1. basic course information final exam wed. may 9, 10-11:50halsmith/476sp2007.pdf0 ¡c b c 0 ¡a ¡b...
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1. Basic Course Information
MAT476 Partial Differential Equations Spring 2007Text: Partial Differential Equations of Mathematical Physicsand Integral Equations, Guenther and LeeInstructor: Hal Smith URL: http://math.la.asu.edu/∼halsmithOffice: PSA 631, Phone: 965-3743, messages: 965-3951, e-mail:[email protected] Hours: 12:40 Mon. & Wed. or by appt.
Final Exam Wed. May 9, 10-11:50 See Dept. web page for oldexams!
The course will cover the standard solution methods for Laplace’sequation, the heat equation, the wave equation, and their close rela-tives. These include separation of variables, Fourier series and Fouriertransform methods, and the method of characteristics. Green’s func-tions, energy methods, and maximum principles will be emphasized.
Homework will be assigned and collected regularly. Midterm andFinal exams will be given. The course grade will be determined by:1/3 for each exam and 1/3 homework.
You may find materials on Prof. Don Jones web page useful:http://math.la.asu.edu/∼dajones/dajones.html
Note previous final exams for this course posted on:http://math.asu.edu/grad/diffeqs.html
Other excellent references:
1. Partial Differential Equations, F. John
2. Mathematics Applied to Deterministic Problems in the Natural Sci-ences, Lin and Segel.
2. Homework assignments
Assignment #1 (Due Friday, Jan 26): Text 1.3: 1,2,4; 1.4: 1;1.8: 3;exercise 0.3 and exercise 0.4 next pages.
Assignment #2 (Due Friday, Feb 9) Text 2.1: 5; Exercise 0.6 andexercise 0.5 (see text below).Also solve each of the following:
−xux + yuy = xyu, u(x, 0) = x
(y + u)ux + yuy = x− y, u(x, 1) = 1 + x
1
2
ux + uy + zuz = u3, u(x, y, 1) = h(x, y)
Assignment #3 (Due Friday, Feb 16) Text 3.1: 1,3,4,7-10,16; 3.2:1,2,9
Assignment #4 (Due Friday, Feb 23) Text 3.3: 1,3,4Assignment #5 (Due Friday, Mar 2) Text 4.1: 6,7; 4.2: 1,3,9,10Also, give a careful argument supporting the displayed estimate
on middle pg 91 of text. You will need to use the triangle inequality
and the inequality | ∫ b
ag(x)dx| ≤ ∫ b
a|g(x)|dx.
Assignment #6 (Due Friday, Mar 9) Text 4.3: 2,6,7Assignment #7 (Due Friday, Mar 23) Text 5.1: 1,3; 5.2 1-5.Assignment #8 (Due Friday, Mar 30) exercise 0.7, Text 5.3: 2, 5.6:
1,2 (read 5.6!)Assignment# 9 (Due Friday, Apr 6) Text 5.1: 5; 8.1 1,10; 8.2 10.Assignment# 10 (Due Friday, Apr 13) Text 8.3: 1,4,11. Also show
that the unique solution u of u′′ = h(x), 0 ≤ x ≤ 1 satisfying u(0) =
0 = u(1) can be expressed as u(x) =∫ 1
0G(x, y)h(y)dy for a Green’s
function G. Give an explicit formula for G!Assignment# 11 (Due Friday, Apr. 20) Text 8.3 verify displayed
equation for normal derivative of G above (3-22) page 313; 8.4: 3,4,11.Do exercise 0.2.
Exercise 0.1. Let f : D → R be a real-valued function on an opensubset D ⊂ R2. Suppose that f has continuous partial derivatives inD (or more generally, suppose that f is differentiable in D) and
∂f
∂x(x, y) = 0 =
∂f
∂y(x, y), (x, y) ∈ D
Prove that f is constant in D: ∃c ∈ R such f(x, y) = c, (x, y) ∈ D?Check your advanced calculus text.
Exercise 0.2. Let f : (a, b) → R be a real-valued function suchthat f ′(x) exists for x ∈ (a, b), f ′(x0) = 0 for some x0, and f ′′(x0) > 0.Use the definition of f ′′(x0) as a limit to show that there exists δ > 0such that f ′(x) > 0 for x0 < x < x0 + δ. Then use the Mean ValueTheorem to show that f(x) > f(x0) for x0 < x < x0 + δ. Use thisresult to argue that if a differentiable function f has a local maximumat x0 and if f ′′(x0) exists, then f ′′(x0) ≤ 0. Use the previous resultto argue that if u is a differentiable function on an open set D ⊂ R3
which attains a local maximum at x0 and if ∂2u∂x2
j(x0), j = 1, 2, 3 exist,
then 4u(x0) ≤ 0.
Assignment# 12 (Due Friday, Apr. 27) Text 9-2:7; 10-2: 8;10-3:1; Also show that eigenvalues of the Laplacian are nonnegative for
3. ADDITIONAL NOTES 3
the Neumann boundary condition (∂u∂ν
= 0) and that eigenfunctionscorresponding to different eigenvalues are orthogonal as we did in classfor the Dirichlet case.
3. Additional Notes
3.1. Vector calculus. In vector calculus we learn the identities
∇×∇φ = 0(1)
∇ · (∇× E) = 0
where φ : R3 → R is a C2 scalar function and E : R3 → R3 is a C2
vector-valued function E = (E1, E2, E3). Recall
∇φ = (∂φ
∂x1
,∂φ
∂x2
,∂φ
∂x3
)
is called the gradient of φ, ∇·E is called the divergence of E and ∇×Eis the curl of E. For divergence and curl,
∇ = (∂
∂x1
,∂
∂x2
,∂
∂x3
)
is a differential operator that acts just like the corresponding vectorproduct:
∇ · E =3∑
i=1
∂Ei
∂xi
(2)
∇× E = (∂E3
∂x2
− ∂E2
∂x3
,∂E1
∂x3
− ∂E3
∂x1
,∂E2
∂x1
− ∂E1
∂x2
)(3)
Exercise 0.3. If E : R3 → R3 is C2, verify that
∇× (∇× E) = ∇(∇ · E)−∇2E
where ∇2 =∑3
i=1∂2
∂x2i, called the Laplacian, operates componentwise.
Recall that the Jacobian of E is the matrix
J =
∂E1
∂x1
∂E1
∂x2
∂E1
∂x3∂E2
∂x1
∂E2
∂x2
∂E2
∂x3∂E3
∂x1
∂E3
∂x2
∂E3
∂x3
From this we see that the divergence of E, ∇ ·E is just the trace of J(sum of diagonal elements of J). Any matrix may be written as a sumof a symmetric matrix and an antisymmetric matrix:
J =1
2(J + JT ) +
1
2(J − JT )
4
Observe that the first summand is symmetric and the second is antisymmetric-its transpose equals its negative. If∇×E = (a, b, c) are the componentsof the curl of E, we see that the non-zero components of the antisym-metric part of J are:
1
2
0 −c bc 0 −a−b a 0
3.1.1. Maxwell’s Equations. Maxwell’s equations for the electric fieldvector E and magnetic field vector B in free space where charged par-ticles and currents are absent are:
∇ · E = 0
∇× E = −∂B
∂t(4)
∇ ·B = 0
c2∇×B =∂E
∂t
where c is the speed of light.Taking the curl of both sides of the second equation, applying the
identity of exercise 0.3, and using the first then the fourth of Maxwell’sequations, we get
−∇2E = −∇× (∂B
∂t)
= − ∂
∂t(∇×B)
= −c−2∂2E
∂t2
Therefore, E = E(x1, x2, x3, t) satisfies the wave equation:
(5)∂2E
∂t2= c2∇2E = c2
3∑i=1
∂2E
∂x2i
Since E = (E1, E2, E3), each component Ei satisfies the wave equation
(6)∂2u
∂t2= c2
3∑i=1
∂2u
∂x2i
A parallel computation shows that B also satisfies (5).
