1. ## Laplacian solution

I would like someone to verify this for me. (I can't find it in my text for some reason.)

Given a solution $\displaystyle \nabla ^2 S(r, \theta) = 0$ of the Laplacian and given that S is constant on some closed boundary, I believe then we know that S is constant throughout the whole of the enclosed area, is that right?

Thanks.

-Dan

2. Originally Posted by topsquark
I would like someone to verify this for me. (I can't find it in my text for some reason.)

Given a solution $\displaystyle \nabla ^2 S(r, \theta) = 0$ of the Laplacian and given that S is constant on some closed boundary, I believe then we know that S is constant throughout the whole of the enclosed area, is that right?
Let me see if I understand the problem:

If $\displaystyle U$ is an open set in $\displaystyle \mathbb{R}^2$ with a smooth boundary $\displaystyle \partial U$. If $\displaystyle u: \bar U\mapsto \mathbb{R}$ is a continous $\displaystyle \mathcal{C}^2$-function so that $\displaystyle u(\bold{x}) = k$ for some $\displaystyle k\in \mathbb{R}$ and all $\displaystyle \bold{x}\in \partial U$ (i.e. $\displaystyle u(\bold{x})$ is constant on the boundary). And if $\displaystyle \nabla^2 u (\bold{x}) = 0$ for all $\displaystyle \bold{x}\in U$. Then $\displaystyle u$ must be a constant function.

This is true by the maximum principle of harmonic functions. The maximum principle states that $\displaystyle u$ attains its maximum and minimum on the boundary. But since the function on the boundary is konstant it means $\displaystyle k\leq u(\bold{x}) \leq k$ for all $\displaystyle x\in \bar U$. And so $\displaystyle u$ is constant.

Though, I am not sure if the maximum principle has any limitations. If $\displaystyle U$ happens to be bounded then it applies. What happens if $\displaystyle U$ is the right-half plane? I do not know in that case. If the maximum principle still applies then the argument still works.

3. Originally Posted by ThePerfectHacker
Let me see if I understand the problem:

If $\displaystyle U$ is an open set in $\displaystyle \mathbb{R}^2$ with a smooth boundary $\displaystyle \partial U$. If $\displaystyle u: \bar U\mapsto \mathbb{R}$ is a continous $\displaystyle \mathcal{C}^2$-function so that $\displaystyle u(\bold{x}) = k$ for some $\displaystyle k\in \mathbb{R}$ and all $\displaystyle \bold{x}\in \partial U$ (i.e. $\displaystyle u(\bold{x})$ is constant on the boundary). And if $\displaystyle \nabla^2 u (\bold{x}) = 0$ for all $\displaystyle \bold{x}\in U$. Then $\displaystyle u$ must be a constant function.

This is true by the maximum principle of harmonic functions. The maximum principle states that $\displaystyle u$ attains its maximum and minimum on the boundary. But since the function on the boundary is konstant it means $\displaystyle k\leq u(\bold{x}) \leq k$ for all $\displaystyle x\in \bar U$. And so $\displaystyle u$ is constant.

Though, I am not sure if the maximum principle has any limitations. If $\displaystyle U$ happens to be bounded then it applies. What happens if $\displaystyle U$ is the right-half plane? I do not know in that case. If the maximum principle still applies then the argument still works.
That works for me. The region in question is an annulus about the origin, so I would say it fits.

Here's the problem as it is stated.
Originally Posted by nairbdm
The steady state temperature of an annular plate is given by the solution to (del^2)T=0 where T=T(r,theta) --- use the Laplacian in polar coordinates.

1. Find the general solution for T. --- Hint - for the angular part, don't forget single-valuedness so that T(r,theta)=T(r, theta+n2pi) for any integer n. The radial equation will not be a Bessel equation. Here, be careful with m=0 as a special case: remember that a linear second order ODE has two independent solutions.

2. Find the exact solution T(r,theta) in the presence of the following boundary conditions. The inside edge of hte plate has r=a and is kept at a temperature T(a,theta)=To. the outer edge has r=2a and is kept at a temperature of T(2a, theta)=2To. (Hint - the boundary conditions mean there is no dependence on the angle theta. So, which m values will survive from the general solution?)
I solved the Laplacian for T and discovered to my dismay that I could not fit both boundary conditions. I now think that this is an ill-posed problem: The constant T on the outer radius demands that T be the same constant throughout the area of the annulus. So T on the inner radius would have to match T on the outer radius, which it does not.

(Please correct me if I'm wrong! )

-Dan

4. Originally Posted by topsquark
(Please correct me if I'm wrong! )
Actually what I wrote does not fit your problem. The boundary here is not smooth, it is made out of two pieces. But I am sure that is okay, I very certain what I wrote above easily generalizes to what you are doing. However, there is still another problem. It can be constant on the outer boundary and on the inner boundary as long as the constant differ. If they happen to be the same then the constant solution is the only solution.

Maybe you can argue that a solution to such an equation can be extened to include the whole disk, in such a case you should be constant on the boundary of the full disk. Which will mean it is constant. But I am not sure how to do that, or if that is even possible.

I do not know much about harmonic functions, , I would like to.

5. Originally Posted by ThePerfectHacker
Actually what I wrote does not fit your problem. The boundary here is not smooth, it is made out of two pieces. But I am sure that is okay, I very certain what I wrote above easily generalizes to what you are doing. However, there is still another problem. It can be constant on the outer boundary and on the inner boundary as long as the constant differ. If they happen to be the same then the constant solution is the only solution.

Maybe you can argue that a solution to such an equation can be extened to include the whole disk, in such a case you should be constant on the boundary of the full disk. Which will mean it is constant. But I am not sure how to do that, or if that is even possible.

I do not know much about harmonic functions, , I would like to.

Okay, so I have a more detailed question then.

$\displaystyle T(r, \theta) = \sum_{m = 0}^{\infty} ( C_m r^m + D_m r^{-m} )( A_m~cos(m \theta) + B_m~sin( m \theta))$

The boundary conditions are
$\displaystyle T(a, \theta ) = T_0$
and
$\displaystyle T(2a, \theta) = 2T_0$

Try as I may I cannot find a set of $\displaystyle A_m, B_m, C_m, D_m$ to fit these two conditions independently.

Any ideas? (Or is this yet another case where separation of variables does not give the most general solution?)

-Dan

6. Originally Posted by topsquark
Any ideas?
Yes. I do not know what it will lead to.
Let $\displaystyle u_1$ be a harmonic function which is zero on the smaller radius and satisfies the boundary value problem on the outer radius.
Let $\displaystyle u_2$ be a harmonic function which is zero on the outer radius and satisfies the boundary value problem on the inner radius.
If you can solve those then $\displaystyle u_1+u_2$ will be a solution to this problem.

7. (sigh)

I have a working solution. If we ignore my previous work on separation of variables and assume a radially based solution to the Laplacian I get that
$\displaystyle T(r, \theta) = -\frac{T_0}{ln(2)}~ln \left ( \frac{a}{2r} \right )$
solves the problem.

What I still don't understand is why separation of variables doesn't give this solution.

-Dan