chapter ii transport and laplace … diff eq...contents chapter i calculus of variations 1-39...
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CONTENTS ------------------------------------------------------------------------------------------------------
CHAPTER I CALCULUS OF VARIATIONS 1-39
CHAPTER II TRANSPORT AND LAPLACE EQUATIONS 40-73
CHAPTER III HEAT AND WAVE EQUATIONS 74-94 CHAPTER IV ANALYTICAL MECHANICS – I 95-131 CHAPTER V ANALYTICAL DYNAMICS-II 132-144
CHAPTER VI ANALYTICAL MECHANICS-III 145-160
CHAPTER VII ANALYTICAL MECHANICS-IV 161-179
CHAPTER VIII NONLINEAR FIRST-ORDER PDE 180-219
CHAPTER IX REPRESENTATION OF SOLUTIONS 220-253
CHAPTER X ATTRACTION AND POTENTIAL-I 254-274
CHAPTER XI ATTRACTION AND POTENTIAL-II 275-290
CALCULUS OF VARIATIONS
5
y
x
(x0, y0)
(x1, y1)
c
Chapter-1 Calculus of Variations
1.1 INTRODUCTION By a functional, we mean a correspondence which assigns a definite real number to each function/curve belonging to some class.
That is, a functional is a kind of function where the independent variable is itself a function. Thus the domain of a functional is a set of admissible functions, rather than a region of a coordinate space.
Examples of Functionals
(1) consider the set of all rectifiable plane curves between two given points (x0, y0) and (x1, y1). Let this family be denoted by A. The length of a curve y(x) ∈A is a functional. This length is given by
J[y] = l[y(x)] = � ��
���
�+1x0x
2
dxdy
1 dx,
y(x)∈A.
(2) The area “S” of a surface z = z(x, y) bounded by a given curve C is a functional. This area “S” is determined by the choice of the surface S, z = z(x, y), as
J[z(x, y)] = �� ���
����
�
∂∂+�
�
���
�
∂∂+
D
22
yz
xz
1 dx dy,
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
6
A B
where D is the projection of the surface, z = z(x, y), bounded by the curve C,
on the xy-plane.
Functionals, called variable quantities, play an important role in many problems arising in analysis, geometry, mechanics, etc. The first important results in this area due to Euler (1707-1783). Nevertheless, up to now, the “calculus of functionals” still does not have methods of a generality comparable to the methods of classical analysis calculus of functions.
The most developed branch of the ‘Calculus of functionals” is concerned with finding the maxima and minima of functionals, and is called the “Calculus of variations”.
Actually, it would be more appropriate to call this branch/subject the “calculus of variations in the narrow sense”, since the significance of the concept of the “variation of a functional” is by no means confined to its applications to the problem of determining the extrema of functionals.
The aim of “calculus of variation” is to explore methods for finding the maximum or minimum of a functional defined over a class of functions. Several physical laws can be deducted from concise mathematical principles to the effect that a certain functional is a given process attains/assumes a maximum or minimum. In mechanics, we have the principle of least action, the principle of conservation of linear momentum, and the principle of conservation of angular momentum. In addition, we have the principle of castigliano in the theory of elasticity.
The history of the calculus of variations (CV) can be traced back to the year 1696 when John Bernoulli formulated the problem of the brachistochrone (shortest time). In this problem one has to find the curve connecting two given points, A and B, that do not lie on a vertical line, such that a particle sliding down this curve under the influence of gravity alone from the point A reaches point B in the shortest time.
We shall see later on that the curve of quickest descent will not be the straight-line connecting the points A and B, though this is the shortest distance between the points.
Apart from Bernoulli, this problem was independently solved by Leibnitz, Newton and L’Hospital. However, the development of “Calculus of Variations” as an independent Mathematical discipline, along with its own methods of investigation, was due to the pioneering studies of Euler during the period 1707-1783.
CALCULUS OF VARIATIONS
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Apart from the above described problem, three other problems, stated below, were the motivating one for the developed of the subject CV.
Problem of Geodesics
In this problem, it is required to determine the line of shortest length connecting two given points A(x0, y0, z0) and B(x1, y1, z1) on a surface S given by
ϕ(x, y, z) = 0.
This problem is a typical problem “Variational problem with a constraint”. Here, we are required to minimize the arc length given by the functional
J[y, z] = dxdxdz
dxdy
11x0x
22
� ��
���
�+��
���
�+
Subject to the constraint ϕ(x, y, z) = 0.
This problem was first solved by Jacob Bernoulli in 1698, but a general method of such category of problems was given by Euler.
A geodesic on a given surface is a curve, lying on that surface, along which distance between two points is minimum. On a plane, a geodesic is a straight line.
The Problem of Minimum Surface of Revolution
A curve y = y(x) ≥ 0 is rotated about the x-axis through an angle 2π. The resulting surface bounded by the planes
x = a and x = b
has the area
J[y] = 2π dxdxdy
1yba
2
� ��
���
�+
The determination of a particular curve
y = y(x)
which minimizes J[y] is a variational problem.
The Isoperimetric Problem
This problem is : “Among all closed curves of a given length l, find the curve enclosing the greatest area”.
This problem was solved by Euler. The required curve turns out to be a circle. The solution of this problem was known ever is ancient Greece.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
8
1.2 FUNCTION SPACES In the study of functions of n variables, it is convenient to use geometric language, by regarding a set of n numbers (y1, y2,…, yn) as a point in the n-dimensional space.
Linear space. Let L be a non-empty set, consisting of elements x, y, z, of any kind, for which the operations of addition and multiplication by real numbers α, β, … are defined and obey the following axioms :
(i) x + y = y + x
(ii) x + (y + z) = (x + y) + z ;
(iii) there exists an element ‘o’, called the zero element, such that
x + 0 = x = 0 + x for all x∈L,
(iv) For each x∈L, there exists an element “−x” in L such that
x +(−x) = 0 = (−x) + x; (v) 1. x = x;
(vi) α(βx) = (αβ)x
(vii) (α + β)x = αx + βx;
(viii) α(x + y) = αx + αy.
Normed Linear Space
A linear space L is said to be a normed linear space, if each x∈L is assigned a non-negative number ||x||, called the norm of x, such that (i) ||x|| = 0 iff x = 0 ;
(ii) ||x+y|| ≤ ||x|| + ||y||,
(iii) ||αx|| = |α| ||x||.
Function Spaces
Linear spaces whose elements are functions are called function spaces.
In studying functionals of various types, it is reasonable to use various function spaces. The concept of continuity plays an important role for functionals, just as it does for the ordinary functions considered in classical analysis. In order to formulate this concept for functionals, we must somehow introduce a concept of “closeness” for elements in a function space. This is most conveniently done by introducing the concept of the norm of a function. The following normed linear spaces are important for our subsequent studies,
CALCULUS OF VARIATIONS
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Examples of Normed Linear Spaces of Function
(1) The space C [a, b] consisting of all continuous functions defined on a closed interval [a, b], is a normed linear space with
||y||0 = bxa
max≤≤
|y(x)|
(2) The space D1[a, b] consisting of all functions y(x) defined on the closed interval [a, b] which are continuous and have continuous first derivative, is a normed linear space with the norm ||y||1 =
bxamax
≤≤|y(x)| +
bxamax
≤≤ |y′(x)|.
Remark. Two functions, y and z, in D1 are regarded as close together if both the functions themselves and their first derivatives are close together, since
||y−z||1 < ∈
implies that
|y(x) − z(x)| < ∈ and |y′(x) − z′(x)| < ∈
for all x∈[a, b]
(3) The space Dn[a, b], consisting of all functions y(x) defined on the closed interval [a, b] which are continuous and have continuous derivatives upto order n inclusive (where n is a fixed positive integer), is a normed linear space with norm
||y||n = �=
n
0i bxamax
≤≤|y(i)(x)|,
where yi(x) = (d/dx)i y(x) and y(0)(x) = y(x).
Remark. Two functions in Dn are regarded as close together if the values of the functions themselves and of all their derivatives upto order n inclusive are close together.
Similarly, we can introduce spaces of functions of severable variables − the space of continuous functions of n variables, the space of functions of n variables with continuous first derivative, etc. Continuity of functionals
After introducing norm on function spaces, it is natural to talk about continuity of functionals defined on a function space L.
Definition. The function J[y] is said to be continuous at the point y ∈L if for any ∈>0, there is a δ>0 such that
|J[y] − J[ y ]| < ∈ provided
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
10
||y− y ||<δ.
Remark. So far, we have talked about linear spaces, and functionals defined on them. However, in many variational problems, we have to deal with functionals defined on sets of functions which do not form linear spaces.
The set of functions/curves satisfying the constraints of a given variational problem, called the admissible functions is in general not a linear space.
1.3 THE CONCEPT OF A VARIATION/DIFFERENTIAL OF A FUNCTIONAL
First, we give some preliminary definitions and facts.
Definition. Let N be a normed linear space of functions. Let each h∈N be assigned a real number φ[h]. That is, let ϕ[h] be a functional defined on N. Then ϕ[h] is said to be a linear functional if (i) ϕ[αh] = αϕ[h] for any h∈N and any real number α ;
(ii) ϕ[h1 + h2] = ϕ[h1] + ϕ[h2] for any h1 and h2 in N ;
(iii) ϕ[h] is continuous for all h∈N.
Example. (1) The integral
ϕ[h] = �b
ah(x) dx
defines a linear functional on the normed linear space C[a, b]
(2) The integral
ϕ[h] =�b
aα(x) h(x) dx
where α(x) is a fixed member of space C[a, b], defines a linear functional on
C[a, b]
Lemma 1. If α(x) is continuous in [a, b], and if
�b
aα(x) h(x) dx = 0
for all h(x) in C[a, b] such that h(a) = h(b) = 0, then
α(x) = 0 for all x in [a, b]
Proof. Suppose the function α(x) is non-zero, at some point, say C in [a, b]. Then, there exists some interval [x1, x2], around c and contained in [a, b], such that α(x) has the same sign in [x1, x2]. Without loss of generality it is assumed that
CALCULUS OF VARIATIONS
11
α(x)>0 in [x1, x2] ⊂ [a, b] …(1) Set
h(x) = � −−
othersie0
]x,x[inxfor)xx()xx( 2121 …(2)
Since h(x) is continuous and h(x1) = h(x2) = 0, so, h(x) ∈ C[a, b]. However,
�b
aα(x) h(x)dx = �
2x
1xα(x) (x−x1) (x2−x) {Θ h(x) = 0 in [a, x1] and [x2, b]
> 0 , …(3)
since the integrand is positive in the open interval (x1, x2). This is a contradiction to the hypothesis in the statement of lemma. This contradiction proves the lemma 1.
Remark. The lemma still holds if we replace the word ‘C[a, b]’ by ‘Dn[a, b]’ in the statement of the lemma. In that situation, we use the same proof with
h(x) = � −− +
otherwise0
]x,x[inxfor)]xx)(xx[( 211n
21
Lemma 2. Statement. If α(x) is continuous in [a, b], and if
�b
aα(x) h′(x) dx = 0
for every function h(x)∈ D1(a, b) such that h(a) = h(b) = 0, then
α(x) = c for all x in [a, b]
where c is a constant.
Proof. Let c be the constant defined by the condition
�b
a[α(x)−c] dx = 0 …(1)
Let
h(x) = �x
a[ α( ξ)−c] dξ …(2)
Then h(x) is differentiable and
h′(x) = α(x) −c, in [a, b] …(3)
by the fundamental theorem of integral calculus. So
h(x) ∈ D1(a, b) …(4)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
12
Also, from equations (1) and (2).
h(a) = h(b) = 0. …(5)
That is, h(x) satisfies all the conditions of the lemma. Hence, by hypothesis
�b
aα(x) h′(x) = 0 …(6)
Now
�b
a[α(x)−c]2dx
= �b
a[α(x)−c]h′(x)
= �b
aα(x) h′(x) dx −c�
b
ah′(x) dx
= 0 − c [h(b) − h(a)] = 0. This gives
�b
a[α(x)−c]2 dx = 0 in [a, b].
It follows that
α(x) −c= 0 for all x in [a, b]
or α(x) = c for all x in [a, b].
This completes the proof of Lemma 2
Lemma 3. Statement. If α(x) and β(x) are continuous in [a, b], and if
�b
a[α(x) h(x) + β(x) h′(x)] dx = 0
for every function h(x) ∈ D1(a, b) such that
h(a) = h(b) = 0,
then β(x) is differentiable, and
β′(x) = α(x) for all x in [a, b]
Proof. Set
A(x) = �αx
a(ξ) dξ, for x∈[a, b] …(1)
CALCULUS OF VARIATIONS
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Then
A(a) = 0 and A′(x) = α(x) for all x∈[a, b] …(2)
Now
�b
aα(x) h(x)dx = { } dxd)()x('hd)()x(h
b
a
x
a
ba � � �
���
�
���
�
ξξα−ξξα�
= − �b
aA(x) h′(x) dx, …(3)
since h(a) = h(b) = 0. The given condition
�b
a[α(x) h(x) + β(x) h′(x)] dx = 0, …(4)
and result in (3), lead to
�b
a[−A(x) + β(x)] h′(x) = 0, …(5)
for every function h(x) ∈ D1(a, b) such that h(a) = h(b) = 0. The lemma 2 applied to relation (5) given at once
−A(x) + β(x) = constt. In [a, b]
i.e., (since A(x) is differentiable)
β′(x) = A′(x) in [a, b] …(6)
Equations (2) and (6) yield
β′(x) = α(x) in [a, b]
This completes the proof of Lemma 3. We now introduce the concept of the variation/differential of a functional. Let J[y] be a functional defined on some normed linear space. Let ∆J[y] = J[y + h] − J[y] …(1)
be its increment corresponding to the increment
h = h(x) …(2)
of the “independent variable” y = y(x). If y is fixed, ∆ J[h] is a functional of h and it is a nonlinear functional, in general. Suppose that
∆J[y] = ϕ[h] + ∈||h||, …(3)
where
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
14
ϕ [h] = a linear functional, …(4)
and
∈→0, …(5)
as ||h||→0.
Then the functional J[y] is said to be differential, and the principal linear part of the increment ∆J[h], i.e., φ[h], is called the variation/ differential of J[y]. It is denoted by δJ[h]. That is, δJ[h] = φ[h]. …(6)
Theorem 1. The differential of a differentiable functional is unique.
Proof. Before proving the main theorem, we state and prove a lemma.
Statement of lemma. If differential ϕ[h] of a functional J[y] is a linear
functional and if
||h||]h[φ →0 …(1)
as ||h||→0, then
ϕ[h] = 0 for all h. …(2)
Proof of lemma. If possible, suppose that
ϕ[h0] ≠ 0 for some h0 ≠ 0. …(3)
Define
hn =||h||]h[
,n
h
0
00 φ=λ ≠ 0. …(4)
Then ||hn||→0 as n→∞, …(5)
but
||n/h||]n/h[
lim||h||]h[
lim0
0
nn
n
n
φ=
φ∞→∞→
= ||h||]h[
lim0
0
n
φ∞→
, since ϕ is linear
= λ
≠ 0. …(6)
This is contrary to hypothesis in (1). Hence, the result (2) holds.
CALCULUS OF VARIATIONS
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Proof of the main theorem
Now, suppose that, if possible, the differential of the functional J[y] is not unique. Then, we can write
∆J[y] = ϕ1[h] + ∈1 ||h||, …(7)
and
∆J[y] = ϕ2[h] + ∈2 ||h||, …(8)
where ϕ1[h] and ϕ2[h] are linear functionals, and
∈1, ∈2→0, …(9)
as ||h||→0. Here
∆J[y] = J[y+h] − J[y]. …(10)
From equations (7) and (8) imply
ϕ1[h] −ϕ2[h] = (∈2−∈1) ||h||
or
||h||
]h[]h[ 21 φ−φ=∈2−∈1
→0 …(11) as ||h||→0. Hence, by above lemma, the functional
ϕ1[h] − ϕ2[h]
vanishes identically. This gives
ϕ1[h] = ϕ2[h] , for all h …(12)
implying that the differential of the differentiable functional J[y] is unique. This completes the proof.
Definition (Extremum). The functional J[y] is said to have a extremum for
y = y if
J[y] − J[ y ]
does not change its sign in some neighbourhood of the curve y = y (x).
Definition (Weak Extremum). The functional J[y] is said to have a weak extremum for y = y if there exists on ∈>0 such that J[y] − J[ y ]
has the same sign for all y in the domain of definition of the functional which
satisfy the condition
||y− y ||1<∈,
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
16
where || ||1 denote the norm in the space D1.
Definition (Strong Extremum). The functional J[y] is said to have a strong extremum for y = y if there exists an ∈>0 such that
J[y] − J[ y ]
Has the same sign for all y in the domain of definition of the functional which
satisfy the condition
||y− y || M∈,
where || ||0 denotes the norm in the space C [a, b].
Note. Every strong extremum is simultaneously a weak extremum.
(1) However, the converse is not true in general.
(2) Finding a weak extremum is simpler than finding a strong extremum.
Theorem 2. Statement. A necessary condition for the differentiable functional J[y] to have an extremum for y = y is that its variation vanishes for y = y .
Proof. We are required to prove that
δJ[y] = 0 …(1)
for y = y and all admissible h.
According to the definition of the variation δJ[h] of J[y], we have
∆J[h] = δJ[y] + ∈ ||h|| …(2) where
∈→0 …(3)
as ||h||→0, and
∆J[h] = J[y + h] − J[y]. …(4)
Thus, for sufficiently small ||h||, the sign of ∆ J[h] will be the same as the sign of the variation δJ[h]. To be explicit, suppose that J[y] has a minimum for y = y . If possible suppose that
δJ[h0] ≠ 0, …(5)
for some admissible h0. Then, for any α>0, no matter however small it may
be, we have
δJ[−α h0] = −δJ [α h0]. …(6)
Hence, (2) can be made to have either sign for sufficiently small ||h||. But this is impossible, since by hypothesis, J[y] has a minimum for y = y , i.e.,
CALCULUS OF VARIATIONS
17
∆J[h] = J[ y +h] − J[ y ] ≥ 0, …(7)
for all sufficiently small ||h||. This contradiction completes the proof of the theorem. 1.4 EULER’S EQUATION −−−− SIMPLEST VARIATIONAL PROBLEM Theorem 3. Let
J[y] = �ba F (x, y, y′) dx …(1)
Be a functional defined on the set of functions y(x) which has continuous first derivative in [a, b] and satisfy the boundary conditions
y(a) = A, y(b) = B. …(2)
Prove that a necessary condition for J[y] to have an extremum for a given function y(x) is that y(x) satisfies the differential equation
Fy − 0)F(dxd
'y = . …(3)
Proof. Let h = h(x) be the increment given to y(x). Then, in order for the function “y + h” to satisfy the boundary conditions in (2), we must have
h(a) = 0, h(b) = 0 …(4)
Now ∆J[h] = J[y + h] − J[y]
= �ba F (x, y + h, y′+h′)dx − �
ba F (x, y, y′)dx
= �ba F[ (x, y +h, y′+h′) − F(x, y, y′)]dx, …(5)
Using Taylor’s theorem, we write
∆ J[h] = �ba yhF[ (x, y, y′) + h′ Fy′(x, y, y′)]dx +…….., …(6)
where the subscripts denote partial derivative w.r.t. the corresponding arguments, and dots denote terms of order higher than 1 relative to h and h′. The integral in the right-hand side of (6) represents the principal linear part of the increment ∆J[h]. Hence, the variation/ linear part of the increment ∆J[h]. Hence, the variation/differential δJ of J[y] is, by definition,
δJ = �ba yhF[ (x, y, y′) + h′ Fy′(x, y, y′)]dx …(7)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
18
We know that a necessary condition for J[y] to have an extremum for
y = y(x) is that
δ J = 0, …(8)
for all admissible h. Equations (7) and (8) imply
�ba yhF[ +h′ Fy’] dx = 0, …(9)
for all admissible h.
The use of lemma 3 an relation (9) imply
Fy = )F(dxd
'y
i.e., Fy − )F(dxd
'y = 0. …(10)
This completes the proof of the theorem
Definition 1. Equation (10) is known as Euler’s equation.
Definition 2. The integral curves of Euler’s equation are called extremals.
Remark. Euler’s equation is, in general, a second-order ordinary differential equation, and its solution will, in general, depend on two arbitrary constants, which are determined from the boundary conditions.
y(a) = A, y(b) = B .
Special Cases
We now consider some special cases where Euler’s equation can be reduced to a first-order differential equation, or where its solution can be obtained entirely by evaluating integrals.
Case I. Suppose the integrand does not depend on y.
In this case, the functional under consideration is of type
J[y] = �ba F (x, y′)dx, …(1)
where F does not contains y explicitly. In this case,
Fy = 0. …(2)
Consequently, the Euler’s equation becomes
dxd
(Fy’) = 0 …(3)
which has the first integral
CALCULUS OF VARIATIONS
19
Fy′ = C, …(4)
where C is a constant.
Equation (4) is a first-order ordinary differential equation. Solving (4) for y′, we obtain an equation of the form
y′ = f(x, c),
from which y can be found by integration
Case 2. Suppose the integrand does not depend on x.
In this case, we have functional as
J[y] = �ba F (y, y′) dx. …(1)
Now
Fy − �
���
���
���
����
����
�+��
���
����
����
�−=
dx'dy
)F('dy
ddxdy
.)F(dyd
F)F(dxd
,y'yy'y
= Fy −y′ Fy’y−y′′ Fy′y. …(2) So, the Euler’s equation is
Fy−y′ Fy’y − y′′ Fy′y′ = 0.
Multiplying by y′, we obtain
y′Fy−(y′)2 Fy’y−y′y′′ Fy′y′ = 0
� [ ] ,0F'yFdxd
'y =− …(3)
where has the first integral
F−y′ Fy′= c
where c is a constant. Euler’s equation (4) is of first-order.
Case 3. Suppose the integrand does not depend on y′. In this case, the function is as
J[y] = �ba F(x, y) dx, …(1)
so
Fy′ = 0. …(2)
Hence, Euler’s equation becomes
Fy(x, y) = 0 …(3)
This equation is not a differential equation, but an algebraic equation in x and y. Its solution consists of one or more curves y = y(x).
Case 4. When functional J[y] is of the form
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
20
J[y] = �ba [f(x, y) 2)'y(1+ ] dx. …(1)
This functional represents the integral of a function f(x, y) with respect to the
arc length s, where
ds = 2)'y(1+ dx.
In ths case,
F(x, y, y′) = f(x, y) 2)'y(1+ …(2)
and, so
[ ]��
�
��
�
�
+−+=��
�
����
�
∂∂−
∂∂
2
2y
)'y(1
'y)y,x(f
dxd
)'y(1)y,x(f'y
Fdxd
yF
= (fy)�
��
−��
�
�
��
�
�
+∂∂+
++
dxdy
)'y(1
'yf
y)'y(1
'y)f()'y(1
22x2
+��
���
��
�
�
��
�
�
+∂∂
dx'dy
.)'y(1
'yf'y 2
= (fy)��
�
��
�
�
+−
+−+
2y2y2
'y1
'y)f('y
'y1
'y)f('y1
������
�
�
������
�
�
+
+−+
−
2
2
22
'y1
'y1
'y'y1
)f)(''y(
=(fy) 2/322
2
y2x2
)'y1(1
).f)(''y('y1
'y)f(
'y1
'y)f('y1
+−
+−
+−+
= 2/322
x
2
y
)'y1(f''y
'y1
f'y
'y1
)f(
+−
+−
+
= 2'y1
1
+[fy − y′ fx −y′′f] …(3)
So, Euler’s equation becomes
fy − y′fx − y′′f = 0.
CALCULUS OF VARIATIONS
21
which is a differential equation of order 2.
ILLUSTRATIVE EXAMPLES Example 1. Solve the variational problem
J[y] = dxx
'y121
2
���
���
�
�� +
…(1)
y(1) = 0, y(2) = 1. …(2)
Solution. We note that the integrand in the given functional does not depend
on y explicitly, and
F(x, y, y′) = x
'y1 2+ …(3)
Euler’s equations for such case is of the form
=∂∂
'yF
c, …(4)
where c is a constant. From equations (3) and (4), we find (exercise)
y′ = 22xc1
cx
− …(5)
Integrating (5), it follows that (exercise)
y = 122 cxc1
c1 +−
or (y−c1)2 + x2 = ,c1
2
…(6)
where c1 is a constant. The curve (6) represents a circle with centre (0, c1), lying on the y-axis, and radius 1/c. Using the boundary conditions (2), we find (exercise)
c = 5
1, c1 = 2. …(7)
So, the require curve is
x2 + (y−2)2 = 5. …(8)
Example 2. Among all the curves joining two given points (x0, y0) and (x1, y1), find the one which generates the surface of minimum area when rotated about the x-axis.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
22
Solution. We know that the area of the surface of revolution generated by rotating the curve y = y(x) about the x-axis is
2π � +1x0x
2 dx'y1y
So, the variational problem is
J[y] = 2π � +1x0x
2 dx'y1y , …(1)
with boundary conditions
y(x0) = y0, y(x1) = y1. …(2)
In this variational problem the integrand does not depend explicitly on x, and
F(y, y′) = 2πy ,'y1 2+ …(3)
and corresponding Euler equation is
y −y′(Fy′) = constt. …(4)
We find (exercise)
2'y1
y
+ = c, c = constt. …(5)
we put y′ = sinh t, …(6)
then (5) imply (exercise)
y = c cosh t. …(7)
Equation (6) and (7) give (exercise)
dx = c dt.
Integrating it, we obtain
x = ct + c1, …(8)
where c1 is a constant. Eliminating t from equations (7) and (8), we find
y = c cosh ��
���
� −c
cx 1 . …(9)
The values of the arbitrary constants c and c1 are determined by the given
conditions in (2).
The required curve is catenary passing through the two given points. The surface generated by rotation of the catenary is called a catenoid.
Example 3. Minimize the functional
CALCULUS OF VARIATIONS
23
J[y] = �ba (x−y)2dx. …(1)
Solution. In this example, the integrand does not contain y′ explicitly and
F(x, y) = (x−y)2 …(2)
The corresponding Euler’s equation is
,0yF =
∂∂
…(3)
which leads to
x−y = 0. …(4)
The required curve (4) is a straight line. Further, the functional (1) vanishes
along this line.
This completes the solution
1.5 THE CASE OF SEVERABLE VARIABLES Now, we consider further generalization of the simplest variational problem. First we consider the case of n dependent functions. Let
J[y1, y2,…, yn] = �ba F(x, y1, y2,…, yn, '
n'1 y,...,y )dx …(1)
Be a functional which depends on n continuously differentiable functions
y1(x), y2(x), …, yn(x)
satisfying the boundary conditions
yi(a) = Ai, yi(b) = Bi, 1≤ i ≤ n. …(2)
Here, we are looking for an extremum of the functional (1) defined on the set of the set of smooth curves joining two fixed points in (n+1)-dimensional Euclidean space Rn+1.
The problem of finding geodesics (shortest curves joining two points of some manifold) is of this type. The same kind of problem arises in geometric optics, in finding the paths along which light rays propagate in an inhomogeneous media. According to Fermat’s principle, light goes from a point, say P0, to a point, say P1, along the path for which the transit time is the smallest.
Theorem. Prove that a necessary condition for the curve
yi = yi(x), (i =1, 2,…, n)
to be an extremal of the functional
J = �ba F (x, y1, y2,…, yn, '
n'2
'1 y,...,y,y ) dx …(1)
is that the functions yi(x) satisfy the equations
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
24
0yF
dxd
yF
ii=��
�
����
�
∂∂−
∂∂
', 1 ≤ i ≤ n. …(2)
Proof. First of all, we calculate the variation δJ of the given functional J in (1). We replace each yi(x) by a varied function yi(x) + hi(x).
By definition, the variations δJ of the functional J[y1,., yn] is linear in hi and hi′, 1 ≤ i ≤ n, and which differs from the increment
∆J = J[y1 + h1, y2 + h2, ., yn + hn] − J[y1, y2,.,yn], …(3)
by a quantity of order higher than 1 relative to hi and hi′ ; i = 1, 2,…, n. Since both yi(x) and yi(x) + hi(x) satisfy the boundary conditions
yi(a) = Ai, yi(b) = Bi, 1 ≤ i ≤ n. …(4)
Therefore, we must have
hi(a) = hi(b) = 0, for each i. …(5)
Using Taylor’s theorem, we obtain
∆J = �ba [ F(x, y1 +h1,.., yi+hi,…, yn + hn ,
),..,,..., ''''''nnii11 hyhyhy +++
− F(x, y1, …, yi,…, yn, )]y,...,y,.,y 'n
'i
'1 dx
= � +���
�� ��
�
����
�
∂∂+
∂∂
=
ba
n
1i i
'i
ii ......dx
'yF
hyF
h …(6)
where the dots denote terms of order higher than 1 relative to hi, hi′ (i = 1, 2, ., n) The integral on the right of (6) represents the principal linear part of the increment ∆J. Hence, by definition, the variation δJ of J is
δJ = ����
�� ��
�
����
�
∂∂+
∂∂
=
ba
n
1i i
'i
ii dx
'yF
hyF
h …(7)
Since all the increments hi(x) are independent, we can choose one of them quite arbitrarily (as long as the boundary conditions are satisfied), setting all the others equal to zero. Therefore, the necessary condition
δJ = 0, …(8)
for an extremum implies
CALCULUS OF VARIATIONS
25
� =���
����
�
∂∂+
∂∂b
ai
'i
ii ,0
'yF
hyF
h …(9)
for each i = 1, 2,…, n. There are now n conditions in (9). Using lemma 3, we
obtain
���
����
�
∂∂=
∂∂
'ii y
Fdxd
yF
, 1 ≤ i ≤ n
or
iyF −
dxd
( 'iyF ) = 0, 1 ≤ i ≤ n. …(10)
Equations (10) are called Euler’s equations. We note that (10) is a system of n second order, in general ordinary differential equations. Solution of (10) contains, in general, 2n arbitrary constants, which are determined from the boundary conditions in (4). This completes the proof.
Definition. Two functionals are said to be equivalent if they have the same
extremals.
Example. Find the external of the functional
J[y, z] = �π 2
0
/(y′2 + z′2 + 2yz)dx
y(0) = 0, y (π/2)= 1, z(0) = 0, z(π/2)= −1.
Solution. Taking
y1(x) = y(x), y2(x) = z(x), …(1)
and F[y1, y2] = (y1′)2 + (y2′)2 + 2 y1 y2, …(2)
Euler’s equations
,0'y
Fdxd
yF
ii
=���
����
�
∂∂−
∂∂
i = 1, 2, …(3)
become
2z −dxd
(2y′) = 0,
2y − dxd
(2z′) = 0.
This gives z = y′′, …(4)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
26
y = z′′. …(5)
Equations (4) and (5) imply
,0ydx
yd4
4
=− …(6)
solution of (6) is
y(x) = c1 ex + c2 e−x + c3 sin x + c4 cosx, …(7)
where c1, c2, c3, c4 are constants. Equations (4) and (7) give
z(x) = c1 ex + c2e−x − c3 sinx − c4 cos x. …(8)
Using the given boundary conditions
���
−=π==π=
,)/(,)()/(,)(
12z00z12y00y
…(9)
we obtain (exercise)
c1 = c2 = 0, c3 = 1, c4 = 0. …(10)
Hence, an extremum of the given functional is given by
y(x) = sinx,
z(x) = −sin x. …(11)
1.6 THE PROBLEM OF GEODESICS
Suppose we have surface σ specified by a vector equation
rrρρ
= (u, v).
The shortest curve (of minimum length) lying on the surface σ and connecting
two points A and B of surface, σ, is called the geodesic connecting the two
points.
The equations for the geodesics of σ are the Euler equations of the corresponding variational problem−namely, the problem of finding the minimums distance (measured along surface σ) between two points of the surface σ.
Euler’s Equations on Geodesics
A curve lying on the surface
rrρρ
= (u, v), …(1)
can be specified by the equations
u = u(t),
CALCULUS OF VARIATIONS
27
v = v(t), t being a parameter. …(2)
Let vu)v,u(vu)v,u(vu)v,u( r.rG,r.rF,r.rE
ρρρρρρ=== …(3)
These quantities are called the coefficients of the first fundamental form of the surface (1). The arc length between the points A(t1) and B(t2), corresponding to the parameter t, is given by (using results from Differential Geometry by C.E. Wealtherburn)
J[u, v] = dt'Gv'v'Fu2'uE1t0t
22� ++ …(4)
Euler’s equations for the functional (4) are
0GvvFu2Euudt
dGvvFu2Eu
u2222 =� �
��
� ++∂∂−++
∂∂
'''''
]''''[
0GvvFu2Euvdt
dGvvFu2Eu
v2222 =� �
��
� ++∂∂−++
∂∂
'''''
''''
These become
0'Gv'v'Fu2'Eu
)'Fv'Eu(2dtd
'Gv'v'Fu2'Eu
'vG'v'uF2'uE2222
2uu
2u =
��
�
���
�
++
+−��
�
���
�
++
++ …(5)
and 0GvvFu2Eu
FvFu2dtd
GvvFu2Eu
vGvuF2uE2222
2vv
2v =
��
�
���
�
++
+−��
�
���
�
++
++
''''
)''(
''''
'''' …(6)
Remark. The concept of a geodesic can be defined not only for surfaces, but also for higher-dimensional manifolds. Finding the geodesics of an n-dimensional manifold reduces to solving a variational problem for a functional depending on n functions.
Example 1. Find the geodesics of the circular cylinder
(r =ρ
a cos ϕ, a sin ϕ, z)
Solution. The variables ϕ and z play the role of the function u and v in the
above article. Now
(r =ρ
a cos ϕ, a sin ϕ, z). …(1)
Then �
�
=
−=
).1,0,0(r
),0,�cosa,�sina(r
z
�
ρ
ρ …(2)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
28
Therefore
����
�
==
==
==
1r.rG
0r.rF
ar.rE
zz
z�
2��
ρρ
ρρ
ρρ
…(3)
The arc length between two points A(t1) and B (t2) lying on the cylinder (1) is given by the functional
J[ϕ, z] = � ++2t1t
22 dt'Gz'z'�F2'�E …(4)
or
J[ϕ, z] = � +2t1t
222 'z'�a dt. …(5)
Euler’s equations for the functional (5) are
0 − ,0'z'�a
'�adtd
222
2
=��
�
��
�
�
+ …(6)
0 − .''
'0
za
adtd
222
2=
��
�
��
�
�
+φ
φ …(7)
These equations, on integration, yield
22221222
2
c'z'�a
'z,c
'z'�a
'�a =+
=+
Dividing the second of these equations by the first, we obtain
z′/ϕ′ = constt.
or c�d
dz =
which has the solution
z = c1ϕ + c2. …(8)
Equation (8) represents a two-parameter family of helical lines lying on the cylinder (1). Thus, a geodesic on cylinder (1) is a helix.
Example 2. Find the geodesics of the sphere
rρ
= (a sin θ cosφ, a sin θ sinφ, a cosθ)
Solution. On the surface of a given sphere
CALCULUS OF VARIATIONS
29
rρ
= (a sin θ cosφ, a sin θ sinφ, a cosθ) …(1)
we find (exercise)
E = a2, F = 0, G = a2 sin2θ. …(2)
The variational functional is (exercise)
J = � �sin'� 22 + dφ, θ′ = .�d�d
…(3)
Here, the integrand is
F = F(θ, θ′) = �sin'� 22 + = independent of φ. …(4)
So, the corresponding Euler’s equation is
F − θ′ Fθ′ = constt. = c
� c�sin'�
'��sin'�
22
222 =
+−+
� c�sin'�
)�(sin22
2
=+
� sin4θ = c2(θ′2 + sin2θ)
� c2θ′2 = sin4θ − c2 sin2θ
� 2
222
c)c�(sin�sin
�d�d −
θ−−
θ=θ−θ
=θφ
222
2
222 cc1
ecc
ecc1
cdd
cot)(
cos
cossin
Integrating
ϕ = cos−1 'cc1
�cotc2
+��
�
�
��
�
�
−
� cos (φ − c′) = ��
�
�
��
�
�
−
θ2c1
c cot
� c1 cotθ = cosϕ cos c′ + sin ϕ sin θ′
� c1 cosθ = sinθ sinϕ sin c′ + sinθ sinϕ sin c′
� z = Ax + By. …(5)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
30
This is the equation of the plane passing through the centre (0, 0, 0) of the sphere and intersecting the sphere along a great circle.
Thus, the shorten curve, i.e., geodesic on a sphere is n are of a great circle.
Example 3. Find the geodesic on the plane.
Solution. The geodesic on the plane is an extremal of the functional
J[y] = � +1x0x
2 dx'y1 …(1)
The integrand F does not contain y explicity. Hence, the corresponding Euler’s
equation is
Fy′ = c …(2)
i.e.,
c'y2.)'y1(21 2
12 =+
−
� y′ = c 2'y1+
y′2 = c2(1+y′2)
y′2(1−c2) = c2
� y′ = A
� y(x) = Ax + B. …(3)
This is the equation of a straight line in the plane. Thus, geodesics in a plane
are straight lines.
1.7 FUNCTIONALS DEPENDING ON HIGHER-ORDER
DERIVATIVES Theorem. Statement. Among all functions y(x) belonging to the space Dn(a, b) and satisfying the conditions y(i)(a) = Ai, y(i)(b) = Bi, 0 ≤ i ≤ n, …(1)
find the function for which the functional
J[y] = �ba F (x, y, y′, y′′,…, y(n)) dx, …(2)
has an extremum.
CALCULUS OF VARIATIONS
31
Solution. First, we state the general result which states that a necessary condition for a functional J[y] to have an extremum is that its variation vanish, i.e.,
δJ = 0. …(3)
We replace y(x) by the “varied” function “y(x) + h(x)”, where h(x) belongs to Dn(a, b) and satisfy the boundary conditions (1). For this, we must have
h(i)(a) = h(i)(b) = 0 for i = 0, 1, 2,…, n.. …(4)
we know that by the variation δJ of the functional J[y], we mean the expression which is linear in h, h′,…, h(n), and which differs from the increment
∆J = J[y + h] − J[y], …(5)
by a quantity of order higher than 1 relative to h, h′,…, h(n). Next, we use Taylor’s theorem to obtain
∆J = �ba (F[ x, y + h, y′ + h′,…, y(n) + h(n)) − F(x, y′,…y(n))]dx
= �ba [ h Fy + h′ Fy′ +…+ h(n) Fy(n)]dx +…., …(6)
where the dots denote terms of order higher than 1 relative to h, h′, …, h(n). The last integral in (6) represents the principal linear part of the increment ∆J. Therefore, by definition of the variation of J[y], we write
δJ = �ba [ hFy+h′ Fy′ +…+ h(n) Fy(n)]dx. …(7)
The necessary condition (3) for an extremum implies that
�ba [ h Fy+h′ Fy′ +…+ h(n) Fy(n)] dx = 0. …(8)
Integrating (8) by parts repeatedly and using boundary condition (4), we find
that (exercise)
���
�
���
�−+++−b
a yn
nn
y2
2
yy nFdx
d1f
dx
dF
dxd
F )()(...)()( )(''' = 0, …(9)
for any function h(x) which has continuous derivatives and satisfies the boundary condition in (4). It follows from lemma 1 that
Fy− 0)F(dxd
)1(...)F(dxd
)F(dxd
'n'yn
nn
''y2
2
'y =−+++ …(10)
Equation (10) is called Euler’s equation. Equation (10) is an ordinary differential equation of order 2n, its general solution contains 2n arbitrary constants, which can be determined from the 2n boundary conditions in (4).
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
32
�ba h′(Fy′) dx = [ ] � �
�
���
�− ba 'y
ba'y )F(
dxd
h)F(h dx
= � �
���
�ba 'y ,dx)F(
dxd
h
�ba h′′(Fy′′) dx = [ ] { }�− b
a ''yba''y dxF
dxd
'h)F('h
= −��
�
���
��−
���
� b
a ''y2
2b
a''y dx)F(
dxd
h)F(dxd
h
= (−1)2 �ba ''y2
2
,dx)F(dxd
h
in general,
� �−=b
a
b
a yk
kk
kyk dxF
dx
dxh1dxFh k }{)()(}{ )()(
)(
1.8 THE CONCEPT OF VARIATIONAL DERIVATIVE Let J[y] be a functional depending on the function y(x), and suppose we give y(x) an increment h(x) which is different from zero only in the neighbourhood of a point x0.
Let ∆σ denote the area lying between the curve y = y(x) and y = y(x) + h(x). Consider the ratio
�
]y[J]hy[J∆
−+ …(1)
of the increment
∆J = J[y+h] −J[y] …(2)
to the area ∆σ
Let the area
∆σ→0 …(3)
in such a way that
max |h(x)|→0 …(4)
as well as the length of the interval in which h(x) is non zero, goes to zero. Then, if the ratio (1) converges to a limit as ∆σ→0, this limit is called the variational derivative of the functional J[y] at the point x0(for the curve y = y(x)), and is denoted by
CALCULUS OF VARIATIONS
33
0xxy�
J�
=
…(5)
Remark. In the light of above, we write
∆J = J[y+h] − J[y] = ��
���
�
��
∈+= 0xxy�
J� ∆σ, …(6)
where
∈→0,
as ∆σ→0.
(2) The variation/differential of a functional J[y] at the point x = x0, in terms of the variational derivative, is given by the formula
δJ =��
���
�
��
= 0xxy�J� ∆σ. …(7)
1.9 VARIATIONAL PROBLEMS WITH SUBSIDIARY CONDITIONS
(THE ISOPERIMETRIC PROBLEM) Theorem. Given the functional
J[y] = �ba F (x, y, y′) dx,
Let the admissible curves satisfy the conditions
y(a) = A, y(b) = B,
K[y] = �ba G (x, y, y′)dx = l ,
where K[y] is another functional, and let J[y] have an extremum for y = y(x). Then, if y = y(x) is not an extremal of K[y], there exists a constant λ such that y = y(x) is an extremal of the functional
�ba F( +λG)dx.
Proof. Let J[y] = �ba F (x, y, y′)dx, …(1)
have an extremum for the curve y = y(x), subject to the conditions
y(a) = A, y(b) = B, …(2)
K[y] = �ba G (x, y, y′)dx = l . …(3)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
34
We choose two points x1 and x2 in the interval [a, b], where x1 is arbitrary and x2 satisfies a condition to be stated later on, but is otherwise arbitrary.
We give y(x) an increment
δ1y(x) + δ2y(x),
where
δ1y(x) is non-zero only in the neighbourhood of x1, (4a)
and
δ2(x) is nonzero only in a neighbourhood of x2 …(4b)
Let y*(x) = y(x) + δ1y(x) + δ2y(x). …(5)
We now require that the “varied” curve y = y*(x) satisfy the condition
K[y*] = K[y]. …(6)
Using variational derivatives, we can write the increment ∆J of the functional J in the form
∆J = ,�y�F�
�y�F�
22
2xx11
1xx
∆��
���
�
��
∈++∆��
���
�
��
∈+==
…(7)
where
∆σ1 = �ba [ δ1y(x)]dx,
∆σ2 = �ba [ δ2y(x)]dx, …(8)
and ∈1, ∈2→0 …(9) as ∆σ1, ∆σ2→0. …(10)
Writing ∆K in a form similar to (7), we obtain
∆K = K[y*] −K[y]
= ,�y�G�
�y�G�
2'2
2xx1
'1
1xx
∆��
���
�
��
∈++∆��
���
�
��
∈+==
…(11)
where
∈1′, ∈2′→0 …(12a)
as
∆σ1, ∆σ2→0. …(12b)
Next, we choose the point x2 to be a point for which
CALCULUS OF VARIATIONS
35
.0y�G�
2xx
≠=
…(13)
Such a point exists, since by hypothesis y = y(x) is not an extremal of the functional K. The condition of the point x2, given in (13), is the condition which we had mentioned earlier. With this choice of point x2, Equations (6) and (11) imply
∆σ2 = −
��
�
��
�
�
��
��
�
∈+
=
= '
y�G�
y�G�
2xx
1xx ∆σ1, …(14)
where
∈′→0 as ∆σ1→0.
We set
λ = −
2xx
2xx
y�G�
y�F�
=
= . …(15)
Using (14) and (15) into (7), we obtain
∆J =
��
�
��
�
�
��
��
�
+��
���
�
��
+−∆��
���
�
��
∈+
=
=
==
'�
y�G�
y�G�
�y�F�
�y�F�
2xx
1xx2
2xx11
1xx
∆σ1
= ��
���
�
��
+== 1xx1xx y�
G��
y�F� ∆σ1 + ε ∆σ1, …(16)
where
∈→0
as
∆σ1→0.
This expression for ∆J explicitly involves variational derivatives only at the point x = x1 and the increment h(x) is now first δ1y(x). The “compensating increment” δ2y(x) has been taken into account automatically by using the
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
36
condition ∆K = 0. Thus, the first term in the right-hand side of (16) is the principal linear part of ∆J. So, the variation of the functional J at the point x1 is
δJ = ��
���
�
��
+== 1xx1xx y�
G��
y�F� ∆σ1. …(17)
We know that a necessary condition for an extremum is that
δJ = 0. …(18)
Since ∆σ1 is nonzero while x1 is arbitrary, we finally obtain
0yG
yF =
δδλ+
δδ
i.e.,
0)G(dxd
G�)F(dxd
F 'yy'yy =���
� −+
���
� − …(19)
This shows that y = y(x) is an extremal of the functional
�ba F( +λ G)dx,
where λ is given by (15).
This completes the proof.
Remarks (1) The general solution of differential equation (19) will contain two arbitrary constants in addition to the parameter λ. We shall determine these three quantities from two boundary conditions
y(a) = A,
y(b) = B
and the subsidiary condition
K[y] = l .
Remark (2). The above theorem/result generalizes immediately to the case of functionals depending on several functions.
Suppose we are looking for an extremum of the functional
J[y1, y2,.., yn] = �ba F (x, y1,…, yn, '
n'1 y,...,y )dx, …(20)
Subject to the conditions
yi(a) = Ai,
yi(b) = Bi, 1 ≤ i ≤ n, …(21)
and
CALCULUS OF VARIATIONS
37
�ba kG (x, y1, y2,…, yn, '
n'1 y,...,y )dx = lk …(22)
for k = 1, 2,…, m.
In this case a necessary condition for an extremum is that
0G�F'ydx
dG�F
y
m
1kkk
i
m
1kkk =
���
�
��
���
��+
∂∂−�
�
���
��+
∂∂
== …(23)
for i = 1, 2, …, n.
The 2n arbitrary constants appearing in the differential equation system (23) and the values of m parameters λ1, λ2,…, λm, sometimes called Largange multipliers & are determined from the boundary conditions (21) and subsidiary conditions (22). Here, the number of Lagrange multiplier equals the number of conditions of constraint. Example 1. Among all curves of length l in the upper half-plane passing through the points (−a, 0) and (a, 0) find the one which together with the interval [−a, a] encloses the largest area. Solution. We have to find the function
y = y(x)
for which the integral
J[y] = �−aa y(x)dx …(1)
takes the largest value subject to the conditions
y(−a) = 0, y(a) = 0, …(2)
K[y] = � +−aa
2'y1 dx = l. …(3)
We form the functional
J*[y] = J[y] + λ K[y]
= � ++−aa
2 ]'y1�y[ dx …(4)
The corresponding Euler’s equation is
0'y1
'y2.21
.�
dxd
]'y1�y[dyd
2
2 =
����
�
����
�
�
+−++
1− λ 0'y1
'ydxd
2=
��
�
�
��
�
�
+ …(5)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
38
Integrating, we obtain
x − 12c
'y1
'y� =+
or (x−c1) = 2'y1
'y�
+
or (x−c1)2 (1+ y′2) = λ2 y′2
y′ = 2
12
1
)cx(�
cx
−
−. …(6)
Integrating (6), we obtain
y(x) = dx)cx(�
cx2
12
1
−−
−�
= 22
12 c)cx(� +−−
or (x−c1)2 + (y−c2)2 = λ2. …(7)
It is a family of circles. The values of c1, c2 and λ are determined from the
given conditions in (2) and (3).
We find (−a − c1)2 + c22 = λ2
(a−c1)2 + c22 = λ2
� c1 = 0. …(8)
Then, we have
c2 = − 22 a� − …(9)
So, solution (7) now becomes
x2 + (y + 2222 �)a� =− . …(10)
This gives
y = 2222 a�x� −−−
and y′ = 22 x�
x
−
− …(11)
CALCULUS OF VARIATIONS
39
Now, the condition (3) implies
l = dxx�
x1a
a 22
2
�−
+−
= dxx�
�aa 22
�−
−
= 2λ sin−1 (a/λ).
This gives
a/λ = sin(e/2λ). …(12)
Equation (12) is a transcendal equation for λ. Solving it, we find a definite/certain value, say λ = λ0. Then, solution curve (10) becomes
x2 + ( ) 20
222
0 �a�y =−+ …(13)
The result (13) is the required form.
Example 2. Find the extremal of the functional
J[y] = �π0
y′2 dx
Subject to the conditions
y(0) = 0, y(π) = 0,
�π0
y2 dx = 1.
Solution. We form an auxiliary function
J*[y] = �π0
F(x, y, y′)dx …(1)
where
F(x, y, y′) = y′2 + λ y2, λ being a parameter. …(2)
Euler’s equation for (1) is
Fy-dxd
(Fy′) = 0
i.e., 2λy −dxd
(2y′) = 0
or y′′ − λy = 0. …(3)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
40
First of all, we claim
λ < 0. …(4)
If possible, consider the case when λ ≥ 0, then the general solution of second
order ODE (3) is
y(x) = c1x�
2x� ece −+ …(5A)
Use of boundary conditions
y(0) = 0, y (π)= 0 …(5)
give (exercise)
c1 = c2 = 0, …(5B)
and y(x) ≡ 0.
