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Given that

$yu_x+xu_y=xy$ , $x\geqslant0$, $y\geqslant0$ with $u(0,y)=e^{-y^2}$ for $y>0$ , and $u(x,0)=e^{-x^2}$ for $x>0$.

My doubt is that how to use initial values in this case?

The answer given is

$ \left\{ \begin{aligned} \frac12y^2 + e^{-(x^2-y^2) } \quad for \quad x>y\\ \frac12 x^2 + e^{-(y^2-x^2) } \quad for \quad x<y \end{aligned} \right.$

Please guide me.

I got the solution.Let me answer it.

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  • $\begingroup$ Ok answer it then. $\endgroup$ – Kushal Bhuyan Dec 8 '15 at 14:03
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$yu_x+xu_y=xy$ , $x\geqslant0$, $y\geqslant0$ with $u(0,y)=e^{-y^2}$ for $y>0$ , and $u(x,0)=e^{-x^2}$ for $x>0$.

Dividing it by $xy$, we get, ${u_x\over x}$+ ${u_y\over y}$=1

${dx\over dt} = { 1\over x}$ with $x(0)=0$ $\implies$${ x^2\over 2}=t$

${dy\over dt}={1\over y}$ with $y(0)=y_0$ $\implies$ ${ y^2\over 2}=y_0+t$ $\implies$ ${ y^2\over 2}-t=y_0 $ $\implies$ ${ y^2\over 2}- { x^2\over 2}=y_0$

${ du\over dt}=1$ with $u(0)=f(y_0)$ $\implies$ $u=t+f(y_0)$ $\implies$ $u= { x^2\over 2}+f( -{ x^2\over 2}$ +${ y^2\over 2} )$

Now $u(x,0)={ x^2\over 2}+ f( -{ x^2\over 2})$

$\implies$ $e^{-x^2}= { x^2\over 2}+ f( -{ x^2\over 2})$

put $-{ x^2\over 2}=t$ ,we get $e^{2t}-t=f(t)$ $\implies$ $u={ x^2\over 2} +e^{2(-{ x^2\over 2} + { y^2\over 2})} -{ x^2\over 2}+{ y^2\over 2}$

$\implies$ $u= { y^2\over 2}+ e^{ -({ x^2 } - { y^2 })} $

Also $u(0,y)=e^{-y^2}=f( { y^2\over 2}) $

Put ${y^2\over2}=t$, we get $e^{-2t}= f( t )$
$\implies$ $u={x^2\over 2}+e^{ -2(-{ x^2\over2 } + { y^2 \over 2})} $

$\implies$ $u={x^2\over 2}+e^{ ( { x^2 } - { y^2 })} $

Or $u={x^2\over 2}+e^{ -({ y^2 } - { x^2 })} $

I still have a doubt. Why are $x>y$ and $y>x$ written in the final answer?

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