wave equations
DESCRIPTION
Partial Differential EquationTRANSCRIPT
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 1
Subject: Math
Lesson: Wave Equations
Course Developer: Dr. Preeti Jain
College/Department: A.R.S.D. College, University of Delhi
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 2
Table of Contents:
Chapter : Wave Equations
1: Learning Outcomes
2: Introduction
3: Homogeneous Wave Equation
4: Initial Boundary Value Problem
5: Non- Homogeneous Boundary Conditions
6: Vibration of Finite Strings with Fixed Ends
7: Non- homogeneous Wave Equations
8: Riemann Method
9: Goursat Problem
10: Spherical Wave Equation
Summary
Exercises
Glossary
References/ Further Reading
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1. Learning Outcomes:
After you have read this lesson , you will be able to understand, define
and simplify the homogeneous wave equation, non- homogeneous
wave equation, non- homogeneous boundary conditions, initial
boundary value problem, finite string problem with fixed ends,
Riemann problem, Goursat problem and spherical wave equation. You
should be able to differentiate between homogenous wave equations
and non- homogeneous wave equations. You will also be able to
understand why it is said that the solution of finite vibrating string
problem with fixed ends is more complicated then the problem of
infinite vibrating string. Moreover, you should be able to apply the
knowledge to solve various initial boundary problems which arises in
many practical problems.
2. Introduction:
Differential equations enables us to solve various problems which we
come across in many mathematical problems. Our main concern here
is on boundary value problems. Boundary value problems are
problems in differential equations in which certain conditions (which
we can also call as constraints) are imposed on the equations. After
finding the general solution of these problems, we apply these
additional condition to get a solution which also satisfies the boundary
conditions. We also come across the boundary value problems in many
physical quantities. Generally, these conditions are specified at the
extremes. Here, when we are dealing with wave equations, we will be
considering both the cases homogeneous wave equation and non-
homogeneous wave equation. If we define (initialize) the independent
variable at the lower boundary of the domain, we can often term
homogeneous wave equations as initial boundary value problems.
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3. Homogeneous Wave Equation:
The equation 2 0tt xxu c u is the standard example of hyperbolic
equation. There are two real characteristic slopes at each point ,x t .
This equation is popularly known as wave equation in one dimension
and describes the propagation (bi-directional) of waves with finite
speed c . The characteristics are therefore, curves in the real domain
of the problem. Oscillatory (not always periodic) behaviour in time.
Here time reversible is permissible as it reverses the direction of wave
propagation.
Now, we will obtain the solution of one dimensional wave equation in
free space. For this, consider the Cauchy problem of an infinite string
with the initial conditions
2 0 , , 0tt xxu c u x R t (3.1)
,0 ,u x f x x R (3.2)
,0 ,tu x g x x R (3.3)
To solve equation (3.1), we first reduce it into canonical form. The two
characteristic coordinates
, ,x ct x ct
transforms equation (3.1) into 0.u
After performing two straightforward integration, we get
,u ,
where and are arbitrary functions to be determined, (provided
they are differentiable twice). Thus, the general solution of wave
equation in terms of original variables x and t is
,u x t x ct x ct , (3.4)
,0u x x x f x , (3.5)
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,0tu x c x x g x . (3.6)
Integrating the last equation and then simplifying for and , we get
0
1 1
2 2 2
x
x
Kx f x g d
c , (3.7)
0
1 1
2 2 2
x
x
Kx f x g d
c , (3.8)
0x and K called the arbitrary constants.
The solution of wave equation is, therefore, given by
1 1
,2 2
x ct
x ctu x t f x ct f x ct g d
c
. (3.9)
Solution shown in equation (3.9) is known as DAlembert solution of
the Cauchy problem for one dimensional wave equation.
if double derivative of f and derivative of g exist then by direct
substitution it is evident that ,u x t satisfies the equation (3.1). The
DAlembert solution describes two distinct waves- one moves to right
direction and other on the left both with speed c. Physically,
x ct represents a propagating wave progressing in the negative x-
x
y
0 0,x t
R
O
Figure 1
Range of influence
Domain of dependence
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Institute of Lifelong Learning, University of Delhi pg. 6
direction and x ct represents a propagating wave progressing in
the positive x- direction. These two waves do not interact with
themselves nor with each other and hence the solutions are
superposed. Waves do not change their shape as they propagate. The
results depend continuously on the initial data and boundary data, so
the equation is well posed and the solution is unique.
As one can see, the solution of equation(3.1), consists of terms of
f x ct and integral of g. Equation (3.9), which is a solution of
equation (3.1), further suggests that the solution does not depend on
values of function f x which is independent to initial data. And the
end points of the interval are ,x ct x ct which is known as domain of
dependence of ,x t . Further, the solution depends on the integral
value of g between the limits ( x ct ).
Now, the value of u at 0 0,x t does not depend on all the values of u
and tu at 0,t but only on the values in the interval
0 0 0 0,x ct x ct (domain of dependence of 0 0,x t ). If we change f away
from these two points 0 0x ct and 0 0x ct , and g outside this interval,
value of u does not changes at 0 0,x t . Thus, at the point 0 0,x t the
range of influence is the infinite sector R as shown in figure 1.
Value Addition: What we learnt
From this section, we learnt that waves do not change their shape as
they propagate. They only vary in speed and they propagate with
constant speed c in both directions.
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Example 1: Solve the partial differential equation 2 3 0,xx xy yyu u u
with the given initial conditions
,0 sin ,u x x ,0yu x x .
Solution: The given equation is hyperbolic in nature. The two
characteristic coordinates takes the form
3
yx , x y .
