Show Laplace operator is rotationally invariant 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
$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2}({\sin}^2 \theta + {\cos}^2 \theta) + \frac{\partial^2 f}{\partial v^2}({\sin}^2 \theta + {\cos}^2 \theta)$$
 A: In a paper   Discrete spherical means of directional derivatives and Veronese maps (arXiv:1106.3691) get get the identity:
$$ \Delta f = \frac{2}{\pi}\int_0^\pi d\phi \frac{\partial^2 f}{\partial\mathbf{e}_\phi^2} $$
The Laplacian is the average of the second directional derivative in all directions. $\mathbf{e}_\phi = (\cos \phi, \sin \phi)$ and 
$$ \frac{\partial}{\partial\mathbf{e}_\phi} = \cos \phi \frac{\partial}{\partial x}+ \sin \phi \frac{\partial}{\partial y} = \nabla \cdot (\cos \phi, \sin \phi)$$
Directional derivatives are ways of taking dervatives in directions other than $x$ and $y$ axes. 

If this seems too much, let's just try 3 directions (then try 5, 7, or more):
$$ \Delta f = \frac{1}{3} \bigg[ 
\frac{\partial^2 f}{\partial x^2} +
\big( \underbrace{\cos \tfrac{2\pi}{3}\cdot \frac{\partial^2 f}{\partial x^2} + 
\sin \tfrac{2\pi}{3}\cdot \frac{\partial^2 f}{\partial y^2}}_{\phi = 2\pi/3}\big) +
\big(\underbrace{\cos \tfrac{4\pi}{3}\cdot \frac{\partial^2 f}{\partial x^2} + 
\sin \tfrac{4\pi}{3}\cdot \frac{\partial^2 f}{\partial y^2}}_{\phi=4\pi/3}\big) \bigg]$$
A: Off the top of my head, I think an easier way to do this would be to write it all in vector/matrix notation. Under a rotation, $\vec\nabla f\rightarrow R\cdot \vec\nabla f$ with $R$ being your rotation matrix. Then the Laplacian would transform like $\nabla^{2}f=\vec\nabla\cdot \vec\nabla f\rightarrow \vec\nabla\cdot R^{T}R\cdot \vec\nabla f$. Rotation matrices satisfy $R^{T}R=1$ so that should do it. There is a bit more rigor to this but this should be a good starting point.
A: $$ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial f}{\partial u} \sin \theta+ \frac{\partial f}{\partial v} \cos \theta$$
$$\frac{\partial^2f}{\partial y^2}=\frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial y} \right)=\frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial u} \sin \theta + \frac{\partial f}{\partial v} \cos \theta \right)= \frac{\partial}{\partial y} \frac{\partial f}{\partial u} \sin \theta + \frac{\partial}{\partial y} \frac{\partial f}{\partial v} \cos \theta$$
Now the problem is to compute $\frac{\partial}{\partial y} \frac{\partial f}{\partial u}$ and $\frac{\partial } {\partial y} \frac{\partial f}{\partial v}$. We can think of $\frac{\partial}{\partial y} \frac{\partial f}{\partial u}$ as $\frac{\partial}{\partial u} \frac{\partial f}{\partial y}$ and similarly for the latter. 
$$\frac{\partial}{\partial u} \frac{\partial f}{\partial y} = \frac{\partial}{\partial u} \left( \frac{\partial f}{\partial u} \sin \theta + \frac{\partial f}{\partial v} \cos \theta \right) = \frac{\partial^2 f}{\partial u^2} \sin \theta + \frac{\partial f}{\partial u \partial v} \cos \theta$$
$$\frac{\partial}{\partial v} \frac{\partial f}{\partial y} = \frac{\partial}{\partial v} \left( \frac{\partial f}{\partial u} \sin \theta + \frac{\partial f}{\partial v} \cos \theta \right) = \frac{\partial^2 f}{\partial u \partial v} \sin \theta + \frac{\partial^2 f}{\partial v^2} \cos \theta$$
Now plug this back in to $\frac{\partial^2f}{\partial y^2}$ giving, 
$$\frac{\partial^2f}{\partial y^2}= \left (\frac{\partial^2 f}{\partial u^2} \sin \theta + \frac{\partial f}{\partial u \partial v} \cos \theta \right) \sin \theta + \left (\frac{\partial^2 f}{\partial u \partial v} \sin \theta + \frac{\partial^2 f}{\partial v^2} \cos \theta \right) \cos \theta $$
Repeat the process for $\frac{\partial^2 f}{\partial x^2}$, add together, and you will arrive at the desired result. 
A: 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.
A: Consider a function $f: \mathbb{R}^2 \to \mathbb{R}$, say $(x,y) \mapsto f(x,y)$
Rotate $(x,y)$ by $\theta$ to get $(u,v)$ :
$$
\begin{bmatrix}
u \\
v
\end{bmatrix} =
\begin{bmatrix}
\cos \theta & -\sin\theta \\
\sin \theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
$$
In reverse, we get:
$$
\begin{bmatrix}
x \\
y
\end{bmatrix} =
\begin{bmatrix}
u\cos \theta +v\sin\theta \\
-u\sin \theta + v\cos\theta
\end{bmatrix}
$$
We can define a new function $\tilde{f}:\mathbb{R}^2\to \mathbb{R}$ that
$$
\tilde{f}(u,v):=f(x(u,v),y(u,v))
$$
If Laplace’s Operator is rotationally invariant, then $\nabla^2 \tilde{f}(u,v)= \nabla^2f(x,y)$.  This can be proved by the following:
$$\begin{aligned}
&\nabla^2 \tilde{f}(u,v)\\
=&\tilde{f}_{uu}+\tilde{f}_{vv} \\
=&{\partial \over \partial u} \left(f_x {\partial x \over \partial u} + f_y {\partial y \over \partial u} \right) +\tilde{f}_{vv} \\
=&f_{xx} \left({\partial x \over \partial u}\right)^2 + f_{xy} {\partial y \over \partial u}{\partial x \over \partial u}+f_x {\partial^2 x \over \partial u^2} \\
&f_{yy} \left({\partial y \over \partial u}\right)^2 + f_{yx} {\partial x \over \partial u}{\partial y \over \partial u}+f_y {\partial^2 y \over \partial u^2}\\
&+\tilde{f}_{vv}\\
=&f_{xx} \cos^2\theta -2\sin\theta\cos\theta f_{xy} +
\sin^2\theta f_{yy} +\tilde{f}_{vv} \\
=&f_{xx} \cos^2\theta -2\sin\theta\cos\theta f_{xy} +
\sin^2\theta f_{yy} \\
&f_{xx} \sin^2\theta +2\sin\theta\cos\theta f_{xy} +
\cos^2\theta f_{yy} \\
=&f_{xx}+f_{yy} \\
=&\nabla^2f(x,y)
\end{aligned}$$
