# Solve $u_{rr}+r^{-1}u_r+r^{-2}u_{\theta\theta}=0.$

I have two BVP's and for both of them the solution is similar, however there is one thing I can't explain. First of all let me state the BBVP's. The first one is

$$(1)= \begin{cases} &u_{rr}+r^{-1}u_r+r^{-2}u_{\theta\theta} = 0&&,1

And the second one is

$$(2)= \begin{cases} u_{rr}+r^{-1}u_r+r^{-2}u_{\theta\theta} = 0&&,0

In both cases, one can let $$u(r,\theta)=R(r)\Theta(\theta)$$ and after plugging this into the PDE, one can write

$$r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{\Theta''}{\Theta}=\lambda, \ \text{where} \ \lambda=\text{constant.}\tag3$$

Since the $$\Theta-$$part is easier we start with that one. Since we are working in disks/annuluses we need the solution to be $$2\pi-$$periodic. Thus the solution has the form (up to constant multiples) $$\Theta=e^{in\theta},$$ where $$n\in\mathbb{Z}.$$ Plugging this onto the ODE, $$\Theta''=-\lambda\Theta$$ gives $$\lambda=n^2.$$ Thus

$$\Theta_n(\theta)=e^{in\theta}. \tag4$$

Using $$(3)$$ we can solve for $$R(r)$$ since $$r^2R''+rR'=\lambda R=n^2R,$$ re arranging we simply have the Euler equation

$$r^2R''+RR'-n^2R=0.$$

Thus far, both $$(1)$$ and $$(2)$$ can be solve thid way. But here comes the differents. In the notes for $$(1)$$, my professor says that:

We need to examine the solutions of the Euler equation for two cases: $$n=0$$ and $$n\neq 0.$$

But in the notes for $$(2)$$ he does not do this and only considers $$n\neq 0.$$

What in the problem statement makes this difference?

There is no difference. Perhaps he just neglected to mention $$n=0$$ for (2), or thought you remembered it from (1). But problem (2) also needs a boundary condition at $$r=0$$, which you didn't mention. Assuming the boundary condition is that $$\lim_{r \to 0} u(r,\theta)$$ exists, your solution should end up as
$$u(r,\theta) = \frac{1}{2} + r \cos(\theta) - \frac{r^2}{2} \cos(2\theta)$$
and the term $$1/2$$ comes from $$n=0$$.
• I don't think he neglected to mention it because then the full solution he provided would have differed. I did not mention the boundary condition because there is none :S This is what my prof got as his final answer for $(1)$ and $(2)$ respectively: $$u(r,\theta)=\frac{2\ln(r)}{3\ln(2)}+\sum_{n\in\mathbb{Z}\setminus {0}}e^{in\theta}a_n(r^n+r^{-n}),$$ where $$a_n=\frac{1}{2\pi(2^n-2^{-n})}\int_{-\pi}^{\pi}\left(1-\frac{\theta^2}{\pi^2}\right)e^{-in\theta} \ d\theta.$$ Mar 4, 2019 at 22:27
• And for $(2)$ it's $$u(r,\theta)=\sum_{n\in Z}c_nr^{|n|}e^{in\theta},$$ where $$c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}(\sin^2(\theta)+\cos(\theta))e^{in\theta} d\theta.$$ Both of these seem to differ from your answer. Mar 4, 2019 at 22:28