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'));
 

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

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