3. ADDITIONAL NOTES 5
3.1.2. Conservative vector fields. We also learn from vector calculusthat if ∇ × E = 0 for some C1 vector function, not necessarily theelectric field vector, then there is a scalar function φ such that
∇φ = E
It is worth noting that φ must satisfy the system of partial differentialequations:
(7)∂φ
∂x1
= E1,∂φ
∂x2
= E2,∂φ
∂x3
= E3
We solve these equations using Stokes Theorem∮
∂S
E · dr =
∫ ∫
S
(∇× E) · ndS
where S is a surface in R3, oriented by the unit normal vector n, withsmooth boundary ∂S oriented by the right-hand rule compatible withn (see your vector calculus text). Because ∇×E = 0, Stokes Theoremsays that the line integral of E around every closed curve is zero andtherefore E is conservative, meaning that the line integral
∫ p
q
E · dr
along any path from point q to point p is the same. It is easy to seethat if p = (x1, x2, x3) and q is a fixed point (e.g., q = (0, 0, 0)), then
φ(p) =
∫ p
q
E · dr
satisfies ∇φ = E and φ(q) = 0. If E were defined on a sub-domain Ωof R3, then we can draw the same conclusion provided that Ω is simplyconnected: every closed curve in Ω is homotopic in Ω to a single point.
Exercise 0.4. Show that φ, defined by the integral above, satisfiesthe system (7). Hint: check out your vector calculus or physics text.
Thus we see that for a C1 vector function E in R3, E = ∇φ forsome C2 scalar function φ if and only if ∇× E = 0 in R3.
In electrostatics, one assumes that the electric and magnetic fieldvectors do not depend on time. In that case, the second of (4) gives∇ × E = 0 so E is conservative. It follows that E is the gradient ofsome function φ, called the electrostatic potential:
E = ∇φ
6
By the first of equations (4), we see that φ satisfies Laplace’s equation
0 = ∇2φ =3∑
i=1
∂2φ
∂x2i
3.1.3. Divergence-free vector fields. What can we say about a C1
vector function E on R3 if we know that
∇ · E = 0.
From Maxwell’s equations we see that both the electric field and themagnetic field vector are divergence free.
By the second identity of (1), we know that the divergence of a curlis zero. Could it be true that since the divergence of E is zero, thenE must be a curl of some other vector function? You will prove thefollowing result
Proposition 1. Let E : R3 → R3 is a C1 vector-valued functionsatisfying
∇ · E = 0
in R3. Then there exists a C1 vector-valued function A : R3 → R3 suchthat
(8) E = ∇× A
A can be viewed as a vector potential function for E.Notice that A = (A1, A2, A3) must satisfy the system of partial
differential equations
∂A3
∂x2
− ∂A2
∂x3
= E1
∂A1
∂x3
− ∂A3
∂x1
= E2(9)
∂A2
∂x1
− ∂A1
∂x2
= E3
The following result shows that (8) does not have a unique solution.
Lemma 1. If (8) has a solution A(x1, x2, x3) and if φ is any C2
scalar function, then A +∇φ is also a solution.If (8) has two solutions A(x1, x2, x3) and B(x1, x2, x3), then there
is a scalar function φ such that
A−B = ∇φ
Proof. The first assertion follows from (1). The second assertionfollows immediately from the fact that ∇× (A−B) = 0 and our vectorcalculus review above. ¤
3. ADDITIONAL NOTES 7
Corollary 1. If (8) has a C2 solution A = (A1, A2, A3), then ithas a C1 solution B with B1 ≡ 0 in R3.
Proof. By Lemma, B = A − ∇φ is a solution for any C2 scalarfunction φ. We can choose a C2 scalar function φ such that
∂φ
∂x1
= A1(x1, x2, x3)
Clearly, then B1 = 0. ¤In view of the Corollary, we may as well try to solve (8) for A =
(0, A2, A3). We need only find two functions rather than three! Theequations for A2, A3 from (9) are:
∂A3
∂x2
− ∂A2
∂x3
= E1
∂A3
∂x1
= −E2(10)
∂A2
∂x1
= E3
Exercise 0.5. Solve the system of equations (10). Hint: first solvethe last two equations to find A2 and A3 and then show that the firstequation automatically holds-well, almost!
8
3.2. Flow of Ideal Gases. This brief section is devoted to flesh-ing out Chapter 1, section 7 of our text. It presumes a substantialknowledge of the theory of ordinary differential equations. I suggestnot to read it!
We follow the time evolution of a bounded volume of gas particleswith density ρ(x, t) as it moves under the velocity field v(x, t). Recallthat conservation of mass gives
(11) ρt +∇ · (ρv) = 0
If Vt0 denotes this volume at t = t0, let Vt be the correspondingvolume of particles at time t. In order to express Vt, we need theparticle trajectory x(t, t0, x0) which passes through x0 at time t0. Itsatisfies the ODE
x′ = v(x, t), x(t0) = x0
Then
Vt = x(t, t0, •)(Vt0) = x(t, t0, x0) : x0 ∈ Vt0It will be useful to express integrals such as
∫Vt
f(y)dy as integrals
over the original domain Vt0 by the change of variables y = g(x0) ≡x(t, t0, x0). Recall this formula is given by
∫
Vt
f(y)dy =
∫
g(Vt0 )
f(y)dy =
∫
Vt0
(f g)(x0)| det Jg(x0)|dx0
where Jg is the Jacobian matrix of g. Consequently, we need an ex-pression for Jg. This comes from basic ODE theory since
Jg = X(t) ≡ ∂x(t, t0, x0)
∂x0
satisfies the matrix differential equation
X ′ = A(t)X, X(t0) = I
where
A(t) =∂v(x(t, t0, x0), t)
∂x
By Abel’s Theorem
det Jg = det X(t) = det X(t0) exp
(∫ t
t0
∑i
aii(s)ds
)
= exp
(∫ t
t0
(∇x · v)(x(s, t0, x0), s)ds
)
3. ADDITIONAL NOTES 9
Suppose f(y, t) is a vector function. Then∫
Vt
f(y, t)dy =
∫
Vt0
f(x(t, t0, x0), t) exp
(∫ t
t0
(∇x · v)(x(s, t0, x0), s)ds
)dx0
Differentiating at t = t0 we get
d
dt|t=t0
∫
Vt
f(y, t)dy =
∫
Vt0
[(∇f) · v + ft + (∇x · v)f ]dx0
=
∫
Vt0
∇ · (fv) + ftdx0
=
∫
Vt0
ftdx0 +
∫
∂Vt0
fv · ndS
We do not need the last expression as the middle one is great.Now we can express the i-th component of the momentum of this
