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  "body": "\n## Definition\n\nLet $a \u003c b$, let $\\Delta = \\set{(t, s) : a \\le s \\le t \\le b}$, let $K \\colon \\Delta \\to \\C$ be a [[kernel|GSMBXT7FPK]], let $f \\colon [a, b] \\to \\C$ be a given function, and let $\\lambda \\in \\C$.\nA *linear Volterra equation of the second kind* is the equation\n$$x(t) = f(t) + \\lambda \\int_a^t K(t, s) \\, x(s) \\, ds, \\quad t \\in [a, b],$$\nfor an unknown function $x$ on $[a, b]$.\nThe unknown appears outside the [[integral|0W29TBTY12]] with coefficient $1$, and the upper limit of integration is the variable $t$.\nThe equation is *homogeneous* when $f = 0$.\n\n## Remarks\n\nExample: $x(t) = t + \\int_0^t (s - t) \\, x(s) \\, ds$ on $[0, 1]$, with $K(t, s) = s - t$ and $\\lambda = 1$.\nThe function $x = \\sin$ solves it: $\\int_0^t (s - t) \\sin s \\, ds = \\sin t - t$, so the right side is $\\sin t$.\nThis is the integral form of $x'' = -x$, $x(0) = 0$, $x'(0) = 1$.\nNon-example: $x(t) = f(t) + \\int_a^b K(t, s) \\, x(s) \\, ds$ has the fixed upper limit $b$, so it is a Fredholm equation, not a Volterra one.\nThe usual standing hypotheses are that $K$ is continuous on $\\Delta$ and $f$ is continuous on $[a, b]$; then the [[unique solvability theorem|WG4GVC9HR8]] applies.\nSetting $K(t, s) = 0$ for $s \u003e t$ shows that a Volterra equation is a Fredholm equation with a kernel that vanishes above the diagonal.\n",
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