Therefore,
2xxu u u u ,
1 2
9 3yyu u u u ,
1 2
3 3xyu u u u ,
The canonical form (we get after substitution) is 0u .
Therefore,
,3
yu x y x x y
, (3.10)
,0 sinu x x x x , (3.11)
1
,03
y y yu x x x x ,
Integrating, the above eq. we get
0
0
21
3 2
xx
xx
x x d
(3.12)
where 0x is an arbitrary point.
Solving equation (3.11) and (3.12), we get
0
23 3sin
4 4 2
x
x
x x
,
0
21 3sin
4 4 2
x
x
x x
,
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Institute of Lifelong Learning, University of Delhi pg. 8
21 3
, sin sin .4 4 3 3
y yu x y x y x xy
Value Addition: Activity 1
Find the solution of the following partial differential equations with the
given initial condition:
1. 0, , 0,tt xxu u x R t ,0 sin3 , ,0 cos3 .tu x x u x x
2. 9 0, , 0,tt xxu u x R t 3,0 , ,0 cos .tu x x u x x x
4. Initial Boundary Value Problem:
Two initial conditions where the wave starts from and two boundary
conditions (how the wave and boundary interact, the wave might be
scattered or absorbed are required for the IBVP). We mention some
conditions on the function constituting a solution of the BVP.
1. The solution must be continuous in the region for which the
problem is posed, up to the boundary of the region.
2. The solution must have a continuous second derivatives within
the region and satisfy the given equation and also satisfy the
given boundary conditions of the region.
3. The solution (if the region is three- dimensional and infinite)
approaches to zero as we display a given point to an infinte
distance along an arbitrary ray C.
The purpose of this section is to make you familiar with the effect of
boundary conditions on the solution.
Value Addition: Did you know
In an initial boundary value problem, an initial condition is imposed
over the problem while in a boundary value problem two conditions
are imposed one at initial point and other at the end point.
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4.1. Semi- Infinite String with Fixed End:
the purpose of this segment is to make you aware, how to solve the
problem of semi- infinite vibrating string with a fixed end, i.e.,
2 0, 0 , 0tt xxu c u x t ,
,0 ,u x f x 0 ,x
,0tu x g x ,0 ,x
0, 0,u t 0 .t (4.1)
Now, x ct represents a straight line passing through the origin which
divides the region into two halves. Here, we obtain two waves which
propagates in both directions. One in the positive x- direction with the
velocity c which we obtain from x ct and one in the negative x-
direction with the velocity c which we obtain from x ct . Therefore,
for x ct solution is determined by equation (3.9). For x ct as shown
in figure 2, some part of the solution still lies in the first quadrant for
the interval ,x ct x ct as from D Alembert formula
,u x t x ct x ct . (4.2)
Applying boundary conditions, we get
x
y
0 0,x t
x ct
O
Figure 2
x ct
x ct
0 0 ,0x ct 0 0 ,0x ct
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0, 0u t ct ct .
Hence,
ct ct
Replacing ct by x- ct, we obtain
0
1 1
2 2 2
ct x Kx ct f ct x g d
c
.
The solution is therefore, divided into two parts and is given by
1 1
,2 2
x ct
x ctu x t f x ct f x ct g d
c
for ,x ct (4.3)
1 1
,2 2
x ct
ct xu x t f x ct f ct x g d
c
for .x ct (4.4)
The solution will exist if f and g also satisfies these additional
conditions, 0 0 0 0f f g , along with the conditions of
differentiability on them (i.e., f should be twice continuously
differentiable and g should be continuously differentiable).
Example 2: Solve the initial boundary-value problem
4 0, 0 , 0tt xxu u x t
4,0 ,u x x 0 ,x
,0 0,tu x 0 ,x
0, 0,u t 0.t
Solution: The given problem is the problem of Semi- Infinite String
with a Fixed End.
Here 4x is twice continuously differentiable, 0g x and
0 0 0 0f f g . Thus, all the conditions of the problem are
satisfied. Hence, solution is given by equation(4.3) and (4.4), i.e.,
For 2 ,x t 4 41
, 2 22
u x t x t x t
,
For 2 ,x t 4 41
, 2 22
u x t x t t x
.
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Also, 0, 0,u t is satisfied by ,u x t for 2 ,x t ( 0t ).
Value Addition: Activity 2
Solve the following initial boundary-value problems
4 0, 0 , 0, 0, 0, 0,tt xxu u x t u t t
( ,0) sin , ,0 0, 0 .tu x x u x x
4.2. Semi-Infinite String with a Free End:
Now, our purpose is to makes you familiar with the problem of Semi-
Infinite String with a Free End. The differential equation for the
problem is as follows:
2 0, 0 , 0tt xxu c u x t ,
,0 ,u x f x 0 ,x
,0tu x g x ,0 ,x
0, 0xu t ,0 ,t (4.5)
As explained above here, the solution for x ct is given by
equation(3.9). To compute solution for x ct (the portion which lies in
first quadrant 0t ), consider equation(3.4) from the DAlembert
solution
,u x t x ct x ct ,
Applying boundary conditions on it, we get
0, 0.xu t ct ct
After integrating we get
ct ct K ,
where K is a constant of integration
Thus,
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0
1 1
2 2 2
ct x Kx ct f ct x g d
c
.
The solution is given by
1 1
,2 2
x ct
x ctu x t f x ct f x ct g d
c
for ,x ct (4.6)
0 0
1 1,
2 2
x ct ct x
u x t f x ct f ct x g d g dc
for x ct .
(4.7)
The solution will exist if f and g also satisfies these additional
conditions, 0 0 0f g , along with the conditions of differentiability
on them.
