PDE Full Solutions

7/22/2019 PDE Full Solutions

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HOMEWORK 1 SOLUTIONS

Section1.1 : #2, 3, 11, 12

#2) Which of the following operators are linear? (This is accomplished bychecking ifLu= cLu andL(u+v) = Lu+ LV are satisfied).

(a)Lu= ux+xuy(b)Lu= ux+uuy(c)Lu= ux+u2y(d)Lu= ux+uy+ 1(e)Lu= 1 +x2 cos(y)ux+uyxy arctan(xy )uAnswer: Only (a) and (e) are Linear. (b), (c), and (d) do not satisfy

L(cu) = cL(u). #3) For each of the equations below state the order and type.

Equation Order Type(a) 2 Linear Inhomogeneous(b) 2 Linear Homogeneous(c) 3 Non-Linear(d) 2 Linear Inhomogeneous(e) 2 Linear Homogeneous(f) 1 Non-Linear(g) 1 Linear Homogeneous(h) 4 Non-Linear

#11 Verify thatu(x, y) = f(x)g(y) is a solution for all pairs of differentiable

functionsf and g of one variable to the equation:

uuxy = uxuy

Answer:

ux= f(x)g(y)

uy =f(x)g(y)

uxy=f(x)g(y)

Then we have:

uuxy = f(x)g(y)f(x)g(y) = f(x)g(y)f(x)g(y) = uxuy

#12 Verify by substitution that:un(x, y) = sin(nx) sinh(ny)

is a solution ofuxx+uyy = 0 for every n >0.

Answer: (un)xx =n2 sin(nx) sinh(ny) and (un)yy = n2 sin(nx) sinh(ny). Whenwe add these together, we get (un)xx+ (un)yy = 0.

1


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Section 1.2 : #1, 3, 7, 8

#1 Solve the first order equation with initial condition:

2ut+ 3ux= 0

u(x, 0) = sin(x)

Answer: Following the Geometricor Coordinate Method, we get:

u(x, t) = f(2x 3t)We solve for f(x) using the initial condition.

u(x, 0) = f(2x) = sin(x)

Then f(x) = sin(x2 ), which means

u(x, t) = sin(x 32

t)

#3 Solve (1 +x2)ux+uy = 0 and sketch some characteristic curves.Answer: Following the Geometric Method:

dy

dx=

1

1 +x2

y= arctan(x) +c

c= y arctan(x)u(x, y) = f(c) = f(y arctan(x))

The characteristic curves are graphs of the form y = arctan(x) + c forany value ofc.

#7 Solve aux+buy+ cu= 0.Answer: Following the Coordinate Method:

Let x =ax+by andy =bx ay. Then we have:aux+buy = (a

2 +b2)ux

(a2 +b2)ux+ cu= 0

ux

u =

ca2 +b2

ln(u) = ca2 +b2

x +f(y)

u(x, y) = f(y)ecx

a2+b2

u(x, y) = f(bx ay)ec(ax+by)

a2+b2

Alternative Solution: Rewrite the equation as:

aux

u +b

uy

u +c= 0


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Then let v = ln uand solve:

avx+bvy+ c= 0

This has the homogenous solution vh= f(bxay) and a particular solutionvp=

cxa

. Then

v(x, y) = f(bx ay) +cxa

u(x, y) = f(bx ay)ecxa

Note: This solution does not appear valid for a = 0, but with an appro-

priate choice off(bxay) it is equivalent to the solution reached using theCoordinateMethod.

#8 Solve ux+uy+ u= ex+2y withu(x, 0) = 0.Answer: One possible approach includes using the solution to the pre-

vious problem to find the solution to the homogenous equation and then

finding a particular solution.

Heres another approach using a change of coordinates and an integrating

factor.

Let x =x+y and y =x y. The equation to solve becomes:2ux+ u= e

3xy

2

ux+1

2u=

1

2e3xy

2

Multiply the equation by the integrating factor ex

2 to get:

ex

2 ux+12

uex

2 = 12

e4xy

2

ex

2 ux

=1

2e4xy

2

ex

2 ux

dx =

1

2e4xy

2 dx

ex

2 u=1

4e4xy

2 +f(y)

u(x, y) =1

4e3xy

2 +ex

2 f(y)

u(x, y) =1

4ex+2y +e

xy2 f(x y)

Using the initial condition to solve forf, we have:

0 = u(x, 0) =1

4ex +e

x2 f(x)

Then f(x) = 14 e3x2 , and we get:

u(x, y) =1

4ex+2y 1

4ex2y

.


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Section 1.3 : #7, 8

#7 Derive the equation for heat flow in a ball where the temperature de-

pends only on the spherical coordinate r =

x2 +y2 +z2.

Answer: We start with the general heat equation:

cut = (u)In this problem, we are told that c, , and are constants, so

ut = Cu= C(uxx+uyy + uzz)

Using the chain rule, we rewrite uxx in terms of derivatives in terms ofr .

uxx= (urrx)x= urxrx+urrxx= urrr2x+urrxx

Also,

r=

x2 +y2 +z2

rx= x

r

rxx=1

r x

2

r3

Then

uxx= urrx2

r2 +ur

1

r x

2

r3

The expressions for uyy and uzz are similar. When added together, we get

ut = C

urr+

2urr

#10 Given f(x) is continuous and|f(x)| 1|x|3+1 for all x. Let B (R) be a

ball of radius R. Thenall space

f dx = limR

B(R)

f(x)dx = limR

B(R)

f(x) n dS

limR

B(R)

|f(x) n |dS= limR

B(R)

|f(x)||n |dS

= limR

B(R)

|f(x)|dS limR

B(R)

1

|x|3 + 1 dS

= limR

B(R)

1

R3 + 1dS= lim

R

4R2

R3 + 1= 0

Similarly,all space

f dx limR

B(R)

|f(x) n |dS 0

Then by the squeeze theorem:all space

f dx = 0


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Section1.5 : #1, 2, 3, 4(a) (b)

#1) Solve the boundary value problem:

uxx+u= 0

u(0) = 0 and u(L) = 0.

Answer: The solution to the ODE is:

u(x) = A sin(Bx+C)

IfA = 0, then the solution is u(x) 0.

If A = 0, then we can solve for the variables by substituting the general

solution into the ODE.

uxx+u= (A AB2)sin(Bx+C) = 0

A(1 B2) = 0

B= 1

Using the boundary conditions we have:

u(0) =A sin(C) = 0

Then C= n with n an integer. Also,

u(L) = A sin(BL+C) = 0

BL +C=mL+n = m

L= m

So, ifL = m for some integer m, then u(x) = A sin(x + n) is a solution.

IfL =m, then u(x) 0 is the only solution.

#2) Consider the problem

u(x) +u(x) = f(x)

u(0) =u(0) =1

2[u(l) +u(l)]

(a) Does the following problem have a unique solution?(b) Does a solution necessarily exist?

Answer: (a) The problem does not have a unique solution.

Proof. Letu1and u2be solutions to the problem. Let v(x) = u1(x)u2(x).

Then v(x) is a solution of

u(x) +u(x) = 01


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u(0) =u(0) =1

2[u(l) +u(l)]

Then v(x) = A+Bex, and

v(0) =v(0) =1

2[v(l) +v(l)]

B= A+B =1

2[Bel +A+Bel]

B= A+B = A

2A= 2B

v(x) = B(ex 2), for any real number B

Then ifu(x) is a solution of the initial problem, u2(x) = u1(x) + ex 2 is

also a solution.

(b)A solution does not necessarily exist unless f(x) satisfies the condition l0

f(x)dx= 0

Proof. We integrate both sides of the ODE and use the conditions on u at

x= 0 and at x = l. l0

[u(x) +u(x)]dx=

l0

f(x)dx

u(l) +u(l) u(0) u(0) =

l0

f(x)dx

2u(0) u(0) u(0) = l0 f(x)dx

0 =

l0

f(x)dx

#3 Solve the boundary value problem

u(x) = 0 if 0< x


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Case 1 (+k): The equations from the boundary conditions are

A= BkBk2 = 0

Then eitherB = 0 or k = 0, which means that A = 0. Thenu(x) = B with

B = 0 ifk = 0.

Case 2 (k): The equations from the boundary conditions are

A= Bk

Bk(k 2) = 0

Then either B = 0, k = 0, or k = 2. Again, ifB = 0 or k = 0, then A = 0.

Ifk = 2, then A = 2B

u(x) =

B ifk = 0 and B R,0 ifk = 0 andk = 2,

2Bx+B ifk = 2 .

The only degree 1 solution occurs when k = 2 in the k case.

#4 Consider the Neumann problem:

u= f(x, y, z) in D ,

nu= 0 on boundary ofD .

(a) What can we add to any solution to get another solution?

(b) Show that

Df(x,y,z)dxdydz= 0.

Answer: (a) We can add any solution to the homogeneous Neumann prob-

lem to any solution of the above Neumann problem. This include constants

and ax +by+cz with a,b,c n= 0.

(b) Using the divergence theorem and the information given in the PDE,

we have: D

f(x,y,z)dxdydz=

D

udxdydz

=

D udxdydz

=

D

n udS

=

D

nudS

=

D

0dS= 0


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Section 1.6 : #1, 4

#1 What are the types of the following equations?

(a) uxx uxy+ 2uy+ uyy 3uyx+ 4u= 0

Answer: Remember that uxy = uyx, then the equation becomes:

uxx 4uxy+ uyy + 2uy+ 4u= 0

D =

4

2

2 (1)(1) = 3> 0

Then the equation is Hyperbolic.

(b) 9uxx+ 6uxy+ uyy +ux= 0

Answer: D =6

2

2(9)(1) = 99 = 0. Then the equation is Parabolic.

#4 What is the typeof the equation

uxx 4uxy+ 4uyy = 0?

Answer: D = (2)2 (1)(4) = 0. It is Parabolic.

Show by direct substitution that u(x, y) = f(y+ 2x) + xg(y+ 2x) is a

solution for arbitrary functions f and g. Answer: Compute the partial

derivatives and then substitute.

ux= 2f(y+ 2x) +g(y+ 2x) + 2xg(y+ 2x)

uxx= 4f(y+ 2x) + 2g(y+ 2x) + 2g(y+ 2x) + 4xg(y+ 2x)

= 4f(y+ 2x) + 4g(y+ 2x) + 4xg(y+ 2x)

uxy = 2f(y+ 2x) +g(y+ 2x) + 2xg(y+ 2x)

uy = f(y+ 2x) +xg(y+ 2x)

uyy = f(y+ 2x) +xg(y+ 2x)

Then substituting, we have:

uxx 4uxy+ 4uyy = 4f(y+ 2x) + 2g(y+ 2x) + 2g(y+ 2x) + 4xg(y+ 2x)

8f(y+ 2x) 4g(y+ 2x) 8xg(y+ 2x)

+ 4f(y+ 2x) + 4xg(y+ 2x)

= 0


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Section 2.1 : #8, 9

#8 Consider the spherical wave equation:

utt = c2

urr +

2

rur

(a) Change variables by making the substitution v = ru.

