[proofplan] We prove the two first variation formulae directly. For the area functional, differentiating the induced metric gives the variation of the Riemannian volume measure; when the variation field is normal, the resulting trace is exactly $-h(V,H_F)$. For the Dirichlet energy, differentiating $|du_t|^2$ gives a term involving the covariant derivative of the variation field, and compact-support [integration by parts](/theorems/2098) moves this derivative onto $du$, producing $-\tau_g(u)$. In both cases, the equivalence with the vanishing of the Euler-Lagrange field follows from the arbitrariness of compactly supported variation fields and the exponential map construction of variations. [/proofplan]

[step:Differentiate the induced volume measure along a normal variation] Let $\Phi:(-\varepsilon,\varepsilon)\times \Omega \to N$ be the smooth map defined by $\Phi(t,x)=F_t(x)$. This is a smooth normal variation of $F$ with compact support in $\Omega$: after shrinking $\varepsilon$ if necessary, there is a compact set $K\subset\Omega$ such that $F_t=F$ on $\Omega\setminus K$ for all $|t|<\varepsilon$. Its variation field is $V:\Omega \to F^*TN$, defined by $V(x)=\left.\frac{\partial F_t(x)}{\partial t}\right|_{t=0}$. For each $t$, write $g_t:=F_t^*h$ and let $\mu_t$ be the Riemannian measure associated to $g_t$. Fix a point $p\in \Omega$, and choose a local $g_F$-orthonormal frame $e_1,\dots,e_m$ on a neighbourhood of $p$. For smooth vector fields $X,Y$ on this neighbourhood, differentiating $g_t(X,Y)=h(dF_t(X),dF_t(Y))$ at $t=0$ gives the following identity. Here and below, the notation $\nabla^N_X$ along $F$ means $\nabla^N_{dF(X)}$, and along $F_t$ means $\nabla^N_{dF_t(X)}$. \begin{align*} \left.\frac{\partial}{\partial t}\right|_{t=0}g_t(X,Y)=h(\nabla^N_X V,dF(Y))+h(dF(X),\nabla^N_Y V). \end{align*} The [determinant differentiation formula](/page/Determinant) for a one-parameter family of positive-definite matrices, applied in a local frame to the Riemannian volume density, gives \begin{align*} \left.\frac{\partial}{\partial t}\right|_{t=0}d\mu_t=\frac{1}{2}\operatorname{tr}_{g_F}\left(\left.\frac{\partial g_t}{\partial t}\right|_{t=0}\right)d\mu_F=\sum_{i=1}^m h(\nabla^N_{e_i}V,dF(e_i))\,d\mu_F. \end{align*} Since the variation is normal, $h(V,dF(e_i))=0$ for each $i$. Differentiating this identity in the direction $e_i$ yields \begin{align*} 0 = e_i\bigl(h(V,dF(e_i))\bigr) = h(\nabla^N_{e_i}V,dF(e_i)) + h(V,\nabla^N_{e_i}dF(e_i)). \end{align*} Therefore \begin{align*} \sum_{i=1}^m h(\nabla^N_{e_i}V,dF(e_i)) = -\sum_{i=1}^m h(V,\nabla^N_{e_i}dF(e_i)). \end{align*} Only the normal component of $\nabla^N_{e_i}dF(e_i)$ contributes to the last [inner product](/page/Inner%20Product) because $V$ is normal. Hence, by the definition of the mean curvature vector, \begin{align*} \sum_{i=1}^m h(\nabla^N_{e_i}V,dF(e_i)) = -h(V,H_F). \end{align*} Because the variation itself has compact support, $F_t=F$ on $\Omega\setminus K$ for all sufficiently small $|t|$. Hence the local volume density and its difference quotient agree with the unvaried density outside $K$, so all variation of the integral is supported in the fixed compact set $K$. The maps $\Phi$, $F$, the metric $h$, and the chosen local frames are smooth; after covering $K$ by finitely many coordinate charts, the coordinate expressions for the first $t$-derivatives of the volume densities are bounded uniformly for $|t|$ sufficiently small. Therefore the ordinary parameter differentiation theorem in these finitely many charts justifies differentiating under the integral sign at $t=0$. Integrating over $\Omega$ with respect to $\mu_F$ gives the [first variation formula](/theorems/2728) for compactly supported normal variations: \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\int_\Omega 1\,d\mu_t(x) = -\int_\Omega h(V,H_F)\,d\mu_F(x). \end{align*} [/step]

