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Prove that Laplace's equation $\Delta u = 0$ is rotation invariant; that is, if $O$ is an orthogonal $n\times n$ matrix and we define

$$ v(x) := u(Ox) \quad (x \in \mathbb{R}^n)$$ then $\Delta v = 0$.

I wanted to see if what I had below was correct and complete. Any feedback on rigor is greatly appreciated.

In cartesian coordinates, we have $\Delta u = \sum_{i = 1}^n \frac{\partial^2 u}{\partial x_i^2}$. Letting the $i$th row and $j$th column of the matrix $O$ be $a_{ij}$ we have

\begin{equation} \frac{\partial v}{\partial x_j} = \sum_{i =1}^n\frac{\partial u}{\partial x_i}a_{ij} \end{equation}

Thus, \begin{equation} \frac{\partial^2 v}{\partial x_j^2} = \sum_{i =1}^n\sum_{k = 1}^n\frac{\partial^2 u}{\partial x_i\partial x_k}a_{ij}a_{kj} \end{equation}

Therefore: \begin{align} \Delta v &= \sum_{j = 1}^n \frac{\partial^2 v}{\partial x_j^2}\\ &= \sum_{j = 1}^n \left(\sum_{i =1}^n\sum_{k = 1}^n\frac{\partial^2 u}{\partial x_i\partial x_k}a_{ij}a_{kj}\right)\\ &= \sum_{i =1}^n\sum_{k = 1}^n\left(\sum_{j = 1}^n \frac{\partial^2 u}{\partial x_i\partial x_k}a_{ij}a_{kj}\right)\\ &= \sum_{i =1}^n\sum_{k = 1}^n \frac{\partial^2 u}{\partial x_i\partial x_k}\left(\sum_{j = 1}^na_{ij}a_{kj}\right)\\ &=\sum_{i =1}^n\sum_{k = 1}^n \frac{\partial^2 u}{\partial x_i\partial x_k}\delta_{i, k}\\ &= \sum_{i =1}^n\frac{\partial^2 u}{\partial x_i^2} = 0 \end{align}

Math_Day
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1 Answers1

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A coordinate-free proof uses that $\triangle u = {\rm tr}(D\nabla u)$, where $\nabla u:\Bbb R^n \to \Bbb R^n$ is the gradient function of $u$ and the linear map $D(\nabla u)(x):\Bbb R^n \to \Bbb R^n$ is the Hessian of $u$ at $x$.

The chain rule says that $$\langle \nabla(u\circ O)(x) ,v\rangle = D(u\circ O)(x)(v) = Du(Ox)\circ DO(x)(v) = Du(Ox)(Ov) = \langle \nabla u(Ox),Ov\rangle = \langle O^\top \nabla u(Ox),v\rangle,$$ so $\nabla(u\circ O)(x) = O^\top \nabla u(Ox)$. Apply $D$ to get $$D\nabla (u\circ O)(x)(v) = O^\top D\nabla u(Ox)O(v).$$ Thus $$\triangle(u\circ O)(x) = {\rm tr}(O^\top D\nabla u(Ox) O) = {\rm tr}(OO^\top D\nabla u(Ox)) = {\rm tr}(D\nabla u(Ox)) = (\triangle u)(Ox),$$by cyclic invariance of trace and $OO^\top = {\rm Id}$. So $\triangle(u\circ O) = (\triangle u)\circ O$. The conclusion is that $u$ is harmonic if and only if $u\circ O$ is harmonic for one (and hence every) orthogonal map $O$.

Ivo Terek
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