# Definition of the weak derivative involving the mean curvature

Definition 2.11 of the weak derivative:

A function $$f \in L^1(\Gamma)$$ has the weak derivative $$v_i=D_if \in L^1(\Gamma)$$, $$i \in \{1,...,n+1\}$$, if for every function $$\phi \in C^1_0(\Gamma)$$ we have the relation $$\int_\Gamma f D_i\phi \,dA =- \int_\Gamma\phi v_i \,dA+ \int_\Gamma f\phi Hv_id\,A$$ where $$\Gamma$$ is a hypersurface in $$\mathbb R^{n+1}$$ and $$H$$ is its mean curvature.

Question:

I am familiar with the "standard" definition of the weak derivative (https://en.wikipedia.org/wiki/Weak_derivative). How are these two different defintions to be reconciled and what role does the mean curvature get into the equation? Some explanation/interpretation would be much appreciated.

• it seems like in the pdf you link the last surface integral doesn't have a $v_i$ but a $\nu_i$, where $\nu$ seems to be the surface normal. – 0x539 Jan 14 at 19:04
• Are you happy with Theorem 2.10? The paper says "The formula for integration by parts on Γ leads to the notion of a weak derivative" i.e. Theorem 2.10 justifies Deﬁnition 2.11 - the derivative of a $C^1$ function is the weak derivative. A weak derivative is just something that makes integration by parts work. – Dap Jan 14 at 21:22
• @Dap So what is the reason that it differs from the conventional definition for a weak derivative? – Tesla Feb 15 at 14:54
• @Tesla: because it's using partial derivatives in the ambient space, which is a different setting to just using derivatives in $\mathbb R^n$ to define weak derivatives on functions on $\mathbb R^n.$ So for the same reason that Theorem 2.10 is different from conventional Stokes theorem/integration by parts. – Dap Feb 15 at 16:54
• @Dap It's been a while, but I picked it up again. So how does one exactly get the partial derivatives in the ambient space? – Tesla May 14 at 13:42