collection of particles as
Mi(t) =
∫
Vt
(ρvi)(y, t)dy
Differentiating with respect to t, evaluating at t = t0, and using(11) gives
Mi(t0) =
∫
Vt0
[∇ · (ρviv) + (ρvi)t]dx0
=
∫
Vt0
[∇ · (ρviv) + ρtvi + ρvit]dx0
=
∫
Vt0
[∇ · (ρviv)− (∇ · ρv)vi + ρvit]dx0
=
∫
Vt0
ρ[(v · ∇)vi + vit]dx0
This must equal the i-th component of the forces acting on Vt0 whichare given by
−∫
Vt0
ρdx0g −∫
∂Vt0
pndS = −∫
Vt0
ρdx0g −∫
Vt0
∇pdx0
See our text, p denotes pressure, g denotes gravitational constant vec-tor. Since Vt0 was arbitrary, we get the partial differential equation:
ρ[vt + (v · ∇)v] = −∇p− ρg
10
3.3. Equations of Demography. Let u(a, t) be the age-densityof a (human or animal) population. Here a ≥ 0 denotes age and t ≥ 0denotes time. By this we mean that, given an age range a1 < a < a2
and time t then∫ a2
a1
u(a, t)da = number of individuals with ages a1 < a < a2 at time t
The total size of the population at time t would then be∫ ∞
0
u(a, t)da
If you wanted to know the size of the cohort of individuals whichat time t0 have ages between a1 and a2 as they age, i.e., for t > t0, youwould write: ∫ a2+(t−t0)
a1+(t−t0)
u(a, t)da
Notice that when t = t0 in the above expression, we get the size ofthis cohort at time t0. Differentiating this expression, using Leibnizformula and the fundamental theorem of calculus, we get
d
dt
∫ a2+(t−t0)
a1+(t−t0)
u(a, t)da =
∫ a2+(t−t0)
a1+(t−t0)
ut(a, t)da
+u(a2 + (t− t0), t)− u(a1 + (t− t0), t)
=
∫ a2+(t−t0)
a1+(t−t0)
[ut(a, t) + ua(a, t)]da
The only change can be caused by death since no one is born into thiscohort. If
µ(a, t) = per capita death rate at time t of individuals of age a
then
µ(a, t)u(a, t)da = death rate at time t of individuals of age a
and the time derivative computed above must be
d
dt
∫ a2+(t−t0)
a1+(t−t0)
u(a, t)da = −∫ a2+(t−t0)
a1+(t−t0)
µ(a, t)u(a, t)da
Consequently, we find that∫ a2+(t−t0)
a1+(t−t0)
ut(a, t) + ua(a, t) + µ(a, t)u(a, t)da = 0
3. ADDITIONAL NOTES 11
Evaluating this at t = t0, we get∫ a2
a1
ut(a, t0) + ua(a, t0) + µ(a, t0)u(a, t0)da = 0
and since a1, a2 and t0 were arbitrary, u must satisfy
ut(a, t) + ua(a, t) + µ(a, t)u(a, t) = 0, 0 < a < ∞, t > 0
It seems reasonable that we should be given the age-density of thepopulation at time t = 0:
u(a, 0) = u0(a), a > 0
In addition, we must account for births of individuals of age a = 0.Suppose that we are given the number of births, B(t), at time t. Thenwe must have
u(0, t) = B(t), t > 0
We are lead to hope that the following is a well-posed problem for u
ut(a, t) + ua(a, t) + µ(a, t)u(a, t) = 0, 0 < a < ∞, t > 0
u(a, 0) = u0(a), a > 0(12)
u(0, t) = B(t), t > 0
Exercise 0.6. Solve (12) when u0(a) = 10e−a, µ(a, t) = a andB(t) = 1. Is your solution continuous? If not, where is it not continu-ous?
Equations (12) are unsatisfactory because we cannot predict births-they are simply assumed given. A better alternative is to prescribe abirth rate:
β(a, t) = per capita birth rate at time t of individuals of age a
Then we have a formula for B(t):
B(t) =
∫ ∞
0
β(a, t)u(a, t)da
Putting this into (12) yields a more challenging problem!
12
3.4. Second Order Constant Coefficients. Consider the linearsecond order differential equation
(13) auxx + 2buxy + cuyy = hu + dux + euy + f(x, y)
for the function u = u(x, y). Here, the coefficients a, b, c, d, e, h are realconstants, with a, b, c not all zero.
Following the same reasoning used to show that the quadratic form
(14) ax2 + 2bxy + cy2 = f + dx + ey
represents an ellipse, a hyperbola, or parabola, we will obtain standardforms for (13).
Let A be the symmetric matrix defined by
A =
(a bb c
)
Then the left side of (14) can be expressed as Az · z where z = (x, y)T
and superscript T denotes transpose. If e1 = (1, 0) and e2 = (0, 1),note that a = Ae1 · e1, b = Ae1 · e2, c = Ae2 · e2.
For first order equations, we have to give the value of u along certaincurves and then the PDE will allow us to calculate u nearby. For secondorder equations we should expect to have to give u and perhaps onemore piece of info along a curve.
Suppose we give the value of u along the x-axis: u(x, 0) = g(x).From this we may immediately compute ux(x, 0) = g′(x) so we get onederivative for free. Does the equation (13) together with the auxiliarycondition u(x, 0) = g(x) allow us to compute all partial derivatives ofu, say at (0, 0)? If so, then we might expect to compute u near (0, 0)from its Taylor series expansion. It seems clear that we have no way ofcomputing uy(x, 0) from anything that we are given. Hence we are leadto conclude that we must specify both u and the value of its derivativenormal to the curve, namely uy(x, 0), along the x-axis:
u(x, 0) = g(x), uy(x, 0) = k(x)
where g, k are given functions. From this information, we may computethe derivatives:
ux(x, 0) = g′(x), uxx(x, 0) = g′′(x), uyx(x, 0) = k′(x)
But we need help from (13) to compute uyy(x, 0)! It gives
cuyy(x, 0) = hg(x) + dg′(x) + ek(x) + f(x, 0)− 2bk′(x)− ag′′(x)
Thus, if c = Ae2 · e2 6= 0 we may compute all derivatives of u along thex-axis up to the second order. It is not so hard to show that in fact we
3. ADDITIONAL NOTES 13
may compute all partial derivatives of u if c 6= 0. Therefore, u shouldbe completely determined, at least near the x-axis, if c 6= 0.
We are in big trouble though if c = Ae2 · e2 = 0! This suggests thatvectors v satisfying Av ·v = 0 may be in some way special for (13). Wecome back to this point later.