[step:Conclude that area stationarity is equivalent to zero mean curvature] Assume first that $H_F=0$ on $\Omega$. For every compactly supported normal variation field $V:\Omega\to \nu_FM$, the formula from the preceding step gives \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\int_\Omega 1\,d\mu_{F_t}(x) = -\int_\Omega h(V,H_F)\,d\mu_F(x) = 0. \end{align*} Thus $F$ is stationary under all compactly supported normal variations. Conversely, assume $F$ is stationary under all compactly supported normal variations. Let $p\in \Omega$. Suppose, for contradiction, that $H_F(p)\ne 0$. By continuity of $H_F$, there is an open neighbourhood $U\subset \Omega$ of $p$ such that $|H_F|_h^2>0$ on $U$. Choose a smooth function \begin{align*} \psi:\Omega &\to [0,\infty) \end{align*} with compact support in $U$ and $\psi(p)>0$. Define the compactly supported normal section $V:\Omega \to \nu_FM$ by \begin{align*} V(x)=\psi(x)H_F(x). \end{align*} Because $\operatorname{supp}V$ is compactly contained in $\Omega$, the set $F(\operatorname{supp}V)\subset N$ is compact. The [exponential map](/page/Exponential%20Map) is defined on an open neighbourhood of the zero section in $TN$, so compactness gives a number $\delta>0$ such that $(F(x),tV(x))$ lies in this neighbourhood for every $x\in \operatorname{supp}V$ and every $|t|<\delta$. Define $F_t:\Omega \to N$ for $|t|<\delta$ by \begin{align*} F_t(x)=\exp^N_{F(x)}(tV(x)). \end{align*} This map equals $F$ outside $\operatorname{supp}V$. Choose an [open set](/page/Open%20Set) $K_0\subset \Omega$ with compact closure such that $\operatorname{supp}V\subset K_0$. On $\overline{K_0}$, the map $dF:T\overline{K_0}\to TN$ is injective on each tangent space, and injectivity of a [linear map](/page/Linear%20Map) is an open condition in finite dimensions. Since $dF_t\to dF$ uniformly on $\overline{K_0}$ as $t\to 0$, there is $0<\delta_1\leq \delta$ such that $dF_t$ remains injective on $\overline{K_0}$ for every $|t|<\delta_1$. Outside $\operatorname{supp}V$, $F_t=F$, so $dF_t=dF$ there. Therefore $F_t$ is an immersion on all of $\Omega$ for $|t|<\delta_1$, and $V$ is realised by an admissible normal variation. Stationarity and the first variation formula give \begin{align*} 0 = -\int_\Omega h(V,H_F)\,d\mu_F(x) = -\int_\Omega \psi |H_F|_h^2\,d\mu_F(x). \end{align*} The integrand $\psi |H_F|_h^2$ is non-negative and is positive on a nonempty open subset of $U$, so the integral is positive, a contradiction. Therefore $H_F(p)=0$. Since $p\in \Omega$ was arbitrary, $H_F=0$ on $\Omega$. [/step]

[step:Differentiate the Dirichlet energy and integrate by parts]Let $\Psi:(-\varepsilon,\varepsilon)\times \Omega \to N$ be the smooth map defined by $\Psi(t,x)=u_t(x)$. This is a smooth variation of $u$ with compact support in $\Omega$: after shrinking $\varepsilon$ if necessary, there is a compact set $K\subset\Omega$ such that $u_t=u$ on $\Omega\setminus K$ for all $|t|<\varepsilon$. Its variation field is $W:\Omega \to u^*TN$, defined by \begin{align*} W(x)=\left.