Help factorizing function

[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?

Thanks in advance

[mathman]

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

[picard]

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

[HallsofIvy]

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.

[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...

[HallsofIvy]

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.

[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...

[Hurkyl]

Hi,
I need some help factorizing the following:

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

into A(a) X(x)
(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).

[picard]

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

Cheers