Taken from Mathematica Symbolic PDE document
quasilinear first-order PDE, scalar conservation law
Solve for \(u(x,y)\) \[ u_x + u u_y = 0 \]
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, y], {x}] + u[x, y]*D[u[x, y], {y}] == 0; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[pde, u[x, y], {x, y}]], 60*10]];
\[\left \{\left \{u(x,t)\to x^4-12 t^2\right \}\right \}\] Implicit solution
Maple ✓
restart; interface(showassumed=0); pde := diff(u(x, y), x) + u(x,y)*diff(u(x, y),y) =0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime')); sol:=DEtools:-remove_RootOf(sol);
\[-y+xu \left ( x,y \right ) +{\it \_F1} \left ( u \left ( x,y \right ) \right ) =0\]
Hand solution
Solve for \(u\left ( x,y\right ) \) in \(u_{x}+u\ u_{y}=0.\) Using the Lagrange-Charpit method, the characteristic equations are\[ \frac {dx}{1}=\frac {dy}{u}=\frac {du}{0}\] From the first pair of equation we obtain\[ u=\frac {dy}{dx}\] But \(du=0\) or \(u=C_{2}\). Hence the above becomes
\begin {align*} \frac {dy}{dx} & =C_{2}\\ y & =xC_{2}+C_{1}\\ C_{1} & =y-xC_{2} \end {align*}
Since \(C_{2}=F\left ( C_{1}\right ) \) where \(F\) is arbitrary function, then \[ u\left ( x,y\right ) =F\left ( y-ux\right ) \]
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