[proofplan] We differentiate the Dirichlet energy of the time-slice map $u_t$ and express the result using the pullback connection on $u_t^*TN$. The key identity is that the covariant time derivative of $du_t$ equals the spatial covariant derivative of the variation field $\partial_t u$, because the Levi-Civita connection on $N$ is torsion-free. An [integration by parts](/theorems/2098) on the closed manifold $M$ then converts the pairing with $\nabla \partial_t u$ into the negative pairing with the tension field. Finally, the heat-flow equation $\partial_t u=\tau_g(u_t)$ turns the first variation identity into the energy dissipation identity. [/proofplan]

[step:Declare the time slices, variation field, and pullback connection] For each $t \in [0,T)$, define the smooth time-slice map $u_t: M \to N$ by $u_t(x)=u(x,t)$ for $x \in M$. Let $\mathfrak{X}(M)$ denote the space of smooth vector fields on $M$. Let $V_t \in \Gamma(u_t^*TN)$ denote the variation field \begin{align*} V_t(x) := \partial_t u(x,t) \in T_{u_t(x)}N. \end{align*} Let $\nabla^M$ be the [Levi-Civita connection](/page/Levi-Civita%20Connection) of $(M,g)$. Let $\nabla^N$ be the Levi-Civita connection of $(N,h)$, and let $\nabla^u$ denote the [pullback connection](/page/Pullback%20Connection) induced by $\nabla^N$ on the bundle $u_t^*TN \to M$. For a vector field $X \in \mathfrak{X}(M)$ and a section $W \in \Gamma(u_t^*TN)$, $\nabla_X^u W$ denotes the covariant derivative of $W$ in the spatial direction $X$. The differential $du_t$ is a smooth $u_t^*TN$-valued one-form on $M$, and its covariant derivative is the $u_t^*TN$-valued two-tensor defined by \begin{align*} (\nabla^u du_t)(X,Y) := \nabla_X^u(du_t(Y))-du_t(\nabla_X^M Y) \end{align*} for $X,Y \in \mathfrak{X}(M)$. The notation $\nabla_t^u$ denotes covariant differentiation along the curve $s \mapsto u(x,s)$ in $N$ at $s=t$, applied pointwise in $x \in M$. For a smooth vector field $Y \in \mathfrak{X}(M)$, $\operatorname{div}_g Y$ denotes its Riemannian divergence with respect to $g$, defined in a local $g$-orthonormal frame $(e_1,\dots,e_m)$ by \begin{align*} \operatorname{div}_g Y := \sum_{i=1}^m g(\nabla^M_{e_i}Y,e_i). \end{align*} The [tension field](/page/Tension%20Field) of $u_t$ is the section $\tau_g(u_t) \in \Gamma(u_t^*TN)$ defined by taking the $g$-trace of the second covariant derivative: \begin{align*} \tau_g(u_t) := \operatorname{tr}_g \nabla^u du_t. \end{align*} Equivalently, if $(e_1,\dots,e_m)$ is a local $g$-orthonormal frame on an [open set](/page/Open%20Set) in $M$, where $m=\dim M$, then \begin{align*} \tau_g(u_t)=\sum_{i=1}^m (\nabla^u du_t)(e_i,e_i). \end{align*} [/step]

[step:Differentiate the energy under the integral sign]Fix $t \in (0,T)$. Since $M$ is compact and $u$ is smooth on $M \times [0,T)$, the function \begin{align*} (x,s) \mapsto |du_s(x)|_{g^* \otimes u_s^*h}^2 \end{align*} is smooth on $M \times [0,T)$, so differentiating under the integral sign is valid. Using a local $g$-orthonormal frame $(e_1,\dots,e_m)$, the energy is locally represented by \begin{align*} E[u(s)] = \frac{1}{2}\int_M \sum_{i=1}^m h(du_s(e_i),du_s(e_i))\,d\operatorname{vol}_g. \end{align*} Metric compatibility of $\nabla^N$ gives \begin{align*} \frac{d}{dt}E[u(t)] = \int_M \sum_{i=1}^m h(\nabla_t^u du_t(e_i),du_t(e_i))\,d\operatorname{vol}_g. \end{align*}[/step]

