5.3.1 Spherical coordinates

5.3.1.1 [414] No I.C. no B.C.

5.3.1.1 [414] No I.C. no B.C.

problem number 414

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["Global`*"]; 
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]];
 

\[\left \{\left \{u(r,\theta ,\phi ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {\sqrt {2} e^{-\frac {1}{2} \sqrt {c_{10}} (2 \phi +\pi )-t \sqrt {c_{11}}} \left (J_{\frac {1}{2} \sqrt {\frac {4 c_9}{c^2}+1}}\left (\frac {r \sqrt {-c_{11}}}{\sqrt {c^2}}\right ) c_1+Y_{\frac {1}{2} \sqrt {\frac {4 c_9}{c^2}+1}}\left (\frac {r \sqrt {-c_{11}}}{\sqrt {c^2}}\right ) c_2\right ) \left (e^{2 \phi \sqrt {c_{10}}} c_5+c_6\right ) \left (e^{2 t \sqrt {c_{11}}} c_7+c_8\right ) \left (c_4 \, _2F_1\left (\frac {1}{4} \left (-\frac {\sqrt {c^2+4 c_9}}{\sqrt {c^2}}+2 \sqrt {-c_{10}}+1\right ),\frac {1}{4} \left (\frac {\sqrt {c^2+4 c_9}}{\sqrt {c^2}}+2 \sqrt {-c_{10}}+1\right );\frac {1}{2};\cos ^2(\theta )\right )+c_3 \cos (\theta ) \, _2F_1\left (\frac {1}{4} \left (-\frac {\sqrt {c^2+4 c_9}}{\sqrt {c^2}}+2 \sqrt {-c_{10}}+3\right ),\frac {1}{4} \left (\frac {\sqrt {c^2+4 c_9}}{\sqrt {c^2}}+2 \sqrt {-c_{10}}+3\right );\frac {3}{2};\cos ^2(\theta )\right )\right ) \sin ^{i \sqrt {c_{10}}}(\theta )}{\sqrt {r}} & c\neq 0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple

restart; 
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);
 

\[u \left ( r,\theta ,\phi ,t \right ) ={\frac { \left ( -1 \right ) ^{i/2\sqrt {{\it \_c}_{{3}}}}{{\rm e}^{-\sqrt {{\it \_c}_{{3}}}\phi -\sqrt {{\it \_c}_{{4}}}t}} \left ( \sin \left ( \theta \right ) \right ) ^{i\sqrt {{\it \_c}_{{3}}}} \left ( {{\rm e}^{2\,\sqrt {{\it \_c}_{{4}}}t}}{\it \_C7}+{\it \_C8} \right ) \left ( {{\rm e}^{2\,\sqrt {{\it \_c}_{{3}}}\phi }}{\it \_C5}+{\it \_C6} \right ) }{\sqrt {r}} \left ( {\it \_C1}\,\BesselJ \left ( 1/2\,\sqrt {{\frac {{c}^{2}-4\,{\it \_c}_{{1}}}{{c}^{2}}}},{\frac {\sqrt {-{\it \_c}_{{4}}}r}{c}} \right ) +{\it \_C2}\,\BesselY \left ( 1/2\,\sqrt {{\frac {{c}^{2}-4\,{\it \_c}_{{1}}}{{c}^{2}}}},{\frac {\sqrt {-{\it \_c}_{{4}}}r}{c}} \right ) \right ) \left ( \hypergeom \left ( [1/4\,{\frac {2\,\sqrt {-{\it \_c}_{{3}}}c+\sqrt {{c}^{2}-4\,{\it \_c}_{{1}}}+3\,c}{c}},1/4\,{\frac {2\,\sqrt {-{\it \_c}_{{3}}}c-\sqrt {{c}^{2}-4\,{\it \_c}_{{1}}}+3\,c}{c}}],[3/2],1/2\,\cos \left ( 2\,\theta \right ) +1/2 \right ) \left | \cos \left ( \theta \right ) \right | \sqrt {2}{\it \_C4}+\hypergeom \left ( [1/4\,{\frac {2\,\sqrt {-{\it \_c}_{{3}}}c-\sqrt {{c}^{2}-4\,{\it \_c}_{{1}}}+c}{c}},1/4\,{\frac {2\,\sqrt {-{\it \_c}_{{3}}}c+\sqrt {{c}^{2}-4\,{\it \_c}_{{1}}}+c}{c}}],[1/2],1/2\,\cos \left ( 2\,\theta \right ) +1/2 \right ) {\it \_C3} \right ) }\]

____________________________________________________________________________________