5.3.1 Spherical coordinates

5.3.1.1 [415] No I.C. no B.C.

problem number 415

Added Jan 10, 2019.

Solve for $u(r,\theta ,\varphi ,t)$ the wave PDE in 3D $$u_{tt}=c^2{\mathrm{\nabla }}^{2}u$$ Using the Physics convention for Spherical coordinates system.

Figure 5.109: PDE specification

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]];
 

Failed

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);
 

$$u(r,\theta ,\varphi ,t)=\frac{\sqrt{2}{\mathrm{e}}^{1/2(-\pi -2\varphi )\sqrt{{\mathit{\_}\mathit{c}}_3}-\sqrt{{\mathit{\_}\mathit{c}}_4}t}{(\sin \left(\theta \right))}^{i\sqrt{{\mathit{\_}\mathit{c}}_3}}({\mathrm{e}}^{2\sqrt{{\mathit{\_}\mathit{c}}_4}t}\mathit{\_}\mathit{C}\mathit{7}+\mathit{\_}\mathit{C}\mathit{8})({\mathrm{e}}^{2\sqrt{{\mathit{\_}\mathit{c}}_3}\varphi }\mathit{\_}\mathit{C}\mathit{5}+\mathit{\_}\mathit{C}\mathit{6})}{\sqrt{r}}(\mathrm{BesselJ}(1/2\sqrt{\frac{c^2+4{\mathit{\_}\mathit{c}}_1}{c^2}},\frac{\sqrt{-{\mathit{\_}\mathit{c}}_4}r}{c})\mathit{\_}\mathit{C}\mathit{1}+\mathrm{BesselY}(1/2\sqrt{\frac{c^2+4{\mathit{\_}\mathit{c}}_1}{c^2}},\frac{\sqrt{-{\mathit{\_}\mathit{c}}_4}r}{c})\mathit{\_}\mathit{C}\mathit{2})(\mathrm{hypergeom}([1/4\frac{2\sqrt{-{\mathit{\_}\mathit{c}}_3}c+\sqrt{c^2+4{\mathit{\_}\mathit{c}}_1}+3c}{c},-1/4\frac{-2\sqrt{-{\mathit{\_}\mathit{c}}_3}c+\sqrt{c^2+4{\mathit{\_}\mathit{c}}_1}-3c}{c}],[3/2],1/2\cos \left(2\theta \right)+1/2)\cos \left(\theta \right)\mathit{\_}\mathit{C}\mathit{3}+\mathrm{hypergeom}([-1/4\frac{-2\sqrt{-{\mathit{\_}\mathit{c}}_3}c+\sqrt{c^2+4{\mathit{\_}\mathit{c}}_1}-c}{c},1/4\frac{2\sqrt{-{\mathit{\_}\mathit{c}}_3}c+\sqrt{c^2+4{\mathit{\_}\mathit{c}}_1}+c}{c}],[1/2],1/2\cos \left(2\theta \right)+1/2)\mathit{\_}\mathit{C}\mathit{4})$$

____________________________________________________________________________________