# Semi Linear PDE

• Feb 14th 2011, 07:39 AM
bugatti79
Semi Linear PDE
$
\displaystyle -2xy \frac{\partial u}{\partial x} +4x \frac{\partial u}{\partial y}=4xy-yu
$
.

$u(x,0)=2 \sqrt x$

$
\displaystyle \frac{dy}{dx}=\frac{4x}{-2xy}=\frac{-2}{y} \implies \frac{y^2}{2}=-2x+k
$

$
\displaystyle \frac{du}{dx}=\frac{4xy-yu}{-2xy}=-2+\frac{u}{2x}
$
. Solve using an integrating factor where $p=\frac{-1}{2x}$ giving

$
\displaystyle u x^{\frac{-1}{2}}= \int -2 x^{\frac{-1}{2}} dx=-4x^{\frac{1}{2}}+f(\frac{y^2}{2}+2x)
$
Therefore

$
\displaystyle u(x,y)=-4x+x^{\frac{1}{2}} f(\frac{y^2}{2}+2x)
$

Using intial conditions gives $\displaystyle 2 \sqrt x=-4x+x^{\frac{1}{2}}f(2x)$. Let $\displaystyle t=2x \implies \frac{t}{2}=f(t)$ therefore

$
\displaystyle f(t)=\{2\sqrt {\frac{t}{2}}+2t}\}({\frac{t}{2})^\frac{-1}{2}
$
. Therefore

$
\displaystyle u(x,y)=-4x+x^{\frac{1}{2}} \{2\sqrt {\frac{\frac{y^2}{2}+2x}{2}}+2(\frac{y^2}{2}+2x)\} (({\frac{y^2}{2}+2x)/2)^{\frac{-1}{2}}
$

Could anyone check this is right so far? the above equation doesnt look correct.
Thanks
• Feb 15th 2011, 03:55 AM
Jester
It's correct but why not simple your $f(t)$ a little. Such as

$f(t) = 2 + \sqrt{8t}$.