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

problem number 85

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


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

Maple


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

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