# Neumann boundary condition in spherical coordinates

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).

• 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. – Gert Sep 19 at 15:44
• You'll also need an initial condition. – Gert Sep 19 at 19:23
• I know an initial condition is needed, but I omitted it for now. – Miguel Ibáñez Sep 19 at 22:37
• I edited my question in order to explain better what I did. Thank you very much. – Miguel Ibáñez Sep 19 at 22:55
• Much better but have you developed $G(\theta)$? – Gert Sep 19 at 22:59