Non trivial solution of Fredholm integral equation of second kind with constant kernel

Question

Let us consider the following integral equation$$f(x) + \lambda \int_0^1 {K(s,x)f(s)ds = 0,{\text{ x}} \in {\text{(0}}{\text{,1)}}{\text{.}}} $$ I'm looking of the values of $\lambda$ so that the above equation has only $f=0$ as solution with a constant kernel. Suppose that $K(s,x)=K$, we obtain $$f(x) + \lambda K\int_0^1 {f(s)ds = 0,{\text{ x}} \in {\text{(0}}{\text{,1)}}{\text{.}}} $$ By taking the integral over $(0,1)$, we get $$(1 + \lambda K)\int_0^1 {f(s)ds = 0} $$. for all $f$. Now, if $\lambda$ is different of $-1/K$, then $$\int_0^1 {f(s)ds = 0} $$. I don't see how this can be helpful. Any suggestions?. Thank you.

Answer 1

Your step of taking the integral is too crude, at least initially. When you have $$ f(x)+\lambda K\int_0^1f(s)\,ds=0, $$ you can write this as $$ f(x)=-\lambda K\int_0^1f(s)\,ds $$ to conclude that $f$ is constant. If $\lambda=0$, you get $f=0$. If $\lambda\ne0$ and $\lambda\ne-1/K$, your trick of integrating again gives you that $\int_0^1 f=0$, so $f=0$.

When $\lambda=-1/K$ the solution is not unique, as any constant $f$ will be a solution.