Unique solvability of Volterra equations of the second kind

Remarks

Existence is given by the convergence of the Neumann series for Volterra kernels. For uniqueness, let $y$ be the difference of two continuous solutions; then $y(t) = \lambda \int_a^t K(t, s) \, y(s) \, ds$. With $M = \max \abs{K}$ and $C = \max \abs{y}$, induction gives $\abs{y(t)} \le C \frac{(\abs{\lambda} M (t - a))^n}{n!}$ for every $n$, so $y = 0$. In particular, the homogeneous equation has only the zero solution, so a continuous Volterra kernel has no characteristic values. Example: $x(t) = \cos t - \int_0^t x(s) \, ds$ has the unique solution $x(t) = \tfrac12 (\cos t - \sin t + e^{-t})$; it solves $x' = -\sin t - x$ with $x(0) = 1$. The continuity of $K$ can be weakened to boundedness with measurability, or to a weak singularity such as $\abs{K(t, s)} \le C (t - s)^{-\alpha}$ with $0 \le \alpha < 1$, with the same conclusion. For the equation of the first kind the statement fails: $\int_0^t x(s) \, ds = 1$ has no continuous solution, because the left side vanishes at $t = 0$.

Going deeper

For a student

For a Fredholm equation of the second kind, the Fredholm alternative says that solvability fails exactly at the characteristic values of the kernel. A Volterra kernel has none, because the Volterra integral operator is quasinilpotent: its spectrum is $\set{0}$. The theorem covers the integral form of every linear initial value problem $y^{(n)} + a_{n-1}(t) y^{(n-1)} + \cdots + a_0(t) y = g(t)$ with continuous coefficients, and thus gives global existence and uniqueness for those.