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Math Help - [SOLVED] Linear, homogenous PDE

  1. #1
    Niles
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    [SOLVED] Linear, homogenous PDE

    Hi all

    Please take a look at the example at the bottom (at eq. 18:18, at page 619):

    Mathematical Methods for Physics and ... - Google Bogsøgning

    Q1: In case (ii), why do they add g(x^2+y^2)?
    Q2: Why do they not add it in case (i)?
    Q3: In case (ii), if I chose f(p) = p + 1 instead of f(p) = 2p, then the solution would be u(x,y) = 1+ x^2 +y^2 + g(x^2 + y^2), right?

    I hope you can help me; I find this really hard to understand, and I've spent hours trying to find out, but I find that the book is poorly written. They don't emphasize the important things at all, and the reader is left behind with so many questions.

    Thanks in advance.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Niles View Post
    Hi all

    Please take a look at the example at the bottom (at eq. 18:18, at page 619):

    Mathematical Methods for Physics and ... - Google Bogsøgning

    Q1: In case (ii), why do they add g(x^2+y^2)?
    Q2: Why do they not add it in case (i)?
    I think the short answer is that the boundary conditions are very different in the two cases. In case (i) you are given the value of u(x,0) for all x, but in case (ii) you only have the value at one point, u(1,0). In case (i), the boundary condition is strong enough to determine the solution uniquely. But in case (ii) there is not enough information for that, and the general solution has to include an arbitrary function of x^2+y^2.
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