It is standard practice to seek new coordinates that will simplify adifferential equation. Lets try an orthogonal transformation given by
(15) (x, y)T = V (r, s)T
where
V = vijis an orthogonal matrix whose columns v1 = (v11, v21)
T and v2 =(v12, v22)
T give an orthonormal basis. Then V −1 = V T .We mat write the change of variables more simply as:
x = rv11 + sv12
y = rv21 + sv22
Its inverse is given by
(r, s)T = V T (x, y)T
or,
r = xv11 + yv21
s = xv12 + yv22
Let
U(r, s) = u(rv11 + sv12, rv21 + sv22) = (u V )(r, s)
where now we abuse notation by regarding V as a mapping V : R2 →R2 by the rule (r, s) → (x, y) = (rv11 + sv12, rv21 + sv22). Then
u(x, y) = U(xv11 + yv21, xv12 + yv22) = (U V T )(x, y)
with the same abuse of notation. By the chain rule:
Du = DU DV T
or, more simply,
ux = Urrx + Ussx = Urv11 + Usv12(16)
uy = Urry + Ussy = Urv21 + Usv22
from which we compute:
uxx = [Urrrx + Urssx]v11 + [Usrrx + Usssx]v12
= Urrv211 + 2Ursv11v12 + Ussv
212
14
and
uxy = [Urrry + Urssy]v11 + [Usrry + Usssy]v12
= Urrv11v21 + Urs[v11v22 + v12v22] + Ussv12v22
and
uyy = [Urrry + Urssy]v21 + [Usrry + Usssy]v22
= Urrv221 + 2Ursv21v22 + Ussv
222
Hence
auxx + 2buxy + cuyy = a[Urrv211 + 2Ursv11v12 + Ussv
212]
+ 2b[Urrv11v21 + Urs[v11v22 + v12v22] + Ussv12v22]
+ c[Urrv221 + 2Ursv21v22 + Ussv
222]
= Urr[av211 + 2bv11v21 + cv2
21]
+ Urs[2av11v12 + 2b(v11v22 + v12v22) + 2cv21v22]
+ Uss[av212 + 2bv12v22 + cv2
22]
= Urr[Av1 · v2] + Urs[Av1 · v2] + Uss[Av2 · v2]
We have considerable freedom in choosing the orthogonal matrixV . However, one choice stands out. Let vi, i = 1, 2 be unit-lengtheigenvectors for A corresponding to eigenvalues λi.
Avi = λivi, i = 1, 2
Since A is symmetric, v1 · v2 = 0. They give an orthonormal basis forR2 and V diagonalizes A
AV = V
(λ1 00 λ2
)
This V leads to
Av1 · v1 = λ1, Av1 · v2 = 0, Av2 · v2 = λ2
Therefore,auxx + 2buxy + cuyy = λ1Urr + λ2Uss
In the new coordinates
f(x, y) = f(rv11 + sv12, rv21 + sv22) ≡ F (r, s)
dux + euy + hu = d(Urv11 + Usv12) + e(Urv21 + Usv22) + hU
= dUr + eUs + hU
Hence, in the new coordinates, (13) becomes:
(17) λ1Urr + λ2Uss = F (r, s) + dUr + eUs + hU
This does not seem impressive for all the work but let’s continue.
3. ADDITIONAL NOTES 15
There are three cases:
(a) Elliptic case: λ1λ2 > 0.(b) Hyperbolic case: λ1λ2 < 0.(c) Parabolic case: λ1 = 0 6= λ2
In any case, we may assume, after possibly multiplying (13) by −1 sothe signs of a, b, c are reversed, that λ2 > 0, i = 1, 2.
In the elliptic case, where both eigenvalues are positive, we makeone further change of variables:
r =√
λ1X, s =√
λ2Y
andw(X,Y ) = U(r, s)
The resulting equation for w is:
(18) wXX + wY Y = G(X, Y ) + dwX + ewY + hw
where G(X,Y ) = F (√
λ1X,√
λ2Y ). If the entire right-hand side van-ishes, we get Laplace’s equation:
wXX + wY Y = 0
If all but G vanishes on the right side, we get Poisson’s equation:
wXX + wY Y = G
In the hyperbolic case, λ1 < 0 < λ2, we change variables:
r =√|λ1|X, s =
√λ2Y
andw(X,Y ) = U(r, s)
The resulting equation for w is:
(19) −wXX + wY Y = G(X,Y ) + dwX + ewY + hw
where, hopefully, the reader can express G in terms of F as we didin the elliptic case. If the entire right side vanishes, this is the waveequation:
−wXX + wY Y = 0
We could change variables again by
ξ = X − Y, η = X + Y
andW (ξ, η) = w(X, Y )
to obtain another standard form of the wave equation in two variables:
(20) Wξη = H(ξ, η) + dWX + eWY + hW
16
In the parabolic case, the equation for U(r, s) becomes
(21) Uss = λ−12 [F (r, s) + dUr + eUs + hU ]
3.5. Energy Method for Wave Equation. We show how theenergy method can be used to establish uniqueness of solutions of thewave equation:
0 = utt − c2uxx, −∞ < x < ∞, t > 0
u(x, 0) = f(x), −∞ < x < ∞ut(x, 0) = g(x), −∞ < x < ∞
By superposition, it suffices to show that u = 0 is the only solutionwhen f = g = 0.
Fix a point P = (x′, t′) with t′ > 0. The characteristics through Ptake the form
x− ct = x′ − ct′ and x + ct = x′ + ct′
Together with the interval [x′ − ct′, x′ + ct′] on the x-axis in the (x, t)-plane, the characteristics bound an isosceles triangle T with P at itsapex.
3. ADDITIONAL NOTES 17
Multiplying the wave equation by ut and integrating over T we have
0 =
∫ ∫
T
ututt − c2utuxxdxdt
=
∫ t′
0
(∫ x′+ct′−ct
x′−ct′+ct
ututtdx− c2
∫ x′+ct′−ct
x′−ct′+ct
utuxxdx
)dt
=
∫ t′
0
∫ x′+ct′−ct
x′−ct′+ct
[u2
t
2+ c2u2
x
2
]
t
dxdt
−c2
∫ t′
0
[(utux)(x′ + ct′ − ct, t)− (utux)(x
′ − ct′ + ct, t)]dt
=1
2
∫ x′
x′−ct′[u2
t + c2u2x](x,
x− x′ + ct′
c)dx +
1
2
∫ x′+ct′
x′[u2
t + c2u2x](x,
x′ − x + ct′
c)dx
−c2
∫ t′
0
[(utux)(x′ + ct′ − ct, t)− (utux)(x
′ − ct′ + ct, t)]dt
=c
2
∫ t′
0
[u2t + c2u2
x](cs + x′ − ct′, s)ds +c
2
∫ t′
0
[u2t + c2u2
x](x′ + ct′ − cs, s)ds
−c2
∫ t′
0
[(utux)(x′ + ct′ − ct, t)− (utux)(x
′ − ct′ + ct, t)]dt
=1
2
∫ t′
0
[c(u2t + c2u2
x) + 2c2utux]|(x′+ct−ct′,t)dt
+1
2
∫ t′
0
[c(u2t + c2u2
x)− 2c2utux|](x′+ct′−ct,t)dt
=c
2
∫ t′
0
(ut + cux)2(x′ + ct− ct′, t)dt +
c
2
∫ t′
0
(ut − cux)2(x′ + ct′ − ct, t)dt
Integration by parts is followed by a change of order of integration andfinally by a change of variable to obtain the final expression. It followsthat
0 = ut + cux, (x, t) = (x′ + ct− ct′, t), 0 ≤ t ≤ t′
0 = ut − cux, (x, t) = (x′ + ct′ − ct, t), 0 ≤ t ≤ t′
In particular,
ut(x′, t′) = ux(x
′, t′) = 0
and as P was arbitrarily chosen, we find that these derivatives vanishfor all (x′, t′) with t′ > 0. It follows that u is a constant and identicallyzero by the initial conditions.
