I am just looking for a more intuitive proof that the $\nabla^2$ operator is rotationally invariant in Euclidean space. That is: If $u(x)$ solves $\nabla ^2 u=0$, then $v(x)=u(Rx)$ solves it too, where $R$ is the usual rotation matrix ($RR^T=Id$). Here $x\in \mathbb R^n$.

I was able to prove it using brute force index gymnastics, but I am sure there something prettier out there.


To expand on Daniel Fischer's comment, the solutions to the equation $\nabla^2u=0$ are precisely the integrable functions $u$ satisfying $$u(x)=\frac{1}{V(B(x,r))}\int_{B(x,r)} u(y)dy$$ for any $x$ and any $r>0$, where $B(x,r)$ is the ball of radius $r$ around $x$ and $V$ denotes volume. Let $R$ be a rotation about $0$. Then we have $$\begin{align*} u(Rx) &=\frac{1}{V(B(Rx,r))}\int_{B(Rx,r)} u(y)dy\;\; \text{because $u$ satisfies MVP}\\ &=\frac{1}{V(B(Rx,r))}\int_{RB(x,r)} u(y)dy\;\; \text{because}\; RB(x,r)=B(Rx,r)\\ &=\frac{1}{V(B(Rx,r))}\int_{B(x,r)} u(Ry)dy \end{align*}$$ as desired.


Just treat the rotation like a function: $v(x) = (u \circ R)(x)$. Then use the chain rule. Let $a$ be an arbitrary vector. The chain rule is then:

$$a \cdot \nabla v|_x = [(a \cdot \nabla) R(x)] \cdot \nabla u|_{R(x)}$$

$R$ is a linear function, and as such, $(a \cdot \nabla) R(x) = R(a)$. For brevity, let $R(x) = x'$, and we get

$$a \cdot \nabla v|_x = R(a) \cdot \nabla u|_{x'} \implies \nabla (u \circ R)|_x = R^T(\nabla u)|_{x'}$$

(You may be thinking, "What on earth does it mean for a rotation matrix to act on $\nabla$? Well, you just treat the partial derivatives as if they were components of a vector.)

A similar line of logic would show that, for a vector field $F$,

$$\nabla \cdot (F \circ R)|_x = R^T(\nabla) \cdot F |_{x'}$$

So the Laplacian would take the form

$$\nabla^2 (u \circ R)|_x = R^T(\nabla) \cdot R^T( \nabla) u|_{x'}$$

But the right-hand side reduces to $R^T R(\nabla) \cdot \nabla u|_{x'}$, and we know that $R^T R = 1$, leaving us with $\nabla^2 u|_{x'} = \nabla^2 v |_x$. If $\nabla^2 u = 0$ everywhere, then $\nabla^2 v = 0$ everywhere as well.


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