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. $$G=\frac{1}{F}$$ or$$G=\frac{1}{F^2}$$ 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
$$K(x,t)=\frac{-f'(t)-\frac{f(R)}{x-R}}{g(t)}$$

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

 

[WastedGunner]

Ah, I see, very simple.

And thanks for the advice.