# Fredholm Integral Equation

1. Apr 10, 2013

### LeonhardEuler

Hello Everyone. An interesting equation has come in my thesis research, and I was wondering whether anyone had any useful information about it. It is this equation:
$$\int_{a_1}^{b_1}...\int_{a_n}^{b_n}P(x_1,...x_n)K(x_1,...x_n,s_1...s_n)dx_1...dx_n=C$$
K is a known function of the x's and s's. C is an unknown constant. P is a probability distribution and so subject to
$$\int_{a_1}^{b_1}...\int_{a_n}^{b_n}P(x_1,...x_n)dx_1...dx=1$$
$$P(x_1,...x_n)\ge 0$$
http://en.wikipedia.org/wiki/Fredholm_integral_equation
So I see that this is a Fredholm integral equation of the first kind. However, none of the theorems they present have any clear relevance to helping solve this equation, and there is nothing about how to impose the probability conditions.

It would be great if there was a general method for a numerical solution of these equations, and it would be good to know about analytic solutions in certain cases. A general analytic solution is probably too much to hope for. A way to transform this to a differential equation would also be good, since I know more about those.

2. Apr 10, 2013

### mikeph

I'm a little confused at this specific equation; are you sure C is a constant and not dependant on s1, s2, ... sn? If you are, then how can K depend on s1, s2, ... sn?

3. Apr 10, 2013

### LeonhardEuler

Yes, I am sure, and this is a question I get from a lot of people I have shown this to. Here is an example of a solution to the equation in one dimension:
$$F(x,s)=\sqrt{\frac{2}{\pi}}\frac{(b-a)e^{-\frac{(x-s)^2}{2\sigma^2}}}{\sigma[erf(\frac{s-a}{\sqrt{2}\sigma})+erf(\frac{b-s}{\sqrt{2}\sigma})]}$$
$$P(x)=\frac{1}{b-a}$$
$$C=1$$
Given any K and P that don't solve the problem because they give f(s) instead of C, you can always generate a solved version of the problem (though not the one you are trying to solve) by dividing K by f(s).

4. Apr 10, 2013

### mikeph

I think you've lost me, what's F, a, b?

The last sentence also went over my head; am I right in saying that because C is a constant, dK/ds1 = dK/ds2 = ... = dK/dsn = 0? I don't think you'd be asking if it were, but I can't comprehend how the RHS is constant yet the LHS depends on a set of variables.

On a more fundamental level, it seems like you have a very large space of unknowns (a continuous function P) and a single scalar known value C. It would seem to be vastly underdetermined.

5. Apr 10, 2013

### LeonhardEuler

Really sorry about that. I meant
$$K(x,s)=\sqrt{\frac{2}{\pi}}\frac{(b-a)e^{-\frac{(x-s)^2}{2\sigma^2}}}{\sigma[erf(\frac{s-a}{\sqrt{2}\sigma})+erf(\frac{b-s}{\sqrt{2}\sigma})]}$$
And a and b are the limits of integration in the original problem, with the subscripts dropped because this is a 1D case.

Let me clarify what I meant before with the last sentence. Suppose
$$\int_{a_1}^{b_1}...\int_{a_n}^{b_n}P(x_1,...x_n)K(x_1,...x_n,s_1...s_n) dx_1...dx_n=f(s_1,...s_n)$$
Then define
$$K_{new}(x_1,...x_n,s_1,...s_n)=\frac{K(x_1,...x_n,s_1,...s_n)}{f(s_1,...s_n)}$$
Then
$$\int_{a_1}^{b_1}...\int_{a_n}^{b_n}P(x_1,...x_n)K_{new}(x_1,...x_n,s_1...s_n) dx_1...dx_n=1$$
From this you see that the derivatives of K with respect to the s variables are not necessarily 0. Unless I'm still unclear. Thanks for taking the time to look at this.

6. Apr 10, 2013

### LeonhardEuler

Also, here is another simple example of the same counter intuitive kind of thing. Suppose
$$\int_{0}^{\infty}f(x,s)dx=1$$
for all s>0. You might think that
$$\frac{\partial f}{\partial s}=0$$
But suppose
$$f(x,s)=se^{-sx}$$
Then
$$\int_{0}^{\infty}se^{-sx}dx=1$$
but
$$\frac{\partial f}{\partial s}=-s^2e^{-sx}+e^{-sx}$$