Then �π0
y2 dx = 0 ≠ 1, …(5C)
which is a violation given condition. Hence, our claim in (4) is valid.
Consequently, solution of ODE (4) is
y(x) = c1 sin �− x + c2 sin �− x …(6)
The boundary condition y(0) = 0 gives
c2 = 0, …(7)
and boundary condition
y (π) = 0
implies
λ = −k2 …(8)
for k = 1, 2, 3,…
Thus, a solution of ODE satisfying two boundary condition is
y(x) = c1 sink x, …(9)
where c1 is a non-zero constant and yet to be determined.
The condition
�π0
y2dx = 1, …(10)
gives (exercise)
c1 = π± /2 . …(11)
CALCULUS OF VARIATIONS
41
Hence, extremals are
y(x) = π
± 2sin kx, …(12)
where k = 1, 2, 3,… . …(13)
Example 3. Find an extremal of the functional
J[y, z] = �10 [y′2 + z′2 − 4xz′ − 4z]dx,
y(0) = 0, y(1) = 1
z(0) = 0, z(1) = 1
subject to the condition
�10 [y′2 − xy′−z′2] dx = 2.
Solution. We form an auxiliary functional
J*[y, z] = �10 F(x, y, z, y′, z′)dx …(1)
where
F(x, y, z, y′, z′) = (y′2 + z′2 −4xz′ − 4z) +λ(y′2−xy′−z′2), …(2)
in which λ is a parameter.
The system of Euler’s equations are
0 +dxd
(2y′ + 2λy′ − λx) = 0 …(3)
4 +dxd
(2z′ − 4x −2λz′) = 0 …(4)
Solving these equations (exercise), we obtain
y(x) = 21
2
c)�1(4
xc2x� ++
+ …(5)
z(x) = ,c)�1(2
xc4
3 +−
…(6)
where c1, c2, c3, c4 are constants of integration. Using the boundary conditions
���
====
1)1(z,0)0(z1)1(y,0)0(y
…(7)
we find (exercise)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
42
c1 2
4�3 +, c2 = 0, c3 = 2(1−λ), c4 = 0. …(8)
Hence, solution of Euler’s system is
y(x) = )�1(4
x)4�3(x� 2
+++
…(9)
z(x) = x. …(10)
To find λ, we substitute the value of y(x) and z(x) from equations (9) & (10)
into the given condition
�10 (y′2−xy′−z′2)dx = 2, …(11)
we find (exercise), two value of λ, namely
λ1 = −1112
�,1110
2 −= …(12)
The actual substitution of λ, y and z in (2), we find that λ2 does not satisfy it, but λ1 does. Hence, the desired extremal is determined by the equations
.)(
,)(
xxz2
x5x7xy
2
=
−= …(13)
1.10 FINITE SUBSIDIARY CONDITIONS We now consider a problem which can be stated a follows :
Problem. Find the function yi(x) for which the functional
J[y1, y2,…, yn] = �ba F(x, y1, …, yn, 1
n'1 y,...,y ) dx, …(1)
has an extremum, where the admissible functions satisfy the boundary
conditions
yi(a) = Ai, yi(b) = Bi, 1 ≤ i ≤ n, …(2)
and m “finite” subsidiary conditions (m < n)
gk(x, y1,…, yn) = 0, 1 ≤ k ≤ m. …(3)
Note. We note that in the above problem, the functional (1) is not considered for all curves satisfying the boundary conditions (2), but only for those which lie in the (n−m) −dimensional manifold defined by the systems (3).
Remark. For simplicity, we restrict ourselves to the case
CALCULUS OF VARIATIONS
43
n = 2 and m = 1.
Theorem. Given the functional
J[y, z] = �ba F(x, y, z, y′, z′)dx, …(1)
let the admissible curves lie on the surface
g(x, y, z) = 0, …(2)
and satisfy the boundary conditions
y(a) = A1, y(b) = B1, …(3)
z(a) = A2, z(b) = B2, …(4)
and moreover, let J[y, z] have an extremum for the curve
y = y(x), z = z(x) …(5)
Then, if gy and gz do not vanish simultaneously at any point of the surface (2), there exists a function λ(x) such that (5) is an extremal of the functional
�ba [ F+λ(x) g]dx. …(6)
Proof. We are required to prove that (5) satisfies the differential equations
Fy + λgy −dxd
(Fy′) = 0, …(7)
Fz + λgz −dxd
(Fz′) = 0. …(8)
Let J[y, z] have an extremum for the curve (5), subject to the conditions (2) to (4). Let x1 be an arbitrary point of the interval [a, b]. Next, we give y(x) an increment δy(x) and z(x) an increment δz(x), where both δy(x) and δz(x) are non-zero only in a neighbourhood, say [α, β] ⊂ [a, b], of x1. Using the notion of variational derivatives, we can write the corresponding increment
∆J = J[y + δy, z + δz] − J[y, z], …(9) in the form
∆J = ,22xx
11xx 11
zF
yF σ∆
��
���
�
��
∈+δδ+σ∆
��
���
�
��
∈+δδ
== …(10)
where
∆σ1 = �ba � y(x)dx, ∆σ2 �
ba � z(x) dx, …(11)
and
∈1, ∈2→0
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
44
as
∆σ1, ∆σ2→0. …(12)
We now require that the “varied” curve
y = y*(x) = y(x) +δy(x),
z = z*(x) = z(x) + δz(x) …(13)
satisfy the condition (2), i.e.,
g(x, y*, z*) = 0. …(14)
Then 0 = �ba [G(x, y*, z*) −g(x, y, z)]dx
= �ba dx]z�gy�g[ zy +
= { } ,''22xxz11xxy 11
gg σ∆∈++σ∆���
� ∈+ ==
…(15)
where ∈1′, ∈2′ →0 as ∆σ1, ∆σ2→0, and the overbar indicates that the corresponding derivatives are evaluated along certain intermediate curves. By hypothesis, either
1xxz
1xxy gorg ==
is nonzero. If
1xxzg = ≠ 0, …(16)
we can write the condition (16) in the form
∆σ2 = −��
��
�
�
�
�
∈+=
= '1
1
xxz
xxy
g
g∆σ1, …(17)
where ∈′→0 as ∆σ1→0. Substituting (18) into the formula (10) for ∆J, we
obtain
∆J = ��
��
�
�
�
�
���
����
�−
== 1xxz
y
1xx z�F�
g
g
y�F� ∆σ1 + ∈ ∆σ1 …(18)
where ∈→0 as ∆σ1→0. The first term in the right side of (19) is the principal linear part of ∆J. Hence, by definition, the variation δJ of the functional J at the point x1 is
CALCULUS OF VARIATIONS
45
δJ = ��
��
�
�
�
�
���
����
�−
== 1xxz
y
1xx z�F�
g
g
y�F� ∆σ1 …(19)
we know that a necessary condition for an extremum of the functional J is that
δJ = 0. …(20)
Since ∆σ1 is non-zero while x1 is arbitrary, equations (20) and (21) imply
,0z�F�
g
g
y�F�
z
y =���
����
�−
or 0)F(dxd
Fg
g)F(
dxd
F 'zzz
y'yy =�
���
� −−� �
��
� −
or z
'zz
y
'yy
g
)F(dxd
F
g
)F(dxd
F −=
− …(21)
Along the curve
y = y(x),
z = z(x)
the common value of the ratios (22) is some function of x, say −λ(x). Then (22) reduces to system of differential equations
Fy + λgy −dxd
(Fy′) = 0
Fz + λgz−dxd
(Fy′) = 0,
which are precisely equations (7) and (8). This completes the proof of
theorem.
Remark 1. If the functional J has an extremum for a curve γ, subject to the
condition
g(x, y, z, y′, z′) = 0, …(∗)
and if the derivatives gy′ and gz′ do not vanish simultaneously along γ, then there exists a function λ(x) such that γ is an integral curve of the system of differential equations
Φy−dxd
(Φy′) = 0,
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
46
Φz −dxd
(Φz′) = 0,
where
Φ = F + λG.
Remark 2. If we assume that the condition (2) does not hold everywhere, but only at some fixed point
g(x1, y, z) = 0, …(∗∗)
we obtain a condition whose left-hand side can be regarded as a functional of y and z. Thus, the condition (2) can be regarded as an infinite set of conditions, each of which is a functional.
Example 1. Among all curves lying on the sphere x2 + y2 + z2 = a2, and passing through two given points (x0, y0, z0) and (x1, y1, z1), find the one which has the least length.
Solution. The length of the curve
y = y(x), z = z(x) …(1)
is given by the integral
J[y, z] = dx'z'y11x0x
22� ++ …(2)
The curve (1) lies on the sphere
x2 + y2 + z2 = a2. …(3)
we form the auxiliary functional
J* = � ++1
0
x
x22 zy1 ''( + λ(x) (x2 + y2 + z2)]dx …(4)
The other boundary conditions are
���
====
,z)x(z,z)x(z
y)x(y,y)x(y
1100
1100 …(5)
The Euler’s equations, corresponding to function (4), are
2λ(x) y− 0'z'y1
'ydxd
22=
��
�
�
��
�
�
++, …(6)
2λ(x)y − 0'z'y1
'zdxd
22=
��
�
�
��
�
�
++. …(7)
CALCULUS OF VARIATIONS
47
Solving these equations (6) and (7), we obtain a family of curves depending on four constants, whose values are determined from the boundary conditions in (5).
Example 2 . Find the shortest distance between the points A(1, −1, 0) and B(2, 1, −1) lying on the surface 15x−7y +z −22 = 0.
Solution. In this question, we have to find the minimum of the functional
J[y, z] = ,dx'z'y121
22� ++ …(1)
subject to the conditions
���
−===−=
1)2(z,0)1(z1)2(y,1)1(y
…(2)
provided
g(x, y, z) ≡ 15x −7y +z − 22 = 0. …(3)
To achieve this end, we form an auxiliary functional
J*[y, z] = �2
1 F(x, y, z, y′, z′)dx …(4)
where
F = 22 'z'y1 ++ + λ(x) [15x−7y +z −22]. …(5)
The corresponding Euler’s equations are
0 + λ(x) {−7} − 0'z'y1
'ydxd
22=
��
�
��
�
�
++ …(6)
0 + λ(x). {1} − 0'z'y1
'zdxd
22=
��
�
��
�
�
++ …(7)
Combined together, we get
0'z'y1
'z'ydxd
22=
��
�
��
�
�
++
+
Integrating, we find
122c
'z'y1
'z7'y =++
+ …(8)
From (3), we write
z′ = 7y′ −15. …(9)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
48
From equation (8) and (9), and then integrating, we obtain (exercise)
y(x) = αx +β. …(10)
Using boundary conditions in (2), we find (exercise)
α = 2, β = −3, y(x) = 2x−3. …(11)
From equation (9) and (11), we have
z′ = −1
giving z(x) = c−x.
The B.C.′s in (2), give
z(x) = 1−x …(12)
Putting y(x) and z(x) from equations (11) and (12) into equation (6), we find
λ(x) ≡ 0. …(13)
The desired shortest distance is (exercise)
l = � =++21
22 6'z'y1
The Books Recommended for Chapter I
1. I.M. Gelfand Calculus of Variations, and S.V. Fovmin Prentice Hall.
TRANSPORT AND LAPLACE EQUATIONS
49
Chapter-2
Transport and Laplace Equations
2.1 INTRODUCTION Many physical problems in science, engineering and geometry can be modeled mathematically by partial differential equations (PDE). A partial differential equation is an equation involving an unknown function of two or more variables and certain of its partial derivatives.
Before writing symbolically a typical PDE, we first present the notation / symbol to be used consequently.
2.1.1 Geometric Notation
(i) Rn = n – dimensional real Euclidean space,
(ii) R1 = R = real line.
(iii) ei = ith standard coordinate vector
= (0, 0, ……, 0, 1, 0,…….0).
(iv) A typical point x in Rn is
x = (x1, x2,……, xn).
Sometimes, we will also regard x as a row or column vector.
(v) Rn+ = open upper half – space
= { x = (x1, x2,……,xn) ∈ Rn | xn > 0}.
(vi) R+ = { x ∈ R | x > 0}
(vii) U, V, W etc are usually open subsets of Rn.
(viii) ∂ U = boundary of U
(ix) U = closure of U
= U ∪ ∂U.
(xi) A typical point in Rn+1 will often be denoted as
(x, t) = (x1, x2,…., xn, t),
and we usually interpret
t = xn+1 = time.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
50
(xii) A point x ∈ Rn will sometimes be written as
x = (x1, xn)
for x1 = (x1, x2,…., xn-1) ∈ Rn-1.
(xiii) B(x, r) = closed ball in Rn with center at x and having radius r, r > 0.
= { y ∈ Rn | |y – x| ≤ r}.
(xiv) B0(x, r) = open ball in Rn with centre at x and radius r =
= { y ∈ Rn | |y – x| < r}
(xv) For a = (a1, a2,….., an) and b = (b1, b2,……., bn)
a . b =�=
n
i 1
ai bi
| a | = 21
1
2��
���
��
=
n
iia = 22
22
1 .......... naaa +++ ≡ Euclidean norm of a
(xvi) Cn = n – dimensional complex space,
(xvii) C1 = C = complex plane.
(xviii) α(n) = volume of unit ball B(0, 1) in Rn
=
��
��
�
���
��
�
��
���
� +Γ 12
2/
n
nπ ,
In particular for n = 3 ,
α(3) = π34
for r = 1
(xix) n α(n) = surface area of unit sphere B(0, 1) in Rn
= ∂B(0, 1).
2.1.2. Notation for Functions
If u : U → R is a real valued function with domain U ⊂ Rn, we write
u(x) = u(x1, x2,……, xn), for x ∈ U.
Definition: Function u is called smooth when u is infinitely differentiable
If u and v are two functions, then we write
TRANSPORT AND LAPLACE EQUATIONS
51
u ≡ v (read : u is identically equal to v)
when functions u and v agree for all values of their arguments.
(i) We write
u : = v
to define u as equaling v
(ii) u+ = max (u, 0) , u+ ≥ 0
(iii) u- = − min(u, 0) , u- ≥ 0
(iv) u = u+− u-,
(v) | u | = u++ u-.
(vi) The sign function is defined as
sgn (x) = ��
�
�
<−=>
.0100
01
xif
xif
xif
(vii) If u : U → Rm , U ⊂ Rn ,we write
u (x) = (u1(x) , u2(x) , .... , um(x)) for x ∈ U
Here ,uk is the kth component of u for k = 1,2,…., m. Further
uk : U → R.
(viii) The function
χE(x) = � �
∉∈
Exif
Exif
01
is called the Indicator Function of E.
(ix) A function u : U → R is called Lipschitz continuous if
| u(x) – u(y) | ≤ C | x – y | for al x , y ∈ U
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
52
and for some constant C. Here, on the left there is norm in R, and on the right, there is norm in Rn.
2.1.3. Notation for Derivatives
Let u : U → R, x ∈ U ⊂ Rn. We write
(i) ��
���
� −+→
=∂∂
hxuhexu
hx
xu i
i
)()(0
lim)( , provided this limit exists, h ∈ R.
(ii) We usually write ixu for
ixu
∂∂
.
(iii) ixu
jx = ji xx
u∂∂
∂2
, ixu
jx kx =kji xxx
u∂∂∂
∂3
, etc.
2.1.4. Multiindex Notation
(1) A vector / n-tuple of the form
α = (α1, α2,………., αn), αi is a non-negative integer for each i,
is called a multiindex. Its order is defined as
| α | = α1 + α2 + ……..+αn = �=
n
i 1
αi
Note : | α | ≥ 0 and | α | is a non – negative integer.
Also, we define
α! = α1! α2! ……. αn!
(2) For x ∈ Rn, we define
xα = x1 1α x2 2α ……. x3 nα
(3) We employ the symbol
Du
to denote the gradient vector of the function u.
(4) Given a multiindex α = (α1, α2,……, αn) , we define
TRANSPORT AND LAPLACE EQUATIONS
53
Dα u(x) = .................
)(||
uxu n
n2
21
1nn
22
11
xxxxxx
αααααα
α∂∂∂=
∂∂∂∂
=���
�
�
���
�
�
���
����
�
∂∂
α
=∏
i
i
n
1i xu
In particular, if α = 0, then Dα is the identity operator.
(5) If k is a non – negative integer, we define
Dk u(x) : = {Dk u(x) : |α| = k}. (*)
Thus, Dk u(x) is the set of all partial derivatives of order k. Assigning some ordering to the various partial derivatives in (*), we can also regard
Dk u(x) as a point in knR - space.
(6) We define
| Dk u | = 21
||
2|| ���
����
��
=k
uDα
α
(7) Special cases:
(a) When k = 1, Du is a point in Rn – space and we arrange the elements of Du in a vector of the form
Du = (u1x ,u
nxx u.......,2
) = gradient vector
In particular, for n = 3,
Du = ((u1x ,u
32, xx u )
(b) When k = 2, D2u can be regarded as an element of 2nR - space, and the
elements of D2u are being arranged in a matrix
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
54
D2u =
���������
�
�
���������
�
�
∂∂∂
∂∂∂
∂∂∂
−−−−−−−−−−−−−−−∂∂
∂∂∂
∂
∂∂∂
∂∂∂
∂∂
nnnn
n
xxu
xxu
xxu
xxu
xxu
xxu
xxu
xu
2
2
2
1
2
22
2
12
2
1
2
21
2
21
..........
...................
.............
This matrix is called the HESSIAN MATRIX.
For n = 2 (i.e., in two dimensional space),
u = u(x, y) and D2u =
222
22
2
2
2
����
�
�
����
�
�
∂∂
∂∂∂
∂∂∂
∂∂
yu
xyu
yxu
xu
(c) tr(D2 u) = �=
n
i 1ixu
ix
= Laplacian of u
= ∆ u.
(d) For a function of two variables u = u(x, y) ,
x = (x1, x2,….., xn) , and
y = (y1, y2, ….., yn) ,
Dx u = (1xu ,
2xu …..,nxu ) ,
Dy u = (1yu ,
2yu ….,nyu ) .
2.1.5 Vector – valued Functions
(i) Let U ⊂ Rn and m > 1. Let
u : U → Rm
be a vector – valued function and
u = (u1, u2,….., um).
We define
Dαu = (Dα u1, Dα u2, …..Dα um)
TRANSPORT AND LAPLACE EQUATIONS
55
for each multi-index α.
We note that Dα ui are defined earlier under the heading “Notation for
Derivatives”.
(ii) For a non – negative integer k, we define
Dku = { Dαu : |α| = k}
and
| Dku | = norm in m – dimensional space
= 21
||
2|| ���
����
��
=k
uDα
α
as defined earlier for scalar – valued functions.
2.1.6 Measures and Integrals
(i) The integral of a function f : U ⊆ Rn → R, over a subset U ⊆ Rn, with respect to Lebesgue measure is denoted by
�U
f (x) dx or simply �U
f .
Note: If no subscript occurs on the integral sign , the region of integration is understood to be Rn.
(ii) Let � be a smooth (n – 1) dimensional surface in Rn, we write
�� dsf
for the integral of f over � , with respect to (n – 1) – dimensional surface measure.
(iii) If C is a curve in Rn, we denote by
�C dtf ,
the integral of f over C w.r.t. arc length.
(iv) The convolution of the function f and g, denoted by
f * g,
is given by
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
56
(f * g) (x) = � f (x – y) g(y) dy
= � f(y) g(x – y) dy
= ( g * f) (x),
provided the integrals exists.
2.1.7 Function Spaces
(1) C(U) = { u| u : U →→→→ R is continuous }
C( U ) = { u : u ∈∈∈∈ C(u) and is uniformly continuous }
(2) Ck(U) = { u : U → R, U ⊆ Rn | u is k – times continuously differentiable}
Ck( U ) = { u ∈ Ck(U) | Dαu is uniformly continuous for all | α | ≤ k}
Thus, if u ∈ Ck( u ), then Dαu continuously extends to U for each multiindex α such that | α | ≤ k.
(3) C∞ (U) = { u : U → R | u is infinitely differentiable}
= Ι∞
=0k
Ck (U)
C∞( U ) = Ι∞
=0k
Ck( U ).
…(4)
Lp(U) = { u : U → R : u is Lebesgue measurable, || u ||)(ULp < ∞ }
where
|| u || Lp(U)=p1
p dxf ���
����
��∪
|| . 1 ≤ p < ∞.
2.1.8 Notation for Matrices
(1) A = (aij)
= a matrix A which is an m × n matrix with (i, j)th entry aij.
A = diag(d1, d2,…., dn)
= a diagonal matrix.
TRANSPORT AND LAPLACE EQUATIONS
57
(2) Mm×n = space of real m × n matrices
Sn×n = space of real symmetric n × n matrices
(3) tr A = trace of A = a11 + a22 +…..+ ann
= sum of diagonal elements
(4) det A = determinant of the matrix A
(5) cof A = cofactor matrix of A
= Transpose of (Adj A)
= (Adj A)T
AT = transpose of the matrix A
(6) If A = (aij), B = (bij) are m × n matrices, then
A : B = ��= =
m
i
n
j1 1
aij bij
| A | = norm of matrix A
= (A : A)1/2
=21
n
1i
n
1j
2ija �
�
���
���= =
)(
= [(a11)2 + (a12)2 +……+ (a1n)2 + (a21)2 + (a22)2 +….+ (a2n)2
+ ……..+ (ann)2 ]1/2
(7) If A = (aij) ∈ Sn×n and x = (x1, x2, …., xn) ∈ Rn, then
x . A x = ��= =
m
i
n
j1 1
aij xi xj
= �=
m
ji 1,
aij xi xj
= Quadratic Form corresponding to (aij) .
(8) Let A ∈ Sn×n. If
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
58
x . A x ≥ θ | x |2 for all x ∈ Rn, and some real number θ, then, we write A ≥ θ. I
(9) For A ∈ Mn×n, y ∈ Rn, we sometimes write
y A = AT y.
2.2 TRANSPORT EQUATION
The transport equation with constant coefficients is the PDE
Ut + b. D u = 0 in Rn × [0, ∞) , (1)
where
b = (b1, b2,…., bn) is a fixed vector in Rn,
and
u : R × [0, ∞] → R
is the unknown function, and
u = u(x, t).
Note. Here x = (x1, x2, ….., xn) ∈ Rn is a typical point in space, and t ≥ 0
denotes a typical time variable.
We write D u = Dx u = (
1xu ,2xu …..,
nxu ) (2)
for the gradient of the scalar function u with respect to the spatial variable x.
Initial – Value Problem
Let us consider the homogeneous linear initial – value problem
Ut + b . D u = 0 in Rn × [0, ∞) (1)
u = g on Rn × {0 = t} (2)
where g : Rn → R is known.
The problem is to compute u = u(x ,t).
Solution. Let (x, t) be any given (hence fixed) point in Rn × [0, ∞).
The line through (x, t) with direction (b, 1 ) is represented parametrically by
TRANSPORT AND LAPLACE EQUATIONS
59
�
+=+=stst
bsxsx
)()(
, s∈R (3)
i.e., by (x + s b, t + s) for s ∈ R.
This line hits the plane
Γ : Rn × {t = 0} (4)
at the point (x – t b, 0) , when
s = − t. (5)
Since u is constant on the line and
u(x – t b, 0) = g(x – t b), (6)
by virtue of given initial condition (2), we deduce that
u(x, t) = g(x – t b) (7)
for x ∈ Rn and t ≥ 0.
So, if the given initial – value problem has a sufficiently regular solution, u = u(x, t), it must certainly be given by (7) above. Conversely, if g is C1, then u = u(x, t) defined by (7) is indeed a solution of the given initial – value problem. Verification:
From, (7), we find
Ut = − b . D(ξ), where ξ = x – t b
D u = D(ξ)
Hence ut + b. D u = [ −b . D(ξ)] + b . [D(ξ)]
= 0 (8)
and, for t = 0,
u(x, 0) = g(x) on Rn (9)
This completes the result.
Remark : If g is not C1, then there is no C1 solution of the given initial – value problem. But even in this case, formula (7) certainly provides a strong, and in fact, the only reasonable candidate for a solution.
We may thus formulary declare
u(x, t) = g(x – t b), x ∈ Rn, t ≥ 0. (10)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
60
to be a weak solution of IVP, even should g not be C1. This all makes sense even if g, and thus u, are discontinuous.
Non homogeneous problem
Problem: Consider the non – homogeneous initial – value problem
)2(}0t{Ringu
)1(),0[RinfuD.bun
nt
=×=
∞×=+
in which b = (b1, b2, ….., bn) ∈ Rn is a fixed vector, and
u : R × [ 0, ∞) → R
is the unknown function, and
u = u(x, t),
x = (x1, x2,……, xn) ∈ Rn is a point in space,
t ≥ 0 denotes a typical time variable,
Du = Dxu = (1xu ,
2xu …..,nxu )
denote the gradient of u with respect to the spatial variable x,
g : Rn → R
is known, f : Rn × [0, ∞) → R
is known. The problem is to compute u = u(x . t).
Solution: Let (x, t) be any given, hence fixed, point in Rn × [0, ∞). Define a
function
),()(
:
stbsxusz
RRz
++=
→ (3)
for all s ∈ R. Then
z&(s) = b . D u(x + s b, t + s) + ut (x + s b, t + s)
= f(x + s b, t + s) (4)
using (1).
Now, using (2), (3) and (4), we find
u(x, t) – g(x – b t) = z(0) – u(x – b t, 0)
= z(0) – z(- t)
TRANSPORT AND LAPLACE EQUATIONS
61
= �−
0
t
z&(s) ds
= �−
0
t
f(x + s b, t + s) ds
= �t
0
f(x + (s – t) b, s) ds (5)
This gives
u(x, t) = g(x – b t) + �t
0
f(x + (s – t) b, s) ds (6)
for x ∈ Rn, t ≥ 0
as solution of the given non – homogeneous initial – value problem.
2.3 LAPLACE’S EQUATION
Problem: Laplace’s equation is
∆u = 0, (1)
and Poisson’s equation is
∆u = − f . (2)
In equation (2), the minus sign is taken so that the notation is consistent with notation for general second – order elliptic operators. In both equations (1) and (2),
x ∈ U ⊆ Rn, U is an open set
and the unknown function is
u : U → R , U = closure of U
u = u(x).
In equation (2),
f : U → R
is given. Further
∆u = Laplacian of u
=�=
n
i 1ii xxu .
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
62
Definition : A function u ∈ C2 is called harmonic function if u satisfies the
Laplace’s equation
∆u = 0.
Physical Interpretation
Laplace’s equation comes up in a wide variety of physical contexts – such as when u denotes the chemical concentration / temperature / electrostatic potential.
Laplace’s equation arises as well in the study of analytic functions.
Fundamental Solution of Laplace’s Equation
We attempt to find a solution of the given Laplace equation
∆u = 0 (1)
by searching radial solutions of the form
u(x) = v(r) (2)
where
r = | x |
= 21
222
21 )......( nxxx +++ , (3)
and v is to be selected, if possible, so that
∆ v = 0, (4)
holds.
First, we note that
21=
∂∂
ixr
( 21
222
21 ).....
−+++ nxxx (2xi)
= rxi , (x ≠ 0) (5)
Thus, we have
uxi = v′(r) ��
���
�
∂∂
ixr
= v′(r) ��
���
�rxi , (6)
TRANSPORT AND LAPLACE EQUATIONS
63
and
uii xx = v′′(r)
2
��
���
�
rxi + v′(r)
�
� �
− 3
21rx
ri (7)
for i = 1, 2,…, n. So,
∆u = �=
n
i 1
uxi xi
= v″(r) + ��
���
� −r
n 1v′(r). (8)
Hence
∆ u = 0
iff
v′′ + ��
���
� −r
n 1 v′ = 0. (9)
If v′ ≠ 0, we deduce
r
nvv −= 1
'"
or ( )r
nv
drd −= 1
'log .
or log v′(r) = (1 – n) log r + constt.
or v′(r) = 1−nra
, (10)
for some constant a . Consequently, if r > 0, we obtain
v(r) = ��
�
�
≥+
=+
− 3,
2,log
2 ncr
b
ncrb
n
(11)
where b and c are constants. Let
Φ(x) =
���
��
�
≥−
=−
− 3,||
1)()2(
1
2,||log21
2 nxnnn
nx
nα
π (12)
for x ∈ Rn, x ≠ 0. α(n) = volume of unit ball in Rn.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
64
Then Φ(x) is a solution of the given Laplace equation (1) and is called the Fundamental Solution of Laplace’s Equation.
Note: This fundamental solution is radial.
2.4 FUNDAMENTAL SOLUTION OF POISSON’S EQUATION
Let Φ(x) be the fundamental solution of Laplace’s equation
∆u = 0, (1) where
Φ(x) =
���
��
�
≥−
=−
− 3,||
1)()2(
1
2,||log21
2 nxnnn
nx
nα
π (2)
and x ∈ Rn, x ≠ 0.
So, the mapping
x → Φ(x) , x ≠ 0 , (3)
is harmonic.
If we shift the origin to a new point y, the PDE (1) is unchanged, so the
mapping
x → Φ(x – y) (4)
is also harmonic as a function of x, x ≠ y.
Now, we consider the Possion’s equation
∆u = − f, (5)
where
f : Rn → R. (6)
we note that the mapping
x → Φ(x – y) f(y), (7)
for x ≠ y, is harmonic for each point y ∈ Rn, and so is the sum of finitely many such expressions built / constructed for different points y. Consider convolution u(x) = �
nR
Φ(x – y) f(y) dy . (8)
TRANSPORT AND LAPLACE EQUATIONS
65
From equations (2) and (8), we write
u(x) =
��
�
��
�
≥−α−
=−π
−
�
�
− )(,||)(
)()(
)()(|)log(|
3ndyyxyf
n2nn1
2ndyyfyx21
n
n
R2n
R . (9)
For simplicity, we assume that the function f, given in Possion’s equation (5), is twice continuously differentiable with compact support. Now, we shall show that, u(x) defined by (9) satisfies (i) u ∈ C2(Rn)
(ii) ∆u = −f in Rn.
Consequently, the function in (9) provided us with a formula for a solution of Possion’s equation (5).
Proof of (i):
We have
u(x) = �nR
Φ(x – y) f(y) dy = �nR
Φ(y) f(x – y) dy (10)
Hence
hxuhexu i )()( −+
= �nR
Φ(y) dyh
yxfyhexf i��
���
� −−−+ )()( (11)
where h ≠ 0 is a real number and ei ∈ Rn,
ei = (0, 0, …,0, 1, 0, ….., 0)
with 1 in the ith slot.
But
h
yxfyhexf i )()( −−−+
→ ix
f∂∂
(x−y) (12)
uniformly on Rn as h → 0. Thus, on taking h → 0 in (11) and making use of
result in (12), we write
��
� �
−∂∂Φ=
∂∂
nR ii
dyyxxf
yxxu
,)()()( (13)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
66
for i = 1, 2, 3,…, n.
Similarly
� �
��
��
� �
−∂∂
∂Φ=∂∂
∂nR jiji
dyyxxxf
yxxxu
,)()()(22
(14)
for i, j = 1, 2, …, n.
As the expression on the right hand side of (14) is continuous in the variable x, we see that
u ∈ C2(Rn) (15)
This proves (i).
Proof of (ii) :
This function Φ(x), defined in (9), blows up at x = 0 , we will need for subsequent calculations to isolate this singularity (x = 0) inside a small ball. So, fix ∈ > 0. Let B(0, ∈) denote an open ball at x = 0 with radius ∈. Then, from equation (10), we obtains
∆u(x) = �∈
Φ),0(
)(B
y ∆x f(x – y) dy + �∈−
Φ),0(
)(BR n
y ∆x f(x – y) dy
= I∈ + J∈ , say, (16)
where
I∈ = �∈
Φ),0(
)(B
y ∆x f(x – y) dy, (17)
J∈ = �∈−
Φ),0(
)(BR n
y ∆x f(x – y) dy, (18)
Now
| I∈ | ≤ |)(|),0(�
∈
ΦB
y | ∆x f(x – y) | dy ,
≤ C || D2 f ||)( nRL∞ �
�
�
�
��
�
�Φ�
∈dyy
0B
|)(|),(
TRANSPORT AND LAPLACE EQUATIONS
67
≤��
� �
≥∈
=∈∈
)(
)(|log|
3nC
2nC2
2
(19)
Also, by integration by parts, we get
J∈ = �∈−
Φ),(
)(0BRn
y ∆y f(x – y) dy
= − �∈− ),0(BR n
D Φ(y) . Dy f(x – y) dy
+ �∈∂ ∂
∂Φ),0(
)(B v
fy (x – y) dS(y), using divergence then
= K∈ + L∈. (20)
v indicating the inward pointing unit normal along the boundary ∂B(0, ∈) of the ball B(0, ∈).
Further
| L∈ | ≤ || D f ||)( nRL∞ �
�
�
�
��
�
�Φ�
∈−
)()(),(
ydsy0BRn
≤ � �
≥∈=∈∈
3nC2nC ,|log|
. (21)
We continue by integration by parts once again in the term K∈, to obtain /
discover
k∈ = �∈−
∆Φ),(
)(0BRn
y f(x – y) dy
– �∈−
Φ),(
)(0BR n
y f(x – y) dS(y)
= �∈∂ ∂
Φ∂
),0(B v(y) f(x – y) dS(y), (22)
since the function Φ is harmonic away from the origin (x ≠ 0).
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
68
Now
����
�
�
∈−=−=
≠−=Φ
,||
,0,||)(
1)(
yyy
v
yyy
nnyD nα
(23)
on the boundary ∂ B(0, ∈). Consequently
v∂Φ∂
(y) = v . D Φ(y)
= 1)(1
−∈nnα, (24)
on the boundary ∂B(0, ∈).
Since n α(n) ∈n-1 is the surface area of the n – dimensional sphere ∂B(0, ∈), we
have
K∈ = − 1)(1
−∈nnα �∈∂ ),0(B
f(x – y) dS(y)
= − ),( ∈∂
�0B
f(y) dS(y)
→ − f(x), as ∈ → 0. (25)
Here, a slash through an integral denote an average value.
Combining now equations (16) – (25), and letting ∈ → 0, we find
∆ u(x) = − f(x), (26)
as asserted earlier.
Thus, u(x), given by (9), in a solution of (26). This completes the solutions of Poisson’s equation.
Remark: (i) We sometimes write
∆ Φ = − δ0
in Rn, δ0 denoting the Dirac measure on Rn giving unit mass to the point x = 0.
Adopting this notation, we formally compute
∆u(x) = �nR
[∆x Φ(x – y)] f(y) dy
TRANSPORT AND LAPLACE EQUATIONS
69
= − �nR
δx f(y) dy
= − f(x),
x ∈ Rn, in accordance with above theorem.
Remark (ii). The above theorem (Solving Poisson’s equation) is in fact valid under for less stringent smoothness requirements for f.
2.5 MEAN – VALUE FORMULAS FOR LAPLACE’S EQUATION
Let U ⊂ Rn be an open set. Let
u : U → R
be a harmonic function. We define
(i) average of f over the ball B(x, r)
= �),()(
1
rxBn dyf
rnα
= �∂ ),(
.rxB
dsf
where
α(n) = volume of unit ball B(0, 1) in Rn
= ,1
2
2/
��
���
� +Γ n
nπ
n α(n) = surface area of unit sphere ∂B(0, 1) in Rn
Note: For x ∈ U ⊂ Rn, r = | x | ,
we shall now derive the important mean – value formulas, which declare that
“u(x) equals both the average of u over the sphere ∂∂∂∂B(x, r) and the average of u over the entire ball B(x, r), provided
B(x, r) ⊂⊂⊂⊂ U”.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
70
Theorem (Mean – value formulas for Laplace’s equation)
Statement : If u ∈ C2 (U) is harmonic, then
u(x) = �∂ ),( rxB
u ds = �),( rxB
u dy,
for each ball B(x, r) ⊂ U.
Proof of Part – I
Set
φ(r) = �∂ ),( rxB
u(y) dS(y)
= �∂ )1,0(B
u(x + r z) dS(z). (1)
Then
φ′(r) = �∂ )1,0(B
z . Du(x + r z) dS(z), (2)
and consequently, using Green’s formula, we compute
φ′(r) = �∂ ),( rxB
��
���
� −r
xy. Du(y) d S(y)
= �∂ ),( rxB v
u∂∂
dS(y)
= nr�
),( rxB
∆u(y) dy
= 0. (3)
Hence φ is constant, and so
φ(r) = 0
lim→t
φ(t)
TRANSPORT AND LAPLACE EQUATIONS
71
= 0
lim→t �
��
�
���
��
∂ ),(
)()(txB
ydSyu
= u(x). (4)
Equations (1) and (4) prove the part – I, i.e. ,
u(x) = �∂ ),( rxB
u(y) dS(y) = average of u over the sphere ∂B(x, r). (5)
Proof of Part – II : We observe that by employing polar coordinates, one gets
� � � ��
�
���
�=
ξ∂),( ),(rxB
r
0 xB
ddsudyu
= u(x) �r
0
[n α(n) ξn-1] dξ
= α(n) rn u(x). (6) Hence
u(x) = �),()(
1
rxBn dyu
rnα
= �),( rxB
f dy
= average of u over the entire ball B(x, r). (7)
This complete the proof of both the mean – value formulas for Laplace’s equation.
Theorem (Converse of mean – value property for Laplace’s equation):
Statement: If u ∈ C2(U) satisfies the mean formula
u(x) = �∂ ),( rxB
u dS
for each ball B(x, r) ⊂ U, then prove that
u : U → R
is harmonic.
Proof: If possible assume that
∆U ≠ 0 in U ⊆ Rn (1)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
72
Then, there exists some open ball
B(x, r) ⊂ U (2)
such that
∆u > 0, within B(x, r) . (3)
Set
φ(r) = �∂ ),( rxB
u(y) dS(y) (4)
Then, as proved earlier (exercise)
φ′(r) = nr�
),( rxB
∆u(y) dy . (5)
Using (3) and (5), we get
φ′(r) > 0 . (6)
From the hypothesis and equation (4), it follows that
u(x) = φ(r) = constant (7)
This contradicts (6). Hence, the result follows. This completes the proof.
2.6. ENERGY METHODS
Definition (Energy functional) :
It is defined as
I[w] = � ��
���
� −U
2 dxfwwD21
|| (1)
where w belongs to the admissible set.
A = { w ∈ C2 (U )| w = g on ∂ U}. (2)
and
∆w = − f in U. (3)
Theorem (Dirichlet’s principle):
Statement: Assume u ∈ C2 ( U ) solves the boundary – value problem
��
�
∂=−=∆
UonguUinfu
(*)
TRANSPORT AND LAPLACE EQUATIONS
73
where U is open and bounded subset of Rn and its boundary ∂U is C1. Prove that
],[min][ wIuI
Aw∈=
(**)
where I[w] is the energy functional and w belongs to the admissible set
A = {w ∈ C2 (U )| w = g on ∂U} (***)
Conversely, if u ∈ A satisfies (**), then u solves the boundary value problem
(*).
Proof (Part – I) : Choose w ∈ A. Then
W = g on ∂ U (1)
Let u ∈ C2 (U) solves the BVP (*). Then
Uinfu −=∆ (2)
Uongu = (3)
Now
�U
( ∆u + f) (u – w) dx
= 0 (4)
by virtue of (2). This gives
�U
[ (∆u) (u – w) + f(u – w)] dx = 0
An integration by parts yields (using Green’s formula)
�U
[ D u. D(u – w)] dx
= − �U
( ∆u) (u – w) dx + �∂
��
���
�
∂∂
U vu
(u – w) dS
= �U
( f) (u – w) dx + 0,
using (1), (2) and (3).
This implies
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
74
�U
[ D u . D(u – w) – f(u – w)] dx = 0. (5)
as u = w = g on ∂ U, and hence there is no boundary term Equation (5) gives
�U
[ | D u |2 – u f] dx = �U
[ D u. D w – w f] dx. (6)
we know the estimates
| D u . D w | ≤ | D u | | D w |
≤ 21
| D w |2 + 21
| D w |2 (7)
by virtue of Cauchy – Schwarz and Cauchy inequalities. From (6) and (7), we write
�U
{ | D u |2 – u f } dx ≤ 21�U
| D u |2 + 21�[ | D w |2 – w f] dx (8)
By definition, the energy functional is given by
I[w] = � ��
���
� −U
2 fwDw21
|| dx (9)
Hence, relation (8) concludes
I[u] ≤ I[w], w ∈ A. (10)
Since u ∈ A, it follows that
I[u] = Aw∈
min I[w] . (11)
This proves part – I .
Proof of Part – II : Conversely, assume that the conclusion (**) of the
statement of the theorem holds.
Let v ∈ ∞cC (U) be any but fixed function. Let
λ(τ) = I[u + τ v], τ ∈ R (12)
where the energy function I is defined above in equation (9). Since u + τ v ∈ A
TRANSPORT AND LAPLACE EQUATIONS
75
for each τ, the scalar function λ(τ) has a minimum at zero, by virtue of
assumption in (**). So λ′(0) = 0, (13)
provided this derivative of λ(τ) at τ = 0 exists. But
λ(τ) = I[u + τ v]
= � ��
���
� τ+−τ+U
2 dxfvuDvDu21
)(||
= � ��
���
�τ+−τ+τ+
U
22
2 dxfvuDvDuDv2
Du21
)(.|||| (14)
Equation (13) and (14) give at once
�U
( D u . D v – v f) dx = 0 .
This gives
�U
( – ∆u – f ) v dx = 0 . (15)
This identity is valid for each function v ∈ ∞cC (U). So we must have
− ∆ u – f = 0 in U
or
∆ u = − f in U. (16)
This shows that u solves the given boundary – value problem. Hence, the proof of the converse of Dirichlet’s principle is complete.
This completes fully the Dirichlet’s principle.
Note (i) : In other words, the Dirichlet’s principle states that
If u ∈ A, then P D E
∆ u = − f in U
u = g on ∂ U
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
76
is equivalent to the statement that the solution function u = u(x, t) minimizes the associated energy functional
I[ . ].
Note (ii) : Dirichlet’s principle is an instance of the calculus of variations applied to Laplace Equation.
Theorem (Uniqueness theorem)
Statement : Prove that there exist at most one solution u ∈ C2(U ) of the boundary – value problem,
∆ u = − f in U
u = g on ∂ U
where U is bounded, open, and ∂ U is C1.
Proof: If possible assume that, in addition to u, there is another solution, sayu of the given boundary-value problem. Set w = u −u in U . (1)
Since u and u are solutions of the given boundary value problem, so
∆ u = − f in U (2)
u = g on ∂ U (3)
∆u = − f in U (4)
u = g on ∂ U (5) Now, in U,
∆ w = ∆ u − ∆u
= (− f) – (−f)
= 0 in U , (6)
and
w = 0 on ∂U . (7)
From Green’s formula, we write
�U
| D w |2 dx = �U
(D w . d w) dx
= − �U
w(∆ w) dx + �∪∂
��
���
�
∂∂
vw
w dS
TRANSPORT AND LAPLACE EQUATIONS
77
= 0 + 0
= 0, (8)
using (6) and (7). Equation (8) shows that
D w ≡ 0 in U (9)
Since w = 0 on the boundary ∂ U and w is constant in U, it follows that
w = 0 in U
or
u =u in U. (10)
This proves uniqueness theorem.
2.7 PROPERTIES OF HARMONIC FUNCTION
We now present a sequence of interesting deductions about harmonic functions, all based upon the mean – value formulas. Assume for the following that U ⊂ Rn is open and bounded.
Theorem: (Strong maximum principle).
Statement : Suppose u ∈ C2(U) ∩ C(U) is harmonic within U.
(i) Then uu
UU ∂= maxmax
(ii) Furthermore, if U is connected and there exists a point x0 ∈ U such that
u(x0) = uU∂
max ,
then u is constant within U.
Proof: Suppose there exists a point x0 ∈ U with
u(x0) = M = maxu u . (1)
Then for
0 < r < dist(x0, ∂U),
the mean – value property asserts
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
78
M = u(x0) = �),( 0 rxB
u dy
≤ M. (2)
As equality holds only
if u ≡ M within B(x0, r), (3) we see
u(y) = M (4)
for all y ∈ B(x, r). Hence the set
{x ∈ U | u(x) = M }
is both open and relatively closed in U, and thus equals U if U is connected. This proves assertion (ii), from which (i) follows.
Note : Assertion (i) is the maximum principle for Laplace’s equation and (ii) is the strong maximum principle. Replacing u by –u, we recover also similar assertions with “min” replacing “max”.
Remark (i) : The strong maximum principle asserts in particular that if U is connected and
u ∈ C2(U) ∩ C )(U
satisfies
� �
∂==∆
,00
Uonu
Uinu
where g ≥ 0, then u is positive everywhere in U if g is positive somewhere on ∂∂∂∂U.
Remark (ii) : An important application of the maximum principle is establishing the uniqueness of solutions to certain boundary – value problems for Poisson’s equation.
Theorem: (Uniqueness).
Statement : Let g ∈ C(∂U), f ∈ C(U). Then there exists at most one solution u ∈ C2(U) ∩ C )(U of the boundary – value problem
TRANSPORT AND LAPLACE EQUATIONS
79
� �
∂==∆−
.Uongu
Uinfu (1)
Proof: If u and (u ) both satisfy (1) , apply theorem above to the harmonic functions
w = ± (u − u ) .
Local Estimates for Harmonic Functions
Next we employ the mean – value formulas to derive careful estimates on the various partial derivatives of a harmonic function. The precise structure of these estimates will be needed below, when we prove analyticity.
Theorem (Estimates on derivatives)
Statement : Assume u is harmonic in U. Then
| Dα u(x0) | ≤ knk
rC
+ || u )),((
||rxBL 0
1 (1)
for each ball B(x0, r) ⊂ U and each multi-index α of order | α | = k.
Here
C0 = ,)(n
1α
)()(
nnk2
Ck1n
k α=
+ (k = 1,……) (2)
Proof 1: We establish (1) and (2) by induction on k . The case k = 0 being immediate from the mean – value formula. For k = 1, we note upon differentiating Laplace’s equation that
ixu (i = 1, …., n) is harmonic. Consequently
| ixu (x0) | = | � )2/,( 0 rxB ixu dx |
= irxBn
n
uvrn �∂ 2/,( 0)(
2|α
dS | (3)
≤ .||||2
))2
,0((r
xBLu
rn
∂∞
Now if x ∈ ∂B(x0, r/2), then B(x, r/2) ⊂ B(x0, r) ⊂ U, and so
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
80
|u(x)| ≤ )),(
||||)( rxBL
n
01u
r2
n1
��
���
�
α .
Combining the inequalities above, we deduce
| Dαu(x0) | ≤ )),((1
1
01||||
1)(
2rxBLn
n
urn
n+
+
α (4)
if | α | = 1. This verifies (1) and (2) for k = 1 .
2. Assume now k ≥ 2 and (1) and (2) is valid for all balls in U and each multiindex of order less than or equal to k – 1. Fix B(x0, r) ⊂ U and let α be a multiindex with | α | = k. Then Dα u = (Dβ u)xi for some i ∈ {1,…., n} , | β | = k – 1. By calculations similar to those in (3), we establish that (exercise)
| Dα u(x0) | ≤ .||||)),(( 0 k
rxBL
uDrkn
∂∞
β (5)
If x ∈ ∂B(x0, kr
), then
B(x, k
k 1−r) ⊂ B(x0, r) ⊂ U.
Thus (1) , (2) for k – 1 imply
| Dβu(x) | ≤ )),((1
11
01||||
1)(
))1(2(rxBLkn
kn
u
rk
kn
kn−+
−+
��
���
� −−
α. (6)
Combining the two previous estimates yields the bound
| Dαu(x0) | ≤ .||||)(
)2()),((
1
01 rxBLkn
kn
urnnk
+
+
α (7)
This confirms (1), (2) for | α | = k.
Liouville’s Theorem.
Next we see that there are no nontrivial bounded harmonic functions on all of Rn.
Theorem (Liouville’s Theorem)
TRANSPORT AND LAPLACE EQUATIONS
81
Statement : Suppose u : Rn → R is harmonic and bounded. Then u is constant.
Proof: Fix x0 ∈ Rn, r > 0, then
| Du(x0) | ≤ )),((1
1
01||||
rxBLn urC
+
≤ )(
||||)(
nL1 u
rnC
ℜ∞α
→ 0 ,
as r → ∞. Thus
Du ≡ 0,
and so
u is constant. This proves the Liouville’s Theorem.
2.8 GREEN’S FUNCTION
Assume now U ⊂ Rn is open, bounded, and ∂U is C1. We propose next to obtain a general representation formula for the solution of Poisson’s equation
−∆u = f in U, (*)
subject to the prescribed boundary condition
u = g on ∂U. (**)
Derivation of Green’s function.
Suppose first of all u ∈ C2(U) is an arbitrary function. Fix x ∈ U, choose ∈ > 0 so small that B(x, ∈) ⊂ U, and apply Green’s formula on the region V∈ = U – B(x, ∈) to u(y) and Φ(y – x). We thereby compute
�∈V
u(y) ∆Φ(y – x) - Φ(y – x) ∆u(y) dy
= �∈∂V
u(y) v∂Φ∂
(y – x) - Φ(y – x) vu
∂∂
(y) dS(y), (1)
v denoting the outer unit normal vector on ∂V∈. Recall next
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
82
∆Φ(x – y) = 0 for x ≠ y.
