Prove that the Laplace's equation $\Delta u=0$ is rotation invariant on ${\bf R}^n$.
I know this is a very basic question, but I am not very sure for some steps. I wrote down my working for this problem in detail as possible as I can.
My attempt :
Fix an $n\times n $ orthogonal matrix $O=(a_{ij})$ with real entries , we try to show that $\Delta\big(u(O\cdot x)\big)=0$
Define $v(x)=u(O\cdot x)$.
Let $\xi=O\cdot x$ , then for all $j=1,2,...,$ one has $\xi_{j}=a_{j1}x_{1}+a_{j2}x_{2}+\cdots+a_{jn}x_{n}=\displaystyle\sum_{i=1}^{n}a_{ji}x_{i}$ Thus for each $i=1,2,...,$ we have $$\frac{\partial v}{\partial x_{i}}=\sum_{j=1}^{n}\frac{\partial u}{\partial \xi_{j}}\frac{\partial \xi_{j}}{\partial x_{i}}=\sum_{j=1}^{n}\frac{\partial u}{\partial \xi_{j}}a_{ji}$$ Therefore, for all $i=1,2,...,n$ , we obtain $$\frac{\partial^{2} v}{\partial x_{i}^{2}}=\sum_{j=1}^{n}\sum_{l=1}^{n}\frac{\partial^{2}u}{\partial \xi_{j}\partial\xi_{l}}\frac{\partial \xi_{l}}{\partial x_{i}}a_{ji}=\sum_{j=1}^{n}\sum_{l=1}^{n}\frac{\partial^{2}u}{\partial \xi_{j}\partial\xi_{l}}a_{li}a_{ji}$$
Hence, we have \begin{align} \Delta v&=\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{l=1}^{n}\frac{\partial^{2}u}{\partial \xi_{j}\partial\xi_{l}}a_{li}a_{ji}\\ &=\sum_{j=1}^{n}\sum_{l=1}^{n}\frac{\partial^{2}u}{\partial \xi_{j}\partial\xi_{l}}\sum_{i=1}^{n}a_{li}a_{ji}\\ &=\sum_{j=1}^{n}\sum_{l=1}^{n}\frac{\partial^{2}u}{\partial \xi_{j}\partial\xi_{l}}\sum_{i=1}^{n}O_{li}(O^{T})_{ij}\\ &=\sum_{j=1}^{n}\sum_{l=1}^{n}\frac{\partial^{2}u}{\partial \xi_{j}\partial\xi_{l}}(OO^{T})_{lj}\\ &=\sum_{j=1}^{n}\sum_{l=1}^{n}\frac{\partial^{2}u}{\partial \xi_{j}\partial\xi_{l}}I_{lj}\\ &=\sum_{j=1}^{n}\sum_{l=1}^{n}\frac{\partial^{2}u}{\partial \xi_{j}\partial\xi_{l}}\delta_{lj}\\ &=\sum_{l=1}^{n}\frac{\partial^{2} u}{\partial \xi_{l}^{2}}\\ &=0 \end{align}
where $I$ is a $n\times n$ identity matrix and $\delta_{lj}=1$ as $l=j$ , $\delta_{lj}=0$ as $l\ne j$
Can anyone check my attempt for validity?
You should really use mutlidimensional calculus: Note that $\Delta u = \mathrm{trace}(\mathrm{Hess} \, u)$. Now we define $v(x) = u(Ux)$ for some orthogonal matrix $U \in O(n)$. Then, (using row vectors for $v,u$) $$D_xv = (D_{Ux} v) U = (D_x u) U$$ (if we were to use column vectors, one will obtain $D_x v=U^T (D_x u)$)
and furthermore $$\mathrm{Hess}_x v = U^T \left(\mathrm{Hess}_{Ux} v \,\right) U = U^T \left(\mathrm{Hess}_{x} u \,\right) U.$$ Since the trace is cyclical, we get $\Delta v =\mathrm{trace} \left( \mathrm{Hess}_x \, v\right) = \mathrm{trace} \left(\mathrm{Hess}_{x} u \,\right) = \Delta u =0.$
However, your argument is fine and verify also the above-mentioned formulas implicitly.