# Solving Laplace's equation in an annulus

1. Dec 7, 2008

### maddogtheman

1. The problem statement, all variables and given/known data
Solve Laplace's equation $$\nabla^2 u(r,\vartheta) =0$$ in an annulus with inner radius $$r_1$$ and outer radius $$r_2$$. (a) For boundary conditions take $$u(1,\vartheta) = 3$$ and $$u(2,\vartheta) = 5$$. (b) What is the solution using this second set of boundary conditions $$u(1,\vartheta) = cos(\vartheta)$$ and $$u(2, \vartheta) = sin(\vartheta)$$?

2. Relevant equations

3. The attempt at a solution

I's assuming I'm going to get Bessel functions out of this but I don't know what the exact solution is and I'm having an even harder time applying the boundary conditions.

2. Dec 8, 2008

### gabbagabbahey

Why assume that you're going to get Bessel functions as solutions? Surely you've been taught at least one method you could use to actually solve Laplace's equation in this case. Perhaps you've learned the technique called 'separation of variables'?

3. Dec 8, 2008

### maddogtheman

Yeah I have and I think that leave's me with $$u(r,\vartheta) = \sum_{n=0}^\infty (A_n r^n + B_n r^{-n-1})(C_n cos(n\vartheta) + D_n sin(n\vartheta))$$ but I don't know how to apply the boundary conditions correctly

4. Dec 8, 2008

### gabbagabbahey

Close, I think you should end up with a general solution of $$u(r,\vartheta) = \sum_{n=0}^\infty (A_n r^n + B_n r^{-n})(C_n cos(n\vartheta) + D_n sin(n\vartheta))$$ (note the power of r is -n not -n-1).

Do you understand how this solution is obtained, and why it is an infinite sum of this form?

If so, then start applying the boundary conditions in (a)....what does the fact that $u(1,\theta)=3$ for all values of theta tell you about $$(C_n cos(n\vartheta) + D_n sin(n\vartheta))$$?

5. Dec 8, 2008

### maddogtheman

Yeah I think you're right for some reason the -n-1 power is used in our class tutorial. Yes I understand how it was found and the principal of linear superposition. The second part of the expression is a constant because u is theta independent for the first set of boundary conditions.

6. Dec 8, 2008

### gabbagabbahey

Okay, good.... now for what values of 'n' can that part actually be a constant? Clearly, you require $C_n=D_n=0$ for values of n that don't allow $(C_n cos(n\vartheta) + D_n sin(n\vartheta))=\text{constant}$.

7. Dec 8, 2008

### maddogtheman

Well n=0 has to be one right? I can't think of any others

8. Dec 8, 2008

### gabbagabbahey

Right, n=0 is the only one that can have nonzero C_n and D_n....so what does your expression for u(r,theta) simplify to now?

9. Dec 8, 2008

### maddogtheman

$$u(r,\vartheta) = A_0 + B_0$$ right?

10. Dec 8, 2008

### maddogtheman

Ok I think I realized my problem. My tutorial solved a slightly different DE for R(r) which resulted in the diffence of powers. Also for this problem I think the n=0 case takes on the solution A_0 + B_0*ln(r) which would give a final solution u(r,theta)=3+2ln(r)/ln(2). I'm going to bed, let me know if this is correct. I know how to apply the boundary conditions in (b). Thanks for the help!

11. Dec 8, 2008

### gabbagabbahey

hmm....actually, your general solution $$u(r,\vartheta) = \sum_{n=0}^\infty (A_n r^n + B_n r^{-n})(C_n cos(n\vartheta) + D_n sin(n\vartheta))$$ is incorrect. I think you should actually solve Laplace's equation using separation of variables in order to get the correct general solution.

In the end you should find that the general solution is $$u(r,\vartheta) =A_0+B_0\ln(r)+ \sum_{n=1}^\infty r^n(A_n cos(n\vartheta) + B_n sin(n\vartheta))+r^{-n}(C_n cos(n\vartheta) + D_n sin(n\vartheta))$$.

I'm going to bed now, but I strongly recommend you actually go about solving Laplace's equation for yourself.

When you apply the boundary conditions for part (a), you will see that $A_n=B_n=C_n=D_n=0$ and you only need to solve for A_0 and B_0.

For part (b) you will need to use Fourier's method to determine the coefficients A_n, B_n, C_n, and D_n.

Give it a try and I'll check on your progress in the morning.

12. Dec 8, 2008

### gabbagabbahey

That looks more like it

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