18
3.6. Nonhomogeneous Wave Equation.
F (x, t) = utt − c2uxx, −∞ < x < ∞, t > 0
u(x, 0) = 0, −∞ < x < ∞ut(x, 0) = 0, −∞ < x < ∞
F (x, t) = vt − cvx
v(x, t) = ut + cux
ut(x, 0) = 0 = v(x, 0), −∞ < x < ∞Method of characteristics gives
v(x, t) =
∫ t
0
F (x + ct− cη, η)dη
and
u(x, t) =
∫ t
0
∫ ξ
0
F (x− ct + 2cξ − cη, η)dηdξ
=
∫ ∫
RT
F G(ξ, η)dηdξ
=
∫ ∫
G(RT )
F G G−1(x′, t′)| det DG−1|dx′dt′
=1
2c
∫ ∫
T
F (x′, t′)dx′dt′
where
(x′, t′)∗ = G(ξ, η) = (x−ct+2cξ−cη, η)∗ =
(2c −c0 1
)(ξη
)+
(x− ct
0
)
maps the right triangle RT = (ξ, η) : 0 ≤ η ≤ ξ ≤ t onto the isoscelestriangle T (x, t) formed by the characteristic curves x′ − ct′ = x − ctand x′ + ct′ = x + ct and the segment t′ = 0 and x− ct ≤ x′ ≤ x + ctin the x′ − t′-plane. So
(22) u(x, t) =
∫ t
0
∫ x+c(t−t′)
x−c(t−t′)F (x′, t′)dx′dt′
As an example, suppose that some nonzero forcing occurs near x =x0 at time near t = t0 > 0, e.g.:
F (x, t) = 1
πr2 , (x− x0)2 + (t− t0)
2 < r2
0, (x− x0)2 + (t− t0)
2 > r2
where r is a very small constant: r << t0. Suppose that you are atposition x′ > x0. When will you know about this event? Equation (22)
3. ADDITIONAL NOTES 19
implies that u = 0 until such time as the triangle T (x′, t) includes apoint of the disk (x− x0)
2 + (t− t0)2 < r2. Hence, for very small r > 0
u(x, t) ≈ 0, t < t0 + x′−x0
c
1, t > t0 + x′−x0
c
This seems strange! The wave equation describes sound waves (c =1100 ft/sec). We imagine the solution above as the result of a firecrackerburst at x = x0 when t = t0. Our solution shows that we do nothear anything for a while and then we hear forever after, and with nodiminution, the result of the explosion!
3.7. Midterm Exam Solutions. 1. Find the solution, if it exists,of
ux + ut = x + t2, u(x, 0) =1
1 + x2
Characteristics:
x = τ + s, t = τ
sodu
dτ= τ + s + τ 2
thus
u = τ 2/2 + sτ + τ 3/3 +1
1 + s2= t2/2 + t(x− t) + t3/3 +
1
1 + (x− t)2
2. Consider the wave equation :
utt = c2uxx, −∞ < x < ∞, t > 0
u(x, 0) = f(x), −∞ < x < ∞ut(x, 0) = g(x), −∞ < x < ∞
a. State d’Alembert’s solution.
b. Let u1(x, t) be a solution of the initial value problem correspondingto initial data f1(x) and g1(x) as above and let u2(x, t) be a solution ofthe initial value problem corresponding to initial data f2(x) and g2(x).Suppose that
|f1(x)− f2(x)| ≤ ε, |g1(x)− g2(x)| ≤ ε, −∞ < x < ∞Show that
|u1(x, t)− u2(x, t)| ≤ (1 + T )ε, 0 ≤ t ≤ T, −∞ < x < ∞
20
c. Let
f(x) = cos(x), x ∈ [−π/2, π/2]0, x /∈ [−π/2, π/2]
and suppose g(x) = 0 for all x. Find all t such that u(8π, t) > 0.Justify your answer!
Since
u(8π, t) =f(8π + ct) + f(8π − ct)
2we need to consider each summand:
f(8π + ct) > 0
⇔ −π/2 < 8π + ct < π/2
⇔ −17π
2< ct < −15π
2
⇔ −17π
2c< t < −15π
2c
and
f(8π − ct) > 0
⇔ −π/2 < 8π − ct < π/2
⇔ −17π
2< −ct < −15π
2
⇔ −17π
2c< −t < −15π
2c
⇔ 15π
2c< t <
17π
2c
Note that both summands are positive or zero so there are no cancela-tions. So
u(8π, t) > 0 ⇔ −17π
2c< t < −15π
2cor
15π
2c< t <
17π
2c
3. Let f(x) be a 2L-periodic function on −∞ < x < ∞.
a. Using the relations
0 =
∫ L
−L
sin(mπx
L) sin(
nπx
L)dx =
∫ L
−L
cos(mπx
L) cos(
nπx
L)dx, n 6= m
0 =
∫ L
−L
sin(mπx
L) cos(
nπx
L)dx, all m,n
L =
∫ L
−L
sin2(mπx
L)dx =
∫ L
−L
cos2(mπx
L)dx, m ≥ 1.
3. ADDITIONAL NOTES 21
show how we formally obtain the coefficients an, bn in the Fourier series
(23) f(x) =a0
2+
∞∑n=1
an cos(nπx
L) + bn sin(
nπx
L)
b. What does it mean that the Fourier series (23) converges to f(x)uniformly on the interval [a, b]? Give a mathematically precise defini-tion!
If
Sm(x) =a0
2+
m∑n=1
an cos(nπx
L) + bn sin(
nπx
L)
denotes the finite sum of the first m + 1 terms (including a0) of (23),then the Fourier series converges uniformly on [a, b] means that forevery ε > 0 there exists a positive integer N such that
|f(x)− Sm(x)| < ε, m ≥ N, a ≤ x ≤ b.
4. Consider the ODD 4-periodic extension of the function
f(x) = 1, 0 ≤ x ≤ 12− x, 1 < x ≤ 2
a. Give its Fourier series and the formula for all coefficients. Evalu-ate only those coefficients that are zero due to even-ness or odd-nessconsiderations.
f(x) =∞∑
n=1
bn sin(nπx
2)
where
bn =
∫ 2
0
f(x) sin(nπx
2)dx
b. Give the sum of the series for each x ∈ [0, 2], carefully justifyingyour answers. Be sure to explicitly verify ALL hypotheses of any theo-rems you use. Where does the series converge uniformly? Justify youranswer as above.
22
After CAREFULLY checking that f and f ′ are piecewise continu-ous, in particular noting the values of
f(0−), f(0+), f(2−), f(2+), f ′(0−), f ′(0+), f ′(1−), f ′(1+), f ′(2−), f ′(2+)
and verifying they are finite, we can apply Dirichlet’s Theorem. TheFourier series converges to:
f(0−) + f(0+)
2= 0 at x = 0
and to f(x) at each x ∈ (0, 2]. It converges uniformly to f on anyinterval [a, b] where 0 < a < b < 4 and all translates by 4z for integerz.
3.8. On Term-by-Term Differentiation of Series. The sepa-ration of variables technique applied to linear second order pdes ulti-mately leads to formal series “solutions” whose validity must be veri-fied. For this our text supplies Proposition 2-3, page 59:
Proposition 2. Let gn(x) be continuously differentiable on an in-terval I and assume the following:
(i)∑∞
n=0 gn(a) converges at some point a ∈ I;(ii)
∑∞n=0 g′n(x) is uniformly convergent on I.
