[proofplan] We compute the second derivative of area on a relatively compact smooth domain in $\Sigma$ containing the support of the variation, so that the [First Variation Formula](/page/First%20Variation%20Formula) applies without [boundary regularity](/theorems/99) ambiguity. Since $\Sigma$ is minimal and the variation is supported away from the auxiliary boundary, the first variation vanishes at $t=0$, so only the linearization of the mean curvature contributes. With the convention $A(X,Y)=g(\nabla^M_X\nu,Y)$ and $H=\operatorname{tr}_\Sigma A$, we use the sign convention $\frac{d}{dt}\mathcal H^m(\Sigma_t)=\int_{\Sigma_t}H_t f_t\,d\mathcal H^m_t$ and then compute directly that $\partial_t H_t|_{t=0}=-\Delta_\Sigma\phi-\left(|A|^2+\operatorname{Ric}_M(\nu,\nu)\right)\phi$. Finally, compact support permits [integration by parts](/theorems/2098) on $\Sigma$, converting the Laplacian term into the Dirichlet energy of $\phi$. [/proofplan] [step:Fix the sign conventions and write the first variation formula] Let $V: \Sigma \to TM|_\Sigma$ denote the initial variational vector field \begin{align*} V(p)=\left.\frac{\partial F}{\partial t}\right|_{t=0}(p)=\phi(p)\nu(p). \end{align*} By the compact support hypothesis in the statement, there is a compact set $K\subset\Sigma$ such that $F_t(p)=p$ for every $p\in\Sigma\setminus K$ and every sufficiently small $|t|$. Choose a relatively compact smooth domain $\Omega\subset\Sigma$ containing $K$ in its interior. After decreasing the time interval if necessary, $F_t(p)=p$ for all $p\in\Sigma\setminus\Omega$ and for all sufficiently small $|t|$. For each sufficiently small $t$, write $\Sigma_t:=F_t(\Sigma)$ and $\Omega_t:=F_t(\Omega)$. Choose $\nu_t:\Omega_t\to TM|_{\Omega_t}$ to be the unique smooth unit normal field satisfying $\nu_0=\nu|_\Omega$ and agreeing with the transported orientation for $|t|$ small. Let $H_t:\Omega_t\to\mathbb R$ denote the scalar mean curvature with respect to $\nu_t$, with sign fixed by the first variation convention \begin{align*} \left.\frac{d}{dt}\right|_{t=s}\mathcal H^m(\Omega_t) = \int_{\Omega_s} H_s f_s\,d\mathcal H^m_s, \end{align*} where $f_s:\Omega_s\to\mathbb R$ is the normal speed function defined by \begin{align*} f_s(q)=g\left(\left.\frac{\partial F}{\partial t}\right|_{(s,p)},\nu_s(q)\right), \qquad q=F_s(p). \end{align*} The [First Variation Formula](/page/First%20Variation%20Formula) is applied only to the smooth relatively compact domains $\Omega_t$; no regularity assumption on an arbitrary compact support set is needed. Because $\Sigma$ is minimal, $H_0=0$ on $\Omega\subset\Sigma$. Differentiating the [First Variation Formula](/page/First%20Variation%20Formula) at $s=0$ therefore gives \begin{align*} \left.\frac{d^2}{dt^2}\right|_{t=0}\mathcal H^m(\Omega_t) = \int_\Omega \left.\frac{\partial H_t}{\partial t}\right|_{t=0}\phi\,d\mathcal H^m. \end{align*} The derivative of the area element and the derivative of the speed do not contribute at $t=0$, because both are multiplied by $H_0=0$. [/step] [step:Compute the linearized mean curvature in the normal direction] We use the convention \begin{align*} A(X,Y)=g(\nabla^M_X\nu,Y) \end{align*} for tangent vector fields $X,Y\in\mathfrak X(\Sigma)$, so that $H=\operatorname{tr}_\Sigma A$. Under the normal variation with speed $\phi$, the scalar mean curvature satisfies \begin{align*} \left.