Answer: First solve for u, then take partial derivatives.

u=1

rv

utt =1

rvtt

urr =1

rvrr

2

r2vr+

2

r3v

Substituting into the spherical wave equation, we get

1

rvtt =

1

rc2vrr

vtt = c2vrr

(b) Solve for v using the fact that utt = c2uxx has a solution of the form

u(x, t) = f(x+ct) +g(x ct)

Answer: All that has changed here are the variables, so

v(r, t) = f(r+ct) +g(r ct)

is a solution ofvtt = c2vrr, which means

u(r, t) =1

rf(r+ct) +

1

rg(r ct)

is a solution of the spherical wave equation.

(c) Solve the spherical wave equation with initial conditions

u(r, 0) = (r)

ut(r, 0) = (r)

with both and even functions ofr .

Answer: First, rewrite the initial conditions in terms of v, and then

solve using the solution to the initial value problem wave equation given by

(8) in the book.

v(r, 0) = r(r) = L(r)

vt(r, 0) = r (r) = M(r)


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Then from the solution to the IVP, we have

v(r, t) =

1

2 [L(r+ct) +L(r ct)] +

1

2c r+ct

rct M(s)ds

v(r, t) =1

2[(r+ct)(r+ct) + (r ct)(r ct)] +

1

2c

r+ctrct

s(s)ds

u(r, t) = 1

2r[(r+ct)(r+ct) + (r ct)(r ct)] +

1

2rc

r+ctrct

s(s)ds

#9 Solve the initial value problem

uxx 3uxt 4utt = 0,

u(x, 0) = x2,

ut(x, 0) = ex.

Answer: By factoring the operator acting on u, we have

4(t1

4x)(t+x)u= 0.

Noticing the similarities of this to the wave equation, which factors like

(t cx)(t+cx)u= 0,

you may guess that the solution were looking for has the form

u(

x, t) =

f(

x

t) +

g(

x+

1

4t)

.

One way to show this is to let v = (x+t)u. Then we have

ut+ux= v

vt1

4vx= 0

Then ux+ut = v(x, t) = h(x+ 1

4t).

ux+ut = h(x+1

4t) has a homogeneous solution of the form uh= f(xt).

It also has a particular solution up= g(x+ 1

4t) with g (s) = 4

5h(s). Then

u(x, t) = f(x t) +g(x+1

4t)

ut(x, t) = f(x t) +

1

4g(x+

1

4t)

Considering the initial conditions, we have

u(x, 0) = f(x) +g(x) = x2

ut(x, 0) = f(x) +

1

4g(x) = ex


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Solving for f and g , we integrate

exdx= f(x) +1

4

g(x)dx

ex =f(x) +1

4g(x) +c

ex =g(x) x2 +1

4g(x) +c

So, we have

g(x) =4

5

ex +x2

+c

f(x) =4

5 ex +

1

5x2 c

u(x, t) =4

5

ex+

1

4t ext

+x2 +

1

4t2


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Section 2.1 : #2, 7

#2) Solve

utt= c2uxx

u(x, 0) = log(1 +x2)

ut(x, 0) = 4 +x

Answer: Using the solution to the initial value problem wave equation, we

have

u(x, t) =

1

2

log(1 + [x+ct]2

) + log(1 + [xct]2

)

+

1

2c

x+ct

xct 4 +sds

After combining the logarithms and evaluating the integral

u(x, t) =1

2log

(1 + [x+ct]2)(1 + [xct]2)

+ 4t+xt

#7) If both and are odd functions ofx, show that solution u(x, t) of

the wave equation is also an odd function ofx for all t.

Answer: Show u(x, t) = u(x, t).

u(x, t) =1

2[(x+ct) +(xct)] +

1

2c

x+ct

xct

(s)ds

u(x, t) =1

2[(x+ct) +(xct)] +

1

2c

x+ct

xct

(s)ds

=1

2[(xct)(x+ct)]

1

2c

xct

x+ct

(s)ds

(let s =r, then ds=dr and we change the limits of integration)

u(x, t) =1

2[(xct)(x+ct)]

1

2c

x+ct

xct

(r)(dr)

=1

2 [(xct) +(x+ct)]

1

2c

x+ct

xct

(r)dr

=u(x, t)

Section 2.2 : #1, 2, 3

#1) Use energy conservation of the wave equation to prove that the only

solution with 0 and 0 is u 0.1


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Answer: We know u(x, 0) 0 and ut(x, 0) 0.

E(x, t) =

1

2

(ut(x, t))

2

+T(ux(x, t))

2dx

E(x, 0) =1

2

(ut(x, 0))

2 +T(xu(x, 0))2

dx

=1

2

0)2 +T(x(0))

2

dx

E(x, 0) =1

2

0dx= 0

Then E(x, t) 0, since tE= 0.

0 =1

2

(ut(x, t))

2 +T(ux(x, t))2

dx(ut(x, t))2 +T(ux(x, t))

2 0

Then by the vanishing theorem

(ut(x, t))2 +T(ux(x, t))

2 0

Then

ut(x, t)0

ux(x, t)0

Then u(x, t) is a constant. However, u(x, 0)0, then u(x, t)0.

#2) For a solution u(x, t) of the wave equation with c = 1, the energy

density e, and the momentum density p are defined as

e=1

2

u2t+u2

x

p= utux

(a) Show that te= xp and tp= xe.

(b) Show that both e(x, t) and p(x, t) satisfy the wave equation.

Answer: (a) We take the partial derivatives using the chain rule and use

the fact that utt= uxx.

te=1

2t

u2t+u2

x=1

2(2ututt+ 2uxuxt)

=ututt+uxuxt

=utuxx+uxutx

=x(utux) = xp


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Do the same for the next part

tp= t(utux)

=uttux+utuxt

=uxxux+utuxt

=1

2(u2

x)x+

1

2(u2

t)t

=1

2

u2x

+u2t

x

= xe

(b) Here are two methods to prove this, the second being the fastest.

To show p satisfies the wave equation we take the second partial deriva-

tives with respect to t and x, and then show that they are equal to each

other using utt= uxx.

ptt= (utux)tt

= (uttux+utuxt)t

=utttux+ 2uttuxt+utuxtt

= (utt)tux+ 2uxxutx+ut(utt)x

=uxxtux+ 2uxxutx+utuxxx

=utxxux+ 2uxxutx+utuxxx

= (utxux+utuxx)x

= (utux)xx= pxx

To show e satisfies the wave equation we make use of part (a).

ett= (et)t= (px)t= pxt= ptx= (pt)x= (ex)x= exx

#3) Show that the wave equation has the following invariance properties.

(a) Any translate u(xy, t), where y is fixed is also a solution.

Answer: We know utt(x, t) = c2uxx(x, t) for any pair (x, t).

2u(xy, t)

x2 =

x[ux(xy, t)x(xy)]

=

x[ux(xy, t)1]

=uxx(xy, t)x(xy) = uxx(xy, t)

2u(xy, t)

t2 =

t[ut(xy, t)t(t)]

=utt(xy, t)

Then

2u(xy, t)

t2 =utt(xy, t) = c

2uxx(xy, t) = c2

2u(xy, t)

x2


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(b) Any derivative, say ux, is also a solution.

Answer: We can change the order of partial derivatives, so

2

t2ux(x, t) = uxtt(x, t)

=uttx(x, t)

= (utt(x, t))x

= (c2uxx(x, t))x

=c2 2

x2ux(x, t)

(c) The dilated function u(ax, ay) is also a solution.

Answer: Use the chain rule and utt(ax, at) = c2uxx(ax, at).

2

t2u(ax, at) = a2utt(ax, at)

=a2c2uxx(ax, at)

=c2 2

x2u(ax, at)

Section 2.3 : #2, 4

#2) Consider a solution of the diffusion equation in

{0 x , 0 t T. Then

M(T)< M(S). So, M(T) cannot be decreasing.

(b)) Let m(T) be the minimum of the solution in the rectangle

{0 x , 0 t T}.

Doesm(T) increase or decrease as a function of time?

Answer: m(T) is decreasing as a function of time. By the maximum

principle the minimum occurs on the bottom or lateral sides of the rectan-

gle. Suppose it occurs on the lateral side for some time t= S > T. Then

m(T)> m(S). So, m(T) cannot be increasing.


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#4) Consider the diffusion equation

ut= uxx in{0< x


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Answer: Using the energy method and u(1, t) = u(0, t) = 0.

0 = utuxx

= (utuxx)u

=uut= uuxx

0 =1

2(u2)t(uux)x+ (ux)

2

0 =1

2

10

(u2)tdx

10

(uux)xdx+

10

(ux)2dx

=1

2

t

10

u2dxuux|10dx+

10

(ux)2dx

0 =1

2

t

10

u2dx+

10

(ux)2dx

Then we have

t

10

u2dx 2

10

(ux)2dx 0

However, if 10

(ux)2dx= 0

Then ux 0 by the vanishing theorem. Then u(x, t) = f(t), but we know

that u(x, 0) = 4x(1x). Then 10

(ux)2dx >0

which means that

t

10

u2dx


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Section 2.4 : #2, 4, 9, 11, 15

#2) Solve the diffusion equation with the given initial condition:

ut = kuxx in{< x


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and let p= (y+2ktx)

4kt . Then dy =

4ktdp. After substituting and remem-

bering to change the limits of integration, we have

u(x, t) = 1

ektx

2ktx4kt

ep2

dp

= 1

ektx

0

ep2

dp2ktx

4kt

0

ep2

dp

= 1

ektx

2

2Erf

2ktx

4kt

=1

2ektx

1 Erf

2ktx

4kt

#9) Solve the diffusion equation with initial condition:ut= kuxx

u(x, 0) = x2.

After differentiating both sides with respect to xthree times, we have

(uxxx)t = k(uxxx)xx

uxxx(x, 0) = 0.

Then uxxx(x, t) = 0 is a solution to the above equation, but by uniqueness

of solutions, we have:

uxxx(x, t)

0.

Integrating both sides with respect tox three times, yields

u(x, t) = A(t)x2 +B(t)x+C(t).

