#### 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

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

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\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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