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Math Help - Finding all solutions for a given PDE

  1. #1
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    Finding all solutions for a given PDE

    The problem is stated as follows:

    ---

    a) Transform the differential equation

    \frac{\partial^2 f}{\partial x^2} - y^2 \frac{\partial^2 f}{\partial y^2} - y \frac{\partial f}{\partial y}=4y^4,\,

    with the substitution u=ye^x and v=ye^{-x} in the region y>0.

    b) Determine all solutions of class C^2 (that is, solutions with continuous second (partial) derivatives) for the differential equation in this region. The solutions should be expressed as functions of the variables x and y.

    ---

    My question is about part b), since I don't know how I should get started on that. I'll show some of my work on part a) for some context:

    \begin{cases}<br />
u = ye^x\\<br />
v = ye^{-x}<br />
\end{cases}<br /> <br />
\Leftrightarrow<br /> <br />
\begin{cases}<br />
x = \ln(\sqrt{\frac{u}{v}})\\<br />
y = \sqrt{uv}<br />
\end{cases}

    (since y>0).

    Also, by the looks of the PDE, we have f(x,y) = f(x(u,v),y(u,v)) = g(u,v), so we can use the chain rule:

    \begin{cases}<br />
\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}\\<br />
\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial v}<br />
\end{cases}<br />

    By finding the partials of x and y with respect to u and v, we get

    \begin{cases}<br />
\frac{\partial f}{\partial u} = \frac{1}{2u} \frac{\partial f}{\partial x} + \frac{\sqrt{v}}{2\sqrt{u}} \frac{\partial f}{\partial y}\\<br />
\frac{\partial f}{\partial v} = -\frac{1}{2v} \frac{\partial f}{\partial x} + \frac{\sqrt{u}}{2\sqrt{v}} \frac{\partial f}{\partial y}<br />
\end{cases}<br /> <br />
\Leftrightarrow<br /> <br />
\begin{cases}<br />
\frac{\partial f}{\partial x} = u \frac{\partial f}{\partial u} - v \frac{\partial f}{\partial v}\\<br />
\frac{\partial f}{\partial y} = \sqrt{\frac{u}{v}} \frac{\partial f}{\partial u} + \sqrt{\frac{v}{u}} \frac{\partial f}{\partial v}<br />
\end{cases}.

    Differentiating again, and substituting the relevant partials back into the original PDE, we eventually, after a few pages of work, get

    \frac{\partial^2f}{\partial u\, \partial v} = -uv

    (assuming I didn't mess up somewhere).


    I'm not sure what to make of that, other than how we have a function that depends on u and v, and consequently, on x and y. But, I guess that was kind of the premise for the problem. How do I approach part b)?

    I suppose I could integrate with respect to one variable at a time and get f(u,v) = -\frac{u^2 v^2}{4} + Cu +D or f(u,v) = -\frac{u^2 v^2}{4} + Cv +D (C and D being integration constants), but I suspect that won't cover all possible cases.
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  2. #2
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    If f_{uv} = - uv (which I agree with), then integrating wrt v gives

    f_u = -\dfrac{1}{2}u v^2 +A'(u) and then integrating wrt u gives

    f = - \dfrac{1}{4} u^2 v^2 + A(u) + B(v)

    where A and B are arbitrary functions of their arguments. Then back substitute giving

    f = - \dfrac{1}{4} y^4 + A\left(ye^x\right) + B\left(ye^{-x}\right).
    Last edited by Jester; February 10th 2011 at 10:34 AM. Reason: fixed latex
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  3. #3
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    Aah, of course you wouldn't get constants but rather arbitrary functions of the other variable when integrating. Thank you!

    (I'm hoping the missing minus sign after the second integration was a LaTeX mishap rather than a lack of understanding on my part.)
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