Next, use the initial condition and the diffusion equation to determine the

functionsA(t), B (t), and C(t).

u(x, 0) = A(0)x2 +B(0)x+C(0)

=x2

Then A(0) = 1, B (0) = 0, and C(0) = 0. Furthermore,

kuxx = ut

k(2A(t)) = A(t)x2 +B(t)x+C(t)

A(t) = 0A(t) = constantB(t) = 0B(t) = constantC(t) = k(2A(t))


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Since A(t) is a constant, and A(0) = 1, we know A(t) = 1. Similarly

B(t) = 0. Then we have C(t) = 2k. Since C(0) = 0, we know that

C(t) = 2kt. Then the solution is given by

u(x, t) = x2 + 2kt

#11) Consider the diffusion equation on the whole line withu(x, 0) = (x)

(1) If(x) is an odd function ofx, show that the solution u(x, t) is also.

Proof. Let w(x, t) = u(x, t) + u(x, t). Then w(x, t) satisfies the dif-fusion equation, and

w(x, 0) = u(

x, 0) +u(x, 0) = (

x) +(x) =

(x) +(x) = 0.

Then we have the initial condition diffusion equationwt = kwxxw(x, 0) = 0

By uniqueness, w(x, t)0.Thenu(x, t) +u(x, t) = 0

u(x, t) =u(x, t).Thus,u(x, t) is an odd function ofx.

(2) If(x) is an even function ofx, show that the solution u(x, t) is also.

Proof. Let w(x, t) = u(x, t)u(x, t). Then w(x, t) satisfies the dif-fusion equation, and

w(x, 0) = u(x, 0)u(x, 0) = (x)(x) = (x)(x) = 0.Then we have the initial condition diffusion equation

wt = kwxxw(x, 0) = 0

By uniqueness, w(x, t)0.Thenu(x, t)u(x, t) = 0

u(

x, t) = u(x, t).

Thus,u(x, t) is an even function ofx.

(3) Show that the analogous statements are true for the wave equation.

Proof. The analogous statement for the wave equation involves spec-

ifying the oddness/evenness of both (x) and (x). Otherwise, the

proof is the same.


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4

If(x) and (x) are odd functions ofx, show that u(x, t) is an

odd function ofx. Let w(x, t) = u(x, t) + u(x, t). Thenw(x, t)satisfies the wave equation, and we have the initial conditionwave equation

wtt = c

2wxx,

w(x, 0) = 0,wt(x, 0) = 0.

By uniqueness, w(x, t)0.Thenu(x, t) +u(x, t) = 0

u(x, t) =u(x, t).Thus,u(x, t) is an odd function ofx.

If(x) and (x) are even functions ofx, show that u(x, t) is aneven function ofx. Letw(x, t) = u(x, t)u(x, t). Thenw(x, t)satisfies the wave equation, and we have the initial condition

wave equation

wtt = c2wxx,

w(x, 0) = 0,wt(x, 0) = 0.

By uniqueness, w(x, t)0.Thenu(x, t)u(x, t) = 0

u(

x, t) = u(x, t).

Thus,u(x, t) is an even function ofx.

#15) Use the energy method to prove the uniqueness of the diffusion prob-lem with Neumann boundary conditions:

utkuxx = f(x, t) for{0< x < , 0< t


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5

Then

0 = 0

w= (wt

kwxx)w

0 =1

2(w2)t(kwxw)x+k(wx)2

Integrating both sides with respect tox, we have

0 =1

2

0

(w2)tdx(kwxw)|0+ 0

k(wx)2dx

0 =1

2

t

0

w2dx+

0

k(wx)2dx

Then

t

0

w2dx=2

0

k(wx)2dx0

Then0 w

2(x, t)dx is decreasing with respect to t. Since t >0, we have

0 0

w2(x, t)dx 0

w2(x, 0)dx= 0.

Then

w2(x, t)0w(x, t)0

u1(x, t)u2(x, t)0u1(x, t)u2(x, t).

Then, there is a unique solution.


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Section 3.1 : #1, 3, 4

#1) Solve the diffusion equation on the half-line with the Dirichlet bound-ary condition:

ut= kuxx on{0< x


24/178

2

#3) Solve the diffusion equation on the half-line with the Neumann bound-ary condition:

wt= kwxx on{0< x


25/178

3

will be satisfied if we make sure that vx(x, t) 2v(x, t) is an odd functionofx. Let

f(x) =

x for x >0x+ 1 e2x for x


26/178

4

solution is unique, it must be given by

u(x, t) = 1

4kt

e(xy)2

4kt f(y)dy.

Section 3.2 : #1, 6

#1) Solve the Neumann problem for the wave equation on the half-line:utt= c

2uxx on x, t(0,)u(x, 0) = (x)

ut(x, 0) = (x)

ux(0, t) = 0

Solution: Using the method of even extensions, consider the wave equation

on the whole linevtt= c

2vxx onx, t(0,)

v(x, 0) = even(x) =

(x) x0(x) x0

vt(x, 0) = even(x) =

(x) x0(x) x0

The guarantees that v(x, t) is an even function, and therefore vx(0, t) = 0,

andu(x, t) = v(x, t) whenx >0. The solution to the wave equation on the

whole line is given by

v(x, t) =1

2

[even(x+ct) +even(x

ct)] +

1

2c x+ct

xct

even(y)dy

Whenx ct >0, even(x ct) = (x ct). Then we have

u(x, t) =1

2[(x+ct) +(x ct)] + 1

2c

x+ctxct

(y)dy

When 0< x < ct, x ct 0 and a > c.


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5

Solution: Consider the function w(x, t) defined on the whole line by

w(x, t) =

ut(x, t) +aux(x, t) , for x >0;

0 , for x = 0;ut(x, t) aux(x, t) , for x 0;0 , for x = 0;

ut(x, 0) aux(x, 0) , for x 0;0 , for x = 0;

V axu(x, 0) , for x 0;0 , for x = 0;

V ax(0) , for x 0;0 , for x = 0;V , forx 00, forx = 0V, forx ct;0, for 0< x < ct.

Then for x > ct, we have

ut(x, t) +aux(x, t) = V .


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6

The homogeneous solution to this is uh(x, t) =f(at x), and a particularsolution is up(x, t) = V t. Then

u(x, t) = V t +f(at x)u(x, 0) = f(x) = 0

Then f(x)0. Then u(x, t) = V t for x > ct.When 0< x < ct, we have

ut(x, t) +aux(x, t) = 0

Then u(x, t) = h(at x). We know this must satisfy the wave equation.

utt(x, t) = c2uxx(x, t)

a2h(at

x) = c2h(at

x)

(a2 c2)h(at x) = 0

Then

h(at x) = 0h(at x) = Kh(at x) = K(at x)

Then u(x, t) = K(at x) for 0< x < ct. To solve for K, we ensure that ucontinuous at x= ct. Then

V t = u(ct,t)=K(at ct)

K= V

a cThen we have for a final solution to the initial problem

u(x, t) =

V t, forx > ct;atxac

V , for 0< x < ct.

Section 3.3 : #1, 3

#1) Solve the inhomogeneous diffusion equation on the half-line with theDirichelt boundary condition:

utkuxx= f(x, t) on{0< x


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7

Solution: Using the method of reflection, consider the inhomogeneous

diffusion equation on the whole line

vtkvxx= fodd(x, t) =

f(x, t), x >0;0, x= 0;

f(x, t), x 0;0, x= 0;

(x), x 0. The solution for v(x, t) was derived in

3.3. The result is

v(x, t) =

S(x y, t)odd(y)dy+ t0

S(x y, t s)fodd(y, s)dyds

=

0

S(x y, t)(y)dy+ 0

S(x y, t)[(y)]dy

+

t0

0

S(x y, t s)f(y, s)dyds+ t0

0

S(x y, t s)[f(y, s)]dyds

=

0

[S(x y, t) S(x+y, t)](y)dy+ t0

0

[S(x y, t s) S(x+y, t s)]f(y, s)dyds

#3) Use the subtraction method to solve the inhomogeneous Neumanndiffusion problem on the half-line:

wt= kwxx on{0< x


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8

Then u(x, t) will be an even function of x, which guarantees u satisfies

the boundary condition ux(0, t) = 0. Also, v(x, t) = u(x, t) for x > 0.

The solution to the inhomogeneous diffusion problem on the whole line isderived in 3.3, and it is given by equation (2). Then for x >0,

v(x, t) =

S(x y, t)even(y)dy+ t0

S(x y, t s)feven(y, s)dyds

=

0

[S(x y, t) +S(x+y, t)][(y) yh(0)]dy

+

t0

0

[S(x y, t s) +S(x+y, t s)][yh(s)]dyds

Then we have

w(x, t) = 0

[S(x y, t) +S(x+y, t)][(y) yh(0)]dy

t0

0

[S(x y, t s) +S(x+y, t s)]yh(s)dyds+xh(t)


31/178


Section

3.2 : #8

,10

,11

#8) For the wave equation in the finite interval (0 , ) with Dirichlet con-ditions, explain the solution formula within each diamond-shaped region.

Solution: The equation and boundary conditions we are considering is

equation below defined for x (0, ).utt = c

2uxx

u(x, 0) = (x)

ut(x, 0) = (x)

u(0, t) = u(, t) = 0

We can extend this to the whole line by extending and in a way suchthat their extensions are odd functions ofx aroundx = 0 andx = . Define

extand extby

ext(x) =

(x) for 0< x < ;(x) for < x


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2

CASE 3: (n, n + 1) with n even:

u(x, t) =

1

2 (x ct + n) 1

2 (x ct + (n + 2)) + 1

2c xct+(n+2)xct+n (y)dy

CASE 4: (n, n) with n odd:

u(x, t) = 12

(x+ct(n1)) 12

(xct+(n+1))+ 12c

xct+(n+1)

x+ct(n1)

(y)dy

CASE 5: (n + 1, n) with n odd:

u(x, t) =1

2(x ct + (n + 1)) 1

2(x ct + (n + 1)) + 1

2c

xct+(n+1)

xct+(n+1)

(y)dy

CASE 6: (n, n + 1) with n odd:

u(x, t) = 12

(x + ct (n1)) +12

(x + ct (n + 1)) + 12c x+ct(n+1)x+ct(n1)

(y)dy

#10) Solve the wave equation on the finite interval x (0, 2 ) with thegiven boundary conditions:

utt = 9uxx in{0< x < 2

, 0< t < }u(x, 0) = cos(x)

ut(x, 0) = 0

ux(0, t) = 0

u(

2, t) = 0

Solution: Consider the wave equation on the whole line

vtt = 9vxx

v(x, 0) = cos(x)

vt(x, 0) = 0

Since cos(x) is even aroundx = 0, and odd aroundx = 2 , we will have that

v(x, t) is an even function ofx aroundx = 0, andv(x, t) is an odd function

ofx around x = 2 . This will guarantees that vx(0, t) = 0 and v(

2 , t) = 0.