\frac{\partial H_t}{\partial t}\right|_{t=0} = -\Delta_\Sigma \phi- \left(|A|^2+\operatorname{Ric}_M(\nu,\nu)\right)\phi, \end{align*} where $\Delta_\Sigma=\operatorname{div}_\Sigma\nabla^\Sigma$. We verify this identity in the guided computation below. [guided] We verify the variation formula at a fixed point $p\in\Sigma$. The calculation below uses only the initial velocity field $V=\phi\nu$ at $t=0$: all differentiated quantities are evaluated at $t=0$, and the possible $t$-dependence of the speed function away from $t=0$ does not enter the linearization of $H_t$. Choose a local tangent frame $e_1,\dots,e_m$ near $p$ that is orthonormal at $p$ and satisfies $\nabla^\Sigma_{e_i}e_j(p)=0$ for all $1\le i,j\le m$. Extend the frame along the variation by pushforward: $e_i(t)=dF_t(e_i)$. The pushed-forward frame is usually not orthonormal for $t\ne0$, so we must differentiate the inverse induced metric as well as the second fundamental form. This inverse-metric variation is not negligible; it is one of the terms that produces the $-|A|^2\phi$ contribution. Let $g_{ij}(t)=g(e_i(t),e_j(t))$ denote the induced metric matrix on $\Omega_t$ in the moving frame, and let $g^{ij}(t)$ denote the entries of its inverse matrix. Let \begin{align*} h_{ij}(t)=g(\nabla^M_{e_i(t)}\nu_t,e_j(t)) \end{align*} denote the components of the second fundamental form of $\Omega_t$ with respect to $\nu_t$. The scalar mean curvature is therefore \begin{align*} H_t=\sum_{i,j=1}^m g^{ij}(t)h_{ij}(t). \end{align*} At $t=0$ and $p$, $g^{ij}(0)=\delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta. The normal velocity is $V=\phi\nu$. Differentiating the orthogonality relation $g(\nu_t,e_i(t))=0$ at $t=0$ gives \begin{align*} g(\nabla^M_V\nu,e_i)+g(\nu,\nabla^M_{e_i}V)=0. \end{align*} Since $\nabla^M_{e_i}V=e_i(\phi)\nu+\phi\nabla^M_{e_i}\nu$ and $g(\nabla^M_{e_i}\nu,\nu)=0$, this becomes \begin{align*} g(\nabla^M_V\nu,e_i)=-e_i(\phi). \end{align*} Differentiating $g(\nu_t,\nu_t)=1$ also gives $g(\nabla^M_V\nu,\nu)=0$, so $\nabla^M_V\nu$ is tangent. Hence \begin{align*} \nabla^M_V\nu=-\nabla^\Sigma\phi. \end{align*} Now we differentiate the two factors in $H_t=\sum_{i,j=1}^m g^{ij}(t)h_{ij}(t)$. First, differentiating the induced metric gives \begin{align*} g'_{ij}(0)=g(\nabla^M_{e_i}V,e_j)+g(e_i,\nabla^M_{e_j}V). \end{align*} Using $V=\phi\nu$, $g(\nu,e_i)=0$, and $A_{ij}=g(\nabla^M_{e_i}\nu,e_j)$, this becomes \begin{align*} g'_{ij}(0)=2\phi A_{ij}. \end{align*} Since the derivative of an inverse matrix satisfies $(g^{ij})'(0)=-g'_{ij}(0)$ at a point where $g_{ij}(0)=\delta_{ij}$, we obtain \begin{align*} (g^{ij})'(0)=-2\phi A_{ij}. \end{align*} Consequently the inverse-metric part contributes \begin{align*} \sum_{i,j=1}^m (g^{ij})'(0)h_{ij}(0)=-2\phi |A|^2. \end{align*} This is the term that would be missed if the pushed-forward frame were treated as orthonormal away from $t=0$. Next we differentiate $h_{ij}(t)$. Work on the swept-out map $F:(-\varepsilon,\varepsilon)\times\Omega\to M$ and extend $V$ to the velocity field $\partial_tF$ along this image, while extending each $e_i$ by the pushed-forward field $e_i(t)=dF_t(e_i)$. These fields are $F$-related to the coordinate vector field $\partial_t$ and to the time-independent vector field $e_i$ on $\Omega$, respectively, so their Lie bracket along the variation satisfies $[V,e_i(t)]=0$. Using the curvature convention \begin{align*} R^M(X,Y)Z=\nabla^M_X\nabla^M_YZ-\nabla^M_Y\nabla^M_XZ-\nabla^M_{[X,Y]}Z, \end{align*} we get \begin{align*} h'_{ij}(0) = g(\nabla^M_{e_i}\nabla^M_V\nu,e_j) + g(R^M(V,e_i)\nu,e_j) + g(\nabla^M_{e_i}\nu,\nabla^M_{e_j}V). \end{align*} Tracing with $\delta_{ij}$ gives three terms. Since $\nabla^M_V\nu=-\nabla^\Sigma\phi$ is tangent and $\nabla^\Sigma_{e_i}e_i(p)=0$, the first traced term is \begin{align*} \sum_{i=1}^m g(\nabla^M_{e_i}\nabla^M_V\nu,e_i)=-\Delta_\Sigma\phi. \end{align*} For the curvature term, $V=\phi\nu$. With the Ricci convention $\operatorname{Ric}_M(\nu,\nu)=\sum_{i=1}^m g(R^M(e_i,\nu)\nu,e_i)$ and the skew-symmetry of $R^M$ in the first two slots, we obtain \begin{align*} \sum_{i=1}^m