# How to solve nabla^2 T = 0

1. Nov 19, 2009

### mcfc

I need to solve:
$\nabla ^2 T = 0$ with T=T(r) and r=a/T=T1 and r=b/T=T2

Can anyone offer advice as to the solution?

2. Nov 19, 2009

### flatmaster

Re: Laplacian

I don't quite understand what you mean with your boundry conditions, do you mean...

when r=a, T=T1
when r=b, T=T2

Is this what you mean?

3. Nov 19, 2009

### mcfc

Re: Laplacian

Hi,

Yes, sorry for the confusion

4. Nov 19, 2009

### Phyisab****

Re: Laplacian

Where is your attempt at a solution? You need to crack open a textbook and flip to the section called "Laplace's equation".

5. Nov 19, 2009

### mcfc

Re: Laplacian

My result is :
$T= \frac{1}{2} Ar + \frac{B}{ r}$
A, B constants. Also in this example, I'm not sure how to apply the boundary conditions....
But the result I saw(without proof) involved logarithms...?

6. Nov 19, 2009

### Phyisab****

Re: Laplacian

Well that's not the right solution. It's a pretty simple problem. Either you started with the wrong formula for the Laplacian in polar coordinates, or you integrated it wrong. You apply the boundary conditions the same way you would in any other situation.

7. Nov 19, 2009

### mcfc

Re: Laplacian

$\nabla ^2 T = \frac {\partial}{\partial r}( {1 \over r}\frac{\partial rT}{\partial r})=0$
is the polar form used,

which implies ${1 \over r}\frac{\partial rT}{\partial r}= A$ and integrate to get my (incorrect) result above...
What am I missing!!?

8. Nov 19, 2009

### Phyisab****

Re: Laplacian

Ah you just wrote the polar Laplacian wrong. It should be r(dT/dr)=A at your last step.

9. Nov 19, 2009

### Phyisab****

Re: Laplacian

$$\frac{\partial T}{\partial r} = \frac{A}{r}$$

$$T(a)= Aln(a) + B = T_{1}$$
$$T(b)= Aln(b) + B = T_{2}$$

Last edited: Nov 19, 2009
10. Nov 19, 2009

### mcfc

Re: Laplacian

Thanks.

I've been confused because I've seen two forms of the polar laplacian:
1)$\frac {\partial}{\partial r}( {1 \over r}\frac{\partial T r}{\partial r})$

and

2)${1 \over r} \frac{\partial}{\partial r}(r \frac{\partial T}{\partial r})$

how are these equivalent?

11. Nov 19, 2009

### Phyisab****

Re: Laplacian

$$\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial T}{\partial r}\right)=0$$

the factor of $$\frac{1}{r}$$ comes from the laplacian with theta dependence. Here you are independent of angle. Just multiply by r and it is gone.