# Most General solution of Laplace's equation

• Jan 16th 2009, 05:35 AM
Jason Bourne
Most General solution of Laplace's equation
Laplace's equation written in plane polar coordinates $\displaystyle (r,\theta)$ is

$\displaystyle T_{rr} + \frac{1}{r}T_r + \frac{1}{r^2}T_{\theta \theta} = 0$

I need to find the most general solution $\displaystyle T(r)$ which is a function of $\displaystyle r$ only.

Can anyone see what this would be?
• Jan 16th 2009, 08:22 AM
HallsofIvy
Quote:

Originally Posted by Jason Bourne
Laplace's equation written in plane polar coordinates $\displaystyle (r,\theta)$ is

$\displaystyle T_{rr} + \frac{1}{r}T_r + \frac{1}{r^2}T_{\theta \theta} = 0$

I need to find the most general solution $\displaystyle T(r)$ which is a function of $\displaystyle r$ only.

Can anyone see what this would be?

If T is a function of r only, then $\displaystyle T_{\theta\theta}= 0$ so the equation becomes the ordinary differential equation $\displaystyle T_{rr}+ \frac{1}{r}T_r= 0$. Let $\displaystyle u= T_r$ and we have $\displaystyle u_r+ \frac{1}{r}u= 0$ which is a separable first order differential equation. $\displaystyle \frac{du}{u}= -\frac{dr}{r}$. Can you finish it from there?
• Jan 16th 2009, 09:09 AM
Jason Bourne
Quote:

Originally Posted by HallsofIvy
If T is a function of r only, then $\displaystyle T_{\theta\theta}= 0$ so the equation becomes the ordinary differential equation $\displaystyle T_{rr}+ \frac{1}{r}T_r= 0$. Let $\displaystyle u= T_r$ and we have $\displaystyle u_r+ \frac{1}{r}u= 0$ which is a separable first order differential equation. $\displaystyle \frac{du}{u}= -\frac{dr}{r}$. Can you finish it from there?

Thanks. So $\displaystyle T = Aln(r) + B$

A,B constants?
• Jan 16th 2009, 11:14 AM
HallsofIvy
Yes, that is correct.