Gradient of a function defined on moving curves My question could be summarized as follows: how to differentiate in time a function defined on moving curves, in the following sense.
Let $h(t,\cdot): \mathbb R \to \mathbb R$ smooth and let $u(t,\cdot):  G_t \to \mathbb R$ smooth for each $t$, where $G_t = \text{Graph}(h(t,\cdot))$.
Thus $u$ is a family of functions defined on moving graphs. Now let $$w(t,x) = u(t, (x,h(t,x)) )$$
defined on the fixed $\mathbb R_+ \times \mathbb R$. Since the graphs move smoothly and the $u$ are smooth on each graph, it is tempting to give a meaning to $\partial_t w$.
Formally, one should have
$$\partial_t w(t,x) = \partial_t u(t,(x,h(t,x))) + (\nabla_{G_t} u) (t,(x,h(t,x))) \cdot \begin{pmatrix} 0 \\ \partial_t h(t,x) \end{pmatrix}$$
But of course, as the $u(t,\cdot)$ are not defined on the same sets, the chain rule and specifically the second term doesn't make sense directly here. Two possible ideas, by extending $u$ in neighborhoods of the graphs are:


*

*At each time extend $u$ by $\bar u$ constantly along the normals to $G_t$ in a small tubular neighbourhood. One can then compute this second term and it yields
$$\nabla_{G_t} u(t,(x,h(t,x))) \cdot P_{TG_t(x)} \begin{pmatrix} 0 \\ \partial_t h(t,x) \end{pmatrix}$$
where $\nabla_{G_t}$ is the tangential gradient on $G_t$ and $P_{TG_t(x)}$ is the projection onto the tangent space of the graph at $(t,x)$.

*Extend $u$ by $\tilde u$ constantly along vertical lines in a small neighborhood of $G_t$. But then, clearly the second term is $0$.


Is there any natural idea on how to compute that quantity ? The normal extension seems to make more sense, but I don't understand why.
Thank you.
 A: If I understood your question correctly, the problem is that your assumptions don't guarantee that $\partial_t w(t,x)$ even exists and so you can't expect some sort of a chain rule to give you the partial derivative.
In your situation, you have a family $u_t \colon G_t \rightarrow M$ of smooth functions defined on a time-dependent domain. In order for the statement "$u_t$ is smooth" to make sense, the domains $G_t$ should be smooth manifolds. Now, if you want to differentiate $u_t$ in the $t$-direction, you want two things:


*

*The manifolds $G_t$ should vary smoothly with $t$. To make sense of this statement, one usually takes $G_t$ to be submanifolds of some ambient space $X$ and which can be described as the images $\sigma_t \colon G \rightarrow X$ of a fixed manifold $G$ under a time-dependent parametrization $\sigma_t$. Then $G_t = \sigma_t(G)$ and the statement that $G_t$ vary smoothly (inside $X$) is translated to the fact that the the map $\sigma \colon I \times G \rightarrow X$ is a smooth map (where $\sigma(t,p) = \sigma_t(p)$).

*The functions $u_t$ should vary smoothly with $t$ as the domain $G_t$ varies smoothly. Since the domains $G_t$ are changing inside $X$, to make sure that $u_t$ varies smoothly in the $t$ direction, the easiest thing to do is to consider a smooth time-dependent map $u \colon I \times X \rightarrow M$ where each $u_t \colon X \rightarrow M$ is defined on the whole of the ambient space $X$. This makes sure that as $G_t$ changes inside $X$, $u_t|_{G_t}$ change smoothly along $G_t$.


Under the setting above, you are have a map $\Phi(t,p) = u(t, \sigma(t,p)) \colon I \times G \rightarrow M$ which is guaranteed to be smooth and so $\partial_t \Phi$ makes sense and can be calculated using the regular chain rule.

In your case, $G_t$ are smooth one-dimensional curves in $X = \mathbb{R}^2$ which you describe as graphs and so you have a smooth map $\sigma \colon I \times \mathbb{R} \rightarrow \mathbb{R}^2$ given by $\sigma(t,x) = (x, h(t,x))$. For a fixed $t$, the map $\sigma_t$ parametrizes a single curve in $\mathbb{R}^2$ and since $h$ is smooth, the map $\sigma$ is also smooth. This handles part $(1)$. However, if all you know is that each $u_t \colon G_t \rightarrow \mathbb{R}$ is a smooth function for a fixed $t$ then there's no reason that the composition $u(t,(x,h(t,x)))$ will be differentiable in the $t$-direction.
For example, let $\varphi \colon I \rightarrow \mathbb{R}$ be some arbitrary function and set $h(t,x) := t$ and $u_t(x, t) = x \cdot \varphi(t)$. In this case, the one-dimensional curves $G_t = \operatorname{Graph}(h_t)$ are just the lines $ \{ (x, t) \, | \, x \in \mathbb{R} \}$. The graphs $G_t$ vary smoothly with $t$ and each $u_t \colon G_t \rightarrow \mathbb{R}$ is definitely a smooth map but the composition $u(t,(x,h(t,x)) = x\varphi(t)$ won't be differentiable in the $t$-direction if $\varphi$ isn't differentiable.
