Volterra equation of the second kind

Remarks

Example: $x(t) = t + \int_0^t (s - t) \, x(s) \, ds$ on $[0, 1]$, with $K(t, s) = s - t$ and $\lambda = 1$. The function $x = \sin$ solves it: $\int_0^t (s - t) \sin s \, ds = \sin t - t$, so the right side is $\sin t$. This is the integral form of $x'' = -x$, $x(0) = 0$, $x'(0) = 1$. Non-example: $x(t) = f(t) + \int_a^b K(t, s) \, x(s) \, ds$ has the fixed upper limit $b$, so it is a Fredholm equation, not a Volterra one. The usual standing hypotheses are that $K$ is continuous on $\Delta$ and $f$ is continuous on $[a, b]$; then the unique solvability theorem applies. Setting $K(t, s) = 0$ for $s > t$ shows that a Volterra equation is a Fredholm equation with a kernel that vanishes above the diagonal.