We observe also that
| � ∈∂ ),( xBΦ(y – x)
vu
∂∂
(y) dS(y) | ≤ C ∈n-1 ),0(
max∈∂B
| Φ | = 0(1)
as ∈ → 0. Furthermore
� ∈∂ ),( xBu(y)
v∂Φ∂
(y – x) dS(y) = � ∈∂ ),( xBu(y) dS(y) → u(x)
as ∈ → 0. Hence our sending ∈ → 0 in (1) yields the formula:
u(x) = �∂UΦ(y – x)
vu
∂∂
(y) – u(y) v∂Φ∂
(y – x) dS(y)
− �U Φ(y – x) ∆u(y) dy. (2)
This identity is valid for any point x ∈ U and any function u ∈ C2 )(U .
Now formula (2) would permit us to solve for u(x) if we knew the values of ∆u within U and the values of u, ∂u / ∂v along ∂U. However for our application to Poisson’s equation with prescribed boundary values for u, somehow modify (2) to remove this term.
The idea is now to introduce for fixed x a corrector function
φx = φx(y),
solving the boundary – value problem:
� �
∂−Φ==∆
.)(
0
Uonxy
Uinx
x
φφ
(3)
Let us apply Green’s formula once more, now to compute
− �U φx(y) ∆u(y) dy = �∂Uu(y)
v
x
∂∂φ
(y) - φx(y) vu
∂∂
(y) dS(y)
= �∂Uu(y)
v
x
∂∂φ
(y) - φ(y -x) vu
∂∂
(y) dS(y)
We introduce next this.
TRANSPORT AND LAPLACE EQUATIONS
83
Definition: Green’s function for the region U is
G(x, y) = Φ(y – x) − φx(y)
for x, y ∈ U, x ≠ y . Adopting this terminology and adding (2) to (4), we find
u(x) = − �∂Uu(y)
vG
∂∂
(x, y) dS(y) − �U G(x, y) ∆u(y) dy (x ∈ U), (5)
where
vG
∂∂
(x, y) = Dy G(x, y) . v(y) (6)
is the outer normal derivative of g with respect to the variable y. Observe that the term ∂u / ∂v does not appear in equation (5). We introduces the corrector φx precisely to achieve this.
Suppose now u ∈ C2(U ) solves the boundary – value problem
� �
∂==∆−
,Uongu
Uinfu (7)
for given continuous functions f, g. Plugging into (5), we obtain the following theorem.
Theorem: (Representation formula using Green’s function).
Statement : If u ∈ C2(U ) solves problem, then
u(x) = − �∂Ug(y)
vG
∂∂
(x, y) dS(y) + �U f(y) G(x, y) dy (x ∈ U). (8)
Here we have formula for the solution of the boundary – value problem (7), provided we can construct Green’s function G for the given domain U. This is in general a difficult matter, and can be done only when U has simple geometry. Subsequent subsections identify some special cases for which an explicit calculation of G is possible.
Remark: Fix x ∈ U. Then regarding G as a function of y, we may symbolically write
� �
∂==∆−
,0 UonG
UinG xδ
δx denoting the Dirac measure giving unit mass to the point x.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
84
Before moving on to specific examples, let us record the general assertion that G is symmetric in the variables x and y .
Theorem: (Symmetry of Green’s function)
Statement : For all x, y ∈ U, x ≠ y, we have
G(y, x) = G(x, y). (9)
Proof: Fix x, y ∈ U, x ≠ y. Write
v(z) = G(x, z),
w(z) = G(y, z), (10)
for z ∈ U . Then
∆v(z) = 0 (z ≠ x), (11)
∆w(z) = 0 (z ≠ y) (12)
and
w = v = 0 (13)
on ∂U. Thus our applying Green’s identity on
V = U – [B(x, ∈) ∪ B(y, ∈)] (14)
for sufficiently small ∈ > 0 yields
� ∈∂ ∂∂−
∂∂
),( xB vw
wvu
v dS(z) = � ∈∂ ∂∂−
∂∂
),(yB vu
vvw
w dS(z), (15)
v denoting the inward pointing unit vector field on ∂B(x, ∈) ∪ ∂B(y, ∈). Now w is smooth near x . So
| � ∈∂
−
∈∂∈≤
∂∂
),(
1
),(sup
|xB
n
xBCdSv
vw
| v |
= o(1) (16)
as ∈ → 0.
On the other hand, v(z) = Φ(z – x) - φx(z),
where φx is smooth in U. Thus
TRANSPORT AND LAPLACE EQUATIONS
85
�� ∈∂∈∂ ν∂Φ∂
∈→=
ν∂∂
∈→ ),(),(
limlimxBxB 0
dSwv
0(x – z) w(z) dS
= w(x) .
Thus the left – hand side of (15) converges to w(x) as ∈ → 0. Likewise the right hand side converges to v(y). Consequently
G(y, x) = w(x) = v(y) = G(x, y) . This completes the proof. The Books Recommended for Chapter II 1. L.C. Evans Partial Differential Equations, Graduate Studies
in Mathematics, Volume 19, AMS, 1998.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 86
Chapter-3 Heat and Wave Equations 3.1 INTRODUCTION
The heat equation is
ut −∆u = 0 (∗)
and the non-homogeneous heat equation is
ut − ∆u = f, (∗∗)
where t > 0 and x∈U, U⊂Rn is open. The unknown function
u = u(x, t) is
u : U [0, ∞) →R. (∗∗∗)
The Laplician ∆ is taken with respect to the spatial variables x = (x1, x2,…, xn),
and
∆u = ∆x u
= �=
n
1iixixu . (∗∗∗∗)
In equation (∗∗), the function
f : U×[0, ∞)→R
is given.
Remark (1). The heat equation is also known as the diffusion equation.
Remark (2). In typical applications, the heat equation describes the evolution in time of the density u of some quantity such as heat, chemical concentration, etc.
Remark (3). The heat equation appears as well in the study of BROWNIAN
MOTION.
3.2. FUNDAMENTAL SOLUTION OF HEAT EQUATION Article :- Derivation of the fundamental solution of the heat equation ut − ∆u = 0, in U ×[0, ∞)
HEAT AND WAVE EQUATIONS
87
where U⊂Rn is open.
Solution. We observe that the heat equation
ut −∆u = 0 in U ×[0, ∞) …(1)
involves one derivative w.r.t. the time variable t, but two derivatives w.r.t. the space variables x1, x2,…, xn . Consequently, we see that if u = u(x, t) solves (1), then so does u(λx, λ2t) for λ∈R. This scaling indicates that the ratio
,t
r 2
2n
21 xxxr ++== ...|| …(2)
is important for the heat equation. It also suggests that we seek a solution of heat equation (1) of the form
u = u(x, t) = v ���
����
�=��
�
����
�
t|x|
vt
r 22
, …(3)
for t > 0 and x∈Rn, for some function v as yet undetermined.
However, it is quicker to try a solution u having the special structure
u(x, t) = ��
���
��� tx
vt1
…(4)
for x∈Rn, t > 0. Here, α and β are constants and the function
v : Rn→R …(5)
must be found.
Inserting (4) into heat equation (1), and thereafter comprising, we obtain
αt−(α+1) v(y) + βt−(α+1)y. D v(y) + t−(α+2β) ∆v(y) = 0, …(6) where
y = t−βx = �tx
…(7)
Canceling t−(α+1) from equation (6), we find
α v(y) + β y. D(y) + t−(2β−1) ∆ v(y) = 0. …(8)
In order to transform (8) into an expression involving the variable y alone, we
take
β = 21
. …(9)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 88
Then, equation (8) reduces to
αv + +Dy.y21 ∆v = 0. …(10)
We simplify further by guessing v to be radial, i.e.,
v(y) = w(|y|), …(11)
for some
w : R→R. …(12)
Then (left as an exercise)
∆v = r
1n −w′, …(13)
and equation (10) now becomes
αw + ,0'wr
1n''w'wr
21 =−++ …(14)
for r = |y| and ′ =drd
.
Now, if we set
α = n/2, …(15)
then equation (14) simplifies to read
(rn−1 w′)′ + 21
(rn w)′ = 0. …(16)
On integration, one obtains
rn−1 w′ +21
rn w = constant = a. …(17)
We assume that w and w′ tend to zero as r→∞. Under these conditions, we
find
a = 0. …(18)
Hence, equations (17) and (18) imply
w′ = −21
r w. …(19)
Integrating (19), we obtain
w = b 42re− , …(20)
HEAT AND WAVE EQUATIONS
89
for some constant b. Combining equations (4), (9), (11), (15) and (20), we
conclude
u(x, t) = �
��
−t4
|x|exp
tb 2
2/n, …(21)
solves the heat equation (1).
We define Φ(x, t) =
���
���
�
<∈
>∈−
)0t,Rx(;0
)0t,Rx(;e)t�4(
1
n
nt4
2|x|
2/n …(22)
The function Φ(x, t) called the fundamental solution of the heat equation (1).
Remarks. φ is singular at the point (0, 0). (2) We will sometimes write
Φ(x, t) = Φ(|x|, t) …(23)
to emphasize that the fundamental solution is radial in the variable x.
Theorem. (Integral of fundamental solution of heat equation)
Statement. For each time t > 0,
�nR
Φ(x, t) dx = 1.
Proof. We find
�nR
Φ(x, t) dx = 2/n)t�4(
1�nR
exp[−|x|2/4t]dt
= 2/n�
1�nR
exp[−|z|2]dz
= 2/n�
1∏ �=
∞
∞−
n
1iexp[−|zi|2] dzi
= 1. This proves the theorem.
Article. Solve the initial value (or Cauchy) problem
ut −∆u = 0 in Rn×(0, ∞) …(1)
u =g on Rn×{t = 0} …(2)
associated with homogeneous heat equation.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 90
Solution. Let Φ(x, t) = ��
���
π
−t4
x
2n
2
et41
||
/)((x∈Rn, t>0) …(3)
be the fundamental solution of the heat equation (1). We note that the function
(x, t) → Φ(x, t) …(4)
solves the heat equation away from the singularity at (0, 0), and thus so does
(x, t) → Φ(x−y, t) for each fixed y∈Rn. …(5)
Consequently, consider the convolution
u(x, t) = �nR
Φ(x−y, t) g(y)dy
= t4yx
R2n
2
n
et41
||
/)(
−−
�πg(y)dy …(6)
for x∈Rn, t>0. (A) First, we shall show that u∈C∞ (Rn×(0, ∞)).
Since the function ��
��
�
−t4
2|x|
2/ne
t1
is infinitely differentiable, with
uniformly bounded derivatives of all order, on Rn×[δ, ∞) for each δ>0, we
see that
u∈⊂∞(Rn×(0, ∞)). …(7)
(B) Furthermore, from equation (6), we write
ut(x, t) −∆u(x, t) = �nR
[(Φt−∆xΦ)(x−y, t)]g(y) dy
= 0 ,
for all x∈Rn and t>0, since the fundamental solution Φ(x, t) itself solves the
heat equation. Thus,
ut(x, t) −∆u(x, t) = 0 …(8)
in Rn×(0, ∞).
(C) Let x0 ∈Rn be a fixed point. Let ∈>0 be given. Choose δ>0 such that
|g(y)−g(x0)| <∈ …(9)
HEAT AND WAVE EQUATIONS
91
whenever
|y−x0|<δ for y∈Rn. …(10)
We know that
�nR
Φ(x, t) dx = 1 …(11)
for each time t>0 .
Then, if
|x−x0|< δ/2, …(12)
we have, using equations (6) and (11),
|u(x, t) −g(x0)| = | �nR
Φ(x−y,t) {g(y)−g(x0)}dy|
≤ � Φ)�,0x(B
(x−y, t) |g(y)−g(x0) |dy
+ � − )�,0x(BnR Φ(x−y t) |g(y)−g(x0)|dy
= I + J, say …(13)
where I = � )�,0x(B Φ(x−y, t) |g(y)−g(x0)| dy, …(14)
J = � − )�,0x(BnR Φ(x−y, t) |g(y)−g(x0)|dy …(15)
Now, owing to inequality (9) and relation (11), we find
I ≤ ∈ � )�,0x(B (x−y, t) dy = ∈
This implies
I ≤ ∈. …(16)
Furthermore, if
|x−x0| ≤ δ/2 and |y−x0| ≥ δ, …(17) then |y−x0| ≤ |y−x| + |x−x0|
≤ |y−x| + δ/2
≤ |y−x| +21
|y−x0|
or |y−x| ≥ 21
|y−x0|. …(18)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 92
Consequently,
J ≤ � −∞ )�,0x(BnRL||g||2 Φ(x−y, t)dy
≤ dy|yx|t4
1exp
tC 2
)�,0x(BnR2/n �
��
−−� −
≤ dy|xy|t16
1exp
tC 20
)�,0x(BnR2/n �
��
−−� − , using (18)
= )drr(rt16
1exp
tC 1n2
�2/n−∞
�
��
−�
→ 0 …(19)
as t→0+.
Hence, if |x−x0| < δ/2 and t > 0 is small enough, then
|u(x, t) −g(x0)| < 2∈, …(20)
Using equations (13), (16) and (19). The relation (20) implies
)(),(lim
,),(),(
0
0tRx0xyx
xgtxun
0=
+→∈→
…(21)
for each point x0∈Rn.
Thus, we have shown that u(x, t), given by (6), is the solution of the initial-value problem constiting of equations (1) & (2). This complete the proof. 3.2 MEAN-VALUE FORMULA FOR THE HEAT
EQUATION Let U⊂Rn be open and bounded. We fix a time T > 0.
Definition. The parabolic cylinder is defined as
UT = U×(0, T],
and the parabolic boundary of UT is denoted by ΓT and is defined as
ΓT = )U( T −(UT).
Interpretation. We interpret UT as being the parabolic interior ofU×[0, T]. We must note that UT includes to top U×{t = T}. The parabolic boundary ΓT comprises the bottom and vertical sides of U×[0, T],
but not the top.
HEAT AND WAVE EQUATIONS
93
Definition (Heat ball)
For fixed x∈Rn, t∈R and r > 0, we define
E(x, t; r) = ���
��� ≥−−Φ≤∈ +
n1n
r1
)st,yx(andtsR)s,y( .
Note. E(x, t; r) is a region in space-time. Its boundary is a level set of fundamental solutions Φ(x−y, t−s) for the heat equation. The point (x, t) is at the center of the top. E(x, t; r) is called a heat ball.
Theorem. (A mean-value property for the heat equation)
Statement. Let u∈C12(UT) solve the heat equation
ut − ∆u = 0 in Rn×(0, ∞). …(1) Prove that
u(x, t) = dsdystyx
syur41
2
2
rtxEn )(
||),(
),,( −−
�� …(2)
for each heat ball E(x, t; r) ⊂ UT.
Proof. Formula (2) is a mean-value formula for heat equation. We find that the right hand side of (2) involves only u(y, s) for times s ≤ t. This is reasonable, as the value u(x, t) should not depend upon future times. We may assume upon translating the space and time coordinates that
x = 0, t = 0. …(3)
We write
E(r) = E(0, 0; r). …(4)
and set
φ(r) = dsdysy
syur1
2
2
rEn
||),(
)(�� …(5)
= dsdysy
srryu2
22
1E
||),(
)(�� …(5A)
We calculate
φ′(r) = dsdys|y|
ur2s
|y|uy
)1(E
2
s2
2n
1i iyi����
���
��
���
���
����
�+���
����
��=
= 1nr
1+ dsdy
s|y|
u2s
|y|uy
)r(E
2
s2
2
iyi����
���
��
���
���
����
�+���
����
�
= A + B, say …(6) We introduce the useful function
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 94
ψ = − rlogns4|y|
)s�4log(2n 2
++− …(7)
Then
ψ = 0, on ∂E(r), …(8)
since, Φ(y, −s) = r−n on ∂E(r), …(9)
be definition of heat ball. Now, we utilize (7) to write
B = {�� �=+
)r(E
n
1iiyis1n
dsdy�yu.4r
1
= − dsdyyu4un4r
1
rE
n
1iisys1n i�� ����
��� ψ+ψ
=+
)(
)( , …(10)
there is no boundary term, since ψ = 0 on the boundary ∂E(N), by virtue of (8). Integrating by part w.r.t. ‘s’, we discover
B = �����
���
�+−=+
)r(E
n
1isiiys1n�yu4�un4
r1
dy ds
= ����
���
��
���
� ���
����
�−−+−
=+)r(E
n
1i2
2
iiys1n s4|y|
s2n
yu4�un4r
1dy ds
= �� ����
��� −ψ−
=+
)(rE
n
1iiys1n
yusn2
un4r
1i
dy ds − A .
This implies
A + B = �����
���
�−∆−=+
)r(E
n
1iiiy1n
yusn2
�un4r
1dy ds,
Since u solves the heat equation. So
φ′(r) = � �
��
���
��� −−��
= +
n
1iiiyiyiy1n
dsdyyusn2
�un4r
1 = 0,
…(11)
by virtue of (7). Equation (11) gives
φ(r) = constant.
or φ(r) = 0t
lim→
φ(t)
HEAT AND WAVE EQUATIONS
95
= u(0, 0) ���
���
��→ )t(E 2
2
n0tdsdy
s|y|
t1
lim
= 4 u(0, 0), …(12) since,
.4dsdys
|y|dsdy
s|y|
t1
)1(E 2
2
)r(E 2
2
n=��=�� …(13)
From equation (4) and (12), we write
u(x, t) = 41 φ(r) …(14)
From equation (5) and (14), we have
u(x, t) = ��−−
)r;t,x(E 2
2
ndsdy
)st(|yx|
)s,y(ur41
…(15)
This completes the proof of mean-value formula for the heat equation.
3.4 ENERGY METHODS FOR HEAT EQUATIONS
Theorem. (Uniqueness theorem for heat equation)
Statement. Prove that there exists atmost one solution u∈C12(UT) of the
problem
ut −∆u = f in UT,
u = g on ΓT,
where U⊂ Rn is open and bounded, and ∂U is C1. The terminal time T >0 is
given.
Proof. Let u and u be two solutions of the above problem. Then
ut −∆u = f in UT, …(1)
u = g on ΓT, …(2)
u t − ∆ u = f in UT, …(3)
u = g on ΓT. …(4) Let w = u− u …(5)
Then equations (1) to (5) yield
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 96
wt − ∆w = (ut− tu ) − (∆u −∆ u )
= (ut −∆u) − ( tu −∆ u )
= f −f
= 0 in UT …(6)
Also w = u − u
= g − g
= 0 on ΓT. …(7)
Set e(t) = �U w2(x, t)dx, 0 ≤ t ≤ T. …(8)
Then
�=≡U
2)t(edtde & w wt dx
= 2 �U w ∆w dx, using (6)
= −2 �U |Dw|2 dx
≤ 0. …(9)
So e(t) is a decreasing function, and so
e(t) ≤ e(0) = 0, for 0 ≤ t ≤ T. …(10)
�U w2(x, t)dx = 0 for all 0 ≤ t ≤ T ,
w ≡ 0 in UT ,
u = u in UT .
Hence, the solution is unique. This completes the proof.
3.5 PROPERTIES OF SOLUTIONS
First we employ the mean-value property to give a quick proof of the strong maximum principle. Theorem. (Strong maximum principle for the heat equation). Statement : Assume u∈ )U(C)U(C TT
21 ∩ solves the heat equation in UT.
(i) Then .umaxumax
TTU Γ=
HEAT AND WAVE EQUATIONS
97
Rn
t
Strong maximum principle for the heat equation
(ii) Furthermore, if U is connected and there exists a point (x0, t0) ∈ UT such
that
u(x0, t0) = ,umaxTU
then u is constant in
0tU .
Assertion (i) is the maximum principle for the heat equation and (ii) is the strong maximum principle. Similar assertions are valid with “min” replacing “max”.
Remark. So if u attains its maximum (or minimum) at an interior point, then u is constant at all earlier times. This accords with our strong intuitive interpretation of the variable t as denoting time : the solution will be constant on the time interval [0, t0] provided the initial and boundary conditions are constant. However, the solution may change at times t > t0, provided the boundary conditions alter after t0. The solution will however not respond to changes in boundary conditions until these changes happen.
Take note that whereas all this is obvious on intuitive, physical grounds, such insights do not constitute a proof. The task is to deduce such behaviour from the PDE.
Proof. 1. Suppose there exists a point (x0, t0)∈ UT with
u(x0, t0) = M = TUmax u.
Then for all sufficiently small r > 0,
E(x0, t0; r) ⊂ UT ,
and we employ the mean-value property to deduce
(x0, t0)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 98
M = u(x0, t0)
= dyds)st(
|yx|)s,y(u
r41
20
20
)r;0t,0x(En −−
��
≤ M, since
1 = .dyds)st(
|yx|
r41
20
20
)r;0t,0x(En −−
��
Equality holds only if u is identically equal to M within E(x0, t0; r).
Consequently
u(y, s) = M for all (y, s) ∈ E(x0, t0; r).
Draw any line segment L in UT connecting (x0, t0) with some other point (y0, s0) ∈ UT, with s0 < t0. Consider r0 = min {s ≥ s0 | u(x, t) = M for all points (x, t) ∈ L, s ≤ t ≤ t0}.
Since u is continuous, the minimum is attained. Assume r0 > s0. Then
u(z0, r0) = M
for some point (z0, r0) on L ∩ UT and so
u ≡ M on E(z0, r0; r) for all sufficiently small r > 0.
Since E(z0, r0; r) contains L ∩ {r0 − σ ≤ t ≤ r0} for some small σ > 0, we have a
contradiction. Thus
r0 = s0,
and hence u ≡ M on L .
2. Now fix any point x∈U and any time 0 ≤ t < t0. There exists points {x0, x1,…,xm = x} such that the line segments in Rn connecting xi−1 to xi lie in U for i = 1,…,m. (This follows since the set of points in U which can be so connected to x0 by a polygonal path is nonempty, open and relatively closed in U.) Select times t0 > t1 >…> tm = t. Then the line segments in Rn+1 connecting (xi−1, ti−1) to (xi, ti) (i = 1,…, m) lie in UT. According to Step 1,
u ≡ M
HEAT AND WAVE EQUATIONS
99
on each such segment and so
u(x, t) = M.
This completes the proof.
Remark. The strong maximum principle implies that if U is connected and u∈ )U(C)U(C TT
21 ∩ satisfies
��
��
�
=×=×∂=
=∆−
}0t{Uongu]T,0[Uon0u
Uin0uu Tt
where g ≥ 0, then u is positive everywhere within UT if g is positive somewhere on U. This is another illustration of infinite propagation speed for disturbances. An important application of the maximum principle is the following
uniqueness assertion.
Theorem. (Uniqueness on bounded domains). Statement. Let g∈C(ΓT), f ∈ C(UT). Then there exists at most one solution u ∈ )U(C)U(C TT
21 ∩ of the initial/boundary-value problem
���
Γ==∆−
.T
Tt
onu
Uinfuu
g …(1)
Proof. If u and u~ are two solutions of (1), apply previous theorem to
w = + (u− u~ )
to get the result.
3.5 WAVE EQUATION The wave equation is
utt −∆u = 0 …(∗)
and the non-homogeneous wave equation is
utt − ∆u = f. …(∗∗)
Here t > 0 and x∈U, U⊂ Rn is open. The unknown function is
u = U ×[0, ∞)→R, …(∗∗∗)
u = u(x, t),
and the Laplacian ∆ is taken w.r.t. the spatial variables.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 100
x = (x1, x2,…, xn)
In equation (∗∗)
f : U×[0, ∞)→R …(∗∗∗∗) is given.
Generally, we use the abbreviation
� u = utt − ∆u. …(∗∗∗∗∗)
Remark. The wave equation is a simplified model for a vibrating
string (n = 1),
membrance (n = 2),
elastic solid (n = 3).
In each of the above, u(x, t) represents the displacement in some direction of
the point x at time t ≥ 0.
Solutions by Special Means
Article d′′′′. Alembert’s formula (for n = 1)
We consider the initial-value problem for the one-dimensional wave equation in all of R: utt − uxx = 0 in R×(0, ∞) …(1)
u = g , ut = h on R × {t = 0}, …(2)
where g, h are given functions.
We desire to derive a formula for u = u(x, t) in terms of known functions g and h. The two initial conditions in (2) imply that the displacement u(x, 0) and the velocity ut(x, 0) are known. The PDE (1) can be factored to write
��
���
�
∂∂−
∂∂
��
���
�
∂∂+
∂∂
xtxtu = 0. …(4)
Set
v(x, t) = ��
���
�
∂∂−
∂∂
xtu(x, t). …(5)
Then equations (4) says
ut(x, t) + vx(x, t) = 0 ; x∈R, t > 0. …(6)
Equation (6) is a homogeneous transport equation with constant coefficients
(b = 1). Let
HEAT AND WAVE EQUATIONS
101
v(x, 0) = a(x). …(7)
We know that the fundamental solution of the initial-value problem consisting of transport equation (6) and condition (7) is v(x, t) = a(x−t), x∈R, t ≥ 0. …(8)
Combining equations (5) and (8), we obtain
ut(x, t) − ux(x, t) = a(x−t) in R × (0, ∞) …(9)
Also u(x, 0) = g(x) in R, …(10)
By virtue of initial condition (2). Equations (9) and (10) constitute the non-homogeneous transport problem. Hence, its solution is u(x, t) = g(x + t) + �
t0 a (x + (s−t) (−1)−s) ds
= g (x + t) + �+−
txtx2
1a(y)dy. …(11)
The second initial condition in (2) imply
a(x) = v(x, 0)
= ut(x, 0) − ux(x, 0)
= h(x) − g′(x), x∈R. …(12)
Substituting (12) into equation (11), we obtain
u(x, t) = g(x + t) + �+−
txtx2
1[h(y) − g′(y)]dy
= 21
[g(x+t) + g(x−t)] + �+−
txtx2
1h(y)dy, …(13)
for x∈R, t ≥ 0.
This is the d’ Alembert’s formula. We have derived (13) assuming u is a (sufficiently smooth) solution of (1). Application of D’ Alembert’s Formula
Initial/boundary-value problem on the half-line. R+ = {x > 0}
Example. Consider the problem utt − uxx = 0 in R+ ×(0, ∞)
u = g, ut = h on R+ ×{t = 0} …(1)
u = 0 on {x = 0} × (0, ∞),
where g, h are given, with
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 102
g(0) = 0, h(0) = 0. …(2)
Solution. We convert the given problems on the half-line into the problem on whole of R. We do so by extending the functions u, g, h to all of R by odd reflection method as below we set
���
≥≤−−≥≥
=,,),(
,),(),(~
0t0xfortxu
0t0xfortxutxu …(3)
���
≤−−≥
=,0xfor)x(g
0xfor)x(g)x(g~ …(4)
���
≤−−≥
=.0xfor)x(h
0xfor)x(h)x(h
~ …(5)
Now, problem (1) becomes
��
=×==
∞×=
}0t{Ronh~
u~,g~u~),0(Rinu~u~
t
xxtt …(6)
Hence, d’ Alembert’s formula for one-dimensional problem (6) implies
�+−++= +−
txtx )y(h
~21
)]tx(g~)tx(g~[21
)t,x(u~ dy . …(7)
Recalling the definitions of h~
,g~,u~ in equations (3)−(5), we can transform equation (7) to read for x ≥ 0, t ≥ 0. u(x, t)
=
���
���
�
≤≤+−−+
≥≥+−++
�
�
+
+−
+
−
tx
tx
tx
tx
tx0fordyyh21
xtgtxg21
0txfordyyh21
txgtxg21
;)()]()([
;)()]()([ …(8)
Formula (8) is the solution of the given problem on the half-line R+ = {x > 0}.
Remark. If h ≡ 0 (9) in R+ ×{t = 0}, then the solution of the corresponding problem, as given by (8), is
u(x, t) =
���
���
�
≤≤−−+
≥≥−++
.ttxfor)];xt(g)tx(g[21
,0txfor)];tx(g)tx(g[21
…(10)
HEAT AND WAVE EQUATIONS
103
The formula (10) shows that the initial displacement, u(x, 0) = g(x), splits into two parts − one moving to the right with speed one (c = 1) and the other to the left with speed one. The latter part reflects off the point x = 0, where the vibrating string is held
fixed.
Article. Derive Kirchloff’s formula for the solution of three-dimensional (n = 3) initial-value problem utt − ∆u = 0 in R3× (0, ∞) …(1)
u = g on R3×{t = 0} …(2)
ut = h on R3×{t = 0}. …(3)
Solution. Suppose u ∈ C2 (R3 × [0, ∞)) solves the above initial-value problem.
We know that
U(x; r, t) = ),( rxB∂
� u(y, t) dS(y) …(4)
Defines the average of u(⋅, t) over the sphere ∂B(x, r). Similarly,
G(x; r) = ),( rxB∂
� g(y)dS(y) …(5)
H(x; r) = ),( rxB∂
� h(y) d S(y). …(6)
For fixed x, we hereafter regard U as a function of r and t only. Next, set
,UrU~ = …(7)
.HrH~
,GrG~ == …(8)
we now assert that U~
solve
��
�
��
�
�
∞×==
=×=
=×=
∞×=−
+
+
+
),(}{~
}{~~
}{~~
),(~~
00ron0U
0tRonHU
0tRonGU
0Rin0UU rrtt
…(9)
We note that the transformation in (7) and (8) converts the three-dimensional wave equation into the one-dimensional wave equation. From equation (7), we find
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 104
tttt UrU~ =
= r ∆U
= r ,Ur2
U rrr �
��
+ Laplacian for n = 3
= r Urr + 2Ur
= (U + r Ur)r
= ( rr )U~
= .U~ rr …(10)
The problem (9) is one the half-line R+ = {r ≥ 0}. The d’ Alembert’s formula for the same, for 0 ≤ r ≤ t, is
�+−−+= ++−trtr .dy)y(H
~21
)]tr(G~
)tr(G~
[21
)t,r;x(U~
…(11)
From (4), we find
u(x, t) = +→0r
lim U(x; r, t), …(12)
Equations (7), (8), (11) and (12) imply that
u(x, t) = +→0r
lim �
��
r)t,r;x(U
~
= +→0r
lim �
��
�+−−+ +
−rtrt dy)y(H~
r21
r2)rt(G
~)rt(G
~
= ).t(H~)t('G~ + …(13)
Owing then to (5) and (6), we deduce from (13)
u(x, t) = { }�∂∂+
���
���
�∂∂
),(),()()()()(
txBtxBySdyhtydSygt
t …(14)
But
� �∂ ∂+=
),( ),().()()()(
txB 10BzSdtzxgySdyg …(15)
Hence
)()}.({)()(),(),(
zSzdtzxDgySdygt 10BtxB
+�=���
���
�∂∂
∂∂
HEAT AND WAVE EQUATIONS
105
= ��
���
� −�
∂ txy
yDgtxB
)}.({),(
d S(y). …(16)
Now, equation (14) and (16) conclude
u(x, t) = ),( txB∂
� [g(y) + {Dg(y)}. (y−x) + t h(y)] d S(y) …(17)
for x∈R3, t > 0.
The formula (17) is called KIRCHHOFF’s formula for the solution of the initial-value problem (1)−(3), in 3D. Nonhomogeneous Problem
We next investigate the initial-value problem for the nonhomogeneous wave
equation
��
���
=×==
∞×=∆−
}.0t{Ron0u,0u
),0(Rinfuun
t
ntt …(1)
Motivated by Duhamel’s principle, we define u = u(x, t; s) to be the solution of
��
���
=×⋅=⋅=⋅
∞×=⋅∆−⋅
}.st{Ron)s;(f)s;(u,0)s;(u
),s(Rin0)s;(u)s;(un
t
ntt …(2)
Now set u(x, t) : = �
t0 u(x, t; s)ds (x∈Rn, t ≥ 0). …(3)
Duhamel’s principle asserts this is solution of
��
���
=×==
∞×=∆−
}.0t{Ron0u,0u
),0(Rinfuun
t
ntt …(4)
Theorem. (Solution of nonhomogeneous wave equation). Statement. Assume n ≥ 2 and f ∈ C[n/2]+1 (Rn ×[0, ∞)). Define u
by (3). Then
(i) u ∈ C2 (Rn ×[0, ∞)),
(ii) utt − ∆u = f in Rn × (0, ∞),
and
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 106
(iii)
0t,nRx)0,0x()t,x(
lim
>∈→
u(x, t) = 0,
0t,nRx)0,0x()t,x(
lim
>∈→
ut(x, t) = 0 for each point x0 ∈ Rn.
Proof. 1. If n is odd, 2
1n1
2n +=+�
��
. Also u(⋅,⋅;s) ∈ C2(Rn ×[s, ∞)) for each
s ≥ 0, and so u∈C2 (Rn × [0, ∞)). If n is even, 2
2n1
2n +=+�
��
. Hence
u ∈ C2 (Rn ×[0, ∞)). 2. We then compute :
ut(x, t) = u(x, t; t) + �t0 ut(x, t; s)ds
= �t0 ut(x, t; s)ds,
utt(x, t) = ut(x, t; t) + �t0 utt(x, t; s)ds
= f(x, t) + �t0 utt(x, t; s)ds.
Furthermore
∆u(x, t) = �t0 ∆u(x, t; s)ds
= �t0 utt(x, t; s)ds.
Thus utt(x, t) − ∆u(x, t) = f(x, t) (x ∈ Rn, t > 0),
and clearly
u(x, 0) = ut(x, 0) = 0 for x ∈Rn.
Examples. (i) Let us work out explicitly how to solve (4) for n = 1. In this case d’ Alembert’s formula gives
u(x, t; s) = ,),( dysyf21 stx
stx�−+
+−
dydssyf21
txustx
stx
t
0),(),( ��
−+
+−=
That is,
u(x, t) = ��+−
sxsx
t02
1f(y, t−s)dy ds (x∈R, t ≥ 0). …(5)
(ii) For n = 3, Kirchhoff’s formula implies
HEAT AND WAVE EQUATIONS
107
u(x, t; s) = (t −s) ),( stxB −∂
� f(y, s) dS;
so that
u(x, t) = � ��
���
��−
−∂
t
0 stxBdsdSsyfst ),()(
),(
= dSds)st()s,y(f
�41
)st,x(Bt0 −�� −∂
= .dSdrr
)rt,y(f�4
1)r,x(B
t0
−�� ∂
Therefore
u(x, t) = � −−−
)t,x(B dy|xy|
|)xy|t,y(f�4
1 (x ∈ R3, t ≥ 0) …(6)
solves (4) for n = 3.
The integrand on the right is called a retarded potential.
3.6 ENERGY METHODS There is the necessity of making more and more smoothness assumptions upon the data g and h to ensure the existence of a C2 solution of the wave equation for larger and larger n. This suggests that perhaps some other way of measuring the size and smoothness of functions may be more appropriate. Indeed we will see in this section that the wave equation is nicely behaved (for all n) with respect to certain integral “energy” norms. Uniqueness
Let U ⊂ Rn be a bounded, open set with a smooth boundary ∂U, and as usual set UT = U × (0, T], ΓT = TU −UT, where T > 0. We are interested in the initial/boundary-value problem
��
��
�
=×=Γ=
=∆−
}.0t{Uonhu
ongu
Uinfuu
t
T
Ttt
…(1)
Theorem. (Uniqueness for wave equation).
Statement. There exists at most one function u∈C2 )U( T solving (1).
Proof. If u~ is another such solution, then w : = u − u~ solves
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 108
Cone of dependence
��
��
�
=×=Γ=
=∆−
}.0t{Uon0w
on0w
Uin0ww
t
T
Ttt
Define the “energy”
e(t) = �U21 2
tw (x, t) + |Dw(x, t)|2dx (0 ≤ t ≤ T).
We compute
��
���
� =⋅⋅+�=dtd
dxDwDwww)t(e ttttU&
= �U wt(wtt − ∆w)dx = 0.
There is no boundary term since w = 0, and hence wt = 0, on ∂U × [0, T]. Thus for all 0 ≤ t ≤ T, e(t) = e(0) = 0, and so wt, Dw ≡ 0 within UT. Since w ≡ 0 on U × {t = 0}, we conclude w = u − u~ ≡ 0 in UT. Domain of dependence
As another illustration of energy methods, let us examine again the domain of dependence of solutions to the wave equation in all of space. For this, suppose u ∈ C2 solves utt − ∆u = 0 in Rn ×(0, ∞).
Fix x0 ∈ Rn, t0 > 0 and consider the cone
C = {(x, t) |0 ≤ t ≤ t0, |x−x0| ≤ t0 − t}. The Books Recommended for Chapter III 1. L.C. Evans Partial Differential Equations, Graduate Studies
in Mathematics, Volume 19, AMS, 1998.
(x0, t0)
B(x0, t0−t)
ANALYTICAL MECHANICS – I 109
Chapter-4
Analytical Mechanics – I
4.1 INTRODUCTION In the literature on mechanics there is no single generally accepted interpretation of term “analytical mechanics”. Some writers identify analytical mechanics with theoretical mechanics. Some authors maintain that an exposition in generalized coordinates constitutes the distinguishing feature of analytical mechanics. According to Gantmacher, analytical mechanics is characterized both by a specific system of presentation and also by a definite range of problems investigated. In analytical mechanics, general principles (differential or integral) serve as the foundation and then the basic differential equations of motion are derived from these principles analytically. 4.2. FREE AND CONSTRAINED SYSTEM The motion is studied of a system of particles Pk , (k = 1, 2,…,N), relative to some inertial (Galilean) system of coordinates. There are some restrictions on the positions and velocities of the particles of the system. These restrictions may be of a geometrical or kinematical nature. Such restrictions are called constraints. Systems with such constrains are termed as constrained systems. If there are no constraint in the system, then the system is called a free system.
Let t denotes time, krρ
, (k = 1, 2,…, N), be the radius vectors taken from a
single pole (that is stationary in the given system of coordinates) for all system of particles Pk and kk vr
ρ&ρ = (k = 1, 2,…,N) (1)
denote velocities of all points Pk of the system. Here dot (⋅) represents differentiations with respect to time t . Analytically, a constraint is expressed by the equation f(t, kk r,r &ρρ
) = 0. (2)
In the general case, constraint (2) is called differential or kinematical . In (2), f(t, )r,r kk
&ρ is an abridged notation for the function
f(t, ).r,...,r,r,rr,r N21N,...,21&ρ&ρ&ρρρρ
Such abbreviated notation will be used throughout the chapters on analytical mechanics.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 110
If the velocities kr&ρ
do not enter into the constraint equation (2), the constraint is termed finite or geometric. Analytically, it is written as, f(t, )rk
ρ= 0 (3)
Given a finite constraint of type (3), a system cannot occupy an arbitrary position in space at every given instant of time. Finite constrains impose restrictions to possible positions of the system at time t. But with a differential constraint alone, the system may occupy any arbitrary position in space at any time t. However, in this position the velocities of the particles of the system cannot any longer be arbitrary, since the differential constraint imposes restrictions on these velocities. From now on, we shall confine our consideration solely to such differential constraints whose equations contain the velocities of the particles in linear form :- Drrrrl NN332211 +++++ &ρρ&ρρ&ρρ&ρρ
....... lll = 0 or
�=
N
1kkk r&
ρρ.l + D = 0 (4)
where kk r&ρρ
.l is the scalar product of the vectors kk rand &ρρl and the vectors kl
ρ and
the scalar D are specified functions of time t and of all �rρ
(µ = 1, 2,…, N). It is
assumed here that the vectors klρ
cannot all vanish at the some time. Each finite constraint of type (3) implies, as a consequence, a differential constraint whose equation is obtained by termwise differentiation of equation (3) :
0tf
r.rfN
1kk
k
=∂∂+� ��
�
����
�
∂∂
=
&ρρ , (5)
where kzjyixr kkkk ++=ρ
, (6)
and k,j,i are mutually orthogonal unit vectors of the co-ordinate axes. Then
kzf
jyf
ixf
rf
kkkk ∂∂+
∂∂+
∂∂=
∂∂ρ , (k = 1, 2,…, N) (7)
or
krfρ∂
∂= gradk f , (k = 1, 2…N) (8)
But differential constraint (5) is not equivalent to the finite constraint (3). It is equivalent to the finite constraint f(t, kr
ϖ) = c, (9)
ANALYTICAL MECHANICS – I 111
where c is an arbitrary constant. For this reason, the finite constraint (9) is
called integrable.
In rectangular Cartesian co-ordinates, we write
kzjyixr kkkkˆˆˆ ++=ρ
, (10)
kCjBiA kkkkˆˆˆ ++=l
ρ, (11)
and kzjyixr kkkk &&&&ρ ++= , (12)
where Ak, Bk and Ck , (k = 1, …N), are scalar functions of t, x1, y1, z1,…, xN, yN, zN. Then the above constraint equations are now written as : f(t, xk, yk, zk, kkk z,y,x &&& ) = 0 (13)
f(t, xk, yk, zk) = 0 (14)
� ++=
N
1kkkkkkk )zCyBxA( &&& + D = 0 (15)
� =∂∂+��
�
����
�
∂∂+
∂∂+
∂∂
=
N
1kk
kk
kk
k
0tf
zzf
yyf
xxf &&& (16)
4.3 CLASSIFICATION OF CONSTRAINTS If t is not expressed explicitly in the constraint equation, i.e.,
,0tf =
∂∂
(17)
then the constraint is termed as stationary.
Note (1) : If the differential constraint (5) is stationary, then differential equation (5) is linear and homogeneous in the velocities.
Note (2) : By analogy, the differential constraint (4) or (15) is termed
stationary if
D = 0 (18)
and vectors klρ
in equation (3) and, respectively, the coefficients Ak, Bk, and Ck in equation (15) are not explicit function of time t.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 112
Illustration 1. A particle is constrained to move over a surface. Let the equation of this surface be given in the form
f )r(
ρ= 0 (19)
or f(x, y, z) = 0 (20)
This is a finite stationary constraint.
If the surface is moving or undergoing deformation, then the time t enters into the equation of the surface explicitly and its equation is of the form f(t, r
ρ) = 0 (21)
or f(t, x, y, z) = 0 (22)
In this case, the constraint is finite but non-stationary.
System of particles Definition 1. A system of particles is called holonomic if the particles of the system are not subjected to differential nonintegrable constraints. Thus, a holomic system is any free system of particles and also any constrained system with finite or differential but integrable constraints. All constraints in a holonomic system may be written in closed form. Definition 2. A system of particles is called nonholonomic if there are differentiable integrable constraints.
Nonintegrable differential constraints are themselves frequently called nonholonomic. Sometimes, integrable differential constraints are termed seminomic if only stationary constraints are imposed. Otherwise, it is called sclernomic.
Illustration 2 : Two particles are connected by a rod of constraint length l. Then the constraint equation is of the form
( 2
21 )rrρρ
− − l2 = 0 (23) or (x1−x2)2 + (y1−y2)2 + (z1−z1)2 − l2 = 0 (24)
Here 21 randrρρ
are the position vectors of the end points of the given rod. This is a holonomic scleronomic system. Note : We note that a rigid body may be regarded as a system of particles equidistant from one another, that is, subjected to constraints of type (23).
ANALYTICAL MECHANICS – I 113
Thus a free rigid body is a special case of a constrained holonomic scleronomic system of particles.
Illustration 3. Two particles are connected by a rod of variable length l = f(t). The constraint equation for this is
( 2
21 )rrρρ
− − f2(t) = 0 , (25) or (x1−x2)2 + (y1−y2)2 + (z1−z1)2 − f2(t) = 0 . (26)
This system is a holonomic sheonomic system.
Illustration 4 . Two particles in a plane are connected by a rod of constant length l and are constrained to move in such a manner that the velocity of the middle of the rod is in the direction of the rod. The constraint equations for this are z1 = 0, z2 = 0, (27)
(x1−x2)2 + (y1−y2)2 − l2 = 0, (28)
21
21
21
21
yyyy
xxxx
−+
=−+ &&&&
, (29)
since, the velocity of the centre of the rod is ��
���
� ++2
yy,
2xx 2121 &&&&
and direction
ratios of the rod are < >++2
yy,
2xx 2121 . This system is a nonholonomic
system because equation (29) defines a nonintegrable differential constraint. Unilateral Constraints : The constraints discussed earlier are called bilateral constraints. The constraints of the form f(t, kk r,r &ρρ
) ≥ 0 (30)
are called unilateral constraints. If in condition (30), we have an equal sign, it is said that the constraint is taut. Illustration 5. Consider two particles connected by a thread of length l. Then the constraint equation is expressed by the inequality l2 − 2
21 )rr(ρρ
− ≥ 0, (31)
and is unilateral constraint.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 114
Remark. The motion of a system of particles on which a unilateral constraint is imposed may be divided into portions so that in certain portions the constraint is taut and the motion occurs as if the constraint were bilateral, and in other portions the constraint is not taut and the motion occurs as if there were no such constraint.
In other words, in certain portions a unilateral constraint is either replaced by a bilateral constraint or is eliminated altogether.
So we shall hence forth consider only bilateral constraints. 4.4. POSSIBLE AND VIRTUAL DISPLACEMENT
On a material system, let us impose the following d finite constraints.
f1(t, 0rrr N21 =),.......,,ρρρ
f2(t, 0rrr N21 =),,.........,ρρρ
f3(t, 0rrr N21 =).,,.........,ρρρ
Μ
fd(t, 0rrr N21 =)..,..........,ρρρ
(1)
or fα(t, kr
ρ) = 0, (α = 1, 2,…d) (2)
and following g differential constraints :
���
����
��
=β
N
1kkk v
ρρ&l + Dβ = 0 , (3)
or
�����
=+⋅++⋅+⋅
=+⋅++⋅+⋅
0Dvvv
0Dvvv
gNgN22g11g
1NN1212111
ρρρρρρ
ρρρρρρ
lll
lll
...
............................................................
...........................................................
...
. (4)
We replace the finite constraints (2) by the differential constraints by
differentiating them. We have
0t
fV.
rf �
N
1kk
k
� =∂
∂+���
����
��
∂∂
=
ρρ , (α = 1, 2,…d) (5)
ANALYTICAL MECHANICS – I 115
Definition (Possible velocities). The system of vectors kvρ
are called possible velocities for a certain instant of time t and for a certain possible (at that instant) position of the system if these vectors kv
ρ satisfy the above “d + g”
linear equations in (3) and (5).
Thus, possible velocities are velocities permitted by the constraints.
For every possible position of the system at time t there exists an infinity of systems of possible velocities. One of these system of velocities is realized in the actual motion of the system at time t.
Definition (Possible displacements). Consider the system of infinitely small displacements
dtVrd
dtVrd
dtVrd
NN
22
11
ρρΜ
ρρ
ρρ
=
=
=
(6)
or dtVrd kk
ρρ= (k = 1, 2,…N), (7)
where kvϖ (k = 1, 2,….N) are the possible velocities. The infinitesimal displacements krd
ρ are called possible infinitesimal displacements or simply
possible displacements.
Remark : Multiplying equation (3) and (5) termwise by dt, and then using
equation (7), we find
���
����
��
=β k
N
1kk rd
ρρ&l Dβ dt = 0 , (8)
for β = 1, 2,…, g , and
0dtt
frd.
rf �
k
N
1k k
� =∂
∂+���
����
��
∂∂
=
ρϖ , (9)
for α = 1, 2,…,d. Equation (8) and (9) determine the possible displacements.
Virtual Displacements
Let us take two systems of possible displacements at one and the same instant of time and for one and the same position of the system : dtvdrd kk
ρρ= ,
and
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 116
dtvr'd kk &ϖϖ = , (k = 1, 2,…N), (10)
Here, both possible displacements, kk r'anddrdρϖ
satisfy the equations (8) and (9).
Therefore,
0dtt
frd.
rf �
k
N
1k k
� =∂
∂+���
����
��
∂∂
=
ρϖ (11)
0dtt
fr'd.
rf �
k
N
1k k
� =∂
∂+���
����
��
∂∂
=
ρϖ (12)
and
���
����
��
=β k
N
1kk rd
ρρ&.l + Dβ dt = 0 (13)
���
����
��
=β k
N
1kk rd
ρρ& '.l + Dβ dt = 0 (14)
Subtract equation (11) from (12) and (13) from (14) we have
,0)rdr'd(.rf
kk
N
1k k
� =−�∂∂
=
ρρρ (15)
and
0rdrd kk
N
1kk =−�
=β )'(
ρρρl (16)
for β = 1, 2,…, g. We denote
δ ,rdr'dr kkkρρρ
−= (17)
k = 1, 2,…, N.
Then equations (15) and (16) become
0r�..rf
k
N
1k k
� =���
����
��
∂∂
=
ρϖ (18)
0rkk
N
1k=��
�
����
� δβ=�
ρρ..l , (19)
for β = 1, 2,…, g. The displacements δ kr
ϖsatisfy the homogeneous relations (18) and (19) are
called virtual displacements.
ANALYTICAL MECHANICS – I 117
uρ
P rdρ
vρ
uρ
1vρ S
P dtvrd
ρρ=
vρ
Any system of vectors δ krϖ
satisfying equations (18) and (19) is a system of virtual displacements. Remark 1 : We can say that virtual displacements are displacements of points of a system from one possible position of the system at time t to another infinitely close possible position of the system. Remark 2 : In the case of stationary constraints, virtual displacements coincide with possible displacements. Illustrations 1 : Consider a particle in motion on a fixed surface. In this case, any vector v
ρ constructed from the point P and tangent to the
surface at P will constitute a possible velocity. The corresponding possible displacement vrd
ρρ= dt
lies in the plane tangent to the given fixed surface. The difference δ rdr'dr
ρρρ−=
of the two tangent vectors is also a vector tangent to the surface at the same
point P.
Thus, any vector constructed from P and lying in the tangent plane may be regarded as a certain rd
ρ and as a certain r�ρ .
Here, the constraint is stationary and the virtual displacements coincide with the possible displacements. Illustrations 2 : The constraint is a surface S which is itself in motion (as a rigid body) with a certain velocity u
ρ relative to the original system of
coordinates.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 118
In this case, the possible velocity vρ
is obtained from an arbitrary vector 1vρ
that is tangent to the surface by adding to it the velocity u
ρ. That is,
.uvv 1
ρρρ += Therefore,
dtvrdρρ
=
= .dtudtv1ρρ +
Similarly, for another possible displacement
dtudtvr'd '1
ρρρ+= .
