[proofplan] The proof tests the weak parabolic equation against the solution itself. Since $u$ has only weak time regularity, this testing is justified through the Lions-Magenes energy identity, which identifies the dual pairing of $\partial_t u$ with $u$ as the time derivative of the $L^2$ norm. Ellipticity gives the coercive gradient term, and the forcing term is controlled by $H^{-1},H^1_0$ duality followed by [Young's inequality](/theorems/244) with parameter $\theta$. Integrating the resulting differential inequality gives the estimate. [/proofplan] [step:Use the Lions-Magenes energy identity to test by the solution] By the Lions-Magenes energy identity for the Gelfand triple $H^1_0(U)\subset L^2(U)\subset H^{-1}(U)$, applied to $u\in L^2(0,T;H^1_0(U))$ with $\partial_t u\in L^2(0,T;H^{-1}(U))$, the representative of $u$ belongs to $C([0,T];L^2(U))$ and, for every $t\in[0,T]$, \begin{align*} \frac{1}{2}\|u(t)\|_{L^2(U)}^2-\frac{1}{2}\|u_0\|_{L^2(U)}^2=\int_0^t \partial_t u(\tau)(u(\tau))\,d\mathcal{L}^1(\tau). \end{align*} We use the weak formulation in its integrated form. Namely, for every map $w:(0,T)\to H^1_0(U)$ with $w\in L^2(0,T;H^1_0(U))$, \begin{align*} \int_0^t \partial_t u(\tau)(w(\tau))\,d\mathcal{L}^1(\tau)+\int_0^t\int_U A(x,\tau)\nabla u(x,\tau)\cdot\nabla w(x,\tau)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(\tau)=\int_0^t f(\tau)(w(\tau))\,d\mathcal{L}^1(\tau). \end{align*} This integrated identity follows from the pointwise-in-time weak formulation first for simple maps $w(\tau)=\sum_{j=1}^m \mathbb{1}_{E_j}(\tau)v_j$, where $E_j\subset(0,T)$ is Lebesgue measurable and $v_j\in H^1_0(U)$, and then for all of $L^2(0,T;H^1_0(U))$ by density of such simple maps and boundedness of the three terms. Since $u\in L^2(0,T;H^1_0(U))$, we may take $w=u$. This gives \begin{align*} \frac{1}{2}\|u(t)\|_{L^2(U)}^2-\frac{1}{2}\|u_0\|_{L^2(U)}^2+\int_0^t\int_U A(x,\tau)\nabla u(x,\tau)\cdot\nabla u(x,\tau)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(\tau)=\int_0^t f(\tau)(u(\tau))\,d\mathcal{L}^1(\tau). \end{align*} [guided] The natural [test function](/page/Test%20Function) is $u(\tau)$ itself, because this is the choice that produces both the time derivative of the energy and the coercive gradient term. The only point requiring care is that $u$ is not assumed classically differentiable in time. The Lions-Magenes energy identity for the Gelfand triple $H^1_0(U)\subset L^2(U)\subset H^{-1}(U)$ applies because $u\in L^2(0,T;H^1_0(U))$ and $\partial_t u\in L^2(0,T;H^{-1}(U))$. It gives a representative $u\in C([0,T];L^2(U))$ and, for every $t\in[0,T]$, \begin{align*} \frac{1}{2}\|u(t)\|_{L^2(U)}^2-\frac{1}{2}\|u_0\|_{L^2(U)}^2=\int_0^t \partial_t u(\tau)(u(\tau))\,d\mathcal{L}^1(\tau). \end{align*} Now we use the weak formulation in a way that permits a time-dependent test function. For every map $w:(0,T)\to H^1_0(U)$ with $w\in L^2(0,T;H^1_0(U))$, the integrated weak formulation states \begin{align*} \int_0^t \partial_t u(\tau)(w(\tau))\,d\mathcal{L}^1(\tau)+\int_0^t\int_U A(x,\tau)\nabla u(x,\tau)\cdot\nabla w(x,\tau)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(\tau)=\int_0^t f(\tau)(w(\tau))\,d\mathcal{L}^1(\tau). \end{align*} Why is this stronger-looking formulation available? Starting from the pointwise-in-time weak formulation, we first multiply by indicators of measurable time sets and fixed test functions $v_j\in H^1_0(U)$, then add finitely many such identities. This proves the integrated identity for simple maps $w(\tau)=\sum_{j=1}^m \mathbb{1}_{E_j}(\tau)v_j$. Such simple maps are dense in $L^2(0,T;H^1_0(U))$, and the three terms are continuous with respect to this norm: the time-derivative term by $\partial_t u\in L^2(0,T;H^{-1}(U))$, the elliptic term by boundedness of $A$, and the forcing term by $f\in L^2(0,T;H^{-1}(U))$. Therefore the identity holds for $w=u$, since $u\in L^2(0,T;H^1_0(U))$. Replacing the time-derivative term by the Lions-Magenes identity gives