# Solving $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$

Let $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$, how would you go about solving this?

So far, I have show it is hyperbolic everywhere except for the line $y=x$ and have been attempting to find the characteristic variables by solving:

$$\frac{d\phi}{dx}+\frac{x}{y}\frac{d\phi}{dy}=0$$ and $$\frac{d\psi}{dx}+\frac{d\psi}{dy}=0$$

I have $\psi=x-y$, but am stuck on the first pde. Once I have these I plan to $u_{\phi \psi}=0$, the canonical form, to find the general solution. I would appreciate letting me know i'm on the right lines and any help very much, thanks!

EDIT: $\phi=y^2-x^2$ is a solution, so by the chain rule I end up with $(x^2+y^2)u_{\phi \psi}=0$ is this correct? As the solution is meant to be $$u=\frac{1}{y-x}f(y^2-x^2)+g(y-x)$$

this seems to give

$$u=f(y^2-x^2)+g(y-x)$$

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$u(x,y) = x^3 - y^3$ also solves the equation, so your final result can't be right. –  Zarrax Nov 22 '11 at 14:34
Your substitution method has problems. Please follow the substitution method in math.stackexchange.com/questions/88825. –  doraemonpaul Jul 8 '12 at 22:07
Reminder that $u=\frac{1}{y-x}f(y^2-x^2)+g(y-x)$ should not simplify to $u=f(y^2-x^2)+g(y-x)$ . The meanings of them are different. –  doraemonpaul Aug 19 '12 at 2:08
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. –  doraemonpaul Sep 10 '12 at 1:30

Try let $v=x+y$ , $w=x-y$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial v}\dfrac{\partial v}{\partial x}+\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial x}=\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}\right)=\dfrac{\partial}{\partial v}\left(\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}\right)\dfrac{\partial v}{\partial x}+\dfrac{\partial}{\partial w}\left(\dfrac{\partial u}{\partial v}+\dfrac{\partial u}{\partial w}\right)\dfrac{\partial w}{\partial x}=\dfrac{\partial^2u}{\partial v^2}+\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}=\dfrac{\partial^2u}{\partial v^2}+2\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}$

$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial v}\dfrac{\partial v}{\partial y}+\dfrac{\partial u}{\partial w}\dfrac{\partial w}{\partial y}=\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}$

$\dfrac{\partial^2u}{\partial xy}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)=\dfrac{\partial}{\partial v}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)\dfrac{\partial v}{\partial x}+\dfrac{\partial}{\partial w}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)\dfrac{\partial w}{\partial x}=\dfrac{\partial^2u}{\partial v^2}-\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial vw}-\dfrac{\partial^2u}{\partial w^2}=\dfrac{\partial^2u}{\partial v^2}-\dfrac{\partial^2u}{\partial w^2}$

$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)=\dfrac{\partial}{\partial v}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)\dfrac{\partial v}{\partial y}+\dfrac{\partial}{\partial w}\left(\dfrac{\partial u}{\partial v}-\dfrac{\partial u}{\partial w}\right)\dfrac{\partial w}{\partial y}=\dfrac{\partial^2u}{\partial v^2}-\dfrac{\partial^2u}{\partial vw}-\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}=\dfrac{\partial^2u}{\partial v^2}-2\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}$

$\therefore y\left(\dfrac{\partial^2u}{\partial v^2}+2\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}\right)+(x+y)\left(\dfrac{\partial^2u}{\partial v^2}-\dfrac{\partial^2u}{\partial w^2}\right)+x\left(\dfrac{\partial^2u}{\partial v^2}-2\dfrac{\partial^2u}{\partial vw}+\dfrac{\partial^2u}{\partial w^2}\right)=0$

$2(x+y)\dfrac{\partial^2u}{\partial v^2}-2(x-y)\dfrac{\partial^2u}{\partial vw}=0$

$v\dfrac{\partial^2u}{\partial v^2}-w\dfrac{\partial^2u}{\partial vw}=0$

Let $z=\dfrac{\partial u}{\partial v}$ ,

Then $v\dfrac{\partial z}{\partial v}-w\dfrac{\partial z}{\partial w}=0$

This belongs to a PDE of the form http://eqworld.ipmnet.ru/en/solutions/fpde/fpde1104.pdf

$z=\dfrac{\partial u}{\partial v}=c\left(\int\dfrac{1}{v}~dv+\int\dfrac{1}{w}~dw\right)=c(\ln v+\ln w)=c(\ln vw)=C(vw)$

$u=\int C(vw)~dv=\int C(v)~d\left(\dfrac{v}{w}\right)=\int\dfrac{C(v)}{w}dv=\dfrac{C_1(v)+C_2(w)}{w}=\dfrac{C_1(x+y)+C_2(x-y)}{x-y}$

Note that when the independent variables in the arbitrary functions are more simple, the general solutions are more general.

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