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  "body": "\n## Example\n\nLet $n \\ge 2$, let $B = \\set{x \\in \\R^n : \\abs{x} \u003c 1}$ be the unit [[open ball|0YE5868HZB]], and fix a point $\\zeta$ with $\\abs{\\zeta} = 1$.\nConsider the function\n$$u(x) = \\frac{1 - \\abs{x}^2}{\\abs{x - \\zeta}^n}, \\qquad x \\in B,$$\nwhich is the [[Poisson kernel|TCYSWCTYYM]] of the ball with pole $\\zeta$, up to a constant factor.\nIt is positive on $B$, because $\\abs{x} \u003c 1$ and $x \\ne \\zeta$.\nTo check that it is harmonic, write $f(x) = 1 - \\abs{x}^2$, $g(x) = \\abs{x - \\zeta}^{-n}$ and $r = \\abs{x - \\zeta}$.\nThen $\\Delta f = -2n$, $\\nabla f = -2x$, $\\nabla g = -n r^{-n-2} (x - \\zeta)$ and $\\Delta g = 2n r^{-n-2}$, where $\\Delta$ is the [[Laplacian|0E0B6RA8K0]].\nThe product rule $\\Delta(fg) = f \\Delta g + 2 \\nabla f \\cdot \\nabla g + g \\Delta f$ gives\n$$\\Delta u = 2n r^{-n-2} \\left( 1 - \\abs{x}^2 + 2 x \\cdot (x - \\zeta) - r^2 \\right).$$\nSince $r^2 = \\abs{x}^2 - 2 x \\cdot \\zeta + 1$, the bracket equals $0$, so $\\Delta u = 0$.\n\n## Remarks\n\nAs $x$ tends to a boundary point $\\eta \\ne \\zeta$, the numerator tends to $0$ and the denominator stays away from $0$, so $u(x) \\to 0$.\nAs $x \\to \\zeta$ along the radius, $u(t\\zeta) = (1 + t)/(1 - t)^{n-1}$ tends to $+\\infty$.\nSo $u$ is a positive harmonic function on the ball that vanishes on the whole boundary sphere except at one point.\n\n## Explanations\n\n### L1 human\n\nThis function is the steady temperature in a ball when all the heat enters through one point $\\zeta$ of the surface.\nIt is zero near the rest of the surface and blows up at $\\zeta$.\nEvery positive harmonic function on the ball is an average of such functions with different poles.\n",
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