How to solve this linear second order PDE in spherical coordinates

Question

I have this simple looking linear PDE which I can't seem to solve:$$\psi_{rr}+\frac{\psi_{\theta \theta}}{r^2}-\cot \theta \frac{\psi_\theta}{r^2}=0$$ where $\psi (r,\theta)$ is the Stokes stream function of fluid dynamics and $(r,\theta)$ are spherical coordinates (without the azimuthal coordinate $\phi$). Any ideas?

Answer 1

For the velocity field with axial symmetry in spherical coordinates we have $$\vec v(r,\theta)=u(r,\theta)\hat r+v(r,\theta)\hat \theta+w(r,\theta)\hat\phi $$where the Stokes stream function $\psi(r,\theta)$ (see Wikipedia) is defined by$$u(r,\theta)=+ \frac{1}{r^2 \sin \theta}\frac {\partial \psi}{\partial \theta}$$and$$v(r,\theta)=-\frac{1}{r \sin \theta}\frac {\partial \psi}{\partial r}$$ Assume a separation of variable form for the solution to the above PDE for $\psi(r,\theta)$ to be $\psi(r,\theta) = \sum_{n} R_n(r) T_n(\theta)$ with $0\le \theta\le \pi$ with $n = 1,2,3,...$ so that $u(r,\theta), v(r,\theta)$, and $\psi (r,\theta) $ do not diverge for any values of $\theta$ and $T_k(\theta)$. The second order PDE for $\psi$ above reduces to a pair of ODEs for $R_n(r)$ and $T_n(\theta)$ which gives $$\frac{d^2 R_n(r)}{dr^2}-\frac{n(n+1)R_n(r)}{r^2} = 0$$ and $$\frac{d^2 T_n(\theta)}{d \theta^2}-\cot \theta\frac{d T_n(\theta)}{d \theta}+n(n+1)T_n(\theta) = 0$$ The solutions are$$R_n(r)=c_1 r^{n+1}+\frac {c_2}{r^n}$$ and $T_1 (\theta)=\sin^2 \theta $ , $T_2(\theta)=\cos \theta\sin^2\theta$ , $T_3(\theta)=\cos^4 \theta-\frac{6}{5} \cos^2 \theta+\frac{1}{5}$ , $T_4(\theta)=\cos^5 \theta-\frac{10}{7}\cos^3 \theta+\frac{3}{7}\cos \theta$ , etc. , giving for the first term of the series for $\psi (r,\theta)$ $$\psi_1(r,\theta)=\left (c_1 r^2+\frac {c_2}{r}\right) \sin^2 \theta$$So using the Stokes stream function above we have the first term of $u_1(r,\theta)$ and $v_1(r,\theta)$ $$u_1(r,\theta)=+2 c_2 \frac{\cos \theta}{r^3}$$ and$$v_1(r,\theta)=+c_2 \frac{\sin\theta}{r^3}$$ for $r$ far from the origin , while for $r$ near the origin we have $$u_1(r,\theta)=+2 c_1 \cos \theta$$ and $$v_1(r,\theta)=-2 c_1 \sin \theta$$

The $T_n(\theta)$ functions could be multiplied by constants. At this stage there are no boundary conditions applied except that we must have the near origin solution match the far solution at $r = a$ for continuity. The $T_n(\theta)$ used here are not the Chebyshev polynomials which have the same symbol . I chose $T_n(\theta)$ here because $T$ is the first letter of the Greek symbol $theta$ .
These $T_n(\theta)$ functions are orthogonal with a weight function $W(x) = \frac {1}{1-x^2}$ where $x = \cos \theta$ . $$\int_{-1}^{+1}T_m(x)T_n(x) \frac{1}{1-x^2}dx = K_n\; \delta_{mn}$$where $K_n$ are constants . The general solution for all integers $n$ is $$T_n(\cos \theta) = \;_2F_1\left(-\frac {n}{2}-\frac {1}{2},\frac{n}{2},\frac{1}{2},\cos^2\theta\right)$$ (multiplied by constants) where $_2F_1$ is the hypergeometric function.