4.2.1 Spherical coordinates

   4.2.1.1 [317] In a sphere

4.2.1.1 [317] In a sphere

problem number 317

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

Failed

Maple

\[F \left ( r,\theta ,\phi \right ) ={\frac{\sqrt{2} \left ( - \left ( \sin \left ( \theta \right ) \right ) ^{2} \right ) ^{1/2\,\sqrt{{\it \_c}_{{2}}}} \left ({\it \_C5}\,\sin \left ( \sqrt{{\it \_c}_{{2}}}\phi \right ) +{\it \_C6}\,\cos \left ( \sqrt{{\it \_c}_{{2}}}\phi \right ) \right ) \left ({\it \_C1}\,{r}^{1/2\,\sqrt{1+4\,{\it \_c}_{{1}}}}+{\it \_C2}\,{r}^{-1/2\,\sqrt{1+4\,{\it \_c}_{{1}}}} \right ) \left ( \cos \left ( \theta \right ){\it csgn} \left ( \sin \left ( \theta \right ) \right ) \hypergeom \left ( [1/2\,\sqrt{{\it \_c}_{{2}}}+1/4\,\sqrt{1+4\,{\it \_c}_{{1}}}+3/4,1/2\,\sqrt{{\it \_c}_{{2}}}-1/4\,\sqrt{1+4\,{\it \_c}_{{1}}}+3/4],[3/2],1/2\,\cos \left ( 2\,\theta \right ) +1/2 \right ){\it \_C3}+{\it \_C4}\,\hypergeom \left ( [1/2\,\sqrt{{\it \_c}_{{2}}}+1/4\,\sqrt{1+4\,{\it \_c}_{{1}}}+1/4,1/2\,\sqrt{{\it \_c}_{{2}}}-1/4\,\sqrt{1+4\,{\it \_c}_{{1}}}+1/4],[1/2],1/2\,\cos \left ( 2\,\theta \right ) +1/2 \right ) \right ) }{\sqrt{r}}}\]

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