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

$$(1)= \begin{cases} &u_{rr}+r^{-1}u_r+r^{-2}u_{\theta\theta} = 0&&,1<r<2,|\theta|\leq \pi,\\ &u(1,\theta) = 0&&, \theta\leq \pi\\ &u(2,\theta) = 1-\frac{\theta^2}{\pi^2}&&, \theta\leq \pi \end{cases} $$

And the second one is

$$(2)= \begin{cases} u_{rr}+r^{-1}u_r+r^{-2}u_{\theta\theta} = 0&&,0<r<1,|\theta|\leq \pi,\\ u(1,\theta) = \sin^2\theta + \cos\theta&&,\theta\leq \pi\\ \end{cases} $$

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


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?


1 Answer 1


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 $n=0$.

  • $\begingroup$ 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 $$a_n=\frac{1}{2\pi(2^n-2^{-n})}\int_{-\pi}^{\pi}\left(1-\frac{\theta^2}{\pi^2}\right)e^{-in\theta} \ d\theta.$$ $\endgroup$
    – Parseval
    Mar 4, 2019 at 22:27
  • $\begingroup$ 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. $\endgroup$
    – Parseval
    Mar 4, 2019 at 22:28
  • $\begingroup$ @Parseval The given solution for (2) and Robert's solution are in fact the same. $\endgroup$
    – Dylan
    Mar 5, 2019 at 9:24

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