Solving the Partial Derivative Equation: xy*z^2

[tehno]

$$\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}=xyz^2$$

People,do we know how to solve this?
I'm looking for the explicite solution z=f(x,y) so far I'm unable to solve this .Thinking of it for a half a day without much of the progress.Even though I haven't tryed all dirty tricks yet.

 

[tehno]

Expert help would be appreciated.

 

[arildno]

Well, you might see how separation of variables can help you:
$$z=g(x)h(y)\to{g}'gh'h=xy(gh)^{2}\to\frac{g'}{gx}=\frac{yh}{h'}$$

Since LHS is a function of x, whereas RHS is a function of y, we get:
$$g'(x)=Kxg(x), h'(y)=\frac{1}{K}yh(y)\to{g}(x)=Ae^{\frac{Kx^{2}}{2}},h(y)=Be^{\frac{y^{2}}{2K}}$$
with A, B, K constants.

Note that it by no means follows from this that all solutions of your PDE must be on this form!

For example, let f(x,y)=G(xy). Then:
$$\frac{\partial{f}}{\partial{x}}=yG',\frac{\partial{f}}{\partial{y}}=xG'\to{(G')^{2}}=G^{2}\to{G'}=\pm{G}, z_{+}(x,y)=Ae^{xy},z_{-}(x,y)=Be^{-xy}$$

These radically different special solutions should tell you that to arrive at some simple, generally valid formula for the solutions is impossible. That's usually the case with non-linear PDE's in particular.

 

[tehno]

Thanks arildno for the inputs.
Neverthless,I have good news:I managed today to find "all" PDE solutions as representation in formal explicite expression z=f(x,y).
Indeed ,as you indicate,that's really broad spectrum of the functions .

 

[arildno]

I dis-assume it is a "simple" formula..

 

[tehno]

I dis-assume it is a "simple" formula..

These radically different special solutions should tell you that to arrive at some simple, generally valid formula for the solutions is impossible. That's usually the case with non-linear PDE's in particular.

But I arrived at the generally valid (and explicite!) formula that describes all the solutions of this PDE.
For the first time I effectively find this forum helpful for me,becouse your suggestion of thinking in "exponential terms" was the idea (there's no a classical linear superpositon ,however).
The method of the separation of variables is natural there,but I was blind or something haven't had noticed it before.

The most important was the first step :to write PDE as

$$z^{-1}\frac{\partial z}{\partial x}\cdot z^{-1}\frac{\partial z}{\partial y}=xy$$
Other work I will leave as an exercise.
Here's my general formula:

$$z=e^{\frac{C^2x^2+y^2}{2C}+ \phi (C)}$$


where $\phi$ is some derivabile arbitrary function such that:

$\phi '(C) +\frac{x^2}{2}-\frac{y^2}{2C^2}=0$

 

[arildno]

Oh, really?
As far as I know,
$$\frac{1}{z}\frac{\partial{z}}{\partial{x}}=\frac{\partial}{\partial{x}}(\log(|z(x,y)|)+f(y))$$
where f is an arbitrary function of y
Did you take that into account as well?


EDIT:

Personally, I'd have thought some progress might be made to make a switch to polar, or hyperbolic coordinates, but I haven't done so myself..

 

[tehno]

I collected all the solutions by my formula, neglecting variable interchangable ones,becouse I solved PDE.Try to solve PDE.You will see.

 

[arildno]

Hmm...since I'm not very clever, I won't be able to do so.
Perhaps you can post your solution in a while?

 

[tehno]

All righty.I think ,at least,I owe you that so here is the method.

Write PDE as:
$$\frac{\partial z}{z\partial x}\cdot \frac{\partial z}{z\partial y}=xy$$

Let's search for the solutions of this PDE in implicite form.
In order to do that note:

$$(ln\ z)'=\frac{1}{z}$$.

Having on mind this and by introducing substitutions:

$$\frac{\partial u}{\partial x}=p,\frac{\partial u}{\partial y}=q$$

our PDE turns into:

$$p\cdot q=xy$$

Voila!
The separation of the variables occurs.We have:

$$\frac{p}{x}=\frac{y}{q}=C \rightarrow du=Cxdx+\frac{ydy}{C}\rightarrow u=\frac{Cx^2}{2}+\frac{y^2}{2C}+C_{1}$$


Therefore the complete integral of PDE is:

$$ln\ z=\frac{Cx^2}{2}+\frac{y^2}{2C} +C_{1}$$


General solution of first order PDE in variables ,knowing one solution H
is obtained per definition from the system:


$$H(x,y,z,a,b)=0$$
$$\frac{\partial H}{\partial a} + \frac{\partial H}{\partial b}\cdot f'(a)=0$$

So,under assuption that between parameters a and b exists functional relation f(a)=b ,where f is an arbitrary derivabile function the general solution ,after solving the system can be written as:

$$ln\ z=\frac{Cx^2}{2}+\frac{y^2}{2C}+2C + f(C)$$

$$f'(C) +\frac{x^2}{2}-\frac{y^2}{2C^2}=0$$

Where f is arbitrary function and C real constant.


This solution isn't equivalent in notation with the one given previously.
However, I stressed that I was looking for only solutions given in explicite form z=f(x,y).Hence..