Find $f$ explicitly when $f_n$ is defined recursively and $\lim_{n\to\infty} f_n = f.$

Question

Given that $f_1(x) = 0$ and \begin{equation} f_{n+1}(x) = e^{-2x} + \int_0^x e^{-2t}f_n(t) \; dt, \; \text{ where }n = 1,2,\dots \end{equation} identify $f(x)$ explicitly where $\lim_{n\to\infty} f_n(x) = f(x).$

I have shown that $f(x)$ exists and is unique, but I am unable to find the limiting function.

Answer 1

the idea

If the limit exists it will satisfy :

$$f_{\infty}(x)=e^{-2x}+\int_0^xe^{-2t}f_{\infty}(t)dt$$

if you derive this equation, you get:

$$f'_{\infty}(x)=-2e^{-2x}+e^{-2x}f_{\infty}(x)$$

hoping that I did not make mistake, the following lines might be correct

setting $f_{\infty}(x)=e^{\frac{-1}{2}e^{-2x}}g(x)$, you have

$$f'_\infty(x)=e^{-2x}f_\infty(x)+e^{\frac{-1}{2}e^{-2x}}g'(x)$$

so $$g'(x)=-2e^{-2x+\frac{1}{2}e^{-2x}}$$

ang thus :

$$f_{\infty}(x)=e^{\frac{-1}{2}e^{-2x}}\left(e^{\frac{1}{2}}-\int_0^{x}2e^{-2u+\frac{1}{2}e^{-2u}}du\right)=3e^{\frac{1}{2}(1-e^{-2x})}-2$$

you can now verify that it satisfies the first equation.