Boundary version of the fundamental lemma of calculus of variations

Question

I was looking for a detailed version with proof of the following statement: $$\int_{\partial D} (f\varphi +X \cdot \operatorname{grad}\varphi ) \,\,\mathrm d a =0 ~~~~\forall \varphi:D \to \mathbb R ~~~~~\Longrightarrow~~~~~ f=0,~X=0 ~~\text{on}~ \partial D .$$ Here $D$ is a three-dimensional domain. It is some sort of boundary version of the fundamental lemma of calculus of variations with the derivatives of the test function. A simple reference would be ok. Thanks!

Answer 1

$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\n}{\mathbf{n}}$ $\newcommand{\pl}{\partial}$ $\DeclareMathOperator{\dv}{div}$ $\newcommand{\vp}{\varphi}$

For simplicity, let me assume that $D \subseteq \R^n$ is a bounded smooth domain, so that $\pl D$ is a closed submanifold. Let me also assume that $f \colon \pl D \to \R$ and $X \colon \pl D \to \R^n$ are smooth. Our assumption is that $$ \int_{\pl D} f \vp + X \cdot \nabla \vp = 0 \quad \text{for all smooth } \vp, $$ where $\vp$ is defined on all of $\overline{D}$ (or even $\R^n$) and $\nabla \vp$ is the full gradient.

Let us denote by $\n$ the unit outer normal to $\pl D$ and decompose $X$ and $\nabla \vp$ into their tangent and normal parts: $$ X = X^\top + X^\perp \cdot \n, \quad \nabla \vp = \nabla_{\pl D} \vp + \pl_\n \vp \cdot \n, $$ so that $X \cdot \nabla \vp$ becomes $$ X \cdot \nabla \vp = X^\top \cdot \nabla_{\pl D} \vp + X^\perp \cdot \pl_\n \vp. $$ The first part can be integrated by parts over $\pl D$, yielding $$ \int_{\pl D} X^\top \cdot \nabla_{\pl D} \vp = - \int_{\pl D} \dv_{\pl D} X^\top \cdot \vp. $$ Our main assumption then translates to the statement that for each smooth $\vp$ we have $$ 0 = \int_{\pl D} f \vp + X \cdot \nabla \vp = \int_{\pl D} (f - \dv_{\pl D} X^\top) \cdot \vp + \int_{\pl D} X^\perp \cdot \pl_\n \vp. $$ I claim that $\vp$ and $\pl_\n \vp$ can be chosen as arbitrary smooth functions on $\pl D$, so our starting assumption is equivalent to $$ f = \dv_{\pl D} X^\top, \quad X^\perp = 0 \quad \text{on } \pl D. $$ To justify the claim, let us choose arbitrary smooth functions $a,b \colon \pl D \to \R$. The formula $$ \vp(x+t\n) = a(x)+tb(x) $$ gives us a well-defined function on a tubular neighborhood of $\pl D$, which can be extended elsewhere, if needed. Of course, $\vp = a$ and $\pl_\n \vp = b$ on $\pl D$.

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What about curvature? The integration by parts formula $\int_{\pl D} X^\top \cdot \nabla_{\pl D} \vp = - \int_{\pl D} \dv_{\pl D} X^\top \cdot \vp$ is a consequence of the formula $\int_{\pl D} \dv_{\pl D} V = 0$ valid for tangent fields $V$ on $\pl D$. If in doubts, this can be further translated to an intrinsic formula (Gauss-Green, Stokes) for abstract manifolds, e.g. by translating fields and divergences into forms and differentials.

The mean curvature pops out when one applies $\dv_{\pl D} V$ (understood as $\operatorname{tr}(DV \circ (T_x \pl D)_\#)$) to a general (not tangent) vector field. The result is indeed $$ \int_{\pl D} \dv_{\pl D} V = - \int_{\pl D} V \cdot \mathbf{h}, $$ where $\mathbf{h}$ is the mean curvature vector. See the definition of first variation here (sorry for unnecessary generality).