# Fredholm Integral Equations

• October 5th 2008, 09:36 AM
apalmer3
Fredholm Integral Equations
Hello All!

I am currently in an Applied Analysis class, and I'm trying to do a little research outside of the classroom to try and understand what my teacher is trying to say.

So, I'm supposed to understand how to solve Fredholm Integral Equations (inhomogeneous and of the second kind). Would anybody be willing to help me understand?

Here's an example problem from the notes:

u(x) = cos(x) + $\lambda$ $\int$sin(x-y)u(y)dy

The limits of the integral are from 0 to $\pi$.

I'm supposed to solve for u(x), and I'm not exactly sure how. I know that, if it were a Volterra equation, I could solve it by convolution. Is it the same for a Fredholm equation? Or is there a better approach?

Thanks so very very much!
Abigail
• October 8th 2008, 10:03 AM
shawsend
Figure you got this already but let me solve it just for fun ok.

$u(x)=\cos(x)+\lambda\int_0^{\pi} \sin(x-y)u(y)dy$

$u(x)=\cos(x)+\lambda\left(\int_0^{\pi}\sin(x)\cos( y)u(y)dy-\int_0^{\pi}\sin(y)\cos(x)u(y)dy\right)$

$u(x)=\cos(x)+\lambda\left(\sin(x)\int_0^{\pi}\cos( y)u(y)dy-\cos(x)\int_0^{\pi}\sin(y)u(y)dy\right)$

Letting:

$a=\int_0^{\pi}\cos(y)u(y)dy;\quad b=\int_0^{\pi}\sin(y)u(y)dy$

we have:

$u(x)=\cos(x)+\lambda\left(a\sin(x)-b\cos(x)\right)$

Then:

$a=\int_0^{\pi}\cos(y)\bigg(\cos(y)+\lambda(a\sin(y )-b\cos(y)\bigg)dy$

$b=\int_0^{\pi}\sin(y)\bigg(\cos(y)+\lambda(a\sin(y )-b\cos(y)\bigg)dy$

Solving for a and b:

$a=\frac{2\pi}{4+\pi^2\lambda^2}$

$b=\frac{\pi^2\lambda}{4+\pi^2\lambda^2}$

We can now substitute these values into the expression:

$u(x)=\cos(x)+\lambda\left(a\sin(x)-b\cos(x)\right)$

See "A First Course in Integral Equation" by Abdul-Majid Wazwaz. It's a good book that's easy to read.