# Help with 1st order quasilinear PDE

1. Sep 24, 2009

### BobbyBear

I have to solve:

$$x(y^2 - z^2) \frac{\partial z}{\partial x} + y(z^2 - x^2) \frac{\partial z}{\partial y} + z(x^2 - y^2)$$

So, I write out the characteristic system of ODEs:

$$\frac{dx}{x(y^2 - z^2)} = \frac{dy}{y(z^2 - x^2)} = \frac{dz}{z(x^2 - y^2)}$$

Now, the variables aren't seperated so I can't integrate two pairs seperately, so what I did was use the componendo et dividendo rule for fractions to write :

$$\frac{dx}{x(y^2 - z^2)} = \frac{dy}{y(z^2 - x^2)} = \frac{dz}{z(x^2 - y^2)} = \frac{yzdx+xzdy+xydz}{xyz(y^2 - z^2)+xyz(z^2 - x^2)+xyz(x^2 - y^2)} = \frac{yzdx+xzdy+xydz}{0}$$

Thus
$$yzdx+xzdy+xydz=0$$

which is in integrable Pfaffian Equation, and its integration yields:

$$xyz=C_1$$

ie. I have one of the constants of integration of the characteristic system of ODEs.
But now I'm stuck because I don't know how to obtain the other :S

I thought to use one pair of the ODE system, eg,

$$\frac{dx}{x(y^2 - z^2)} = \frac{dy}{y(z^2 - x^2)}$$

and substitute
$$z=C_1/(xy)$$

from the first integration, so that I'd be left with an ode:

$$\frac{dx}{x(y^2 - C_1^2/(xy)^2)} = \frac{dy}{y(C_1^2/(xy)^2) - x^2)}$$

from which in theory we could obtain a second integration constant, but this is too hard to solve (I think), so there must be an easier way to get the second integration constant, ie.

$$g(x,y,z)=C_2$$

And the general solution of the PDE would be

$$F[xyz,g(x,y,z)]=0,$$

with F an arbitrary function.

Thank you :)

2. Sep 25, 2009

### kosovtsov

It seems to me that the general solution in implicit form to your PDE is as follows

\int_{-\infty}^ \infty F(z^2-x^2-y^2,c)(\frac{xy}{z})^c dc=0 ,

where F(a,c) is an arbitray function.

3. Sep 25, 2009

### BobbyBear

Oh dear, how in the world did you get that?