Linear partial differential eqn.

Posted September 22nd, 2007 by vic777

General questions

How can I reduce the following eq. to canonical form?
$u_{xx} + yu_{yy} +\frac{1}{2}u_y + 4yu_x = 0$

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Answer

On September 22nd, 2007 Structure says:

This equation is of different type in different regions.
In half plane y<0 is of hyperbolic type and in y>0 is of elliptic type.
Let us begin with y<0
Characteristic equation is $y'^2+y=0$ so $y'(x)=-\sqrt{-y}$ and $y'(x)=\sqrt{-y}$
Write in form
$-\frac{y'(x)}{\sqrt{-y}}=1$
and you have $2\sqrt{-y}=x+c_1$
and
$-\frac{y'(x)}{\sqrt{-y}}=-1$
and you have $2\sqrt{-y}=-x+c_2$
take
$\xi(x,y)=x+2\sqrt{-y}$
$\eta(x,y)=x-2\sqrt{-y}$
and
$u(x,y)=v(\xi (x,y),\eta (x,y))=v(x+2\sqrt{-y},x-2\sqrt{-y})$
$\frac{\partial u}{\partial x}(x,y)=\frac{\partial v}{\partial \xi}(x+2\sqrt{-y},x-2\sqrt{-y})+\frac{\partial v}{\partial \eta}(x+2\sqrt{-y},x-2\sqrt{-y})$
Replacing into the equation we get
$4\frac{\partial^2 v}{\partial \xi\partial\eta}$
$\frac{\partial u}{\partial y}(x,y)=-\frac{1}{\sqrt{-y}}\frac{\partial v}{\partial \xi}(x+2\sqrt{-y},x-2\sqrt{-y})+\frac{1}{\sqrt{-y}}\frac{\partial v}{\partial \eta}(x+2\sqrt{-y},x-2\sqrt{-y})$

$\frac{\partial ^2 u}{\partial x^2}=\frac{\partial ^2 v}{\partial \xi^2}+2\frac{\partial ^2 v}{\partial \xi \partial \eta}+\frac{\partial ^2 v}{\partial \eta^2}$

$\frac{\partial ^2 u}{\partial y^2}=-\frac{1}{y}\frac{\partial ^2 v}{\partial \xi^2}+2\frac{\partial ^2 v}{\partial \xi \partial \eta}+\frac{\partial ^2 v}{\partial \eta^2}+\frac{1}{2y\sqrt{-y}}\frac{\partial v}{\partial \xi}-\frac{1}{2y\sqrt{-y}}\frac{\partial v}{\partial \eta}$

Replacing into the equation we get
$4\frac{\partial^2 v}{\partial \xi\partial\eta}+4y(\frac{\partial v}{\partial \xi}+\frac{\partial v}{\partial \eta})=0$
But $\xi-\eta=4\sqrt{-y}$
so $y=-\frac{1}{16}(\xi-\eta)^2$

Finally
$16\frac{\partial^2 v}{\partial \xi\partial\eta}-(\xi-\eta)^2(\frac{\partial v}{\partial \xi}+\frac{\partial v}{\partial \eta})=0$

Case 2
y>0
$y'^2+y=0$
$y'=i\sqrt{y}$
$\frac{y'}{\sqrt{y}}=i$
$2\sqrt{y}-xi=c$
Now we can take the real and the imaginary part for variable.
$u(x,y)=v(x,2\sqrt{y})$
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial \xi}$
$\frac{\partial u}{\partial y}=\frac{1}{\sqrt{y}}\frac{\partial v}{\partial \eta}$
$\frac{\partial ^2 u}{\partial x^2}=\frac{\partial ^2v}{\partial \xi^2}$
$\frac{\partial ^2u}{\partial y^2}=\frac{1}{y}\frac{\partial ^2v}{\partial \eta^2}-\frac{1}{2y\sqrt{y}}\frac{\partial v}{\partial \eta}$
Replacing into equation we get
$\frac{\partial^2 v}{\partial \xi^2}+\frac{\partial^2 v}{\partial \eta^2}+4y\frac{\partial v}{\partial \xi}=0$
Now it rest to replace y