(NOTE: you also need to be sure that vt(x, 0) is both even around x = 0,

and odd around x = 2 ).

Then u(x, t) = v(x, t) for 0< x < 2 .

u(x, t) =1

2[cos(x + 3t) + cos(x 3t)] + 1

2c

x+3tx3t

0ds

= cos(x) cos(3t)


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3

#11) Solve the wave equation on the finite interval x (0, ) with the givenboundary conditions:

utt = c2uxx in{0< x < , 0< t < }u(x, 0) = 0

ut(x, 0) = x

u(0, t) = u(, t) = 0

Solution: Consider the wave equation on the whole line

vtt = c2vxx

v(x, 0) = 0

vt(x, 0) = ext(x) = x for < x <

with period 2.

Since ext(x) is odd around x = 0 and x = , then v(x, t) will be as well.

Then v(0, t) = v(, t) = 0. Then u(x, t) = v(x, t) for x (0, ).

u(x, t) = 1

2c

x+ctxct

ext(y)dy

For any point (x, t), in a diamond-shaped region , you can find u(x, t) by

using the results of problem 3.4.8 given above. For any initial point or

point on the boundary, you can check that this solution will satisfy the

given conditions.

Section 3.4 : #1, 6, 11, 12

#1) Solveutt = c

2uxx+ xt

u(x, 0) = 0

ut(x, 0) = 0

Solution: From equation (3), we have:

u(x, t) = 1

2c

t

0

x+c(ts)

xc(ts)

ysdyds

= 1

4c t

0

y2s

|x+c(ts)xc(ts)ds

=

t0

xts xs2ds

=1

6xt3

#6) Derive the formula for the inhomogeneous wave equation by factoringthe operator.


34/178

4

(a) We have

utt

c2uxx = f(x, t)

(t cx)(t+ cx)u= f(x, t)

Let

v(x, t) = (t+ cx)u(x, t)

f(x, t) = (t cx)v(x, t)

(b) We can solve for u in terms ofv by using a change of coordinates.

Let

t =t + cx

x

=ct

x

Then

t= t + cx

1 + c2

x= ct x

1 + c2

Then

ut = u

x

x

t+

u

t

t

t=

1

1 + ccv

Integrating both sides with respect tot, we have

u(x, t) =

1

1 + c2 tcx

v(x, r)dr

= 1

1 + c2

t

cx

v

cr x1 + c2

,r + cx

1 + c2

dr

Let

s= r + cx

1 + c2

Then we have

u(x, t) =

t

0

v (x ct + cs,s) ds

(c) Similarly, we can solve for v in terms of fby using a change of

coordinates. The result is

v(x, t) =

t0

f(x + ct cr, r)dr

(d) Substituting, part (c) into part (d), we have

u(x, t) =

t0

s0

f(x ct + 2cs cr, r)drds= 12c

f


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5

(e) Part (d) is a particular solution to the wave equation. To obtain

the full solution, we need to add the homogeneous solution. Then

u(x, t) = 12

[(x + ct) + (x ct)] + 12c

x+ct

xct

(y)dy+ 12c

f

#11) Show by direct substitution that u(x, t) = h t xc

for x < ct and

u(x, t) = 0 for x ct solves the wave equation on the half-line with zeroinitial data and boundary condition u(0, t) = h(t).

utt = c2uxx

u(x, 0) = 0

ut(x, 0) = 0

u(0, t) = h(t)

Solution: When x= 0, we have x < ct, then u(0, t) = h(t). When t= 0,

we have x > c t, then u(x, t) 0. Then u(x, 0) = 0 and ut(x, 0) = 0.Clearly, for x > ct, u(x, t) = 0 satisfies the wave equation. It only remains

to check that u(x, t) satisfies the wave equation for x < ct.

utt = h(t x

c)

=c21

c2h(t x

c)

=c2uxx

#12) Solve the following using Greens Theoremvtt c2vxx = f(x, t) in 0< x <

v(x, 0) = (x)

vt(x, 0) = (x)

v(0, t) = h(t)

Solution: For x > ct, the domain of dependence does not intersect the

boundary x = 0. Then we can use Greens Theorem exactly as done on

pages 73 74. The result is the same as for the whole line

u(x, t) =1

2[(x + ct) + (x ct)] + 1

2c

x+ct

xct

(y)dy+ 1

2c

fdxdt

For x < ct, the domain of dependence reflects off of the boundary x = 0.Then instead of being a triangle, like before, the domain of dependence

is a quadrilateral. You should draw the domain of dependence. The four

vertices are

(0, t0 x0c

) (x0, t0)

(ct0 x0, 0) (x + ct0, 0)


36/178

6

Then define the four lines that go along the edges of the domain of depen-

dence to be

L0 = the line from (ct0 x0, 0) to (x + ct0, 0)L1 = the line from (x + ct0, 0) to (x0, t0)

L2 = the line from (x0, t0) to (0, t0 x0c

)

L3 = the line from (0, t0 x0c

) to (ct0 x0, 0)

Then by Greens theorem, we have

fdxdt=

L0+L1+L2+L3

(c2vxdt vtdx)

Compute each line integral separately, and then add them together.L0

(c2vxdt vtdx) = x0+ct0ct0x0

(x)dx

L1

(c2vxdt vtdx) = cL1

dv

=cv(x0, t0) c(x0+ ct0)L2

(c2vxdt vtdx) = cL2

dv

= cht0 x0

c + cv(x0, t0)L3

(c2vxdt vtdx) = cL3

dv

=c(ct0 x0) ch

t0 x0c

After adding these together, we have

fdxdt= 2cv(x0, t0)+c(ct0x0)c(x0+ct0) x0+ct0ct0x0

(x)dx2ch

t0 x0c

Then solving for v, we have

v(x0, t0) =1

2[(x0+ct0)

(ct0

x0)]+

1

2c x0+ct0

ct0x0

(x)dx+1

2c

fdxdt+ht0 x0

c

Then the full solution for v is given by

v(x, t) =

12 [(x + ct) (ct x)] + 12c

x+ctctx

(y)dy+ 12c

fdxdt + h

t xc

ifx < ct

12 [(x + ct) + (x ct)] + 12c

x+ct

xct (y)dy+ 12c

fdxdt ifx > ct


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7

Section 4.1 : #4, 6

#4) Write the series expansion solution of a wave in a resistant medium

that satisfies

utt = c2uxx rut for 0< x <

u(x, 0) = (x)

ut(x, 0) = (x)

u(0, t) = u(, t) = 0

where r is a constant such that

0< r 0

Then we have two ODEs

X(x) + 2X(x) = 0

T(t) + rT(t) + c22T(t) = 0

Solving the first ODE, we have

X(x) = C1cos(x) + C2sin(x)

Since X(0) =X() = 0, we have

= n

which meansn =

n

2Then

Xn(x) = sinnx

forn N.

The second ODE is now

T(t) + rT(t) + nc2

T(t) = 0.

This has the characteristic equation

s2 + rs +nc

2= 0

Solving this for s, we have

s=r

r2 4 nc

22


38/178

8

Since 0< r < 2c

, s is a complex number. Then

T(t) = ARe

e

rt+t

r24(nc )2

2

+ BIm

e

rt+t

r24(nc )2

2

Tn(t) =

Ancos

t

r2

4nc

2+ Bnsin

t

r2

4nc

2ert

2

un(x, t) =

Ancos

t

r2

4nc

2+ Bnsin

t

r2

4nc

2ert

2 sinnx

u(x, t) =

n=1

Ancos

t

r2

4nc

2+ Bnsin

t

r2

4nc

2ert

2 sinnx

where

(x) =

n=1

Ansin

nx

(x) =

n=1

Bn

r2

4

nc

2 r

2An

sinnx

#6) Consider the following differential equation on a finite interval withgiven boundary conditions. Separate variables and show that there are an

infinite number of solutions for the given initial condition.

tut = uxx+ 2u

u(0, t) = u(, t) = 0

u(x, 0) = 0

Lettingu(x, t) = X(x)T(t), we have

tX(x)T(t) = X(x)T(t) + 2X(x)T(t)

Then separating the variables,

tT(t)

T(t) 2 = X

(x)

X(x) =

Then we have two ODEs

X(x) + X(x) = 0

tT(t) + ( 2)T(t) = 0The first ODE, yields

X(x) = sin(

x)

Since X(0) =X() = 0, we have

Xn(x) = sin(nx) for n N


39/178

9

where n = n2. The second ODE is separable, and we can rewrite it as

T(t)

T(t)

=2

t T(t)

T(t)dt =

2

t dt

ln(T(t)) = (2 )ln(t) + CT(t) = C t2

Tn(t) = Cnt2n2

Then

u(x, t) =

n=1

Cnt2n2 sin(nx)

Since u(x, 0) = 0, we require that Cn = 0 for n 2. Thenu(x, t) = C1t sin(nx)

However, we can choose any value we want for C1, and u(x, t) will still

satisfy the differential equation, the boundary conditions, and the initial

condition.


40/178

HOMEWORK 9

6.1.4. We look for a solution u = u(r). Thus u must satisfy

urr +2

rur = 0 (r

2ur)r = 0 r2ur = c1 u=

c1r

+c2.

From the boundary conditions

u(a) =A c1a

+c2 = A

u(b) = B c1b +c

2 = B.

Solving forc1, c2 we obtain

c1 = (B A)(1/a 1/b)1, c2 = B + (B A)(1/a 1/b)

1/b.

6.1.11. Integrate u= f to obtain

(1)

D

udx=

D

f dx.

From the divergence theorem

(2) D

udx= D

div(u)dx= D

u ndS= D

u

n

dS.

Combining (1), (2) and the boundary condition un

= g on D, we obtain thedesired equality.

1


41/178

HOMEWORK 10

6.2.1. Following the hint, guess that

u(x, y) = Ax2 + By2 + Cxy + Dx + Ey + F.

Then,

(1) u= 0 A= B.

(2) ux(0, y) = a CY + D= a.

(3) ux(a, y) = 0 2Aa + CY + D= 0.

(4) uy(x, 0) = b Cx + E= b.

(5) uy(x, b) = 0 2Bb + Cx + E= 0.

From (2) and (3) one immediately gets 2Aa a= 0 A= 1/2.From (4) and (5) one immediately gets 2Bb + b= 0 B= 1/2.This is consistent with (1).Since (2),(3), (4) and (5) must hold for all x, y, one must choose C= 0 which

impliesD = a,E=b.Thus

u(x, y) = 1/2x2 1/2y2 ax + by+ F.

Note, this is a Neumann problem hence there is uniqueness up to additive con-stants.

6.3.1.

