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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:

- 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 .

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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..

But I arrived at the generally valid (and explicite!) formula that describes all the solutions of this PDE.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.

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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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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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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