# Homework Help: Need help with a couple of BVP's.

1. Oct 24, 2006

### Grogs

I've got a couple of BVP's I'm working on that are giving me some problems. First, I'm asked to solve the 1-D Heat Equation:

$$k \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}$$

in a thin rod of length L subject to the following:

BC1) u(0, t) = 0
BC2) u(L, t) = 0
IC) 2 different problems. For the first, $u(x,0) = 1 - 2 cos(\frac{9 \pi x}{L})$. For the second, $u(x,0) = 3 cos(\frac{9 \pi x}{2 L})$.

I can separate this problem into:

$$\frac{X''}{X} = \frac{T'}{k T} = - \lambda$$

easily enough and apply the 2 BC's to get:

$$u_{n} (x,t) = A_{n} \sin(\frac{n \pi x}{L}) \exp(\frac{-k n^2 \pi^2 t}{L}) , n=1,2,3,...$$

My problem is with applying the Initial condition and solving for u(x,t):

$$u(x,t) = \sum \limits_{n=1} ^ {\infty} A_{n} \sin(\frac{n \pi x}{L}) = IC$$

My book shows me how to do a half-range expansion here, but according to my professor, I shouldn't need to. He even sent us an e-mail changing one of the IC's "so it would be an eigenfunction." How do I get there from here? Is there some easy trig identity to convert those IC's to a sin function I'm just missing?

My 2nd question involves steady state heat conduction in a 2-D square plate of side L:

$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$

Subject To:

BC1) $\frac{\partial u}{\partial x} = 0$ at x = 0
BC2) $\frac{\partial u}{\partial x} = 0$ at x = L
BC3) $\frac{\partial u}{\partial y} = 0$ at y = 0
BC4) u(x,L) = 5

I separate the Laplacian into:

$$\frac{X''}{X} = \frac{-Y'}{Y} = - \lambda$$

and solve using the first 3 BC's, I get $X = c_{1}$ and $Y = c_{4}$ for $\lambda = 0$ and

$$X(x) = c_{2} \sin(\frac{n \pi x}{L})$$

$$Y(y) = c_{4} \cosh(\frac{n \pi y}{L})$$

for $\lambda > 0$.

Using superposition, I end up with:

$$u(x,y) = A_{0} + \sum \limits_{n=1} ^ {\infty} A_{n} \sin(\frac{n \pi x}{L}) \cosh(\frac{n \pi y}{L})$$

When I apply the final BC, u(x,L) = 5, I end up with a solution of u(x,y) = 5. I guess I'm just looking for some sanity checking here. I can certainly see how u = 5 works here, but is it the only solution?

Grogs

Last edited: Oct 24, 2006
2. Oct 24, 2006

### HallsofIvy

(To get the sigma to show use \sum with a small "s".)
For your first problem, there is a difficulty with your initial conditions: you don't have enough! Since the equation is second order in t, you need two initial conditions. Perhaps you were also told that $u_t(x,0)= 0$? That is, that the thin rod was deformed to the given initial position and then released "from rest".
The "complication" is that, since u= 0 at both endpoints, you have written u(x,t) as a sine series in x which is always an odd function by your initial conditions are even functions. That isn't really a problem since your interval is from 0 to L but notice that, according to the first initial condition, u(0, 0)= 1- 2cos(0)= -1 and for the second initial condition, u(0,0)= 3cos(0)= 3, while, according to the boundary conditions, u(0,0)= u(0,t)= 0! You necessarily have a discontinuity at the boundary points.

Again, you can't expect the solution to be continues at y= L because the initial conditions aren't. Yes u(x,y)= 5 is the only solution.

3. Oct 24, 2006

### Grogs

Hmm, interesting. I see where you're coming from with the cosines. There's a mathematical discontinuity between 2 cos(0) and 0 at x = 0. Looking at that leads me to believe there's no way this problem can be solved in the same way as an IC of say, $3 \sin(\frac{5 \pi x}{L}$. So would using the half-range expression:
$$A_{n} = \frac{2}{L} \int \limits_{0} ^ {L} sin(\frac{n \pi x}{L}) f(x) dx$$