I have the following mixed boundary value problem
$$ \begin{array}{l} \Delta \tau(\theta) = -1, \quad 0\leq\theta <\theta_{\max}, \\ \tau(\theta_{\max}) = 0, \\ \left. \dfrac{\partial \tau(\theta)}{\partial \theta} \right|_{\theta = 0} =0. \end{array} $$ Where $\Delta$ is the laplace operator in spherical coordinates. This can easily be integrated. Namely $r = sin \theta$. Also note that there is no radial dependence and azimuthal symmetry
\begin{align*} & \Delta \tau(r)=\frac{\sqrt{1-r^2}}{r} \frac{\partial}{\partial r} \Big( r \sqrt{1-r^2} \frac{\partial}{\partial r} \Big) \tau = -1, \Longleftrightarrow \frac{\partial \tau(r)}{\partial r} = \frac{1}{r} + \frac{C_1}{r \cdot \sqrt{1-r^2}}, \Longleftrightarrow\\ &\tau(r) = \log r + \frac{C_1}{2} \log \left( \frac{1-\sqrt{1-r^2}}{1+\sqrt{1-r^2}} \right) + C_2, \Longleftrightarrow \\ &\tau(\theta) = \log (\sin \theta) + \frac{C_1}{2} \log \left( \frac{1-\cos\theta}{1+\cos \theta}\right) + C_2. \end{align*}
How can I deal with the singularity at $\theta = 0$? Physically, the solution for $\tau(\theta)$ must be finite even in $\theta = 0$
Mathematica gives the solution
$$ \tau(\theta) = \log(1+\cos\theta) -\log(1-\cos \theta_{\max}) $$
This solution can be "derived" by ignoring the singularities. No idea