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Show all of your work. Full credit may not be given for an answer alone. You may use the backs of the pages for scratch work. Do not unstaple or remove pages.

This is a non-calculator exam. One side of one sheet of US letter-size paper on which definitions, theorems, and formulas are handwritten by you may be used. Worked-out examples are not allowed on the formula sheet. You will turn in your formula sheet with the exam.

By taking this exam you are agreeing to abide by these rules and Rutgers University's Academic Integrity Policy.

 Problem Possible Points Number Points Earned 1 10 2 10 3 15 4 15 5 15 6 10 7 25 Total 100

1. (10 points) Show that the function

is harmonic (i.e., ) where it is defined.

Proof. [Solution]This is easy if you remember the derivative of arctan, and I make no apologies for using this function. We have

Cleary

2. (10 points) Formulate a boundary value problem (partial differential equation, boundary conditions, and initial conditions) for the motion of an elastic string of length , fastened at the left end, and attached with a ring to a pole at the right end. The ring allows the string to slide up and down the pole, so that end is free. The initial position of the string is a function .

Just set up the problem and stop!

Proof. [Solution]We expect the string to satisfy the wave equation. So the displacement will be

for some constant , for all , . The fact that the left end is fixed gives us

for all , and the fact that the right end is free gives us

for all . Initial conditions are

for . I did not mention initial velocity; in general we should have

for some function and .

3. (15 points) Solve the following initial value problem for the wave equation in an infinitely long string:

Proof. [Solution]Since we have an infinitely long string, Fourier methods do not apply. Instead, we have to use the d'Alembert methods of Section 16.4. We have

Using our data , , we have

Now

So

4. (15 points) Solve the following boundary-value problem for the wave equation:

Proof. [Solution]Again, I said nothing about initial velocity; most assumed wisely that it was zero. We know that the general solution is

where

Now as we have done in homework and lecture several times:

Likewise

So

You could go on, but why?

5. (15 points) Solve the following boundary-value problem for the potential equation:

Proof. [Solution]Notice that we have zero on all three sides of the rectangle except the top. Thus our general solution is (remember )

where

But since , we have without even doing the integral the result

So the series has only a single term:

6. (10 points) Consider heat conduction in a thin rod with insulated ends:

Remember is a positive constant. Show that the integral

is a strictly decreasing function of as long as .

Hint 1   Show .

Proof. [Solution]I should have given the hint not to solve the problem for ! Instead, just differentiate under the integral sign, apply the differential equation, and integrate by parts.

Still, many of you tried to go the route of solving for directly. You get

where the are some constants determined by initial data. (Yes, we know that they are the Fourier cosine coefficients of the initial position, and is actually off by a factor of , but none of that matters.) Then

Therefore

7. (25 points) We will solve the following boundary-value problem for the heat equation:

1. (5 points) Transform the problem into , where satisfies the heat equation with homogeneous boundary conditions:

What differential equation and boundary conditions must satisfy? What initial conditions must satisfy?

Proof. [Solution]A word to the wise: read the problem and answer what it asks. Many people missed the chance for points because they rushed on to part (c) without reading the directions.

All we have to do is apply both sets of equations to the identity

Since and , we have

Since , we must have

Since and , we must have

Since , we must have

2. (5 points) Solve for .

Proof. [Solution]The fact that means that

for some constants and . Since and we need it to be zero, we have and thus is a constant. Since and we need it to be , we must have

3. (10 points) Find the special solutions for . Write down the general solution.

Proof. [Solution]We seek special solutions of the form

Then if we are to satisfy the heat equation we must have

 constant

So we have a problem for as well as one for . The problem is

There is a solution to the differential equation for each , but that solution may or may not fit the boundary conditions. For example, if , then is a straight line. Since , is a constant, and since , that constant is zero. So the only linear solution is the trivial solution.

If , then for some , and

for some constants and . But

so , and

so . Thus the only hyperbolic solution is the trivial solution.

If , then for some , and then

for some constants and . But

so , and

so we must have that is an odd multiple of (not just any multiple, because the set of multiples of includes the set of multiples of , which we definitely do not want). Thus for each we have a special solution

The problem is easier. Since , we have

So our special solution is

The general solution is therefore

4. (5 points) How would one find the coefficients in the general solution? You may find the following fact useful: If and are integers, then

Proof. [Solution]If we take the general solution and plug in , we have

Now pick a positive integer , multiply by , and integrate between 0 and . We have

Using the fact, we have

Thus for any ,

Matthew Leingang 2002-11-25