# Fredholm Integral Eqn of the 2nd kind

1. Jun 3, 2010

### 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

2. Jun 4, 2010

### 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.

3. Jun 4, 2010

### 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?

4. Jun 4, 2010

### 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

5. Jun 5, 2010

### 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.