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  "body": "\n## Definition\n\nLet $T \u003e 0$ or $T = \\infty$, let $k \\colon (0, T) \\to \\C$ be locally integrable, let $f \\colon [0, T) \\to \\C$ be a given function, and let $\\lambda \\in \\C$.\nA *Volterra equation of convolution type* is a [[Volterra integral equation|DND8CBEVXT]] whose kernel is a [[difference kernel|B5AB451TB7]] $K(t, s) = k(t - s)$.\nIn the second kind it reads\n$$x(t) = f(t) + \\lambda \\int_0^t k(t - s) \\, x(s) \\, ds, \\quad 0 \\le t \u003c T,$$\nand in the first kind it reads\n$$\\int_0^t k(t - s) \\, x(s) \\, ds = f(t), \\quad 0 \\le t \u003c T.$$\n\n## Remarks\n\nThe integral is the [[convolution|2Z29JN3VES]] $(k * x)(t) = \\int_0^t k(t - s) \\, x(s) \\, ds$ of functions on $[0, \\infty)$, so the second kind is written $x = f + \\lambda \\, k * x$.\nExample: $x(t) = 1 + \\int_0^t e^{-(t - s)} x(s) \\, ds$, with $k(u) = e^{-u}$.\nThe function $x(t) = 1 + t$ solves it: $\\int_0^t e^{-(t - s)} (1 + s) \\, ds = t$, by the substitution $u = t - s$ and $\\int_0^t e^{-u}(1 + t - u) \\, du = t$.\nNon-example: $x(t) = 1 + \\int_0^t t s \\, x(s) \\, ds$ has kernel $K(t, s) = ts$, which is not a function of $t - s$ alone.\nThe [[Laplace transform method|EHGECREYB4]] solves convolution equations on $[0, \\infty)$ when $k$ and $f$ have Laplace transforms.\n",
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