Stability of Volterra equations

Remarks

If the equation is stable and $1 \le p \le \infty$, then $f \in L^p(\R_+)$ gives $x \in L^p(\R_+)$ with $\norm{x}_p \le (1 + \norm{r}_1) \norm{f}_p$, by Young's inequality in the Lp spaces. If moreover $f$ is bounded and $f(t) \to f_\infty$ as $t \to \infty$, then $x(t) \to \big(I - \int_0^\infty r(s) \, ds\big) f_\infty$, by dominated convergence.

Example: $n = 1$ and $k(t) = a e^{-t}$ with $\operatorname{Re} a > -1$. The resolvent is $r(t) = a e^{-(1 + a) t}$. Check: $(k * r)(t) = a^2 e^{-t} \int_0^t e^{-a s} \, ds = a e^{-t} - a e^{-(1 + a) t}$ for $a \ne 0$, so $r + k * r = k$, and the case $a = 0$ is trivial. Since $\operatorname{Re}(1 + a) > 0$, $r$ is integrable and the equation is stable.

Non-example: $n = 1$ and $k(t) = -1$. Then $r(t) = -e^t$, since $-e^t + \int_0^t e^s \, ds = -1$, and $r$ is not integrable. For $f = 1$ the solution is $x(t) = 1 + (e^t - 1) = e^t$, which is unbounded although $f$ is bounded.

For nonlinear Volterra equations one speaks instead of stability of a particular solution, in the sense of Lyapunov.