\frac{\partial u_t(x)}{\partial t}\right|_{t=0}. \end{align*} Let $\nabla^{u^*TN}$ denote the pullback connection on $u^*TN$. Fix a point $p\in \Omega$, and choose a local $g$-orthonormal frame $e_1,\dots,e_m$ on a neighbourhood of $p$. Since the Levi-Civita connection on $N$ is torsion-free, covariant derivatives commute in the variation direction and domain directions: \begin{align*} \nabla^N_{\partial_t}du_t(e_i)\big|_{t=0} = \nabla^{u^*TN}_{e_i}W. \end{align*} Therefore \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\frac{1}{2}|du_t|_{g,h}^2 = \sum_{i=1}^m h(\nabla^{u^*TN}_{e_i}W,du(e_i)). \end{align*} Define the compactly supported vector field $Z$ on $\Omega$ by requiring \begin{align*} g(Z,X)=h(W,du(X)) \end{align*} for every smooth vector field $X$ on $\Omega$. The divergence of $Z$ is \begin{align*} \operatorname{div}_g Z = \sum_{i=1}^m e_i\bigl(h(W,du(e_i))\bigr) - \sum_{i=1}^m h(W,du(\nabla^g_{e_i}e_i)). \end{align*} Expanding the derivative using compatibility of $\nabla^N$ with $h$ gives \begin{align*} \operatorname{div}_g Z = \sum_{i=1}^m h(\nabla^{u^*TN}_{e_i}W,du(e_i)) + h\left(W,\sum_{i=1}^m \left(\nabla^{u^*TN}_{e_i}du(e_i)-du(\nabla^g_{e_i}e_i)\right)\right). \end{align*} The second term is $h(W,\tau_g(u))$. Hence \begin{align*} \sum_{i=1}^m h(\nabla^{u^*TN}_{e_i}W,du(e_i)) = \operatorname{div}_g Z-h(W,\tau_g(u)). \end{align*} Because $Z$ is compactly supported in $\Omega$, the [Divergence Theorem](/page/Divergence%20Theorem) applied on a relatively compact open set containing $\operatorname{supp}Z$ gives the compact-support [integration by parts](/theorems/210) identity \begin{align*} \int_\Omega \operatorname{div}_g Z\,d\mu_g(x)=0. \end{align*} The compact-support hypothesis on the variation gives $u_t=u$ on $\Omega\setminus K$ for all sufficiently small $|t|$, so the energy density and its difference quotient are unchanged outside the fixed compact set $K$. Smoothness of $\Psi$, $g$, and $h$ implies that, after covering $K$ by finitely many coordinate charts, the coordinate expressions for the first $t$-derivatives of the energy densities are bounded uniformly for $|t|$ sufficiently small. Therefore the ordinary parameter differentiation theorem in these finitely many charts justifies differentiating under the integral sign at $t=0$. Thus the first variation formula for Dirichlet energy is \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\frac{1}{2}\int_\Omega |du_t|_{g,h}^2\,d\mu_g(x) = -\int_\Omega h(W,\tau_g(u))\,d\mu_g(x). \end{align*}[/step]

[guided]We compute the energy variation by differentiating the integrand first, then using integration by parts to move the derivative off the variation field. The variation is the smooth map $\Psi:(-\varepsilon,\varepsilon)\times \Omega \to N$ defined by $\Psi(t,x)=u_t(x)$, with $u_0=u$, and there is a compact set $K\subset\Omega$ such that $u_t=u$ outside $K$ for all sufficiently small $|t|$. Its variation field is the section $W:\Omega \to u^*TN$ defined by $W(x)=\left.\frac{\partial u_t(x)}{\partial t}\right|_{t=0}$. The compact support of the variation is the hypothesis that will remove all boundary terms and localize differentiation under the integral sign. Fix a local $g$-orthonormal frame $e_1,\dots,e_m$. The energy density is \begin{align*} \frac{1}{2}|du_t|_{g,h}^2 = \frac{1}{2}\sum_{i=1}^m h(du_t(e_i),du_t(e_i)). \end{align*} Differentiating this expression with respect to $t$ and using compatibility of the Levi-Civita connection of $N$ with $h$ gives \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\frac{1}{2}|du_t|_{g,h}^2 = \sum_{i=1}^m h\left(\left.\nabla^N_{\partial_t}du_t(e_i)\right|_{t=0},du(e_i)\right). \end{align*} To justify the commutation identity, regard $\partial_t$ and the lifted vector field $e_i$ as vector fields on the product $(-\varepsilon,\varepsilon)\times \Omega$. They commute, so $[\partial_t,e_i]=0$. Applying the torsion-free identity for the Levi-Civita connection of $N$ to the map $\Psi$ gives \begin{align*} \nabla^N_{\partial_t}d\Psi(e_i)-\nabla^N_{e_i}d\Psi(\partial_t)=d\Psi([\partial_t,e_i])=0. \end{align*} Since $d\Psi(e_i)=du_t(e_i)$ and $d\Psi(\partial_t)=\partial_t u_t$, evaluating at $t=0$ gives \begin{align*} \left.