2.1.12 $$x u_y + y u_x = -4 x y u$$ and $$u(x,0)=e^{-x^2}$$

problem number 12

Taken from Mathematica help pages

Solve for $$u(x,y)$$ $x u_y + y u_x = -4 x y u$ with initial value $$u(x,0)=e^{-x^2}$$

Mathematica

$\left \{\left \{u(x,y)\to e^{-x^2-y^2}\right \}\right \}$

Maple

$u \left (x , y\right ) = {\mathrm e}^{-x^{2}-y^{2}}$

Hand solution

Solve $xu_{y}+yu_{x}=-4xyu$ with $$u\left ( x,0\right ) =e^{-x^{2}}$$.

Solution

Let $$u\equiv u\left ( x\left ( y\right ) ,y\right )$$. We’ve taken $$y$$ as the independent variable for $$x\left ( y\right )$$ here, since the initial conditions has $$y\left ( 0\right )$$ in it. The PDE can be written as\begin {equation} u_{y}+\frac {y}{x}u_{x}=-4yu\tag {1} \end {equation} Then\begin {equation} \frac {du}{dy}=\frac {\partial u}{\partial x}\frac {dx}{dy}+\frac {\partial u}{\partial y}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dy} & =-4yu\tag {3}\\ \frac {dx}{dy} & =\frac {y}{x}\tag {4} \end {align}

Solving (3) gives\begin {align} \ln \left \vert u\right \vert & =-\frac {4y^{2}}{2}+C_{1}\nonumber \\ u & =C_{1}e^{-2y^{2}}\tag {5} \end {align}

At $$y=0$$, using initial conditions the above becomes$e^{-x\left ( 0\right ) ^{2}}=C_{1}$ (5) becomes\begin {align} u & =e^{-x\left ( 0\right ) ^{2}}e^{-2y^{2}}\nonumber \\ & =e^{-x\left ( 0\right ) ^{2}-2y^{2}}\tag {5A} \end {align}

All what is left is to ﬁnd $$x\left ( 0\right )$$ to ﬁnish the solution. From (4)\begin {equation} \frac {x^{2}}{2}=\frac {y^{2}}{2}+C_{2}\tag {6} \end {equation} At $$y=0$$$\frac {x\left ( 0\right ) ^{2}}{2}=C_{2}$ Hence (6) becomes\begin {align*} \frac {x^{2}}{2} & =\frac {y^{2}}{2}+\frac {x\left ( 0\right ) ^{2}}{2}\\ x\left ( 0\right ) ^{2} & =x^{2}-y^{2} \end {align*}

Substituting the above in (5A) gives\begin {align*} u\left ( x\left ( y\right ) ,x\right ) & =e^{-\left ( x^{2}-y^{2}\right ) -2y^{2}}\\ & =e^{-x^{2}-y^{2}} \end {align*}

The following is a plot of the above solution showing the initial conditions are red line

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