So, the virtual displacement is
δ rdr'drρρρ
−= = dt)vv( 2
'1
ρρ − .
Degree of Freedom
The vector ,r� kρ
in Cartesian co-ordinates, is characterized by three projections on the axes δxk, δyk, δzk (k = 1, 2,…,N) and the equations (18) and (19) which define the virtual displacements may be written in the following form :
� ���
����
�
∂∂
+∂∂
+∂∂
=
N
1kk
k
�k
k
�k
k
� z�zf
y�yf
x�xf
= 0 , (20)
for α = 1, 2,…,d and
� =++=
N
1kkk�kk�kk� 0)z�Cy�Bx�A( (21)
for β = 1, 2,…,N .
If the above d + g equations in (20) and (21) are independent, then out of the 3N virtual increments δxk, δyk, δzk, there will be (3N − d − g) independent virtual increment. Let n = 3N − d − g (22)
then n is called the number of degrees of freedom of the given system of particles. 4.5. POSSIBLE ACCELERATION Let the corresponding forces ),.....,,(, N21kFk =
ρ, be impressed at the points Pk
of the system. Here kFρ
is the resultant of all forces applied directly to the
ANALYTICAL MECHANICS – I 119
particle Pk , (k = 1, 2,…,N). If the constraint were absent, then by Newton’s second law we would have the relations kkk wmF
ρρ= , (23)
for k = 1, 2,…,N , between the masses mk, the accelerations wk and the forces
Fk.
Given constraints, the accelerations
kk
k Fm1
wρρ = (24)
(at a given instant of time t, in a given position of the particles of the system ,rk and for given velocities kv
ρ) may prove incompatible with the constraints.
Differentiating the equations (3) and (5) termwise w.r.t. time, we get
0t
fdtd
v.rf
dtd
w.rf �
N
1kk
k
�N
1kk
k
� =��
���
�
∂∂
+���
����
�� ��
�
����
�
∂∂
+���
����
��
∂∂
==
ρρ
ρρ , (25)
for α = 1, 2,…, d, and
0Ddtd
vdtd
wN
1kkkk
N
1kk =+��
�
����
���
���
�+���
����
�β
=β
=β ��
ρρρρ& .. ll , (26)
for β = 1, 2,…,g.
The left hand sides in relations (25) and (26) are linearly dependent on the accelerations kw
ρ. These left hand sides are also dependent on t, kk v,r
ρρ(k = 1,
2,…N). Equations (25) and (26) are analytic expressions for the restrictions imposed by the constraints on the accelerations kw
ρ of the particles of the
system. The accelerations (24), i.e.,
kk
k Fm1
wρρ = ,
may not satisfy above relations (25) and (26). Then the materially effected constraints will act on the particles Pk of the system with certain supplementary forces kR
ρ, (k = 1, 2,…,N). These forces are called the reaction forces of the
constraint. The reactions that arise are such that the accelerations determined from the equations mk kkk RFw
ρρρ += , (k = 1, 2,…., N) (27)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 120
are already permitted by the constraints. Unlike the reactions kRρ
(k = 1,
2,…N), the pressigned forces kFρ
(k = 1, 2,….,N) are called effective forces. Effective forces are ordinarily specified as known functions of the time, position and velocities of the particles of the system, i.e., ),,( kkkk vrtFF
ρρρρ= . (k = 1, 2,…N) (28)
Basic Problem : The basic problem of the dynamics of a constrained system consists in the following : Given effective forces kF
ρ= ),,( kkk vrtF
ρρρ and the initial positions
ok
ok vvelocitiesinitialtheandr
ρρ of the particles of the system both are
compatible with constraints it is required to determine the motion of the system and the reactions of the constraints kR
ρ (k = 1, 2,…N).
It nothing is known about the nature of the constraints except the defining equations (2) and (3) and, consequently, nothing in known about the reaction
kRρ
produced by these constraints, then the above problem is indeterminate, since the number of scalar quantities (xk, yk, zk, Rkx, Rky, Rkz) that have to be determined is greater than the number of available scalar equations (6 N > 3N + d+ g)
For the basic problem of dynamics to become determinate, it is necessary to have some kind of additional
6N − (3N + d + g)
= 3N − d − g
= n (29)
independent relations between sought-for quantities. These relation can be obtained if we confine ourselves to the following important class of ideal constraints.
Ideal Constraints : If the sum of the works of the reactions of constraints on any virtual displacements is equal to zero, constraints are termed ideal. That is, for ideal constraints, 0r�R...r�Rr�R NN2211 =⋅++⋅+⋅
ρρρρρρ
or
0r�RN
1kkk =� ⋅
=
ρρ . (30)
In Cartesian co-ordinates equal (30) may be rewritten as :
ANALYTICAL MECHANICS – I 121
(R1x δx1 + R1y δy1 + R1z δz1)
+ (R2x δx2 + R2y δy2 + R2z δz2) + …..+ (RNx δxn + RNy δyN + RNz δzN) = 0, or
�=
N
1k(Rkx δxk + Rky δyk + Rkz δzk) = 0. (31)
Among the 3N quantities δxk, δyk, δzk, there are n independent ones (n = 3N − d − g is the degree of freedom of the system). It is therefore possible in (31) to express 3N−n dependent increments δxk, δyk, δzk in terms of n independent increments and equate to zero the co-efficients of these independent increments. We then obtain the n relations still lacking and need to make determinate the basic problem of the dynamics mentioned above. 4.6 LAGRANGE’S EQUATIONS OF THE FIRST KIND
We assumed that all constraints imposed on a system of particle are ideal. If mk is the mass of the kth particle, kw
ρ is its acceleration, and kk RandF
ρρare,
respectively, the resultant of the effective forces and the resultant of the forces of reaction operating on this particle (k = 1, 2,…N), then for particles of a constrained system, we have kkkk RFwm
ρρρ += . (k = 1, 2…N) (1) Since constraints are ideal, for any position of the system under any virtual
displacements, we have
0r�RN
1kkk =� ⋅
=
ρρ, (2)
Eliminating Rk from equations (1) and (2), we obtain
0r�).wmF kkk
N
1kk =� −
=
ρρρ . (3)
This is known as the general equation of dynamics.
It states that, given a system in motion, at any instant of time the sum of the works of the effective forces and the forces of inertia on any virtual displacements is zero.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 122
Thus, general equation of dynamics always holds for any motion that is compatible with constraints and that corresponds to the specified effective forces kF
ρ, (k = 1, 2,…,N).
Derivation of Lagrange’s Equations of the First Kind
Let us find the expressions for the reaction forces kRρ
by means of the undetermined multipliers of Lagrange. The relations defining the virtual displacements of particles of a system are
0r�.rf
k
N
1k k
� =���
����
��
∂∂
=
ρρ , (4)
for α = 1, 2, …d, and
,0rkN
1kk =��
�
����
� δ�=
βρρ
l (5)
for β = 1, 2, …, g.
Multiplying the equations (4) and (5) termwise by arbitrary scalar multipliers (−λα) and (−µβ) and adding termwise the resulting equations to equation (2), we get,
0rrf
R kk
g
1
d
1 kk
N
1k=�
�
����
�µ−
∂∂
λ− ββ=β=α
αα
=���
ρρρ
ρ.l . (6)
In expanded form in Cartesian co-ordinates, we have
kk��
g
1�
d
1� k
��kx
N
1kx�A�
xf
�R ���
����
��−�
∂∂
−�===
+ kk��
g
1�
d
1� k
��ky
N
1ky�A�
yf
�R ���
����
��−�
∂∂
−�===
+ kk��
g
1�
d
1� k
��kz
N
1kz�A�
zf
�R ���
����
��−�
∂∂
−�===
= 0. (7)
The undetermined multipliers λα and µβ may be chosen so that all the scalar coefficients in (7) and, hence, all the vector co-effecients in (6) vanish. This gives
ANALYTICAL MECHANICS – I 123
k��
g
1�
d
1� k
��k l�
rf
�Rρ
ρρ
�+�∂∂
===
, (8)
for k = 1, 2,…N. Expressions (8) is a general expression for the reaction forces of ideal constraints in terms of the undermined multipliers of Lagrange λα, µβ (α = 1, 2,…d ; β = 1 2, …g). Putting the expressions for kR
ρ into equation (1), we get
���
����
��+��
�
����
��
∂∂
+===
k��
g
1�
d
1� k
��kkk l�
zf
�Fwmρρρ
, (9)
for k = 1, 2,…N. The constraint equations for above equations are
fα( 0)rk =ρ
, (10)
for α = 1, 2,…, d, and
0DrN
1kkk =+�
=ββ &l , (11)
for β = 1, 2,…, g.
Equations (9) are called Lagrange equations of the first kind.
Remark : By replacing each vector equation in (9) by three scalar equations, equations (9) to (11) constitute a set of (3N + d + g) scalar equations in (3N + d + g) unknown scalar quantities xk, yk, zk, λα, βµ .
Integrating this set of equations, we get the final equations of motion and, at the same time, from equation (8) we get the magnitude of the reaction forces of constraints. However, integration of such a set of equations is very cumbersome due to the large number of equations. That is why the Lagrange equations of the first kind find little use in actual practice.
Example. Two ponderable particles M1 and M2 of identical mass m = 1 are joined by a rod of invariable length l and negligibly small mass. The system is constrained to move in the vertical plane and only in such manner that the velocity of the midpoint of the rod is directed along it. Determine the motion of the particles M1 and M2
Solution. Let (x1, y1) and (x2, y2) be the co-ordinates of the particles M1 and M2. Then, the constraint equations are
−−+− 212
211 )yy()xx[(
21
l2] = 0, (1)
and
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 124
(x2 − x1) )xx()yy( 1212 &&&& +−+ (y2 − y1) = 0. (2)
The Lagrange equations with undermined multipliers λ and µ are
mk � �=α =β
ββα
α µ+∂∂
λ+=d
1
g
1k
kkk r
fFw l
ρρ
ρρ (3)
for k = 1, 2,…, N. These equations give
1x&& = −λ(x2 − x1) − µ(y2 − y1),
1y&& = −g −λ (y2−y1) + µ(x2−x1), (4)
and 2x&& = λ(x2−x1) − µ(y2−y1),
2y&& = −g + λ(y2−y1) + µ(x2−x1). (5)
Equations in (4) are rewritten as
λ(x2−x1) + µ(y2−y1) + 1x&& = 0, (6)
λ(y2 −y1) −µ(x2 −x1) + gy1 +&& = 0. (7)
Solving equations (6) and (7) for λ and µ, we get
λ =2
122
12
112112
yyxx
xxxgyyy
)()(
)())((
−−−−−++− &&&&
. (8)
Using equation (1), we obtain
λ = ( )1121122122y)yy(x)xx(
l1
)yy(eg &&&& −+−−−− (9)
Similarly, we shall find (left as an exercise)
µ = ( )1121122122yxxxyy
1xx
g &&&& )()()( −−−−−ll
. (10)
It is clear that equation (3) can be obtained from equation (4) if are replace λ by “−λ” and 11 y,x &&&& by .y,x 2 &&&& The values of λ and determined from equation are (exercise)
λ = [ ]2122122122yyyxxx
1yy
g &&&& )()()( −+−+−el
, (11)
ANALYTICAL MECHANICS – I 125
µ = [ ]2122122122yxxxyy
1xx
g &&&& )()()( −−−−−ll
. (12)
Equating the approximate expressions for µ and λ in the above formulae, we
find
)yy()yy)(xx( 121212 &&&&&&&& −−−− (x2 − x1) = 0 (13)
)yy()xx)(xx( 121212 &&&&&&&& −+−− (y2−y1) + 2g (y2−y1) = 0. (14)
Next, we introduce the following abbreviated notation :
u = x2−x1,
v = y2 − y1,
P = 21 xx && + , Q = 21 yy && + (15)
Then we write
u2 + v2 = l2 , (16)
,0vuvu =− &&&& (17)
P v− Qu = 0 , (18)
0gv2vQuP =++ && . (19)
Equations (16) and (17) show that in a u, v-plane a particle with coordinates (u, v) moves in a circle with radius l and with centre at the origin. Its acceleration will all the time be directed towards the centre. The motion of particle will then be uniform. For this reason, we write u = l cosφ,
v = l sinφ (20)
Since motion is uniform so change in φ is uniform, i.e., rate of change of φ is
constant. Let
�&= α . (constant) (21)
Then
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 126
φ = αt + β (22)
According to (18), we may put
P = vf
Quf
ll=, (23)
Substituting these values in (19), are write
0vg2fvvvffuuuf 22 =++++ l&&&& . (24)
Using (20) and (21), we obtain
sing2f −=& φ. (25)
Then
dtddtdf
ddf
φ=
φ
α
= f&
α
−= g2 sinφ.
This implies
f = �
g2cos φ + 2γ . (26)
Consequently, we get
P = 2 ��
���
� φα
+γ cosg
cosφ
= 21 xx && + , (27)
and
Q = 2 ��
���
� φα
+γ cosg
sinφ
= 21 yy && + . (28)
Integrating we get
x1 + x2 = �P dt
ANALYTICAL MECHANICS – I 127
= P21� dφ
= αγg
sinφ +2�
gsin φ cos φ +
2�
g φ + 2δ , (29)
and
y1 + y2 = −�
�2cos φ −
2�
gcos2φ + 2∈. (30)
Finally, we obtain
x1 = �
�sinφ +
2�2g
sin φ cos φ + 2�2
g φ −2l
cos φ+ δ (31)
y1 = −�
�cosφ −
2�2g
cos2φ −2l
sin φ + ∈ (32)
x2 = �
�sinφ +
2�2g
sin φ cos φ + 2�2
g φ + 2l
cos φ+ δ (33)
y2 = −�
�cosφ −
2�2g
cos2φ −2l
sin φ + ∈ (34)
φ = αt + β (35)
where α, β, γ, δ and ∈ are arbitrary constants.
4.7 INDEPENDENT COORDINATES AND GENERALIZED FORCES
Let us consider a holonomic system of N particles Pk with radii
vectors
kzjyixr kkkk ++=ρ (1)
for k = 1, 2,…, N, and with finite constraints
fα(t, )rkρ
= 0, (2)
with α = 1, 2,…, d. In Cartesian form, equivalently, we write
fα(t, xk, yk, zk) = 0 . (3)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 128
We shall assume that d functions fα of 3N arguments xk, yk, zk are independent. Here, t is regarded as a parameter. We can therefore express d co-ordinates of equations (3) as the functions of remaining 3N − d coordinates. The time t and there 3N − d co-ordinates are regarded as independent quantities that define the position of the system at time t. All the 3N Cartesian coordinates may be expressed in the form of functions of n = 3N−d (4)
independent parameters
q1, q2 …qn (5)
and time t. That is,
xk = φk(t, q1, q2…qn),
yk = ψk(t, q1, q2,…, qn),
zk = χk(t, q1, q2,…, qn), (6)
for k = 1, 2,…, N. When these functions are put in the constraint equations (1), the latter become identities. We will assume that any position of the system that is compatible with constraints at the given instant of time may be obtained from the equations (6) for certain values of the quantities q1, q2,…, qn. In vector form, equation (6) can be written as
kk rrρρ
= (t, q1, q2,…, qn) (7)
for k = 1, 2,…, N. The scalar functions in (6) and vector functions in (7) both are assumed continuous and differentiable. The minimal number of quantities qi with the aid of which formulas (6) can embrace all possible positions of a holonomic system coincides with the number of degrees of freedom of the system n = 3N − d .
The quantities q1, q2…qn in formula (6) or (7) (where n is the number of degree of freedom) are called the independent generalized coordinates of the system.
For each instant of time t, a one to one corespondent is established between the possible states of the system and the points of a certain region in the n-dimensional coordinate space (q1, q2…qn). To each position of the system at
ANALYTICAL MECHANICS – I 129
time t, there corresponds a point in the space (q1, q2…qn) that describes this position of the system. The motion of a point in the coordinate space (q1, q2…qn) corresponds to the motion of the system.
If all constraints are stationary, then the time t does not appear explicity in equations (3). It is then always possible to choose coordinates q1, q2…qn such that time t does not enter the equations (6) either.
From now on, it is assumed that for a scleronomic system the independent coordinates q1, q2,…, qn are chosen in precisely that way. Then, for a scleronomic system, the formulas (6) and (7) take on the form xk = φk(qi),
yk = ψk(qi),
zk = χk(qi) , (8)
or kk rr
ρρ= (qi), (9)
for = 1, 2,…, N.
Generalized Forces :- To every coordinate qi, there corresponds a generalized force Qi for i = 1, 2,…, N. The generalized forces are determined as follows.
Consider the elementary work of effective forces on virtual displacements
δA = � ⋅=
N
1kkk r�F
ρρ. (1)
But the virtual differentials of the function )q,t(r ikρ
are the virtual
displacements kr�ρ
:
δ �∂∂
==
n
1i i
kk q
rr
ρρ
δqI , (2)
for k = 1, 2,…, N.
Substitute the expressions (2) into the right-hand side of formula (1) and express the elementary work of the effective forces on the virtual displacements in terms of arbitrary elementary increments δqi of the independent coordinates qi , (i = 1, 2,…m) :
δA = � �= =
�
�
�δ
∂∂N
1k
n
1ii
i
kk q
qr
Fρρ
.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 130
= � ���
����
�
∂∂
⋅�= =
n
1ii
i
kN
1kk q�
qr
Fρρ
= �=
n
1iii ,q�Q (3)
where
Qi = ���
����
�
∂∂
⋅�= i
kN
1kk q
rF
ρρ , (4)
for i = 1, 2,…n. Qi are called the generalized forces, which are coefficients of δδδδqi.
It will be noted that for practical purposes formula (4) is by far not always used to find the quantity Qi. Instead, the system is given an elementary virtual displacement such that only the ith coordinate qi receives a certain increment while the remaining independent coordinates do not change. After that the work of effective forces δAi is calculated on just such a specially chosen displacement. Then δAi = Qiδqi , (5) or
Qi = i
i
q�A�
. (6)
Theorem : Prove that the position of a holonomic system is an equilibrium position if and only if all the generalized forces in this position are zero.
Proof. Let a certain position of the system be a position of equilibrium. According to the principle of virtual displacements, this is possible if and only if
δA = 0 , (1)
or
�=
n
1iiQ δqi = 0. (2)
Here, Qi are generalized forces. But the increments δqi in the independent/generalized coordinates qi are arbitrary. Therefore, the equation (2) is equivalent to the following system of equations Qi = 0, (3)
for i = 1, 2,…, n.
ANALYTICAL MECHANICS – I 131
This proves the theorem.
Illustration 1. Consider a rigid body which is constrained to move
translationally along the x-axis.
The abscissa x of some point of the given rigid body may be taken as the only independent coordinate. Here n = 1 and δA = X δx ,
where X is the sum of the projections, on the x-axis, of all effective forces
acting on the body. SO
Q = X ,
is the generalized force for the single independent coordinate x.
Illustration 2. Consider a rigid body which is constrained to rotate about a certain fixed axis, say u. In this problem, the angle of rotation, say φ, may be taken as the only independent coordinate. Then δA = Luδφ ,
where Lu is the total moment of all effective forces about the axis of rotation.
We find
Q = Lu ,
as the generalized force.
Illustration 3. Consider a free rigid body.
It has six degrees of freedom. For this, we take the three coordinates xA, yA, zA of some point A of the body as the independent coordinates and the three Eulerian angles ψ, θ, φ that define the rotation. Then δA = Qx δx + Qy δy + Qz δz + Qψ δψ + Qθ δθ + Qϕ δϕ . (1)
To determine Qx, we impart to the body an elementary displacement along the x-axis. Then δyA = δzA = 0,
δψ = δθ = δφ = 0. (2)
Equations (1) and (2) imply
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 132
δA = Qx δxA. (3)
Let X, Y, Z be the projections of the stationary axes x, y, z of the principal vector of all effective forces acting on the body. Then Qx = X. (4)
Similarly
Qy = Y,
Qz = Z. (5)
We now impart to our body an elementary displacement such that only the angle ψ changes, while the other quantities remain invariable. Then δA = Qψ δψ. (6)
Let Lψ be the total moment of all effective forces about the Az1-axis (parallel to oz− axis), about which a rotation through an angle ψ is performed. Then Qψ = Lψ (7)
In quite analogous fashion,
Qθ = Lθ,
Qφ = Lφ, (8)
where Lθ and Lϕ are the total moments of the effective forces.
4.8. LAGRANGE’S EQUATIONS OF THE SECOND KIND
We know that the general equation of dynamics is
� −=
N
1kkkkk r�).wmF(
ρρρ= 0. (1)
Expression for the elementary work of effective forces is
δA = �=
N
1kkk r�.F
ρρ
= �=
n
1iii q�Q , (2)
ANALYTICAL MECHANICS – I 133
where
Qi = �∂∂
=
N
1k i
kk ,
qr
.Fρ
(3)
for i = 1, 2,…,n.
The elementary work of the inertial forces “−mk kwρ
”, (k = 1, 2,….N), is
δB = − � ⋅=
N
1kkkk r�wm
ρρ
= −�=
δn
1iii qZ (4)
where, by analogy with expression (3),
Zi = �∂∂
⋅=
N
1k i
kkk q
rwm
ρρ
= ���
����
�
∂∂
⋅� ���
����
�
= i
kN
1k
kk q
rdtrd
mρ&ρ
= � ���
����
�
∂∂
−�
�
��
∂∂
⋅==
N
1k i
kkk
N
1k i
kkk ,
qr
dtd
rmqr
rmdtd &ρ
ρ&ρ
for i = 1, 2,…, n.
But the velocity kr&ρ
is also given as
1kk q(r[dtd
rρ&ρ = , q2, …, qn, t)]
= �∂∂
+���
����
�
∂∂
=
n
1i
ki
i
k
tr
qqr
ρ&
ρ . (6)
This shows that the velocity kr&ρ
is linearly dependent on n21 q,...,q,q &&& . From formula (6), we find that
,qr
qr
i
k
i
k
∂∂
=∂∂
ρ
&&
(7)
for i = 1, 2,…n and k = 1, 2,…N. On the other hand, from the same equation (6), we obtain
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 134
tq
rq
qqr
qr
i
k2n
1kk
ki
k2
i
k
∂∂∂
+���
����
��
∂∂∂
=∂∂
=
ρ&
ρ&ρ
= ���
����
�
∂∂
i
k
qr
dtd
, (8)
for i = 1, 2,…, n and k = 1, 2,…, N.
Using equation (7) and (8) in equation (5), we obtain the following expression for Zi
Zi = �∂∂⋅−�
�
��
∂∂
⋅==
N
1k ikk
N
1k i
kkk q
r)rm(
qr
rmdtd &ρ&ρ
&
&ρ&ρ , (9)
for i = 1, 2,…,n. Let T denote the kinetic energy of the system. Then
T = �
�
�� ⋅=
N
1kkkk )rr(m
21 &ρ&ρ (10)
� ���
����
�
∂∂
=∂∂
=
N
1k i
kkk
i
,qr
).rm(qT &ρ
&ρ (11)
and
�=
���
����
�
∂∂
=∂∂ N
1k i
kkk
i qr
rmqT
&
&ρ&ρ&
).( . (12)
Equations (9), (11) and (12) imply
Zi = ,qT
qT
dtd
ii ∂∂−��
�
����
�
∂∂
(13)
for i = 1, 2,…n.
The general equation of dynamics (1) gives as
δA + δB = 0 (14)
Equations (14), (2) and (4) imply
�=
n
1i(Qi − Zi) δqi = 0 (15)
ANALYTICAL MECHANICS – I 135
Since qi are the independent coordinates and, for this reason, the δqi are absolutely arbitrary increments in the coordinates. It follows that equation (15) can hold when and only when all the coefficients of δqi in equation (15) are equal to zero. Therefore, the general equation of dynamics (15) is equivalent to the following set of equations Zi = Qi , (16)
for i = 1, 2,…,n. Equations (13) and (16) give
iii
QqT
qT
dtd =��
�
����
�
∂∂−��
�
����
�
∂∂&
, (17)
for i = 1, 2,…,n. The above equations in (17) are called the Lagrange equations of the second kind or Lagrange equation in independent co-ordinates. Definition : The quantities iq&, (i = 1, 2,…, n), are called generalized
velocities.
We note that the velocities of the points of the system )rv( kk&ρρ = are expressed
in terms of the generalized velocities )q,...,q,q( n21 &&& and also in terms of independent co-ordinates (q1, q2,…, qn) and the time ‘t’ by means of the formula (6).
Definition : The quantities ,q i&& (i = 1, 2,…n), are called generalized accelerations.
Remark 1 : After performing the operation ,dtd
the left-hand sides of the
Lagranges equations (17) contain the time t, the generalized coordinate qi, the generalized velocities iq& and the generalized acceleration iq&& , (i = 1, 2,… n). The generalized forces Qi,(i = 1, 2,…,n), on the right-hand side of the Lagrange equations (17) are ordinarily specified as functions of t, qk, kq& , (k = 1, 2,…n). That is,
Qi = Qi(t, qk, kq& )
for i = 1, 2,…,n.
Remark 2. The Lagrange equations (17) form a set of n ordinary differential equations, each of the second order in n unknown functions qi of the independent variable t. The order of this system of differential equations is g n . Note that the set of differential equations determining the motion of a holonomic system with n degrees of freedom cannot be of order less than g n,
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 136
since by virtue of the arbitrariness of the initial values of the quantities qi and iq&, (i = 1, 2,…,n), the solution of the system must contain at least g n arbitrary
constants. Thus the set of Lagrange equations in independent coordinates has the lowest possible order. Remark 3. In the case of a constrained system, the reaction forces do not enter into Lagrange’s equations. This is an essential advantage of Lagrange’s equation. After Lagrange’s equations have been integrated and the functions qi(t) found, ),t(rr kk
ρρ= )r,r,t(Fandrw,rv kkkkkkk
&ρρ&&ρρ&ρρ == are determined consequently. After that the unknown reaction forces are determined from the formulas ,FwmR kkkk
ρρρ−=
for k = 1, 2,…, N.
Remark 4 : In the case of a free system of particles, Lagrange’s equations (17) are a compact notation of the equations of motion in an arbitrary system of coordinates. Illustration 1 : Consider a rigid body in rotation about a stationary axis u. In this situation, there is one independent coordinate, the angle of rotation ϕ, to represent the motion. So, we take
qi = φ . (1)
The appropriate generalized force Q for the present motions is equal to
moment of rotation Lu ,i.e.,
Q = Lu. (2)
The total kinetic energy is given by
T = 2u �I
21 & , (3)
where Iu is the moment of inertia of the body about the axis of rotation.
The Lagrange equation for the present problem becomes
.Q�
T�
Tdtd =
∂∂−��
�
����
�
∂∂
& (4)
we find
�I�
Tu &
&=
∂∂
(5)
ANALYTICAL MECHANICS – I 137
y
y
ω P(t)
x
v
and
.0�
T =∂∂
(6)
Lagrange equation (4) now takes the form
Iu uL� =&& . (7)
This is the differential equation of the rotation of a rigid body about a stationary axis. Illustration 2 (Circular Motion). Suppose that a moving particle is describing a circle of constant radius r about the centre 0 with angular velocity ω. Then )jsini(cosrr +=ρ
(1)
is the position vector of the particle. Let r Landi be the unit vectors in the radial (r increasing) and transversal (θ increasing) directions. Then ri = cosθ) j)(sini + (2)
jii ˆ)(cosˆ)sin(ˆ θ+θ−=θ . (3)
The velocity and acceleration are
dtrd
vρρ =
j
o
r
i
θ
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 138
l
m θ
= ]jcosi)sin[(r +−&
= θθ ir& (4)
and
dtvd
fρρ
=
= r .ˆ)(ˆ)( θθ+θ−= irir r2 &&& (5)
If rv , vθ and fr, fθ are components in radial and transversal direction, then
vr = 0, vθ = r&, (6)
fr = −r&2, fθ = r&&. (7)
The quantity ω =& is the instantaneous rate of change of θ and is the angular velocity of the moving point at P.
Illustration 3 : Let us consider a simple pendulum. Assume that a particle of mass m is attached to a massless rod that is free to rotate in a vertical plan about a frictionless pin.
The motion of this single-degree-of-freedom system may be described by the generalized coordinate θ (shown in the figure given below).
The kinetic energy of this system is given in terms of the generalized velocity &as
T = 2mv21
= 22m21 θ&l . (1)
The generalized force associated with the rotational coordinate of a pendulum is Qθ = −mgl sinθ. (2)
ANALYTICAL MECHANICS – I 139
O
l1
φ1 A
l2
φ2
m1g B
m2g z
The equation of motion based on the Lagrangian formation is
T
Tdtd
∂∂−�
�
���
�
∂∂
& = Qθ. (3)
This gives
lg+θ&& sinθ = 0. (4)
Example : A double simple pendulum is in motion in a vertical plane. Find the Langrangian equations of motion. Solution : Let OA and AB be the rods hinged at 0 and A, making angles φ1 and φ2 with the vertical at any time t. Let mass of rod OA bc m1 and mass of rod AB be m2. Let OA = l1 and AB = l2 (see figure below).
Let z1 = l1 cos φ1 (1)
z2 = l1 cos ϕ1 + l2 cos ϕ2. (2)
We know that the elementary work equation is
δA = m1g δz1 + m2 gδz2 (3)
Form equations (1) and (2), we find
δz1 = −l1 sinφ1, δφ1, (4)
δz2 = −l1 sinφ1 δφ1 − l2 sinφ2 δφ2. (5)
Equations (3) to (5) yield
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 140
δA = [−(m1 + m2) gl1 sinφ1] δφ1 [m2gl2 sinφ2] δφ2 (6)
This gives
Qφ1 = −(m1 + m2) gl1 sinφ1, (7)
and
Qφ2 = −m2g l2 sinφ2. (8)
The K.E. of rod OA is
T1 = 21
211m
21 φ&l , (9)
and K.E. of rod AB is
T2 = cos(( 2122
22
21
212 qm
21
llll +φ+φ && φ1 − φ2) )�� 21 && . (10)
The total K.E. of the system is
T = T1+ T2 =21
(m1 + m2) l12
2121221 m φφ+φ &&& ll cos (φ1 − φ2) + 2
2222m
21 φ&l .
(11)
Then the first Lagrange’s equation of motion for the generalized co-ordinate φ1 is
11
TTdtd
φ∂∂−
φ∂∂& = Qφ1. (12)
On simplification of (12), one gets
12121 mm
dtd φ+ &l)[( + m2 l1l2 2φ& cos (φ1 − φ2)]
+ m2 l1l2 21 φφ && sin (φ1 −φ2)
= (m1 +m2) gl1 sinφ1. (13)
The second Lagrange’s equation of motion for the generalized coordinate φ2 is
ANALYTICAL MECHANICS – I 141
22
TTdtd
φ∂∂−
φ∂∂& = Qφ2. (14)
This gives
1212mdtd φ&ll[ cos (φ1 − φ2) + m2 l2
2 ]�2&
+ m2 l1l2 21 φφ && sin (φ1 − φ2)
= −m2 g l2 sin φ2. (15) Special Case : When
m1 = m2 = m,
l1 = l2 = l . (16)
After simplification, equations (13) and (15) reduce to
2 21 φ+φ &&&& cos (φ1 − φ2) + 21φφ && sin (φ1 − φ2) +2 ��
���
�
lg
sin φ1 = 0, (17)
22 φ+φ &&&& cos (φ1 − φ2) + 21φφ && sin (φ1 − φ2) + ��
���
�
lg
sin φ2 = 0, (18)
Further, for small oscillations,
sin φ1 = φi, cos φi = 1 . (19)
Neglecting the second and higher order terms, equation (17) and (18) become
2 0g
221 =��
���
�+φ+φl
&&&& , (20)
0g
21 =��
���
�+φ+φl
&&&& . (21)
4.9. UNIQUENESS OF SOLUTION We have seen that in order to form the lagrange equations of motion for a holonomic system, it is necessary first to find the expression for the kinetic energy T as a function of the time t, the generalized velocities 1q& (i = 1, 2,…n). Let us do this in the general form :
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 142
T = �=
N
1k
2kk rm
21 &ρ . (1)
We know that
� ���
����
�
∂∂
+∂∂
==
n
1i
ki
i
kk t
rq
qr
rρ
&ρ
&ρ . (2)
Equations (1) and (2) give
T = � ���
����
��
∂∂
+∂∂
= =
N
1k
2n
1i
ki
i
kk t
rq
qr
m21
ρ&
ρ.
= � ��
���
�
∂∂+�
��
����
����
��
∂∂
= =
N
1k
2k
2n
1ii
i
kk t
rq
qr
m21
ρ&
ρ
+ �=
��
�
∂∂
⋅���
���
∂∂N
1k
ki
i
kk t
rq
qr
m2ρ
&ρ
= � ��
���
�
∂∂
�+���
����
��
∂∂
= ==
N
1k
2k
N
1kk
2n
1ii
i
kk t
rm
21
qqr
m21
ρ&
ρ
+ � � ���
����
�
∂∂
∂∂
= =
N
1ki
n
1i
k
i
kk q
tr
.qr
m &ρρ
. (3)
Let aiK ai, a0 be functions of t, q1, q2, …, qn defined by the following equations
aiK = �=
���
����
�
∂∂
∂∂N
1k K
k
i
kk q
rqr
mρρ
. (4)
ai = � ���
����
�
∂∂
∂∂
=
N
1k
k
i
kk t
r.
qr
mρρ
, (5)
a0 = � ��
���
�
∂∂
=
N
1k
2k
k tr
m21
ρ (6)
where i, K = 1, 2,…,n. From equation (4), we find
aiK = aKi . (7)
ANALYTICAL MECHANICS – I 143
Using relations (4)−(6) in equation (3) we find
T = T2 + T1 + T0 (8)
where
T2 = �=
n
1KiKiiK qqa
21
,.&& ,
T1 = �=
n
1iiiqa .&
T0 = a0. (9)
We also know that, in the case of a sclaronomic system, the time does not explicitly enter into the relation between .qandr ik &ρ
For this reason
.0trk =
∂∂
ρ (10)
for k = 1, 2,…,N. Consequently, we get
a0 = 0,
ai = 0; (i = 1, 2, …n) (11) and
T = T2 = �=
n
1KiKiiK qqa
21
,
&& . (12)
Thus, we see that the kinetic energy of a scleronomic system appears in the form of a homogenous function of the second degree of the generalized velocities. Remark 1 : It will be noted that in an arbitrary (Scleronomic or rheonomic) holonomic system, the homogeneous quadratic form T2 is always nondegenerate. That is, a determinant made up of its coefficients is different from zero, or . det (aiK) ≠ 0, (12)
for i, K = 1 to n.
For this, if possible, assume that
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 144
det(aiK) = 0. (13)
Then the system of n homogeneous linear equations
�=
n
1KaiK λK = 0 (14)
for i = 1, 2,…,n has a real non-zero solution for λK. Multiplying the set of equations (14), termwise, by λi then summing with respect to i from 1 to n and using formula (4), we get
0 = �=
n
1Ki,aiK λi λK
= � �= =
���
����
�
∂∂
∂∂n
1Ki
N
1k K
k
i
kk q
rqr
m,
.ρρ
(λi λK)
= � �= =
�
�
�λλ��
�
����
�
∂∂
∂∂N
1kKi
n
1Ki K
k
i
kk q
rqr
m,
.ρρ
= � ��= ==
�
�
����
����
�
∂∂
��
����
�
∂∂
λN
1k
n
1K K
KK
n
1i i
kik q
rqr
mρρ
..
= � �= =
���
����
�
∂∂
λN
1k
2n
1i i
kik q
rm .
ρ. (15)
This implies
� =∂∂
=
N
1k i
ki 0
qr
�
ρ , (16)
for k = 1, 2,…, N. These N vector equations may be replaced by 3N scalar
equations
� =∂∂
=
n
1ii
i
ki 0�
qx
� ,
� =∂∂
=
n
1ii
i
ki 0�
qy
� ,
ANALYTICAL MECHANICS – I 145
� =∂∂
=
n
1ii
i
ki 0�
qz
� (17)
for k = 1, 2,…, N. The equations (17) show that in the following Jacobian functional matrix.
J =
������������������
������������������
�
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
n
N
1
N
n
N
1
N
n
N
1
N
n
1
1
1
n
1
1
1
n
1
1
1
qz
..........qz
qy
..........qy
qx
..........qx
qz
..........qz
qy
..........qy
qx
..........qx
ΜΜ , (18)
The columns are linearly dependent. So, the rank, say ρ, of this functional matrix J is less than n. That is, ρ > n. Then among the 3N functions x1, y1, z1, x2, y2, z2,…xN, yN, zN of the n arguments q1, q2…qn (t is regarded as a parameter) there are ρ independent quantities in terms of which all the remaining Cartesian coordinates of the points of the system may be expressed. This is a contradiction to the fact that the minimal number of independent coordinates of the system is equal to the number of degrees of freedom n. This contradiction establishes the claim in (12). Remark 2. Since T2 ≥ 0, (19)
always, it follows from inequality (12) that the quadratic form
T2 = �=
n
1k,ikiik qqa
21 && (20)
is positive definite, and
T2 = 0
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 146
only when
iq&= 0 for each i.
Therefore, from the theory of matrices, we have
a11 > 0,
,0aaaa
2221
1211 >
…………………,
nn2n1n
n22221
n11211
a......
a......
a.....
aaaaaa
> 0. (21)
Remark 3 : Putting the expression (1) for kinetic energy into the Lagrange
equations of motion
iii
QqT
qT
dtd =
∂∂−��
�
����
�
∂∂&
(22)
for i = 1, 2,…, n; we get
kik
n
1kqa &&�
=+ (sum f the terms not involving second derivatives of the
coordinates w.r.t. time)
= Qi (t, qi, jq& ), (23)
for i = 1, 2,…, n. The right-hand sides likewise do not contain second derivatives, since, in the general case, they are functions of the quantities t, qj,
jq& for i = 1, 2,………..n. By virtue of (12), it follows that the equations (23) may be solved for the second derivatives and represented in the form iq&& = Gi (t, qk, kq& ) (24)
ANALYTICAL MECHANICS – I 147
for i = 1, 2,…,n. But, then, as we know from the theory of differential equations, for certain assumptions relative to the right hand sides Gi (the functions Gi, 1 ≤ i ≤ n, have continuous first order partial derivatives) then there is one and only one solution of the Lagrange equations for arbitrary pre-assigned initial quantities qi, iq& with t = t0 for i = 1, 2,…,n. Thus, the motion of a holonomic system is uniquely determined by specifying the initial position
oiq and initial velocities o
iq& . 4.10. THEOREM ON VARIATION OF TOTAL ENERGY
EQUATION FOR CONSERVATIVE FIELDS If the generalized forces do not depend on the generalized velocities, i.e.,
Qi = Qi(t, q1, q2…qn) (1)
for i = 1, 2,…,n and there exists a function U(t, q1, q2…qn) such that
Qi = −iq
U∂∂
(2)
for i = 1, 2,…,n; then the forces Qi are called potentials and the function U is the potential of the forces or the potential energy. We know that the elementary work of the forces Qi is given by
δA = �=
n
1iiQ δqI . (3)
From equations (2) and (3) , we find
δA = −δU . (4)
Let us now consider the general case when in addition to the potential forces determined by the potential U, the system is acted upon also by non-potential forces )q,q,t(Q
~Q~
jjii &= , (5)
for i = 1, 2,…,n. Then the total generalized force is
Qi = − ii
Q~
qU +
∂∂
, (6)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 148
and the Lagrange equations of motion assume the following form ;
iii
Q~
qU
qT
qT
dtd +
∂∂−=
∂∂−��
�
����
�
∂∂&
(7)
for i = 1, 2,…, n. We now consider the total energy E, which is equal to the sum of the K.E. and potential energy, is given by E = T + U . (8)
To compute the derivative dtdE
, we first evaluate .dtdT
We find
�∂∂+��
�
����
�
∂∂+
∂∂=
=
n
1ii
ii
i tT
qqT
qqT
dtdT &&
&&
= �∂∂+�
�
����
����
�
∂∂−
∂∂+��
�
����
��
∂∂
==
n
1ii
ii
n
1ii
i
.tT
qqT
dtd
qT
qqT
dtd &
&&
& (9)
The Euler’s formula for a homogeneous function gives
� =∂∂
=
n
1i2i
i
2 T2qqT &&
(10)
� =∂∂
=
n
1i1i
i
1 TqqT &&
. (11)
We know that
T = T2 + T1 + T0 (12)
From equations (9) to (12) and Lagrange’s equation of motion (7), we find
�=
���
����
�−
∂∂+
∂∂++=
n
1iii
i12 qQ
qU
tT
TT2dtd
dtdT &~
)( . (13)
From (12), we find the value of T2 in term of T, T1 and T0.
ANALYTICAL MECHANICS – I 149
T2 = T−T1 − T0 . (14)
Using equation (14) in (13), we obtain
dtdU
tT
)T2T(dtd
dtdT
2dtdT
01 +∂∂++−=
− �−∂∂
=
n
1iii qQ
~tU & . (15)
From equation (8), we find
dtdU
dtdT
dtdE += . (16)
Using relation (15) in equation (16), we write
�∂∂+
∂∂−++=
=
n
1i01ii t
UtT
)T2T(dtd
qQ~
dtdE & . (17)
Formula (17) determines the total energy of an arbitrary holonomic system.
Further,
�=
n
1iii qQ
~ & = �=
n
1i
ii dt
qdQ~ &
= dt
qdQ~n
1iii�
=&
= ,dtA~�
(18)
where A~� is the elementary work of the nonpotential forces .Q
~i
It is called the power of the non potential forces iQ~ .
For a conservative system, we have
(i) a scalernomic system,
(ii) a system where all forces are potential, and
(iii) the potential energy U is not explicitly dependent on the time.
Thus, for a conservative system equation (17) leads to
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 150
0dtdE = . (19)
This gives E = constant = h , say, (20) for a conservative system. This shows that the total energy of a conservative system does not change when the system is in motion. The Books Recommended for Chapters IV, V, VI and VII 1. F. Gantmatch Lectures in Analytical Mechanics,
MIR Publications, Moscow, 1975.
2. Louis N. Hand and J.D. Finch Analytical Mechanics,
Cambridge University Press, 1998.
3. J.S. Torok Analytical Mechanics,
John Wiley and Sons, 2000.
ANALYTICAL DYNAMICS-II 151
Chapter-5 Analytical Dynamics-II 5.1. LAGRANGE’S EQUATIONS FOR POTENTIAL
FORCES Let the generalized forces Qi be potential, i.e., there exist a force potential
(potential energy)
U = U(t, qi) (1)
such that
Qi = − ,q
U
i∂∂
(2)
for i = 1, 2,…,n.
We define
L = T − U. (3)
The function L is called the Lagrangian function or the kinetic potential. Since the potential energy U does not depend on the generalized velocity iq& , so
ii q
T
q
L
∂∂
=∂∂&
, (4)
iii q
U
q
T
q
L
∂∂
−∂∂
=∂∂
, (5)
for i = 1, 2,…, n. We know that the Lagrange’s equation of motion in terms of kinetic energy T are given by
iii q
U
q
T
q
T
dt
d
∂∂
−=∂∂
−���
����
�
∂∂&
. (6)
from equations (4) to (6), we obtain
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 152
0q
L
q
L
dt
d
ii
=∂∂
−���
����
�
∂∂&
(7)
for i = 1, 2, …, n.
Let us now consider the case when in place of the ordinary potential U(t, qi), there exists a generalized potential V = V(t, qi, iq& ) (8) in term of which the generalized forces Qi are expressed by means of the
following formulas
Qi = ii q
V
q
V
dt
d
∂∂
−���
����
�
∂∂&
(9)
for i = 1, 2, …., n.
In this case, the Lagrange’s equations
=∂∂
−���
����
�
∂∂
ii q
T
q
T
dt
d& ii q
V
q
V
dt
d
∂∂
−���
����
�
∂∂&
becomes
ii q
L
q
L
dt
d
∂∂
−���
����
�
∂∂&
= 0 . (10)
These equation in (10) are of the same type/form as in (7).
Remark I: From formulas (9), it follows that
Qi = �=
���
����
�
∂∂∂n
1KK
Ki
2
qqqV &&&&
+ (*) (11)
for i = 1, 2,….., n. (*) denotes the sum of the terms that do not contains generalized accelerations q&&, K = (1, 2,…,n) Inasmuch as in mechanics we consider only the core when the generalized forces Qi are not explicitly dependent on the generalized accelerations but depend solely on the time, on the coordinates and on the generalized velocities.
ANALYTICAL DYNAMICS-II 153
Qi = Qi(t, qK, Kq& ) (12)
for i = 1, 2,…,n .
It then follows according to formulas (11) that all partial second order derivatives of V with respect to the generalized velocities must be identically equal to zero. This implies that the general potential V, in this case, depends linearly on the generalized velocities. Therefore, we can write
V = i
n
1ii qU &�
=+ U = V1 + U , (13)
where Ui and V are functions of coordinates q1, q2,…, qn and of time t.
Substituting expression (13) for V into formula (9), we get
Qi = �
��
+∂∂− �
=UqU
qdtdU
K
n
1KK
i
i &
= − �= ∂
∂+��
�
����
�
∂∂
−∂∂
+∂∂ n
1K
iK
i
K
K
i
i tU
UqU
qU & . (14)
The formulas (14) show that when the linear part V1 of the generalized potential does not depend explicitly on the time variable t, the generalized forces Qi are made up out of potential forces
− iq
U
∂∂
(15)
and gyroscopic forces
�=
γ=n
1KKiKi qQ &~
, (16)
where
γiK = −γKi = i
K
K
i
qU
qU
∂∂
−∂∂
(17)
for i, K = 1, 2,…, n. 5.2 LAGRANGIAN AND HAMILTONIAN VARIABLES
If the kinetical potential or the Lagrangian function L = L(t, qi, )q i& is known, then the differential equations of motion of a system can be written. The
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 154
variables t, qi, iq&(i = 1, 2,…n), in terms of which the Lagrangian function is expressed, are called Lagrangian variables. For the basic variables that characterize the state of a system, Hamilton proposed the quantities t, qi, pi (i = 1, 2,…,n), where pi are the generalized momenta defined by the following equalities
pi = iq
L&∂
∂ (1)
for i = 1, 2,…,n. The variables t, qi, pi are called Hamiltonian variables. Classical systems in which the forces have the ordinary potential U(t, qi) or the generalized potential V(t, qi, iq&) will be called natural. For such systems the Lagrangian function L is a quadratic function of the generalized velocities. For natural system we have
iKKi
2
Ki
2
aqqT
qqL =
∂∂∂=
∂∂∂
&&&& (2)
for i, K = 1, 2,…,n.
We notice that the Jacobian of the Right side of the equations (1) w.r.t. the variables iq& is the Hessian of the function L. We assume that the Hessian of the function L w.r.t. to the generalized velocities iq& is not identically zero. Therefore
det ���
����
�
∂∂∂
Ki
2
qqL&&
≠ 0 (3)
it follows that equation (1) can be solved for qi, and we write
iq& = Φi(t, qK, pK) (4)
for i = 1, 2,…, n.
Thus, Hamilton’s variables may be expressed in terms of the Lagrange variables and vice versa. Consequently, the state of the system may be described both as a system of values of the Lagrange variables and as a system of values of Hamiltonian variables.
ANALYTICAL DYNAMICS-II 155
We know that in the case of a natural system, Lagrangian function L is a quadratic function of the generalized velocities. By virtue of equation (1), the generalized momenta pi are linearly expressible in terms of the generalized velocities :
pi = �=
+n
1KiKiK cqa & (5)
for i = 1, 2,…,n. Solving the linear system (5) for iq&, we get linear
expressions for iq& of the type
iq& = �=
n
1KKiK pb + bi (6)
for i = 1, 2,…,n. Here, bik and bi are functions of t, q1, q2,…, qn.
If in a natural system the forces Qi have an ordinary potential U(t, qi), it
follows from the equation
L = T − U (7)
pi = iq
T&∂
∂ (8)
If forces Qi have a generalized potential, then we have
pi = ii
Uq
T−
∂∂&
(9)
Let F = F(t, qi, iq&) , (10)
be any function of Lagrangian variables. After substitution of the expression (4) or (6) into (10) in place of the generalized velocities iq&, the function (10) is converted into a certain functions, say, F (t, qi, pi), of the Hamiltonian variables. We call the function F (t, qi, pi) the associated function of the function F(t, qi, iq&). Hamilton (1834) introduced the function H(t, qi, pi) defined by the equation
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 156
H = � −=
n
1iii LqP & , (11)
where L is the associated function of L, in the sense described above.
The function H is called Hamiltonian function.
With the help of Hamiltonian function H, the equations of motion may be written in the form of the following system of 2n ordinary differential equations of the first order
,p
H
dt
dq
i
i
∂∂
= (12a)
i
i
q
H
dt
dp
∂∂
−= (12b)
for i = 1, 2,…,n.
These equations (12a, b) are called canonical equations or Hamilton’s
equations.
DONKIN’S THEOREM : The derivation of the canonical equations of Hamilton will follows from the DONKIN’S Theorem : Statement : Given a certain function X(x1 , x2,…, xn), the Hessian of which is different from zero. Let there also be a transformation of the variables “generated” by the function X(x1, x2…xn) :
yi = ,x
X
i∂∂
for i = 1, 2,…,n. Then, there exists a transformation which likewise generates some function Y(y1, y2…yn) :
xi = iy
Y
∂∂
for i = 1, 2,…,n. If the function X contains the parameters α1, α2,…, αm , i.e.,
X = X(x1, x2………xn ; α1, α2…….αm),
then Y also contains these parameters, i.e.,
ANALYTICAL DYNAMICS-II 157
Y = Y(y1, y2………yn ; α1, α2…….αm) and
jj
XY
α∂∂
−=α∂
∂
for j = 1, 2,…,n.
