## Homework 14 collaboration area

Question Page 560, Problem 11:

Are the boundary conditions for this problem u(0,t)=U1 and u(L,t)=U2 or u'(0,t)=U1 and u'(L,t)=U2? I think it is the u' option, but I'm not sure why?

Answer: The problem says that it is fixed at U1 and U2 for all time. I interpret this as X(0)=U1 and X(L)=U2, and X'(0)=X'(L)=0 since they are not changing at those points. Apply these to the X equation after you separate your variables. The T equation will come out the same as before.

RESPONSE: I don't think that X'(0)=X'(L)=0 is correct as problem 10 only mentions that the temperatures are fixed. Making the gradients zero means that the ends are insulated. Now we can have a fixed temperature even if flux is non-zero (if flux in is same as flux out). The part that i am confused about is that while solving X equation we will have non-zero solutions for all values of lambda .. so do we solve three general cases according to lambda values. But the solution at the book's back only mentions a single solution !!.

Hint from Bell: That response is right on. First you need to find the steady state solution to the problem. (If you can't guess that, you can note that the derivative in t would be zero at steady state, so the heat equation becomes u"(x)=0 and the problem is a simple ODE problem.) The key to solving the more general problem is to subtract the steady state solution from the solution you are after. That difference will satisfy homogeneous (=zero) boundary conditions and you can use the formulas from the book to write the difference. (Note: be careful about what the intitial conditions are for the difference.)

Question Page 585, Prob 6:

Can anyone provide some direction on how to start this problem? I'm not really sure how to get started on it.

Answer: I'm not sure about part b, but for part a, differentiate the two given solutions and plug them into formula five to verify that they're solutions. In part c, you're given f(theta), and to solve, you find the fourier coefficients and plug them into (20). Since f(theta) is odd, this is pretty simple.

Any thoughts on part b of Page 585, Problem 6?

Question Page 568, Prob 2:

What are the limits of integration for A(p) and B(p) in this problem. I think it is -infinity to infinity, but I'm not sure if this is correct.

Answer: see page 508 for a Fourier Integral refresher. You are right that it is -inf to inf. Sometimes the integrals are zero because the functions inside are odd, and sometimes you can reduce an integral from minus infinity to infinity to two times one from zero to infinity if the function inside is even.

Question Page 562, Prob 31: I am confused regarding the boundary conditions as the problem says that the faces are insulated. However the sides are at 0. So should the B.Cs be X(0)=0, X(24)=0,Y(0)=0 or X'(0)=0, X'(24)=0, Y'(0)=0?

I actually see this problem as being over-specified. Per what the problem says, Y(0)=0, Y'(0)=0, Y'(a)=0, X(0)=0, X(a)=0, X'(0)=0, and X'(a)=0 in addition to the fact that u(a,a) = 20. To me that looks like eight boundary conditions when only four are needed. --Bryce

Another Question Page 562, Problem 31: What's with the 2n-1 format of the solution? Why isn't it left in n*pi*x/a format? Does this have something to do with the sinh terms? Am I missing something when evaluating the X terms, because having lambda = n*pi/a as in Bell's example seems to satisfy the conditions just fine to me. Why would we need odd multiples of pi?

Answer: When you solve for An* all even n terms are 0, so only odd n terms are non-zero. Hence the 2n-1 term for the U(x,y) solution.

Question Page 560, Prob 7: I am having an issue getting the solution in the back of the book. When I evaluate the integral for Bn using integration by parts, I get Bn = 4/(n^2 pi*2) * sin(n*pi/L). For even values of n, Bn is zero, so I'm not sure where the second term in the answer in the book comes from. Anybody know what I'm doing wrong?

RESPONSE: You probably are not integrating properly. In integration by parts 'uv' will give a 'cos' term and the vdu will give a sine term. (Cos(n*pi/2) is not always zero)

## Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood