Solutions of the Laplace equation How do I find solutions $u=f(r)$ of the two-dimensional Laplace equation $u_{xx}+u_{yy}=0$ that depend only on the radial coordinate $r= \sqrt{x^2+y^2}$
 A: You should also use the polar coordinate version of the Laplace operator 
$$
\Delta u 
= {\partial^2 u \over \partial r^2}
+{1 \over r} {\partial u \over \partial r}
+ {1 \over r^2} {\partial^2 u \over \partial \theta^2}
$$
and trying for $u(r, \phi) = f(r)$. This reduces to solving the ODE
$$
0 = g'(r) +{1 \over r} {g}
$$
for $g(r) = f'(r)$.
A: As @CameronWilliams pointed out, the easiest way to do this is to switch to polar coordinates.    But, here is a brute force way to proceed.  
Let $f(x,y)=g(x^2+y^2)$.  Then, $\frac{\partial f}{\partial x}=2xg'(x^2+y^2)$.
Taking a second partial yields $\frac{\partial^2 f}{\partial x^2}=(2x)^2g''(x^2+y^2)+2g'(x^2+y^2)$.
One can easily see then (replace $x$ with $y$) that $\frac{\partial^2 f}{\partial y^2}=(2y)^2g''(x^2+y^2)+2g'(x^2+y^2)$.
Thus, 
$$\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=4(x^2+y^2)g''(x^2+y^2) + 4g'(x^2+y^2)=0$$
Then, we may write 
$$\frac{g''(x^2+y^2)}{g'(x^2+y^2)}=\left(\log g'(x^2+y^2)\right)'=-\frac{1}{x^2+y^2}$$
where again the "prime" is on $x^2+y^2$.  Integrating once reveals that $\log g'(x^2+y^2) = -\log(x^2+y^2) +C \Rightarrow g(x^2+y^2) = A\log (x^2+y^2) +B$.
A: ..and here's how you arrive at the polar coordinate version:
$$ 
\begin{align}
& u := f(r(x,y)), \\
& u_x = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}, \\
& u_{xx} = \left(\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}\right)_x = \frac{\partial^2 f}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 + \frac{\partial f}{\partial r}\frac{\partial^2 r}{\partial x^2}, \\
& u_{xx} + u_{yy} = \frac{\partial^2 f}{
\partial r^2}\left(\left(\frac{\partial r}{\partial x}\right)^2 + \left(\frac{\partial r}{\partial y}\right)^2\right) + \frac{\partial f}{
\partial r}\left(\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2}\right), \\
& u_{xx} + u_{yy} = \frac{\partial^2 f}{
\partial r^2}\left(\frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2}\right) + \frac{\partial f}{
\partial r}\left(\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2}\right), \\
& u_{xx} + u_{yy} = \frac{\partial^2 f}{\partial r^2} + \frac{\partial f}{
\partial r} \left( - \frac{x^2 + y^2}{(x^2 + y^2)^{3/2}} + \frac{2}{\sqrt{x^2 + y^2}} \right), \\
& u_{xx} + u_{yy} = \frac{\partial^2 f}{\partial r^2} + \frac{\partial f}{
\partial r} \frac{1}{r} = 0.
\end{align}
$$