Proof. Let the generating function Y of the inverse transformation be connected with the generating function X of the direct transformation
yi = ix
X
∂∂
(1)
for i = 1, 2,…,n . By the formula (known as Legendre transformation)
Y = �=
n
1iii yx −X . (2)
The Hessian of the function X coincides with the Jacobian of the right hand side of equations (1), and by virtue of the hypotheses given in the statement of the theorem, we have
det��
�
�
��
�
�
∂∂∂
ki
2
xx
X≠ 0 , (3)
for this reason it is possible to express the variables x1, x2,…, xn in terms of y1,
y2,…, yn .
We write
xi = fi (y1, y2,…,yn) (4)
for i = 1, 2,…,n. Now we replace the variables xi appearing in formula (2) by the expressions in
(4). Then
���
����
� −∂∂=
∂∂
�=
n
1KKK
ii
Xyxyy
Y
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 158
= ��== ∂
∂∂∂−+
∂∂ n
1K i
K
K
n
1KiK
i
K
yx
xX
xyyx
. (5)
By virtue of equations (1), the two sums on the right hand side of equation (5) cancel. Hence, we obtain
ii
xy
Y=
∂∂
, (6)
for i = 1, 2,…,n.
This proves the first part of the Donkin’s theorem.
2nd Part : Now, let X contain parameters α1, α2…αm in addition to the variables x1, x2…xn. Then these parameters occur in the direct transformation (1) and, consequently, in the reverse one as well : xi = fi(y1, y2…yn ; α1, α2…αm) (7)
for i = 1, 2,…,n.
The function Y is determined by equation (2) in which the xi are replaced by
fi(y1, y2…yn; α1, α2…αn),
and so (regarding y1, y2,…, yn as constants) we obtain
��
���
�� −
α∂∂
=α∂
∂=
n
1iii
jj
XyxY
= �α∂
∂−��
�
�
��
�
�
α∂∂
∂∂
−� ��
�
�
��
�
�
α∂∂
==
n
1i jj
i
i
n
1ii
j
i Xx
x
Xy
x
= �α∂
∂−
α∂∂
∂∂
� −∂∂
α∂∂
==
n
1i jj
i
i
n
1i ij
i Xx
x
X
x
Xx
= − ,X
iα∂∂
(8)
for j = 1, 2,…,n.
This completes the proof of Donkin’s theorem.
ANALYTICAL DYNAMICS-II 159
5.3 HAMILTON CANONICAL EQUATIONS
To derive these equations, we use Donkin’s theorem to make transition from the Lagrangian variables to the Hamiltonian variables. For this, by the function X is replaced by L, the variables x1, x2,…, xn by n21 q,...,q,q &&& , the parameters α1, α2 … αm by q1, q2,…, qn and t, the variables y1, y2,…, yn by p1, p2,…, pn and the function
Y = � −=
n
1iii Xyx (1)
in the Donkin’s theorem. We know that
H = �=
−n
1iii Lqp
))& , (2)
and
pi = iq
L&∂
∂ , (3)
for i = 1, 2,…,n.
Hence by Donkin’s theorem, it is concluded that
i
i p
Hq
∂∂
=& , (4)
ii q
HqL
∂∂−=
∂∂
, (5)
and
t
H
t
L
∂∂
−=∂∂
(6)
for i = 1, 2,…, n. Lagrange equations of motion are
0q
L
q
L
dt
d
ii
=∂∂
−���
����
�
∂∂&
, (7)
for i = 1, 2,…,n. Using (3) in (7), we write
i
i q
L)p(
dt
d
∂∂
= . (8)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 160
Hence, from (5) to (8), we obtain
ii
i
ii
qH
qL
dtdp
pH
q
∂∂−=
∂∂=
∂∂= ,&
or
,i
i
pH
dtdq
∂∂=
i
i
qH
dtdp
∂∂−= , (9)
for i = 1, 2,…,n.
Equations in (9) are the required canonical equations of Hamilton.
5.4 ROUTH’S EQUATIONS
For the basic variables characterizing the state of a system at a time t, Routh proposed taking a part of the Lagrangian variables and a part of the Hamiltonian variables. The Routh variables are the quantities t, qi, qα, iq&, pα , (1)
for i = 1, 2,…, m ; α = m + 1,…,n, m is an arbitrary fixed number less than n.
The Lagrangian variables can be replaced by Routh variables if we express all the αq& in terms of pα, by the relations
pα = α∂
∂q
L&
, (2)
for α = m + 1, m + 2,…,n.
Suppose that the Hessain of the function L of the generalized velocities αq& is different from zero. That is,
det��
�
�
��
�
�
∂∂∂
βα qq
L2
&& ≠ 0. (3)
ANALYTICAL DYNAMICS-II 161
Then, by applying the Donkin theroem, we get a transformation that is inverse to the transformation (2), namely
α
α ∂∂=pR
q& (4)
for α = m + 1, m + 2,…,n
and R = R(t, qi, qα, ,q i& pα)
is the Routh function defined by the equation
R = �+=α
αα −n
1mLqp))
& . (5)
The sign .) signifying that all the αq& are expressed in terms of pα.
The variables
t, qi, qα, iq&
(i = 1, 2,…, m ; α = m + 1, m + 2, …,n) are now regarded as parameters. Consequently, Donkin theorem gives
,ii q
LqR
∂∂−=
∂∂
ii qL
qR
&& ∂∂−=
∂∂
(6)
for i = 1, 2,…,m ; and
αα ∂
∂−=
∂∂
q
L
q
R (7)
for α = m + 1, m + 2,…, n , and
.t
L
t
R
∂∂
−=∂∂
(8)
We know that the Lagrange equations for the coordinates qi are
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 162
0q
L
q
L
dt
d
ii
=∂∂
−���
����
�
∂∂&
. (9)
Using (6) equations (9) may be written as
0q
R
q
R
dt
d
ii
=∂∂
−���
����
�
∂∂&
, (10)
for i = 1, 2,…,m.
Lagrange’s equations of the coordinates qα are
α
α
∂∂
=q
L
dt
dp (11)
for α = m + 1, m + 2,…, n.
Using (4) and (7), equation (11) become
α
α
∂∂
−=q
R
dt
dp (12)
and
α
α
∂∂
=p
R
dt
qd, (13)
for α = m + 1, m + 2,…, n.
Equations (10), (12) and (13) form a set of following Routh equations :
,0q
R
q
R
dt
d
ii
=∂∂
−���
����
�
∂∂&
,q
R
dt
dp
α
α
∂∂
−=
α
α
∂∂
−=p
R
dt
dq , (14)
for i = 1, 2,…, m and α = m + 1, m + 2,…,n. These equations consist of m second-order differential equations of the Lagrange type and 2(n−m) first order differential equations of the Hamiltonian
ANALYTICAL DYNAMICS-II 163
type. The Routh function in the first equations play the role of the Lagrangian function, while those in the latter equations play the role of the Hamiltonian function. 5.5 CYCLIC COORDINATES
The Lagrangian of any physical system is generally expected to have explicit dependence on all the generalized coordinates qi, all the generalized velocities
iq& and time t, that is L = L(q1, q2… qn, )t,q...q,q n21 &&&
where n is the total number of generalized coordinates. If some of the generalized coordinates do not appear explicitly in the expression for the Lagrangian, these coordinates are called cyclic (ignorable). Any change in these coordinates do not affect the Lagrangian. 5.6 POISSON BRACKETS Poisson introduced a special term, called the Poisson bracket, for the following expression composed of the partial derivatives of two arbitrary functions φ(t, qi, pi) and ψ(t, qi, pi) :
(φ ψ) = � ���
����
�
∂ψ∂
∂φ∂
−∂
ψ∂∂
φ∂=
n
1i iiii qPPq. (1)
Remark : For the functions φ(t, qi, Pi), ψ(t, qi, pi), χ(t, qi, pi) and constant c, the following properties are satisfied by Poisson bracket : (1) (φ ψ) = −(ψ φ)
(2) (c φ ψ) = c(φ ψ)
(3) (φ + ψ χ) = (φ χ) + (ψ χ)
(4) ���
����
�
∂ψ∂
φ+���
����
�ψ
∂φ∂
=φψ∂∂
tt)(
t.
POISSON’S IDENTITY : For the functions φ(t, qi pi) ψ(t, qi, pi) and χ (t, qi, pi); the following property holds ((φ ψ)χ) + ((ψ χ)φ) + ((χ φ)ψ) = 0
This property is known as Poission’s identity.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 164
Definition : A function f(t, qi, pi) is called the integral of the equations of
motion
,i
i
pH
dtdq
∂∂=
i
i
qH
dtdp
∂∂−= , (1)
for i = 1, 2,…,n ; if for any motion of the given system this function retains a
constant value say C :
f(t, qi, pi) = C (2)
Remark : The necessary and sufficient condition for the function f(t, qi, pi) to be integral of the equations of motion (1) is that
t
f
dt
df
∂∂
= + ( f H) = 0. (3)
5.7 JACOBI-POISSON THEOREM Statement : If f and g are integrals of equations of motion, then (f g) is also an integral of these equations. Proof : Since f and g are integrals of equations of motion then
0)Hf(t
f=+
∂∂
, (1)
t
g
∂∂
+ (g H) = 0 . (2)
Now to prove that (f g) is also an integral of the same equations of motion, it is
required to prove that
t∂
∂(f g) + ((fg)H) = 0 . (3)
Now
ANALYTICAL DYNAMICS-II 165
���
����
�
∂∂
+���
����
�
∂∂
=∂∂
t
gfg
t
f)gf(
t (4)
From equations (1) and (2), we write
)Hf(t
f−=
∂∂
, (5)
t
g
∂∂
= − (g H). (6)
Using (5) and (6) in (4), we write
t∂
∂(f g) = −((f H)g) − (f(gH)) ,
t∂
∂(f g) = ((H f) g) + ((g H) f) . (7)
From (3) and (7), we write
t∂
∂(f g) + ((f g)H) = ((H f)g) + ((gH) f) + ((f g)H) = 0 .
Thus
t∂
∂(f g) + ((f g)H) = 0 .
This completes the proof of the theorem.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 166
Chapter-6
Analytical Mechanics-III
6.1 HAMILTON’S PRINCIPLE Definition : We consider an arbitrary holonomic system with independent coordinates q1, q2,…, qn and the Lagrangian function L = L(t, qi, iq&), 1 ≤ i ≤ n. The integral
W = �1
0
t
tLdt …(1)
is called the “Hamilton action” during a time interval (t0, t1). The expression Ldt is called the elementary Hamilton action.
Note : Since the function L is of the form
L = L (t, qi, iq&), …(2)
it is necessary, in order to compute the Hamilton action W, to specify the
functions
qi = qi(t) …(3)
for i = 1, 2,…, n; in the time interval [t0, t1]. This imply that the Hamilton action W is a functional dependent on the motion of the system.
Remark : Let 0iq be a given initial position of the system at time t = t0 and qi′
be a given final position, which it occupies at time t = t1. We fix the initial and terminal instants of time t0 and t1, and the initial and terminal positions of the system. The motions are otherwise arbitrary.
In the extended (n + 1) −dimensional coordinate space, where the quantities qi and the time t are the coordinates, this motion is depicted by some Curves or paths. We shall consider all possible such paths, passing through two specified points of space M0 (t0, qi
o) and M1(t1, qi′) as shown in figure below.
t
qi
M1
M0
ANALYTICAL MECHANICS-III
167
That is, we consider all possible motions that translate the systems from initial position qi
o to final position qi′. Straight path and Circuitous paths
Suppose that among the paths considered above, there is a path along which the system can move for a specified function L, i.e., in a given field of force. Such a path is called “straight path”. In the above figure, the “straight path” is depicted by a solid-line. For a straight path, the functions, qi = qi(t), satisfy the Lagrange equations of motion, namely,
,0q
L
q
L
dt
d
ii
=∂∂
−���
����
�
∂∂&
…(4)
for i = 1, 2,…,n. All other paths passing through the points M0 and M1 are termed as “circuitous paths”. Statement of Hamilton’s principle (1834-35)
“The Hamilton action W has a stationary value for the straight path as compared with the circuitous paths”. Proof Hamilton’s principle : Let us consider an arbitrary one-parameter
family of paths
qi = qi (t, α), …(5)
where α is a parameter, −γ ≤ α ≤ γ, t0 ≤ t ≤ t1 and i = 1, 2,…,n. For α = 0, one obtains a given straight path and for α ≠ 0, the paths are circuitous. Further, let us assume that all these paths in (5) have a common initial point M0 and a common end point M1. That is, qi(t0, α) = qi
0,
…(6a)
qi(t1 α) = qi′, …(6b)
for −γ ≤ α ≤ γ and i = 1, 2,…, n. The Hamilton action as computed along a path of this one-parameter family is a function of the parameter α, and is denoted by W(α), defined as
W(α) = � α1
0
t
tii q),,t(q,t(L & (t, α)) dt. …(7)
Now we compute the variation δW of the Hamilton action W. We defined
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 168
δW = � δ1
0
t
tdtL
= dtqq
Lq
q
L1
0
t
t
n
1ii
ii
i�
��
���
� ��
�
����
�δ
∂∂
+δ∂∂
=&
& …(8)
We note that
δ �
��
αδ= )},t(q{
dt
dq ii&
= δα��
���
��
���
�
��
α
α∂∂
)},t(q{dt
di
= ��
���
��
���
δα�
��
α
α∂∂
),t(qdt
di
= dt
d(δqi). …(9)
Now
dtqq
L1
0
t
t
n
1ii
i� �
��
� δ
∂∂
=&
&
= dt)q(dt
d
q
L1
0
t
t
n
1ii
i�
��
���
� �
�
����
�
∂∂
=&
&
= � ���
���
��
����
�
∂∂
−��
���
� �
�
����
�
∂∂
=
=
==
1
0
1
0
t
t
n
1ii
i
tt
tt
n
1ii
i
dtqq
L
dt
dq
q
L&&
= − ,dtqq
L
dt
d1
0
t
t
n
1ii
i� �
��
���
��
����
�
∂∂
= & …(10)
since the variations δqi are zero at t = t0 and t = t1 by virtue of the assumption that the straight path and all the circuitous paths pass through M0 and M1 in extended co-ordinate square. Combining equations (8) and (10), one writes
δW = ���
���
��
���� δ
��
���
���
����
�
∂∂
−∂∂
=
1
0
t
t
n
1ii
ii
dtqq
L
dt
d
q
L&
…(11)
ANALYTICAL MECHANICS-III
169
For a straight path (i.e., α = 0), the functions qi = qi(t) satisfy the Lagrange equations of motion, namely,
,0q
L
q
L
dt
d
ii
=∂∂
−���
����
�
∂∂&
…(12)
for i = 1, 2, …,n. From equations (11) and (12), it is concluded that δW = 0, …(13)
for the straight path. This proves that the Hamilton actions W has a stationary value for the straight path. Hence, the proof of Hamilton’s principle is complete.
Remark : The converse of Hamilton’s principle is true. That is, if
δW = 0 …(14)
for some path, then the paths straight.
Remark 2 : Since, from the Hamilton principle there follows the Lagrange equations of motion (using equation 11) and vice versa, Hamilton’s principle may be placed at the foundation of the dynamics of holonomic systems.
Remark 3 : The variational principle characterizes the entire straight path as a whole. It formulates the stationary property of a certain functional (Hamilton action), which property distinguishes the straight-line path from among other kinetically possible paths (circuitous paths). The variational principle has a more surveyable and compact form and if frequently used as foundation for new non-classical domains of mechanics.
Remark 4 : The value of the Hamilton actions W is least for a straight-line
path.
6.2 POINCARE −−−− CARTAN INTEGRAL INVARIANT We shall now derive a formula for the variation of Hamilton action
W = �1
0
t
t,dtL …(1)
in the general case when the initial and terminal instances of time, just like the initial and terminal coordinates, are not fixed but are functions of a parameter, say α, of the type t0 = t0(α), t1 = t1(α), …(2)
and
)(qq),(qq 1i
1i
0i
0i α=α= . …(3)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 170
Using the Leibnitz rule for differentiation under integral sign, the differentiation of (1) with respect to parameter α gives
δW = δ ���
����
��α
α
)(t
)(t
1
0
dtL
L1δt1 −L0δt0 + dtqq
Lq
q
L)(t
)(t
n
1ii
ii
i
1
0
� � ���
����
�δ
∂∂
+δ∂∂α
α =&
& …(4)
Integrating by parts, we write
dtqdt
d
q
Ldtq
q
Li
)(t
)(t i
)(t
)(ti
i
1
0
1
0���
����
�δ� ��
�
����
�
∂∂
=� ���
����
�δ
∂∂ α
α
α
α &&
&
= � ���
����
�
∂∂
−�
��
δ
∂∂ α
α
α
α
)(t
)(t i
)(t
)(ti
i
1
0
1
0q
L
dt
dq
q
L&&
(δqi) dt
= � ���
����
�
∂∂
−δ−δα
αα=α=
)(t
)(t i)(tti
0i)(tti
1i
1
001 q
L
dt
d}q(p}q{p
& (δqi) dt, …(5)
where we have used the facts that
pi = ,q
L
i&∂∂
…(6a)
and
),q(dt
dq ii δ=δ& …(6b)
for i = 1, 2,…,n. Here, pi is the generalized momenta of the system. Further, we have used the notation [ ] )(ttii pp αλ=
λ =
= pi(tλ(α), α), …(7)
for λ = 0, 1. We note that
{ } δα�
��
α
α∂∂
=δλ
λ=
=tt
itti ),t(qq …(8)
for λ = 0, 1. On the other hand, for the variations of the terminal coordinates
ANALYTICAL MECHANICS-III
171
],),(t[qq 1'i
'i αα=
we have the formulas
δα�
��
α∂α∂
+δ=δ= 1tt
i1
'i
'i
),t(qtqq &
= [ ] .qtq1tti1
'i =δ+δ&
This implies
[ ] 1'i
'itti tqqq
1δ−δ=δ = & . …(9)
Similarly,
[ ] .tqqq 00i
0itti 0
δ−δ=δ = & …(10)
First, we substitute the expressions for [ ]λ=δ ttiq from equations (9) and (10)
into right side of equation (5) and then use the value of leftside of (5) into equation (4) to get the following expression for δW.
δW = ( )� δ+δ−δ=
n
1i111
'i
'i
'i tLtqqp &
− 0
n
1i00
0i
0i
0i tL)tqq(p δ� −δ−δ
=&
+ ���
���
��
���
��
���
���
����
�
∂∂
−∂∂α
α =
)(t
)(ti
n
1i ii
1
0
dtqq
L
dt
d
q
L&
…(11)
we know that
HLqpn
1iii =−�
=& , …(12)
so that
�=
λλn
1iii qp & − Lλ = Hλ, …(13)
for λ = 0, 1. Using (13) in (11), we finally obtain
δW = 1
0
n
1iii tHqp �
��
δ−δ�=
+ ���
���
��
���� δ
��
���
���
����
�
∂∂
−∂∂
=
1
0
t
t
n
1ii
ii
dt,qq
L
dt
d
q
L&
…(14)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 172
t
q1
C0
C1
p1
(For n = 1)
where
��==
δ−δ=�
��
δ−δn
1i11
1i
1i
1
0
n
1iii tHqptHqp
− .tHqpn
1i00
0i
0i� δ+δ
= …(15)
In place of the (n + 1) −dimensional extended coordinate space, we take the (2n+1)−dimensional extended phase space in which the quantities qi, pi and t will be the coordinates of the point.
In this space we take an arbitrary closed curve C0 given by the equations
qi = 0iq (α),
pi = 0iq (α),
t = t0(α), …(16)
for i = 1, 2,…,n and 0 ≤ α ≤ l. Here, we have one and the same point of the curve C0 for α = 0 and α = l. Taking each point on the curve C0 as the initial one, we draw the appropriate straight-line path. Such a path is uniquely determined by a system of Hamilton’s canonical equations and initial-conditions. We obtain a closed tube of straight paths qi = qi(t, α),
pi = pi(t, α), …(17)
qi(t, 0) = qi(t, l),
pi(t, 0) = pi(t, l), …(18)
for i = 1, 2,…,n and 0 ≤ α ≤ l.
ANALYTICAL MECHANICS-III
173
On this tube, we choose arbitrarily a second closed curve C1 around the tube that has only one common point with each generatrix. The equations of curve C1 may be written in the following form
qi = 1iq (α),
pi = 1iq (α),
t = t1(α). …(19)
Now we shall examine the Hamilton action W along the generatrix of the tube from the curve C0 to the curve C1. In this case, for any α, the generatrix is a straight-line path and by virtue of Lagrange equations of motion
0q
L
dt
d
q
L
ii
=���
����
�
∂∂
−∂∂
&, …(20)
for i = 1, 2,…,n.
Consequently, the variations of Hamilton action, δW, for this case takes the following simplest form:
δW = ,tHqp1
0
n
1iii �
��
� δ−δ=
…(21)
by virtue formula in (14). This gives
W′(α) δα = .tHqp1
0
n
1iii �
��
� δ−δ=
…(22)
Integrating (22) with respect to α from α = 0 to α = l, we get
0 = w(l) −w(0) = � � �
��
δ−δ=
1
0
1
0
n
1iii tHqp
= � �� ����
��� δ−δ−
���
��� δ−δ
==
11
0
n
1i00
0i
0i
0
n
1i11
'i
'i
tHqptHqp
= ����
���� δ−δ−�
���
���� δ−δ
== 01 C
n
1iii
C
n
1iii tHqptHqp .
This gives
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 174
����
���� δ−δ=�
���
���� δ−δ
== 01 C
n
1iii
C
n
1iii tHqptHqp . …(23)
The relation (23) proves that the line-integral
I = ����
���� δ−δ=C
n
1iii tHqp …(24)
taken along any close contour C does not change its value in the case of an arbitrary displacement of the contour along a tube of straight-line paths. This imply that the integral I is invariant.
Definition : The integral I, defined above in (24), is called the POINCARE-CARTAN INTEGRAL INVARIANT.
Remarks 1 : The invariance of the Poincare−Cartan integral may be placed at the foundation of mechanics since from this invariance it follows that the motion of a system obeys Hamilton’s canonical equations.
Remark 2 : The converse proposition of the above result is true. That is, if the Poincare−Cartan I is an integral invariant with respect to the straight-line paths defined by the set of following first-order differential equations
=dt
dq i Qi(t, qk, pk),
dt
dp i = Pi(t, qk, pk), …(25)
for i = 1, 2,…,n ; then the following relations hold between the function H and
the functions Qi, Pi :
Qi = ip
H
∂∂
Pi = −iq
H
∂∂
…(26)
for i = 1, 2,…,n.
Remark 3 : In the Poincare−Cartan integral (24), the time t enters as the coordinate qi, and the role of the corresponding momentum is played by the quantity, −H, i.e., the energy taken with opposite sign. This is a far−reaching analogy. We change the variables in the integral I by introducing a new variable z connected with the old variables by the relation
z = −H(t, q1, q2, …, qn, p1, p2,…,pn) …(27)
Using this relation (27), we express p1 in terms of other variables. Let
ANALYTICAL MECHANICS-III
175
p1 = −K(t, q1, q2,…, qn ; z; p2, p3,…, pn). …(28)
Using this relation (27), and (28), the expression (24) now becomes
I = �C
{zδt + p2 δq2 + p3 δq3 +…+ pn δqn − K δq1}. …(29)
Thus, in the new variables, the integral I has the aspect of the Poincare − Cartan integral, but the role of the time is now played by the variable q1 and in place of the earlier energy H we have the momentum p1 taken with reversed sign, i.e., K. Thus, the motion of a system in the new variables is described by the following Hamiltonian system of differential equations.
tK
dqdz
,zK
dqdt
11 ∂∂−=
∂∂= , …(30)
j1
j
j1
j
qK
dq
dp,
pK
dq
dq
∂∂−=
∂∂= , …(31)
for i = 2, 3,…,n; q1 being the independent variable.
Illustration : In the case of a linear oscillator for which
H = .2
cq
m2
p 22
+ …(1)
To form the canonical equations taking q for the independent variable, we put
z = −��
�
�
��
�
�+
2
cq
m2
p 22
. …(2)
Equation (2) implies
p = )cqz2(m 2+− . …(3)
Thus, we have,
K = − )cqz2(m 2+− . …(4)
The corresponding canonical equations are
,q)c/z(2
c/m
dq
dt2−−
= …(5)
0dq
dz= . …(6)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 176
Solving (6), we get
z = constant
= −h, say. …(7)
From equations (5) and (7), we find
α−−
�=�2q)c/h2(
dqdt
m
c
This gives
wt + α = sin−1
��
�
�
��
�
�
h2
cq …(8)
where w = m/c , …(9)
and α is the constant of integration. Equation (8) implies
q = A sin (wt + α) …(10) where A = c/h2 .
The solution of canonical equations (5) and (6) consists of relations in (7) and (10).
6.3 WHITTAKER’S EQUATIONS We consider a generalized conservative system for which the Hamiltonian function H is not explicitly dependent on time, i.e., H = H(qi, pi) …(1) and
.0t
H=
∂∂
…(2)
We know that Hamilton’s canonical equations are
i
i
i
i
q
H
dt
dp,
p
H
dt
dq
∂∂
−=∂∂
= …(3)
for i = 1, 2,…,n. Differentiating (1), we write
� ���
����
�
∂∂
+∂∂
==
n
1i
i
i
1
i dt
dp
p
H
dt
dq
q
H
dt
dH
= 0, …(4)
ANALYTICAL MECHANICS-III
177
by virtue of (2) and (3). Integrating (3), we get
H(qi, pi) = constant
= h, say, …(5)
during the motion of the system. The function H is called the generalized total energy and relation (5) is termed as the generalized integral of energy.
We consider an ordinary 2n−dimensional phase space in which the quantities qi and pi, 1 ≤ i ≤ n, are the coordinates of a point. We confine ourselves to only those points of the phase space whose coordinates satisfy the equation (5) with fixed value of the constant h, say h0. In other words, we confine ourselves solely to those states of the system to which the given magnitude of the total energy corresponds :
H = H(qi0, pi
0) = h0. …(6)
The basic integral invariant I for a generalized conservative system is
I = � �
��
� δ=
n
1iii qp . …(7)
We solve equation (5) for one of the momenta, say p1, to get
p1 = −K (q1, q2,…, qn, p2, p3,…, pn, h0), …(8)
and put the expression (8) for p1 into the integral (7). We have
I = �
��
δ−δ� �
=
n
2j1jj qKqp . …(9)
But the integral invariant (9) again has the form of the Poincare−Cartan integral if it is assumed that the basic coordinates and momenta are the quantities qj and pj for j = 2, 3,…,n, and the variable q1 plays the role of time variable; and instead of H we have the function K. Consequently, the motion of a generalized conservative system satisfy the following Hamiltonian system of 2n−2 differential equations.
,pk
dq
dq
j1
j
∂∂=
j1
j
qk
dq
dp
∂∂−= , …(10)
for j = 2, 3,…,n. These differential equations in (10) were obtained by Whittaker. For this reason, equations (10) are called Whittaker Equations.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 178
6.4 JACOBI EQUATIONS We know that for an ordinary conservative system, Whittaker’s equations are
j1
j
j1
j
q
k
dq
dp,
p
k
dq
dq
∂∂
−=∂∂
= , …(1)
and H = constant = h. …(1a)
for j = 2, 3,…n. Integrating the system (1), we find the qj and pj as functions of variable q1 and 2n−2 arbitrary constants c1, c2,…, c2n−2.
Moreover, the integrals of Whittaker’s equation will contain an arbitrary constant h0, representing the given magnitude of the total energy of the system. Thus,
qj = ϕj (q1, h0, c1, c2,…, c2n−2), …(2)
pj = ψj(q1, h0, c1, c2,…, c2n−2), …(3)
for j = 2, 3,…, n. We know that
p1 = −K(q1, q2,…, qn, p2, p3,…, pn, h0). …(4)
Substituting (2) and (3) into equation (4), we find
p1 = ψ1(q1, h0, c1, c2,…, c2n−2). …(5)
The dependence of the coordinates on the time t is obtained from the following Hamilton equation
.p
H
dt
dq
1
1
∂∂
= …(6)
Integrating (6), we obtain
t = )pH(
dq
1
1
∂∂� + c2n−1, …(7)
where c2n−1 is the constant of integration, and all the variables in the partial derivative (∂H/∂p1) are expressed in terms of q1 with the help of the equations (2) to (4)
ANALYTICAL MECHANICS-III
179
1
j'j dq
dqq = , …(8)
for j = 1, 2,…,n.
Then q1′ = 1.
Let P = P(q1, q2,…, qn, q2′, q3′,…, qn′)
= �=
n
2j
'jj qp −K …(9)
The Hamiltonian system (1), by eliminating K and pj from equations (1), (8) and (9), is equivalent to the system of equations of the Lagrangian type :
,0q
P
q
P
dt
d
j'j
=∂∂
−��
�
�
��
�
�
∂
∂ …(10)
for j = 2, 3,…,n.
The system (10) contains (n−1) second-order equations.
Next, we transform the expression for the function P by using (4) for p1. We
write
P = �=
n
2j
'jj qp + p1. …(11)
Now P = p2 q2′ + p3 q3′ +…+ pn qn′ + p1
= p2 11
nn
1
33
1
2 pdq
dqp...
dq
dqp
dq
dq++++
= 1nn33221
p)qp...qpqp(q
1++++ &&&
&
= ���
����
��=
n
1iii
1
qpq1 &&
= ),HL(q
1
1
+&
…(12)
because
� =−=
n
1iii .HLqp & …(13)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 180
We know that for a conservative system.
L = T−U,
H = T + U, …(14)
where T is the kinetic energy and U is the potential energy. The kinetic energy
T may be written as
T = k
n
1k,iiik qqa
2
1&&�
=. …(15)
Let
G(q1, q2, qn, q2′, q3′,…, qn′) = �=
n
1k,i
'k
'iik qqa
21
…(16)
Equations (12) to (16) yield
P = 1q
T2&
…(17)
H−U = T = .Gq 21& …(18)
From equations (1a) and (18), we find
G
Uhq1
−=& , …(19)
and from equations (1a), (17) and (19), we get the following expression for the
function P.
P = 2 )Uh(G − . …(20)
Definition : The differential equations (10), in which the function P is of the form (20) and which therefore belong to the ordinary conservative systems (natural) are refereed to as JACOBI EQUATIONS.
Remark : Integrating Jacobi’s equations, we determine all the trajectories in the coordinate space (q1, q2,…, qn). qj = ϕj (q1, h, c1, c2,…, c2n−2). …(21)
The relation between the coordinates and the time variable is established from equation (19) by integration. We find
t = 1n21 CdqUh
G−+
−� . …(22)
ANALYTICAL MECHANICS-III
181
qi′ q3
q1
qi0
q2
6.5 PRINCIPAL OF LEAST ACTION The Jacobi equations are
,0qP
'qP
dqd
jj1
=∂∂−
��
�
�
��
�
�
∂∂
…(1)
for j = 2, 3,.., n, and
1
j'j dq
dqq = . …(2)
Here, q1 is the independent variable and plays the role of the time. The Lagrangian action, denoted by W*, is defined as
W* = .dqP'1
01
q
q1� …(3)
Here, all the motions of a generalized conservative system that transfer the system from a given initial position 0
iq to a specified terminal position 'iq
(figure below). The instants of time t0 and t1 are not fixed and may vary when passing from the straight-line path to circuitous paths. Statement (Principal of least action). The variation of the Lagrange action W* is zero for the straight-line path.
Proof : We note that Jacobi equations (1) are Lagrangian type equations. So
by Hamilton principle,
δW* = 0
for the straight-line path.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 182
Remark 1 : We have
W* = � +1
0
t
tdt)HL(
= � �+1
0
1
0
t
t
t
tdthdtL
= W + h(t1 − t0).
Remark 2 : For an ordinary system, we have the Lagrange action W* in the
following form
W* = 2 �1
0
t
tdtT
= ����
����=
1
0
t
t
N
1k
2kk dtvm
ρ
= ���
��
�
�
=
N
1kk
2k
s
sk dsvm
'k
0k
ρ.
6.6 LEE HWA −−−− CHUNG’S THEOREM
Poincare introduced for the first time, the following integral
I1 = � �=
n
1ipi δqi
…(1)
Round the contour C consisting of simultaneous states of a system. Poincare’s integral invariant I1 does not change its value if the contour C is displaced along the tube of straight-line paths to the contour C′, which again consists of simultaneous states.
It is convenient toe consider the integral I1 in the ordinary non extended 2n−dimensional phase space (q1, p1, q2, p2,…, qn, pn). In this the contour D and D′ (figure below) bounding the tube of straight-line paths.
ANALYTICAL MECHANICS-III
183
q1
pn
p1
D
D′
Here,
� � δ=� � δ== 'D
n
1iii
D
n
1iii qpqp .
The Poincare integral invariant I1 is called the universal integral
invariant.
In 1947, the Chinese scientist Lee Hwa−Chung proved the uniqueness of universal integral invariants. He demonstrated that any other universal integral invariant differs by a constant factor from I1. Statement of Lee−−−−Hwa Chung Theorem :
If
I′ = � �=
n
1i[ Ai(t, qk, pk) δqi + Bi(t, qk, pk) δpi]
is a universal relative integral invariant, then
I′ = c I1,
where C is a constant, and I1 is the Poincare integral.
Note : The term ‘relative’ means that the domain of integration is a closed contour.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
184
Chapter-7
Analytical Mechanics-IV
7.1 CANONICAL TRANSFORMATIONS
Definition :- The transformation of 2n-dimensional space
ii q~q~ = (t, qk, pk), …(1)
ii p~p~ = (t, qk, pk), …(2)
for i = 1, 2, 3,…, n, with the condition that
)p,q,...,p,q,p,q()p~,q~,...,p~,q~,p~,q~(
nn2211
nn2211
∂∂
≠ 0, …(3)
is called canonical if it carries any Hamiltonian system
,pH
dtdq
i
i
∂∂=
,qH
dtdp
i
i
∂∂−= …(4)
for i = 1, 2,…,n, again into another Hamiltonian system
,p~H~
dtq~d
i
i
∂∂=
,q~H~
dtp~d
i
i
∂∂−= …(5)
for i = 1, 2,…,n.
ANALYTICAL MECHANICS-IV
185
Remark 1 :- In the transformation (1, 2), the time variable t is considered as a parameter
Remark 2 :- In equation (5), is another Hamiltonian function.
Remark 3 :- The importance of studying canonical transformations is due to the fact that these transformations permit replacing a given Hamiltonian system (4) by another Hamiltonian system (5) in which the function H~ is of a simpler structure than H.
Remark 4 :- Canonical transformations are some time also called contact transformations
Result :- The set of all canonical transformation form a group. If in a phase space, we perform two canonical transformation in succession, the resulting transformation will again be canonical. Further more, a transformation that is inverse to a certain canonical transformation will always be canonical. The identity transformation
ii qq~ = ,
ii pp~ = ,
for i = 1, 2,…,n is canonical. Hence the result.
Example 1 :- Consider the transformation
ii q�q~ = ,
ii p�p~ = ,
for i = 1, 2,…,n, α ≠ 0, β ≠ 0, and α and β are constant. This transformation is canonical and it transforms the system (4) into the system (5) with
H~ = αβH.
Example 2 :- Consider the transformation
ii q�q~ = ,
ii p�p~ = ,
for i = 1, 2, …,n, α ≠ 0, β ≠ 0 and α, β are some constant. This transformation is canonical and it transform the system (4) into the system (5) with
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
186
H~ = −αβH.
Result :- A necessary and sufficient condition for the transformation
ii q~q~ = (t, qk, pk),
ii p~p~ = (t, qk, pk),
for i = 1, 2,…,n with
)p,q,...,p,q,p,q()p~,q~,...,p~,q~,p~,q~(
nn2211
nn2211
∂∂
≠ 0
to be canonical is the existence of a generating function F and some constant C for which
����
���� −=−
= =
n
1i
n
1iiiii t�H)q�p(Ct�H
~)q~�p~( −δF
is identically satisfied by virtue of the above transformation.
Note 1 :- The constant C is called the valence of the canonical transformation under consideration. The canonical transformation will be called univalent if
C = 1
for it.
Note 2 :- δF = t�tF
p�pF
q�qF
ii
ii
n
1i ∂∂+��
���
∂∂+
∂∂
�=
Note 3 :- In the literature, only univalent canonical transformations are frequently considered, and many authors erroneously hold that these transformations exhaust all the transformations that carry Hamiltonian systems again into Hamiltonian systems.
7.2 FREE CANONICAL TRANSFORMATION
Definition :- A canonical transformation is called a free canonical transformation if the inequality
)p,...,p,p()q~,...,q~,q~(
n21
n21
∂∂
≠ 0 …(1)
ANALYTICAL MECHANICS-IV
187
holds additionally.
Note 1 :- The inequality (1) for a free transformation ensures the independence of the quantities
t, q1, q2 ,…, qn, ,q~,...,q~,q~ n21
which can now be taken as the basic variables. Further the generalized momenta p1, p2,…, pn can now be expressed in terms of 2n+1 quantities t, qi, iq~ for i = 1, 2,…, n consequently the generating function F for a free canonical transformation is represented as
F(t, qi, pi) = S(t, qi, )q~i …(2)
For univalent (c = 1) free canonical transformation we obtain the following formulas
ii
pqS =
∂∂
, …(3)
iq~
S∂∂
= −pi, …(4)
tS
HH~
∂∂+= …(5)
in the equation (3) and (4), i = 1, 2,…,.
Example :- The canonical transformation
ii p�q~ =
ip~ = β qi
for i = 1, 2,…,n α ≠ 0, β ≠ 0 is free
Remark :- For a natural system, the coordinates q1, q2,…, qn, define the position of the system, and together with the momenta p1, p2, …, pn they define the state of the system, that is the positions and velocities of its points. This specificity of the coordinates is lost in a general-type canonical transformation. The quantities n21 q~,...,q~,q~ no longer define the position of the system, and only together with the n21 p~,...,p~,p~ do they define the state of the system. The variables n21 q~,...,q~,q~ will as before define the position of the system only in
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
188
the particular case of a point canonical transformation for which the functions iq~ (t, qk, pk) actually do not contain the momenta
ii q~q~ = (t, qk)
for i = 1, 2,…,n.
Note that subsequently the transformation of an arbitrary Hamiltonian system into a system with the function H of simple structure may be effected with the aid of a free canonical transformation. But a free canonical transformation is not a point transformation. Thus, non point canonical transformations play an essential role in the theory of Hamiltonian systems.
7.3 THE HAMILTON−−−−JACOBI EQUATION
Let there be given a holonomic system whose motion obeys the canonical equations of Hamilton :
i
i
pH
dtdq
∂∂= , …(1)
i
i
qH
dtdp
∂∂−= , …(2)
for i = 1, 2, …,n. We shall try to find/determine a free univalent canonical transformation such that in the transformed Hamiltonian system
i
i
p~H~
dtq~d
∂∂= , …(3)
i
i
q~H~
dtp~d
∂∂−= , …(4)
for i = 1 2,…,n and the function H~ will be identically zero, i.e.,
H~ ≡ 0. …(5)
Using (5), equations (3) and (4) reduce to
0dtp~d
,0dtq~d ii == , …(6)
for i = 1, 2,…,n. Integrating (6), we get
ANALYTICAL MECHANICS-IV
189
,�p~,�q~ iiii == …(7)
where αi and βi are 2n arbitrary constants. This implies that the new variables are also identically constant.
Knowing the canonical transformation, i.e., the relation between qi, pi and ii p~,q~ ; we can express all the qi and pi as functions of the time t and of the 2n
arbitrary constants αk, βk (k = 1, 2,…,n). That is, we can find the final equations of motion of the given holomonic system completely, i.e., all the solutions of the system (1) and (2).
We know that for such a univalent free canonical transform, there exists a generating function S = S(t, qi, )q~i for which
,tS
HH~
∂∂+= …(8)
,pqS
ii
=∂∂
ii
p~q~S −=
∂∂
…(9)
for i = 1, 2,…,n.
Using (5), (8) and (9) ; we write
0qS
,q,tHtS
ii =��
���
∂∂+
∂∂
. …(10)
The partial differential equation in (10) is called the Hamilton-Jacobi equation
The form of this equation is very simple to write down. Given a specific Hamiltonian function H = H(t, qi , pi), the momentum components are formally replaced by the partial derivative of S as in equation (9). The result is a first order partial differential equation. By assumption, the new generalised coordinates iq~ are constants. Hence, the form of the generating function is
S = S(t, q1, q2,…, qn, α1, α2,…, αn) …(11)
dependent on the original coordinates and possibly time. So S is a function of n + 1 variables and n parameters. The solution of the Hamilton-Jacobi
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
190
equation (10) is equivalent to finding the solution of the original canonical equations of motion. Besides the Hamilton-Jacobi equation (10), the condition
det��
��
�
∂∂∂
ji
2
�qS
≠ 0, …(12)
must hold for the generating function. As soon as the generating function S (t, qi, αi) is found, the formulas (9) will define the required/desired univalent free canonical transformation
Definition :- The solution of partial differential equation (10) of Hamilton-Jacobi containing n arbitrary constants α1, α2,…, αn is called the complete integral of this equation if the condition (12) is fulfilled.
In the above, we have proved the following theorem :
Theorem (Jacobi’s Theorem)
Statement : If S(t, qi, αi) is some complete integral of the Hamilton−Jacobi equation
0qS
q,tHtS
ii =��
���
∂∂+
∂∂
,
then the final equations of motion of a holomonic system with the given function H may be written in the form
ii
ii
��
S,p
qS =
∂∂=
∂∂
for i = 1, 2,…, n; and αi and βi are arbitrary constants.
Remark :- A knowledge of the complete integral of the partial differential equation (10) relieves us of the necessity of integrating the system of ordinary differential equations in (1) and (2).
Conservative System
When the Hamiltonian function H does not depend explicitly on time, then the Hamiltonian itself is a constant of motion.
H(qi, pi) = h …(1)
ANALYTICAL MECHANICS-IV
191
in which h is an energy constant. In this case the Hamilton−Jacobi equations reduces to
,0qS
,qHtS
ii =��
���
∂∂+
∂∂
…(2)
in which
S = S(t, q1, q2,…, qn, α1, α2,…, αn) …(3)
is the generating function for free canonical transformation with n parameters α1, α2,…, αn. From equations (1) and (2)
0htS =+
∂∂
. …(4)
We assume the following special form of S for time dependence of S.
S = −ht + V(q1, q2,…, qn, α1, α2,…,αn). …(5)
Substituting of (5) into equation (4) gives
H hqV
,qi
i =��
���
∂∂
. …(6)
Note 1 : Equation (6) is called the reduced Hamilton – Jacobi equation.
This is a first order partial differential equation in n dependent variables.
Note 2: Since the solution of (6) already dependents on the energy parameter h, there is no loss in assigning one of the parameters, say αn ; the constant h that is, αn = h, so
V = V(q1, q2,…, qn; α1, α2,…, αn−1, h). …(7)
Note 3: We have the following final equations of motion of a generalised conservative system.
ii
pqV =
∂∂
, 1 ≤ i ≤ n, …(8)
j�
V∂∂
= βj, 1 ≤ j ≤ n−1 …(9)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
192
hV
∂∂
= t + γ …(10)
where αj, βj, h and γ are arbitrary constants
7.4 METHOD OF SEPARATION OF VARIABLES
The Hamilton − Jacobi theory offers an elegant method by which the canonical equations can be solved by reducing the dynamical problem to that of finding a solution to a partial differential equation. Unfortunately, there is no general technique for construction of complete solutions to partial differential equations.
The most practical method, which works for certain classes of differential equations, is the method of separation of variables. In this method, we search for solutions in which the independent variables of the given PDE are grouped together so that the original problem converts into a collection of problems involving only ODE.
Under certain circumstances, it is feasible to assume certain convenient forms for the solution to the Hamilton−Jacobi equation. We consider the conservative Hamiltonian for which the generating function S is
S = −h t + V, …(1)
where
H ��
���
∂∂
ii q
v,q = h. …(2)
Let
H(qi, pi) = G[f1(q1, p1),…, fn (qn, pn)] . …(3)
Here, the variables in the expression for the function H are separated. Equations (2) and (3) imply
G ��
���
���
���
∂∂
��
���
∂∂
nnn
111 q
V,qf,...,
qV
,qf = h, …(4)
since
pi = iq
V∂∂
…(5)
ANALYTICAL MECHANICS-IV
193
we now put
fi ��
���
∂∂
ii q
V,q = αi …(6)
for i = 1, 2, …, n. The constants in (6) are otherwise arbitrary but by (4), must satisfy the relation.
G(α1, α2,…, αn) = h. …(7)
Solving (5) for iq
V∂∂
, we find
iq
V∂∂
= Fi(qi, αi) …(8)
for i = 1, 2,…, n Consequently, we obtain
V = [ ]iiii
n
1idq)�,q(F��
= …(9)
and then
S = − G(α1, α2,…, αn)t + [ ]iiii
n
1idq)�,q(F��
= …(10)
Remark 1. We find
i
i
ii
2
�
F�qS
∂∂=
∂∂∂
…(11)
for i = 1, 2,…, n, and
0�qS
ki
2
=∂∂
∂ …(12)
for i ≠ k and i, k, = 1, 2,…n. Hence, the condition.
det ��
���
∂∂∂
ki
2
�qS ≠ 0, …(13)
reduces (for this method) to
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
194
∏ ��
���
∂∂
=
n
1i i
i
�
f≠ 0. …(14)
Remark 2
1
i
i
i
i
pf
�
F−
��
���
∂∂
=∂∂
≠ 0, …(15)
for i = 1, 2,…,n.
Remark 3. The formula (10) defines a complete integral of the reduced Hamilton −Jacobi equations for a conservative system.
7.5 LAGRANGE BRACKETS
Let φ and ψi be 2n functions of the two variables p and q, for i = 1, 2,…,n. We define
[q p] = )p,.q(
)�,�( jjn
1j ∂∂
�=
= � ��
���
�
∂∂
∂∂
−∂
∂∂
∂
=
n
1j
jjjj
q
�
p
�
p
�
q
�. …(1)
[q p] are called Lagrange brackets.
We find
[q p] = −[p q], …(2)
which is an antisymmetric property.
Note : Comparing Lagrange brackets with Poisson brackets, we find that there were two functions φ and ψ of 2n variables qi, pi for Poisson brackets, whereas, there are 2n functions φi, ψi of two variables p, q for Lagrange brackets.
Canonical character of a transformation in terms of Lagrange brackets
We shall now derive the necessary and sufficient conditions that must be satisfied by 2n independent functions
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195
iq~ = ϕi(t, qk, pk),
ip~ = ψi(t, qk, pk), …(1)
for i, k = 1, 2,…,n, so that the transformation defined above in (1) should be canonical.
We know that the necessary and sufficient condition for the transformation (1) to be canonical is the existence of a generating function F = F(t, qi, pi) and some constant c for which the following identity hold.
��
���
�� −=−�==
n
1iiii
n
1ii t�Hq�pct�H
~q~�p~ − δF(t, qi, pi) …(2)
We assume that the transformation (1) is canonical. Then, the identity (2) holds. We take an arbitrary fixed value t = t (so δ t = 0) in (20. We write
��==
=δn
1ii
n
1ii q~p~ pi δqi − δF ( t , qi, pi) …(3)
But (3) is a defining identity for a transformation that does not contain the time explicity,
=iq~ φi( t , qk, pk), ip~ = ψi( t , qk, pk) …(4)
for i = 1, 2,…, n.
Hence, formulas (4) define a canonical transformation with valence C which is independent of the chosen value of t = t
On the contrary, let it now be given that all transformations obtained from the transformation (1) by replacing the variable t by various fixed values of t are canonical and with one and the same valence c. Then, defining the function H~ by the equation
�= ∂
∂+
∂∂+=
n
1i
ii t
q~p~
tF
CHH~ , …(5)
we get equation (2) from equations (3) and (5). Thus, we find that the transformation (1) that depends on the time t is canonical
Hence, for the time-dependent transformation (1) to be canonical it is necessary and sufficient that all the time-independent transformations obtained from the transformation (1) by replacing t with an arbitrary value of t be
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196
canonical and with one and the same valence c. For this reason, when establishing tests for canonical character, we can confine ourselves to canonical transformations that do not contain the time variable t explicitly :
iq~ = φi(qk, pk), …(6)
ip~ = ψi(qk, pk), …(7)
with
)p,...,p;q,...,q,q()p~,...,p~,q~,...q~,q~(
n1n21
n1n21
∂∂
≠ 0, …(8)
for i = 1, 2,…,n.
For the above canonical transformation (6) to (8), the defining identity (2) is now written as
��
���
� δ=δ ��==
k
n
1kkk
n
1kk qpCq~p~ − δK (qk, pk) …(9)
From (7), we write
� ��
���
�
∂∂
+∂∂
==
n
1ii
i
ki
i
kk p�
pq~
q�qq~
q~� . …(10)
let
Φi = �=
−∂∂n
1ki
i
kk Cp
qq~
p~ , …(11)
Ψi = �∂∂
=
n
1k i
kk q
q~p~ , …(12)
for i = 1, 2,…,n. Using (10)−(12) in (9), we obtain
�=
n
1i(Φi δqi + Ψi φpi) = −δK (qk, pk). …(13)
The conditions that the left-hand side of (13) is differential are
i
k
k
i
qq ∂Φ∂
=∂Φ∂
, …(14a)
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197
i
k
k
i
pp ∂Ψ∂
=∂Ψ∂
, …(14b)
i
k
k
i
qp ∂Ψ∂
=∂Φ∂
, …(14c)
where i, k = 1, 2,…,n. Substituting the expressions for Φi and Ψi from equations (11) and (12) into equations (14a−c), we obtain
� ��
���
�
∂∂
∂∂
−∂∂
∂∂
=
n
1j i
j
k
j
k
j
i
j
q
p~
q
q~
q
p~
q
q~ = 0, …(15a)
�=
��
���
�
∂∂
∂∂
−∂∂
∂∂n
1j i
j
k
j
k
j
i
j
p
p~
p
q~
p
p~
p
q~ = 0, …(15b)
�=
��
���
�
∂∂
∂∂
−∂∂
∂∂n
1j i
j
k
j
k
j
i
j
q
p~
p
q~
p
p~
q
q~= C δik, …(15c)
for i, k = 1, 2,…,n.
