# Show that a harmonic function $u(r, \theta)$ dependent only on $r$ has the form $u(r, \theta) = a \log r + b$

What I have done so far is this:

I've shown that if $u(r, \theta)$ is a harmonic function dependent only on $r$ then Laplace's equation becomes $u_{rr} + \frac{1}{r}u_r = 0$

I've also shown that $u(r, \theta) = a \log r$ satisfies $u_{rr} + \frac{1}{r}u_r = 0$

Now I think the next step is to suppose that $v(r, \theta)$ is another harmonic function dependent only on $r$. From the above, we know that $v_{rr} + \frac{1}{r}v_r = 0 \implies v_{rr} + \frac{1}{r}v_r = u_{rr} + \frac{1}{r}u_r$. This is where I am stuck. Any tips?

## 2 Answers

You can think of $\theta$ as a parameter, so that the PDE is actually an ODE $$u''(r)+\frac{1}{r} u'(r)=0$$ Setting $v=u'$ the equation becomes first ordered $$v'+\frac{1}{r}v=0$$ You can easily solve this using separation of variables to find the most general solution.

• I took ODE 4 years ago. We never discussed any ODE techniques in class, so I'm a little lost on how to solve this. – Max Jun 13 '13 at 8:38
• The ODE is $\frac{dv}{dr}=-\frac{1}{r} v$ separate the variables to get $\frac{dv}{v}=-\frac{dr}{r}$. Integrate this equation to find $v(r)$ – user1337 Jun 13 '13 at 10:57

Hint: You already know a PDE for u(r,θ), so just solve the PDE and then you can get the general answer.