(a) By the maximum principle u achieves its maximum on r = 2. Hence itsuffices to find the maximum value ofu = 3sin 2+ 1. Clearly this max value is 4,as the sine function has 1 as its max value.

(b) By the MVP, (a= 2)

u(0) = 14

2

0

(3sin2+ 1)2d= 1.

6.4.6. We look for a separated solution u = R. Then

(1) + = 0 (0) = () = 0.

(2) r2R + rR R= 0 R(0)f inite,R(1) = sin sin2.

From (1) we have (Dirichlet problem)1


42/178

2 HOMEWORK 10

= n2, n= sin n,n= 1, 2, . . .

Substituting= n2 in (2) and looking for a solution of the form r, we obtain= n.Thus,

Rn= Crn + D/rn.

We chooseD = 0, so thatR is finite at zero. Then our separated solutions looklike,

un= Anrn sinn.

Summing them up we get,

u=n=1

Anrn sinn.

By imposing the boundary condition at 1 we have,

sin sin2=n=1

Ansinn.

This immediately implies

A1= , A2= 1, An= 0 n = 1, 2.

Thus,

u= r sin r2 sin2.

7.1.5.

Let v = u w. Then

(1) E[w] =E[u]

D

u vdx +

D

|v|2dx +

D

hvdS.

Now apply G1 tou and v :

D

vu

ndS=

D

u vdx +

D

vudx.

Since u= 0 and un

=h on D, we get

(2)

D

hdS=

D

u vdx.

Combining (1) and (2) we get,

E[w] =E[u] +

D

|v|2dx E[u].


43/178

HOMEWORK 11

7.2.2. Apply G2 with u = , v= 1/(4|x|), over the domain D = B\ B.(B= unit ball centered at zero, B = ball of radius centered at zero.)Then v= 0 in D and we get

D

1

4|x|dx=

D

(v

n v

n)dS=

B

(v

n v

n)dS+

B

(v

n v

n)dS.

The first addend in the sum above is zero because 0 outside ofB , and it issmooth in R3.

As 0 we then obtain (r= |x|,/n= /r)

B

1

4|x|dx= lim

0

B

v

n v

n

dS= lim

0

r=

r

1

4r

+

1

4r

n

dS=

= lim0

r=

1

4r2

+

1

4r

n

dS= lim

0

1

42

r=

dS+ 1

4

r=

n

dS=

= lim0 +

n =(0),

where for any function f, we denote by f the average offon the boundary of theballB.

7.3.2. For notational simplicity I will take x0 = 0. Apply G2 to u, v withv = (4|x|)1, over the domain D = D \ B (B = ball of radius centered atzero.)

Then v= 0, u= f in D, and we get

D

1

4|x|f dx=

D

(uv

n v

u

n)dS=

D

(u vn

v un

)dS+B

(u vn

v un

)dS.

As 0, using the same argument as in the previous problem, we then obtain

(1)

D

1

4|x|f dx=

D

(hv

n v

u

n)dS u(0),

where we also used that u = h on D.Now lets write G(x, 0) =v(x) +H(x) with H harmonic in D . We apply G2 to

u, Hin the domain D to obtain1


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2 HOMEWORK 11

(2) D

Hf dx= D

(hH

n

Hu

n

)dS,

where we also used that u = h on D.Adding (1) and (2), and using that G 0 on D we obtain

D

Gfdx=

D

hG

ndS u(0),

which gives the desired claim.

7.4.9. We have

G(x, x0) = 1

4|x x0|+

1

4|x x0|,

where x

0 is the reflection of x0 across the plane ax+ by +cz = 0. Thus, setn= (a,b,c), the normal vector to the plane, we have

x0

= x0 2nax0+by0+cz0

a2 +b2 +c2 .

Property 1 : the only singularity ofG is at x0, since x0 is not in D.

Property 2: when x D, then |x x0|= |x x0|. Hence G(x, x0) = 0.

Property 3: G(x, x0) + 1

4|xx0| = 1

4|xx0|which is harmonic everywhere in

D, since x0 is not in D.

7.4.22. Following the hint, consider the function v = u/y, which solves

v= 0 in {y >0}, v= h on {y= 0}.

Thus, according to formula (4) page 183 (modified for the 2D case),

v(x, ) =

+

h()

( x)2 +2d

(I am calling (x0, y0) with (x, ) and (x, 0) with (, 0). This is just for elegancepurposes.)

Integrating ind betweeny and +we get (we can assume that u behaves likea constant at ):

u(x, y) = C1

y

+

h()( x)2 +2

dd

=C limR+

1

Ry

+

h()

( x)2 +2dd

But

Ry

+

h()

( x)2 +2dd


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HOMEWORK 11 3

=

+

h()

Ry

( x)2 +2dd=

=12

+

h()[log(( x)2 +R2) log(( x)2 +y2)]d=

= 1

2

+

h()[log(( x)2 +R2) log R2 log(( x)2 +y2)]d

Here we used that in order for the Neumann problem to have a solution onemust have (problem 6.1.11)

+

h()d= 0.

Thus,

Ry

+

h()

( x)2 +2 dd

=1

2

+

h()[log(( x)2/R2 + 1) log(( x)2 +y2)]d

Hence,

u(x, y) =C limR+

1

Ry

+

h()

( x)2 +2dd

=C+ 1

2

+

h() log(( x)2 +y2)d.


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Partial Differential Equations Math 442 C13/C14

Fall 2009Homework 1 Solutions

1. Determine which of the following operators are linear:

(a) Lu= uxx+uxy

(b) Lu= uux

(c) Lu= 4x2uy 4y2uyy

Solution: We will see that the first and third are linear and the second is not. For example, wecompute:

(a)

L(u+v) = (u+v)xx+ (u+v)xy = uxx+vxx+uxy+vxy

= (uxx+uxy) + (vxx+vxy) = Lu+Lv,

and

L(u) = (u)xx+ (u)xy = uxx+uxy = (uxx+uxy) = Lu.(b) L(u) = (u)(u)x = u(ux) =

2ux and this is not equal to Lu if2 =.

(c)

L(u+v) = 4x2(u+v)y 4y2(u+v)yy = 4x

2(uy+vy) 4y2(uyy +vyy)

= 4x2uy 4y2uyy + 4x

2vy 4y2vyy = Lu+Lv,

and

L(u) = 4x2(u)y 4y2(u)yy = 4x

2uy 4y2ux= (4x

2uy 4y2uyy) = Lu.

2. (Strauss, 1.1.4) Show that the difference of two solutions toLu = g is a solution toLu = 0, whenL isany linear operator.

Solution: Assume thatLu = g andLv = g. Define w = u v, thenLw= L(u v) =Lu Lv= g g= 0,

where the second equality comes from L being linear.

3. Solve:

2ux+ 3ut= 0,

u(x, 0) = x2.

Solution: From the formula derived in class, we know the solution is of the form

u(x, t) = f(3x 2t),

for some undetermined functionf. Using the initial condition, we obtain

u(x, 0) = f(3x) = x2,

so that f(s) = s2/9, oru(x, t) = (3x 2t)2/9.

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4. Consider the heat equation with initial condition given as

ut = uxx,

u(x, 0) = x+,

where, are real numbers. Make an educated guess for the solution to this PDE and check that it is

correct. Interpret this as a statement about the evolution of temperature in a 1D object.Solution: There are at least two ways to come up with a guess.

If we think of the temperature of an object as being a linear function, notice then that at every point,we expect the heat coming into the point to be the same as the heat leaving, so while there is a heatflux at every point, the actual temperature stays fixed. Thus we expect this temperature not to change,and we would guess u(x, t) = x+.

Another approach is to notice that if we plug the initial condition into the right hand side of theequation, then we get zero, which means that ut is also zero, which means that u should not changein time. Then we would guess that u(x, t) = x+for all t.

Either way, though, we can check thatu(x, t) = x +is a solution to the system. Clearly it satisfiesthe initial conditionu(x, 0) =x +, and moreover it satisfies the PDE (ut = 0 and uxx = 0). So itis a solution. (We will show later in class that it is the onlysolution.)

5. Determine the type of the following equations:

(a) uxx+uxy+uyy + 3uy = 0,

(b) 9uxx uy = 0.

(c) Now, for the equation uxx+ 3uxy+ uyy = 0, determine which values of make the equationelliptic.

Solution:

(a) Using the notation of class (or the book), we have11 = 1, 12 = 1

2, 22 = 1, and thus 1122 = 1

and 212 = 1

4, so the equation is elliptic.

(b) We have11 = 9 and12 = 22 = 0, and thus we have 0 9 = 02, so that the equation is parabolic.

In fact, writing this equation as uy = 9uxx and thinking ofy as time, we have the heat equationexactly.

(c) Here we have 11 = 1, 12 = 3

2. So we compare 1 and (3/2)2 = 9/4. Thus if > 9/4, the

equation is elliptic.

6. We define the operator x by the equation xu = u

x, xy by xyu =

2uxy

, and similarly for other

independent variables. Moreover, when we concatenate operators, we take it to mean composition, i.e.

LM u:= L(M(u)).

(a) Show that2x= xx.

(b) Show that

(x+y)2

=2

xx+ 2xy+2

yy ,i.e. that they multiply like polynomials.

(c) From this, prove that if,, are complex numbers, then there exist1, 2 complex such that

xx+xy+ yy = (x 1y)(x 2y).

Compute1, 2 in terms of,, .

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Solution: We compute the effect of each operator on a function and see that it is the same.

(a) We check:

2xu= xxu= xu

x=

2u

x2 =xxu.

This is true for any function u and we are done.

(b) Now we compute:

(x+y)2u= (x+y)(x+y)u

= (x+y)(ux+uy)

=x(ux+uy) +y(ux+uy)

=2uxx+uyx+uxy+2uyy

=2uxx+ 2uxy+2uyy

= (2xx+ 2xy+2yy)u.

(c) There are a couple of ways to do this, but the most straightforward might be to compute theright-hand side first, and we have

(x 1y)(x 2y)u= (x 1y)(ux 2uy)

= [x(ux 2uy) 1y(ux 2uy)]

= [uxx (1+2)uxy+12uyy ] .

So this will work, if we have

(1+2) = ,

12 =

Solving the second equation for1 gives1 = /2; plugging this into the first gives

2

+2 = ,

or2

2+2+= 0.

(Notice this similarity of this equation and the original left-hand side of the equation.) We thenhave two solutions for this quadratic polynomial, namely

2 =

2 4

2 ,

and its not hard to see that if we pick 2 to be either of these roots, this will make1 the otherone.

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Fall 2009Homework 2 due September 18

1. (Strauss 2.1.1.) Solve utt= c2uxx, u(x, 0) = e

x, ut(x, 0) = sin x.