Here δik is the substitution tensor. Using Lagrange brackets, the above conditions in (15a−c) can be written as
[qi qk] = 0,
[pi pk] = 0,
[qi pk] = C δik …(16)
The equalities (16) express the necessary and sufficient conditions for the transformation (6) and (7) to be canonical.
7.6 JACOBIAN MATRIX OF A CANONICAL TRANSFORMATIONS
Let
Q = k
i
∂∂
…(1)
be the Jacobian matrix of order n. Let
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198
P = ,pp
k
i
∂∂
,pq~
Rk
i
∂∂
=
k
i
qp~
S∂∂
= …(2)
be other Jacobian matrices, each of order n. Let
M = ��
���
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
PSRQ
pp~
.....................qp~
...................................
...........................qp~
pq~
...pq~
qq~
...qq~
..................................
pq~
...pq~
qq~
...qq~
n
n
1
n
1
1
n
n
1
n
n
n
1
n
n
1
1
1
n
1
1
1
…(3)
be the Jacobian matrix of the canonical transformation
),p,q(q~ kkii φ=
ip~ = ψi(qk, pk). …(4)
Let E be a unit matrix of order n. Let
J = ��
���
� −OEEO
…(5)
be a matrix of order 2n.
Then
det J = 1 . …(5A)
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199
Since the transformation (4) is canonical, so the following conditions, in terms of Lagrange brackets, hold :
[qi qk] = 0,
[pi pk] = 0,
[qi pk] = C δik …(6)
Using (6), it can be checked (left as an exercise) that
M′ JM = C J. …(7)
For a univalent canonical transformation (c = 1), equation (7) becomes
M′ JM = J. …(8)
Definition 1: The matrix M that satisfies (7) is called a GENERALIZED − SIMPLICIAL matrix.
Definition 2: The matrix M for which (8) hold is called SIMPLICIAL.
For such matrices
det M = + Cn .
That is, simplicial matrices are non-singular.
Result :- All generalized−simplicial matrices ( for c ≠ 0) form a group and
det M = + Cn.
Note :- In view of the above, the test for the canonical character of the transformations may be stated as
“For a certain transformation
)p,q,t(p~p~),p,q,t(q~q~ kkiikkii ==
to be canonical, it is necessary and sufficient that the Jacobian matrix M corresponding to this transformation should be generalized−simplicial with constant valence C”.
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200
7.7 CONDITION OF CANONICAL CHARACTER OF A TRANSFORMATION IN TERMS OF POISSON BRACKETS
We know that the condition of canonicity of a transformation
),p,q,t(q~q~ kkii =
)p,q,t(p~p~ kkii = …(1)
(i, k = 1, 2,…) is that
M′ JM = C J , …(2)
where C ≠ 0 is the valence of the canonical transformation, M is a 2n×2n generlalized − simplicial matrix and
J = ��
���
−OEEO
…(3)
in which E is a unit matrix of order n. Matrix M is non-singular. Equation (2) gives
(M′)−1 (M′ JM) M−1 = (M′)−1 (c J) M−1
or J = C [(M′)−1 J M−1]
or (M′)−1 J M−1 = .JC1
…(4)
From equation (3), we find (exercise)
J−1 = −J. …(5)
Taking inverse of (4) both sides and using (5), we write
M J M′ = C J. …(6)
The equality (6) may be considered as obtained from equality (2) on replacing the Jacobian matrix M by its transpose M′. In view of definition of M,
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201
M =
����
����
�
∂∂
∂∂
∂∂
∂∂
k
i
k
i
k
i
k
i
pp~
qp~
pq~
qq~
, …(7)
the above substitution reduces to replacing the derivatives
k
i
k
i
k
i
k
i
pp~
,qp~
,pq~
,qp~
∂∂
∂∂
∂∂
∂∂
,
respectively, by the derivatives
i
k
i
k
i
k
i
k
pq~
,pq~
,qq~
,qq~
∂∂
∂∂
∂∂
∂∂
.
That is in each derivative the letters and indices above and below are interchanged. He know that in terms of Lagrange brackets, the equation (2) is equivalent to the following system of equalities :
[qi qk] = 0,
[pi pk] = 0,
[qi pk] = C δik …(8)
for i , k = 1, 2,…,n. In view of the discussion in the proceeding paragraph, the equation (6) will be equivalent to the following system of equalities ;
[qi qk]* = 0,
[pi pk]* = 0,
[qi pk]* = C δik …(9)
Here, the asterisk (*) indicates that the above mentioned interchange of derivatives is to be performed within Lagrange brackets. That is,
[qi qk]* = *
q
q~
q
p~
q
p~
q
q~n
1j k
j
i
j
k
j
i
j
���
�
���
�� �
�
���
∂∂
∂∂
−∂∂
∂∂
=
= � ��
��
�
∂∂
∂∂
−∂∂
∂∂
=
n
1j j
k
j
i
j
k
j
i
qq~
pq~
pq~
qq~
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= ),q~q~( ki …(10a)
where )q~q~( ki are Poisson brackets of the function iq~ and kq~ with respect to the independent variables q1, p1, q2, p2, Poisson brackets. Similarly,
[pi pk]* = ),p~p~( ki
)p~q~(]*pq[ kiki = …(10b)
Hence, the conditions of the canonicity of transformation (8) may be written in the following form in terms of Poisson brackets.
,0)q~q~( ki =
,0)p~p~( ki =
)p~q~( ki = C δik …(11)
for i, k = 1, 2,…,n.
7.8 INVARIANCE OF THE POSISSON BRACKETS IN A CANONICAL TRANSFORMATIONS
Consider two function
φ = φ(t, qi, pi) ψ = ψ(t, qi, pi) …(1) Let kk q~q~ = (pi, qi) kk p~p~ = (pi, qi) …(2) be the canonical transformation and its inverse be pi = pi )p~,q~( kk qi = qi )p~,q~( kk …(3)
Substituting the function pi and qi in (1) interm of kk p~,q~ with the help of (3), we can regard these same functions φ and ψ as functions of the variable
kk p~,q~ . Accordingly, the Poisson brackets of φ and ψ may be evaluated both with respect to the variable qi and pi and with respect to the variables ii p~,q~ .
Let (φ ψ) denote the Poissons brackets with respect to variables qi, pi and )��( denote the same with respect to variables .p~,q~ ii
We assert that (left as an exercise to the readers)
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203
(φ ψ) = C )��( , …(4)
where C is the valence of the canonical transformations
The converse of the above assertion also holds. That is, if for any two functions φ and ψ, the identity (4) is fulfilled for one and the same constant C ≠ 0, then the transition from the 2n variables qi, pi to the 2n variables ii p~,q~ is accomplished by a canonical transformation with valence C.
In particular, for a univalent canonical transformation (C = 1), we have
(φ ψ) = )��( . …(5)
This proves that the Poisson brackets are invariant under univalent canonical transformations.
This property of univalent canonical transformations singles out these transformations from among all possible transformations of phase space.
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Chapter-8
Nonlinear First-Order PDE
8.1 INTRODUCTION
We shall study PDE of the form
F(Du, u, x) = 0, …(∗)
where x∈U⊂ Rn, U is open, and
u : U →R ,
is the unknown, u = u(x). The function F is given.
Notation : We write
F = F(p, z, x)
= F(p1, p2,…,pn, z, x1, x2,…, xn)
for
p∈Rn, z∈R, x∈U.
Thus
“p” is the name of the variable for which we substitute the gradient Du(x), and “z” is the variable for which we substitute u(x).
We also assume hereafter that F is smooth, and set
DpF = ( )F...,F,Fnp,2p1p
DzF = Fz
DxF = ( )F...,F,Fnx,2x1x
We are concerned with discovering solutions u of the PDE (∗) in U, usually subject to the boundary condition
u = g on Γ, …(∗∗)
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205
where Γ is some given subset of ∂U and
g : Γ→R …(***)
is prescribed.
8.2 COMPLETE INTEGRALS
Consider the nonlinear first-order PDE
F(Du, u, x) = 0. …(1)
Suppose first A ⊂ Rn is an open set. Assume for each parameter
a = (a1, a2,…, an) ∈A,
we have a C2 solution
u = u(x ; a) …(2)
of the PDE (1). We write
)uD,uD( 2xaa =
�����
�
�
�����
�
�
nnn1n
2n212
1n111
axaxa
axaxa
axaxa
u...uu
u...uu
u...uu
Μ . …(3)
Definition : A C2 function u = u(x ; a) is called a complete integral in U×A, provided
(i) u(x ; a) solves the PDE (1) for each a∈A ,
(ii) rank )uD,uD( 2xaa = n ,
for x∈U, a∈A .
Remark : The condition (ii) above ensures u = u(x ;a)
“depends on all the n independent parameters a1, a2,…, an”.
Example 1. The Clairaut’s equation is the PDE (nonlinear)
x⋅Du + f(Du) = u, …(1)
where
f : Rn→R is given.
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A complete integral of PDE (1) is
u(x ; a) = a⋅ x f(a) , …(2)
for all x∈U and a∈Rn.
Example 2 : The eikonal PDE is (non-linear)
|Du| = 1. …(3)
A complete integral of PDE (3) is
u(x ; a, b) = a ⋅ x + b …(4)
for all x∈U ⊂ Rn, a∈∂B(0, 1), b∈R.
Example 3 : The Hamilton−Jacobi equation from mechanics is, in its simplest form,
the PDE (nonlinear)
ut + H(Du) = 0, …(5)
where
H : Rn → R
is given.
In PDE(5), u depends on x = (x1, x2,…, xn) ∈Rn and t∈R.
A complete integer of PDE (5) is
u(x, t; a, b) = a⋅x − t H(a) + b, …(6)
for x∈Rn, t ≥ 0 and a∈Rn, b∈R.
New Solutions as envelopes of Complete Integrals
Definition : Let u = u(x; a) be a C1 function of x∈U, a∈A, where U⊂Rn and A⊂Rm are open sets. Consider the vector PDE
Da u (x ; a) = 0, …(1)
for x∈U and a∈A.
Suppose that we can solve (1) for the parameter a as a C1 function of x, of the form
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207
a = φ(x). …(2)
Then
Dau(x; φ(x)) = 0,
for all x∈U. …(3)
We then call
v(x) = u(x; φ(x)),
x∈U …(4)
the envelope of the functions
{u(⋅ ; a)}a∈A.
Result : By forming envelopes of complete integrals (or of other m-parameter families of solutions), we construct new solutions of given nonlinear first order PDE.
Such solution is called a Singular Integral of given equation
F(Du, u, x) = 0.
Example : Consider the nonlinear first order PDE
u2(1 + |Du|2) = 1. …(1)
A complete integral of PDE (1) is
u(x ; a) = + (1−|x−a|2)1/2, …(2)
for
|x−a| < 1. …(3)
we compute
Dau = 2/1|)ax|1(
)ax(−−−±
. …(4)
The vector equation
Dau = 0 , …(5)
gives
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208
a = x ≡ φ(x), say. …(6)
Thus
v(x) = u(x, φ(x))
or
v(x) = + 1 , …(7)
are singular integrals of nonlinear PDE (1).
8.3 CHARACTERISTICS METHOD
Consider basic nonlinear first-order PDE
F(Du, u, x) = 0 in U, …(1)
subject to the boundary condition
u = g on Γ, …(2)
where Γ ⊆ ∂ U and
g : Γ→R …(3)
are given.
We hereafter suppose that F, g are smooth functions.
PLAN :- We develop next the method of characteristic, which solves (1) and (2) by converting the PDE into an appropriate system of first order ODE.
Method
Suppose u solves (1), (2) and fix any point x∈U. We would like to calculate u(x) by finding some curve lying within U, connecting the point x with a point x0∈Γ and along which we can compute u.
Since the boundary condition (2) says that the function u is known on Γ, the value of u at the one end x0 becomes known.
We hope then to be able to calculate u all along the curve, and so in particular at the point x.
Let us suppose that this curve is described parametrically by the function
x (s) = (x1(s), x2(s),…, xn(s)) …(4)
the parameter s lying in some subinterval of R.
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209
Assuming u is a C2 solution of PDE (1), we define also
z(s) = u( x (s)). …(5)
We also set
p (s) = Du( x (s)), …(6)
i.e.,
p (s) = (p1(s), p2(s),…, pn(s)), …(7)
where
pi(s) = ixu ( x (s)), …(8)
for i = 1, 2,…, n.
The function z(⋅) gives the value of u along the curve and p (⋅) determines the values of the gradient Du.
We must choose the function x (⋅) in such a way that we can compute z(⋅) and p (⋅).
For this, we differentiate (8) and write
),s(x))s(x(u)s(p jn
1jjxix
i && �=
= …(9)
for i = 1, 2,…,n.
Here, dot (⋅) signifies d/ds. The right side of (9) involves the second derivatives of u. On the other hand, we can also differentiate the PDE (1) with respect to xi, to get
ix
n
1jixjx
j
u)x,u,Du(zF
u)x,u,Du(pF
��
��
∂∂+
��
�
��
��
∂∂
=
+ )x,u,Du(xF
i∂∂
= 0, …(10)
for i = 1,2,…,n.
We are able to employ this identity (10) to get rid of the “dangerous” second derivative terms in (9), provided we first set
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210
))s(x),s(z),s(p(pF
)s(xj
j
∂∂=& , for j = 1, 2,…,n. …(11)
Assuming now (11) holds, we evaluate (10) at x = x (s), obtaining from (5), (6), the identity,
�= ∂
∂n
1jjxix
j
))s(x(u))s(x),s(z),s(p(pF
+ )s(p))s(x),s(z),s(p(zF i
∂∂
+ 0))s(x),s(z),s(p(xF
i
=∂∂
, …(12)
for i = 1, 2,…,n.
Substituting from (11) and (12) into equation (9) we write
))s(x),s(z),s(p(xF
)s(pi
i
∂∂−=&
− ),s(p))s(x),s(z),s(p(zF i
∂∂
…(13)
for i = 1, 2,…,n.
Finally, we differentiate (5) and obtain
)s(x))s(x(xu
)s(z jn
1j j
&& �= ��
�
��
��
∂∂=
= ���
�
��
��
∂∂
=
n
1j j
j ,))s(x),s(z),s(p(pF
)s(p …(14)
by using equations (8) and (11).
We now summarize equations (11), (13) and (14) and rewrite in vector notation
)s(p))s(x),s(z),s(p(FD))s(x),s(z),s(p(FD)s(p zx −−=& , …(15A)
)s(p)}.s(x),s(z),s(p(FD{)s(z p=& , …(15B)
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211
)).s(x),s(z),s(p(FD)s(x p=& …(15C)
The system (15) consists of (2n+1) first order ODE. It comprises the characteristic equations of the given nonlinear first order PDE (1).
The functions
))(p),...,(p),(p()(p n21 ⋅⋅⋅=⋅ ,
z )(⋅ ,
)),(x),...,(x),(x()(x n21 ⋅⋅⋅=⋅
are called the characteristics.
Remark 1 : In the above, we have proved the following theorem regarding the “Structure of Characteristic ODE”.
Statement : Let u∈C2(U) solve the nonlinear first-order PDE (1) in U. Assume )(x ⋅ solves the ODE (15c). Then )(p ⋅ solves the ODE (15a) and )(z ⋅ solves the ODE (15b), for those s such that )s(x ∈U.
Remark 2 : We still need to discover appropriate initial conditions for the system of ODE (15), in order that this theorem be useful.
Remark 3 : The form of the full characteristic equations can be quite complicated for fully nonlinear first order PDE, but sometimes a remarkable mathematical structure emerges.
Question 1. Derive the characteristic equations for the linear and homogeneous PDE
F(Du, u, x) = b (x) ⋅ Du(x) + c(x) u(x) = 0, …(1)
for x∈U. Hence, solve the problem
x1 uuxu1x22x =− in U
u = g on Γ,
where U is the quadrant {x1 > 0, x2 > 0} and
Γ = {x1 > 0, x2 = 0} ⊆ ∂ U.
Solution Part I.
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212
We write
nR)x(Dup ∈= , …(1)
R)x(uz ∈= . …(2)
Then , the given equation can be written as
F(p, z, x) = z)x(cp)x(b +⋅ = 0 …(3)
So
DpF = )x(b . …(4)
The characteristic equation (15c) of the Article now becomes
)),s(x(b)s(x =& …(5)
which is an first order ODE involving only the function x (⋅), and can be solved easily.
The characteristic equation (15b) now becomes
)s(p))}.s(x(b{)s(z =& …(6)
Equations (1), (2) and (6) simplify to
)s(z))s(x(c)s(z −=& . …(7)
This is a first order linear ODE in z(⋅), once the function x (∗) is known from ODE (5).
Thus, equations (5) and (7) comprise
)),s(x(b)s(x =& …(8a)
),s(z))s(x(c)s(z −=& …(8b)
the characteristic equations for the linear first order PDE (3).
Part −−−−II : Comparing the given PDE with standard PDE (3), we find
),x,x(x 21=
b = (−x2, x1),
c = −1. …(9)
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213
Thus, the system (8a, b) for this present problem consists of
,xx 21 −=&
,xx 12 =& …(10)
and
.zz =& …(11)
Solving the system of two ODE in (10), we find (exercise)
scosx)s(x 01 = ,
ssinx)s(x 02 = . …(12)
Solution of (11) is
z(s) = z0 es
= g(x0) es, …(13)
where x0 ≥ 0 and 0 ≤ s ≤ π/2 .
Fix a point (x1, x2) ∈U. We select s > 0, x0 > 0 so that
(x1, x2) = (x1(s), x2(s)) .
This gives x0 = (x12 + x2
2)1/2,
s = tan−1 (x2/x1). …(14)
Therefore
u(x1, x2) = u(x1(s), x2(s))
= z(s)
= g(x0) es
= g((x12 + x2
2)1/2) exp [tan−1 (x2/x1)] , …(15)
as the solution of the given boundary-value problem.
Article : Derive the characteristic equations for the quasilinear PDE of the form
F(Du, u, x) = b (x, u (x)) ⋅ Du(x) + c(x, u(x)) = 0.
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214
Hence, solve the boundary-value problem
22x1x uuu =+ in U ,
u = g on Γ,
where U is the half-space {x2 > 0} and Γ = {x2 = 0} = ∂U.
Solution : Part I.
For
��
�
∈=∈=R))s(x(u)s(z
R))s(x(Du)s(p n
, …(1)
the given nonlinear PDE is
F .0)z,x(cp)z,x(b)x,z,p( =+⋅= …(2)
We obtain
DpF = b (x, z). …(3)
Hence, characteristic equation (15c) now becomes
))s(z),s(x(b)s(x =& . …(4)
Characteristic equation (15b) becomes
)s(p))}s(z),s(x(b{)s(z ⋅=&
= − c )),s(z),s(x( …(5)
Using (2). So, the characteristic equations for the quasi linear first order PDE consists of two ODE (4) and (5).
Part-II. In this example,
),1,1(b =
c = −z2,
x = (x1, x2) …(6)
ODE equations (4) and (5) become
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215
���
�
=
=
1x
1x2
1
&
&, …(7)
and
2zz =& . …(8)
Solving (7), we get
x1(s) = x0 + s,
x2(s) = s. …(9)
Solving (8), we get (exercise)
z(s) = 0
0
sz1z
−
= )x(sg1
)x(g0
0
−, …(10)
where x0∈R, s ≥ 0, provided the denomination is not zero.
For a point (x1, x2) ∈U, we select s > 0 and x0∈R so that
(x1, x2) = (x1(s), x2(s))
= (x0 + s, s) ,
i.e., x0 = x1 − x2,
s = x2. …(11)
Then
u(x1, x2) = u(x1(s), x2(s))
= z(s)
=)x(sg1
)x(g0
0
−.
)xx(gx1
)xx(g
212
21
−−−
= …(12)
is the solution, provided the denominator is non-zero.
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216
8.4 CHARACTERISTIC FOR THE HAMILTON−−−−JACOBI EQUATION
The general Hamilton−Jacobi PDE is
G(Du, ut, u, x, t) = ut + H(Du, x) = 0, …(1)
where
Du = Dxu = ),u,...,u,u(nx2x1x
H : Rn→R,
x = (x1, x2,…, xn) ∈Rn,
t∈R.
We set
t = xn+1 ,
q = (p, pn+1)
z = u(x) ,
y = (x, t). …(2)
We have
G(q, z, y) = pn+1 + H(p, x) …(3)
So, DqG = (DpH(p, x), −1), …(4)
DyG = (Dx H(p, x), 0), …(5)
Dz G = 0. …(6)
The characteristic equation (15c) of the main article yields
));s(x),s(p(pH
)s(xi
i
∂∂=&
1)s(x 1n =+& , …(7)
for i = 1,2,.., n .
NONLINEAR FIRST-ORDER PDE
217
In particular, we can identity the parameter s with the time t.
The characteristic equation (15a) for Hamilton − Jacobi PDE (1) reads as
)),s(x),s(p(xH
)s(pi
i
∂∂−=&
.0)s(p 1n =+& …(8)
)ni1( ≤≤ .
The characteristic equation (15b) is now
���
�
−⋅=
+⋅= +
),s(x),s(p(H)s(p))s(x),s(p(HD
),s(p)s(p))s(x),s(p(HD)s(z
p
1np&
…(9)
using equations (1) and (3).
In summary, the characteristic equations for the given Hamilton−Jacobi PDE(1) are the following set of ODE.
)),s(x),s(p(HD)s(p x−=& …(10A)
)),s(x),s(p(H)s(p))}.s(x),s(p(HD{)s(z p −=& …(10B)
)),s(x),s(p(HD)s(x p=& …(10c)
for
)),(p),...,(p),(p()(p n21 ⋅⋅⋅=⋅
))(x),...,(x),(x()(x n21 ⋅⋅⋅=⋅ ,
and
)(z ⋅ .
Definition : Equalities (10A) and (10C), i.e.,
)x,p(HDx p=& , …(11A)
)x,p(HDp x−=& , …(11B)
are called Hamilton’s Equations.
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218
Note 1 : Obverse that the ODE (10B) for z(⋅) is trivial, once x (⋅) and p (⋅) have been found by solving Hamilton’s equations (11A, B).
Note 2 : The initial-value problem for the Hamilton−Jacobi equation does not, in general, have a smooth solution u lasting for all times t > 0.
Initial-value problem for the Hamilton−−−−Jacobi Equations.
Problem : Consider the initial-value problem for the Hamilton-Jacobi equation :
ut + H(Du) = 0 in Rn ×(0, ∞) …(1)
u = g on Rn × {t = 0}. …(2)
Here
u : Rn × [0, ∞) → R ,
is the unknown, u = u(x, t), and
Du = Dxu = )u,...,u,u(nx2x1x . …(3)
The Hamiltonian
H : Rn → R …(4)
and the initial function
g : Rn → R, …(5)
are given.
Remark : We have derived earlier two Hamilton’s ODE. In the next section, we shall derive them from a variational principle.
8.5 DERIVATION OF HAMILTON’S ODE FROM A VARIATIONAL PRINCIPLE
Assume that
L : Rn × Rn → R , …(1)
is a given smooth function.
We call L to be Lagrangian.
We write
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219
L = L(q, x) = L(q1, q2,…, qn, x1, x2,…, xn) …(2)
for q ∈ Rn and x∈Rn and
��
�
=
=
).L,...,L,L(LD
),L,...,L,L(LD
nx2x1xx
nq2q1qq …(3)
Now fix two points x, y ∈Rns and a time t > 0.
We introduce the action functional
I �=⋅t
0ds))s(w),s(w(L)](W[ & , …(4)
in which
.ds
)s(wd)s(w =&
Here, the functional (4) is defined for functions
w (⋅) = (w1(⋅), w2(⋅),…, wn(⋅)) , …(5)
belonging to the admissible class
A = ( ){ }.x)t(w,y)0(wR];t,0[C)(w n2 ==∈⋅ . …(6)
Thus a C2 curve )(w ⋅ belongs to A if it starts at the point y at time 0, and reaches the point x at time t.
According to the calculus of variations, we shall find a curve x (⋅) ∈A such that
)](w[Imin)](x[IA)(w
⋅=⋅∈⋅
. …(7)
That is, we are seeking a function )(w ⋅ which minimizes the functional I[⋅], given in equation (4), among all admissible functions/candidates )(w ⋅ in class A.
Theorem (Euler−−−−Lagrange equations)
Statement : Prove that any minimizer )(x ⋅ belonging to the admissible class
A = { )(w ⋅ ∈C2 ([0, t]; Rn): w (0) = y, w (t) = x} …(1)
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220
of the action functional
I[ w (⋅)] = �t
0ds))s(w),s(w(L & , …(2)
solves the system
− { } ,0))s(x),s(x(LD))s(x),s(x(LDdsd
xq =+ && (0 ≤ s ≤ t) …(3)
of Euler−Lagrange ordinary differential equations.
Proof : Choose a smooth function
v : [0, t] → Rn …(4)
satisfying
v (0) = v (t) = 0, …(5)
and
v = (v1, v2,…, vn).
For τ∈R, define
)(w ⋅ = ).(v�)(x ⋅+⋅ …(6)
Then )(w ⋅ belongs to the admissible class A and )(x ⋅ being the minimizer of the action functional, we write
I )].(w[I)](x[ ⋅=≤⋅ …(7)
Therefore, the real-valued function
i : R → R …(8)
defined by
i(τ) = I )](v�)(x[ ⋅+⋅ , …(9)
has a minimum at
τ = 0. …(10)
Consequently,
i′(0) = 0, …(11)
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221
provided i′(0) exists.
Now, we shall compute this derivative explicitly. We find
i(τ) = � ++t
0)]s(v�)s(x),s(vt)s(x[L && ds,
so that
i′(τ) = ( ) ( ){ }� � ��
���
� τ+++τ+τ+=
t
0
n
1i
iixiq vvx,vxLvvx,vxL &&&&& i ds .
Setting τ = 0 and using the relation (11), we find
0 = { }� ��
���
�� +
=
n
1i
t
0
iix
iiq dsv)x,x(Lv)x,x(L &&& . …(12)
Integrating by parts in the first term inside the integral and using conditions in (5), we find
0 = ( )� ��
���
��
�
�� +−
=
n
1i
t
0
iixiq dsv)x,x(L)x,x(L
dsd && . …(13)
This identity (13) is valid for all smooth functions v = (v1, v2,…,vn) satisfying the boundary conditions (5), so we must have
( ) ,0)x,x(L)x,x(Ldsd
ixiq =+− && …(14)
for all i = 1, 2,…,n, and, 0 ≤ s ≤ t. Hence, in vector form,
( ) ,0))s(x),s(x(LD))s(x),s(x(LDdsd
xq =+− && …(15)
for 0 ≤ s ≤ 1.
This completes the proof
Note (1) Equation (15) is a vector equation. It consists of n coupled second-order ODE.
Note (2) It is of course possible that a curve x (⋅) ∈A may solve the EL-equations without necessarily being a minimizer. In such a case, we say that solution x (⋅) is a critical point of the action functional I[⋅].
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
222
So every minimizer of a functional is a critical point, but a critical point need not be a minimizer.
Example : If, we take
L(q, x) = 2|q|m21 − φ(x)
where m>0, the corresponding Euler−Lagrange equation is
m ))s(x(f)s(xρ&& =
for
fρ
= −Dφ.
This is Newton’s law for the motion of a particle of mass m moving in the force field f generated by the potential φ.
8.6 DERIVATION OF HAMILTON’S ODE
Assume that the C2 function )(x ⋅ is a critical point of the action functional. Thus, it solves the Euler−Lagrange equations (or vector equation)
( ) ,0))s(x),s(x(LD))s(x),s(x(LDdsd
xq =+− && 0 ≤ s ≤ t . …(1)
First we set
)),s(x),s(x(LD)s(p q&= 0 ≤ s ≤ t. …(2)
p (⋅⋅⋅⋅) is called the generalized momentum corresponding to the position x (⋅⋅⋅⋅) and velocity x&(⋅⋅⋅⋅).
We now make the following important hypothesis.
Hypothesis : Suppose for all x, p∈Rn that the equation
p = DqL(q, x) , …(3a)
can be uniquely solved for q as a smooth functions of p and x,
q = q(p, x). …(3b)
Definition : The Hamiltonian H associated with the Lagrangian L is defined to be
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223
H(p, x) = p q (p, x) −L(q (p, x), x) , …(4)
for p, x ∈Rn. The function q (⋅, ⋅) is defined implicitly by (3).
We now convert the Euler−Lagrange equations into Hamilton’s equation.
We rewrite the Euler−Lagrange equations in terms of p (⋅) and x (⋅). For this purpose, we state and prove a theorem.
Statement : The functions x (⋅) and p (⋅) satisfy Hamilton’s equations :
��
�
−=
=
)),s(x),s(p(HD)s(p
))s(x),s(p(HD)s(x
x
p
&
& …(5)
for 0 ≤ s ≤ t.
Furthermore, the mapping
s→H( ))s(x),s(p …(6)
is constant.
Proof : Let us hereafter write
))(q),...,(q),(q()(q n21 ⋅⋅⋅=⋅ . …(7)
From equation (4), we compute, for 1 ≤ i ≤ n,
�∂∂−
∂∂
∂∂−
∂∂=
∂∂
=
n
1k ii
k
ki
k
ki
)x,q(xL
)x,p(xq
)x,q(qL
)x,p(xq
p)x,p(xH
ix
L∂∂−= (q, x), …(8)
using (3). Also
�∂∂
∂∂−
∂∂+=
∂∂
=
n
1k i
k
ki
k
ki
i
)x,p(pq
)x,q(qL
)x,p(pq
p)x,p(q)x,p(pH
= qi(p, x) …(9)
by using again (3).
Thus
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
224
))s(x),s(p(q))s(x),s(p(pH i
i
=∂∂
= )s(x i& ; …(10)
and likewise
))s(x)),s(x),s(p(q(xL
))s(x),s(p(xH
ii ∂∂−=
∂∂
= − ))s(x),s(x(xL
i
&∂∂
= − ���
����
�
∂∂
)),s(x),s(x(qL
dsd
i
& ,
= − )s(p i& . …(11)
Equations (10) and (11) are required Hamilton’s ODE.
These equations comprise a coupled system of 2n first order ODE for xi(⋅) and pi(⋅)
for 1 ≤ i ≤ n.
Finally, we observe
� ��
���
�
∂∂+
∂∂=
=
n
1i
i
i
i
i
xxH
ppH
))s(x),s(p(Hdsd &&
=� ��
���
����
����
�
∂∂−
∂∂+��
�
����
�
∂∂−
∂∂
=
n
1i iiii
,pH
xH
xH
pH
= 0 , (12)
using (10) and (11). Equation (12) shows that the mapping (6) is constant.
This completes the proof.
8.7 LEGENDRE TRANSFORM
Now we try to find a connection between the Hamilton−Jacobi PDE and the calculus of variations problem−minimizing of the action functional.
To simplify further, we also drop the x-dependence in the Hamiltonian so that
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225
H = H(p). ….(1)
We hereafter suppose that the Lagrangian
L : Rn → R ,
satisfies the following conditions :
(a) the mapping
q → L(q) …(2)
is convex and
(b) �
��
∞→ |q|)q(L
lim|q|
= + ∞. …(3)
Result : The convexity of the mapping in (2) implies L is continuous.
Definition : The Legendre transform of L is denoted by L*(p) and is defined as
L*(p) = nRq
sup∈
{p⋅q − L(q)}, …(4)
for p∈Rn.
Theorem : Show that the Hamiltonian H can be obtained from the Lagrangian L. Establish the relation.
Proof : Suppose that the Lagrangian L satisfies the conditions (2) and (3). Let L* denote the Legendre transform of L, defined in (4).
We note that in view of (3), the “sup” in the definition of L* in (4) is really a “max”. That is, there exists some q*∈Rn for which
L*(p) = p⋅q*−L(q*), …(5)
and the mapping
q → p⋅q − L(q) …(6)
has a maximum at q = q*.
But then
p = DL(q*), …(7)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
226
provided L is differentiable at q*. Hence the equation
p = DL(q)
is solvable for q in terms of p, as
q* = q(p) . …(8)
Therefore, from equations (5) and (8), we write
L*(p) = p q (p) − L(q(p)). …(9)
From, definition of the Hamiltonian associated with the Lagrangian L, we write
H(p) = p⋅ q (p) − L( q (p)), for p∈Rn …(10)
In equation (10), the x-dependence in the Hamiltonian is dropped for simplification, as the variable x is not appearing.
From equations (9) and (10), we write.
H(p) = L*(p) for p∈Rn .
Hence
H = L* . …(11)
This gives the formula to obtain the Hamiltonian H from the Lagrangian L, under certain conditions.
Theorem : Prove that L = H*, under certain assumptions.
Proof : This theorem gives us a formula to compute L, when H is given. It states that L is the Legendre transform of H. We have already checked that H is the Legendre transform f L under certain conditions.
Thus, we shall say that H and L are dual convex functions.
To prove the result, we assume that the Lagrangian
L : Rn→R,
L = L(q) for q∈Rn , …(1)
satisfies the conditions
a) the mapping
q → L(q) …(2)
NONLINEAR FIRST-ORDER PDE
227
is convex and
b) |q|)q(L
limq ∞→
= + ∞. …(3)
we know that the Legendr transform, L*(p), is defined as
L*(p) = nRq
sup∈
{p⋅q − L(q)} …(4)
for p∈Rn. We also know that the Hamiltonian H is given by the formula
H = L*. …(5)
To achieve the desired result, we shall show that
i) the mapping
p →H(p) …(6)
is convex and
ii) |p|)p(H
lim|p| ∞→
= + ∞, …(7)
iii) L = H*. …(8)
For each fixed q, the function
p→p ⋅q −L(q) …(9)
is linear, and consequently, the mapping
p→H(p) = L*(p),
= nRq
sup∈
{p⋅q − L(q)}, …(10)
is convex, using (4) and (5).
Indeed, if 0 ≤ τ ≤ 1, p and p ∈Rn, then
H(τp + (1−τ) p ) = q
sup {(τp + (1−τ) p ) ⋅q − L(q)}
≤ τq
sup {p ⋅q − L(q)}
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
228
+ (1−τ) q
sup { p ⋅ q − L(q)}
= τ H(p) + (1−τ) H( p ) …(11)
This proves part (i) in (6) that the mapping is convex.
To prove (ii), fix any λ > 0 and p ≠ 0. Then
H(p) = nRq
sup∈
{p⋅q − L(q)},
using (10) or (4) and (5).
≥ λ |p| − L ���
����
�λ
|p|p
, on taking q = |p|
pλ ∈ Rn
≥ λ |p| − ),0(BLmax
λ
Thus,
lim |p|)p(H
inf|p| ∞→
≥ λ for all λ > 0. …(12)
This proves (ii) in (7).
To prove (iii) in (8), equation (10) gives
H(p) + L(q) ≥ p⋅q , …(13)
for all p, q ∈Rn. Consequently,
L(q) ≥ nRp
sup∈
{p ⋅q − H(p)}
= H*(q) .
This gives
L(q) ≥ H*(q) for all q∈Rn. …(14)
On the other hand,
H*(q) = nRp
sup∈ �
��
−⋅−∈
)}r(Lrp{supq,pnRr
,
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229
= nRp
sup∈ �
�
∈ nRrinf {p⋅(q−r) + L(r)}
�
, …(15)
by definition of Legendre transform and properties of sup and inf. Since the mapping
q→L(q)
is convex, so there exists s∈Rn such that
L(r) ≥ L(q) + s⋅ (r−q), for r ∈Rn . …(16)
Taking p = s in (15) and using (16), we compute
H*(q) ≥nRr
inf∈
{s⋅(q−r) + L(r)}
= L(q) .
This gives
H*(q) ≥ L(q) for all q∈Rn. …(17)
From equations (14) and (17), we find
L(q) = H*(q) for all q∈Rn
Hence
L = H*. …(18)
This proves part (iii) in equation (8). Hence, the proof of the theorem is complete.
8.8 HOPF−−−−LAX FORMULA
Consider the initial-value problem for the Hamilton - Jacobi equation
ut + H(Du) = 0 in Rn × (0, ∞) …(1)
u = g on Rn × {t = 0}. …(2)
We know that the calculus of variations problem with Lagrangian L leads to Hamilton’s ODE for the associated Hamiltonian H. Since these ODE are also the characteristic equations of the Hamilton-Jacobi PDE, we infer there is probably a direct connection between this PDE and the calculus of variations.
Theorem : (Hopf-Lax formula)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
230
Statement : If x∈Rn and t > 0, then prove that the solution u = u(x, t) of the minimization problem
u(x, t) = inf�
��
==� + x)t(w,y)0(w)y(gds))s(w(Lt
0
& ,
the infimum taken over all C1 functions
w : [0, t] → Rn
satisfying
w (t) = x,
is
u(x, t) = �
�� +�
�
���
� −∈
)y(gt
yxLtmin
nRy.
Proof : Fix any y ∈Rn and define
w (s) = y ),yx(ts −+ …(1)
for 0 ≤ s ≤ t. Then
w (0) = y …(2)
and
w (t) = x. …(3)
It is given that
u(x, t) = inf �
��
==� + x)t(w,y)0(w)y(g))s(w(Lt
0
& . …(4)
It implies that
u(x, t) ≤ �t
0))s(w(L & ds + g(y), …(5)
by definition of infimum. Equations (1) and (5) yield
u(x, t) ≤ � ��
���
� −t
0 tyx
L ds + g(y)
NONLINEAR FIRST-ORDER PDE
231
= t L ��
���
� −t
yx+ g(y) .
This gives
u(x, t) ≤ �
�� +�
�
���
� −∈
)y(gt
yxLtinf
nRy. …(6)
On the other hand, if w (⋅) is any C1 function satisfying the condition
,x)t(w = …(7)
we have, by Jensen’s inequality (exercise),
L �� ≤���
����
� t
0
t
0
.ds))s(w(Lt1
ds)s(wt1 && …(8)
If we write
w (0) = y, …(9)
we find
t L �≤+��
���
� − t
0))s(w(L)y(g
tyx & ds + g(y),
and consequently,
)}y(gds))s(w{(inf)y(gt
yxLtinf
wnRy+≤
�
�� +�
�
���
� −∈
& ,
= u (x, t), …(10)
by definition (4). Equations (6) and (10) yield the desired Hopf-Lax formula for the given variational problem stated in the statement of the theorem.
This completes the proof of Hopf-Lax formula.
Remark : We propose now to investigate the sense in which u so defined above (as a minimization problem) actually solves the initial-value problem for the Hamilton-Jacobi PDE.
ut + H(Du) = 0 in Rn×(0, ∞) …(1)
u = g on Rn ×{t = 0}. …(2)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
232
Recall we are assuming H is
i) smooth , …(3)
ii) convex, and
iii) |p|)p(H
lim|p| ∞→
= + ∞. …(4)
We henceforth suppose also
g : Rn → R , …(5)
is Lipschitz continuous, i.e.,
Lip (g) = <�
��
−−
≠∈ |yx|
|)y(g)x(g|sup
yxnRy,x
∞. …(6)
Our ultimate goal is showing “Hopf−Lax formula” provides a reasonable “weak solution” of the initial-value problem (1) for the Hamilton-Jacobi PDE.
First, we record some preliminary observations/properties of the function u = u(x, t) defined earlier by the Hopf−Lax formula.
Lemma 1 : (known as a functional identity)
Statement : For x∈Rn and 0 ≤ s ≤ t, we have
u(x, t) = .)s,y(ustyx
L)st(minnRy �
�� +�
�
���
�
−−−
∈
In other words, to compute u (⋅, t), we can calculate u at time s and then use u(⋅, s) as the initial condition on the remaining time interval [s1, t].
Proof of Lemma 1 : For y∈Rn and 0 < s < t and choose z∈Rn so that
u(y, s) = s L ��
���
� −s
zy + g(z). …(1)
by virtue of Hopf-Lax formula.
Further
��
���
� −��
���
�+��
���
�
−−
��
���
� −=−s
zyts
styx
ts
1t
zx, …(2)
NONLINEAR FIRST-ORDER PDE
233
and 0 < .1ts < Since L is convex, so we have
L ��
���
� −��
���
�+��
���
�
−−
��
���
� −≤��
���
� −s
zyL
ts
styx
Lts
1t
zx. …(3)
Thus, combining with Hope-Lax formula
u(x, t) = �
�� +�
�
���
� −∈
)y(gt
yxLtmin
nRy, …(4)
we write
u(x, t) ≤ t L )z(gt
zx +��
���
� −
≤ (t − s) L ��
���
� −+��
���
�
−−
szy
Lsstyx
+ g(z)
= (t −s) L ��
���
�
−−
styx
+ u (y, s), …(5)
using the relation (1). The inequality (5) is true for each y∈Rn. Therefore, relation (5) gives
u(x, t) ≤ �
�� +�
�
���
�
−−−
∈)s,y(u
styx
L)st(minnRy
. …(6)
Now, it remains to prove that (to complete the proof of Lemma)
�
�� +�
�
���
�
−−−
∈)s,y(u
styx
L)st(minnRy
≤ u(x, t) . …(7)
To prove (7), we now choose w such that
u(x, t) = t L ��
���
� −t
wx+ g(w), …(8)
and set
y = wts
1xts
��
���
� −+ . …(9)
Then
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
234
s
wystyx
twx −=
−−=−
. …(10)
Consequently
(t −s) L ��
���
�
−−
styx
+ u(y, s)
≤ (t −s) L ��
���
� +��
���
� −+��
���
� −)w(g
swy
Lst
wx,
= t L ��
���
� −t
wx+ g(w), …(11)
= u(x, t),
using (8) and (10). Hence
�
�� +�
�
���
�
−−−
∈)s,y(u
styx
L)st(minnRy
≤ u(x, t). …(12)
Results (6) and (12) combine together prove the desired result.
This completes the proof of Lemma
Lemma 2 : (Lipschitz Continuity)
Statement : The function u is Lipschitz continuous in Rn×[0, ∞) and
u = g on Rn×{t = 0}.
Proof : For t > 0 and x, x ∈Rn. Choose y∈Rn such that
t L ��
���
� −t
yx+ g(y) = u(x, t), …(1)
using Hopf-Lax formula. Then
u( x , t) − u(x, t) = )y(gt
yxLt)z(g
tzx
LtInfz
−��
���
� −−�
�� +�
�
���
� −
≤ g( x −x +y) −g(y), on taking z = x −x +y
≤ Lip (g) {| x −x|},
NONLINEAR FIRST-ORDER PDE
235
as g is Lip continuous. Hence
u( x , t) −u(x, t) ≤ Lip (g) | x −x|. …(2)
Interchanging the roles of x and x, we write
u(x, t) − u ( x , t) ≤ Lip(g) | x −x|. …(3)
Combining (2) and (3), we write
|u( x , t) −u(x, t)| ≤ Lip (g) | x −x|. …(4)
Next select x∈Rn, t > 0. Choose y = x in Hopf-Lax formula, we discover
u(x, t) ≤ t L(0) + g(x). …(5)
Furthermore, by Hopf-Lax formula,
u(x, t) = �
�� +�
�
���
� −∈
)y(gt
yxLtmin
nRy
≥ g(x) +�
��
��
���
� −+−−∈ t
yxLg|yx|)g(Lipmin
nRy,
= g(x) −t { })z(L|z|)g(LipmaxnRz
−∈
; on taking x−y = tz,
= g(x) − t �
�� −
∈∈)}z(Lz.w{maxmax
nRz))g(Lip,0(Bw
= g(x) −t ))g(Lip,0(B
max H. …(6)
Inequalities (5) and (6) together imply
|u(x, t) −g(x)| ≤ C t …(7)
for
C = max .|H|max|,)0(L|))g(Lip,0(B �
�� …(8)
Finally select x∈Rn, o < t < t. Then
Lip (u(⋅, t)) ≤ Lip (g), …(9)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
236
by virtue of inequality (4) above. Consequently Lemma 1 and calculations like those employed above imply
|u(x, t) − u(x, t )| ≤ C |t − t |, …(10)
for the constant C defined in (8).
Inequalities (4) and (10) proves the fact that the function u is Lipschitz continuous in Rn×[0, ∞). Moreover, inequality (7) proves that
u = g on Rn ×{t = 0}.
This completes the proof of Lemma 2.
Result : By Rademacher’s theorem (proof out of course), it is asserted that a Lipschitz function is differentiable almost every where.
Consequently, by Lemma 2, our function u = u(x, t) defined above by the Hopf-Lax formula is differentiable almost everywhere in Rn×[0, ∞).
The next theorem concludes u, in fact, as defined by Hopf-Lax formula, solves the Hamilton-Jacobi PDE wherever u is differentiable.
Theorem : (Solving the Hamilton-Jacobi equation).
Statement : Suppose x∈Rn, t > 0, and u = u(x, t) defined by the Hopf-Lax formula is differentiable at a point (x, t) ∈ Rn ×(0, ∞).
Then
ut(x, t) + H(Du(x, t)) = 0.
Proof : Fix q ∈Rn, h > 0. Owing to Lemma 1,
u(x + hq, t + h) = �
�� +�
�
���
� −+∈
)t,y(uh
yqhxLhmin
nRy
≤ h L(q) + u(x, t).
Hence
h
)t,x(uht,qhx(u −++ ≤ L (q) .
Taking h→ 0 +, we write
q ⋅ Du(x, t) + ut(x, t) ≤ L(q). …(1)
NONLINEAR FIRST-ORDER PDE
237
Since
H = L*, …(2)
therefore,
ut(x, t) + H(Du(x, t)) = ut(x, t) + nRq
max∈
{q ⋅ Du(x, t) − L(q)}
≤ 0, …(3)
because the inequality (1) is valid for all q∈Rn.
In order to prove the required result, it is now enough
to show that
ut(x, t) + H(Du(x, t)) ≥ 0. …(4)
To prove (4), we choose z such that
u(x, t) = t L ��
���
� −t
zx+ g(z). …(5)
Fix h > 0 and set
s = t −h,
y = zts
1xts
��
���
� −+ . …(6)
Then
,s
zyt
zx −=− …(7)
and thus
u(x, t) −u(y, s) ≥ ��
���
� +��
���
� −)z(g
tzx
Lt
− ��
���
� +��
���
� −)z(g
szy
Ls
= (t−s) L ��
���
� −t
zx, …(8)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
238
using (7). This gives
,t
zxL
h
ht,zth
xth
1u)t,x(u
��
���
� −≥���
����
� −+��
���
� −− …(9)
using (6). Letting h→0+ in (9), we compute
.t
zxL)t,x(u)t,x(Du.
tzx
t ��
���
� −≥+��
���
� −
Consequently
ut(x, t) + H(D(u, t)) = ut(x, t) +nRq
max∈
{q ⋅ Du(x, t) − L(q)}
≥ ut(x, t) + ��
���
� −−⋅��
���
� −t
zxL)t,x(Du
tzx
≥ 0. …(10)
This prove (4) and hence the theorem
We summarize the above results in the form of following theorem :
Theorem (Hopf−−−−Lax formula as solution). The function u (x,t) defined by the Hopf-Lax formula is Lipschitz continuous, is differentiable a. e. in Rn×(0, ∞), and solves the initial-value problem.
ut + H(Du) = 0 a.e. in Rn×(0, ∞) ,
u = g on Rn×{t = 0} .
8.9 WEAK SOLUTIONS, UNIQUENESS
Semiconcavity
In view of Theorem above it may seem reasonable to define a weak solution of the initial-value problem to be a Lipschitz function which agrees with g on Rn×{t = 0}, and solves the PDE a.e. on Rn×(0, ∞). However, this turns out to be an inadequate definition, as such weak solutions would not in general be unique.
Example : Consider the initial-value problem
NONLINEAR FIRST-ORDER PDE
239
��
=×=∞×=+
}.0t{Ron0u
),0(Rin0|u|u 2xt …(1)
One obvious solution is
u1(x, t) ≡ 0. …(2)
However the function
u2(x, t) =��
��
≤≤−−−≤≤−
≥
0xtiftxtx0iftx
t|x|if0
…(3)
is Lipschitz continuous and also solves the PDE a.e. (everywhere, in fact, except on the lines x = 0, + t). It is easy to see that actually there are infinitely many Lipschitz functions satisfying (1).
This example shows we must presumably require more of a weak solution than merely that it satisfy the PDE a.e. We will look to the Hopf-Lax formula for a further clue as to what is needed to ensure uniqueness.
The following lemma demonstrates that u inherits a kind of “one-sided” second-derivative estimate from the initial function g.
Lemma 3: (Semiconcavity). Suppose there exists a constant C such that
g(x + z) −2g(x) + g(x−z) ≤ C|z|2 …(1)
for all x, z ∈Rn. Define u by the Hopf-Lax formula. Then
u(x + z, t) −2u(x, t) + u(x −z, t) ≤ C |z|2 …(2)
for all x, z ∈Rn, t > 0.
