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  "body": "\n## Statement\n\nLet $m \\in \\set{2, 3}$, and let $D \\subset \\R^m$ be a bounded domain whose boundary $\\partial D$ is a connected $C^2$ curve or surface with outward unit normal $\\nu$.\nLet $\\Phi$ be the [[fundamental solution of the Laplacian|BSJ8X2XRN9]] with $-\\Delta \\Phi = \\delta_0$, so $\\Phi(x) = \\frac{1}{2\\pi} \\ln \\frac{1}{\\abs{x}}$ for $m = 2$ and $\\Phi(x) = \\frac{1}{4\\pi \\abs{x}}$ for $m = 3$.\nFor a continuous density $\\varphi$ on $\\partial D$ put\n$$(K\\varphi)(x) = \\int_{\\partial D} \\frac{\\partial \\Phi(x - y)}{\\partial \\nu(y)} \\, \\varphi(y) \\, ds(y), \\quad x \\in \\partial D.$$\nLet $f \\colon \\partial D \\to \\R$ be continuous.\nThen:\n\n1. The [[integral equation of the second kind|8KJCN49V64]] $\\varphi - 2K\\varphi = -2f$ has exactly one continuous solution $\\varphi$ on $\\partial D$.\n2. For this $\\varphi$, the [[double layer potential|EG8Z3FWV9A]] $u(x) = \\int_{\\partial D} \\frac{\\partial \\Phi(x - y)}{\\partial \\nu(y)} \\varphi(y) \\, ds(y)$, $x \\in D$, extends continuously to $\\overline{D}$, is [[harmonic|0DEB0JMN0W]] in $D$, and equals $f$ on $\\partial D$.\n\nSo $u$ is the unique solution of the interior [[Dirichlet problem|8D56GQ4AG8]] with boundary values $f$.\n\n## Remarks\n\nPart 2 follows from the [[jump relations for layer potentials|J58YA8QNNX]]: the limit of the double layer potential from inside is $K\\varphi - \\frac12 \\varphi$, and this equals $f$ exactly when $\\varphi - 2K\\varphi = -2f$.\nFor part 1, $K$ is a [[compact operator|M2NEP1K5AE]] on $C(\\partial D)$ because its kernel is continuous for $m = 2$ and weakly singular for $m = 3$.\nBy the [[Fredholm alternative|4HH8C9V8JQ]] it suffices to show that $\\varphi - 2K\\varphi = 0$ forces $\\varphi = 0$.\nFor such $\\varphi$ the double layer potential vanishes in $D$ by the maximum principle.\nIts normal derivative is continuous across $\\partial D$, so it solves an exterior Neumann problem with zero data and vanishes outside as well.\nThe jump of the potential across $\\partial D$ is $\\varphi$, so $\\varphi = 0$.\n\nExample: on the unit disc in $\\R^2$ one has $\\frac{\\partial \\Phi(x - y)}{\\partial \\nu(y)} = -\\frac{1}{4\\pi}$ for $x \\ne y$ on the unit circle, so $K\\varphi = -\\frac{1}{4\\pi} \\int_{\\partial D} \\varphi \\, ds$.\nFor $f = 1$ the equation becomes $\\varphi + \\frac{1}{2\\pi} \\int \\varphi \\, ds = -2$, whose solution is the constant $\\varphi = -1$, and then $u = 1$ in $D$.\n\nThe method turns a problem on the domain $D$ into an equation on its boundary, one dimension lower, which is the basis of boundary element methods.\n\n## Explanations\n\n### L1 human\n\nTo find a harmonic function inside a region with given boundary values, one can place a layer of dipoles on the boundary and look for its strength.\nThe strength must satisfy an integral equation on the boundary alone.\nThis equation of the second kind always has exactly one solution, which proves that the Dirichlet problem is solvable and gives a way to compute the solution.\n\n### L2 human\n\nThe theorem is the classical route, going back to the work of Neumann and Fredholm, from potential theory to [[Fredholm theory|242TFD41KZ]].\nIts proof uses the jump relations, the compactness of $K$ and the Fredholm alternative, and uniqueness for the exterior Neumann problem.\nFor the interior Neumann problem one uses a single layer potential instead, see the [[boundary integral equation for the Neumann problem|CN7NP6R1AQ]].\nFor the exterior Dirichlet problem the corresponding equation $\\varphi + 2K\\varphi = 2f$ is not always uniquely solvable, and the ansatz has to be modified.\n",
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