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

Question

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?

Answer 1

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.

Answer 2

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