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How to solve the following integral-differential equations?

  1. Dec 12, 2012 #1
    hello, everyone. Recently I confront with the integral differential equations(ide's) in my research. Anyone can help to solve h(t,s), A1(t,s), A2(t,s), B1(t,s), B2(t,s)?

    The file attached as following. Thanks a lot. Numerical method is also okay!

    Attached Files:

  2. jcsd
  3. Dec 12, 2012 #2


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    Some of your equations look like they will be simplified if you Laplace transform them in s or t (or both).

    The key rule is the convolution rule:

    $$\mathcal L \left[ \int_0^t dt'~K(t-t')f(t')\right] = \hat{K}(s)\hat{f}(s).$$

    Since some of your integrals are of this convolution form, Laplace transforming your equations looks like it may leave you with an algebraic equation for the Laplace transform of the function(s) you want to solve for.
  4. Dec 13, 2012 #3
    Hi, Dear Mute:
    Thanks for your the key rule: Laplace transform of the convolution form. However, my beta function only has numerical values(obtained by matlab), not a closed form(because the upper and lower limit of the integration). So how can I get a Laplace transform for numerical values of the beta function(by any command in other package)? Otherwise, how can I do the inverse Laplace transform of this complex algebraic equation(because the Laplace transform of beta is very complex, I think), by the command residue in mathematica or others?
  5. Dec 16, 2012 #4


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    As the Laplace transform is just ##\mathcal L [f(t)](y) = \int_0^\infty dt~\exp(-yt)f(t)##, you could easily solve for the Laplace transform numerically. The tricky part would probably be inverse transforming numerically.

    Alternatively, you could fit a functional form to your ##\beta## function in terms of standard functions and compute the Laplace transform of the fit if you want to try and solve the equations analytically.

    I suppose you could also combine both approaches and calculate the Laplace transform numerically and then fit a functional form to the Laplace transform of your ##\beta##.
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