utt − uxx − 2 xux = 0 with u(x, 0) = 0,ut(x, 0) = g(x)

Problem 2.4.18 from Peter Olver, Intoduction to Partial diﬀerential equations, 4th edition.

With \(u(x,0)=0,u_t(x,0)=g(x)\). Note, in the book, it says to assume \(g(x)\) is even function. In the code below, this assumption is not used. When I ﬁnd the correct way to implement this assumption in CAS, will have to re-run these.

Mathematica ✗

ClearAll["Global`*"]; 
pde =  D[u[x, t], {t, 2}] - D[u[x, t], {x, 2}] - 2/x*D[u[x,t],x] == 0; 
ic = {u[x,0]==0,Derivative[0,1][u][x,0]==g[x]}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[x, t], {x, t}], 60*10]];

Failed

Maple ✓

restart; 
pde := diff(u(x,t),t$2)- diff(u(x,t),x$2) - 2/x*diff(u(x,t),x)=0; 
ic:= u(x,0)=0, D[2](u)(x,0)=g(x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,t)) ),output='realtime'));

\[u \left (x , t\right ) = \sum _{n=0}^\infty \frac {t^{2 n +1} \left (\textbf {proc} (U) \\ \textbf {option} \,operator,\,arrow; \\ \mapleIndent {1} \mathit {diff} (\mathit {diff} (U,\,x),\,x) + 2 \ast x\hat {~}{-1} \ast \mathit {diff} (U,\,x)\\ \textbf {end\ proc};\right )^{\left (n \right )}\left (g \left (x \right )\right )}{\left (2 n +1\right )!}\]

2.9.3 \(u_{tt} - u_{xx} - \frac {2}{x} u_x = 0\) with \(u(x,0)=0,u_t(x,0)=g(x)\)