How to Separate Charpit Equations

Question

I've been attempting to solve this non-linear PDE

$$4\Omega x^2 y^2 \frac{\partial z}{\partial y} -x^2 y (\frac{\partial z}{\partial x})^2 + 2x^2 y^2 E-N^2=0$$

using Charpit's method. The variables $\Omega$, $E$, and $N$ are constants. I've derived the relation between Charpit's auxillary equations,

$$\frac{x}{N^2}dp=-\frac{y}{N^2}dq=-\frac{1}{x^2 y p}dx=\frac{1}{2\Omega x^2 y^2} dy$$

but I have been unable to separate these to obtain the second function relating $p$ and $q$. Most of the examples I have seen are separated very easily with terms such as

$$\frac{dp}{p}=\frac{dq}{q}$$

or something similar.

A little background on Charpit's method and the Method of Characteristics...

We begin by defining the primary non-linear PDE

$$F(x,y,z,p,q)=4\Omega x^2 y^2 \frac{\partial z}{\partial y} -x^2 y (\frac{\partial z}{\partial x})^2 + 2x^2 y^2 E-N^2=0$$

Then, by expanding the total derivatives

$$\frac{dF}{dx}=0$$ and $$\frac{dF}{dy}=0$$

in terms of partial derivatives, along with the definitions

$$\frac{dx}{dt}\equiv\frac{\partial F}{\partial p}$$ and $$\frac{dy}{dt}\equiv\frac{\partial F}{\partial q}$$

we get the five Charpit Equations:

$$\frac{dx}{dt}=\frac{\partial F}{\partial p}\\ \frac{dy}{dt}=\frac{\partial F}{\partial q}\\ \frac{dp}{dt}=-\frac{\partial F}{\partial x}-p\frac{\partial F}{\partial z}\\ \frac{dq}{dt}=-\frac{\partial F}{\partial y}-q\frac{\partial F}{\partial z}\\ \frac{dz}{dt}=p\frac{dx}{dt}+q\frac{dy}{dt}$$

By eliminating $dt$ they can all be set equal to each other (see Charpit's auxillary equations mentioned above), and any two (or more) can be used to integrate a total derivative relating $p$ and $q$. This new relation is then substituted into the original differential equation, which can then be written as

$$dz=p(x,y)dx+q(x,y)dy$$

which is a total differential which can be directly integrated to find the solution z=z(x,y).

Answer 1

N.B.: I checked your calculations a couple of times, and I got a different sign in the Charpit equations:

$$ -\frac{x}{N^2}dp=-\frac{1}{x^2 y p}dx=\frac{1}{2\Omega x^2 y^2} dy $$ (Notice the minus sign in the leftmost expression. Also, because you can solve the equation $F=0$ for $q$, you don't need to worry about it. It would be enough to solve the reduced Charpit equations in $xyp$-space.)

No general method to integrate the flow of a vector field in dimension above $2$ is known (or likely to be found, because of chaos theory). However, there are some tricks to try that work in many cases. A powerful ansatz is to exploit symmetry when you notice it. In this particular, case, inspection shows that the reduced Charpit equations are invariant under the following scaling: $$ (x,y,p) \longmapsto (\lambda x, \lambda^{-4}y,\lambda p). $$ This means that you can make a change of variables that will 'filter out' this dependence. A little experimentation shows that, if you set $k = 2N/\Omega$ and introduce new variables $(u,v)$ so that $$ y =\frac{k^2}{x^4u}\qquad\text{and}\qquad p = \tfrac12\Omega\,xv $$ then your equations are expressed as $$ \mathrm{d}u = \frac{4u(1{-}v)}{v}\,\frac{\mathrm{d}x}{x}\,,\qquad \mathrm{d}v = \frac{(u{-}v^2)}{v}\,\frac{\mathrm{d}x}{x}\,. $$ Thus, you get the relation $$ 4u(1{-}v)\,\mathrm{d}v + (v^2{-}u)\,\mathrm{d}u=0. $$ In particular, finding integral curves of the original equation requires that you find integral curves of this relation in the $uv$-plane. There is a $1$-parameter family of such curves, and they can be understood by looking at the corresponding phase portrait in the plane. (This does not appear to be an explicitly integrable case, though. However, I note that $u=0$ is an integral curve of the system and that $(u,v)=(1,1)$ and $(u,v)=(0,0)$ are singular points.)

Once you have such an integral curve, you can then use it to compute $x$ as a function along this curve by integration (i.e., quadrature).

I'm afraid that you won't be able to integrate these equations explicitly, but you can understand the asymptotics of the integral curves using the above phase portrait and integrations. This will give you qualitative features of the solutions of the above equations for given initial conditions, given sufficient patience.