Expressing the 2D Laplacian using information given on a parametric curve

The two coordinate systems $(x,y)$ and $(\tau,n)$ are both orthogonal, which is the only important thing. So there exists an (orthogonal) rotation transformation $$ \begin{cases} \tau = \cos(\alpha)\,x-\sin(\alpha)\,y \\ n = \sin(\alpha)\,x+\cos(\alpha)\,y \end{cases} $$ We want to know how the derivatives of the solution $\,u\,$ transform. $$ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial \tau}\frac{\partial \tau}{\partial x} + \frac{\partial u}{\partial n}\frac{\partial n}{\partial x} \\ \frac{\partial u}{\partial y} = \frac{\partial u}{\partial \tau}\frac{\partial \tau}{\partial y} + \frac{\partial u}{\partial n}\frac{\partial n}{\partial y} $$ It follows, with operator notation: $$ \frac{\partial}{\partial x} u = \left[ \cos(\alpha)\frac{\partial}{\partial \tau} +\sin(\alpha)\frac{\partial}{\partial n} \right] u\\ \frac{\partial}{\partial y} u = \left[ -\sin(\alpha)\frac{\partial}{\partial \tau} +\cos(\alpha)\frac{\partial}{\partial n} \right] u $$ So we have: $$ \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = \left(\frac{\partial}{\partial x}\right)^2 + \left(\frac{\partial}{\partial y}\right)^2 = \\ \left[\cos(\alpha)\frac{\partial}{\partial \tau} +\sin(\alpha)\frac{\partial}{\partial n} \right]^2 + \left[ -\sin(\alpha)\frac{\partial}{\partial \tau}+\cos(\alpha)\frac{\partial}{\partial n} \right]^2 = \\ \left[\cos^2(\alpha)\left(\frac{\partial}{\partial \tau}\right)^2 + 2\sin(\alpha)\cos(\alpha)\frac{\partial}{\partial \tau}\frac{\partial}{\partial n} +\sin^2(\alpha)\left(\frac{\partial}{\partial n}\right)^2\right] + \\ \left[\sin^2(\alpha)\left(\frac{\partial}{\partial \tau}\right)^2 -2\sin(\alpha)\cos(\alpha)\frac{\partial}{\partial \tau}\frac{\partial}{\partial n} +\cos^2(\alpha)\left(\frac{\partial}{\partial n}\right)^2\right] = \\ \left(\frac{\partial}{\partial \tau}\right)^2 + \left(\frac{\partial}{\partial n}\right)^2 = \frac{\partial^2}{\partial \tau^2} + \frac{\partial^2}{\partial n^2} $$ Therefore $\Delta u$ is the same in the (global) plane coordinates as well as in the (local) curve coordinates.