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]];

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

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'));

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

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