Proof that laplace's equation is rotationally invariant using chain rule Suppose $(x, y)$ and $(p, q)$ are coordinates in the plane related by rotation around a fixed point $(a, b)$, as follows:
$$\begin{bmatrix} p\\ q\end{bmatrix} = \begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix} \begin{bmatrix} x-a \\ y-b \end{bmatrix}$$
where $t$ is the rotation angle. Applying the chain rule show that $u(p(x,y), q(x,y)) $ satisfies $u_{xx}+u_{yy}=0$ iff $u_{pp}+u_{qq}= 0$. 
Where would I use chain rule in this problem? I am kind of confused because $u$, $p$, and $q$ are all functions of two variables.
 A: 1) Since $u(x,y) = u(p(x,y), q(x,y))$, you have (by the chain rule)
$$
\frac{\partial u}{\partial x} = \frac{\partial u}{\partial p} \frac{\partial p}{\partial x} + 
\frac{\partial u}{\partial q} \frac{\partial q}{\partial x}.
$$
The second derivative:
$$
\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(\frac{\partial u}{\partial x} \right) = \frac{\partial^2 u}{\partial p^2} \left(\frac{\partial p}{\partial x} \right)^2+
\frac{\partial u}{\partial p} \frac{\partial^2 p}{\partial x^2}+
2 \frac{\partial^2 u}{\partial p \partial q} \frac{\partial p}{\partial x} \frac{\partial q}{\partial x} + 
\frac{\partial^2 u}{\partial q^2} \left(\frac{\partial q}{\partial x} \right)^2+
\frac{\partial u}{\partial q} \frac{\partial^2 q}{\partial x^2}.
$$
The similar equalities you can obtain for $u_y$ and $u_{yy}$.
2) Note that
$$
p = (x-a) \, \cos t - (y-b) \, \sin t,\\
q = (x-a) \, \sin t + (y-b) \, \cos t
$$
Therefore,
$$
\frac{\partial p}{\partial x} = \cos t, \quad \frac{\partial^2 p}{\partial x^2} = 0,\\
\frac{\partial q}{\partial x} = \sin t, \quad \frac{\partial^2 q}{\partial x^2} = 0.
$$
Substituting these equalities (and the corresponding for $p_y$, $q_y$) to the expression for $u_{xx}$ (and $u_{yy}$), and summing them, you will finally get what you want.