Solution: We use dAlemberts formula with

(x) =ex, (x) = sin x.

Then we have x+ctxct

sin s ds= cos(x+ct) cos(x ct),

and so we have

u(x, t) =1

2(ex+ct +exct) +

1

2c(cos(x+ct) cos(x ct)).

2. (Strauss 2.1.7.) We define an odd function to be any function f such that f(x) = f(x) for all x.Prove that if that the initial conditions , are odd functions, then so is the solution u(x, t) for anyfixed time t.

Solution: There are two completely different solutions:

Again use dAlembert. We know that the solution is

u(x, t) =1

2((x+ct) +(x ct)) +

1

2c

x+ctxct

sin sds.

Now, we compute

u(x, t) =1

2((x+ct) +(x ct)) +

1

2c

x+ct

xct

(s) ds

Using the fact that is odd, we have

(x ct) = ((x+ct)) = (x+ct),

(x+ct) = ((x ct)) = (x ct).

Now consider the integral, and make a change of variables r = s, giving

x+ct

xct

(s) ds=

xctx+ct

(r) dr=

x+ctxct

(r) dr.

(Note in the last equality there are actually three minus signs, since we use the fact that is odd,andwe flip the domain of integration.) Putting all of this together gives u(x, t) = u(x, t) andthus u is odd.

We showed that reversing time gives a solution to the wave equation, and so does reversing space:if we choose v (x, t) = u(x, t), then we see that

vtt(x, t) =utt(x, t), vxx(x, t) = uxx(x, t),

and thus ifu satisfies

utt= c2uxx, u(x, 0) = (x), ut(x, 0) =(x),

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thenv satisfiesvtt= c

2vxx, v(x, 0) = (x), vt(x, 0) =(x).

This is true for any solution to the wave equation. If we now further assume that, are odd,then these two PDE have the same initial data, and therefore by uniqueness, v (x, t) =u(x, t) forall x, t, and thus u(x, t) = u(x, t), and u is odd.

3. We have defined a well-posed problem in class (also see book) typically for PDE, but we can considerif an ODE satisfies these three properties as well. Here you are given a sequence of ODEs and initialconditions; determine which of these problems are well-posed, and which are not1:

(a) dy

dt = 2y, y(0) = 2,

(b) dy

dx= ln x, y(0) = 0.

Solution:

(a) This ODE satisfies the hypotheses of the E-U theorem for ODEs (in fact, the vector field is C)and thus it is well-posed.

(b) We cannot apply the theorem directly, but we can solve this ODE exactly. In fact, the generalsolution can be written asy(x) =x log xx + Cfor some constantC. We see that there is exactlyone solution with y(0) = 0 (in fact, choose C = 0). To show stability, we need to show that ifwe choose two different initial conditions, we can make the solutions close by choose the initialconditions close. So consider the two solutions

y1(x) =x log x x, y2(x) = x log x x+C.

Clearly, y2(0) =C, and by choosing|y2(0)|< , we can make |y1(x) y2(x)|< for anyx.

4. (Strauss 2.2.2.) Let us consider a solution to the wave equation utt = uxx (we have assumed thatc2 = 1). Define the energy density e(x, t) = 12(u

2t +u

2x) and the momentum density p(x, t) = utux.

Show that

(a) e

t = p

x and p

t = e

x ,

(b) e and p both satisfy the wave equation themselves (although with different initial conditions).

Solution:

(a) We havee

t =ututt+uxuxt,

p

x=utxux+utuxx.

Using utt= uxx shows these are equal. (similar for the other)

(b) We have

ex= ututx+uxuxx,

exx= u2tx+ututxx+u2xx+uxuxxx,et= ututt+uxuxt,

ett= u2tt+ututtt+u

2tx+uxuxtt

1Recall the ExistenceUniqueness Theorem which you saw in ODEs

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Then

exx ett= ututxx ututtt+u2xx u

2tt+uxuxxx uxuttx

=ut(uxx utt)t+u2xx u

2tt+ux(uxx utt)x= 0 + 0 + 0 = 0.

(similar forp)

5. (Strauss 2.2.3.) Show the following invariance properties for solutions of the wave equation. Assumethat u(x, t) satisfies the wave equation, then show that each of the transformed solutions also satisfythe wave equation:

(a) translation: u(x , t) for any,

(b) derivative: ux(x, t),

(c) dilation: u(ax, at) for any a

Solution:

(a) Define v (x, t) =u(x , t). We see, using the chain rule, that

vx

(x, t) = ux

(x , t) ddx

(x ) = ux

(x , t).

Similarly,2v

x2(x, t) =

2u

x2(x , t).

We also work out that2v

t2(x, t) =

2u

t2(x , t).

Thereforevtt(x, t) c

2vxx(x, t) =utt(x , t) c2uxx(x , t) = 0.

(b) Define v (x, t) =ux(x, t). Thenvtt= uxtt, vxx= uxxx,

so

vtt c2vxx= uxtt c

2uxxx= uttx c2uxxx= (utt c

2uxx)x = 0

x= 0.

(c) Define v (x, t) =u(ax, at). Then

v

x(x, t) =

u

x(ax, at) a,

2v

x2(x, t) =

x(a

u

x(ax, at)) = a

2u

x2(ax, at) a= a2

2u

x2(ax, at),

v

t(x, t) =

u

t(ax, at) a,

2v

t2(x, t) =

t (au

t (ax, at)) = a2u

t2(ax, at) a= a2

2u

t2(ax, at).

Thus we have

vtt(x, t)c2vxx(x, t) =a

2utt(ax, at)a2c2uxx(ax, at) = a

2(utt(ax, at)c2uxx(ax, at)) = a

2 0 = 0.

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6. (Strauss 2.3.3.) Consider a solution to the diffusion equation ut = uxx for x [0, L] and t > 0.Define

M(T) = maximum ofu(x, t) on the rectangle [0, L] [0, T],

m(T) = minimum ofu(x, t) on the rectangle [0, L] [0, T].

Does M(T) increase or decrease as a function ofT? Does m(T) increase or decrease as a function ofT? Explain why.

Solution: It turns out the answer to this is somewhat complicated as it could depend on the boundaryand initial conditions. Let us first fix the boundary conditions as

u(0, t) = 0, u(L, t) = 0,

for all t and assume that the initial condition (x) is positive for some x [0, L] so that M(0) > 0.The claim here then is that M(T), m(T) are functions which are constant in T.

First, notice that by definition, ifT > T, thenM(T) M(T) (since were taking the maximum overa larger set).

We now want to prove that M(T) M(T), and then we are done. So we prove by contradiction:

assume thatM(T

)> M(T). If this is so, then clearly the maximum inside the rectangle [0, L] [0, T

]must occur for t (T , T]. Since M(T)> M(T) M(0)> 0, this means that this maximum may notoccur on the left- or right-hand edges, but must occur either in the interior of the rectangle, or on thetop edge. But this directly contradicts the Maximum Principle. Since assumingM(T)> M(T) leadsto a contradiction, it must be true that M(T) M(T).

The argument for m(T) is similar: reverse every inequality above and use the Minimum Principle.

Of course, with different boundary conditions things could be more complicated. If we now assumethat u(0, t) = f(t), and this function is increasing (rapidly enough), then its possible that M(T)increases, since M(T) maxt[0,T] f(t) at least. In this case, the final statement would be that M(T)can increase, but no faster than f. Other permutations are left to the reader...

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1. Here we will prove that solutions to the heat equation satisfy (some of) the invariance principles

mentioned in class, or in the book in 2.4. That is, ifu(x, t) is a solution to ut = kuxx for x R, t > 0,then so are

(a) u(x y, t) for any fixed y ,(b) ux, ut,

(c) v(x, t) = u(x y, t)g(y) dy whereg has finite support,

(d) v(x, t) =u(

ax,at) for any a >0.

Solution:

(a) Letv(x, t) = u(x y, t). Thenv

t (x, t) =

u

t (x y, t) 1 =u

t (x y, t),v

x(x, t) =

u

x(x y, t) 1 = u

x(x y, t),

2v

x2(x, t) =

x

v

x(x, t) =

x

u

x(x y, t) =

2u

x2(x y, t).

Thenv

t(x, t) k

2v

x2(x, t) =

u

t(x y, t) k

2u

x2(x y, t) = 0,

since u solves the heat equation.

(b) We compute forux, the other is similar. Denoting v = ux gives

vt = (ux)t = uxt,

vxx= (ux)xx= uxxx.

Thenvt kvxx= uxt uxxx= utx uxxx= (ut uxx)x= 0x= 0.

(c) Since g has compact support, we can exchange derivatives and integration (see e.g. TheoremA.3.2 from Strauss), and thus we have

v

t =

t

u(x y, t)g(y) dy=

u

t(x y, t)g(y) dy

and2v

x2

= 2

x2

u(x

y, t)g(y) dy=

2u

x2

(x

y, t)g(y) dy.

But then

vt kvxx=

u

t(x y, t)g(y) dy

2u

x2(x y, t)g(y) dy

=

u

t(x y, t)

2u

x2(x y, t)

g(y) dy=

0 dy= 0.

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(d) We have

v

t(x, t) = a

u

t(

ax,at),

v

x(x, t) =

a

u

x(

ax,at),

2

vx2

(x, t) = a 2

ux2

(ax, at).

So then

vt kvxx= aut(

ax,at) auxx(

ax, at) = a(ut(

ax,at) uxx(

ax,at)) = a 0 = 0.

2. (Strauss 2.4.1.) Solve the heat equation with initial condition

(x) =

1, |x| < L,0, |x| L.

(You can use the formula for the solution as derived in class, but there is a simpler way to build thissolution using the invariance principles above.)

Solution: We will solve two ways, the first using the formula. We have

u(x, t) =

S(x y, t)(y) dy,

where

S(x, t) = 1

4ktex

2/4kt.

We can then write

u(x, t) = L

S(x

y, t)(y) dy+

L

LS(x

y, t)(y) dy+

L

S(x

y, t)(y) dy

=

L

0 dy+

LL

S(x y, t) dy+L

0 dy

=

LL

S(x y, t) dy,

so we need to evaluate1

4kt

LL

e(xy)2/4kt dy.

Changing variables with s = (x y)/4kt,ds= 1/4ktdy, gives

u(x, t) = 14kt

xL4ktx+L4kt

es2

(4kt) ds

= 1

x+L4kt

xL4kt

es2

ds

=1

2

erf

x + L

4kt

erf

x L

4kt

.

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A completely different method is to use the transformations in Question #1, particularly (a). We knowthat if the initial condition is the Heaviside function H(x), then we get the solution

Q(x, t) =1

2+

1

x/4kt0

ep2

dp=1

2

1 + erf

x

4kt

.

