# 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 $u_xu_y=xy$ ?

What you are trying to do is theoretically possible, but not always possible in practice because the calculus process can be blocked by an integration or by the inversion of a function extremely difficult.

The same hitch is encountered whatever the method used for solving the PDE.

For example, consider the PDE : $$\quad u_xu_y=xy \tag 1$$ $$u_y=\frac{xy}{u_x} \quad\to\quad u_{xy}=\frac{y}{u_x}-\frac{xyu_{xx}}{(u_x)^2}$$ $v=u_x \quad\to\quad v_{y}=\frac{y}{v}-\frac{xyv_{x}}{v^2}$ $$xyv_x+v^2v_y=yv \tag 2$$ The set of characteristic ODEs is : $\quad \frac{dx}{xy}=\frac{dy}{v^2}=\frac{dv}{yv}$

A first set of characteristic curves comes from $\frac{dy}{v^2}=\frac{dv}{yv} \quad\to\quad v^2-y^2=c_1$

A second set of characteristic curves comes from $\frac{dx}{xy}=\frac{dv}{yv}\quad\to\quad \frac{v}{x}=c_2$

The general solution of PDE (2) can be expressed on the form of an implicit equation : $$v^2-y^2=F\left(\frac{v}{x}\right) \tag 3$$ where $F$ is any differentiable function.

Without knowing what is the function $F$, it is not possible to obtain $u_x=v$ and then to continue up to $u(x,y)$.

So, a condition is needed in order to determine the function $F$.

Given a condition, for example $u(x,0)=f(x)$ with a given function $f(x)$, then $v(x,0)=u_x(x,0)=f'(x)$. Putting this condition into (3) theoretically allows to determine the function $F$. As a consequence, the general solution $v(x,y)$ of EDP (2) is obtained.

Then, with $u_x(x,y)=v(x,y)$ and $u_y(x,y)=\frac{xy}{v(x,y)}$ it is theoretically possible to get to the general solution of PDE (1).

But in practice, that is a different kettle of fish. First remember, we saw that we can't answer to your question if no condition is specified. Second, even with a condition such as $u(x,0)=f(x)$, the possibility to determine the function $F$ from (3) depends on the form of the given $f(x)$.

In the cases where the function $F$ can be derived, bringing it into (3) generally don't leads to an explicit form of $u_x(x,y)=v(x,y)$. As well for $u_y(x,y)$. Then, the integration for $u(x,y)$ becomes generally too complicated.

As a conclusion, a general solution for EPD (1) can be obtained, even on implicit form, only in particular cases of $f(x)$, which can be used as textbook cases.