A Fredholm equation with a particular kernel How to solve this kind of Fredholm’s equation?
$$
x(t)+\lambda \int\limits_{0}^{1}\! \big[ts - \min\{t,s\}\big]x(s)ds=t
$$
Thanks for any help.
 A: You can immediately translate this integral equation into an  easy second order linear ODE on $[0,1]$ with boundary conditions $x(0)=0$, $x(1)=1$. (I'm not going to do it for you). Just note that your integrand $ts-\min(t,s)$ is the Green function of the Laplacian in dimension $1$ (i.e. the second derivative) with Dirichlet boundary condition: one has
$$\int_0^1\big(sr-\min(t,s)\big)x(t)ds=u(t)$$
if and only  $$\cases{\ddot u(t)=x(t)\\u(0)=0\\u(1)=0\,.}$$
(The easy computation to check the latter fact starts writing $\int_0^1=\int_0^t+\int_t^1$; then pull the $t$-factors out of the integrals and derive (...). After that, you may want to precise further the above claim in the various functional settings; however for your eigenvalue problem everything is necessarily classic and smooth). 
A: $\newcommand{\la}{\lambda}
$
Let us formalize the question as follows: 

Take any complex number $\la$. Let $K(s,t):=s\wedge t-st$ for $s,t$ in $[0,1]$. Find $x\in L^2[0,1]$ such that 
  \begin{equation*}
 x(t)-\la \int_0^1 K(s,t)x(s)\,ds=t \tag{1}
\end{equation*}
  for almost all $t\in[0,1]$. 

Note that the kernel $K$ is the covariance function of the Brownian bridge, with the eigenfunctions $\sin j\pi\cdot$ and the corresponding eigenvalues $1/(j^2\pi^2)$ for natural $j$; see e.g. Section The Brownian bridge. Also, for $t\in[0,1)$
\begin{equation*}
 t=\sum_{j=1}^\infty b_j\sin j\pi t,\quad\text{where}\quad b_j:=\frac{2(-1)^{j-1}}{j\pi}. 
\end{equation*}
Therefore, equation (1) can be rewritten as 
\begin{equation*}
x(t)=\sum_{j=1}^\infty c_j\sin j\pi t \tag{2} 
\end{equation*}
with 
\begin{equation*}
 c_j-\frac\la{j^2\pi^2}\,c_j=b_j, 
\end{equation*}
so that 
\begin{equation*}
c_j=\frac{b_j}{1-\la/(j^2\pi^2)}; \tag{3}
\end{equation*}
this unique solution $x$ defined by (2)--(3) exists iff $\la\notin\{j^2\pi^2\colon j=1,2,\dots\}$. 
