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  "body": "\n## Statement\n\nLet $a \u003c b$, let $\\Delta = \\set{(t, s) : a \\le s \\le t \\le b}$, let $K \\colon \\Delta \\to \\C$ be continuous, let $f \\colon [a, b] \\to \\C$ be continuous, and let $\\lambda \\in \\C$.\nThen the [[Volterra equation of the second kind|ATPSAHYG7C]]\n$$x(t) = f(t) + \\lambda \\int_a^t K(t, s) \\, x(s) \\, ds, \\quad t \\in [a, b],$$\nhas exactly one continuous solution $x \\colon [a, b] \\to \\C$.\nIt is $x(t) = f(t) + \\lambda \\int_a^t R(t, s; \\lambda) \\, f(s) \\, ds$, where $R$ is the [[resolvent kernel|ARTJV1XCRX]] of $K$.\n\n## Remarks\n\nExistence is given by the [[convergence of the Neumann series for Volterra kernels|3QDDSSH4FB]].\nFor uniqueness, let $y$ be the difference of two continuous solutions; then $y(t) = \\lambda \\int_a^t K(t, s) \\, y(s) \\, ds$.\nWith $M = \\max \\abs{K}$ and $C = \\max \\abs{y}$, induction gives $\\abs{y(t)} \\le C \\frac{(\\abs{\\lambda} M (t - a))^n}{n!}$ for every $n$, so $y = 0$.\nIn particular, the homogeneous equation has only the zero solution, so a continuous Volterra kernel has no [[characteristic values|DC7HMHVREP]].\nExample: $x(t) = \\cos t - \\int_0^t x(s) \\, ds$ has the unique solution $x(t) = \\tfrac12 (\\cos t - \\sin t + e^{-t})$; it solves $x' = -\\sin t - x$ with $x(0) = 1$.\nThe 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 $0 \\le \\alpha \u003c 1$, with the same conclusion.\nFor 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 $t = 0$.\n\n## Explanations\n\n### L1 human\n\nA linear Volterra equation of the second kind with a continuous kernel can always be solved, and in only one way.\nThis holds for every value of the parameter $\\lambda$, with no exceptions.\nThe reason is that the solution is built forward in time, each value being determined by the earlier ones.\n\n### L2 human\n\nFor a Fredholm equation of the second kind, the [[Fredholm alternative|4HH8C9V8JQ]] says that solvability fails exactly at the characteristic values of the kernel.\nA Volterra kernel has none, because the Volterra integral operator is quasinilpotent: its spectrum is $\\set{0}$.\nThe 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.\n",
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  "name": "Unique solvability of Volterra equations of the second kind",
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