g(R^M(V,e_i)\nu,e_i)=-\phi\operatorname{Ric}_M(\nu,\nu). \end{align*} For the last traced term, the identity $\nabla^M_{e_i}V=e_i(\phi)\nu+\phi\nabla^M_{e_i}\nu$ and the tangency of $\nabla^M_{e_i}\nu$ give \begin{align*} \sum_{i=1}^m g(\nabla^M_{e_i}\nu,\nabla^M_{e_i}V)=\phi |A|^2. \end{align*} Therefore the differentiated second-fundamental-form part contributes \begin{align*} \sum_{i=1}^m h'_{ii}(0)=-\Delta_\Sigma\phi-\phi\operatorname{Ric}_M(\nu,\nu)+\phi |A|^2. \end{align*} Combining this with the inverse-metric contribution $-2\phi |A|^2$ yields \begin{align*} \left.\frac{\partial H_t}{\partial t}\right|_{t=0} = -\Delta_\Sigma \phi- \left(|A|^2+\operatorname{Ric}_M(\nu,\nu)\right)\phi. \end{align*} [/guided] [/step] [step:Insert the mean curvature variation into the second derivative of area] Substituting the linearized mean curvature formula into the differentiated [first variation formula](/theorems/2728) gives \begin{align*} \left.\frac{d^2}{dt^2}\right|_{t=0}\mathcal H^m(\Omega_t) = \int_\Omega \phi \left( -\Delta_\Sigma \phi- \left(|A|^2+\operatorname{Ric}_M(\nu,\nu)\right)\phi \right) \,d\mathcal H^m. \end{align*} Because $\phi$ is supported in $\Omega$, extending the integral from $\Omega$ to $\Sigma$ does not change its value, and hence \begin{align*} \left.\frac{d^2}{dt^2}\right|_{t=0}\mathcal H^m(\Omega_t) = -\int_\Sigma \phi\,\Delta_\Sigma\phi\,d\mathcal H^m - \int_\Sigma \left(|A|^2+\operatorname{Ric}_M(\nu,\nu)\right)\phi^2 \,d\mathcal H^m. \end{align*} All integrals are finite because $\phi$ is smooth with compact support and $A$ and $\operatorname{Ric}_M(\nu,\nu)$ are smooth on $\Sigma$. [/step] [step:Integrate by parts using compact support] Because $\phi\in C_c^\infty(\Sigma)$, define the compactly supported tangent vector field $X:\Sigma\to T\Sigma$ by \begin{align*} X(p)=\phi(p)\nabla^\Sigma\phi(p). \end{align*} The divergence identity \begin{align*} \operatorname{div}_\Sigma(\phi\nabla^\Sigma\phi) = |\nabla^\Sigma\phi|^2+\phi\Delta_\Sigma\phi \end{align*} holds on $\Sigma$. Choose a relatively compact smooth domain $\Omega_1\subset\Sigma$ containing $\operatorname{supp} X$ in its interior. Since $X=0$ in a neighbourhood of $\partial\Omega_1$, the boundary term in the [Divergence Theorem](/page/Divergence%20Theorem) on $\Omega_1$ vanishes, and extending by zero outside $\Omega_1$ gives \begin{align*} 0 = \int_\Sigma \operatorname{div}_\Sigma(\phi\nabla^\Sigma\phi)\,d\mathcal H^m = \int_\Sigma |\nabla^\Sigma\phi|^2\,d\mathcal H^m + \int_\Sigma \phi\Delta_\Sigma\phi\,d\mathcal H^m. \end{align*} Therefore \begin{align*} -\int_\Sigma \phi\Delta_\Sigma\phi\,d\mathcal H^m = \int_\Sigma |\nabla^\Sigma\phi|^2\,d\mathcal H^m. \end{align*} Substituting this into the previous expression yields \begin{align*} \left.\frac{d^2}{dt^2}\right|_{t=0}\mathcal H^m(\Omega_t) = \int_\Sigma \left( |\nabla^\Sigma \phi|^2 - \left(|A|^2+\operatorname{Ric}_M(\nu,\nu)\right)\phi^2 \right) \,d\mathcal H^m. \end{align*} The right-hand side is independent of the chosen relatively compact smooth domain $\Omega$ containing the compact set $K$ on which the variation is supported. Indeed, if $\Omega'$ is another such domain, then $F_t$ is the identity on both $\Omega'\setminus K$ and $\Omega\setminus K$ for sufficiently small $|t|$, so the differences \begin{align*} \mathcal H^m(F_t(\Omega))-\mathcal H^m(\Omega) \quad\text{and}\quad \mathcal H^m(F_t(\Omega'))-\mathcal H^m(\Omega') \end{align*} have the same value. Thus the displayed formula computes the second derivative of the area change of the compactly supported variation, independently of the auxiliary smooth domain used to localize the computation. This is the asserted [second variation formula](/theorems/2729). [/step]