\nabla^N_{\partial_t}du_t(e_i)\right|_{t=0}=\nabla^{u^*TN}_{e_i}W. \end{align*} Therefore \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\frac{1}{2}|du_t|_{g,h}^2 = \sum_{i=1}^m h(\nabla^{u^*TN}_{e_i}W,du(e_i)). \end{align*} Now we identify this expression as a divergence minus the tension pairing. Define the vector field $Z$ on $\Omega$ by \begin{align*} g(Z,X)=h(W,du(X)) \end{align*} for every smooth vector field $X$ on $\Omega$. This is the natural vector field because its divergence differentiates exactly the scalar pairing $h(W,du(e_i))$. Since $W$ has compact support in $\Omega$, the vector field $Z$ also has compact support in $\Omega$. Using the definition of divergence in the local orthonormal frame, \begin{align*} \operatorname{div}_g Z = \sum_{i=1}^m e_i\bigl(g(Z,e_i)\bigr) - \sum_{i=1}^m g(Z,\nabla^g_{e_i}e_i). \end{align*} Substituting $g(Z,X)=h(W,du(X))$ gives \begin{align*} \operatorname{div}_g Z = \sum_{i=1}^m e_i\bigl(h(W,du(e_i))\bigr) - \sum_{i=1}^m h(W,du(\nabla^g_{e_i}e_i)). \end{align*} Expanding the derivative of the $h$-pairing yields \begin{align*} \operatorname{div}_g Z = \sum_{i=1}^m h(\nabla^{u^*TN}_{e_i}W,du(e_i)) + \sum_{i=1}^m h(W,\nabla^{u^*TN}_{e_i}du(e_i)) - \sum_{i=1}^m h(W,du(\nabla^g_{e_i}e_i)). \end{align*} Combining the last two sums gives the trace of $\nabla du$: \begin{align*} \sum_{i=1}^m \left(\nabla^{u^*TN}_{e_i}du(e_i)-du(\nabla^g_{e_i}e_i)\right) = \tau_g(u). \end{align*} Hence \begin{align*} \operatorname{div}_g Z = \sum_{i=1}^m h(\nabla^{u^*TN}_{e_i}W,du(e_i)) + h(W,\tau_g(u)). \end{align*} Solving for the differentiated energy density gives \begin{align*} \sum_{i=1}^m h(\nabla^{u^*TN}_{e_i}W,du(e_i)) = \operatorname{div}_g Z-h(W,\tau_g(u)). \end{align*} Finally, the compact support of $Z$ in $\Omega$ eliminates the divergence term. Indeed, the [Divergence Theorem](/page/Divergence%20Theorem) applied on a relatively compact open set containing $\operatorname{supp}Z$ gives \begin{align*} \int_\Omega \operatorname{div}_g Z\,d\mu_g(x)=0. \end{align*} Because $u_t=u$ outside the fixed compact set $K$ for all sufficiently small $|t|$, the energy density and its difference quotient are unchanged outside $K$. Since $\Psi$, $g$, and $h$ are smooth, finitely many coordinate charts covering $K$ give coordinate expressions whose first $t$-derivatives are bounded uniformly for $|t|$ sufficiently small. Hence the ordinary parameter differentiation theorem permits differentiation under the integral sign at $t=0$. Therefore \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\frac{1}{2}\int_\Omega |du_t|_{g,h}^2\,d\mu_g(x)=-\int_\Omega h(W,\tau_g(u))\,d\mu_g(x). \end{align*} This is the Euler-Lagrange identity for the Dirichlet energy: the only possible obstruction to stationarity against compactly supported variations is the tension field.[/guided]

[step:Conclude that energy stationarity is equivalent to vanishing tension field]Assume first that $\tau_g(u)=0$ on $\Omega$. For every smooth compactly supported variation field $W:\Omega\to u^*TN$, the energy first variation formula gives \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\frac{1}{2}\int_\Omega |du_t|_{g,h}^2\,d\mu_g(x) = -\int_\Omega h(W,\tau_g(u))\,d\mu_g(x) = 0. \end{align*} Thus $u$ is stationary for Dirichlet energy under all compactly supported variations. Conversely, assume $u$ is stationary for Dirichlet energy under all compactly supported variations. Let $p\in \Omega$. Suppose, for contradiction, that $\tau_g(u)(p)\ne 0$. By continuity of $\tau_g(u)$, there is an open neighbourhood $U\subset \Omega$ of $p$ such that $|\tau_g(u)|_h^2>0$ on $U$. Choose a smooth function \begin{align*} \psi:\Omega &\to [0,\infty) \end{align*} with compact support