Need help with a couple of BVP's.

[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?

Thanks in advance,

Grogs

 

[HallsofIvy]

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):

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

That should be a sum above, it's just not showing up for some reason.

(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.

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:

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

That should be a sum above, it's just not showing up for some reason.

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?

Thanks in advance,

Grogs

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.

 

[Grogs]

Halls, thanks for the reply.

(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".

I'm not sure where you're getting that. This is the heat equation (u = temperature,) not the wave equation. It's 1st order in t -> 1 IC required. The 2 IC's I posted are for 2 different problems where everything is the same, save for the IC's.

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.

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:

$$A_{n} = \frac{2}{L} \int \limits_{0} ^ {L} sin(\frac{n \pi x}{L}) f(x) dx$$

where f(x) is the IC be the only way to find u(x,y) in this case?