Then∑∞
n=0 gn(x) is differentiable on I and( ∞∑
n=0
gn(x)
)′
=∞∑
n=0
g′n(x)
As we will see, it is useful to know that the hypotheses of theProposition imply the additional fact that the series
∞∑n=0
gn(x)
converges uniformly on bounded subintervals of I. Indeed, if
Sn(x) =n∑
k=0
gk(x)
is the n-th partial sum of the series, we have
Sn(x)− Sm(x) = ([Sn(x)− Sm(x)]− [Sn(a)− Sm(a)]) + [Sn(a)− Sm(a)]
= [S ′n(c)− S ′m(c)](x− a) + [Sn(a)− Sm(a)]
for some c between a and x by the Mean Value Theorem. Let x anda belong to a bounded subinterval J of I of length L. By the uniformconvergence of
∑∞n=0 g′n(x), if ε > 0 then there exists a natural number
3. ADDITIONAL NOTES 23
N such that |S ′n(x)−S ′m(x)| < ε/2L for all x ∈ J, n ≥ N . We can alsoassume N is so large that |Sn(a) − Sm(a)| < ε/2 for n ≥ N . Takingabsolute values in the inequality above, using the triangle inequalityand our estimates results in |Sn(x) − Sm(x)| < ε for n ≥ N , provingthe uniform convergence of
∑∞n=0 gn(x).
Our differential equations are typically of second order so we areoften faced with showing that
∑∞n=0 gn(x) is twice differentiable and
each derivative is given by term-by-term differentiation. The followingresult is stated so as to be maximally useful in this context.
Proposition 3. Let gn(x) be twice continuously differentiable onan interval I and assume the following:
(i)∑∞
n=0 gn(a) converges at some point a ∈ I;(ii)
∑∞n=0 g′n(b) converges at some point b ∈ I;
(iii)∑∞
n=0 g′′n(x) is uniformly convergent on I.
Then∑∞
n=0 gn(x) is twice differentiable on I and for each x ∈ I:( ∞∑
n=0
gn(x)
)′
=∞∑
n=0
g′n(x)
( ∞∑n=0
gn(x)
)′′
=∞∑
n=0
g′′n(x)
Proof. The hypotheses of the previous proposition are satisfiedfor g′n(x) so we may conclude from it that
∑∞n=0 g′n(x) is differentiable
and its derivative is given by term-by-term differentiation. We alsoknow that
∑∞n=0 g′n(x) converges uniformly on bounded subintervals of
I. This allows us to again apply the previous proposition to show that∑∞n=0 gn(x) is differentiable on I and its derivative is given by term-by-
term differentiation. Therefore,∑∞
n=0 gn(x) is twice differentiable. ¤Obviously, the reasoning may be extended to obtain results for
higher order derivatives.
3.9. Heat Equation on a Ring. Consider a very thin ring, acircle, of circumference 2π whose lateral sides are insulated so that noheat may escape. If we parameterize length along the ring by x ∈[−π, π] then the temperature u(x, t) satisfies the initial boundary valueproblem:
ut = auxx, −π < x < π, t > 0
u(−π, t) = u(π, t), t > 0(24)
ux(−π, t) = ux(π, t), t > 0
u(x, 0) = f(x), −π < x < π
24
The boundary conditions express the fact that there is no boundaryto a ring so the temperature and heat flux must agree at x = −π andx = π since they represent the same point of the ring!
Looking for solutions of the form u = X(x)T (t) we find on insertingthis into the PDE that
T ′(t)aT (t)
=X ′′(x)
X(x)= K, a constant
and so X satisfies
X ′′ −KX = 0, −π < x < π
X(−π) = X(π)
X ′(−π) = X ′(π)
and T satisfiesT ′ = aKT
We can show that K ≤ 0 by the following argument. Multiply thedifferential equation for X by X, integrate over [−π, π] to get
K
∫ π
−π
X2dx = −∫ π
−π
(X ′)2dx
Since X(x) cannot be identically zero for all x, we see that K ≤ 0. SetK = −λ2, λ ≥ 0. Then
X = a cos(λx) + b sin(λx)
and forcing X to satisfy the boundary conditions leads to:
2b sin(λπ) = 0
2aλ sin(λπ) = 0
If sin(λπ) 6= 0, then, since λ 6= 0, we must have a = b = 0 which impliesX ≡ 0-an uninteresting result (u = XT = 0!). We conclude that
sin(λπ) = 0 ⇒ λ = n ∈ 0, 1, 2, · · · and hence
X0 = a0/2, Xn = an cos(nx)+bn sin(nx), n ≥ 1, T0 = 1, Tn = e−an2t
and by superposition we have
(25) u(x, t) =a0
2+
∞∑n=1
e−an2t[an cos(nx) + bn sin(nx)]
Setting t = 0
(26) f(x) =a0
2+
∞∑n=1
[an cos(nx) + bn sin(nx)]
3. ADDITIONAL NOTES 25
Extending f(x) as a 2π-periodic function to all x by translation, wesee that
an =1
π
∫ π
−π
f(x) cos(nx)dx, bn =1
π
∫ π
−π
f(x) sin(nx)dx
Now we must verify that (25) solves our initial boundary valueproblem. Since each un is a solution of the PDE, u will be a solutionprovided the series converges and interchange of differentiation withthe infinite sum is verified. For the latter, we see that both ut anduxx involve virtually the same series. We also note that the decayingexponential term should work in favor of convergence for t > 0. Fixx ∈ [−π, π] and t > 0 arbitrarily small and consider (25) as a functionof t ∈ I = t ≥ t. We are going to apply Proposition 2-3, Chapter 3to verify that u is differentiable with respect to t and that its derivativeis:
(27) ut(x, t) = −a
∞∑n=1
n2e−an2t[an sin(nx) + bn cos(nx)]
According to Proposition 2-3, we must show that the above series con-verges uniformly on I and that the series (25) for u converges, say att = t. The series obtained from the ut series by taking absolute valuesis dominated by the series
∞∑n=1
n2e−an2 t|an|,∞∑
n=1
n2e−an2 t|bn|
for t ∈ I. Observe that the similar series corresponding to u (whichdoes not have the factor n2) will be dominated by the series above aswell! So we just have to find conditions for these two series to converge.The effect of the decaying exponential is so strong that we merely haveto assume that there is some M > 0 so that
(28) |an|, |bn| ≤ M, n ≥ 1
26
If this is the case then we may focus on∞∑
n=1
n2e−an2 t =∞∑
n=1
n2e−an2 t/2e−an2 t/2
≤ L
∞∑n=1
(e−at/2)n2
≤ L
∞∑
l=1
(e−at/2)l
= Le−at/2
1− e−at/2< ∞
where we have used the fact that
z → ze−azt/2
is bounded by the constant L = 2e−1
atfor 0 ≤ z < ∞ to conclude that
n2e−an2 t/2 ≤ L.
Proposition 2-3 and our calculations above show that u is differentiableon I and ut is given by (27) on I. Since t > 0 was arbitrary, we concludethat ut is given by (27) for every t > 0.
An entirely parallel argument using the extension of Proposition 2-3 to second derivatives is required to show that uxx exists and is givenby the formula (27) but without the factor a. With t > 0 now fixed, wemust show that the series for u and ux converge for some x and that foruxx converges uniformly for x ∈ [−π, π]. The same argument as for ut
shows the series for uxx converges uniformly and absolutely on [−π, π]and that series dominates the series for u and ux so they also convergeuniformly on [−π, π].
We must also verify that u satisfies the initial conditions (26). Butthis holds by Dirichlet’s theorem if the 2π-periodic extension of f iscontinuous and piecewise smooth. This condition will automaticallyimply that (28) holds.
