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PDE kind

  1. Feb 2, 2007 #1
    [tex]\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}=xyz^2[/tex]

    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:
     
  2. jcsd
  3. Feb 2, 2007 #2
    Expert help would be appreciated.
     
  4. Feb 2, 2007 #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.
     
    Last edited: Feb 2, 2007
  5. Feb 3, 2007 #4
    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 .
    :smile:
     
  6. Feb 3, 2007 #5

    arildno

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    I dis-assume it is a "simple" formula..
     
  7. Feb 4, 2007 #6
    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]
     
    Last edited: Feb 4, 2007
  8. Feb 4, 2007 #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..
     
    Last edited: Feb 4, 2007
  9. Feb 4, 2007 #8
    I collected all the solutions by my formula, neglecting variable interchangable ones,becouse I solved PDE.Try to solve PDE.You will see.
     
  10. Feb 4, 2007 #9

    arildno

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    Hmm...since I'm not very clever, I won't be able to do so.
    Perhaps you can post your solution in a while? :smile:
     
  11. Feb 5, 2007 #10
    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..
     
    Last edited: Feb 6, 2007
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