Integral equation with a derivative of the function inside the integral

  1. [tex] f(x) = 2\int_{0}^{t} sin(8u)f'(t-u) du + 8sin(8t) , t\geq 0 [/tex]
    is this problem solvable? i've never seen an integral equation like this with an f'(t-u)

    i tried to solve it us the convolution theorem and laplace transforms but ended up with

    [tex] s^{2} F(s) + 64F(s)- 16(F(s) - f(0)) =64 [/tex]
    and i havent been given f(0)
     
    Last edited: Oct 6, 2011
  2. jcsd
  3. This is just a hunch, but if you assume f(t) = A sin(at + theta), you will wind up with an equation in terms of sines and cosines. Smart money says that "a" would be 8. Then it's a matter of solving algebraic equations to get the phase and amplitude. Another hunch: It may be easier to convert everything to exponentials and work it from there.
     
  4. vela

    vela 12,777
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    I think you want f(t) on the LHS, right?
    What do you get if you set t=0 in the original equation?

    I didn't bother trying to reproduce what you did, but your approach sounds fine. If you can't figure it out, show your work and we'll be able to provide more guidance.
     
  5. Ray Vickson

    Ray Vickson 6,369
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    You could always set f(0) = c (an unspecified constant) and get a solution in terms of t and c. However, if looks like you CAN get f(0) from the original integral equation, at least if f' does not have a singularity at zero.

    RGV
     
  6. thanks for the replies

    yeah it was meant to be f(t) on the LHS

    Im going to have another look at it and get back to you guys.
     
  7. so i think i solved it, thanks guys.
    heres the solution in case anyone has a similar one in the future.

    [tex]f(t)=2∫t0sin(8u)f′(t−u)du+8sin(8t),t≥0[/tex]

    set t=0
    then f(0)= 0

    the convolution gives us

    [tex](sF(s) - f(0)) \cdot \frac{8}{s^{2}+64} [/tex]

    and the rest is just algebra
    [tex]f(t) = 64te^{8t}[/tex]
     
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