Value Addition:
In a semi infinite string we know the initial point from where the
vibration in string starts and waves move only in the positive direction.
whereas in an infinite string we don't know the origin of the vibration
in the string i.e. it moves from to .
Example 3: Solve the initial boundary-value problem
9 0, 0 , 0tt xxu u x t
,0 0,u x 0 ,x
3,0 ,tu x x 0 ,x
0, 0,xu t 0.t
Solution: The given problem is the problem of Semi- Infinite String
with a Free End. Here 3x is continuously differentiable, 0f x and
0 0 0f g . Thus, all the conditions of the problem are satisfied.
Hence, solution is given by equation (4.6) and (4.7), that is
For 3 ,x t 3
43
3
33
1 1,
6 6 4
x tx t
x tx t
u x t d
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4 41 3 324
x t x t ,
For 3 ,x t 3 3
3 3
0 0
1,
6
x t t x
u x t d d
4 41 3 324
x t t x ,
3 31, 4 3 4 324
xu x t x t t x .
We notice here that 0, 0,xu t is satisfied for 3 .x t
Value Addition: Activity 3
Solve the initial boundary- value problem
16 0, 0 , 0tt xxu u x t
,0 cos ,2
xu x
0 ,x
3,0 ,tu x x 0 ,x
0, 0xu t 0 t .
5. Non- Homogeneous Boundary Conditions:
So far we dealt with problems of differential equations with
homogeneous boundary conditions. This section is to make you aware
with equations whose boundary conditions are non- homogeneous.
Consider a initial boundary-value problem with non homogeneous
boundary conditions. We illustrate this by considering two cases:
Case 1:
2 0, 0, 0,tt xxu c u x t
,0 , 0u x f x x ,
,0 , 0tu x g x x ,
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0, , 0u t p t t , (5.1)
Proceeding, as in the case of homogeneous wave equation for x ct .
For 0 x ct by d Alemberts solution for the Cauchy problem we have
,u x t x ct x ct (5.2)
Applying boundary conditions, we get
0,u t ct ct p t ,
ct p t ct .
Replacing ct by x ct , the preceding relation becomes
x
x ct p t ct xc
,
Thus, from equation (21), we get
,x
u x t p t x ct ct xc
1 1
2 2
x ct
ct x
xf x ct f ct x g d p t
c c
(5.3)
Here g must be cont. differentiable, f and p must be twice
continuously differentiable and 20 0 , 0 0 , 0 0 .f p p g p c f
Example 4: Solve the signal problem governed by the wave equation
2 , 0, 0,
,0 ,0 0, 0,
0, , 0.
tt xx
t
u c u x t
u x u x x
u t U t t
(5.4)
Solution: Characteristic curves are given by .dx
cdt
After integration we get
, .x ct x ct
Therefore, general solution of equation (5.4), is
,u x t x ct x ct (5.5)
Applying boundary conditions, for x ct , we have
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,0 0u x x x ,
,0 0tu x c x c x ,
Solving them we get solution for equation (5.5), that is
, 0u x t .
For x ct , we have
0,u t ct ct U t ,
Replacing ct by x-ct , we get
x
x ct U t ct xc
,
Since 0x , we get
x
x ct U tc
.
Therefore,
, 0 for
for
u x t x ct
xU t ct x x ct
c
(5.6)
,x x
u x t U t H tc c
where H is the Heaviside unit step function.
Case 2:
2 0, 0, 0,tt xxu c u x t
,0 , 0u x f x x ,
,0 , 0tu x g x x ,
0, , 0.xu t q t t (5.7)
Applying boundary conditions on equation (5.2), we get
0, .xu t ct ct q t
After integration, we get
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0
.t
ct ct c q d K (5.8)
where K is the constant of integration.
Replacing ct by x-ct in equation (5.8), we get
0
.x
tcx ct ct x c q d K
The solution of the initial boundary- value problem for x ct , is given
by
0 0
0
1 1,
2 2
x ct ct x
xtc
u x t f x ct f ct x g d g dc
c q d
(5.9)
Here g must be continuously differentiable, and f must be twice
continuously differentiable and
0 0 , 0 0 .f q g q
Value Additions:
After reading this we are able to differentiate between homogeneous
and non- homogeneous Boundary conditions of wave equations.
Homogeneous boundary conditions of wave equations are those in
which right hand side is zero (is not a function of time t) i.e. 0, 0u t
etc. whereas non- homogeneous boundary conditions of wave
equations are those in which right hand side is not zero (is a function
of time t) i.e. 0,u t p t etc.
Value Addition: Activity 4
Solve the initial boundary-value problem
2 , 0 , 0,
,0 , 0 ,
,0 , 0 ,
0, , , , 0.
tt xx
t
u c u x l t
u x f x x l
u x g x x l
u t p t u l t q t t
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6. Vibration of Finite Strings with Fixed Ends:
We will now study problem of vibration of string when the length of
string is finite and the string is fixed between two ends. This problem
requires more concern and labour then that of previous problem. It is
due to the repeated reflection of waves from the boundaries which is
not present in the case of infinite string problem. Let the length of
string be l which is fixed at both ends. we require to determine the
solution of
2 , 0 , 0,
,0 , 0 ,
,0 , 0 ,
0, 0, , 0, 0,
tt xx
t
u c u x l t
u x f x x l
u x g x x l
u t u l t t
(6.1)
We already know that the solution of the wave equation is
,u x t x ct x ct .
