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Equations with integrals,

  1. Jun 27, 2012 #1
    Hi,

    How would one go about solving equations like [itex]∫^{b}_{a}f(s,t)g(s)ds[/itex]=g(t),for f(s,t). Could we turn it into a differential equation somehow?

    Thanks
     
  2. jcsd
  3. Jun 27, 2012 #2

    chiro

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    Hey 2pi.

    Have you tried first using the fundamental theorem of calculus and the chain rule to get rid of the integral expression (assuming s and t are unrelated and orthogonal)?
     
  4. Jun 27, 2012 #3
    Yeah s and t are independent, but isn't the problem the fact that a and b are constants,and we don't know what f is, and as such the FTC doesn't really get us anywhere useful?
     
  5. Jun 27, 2012 #4

    chiro

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    You will get a definition for the partial derivative in terms of f(s,t) and from that, you should be able to get something useful.

    Remember that the integral is done with respect to ds, so you can consider this partial derivative with respect to the fundamental theorem of calculus and through the chain rule, obtain a partial differential relationship for the integral relationship.

    You also take the derivative of the RHS with respect to s (partial derivative) and from this you get 0 (since g(t) will be considered more or less a constant).

    This means you will be left with an expression involving partial with respect to s involving the chain rule of f(s,t)g(s) and the RHS will be zero. This should give you a PDE.

    More information for this kind of problem, look for integro-differential equations either on the internet or in textbooks.
     
  6. Jun 27, 2012 #5
    Would you mind make clear what is known and what is unknown in the equation.
    Is the function g(t) known and then are you searching an unknown function f(s,t) consistant with the équation ?
    Or, is the function f(s,t) known and then are you searching an unknown function g(t) consistant with the équation ?
    Since there are two parameters a and b involved into the relationship, necesserally they will appear in the function that we are looking for.
    So, if f(s,t) is unknown, then a and b will appear in g(t) and, as a matter of fact, the analytical expression of g(t) is g(a,b,t). Rigth or not ?
    If g(t) is unknown, then a and b will appear in f(s,t) and, as a matter of fact, the analytical expression of f(s,t) is f(a,b,s,t). Rigth or not ?
     
  7. Jun 27, 2012 #6
    @JJacquellin
    Sorry I should have been a bit clearer, f(s,t) is unknown. We are given g(t) and g(s). So then f is actually f(a,b,s,t), as you said? I think I can see where this is going but would you mind elaborating on how we can use FTC and the chain rule. Im still not 100% sure.

    Thanks
     
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