what condition should $\phi$ satisfy to make $u(\phi)$ still be harmonic when $u$ is harmonic let $U\subseteq\mathbb{R}^n$ be open and $u\in C^2(U)$ be a solution to $\Delta u(y)=0$ for all $y\in U$. Let $\phi:V\rightarrow U$ be twice continuously differentiable where $V\subseteq\mathbb{R}^n$ is open and $U=\phi(V)$. Let $v:\phi\rightarrow \mathbb{R}$ be defined by $v(x)=u(\phi(x))$ for all $x\in V$.What condition should $\phi$ satisfy in order to make $\Delta v=0$ for all $x\in V$.
what I have done is that just calculate the twice differential of $v$ which is $$v_{x_i}=u_{x_i}(\phi(x))\phi_{x_i}(x)\mbox{ and }v_{x_ix_i}=u_{x_ix_i}(\phi(x))\phi_{x_i}^{2}(x)+u_{x_i}(\phi(x))\phi_{x_ix_i}(x)$$
then $$\Delta v=\sum_{i=1}^nv_{x_ix_i}=\sum_{i=1}^n[u_{x_ix_i}(\phi(x))\phi_{x_i}^{2}(x)+u_{x_i}(\phi(x))\phi_{x_ix_i}(x)]$$
and I don't know how to carry on to find the condition I need, can someone help me with this?
$ \rule[-10pt]{17.5cm}{0.05em}$
then I tried to use chain rule then I got $\nabla v=\nabla u\circ\phi=((\nabla u)\circ\phi)D\phi$ where $D\phi$ is the square matirx of derivativesof $\phi$. But here I don't know how to get the second derivatives.
 A: I know this might not be enough, but if we assume that $\phi$ can be represented by an orthogonal $n\times n$ matrix 
i.e. $\phi(x)=O\cdot x$ where $O$ is an orthogonal matrix
then we want to prove that $\Delta v(x)=0$
Write $O=(a_{i,j})$ and since $\phi(x)=O\cdot x$ so $D\phi(x)=O$
Since $v$ has been defined to be $v(x)=u(\phi(x))$, hence $$Dv(x)=Du(\phi(x))=\cdot D\phi(x)=Du(O\cdot x)\cdot O$$
Since $v_{x_i}=\sum_{j=1}^{n}u_{x_j}(\phi(x))_{a_{j,i}}$, then by the similar calculation, we have $$D(u_{x_j}\circ\phi)(x)=Du_{x_j}(\phi(x))\cdot O$$
Thus $$v_{x_ix_i}(x)=\frac{\partial v_{x_i}(x)}{\partial x_i}=\sum_{j=1}^n\frac{\partial u_{x_j(\phi(x))a_{j,i}}}{\partial x_i}=\sum_{j=1}^na_{j,i}\left[\sum_{k=1}^nu_{x_jx_k}(\phi(x))a_{k,i}\right]=\sum_{j,k=1}^na_{j,i}a_{k,i}u_{x_jx_k}(\phi(x))$$
Then we can calculate the $\Delta$ of $v$
$$\Delta v(x)=\sum_{i=1}^nv_{x_ix_i}=\sum_{i=1}^n\left[\sum_{j,k=1}^na_{j,i}a_{k,i}u_{x_jx_k}(\phi(x))\right]=\sum_{j,k=1}^n\left[u_{x_jx_k}(\phi(x))\sum_{i=1}^na_{j,i}a_{k,i}\right]$$
By assumption, we have that $O$ is orthogonal, i.e. $$\sum_{i=1}^na_{j,i}a_{k,i}=
 \begin{cases}
  1,&\mbox{ if } j=k\\
  0,&\mbox{ if } j\neq k
 \end{cases}$$
and we also assume that $\Delta u=0$, therefore $$\Delta v(x)=\sum_{i=1}^n\left[u_{x_ix_i}(\phi(x))\right]=\Delta u(\phi(x))=0$$