Remark. We say g is semiconcave provided (1) holds. It is easy to check (1) is valid if g is C2 and
nR
sup |D2g| < ∞. Note that g is semiconcave if and only if
the mapping
x→g(x) −2C
|x|2
is concave for some constant C.
Proof : Choose y∈Rn so that
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
240
u(x, t) = t L ��
���
� −t
yx+ g(y). …(3)
Then, putting y + z and y −z in the Hopf-Lax formulas for u(x, + z, t) and u(x−z, t),
we find
u(x + z, t) −2u(x, t) + u(x −z, t)
≤ ��
���
� ++��
���
� −)zy(g
tyx
Lt
��
���
� +��
���
� −− )y(gt
yxLt2
+ ��
���
� −+��
���
� −)zy(g
tyx
Lt
= g(y + z) −2g(y) + g(y−z)
≤ C |z|2, …(4)
by (1). This proves the lemma.
Note : As a semiconcavity condition for u(x, t) will turn out to be important, we pause to identify some other circumstances under which it is valid. We will no longer assume g to be semiconcave, but will suppose the Hamiltonian H to be uniformly convex.
Definition : A C2 convex function
H : Rn→R
is called uniformly convex (with constant θ > 0) if
�=
n
1j,ijpipH (p)ξiξj ≥ θ|ξ|2 for all p, ξ ∈Rn.
We now prove that if g is not semiconcave, the uniform convexity of H forces u(x,t) to become semiconcave for time t > 0. This is a kind of mild regularizing effect for the Hopf-Lax solution of the initial-value problem.
Lemma 4 : Suppose that H is uniformly convex (with constant θ) ad u(x,t) is defined by the Hopf-Lax formula. Then
NONLINEAR FIRST-ORDER PDE
241
u(x + z, t) −2u(x, t) + u(x −z, t) ≤ 2|z|t�
1
for all x, z ∈Rn, t > 0.
Proof : We note first using Taylor’s formula that uniform convexity of H implies
H8�
)p(H21
)p(H21
2pp
2121 −+≤��
���
� +|p1−p2|2. …(1)
Next we claim that for the Lagrangian L we have the estimate
221
2121 |qq|
�81
2qq
L)q(L21
)q(L21 −+�
�
���
� +≤+ , …(2)
for all q1, q2 ∈Rn. Verification is left as an exercise.
Now choose y so that
u(x, t) = t L ��
���
� −t
yx+ g(y). …(3)
Then using the same value of y in the Hopf-Lax formulas for u(x + z, t) and u(x −z, t), we calculate
u(x + z, t) −2u(x, t) + u(x −z, t)
≤ ��
���
� +��
���
� −+)y(g
tyzx
Lt
��
���
� +��
���
� −− )y(gt
yxLt2
+ ��
���
� +��
���
� −−)y(g
tyzx
Lt
= 2t ��
���
���
���
� −−��
���
� −−+��
���
� −+t
yxL
tyzx
L21
tyzx
L21
≤ 2t2
tz2
81θ
t
1θ
≤ |z|2,
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
242
using (2). Hence the lemma.
Now we show that semiconcavity conditions of the sort discovered for the Hopf-Lax solution u(x, t) in Lemmas 3 and 4 can be utilized as uniqueness criteria.
Definition : We say that a Lipschitz continuous function
u : Rn × [0, ∞) → R
is a weak solution of the initial-value problem:
��
��
=×=
∞×=+
}0t{Rongu
),0(Rin0)Du(Hun
nt …(∗)
provided
(a) u(x, 0) = g(x) for x∈Rn,
(b) ut(x, t) + H(Du(x, t)) = 0 a.e. for (x, t) ∈Rn ×(0, ∞), and
(c) u(x +z, t) −2u(x, t) + u(x −z, t) ≤ C ��
���
� +t1
1 |z|2
for some constant C ≥ 0 and all x,z∈Rn, t > 0.
Next we prove that a weak solution the above initial-value problem is unique, the key point being that this uniqueness assertion follows from the inequality condition (c) above.
Theorem (Uniqueness of weak solutions). Assume H is C2 and satisfies the condition
��
��
+∞=∞→ |p|
)p(Hlim
andconvexisH
|p|
(**)
and g is Lipschitz continuous. Then there exists at most one weak solution of the initial-value problem (∗).
Proof : 1. Suppose that u and u~ are two weak solutions of (∗) and write
w = u− u~ . …(1)
Observe now at any point (y, s) where both u and u~ are differentiable and solve our PDE, we have
NONLINEAR FIRST-ORDER PDE
243
wt(y, s) = ut(y, s) − tu~ (y, s)
= −H(Du(y, s)) + H(D u~ (y, s))
= − �1
0drd
H(r Du(y, s) + (1−r)D u~ (y, s)) dr
= − �1
0
r(DH Du(y, s) +(1−r) D u~ (y, s)) dr . (Du(y, s) − D u~ (y, s))
= −b(y, s) ⋅ Dw(y, s) .
Consequently
wt + b ⋅ Dw = 0 a.e. …(2)
2. Write v = φ(w) ≥ 0, where φ : R→[0, ∞) is a smooth function to be selected later. We multiply (2) by φ′(w) to discover
vt + b ⋅ Dv = 0 a.e. …(3)
3. Now choose ε > 0 and define
uε = ηε ∗ u,
�u~ = ηε ∗ u~ , …(4)
where ηε is the standard mollifier in the x and t variables. Then
|Duε| ≥ Lip(u),
|u~D| � ≤ Lip( u~ ), …(5)
and
Duε → Du,
�u~D → D u~ …(6)
a.e., as ∈→0 .
Furthermore inequality (c) in the definition of weak solution implies
D2uε, �2u~D ≤ C ��
���
� +s1
1 I …(7)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
244
for an appropriate constant C and all ε > 0, y∈Rn, s > 2ε. Verification is left as an exercise.
4. Write
bε(y, s) = �1
0DH(r Duε(y, s) + (1−r) �u~D (y, s)) dr. …(8)
Then (3) becomes
vt + bε ⋅ Du = (bε − b) ⋅ Dv a.e.; …(9)
hence
vt + dv (v bε) = (div bε)v + (bε −b) ⋅ Dv a.e. …(10)
5. Now
div bε = dr)u~)r1(ru)(u~D)(r1(rDu(Hkxx
1
0kxxpkp
n
1,k
εεεε
=−+−+� � 1ll
l
≤ C ��
���
� +s1
1 , …(11)
for some constant c, in view of (5), (7). Here we note that H convex implies
D2H ≥ 0.
6. Fix x0∈Rn, t0 > 0, and set
R = max {|DH(p)| | |p| ≤ max (Lip(u), Lip ( u~ ))}. …(12)
Define also the cone
C = {(x, t) |0 ≤ t ≤ t0|x −x0| ≤ R(t0 − t)|. …(13)
Next write
e(t) = �− ))t0t(R,0x(B
v(x, t) dx …(14)
and compute for a.e. t > 0:
)t(e& = �− ))t0t(R,0x(B
vt dx −R �−∂ ))t0t(R,0x(B
v dS
NONLINEAR FIRST-ORDER PDE
245
= �−∂ ))t0t(R,0x(B
−div (v bε) + (div bε)v + (bε −b). Dv dx
−R �−∂ ))t0t(R,0x(B
v dS
= − �−∂ ))t0t(R,0x(B
v(bε⋅ v + R)dS
+ �− ))t0t(R,0x(B
(div bε)v + (bε −b) ⋅ Dv dx
≤ �− ))t0t(R,0x(B
(div bε)v + (bε−b) ⋅ Dv dx
≤ C ��
���
� +t1
1 e(t) + �− ))t0t(R,0x(B
(bε −b) ⋅ Dv dx
by (11). The last term on the right hand side goes to zero as ε→0, for a.e. t0 > 0, according to (5), (6) and the Dominated Convergence Theorem.
��
���
� +≤t1
1C)t(e& e(t) for a.e. 0 < t < t0. …(15)
7. Fix 0 < ε < r < t and choose the function φ(z) to equal zero if
|z| ≤ ε[Lip(u) + Lip( u~ )]
and to be positive otherwise.
Since u = u~ on Rn × {t = 0},
v = φ(w) = φ (u− u~ ) = 0 at {t = ε}.
Thus
e(ε) = 0.
Consequently Gronwall’s inequality and (15) imply
e(r) ≤ e(ε)�ε
��
���
� +r
dss1
1C
e
.0=
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS
246
Hence
|u− u~ | ≤ ε [Lip(u) + Lip( u~ )] on B(x0, R(t0 −r)).
This inequality is valid for all ε > 0, and so
u ≡ u~ in B(x0, R(t0 −r)).
Therefore, in particular,
u(x0, t0) = u~ (x0, t0).
This completes the proof
In light of Lemma 3, 4 and Theorem above, we have the following theorem.
Theorem : (Hopf-Lax formula as weak solution). Suppose H is C2 and satisfies (**), and g is Lipschitz continuous. If either g is semiconcave or H is uniformly convex, then
u(x, t) = �
�� +�
�
���
� −∈
)y(gt
yxLtmin
nRy
is the unique weak solution of the initial-value problem (∗) for the Hamilton-Jacobi equation.
Example 1: Consider the initial-value problem :
��
��
=×=
∞×=+
}.0t{Ron|x|u
),0(Rin0|Du|21
u
n
n2t …(1)
Here
H(p) = 2|p|21
.|q|21
)q(L 2=
The Hopf-Lax formula for the unique, weak solution of (1) is
u(x, t) = .|y|t2
|yx|min
2
nRy �
��
+−∈
…(2)
Assume |x| > t. Then
NONLINEAR FIRST-ORDER PDE
247
Dy|y|
yt
xy|y|
t2|yx| 2
+−=���
����
�+−
(y ≠ 0); …(3)
and this expression equals zero if x = y +|x|
x)t|x(|y,t
|y|y −= ≠ 0.
Thus
u(x, t) = |x|−21
if |x| > t.
If |x| ≤ t,
the minimum in (2) is attained at y = 0. Consequently
u(x, t) = ��
��
≤
≥−
.t|x|ift2|x|
t|x|if2/t|x|2
Observe that the solution becomes semiconcave at time t > 0, even though the initial function g(x) = |x| is not semiconcave. This accords with Lemma 4.
Example 2 : We next examine the problem with reversed initial conditions :
��
��
=×−=
∞×=+
}.0t{Ron|x|u
),0(Rin0|Du|21
u
n
n2t …(1)
Then
u(x, t) = .|y|t2
|yx|min
2
nRy �
��
−−∈
Now
Dy|y|
yt
xy|y|
t2|yx| 2
−−=���
����
�−−
(y ≠ 0),
and this equals zero if x = y − .|x|
x)t|x(|y,t
|y|y += Thus
u(x, t) = −|x| −2t
(x ∈Rn, t ≥ 0). …(2)
The initial function g(x) = −|x| is semiconcave, and the solution remains so for times t > 0. The Books Recommended for Chapter VIII 1. L.C. Evans Partial Differential Equations, Graduate Studies
in Mathematics, Volume 19, AMS, 1998.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 248
Chapter-9 Representation of Solutions
In this chapter, we collect together a wide variety of techniques that are some-times useful for finding certain more-or-less explicit solutions to various partial differential equations, or at least representation formulas for solutions.
9.1 SEPARATION OF VARIABLES
The method of separation of variables tries to construct a solution u to a given partial differential equation as some sort of combination of functions of fewer variables.
In other words, the idea is to guess that u can be written as, say, a sum or product of as yet undetermined constituent function, to plug this guess into the PDE, and finally to choose the simpler functions to ensure u really is a solution.
This technique is best understood in examples.
Example 1 : Let U⊂ Rn be a bounded, open set with smooth boundary. We consider the initial/boundary-value problem for the heat equation
ut−∆u = 0 in U × (0, ∞),
u = 0 on ∂U × [0, ∞),
u = g on U ×{t = 0}, …(1)
where
g : U→R is given. …(2)
We conjecture there exists a solution having the multiplicative form
u(x, t) = v(t)w(x) (x ∈U, t ≥ 0) . …(3)
That is, we look for a solution of (1) with the variables x = {x1,…,xn} ∈U “separated” from the variable t ∈[0, T].
We compute
ut(x, t) = v′(t)w(x), …(4)
REPRESENTATION OF SOLUTIONS 249
∆u(x, t) = v(t) ∆w(x). …(5)
Hence, equations (1), (4) and (5) imply
v′(t)w(x) − v(t)∆w(x) = 0
or
)x(w)x(w
)t(v)t('v ∆= , …(6)
for all x∈U and t > 0 such that w(x), v(t) ≠ 0.
Now observe the left-hand side of (6) depends only on t and the right hand side depends only on x. This is impossible unless each is constant, say
)x(w)x(w
�)t(v)t('v ∆== (t ≥ 0, x∈U). …(7)
Then
v′ = µv, …(8)
and
∆w = µw. …(9)
We must solve these equations (8) and (9) for the unknowns w, v and µ.
Notice first that if µ is known, the solution of (8) is
v (t) = d eµt …(10)
for an arbitrary constant d. Consequently, we need only investigate equation (9).
We say that λλλλ is an eigenvalue of the operator −−−−∆∆∆∆ on U (subject to zero boundary conditions) provided there exists a function w, not identically equal to zero, solving
���
∂==∆−
.Uon0wUinw�w
…(11)
The function w is a corresponding eigenfunction.
If λ is an eigenvalue and w is a related eigenfunction, we set
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 250
µ = −λ , …(12)
and find
u = de−λtw , …(13)
solves the problem
ut − ∆u = 0 in U × (0, ∞)
u = 0 on ∂U × [0, ∞), …(14)
with the initial condition
u(⋅, 0) = d w. …(15)
Thus the function u defined by (13) solves problem (1), provided
g = d w. …(16)
More generally, if λ1,…,λm are eigenvalues, w1,…, wm corresponding eigenfunctions, and d1,…,dm are constants, then
u = k
m
1k
tk�k wed�
=
− …(17)
solves (14), with the initial condition
u(⋅, 0) = � =m
1k kk .wd …(18)
If we can find m, w1,…, etc. such that
� =m
1k kk wd = g, …(19)
we are done.
We can hope to generalize further by trying to find a countable sequence λ1,… of eigenvalues with corresponding eigenfunctions w1, ….. so that
� =∞
=1kkk gwd in U …(20)
for appropriate constants d1,….
Then presumably
REPRESENTATION OF SOLUTIONS 251
u = �∞
=
−
1kk
tk�k wed …(21)
will be the solution of the initial-value problem (1).
Remark 1 : This is an attractive representation formula for the solution, but depends upon
(a) our being able to find eigenvalues, eigenfunctions and constants satisfying (21) and
(b) our verifying that the series in (21) converges in some appropriate sense.
Remark 2 : Note that our solution (13) is determined by the method of separation of variables. The more complicated forms (17) and (21) depend upon the linearity of the heat equation.
Example 2 : Let us turn once again to the Hamilton-Jacobi equation
ut + H(Du) = 0 in Rn × (0, ∞) , …(22)
and look for a solution u having the form
u(x, t) = w(x) + v(t) (x ∈Rn, t ≥ 0). …(23)
Then
0 = ut(x, t) + H(Du(x, t))
= v′(t) + H(Dw(x)) ,
if and only if
H(Dw(x)) = µ = −v′(t) (x∈Rn, t > 0) , …(24)
for some constant µ. Consequently if
H(Dw) = µ, …(25)
v′(t) = −µ, …(26)
for some µ∈R, then
u(x, t) = w(x) −µt + b …(27)
will for any constant b solve (22).
In particular, if we choose
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 252
w(x) = a⋅x …(28)
for some a∈Rn and set
µ = H(a), …(29)
we discover the solution
u = a ⋅ x −H(a)t + b …(30)
already obtained.
9.2 SIMILARITY SOLUTIONS
When investigating partial differential equations it is often profitable to look for specific solutions u, the form of which reflects various symmetries in the structure of the PDE. We have already seen this idea in our derivation of the fundamental solutions for Laplace’s and the heat equations.
Following are some other applications of this important method.
Plane and Traveling Waves, Solitons
Consider first a partial differential equation involving the two variables x∈R, t∈R. A solution u of the form
u(x, t) = v(x−σt), (x∈R, t∈R) …(1)
is called a traveling wave (with speed σ and profile v).
More generally, a solution u of a PDE in the n + 1 variables x = (x1,…xn) ∈Rn, t∈R having the form
u(x, t) = v(y⋅x−σt), (x∈Rn, t∈R) …(2)
is called a plane wave having wavefront normal to y∈Rn, velocity |y|�
, and
profile v.
Exponential Solutions
In view of the Fourier transform, it is particularly enlightening when studying linear partial differential equations to consider complex-valued plane wave solutions of the form
u(x, t) = ei(y⋅x+ωt), …(3)
where
ω ∈C and y = (y1,…,yn) ∈Rn.
REPRESENTATION OF SOLUTIONS 253
ω being the frequency and n1ii}y{ = the wave numbers.
We will next substitute trial solutions of the form (3) into various linear PDE, paying particular attention to the relationship between y and ω forced by the structure of the equation.
Example 1 : (Heat equation).
If u is given by (3), we compute
ut − ∆u = (iω + |y|2) u = 0, …(4)
provided
ω = i|y|2. …(5)
Hence
u = t2|y|xiye −⋅ , …(6)
solves the heat equation for each y ∈Rn.
Taking real and imaginary parts, we discover further that
u1 = t2|y|e− cos (y⋅x) , …(7)
and
u2 = t2|y|e− sin (y⋅x) , …(8)
are solutions as well.
Notice in this example that since ω is purely imaginary, there results a real,
negative exponential term t2|y|e− in the formulas, which corresponds to dissipation.
Example 2 : (Wave equation).
Upon our substituting (3) into the wave equation, we discover
utt − ∆u = (−ω2 + |y|2 u = 0, …(9)
provided
ω = + |y|. …(10)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 254
Consequently
u = ei(y⋅x+|y|t) , …(11)
solves the wave equation.
The pair of functions
u1 = cos (y⋅x + |y|t) , …(12)
and
u2 = sin(y⋅x + |y|t) , …(13)
also solves the same.
Since ω is real, there are no dissipation effects in these solutions.
Example 3 : (Dispersive equations).
We now let n = 1 and substitute
u = ei(yx + ωt)
into Airy’s equation
ut + uxxx = 0. …(14)
We calculate
ut + uxxx = i(ω −y3) u = 0, …(15)
whenever
ω = y3. …(16)
Thus
u = )t3yyx(ie + , …(17)
solves Airy’s equation.
Once again, as ω is real there is no dissipation. Notice however that the velocity of propagation is y2, which depends non-linearly upon the frequency of the initial value iyxe .
Thus waves of different frequencies propagate at different velocities: the PDE creates dispersion.
REPRESENTATION OF SOLUTIONS 255
Likewise, if n ≥ 1 and we substitute
u = ei(y⋅x+ωt)
into Schrodinger’s equation
iut + ∆u = 0, …(18)
we compute
iut + ∆u = −(ω + |y|2)u = 0. …(19)
Consequently
ω = −|y|2, …(20)
and
u = )t2|y|xy(ie −⋅ . …(21)
Again, the solution displays dispersion.
Solitons
We consider next the Korteweg-de Vries (KdV) equation in the form
ut + 6uux + uxxx = 0 in R × (0, ∞), …(22)
this nonlinear dispersive equation being a model for surface waves in water.
We seek a traveling wave solution having the structure
u(x, t) = v(x −σt) (x ∈R, t > 0). …(23)
Then u solves the KdV equation (22), provided v satisfies the ODE
−σv′ + 6vv′ + v′′′ = 0 . ��
��
−== t�xs,dsd
' …(24)
We integrate (24) by first noting
−σv + 3v2 + v′′ = a, …(25)
a denoting some constant.
Multiply this equality by v′ to obtain
−σ vv′ + 3v2v′ + v′′ v′ = av′,
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 256
and so deduce
232
v2�
v2)'v( +−= + av + b , …(26)
where b is another arbitrary constant.
We investigate (26) by looking now only for solutions v which satisfy
v, v′, v′′ → 0 , as s→ + ∞. …(27)
The function u having the form (23), under conditions (27), is called a solitary wave.
Then (25), (26) imply
a = b = 0. …(28)
Equation (26) thereupon simplifies to read
.2�
vv2)'v( 2
2
��
��
+−=
Hence
v′ = + v(σ − 2v)1/2. …(29)
We take the minus sign above for computational convenience, and obtain then this implicit formula for v:
s = − � +−
)s(v
02/1
,c)z2�(z
dz …(30)
for some constant c. Now substitute
z = 2�
sech2θ. …(31)
It follows that
��d
dz −= sech2θ tanh θ , …(32)
and
REPRESENTATION OF SOLUTIONS 257
z(σ−2z)1/2 = 2� 2/3
sech2 θ tanhθ. …(33)
Hence (30) becomes
s = σ2 θ + c, …(34)
where θ is implicitly given by the relation
2�
sech2 θ = v(s). …(35)
We combine (34) and (35) to compute
v(s) = 2�
sech2 ���
���
− )cs(
2�
, (s∈R). …(36)
Conversely, it is routine to check v so defined actually solves the ODE (24).
The upshot is that
u(x, t) =2�
sech2 ���
���
−σ−σ
)ctx(2
, (x∈R, t ≥ 0) …(37)
is a solution of the KdV equation for each c∈R, σ > 0.
A solution of this form is called a soliton.
Note the velocity of the solution depends upon its height.
Remark : The KdV equation is in fact utterly remarkable, in that it is completely integrable, which means that in principle the exact solution can be computed for essentially arbitrary initial data.
Traveling Waves for a Bistable Equation.
Consider next the scalar reaction-diffusion equation
ut − uxx = f(u) in R×(0, ∞), …(38)
where
f : R→R
has a “cubic-like” shape.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 258
Graph of the function f
We assume, more precisely, f is smooth and verifies
(a) f(0) = f(a) = f(1) = 0
(b) f < 0 on (0, a), f > 0 on (a, 1)
(c) f′(0) < 0, f′(1) < 0
(d) �10 )z(f dz > 0 …(39)
for some point 0 < a < 1.
We look for a traveling wave solution of the form
u(x, t) = v(x − σ t), …(40)
the profile v and velocity σ to be determined, such that
u→0 as x→ − ∞, u→1 as x→ +∞. …(41)
Now since
f ′ < 0 at z = 0, 1,
the constants 0 and 1 are stable solutions of the PDE (and since f′ ≥ 0 at z = a, the constant a is an unstable solution).
So we want our traveling wave (40) to interpolate between the two stable states z = 0, 1 at x = µ ∞.
Plugging (40) into (38), we see v must satisfy the ordinary differential equation.
v′′ + σv′ + f(v) = 0 , ��
��
=dsd
' , …(42)
0 a 1
REPRESENTATION OF SOLUTIONS 259
subject to the conditions
,1)s(vlims
=+∞→
,0)s(vlims
=−∞→
±∞→s
lim v′(s) = 0. …(43)
We outline now (without complete proofs) a phase plane analysis of the ODE problem (42), (43). We begin by setting
w = v′ . …(44)
Then (42), (43) transform into the autonomous first-order system
v′ = w
w′ = −σw −f(v), …(45)
with
),0,1()w,v(lims
=∞→
−∞→s
lim (v, w) = (0, 0). …(46)
Now (0, 0) and (1, 0) are critical points for the system (45), and the eigen-values of the corresponding linearizations are
λ0+ = ,
2))0('f4( 2/12 −σ±σ−
.
2
))1('f4( 2/12
1−σ±σ−=λ± …(47)
In view of (39c), ±±10 �,� are real, with differing sign, and thus (0, 0) and (1, 0)
are saddle points for the flow (45).
Consequently as “stable curve”, Ws approaches (1, 0), as drawn. Furthermore, by calculating eigenvectors corresponding to (47) , we see
Wu is tangent to the line w = +0� v at (0, 0)
Ws is tangent to the line w = −1� (v−1) at (1, 0). …(48)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 260
Stable and unstable curves
Note that ±±10 �,� , Wu and Ws depend upon the parameter σ.
Our intention is to find σ< 0 so that
Wu = Ws in the region {v > 0, w > 0}. …(49)
Then we will have a solution of (45), (46), whose path in the phase plane is a heteroclinic orbit connecting (0 0) to (1, 0).
To establish (49), we fix now a small number ε > 0 and let L denote the vertical line through the point (a + ε, 0).
We claim
Ws ∩ L ≠φ,
Wu ∩ L ≠ φ, …(50)
if σ < 0.
To check this assertion, define
E(v, w) = )z(f2
w v
0
2
�+ dz (v, w∈R) …(51)
and compute
dtd
E(v(t), w(t)) = w(t) w′(t) + f(v(t)) v′(t)
= − σw2(t). …(52)
w
Wu Ws
(0, 0) (1, 0) v
REPRESENTATION OF SOLUTIONS 261
As σ < 0, we see that E is nondecreasing along trajectories of the ODE (45). Note also the level sets of E have the shapes illustrated below :
Level curves of E
Consider next the region R, as drawn. The unstable curve enters R from (0, 0) and cannot exit through the bottom, top or left hand side. Using (45), we deduce that Wu must exit R through the line L, at a point (a + ε, w0(σ)). Similarly we argue Ws must hit L at a point (a + ε, w1(σ)). This verifies claim (50).
We next observe
w0(0) < w1(0); …(53)
this follows since trajectories of (45) for σ = 0 are contained in level sets of E.
We assert further that
w0(σ) > w1(σ) …(54)
The region R
The region S
w
(1, 0) v (a, 0) (0, 0)
T
w L
s
v
w L R
v (a,0) (1,0) (0,0)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 262
Provided σ < 0 and |σ| is large enough. To see this, fix β > 0 and consider the region S, as drawn.
Now along the line segment T = {0 ≤ v ≤ a + ε, w = βv},
we have
v�
)v(f�
w)v(fw�
'v'w −−=−−= . …(55)
Since v
)v(f is bounded for 0 ≤ v ≤ a + ε, we see
��
C�
'v'w >−−≥ on T, …(56)
provided σ < 0 and |σ| is large enough.
The calculation (56) shows that Wu cannot exit S through the line segment T, and so
w0(σ) ≥ β(a + ε) if σ = σ(β)
is sufficiently negative.
On the other hand,
w1(σ) ≤ w1(0) for all σ ≤ 0.
Thus we see that (53) will follow once we choose β large enough and then σ sufficiently negative.
Since w0 and w1 depend smoothly on σ, we deduce from (50) and (53) that there exists σ < 0 with
w0(σ) = w1(σ). …(57)
For this velocity σ there consequently exists a solution of the ODE (45), (46).
Hence we have found for our reaction-diffusion PDE (38) a traveling wave of the form (40).
9.3 TRANSFORM METHODS
In this section we develop some of the theory of Fourier and Laplace transforms, which provides extremely powerful tools for converting certain
REPRESENTATION OF SOLUTIONS 263
linear partial differential equations into either algebraic equations or else differential equations involving fewer variables.
Fourier Transform
In this section all functions are complex-valued, and − denote the complex conjugate.
Definitions and Elementary Properties
Definition of Fourier transform on L1.
If u ∈L1(Rn), we define its Fourier transform
yix
R2/n
e)2(1
)y(un
⋅−�π
= u(x) dx (y ∈ Rn) …(1)
and its inverse Fourier transform
y.ix
R2/n
e)2(1
)y(un�π
=( u(x)dx (y∈Rn). …(2)
Since
|e+ix⋅y| = 1
and u∈L1(Rn), these integrals converge for each y∈Rn.
We intend now to extend definitions (1), (2) to functions u∈L2(Rn).
Theorem 1: (Plancherel’s Theorem). Assume u ∈ L1(Rn) ∩ L2(Rn).
Then u,u ( ∈ L2(Rn) and
.uuu)nR(2L)nR(2L)nR(2L
== ( …(3)
Proof 1: First we note that if v, w ∈L1(Rn), then w,v ∈ L∞(Rn). Also
dy)y(w)y(vdx)x(w)x(vnn RR�� = , …(4)
since both expressions equal
y.ix
RR2/n
e)2(1
nn
−��π
v(x) w(y) dxdy.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 264
Furthermore, (exercise)
t4|y|2/n
R
|x|tyix
2
n
2e
tdxe
−−⋅−��
��
π=� (t > 0). …(5)
Consequently if ε > 0 and
vε(x) = ,e2|x|�− …(6)
we have
.)2(
e)y(v
2/n
4|y| 2
ε=
ε−
ε …(7)
Thus (4) implies for each ε > 0 that
.dxe)x(w)2(1
dye)y(w 4|x|
R2/n
|y|
R
2
n
2
n
ε−ε−
�� ε= …(8)
2. Now take u ∈ L1(Rn) ∩ L2(R2) and set
v(x) = u (−x). …(9)
Let
w = u∗v ∈ L1(Rn) ∩ C(Rn) , …(10)
and we find that
vu)�2(w 2/n= ∈ L∞(Rn). …(11)
But
dx)x(ue)2(1
)y(v yix
R2/n
n
−π
= ⋅−�
= )y(u …(12)
and so
22/n |u|)�2(w = . …(13)
REPRESENTATION OF SOLUTIONS 265
Now w is continuous and thus
).0(w)2(dxe)x(w)2(1
lim 2/n4|x|
R2/n0
2
n
π=ε
ε−
→ε � …(14)
Since
22/n |u|)�2(w = ≥ 0,
we deduce upon sending ε→0+ in (8) that w is summable, with
dy)y(wnR� = (2π)n/2 w(0). …(15)
Hence
.dx|u|dx)x(v)x(u)0(wdy|u| 2
RR
2
R nnn��� =−== …(16)
The proof for u( is similar.
Definition of Fourier transform on L2.
In view of the equality (3), we can define the Fourier transform of a function u ∈L2(Rn)
as follows.
Choose a sequence ∞=1kk }u{ ⊂ L1 (Rn) ∩ L2(Rn) with
uk→ u in L2(Rn).
According to (3),
)nR(2Ljk)nR(2Ljk)nR(2Ljk uuuuuu −=−=− ,
and thus ∞=1kk }u{ is a Cauchy sequence in L2(Rn).
This sequence consequently converges to a limit, which we define to be :u
uu k → in L2(Rn).
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 266
The definition of u does not depend upon the choice of approximating sequence ∞
=1kk }u{ . We similarly define .u(
Next we record some useful formulas in the following theorem.
Theorem 2 : (Properties of Fourier transform).
Assume u, v ∈ L2(Rn). Then
(i) ,dyvudxvunn RR�� = …(17)
(ii) Dαu = (i y)α u
for each multi-index α such that
Dαu ∈ L2(Rn), …(18)
(iii) (u ∗ v)^ = (2π)n/2 vu , …(19)
(iv) u = ∨)u( . …(20)
Applications :
The Fourier transform is an especially powerful technique for studying linear, constant-coefficient partial differential equations.
Example 1 : (Bessel potentials).
We investigate first the PDE
−∆u + u = f in Rn, …(21)
where f ∈ L2(Rn).
To find an explicit formula for u, we take the Fourier transform and use (18) to obtain
(1 + |y|2 )y(f)y(u = (y ∈Rn). …(22)
The effect of the Fourier transform has been to convert the PDE (21) into the algebraic equation (22). The solution of (22) is
.|y|1
fu
2+= …(23)
REPRESENTATION OF SOLUTIONS 267
Thus
∨
���
���
+=
2|y|1f
u , …(24)
and so the only real problem is to rewrite the right hand side of (24) into a more explicit form.
Using (19), we see
u = ,)�2(Bf
2/n
∗ …(25)
where
.|y|1
1B
2+= …(26)
We solve for B as follows.
Since
�= ∞ −0
tadtea1
, …(27)
for each a > 0, we have
)|y|1(t
02
2e
|y|11 +−
∞
�=+
dt. …(28)
Thus
B = ∨
���
���
+ 2|y|11
( ) .dtdyee)2(1
n
2
R
|y|tyix
0
t2/n ��
−⋅∞
−
π= …(29)
Now if a, b ∈ R, b > 0, and we set
z = b1/2 x − ,ib2a
2/1 …(30a)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 268
we find
� �=∞∞− Γ
−−
− ,dzeb
edxe
2z2/1
b42a2bxiax …(30b)
à denoting the contour ���
� � −=
2/1b2a
)zIm( in the complex plane. Deforming Γ
into the real axis, we compute
� �Γ
∞
∞−
−− π== ;dxedze 2/1xz 22 …(31)
and hence
�∞
∞−
−−��
��
π= .b
edxe2/1
b4/abxiax 22 …(32)
Thus
∏ ��=
∞
∞−
−−⋅ =n
1jj
tyyix|y|tyix
R
dyedye2jjj
2
n
.
t4|x|2/n 2
et
−��
��
π= …(33)
Consequently, we conclude from (29), (33) that
B(x) = �∞
−−
0
2/n
t4|x|
t
2/ndt
te
21
2
, (x ∈Rn). …(34)
B is called a Bessel potential.
Employing (25), we derive then the formula
u(x) = 2/n
t4|yx|
t
R02/n t
e)4(1
2
n
−−−∞
��π f(y) dydt (x ∈Rn) …(35)
for the solution of (21).
Example 2 : (Fundamental solution of heat equation).
Consider again the initial-vale problem for the heat equation
REPRESENTATION OF SOLUTIONS 269
ut − ∆u = 0 in Rn ×(0, ∞)
u = g on Rn×{t = 0}. …(1)
We establish a new method for solving (1) by computing ,u the Fourier transform of u in the spatial variables x only.
Thus
tu + |y|2 u = 0 for t > 0 ,
u = g for t = 0 . …(2)
Solution of (2) is (exercise)
u = .ge2|y|t− …(3)
Consequently
u =∨
−���
� ge
2|y|t , …(4)
and therefore
u = ,)�2(
Fg2/n
∗ …(5)
where
.eF2|y|t−= …(6)
But then
F = ( ) dye)2(1
en
22
R
|y|tyix2/n
|y|t � −⋅∨−
π= .
t4|x|
2/n
2
e)t2(1 −= …(7)
using (5). We compute
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 270
u(x, t) = �−−
π n
2
R
t4|yx|
2/ne
)t4(1
g(y) dy. (x∈Rn, t > 0). …(8)
The Fourier transform has provided us with new derivation of the fundamental solution of the heat equation.
Example 3 : (Fundamental solution of Schrodinger’s equation).
Let us next look at the initial-value problem for Schrodinger’s equation
iut + ∆u = 0 in Rn × (0, ∞)
u = g on Rn × {t = 0}. …(1)
Here u and g are complex-valued.
If we formally replace t by ‘i t’ in the solution of heat equation, we obtain the formula
u(x, t) = �−−
π n
2
R
t4|yx|i
2/ne
)it4(1
g(y) dy , (x∈Rn, t > 0), …(2)
where we interpret .easi 4�i
21
This expression clearly makes sense for all times t > 0, provided g∈L1(Rn). Furthermore if |y|2 g∈L1(Rn), we can check by a direct calculation that u, given in (2), solves the differential equation is (1).
Let us next rewrite formula (2) as
u(x, t) = .dy)y(gee)it4(
et4|y|i
R
t2yix
2/n
t4|x|i 2
n
2
�⋅−
π …(3)
Since
|e,e| t4
2|y|it4
2|x|i
= 1,
we can check that if g∈L1(Rn) ∩ L2(Rn), then
REPRESENTATION OF SOLUTIONS 271
)nR(2L)nR(2L
g)t,(u =⋅ (t > 0). …(4)
Hence the mapping
g→u(⋅, t)
preserves the L2-norm. Therefore we can extend formula (2) to functions g ∈ L2(Rn), in the same way that we extended the definition of Fourier transform.
Remark. We call
Ψ(x, t) = 4
2|x|i
2/ne
)it�4(1
(x ∈ Rn, t ≠ 0) …(5)
the fundamental solution of Schrodinger’s equation.
Note that formula (2), u = g ∗ Ψ, makes sense for all time t ≠ 0, even t < 0. Thus we in fact have solved the problem.
iut + ∆u = 0 in Rn × (−∞, ∞)
u = g on Rn × {t = 0}. …(6)
In particular, Schrodinger’s equation is reversible in time, whereas the heat equation is not.
Example 4 : (Wave equation).
We next analyze the initial-value problem for the wave equation
utt − ∆u = 0 in Rn × (0, ∞),
u = g, ut = 0 on Rn × {t = 0}, …(1)
where for simplicity we suppose the initial velocity to be zero.
Take as before u to be the Fourier transform of u in the variable x∈Rn. Then (1) gives
u|y|u 2tt + = 0 for t > 0,
0u,gu t == for t = 0. …(2)
This is an ODE for each fixed y ∈Rn.
We look for a solution having the form
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 272
�te�u = (β, γ ∈ C) …(3)
Plugging into (2) gives
γ2 + |y|2 = 0 , …(4)
and so
γ = + i|y|. …(5)
Remembering the initial conditions from (2), we deduce
)ee(2g
u |y|it|y|it −+= . …(6)
Inverting, we find
u(x, t) = .)ee(2g |y|it|y|it
∨−
��
���
� + …(7)
Consequently, we get the formula
u(x, t) = dy)ee(2
)y(g)2(1 |)y|tyx(i
R
|)y|tyx(i2/n
n
−⋅+⋅ +π � , …(8)
for x∈Rn, t ≥ 0.
Laplace Transform
Remember that we write R+ = (0, ∞).
Definition : If u ∈L1(R+), we define its Laplace transform to be
u#(s) =�∞
−
0
ste u(t)dt (s ≥ 0). …(∗)
Whereas the Fourier transform is most appropriate for functions defined on all of R (or Rn), the Laplace transform is useful for function defined only on R+.
In practice this means that for a partial differential equation involving time, it may be useful to perform a Laplace transform in t, holding the space variables x fixed.
REPRESENTATION OF SOLUTIONS 273
Example 1 : (Resolvents and Laplace Transform).
Consider again the heat equation
vt − ∆v = 0 in U×(0, ∞)
v = f on U×{t = 0}, …(1)
and perform a Laplace transform with respect to time :
v#(x, s) = �∞
−
0
st )t,x(ve dt (s > 0). …(2)
We compute
∆v#(x, s) =�∞
− ∆0
st dt)t,x(ve
=�∞
−
0
tst dt)t,x(ve
= s [ ]�∞
∞==
−− +0
t0t
stst vedt)t,x(ve
= sv#(x, s)−f(x). …(3)
Think now of s > 0 being fixed, and write
u(x) = v#(x, s) . …(4)
Then
−∆u + su = f in U . …(5)
Thus the solution of the resolvent equation (5) with right hand side f is the Laplace transform of the solution of the heat equation with initial data f.
Example 2 : (Wave equation from the heat equation).
Next we employ some Laplace transform ideas to provide a new derivation of the solution for the wave equation, based upon the heat equation.
Suppose u is a bounded, smooth solution of the initial-value problem :
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 274
utt − ∆u = 0 in Rn×(0, ∞),
u = g, ut = 0 on Rn×{t = 0}, …(1)
where n is odd and g is smooth, with compact support.
We extend u to negative times by writing
u(x, t) = u(x, −t) if x∈Rn, t < 0. …(2)
Then
utt − ∆u = 0 in Rn × R. …(3)
Next define
v(x, t) = �∞
∞−
−
πt4/s
2/1
2e
)t4(1
u(x, s)ds (x ∈Rn, t > 0). …(4)
Hence
0t
lim→
v = g , …(5)
uniformly on Rn. In addition
∆v(x, t) = �∞
∞−
−
πt4/s
2/1
2e
)t4(1 ∆u (x, s)ds
= �∞
∞−
−
πt4/s
2/1
2e
)t4(1
uss(x, s)ds
= �∞
∞−
−
πt4/s
2/1
2e
)t4(1
us(x, s)ds
= �∞
∞−
−���
���
−
πt4/s
2
2
2/1
2e
t21
t4s
)t4(1
u(x, s)ds
= vt(x, t). …(6)
Consequently v solves this initial-value problem for the heat equation :
vt −∆v = 0 in Rn×(0, ∞),
REPRESENTATION OF SOLUTIONS 275
v = g on Rn×{t = 0}. …(7)
As v is bounded, we deduce that
v(x, t) = .dy)y(g)t4(
1
t4
2|yx|neR
2/n � −−π
…(8)
We equate (4) with (8), recall (2), and set
λ = .t4
1
We obtain the identity
� �∞
−λ−
−
λ−��
��
πλ=
0
)y(g|yx|
R
21n
s 2
n
2e
21
dse)s,x(u dy.
Thus
,dr)r;x(Gre2
)n(ndse)s,x(u 1nr
0
21n
s
0
22 −λ−∞−
λ−∞
�� ��
��
πλα= …(9)
for all λ > 0, where
G(x ; r) = ).y(dS)y(g)r,x(B�∂
…(10)
We will solve (9), (10) for u.
To do so, we write n = 2k + 1 and note
−22 rr e)e(
drd
r21 λ−λ− λ= . …(11)
Hence
dr)r;x(Gredr)r;x(Gre k2rk
0
1nr
0
21n
22 λ−∞
−λ−∞−
λ=λ ��
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 276
= dr)r;x(Gr)e(drd
r1
2)1( k2r
k
0k
k2
���
�
���
���
��
− λ−∞
�
= ,dre))r;x(Gr(rr
1r
21 2r1k2
k
0k
λ−−∞
���
�
���
���
��
∂∂
� …(12)
where we integrated by parts k times for the last equality.
Owing to (9) (with r replacing s in the expression on the left), we deduce from (12),
.dre))r;x(Gr(rr
1r
2
)n(ndre)r,x(u
22 r
0
1k2k
01k21n
r λ−∞
−∞
+−
λ−� � ���
�
���
���
��
∂∂
π
α= …(13)
Upon substituting τ = r2 we see that each side above, taken as a function of λ, is a Laplace transform. As two Laplace transforms agree only if the original functions were identical, we deduce
u(x, t) = )).t,x(Gt(tt
1t
2�
)n(n 1k2k
1kk−
+ ��
��
∂∂
…(14)
Now n = 2k + 1 and
α(n) = .1
2n�
12n� 2
1k2/n
��
��
+Γ=
��
��
+Γ
+
…(15)
Since
à 2/1�21 =��
��
, …(16)
and
Γ(x + 1) = xΓ(x) for x > 0,
we compute
REPRESENTATION OF SOLUTIONS 277
n1k
2/1
1kk �
13.5)...4n)(2n(
1
12n
2
�n2�
)n(n =−−
=��
��
+Γ=
++ . …(17)
We insert this deduction (17) into (14) and simplify :
u(x, t) = ���
�
�
���
��
��
∂∂
∂∂
γ �∂−
−
dSgttt
1t
1
)t,x(B
2n23n
n
(x ∈Rn, t > 0). …(18)
9.4 CONVERTING NONLINEAR INTO LINEAR PDE
Now we describe several techniques which are sometimes useful for converting certain nonlinear equations into linear equations.
Hopf-Cole Transformation
A parabolic PDE with quadratic nonlinearity.
We consider first of all an initial-value problem for a quasilinear parabolic equation :
ut −a∆u + b|Du|2 = 0 in Rn×(0, ∞)
u = g on Rn ×{t = 0}, …(1)
where a > 0.
This sort of nonlinear PDE arises in stochastic optimal control theory.
Assuming for the moment u is a smooth solution of (1), we set
w = φ(u), …(2)
where
φ : R→R …(3)
is a smooth function, as yet unspecific.
We will try to choose φ so that w solves a linear equation. We have
wt = φ′(u)ut, …(4)
∆w = φ′(u)∆u + φ′′(u)|Du|2; …(5)
and consequently (1) implies
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 278
φ′(u)ut = φ′(u) [a∆u−b|Du|2]
=a∆w − [aφ′′(u) + bφ′(u)] |Du|2 ,
or
wt = a∆w, …(6)
provided we choose φ to satisfy
aφ′′ + bφ′ = 0. …(7)
We solve this differential equation (7) by setting
φ = abu
e−
. …(8)
Thus we see that if u solves (1), then
w = abu
e−
…(9)
solves this initial-value problem for the heat equation (with conductivity a):
��
�
�
=×=
∞×=∆−−
}.0t{Ronew
),0(Rin0waw
nabg
nt
…(10)
Formula (9) is the Hopf-Cole transformation.
Now the unique bounded solution of (10) is
w(x, t) = dyee)at4(
1 )y(gab
at4|yx|
R2/n
2
n
−−−
�π (x ∈Rn, t > 0); …(11)
and, since (9) implies
u = − ,wlogba
…(12)
we obtain thereby the explicit formula
u(x, t) = −���
�
�
���
π �−−−
dye)at4(
1log
ba
n
2
R
)y(gab
at4|yx|
2/n (x ∈Rn, t > 0) …(13)
REPRESENTATION OF SOLUTIONS 279
for a solution of quasilinear initial-value problem (1).
Burgers’ Equation with Viscosity.
As a further application, we examine now for n = 1 the initial-value problem for the viscous Burgers’ equation:
ut−auxx + uux = 0 in R × (0, ∞)
u = g on R ×(0, ∞). …(14)
If we set
w(x, t) = dy)t,y(ux
� ∞−
…(15)
and
h(x) = dy)y(gx
� ∞−
…(16)
we have
wt − awxx +21 2
xw = 0 in R×(0, ∞)
w = h on R×{t = 0}. …(17)
This is an equation of the form (1) for
n = 1, b =21 .
So (13) provides the formula
w(x, t) = −2a log .dye)at4(
1
R
a2)y(h
at4|yx|
2/1
2
���
�
�
���
π �−−−
…(18)
Since
u = wx,
we find upon differentiating (18) that
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 280
u(x, t) =
��
∞
∞−
−−−
∞
∞−
−−−−
dye
dyet
yx
a2)y(h
at4|yx|
a2)y(h
at4|yx| 2
(x ∈R, t > 0) …(19)
is a solution of problem (14), where h is defined by (16).
Potential Functions
Another technique is to utilize a potential function to convert a nonlinear system of PDE into a single linear PDE.
We consider as an example Euler’s equations for inviscid, incompressible fluid flow :
(a) ut + u⋅Du = −Dp + f in R3×(0, ∞)
(b) div u = 0 in R3×(0, ∞)
(c) u = g on R3×{t = 0}. …(20)
Here the unknowns are the velocity field u = (u1, u2, u3) and the scalar pressure p . The external force f = (f1, f2, f3) and initial velocity g = (g1, g2, g3) are given. Here D as usual denotes the gradient in the spatial variables x = (x1, x2, x3). The vector equation 20(a) means
� +−=+=
3
1j
iix
ijx
jit fpuuu (i = 1, 2, 3). …(21)
We will assume
div g = 0. …(22)
If furthermore there exists a scalar function
h : R3 × (0, ∞)→R …(23)
such that
f = Dh, …(24)
we say that the external force is derived from the potential h.
We will try to find a solution (u, p) of (20) for which the velocity field u is also derived from a potential, say
REPRESENTATION OF SOLUTIONS 281
u = Dv. …(25)
The flow will then be irrotational as
curl u ≡ 0. …(26)
Now equations (20) (b) and (26) imply
∆v = 0, …(27)
and so v must be harmonic as a function of x, for each tie t > 0.
Thus if we can find a smooth function v satisfying (27) and
Dv (⋅, 0) = g, …(28)
we can then recover u from v by (25).
How do we compute the pressure p? Let us observe that because of (25), we have
u ⋅Du = 21
D(|Dv|2). …(29)
Consequently (20) (a) reads
D ��
��
+ 2t |Dv|2
1v = D(−p + h).
Therefore we may take
vt + 21 |Dv|2 + p = h. …(30)
This is Bernoulli’s law.
But now we can employ (30) to compute p, since v and h are already unknown.
9.5 HODOGRAPH AND LEGENDRE TRANSFORMS
Hodograph Transform
The hodograph transform is a technique for converting certain quasilinear systems of PDE into linear systems, by reversing the roles of the dependent and independent variables.
As this method is most easily understood by an example, we investigate here the equations of steady, two-dimensional, irrotational fluid flow :
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 282
(a) (σ2(u) −(u1)2) )uu(uuu 21x
12x
2111x +− + (σ2(u) −(u2)2) 0u 2
2x =
(b) 0uu 21x
12x =− , …(31)
in R2.
The unknown is the velocity field u = (u1, u2). The function
σ(⋅) : R2→R,
the local sound speed, is given.
The system (31) is quasilinear.
Let us now, however, no longer regard u1 and u2 as functions of x1 and x2:
u1 = u1(x1, x2), u2 = u2(x1, x2), …(32)
but rather regard x1 and x2 as functions of u1 and u2:
x1 = x1(u1, u2), x2 = x2(u1, u2). …(33)
We have exchanged sub and superscripts in the notation to emphasize the interchange between independent and dependent variables.
According to the Inverse Function Theorem, we can invert equations (32) to yield (33), provided
J = 21x
12x
22x
11x
21
21
uuuu)x,x()u,u( −=
∂∂ ≠ 0 , …(34)
in some region of R2.
Assuming now (34) holds, we calculate
��
� �
=−=
−==
.Jxu,Jxu
Jxu,Jxu2u
1x
12u
1x
2u
2x
1u
2x
212
1112 …(35)
We insert (35) into (31), to obtain
{ }��
� �
=−
=−σ+++−σ
.0xx)b(
,0x)u)u(()xx(uuxu)u()a(2u
1u
1u
22
22u
1u21
2u
22
12
11221
…(36)
This is linear system for x = (x1, x2), as a function of u = (u1, u2).
REPRESENTATION OF SOLUTIONS 283
Remark : We can utilize the method of potential functions to simplify (36) further. Indeed, equation (36b) suggests that we look for a single function z = z(u) such that
x1 = 1uz
x2 = .z2u …(37)
Then (36a) transforms into the linear second-order PDE
(σ2(u) − .0z)u)u(�(zuu2z)u1u1u
22
22u1u212u2u
21 =−++ …(38)
Legendre Transform
A technique closely related to the hodograph transform is the classical Legendre transform. The idea is to regard the components of the gradient of a solution as new independent variables.