[guided]We first isolate what is being differentiated. The map $u_t: M \to N$ changes with $t$, so $du_t(e_i)$ is a time-dependent section of the pullback bundle $u_t^*TN$. The energy contains the squared norm of this section, summed over a local $g$-orthonormal frame: \begin{align*} E[u(t)] = \frac{1}{2}\int_M \sum_{i=1}^m h(du_t(e_i),du_t(e_i))\,d\operatorname{vol}_g. \end{align*} Because $M$ is compact and $u$ is smooth, the integrand and its time derivative are continuous on compact subsets of $M \times (0,T)$. Thus differentiation under the integral sign is justified. The factor $\frac{1}{2}$ cancels the factor $2$ from differentiating the square. Since the Levi-Civita connection $\nabla^N$ is compatible with the metric $h$, we have \begin{align*} \partial_t h(du_t(e_i),du_t(e_i)) = 2h(\nabla_t^u du_t(e_i),du_t(e_i)). \end{align*} Therefore \begin{align*} \frac{d}{dt}E[u(t)] = \frac{1}{2}\int_M \sum_{i=1}^m \partial_t h(du_t(e_i),du_t(e_i))\,d\operatorname{vol}_g. \end{align*} Using the metric-compatibility identity above, this becomes \begin{align*} \frac{d}{dt}E[u(t)] = \int_M \sum_{i=1}^m h(\nabla_t^u du_t(e_i),du_t(e_i))\,d\operatorname{vol}_g. \end{align*} This rewrites the derivative of the energy as an $L^2$ pairing between $du_t$ and its covariant time derivative.[/guided]

[step:Commute the time derivative with the spatial derivative] For each local vector field $X \in \mathfrak{X}(M)$, the torsion-free property of the Levi-Civita connection on $N$ gives \begin{align*} \nabla_t^u du_t(X)=\nabla_X^u V_t. \end{align*} Applying this with $X=e_i$ in the differentiated energy identity yields \begin{align*} \frac{d}{dt}E[u(t)] = \int_M \sum_{i=1}^m h(\nabla_{e_i}^u V_t,du_t(e_i))\,d\operatorname{vol}_g. \end{align*} [/step]

[step:Integrate by parts on the closed domain]Define the smooth vector field $Y_t \in \mathfrak{X}(M)$ locally by \begin{align*} Y_t := \sum_{i=1}^m h(V_t,du_t(e_i))\,e_i. \end{align*} At a point where the local orthonormal frame is chosen geodesic, so that $\nabla^M_{e_i}e_j=0$ at that point, metric compatibility gives \begin{align*} \operatorname{div}_g Y_t = \sum_{i=1}^m h(\nabla_{e_i}^u V_t,du_t(e_i)) + h\left(V_t,\sum_{i=1}^m(\nabla_{e_i}^u du_t)(e_i)\right). \end{align*} Both sides are scalar functions, so this identity holds globally by tensoriality. Since $M$ is closed, the [divergence theorem](/page/Divergence%20Theorem) on a compact boundaryless Riemannian manifold gives \begin{align*} \int_M \operatorname{div}_g Y_t\,d\operatorname{vol}_g = 0. \end{align*} Therefore \begin{align*} \int_M \sum_{i=1}^m h(\nabla_{e_i}^u V_t,du_t(e_i))\,d\operatorname{vol}_g = -\int_M h(V_t,\tau_g(u_t))\,d\operatorname{vol}_g. \end{align*} Combining this with the previous step gives the first variation identity \begin{align*} \frac{d}{dt}E[u(t)] = -\int_M h(V_t,\tau_g(u_t))\,d\operatorname{vol}_g. \end{align*}[/step]

