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  "body": "\n## Example\n\nLet $B = \\set{x \\in \\R^3 : \\abs{x} \u003c 1}$ be the unit [[open ball|0YE5868HZB]].\nFor $x \\in B$ with $x \\ne 0$, let $x^* = x / \\abs{x}^2$ be the inverse point of $x$ in the unit sphere; it lies outside $\\overline{B}$.\nDefine, for $y \\in \\overline{B}$ with $y \\ne x$,\n$$G(x, y) = \\frac{1}{4\\pi} \\left( \\frac{1}{\\abs{x - y}} - \\frac{1}{\\abs{x} \\, \\abs{y - x^*}} \\right) \\quad (x \\ne 0), \\qquad G(0, y) = \\frac{1}{4\\pi} \\left( \\frac{1}{\\abs{y}} - 1 \\right).$$\nThis is the [[Green's function|M36SD56MWQ]] of $B$, normalized so that $\\Delta_y G(x, y) = -\\delta_x$.\nCheck of the three properties, for $x \\ne 0$:\n\n1. The first term is the [[fundamental solution|BSJ8X2XRN9]] $\\frac{1}{4\\pi\\abs{x - y}}$, which has the right singularity at $y = x$.\n2. The second term is [[harmonic|0DEB0JMN0W]] in $y$ on $B$, because its singularity $x^*$ lies outside $\\overline{B}$.\n3. For $\\abs{y} = 1$ one has $\\abs{x}^2 \\abs{y - x^*}^2 = \\abs{x}^2 - 2 x \\cdot y + 1 = \\abs{x - y}^2$, so the two terms cancel and $G(x, y) = 0$.\n\nFor $x = 0$, the term $1/\\abs{y}$ is the fundamental solution times $4\\pi$, the constant $1$ is harmonic, and $G(0, y) = 0$ when $\\abs{y} = 1$.\n\n## Remarks\n\nIn $\\R^n$ with $n \\ge 3$ the same construction gives $G(x, y) = \\Phi(x - y) - \\Phi(\\abs{x}(y - x^*))$, where $\\Phi(z) = \\frac{1}{(n-2)\\sigma_{n-1}} \\abs{z}^{2-n}$ and $\\sigma_{n-1}$ is the area of the unit sphere.\nThe outward normal derivative of $-G(x, \\cdot)$ on the sphere is the Poisson kernel of the ball.\nThe second term is the potential of an image charge of strength $1/\\abs{x}$ placed at $x^*$.\n\n## Explanations\n\n### L1 human\n\nTo make the potential of a point charge vanish on a sphere, place a second, opposite charge at the mirror point outside the sphere.\nThe mirror point lies on the same ray, at distance $1/\\abs{x}$ from the centre.\nOn the sphere the two potentials cancel exactly, and inside the ball the sum is the Green's function.\n",
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