# 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 $u_xu_y=xy$ ?

## 1 Answer

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.