Distributional derivatives on hypersurface?

Question

In a paper I was reading, the define a set $Q=(0,T)\times \Omega$, where $\Omega \subset \mathbb{R}^n$ is a bounded domain, and then they write $$\langle \frac{d}{dt}u - \Delta u, \varphi \rangle_{\mathcal D^*(Q), \mathcal D(Q)} = -\langle u, \frac{d}{dt}u \rangle - \langle u, \Delta \varphi\rangle$$ where $\varphi \in C_c^\infty(Q)$ and the duality pairing on the LHS here is between $\mathcal{D}^*(Q)$ and $\mathcal{D}(Q)$. I guess this is just using the definition of the distributional derivatives $D_1, ... ,D_n$ and $D_t$.

Suppose now that $\Gamma$ is a hypersurface and set $S=(0,T)\times \Gamma$. Does the following make sense in some way: $$\langle \frac{d}{dt}u - \Delta_\Gamma u, \varphi \rangle_{\mathcal D^*(S), \mathcal D(S)} = -\langle u, \frac{d}{dt}u \rangle - \langle u, \Delta_\Gamma \varphi\rangle$$ where $\Delta_\Gamma$ is the Laplace-Beltrami operator not the standard Laplacian. How to make this rigourous and precise? I tried searching for distributions on manifolds but that means something else it seems. Any references to this topic are appreciated.

Related: https://math.stackexchange.com/questions/669379/about-a-function-space-on-bigcup-t-in-0-t-gammat-times-t?noredirect=1#comment1406810_669379

Answer 1

This is indeed the definition of distributional derivatives.

The book "Heat Kernel and Analysis on Manifolds" by Grigoryan (I can even access that part as a free preview on books.google.com) contains formula (7.30): locally integrable function $u(t,x)$ satisfies the heat equation on a manifold $N=(0,\infty)\times M$ in a distributional sense if and only if for every $\varphi \in {\cal D}' (N)$ it holds: $$\left( u, \partial_t \varphi + \triangle_\mu \varphi \right)=0.$$ This is transformed (using Fubini's theorem) in the form (7.31) $$\int_{\mathbb{R}_+} \left( u, \partial_t \varphi\right)_{L^2(M)}dt +\int_{\mathbb{R}_+} \left( u, \triangle_\mu \varphi\right)_{L^2(M)}dt=0.$$ The point in using $L^2$ inner product instead of a duality pairing lies in the fact that for every $t>0$ the solution is a smooth function, no matter what the initial data are. Frankly, I don't understand all the technical details, however, that book might be a worthy reference.

Answer 2

What you wrote does make perfect sense, as integration by parts on a sufficiently smooth manifold $S$ is permissible. Note that you have made a little mistake, and this is the correct version $$ \left\langle \frac{d}{dt}u - \Delta_\Gamma u, \varphi \right\rangle_{\mathscr D^*(S), \mathscr D(S)} = -\left\langle u, \frac{d}{dt}\varphi \right\rangle - \langle u, \Delta_\Gamma \varphi\rangle. $$ In particular $$ \langle \Delta_\Gamma u,v \rangle=-\langle \nabla_\Gamma u,\nabla_\Gamma v \rangle, $$ even for $v\in H_0^1(S)$ and $u\in H^{-1}(S)$. A good reference is:

Emmanuel Hebey, Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities.