UNIVERSITY OF WISCONSIN-MADISON

Math 619 - Spring 2015Prof. Andrej Zlatos

Daniel Howard

February 27, 2015

1 STRAUSS - 2.3.3

PART (A) By the strong maximum principle, u(x, t )> 0 is in the interior points 0< x < 1 and0 < t < inf since the minimum value of u, i.e. 0, is attained at the boundary point and at thetwo sides.

PART (B) Following the hint, let (t ) be the maximum of u(x, t ) on 0 x 1 and X (t ) [0,1]such that (t )= u(X (t ), t ).

On the two lateral sides, we know from part a that u = 0 and thus (t )> 0. So we have thatX (t ) (0,1) and from the proof of the maximum principle in Struass:

ux(X (t ), t )= 0 ut (X (t ), t ) 0

When we differentiate with respect to t ,

(t )= uxX (t )+ut = ut 0

and is decreasing in t > 0.

2 STRAUSS - 2.3.4

PART A The maximum value of u(0,x)= 4x(1 x) occurs at x = 1/2 where u(0,1/2)= 1. Theminimum value 0 occurs at both x = 0 and x = 1. We see that the boundary values are 0, sothe strong maximum principle implies that 0< u(t ,x)< 1 for all t > 0 and 0< x < 1.

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PART B Suppose we have a solution u(x, t ). Consider v(t ,x) = U (t ,1 x). We know thatvt = vxx . Also, v(t ,0)= v(t ,1)= 0 and v(0,x)= u(0,1x)= 4(1x)(x)= u(0,x).

By uniqueness, u(t ,x)= v(t ,x)= u(t ,1x).

PART C By the energy method, we have

d

dt

1

2

1xu2dx =k

10u2xdx.

We see that the energy is strictly decreasing as long as 10u2xdx > 0.

Now suppose there exists some t0 such that 10ux(t0,x)

2dx = 0.

This would imply that ux(t0,x) = 0 for all x, and u(t0,x) is constant. Given our boundaryconditions, this constant must be 0. However, this contradicts Part A, where we know 0 0 and 0< x < 1.

3 STRAUSS - 2.3.7

PART A We have for k > 0 and f gut kuxx = f (3.1)vt kvxx = g (3.2)

and u v at x = 0,x = l , and t = 0.Let w = u v . At the initial condition boundaries,

w < 0 f g wt kwxx 0Thus, w 0 on x = 0 and x = l for all t .

Define w =w +et ,> 0.w xx =wxx ,w t =wt et

Thus, w t kw xx =wt et kwxx < 0.On the boundary where t = 0, w e0 = . Thus,

w t kw xx < 0 (3.3)w | (3.4)

If w has a max at (t0,x0) and 0 < x0 < l , t0 > 0, then w t 0,w xx 0 at (t0,x0), which is acontradiction with 3.3.

It follows that w (t ,x) on 0 x l , t 0. This implies that w +et andw(t ,x) (et +1)

for all . As 0w(t ,x) 0 on 0 x l , t 0.

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PART B If vt vxx sin(x) for 0 x pi,0 < t T

6 STRAUSS - 2.4.9

Since uxxx(t ,x) satisfies the same equation with u(x,0)= 0, uniqueness of solutions impies

uxxx(x, t )= 0.

Integrating with respect to x, we have

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

We pug this into ut = kuxx and we get

A(t )x2+B (t )x+C (t )= 2kA(t ).

Matching coefficients, we see A(t ) = 0, B (t ) = 0, and C (t ) = 2kA(t ). Thus we have theconstants A(t )= A,B(t )=B andC (t )= 2kAt +C0.

ooking at t = 0, we have

x2 = Ax2+Bx+C (0)= Ax2+Bx+C0

Matching coefficients again, we see A = 1,B = 0, andC0 = 0. Therefore,

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

7 STRAUSS - 2.4.11

PART A For the diffusion equation, we have ut = kuxx . Suppose u(x, t ) is a solution of theequation. We can replace x withx and we see that u(x, t ) is also a solution of the diffusionequation since the second derviative in x cancels out the negative seen via the chain rule.

If the initial condition u(x,0) = (x) is odd, then both u(x, t ) and u(x, t ) solve theseinitial conditions and the diffusion equation. Since the initial value problem has a uniquesolution, u(x, t )=u(x, t ), which implies that u(x, t ) is odd for all of t .

PART B For the diffusion equation, we have ut = kuxx . Suppose u(x, t ) is a solution of theequation. We can replace x withx and we see that u(x, t ) is also a solution of the diffusionequation since the second derviative in x cancels out the negative seen via the chain rule.

If the initial conditionu(x,0)=(x) is even, then bothu(x, t ) andu(x, t ) solve these initialconditions and the diffusion equation. Since the initial value problem has a unique solution,u(x, t )= u(x, t ), which implies that u(x, t ) is even for all of t .

PART C The argument is the same for the wave equation since ut t = uxx has the same sec-ond derivative in space x and the derivative in time t does not affect the even or odd natureof the initial conditions and solution.

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8 STRAUSS - 2.4.16

We solve ut kuxx +bu = 0 for < x 0. Set v(x, t )= ebtu(x, t ).Then

vt = bebtu+ebtut (8.1)vxx = ebtuxx (8.2)

(8.3)

Thus,vt kvxx = ebt (bu+ut kuxx)= 0

We also have the same intial conition since v(x,0) = (1)u(x,0) = (x). Applying the solutionformula gives

u(x, t )= ebtv(x, t ) (8.4)

= ebt

p4pikt

e(xy)2

4kt (y)dy. (8.5)

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TODODid not give myself enough time for these last two problems this past week.

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TODODid not give myself enough time for these last two problems this past week.

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