Show that Laplace's equation $\Delta u=0$ is rotation invariant In Evans' book, PDE, it is asked to show that if $O$ is a $n\times n$ orthogonal matrix, and we define$$
v(x):=u(Ox)
$$
then $\Delta v=0$. I tried to compute $\Delta v=D_xv\cdot D_xv$ and expand it using the fact that $O^T=O^{-1}$, but I get stuck after a while. Help!!
 A: A function $u \in C^2(\Omega)$ is harmonic iff it satisfies the mean value property. Then note that for any $B_r(x) \subset \Omega$ there holds $Q(B_r(x))=B_r(Qx)$. Then:
$$
\int_{B_r(x)}u(Qy)dy=\int_{B_r(x)}u(Qy)|\det(Q)|dy=\int_{Q(B_r(x))}u(y)dy=\int_{B_r(Qx)}u(y)dy=u(Qx)
$$
A: Another way to see this is to resort to indices:
Let $O$ be the orthogonal matrix. Then
\begin{align}
\Delta v & = \sum_{i=1}^n \frac{\partial ^2}{\partial x_i^2} u(Ax)
\end{align}
Define $y_k = \sum_j A_{kj}x_j$.
Note that
$$ \frac{\partial v}{\partial x_i} = \sum_{j=1}^n \frac{\partial u}{\partial y_j}\frac{\partial y_j}{\partial x_i} = \sum_{j=1}^n \frac{\partial u}{\partial y_j} a_{ji}$$
Then we have
\begin{align}
\Delta v & = \sum_{i}\frac{\partial}{\partial x_i} \left(\sum_{j=1} \frac{\partial u}{\partial y_j}a_{ji}\right)\\
& = \sum_{ij}a_{ji}\frac{\partial }{\partial x_i}\frac{\partial u}{\partial y_j} \\
& = \sum_{ij}a_{ji}\sum_{k}\frac{\partial u}{\partial y_j \partial y_k}\frac{\partial y_k}{\partial x_i} \\
& = \sum_{ij}a_{ji}\sum_{k}\frac{\partial u}{\partial y_j \partial y_k}a_{ki}\\
& = \sum_{jk}u_{y_j,y_k}\sum_i a_{ji}a^T_{ik}\\
& = \sum_{j} u_{jj} = 0
\end{align}
Since $\sum_i a_{ji}a^T_{ik} = 1$ if and only if $j=k$ by orthogonality.
A: Using your notation, let $O=(a_{ij})$ and $y\equiv Ox$. Then $\Delta v(x)=\Delta u(Ox)=\Delta u(y)$, with$$
y_j=\sum_{i=1}^n a_{ji}x_i.
$$
We then have$$
\frac{\partial v}{\partial x_i}=\sum_{j=1}\frac{\partial u}{\partial y_j} \frac{\partial y_j}{\partial x_i}=\sum_{j=1}\frac{\partial u}{\partial y_j} a_{ji}.
$$
Thus\begin{align*}
\begin{pmatrix}
\frac{\partial v}{\partial x_1}\\
\vdots\\
\frac{\partial v}{\partial x_n}\end{pmatrix}&=\begin{pmatrix}
a_{11}&\cdots&a_{n1}\\
\vdots &\ddots &\vdots\\
a_{1n}&\cdots&a_{nn}
\end{pmatrix}\begin{pmatrix}
\frac{\partial u}{\partial y_1}\\
\vdots\\
\frac{\partial u}{\partial y_n}
\end{pmatrix}=O^T\begin{pmatrix}
\frac{\partial u}{\partial y_1}\\
\vdots\\
\frac{\partial u}{\partial y_n}
\end{pmatrix}
\end{align*}
which can be simply rewritten as $D_x\cdot v=O^TD_y\cdot u $. Then, it follows that\begin{align*}
\Delta v&=D_xv\cdot D_xv   \\
&=   (O^TD_yu)\cdot(O^TD_y u)\\
&=  (O^TD_yu)^TO^TD_yu \\
&=   (D_yu)^T(O^T)^TO^TD_yu\\
&= (D_yu)^TOO^TD_yu  \\
&= (D_yu)^TD_yu  \\
&=  (D_yu)\cdot (D_yu) \\
&=   \Delta u(y)\\
&=   0,
\end{align*}
since $O$ is orthogonal and thus $O^T=O^{-1}$.
