Partial integration on hypersurfaces - Why is there no "boundary integral"

Question

In https://homepages.warwick.ac.uk/staff/C.M.Elliott/DziEll13a.pdf on page $299$ the notion of weak derivatives on hypersurfaces is introduced via

For a $2$-dimensional hypersurface $\Sigma \subset \mathbb R^3$ with unit outer normal $\eta,$ mean curvature $\mathcal H$, we call $v$ the weak gradient or $v_i \in L^1(\Sigma)$ the weak partial derivatives of $u \in L^1(\Sigma)$ if for all $\varphi \in C_0^{\infty}(\Sigma)$ we have the relation

\begin{equation} \int_\Sigma u \frac{\partial \varphi}{\partial x_i} \, dA = -\int_\Sigma \varphi v_i\, dA + \int_\Sigma u \varphi \mathcal H \eta_i \, dA. \end{equation}

Now my question is: Why is the second integral not $$-\int_{\partial\Sigma} \varphi v_i\, dA?$$ Shouldn't there always be a boundary term when partially integrating?

Answer 1

Let's start with Theorem 2.10 on page 297

Assume that $\Gamma$ is a hypersurface in $\mathbb{R}^{n+1}$ with smooth boundary $\partial\Gamma$ and that $f\in C^1(\bar\Gamma)$. Then $$\int_\Gamma\nabla_\Gamma f\,\mathrm{d}A = \int_\Gamma fH\nu\,\mathrm{d}A+\int_{\partial\Gamma}f\nu\,\mathrm{d}A.$$

So we "weaken" this and let $f=u\varphi$, $\varphi\in C^\infty_0(\Gamma)$, $u\in L^1(\Gamma)$ and we have $$\require{color}\require{cancel} \int_\Gamma\nabla_\Gamma (u\varphi)\,\mathrm{d}A = \int_\Gamma u\varphi H\nu\,\mathrm{d}A{\color{red}+\cancelto{0}{\int_{\partial\Gamma}u\varphi\nu\,\mathrm{d}A}}. $$ so using Leibniz $\nabla_\Gamma(u\varphi)=(\nabla_\Gamma u)\varphi+u(\nabla_\Gamma\varphi)$ and rearrange, we have $$ \int_\Gamma uD_i\varphi\,\mathrm{d}A = -\int_\Gamma v_i\varphi\,\mathrm{d}A+\int_\Gamma u\varphi H\nu_i\,\mathrm{d}A$$ that you see in Definition 2.11.