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

Question

Given a planar parametric curve $C(t) = \big( x(t), y(t) \big)$ and a scalar function $u(t)$, both with $t \in [0,1]$, we can associate each point on the curve with a scalar value — for some point $P = C(t_P)$, that value would simply be $u(t_P)$. Then, we can express the tangential derivative of $u$ along $C$ as

$$\frac{\partial u}{\partial\tau} = \lim_{\Delta t \to 0} \frac{u(t + \Delta t) - u(t)}{\| C(t + \Delta t) - C(t) \|} = \lim_{\Delta t \to 0} \frac{u(t + \Delta t) - u(t)}{\| \Delta t C'(t) \|} = \frac{1}{\| C'(t) \|} \frac{du}{dt},$$

where $C'(t)$ is the tangent vector to the curve. Note that $\tau(t) = \frac{C'(t)}{\| C'(t) \|}$, in other words, $\| \tau(t) \| = 1$. Likewise, for the second-order tangential derivative we have (updated thanks to Ted Shifrin's comments below)

$$\frac{\partial^2u}{\partial\tau^2} = \frac{1}{\|C'(t)\|} \frac{d}{dt} \left( \frac{1}{\|C'(t)\|} \right) \frac{du}{dt} + \frac{1}{\| C'(t) \|^2} \frac{d^2u}{dt^2}.$$

Now, imagine that we are also given a function $q(t) = \frac{\partial u}{\partial n}$, that is, the normal derivative of $u$ along $C$, with $n(t)$ a $90^\circ$ counter-clockwise rotation of $\tau(t)$. Note that from our computed $\frac{\partial u}{\partial \tau}$ and the given $\frac{\partial u}{\partial n}$, we could compute the gradient $\big( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y} \big)^T$.

My question is whether we can express the 2D Laplacian using this information. In other words, do we have sufficient information to derive the second-order normal derivative $\frac{\partial^2u}{\partial n^2}$ (e.g. using the curvature $\kappa(t)$ of the curve), such that we can express

$$\Delta u = \frac{\partial^2u}{\partial\tau^2} + \frac{\partial^2u}{\partial n^2} = \frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2}$$

Answer 1

There seem to be some slight imprecisions of definition here (in particular I would argue that $\frac{\partial^2 u}{\partial n^2}$ is not well defied), but it's quite clear that what you're asking for is impossible, as the following counterexample demonstrates:

Let $C(t)=(t,0)$ be a parameterization of the $x$-axis, and define $u_1(x,y)=0$ and $u_2(x,y)=y^2$. One can see that $u_1\circ C=u_2\circ C$ and $\frac{\partial u_1}{\partial n}\circ C=\frac{\partial u_2}{\partial n}\circ C$, so all of the "given" information is the same between the two functions. Nonetheless, their Laplacians do not agree along $C$, i.e. $\Delta u_1\circ C\neq\Delta u_2\circ C$. This means that the Laplacian along $C$ is not uniquely determined by the "given" information.

Answer 2

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.