# Necessary condition for the integral equation $u(x) = f(x)+\lambda\int K(x,t)u(t)dt$ to have a continuous solution?

Is the following condition necessary for the integral equation $$u(x) = f(x)+\lambda\int K(x,t)u(t)dt$$ to have a continuous solution: $f(x) \neq 0$, is real and continuous in the interval $[a,b]$?

When $f(x) = 0$, the integral equation will become homogeneous. So I think this should be a necessary condition. Am I right? Please suggest me

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Are there any limits on the integral? What do you know about $K$? –  flavio Dec 20 '12 at 8:48
limits are from a to b and K is kernel of of fredholm integral equation. –  Alka Goyal Dec 20 '12 at 8:51

## 2 Answers

Why would $f$ have to be nonzero or real? If $f = 0$ there is certainly a continuous solution, namely $0$. There may be nonzero solutions as well. You'll probably want to assume $K(x, \cdot)$ to be in $L^1(a,b)$ so the integral exists for all $u \in C[a,b]$. Then under some conditions on $K$ (e.g. if $K$ is continuous) $\int_a^b K(x,t) u(t)\ dt$ will be continuous for all continuous $u$. If so, $f(x) = u(x) - \lambda \int_a^b K(x,t) u(t)\ dt$ will have to be continuous in order for there to be a continuous solution, and will have to be real if $K$ is real and there is a real solution.

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It has a unique continuous solution.It's solution is given by Lioville-Neumann Series.
http://en.wikipedia.org/wiki/Liouville-Neumann_series

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OK thanks sir. but i want to know whether the condition mentioned above is necessary or not. i think it is but not sure. –  Alka Goyal Dec 20 '12 at 9:05
Under appropriate conditions on $K$, the series will converge to a unique continuous solution for sufficiently small $|\lambda|$. It may not converge for all $\lambda$, and there may be $\lambda$ for which there is no solution or a nonunique solution. –  Robert Israel Dec 20 '12 at 9:26