Now, we apply the initial conditions and get
,0 ,u x f x x x 0 ,x l
,0tu x g x c x c x , 0 .x l
Solving for and , we obtain
0
1 1,
2 2 2
Kf g d
c
0 ,x ct l (6.2)
0
1 1,
2 2 2
Kf g d
c
0 x ct l . (6.3)
Hence,
1 1
, ,2 2
x ct
x ctu x t f x ct f ct x g d
c
(6.4)
The solution is determined uniquely by the initial data in the following
region
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0 , , 0.x l x
x ct t x ct l t tc c
When t is large, the solution depends on the boundary conditions.
Thus, we obtain
0, 0u t ct ct , 0t , (6.5)
, 0u l t l ct l ct , 0t . (6.6)
If we take equation (6.5), we get
, 0 0t ct . (6.7)
Replacing by and then using equation (6.2), we get
0
1 1,
2 2 2
Kf g d
c
0 , 0.l l (6.8)
Thus, we see that the range of is extended to .l l
If we take equation (6.6), (where l ct ) we get
2l , l . (6.9)
Replacing by and then using equation (6.3), we get
2
0
1 12 2 ,
2 2 2
l Kl f l g d
c
0 2 2 .l l l l (6.10)
Thus, the range of is extended to 0 2l . Thus, we can obtain
for every 0 and for every .l
Value Addition:
Here the string under consideration is of finite length and there is
repeated reflection of waves from the boundaries.
Example 5: Find the solution of the initial boundary value problem
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, 0 2, 0,
,0 sin , 0 2,2
,0 0, 0 2,
0, 0, 2, 0, 0.
tt xx
t
u u x t
xu x x
u x x
u t u t t
Solution: Here, the problem is vibration of finite string with fixed
ends
1 1
sin , 0 2,2 2 2 2 2
K Kf
1 1
sin , 0 2,2 2 2 2 2
K Kf
Using boundary conditions, we get
1
sin , 0 2,2 2 2
K
1
sin2 2 2
K
, 2 0,
41
2 sin ,2 2 2
Kl
0 4 2,
1
sin2 2 2
K
, 2 4.
Proceeding in this manner, we determine the solution
1
, sin sin2 2 2
x ct x ctu x t
for all x in 0,2 and for all 0t .
Example 6: Find the solution of the initial boundary value problem
4 , 0 1, 0,
,0 0, 0 1,
,0 1 , 0 1,
0, 0, 1, 0, 0.
tt xx
t
u u x t
u x x
u x x x x
u t u t t
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Solution: Here the problem is again of vibration of finite string with
fixed ends. We shall proceed as follows:
2 3
0 0
1 11 , 0 1,
4 2 4 2 3 2
K Kd
2 3
0
1, 0 1,
4 2 3 2
K
Using boundary conditions, we get
2 3
0
1, 0 1 1 0,
4 2 3 2
K
2
2 3
0
12 ,
4 2 3 2
Kl
0 2 1, 1 2.
Proceeding in this manner, we determine the solution
,u x t
The range of is thus extended to 0 2l . Proceeding in the same
way, we can obtain for all 0 and for all .l Thus, the
solution is determined for all 0 x l and 0t .
Value Addition: Activity 5
Solve the initial boundary value problem
2 , 0 , 0,
,0 cos , 0 1,
,0 0, 0 1,
0, 0, 1, 0, 0.
tt xx
t
u c u x l t
xu x x
l
u x x
u t u t t
Solution: ,u x t for all x in 0, l and for all 0t .
7. Non- homogeneous Wave Equations:
The motive of this section is to make you familiar with the non-
homogeneous wave equation. The partial differential equations were
non- homogeneous either because of the presence of a driving term or
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because of a non- homogeneous boundary condition. The generalized
non- homogeneous wave partial differential equation in one dimension
has the form
2 * , ,tt xxu c u h x t (7.1)
Let us derive a mathematical formula, for a particular case when 1c ,
and then generalize it for any value of c.
Consider the Cauchy Problem for the non- homogeneous wave
equation
, ,xx yyu u h x y (7.2)
with the initial conditions
,0u x f x , (7.3)
,0 ,yu x g x (7.4)
Figure 3
Let 0 0 0,P x y be a point of the plane. Since characteristics
constantx y , are two straight lines drawn through the point 0P with
slopes 1 . They intersect x- axis at 1P and 2P with coordinates
0 0 ,0x y and 0 0 ,0x y respectively. Thus, 0 1 2P PP form a triangle
with sides 0 1 2A A A as shown in figure 3, where the whole closed interior
region of the triangle be denoted by D and boundaries by A .
x
y
0 0 0,P x y
O
0 0x y x y
0 0x y x y
1 0 0 ,0P x y 2 0 0 ,0P x y
1A
0A
2A
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Applying Greens theorem on L.H.S. of equation (7.2), we get
, ,xx yy x yR R
A
u u dR u dy u dx h x y dR (7.5)
Now, along 0 0,A dy x varies from 0 0 0 0tox t x t , x y constant along
1A , therefore, dx dy and along 2A , x y constant, therefore dx dy .
Therefore,
0 1 2
x y y x y x yA A A
A
u dy u dx u dx u dx u dy u dx u dy
0 0 0 0 0 0
0 0 0 0 0 0
, ,0
,0 ,
x y x y x y
yx y x y x y
u dx du du
0 0
0 00 0 0 0 0 0,0 2 , ,0
x y
yx y
u dx u x y u x y u x y
(7.6)
From equation (7.6) and equation (7.5), we get
0 0
0 00 0 0 0 0 0
1 1, ,0 ,0 ,
2 2
x y
yx y R
u x y u dx u x y u x y h x y dR
. (7.7)
The point 0 0,x y which we choose is an arbitrary point it can assume
any value. So, replacing it with any point ,x y , and using initial
conditions in equation (7.7), we get
1 1 1
, ,2 2 2
x y
x y Ru x y g d f x y f x y h x y dR
. (7.8)
Equation (7.8), we obtain when we consider 1c . In general 1c .
