2.5.2 Inviscid Burgers with I.C. \(u_x+ u u_y = 0\) and \(u(x,0)=\frac {1}{x+1}\)

problem number 92

Taken from Mathematica Symbolic PDE document

quasilinear first-order PDE, scalar conservation law with initial value

Solve for \(u(x,y)\)

\[ u_x+ u u_y = 0 \]

With \(u(x,0)=\frac {1}{x+1}\)

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, y], {x}] + u[x, y]*D[u[x, y], {y}] == 0; 
ic  = u[x, 0] == 1/(x + 1); 
sol =  AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, ic}, u[x, y], {x, y}]], 60*10]];
 

\[\left \{\left \{u(x,y)\to \frac {y+1}{x+1}\right \}\right \}\]

Maple ✓

restart; 
interface(showassumed=0); 
pde := diff(u(x, y), x) + u(x,y)*diff(u(x, y),y) =0; 
ic  := u(x,0)=1/(x+1); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,y))),output='realtime'));
 

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

Hand solution

Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)

\begin{align} \frac {dx}{ds} & =1\tag {1}\\ \frac {dy}{ds} & =u\tag {2}\\ \frac {du}{ds} & =0\tag {3}\end{align}

With initial conditions at \(s=0\)

\[ x\left ( 0\right ) =t_{1},y\left ( 0\right ) =t_{2},u\left ( 0\right ) =t_{3}\]

We are given that \(u\left ( x,0\right ) =\frac {1}{1+x}\). This initial condition translates to

\begin{equation} t_{3}=\frac {1}{1+t_{1}},t_{2}=0\tag {4}\end{equation}

Equation (1) gives

\begin{equation} x=s+t_{1}\tag {5}\end{equation}

Equation (2) gives

\begin{align} y & =su+t_{2}\nonumber \\ & =su\tag {7}\end{align}

Equation (3) gives

\[ u=t_{3}\]

Hence the solution is

\begin{align*} u & =t_{3}\\ & =\frac {1}{1+t_{1}}\\ & =\frac {1}{1+\left ( x-s\right ) }\\ & =\frac {1}{1+\left ( x-\frac {y}{u}\right ) }\end{align*}

Solving for \(u\) gives

\begin{align*} u\left ( 1+\left ( x-\frac {y}{u}\right ) \right ) & =1\\ u+xu-y & =1\\ u\left ( 1+x\right ) & =1+y\\ u & =\frac {1+y}{1+x}\end{align*}

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