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  "body": "\n## Definition\n\nConsider the [[Wiener–Hopf equation|QYY4NMDY5H]] $\\varphi(x) - \\int_0^\\infty k(x - t)\\, \\varphi(t)\\, dt = f(x)$ for $x \\ge 0$, with $k \\in L^1(\\R)$.\nWrite $\\hat g(\\xi) = \\int_{\\R} g(x)\\, e^{i \\xi x}\\, dx$ for the [[Fourier transform|DPVM34ADZV]].\nFor this convention, the transform of a function supported in $[0, \\infty)$ is [[holomorphic|0NAHXR1FRH]] in the upper half-plane $\\operatorname{Im} \\xi \u003e 0$.\nThe transform of a function supported in $(-\\infty, 0]$ is holomorphic in the lower half-plane.\nThe *Wiener–Hopf method* is the following procedure.\n\n1. Extend $\\varphi$ and $f$ by $0$ to $x \u003c 0$, and let $\\psi(x) = \\int_0^\\infty k(x - t) \\varphi(t)\\, dt$ for $x \u003c 0$ and $\\psi(x) = 0$ for $x \\ge 0$. Then $\\varphi - k * \\varphi = f - \\psi$ on all of $\\R$.\n2. Take Fourier transforms: $\\bigl(1 - \\hat k(\\xi)\\bigr)\\, \\hat\\varphi(\\xi) = \\hat f(\\xi) - \\hat\\psi(\\xi)$, where $\\hat\\varphi$ and $\\hat f$ are upper functions and $\\hat\\psi$ is a lower function.\n3. Factor the symbol as $1 - \\hat k = a_- \\, a_+$. Here $a_+$ and $1/a_+$ are holomorphic and bounded in the upper half-plane. Also $a_-$ and $1/a_-$ are holomorphic and bounded in the lower half-plane.\n4. Split $\\hat f / a_- = g_+ + g_-$ into an upper and a lower part, for example with Cauchy integrals.\n5. The equation becomes $a_+ \\hat\\varphi - g_+ = g_- - \\hat\\psi / a_-$. The left side extends to the upper half-plane and the right side to the lower half-plane, so together they define an entire function. Growth bounds and [[Liouville's theorem|0EHKVRF4J6]] show that it is a polynomial, often $0$.\n6. Then $\\hat\\varphi = (g_+ + P)/a_+$ for that polynomial $P$, and $\\varphi$ is recovered by the inverse Fourier transform.\n\n## Remarks\n\nThe factorization in step 3 exists when $1 - \\hat k$ has no zeros on $\\R$ and its winding number about $0$ is zero.\nOtherwise rational factors such as $\\bigl((\\xi - i)/(\\xi + i)\\bigr)^\\nu$ appear, see [[Wiener–Hopf factorization|BYKP1V2AQN]].\nThe same steps apply to boundary value problems for partial differential equations, where step 1 comes from a transform in one variable along a half-line boundary.\n\nExample: for $k(x) = \\frac{1}{2} e^{-\\abs{x}}$ one has $\\hat k(\\xi) = \\frac{1}{1 + \\xi^2}$, so $1 - \\hat k(\\xi) = \\frac{\\xi^2}{\\xi^2 + 1} = \\frac{\\xi}{\\xi - i} \\cdot \\frac{\\xi}{\\xi + i}$.\nThe factor $\\xi/(\\xi + i)$ is holomorphic in the upper half-plane and $\\xi/(\\xi - i)$ in the lower half-plane.\nBoth vanish at the real point $0$.\nThis zero of the symbol on $\\R$ allows the solution $1 + x$ of the homogeneous equation, which grows linearly.\n\nNon-example: a direct Fourier transform of the equation, without step 1, does not work, because the integral covers only $t \\ge 0$ and the equation holds only for $x \\ge 0$.\n",
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