____________________________________________________________________________________
Added Jan 10, 2019.
Solve for \(u(r,\theta ,\phi ,t)\) the wave PDE in 3D
\[ u_{tt} = c^2 \nabla ^2 u \]
Using the Physics convention for Spherical coordinates system.
Mathematica ✗
ClearAll[u, t, r, theta, phi]; lap = Laplacian[u[r, theta, phi, t], {r, theta, phi}, "Spherical"]; pde = D[u[r, theta, phi, t], {t, 2}] == c^2*lap; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[r, theta, phi, t], {r, theta, phi, t}, Assumptions -> {0 < theta < Pi}], 60*10]];
\[ \text {Failed} \]
Maple ✓
u:='u';t:='t'; theta:='theta';phi:='phi';r:='r'; lap:=VectorCalculus:-Laplacian( u(r,theta,phi,t), 'spherical'[r,theta,phi] ); pde:= diff(u(r,theta,phi,t),t$2)= c^2* lap; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(r,theta,phi,t),'build') assuming 0<theta,theta<Pi),output='realtime')); sol := simplify(sol);
\[ u \left ( r,\theta ,\phi ,t \right ) ={\frac {\sqrt {2}{{\rm e}^{1/2\, \left ( -\pi -2\,\phi \right ) \sqrt {{\it \_c}_{{3}}}-\sqrt {{\it \_c}_{{4}}}t}} \left ( \sin \left ( \theta \right ) \right ) ^{i\sqrt {{\it \_c}_{{3}}}} \left ( {{\rm e}^{2\,\sqrt {{\it \_c}_{{4}}}t}}{\it \_C7}+{\it \_C8} \right ) \left ( {{\rm e}^{2\,\sqrt {{\it \_c}_{{3}}}\phi }}{\it \_C5}+{\it \_C6} \right ) }{\sqrt {r}} \left ( {\it \_C2}\,\BesselY \left ( 1/2\,\sqrt {{\frac {{c}^{2}-4\,{\it \_c}_{{1}}}{{c}^{2}}}},{\frac {\sqrt {-{\it \_c}_{{4}}}r}{c}} \right ) +{\it \_C1}\,\BesselJ \left ( 1/2\,\sqrt {{\frac {{c}^{2}-4\,{\it \_c}_{{1}}}{{c}^{2}}}},{\frac {\sqrt {-{\it \_c}_{{4}}}r}{c}} \right ) \right ) \left ( {\mbox {$_2$F$_1$}(-1/4\,{\frac {-2\,\sqrt {-{\it \_c}_{{3}}}c+\sqrt {{c}^{2}-4\,{\it \_c}_{{1}}}-3\,c}{c}},1/4\,{\frac {2\,\sqrt {-{\it \_c}_{{3}}}c+\sqrt {{c}^{2}-4\,{\it \_c}_{{1}}}+3\,c}{c}};\,3/2;\,1/2\,\cos \left ( 2\,\theta \right ) +1/2)}\cos \left ( \theta \right ) {\it \_C3}+{\mbox {$_2$F$_1$}(-1/4\,{\frac {-2\,\sqrt {-{\it \_c}_{{3}}}c+\sqrt {{c}^{2}-4\,{\it \_c}_{{1}}}-c}{c}},1/4\,{\frac {2\,\sqrt {-{\it \_c}_{{3}}}c+\sqrt {{c}^{2}-4\,{\it \_c}_{{1}}}+c}{c}};\,1/2;\,1/2\,\cos \left ( 2\,\theta \right ) +1/2)}{\it \_C4} \right ) } \]