$ℝ^n$中Laplace算子是旋转不变的

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例如$n=2$时,利用极坐标下的表达式,$$\Delta f(r,θ)=\frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2} + \frac{1}{r}  \frac{\partial f}{\partial r} + \frac{\partial^2 f}{\partial r^2}$$设$θ_1=θ+θ_0$($θ_0$为常数),则有$Δf(r,θ)=Δf(r,θ_1)$.

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$n=3$时,利用球坐标下的表达式,
$$Δf(r,θ,ϕ)={1 \over r^{2}}{\partial  \over \partial r}\!\left(r^{2}{\partial f \over \partial r}\right)\!+\!{1 \over r^{2}\!\sin \theta }{\partial  \over \partial \theta }\!\left(\sin \theta {\partial f \over \partial \theta }\right)\!+\!{1 \over r^{2}\!\sin ^{2}\theta }{\partial ^{2}f \over \partial ϕ ^{2}}$$设$θ_1=θ+θ_0,ϕ_1=ϕ+ϕ_0$($\theta_0,\phi_0$为常数)如何使用链式法则来证明$Δf(r,θ,ϕ)=Δf(r,θ_1,ϕ_1)$呢?分母中带有$θ_0$好像很难化简掉

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sol1.pdf:
If $O$ is an orthogonal $n \times n$ matrix and we define
$$
v(x)=u(O x) \quad\left(x \in \mathbb{R}^{n}\right),
$$
then $\Delta v=\Delta u$.
Solution.
Let $O=\left[o_{i j}\right]$. We compute:
$$
\begin{aligned}
D_{i} v(x) &=\sum_{k=1}^{n} D_{k} u(O x) o_{k i} \\
D_{i j} v(x) &=\sum_{l=1}^{n} \sum_{k=1}^{n} D_{k l} u(O x) o_{k i} o_{l j}
\end{aligned}
$$
Since $O$ is orthogonal, then $O O^{T}=I$ where $I$ is the $n \times n$ identity matrix, thus for all $k, l=1, \ldots, n$
$$
\sum_{i=1}^{n} o_{k i} o_{l i}=\delta_{k l}:= \begin{cases}1 & \text { if } k=l \\ 0 & \text { if } k \neq l\end{cases}
$$
Thus\begin{align*} \Delta v(x) &=\sum_{i=1}^{n} \sum_{l=1}^{n} \sum_{k=1}^{n} D_{k l} u(O x) o_{k i} o_{l i} \\ &=\sum_{l=1}^{n} \sum_{k=1}^{n} D_{k l} u(O x)\left(\sum_{i=1}^{n} o_{k i} o_{l i}\right) \\ &=\sum_{l=1}^{n} \sum_{k=1}^{n} D_{k l} u(O x) \delta_{k l} \\ &=\Delta u(O x) \end{align*}