We can see that our initial condition (x) can be written(x) = H(x + L) H(x L).

(One can either draw this, or check it algebraically: if x > L, then H(x+ L) = H(x L) = 1, ifx < L, then H(x + L) = H(x L) = 0, and ifL < x < L, then H(x L) = 0 but H(x +L) = 1.)From 1(a) above, we know that Q(x+L, t) and Q(x L, t) are both solutions to the heat equation,and clearly they have initial conditions H(x+L) and H(x L), respectively. By linearity, we knowthat

u(x, t) =Q(x + L, t) Q(x L, t) = 12

+1

2erf

x + L

4kt

1

2 1

2erf

x L

4kt

is also a solution, and it clearly satisfies u(x, 0) = (x).

3. (Strauss 2.4.8.) Show that the tails of

S(x, t) = 1

2

ktex

2/4kt

are uniformly small for small times, i.e. that for any >0,

limt0

max|x|>

S(x, t) = 0.

Interpret this in terms of speed of propagation of information for solutions of the heat equation.

Solution: We first compute the inner term, namely

max

|x|>S(x, t).

First, note that for x >0 and any fixed t >0, S(x, t) is monotone decreasing, because

S

x(x, t) =

2x(4kt)3/2

ex2/4kt

S(x, t) = S(, t) = 1

4kte

2/4kt.

By evenness ofS, we also have that maxxS(x, t), so

max|x|>

S(x, t) = 1

4kte

2/4kt.

So it remains to compute

limt0

e2/4kt

4kt

.

This is an indeterminate form, since when we plug in t = 0 we obtain 0/0. The first guess mightbe to try lHopitals rule, but this will actually not work out, because every time we differentiate thenumerator, we will get a higher power oft in the denominator.

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To simplify the expression, rewrite this limit as

limt0

eC1/t

C2

t

and make the change of variabless = 1/t, which then gives

lims

seC1sC2

= lims

sC2eC1s

.

This is still an indeterminate form of/, but now using lHopitals Rule gives

lims

s1/2/2C1C2eC1s

= 0

= 0.

4. (Strauss 2.4.9.) We will write down an exact solution to the heat equation

ut= kuxx,

u(x, 0) = x2

,

but not using the formula derived in class. The idea is as follows.

(a) Show thatuxxx solves the heat equation with initial condition zero,

(b) Use uniqueness to showuxxx(x, t) 0,(c) From this we can deduce thatu(x, t) = A(t)x2 + B(t)x + C(t) for some functionsA, B,C (Why?),

(d) Solve for A, B,C.

Solution:

(a) This part is similar to problem #1. If we writev = uxxx, then

vt = uxxxt= utxxx,vxx= uxxxxx,

and thusvt kvxx= utxxx kuxxxxx= (ut kuxx)xxx= 0xxx= 0.

Moreover, notice that v(x, 0) = uxxx(x, 0) = (x2)xxx= 0.

(b) We know solutions to the heat equation are unique. Moreover, we know that v solves the heatequation withv(x, 0) = 0. However, it is easy to see that if we define w(x, t) 0 for all x, t, thenw satisfies the heat equation and w(x, 0) = 0. Therefore v w and v(x, t) 0.

(c) We know that

uxxx(x, t) = 0,

uxx(x, t) =A(t),

ux(x, t) =A(t)x+ B(t),

u(x, t) =1

2A(t)x2 + B(t)x + C(t),

whereA(t), B(t), C(t) are arbitrary functions oft. Now redefineA to get rid of the 1/2 since itsarbitrary anyway.

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(d) We know thatu satisfies the heat equation, so we have

ut = A(t)x2 + B(t)x+ C(t),

uxx= 2A(t),

and if these are equal as functions, this gives

C(t) = 2A(t), B(t) = 0, A(t) = 0.

Solving the last two are easy (A(t) =A0,B(t) =B0) and then the first becomes

C(t) = 2A0t+ C0.

Putting this together gives

u(x, t) = A0x2 + B0x + (2A0t + C0).

Plugging in the initial condition gives

u(x, 0) = A0x2 + B0x + C0,

which means thatA0 = 1, B0= C0 = 0, so the solution is

u(x, t) =x2 + 2t.

We could, alternately, plug in the initial conditions as soon as we have the equations, namely saythat

A(t) = 0, A(0) = 1,B(t) = 0, B(0) = 0,

C(t) = 2A(t), C(0) = 0,

and directly solve to get A(t) = 1 and C(t) = 2t.

5. Generalize the previous problem to a general initial condition which is a polynomial ofx. (You dontneed to compute anything exactly here, just describe the algorithm which would allow you to obtaina solution.)

Solution: The general idea is as follows. Lets say thatu(x, 0) = p(x), wherep(x) is a polynomial ofdegreen:

p(x) =

nk=0

kxk.

Then notice that if we taken + 1 derivatives ofp we get zero. Therefore, ifu satisfies the heat equation

with initial condition p(x), then n+1u

xn+1 solves the heat equation with initial condition zero. Therefore

we have n+1uxn+1

0, and by the same argument we know

u(x, t) =

nk=0

Ak(t)xk.

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Plugging this into the heat equation gives

ut(x, t) =n

k=0

Ak(t)xk,

uxx(x, t) =n

k=0

k(k

1)Ak(t)x

k2 =n2

k=0

(k+ 2)(k+ 1)Ak+2(t)xk.

Setting these equal gives

An(t) =An1(t) = 0, A

k(t) = (k+ 2)(k+ 1)Ak+2(t),

and plugging in initial conditions gives Ak(0) = k for all k. The first two equations can be solvedeasily:

An(t) = n, An1(t) = n1,

and then the other equations can be solved recursively, e.g.

An2(t) = n(n 1)An(t) =n(n 1)n,

soAn2(t) = n(n 1)nt+ n2,

etc.

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Fall 2009Homework 4 due October 9

1. Consider the boundary value problem

A + A= 0, A(0) + aA(0) = 0, A(L) = 0.

(a) Show that ifa 0, we know e2bL > 1, and thus1 e2bL 0.

But sinceb(1 + e2bL

)> 0 as well, it is not possible to solve (1).(b) If we have a zero eigenvalue, this means we have a solution to A(x) = 0 with those boundary

conditions. However, this means that A(x) = x + , and plugging this into the boundaryconditions gives

+ a= 0,

L + = 0.

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We want to find a nontrivial solution to this system, and if we write it as a matrix equation: 1 aL 1

=

00

,

and this has a nontrivial solution iff the matrix is singular, i.e. if its determinant is zero. So thecondition is that

det

1 aL 1

= 1 aL= 0,

or aL = 1.

(c) If >0, then the solutions to the ODE look like

C1cos(x) + C2sin(x),

where2 =. Plugging in the boundary conditions gives

C2+ aC1 = 0,

C1cos(L) + C2sin(L) = 0.

Solving the first equation givesC2 =

aC1/, and plugging this into the second and doing some

algebra givestan(L) =

a.

The question is, how many roots does this equation have? We cannot answer this analytically, butwe can see from the graph that it has to have infinitely many; see an example picture in Figure 1.

5 5

15

10

5

5

10

15

Figure 1: Graphs of tan(x) and x

To prove this analytically, choose any nonnegative integer k and consider the interval [(k +1/2)/L, (k+ 3/2)/L]. We know from properties of tan that

limt(k+1/2)L +

tan(x) = , limt(k+3/2)L

tan(x) = ,

so for x slightly larger than (k+1/2)/L, tan(x)< x/a and for x slightly smaller than (k+3/2)/L,tan(x) > x/a. By the Intermediate Value Theorem, the functions must be share at least onepoint in the strip. Since the curves intersect in each of these strips, there are infinitely manyintersections.

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Bonus: Repeating some of part (a), we know we need to solve (1) for b, given a positivea, so we have

tanh(2bL) =a

b.

These curves will always intersect as long asa >0. Notice that tanh(0) = 0 and limb tanh(2bL) = 1,whereas limb0+ a/b=

and limb a/b= 0, so the curves must cross. So the condition is that for

anya >0, there is a negative eigenvalue.

2. (Strauss 4.3.2.) Consider the eigenvalue problem with Robin boundary conditions

A + A= 0, A(0) 0A(0) = 0, A(L) + LA(L) = 0.

(a) Show that zero is an eigenvalue if and only if0+ L= 0LL.(b) Compute the eigenfunction corresponding to this eigenvalue.

Solution.

(a) If we have a zero eigenvalue, then we haveA = 0 or

A(x) =x+ .

Plugging in the boundary conditions gives

0= 0,+ L(L + ) = 0.

We want to find a nontrivial solution for , in this equation, or, as above, we need the matrix 1 0

1 + LL L

to have determinant zero, orL+ 0(1 + LL) = 0.

(b) If this determinant is zero, we know that the two equations we have are redundant, so we cansolve either. The simpler to solve is the first, which gives = 0, and of course we will haveone free choice for . So one eigenfunction we can choose is

A(x) = 0x + 1,

and we can of course choose any scalar multiple of this.

3. Solve the equation

ut = kuxx, x [0,), t > 0,

u(x, 0) =

1, x (0, 1),0, x > 1,

u(0, t) = 0.

Solution.We use the formula as derived in class (equation (6) in3.1):

u(x, t) = 1

4kt

0

e(xy)

2/4kt e(x+y)2/4kt

(y) dy,

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Solving the B equation givesBn(t) = Bn(0)e

int,

which oscillates. Notice of course that we can allow Bn(0) to be complex with no extra difficulties.

Thus we haveun(x, t) = Cne

int cosn

Lx ,

and we form a general solution by linear combinations:

u(x, t) =n=0

Cneint cos

nL

x

.

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1. (Strauss 5.2.2.) Show that cos(x) + cos(x) is periodic if is a rational number and compute its

period. What happens if is not rational?Solution: Let us first notice that iff , g are both periodic with period p, then so is their sum:

(f+ g)(x+p) = f(x+p) +g(x+p) = f(x) +g(x) = (f+ g)(x).

So we simply have to show that cos(x), cos(x) share a period. Note that the periods of cos(x) are2n with n Z, and the periods of cos(x) are 2m/, where m Z. The question is then, do thesetwo sets of numbers share an element, i.e. is there an n and an m so that

2n=2m

?

If = p/q, then we have

2n=2mq

p , or n =

mq

p .

This has many solutions, choose, for example, m = p, n= q.

If is irrational, then this does not work; in fact, solving the first equation gives

=m

n

which is only possible if is rational.

