Show that Laplacian is zero Let $u: \mathbb{R}^n \rightarrow \mathbb{R}$ be harmonic, i.e. $\Delta u =0.$ Now, I want to show that $\Delta (u(Ox+b))=0$ for an orthogonal matrix $O$ and a constant vector $b$.
Does anybody know how this can be shown?
Regarding the answer, I received.
mhmm. $D(u(Ox+b)) = Du (Ox+b)D(Ox+b)= Du(Ox+b)O.$ I don't quite see how the divergence of this could look like?
 A: Hint: $\Delta u(x) = \nabla \cdot \nabla u(x)$. Use chain rule and deduce the claim.
A: Both divergence and gradient can be defined in coordinate-free ways, which makes it manifest that they, and therefore their composition the Laplacian, are invariant under translations and orthogonal transformations.
A: Let's use indexed notation to write the $i$'th component of $Ox+b$ as $\sum_j\,O_{ij}x_j+b_i$.
Then, observe that from the chain rule we have
$$\frac{\partial \,u(Ox+b)}{\partial x_k}=\sum_i\,u_iO_{ik}$$
where the notation $u_i$ designates a partial derivative on the $i$'th argument of $u$.  Now, a second differentiation yields
$$\frac{\partial^2 \,u(Ox+b)}{\partial x_k^2}=\sum_j\sum_i\,u_{ij}O_{jk}O_{ik}$$
Now, we sum over $k$ to find
$$\begin{align}
\sum_k\,\frac{\partial^2 \,u(Ox+b)}{\partial x_k^2}&=\sum_k\sum_j\sum_i\,u_{ij}O_{jk}O_{ik}\\\\
&=\sum_j\sum_i\,u_{ij}\,\sum_k\,O_{jk}O_{ik}\tag 1\\\\
&=\sum_j\sum_i\,u_{ij}\,\delta_{ij} \tag 2 \\\\
&=\sum_i\,u_{ii}\\\\
&=0
\end{align}$$
as was to be shown!
Note in going from $(1)$ to $(2)$ we made use of the fact that $O$ is orthogonal.  In $(2)$, $\delta_{ij}$ is the Kronecker Delta, which is $1$ if $i=j$ and is zero otherwise.
