Reduction of a first-kind Volterra equation to the second kind

Remarks

Proof idea: by the Leibniz rule, the derivative of $\int_a^t K(t, s) \, x(s) \, ds$ is $K(t, t) \, x(t) + \int_a^t \partial_t K(t, s) \, x(s) \, ds$. So differentiating the first-kind equation gives the second-kind equation after division by $K(t, t)$. Conversely, a solution of the second-kind equation makes both sides of the first-kind equation have the same derivative, and both vanish at $t = a$ when $f(a) = 0$. The new kernel $-\partial_t K(t, s) / K(t, t)$ is continuous, so unique solvability applies. Example: $\int_0^t e^{t - s} x(s) \, ds = t$ gives $x(t) = 1 - \int_0^t e^{t - s} x(s) \, ds = 1 - t$, and indeed $\int_0^t e^{t - s} (1 - s) \, ds = t$. The hypothesis $K(t, t) \ne 0$ matters: for $K(t, s) = t - s$, differentiation gives another first-kind equation $\int_a^t x(s) \, ds = f'(t)$, and one needs a second differentiation. For the weakly singular kernel $(t - s)^{-1/2}$ the reduction does not apply, and one uses the Abel inversion formula instead.

Going deeper

For a student

The reduction shows that a first-kind Volterra equation with a good kernel is about as hard as one differentiation. The price is that the data $f$ must be differentiable, and errors in $f$ are amplified by the derivative, which is the source of ill-posedness in practice. When $K(t, t) = 0$ but $\partial_t K(t, t) \ne 0$, differentiating twice gives the same kind of reduction, under the corresponding smoothness assumptions and with the conditions $f(a) = f'(a) = 0$.