2. Definef(x) = x3 on the interval [0, 1]. Compute its Fourier sine series and its Fourier cosine series.

Solution: The Fourier sine series coefficients are given by

Bn= 2

L L0

f(x)sin(nx/L) dx,

and in this caseL = 1, so we need to compute

Bn= 2

10

x3 sin(nx) dx.

Integrating by parts several times gives

Bn=(1)n2(6n22)

n33 ,

and then the F.S.S. is

n=1Bnsin(nx).

The Fourier cosine series coefficients are given by

An= 2

10

x3 cos(nx) dx.

Integrating by parts several times gives

An=6 + (1)n(3n22 6)

n44 ,

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and then the F.C.S. isA02

+n=1

Ancos(nx).

3. Consider the function f(x) =x on the interval [, ]. Compute the full Fourier series for f(x). UseParsevals Identity to compute

n=1

1

n2.

Solution: Sincefis odd, we know all of the cosine terms will be zero, so we compute

Bn= 1

x sin(nx) dx=2(1)n+1

n .

Parsevals Identity tells us that

n=1

B2nsin(nx), sin(nx)= f, f ,

orn=1

4

n2=

x2 dx=23

3 .

Dividing givesn=1

1

n2 =

2

6 .

4. Solve the heat equation given by

ut= kuxx, x [0, L], t > 0,u(x, 0) = x,

u(0, t) = u(L, t) = 0.

Solution: We make the Ansatzu(x, t) = A(x)B(t),

and plugging in this gives usA(x)B(t) = kA(x)B(t),

orA(x)

A(x) =

B(t)

kB(t)=.

Thus we first solveA(x) +A(x) = 0, A(0) =A(L) = 0,

and we know the solutions of this are

An(x) = sin(nx/L), n =n22

L2 .

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Then we haveBn= Cne

knt,

and our general solution is

u(x, t) =n

Cneknt sin(nx/L).

Plugging int = 0 gives

x=n

Cnsin(nx/L),

and thus Cn are the Fourier sine series coefficients ofx, or

Cn= 2

L

L0

x sin(nx/L) dx=2(1)n+1L

n .

Therefore we have

u(x, t) =n

2(1)n+1L

n ekn

22t/L2 sin(nx/L).

5. (Strauss 5.3.8.) Let f and g satisfy the same Robin boundary condition at x = 0 and the sameRobin boundary condition at x = L (i.e., we assume that

f(0) +f(0) =g(0) +g(0) =f(L) +f(L) = g (L) +g(L) = 0.)

Prove then that(f(x)g(x)f(x)g(x))|

x=Lx=0 = 0.

Deduce from this that eigenfunctions of a Robin BVP are orthogonal.

Solution: We have

f(0) =f(0), g(0) =g(0), f(L) = f(L), g(L) = g(L).

Plugging this into the left hand side gives

f(L)g(L) f(L)(g(L))(f(0)g(0)f(0)(g(0))) = 0.

6. Prove that iffhas period p, then p+aa

f(y) dy

is independent ofa.

Solution: One way to do this is to write p+aa

=

p0

a0

+

p+ap

, (1)

and we have that p+ap

f(y) dy=

a0

f(yp) dy=

a0

f(y) dy,

(the first is from the u-substitutionz = y+p and the second is from periodicity) and thus the secondand third terms in (1) cancel.

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7. Consider the infinite list of functions

{1, cos(x), cos(2x), . . . , cos(nx), . . . , sin(x), sin(2x), . . . , sin(nx), . . . }.

Show that this is an orthogonal set of functions on the set [, ], i.e. if we define the inner product

f, g:=

f(x)g(x) dx,

then if we choose any two different functions from that list, then their inner product is zero.

Solution: We first notice that

sin(nx)cos(mx) dx= 0

since sin is odd and cos is even and thus their product is odd, and the integral of an odd function ona symmetric interval is always zero.

Now, consider

sin(nx)sin(mx) dx=

cos((nm)x)cos((n+m)x) dx

= sin((nm)x)nm

+sin((n+m)x)n+m

x=x=

,

as long as n =m, and this is clearly zero by periodicity of sin. We also have

cos(nx)cos(mx) dx=

cos((nm)x) + cos((n+m)x) dx

= sin((nm)x)

nm

sin((n+m)x)

n+m

x=

x=

,

as long as n =m, and this is again zero by periodicity of sin.

Finally, to compute the inner products of the functions with themselves, choose n >0, and notice that

sin2(nx) + cos2(nx) = 1,

and thus

sin2(nx) + cos2(nx) dx= 2.

Moreover, since sine and cosine are the same function under a phase shift, we know from the previousproblem that

sin2(nx) =

cos2(nx)

and therefore they each equal . For n = 0 we have

cos2(0x) dx=

dx= 2.

8. Let{fn(x)} be any sequence of functions which converge to f(x) uniformly on [a, b]. Prove then thatfn(x) converge to f in the L2 sense as well. Show a counterexample to demonstrate that the converseis false, i.e. that we can haveL2 converge but not uniform. (The term used for this is that uniformconvergence is strongerthan L2).

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Solution: To prove the first statement, notice that

ba

|fn(x)|2 dx

ba

maxx[a,b]

f(x)

2

dx=

maxx[a,b]

f(x)

2

(ba).

Then if

limn maxx[a,b] f(x) = 0,

then clearly

limn

ba

|fn(x)|2 dx (ba)

limn

maxx[a,b]

f(x)

2= 0

as well. Also, notice that it is nonnegative by definition, and therefore the limit is zero.

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1. LetD = {(x, y) : x2 +y2 :D2?321331WRbRKK q < 91 >< 9C > :G m N; < o 0 with initial condition u(x, 0) = x(5x)and boundary conditions u(0, t) = u(5, t) = 0.

(a) Use the discretization scheme we defined in class (forward difference in time, second centereddifference in space) with x = 1, t= 1/4. Compute two time steps forward (i.e. compute thesolution att = 1/2).

(b) Do the same, except now choose t= 1/8. Compute forward four steps, again computing untiltimet = 1/2.

(c) Now set x= 1/2, and return t to 1/4. Compute two steps forward.

(d) Compare all of the answers obtained above; explain your observations.

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Solution: In all cases, our discretization will be of the form

un+1k = (1 2)unk + (u

nk+1+ u

nk1),

where = t/(x)2.

(a) For this case we have = 1/4, so our scheme is

un+1k =12

unk +14

(unk+1+ unk1).

Our first row is given by 0, 4, 6, 6, 4, 0. The next row up will be given by

0, 3.5, 5.5, 5.5, 3.5, 0

(notice the left and right edges are zero because of BC!), and finally the third row is

0, 3.125, 5, 5, 3.125, 0.

The rest of the cases are similar, except more work. In the other two cases we have = 1/8 and = 1,respectively. You should observe that the first two cases work reasonably, but the third case showssigns of instability.

5. (Strauss 8.2.11.) Write down a discretization scheme for ut = auxx + bu where we use forwarddifference in time, and centered second difference in space. Define = t/(x)2 and find the conditionon for this scheme to be stable. Solution: Using the difference schemes, we have

un+1k unk

t =a

unk+12unk + u

nk1

(x)2 + bunk ,

and solving givesun+1k = (1 2a + bt)u

nk + a(u

nk+1+ u

nk1).

As for stability, we do the standard argument, takeunk =XkTn, and we obtain

(12a + bt) + aXk+1

Xk+

Xk1

Xk =

andTn+1 = Tn, so for stability we need ||< 1. We make the Ansatz

Xk = (ei(x))k,

and furthermore notice that in the limit of small grids the tterm disappears, and we have

= (12a) + a(ei(x) + ei(x))

= (12a) + 2a cos((x))

= 12a(1cos((x))).

Since the 1cos() term is always positive, we cannot have >1, but we could have


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Math 110B: Homework 1

April 11, 2012

All problems are from Strauss, Partial Differential Equations: An Introduction, second edition.

1 6.1.13

A functionu(x) is subharmonicin D if u 0 in D. Prove that its maximum value is attained onbdy D.

Proof. We can complete this exercise by following the proof of the maximum principle for harmonicfunctions on page 155 with a slight modification. Fix > 0 and define v(x) = u(x) +|x|2.Computing in two dimensions as in the text,

v= u + (x2 + y2) 0 + 4 >0,

the only difference being the first sign because our function is subharmonic instead of harmonic.From here, we follow the proof in the text exactly to prove the result.

2 6.2.1

Solve uxx+ uyy = 0 in the rectangle 0 < x < a, 0< y < b with the following boundary conditions:ux= aon x= 0, ux= 0 on x= a, uy =b on y = 0, and uy = 0 on y= b.

Proof. The hint that suggests guessing that the solution is a quadratic polynomial in x and y , i.e.

u(x, y) =Ax2 + Bxy + Cy2 + Dx + Ey + F.

We have the partial derivativesux(x, y) = 2Ax + By + D

uy(x, y) =Bx + 2Cy+ E.

Now we use the boundary conditions to solve for the constants. Since

a= ux(0, y) =By+ D

must hold for all values ofy, we have B= 0 and D = a. Using

0 =ux(a, y) = 2Aa + By + D= 2Aa a,

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we find thatA= 1/2. Similarlyb= uy(x, 0) =E. Solving 0 =uy(x, b) = 2Cb + b gives B =1/2.Hence

u(x, y) =1

2x2

1

2y2 ax + by+ F

solves the equation for any constant F.

3 6.2.3

Find the harmonic function u(x, y) in the square {0 < x < , 0 < y < } with the boundaryconditions uy = 0 for y= 0 and for y = , u = 0 forx = 0, and u = cos

2y forx= .

Proof. We separate variables u(x, y) =X(x)Y(y), and so X/X+ Y/Y = 0. Hence there existsa constant such that Y +Y = 0 on 0 < y < and X X= 0 on 0< x < . We chose as such since the boundary conditions are homogeneous in y. This way, will be a non-negativeeigenvalue and we will be able to find the eigenvalues and eigenfunctions as in the previous chapter.

Y + Y= 0 has solutions of the form

Y(y) =C y+ D

if= 0 andY(y) =A cos(y) + B sin(x)

where2 = if >0.If = 0,Ysolves the ODEY = 0. Integrating twice givesY(y) =C y +D. Using the boundary

condition for y , 0 =Y(0) =Cshows that the first eigenvalue is 0 with the constant eigenfunctionY(y) =D.

If >0, 0 = Y (0) =A sin(0) +B cos(0) = B and hence B = 0. The second boundarycondition

0 =Y() =A sin()

implies that = 1, 2, 3, . . .. Hence we have eigenvaluesn= n2 with eigenfunctionsYn(y) = cos(nx)for n = 0, 1, 2, . . . .

Now we solve the ODE for X. Whenn = 0, we have X = 0 and so X(x) = Cx +D.

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