Help factorizing function

1. Oct 12, 2008

picard

Hi,
I need some help factorizing the following:

$$\frac{g(x)-g(a)}{u(a)^2}=K(x,a)$$

into $$A(a) X(x)$$

ie I want to find $$A(a), X(x)$$ such that their product is the first equation. The reason I want to do this is because K(x,a) is a kernel and it would help a lot if somehow I could write it as:

$$K(x,a)=A(a) X(x)$$

Finally, note that $$g'(a)=a/u(a)^2$$

Any ideas?

2. Oct 12, 2008

mathman

Off hand, I believe what you want can't be done.

3. Oct 12, 2008

picard

Well, that was my first guess as well, but I thought that someone might have an idea...

4. Oct 12, 2008

HallsofIvy

Staff Emeritus
Kernel in an integral equation? While that is not separable, you can write it as a sum of separable terms:
$$\frac{g(x)}{u(a)^2}- \frac{g(a)}{u(a)^2}$$
and work with each part separately.

5. Oct 13, 2008

picard

Yes, I have tried that. However, then the integral equation is not solvable, while if the kernel can be written with only one term as K(x,a)=A(a) X(x) then there exists an analytical solution...

6. Oct 13, 2008

HallsofIvy

Staff Emeritus
I assume, then, that you are talking about a Fredholm integral equation:
$$y(x)= \inf_a^b K(x,t)y(t)dt+ f(t)[/itex] Yes, if K(x,t) is "separable", K(x,t)= A(x)B(t), then the solution is easy. The equation becomes [tex]y(x)= A(x)\int_a^b B(t)y(t)dt+ f(x)[/itex] and [tex]\int_a^b B(t)y(t)dt$$
is a number. Let
$$X= \int_a^b B(t)y(t)dt$$
and you have y(x)= X A(x)+ f(x) so it is just a matter of solving for X. Multiply both sides of that equation by B(x) and integrate from a to b.
$$\int_a^b B(x)y(x)dx= X\int_a^b A(x)B(x)dx+ \int_a^b B(x)F(x)dx$$
The integral on the left is just X again and the other two integrals are of known functions. Let
$$K= \int_a^b A(x)B(x)dx$$
and
$$F= \int_a^b B(x)f(x)dx$$
and X satisfies the equation X= KX+ F which is easily solvable for X.

Now, suppose K(x,t) is not separable but is the sum of two separable functions: K(x,t)= A1(x)B1(t)+ A2(x)B2(t). we can still take the "x" dependence out of the equations:
$$y(x)= A1(x)\int_a^b B1(t)y(t)dt+ A2(x)\int_a^b B2(t)y(t)dt+ f(x)$$
Let
$$X1= \int_a^b B1(t)y(t)dt$$
and
$$X2= \int_a^b B2(t)y(t)dt$$
Now,
$$y(t)= X1A1(x)+ X2A2(x)+ f(x)[/itex] and we only need to find the numbers X1 and X2. If we multiply that equation by B1(x) and integrate: [tex]\int_a^b B1(x)y(x)dx= X1\int_a^b A1(x)B1(x)dx+ X2\int_a^b A2(x)B1(x)dx+ \int_a^b B1(x)f(x)dx[/itex] Again, the left side is just X1 and the integrals on the right are numbers: [tex]X1= K_{11}X1+ K_{12}X2+ F1$$
with the obvious notation. Doing the same with B2(x),
$$X2= K_{12}X1+ K_{22}X2+ F2$$

That is, we have two linear equations to solve for X1 and X2. (And, in fact, the system is symmetric, guarenteeing real solutions.)

The extension to any kernel which is a finite sum of separable functions should be obvious and, in fact, it can be extended to an infinite sum of such functions, giving an infinite series solution.

7. Oct 13, 2008

picard

@HallsofIvy: Thanks for taking the time and effort to write your answer! I really appreciate it. However, you assumed erroneously... My integral equation is a Voltera of the second kind... The thing is that for a seperable kernel with one term (like the one I am looking for) the solution is readily available, see for example:

http://eqworld.ipmnet.ru/en/solutions/ie/ie0214.pdf

However, if K(x,t) is not separable but is the sum of two separable functions then you have to solve a system of ODEs instead of an algebric (simultaneous) system... In my case the system with the ODEs is quite complicated and that's why I decided to ask if somebody knew of a way to factorize my kernel (so that I could use the analytical solution of the link).

Again, thanks and sorry...

8. Oct 13, 2008

Hurkyl

Staff Emeritus
(I'm assuming that g is known not to be a constant function)

A(0) X(0) = 0

A(0) X(x) = (g(x) - g(0)) / u(0)^2 which is nonzero

Therefore, A(0) is nonzero, and so X(0) = 0.

0 = A(x) X(0) = (g(0) - g(x)) / u(x)^2

Therefore, g(0) = g(x) for all x, which contradicts the fact g is nonconstant

Therefore, K(x,a) cannot be written as A(a) X(x).

9. Oct 14, 2008

picard

@Hurkyl: Thanks, at least I now know that there is no solution to this problem and I can move on.

Cheers