in $U$ and $\psi(p)>0$. Define the compactly supported section $W:\Omega \to u^*TN$ by \begin{align*} W(x)=\psi(x)\tau_g(u)(x). \end{align*} Because $\operatorname{supp}W$ is compactly contained in $\Omega$, the set $u(\operatorname{supp}W)\subset N$ is compact. Since the [exponential map](/page/Exponential%20Map) is defined on an open neighbourhood of the zero section in $TN$, compactness gives a number $\delta>0$ such that $(u(x),tW(x))$ lies in this neighbourhood for every $x\in \operatorname{supp}W$ and every $|t|<\delta$. Define $u_t:\Omega \to N$ for $|t|<\delta$ by \begin{align*} u_t(x)=\exp^N_{u(x)}(tW(x)). \end{align*} This gives an admissible compactly supported variation of $u$. Stationarity and the first variation formula give \begin{align*} 0 = -\int_\Omega h(W,\tau_g(u))\,d\mu_g(x) = -\int_\Omega \psi |\tau_g(u)|_h^2\,d\mu_g(x). \end{align*} The integrand $\psi |\tau_g(u)|_h^2$ is non-negative and is positive on a nonempty open subset of $U$, so the integral is positive, a contradiction. Therefore $\tau_g(u)(p)=0$. Since $p\in \Omega$ was arbitrary, $\tau_g(u)=0$ on $\Omega$. Combining the area and energy conclusions proves both first variation principles.[/step]

[guided]We now prove the Dirichlet-energy equivalence as a testing statement. First assume that $\tau_g(u)=0$ on $\Omega$. Let $W:\Omega\to u^*TN$ be the variation field of any smooth variation $u_t:\Omega\to N$ whose variation field is compactly supported in the interior of $\Omega$. The first variation formula gives \begin{align*} \left.\frac{d}{dt}\right|_{t=0}\frac{1}{2}\int_\Omega |du_t|_{g,h}^2\,d\mu_g(x) = -\int_\Omega h(W,\tau_g(u))\,d\mu_g(x) = 0. \end{align*} Thus $u$ is stationary for Dirichlet energy under all compactly supported variations. Conversely, assume $u$ is stationary for Dirichlet energy under all compactly supported variations. Fix $p\in \Omega$ and suppose, for contradiction, that $\tau_g(u)(p)\ne 0$. Since $\tau_g(u):\Omega\to u^*TN$ is smooth, the function $x\mapsto |\tau_g(u)(x)|_h^2$ is continuous. Hence there is an open neighbourhood $U\subset \Omega$ of $p$ such that $|\tau_g(u)|_h^2>0$ on $U$. Choose a smooth function $\psi:\Omega\to [0,\infty)$ with compact support in $U$ and $\psi(p)>0$. Define the compactly supported section $W:\Omega\to u^*TN$ by \begin{align*} W(x)=\psi(x)\tau_g(u)(x). \end{align*} This is the right test field because pairing it with $\tau_g(u)$ produces the non-negative scalar function $\psi |\tau_g(u)|_h^2$. We must also check that this section is realized by an admissible variation. Because $\operatorname{supp}W$ is compactly contained in $\Omega$, the set $u(\operatorname{supp}W)\subset N$ is compact. The [exponential map](/page/Exponential%20Map) is defined on an open neighbourhood of the zero section in $TN$. Compactness therefore gives a number $\delta>0$ such that $(u(x),tW(x))$ lies in this neighbourhood for every $x\in \operatorname{supp}W$ and every $|t|<\delta$. Define $u_t:\Omega\to N$ for $|t|<\delta$ by \begin{align*} u_t(x)=\exp^N_{u(x)}(tW(x)). \end{align*} Outside $\operatorname{supp}W$ this map equals $u$, so the variation is compactly supported in the interior of $\Omega$ and has variation field $W$ at $t=0$. Stationarity and the first variation formula give \begin{align*} 0 = -\int_\Omega h(W,\tau_g(u))\,d\mu_g(x) = -\int_\Omega \psi |\tau_g(u)|_h^2\,d\mu_g(x). \end{align*} The integrand $\psi |\tau_g(u)|_h^2$ is non-negative and is positive on a nonempty open subset of $U$, because $\psi(p)>0$ and $|\tau_g(u)|_h^2>0$ on $U$. Therefore the integral is strictly positive, contradicting the displayed equality. Hence $\tau_g(u)(p)=0$. Since $p\in \Omega$ was arbitrary, $\tau_g(u)=0$ on $\Omega$.[/guided]