the energy balance \begin{align*} \frac{1}{2}\|u(t)\|_{L^2(U)}^2-\frac{1}{2}\|u_0\|_{L^2(U)}^2+\int_0^t\int_U A(x,\tau)\nabla u(x,\tau)\cdot\nabla u(x,\tau)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(\tau)=\int_0^t f(\tau)(u(\tau))\,d\mathcal{L}^1(\tau). \end{align*} This is the exact weak analogue of multiplying a classical parabolic equation by $u$ and integrating over space and time. [/guided] [/step] [step:Use boundedness and ellipticity to control the spatial energy] Let $\Lambda>0$ denote a boundedness constant for $A$, so that $|A(x,\tau)\xi|\leq \Lambda|\xi|$ for every $\xi\in\mathbb{R}^n$ and for $\mathcal{L}^n\otimes\mathcal{L}^1$-almost every $(x,\tau)\in U\times(0,T)$. Since $u\in L^2(0,T;H^1_0(U))$, the map $(x,\tau)\mapsto |\nabla u(x,\tau)|^2$ belongs to $L^1(U\times(0,T))$, and hence \begin{align*} |A(x,\tau)\nabla u(x,\tau)\cdot\nabla u(x,\tau)|\leq \Lambda |\nabla u(x,\tau)|^2 \end{align*} shows that the bilinear energy integrand is integrable on $U\times(0,T)$. For $\mathcal{L}^n\otimes\mathcal{L}^1$-almost every $(x,\tau)\in U\times(0,T)$, uniform ellipticity gives \begin{align*} A(x,\tau)\nabla u(x,\tau)\cdot\nabla u(x,\tau)\geq \theta |\nabla u(x,\tau)|^2. \end{align*} Integrating this pointwise inequality over $U\times(0,t)$ gives \begin{align*} \int_0^t\int_U A(x,\tau)\nabla u(x,\tau)\cdot\nabla u(x,\tau)\,d\mathcal{L}^n(x)\,d\mathcal{L}^1(\tau)\geq \theta\int_0^t\|\nabla u(\tau)\|_{L^2(U)}^2\,d\mathcal{L}^1(\tau). \end{align*} Therefore the energy balance implies \begin{align*} \frac{1}{2}\|u(t)\|_{L^2(U)}^2-\frac{1}{2}\|u_0\|_{L^2(U)}^2+\theta\int_0^t\|\nabla u(\tau)\|_{L^2(U)}^2\,d\mathcal{L}^1(\tau)\leq \int_0^t f(\tau)(u(\tau))\,d\mathcal{L}^1(\tau). \end{align*} [/step] [step:Estimate the forcing term by duality and Young's inequality] For $\mathcal{L}^1$-almost every $\tau\in(0,T)$, the duality between $H^{-1}(U)$ and $H^1_0(U)$ gives \begin{align*} |f(\tau)(u(\tau))|\leq \|f(\tau)\|_{H^{-1}(U)}\|u(\tau)\|_{H^1_0(U)}. \end{align*} Here the $H^{-1}(U)$ norm is the operator norm dual to $H^1_0(U)$ equipped with the gradient norm $\|v\|_{H^1_0(U)}:=\|\nabla v\|_{L^2(U)}$. With this convention, the preceding duality estimate becomes \begin{align*} |f(\tau)(u(\tau))|\leq \|f(\tau)\|_{H^{-1}(U)}\|\nabla u(\tau)\|_{L^2(U)}. \end{align*} Young's inequality with \begin{align*} a=\frac{1}{\sqrt{\theta}}\|f(\tau)\|_{H^{-1}(U)} \end{align*} and \begin{align*} b=\sqrt{\theta}\|\nabla u(\tau)\|_{L^2(U)} \end{align*} gives \begin{align*} \|f(\tau)\|_{H^{-1}(U)}\|\nabla u(\tau)\|_{L^2(U)}\leq \frac{1}{2\theta}\|f(\tau)\|_{H^{-1}(U)}^2+\frac{\theta}{2}\|\nabla u(\tau)\|_{L^2(U)}^2. \end{align*} Consequently, \begin{align*} \int_0^t f(\tau)(u(\tau))\,d\mathcal{L}^1(\tau)\leq \frac{1}{2\theta}\int_0^t\|f(\tau)\|_{H^{-1}(U)}^2\,d\mathcal{L}^1(\tau)+\frac{\theta}{2}\int_0^t\|\nabla u(\tau)\|_{L^2(U)}^2\,d\mathcal{L}^1(\tau). \end{align*} [/step] [step:Absorb the gradient term and obtain the estimate] Combining the previous two estimates gives \begin{align*} \frac{1}{2}\|u(t)\|_{L^2(U)}^2-\frac{1}{2}\|u_0\|_{L^2(U)}^2+\theta\int_0^t\|\nabla u(\tau)\|_{L^2(U)}^2\,d\mathcal{L}^1(\tau)\leq \frac{1}{2\theta}\int_0^t\|f(\tau)\|_{H^{-1}(U)}^2\,d\mathcal{L}^1(\tau)+\frac{\theta}{2}\int_0^t\|\nabla u(\tau)\|_{L^2(U)}^2\,d\mathcal{L}^1(\tau). \end{align*} Subtracting the last gradient term on the right-hand side from both sides yields \begin{align*} \frac{1}{2}\|u(t)\|_{L^2(U)}^2+\frac{\theta}{2}\int_0^t\|\nabla u(\tau)\|_{L^2(U)}^2\,d\mathcal{L}^1(\tau)\leq \frac{1}{2}\|u_0\|_{L^2(U)}^2+\frac{1}{2\theta}\int_0^t\|f(\tau)\|_{H^{-1}(U)}^2\,d\mathcal{L}^1(\tau). \end{align*} Multiplying by $2$ gives \begin{align*} \|u(t)\|_{L^2(U)}^2+\theta\int_0^t\|\nabla u(\tau)\|_{L^2(U)}^2\,d\mathcal{L}^1(\tau)\leq \|u_0\|_{L^2(U)}^2+\frac{1}{\theta}\int_0^t\|f(\tau)\|_{H^{-1}(U)}^2\,d\mathcal{L}^1(\tau). \end{align*} This is the desired weak [parabolic energy estimate](/theorems/603) for every $t\in[0,T]$. [/step]