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Fredholm Integral Eqn of the 2nd kind

  1. Jun 3, 2010 #1

    I need help finding the solution to the homogenous BVP. Normally I could do this but I'm lost on this one. [tex]\frac{d^2}{dt^2}x(t)+\int^{\pi}_{-\pi}\sin(t-s)x(s)ds=0[/tex]. I'm hoping the only solution is the zero solution. If not, I need to know a method to find all solutions. I thought about using the Neumann series but the [tex]\frac{d^2}{dt^2}x(t)[/tex] is messing me up. I don't know what to do. Any help will be appreciated
  2. jcsd
  3. Jun 4, 2010 #2
    it's not the only solution.

    Anything of the form
    [tex] x(t) = A \left( \cos(t) + \pi \sin(t) - \frac{(1 + \pi^2)t}{2 \pi} \right) + B [/tex]

    as for all of the solutions, can't help there.

    i got the one above by taking two more [itex]t[/itex] derivatives.
  4. Jun 4, 2010 #3
    Ok thanks. One question you said you got the solution by taking two l derivatives. Of what exactly. Did you guess at the solution, then checked to see if it worked by taking two derivatives. Or did you used a Fourier series expansion?
  5. Jun 4, 2010 #4
    I don't understand why you're calling it a boundary value problem but if it was an IVP, then the kernel is separable:

    [tex]\frac{d^2 x}{dt^2}+\sin(t)\int_{-\pi}^{\pi} \cos(s) x(s) ds-\cos(t)\int_{-\pi}^{\pi} \sin(s) x(s)dx=0[/tex]


    [tex]\frac{d^2 x}{dt^2}=\beta \cos(t)-\alpha \sin(t)[/tex]

    since the integrals represent constants.

    If I then integrate twice from 0 to t and let [itex]x(0)=x_0[/itex] and [itex]x'(0)=x_1[/itex] then back-substitute into the expressions for [itex]\alpha[/itex] and [itex]\beta[/itex], I get:

    [tex]x(t)=\frac{2\pi^2\left(\sin(t)-t\right) x_1}{1+\pi^2}-\frac{2\pi \cos(t) x_1}{1+\pi^2}+t x_1+x_0+\frac{2\pi x_1}{1+\pi^2}[/tex]

    This is called the "direct computation method" in "A First Course in Integral Equations" by A. Wazwaz
  6. Jun 5, 2010 #5
    It's an BVP since I have boundary conditions. x(-\pi)=0 and x(\pi)=0. I don't know anything about initial conditions. So I don't think I can do what you did.
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