Taken from Maple pdsolve help pages
Solve for \(u( r,\theta ,\phi )\). Where \(\theta \) is the polar angle and \(\phi \) is the azimuthal angle. Hence \(0<\theta <\pi \) and \(-\pi <\phi <\pi \).
\begin {align*} \frac {\partial }{\partial r} \left (r^2 \frac {\partial u}{\partial r} \right ) + \frac {1}{\sin \theta } \frac {\partial }{\partial \theta } \left (\sin \theta \frac {\partial u}{\partial \theta } \right ) + \frac {1}{\sin ^2\theta } \frac {\partial ^2 u}{\partial \phi ^2}=0 \end {align*}
Mathematica ✗
ClearAll["Global`*"]; lap = Laplacian[f[r, theta, phi], {r, theta, phi}, "Spherical"]; sol = AbsoluteTiming[TimeConstrained[DSolve[lap == 0, f[r, theta, phi], {r, theta, phi}], 60*10]];
Failed
Maple ✓
restart; PDE := diff(r^2*diff(F(r,theta,phi),r),r)+ 1/sin(theta)*diff(sin(theta)*diff(F(r,theta,phi),theta),theta)+ 1/sin(theta)^2*diff(F(r,theta,phi),phi$2) = 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(PDE,F(r,theta,phi),'build')),output='realtime'));
\[F \left ( r,\theta ,\phi \right ) ={\frac {\sqrt {2} \left ( - \left ( \sin \left ( \theta \right ) \right ) ^{2} \right ) ^{1/2\,\sqrt {{\it \_c}_{{2}}}} \left ( {\it \_C5}\,\sin \left ( \sqrt {{\it \_c}_{{2}}}\phi \right ) +{\it \_C6}\,\cos \left ( \sqrt {{\it \_c}_{{2}}}\phi \right ) \right ) \left ( {\it \_C1}\,{r}^{1/2\,\sqrt {1+4\,{\it \_c}_{{1}}}}+{\it \_C2}\,{r}^{-1/2\,\sqrt {1+4\,{\it \_c}_{{1}}}} \right ) \left ( \cos \left ( \theta \right ) {\it csgn} \left ( \sin \left ( \theta \right ) \right ) \hypergeom \left ( [1/2\,\sqrt {{\it \_c}_{{2}}}+1/4\,\sqrt {1+4\,{\it \_c}_{{1}}}+3/4,1/2\,\sqrt {{\it \_c}_{{2}}}-1/4\,\sqrt {1+4\,{\it \_c}_{{1}}}+3/4],[3/2],1/2\,\cos \left ( 2\,\theta \right ) +1/2 \right ) {\it \_C3}+{\it \_C4}\,\hypergeom \left ( [1/2\,\sqrt {{\it \_c}_{{2}}}+1/4\,\sqrt {1+4\,{\it \_c}_{{1}}}+1/4,1/2\,\sqrt {{\it \_c}_{{2}}}-1/4\,\sqrt {1+4\,{\it \_c}_{{1}}}+1/4],[1/2],1/2\,\cos \left ( 2\,\theta \right ) +1/2 \right ) \right ) }{\sqrt {r}}}\]
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