# About finding the general solution of first-order totally nonlinear PDE with two independent variables

Recently I feel very interested about finding the general solution of first-order totally nonlinear PDE with two independent variables. However, most PDE books only discussed little about finding the general solution of first-order totally nonlinear PDE with two independent variables, even for http://books.google.com.hk/books?id=hkWDQ57NlksC&pg=PA1&dq=Partial+Differential+Equations+by+Bhamra&hl=zh-CN&sa=X&ei=8mhPUZeYBciaiAe3m4HADw&ved=0CDEQ6AEwAA.

I known that for $F\left(x,y,u,\dfrac{\partial u}{\partial x},\dfrac{\partial u}{\partial y}\right)=0$ , let $p=\dfrac{\partial u}{\partial x}$ and $q=\dfrac{\partial u}{\partial y}$ , the PDE is related to the following system of ODEs

$\begin{cases}\dfrac{dx}{dt}=\dfrac{\partial F}{\partial p}\\\dfrac{dy}{dt}=\dfrac{\partial F}{\partial q}\\\dfrac{du}{dt}=p\dfrac{\partial F}{\partial p}+q\dfrac{\partial F}{\partial q}\\\dfrac{dp}{dt}=-\dfrac{\partial F}{\partial x}-p\dfrac{\partial F}{\partial u}\\\dfrac{dq}{dt}=-\dfrac{\partial F}{\partial y}-q\dfrac{\partial F}{\partial u}\end{cases}$

However, unlike the linear and quasilinear cases, since there also contains $p$ and $q$ , even the above system can perfectly solved for $x(t)$ , $y(t)$ , $u(t)$ , $p(t)$ and $q(t)$ , I still have no concept about the clear route to combine them to get the general solution like http://en.wikipedia.org/wiki/Method_of_characteristics#Example.

So I am thinking some alternatives about finding the general solution of first-order totally nonlinear PDE with two independent variables.

For example the PDE $u_xu_y=xy$ , I found the procedure in Solve PDE using method of characteristics but the procedure also consider the condition $u(x,0)=x$ so it is directly not suitable for finding the general solution of $u_xu_y=xy$ .

So can I for example modify the condition as $u(x,0)=f(x)$ and make some corresponding modifications of the procedure so than I can find the general solution of $u_xu_y=xy$ ?

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