How can I solve Fredholm Integral Equations?

[apalmer3]

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\intsin(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

 

[HallsofIvy]

I consider Fredholm equations to be much simpler than Volterra equations!

In general, a "Fredholm" integral equation is of the form
y(x)= \int_a^b K(x,t)y(t)dt+ u(x)

Start with the simple case: if K(x,t)= A(x)B(t), that is, if the kernel is "separable" you can do the following: Let
X= \int_a^b B(t)y(t)dt[/itex]<br /> C= \int_a^b A(x)B(x)dx[/itex]&lt;br /&gt; and &lt;br /&gt; D= \int_a^b B(x)u(x)dx[/itex]&amp;lt;br /&amp;gt; Notice that these are all numbers (X is a numerical variable).&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; The integral equation can be rewritten as&amp;lt;br /&amp;gt; y(x)= X A(x)+ u(x)&amp;lt;br /&amp;gt; Multiply both sides of the integeral equation by B(x):&amp;lt;br /&amp;gt; B(x)y(x)= X A(x)B(x) + B(x)u(x)&amp;lt;br /&amp;gt; and integrate from a to b:&amp;lt;br /&amp;gt; X= CX+ D, a numerical equation from which we can solve (1-C)X= D so X= D/(1- C).&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; The solution to the integral equation is y(x)+ XA(x)+ u(x)= DA(x)/(1- C)+ u(x).&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; If K(x,t) is NOT separable, you can write it as a sum (finite or infinite) of &amp;amp;quot;separable terms&amp;amp;quot;: A&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;(x)B&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;(x)+ A&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;(x)B&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;(x)+ ... and get a system of linear equations for &amp;amp;quot;X&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;&amp;amp;quot; rather than a single equation. &amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; In your particular case, it helps to remember that sin(x- y)= sin(x)cos(y)- cos(x)sin(y).