Thus, solution of equation (7.1) can be written as ((using
transformation y ct ) here 2*, hh x y c and *gg x c
21 1 1
, * * ,2 2 2
x ct
x ct Ru x t g d f x ct f x ct h x t dR
c
(7.9)
Example 7: Determine the solution of the initial value problem
2 , ,0 , ,0 sin .tt xx tu c u x ct u x x u x x
Solution: The given problem is the problem of non- homogeneous
wave equation. Its solution can be find by substituting
, * and * ,f x g x h x t in equation (7.9), that is,
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Institute of Lifelong Learning, University of Delhi pg. 23
2 01 1 1
, sin2 2 2
x ct ct y x ct
x ct y x ctu x t d x ct x ct x ct dxdy
c c
2
2 0
1 1cos
2 2 2
y x ctctx ct
x ct
y x ct
xx ctx dy
c c
2 0sin sin 1
2 22 2
ctx ctx x ct y y ct dy
c c
2
2
2
0
1 1sin sin 2
2 2 2
ct
yx x ct x cty y cty
c
2 21 1 1
sin sin2 2 2
x x ct xt ct .
Value Addition: Activity 6
Solve the following equations
(i) 2 , ,0 0, ,0 3.tt xx tu c u x u x u x
(ii) 1, ,0 sin , ,0 .xx yy yu u u x x u x x
Solution: (i) 21
, 3 .2
u x t t xt
(ii) 21 1
, sin sin .2 2
u x y x y x y xy y
8. Riemann Method:
This section is to make you aware with the method of solving the
linear hyperbolic equation. Now, we shall discuss the Riemann
method of solving second order partial differential equation in two
independent variables 2
, , , ,x yu
f x y u u ux y
. It yields an exact solution
in terms of the adjoint equation to the original equation. The basic idea
is that the solution of a non- characteristic initial value problem in two
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 24
dimensions can be found if the adjoint equation with the specific
boundary conditions can be solved.
It is assumed that the function , , , ,x yf x y u u u is continuous at al points
of a region D* defined by ,x y for all values of
, , , ,x y u p q concerned. The method due to Riemann, represents the
solution in a manner depending explicitly on the prescribed boundary
conditions. Although this often involves the solution of another
boundary value problem for the Greens function.
We shall assume that the equation has already been reduced to
canonical form
,L u f x y (8.1)
L is called the linear operator,
2
a b cx y x y
(8.2)
where ,a x y , ,b x y , ,c x y and ,f x y are differentiable functions in
some region D*.
Now, let ,v x y be a function (to be determined suitably) having
continuous second order partial derivatives. Then, compute
vL u uM v , where M is the operator defined by the relation (known
as adjoint operator to the operator L ).
2 av bvv
M v cvx y x y
, (8.3)
and
xy xyu v
vu uv v uy x x y
,
x x xvau u av vau , y y yvbu u bv vbu ,
so that
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Institute of Lifelong Learning, University of Delhi pg. 25
x yvL u uM v U U (8.4)
and
yU auv uv , xV buv u v (8.5)
Now, if M = L, we say the operator L is called self- adjoint. Applying,
Greens theorem to equation (8.4), we have
x yD D C
vL u uM v dxdy U U dxdy Udy Vdx , (8.6)
where is the curve RPQP which bounds the region of integration D
which is in D*.
Let be the continuous curve(as shown in figure 4) which is smooth.
Suppose now, that the values of u and xu or yu are defined along the
curve in the xy plane, and we want to obtain the solution of the
equation (8.1) at the point ,P agreeing with these boundary
conditions. Through P draw lines parallel to x- axis and y-axis as
shown in figure 4, cutting curve in Q and R respectively. Take the
curve to be the closed circuit PQRP . By construction 0dx on PR and
0dy on PQ. Thus,
Q R P
P Q RC
Udy Vdx Vdx Udy Vdx Udy (8.7)
Substituting V from equation(8.5) and integrating by parts, we get
Q Q QQ
xPP P PVdx buvdx uv uv dx . (8.8)
Substituting the values from equation (8.7) and (8.8), in equation
(8.6), we get
Q QQ
xPP PD
vL u uM v dxdy buvdx uv uv dx
R P
Q RUdy Vdx Udy (8.9)
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Institute of Lifelong Learning, University of Delhi pg. 26
So far the function v has been arbitrary. Let us choose the function
, ; ,v x y such that it is the solution of the adjoint equation 0M v .
Also the function v should satisfy the following conditions:
1. ,xv b x y v when y (along PQ)
2. ,yv a x y v when x (along PR)
3. 1v when x , y at P.
Such a function if it exists is called Greens function for the problem.
Using, these conditions in equation (8.9), we get
R R
y xP Q Q QD
u uv uv ady bdx uv dy vu dx vf dxdy . (8.10)
Equation (8.10) gives u at the point P when u and xu are given along
the curve . When yu is given along the curve , we use
d uv uv dx uv dyx y
,
Integrating this expression from Q to R ( along the curve ), we get
R
x yR Q Quv uv uv dx uv dy
x
y
,P
R
Q
O
Figure 4: Smooth initial curve
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 27
R R
x y x yQ Qu vdx uv dy uv dx u vdy (8.11)
Substituting Q
uv from equation (8.11) in equation (8.10), we get
R R
x yP R Q QD
u uv uv ady bdx uv dx vu dy vf dxdy (8.12)
Finally, by adding equation (8.12) and (8.10), we obtain the
symmetrical result
1 12 2
R R
x yP R Q Q Qu uv uv uv ady bdx u v dx v dy
1
2
R
x yQ
D
v u dx u dy vf dxdy (8.13)
Thus, the solution of the given equation at any point in terms of the
values of , xu u and yu along a given curve , can be obtained by using
equation (8.10), (8.11) and (8.12), whichever is appropriate. Further,
we observe that the solution at any point , depends on the values
which is popularly known as Cauchy data. Thus, a slight change in the
initial data outside the curve would change the solution only outside
this region.
