Finding Solutions for Laplace's Equation with Radial Dependence

[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?

 

[flatmaster]

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?

 

[mcfc]

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?

Hi,

Yes, sorry for the confusion

 

[Phyisab****]

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

 

[mcfc]

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

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...?

 

[Phyisab****]

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.

 

[mcfc]

$\nabla ^2 T = \frac {\partial}{\partial r}( {1 \over r}\frac{\partial rT}{\partial r})=0$
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!?

 

[Phyisab****]

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

 

[Phyisab****]

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

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

 

[mcfc]

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?

 

[Phyisab****]

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

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.