Once again an example is instructive. We investigate the minimal surface equation
div ,0)|Du|1(
Du2/12
=���
���
+ …(39)
which for n = 2 may be rewritten as
.0u)u1(uuu2u)u1(2212121112 xx
2xxxxxxx
2x =++−− …(40)
Let us now assume that at least in some region of R2, we can invert the relations
p1 =1xu (x1, x2),
p2 = 2xu (x1, x2), …(41)
to solve for
x1 = x1(p1, p2),
x2 = x2(p1, p2). …(42)
The Inverse Function Theorem assures us we can do so in a neighborhood of any point where
J = det D2u ≠ 0. …(43)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 284
Now define
v(p) = x(p)⋅p −u(x(p)), …(44)
where x = (x1, x2) is given by (42) and p = (p1, p2). We find that (exercise)
���
��
�
=
−=
=
,Jvu
,Jvu
,Jvu
1122
2121
221
ppxx
ppxx
pp1xx
…(45)
Upon substituting the identities (45) into (40), we derive for v the following linear equation
0v)p1(vpp2v)p1(112122 pp
21pp21pp
22 =++++ . …(46)
Remark. The hodograph and Legendre transform techniques for obtaining linear out of nonlinear PDE are in practice tricky to use, as it is usually not possible to transform given boundary conditions very easily.
The Books Recommended for Chapter IX 1. L.C. Evans Partial Differential Equations, Graduate Studies
in Mathematics, Volume 19, AMS, 1998.
ATTRACTION AND POTENTIAL-I
285
Chapter-10
Attraction and Potential-I
10.1 LAW OF GRAVITATION This law states that “every particle in the universe attracts every other particle with a force which is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.”
Thus, if m1, m2 denote the masses of two particles and r their distance apart. Then the force of attraction between them is
γ2
21
r
mm,
where γ is known as the gravitation constant.
Remark I :- This law was discovered by Sir Isaac Newton (1642-1727)
Remark II :- Gravitation constant γ measures the attraction of two particles, each of unit mass, at unit distance apart.
Remark III :- To avoid a difficulty in defining the distance between two particles, we may define a material particle as a body so small that, for the purposes of our investigation, the distance between different parts of body may be neglected.
Remark IV : The numerical value of γ is 000,500,15
1 approximately.
Remark V :- If we choose units such that γ = 1. Such units are called astronomical or theoretical units.
Remark VI :- The acceleration f produced by the attraction of a particle of mass m on a particle at a distance r is given by,
f = γ2r
m,
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 286
so that γ = 1, when f, m and r are all unity. Hence, the astronomical unit of mass is the mass of a particle which by its attraction produces unit acceleration at unit distance.
We can find the astronomical unit of mass in grammes by taking the above formula for acceleration, which holds good in all systems of units, and putting r = 1 cm, f = 1cm/sec2, m = 15,500,000 grammes.
If P be a particle of unit mass and Q another particle of mass m. Then the force of attraction
F = γ2)PQ(
1m ×
is called the attraction of Q at P, and act along the line PQ towards Q.
Field of force :
The attraction of a system of particles at a point external to itself is the force of attraction which the system would exert on a particle of unit mass placed at the point. There must be a definite value for this force at every point at which a particle can be placed. Thus we arrive at the conception of a field of force, or region of space with every point of which there is associated a force which is definite in magnitude and direction.
Remark At the point of equilibrium the definite value of force of attraction is zero.
10.2 ATTRACTION OF A SYSTEM OF PARTICLES
Let particles of masses m1, m2, m3,… be situated at points A1 , A2, A3,… whose co-ordinates referred to rectangular axes are (x1, y1, z1), (x2, y2, z2), (x3, y3, z3),
Q (m)
P(unit mass)
z
o
y
Z
P
Y X
A2(x2, y2, z2)
A1(x1, y1, z1)
x
A3
ATTRACTION AND POTENTIAL-I
287
Let P (x, y, z) be any point in space. Let (X, Y, Z) denote the components of the attraction of the given system of particles at point P(x, y, z).
Let r1, r2, r3,…. Denote the distance PA1, PA2, PA3,….
so that
r12 = (x1, −x)2 + (y1−y)2 + (z1−z)2,
r22 = (x2−x)2 + (y2−y)2 + (z2−z)2,
r32 = (x3−x)2 + (y3−y)2 + (z3−z)2,
………………………………..
………………………………..
………………………………..
and the direction cosines of PA1, PA2, PA3……..are
>−−−
<1
1
1
1
1
1
rzz
,r
yy,
rxx
,
>−−−
<2
2
2
2
2
2
rzz
,r
yy,
rxx
,
>−−−
<3
3
3
3
3
3
rzz
,r
yy,
rxx
,
…………………………..
…………………………..
respectively.
The attraction at the point P of mass m1 situated at the point A1 is m1/r12 (on
taking γ=1) and is directed along 1PA .
Therefore, the particle m1 located at A1(x1, y1, z1) exerts at a force at P(x, y, z), whose components parallel to the axes are
X1 = ���
����
� −��
�
�
��
�
�=��
�
����
� −��
�
�
��
�
�=��
�
����
� −��
�
�
��
�
�
1
12
1
11
1
12
1
11
1
12
1
1
rzz
r
mZ,
ryy
r
mY,
rxx
r
m.
The other particles m2, m3,… make like contributions for the attraction at P(x, y, z).
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 288
The principle of superposition of fields of force states that “the force exerted at a point by a system of particles is the vector sum of the forces exerted by each of the particles separately”.
So, by the principle of superposition of fields of force, total attraction at P(x, y, z) due to the given system of particles is
X = ���
���
−k 3
k
kk
r
)xx(m,
Y = ���
���
−k 3
k
kk
r
)yy(m,
Z = ���
���
−k 3
k
kk
r
)zz(m,
where k = 1, 2, 3,… and the summation extends to all the attracting particles.
10.3 POTENTIAL
Let particles of masses m1, m2, m3,… be situated at points A1, A2, A3,… whose co-ordinates referred to rectangular axes are (x1, y1, z1), (x2, y2, z2),… . Let P (x, y, z) be any point of space. Let r1, r2, r3,… denote the distance PA1, PA2, PA3,……, i.e.,
rk2 = (xk−x)2 + (yk−y)2 + (zk−z)2 …(1)
for k = 1, 2, 3,….
Let us now define a function v(x, y, z) by the formula
V(x, y, z) = � ���
����
�
k k
k
rm
. …(2)
The function V defined in (2) is a function related to a system of attracting particles having a definite value at every point P of space external to the particles. It is a function of the co-ordinates (x, y, z) of P and is clearly a single-valued function, in the sense that it cannot have more then one value at each point P; for it represents simply the sum of the masses of the separate particles divided by their respective distances from P. Further, V represents a sum which does not depend on the particular system of axes of reference.
Now, differentiation of equations (1) and (2) with respect to x gives
ATTRACTION AND POTENTIAL-I
289
� ��
���
�
∂∂
��
�
�
��
�
�−=
∂∂
k
k2k
k ,xr
r
mxV
…(3)
and
rk −=∂∂
xrk (xk−x). …(4)
using equation (4) in (3) we obtain
��
���
−�=
∂∂
3k
kk
k r
)xx(mxV
. …(5)
But, we know that the component X of the force of attraction at P is given by
X = ��
���
−� 3
k
kk
k r
)xx(m. …(6)
Equations, (5) and (6) imply
XxV =
∂∂
, …(7)
similarly
YyV =
∂∂
, …(8)
ZzV =
∂∂
, …(9)
where Y and Z are other components of the force of attraction.
Definition. The function V defined by (2) is called the potential of the attracting particles, or the potential of the field of force.
Result (1) :- We have proved that the derivatives of the potential V with regard to x, y, z give the components of attraction at P in the directions of the axes.
Result (2) :- Since the directions of the axes can be chosen arbitrarily, it follows that the space derivative of the potential V in any direction gives the component of attraction in that direction.
For verification of result (2), let ∂/∂s denote a differentiation in a direction dS
whose direction cosines are < l, m, n> or >∂∂
∂∂
∂∂<
sz
,sy
,sx
. Then, by chain rule,
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 290
��
���
�
∂∂
∂∂+�
�
���
�
∂∂
∂∂+�
�
���
�
∂∂
∂∂=
∂∂
sz
zV
sy
yV
sx
xV
sV
.
= lzV
nyV
mxV
∂∂+
∂∂+
∂∂
= l X + mY + nZ
component of the force of attraction in direction < l, m, n>.
Remark 1 :- In the language of vectors the force of attraction say Rρ
, is the gradient of the potential V. That is,
Rρ
= grad V.
Remark 2 :- If the potential V of any given distribution of matter can be determined, the force of attraction R
ρat any point can be found immediately by
taking the gradient of scalar potential V.
Physical Interpretation of Potential V(x, y, z).
The total differential of the potential V(x, y, z) is
dV = dzzV
dyyV
dxxV
∂∂+
∂∂+
∂∂
= X dx + Y dy + Z dz. …(1)
Hence, by integrating along any path from the point P to the point Q, we get
VQ −VP = � ��
���
�
∂∂+
∂∂+
∂∂Q
Pds
sz
Zsy
Ysx
X . …(2)
But the integral in R.H.S. of equation (2) represents the work which the forces of attraction would perform upon a particle of unit mass as it moved along this path from P to Q. This gives us a measure of potential V in terms of work per unit mass. The potential at any point Q exceeds the potential at any other point P by the work which the forces of attraction would perform upon a particle of unit mass as it moves along any path from P to Q.
Remark (1) : The addition of a constant to the potential V will not affect the values of the force components.
Remark 2 :We know that the potential V of the attracting particles is defined by
ATTRACTION AND POTENTIAL-I
291
ds
V = � ���
����
�
k k
k
rm
…(3)
for k = 1, 2, 3,…
From (3) it is clear that potential V vanish at an infinite distance from the attracting matter. So, when the potential is determined by integration from known force components (X,Y, Z), the constant of integration may be so chosen as to make the potential vanish at an infinite distance from the attracting matter.
On this hypothesis, we see that the potential at a given point P due to a given point P due to a given attracting system is the work that would be done by the attractions of the system on a particle of unit mass as it moves along any path from an infinite distance up to the point considered. Hence the definition of potential V as �(m/r) leads to this expression for potential in terms of work done per unit mass.
Result : We have seen that the definition of potential V as
V = �(m/r)
leads tot he expression for V in terms of work done per unit mass.
Now we shall demonstrate the converse.
Let m be the mass of a typical particle of the system situated at the point A. Let
PP′ = ds
be an element of any path from an infinite distance to the point Q, and let
AP = r,
AP′ = r + dr.
Then so far as the field of force depends upon the particle m at A, its value at P is
A(m)
Q P′ P ∞
r
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 292
= ,r
m2
and is directed along PA . The work done by this force on the unit particle as it moves from P to P′ is
= dsdsdr
r
m2 �
�
���
� −��
���
�
= 2r
drm−.
Hence, the total work done by the attraction of the particle A(m) on a unit particle moving from an infinite distance to the point Q is
= − drr
m2
AQ��
���
��∞
= AQm
.
By the principle of superposition of fields, the total work done by the attractions of all the particles of the system is obtained by adding for their separate effects, so that
V = � ���
����
�
AQm
gives the potential at Q. The above formula represents the sum of the masses of the separate particles each divided by its distance from Q. The interchangeability of the two definitions of potential is thus completely established.
Dimensions : Gravitational potential V and potential energy have different physical dimensions. The dimensions of potential energy are those of work, i.e., ML2 T−2, in terms of the fundamental units of mass, space and time. The dimensions of gravitational potential V are obtained below
Dimensions of the potential V.
It is important to remember that we are using astronomical units and omitting the gravitational constant γ, and though this does not affect the argument when potential is defined as work per unit mass. We now use the formula for potential given below :
ATTRACTION AND POTENTIAL-I
293
V = � ��
���
�
rm
…(1)
If we want to find the dimensions of gravitational potential V we must restore the constant γ in (1) and write
V = γ � ��
���
�
rm
…(2)
Because the constant γ has dimensions. By definition of force of attraction, the quantity
F = γ2
1
r
mm …(3)
Represents a force and is therefore has dimensions MLT−2. Equation (3) gives
γ = M−1 L3 T−2 …(4)
Thus, the dimensions of constant γ are
M−1L3 T −1
Hence, the dimensions of γ ��
���
�
rm
are
(M−1 L3 T−1) ��
���
�
LM
= L2T−2. = (LT−1)2.
NOTE :- (1) The potential energy decreases when the work is done and the gravitational potential increases when the work is done.
(2) The dimensions of gravitational potential V are those of the square of a velocity.
10.4 EQUIPOTENTIAL SURFACES AND LINES OF FORCE
Regarding the potential V(x, y, z) of a given attracting system as a function of coordinates x, y, z the equation
V(x, y, z) = constant, …(1)
represents a surface over which the potential V is constant. Such surfaces are called equipotential surfaces. By the definition of potential V(x, y, z), it is clear that only one equipotential surface passes through any point of space, so
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 294
that no two equipotential surfaces can intersect. Also, since the potential V(x, y, z) has a constant value over a equipotential surface, no work would be done by the attraction on a particle moving on such a surface. Therefore, at every point the resultant attraction is normal to the equipotential surface through the point. The observation is also obvious from the relation
R = grad V. …(2)
Definition :- The curve such that the tangent at any point of it is in the direction of the resultant attractive force at that point is called line of force. Remark : The line of force is at right angles to the equipotential surfaces at all their points of intersection. Conversely, a surface which cuts all lines of force at right angles must be an equipotential surface, because at no point on the surface is there a component of force tangential to the surface, so that no work would be done on a particle moving on the surface and there could therefore be no variation in the potential.
Continuous Bodies
We now pass from the attraction components (X, Y, Z) and potential V(x, y, z) of a system of separate particles to the attraction components (X, Y, Z) and potential V(x, y, z) of distributions of matter regarding as continuous bodies.
By the principle of super position we can obtain the attraction components and potential V(x, y, z) of such a body provided that we have a means of summing the contributions of the separate particles.
It is natural to look to integration to effect the summation, but when we consider what the process of integration involves, we find that it does not fit the physical conditions of the problem precisely, and it is only by giving a special interpretation to our conception of 'body' that we can justify the use of integration. Thus, it is usual to represent the potential V of a continuous body by a volume integral, i.e.,
V = �rdv�
,
where dv denotes an element of volume of the body at the distance r from an external point P and ρ denotes the density of matter in dv.
The process of integration implies that the density of the body is continuous. To justify the use of such an integral it is necessary therefore to suppose that it is applied to a hypothetical continuous distribution of matter occupying the same region as the body and having at each point a suitably chosen density; this density being found by considering a small but finite volume surrounding
ATTRACTION AND POTENTIAL-I
295
the point and taking the average through this small volume of the masses of the particles of the real body contained there in.
Attraction of a uniform straight rod
Let m denote the mass per unit length of a uniform rod AB of finite length. It is required to find the components of attraction of the rod AB at an external point, say P.
Let the perpendicular from P to AB meet AB in M, which for simplicity we take on AB produced.
MP = p. …(1)
consider an element QQ′ of the rod AB, where
MQ = x, QQ′ = dx. …(2)
Let
∠MPQ = θ.
Then, from the triangle PMQ, we have
x = p tan θ, …(3)
dx = p sec2θ dθ …(4)
The mass of the element QQ′ of the rod is
m dx = m p sec2θ dθ.
The attraction at P of the element QQ′ of the rod AB is, therefore,
2
2
)PQ(
�d�secmp along PQ. …(5)
X P
Y
θ R′ R
E
D
A Q′ Q M B
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 296
From the triangle PMQ, we have
PQ = p secθ. …(6)
Combing (5) and (6), the attraction at P of the element QQ′ becomes
p�dm
along PQ. …(7)
Let the components of attraction of the rod AB parallel and perpendicular to BA be X and Y.
Let the angles MPA and MPB be α, β.
Then, we have
X = ��
� pm
sinθ dθ
= pm
(cos β − cos α)
= 21
sinpm2
(α+β) sin ��
���
� −2��
, …(8)
and
Y = ��
� pm
cosθ dθ
= pm
(sin α − sin β)
= ��
���
� β+α��
���
� β−α2
cos2
sinpm2
…(9)
Let R be the resultant attraction. Then
R = 22 YX +
= ��
���
� −2��
sinpm2
= ( ) .BPA21
sinpm2
�
��
…(10)
ATTRACTION AND POTENTIAL-I
297
The direction f R is given by
tan−1��
���
�
XY
= tan−1�
��
−−
�sin�sin�cos�cos
= tan−1�
��
��
���
� +2��
tan
= ��
���
� +2��
. …(11)
Thus, the resultant attraction R acts along the bisector of the angle APB and
makes an angle 21
(α+β) with PM.
Remark 1: The component (8), parallel to the rod. AB, can also be written as
X = PAm
PBm − …(12)
in the sense parallel to BA.
Remark 2 : We note that if a circle of centre P and radius PM cuts PA, PQ′, PQ, PB in D, R′, R, E then the attraction at P of the element RR′ of a rod in the form of a circular arc DE of the same line density (mass per unit length) as AB is
= 2p
�mpd
= p�md
= attraction of QQ′.
Hence the circular arc DE exerts the same attraction as the rod AB.
Corollary : If the rod is infinitely long, the angle APB is two right angles and the resultant attraction is
pm2
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 298
and perpendicular to the rod.
If would appear from this result that if the attracted particle were close to the rod, the attraction would be infinite; but this conclusion is not justified because in the foregoing argument we assumed that every point of an element QQ′ of the rod was at the same distance from the point P, and for this to be true when P is close to the rod it would be necessary for the rod to have no thickness. To find the attraction at a point close to a rod of finite thickness it will be necessary to take account of the cross-section of the rod.
Potential of a uniform straight rod
Let m denote the mass per unit length of a uniform rod AB of finite length. It is required to find the potential of rod AB at an external point, say, P.
Let the perpendicular from P to AB meet AB in M, which for simplicity we take on AB produced. Let
MP = p.
Consider an element QQ′ of the rod, where
MQ = x, QQ′ = dx.
Let the angle MPQ be θ.
From the triangle PMQ, we write
x = p tanθ,
dx = p sec2θ dθ,
PQ = p sec θ.
The mass of the element QQ′ of the rod is
mdx = mp sec2θ dθ
P
A
θ
Q′ Q M B
ATTRACTION AND POTENTIAL-I
299
The potential at P is given by the formula
V = �PQ
mdx
= ��
�
2
�secp�d�secmp
= m log �
��
��
���
���
���
�
2PBA
cot2
PABcot …(1)
Let 2l denote the length of the rod AB and,
PA = r, PB = r′,
and r + r′ + 2l = 2s.
Then,
cot )l2s)('rs(
)rs(s.
)l2s)(rs()'rs(s
2PBA
cot2
PAB−−
−−−
−=��
���
���
���
�
= ll
2'rr2'rr
−+++
. …(2)
so, from equations (1) and (2), the potential V is expressed as
V = m log ��
���
�
−+++
ll
2'rr2'rr
. …(3)
Remark 1 : If the ends A, B of rod are foci of an ellipse passing through P and 2a is its major axis, then
V = m log ��
���
�
−+
lala
, …(4)
or V = m loge1e1
−+
. …(5)
where e denotes the eccentricity of the ellipse.
Hence the potential V is constant over any prolate spheroid of which A, B are the foci. That is, family of confocal prolate spheroids are the equipotential surfaces. Since the normal to an ellipse at any point bisects the angle between the focial distances and a resultant attraction at a point is normal to the equipotential surface, it follows that the resultant attraction at P bisects the angle APB.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 300
Remark (2) If the rod AB be of great length and P in the neighbourhood of its centre, then we may put
r + r′ = 2 22 pl + , …(6)
where p is small compared with l. Then, from equation (3), we obtain
V = m log ��
��
�
−+
++
lp
lp22
22
l
l
= 2m log ��
�
�
��
�
� ++p
lp22l
= 2m log
�����
�����
�
���
����
�+
p
2p
22
ll
,
neglecting the term (p/l)2, we write
V = 2m log 2 −2m log p …(7)
By differentiating (7), we get for the attraction of the rod in the direction p increasing,
pm2
pV −=
∂∂
.
10.5. THE ATTRACTION AND POTENTIAL OF A UNIFORM LONG ROD WHOSE CROSS-SECTION IS A CIRCLE
Take a cross-section of the long rod about the middle of its length. Let O be the centre of the cross-section and P any point inside it.
R R′
P O θ
Q Q′
ATTRACTION AND POTENTIAL-I
301
Through the point P draw chords QPR, Q′PR′ making a small angle dθ with one another, and intercepting small arcs QQ′, RR′ on the circle. We can conceive the long rod to be composed of long parallel rods, and take QQ′, RR′ as the cross-sections of two of them. Then if m denotes the mass per unit area of given long uniform rod, mQQ′ and mRR′ denote the mass per unit length of the two rods.
Let ORP|OQP| = = φ.
Using the result for the attraction of a long rod, the attraction at P due to rod through QQ′
= PQ
)'mQQ(2
= 2m. sec φ dθ …(1)
and acts along PQ. Similarly, the attraction at P due to rod through RR′
= PR
RRm2
= 2m sec φ dθ …(2)
and acts along PR.
Hence, these two rods exert equal and opposite attractions at P; and by dividing up the whole long rod into similar pairs of rods we obtain that its resultant attraction at any internal point is zero. Consequently, the potential must be constant at all points inside the long rod sufficiently far from its ends, and is therefore equal to the potential at O. But, the potential due to the rod QQ′ at O is
= 2m QQ′ log(2 l). −2m QQ′ log(OQ) …(3)
where 2l is the length of the rod. Therefore, for the whole potential, we write
V = 2M log 2l − 2M log a, …(4)
a being the radius of the cross-section and
M = 2πma. …(5)
Now we consider the case when the attraction and potential are to determined at an external point. Let P′ be an external point. Let P be its inverse point w.r.t the circle mentioned above. Then
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 302
OP. OP′ = a2. …(6)
Bu the similarity of triangles
OQP|Q'OP| = = φ = R'OP| …(7)
Then, the attraction at the external point P′ of the rod through QQ′ is
= 'Q'P
'QQm2
= Q'P
�sec�dmPQ2
= (2m dθ sec φ)'OP
a …(8)
and acts along P′Q. But the resultant attraction of the rod at P′ is clearly along P′O, and by resolving the attraction of the rod through QQ′ in this direction, we
get �d'OP
am2.
Similarly, other rods give like resultants and the whole attraction at P′ is, therefore, equal to
'OP
M2'OP
am�4 = . …(9)
We observe that this is the same as if the whole mass of the circular rod were condensed into a rod of equal mass along the axis of the cylinder. To find the potential, we may put
r = OP′ …(10) and write
,rM2
drdV −= …(11)
giving
v = −2M log r + C. …(12)
O
R R′
P′
Q′ Q
P
ATTRACTION AND POTENTIAL-I
303
In order that V may take the form (4) when r = a, we must have
C = M log 2l . …(13) So, V = 2M log 2l − 2M log r. …(14)
10.6 ATTRACTION AND POTENTIAL OF A UNIFORM CIRCULAR DISC AT A POINT ON ITS AXIS
Let O be the centre of the disc, P a point on its axis Oz at a distance z from O. Let m denote mass per unit area.
Divide the disc into concentric rings. Let
OQ = x
be the radius and
QQ′ = dx
The breadth of one of these rings. The mass of the ring is 2π m x dx, and the attraction at P of each element is got by dividing the mass of the element by (PQ)2. But the resultant attraction of the ring is in the direction PO, so that its magnitude is
= 2)PQ(
dxmx�2 cosθ,
where θ is the angle OPQ| . But
x = z tanθ
dx = z sec2 θ dθ.
z
P
S
Q′ Q
θ Q
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 304
So, if α is the angle which a radius of the disc subtends at P, we have for the whole attraction of the disc
= �α
���
����
�
θθθθπ
0
22
22
seczcossectanmz2
dθ
= 2π m (1− cosα). …(1)
Remark (1) : For an infinite plate we may put α = π /2, so that the attraction of an infinite plate is 2πm at right angles to itself.
Potential : The potential at point P of the ring of radius x is given by
�π
PQdxmx2
,
that the potential of whole disc of radius a at point P on its axis Oz at a distance z from O is
V = �π
a
0
PQdxmx2
= 2πm � +
a
0
22 xz
dxx
= 2πm { 22 az + −z}. …(2)
NOTE : The formula (2) gives the attraction in the direction PO as
−��
���
��
���
+−=
22 az
z1m�2
dzdV
. …(3)
Remark (1) : The formula (2) is equivalent to
V = 2πm (SP − OP), …(4)
and that this will give the value of V on either side of the disc if, SP, OP denote numerical lengths.
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305
Solid Angles and Its Use
Definition : The solid angle of a cone is measured by the area intercepted by the cone on the surface of a sphere of unit radius having its centre at the vertex of the cone.
Definition : The solid angle subtended at a point by a surface of any form is measured by the solid angle of the cone whose vertex is at the given point and whose base is the given surface.
Let PP′ be a small element of area dS which subtends a solid angle dω at O.
Let the normal to area dS make an acute angle γ with OP, and let OP = r. Then the cross-section at P of the cone which dS subtends at O is dS cos γ, and this cross-section and the small area dw intercepted on the unit sphere (having centre at o) are similar figures, so that
1r
dcosdS 2
=ω
γ …(1)
Then from (1), we find
dω = 2r
�cosdS, …(2)
or
dS = r2 sec γ dω. …(3)
Integrating equation (3) both side we obtain
S = � r2 sec γ dω …(4)
with suitable limits of integration. Hence the area of a finite surface can be represented as an integral over a spherical surface.
cross-section
γ
P dS
P′
dω
O
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 306
10.7 USE OF SOLID ANGLES
There are many applications of the theory of attraction in which calculations are simplified by the use of the solid angle.
Article : Find the component of attraction perpendicular to itself produced by a plane plate of any form.
Let dS be an element of area of the plate at a distance r, from the point O and subtending a solid angle dω at O.
Let m denote the mass per unit area of the plate. Then the attraction of element dS at the point O is given by
= 2r
mdS. …(5)
Resolving attraction in (5) at right angles to the plate, we get
,r
�cosdSm2
…(6)
where θ is the inclination of r to the normal to the plate. From (2) and (6), we conclude that m dω is the contribution of area element dS of the plate to its whole attraction at O at right angles to itself. Further, the attraction of the whole plate in the same direction is m ω, where ω is the solid angle which the plate subtends at O. This completes the article.
Article : Prove that the potential of a solid of uniform density ρ at an external point P can be represented by a surface integral
21 ρ � cos θ dS
over the surface of the solid, where θ is the angle between the inward normal to dS and the line joining dS to P.
plate
O
θ r
dS
dω
ATTRACTION AND POTENTIAL-I
307
Proof :- Let a cone of small solid angle dω and vertex P cut the surface of the solid in elements of area dS1, dS2 at A, B, where the inward normals make angles Q1, Q2 with the line BAP.
Let AP = r1, BP = r2. The mass of an element of volume of the cone at a distance r from the point P is ρr2dω dr.
Hence, the mass of the cone between A and B produces at P a potential equal to
� =ωρ
rdrdr 2
� ρ r dω dr
= ( ) ω−ρ drr21 2
122
= [ ]1122 �cosdS�cosdS�21 + .
If we take the sum for all cones which intersect the solid, we shall get
�ρ21
cos θ dS
as the potential at P, integration is being taken over the surface of the given solid.
The Books Recommended for Chapters X and XI. 1. A.S. Ramsey Newtonian Gravitation, ELBS and Cambridge
University Press.
P dω
θ1
A
B
θ2
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 308
Chapter-11
Attraction and Potential-II 11.1 ATTRACTION AND POTENTIAL AT INTERNAL POINTS
In the previous chapter we confined our attention to the attraction and potential at points external to the attracting matter.
We have now to consider the case of attraction and potential at points inside the attracting matter.
Our definitions of attraction and potential at an external point imply the existence of a separate attracted particle at the point under consideration. Such a particle cannot exist inside a continuous body because two particles cannot occupy the same space simultaneously.
We therefore imagine that there is a small cavity in the body surrounding the attracted particle placed at the chosen point. We assume that we can calculate the attraction and potential at this point by our former rules, since the attracted particle is not in contact with the matter. Then we define the attraction and potential at the same point in the continuous body to the limits to which the attraction and potential at the point in the cavity tend as the cavity decreases in size and ultimately vanishes.
11.2 ATTRACTION AND POTENTIAL OF UNIFORM THIN SPHERICAL SHEEL
Let m be the mass per unit area of a thin spherical shell of radius a and centre O.
(A) ATTRACTION AT AN INTERNAL POINT :-Let p be any internal point. with p as vertex construct a cone of small solid angle dω intersecting the surface in elements QQ′, RR′ of areas dS, dS′. The attractions at P of these elements are [mdS/(QP)2] and [mdS′/(RP)2 in opposite directions. But
Q
dS Q′
R R′ dS′
O P
ATTRACTION POTENTIAL - II 309
dS = (QP)2 (sec/OQP)dω,
and
dS′ = (RP)2 (sec(ORP) dω,
and the angles ORP|,OQP| are equal. So, the elements QQ′, RR′ of the spherical shell exert equal and opposite attractions at the point P. Since the whole shell can be divided into similar pair of elements by taking cones in all directions round the point P, it follows that the resultant attraction of the spherical shell at the internal point P is zero.
(B) Attraction at an external point :
Let P′ be any external point of the spherical shell and P its inverse, so that
OP⋅OP′ = a2
The resultant attraction at P′ is, by symmetry, along P′O and the element QQ′ exerts an attraction {mdS/(QP′)2} along P′Q. Resolving this attraction in the direction P′O, we get
= Q'OP|cos)'QP(
d)OQP|(sec)QP(m2
2 ω.
But the triangles QPP′ and OQP′ are similar, so
'OP
OQ'QP
QP = ,
and the angles OQP. OP′Q are equal. Therefore, the element QQ′ contributes an amount {ma2dw/(OP′)2} to the resultant attraction. By taking cones in all directions round P, we get for the attraction at P.
of the whole shell
=2
2
)'OP(
ma�4
= 2)'OP(
M,
Q
Q′
R
O P
R′
P′
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 310
where M denotes the mass of the spherical shell. It follows that the attraction of the shell at external points is the same as if its mass were collected into a particle at its centre.
(C) Potential at an internal point: Since the attraction is zero throughout the interior of the shell, there can be no variation in the potential, or the potential is constant. The potential at every point in the interior is, therefore, the same as the potential at the centre, i.e., M/a, where M denotes the whole mass, since every element of M is at the same distance a from the centre O.
(D) Potential at an external point : Let OP′ = r. Since the force at distance r is M/r2 in the direction in which r decreases, therefore,
2r
Mdrdv −= .
Integrating, we obtain
V = cr
M +
where c is a constant of integration. But the potential vanishes at an infinite distance, therefore,
c = 0
Hence, V = .r
M
11.3 ATTRACTION OF A THIN UNIFORM SPHERICAL SHELL AT A POINT OF ITSELF
The attraction at a point of itself of a thin layer of matter depends on the shape of the gap in the surface in which the attracted particle is placed. We may define the principal value of the attraction as the limiting value of the attraction at the centre of a circular hole when its radius tends to zero.
Let m be the mass per unit area of the spherical shell, and omitting a small circular element of the shell surrounding a point P, consider the attraction of the rest of the shell at the point P.
Let an element QQ′ of area dS subtend a solid angle dω at P. The attraction of the element dS at P is
Q
Q′
P O
ATTRACTION POTENTIAL - II 311
2)PQ(
dSm
along PQ. Resolving this attraction along the direction PO, the direction of the resultant attraction, we get
2)PQ(
)OPQ|(cosdSm
= 2)PQ(
)OQP|(cosdSm
= m dω.
If now we allow the gap in the spherical shell round P to shrink to vanishing point, resultant we see that we have to take the sum � m dω for all cones on one. Side of the tangent plane at P. So, the resultant attraction is 2πm.
11.4 ATTRACTION AND POTENTIAL OF A UNIFORM SOLID SPHERE
Let a be the radius and ρ the density of the sphere. Such a sphere may be regarded as composed of a series of concentric thin spherical shells, and the required results may be obtained by summation.
(A) Attraction at an interval point : Take a point P at a distance r, 0 < r < a, from the centre. Imagine a thin spherical shell of matter of radii r +∈ and r−∈ to be removed and consider the attraction at a point P in this cavity.
The concentric shells external to the cavity exerts no attraction at the point P, and those internal to the cavity attracts as though their masses were concentrated at the centre O.
Hence, the attraction at P is the limit as ∈→0 of
23 r)r(34 ∈−ρπ
which is equal to π34 ρr. This shows that the attraction of a uniform solid
sphere at an internal point is directly proportional to the distance from the centre.
o r
a
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 312
(b) Attraction at an external point. Since each of the concentric spherical shells attracts at an external point as though its mass were collected at its centre, the same is true of the solid sphere. The attraction of the solid sphere is, therefore, represented by M/r2, where M is its mass and r the distance of an external point from the centre of the solid sphere.
(c) Potential of a uniform solid sphere at an internal point
Adopting the method of finding the attraction of a solid sphere at an internal point, let R denote the radius of a shell external to the cavity. Its mass is 4πρR2dR, so that the potential it produces at a point inside itself is 4πρRdR. Consequently, the potential at P due to all such external shells is given by
�∈+
a
r4πρRdR = 2π{a2 − (r + ∈)2}.
Also, the shells of radius less than that of the cavity produce the same potential as if the mass were collected at 0, i.e.,
).r()r(��34 3 ∈−∈−
Hence, the whole potential at P is the limit as ∈→0 of
[ ]���
��� ∈+−+∈− 222 )r(a
23
)r(��34
= ��32
(3a2 − r2).
(d) At an external point. Since each of the concentric shells produces at an external point a potential equal to its mass divided by the distance of the point from the centre, the same is true for the solid sphere. That is, the potential V at an external point P, OP = r, due to a uniform solid sphere with centre O is
V = M/r,
where M is the total mass of the uniform solid sphere.
Exercise : Deduce the expression for the potential of a uniforms solid sphere from the attraction.
11.5 WORK DONE BY SELF-ATTRACTING SYSTEMS
Let the component particles be m1, m2,…, and let A1, A2,… be their positions in the given system. Let
rst = distance between ms and mt.
ATTRACTION POTENTIAL - II 313
First bring m1 from infinity to its assigned position A1. The work done is zero, for there are no particles of the system near enough to exert attraction on it.
Next, bring the particle m2 from infinity to its position A2. The work done on it
= m2 ×(potential of m1 at A2)
= γ12
21
rmm
Next, bring the particle m3 from infinity to is position A3. The work done on it
= γ ,r
mm�
rmm
23
23
13
13 +
and so on for the other particles of the system.
Hence, the total work done in collecting all the particles from rest at infinity distances from one another to their positions
= γ
��
�+++
��
�++
34
3
24
2
14
14
23
2
13
13
12
21
rm
rm
rm
m�rm
rm
m�rmm
= γ �12
21
rmm
…(1)
When the particles of the system have all been brought to their positions, let V1, V2,… be the potentials of the system at the respective points A1, A2,…,An,… Then
V1 = γ14
4
13
3
12
2
rm�
rm�
rm
++ +…
V2 = γ24
4
23
3
21
1
rm�
rm�
rm
++ +… …(2)
In view of (2), the expression (1) becomes
= 21�m1 V1, …(3)
as in the expression (3), any such term γst
ts
rmm
is twice repeated.
Hence, the work done in bringing the particles of the system from infinity to their respective positions,
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 314
= 21� m1 v1. …(4)
For continuous masses, we write (4) as
21� V dm …(5)
where V is the potential of the body A at any element dm of itself, and the integration is taken throughout the configuration A.
11.6 LAPLACE EQUATION FOR THE POTENTIAL
Let V be the potential of a system of attracting particles at a point, say P(x, y, z), which is not in contact with the particles. Let m be the mass of a particle at A1(a, b, c) of the given system. Let
r2 = (x−a)2 + (y−b)2 + (z−c)2 …(1)
we know that V is given by the formula
V = � ��
���
�
rm
. …(2)
Equation (1) gives
��
���
� −=∂∂
rax
xr
��
���
� −=∂∂
rby
yr
��
���
� −=∂∂
rcz
zr
. …(3)
Differentiating (2) partially w.r.t. x, we obtain
��
� −Σ−=��
���
�
∂∂
��
���
�Σ−=∂∂
32 r)ax(m
xr
rm
xV
. …(4)
Differentiating again (4) partially w.r.t. x, we have (left as an exercise)
��
� −Σ+��
���
�Σ−=∂∂
5
2
32
2
r)ax(m
3rm
xV
. …(5)
Similarly, we shall have (exercise)
ATTRACTION POTENTIAL - II 315
,r
)by(m3
rm
yV
5
2
32
2
��
� −Σ+��
���
�Σ−=∂∂
…(6)
��
� −Σ+��
���
�Σ−=∂∂
5
2
32
2
r)cz(m
3rm
zV
. …(7)
Adding (5)−(7) vertically, we find (exercise)
0zV
yV
xV
2
2
2
2
2
2
=∂∂+
∂∂+
∂∂
. …(8)
This shows that the potential V of a system of attracting particles satisfies the Laplace equation. That is, V is a harmonic function.
Continuous Body
Let V be the potential of a continuous body or bodies at a point P(x, y, z) outside the body or bodies. Let ρ be the density of the element of volume dv at (x′, y′, z′), and
r2 = (x−x′)2 + (y−y′)2 + (z−z′)2. …(1)
We know that the potential V is given by
V = rdv�
� …(2)
Differentiating (2) under the integral sign, we get (exercise)
,dvr
)'xx(�3r�
xV
5
2
32
2
��
� −−�−=∂∂
…(3)
,dvr
)'yy(�3r�
yV
5
2
32
2
��
� −−�−=∂∂
…(4)
dvr
)'zz(�3r�
zV
5
2
32
2
��
� −−�−=∂∂
…(5)
r P(x,y,z)
(x′,y′,z′)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 316
Adding (3) to (5) vertically, we obtain (exercise)
0zV
yV
xV
2
2
2
2
2
2
=∂∂+
∂∂+
∂∂
. …(6)
The Laplace equation (6) is satisfied by the potential of an attracting system at every point P(x, y, z) at which there is no matter.
11.7 POISSON’S EQUATION FOR THE POTENTIAL
Now let the point P(x, y, z) be inside the given attracting matter. Describe a sphere of small radius ∈ and centre (a, b, c) containing the point P, taking ∈ so small that we may regard the density ρ of the matter in this sphere as uniformly distributed.
The matter which produces the potential V at P may now be divided into two parts the matter outside this small sphere and the matter inside this small sphere. Let V0 and Vi denote their respective contributions to the whole potential V at P.
Since the point P is not in contact with the matter which produces the potential V0, therefore,
∇2 V0 = 0. …(1)
Further, Vi being the potential at a point P(x, y, z) inside a small sphere of radius ∈, we have
Vi = 32 πρ (3∈2 − r2), …(2)
where r is the distance of the point P(x, y, z) from the centre (a, b, c), i.e.,
r2 = (x−a)2 + (y−b)2 + (z−c)2. …(3)
From equations (2) and (3), we find (exercise)
2
i2
2i
2
2i
2
z
V
y
V
x
V
∂∂+
∂∂+
∂∂
= −4πρ. …(4)
Equation (4) is known as Poisson equation.
Remark : The relation
V0
∈ Vi
ATTRACTION POTENTIAL - II 317
Attraction = grad V
which we have seen to be true at points outside a system of attracting particles, is also true at all points outside or inside a continuous distribution of matter.
Summarize of our results upto this point
(i) The attraction is the gradient of a potential function V both outside and inside the attracting matter.
(ii) In empty space, the potential V satisfies Laplace equation
∇2 V = 0.
(iii) At any point at which there is matter of volume density ρ, the potential V satisfies Poisson equation
∇2V = −4πρ.
11.8 GAUSS’S THEOREM (SURFACE INTEGRAL OF NORMAL ATTRACTION OVER ANY CLOSED SURFACE) Statement : If N be the normal attraction at any point of the element dS of any closed surface, measured positively along the normal outwards, due to any attracting mass, then
� N dS = −4γπM
where M is the amount of the attracting mass within the surface and the integral is being taken given the whole surface.
Proof : Let O be the position of any element m of the attracting mass within the closed surface. Through O, draw a cone of very small vertical angle and let it cut the given surface in the elements PQ and P′Q′, whose areas are dS and dS′.
The attraction of the mass m at these elements are
γ m OPN|cosOPdS
2 ��
���
� , along PN, …(1)
γ m 'N'OP|cosOP
'dS2' ��
���
� , along P′N′, …(2)
where PN and P′N′ are the outward drawn normals at P and P′, respectively, as shown below.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 318
Through Q and Q′ draw normal sections QM and Q′M′ of this cone. Let dw be the solid angle of the cone. Let Q be the angle between the elements QM and QP. Then
Q + ∠OPN = π. …(3)
Now
dω = 2OQ
QMarea
= 2OQ�cosdS
= − 2OP
)OPN|(cosdS, …(4)
in the limit when PQ is very small. Similarly,
dω = −2'OP
)'N'OP|(cos'dS …(5)
Hence, the total normal attractions for the elements dS and dS′ at points P and P′ are each equal to −γm dw. Hence, the total normal attraction for the whole surface is
−γm �dm
= −γm π4( ),
i.e., for a single particle m at 0, we have
� NdS = −4Vπm. …(6)
Similarly for all other particles of the attracting mass inside the surface. Hence, finally, for the whole mass, we have
P
N dS
M Q
P′ Q′
N′
O
M′ dS′
ATTRACTION POTENTIAL - II 319
� NdS = −4γπm.
This completes the proof of the theorem.
11.9 EQUIPOTENTIAL SURFACES
For any attracting mass M, the potential V at any point P(x, y, z) will be a function of coordinates x, y, z consider the equation
V(x, y, z) = C, …(1)
Where C is a constant. The equation (1) represents a surface such that the potential V(x, y, z) at any point of it for the given attracting mass is constant and equals to C. It is hence called an equipotential surface. By given different values to C, we get a family of equipotential surfaces.
Remark 1 : Only one equipotential surface passes through any point of space, so that no two equipotential surfaces can intersect.
Remark 2 : No work would be done by the attractions on a particle moving on an equipotential surface. Therefore, at every point the resultant attraction is normal to the equipotential surface through the point.
Remark 3 : In the case of rod AB, the equipotential surfaces are ellipsoids of revolution obtained by rotating confocal ellipses, whose foci are A and B, about AB as axis
Remark 4 : In the case of the spherical shells and sphere, the equipotential surfaces are concentric spheres.
Distribution for a given potential
When the potential is given at all points of space, we can determine the corresponding distribution. For, the potential V being known, we can find ∇2V for every point of space. The Poisson equation gives
ρ = −π4
1 ∇2V. …(1)
Equation (1) serves to determine the volume density. When
∇2V = 0, …(2)
the corresponding density of the distribution is zero. That is, there is no attracting mass at all such points. Whenever
∇2V ≠ 0, …(3)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 320
the corresponding density of the distribution is given by (1),
If the form of the potential function V inside any surface S is different from its
form outside, and if there be an abrupt change in the value of dxdV
as we pass
across this surface, then the surface density σ on S is calculated as below. For, let V1 be the potential just inside S and V2 the potential just outside S. Let dn be an element of the outward drawn normal. The surface density σ of S is determined by the relation
σ = −π41
��
�
∂∂
−∂
∂n
Vn
V 12
where the direction of the normal ∂n is from 1 to 2.
Question : The potential outside a certain cylindrical boundary is zero and inside it is
V = x3 − 3x y2 − ax2 + 3ay2.
Find the distribution of matter.
Solution : First we found the boundary. Since the potential V is continuous across the boundary and zero outside, the boundary must be given by the equation
x3 − 3xy2 − ax2 + 3ay2 = 0. …(1)
Equation (1) can be written as
(x−a) (x− 3 y) (x + 3 y) = 0. …(2)
It shows that the section of the given cylindrical boundary is an equilateral triangle OAB of height a
We find
y
o
x = a A P
a M
B
x
OM = a
ATTRACTION POTENTIAL - II 321
2
2
xV
∂∂
= 6x − 2a, …(3)
2
2
yV
∂∂
= −6x + 6a, …(4)
2
2
zV
∂∂
= 0. …(5)
Hence ∇2V = 2
2
2
2
2
2
zV
yV
xV
∂∂+
∂∂+
∂∂
= 4a. …(6)
we know that inside the given cylindrical region, the volume density ρ is given by the formula
ρ = −�4
1 ∇2V …(7)
Equations (6) and (7) give
ρ = −πa
. …(8)
Further, since V is zero outside the region, hence ρ = 0 outside the region.
On the boundary : When the point P lies on the part AB of the boundary, then the surface density is given by
σ = − ��
���
�
∂∂
−∂
∂nV
nV
�41 12 , …(9)
where V1 and V2 are the potentials on two sides of the boundary. Here
V1 = V, V2 = 0. …(10)
and the normal to AB is along x-direction. Hence, from equations (9) and (10), we have
σ = π4
1 ( )�4
1ax2y3x3 ax
22 =−− = (a2 − 3y2)
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 322
= �4
3 (MA2−MP2)
= �4
3AP⋅PB …(11)
For a point Q on AO part of the boundary, we have (left as an exercise)
σ = �4
3OQ ⋅ QA. …(12)
Similarly for the boundary part OB. This shows that a solid cylindrical region
(prism) of uniform density πa
would produce the same external field as a
distribution of matter of surface density ��
���
�
�43
(AP ⋅ PB) on each of the faces of
the prism. Hence the result.
11.10 SURFACE AND SOLID HARMONIC We know that the potential V satisfy the Laplace’s equation
∇2V = 0
or 2
2
2
2
2
2
zV
yV
xV
∂∂+
∂∂+
∂∂
= 0. …(1)
Definition : Any solution of Laplace’s equation which is homogeneous in x, y, z, is called a harmonic function or a spherical harmonic.
Definition : The degree of homogeneity is called the degree of the function.
Remark 1 : We are concerned with the case in which the degree of the function is an integer.
Remark 2 : If V is a harmonic function of degree n, then
.Vzyx t
t
q
q
p
p
∂∂
∂∂
∂∂
is a harmonic function of degree (n−p−q−t).
Surface and Solid Harmonics
In polar co-ordinate(r, θ, φ) the Laplace’s equation (1) may be written as (left as an exercise)
0�
V�sin
1�
V�sin
��sin1
rV
rr 2
2
22 =��
�
����
�
∂∂+�
�
���
�
∂∂
∂∂+�
�
���
�
∂∂
∂∂
…(2)
ATTRACTION POTENTIAL - II 323
We take,
V = rn Sn, …(3)
Where Sn is independent of r and Sn = Sn(θ, φ). Differentiating equation (3) w.r. to r partially, we write
rV
∂∂
= nrn−1. sn
or r2 rV
∂∂
= n rn+1. Sn …(4)
Differentiate (4) w.r. to r both side partially, we obtain
��
���
�
∂∂
∂∂
rV
rr
2 = n(n+1) rn Sn …(5)
From equations (2), (3) and (5), we obtain (exercise)
2n
2
2n
2n
2
�
S
�sin1
�
S�cot
�
S
∂∂+
∂∂+
∂∂
+ n(n+1) Sn = 0 …(6)
We put
cos θ = µ …(7)
in equation (6) and obtain (exercise)
2n
2
2n2
�
S
�11
�
S)�1(
� ∂∂���
����
�
−+
��
�
∂∂
−∂∂
+ n(n+1) Sn = 0 . …(8)
A solution, Sn, of equation (6) is known as a Laplace’s function or a surface harmonic of order n. Since n(n+1) remains unchanged when we write −(n+1) for n, there are two solutions of equation (2) of which Sn is a factor, namely rnsn and r−n−1 Sn.
Definition : The functions rn Sn and r−(n+1) Sn are called solid harmonics of degree n and −(n+1), respectively.
Results 1 : If U is a harmonic function of degree n, then ��
���
�+1n2r
U is a also a
harmonic function.
For example, U = xyz is a harmonic function of the degree 3, therefore, xyz/r7 is also a harmonic function.
Result 2 : If U is a harmonic function of degree −(n+1), then r2n+1U is also a harmonic function.
PARTIAL DIFFERENTIAL EQUATIONS AND MECHANICS 324
Note : x0 = 1 and θ = tan−1(y/x) are both harmonic functions of degree zero.
Consequently r1
andr1
tan−1(y/x) are harmonics of degree −1.
Differentiating these w.r.t. x, y, z; we find )x/y(tanrz
,rz
,ry
,rx 1
3333− etc., as
harmonics of degree −2, and so on.
11.11. SURFACE DENSITY IN TERMS OF SURFACE HARMONIC
We assume that
V1 = �∞
+0
1n
n
ar
Un, r < a …(1)
and
V2 = �∞
+0
1n
n
ra
Un, r > a …(2)
Give the potential of a certain distribution of matter. In equations (1) and (2), Un denotes the sum of a finite number of surface harmonics (one for each particle) and therefore itself a surface harmonic. Hence
∇2V1 = 0, ∇2V2 = 0. …(3)
The matter resides on the surface of the sphere, and its surface density σ is given by the formula
ar
21
rV
rV
4=
��
���
�
∂∂
−∂
∂=σπ . …(4)
Equations (1), (2) and (4) give (exercise)
σ = �+∞
=0n2a�41n2
Un. …(5)
It follows that if we accept the physical argument that an arbitrary distribution of surface density on the surface of a sphere produces the same kind of field of potential as an aggregate of particles distributed over the sphere with the potentials given by (1) and (2), then the arbitrary surface density is expressible in the form (5).
Hence the result.