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Math Help - Differential equation in a distribution space

  1. #1
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    Question Differential equation in a distribution space

    Hi everybody!

    I need your help to solve the following exercise:

    find the general solution of the equation
    x^2*u'(x) = pv(1/x)
    where u'(x) is the derivative of u with respect to x and pv(1/x) stands for 'principal value of 1/x'.

    The answer should be u(x) = c1 + c2*theta(x) + c3*delta(x) - 1/2*fp(1/x^2), where theta(x) is the Heaviside's function and fp(1/x^2) is the 'finite part of 1/x^2'.

    For the sake of clarity, I solved similar equations in distributions spaces making use of Green's functions, Fourier transforms and convolutions, etc., but I can't figure out how to approach this one.

    Thanks in advance for your suggestions!
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  2. #2
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    I hope someone can help me...
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  3. #3
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    Most people here seem to do Pure Mathematics, so you will be hard pressed to find help. Unfortunately, we don't cover Green's Functions until tomorrow, so I can't help yet.
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  4. #4
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    Jul 2010
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    I think it's too late for givin a usefull answer to mambutu, however i write for next viewer:
    First observe that: if we indicate with <br />
 f^{(m)} the m-order derivative of f
    if m<n
    then x^n \delta^{(m)}=0
    In fact (x^n \delta^{(m)}, \phi)=(\delta^{(m)},x^n \phi)=\pm(\delta,(x^n\phi)^{(m)}) = \pm(\delta, Ax^n+...+Bx^{n-m})=0
    So if u want to solve
    x^2 u^{(1)}=0
    we remember that \theta^{(1)}=\delta
    then
    u=c_1 + c_2\theta(x) + c_3\delta(x)
    this is the general solution.
    The particular solution is - 1/2 fp(\frac 1 {x^2})
    The full solution is the sum of both....

    i hope it's clear.
    Last edited by skyscreper; July 25th 2010 at 08:01 AM.
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