2.15.24 Rayleigh \(u_{tt} - u_{xx} = \epsilon (u_t - u_t^3)\)

problem number 133

Added December 27, 2018.

Taken from https://en.wikipedia.org/wiki/List_of_nonlinear_partial_differential_equations

Rayleigh. Solve for \(u(x,t)\)

\[ u_{tt} - u_{xx} = \epsilon (u_t - u_t^3) \]

Mathematica ✗

ClearAll["Global`*"]; 
pde =  D[u[x, t], {t, 2}] - D[u[x, t], {x, 2}] == epsilon*(D[u[x, t], t] - D[u[x, t], t]^3); 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, t], {x, t}], 60*10]];
 

Failed

Maple ✓

restart; 
pde := diff(u(x,t),t$2)-diff(u(x,t),x$2)=epsilon*(diff(u(x,t),t)-diff(u(x,t),t)^3); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,t),'build')),output='realtime'));
 

\[u \left (x , t\right ) = \frac {\textit {\_c}_{1} x^{2}}{2}+c_{1} x +c_{2} +\int \operatorname {RootOf}\left (t +\int _{}^{\textit {\_Z}}\frac {1}{\textit {\_f}^{3} \epsilon -\textit {\_f} \epsilon -\textit {\_c}_{1}}d \textit {\_f} +c_{3} \right )d t +c_{4}\]

Has RootOf

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