[proofplan] Subtract the two weak formulations and study the difference $w := u_1 - u_2$. The difference satisfies a homogeneous variational identity, and testing that identity with $w$ gives vanishing Dirichlet energy. Since $w \in H_0^1(U)$, Poincare's inequality converts the vanishing gradient into vanishing $L^2$ norm, so the full $H_0^1$ norm of $w$ is zero. [/proofplan] [step:Subtract the two weak formulations to obtain a homogeneous identity] Define $w \in H_0^1(U)$ by $w := u_1 - u_2$. This is valid because $H_0^1(U)$ is a [vector space](/page/Vector%20Space) and $u_1,u_2 \in H_0^1(U)$. Let $v \in H_0^1(U)$ be arbitrary. Since both $u_1$ and $u_2$ solve the weak problem with the same datum $f$, we have \begin{align*} \int_U \nabla u_1 \cdot \nabla v \, d\mathcal{L}^n(x) = \int_U f v \, d\mathcal{L}^n(x) \end{align*} and \begin{align*} \int_U \nabla u_2 \cdot \nabla v \, d\mathcal{L}^n(x) = \int_U f v \, d\mathcal{L}^n(x). \end{align*} Subtracting the second identity from the first and using linearity of the weak gradient gives \begin{align*} \int_U \nabla w \cdot \nabla v \, d\mathcal{L}^n(x) = 0. \end{align*} Thus $w$ satisfies the homogeneous weak equation against every [test function](/page/Test%20Function) $v \in H_0^1(U)$. [/step] [step:Test the homogeneous identity with the difference] Since $w \in H_0^1(U)$, we may choose $v := w$ in the homogeneous identity. This gives \begin{align*} \int_U |\nabla w|^2 \, d\mathcal{L}^n(x) = 0. \end{align*} The integrand $|\nabla w|^2$ is nonnegative and belongs to $L^1(U)$ because $w \in H_0^1(U)$. Hence $|\nabla w|^2 = 0$ $\mathcal{L}^n$-a.e. on $U$, and therefore \begin{align*} \|\nabla w\|_{L^2(U)} = 0. \end{align*} [guided] The homogeneous variational identity says that the Dirichlet energy pairing of $w$ with every $H_0^1(U)$ test function vanishes: \begin{align*} \int_U \nabla w \cdot \nabla v \, d\mathcal{L}^n(x) = 0 \end{align*} for every $v \in H_0^1(U)$. The most informative test function is the unknown itself. This is allowed because $w := u_1 - u_2$ lies in $H_0^1(U)$, since $H_0^1(U)$ is closed under subtraction. Taking $v := w$ gives \begin{align*} \int_U \nabla w \cdot \nabla w \, d\mathcal{L}^n(x) = 0. \end{align*} By the Euclidean [inner product](/page/Inner%20Product) identity $\nabla w \cdot \nabla w = |\nabla w|^2$, this becomes \begin{align*} \int_U |\nabla w|^2 \, d\mathcal{L}^n(x) = 0. \end{align*} Because $w \in H_0^1(U)$, each weak partial derivative $\partial_{x_i} w$ belongs to $L^2(U)$, so $|\nabla w|^2$ belongs to $L^1(U)$. Since the integrand is nonnegative, the vanishing of its [Lebesgue integral](/page/Lebesgue%20Integral) implies $|\nabla w|^2 = 0$ $\mathcal{L}^n$-a.e. on $U$. Equivalently, \begin{align*} \|\nabla w\|_{L^2(U)} = 0. \end{align*} [/guided] [/step] [step:Use the zero boundary condition to force the difference to vanish] By the Poincare inequality for $H_0^1$ functions on bounded open sets (citing a result not yet in the wiki: Poincare inequality for $H_0^1$ on bounded open sets), there exists a constant $C_U > 0$, depending only on $U$, such that every $\varphi \in H_0^1(U)$ satisfies \begin{align*} \|\varphi\|_{L^2(U)} \leq C_U \|\nabla \varphi\|_{L^2(U)}. \end{align*} Applying this to $\varphi := w$ gives \begin{align*} \|w\|_{L^2(U)} \leq C_U \|\nabla w\|_{L^2(U)} = 0. \end{align*} Hence $\|w\|_{L^2(U)} = 0$. The $H_0^1(U)$ norm of $w$ is therefore zero: \begin{align*} \|w\|_{H_0^1(U)}^2 = \|w\|_{L^2(U)}^2 + \|\nabla w\|_{L^2(U)}^2 = 0. \end{align*} Thus $w = 0$ in $H_0^1(U)$. Since $w = u_1 - u_2$, we conclude that $u_1 = u_2$ in $H_0^1(U)$. [/step]