2.5.1 Inviscid Burgers \(u_x + u u_y = 0\)

problem number 91

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);
 

\[-y +u \left (x , y\right ) x +f_{1} \left (u \left (x , y\right )\right ) = 0\]

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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