I would be very grateful if someone could explain how you go out finding the form of the solution to this problem?

http://img691.imageshack.us/img691/5733/17042500.png

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- Mar 28th 2011, 08:48 AMLHSLaplaces Equation using polar coordinates
I would be very grateful if someone could explain how you go out finding the form of the solution to this problem?

http://img691.imageshack.us/img691/5733/17042500.png - Mar 28th 2011, 09:19 AMTheEmptySet
Have your tried expanding the solution is a Fourier series?

$\displaystyle \displaystyle \Delta T(x,y)=\left(\frac{\partial T^2}{\partial r^2} +\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial T^2}{\partial \theta^2}\right)=0$

in polar coordinates

Now use separation of variables with $\displaystyle T(r,\theta)=R(r)\Theta(\theta)$ - Mar 28th 2011, 09:27 AMLHS
So by expressing cos^2(theta) as 1/2+1/2 cos(2*theta) I can equate with the general solution to laplace in polars? which is

T(x,y)=A[0]+B[0]log(r)+ (the infinite summation from n=1 of) (A[n]r^n+B[n]r^-n)(C[n]cos(n theta)+D[n]sin(n theta))? - Mar 28th 2011, 09:45 AMTheEmptySet
I think we are on the same page, but just to be sure

if you separate the equations you get

$\displaystyle r^2R''+rR'-\lambda^2R=0$ and

$\displaystyle \Theta''+\lambda^2\Theta=0$

So

$\displaystyle R(r)=c_1r^\lambda+c_2r^{-\lambda}$

Since the solution must be bounded as r goes to infinity $\displaystyle c_1=0$

This gives

$\displaystyle R(r)=c_2r^{-\lambda}$

Solving the theta equation gives

$\displaystyle \Theta(\theta)=d_1\sin(\lambda \theta)+d_2\cos(\lambda \theta)$

So

$\displaystyle T_\lambda(r,\theta)=r^{-\lambda}[d_1\sin(\lambda \theta)+d_2\cos(\lambda \theta)]$

Also note that $\displaystyle \displaystyle \cos^2(\theta)=\frac{1}{2}+\frac{1}{2}\cos(2\theta )$

Use this and orthogonality to get the solution. - Mar 28th 2011, 09:58 AMLHS
Ah thanks, yes that's much clearer now! :)