In summary, we have
Theorem 1. Let f be piecewise smooth and continuous on [−π, π]and satisfy
(29) f(−π) = f(π)
Then (25) is a solution of (24). However, u(x, t) need not be continuousat points (x, 0), −π ≤ x ≤ π. If f ′ is continuous and satisfies
f ′(−π) = f ′(π)
3. ADDITIONAL NOTES 27
and f ′′ is piecewise continuous on [−π, π] then u(x, t) is continuous forall (x, t) ∈ [−π, π]× [0,∞).
Proof. Let f be the 2π-periodic extension of f . Then f is piece-wise smooth because f is. Since f is continuous, the Fourier series for fconverges to f(x) = f(x) for each x ∈ (−π, π) by Dirichlet’s Theorem.
At x = π, it converges to [f(π−)+ f(π+)]/2. But f(π+) = f(−π−) =
f(−π) = f(π), the last by (29), and f(π−) = f(π−) = f(π). Thusthe Fourier series converges to f(π) at x = π by Dirichlet’s Theorem.Similarly, it converges to f(−π) at x = −π.
Note that the series (25) defining u(x, t) was shown to convergeuniformly on [−π, π]× [t,∞) for each t > 0. Proposition 2-1 of Chapter3 implies that u is continuous on [t,∞) for each t > 0. Consequently, itis continuous on [−π, π]×(0,∞) but not necessarily on [−π, π]×[0,∞).For this stronger conclusion, it suffices to show that (25) convergesuniformly and absolutely on [−π, π] × [0,∞). For this, the decaying
exponential factor e−an2t ≤ 1 can be dropped and by the WeierstrasseM-test it suffices to show that there is some M such that
|an|, |bn| ≤ M/n2, n ≥ 1
The usual integration by parts twice yields
an = − 1
n2π
∫ π
−π
f ′′(x) cos(nx)dx, if f ′(−π) = f ′(π) & (29)
and similarly for bn. Since f ′′ is piecewise smooth it is bounded andwe may take M to be an upper bound for f ′′(x±) on [−π, π].
¤
3.10. A Growing and Diffusing Population. A populationgrows exponentially at rate r > 0. Its density satisfies
ut = ru
If the population occupies a one-dimensional habitat 0 < x < L, thenu(x, t) denotes the population density. The meaning is that
∫ b
a
u(x, t)dx = # individuals in a < x < b at time t
If the organisms making up the population move randomly about theirhabitat, then it is reasonable to model this by the equation
ut = duxx + ru
u(0, t) = u(L, t) = 0(30)
u(x, 0) = f(x)
28
The boundary conditions say that the boundary is lethal. Organismscannot survive there. The diffusion constant d > 0 reflects the rate ofthe random movement.
Exercise 0.7. (a) Use the separation of variables technique or achange of variables v = eλtu to find a formal solution of (30). How bigmust the habitat be in order that the population can survive? Whatmight this say about the destruction of habitat caused by humans orhow big to design wild life refuges.
(b) Show that if f(x) ≥ 0, 0 ≤ x ≤ L, then u(x, t) ≥ 0 for 0 ≤ x ≤L and t > 0. Hint: Apply the Maximum Principle to v(x, t) = u(x, t)eλt
for some λ, or use Problem 2, section 5.3.(c) If we change the boundary conditions to ux(0, t) = 0 = ux(L, t),
then the organisms cannot leave the habitat (no flux!). Find a formalsolution of this problem. Integrate the equation with respect to x toshow that the total populations satisfies
∫ L
0
u(x, t)dx = ert
∫ L
0
f(x)dx
3.11. Fundamental Solution of the Heat Equation. We seeka positive, integrable solution of the heat equation
ut = auxx, −∞ < x < ∞, t > 0
which captures the dynamics following a drop of dye at x = 0 at t = 0.This solution should, by symmetry, be an even function of x, have asingle peak near x = 0 which we expect to slowly fall and widen astime progresses. It should go to zero at infinity fast enough that thefollowing calculations hold. Integrate the equation to obtain
d
dt
∫ ∞
−∞u(x, t)dx =
∫ ∞
−∞ut(x, t)dx
= a
∫ ∞
−∞uxx(x, t)dx
= aux(x, t)|x=+∞x=−∞
= 0
Thus, ∫ ∞
−∞u(x, t)dx = constant
Observe that from the equation, a has units length2/time so that
z =x2
at
3. ADDITIONAL NOTES 29
is dimensionless. It seems natural to try a solution of the form
(31) u(x, t) = w(t)h(z) = w(t)h(x2
at)
where w(t) is chosen so that∫ ∞
−∞u(x, t)dx = w(t)
∫ ∞
−∞h(
x2
at)dx = constant, t > 0
We need to find w and h! Notice that:∫ ∞
−∞h(
x2
at)dx = 2
∫ ∞
0
h(x2
at)dx
=
∫ ∞
0
h(z)(at)1/2 dz
z1/2
= (at)1/2
∫ ∞
0
h(z)dz
z1/2
Therefore, we should choose w(t) = 1(at)1/2 . Hence, we seek a solution
of the form
(32) u(x, t) =1
(at)1/2h(
x2
at)
where only h = h(z) needs to be determined. We have
ux =2x
(at)3/2h′(z)
uxx =4x2
(at)5/2h′′(z) +
2
(at)3/2h′(z) =
4
(at)3/2zh′′(z) +
2
(at)3/2h′(z)
ut = −1
2a−1/2t−3/2h(z)− x2
t(at)3/2h′(z) = −1
2
a
(at)3/2h(z)− a
(at)3/2zh′(z)
If we stick the last two terms into the heat equation and then multiply
through by (at)3/2
a, we get
(33) 4zh′′(z) + (z + 2)h′(z) + (1/2)h(z) = 0, z > 0
Try
h(z) = e−rz
Why not? It has worked before! We could get lucky! We find, ondividing out the common factor erz, that
4zr2 − (z + 2)r + 1/2 = z(4r2 − r)− 2r + 1/2 = 0, ∀z > 0
Clearly, we must have
r(4r − 1) = 0, & 0 = −2r + 1/2
30
Fortunately, r = 1/4 satisfies both. In summary
(34) u(x, t) =1
(at)1/2e−
x2
4at
is a positive, integrable solution of the heat equation which has a con-stant integral over −∞ < x < ∞, independent of t > 0.
Equation (34) reminds me of the normal distribution from statistics.Its density function, the infamous bell-curve, is given by
(35) n(x, σ) =1
σ(2π)1/2e−
x2
2σ2
where σ > 0 is the standard deviation and σ2 is the variance. n is aprobability density function so it has the property that
∫ ∞
−∞n(x, σ)dx = 1
From statistics, I recall that 68% of the area under the normal densitycurve lies within one standard deviation of the mean x = 0:∫ σ
−σ
n(x, σ)dx = .68
Comparing (34) and (35), we see that our u is essentially n withstandard deviation σ = (2at)1/2. This suggests multiplying our u by1/(4π)1/2, which is again a positive solution of the heat equation:
(36) u(x, t) =1
(4πat)1/2e−
x2
4at
This function has total integral equal one:∫ ∞
−∞u(x, t)dx = 1
It is called the fundamental solution of the heat equation.The fundamental solution approaches a singularity as t → 0+ since
its value at x = 0 blows up: u(0, t) →∞. In fact, it approximates thefamous Dirac “function” δ(x) as t → 0+:
u(x, t) → δ(x), t → 0+
Recall that δ(x) is supposed to have the properties:
(1) δ(x) = 0 for x 6= 0.
