TITLE: Show Laplace operator is rotationally invariant
QUESTION [11 upvotes]: I'm trying to show the Laplace operator is rotationally invariant. Essentially this boils down to showing
$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}$$
where
$$u = x \cos \theta + y \sin \theta$$
$$v = -x \sin \theta + y \cos \theta$$
I think I'm on the right track by noting that 
$$\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial u}\left(\frac{\partial f}{\partial x}\right)\frac{\partial u}{\partial x} + \frac{\partial}{\partial v}\left(\frac{\partial f}{\partial x}\right)\frac{\partial v}{\partial x}$$
but I'm having difficulty reaching an end game where I show
$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2}({\sin}^2 \theta + {\cos}^2 \theta) + \frac{\partial^2 f}{\partial v^2}({\sin}^2 \theta + {\cos}^2 \theta)$$

REPLY [11 votes]: Here is an answer along the lines of @TeeJay, but with a few more details and in general $\mathbb R^n$ space. So, we need to show that if $u$ is harmonic ($\Delta_y u(y)=0$) then $v(x)=u(Mx)$ is also harmonic ($\Delta_x v(x)=0$) for an orthogonal $M$ (i.e., $M^\top=M^{-1}$). 
First I note that if $D^2_y u(y)$ denotes the Hessian matrix, then
$$
\Delta_y u(y)={\rm tr}\, D^2_y u(y).
$$ 
Note also that $D^2=\nabla\nabla^T$, where $\nabla$ is the column vector of partial derivatives. 
Now the only thing we need is the fact that $\nabla_x u(Mx)=M^\top \nabla_y u(y)|_{y=Mx}$ and that ${\rm tr}\,(AB)={\rm tr}\,(BA)$. 
$$
\Delta_x v(x)={\rm tr}(D^2_x v(x))={\rm tr}(D^2_x u(Mx))=\\
{\rm tr}(\nabla_x\nabla_x^Tu(Mx))={\rm tr}(\nabla_x(\nabla_x u(Mx))^\top)=\\
{\rm tr}(\nabla_x(M^\top \nabla_y u(y))^\top)={\rm tr}(\nabla_x(\nabla_y^\top u(y))M)=\\
{\rm tr}(M^\top\nabla_y\nabla_y^\top u(y)M)={\rm tr}(\nabla_y\nabla_y^\top u(y))\\
={\rm tr}(D^2_y u(y))=\Delta_y u(y)=0
$$
as required.