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

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.

• The function has mean value 0 in integral sense over interval 0 to 1. It removes one degree of freedom. This means you have infinite set of solutions. Any function fulfilling the mean value equation $=0$ will do. – mathreadler Jan 4 at 15:48
• So if $\lambda= - 1/K$ we have only one solution? – Gustave Jan 4 at 15:54
• If the other factor is $0$ then it does not matter what $f$ is, since the product will always be $0$ so then all functions $f$ will satisfy it. – mathreadler Jan 4 at 16:01
• Thanks. I understand, but what I can say about the uniqueness of the trivial solution with respect to $\lambda$? – Gustave Jan 4 at 16:09

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.