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Integral equation with unknown kernel?

  1. Jul 20, 2009 #1
    Hi, all,

    I would to solve an integral equation, here is the form

    [tex]f(x)=\int_{x}^{R}K(x,t)g(t)dt[/tex]

    f(x) and g(t) are known function, R is an constant, how to compute the unknown Kernel
    K(x,t)?

    Thanks a lot
     
  2. jcsd
  3. Jul 20, 2009 #2
    Obviously such a problem has infinitely many solutions.
     
  4. Jul 22, 2009 #3
    Could you state one for general [tex]f[/tex] and [tex]g[/tex]?


    What I would do first is assume we can separate [tex]K(x,t)=F(x)G(t)[/tex]
    And also assume that [tex] f [/tex] and [tex] F [/tex] are differentiable.

    Then you find that [tex]\frac{f(x)}{F(x)} = \int^R_x G(t)g(t)dt[/tex]
    let [tex]\hat{f} = \frac{f(x)}{F(x)} [/tex]

    and taking the derivative with the FTC you get that:

    [tex]\hat{f}' = -G(x)g(x) [/tex]

    After some calculation you can devise the non-linear ODE:

    [tex]F'(x) - \frac{F^2(x)G(x)g(x)}{f(x)} - \frac{f'(x)F(x)}{f(x)} = 0[/tex]

    So with specific choices for [tex]G[/tex] i.e. [tex]G=\frac{1}{F} [/tex] or[tex]G=\frac{1}{F^2}[/tex] you can get certain solutions.

    for example the choice [tex]G=\frac{1}{F} [/tex] leads to the solution:

    [tex] K(x,t) = e^{h(x)-h(t)}[/tex]**

    where [tex] h(y)=\int^y_a \frac{g(s)+f'(s)}{f(s)}ds[/tex] for any [tex]a[/tex] in the shared domain of [tex]f,g,and\ f'[/tex]


    **this is actually: [tex]K(x,t)=e^{\int^x_t\frac{g(s)+f'(s)}{f(s)}ds[/tex]

    which is just one solution to this problem.

    edit: the solution I gave for K can be multiplied by any constant, presumably determined by [tex]R[/tex]
     
    Last edited: Jul 22, 2009
  5. Jul 23, 2009 #4
    I think this is an obvious solution
    [tex]
    K(x,t)=\frac{-f'(t)-\frac{f(R)}{x-R}}{g(t)}
    [/tex]

    By the way, try the [ itex] [ /itex] pair when writing tex inline.
     
  6. Jul 23, 2009 #5
    Ah, I see, very simple.

    And thanks for the advice.
     
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