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  "body": "\n## Statement\n\nLet $a \u003c b$, let $\\Delta = \\set{(t, s) : a \\le s \\le t \\le b}$, and let $K \\colon \\Delta \\to \\C$ be continuous, with continuous partial derivative $\\partial K / \\partial t$ on $\\Delta$ and $K(t, t) \\ne 0$ for all $t \\in [a, b]$.\nLet $f \\colon [a, b] \\to \\C$ be continuously differentiable.\nLet $x \\colon [a, b] \\to \\C$ be continuous.\nConsider the [[Volterra equation of the first kind|H6ZGEDM4VV]]\n$$\\int_a^t K(t, s) \\, x(s) \\, ds = f(t), \\quad t \\in [a, b].$$\nThen $x$ solves it if and only if $f(a) = 0$ and $x$ solves the [[Volterra equation of the second kind|ATPSAHYG7C]]\n$$x(t) = \\frac{f'(t)}{K(t, t)} - \\int_a^t \\frac{1}{K(t, t)} \\frac{\\partial K}{\\partial t}(t, s) \\, x(s) \\, ds, \\quad t \\in [a, b].$$\nConsequently, if $f(a) = 0$, the first-kind equation has exactly one continuous solution, and if $f(a) \\ne 0$ it has none.\n\n## Remarks\n\nProof idea: by the Leibniz rule, the derivative of $\\int_a^t K(t, s) \\, x(s) \\, ds$ is $K(t, t) \\, x(t) + \\int_a^t \\partial_t K(t, s) \\, x(s) \\, ds$.\nSo differentiating the first-kind equation gives the second-kind equation after division by $K(t, t)$.\nConversely, 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 $f(a) = 0$.\nThe new kernel $-\\partial_t K(t, s) / K(t, t)$ is continuous, so [[unique solvability|WG4GVC9HR8]] applies.\nExample: $\\int_0^t e^{t - s} x(s) \\, ds = t$ gives $x(t) = 1 - \\int_0^t e^{t - s} x(s) \\, ds = 1 - t$, and indeed $\\int_0^t e^{t - s} (1 - s) \\, ds = t$.\nThe 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.\nFor the weakly singular kernel $(t - s)^{-1/2}$ the reduction does not apply, and one uses the [[Abel inversion formula|TDMXFY06A2]] instead.\n\n## Explanations\n\n### L1 human\n\nAn equation of the first kind hides the unknown inside an integral.\nDifferentiating with respect to $t$ brings it out, because the derivative of an integral with variable upper limit contains the integrand at that limit.\nIf the kernel does not vanish on the diagonal, one can divide and obtain an equation of the second kind, which is well understood.\n\n### L2 human\n\nThe reduction shows that a first-kind Volterra equation with a good kernel is about as hard as one differentiation.\nThe 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.\nWhen $K(t, t) = 0$ 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 $f(a) = f'(a) = 0$.\n",
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