I'm trying to solve heat equation
$$\nabla^2 u = \frac{1}{k}\frac{\partial u}{\partial t}$$ in the region
$$ a \leq r \leq b, \ \ \ \ 0 \leq \varphi \leq 2\pi, \ \ \ \ 0 \leq \theta \leq \theta_0 $$
with $\theta_0$ a fixed number, with boundary conditions
$$ \frac{\partial u}{\partial r} = 0 \ \ \ \ in \ \ r=a, r=b $$
and $$ \frac{\partial u}{\partial \theta} = 0 \ \ \ \ in \ \ \theta = \theta_0 $$

I have trouble finding the eigenfunctions and eigenvalues associated to $\theta$. Are they the Legendre polynomials? And if they aren't, is it possible to find them analytically?

-- Update --

Firstly, I separed the temporal part from the spatial one, $u = T(t) \phi(r,\theta,\varphi)$, obtaining: $$ \begin{gather} T'' = -\lambda T \\ \nabla^2 \phi = -\lambda \phi \end{gather} $$

Next, I separed again the spatial function $\phi$ into three parts, each one corresponding to each spherical coordinate: $$ \phi = F(\varphi)R(r)G(\theta) $$

  • If we apply periodic conditions to $F$ ($u(\varphi=-\pi) = u(\varphi=\pi)$ and $\frac{\partial u}{\partial \varphi} (\varphi=-\pi) = \frac{\partial u}{\partial \varphi} (\varphi=\pi)$, the eigenfunctions associated to $\varphi$ are $$ \cos(m\varphi), \ \ \sin(m\varphi) $$

with $m$ a non-negative integer.

  • Eigenfunctions associated to radial part $R$ are $$ \frac{J_{n+1/2} (\sqrt{\lambda}r)}{\sqrt{r}}, \ \ \frac{Y_{n+1/2} (\sqrt{\lambda}r)}{\sqrt{r}} $$

with $J_k$ and $Y_k$ Bessel functions of order $k$ of first and second kind, respectively. We should apply Neumann conditions in $r=a$ and $r=b$.

My doubts arrive when dealing with $G$. We need $G$ to be bounded at $\theta = 0$ and also to verify Neumann condition at $\theta = \theta_0$. So I thought we would need Legendre polynomials (or, at least, Legendre associated functions).

  • $\begingroup$ Please show us what work you've done to solve the problem. It's no good guessing whether there are Legendre polynomials involved (there aren't). And yes: an analytical solution does exist. $\endgroup$ – Gert Sep 19 at 15:44
  • $\begingroup$ You'll also need an initial condition. $\endgroup$ – Gert Sep 19 at 19:23
  • $\begingroup$ I know an initial condition is needed, but I omitted it for now. $\endgroup$ – Miguel Ibáñez Sep 19 at 22:37
  • $\begingroup$ I edited my question in order to explain better what I did. Thank you very much. $\endgroup$ – Miguel Ibáñez Sep 19 at 22:55
  • $\begingroup$ Much better but have you developed $G(\theta)$? $\endgroup$ – Gert Sep 19 at 22:59

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