[proofplan] Fix an admissible direction $v \in H^1_0(U)$ and restrict the functional $I$ to the affine line $u+\varepsilon v$. Because the Dirichlet energy is quadratic in $\nabla u$ and the forcing term is linear in $u$, this one-variable function is a quadratic polynomial in $\varepsilon$. Its derivative at $\varepsilon=0$ is exactly the weak residual against $v$, so vanishing of all first variations is equivalent to the weak formulation of $-\Delta u=f$ with zero boundary condition encoded by $H^1_0(U)$. [/proofplan] [step:Verify that the energy and first variation terms are finite] Let $L^2(U)$ denote the space $L^2(U, \mathcal{B}(U), \mathcal{L}^n)$ of square-integrable real-valued functions on $U$, and let $L^1(U)$ denote the space $L^1(U, \mathcal{B}(U), \mathcal{L}^n)$ of integrable real-valued functions on $U$. Let $L^2(U;\mathbb{R}^n)$ denote the space of measurable maps $G:U \to \mathbb{R}^n$ whose components belong to $L^2(U)$. Let $H^1(U)$ denote the [Sobolev space](/page/Sobolev%20Space) of functions $w \in L^2(U)$ whose weak first partial derivatives $\partial_{x_i}w$ belong to $L^2(U)$ for $1 \leq i \leq n$, and let $H^1_0(U)$ denote the closure of $C_c^\infty(U)$ in the $H^1(U)$ norm, where $C_c^\infty(U)$ is the space of smooth compactly supported functions $\varphi:U \to \mathbb{R}$. Let $u \in H^1_0(U)$ and let $v \in H^1_0(U)$. By definition of $H^1_0(U) \subset H^1(U)$, the functions $u$, $v$, and $f$ belong to $L^2(U)$, and the weak gradients $\nabla u$ and $\nabla v$ belong to $L^2(U;\mathbb{R}^n)$. The energy term is finite because $|\nabla u|^2 \in L^1(U)$. The forcing term is finite since \begin{align*} |f(x)u(x)| \leq \frac{1}{2}|f(x)|^2 + \frac{1}{2}|u(x)|^2 \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$, and the right-hand side is integrable on $U$. Thus $I[u]$ is well-defined. The same argument applies to $u+\varepsilon v \in H^1_0(U)$ for every $\varepsilon \in \mathbb{R}$. Similarly, \begin{align*} |\nabla u(x)\cdot \nabla v(x)| \leq |\nabla u(x)|\,|\nabla v(x)| \leq \frac{1}{2}|\nabla u(x)|^2 + \frac{1}{2}|\nabla v(x)|^2 \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$, and \begin{align*} |f(x)v(x)| \leq \frac{1}{2}|f(x)|^2 + \frac{1}{2}|v(x)|^2. \end{align*} Hence both integrals appearing in the weak formulation are finite. [/step] [step:Compute the directional derivative along an arbitrary admissible direction] Fix $v \in H^1_0(U)$ and define the one-variable function \begin{align*} \Phi_v: \mathbb{R} \to \mathbb{R}, \quad \Phi_v(\varepsilon) := I[u+\varepsilon v]. \end{align*} Since $H^1_0(U)$ is a [vector space](/page/Vector%20Space), $u+\varepsilon v \in H^1_0(U)$ for every $\varepsilon \in \mathbb{R}$. Linearity of weak derivatives gives \begin{align*} \nabla(u+\varepsilon v)=\nabla u+\varepsilon \nabla v \end{align*} in $L^2(U;\mathbb{R}^n)$. Therefore \begin{align*} |\nabla(u+\varepsilon v)|^2 = |\nabla u|^2 + 2\varepsilon \nabla u\cdot \nabla v + \varepsilon^2 |\nabla v|^2 \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$, and \begin{align*} f(u+\varepsilon v)=fu+\varepsilon fv. \end{align*} Substituting these identities into the definition of $I$ gives \begin{align*} \Phi_v(\varepsilon)=I[u]+\varepsilon\left(\int_U \nabla u\cdot \nabla v\,d\mathcal{L}^n(x)-\int_U fv\,d\mathcal{L}^n(x)\right)+\frac{\varepsilon^2}{2}\int_U |\nabla v|^2\,d\mathcal{L}^n(x). \end{align*} Thus $\Phi_v$ is a quadratic polynomial in $\varepsilon$, so it is differentiable at $0$, and \begin{align*} \Phi_v'(0)=\int_U \nabla u\cdot \nabla v\,d\mathcal{L}^n(x)-\int_U fv\,d\mathcal{L}^n(x). \end{align*} [guided] Fix an admissible variation $v \in H^1_0(U)$. To test whether $u$ is critical, we do not vary $u$ in all possible ways at once; we restrict the functional to the affine line through $u$ in the direction $v$. Define \begin{align*} \Phi_v: \mathbb{R} \to \mathbb{R}, \quad \Phi_v(\varepsilon) := I[u+\varepsilon v]. \end{align*} The expression is meaningful because $H^1_0(U)$ is closed under addition and scalar multiplication, so $u+\varepsilon v \in H^1_0(U)$ for every $\varepsilon \in \mathbb{R}$. The [weak derivative](/page/Weak%20Derivative) is linear. Therefore the weak gradient of the varied function is \begin{align*} \nabla(u+\varepsilon v)=\nabla u+\varepsilon \nabla v \end{align*} in $L^2(U;\mathbb{R}^n)$. The point of this computation is that the energy term is quadratic in the gradient. Expanding the Euclidean square gives \begin{align*} |\nabla(u+\varepsilon v)|^2 = |\nabla u|^2 + 2\varepsilon \nabla u\cdot \nabla v + \varepsilon^2 |\nabla v|^2 \end{align*} for $\mathcal{L}^n$-a.e. $x \in U$. The forcing term is linear in the function, so \begin{align*} f(u+\varepsilon v)=fu+\varepsilon fv. \end{align*} Substituting both identities into the definition of $I$ yields \begin{align*} \Phi_v(\varepsilon)=I[u]+\varepsilon\left(\int_U \nabla u\cdot \nabla v\,d\mathcal{L}^n(x)-\int_U fv\,d\mathcal{L}^n(x)\right)+\frac{\varepsilon^2}{2}\int_U |\nabla v|^2\,d\mathcal{L}^n(x). \end{align*} This is a genuine quadratic polynomial in the real variable $\varepsilon$; all coefficients are finite by the previous step. Hence $\Phi_v$ is differentiable at $\varepsilon=0$, and differentiating the polynomial gives \begin{align*} \Phi_v'(0)=\int_U \nabla u\cdot \nabla v\,d\mathcal{L}^n(x)-\int_U fv\,d\mathcal{L}^n(x). \end{align*} This formula identifies the first variation of the energy with the weak residual of the equation $-\Delta u=f$ tested against $v$. [/guided] [/step] [step:Show that criticality implies the weak formulation] Assume that $u$ is a critical point of $I$ in the stated sense. Let $v \in H^1_0(U)$ be arbitrary. By the [first variation formula](/theorems/2728) just computed, \begin{align*} 0=\frac{d}{d\varepsilon}\Big|_{\varepsilon=0}I[u+\varepsilon v]=\int_U \nabla u\cdot \nabla v\,d\mathcal{L}^n(x)-\int_U fv\,d\mathcal{L}^n(x). \end{align*} Rearranging gives \begin{align*} \int_U \nabla u\cdot \nabla v\,d\mathcal{L}^n(x)=\int_U fv\,d\mathcal{L}^n(x). \end{align*} Since $v \in H^1_0(U)$ was arbitrary, this is exactly the homogeneous Dirichlet weak formulation of $-\Delta u=f$ in $U$. [/step] [step:Show that the weak formulation implies criticality] Conversely, assume that $u \in H^1_0(U)$ satisfies \begin{align*} \int_U \nabla u\cdot \nabla v\,d\mathcal{L}^n(x)=\int_U fv\,d\mathcal{L}^n(x) \end{align*} for every $v \in H^1_0(U)$. Fix such a $v$. The first variation formula gives \begin{align*} \frac{d}{d\varepsilon}\Big|_{\varepsilon=0}I[u+\varepsilon v]=\int_U \nabla u\cdot \nabla v\,d\mathcal{L}^n(x)-\int_U fv\,d\mathcal{L}^n(x)=0. \end{align*} Since this holds for every admissible direction $v \in H^1_0(U)$, $u$ is a critical point of $I$ in the stated sense. This proves the equivalence. [/step]