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Suppose that we have two functions $f(x)=f_1(x)+...+f_n(x)$ and $K(x,t)$.

If we know that $$F_2(x)+\int_a^b K(x,t)F_1(t)dt=0$$

, where $F_1(x)+F_2(x)=f(x)$.

What is the best way to split $f(x)$ into two parts $F_1(x)$ and $F_2(x)$ such that the above integral equation is satisfied.

Other than trail and error, is there a quick intuitive way to split $f(x)$.

user3741635
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  • Please check your post for typos and correct them where appropriate: Is $f = f_1 + f_2$ or $f = f_1 + f_2 + \dots + f_n$? Is $f_1 = F_1, f_2 = F_2$? Is the integral $\int_a^bK(x,t) F_1(t) dt$? – Hans Engler Dec 09 '21 at 02:50
  • @HansEngler, the equation is $$F_2(x)+\int_a^b K(x,t)F_1(t)dt=0$$ – user3741635 Dec 09 '21 at 02:54
  • @HansEngler, $f_1$ is different from $F_1$. for example if $f=f_1+f_2+f_3$. Then $F_1$ could be $f_2+f_3$ or could be $f_1+f_3$. Basically just different ways to split $f$ to satasifiy the integral equation. Doing this by trial and error takes time if $n>2$. So I was woundering if there is another quicker way. – user3741635 Dec 09 '21 at 03:01

1 Answers1

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This is a Fredholm integral equation of the second kind for the unknown function $F_1$. Namely, since $F_2 = f - F_1$, the function $F_1$ must satisfy $$ F_1(x) = f(x) + \int_a^b K(x,t) F_1(t) \, dt \, . $$

Usually there is a unique $F_1$ that solves the problem, but if the integral operator has the eigenvalue -1, i.e. if there is a non-trivial function $g$ such that $g(x) + \int_a^b K(x,t) g(t) \, dt = 0$ , there may be infinitely many solutions or none at all, depending on $f$.

You can't choose $F_1$, you must solve for it. After $F_1$ has been found, also $F_2 = f-F_1$ is uniquely determined. Of course at that point you can split $F_1 = f_1 + f_2, F_2 = f_3 + f_4 + f_5$ or whatever you like. But $F_1$ and therefore also $F_2$ cannot be chosen arbitrarily.

Hans Engler
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