(2) If a < 0 < b and f is a continuous function then∫ b
af(x)δ(x)dx =
f(0).
3. ADDITIONAL NOTES 31
−5 0 5−1
−0.5
0
0.5
1
1.5
2Snapshots of Fundamental Solution
t=0.03
t=0.5
If u(x, t) solves the heat equation, so does u(x− y, t) for any fixedy. In the case of our fundamental solution, u(x − y, t) → δ(x − y) ast → 0+. If f is any bounded continuous function, then superpositionsuggests that
v(x, t) =
∫ ∞
−∞f(y)u(x− y, t)dy
is also a solution (differentiate under integral) and
v(x, t) =
∫ ∞
−∞f(y)u(x−y, t)dy →
∫ ∞
−∞f(y)δ(x−y)dy = f(x), t → 0+
It can be shown that
v(x, t) =1
(at)1/2
∫ ∞
−∞f(y)
1
(at)1/2e−
(x−y2
4at dy
satisfies the initial value problem
vt = avxx, −∞ < x < ∞, t > 0
v(x, 0) = f(x), −∞ < x < ∞3.12. Eigenvalue Problem for the Laplacian. Let D be a nor-
mal domain in Rn. For the Dirichlet boundary condition, the problemis to find eigenvector-eigenvalue pairs (u, λ), where u is not the zerofunction, which satisfy
0 = 4u + λu, x ∈ D
0 = u(x), x ∈ B
32
As a familiar example, if n = 1, D = (0, L), the eigenvalue problembecomes
0 = u′′ + λu, x ∈ (0, L)
0 = u(0) = u(L)
The eigenvalues and eigenfunctions are
λn = (nπ
L)2, un = sin(
nπx
L), n = 1, 2, · · ·
We use Green’s Identities to show that λ must be positive. Sup-pose that u is an eigenfunction corresponding to a real eigenvalue λ.Multiply the eigenvalue equation by u and integrate to get
0 =
∫
D
(u4u + λu2)dx
= −∫
D
|∇u|2dx + λ
∫
D
u2dx
where we have used Green’s first identity. Since u is not the zerofunction,
∫D
u2dx 6= 0 so
λ =
∫
D
|∇u|2dx/
∫
D
u2dx ≥ 0
If∫
D|∇u|2dx = 0 then ∇u = 0 in D which implies that u is a constant.
As u = 0 on B, it would follow that u = 0 in D contradicting that u isnot the zero function. Thus we conclude that λ > 0.
Now suppose that λ1 and λ2 are distinct eigenvalues with corre-sponding eigenfunctions u1 and u2. We claim that
∫
D
u1(x)u2(x)dx = 0
Indeed, apply Green’s second identity to u1 and u2 to conclude
0 =
∫
D
(u14u2 − u24u1)dx
=
∫
D
−λ2u1u2 + λ1u2u1dx
= (λ1 − λ2)
∫
D
u1u2dx
3. ADDITIONAL NOTES 33
If D = (0, a) × (0, b) ⊂ R2 where a, b > 0, then separation ofvariables gives the eigenvalues and eigenfunctions as
λnm = π2(n2
a2+
m2
b2)
unm = sin(nπx
a) sin(
mπy
b), n,m = 1, 2, 3, · · ·
Clearly, this extends in the obvious way to the case that D = (0, a)×(0, b)× (0, c) ⊂ R3.
3.13. Exercise 0.2. Let f : (a, b) → R be a real-valued functionsuch that f ′(x) exists for x ∈ (a, b), f ′(x0) = 0 for some x0, and supposethat f ′′(x0) > 0.
Since
f ′′(x0) = limx→x0
f ′(x)− f ′(x0)
x− x0
Given ε > 0 we can find δ > 0 such that
−ε <f ′(x)− f ′(x0)
x− x0
− f ′′(x0) < ε
provided −δ < x − x0 < δ. If we choose ε = f ′′(x0)2
, for example, thenthere is a corresponding δ as above. Since f ′(x0) = 0, we conclude that
0 <f ′′(x0)
2<
f ′(x)− f ′(x0)
x− x0
provided −δ < x− x0 < δ. It follows that
f ′(x) > 0, x0 < x < x0 + δ.
Now, if x0 < x < x0 + δ, the Mean Value Theorem implies there existsc, depending on x, such that x0 < c < x and
f(x)− f(x0)
x− x0
= f ′(c)
Since f ′(c) > 0, we have proved that
f(x) > f(x0), x0 < x < x0 + δ
Now suppose that a differentiable function f has a local maximumat x0 and that f ′′(x0) exists. Note that we do not assume that f ′′(x)exists at any other point! We know that f ′(x0) = 0 at an extrema.If f ′′(x0) > 0, then our argument above shows that f(x) > f(x0) forx0 < x < x0 + δ. This is incompatible with our assumption that x0 isa local maximum. Therefore, we conclude that f ′′(x0) ≤ 0.
34
Finally, suppose that u is a differentiable function on an open setD ⊂ R3 which attains a local maximum at x0 = (x0
1, x02, x
03) and that
∂2u∂x2
j(x0), j = 1, 2, 3, exist. Since x0 is an extrema, we have
∂u
∂xj
(x0) = 0, j = 1, 2, 3
These hypotheses imply that f(x) = u(x, x02, x
03) has a local maxi-
mum at x = x01 and that
f ′(x01) =
∂u
∂x1
(x0) = 0, f ′′(x01) =
∂2u
∂x21
(x0)
Our results above imply that f ′′(x01) ≤ 0 so we conclude that
∂2u
∂x21
(x0) ≤ 0
Similarly, for the other variables we get
∂2u
∂x2j
(x0) ≤ 0, j = 1, 2, 3
so
4u(x0) ≤ 0.
3.14. Growing and Diffusing Population on a RectangularHabitat. Consider our population with density function u(x, t) whichsatisfies:
ut = d4u + ru, x ∈ D, t > 0
u(x, t) = 0, x ∈ B, t > 0
u(x, 0) = f(x), x ∈ D
Here, we imagine D as a bounded domain in R2 (a Petri dish?) or R3.Letting v = e−rtu we see that v satisfies the heat equation with the
same initial conditions and boundary conditions. From this fact andthe maximum principle, we have the following conclusions:
(1) If f(x) ≥ 0, x ∈ D, then u(x, t) ≥ 0, x ∈ D, t > 0.(2) If f(x) ≤ g(x), x ∈ D and u is the solution with initial data
f and u∗ is the solution with initial data g, then u(x, t) ≤u ∗ (x, t), x ∈ D, t > 0.
3. ADDITIONAL NOTES 35
The smallest eigenvalue λ1 > 0 of the Laplacian and correspondingeigenfunction u1(x) will play a key role.
0 = 4u + λu, x ∈ D
u(x) = 0, x ∈ B
It is a fact that u1(x) > 0, x ∈ D (but not on B!). For example, if
D = (0, a)× (0, b)
then
λ1 = π2(1
a2+
1
b2), u1(x, y) = sin(
πx
a) sin(
πy
b)
It is easy to check that for any c > 0
u(x, t) = ce(r−λ1d)tu1(x)
is a positive solution.This solution grows if r−λ1d > 0 and decays if r−λ1d < 0. In the
case of the square D described above we find that
r − λ1d > 0 ⇔ √r
d> π
√a2 + b2
ab