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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 $-\\Delta \\Phi = \\delta_0$.\nFor a continuous density $\\psi$ on $\\partial D$ put\n$$(K'\\psi)(x) = \\int_{\\partial D} \\frac{\\partial \\Phi(x - y)}{\\partial \\nu(x)} \\, \\psi(y) \\, ds(y), \\quad x \\in \\partial D.$$\nLet $g \\colon \\partial D \\to \\R$ be continuous.\nThen:\n\n1. The [[integral equation of the second kind|8KJCN49V64]] $\\psi + 2K'\\psi = 2g$ has a continuous solution $\\psi$ if and only if $\\int_{\\partial D} g \\, ds = 0$.\n2. For such a solution $\\psi$, the [[single layer potential|9T6ZEV4125]] $u(x) = \\int_{\\partial D} \\Phi(x - y) \\psi(y) \\, ds(y)$ is [[harmonic|0DEB0JMN0W]] in $D$, and its normal derivative from inside equals $g$ on $\\partial D$.\n\nSo the interior Neumann problem $\\Delta u = 0$ in $D$, $\\partial_\\nu u = g$ on $\\partial D$, is solvable exactly when $\\int_{\\partial D} g \\, ds = 0$, and its solutions differ by constants.\n\n## Remarks\n\nPart 2 follows from the [[jump relations for layer potentials|J58YA8QNNX]]: the normal derivative of the single layer potential from inside is $K'\\psi + \\frac12 \\psi$.\nFor part 1, $K'$ is compact on $C(\\partial D)$ and is the adjoint of the double layer operator $K$ with respect to the pairing $\\int_{\\partial D} \\varphi \\psi \\, ds$.\nThe null space of $I + 2K$ consists of the constants, since $K1 = -\\frac12$ by Gauss's formula for the double layer potential of a constant density.\nBy the [[Fredholm alternative|4HH8C9V8JQ]], $\\psi + 2K'\\psi = 2g$ is solvable if and only if $2g$ is orthogonal to the constants.\n\nExample: on the unit disc in $\\R^2$, $\\frac{\\partial \\Phi(x - y)}{\\partial \\nu(x)} = -\\frac{1}{4\\pi}$ for $x \\ne y$ on the unit circle, so $K'\\psi = -\\frac{1}{4\\pi} \\int \\psi \\, ds$.\nFor $g(\\theta) = \\cos\\theta$ the equation $\\psi + 2K'\\psi = 2g$ has the solution $\\psi = 2\\cos\\theta$, since $\\int_0^{2\\pi} \\cos\\theta \\, d\\theta = 0$; the resulting single layer potential is $u(x) = x_1$ in the disc, whose normal derivative on the circle is $\\cos\\theta$.\nFor $g = 1$ the condition $\\int g \\, ds = 0$ fails, and indeed no harmonic $u$ in $D$ has $\\partial_\\nu u = 1$, because $\\int_{\\partial D} \\partial_\\nu u \\, ds = \\int_D \\Delta u \\, dx = 0$.\n\n## Explanations\n\n### L1 human\n\nHere the flux of a harmonic function through the boundary is prescribed instead of its values.\nSince nothing is created inside the region, the total flux must be zero, and that is the only obstruction.\nOne looks for the solution as the potential of a layer of charges on the boundary, whose density solves an integral equation on the boundary.\n\n### L2 human\n\nThis is the companion of the [[boundary integral equation for the Dirichlet problem|ZYDVFT6JR7]], which uses the double layer potential.\nThe two boundary operators $K$ and $K'$ are adjoint to each other, and the solvability condition comes from the Fredholm alternative applied to this pair.\nThe same equations, with the signs changed, treat the exterior problems.\nNumerically, both equations are discretized by boundary element or Nyström methods, which reduce the dimension of the problem by one.\n",
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