Value Addition:
Green's Theorem: Let C be a piecewise smooth simple closed curve
in the xy-plane and let D denote the closed region enclosed by C.
Suppose M,N,N
x
and
M
y
are real valued continuous functions in an
open set containing D. Then
D C
N Mdxdy Mdx Ndy
x y
where the line integral is taken around C in the counter clockwise direction.
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 28
Example 8: Prove that for the equation 1 04xy
u u Greens function
is 0, ; ,v x y J x y where 0J denotes Bessels function of
the first kind of order zero.
Solution: Here 10, 0, , & , 0.4a b c f x y
14xyL u u u & 1
4xyM v v v .
The Greens function , ; ,v x y satisfies the following conditions:
1 04xyM v v v
0xv when y
0yv when x
1v when , .x y
To find v , we put 2s x y (8.14)
Thus v v s is a function of one variable
Now 0s at , .x y
Taking logarithm of equation (8.14) and differentiating partially with
respect to x and y, we get
, .2 2
s y s x
x s y s
So,
2
2
1
4xy
d v dvv
ds sds
.
Thus, we obtain 2
2
10
4xy
d v dvv v v
ds sds
Or
2
2 2 2
20 0
d v dvs s s vds ds
(8.15)
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Institute of Lifelong Learning, University of Delhi pg. 29
Above equation is Bessels equation of order zero, and so 0v J s is a
solution of equation (8.15)
Hence, 0v J x y .
Also, it satisfies all the properties of the Greens function as
0 0 1 1J v when , ,x y
0 01
2x x
yv J s s J s
x
0xv when y and
0 01
2y y
xv J s s J s
y
0yv when .x
Value Addition: Activity 7
Solve the following problem using Riemann method:
2 2
1 2
0,
, , , .
xx tt
t
x u t u
u x t f x u x t g x
Solution: 3 3
2 2
1 1 1 1,
2 2 4 2
x x
t t
xt xt
f gxu x t f xt tf xt d xt d
t
.
9. Goursat Problem:
This section is to make you aware with the Goursat problem. In
Goursat problem we finds a solution of the hyperbolic partial
differential equation
, , , ,xy x yu F x y u u u (9.1)
which satisfies the following conditions
, ,u x y f x (9.2)
on a curve, say, 0y , (which is characteristic of the given equation)
and
,u x y g x ,
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 30
on a monotonic increasing curve y y x . Further, we assume that
curve intersects the characteristic at the origin.
Here, we are illustrating few examples to make you aware of the
procedure of solving Goursats problem.
Value Additions:
A partial differential equation of second order i.e.,
xx xy yy x yAu Bu Cu Du Eu Fu G
where A, B, C, D, E, F and G are constants, is said to be
(i) Parabolic if 2 4 0B AC
(ii) Hyperbolic if 2 4 0B AC
(iii) Elliptic if 2 4 0B AC .
Example 9: Solve the Goursat problem
3 3 3 3 0xx yy x yxy u x yu y u x u , (9.3)
,u x y f x on 2 2 16y x for 0 4x ,
,u x y g x on 0x for 0 4y ,
where 0 4f g .
Solution: Here the given equation is not in canonical form. So, we
reduce it into canonical form first. In canonical form equation (9.3) can
be written as 0u ,
where,
2 2y x , 2 2y x .
Thus, the general solution is
2 2 2 2,u x y y x y x
Applying the prescribed conditions , we have
2, 16 2 16u x y f x x on 2 2 16y x (9.4)
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 31
2 2,u x y g x y y on 0x (9.5)
We observe that these equations are compatible, since
0 4 16 16f g (9.6)
Now, replacing x by 2 21
162
y x in equation (64) and y by 2 2y x
in equation (9.5), we have
2 2 2 21
16 162
f y x y x
, (9.7)
2 2 2 2 2 2g y x x y x y , (9.8)
but f x on 2 2 16y x .
Therefore, from equation (9.8),
2 2 16 16x y g
0 16f (using (9.6)) (9.9)
Thus,
2 2 2 21, 16 16 0 162
u x y f y x g y x f
2 2 2 21 16 02
f y x g y x f
. (9.10)
Example 10: Solve 2tt xxu c u ,
,u x t f x on t t x ,
,u x t g x on 0x ct ,
where 0 0g f .
Solution: The general solution of the problem, using initial conditions,
is
,u x t x ct x ct , (9.11)
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 32
,u x t f x x ct x x ct x , (9.12)
0 2g x x , (9.13)
Let, s x ct x , the inverse of it is x s . Thus, equation (9.12) may
be written as
f s s s ct s , (9.14)
Replacing 2x by x ct
2
s ct sg s s ct s
. (9.15)
From equation (9.14) and (9.15), we get
02
s ct sf s s g
,
or,
02
x ct ct x ctx ct f x ct g
. (9.16)
Hence, the solution is given by (using equation (9.15) and(9.16) in
(9.11))
, 0 02 2
x ct ct x ctx ctu x t g f x ct g
.
