[proofplan] We compute the Fréchet derivative of the energy functional by splitting it into the quadratic Dirichlet term and the potential term. The quadratic term is differentiated directly along arbitrary directions $v\in H^1_0(\Omega)$, while the derivative of the potential term is supplied by the stated Fréchet differentiability hypothesis for the potential functional $P$. Once the derivative formula for $I'[u](v)$ is obtained, the equivalence follows from the definition of a critical point as vanishing of the continuous linear functional $I'[u]$ on every direction. [/proofplan] [step:Differentiate the quadratic Dirichlet term along an arbitrary variation] Define the Dirichlet energy functional $Q: H^1_0(\Omega) \to \mathbb{R}$ by \begin{align*} Q[w]=\frac{1}{2}\int_\Omega |\nabla w(x)|^2\,d\mathcal{L}^n(x) \end{align*} for $w\in H^1_0(\Omega)$. Let $L^2(\Omega;\mathbb{R}^n)$ denote the [Hilbert space](/page/Hilbert%20Space) of $\mathcal{L}^n$-square-integrable vector fields $F:\Omega\to\mathbb{R}^n$. Fix $u,v\in H^1_0(\Omega)$. For $t\in\mathbb{R}$, the variation $u+tv$ belongs to $H^1_0(\Omega)$ because $H^1_0(\Omega)$ is a [vector space](/page/Vector%20Space). Using linearity of the weak gradient, \begin{align*} \nabla(u+tv)=\nabla u+t\nabla v \end{align*} in $L^2(\Omega;\mathbb{R}^n)$. Since $\nabla u,\nabla v\in L^2(\Omega;\mathbb{R}^n)$, the functions $|\nabla u|^2$ and $|\nabla v|^2$ are in $L^1(\Omega)$, and the [Cauchy-Schwarz inequality](/theorems/432) gives $\nabla u\cdot\nabla v\in L^1(\Omega)$. Hence \begin{align*} Q[u+tv]=\frac{1}{2}\int_\Omega |\nabla u(x)+t\nabla v(x)|^2\,d\mathcal{L}^n(x). \end{align*} Expanding the Euclidean square gives \begin{align*} Q[u+tv]=\frac{1}{2}\int_\Omega |\nabla u(x)|^2\,d\mathcal{L}^n(x)+t\int_\Omega \nabla u(x)\cdot\nabla v(x)\,d\mathcal{L}^n(x)+\frac{t^2}{2}\int_\Omega |\nabla v(x)|^2\,d\mathcal{L}^n(x). \end{align*} To verify that this directional derivative is the Fréchet derivative, let $h\in H^1_0(\Omega)$. Expanding as above with $h$ in place of $tv$ gives \begin{align*} Q[u+h]-Q[u]-\int_\Omega \nabla u(x)\cdot\nabla h(x)\,d\mathcal{L}^n(x)=\frac{1}{2}\int_\Omega |\nabla h(x)|^2\,d\mathcal{L}^n(x). \end{align*} Since $\|\nabla h\|_{L^2(\Omega)}\leq \|h\|_{H^1_0(\Omega)}$, the remainder is bounded by $\frac{1}{2}\|h\|_{H^1_0(\Omega)}^2$, hence is $o(\|h\|_{H^1_0(\Omega)})$ as $h\to 0$ in $H^1_0(\Omega)$. Therefore the Fréchet derivative at $u$ in the direction $v$ is \begin{align*} Q'[u](v)=\int_\Omega \nabla u(x)\cdot\nabla v(x)\,d\mathcal{L}^n(x). \end{align*} [guided] We isolate the part of the energy whose derivative can be computed by algebra. Define $Q: H^1_0(\Omega) \to \mathbb{R}$ by \begin{align*} Q[w]=\frac{1}{2}\int_\Omega |\nabla w(x)|^2\,d\mathcal{L}^n(x) \end{align*} for $w\in H^1_0(\Omega)$. Let $L^2(\Omega;\mathbb{R}^n)$ denote the Hilbert space of $\mathcal{L}^n$-square-integrable vector fields $F:\Omega\to\mathbb{R}^n$. The goal is to find the derivative of $Q$ at $u$ in an arbitrary direction $v\in H^1_0(\Omega)$. Since $H^1_0(\Omega)$ is a vector space, $u+tv\in H^1_0(\Omega)$ for every $t\in\mathbb{R}$. Since the weak gradient is linear, \begin{align*} \nabla(u+tv)=\nabla u+t\nabla v \end{align*} as elements of $L^2(\Omega;\mathbb{R}^n)$. Because $\nabla u,\nabla v\in L^2(\Omega;\mathbb{R}^n)$, the functions $|\nabla u|^2$ and $|\nabla v|^2$ are integrable with respect to $\mathcal{L}^n$ on $\Omega$. Also, Cauchy-Schwarz gives \begin{align*} \int_\Omega |\nabla u(x)\cdot\nabla v(x)|\,d\mathcal{L}^n(x)\leq \|\nabla u\|_{L^2(\Omega)}\|\nabla v\|_{L^2(\Omega)}<\infty, \end{align*} so the cross term is integrable. Now compute $Q[u+tv]$: \begin{align*} Q[u+tv]=\frac{1}{2}\int_\Omega |\nabla u(x)+t\nabla v(x)|^2\,d\mathcal{L}^n(x). \end{align*} For each $x\in\Omega$ at which the weak gradients are represented, the Euclidean identity $|a+tb|^2=|a|^2+2t\,a\cdot b+t^2|b|^2$ gives \begin{align*} |\nabla u(x)+t\nabla v(x)|^2=|\nabla u(x)|^2+2t\nabla u(x)\cdot\nabla v(x)+t^2|\nabla v(x)|^2. \end{align*} Integrating this identity with respect to $\mathcal{L}^n$ on $\Omega$ yields \begin{align*} Q[u+tv]=\frac{1}{2}\int_\Omega |\nabla u(x)|^2\,d\mathcal{L}^n(x)+t\int_\Omega \nabla u(x)\cdot\nabla v(x)\,d\mathcal{L}^n(x)+\frac{t^2}{2}\int_\Omega |\nabla v(x)|^2\,d\mathcal{L}^n(x). \end{align*} The coefficient of $t$ is the first variation along the line $u+tv$, while the remaining $t^2$ term vanishes after division by $t$ and passage to the limit $t\to 0$. To check that this line computation is genuinely the Fréchet derivative, take an arbitrary increment $h\in H^1_0(\Omega)$. The same expansion gives \begin{align*} Q[u+h]-Q[u]-\int_\Omega \nabla u(x)\cdot\nabla h(x)\,d\mathcal{L}^n(x)=\frac{1}{2}\int_\Omega |\nabla h(x)|^2\,d\mathcal{L}^n(x). \end{align*} The right-hand side is bounded by $\frac{1}{2}\|h\|_{H^1_0(\Omega)}^2$, because $\|\nabla h\|_{L^2(\Omega)}\leq \|h\|_{H^1_0(\Omega)}$. After division by $\|h\|_{H^1_0(\Omega)}$, this bound tends to $0$ as $h\to 0$ in $H^1_0(\Omega)$. Therefore \begin{align*} Q'[u](v)=\int_\Omega \nabla u(x)\cdot\nabla v(x)\,d\mathcal{L}^n(x). \end{align*} [/guided] [/step] [step:Combine the Dirichlet and potential variations] Define the potential functional $P: H^1_0(\Omega) \to \mathbb{R}$ by \begin{align*} P[w]=\int_\Omega G(x,w(x))\,d\mathcal{L}^n(x) \end{align*} for $w\in H^1_0(\Omega)$. By the stated hypothesis on the potential term, $P$ is Fréchet differentiable and, for every $v\in H^1_0(\Omega)$, \begin{align*} P'[u](v)=\int_\Omega g(x,u(x))v(x)\,d\mathcal{L}^n(x). \end{align*} Since $I=Q-P$ and both $Q$ and $P$ are Fréchet differentiable at $u$, the linearity of the Fréchet derivative gives, for every $v\in H^1_0(\Omega)$, \begin{align*} I'[u](v)=Q'[u](v)-P'[u](v). \end{align*} Substituting the two derivative formulas, \begin{align*} I'[u](v)=\int_\Omega \nabla u(x)\cdot\nabla v(x)\,d\mathcal{L}^n(x)-\int_\Omega g(x,u(x))v(x)\,d\mathcal{L}^n(x). \end{align*} [/step] [step:Translate vanishing of the derivative into the weak Euler-Lagrange equation] By definition, $u$ is a critical point of $I$ if and only if $I'[u]=0$ as an element of $(H^1_0(\Omega))^*$. This is equivalent to \begin{align*} I'[u](v)=0 \end{align*} for every $v\in H^1_0(\Omega)$. Using the derivative formula from the previous step, this condition is equivalent to \begin{align*} \int_\Omega \nabla u(x)\cdot\nabla v(x)\,d\mathcal{L}^n(x)-\int_\Omega g(x,u(x))v(x)\,d\mathcal{L}^n(x)=0 \end{align*} for every $v\in H^1_0(\Omega)$. Rearranging the equality gives \begin{align*} \int_\Omega \nabla u(x)\cdot\nabla v(x)\,d\mathcal{L}^n(x)=\int_\Omega g(x,u(x))v(x)\,d\mathcal{L}^n(x) \end{align*} for every $v\in H^1_0(\Omega)$. This is exactly the asserted weak Euler-Lagrange equation, and the two conditions are equivalent. [/step]