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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.

:uhh:

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- Thread starter tehno
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People,do we know how to solve this?

I'm looking for the explicite solution

:uhh:

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Expert help would be appreciated.

- #3

arildno

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Well, you might see how separation of variables can help you:

[tex]z=g(x)h(y)\to{g}'gh'h=xy(gh)^{2}\to\frac{g'}{gx}=\frac{yh}{h'}[/tex]

Since LHS is a function of x, whereas RHS is a function of y, we get:

[tex]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}}[/tex]

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:

[tex]\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}[/tex]

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.

[tex]z=g(x)h(y)\to{g}'gh'h=xy(gh)^{2}\to\frac{g'}{gx}=\frac{yh}{h'}[/tex]

Since LHS is a function of x, whereas RHS is a function of y, we get:

[tex]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}}[/tex]

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:

[tex]\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}[/tex]

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.

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Neverthless,I have good news:I managed today to find "all" PDE solutions as representation in formal explicite expression

Indeed ,as you indicate,that's really broad spectrum of the functions .

- #5

arildno

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

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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

[tex]z^{-1}\frac{\partial z}{\partial x}\cdot z^{-1}\frac{\partial z}{\partial y}=xy[/tex]

Other work I will leave as an exercise.

Here's my general formula:

[tex]z=e^{\frac{C^2x^2+y^2}{2C}+ \phi (C)}[/tex]

where [itex]\phi[/itex] is some derivabile arbitrary function such that:

[itex]\phi '(C) +\frac{x^2}{2}-\frac{y^2}{2C^2}=0[/itex]

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- #7

arildno

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Oh, really?

As far as I know,

[tex]\frac{1}{z}\frac{\partial{z}}{\partial{x}}=\frac{\partial}{\partial{x}}(\log(|z(x,y)|)+f(y))[/tex]

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..

As far as I know,

[tex]\frac{1}{z}\frac{\partial{z}}{\partial{x}}=\frac{\partial}{\partial{x}}(\log(|z(x,y)|)+f(y))[/tex]

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..

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- #9

arildno

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Perhaps you can post your solution in a while?

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

Write PDE as:

[tex]\frac{\partial z}{z\partial x}\cdot \frac{\partial z}{z\partial y}=xy[/tex]

Let's search for the solutions of this PDE in implicite form.

In order to do that note:

[tex](ln\ z)'=\frac{1}{z}[/tex].

Having on mind this and by introducing substitutions:

[tex]\frac{\partial u}{\partial x}=p,\frac{\partial u}{\partial y}=q [/tex]

our PDE turns into:

[tex]p\cdot q=xy[/tex]

Voila!

The separation of the variables occurs.We have:

[tex]\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}[/tex]

Therefore the complete integral of PDE is:

[tex]ln\ z=\frac{Cx^2}{2}+\frac{y^2}{2C} +C_{1}[/tex]

General solution of first order PDE in variables ,knowing one solution H

is obtained per definiton from the system:

[tex]H(x,y,z,a,b)=0[/tex]

[tex]\frac{\partial H}{\partial a} + \frac{\partial H}{\partial b}\cdot f'(a)=0[/tex]

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:

[tex]ln\ z=\frac{Cx^2}{2}+\frac{y^2}{2C}+2C + f(C)[/tex]

[tex]f'(C) +\frac{x^2}{2}-\frac{y^2}{2C^2}=0[/tex]

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..

Write PDE as:

[tex]\frac{\partial z}{z\partial x}\cdot \frac{\partial z}{z\partial y}=xy[/tex]

Let's search for the solutions of this PDE in implicite form.

In order to do that note:

[tex](ln\ z)'=\frac{1}{z}[/tex].

Having on mind this and by introducing substitutions:

[tex]\frac{\partial u}{\partial x}=p,\frac{\partial u}{\partial y}=q [/tex]

our PDE turns into:

[tex]p\cdot q=xy[/tex]

Voila!

The separation of the variables occurs.We have:

[tex]\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}[/tex]

Therefore the complete integral of PDE is:

[tex]ln\ z=\frac{Cx^2}{2}+\frac{y^2}{2C} +C_{1}[/tex]

General solution of first order PDE in variables ,knowing one solution H

is obtained per definiton from the system:

[tex]H(x,y,z,a,b)=0[/tex]

[tex]\frac{\partial H}{\partial a} + \frac{\partial H}{\partial b}\cdot f'(a)=0[/tex]

So,under assuption that between parameters a and b exists functional relation

[tex]ln\ z=\frac{Cx^2}{2}+\frac{y^2}{2C}+2C + f(C)[/tex]

[tex]f'(C) +\frac{x^2}{2}-\frac{y^2}{2C^2}=0[/tex]

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

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