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Theorem 3.2.2 (Kelvin transform)97BA27062B9F

Theorem 3.2.2 (Kelvin transform).97BA27062B9F Let $U\subset \mathbb R^ d\setminus \{ a\} $ be open and let $u\in C^2(U)$. Then $Ku\in C^2(T(U))$ and

0E9BCA99B831 3.2.3
\begin{equation} \Delta (Ku)(x)=\left(\frac{\rho }{|x-a|}\right)^{d+2} (\Delta u)(T(x)). \end{equation}

Moreover, $T^2$ and $K^2$ are the identity. Thus the Kelvin transform preserves harmonicity and nonnegativity in both directions.

Proof. Translate $a$ to $0$, and set $r=|x|$, $e=x/r$, $Q=I-2ee^{\mathsf T}$, $y=T(x)=\rho ^2x/r^2$ and $w(x)=\rho ^{d-2}r^{2-d}$. The matrix $Q$ is symmetric and orthogonal, and $Qx=-x$. Differentiating $y_ j=\rho ^2x_ jr^{-2}$ gives

\[ Dy=\rho ^2r^{-2}Q,\qquad \Delta y_ j=-2(d-2)\rho ^2x_ jr^{-4}. \]

For completeness, $\Delta r^\alpha =\alpha (\alpha +d-2)r^{\alpha -2}$ follows by differentiating $\partial _ jr^\alpha =\alpha x_ jr^{\alpha -2}$ and summing. Applying the product rule to $x_ jr^{-2}$ proves the second identity above. The same radial formula yields

\[ \nabla w=(2-d)\rho ^{d-2}r^{-d}x,\qquad \Delta w=0. \]

The chain rule and the orthogonality of $Q$ now give

\begin{align*} \Delta (u\circ T) & =\rho ^4r^{-4}(\Delta u)(y) -2(d-2)\rho ^2r^{-4}x\cdot \nabla u(y),\\ 2\nabla w\cdot \nabla (u\circ T) & =2(d-2)\rho ^ d r^{-d-2}x\cdot \nabla u(y). \end{align*}

In $\Delta (w(u\circ T))$, the terms containing $x\cdot \nabla u(y)$ therefore cancel, leaving $\rho ^{d+2}r^{-d-2}(\Delta u)(y)$. This proves (3.2.3), including $d=2$.

Since $|T(x)|=\rho ^2/|x|$, substitution gives $T(T(x))=x$ and $w(x)w(T(x))=1$. In particular $T$ is a smooth diffeomorphism of $\mathbb R^ d\setminus \{ 0\} $ onto itself, so $T(U)$ is open and $K^2u=u$. The positive weight $w$ preserves signs. The Laplacian identity and involutivity prove the assertion about harmonicity. $\square$

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