ut + uxxx = 0

Third oder PDE. Solve for \(u(x,y)\) \[ u_t + u_{xxx} = 0 \] With initial conditions \begin {align*} u(x,0)&=f(x) \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == -D[u[x, t], {x, 3}]; 
ic  = u[x, 0] == f[x]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[x, t], {x, t}], 60*10]];

\[\left \{\left \{u(x,t)\to \frac {\int _{-\infty }^{\infty }e^{i K[1] \left (t K[1]^2+x\right )} \int _{-\infty }^{\infty } e^{-i x K[1]} f(x) \, dxdK[1]}{2 \pi }\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(x, t), t)=- diff(u(x, t), x$3); 
ic  := u(x,0)=f(x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic],u(x,t))),output='realtime'));

\[u \left (x , t\right ) = \frac {\int _{-\infty }^{\infty }\frac {4 \pi \sqrt {-\frac {x +\zeta }{\left (-t \right )^{\frac {1}{3}}}}\, \BesselK \left (\frac {1}{3}, \frac {2 \sqrt {3}\, \left (-\frac {x +\zeta }{\left (-t \right )^{\frac {1}{3}}}\right )^{\frac {3}{2}}}{9}\right ) f \left (-\zeta \right )}{3 \left (-t \right )^{\frac {1}{3}}}d \zeta }{4 \pi ^{2}}\]

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2.15.4 \(u_t + u_{xxx} = 0\)