Value Addition: Activity 8
Solve the following problems
3 3 3 3 0,xx yy x yxy u x yu y u x u
2 2, on 8 for 0 2,u x y f x y x x
2 2, on 16 for 2 4,u x y g x y x x with 2 2 .f g
Solution: 2 2 2 28 16
, 22 2
y x y xu x t f g f
.
10. Spherical Wave Equation:
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Institute of Lifelong Learning, University of Delhi pg. 33
This section is to make you aware with spherical waves. We are
rewriting wave equation into spherical coordinates in this section.
wave equation is one of the kind of partial differential equations of
second order. In the wave equation the dependent variable
, , ,u u x y z t is a function of three space variable; x, y, z and time
variable t, is
2 0tt xx yy zzu c u u u .
In spherical polar coordinates , ,r , the above equation takes the
form
2 2
2 2 2 2 2
2 1 1 1sin
sin sinr rr
u u uu u
r r r c t
(10.1)
Solution of equation (10.1) is known as spherical symmetric waves if u
depends on r and t. Thus, we have
2
2 2
2 1r rr
uu u
r c t
(10.2)
Introducing a new dependent variable ,U ru r t , equation (10.2)
reduces to
2 .tt rrU c U (10.3)
Equation (10.3) denotes a wave in one dimension.
As obtained earlier, the solution of the equation(10.3) is of the form
1
,u r t r ct r ctr .
As we can see, there are two progressive spherical waves travelling
with constant velocity c in opposite directions (one approaching
towards origin and other going away from origin) .
Here, we are interested only in the solution for outgoing waves
1
,u r t r ctr , (10.4)
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 34
21 1
,ru r t r ct r ct r ctr r .
Here ru is called the radial velocity. In fluid flow, u denotes the
velocity potential so that the total limiting flux through a sphere of
radius r and center at the origin is
20
lim4 , 4rr
Q t r u r t ct
.
1
4
r rr ct c t Q t
c c
(10.5)
Again in fluid flow, if p denotes the pressure at any time t and 0p the
pressure at equilibrium value, the difference between them is given by
0
4t
rp p u Q t
r c
, (10.6)
here we call the density of the fluid.
Now, we shall obtain the solution of equation (10.3), where ,U ru r t
with the boundary conditions
,0u r f r , ,0tu r g r , 0r . (10.7)
As f and g satisfies equation (10.3) they should be continuously
differentiable as given by the dAlembert solution of the Cauchy
problem for the one dimensional wave equation, that is
1 1
2 2
r ct
r ctru U r ct f r ct r ct f r ct g d
c
(10.8)
provided 0.r ct
As for 0, 0t r , which determines the solution ,u r t only up-to the
characteristic r ct in the r- t plane.
However, when r is less than or equal to ct , this solution does not
holds, because in that case f and g are not defined for negative values
of r . If u is finite at 0r (where 0U ) for all 0t , the solution for
r ct exists and can be obtained as follows:
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Institute of Lifelong Learning, University of Delhi pg. 35
For 0ct r
1
,2
U r t ct r ct r (10.9)
which for 0r gives
0ct ct for 0ct (10.10)
Now, (using equation (10.9) and (10.10)), we get
1
r tU U ct rc
1
r ct f r ct f r ct r ct g r ctc
(10.11)
In view of the fact 1
r tU Uc
is constant on each characteristic ct r =
constant , it turns out from equation (10.11) for 0r ,
ct ct f ct f ct t g ct .
After integrating we get
0
10
t
t t f t g dc
,
so from equation (10.10), we get
0
10
t
t t t f t g dc
.
Substituting these values in equation (10.9), we get
1 1
,2
ct r
ct ru r t ct r f ct r ct r f ct r g d
r c
(10.12)
Value Addition:
The relation between rectangular coordinated and spherical polar
coordinate is given by
sin cos
sin sin
cos .
x r
y r
z r
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 36
Summary:
So far we have obtained the solution of wave equation which is given
by equation (3.9),
1 1
,2 2
x ct
x ctu x t f x ct f x ct g d
c
.
The importance of waves lies in the fact that in case of waves, the
signal breaks into two pieces and the waves propagates in opposite
directions at finite speed c (movement is along the characteristic, and
therefore, information is retained). Then we illustrate procedure to find
solution of Semi- infinite string with a fixed end and with a free end
and the solution is given by equations (4.3), (4.4) and (4.6), (4.7)
respectively. Then we discuss non- homogeneous boundary conditions
in differential equations and gave a procedure to simplify differential
equations with these conditions, which is given by equations (5.3) and
(5.9). After that we outline a procedure to solve problem of finite
strings with fixed ends and its solution is given by equation (6.10).
Then we take problem of non- homogeneous wave equation, whose
solution is given by equation (7.9). After that we discuss Riemann
method, which is used for solving Goursat problem and Cauchy
problem for linear hyperbolic partial differential equations given by
equation (8.13). In section 9 we gave procedure to solve Goursat
problem and in the end we gave solution of the spherical wave
equation which is given by equation no (10.8) and (10.12).
Now, we suppose you should be able to answer the following
questions:
True False:
1. Time reversible is not permissible in case of propagation of
waves.
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 37
2. Waves change their shape as they propagate.
3. The partial differential equations were non- homogeneous either
because of the presence of a driving term or because of a non-
homogeneous boundary condition.
4. Riemann method yields an exact solution of the linear hyperbolic
equation in terms of the adjoint equation to the original
equation.
5. Three dimensional wave equation is given by
2 0tt xx yy zzu c u u u where t is time variable and x, y, z are
space variables.
6. The adjoint operator M of operator L is called self-adjoint if M=L.
7. The problem of the finite string is easier than that of the infinite
string.
