Integral equation with unknown kernel?

[benjaminmar8]

Hi, all,

I would to solve an integral equation, here is the form

f(x)=\int_{x}^{R}K(x,t)g(t)dt

f(x) and g(t) are known function, R is an constant, how to compute the unknown Kernel
K(x,t)?

Thanks a lot

 

[daudaudaudau]

Obviously such a problem has infinitely many solutions.

 

[WastedGunner]

Obviously such a problem has infinitely many solutions.

Could you state one for general f and g?What I would do first is assume we can separate K(x,t)=F(x)G(t)
And also assume that f and F are differentiable.

Then you find that \frac{f(x)}{F(x)} = \int^R_x G(t)g(t)dt
let \hat{f} = \frac{f(x)}{F(x)}

and taking the derivative with the FTC you get that:

\hat{f}' = -G(x)g(x)

After some calculation you can devise the non-linear ODE:

F'(x) - \frac{F^2(x)G(x)g(x)}{f(x)} - \frac{f'(x)F(x)}{f(x)} = 0

So with specific choices for G i.e. G=\frac{1}{F} orG=\frac{1}{F^2} you can get certain solutions.

for example the choice G=\frac{1}{F} leads to the solution:

K(x,t) = e^{h(x)-h(t)}**

where h(y)=\int^y_a \frac{g(s)+f'(s)}{f(s)}ds for any a in the shared domain of f,g,and\ f'**this is actually: K(x,t)=e^{\int^x_t\frac{g(s)+f'(s)}{f(s)}ds

which is just one solution to this problem.

edit: the solution I gave for K can be multiplied by any constant, presumably determined by R

 

[daudaudaudau]

Could you state one for general f and g?

I think this is an obvious solution
<br /> K(x,t)=\frac{-f&#039;(t)-\frac{f(R)}{x-R}}{g(t)}<br />

By the way, try the [ itex] [ /itex] pair when writing tex inline.

 

[WastedGunner]

Ah, I see, very simple.

And thanks for the advice.