Fredholm Integral Eqn of the 2nd kind

[gain01]

Hi,

I need help finding the solution to the homogenous BVP. Normally I could do this but I'm lost on this one. \frac{d^2}{dt^2}x(t)+\int^{\pi}_{-\pi}\sin(t-s)x(s)ds=0. 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 \frac{d^2}{dt^2}x(t) is messing me up. I don't know what to do. Any help will be appreciated

 

[qbert]

it's not the only solution.

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

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

i got the one above by taking two more t derivatives.

 

[gain01]

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?

 

[jackmell]

I don't understand why you're calling it a boundary value problem but if it was an IVP, then the kernel is separable:

\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

and:

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

since the integrals represent constants.

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

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}

This is called the "direct computation method" in "A First Course in Integral Equations" by A. Wazwaz

 

[gain01]

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.