8. Non- homogeneous one- dimensional wave equation is given by
2 0tt xxu c u .
9. Physically, x ct in D Alembert solution of the Cauchy
problem represents a progressive wave travelling in the positive
x- direction with speed different from c.
10.Bessels Equation of order zero is given by 2
2 2
20
d y dyx x x y
dx dx .
Solutions
1. False ;Time reversible is permissible as it reverses the direction
of wave propagation.
2. False ; Waves do not change their shapes while propagation.
3. True.
4. True.
5. False; Three dimensional wave equation is 2 0tt xx yy zzu c u u u .
6. True.
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 38
7. False; The problem is more complicated due to the repeated
reflection of waves from the boundaries.
8. False; Non- homogeneous wave equation is 2 ,tt xxu c u h x t .
9. False; It represents a progressive wave travelling in the negative
x- direction travelling with speed c without change of shape.
10.True.
Descriptive Questions:
1. Give an expression for the D Alembert solution of the Cauchy
problem.
2. Differentiate between homogeneous and non- homogeneous wave
equation.
Find the solution of each of the following initial value problems
3. 2 20, ,0 sin , ,0 .tt xx tu c u u x x u x x
4. 2 10, ,0 cos , ,0 .tt xx tu c u u x x u x e
5. 2 2, ,0 5, ,0 .xtt xx tu c u e u x u x x
6. 2 sin , ,0 cos , ,0 1 .tt xx tu c u x u x x u x x
7. 2 , ,0 sin , ,0 0.ttt xx tu c u xe u x x u x
8. 2 22, ,0 , ,0 cos .tt xx tu c u u x x u x x
9. 10 9 0, ,0 , ,0 .xx xy yy yu u u u x f x u x g x
10. 2 2 0 0, 0, ,1 , ,1 .xx tt tx u t u x t u x f x u x g x
Determine the solution of the following initial boundary value problem
11. 16 0, 0 , 0, 0, 0, 0,tt xxu u x t u t t
2( ,0) sin , ,0 , 0 .tu x x u x x x
12. 0, 0 , 0, 0, 0, 0,tt xx xu u x t u t t
( ,0) cos , ,0 0, 0 .2
t
xu x u x x
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Institute of Lifelong Learning, University of Delhi pg. 39
13. 4 0, 0 1, 0,tt xxu u x t
2( ,0) 0, ,0 1 , 0 1,tu x u x x x x
0, 0, 1, 0, 0.u t u t t
14. Solve the characteristic initial value problem
3 0, 0,xx yy xxu x u u x
2
, on 0 for 0 y 2,2
xu x y f y y
2
, on 4 for 2 y 4,2
xu x y g y y where 2 2 .f g
Solutions:
3. 2 2 31sin cos ,3
x ct x t c t 4. cos cos tx cte
,
5. 2 2 3 21 15 23 2x ct x ct xx t c t e e e
c
6. 21cos cos 2 sin sin cosx ct t x x ctc
7. 2sin cos 2txt ex ct 8. 2 2 2 11 cos sinx t c x ctc
9. 99 9 1, 8 9 8 8y
x
x y
yu x t f x g d f x y
10. 1 1, *2 2xt
x
t
xu x t f xt f g dt
where *g x
g xx
.
11. 3 31, sin cos4 4 4 for 424u x t x t x t x t x t
3 31sin cos4 4 4 for 424x t x t t x x t
12. , cos cos for and .2 2x tu x t x t x t
14. 2 2
, 2 4 02 4 2 4
y x y xu x t f g
Glossary:
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 40
G Goursat Problem: In Goursat problem we finds a solution of the
hyperbolic partial differential equation , , , ,xy x yu F x y u u u which
satisfies the following conditions , ,u x y f x on a curve, say,
0y , (which is characteristic of the given equation) and
,u x y g x , on a monotonic increasing curve y y x .
G Green's Theorem: Let C be a piecewise smooth simple closed
curve in the xy-plane and let D denote the closed region
enclosed by C. Suppose M,N,N
x
and
M
y
are real valued
continuous functions in an open set containing D. Then
D C
N Mdxdy Mdx Ndy
x y
where the line integral is taken around C in the counterclockwise
direction.
H Homogeneous wave equation: A wave equation is said to be
homogeneous if the right hand side of the boundary condition
associated with the wave equation is zero (i.e., it is not a
function of time t) i.e. 0, 0u t .
N Non-Homogeneous wave equation: A wave equation is said to
be non-homogeneous if the right hand side of the boundary
condition associated with the wave equation is not zero (i.e., it
is a function of time t) i.e. 0,u t p t etc.
R Riemann Method: Riemann method of solving second order
partial differential equation in two independent variables
2
, , , ,x yu
f x y u u ux y
yields an exact solution in terms of the
adjoint equation to the original equation.
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Wave Equations
Institute of Lifelong Learning, University of Delhi pg. 41
References/ Further Reading:
1. Tyn Myint-U and Lokenath Debnath, Linear Partial Differential
Equation for Scientists and Engineers, Springer, Indian reprint,
2006.
2. Ian N. Sneddon, Elements of Partial Differential Equations,
McGraw-Hill, 1957.
3. Bateman H., Partial Differential Equations of Mathematical
Physics, Cambridge University Press, Cambridge (1959).
4. Debnath L., Non-Linear Partial Differential Equation for
Scientists and Engineers, (Second Edition), Birkhauser Verlag,
Boston (2005).
5. Epstein B., Partial Differential Equations, McGraw- Hill, New
York (1962).
6. Hadamard J., Lectures on Cauchys Problem in Linear Partial
Differential Equations, Dover Publications, New York (1952).