Kelvin transform

Remarks

The map $x \mapsto x^*$ is an involution, so $(\Omega^*)^* = \Omega$ and $K[K[u]] = u$. For $n = 2$, $K[u] = u \circ {}^*$ is the composition with the inversion in the unit circle, $z \mapsto 1/\bar z$. For $u$ twice continuously differentiable, $\Delta(K[u]) = K[\abs{x}^4 \Delta u]$, where $\Delta$ is the Laplacian; so $u$ is harmonic on $\Omega$ if and only if $K[u]$ is harmonic on $\Omega^*$. Example: in $\R^3$, $K[x_1](x) = \abs{x}^{-1} \cdot x_1 \abs{x}^{-2} = x_1 / \abs{x}^3 = -\partial_1(1/\abs{x})$, which is harmonic on $\R^3 \setminus \set{0}$, as a derivative of a harmonic function. Non-example: without the weight, $u(x^*) = x_1 / \abs{x}^2$ is not harmonic on $\R^3 \setminus \set{0}$; a direct computation gives $\Delta(x_1 \abs{x}^{-2}) = -2 x_1 \abs{x}^{-4}$. The transform turns the behaviour of $u$ near infinity into the behaviour of $K[u]$ near $0$. For instance, let $n \ge 3$ and let $u$ be harmonic outside a ball; then $u(x) \to 0$ at infinity if and only if $K[u]$ has a removable singularity at $0$. Inversion in a sphere of centre $a$ and radius $\rho$ gives the analogous transform $\left(\rho / \abs{x - a}\right)^{n-2} u\left(a + \rho^2 (x - a)/\abs{x - a}^2\right)$.