[guided]The differentiated energy contains $\nabla^u V_t$, but the heat-flow equation identifies $V_t$ with the tension field. Thus we want to transfer the spatial derivative from $V_t$ onto $du_t$. This is exactly an [integration by parts](/theorems/210) on the closed manifold $M$. Define the vector field $Y_t \in \mathfrak{X}(M)$ by \begin{align*} Y_t := \sum_{i=1}^m h(V_t,du_t(e_i))\,e_i \end{align*} in a local $g$-orthonormal frame $(e_1,\dots,e_m)$. This definition is independent of the chosen orthonormal frame because it is the vector field metrically dual to the one-form \begin{align*} X \mapsto h(V_t,du_t(X)). \end{align*} We compute its divergence. At a fixed point of $M$, choose the local orthonormal frame to be geodesic at that point, meaning $\nabla^M_{e_i}e_j=0$ at the point. Since divergence and the expression below are tensorial, proving the identity in such a frame proves it everywhere. Metric compatibility of $\nabla^N$ gives \begin{align*} \operatorname{div}_g Y_t = \sum_{i=1}^m e_i\bigl(h(V_t,du_t(e_i))\bigr). \end{align*} Metric compatibility of $\nabla^N$ expands this as \begin{align*} \operatorname{div}_g Y_t = \sum_{i=1}^m h(\nabla_{e_i}^u V_t,du_t(e_i)) + \sum_{i=1}^m h(V_t,\nabla_{e_i}^u(du_t(e_i))). \end{align*} By the definition of the second covariant derivative and of the tension field, this is \begin{align*} \operatorname{div}_g Y_t = \sum_{i=1}^m h(\nabla_{e_i}^u V_t,du_t(e_i)) + h\left(V_t,\sum_{i=1}^m(\nabla_{e_i}^u du_t)(e_i)\right). \end{align*} Therefore \begin{align*} \operatorname{div}_g Y_t = \sum_{i=1}^m h(\nabla_{e_i}^u V_t,du_t(e_i)) + h(V_t,\tau_g(u_t)). \end{align*} The hypothesis that $M$ is closed is used here. Since $M$ is compact and has no boundary, the [divergence theorem](/page/Divergence%20Theorem) gives no boundary contribution: \begin{align*} \int_M \operatorname{div}_g Y_t\,d\operatorname{vol}_g=0. \end{align*} Integrating the divergence identity and rearranging gives \begin{align*} \int_M \sum_{i=1}^m h(\nabla_{e_i}^u V_t,du_t(e_i))\,d\operatorname{vol}_g = -\int_M h(V_t,\tau_g(u_t))\,d\operatorname{vol}_g. \end{align*} Substituting this into the differentiated energy identity yields \begin{align*} \frac{d}{dt}E[u(t)] = -\int_M h(V_t,\tau_g(u_t))\,d\operatorname{vol}_g. \end{align*} This is the [first variation formula for the Dirichlet energy](/page/First%20Variation%20of%20Energy) along the particular variation $t \mapsto u_t$.[/guided]

[step:Substitute the harmonic map heat-flow equation] By definition, $V_t=\partial_t u(\cdot,t)$. Since $u$ solves harmonic map heat flow, we have \begin{align*} V_t=\partial_t u(\cdot,t)=\tau_g(u_t). \end{align*} Substituting this into the first variation identity gives \begin{align*} \frac{d}{dt}E[u(t)] = -\int_M h(\tau_g(u_t),\tau_g(u_t))\,d\operatorname{vol}_g. \end{align*} Using the definition of the pointwise norm induced by $h$, we obtain \begin{align*} \frac{d}{dt}E[u(t)] = -\int_M |\tau_g(u_t)|_h^2\,d\operatorname{vol}_g. \end{align*} Because $\partial_t u(\cdot,t)=\tau_g(u_t)$, the same identity is \begin{align*} \frac{d}{dt}E[u(t)] = -\int_M |\partial_t u(\cdot,t)|_h^2\,d\operatorname{vol}_g. \end{align*} Thus \begin{align*} \frac{d}{dt}E[u(t)] \le 0 \end{align*} for every $t \in (0,T)$, because the integrand $|\partial_t u(\cdot,t)|_h^2$ is nonnegative. Let $0<a<b<T$. Since $t \mapsto E[u(t)]$ is differentiable on $(0,T)$, the one-variable [mean value theorem](/theorems/186) applied on $[a,b]$ gives $E[u(b)] \le E[u(a)]$. Hence $E[u(t)]$ is nonincreasing on every compact subinterval of $(0,T)$. Since $u$ is smooth on $M \times [0,T)$ and $M$ is compact, $t \mapsto E[u(t)]$ is continuous at $t=0$; letting $a \downarrow 0$ gives $E[u(b)] \le E[u(0)]$ for every $b \in (0,T)$. Therefore $t \mapsto E[u(t)]$ is nonincreasing on $[0,T)$. [/step]