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Math Help - General Solution of a PDE

  1. #1
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    General Solution of a PDE

    Find the general solution of the following PDE.

    uxxyy = 0

    I know I have to go from right to left i.e integrate with respect to y, then y then x and x.

    So here it goes

    uxxy = F(x) Where F(x) is a constant in terms of x.

    uxx =yF(x) +F1(x) Where F1(x) is a constant in terms of x.

    ux = yxF(x) + xF1(x) + F2(y) Where F2(y) is a constant in terms of y.

    u = u(x,y) = 1/2 (yx2F(x) ) + 1/2 (x2F1(x) ) + xF2(y) + F3(y) Where F3(y) is a constant in terms of y.

    Is that correct?

    Thanks
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  2. #2
    GJA
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    Re: General Solution of a PDE

    Hi, princessmath.

    Nice work so far. Something did catch my eye. You have the line

    u_{xx}=yF(x) + F_{1}(x)
    .

    We need to be careful at this point, because the next step is to take an antiderivative with respect to x. Since there are functions of x on the right hand side we cannot do this by putting x's in front of them like we were able to put y in front of F(x) in the previous integration.

    Does this make sense? Let me know if there are any questions.

    Good luck!
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  3. #3
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    Re: General Solution of a PDE

    Hey GJA,

    Do you mean I have to change F(x) to F(x^2/2) if we were to integrate wrt x?

    The y in front is fine because the LHS has no y.

    is that what you mean?
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  4. #4
    GJA
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    Re: General Solution of a PDE

    Hi again. I'm glad you're sticking with it on this tricky little problem!

    Let's start from

    u_{xx}=yF(x)+F_{1}(x).

    We want to integrate both sides with respect to x now. But we have a problem because we don't know what F(x) and F_{1}(x) are; all we know is that they are functions of x.

    I think if we use a concrete function for F_{1}(x) what I'm getting at might be a little more clear.

    Note: I'm ignoring F(x) for now because I'm trying to demonstrate why after integrating we don't have xF_{1}(x).

    For example, pretend F_{1}(x)=e^{x}. Then we have

    u_{xx}=yF(x)+e^{x}.

    When you take an antiderivative of the right hand side e^{x} stays e^{x}, it does NOT become xe^{x}=xF_{1}(x).

    Does that help demonstrate what we're getting at?

    Again, the main issue is that we don't know what F(x) and F_{1}(x) are specifically; they can be any functions with x's in it. So, at best, after we take an antiderivative we can write

    u_{x}=y\int F(x)dx+\int F_{1}(x)dx+F_{2}(y),

    where F_{2}(y) is a function of y only, and so is constant with respect to x.

    Again, good work. Keep working hard and asking questions and it will make sense!

    Good luck!
    Last edited by GJA; August 18th 2012 at 08:17 AM.
    Thanks from princessmath
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  5. #5
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    Re: General Solution of a PDE

    Oh I see what you mean. Yeah I never thought about it like that. So it's a function in terms of x. Hmm let me think about it and see if I can produce a final solution.

    Thanks a lot!
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  6. #6
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    Re: General Solution of a PDE

    Can you generalize it to say Integral of F(x) is equal to I(x) where I(x) = S F(x) dx and then will continue the same process? Do you think that is valid?
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  7. #7
    GJA
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    Re: General Solution of a PDE

    If you want to simplify the notation a bit and write

    I(x)=\int F(x)dx

    and

    I_{1}(x)=\int F_{1}(x)dx,

    that's no problem at all
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  8. #8
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    Re: General Solution of a PDE

    I got it!

    So basically I get this general solution where I have something that looks like this.

    U(x,y) = yM(x) + M1(x) + xF2(y) +F3(y)

    Where M(x) = S I(x) and where I(x) = S F(x)
    sames goes for M1 = S I1(x) where I1(x) = S F1 (x).

    Is there a simpler way of doing it?

    Thanks
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