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  "body": "\n## Definition\n\nLet $\\Omega$ be a measurable subset of $\\R^m$, for example a closed interval $[a, b]$.\nLet $k \\colon \\Omega \\times \\Omega \\to \\R$ be a [[kernel|GSMBXT7FPK]], let $g \\colon \\Omega \\times \\R \\to \\R$ be a function, and let $f \\colon \\Omega \\to \\R$ be given.\nA *Hammerstein integral equation* for an unknown function $\\varphi \\colon \\Omega \\to \\R$ is\n$$\\varphi(x) = f(x) + \\int_\\Omega k(x, t)\\, g\\bigl(t, \\varphi(t)\\bigr)\\, dt, \\quad x \\in \\Omega.$$\nWith the linear operator $(K v)(x) = \\int_\\Omega k(x, t)\\, v(t)\\, dt$ and the [[Nemytskii operator|W8G3AZ0DNG]] $(N_g \\varphi)(t) = g(t, \\varphi(t))$, it reads $\\varphi = f + K N_g \\varphi$.\n\n## Remarks\n\nIt is the [[Urysohn equation|WZ4Z26XSVC]] with the product kernel $K(x, t, u) = k(x, t)\\, g(t, u)$.\nMany authors write it as $\\varphi + K N_g \\varphi = f$, which only changes the sign of $g$.\n\nExample: on $[0, 1]$, take $f = 0$, $k(x, t) = x t$ and $g(t, u) = u^2$.\nTry $\\varphi(x) = c x$; the right side is $x \\int_0^1 t \\cdot c^2 t^2\\, dt = \\frac{c^2}{4} x$, so $c = c^2/4$, that is $c = 0$ or $c = 4$.\nSo $\\varphi = 0$ and $\\varphi(x) = 4x$ are two solutions, and the equation has no uniqueness.\n\nNon-example: $\\varphi(x) = \\int_0^1 \\sin\\bigl(x\\, \\varphi(t)\\bigr)\\, dt$ is a Urysohn equation but not a Hammerstein one with a single product kernel, because $\\sin(xu)$ is not of the form $k(x, t)\\, g(t, u)$.\nIndeed, the functions $u \\mapsto \\sin(x u)$ for different $x$ are not all multiples of one function.\n\nExample of origin: on $[0, 1]$ the problem $-u'' = h(x, u)$, $u(0) = u(1) = 0$, with continuous $h$, is equivalent to $u(x) = \\int_0^1 G(x, t)\\, h(t, u(t))\\, dt$, where $G(x, t) = \\min(x, t)\\bigl(1 - \\max(x, t)\\bigr)$ is the Green's function.\n",
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