[proofplan] Subtract the two evolution equations and study the difference $w=u_1-u_2$, which satisfies the same parabolic equation with forcing $g=f_1-f_2$ and initial value $w_0=u_{1,0}-u_{2,0}$. The Hilbert-triple time-regularity identity for functions in $L^2(0,T;V)$ with derivative in $L^2(0,T;V^*)$ converts the equation into a differential inequality for $\|w(t)\|_H^2$. Coercivity controls the $V$-norm of $w$, while an explicit algebraic square estimate absorbs the forcing term into the coercive term. The [Gronwall Inequality](/theorems/872) then bounds the $H$-norm uniformly in time, and reinserting this bound gives the integral $V$-norm estimate. [/proofplan] [step:Subtract the two equations and define the difference variables] Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Let $M\geq 0$, $\alpha>0$, and $\beta\geq 0$ denote the constants from the boundedness and Gårding coercivity hypotheses in the statement. The measurability of $t\mapsto a(t;v,z)$ for each fixed $v,z\in V$ belongs to the ambient parabolic well-posedness framework. The stability estimate below does not use measurability directly after the two evolution identities have been assumed; it uses only the a.e. set on which those identities and the coercivity inequality hold. The boundedness hypothesis justifies the operator definition: for a.e. $t\in(0,T)$ and every $v\in V$, the functional $\mathcal A(t)v:V\to\mathbb R$ satisfies \begin{align*} |\mathcal A(t)v(z)|\leq M\|v\|_V\|z\|_V \end{align*} for all $z\in V$, so $\mathcal A(t)v\in V^*$ and $\|\mathcal A(t)v\|_{V^*}\leq M\|v\|_V$. The estimate below then uses only the difference equation, Gårding coercivity, and the energy identity. Define the map $w:(0,T)\to V$ by $w(t):=u_1(t)-u_2(t)$ for a.e. $t\in(0,T)$; then $w\in L^2(0,T;V)$. Define the map $g:(0,T)\to V^*$ by $g(t):=f_1(t)-f_2(t)$ for a.e. $t\in(0,T)$; then $g\in L^2(0,T;V^*)$. Also set \begin{align*} w_0:=u_{1,0}-u_{2,0}\in H. \end{align*} Since distributional differentiation is linear, \begin{align*} w'=u_1'-u_2'\in L^2(0,T;V^*). \end{align*} Subtracting the two identities in $V^*$ gives, for a.e. $t\in(0,T)$, \begin{align*} w'(t)+\mathcal{A}(t)w(t)=g(t). \end{align*} The continuous representatives satisfy $w\in C([0,T];H)$ and $w(0)=w_0$ in $H$. [/step] [step:Test the difference equation against the solution difference] Write $\langle \ell,v\rangle_{V^*,V}:=\ell(v)$ for the duality pairing between $V^*$ and $V$. We use the standard Hilbert-triple time-regularity identity: if a function belongs to $L^2(0,T;V)$ and its [distributional derivative](/page/Distributional%20Derivative) belongs to $L^2(0,T;V^*)$, then its representative in $C([0,T];H)$ has absolutely continuous squared $H$-norm and satisfies the derivative formula below. Since $w\in L^2(0,T;V)$ and $w'\in L^2(0,T;V^*)$, the map $y:[0,T]\to[0,\infty)$ defined by $y(t):=\|w(t)\|_H^2$ is absolutely continuous and, for a.e. $t\in(0,T)$, \begin{align*} \frac{1}{2}y'(t)=\langle w'(t),w(t)\rangle_{V^*,V}. \end{align*} Testing the equation $w'(t)+\mathcal{A}(t)w(t)=g(t)$ against $w(t)\in V$ gives \begin{align*} \langle w'(t),w(t)\rangle_{V^*,V}+a(t;w(t),w(t))=\langle g(t),w(t)\rangle_{V^*,V}. \end{align*} Using the Gårding coercivity hypothesis with $v=w(t)$, \begin{align*} \frac{1}{2}y'(t)+\alpha\|w(t)\|_V^2\leq \beta\|w(t)\|_H^2+\langle g(t),w(t)\rangle_{V^*,V}. \end{align*} [guided] The goal is to turn the equation in $V^*$ into a scalar inequality. Write $\langle \ell,v\rangle_{V^*,V}:=\ell(v)$ for the duality pairing between $V^*$ and $V$. Since $w'(t)$, $\mathcal A(t)w(t)$, and $g(t)$ are elements of $V^*$ for a.e. $t\in(0,T)$, the natural test object is $w(t)\in V$. Thus, for a.e. $t\in(0,T)$, \begin{align*} \langle w'(t),w(t)\rangle_{V^*,V}+\langle \mathcal A(t)w(t),w(t)\rangle_{V^*,V}=\langle g(t),w(t)\rangle_{V^*,V}. \end{align*} By the definition of $\mathcal A(t)$, the second term is \begin{align*} \langle \mathcal A(t)w(t),w(t)\rangle_{V^*,V}=a(t;w(t),w(t)). \end{align*} The Hilbert-triple time-regularity identity applies because $w\in L^2(0,T;V)$ and $w'\in L^2(0,T;V^*)$. Its conclusion is that $t\mapsto \|w(t)\|_H^2$ is absolutely continuous and \begin{align*} \frac{1}{2}\frac{d}{dt}\|w(t)\|_H^2=\langle w'(t),w(t)\rangle_{V^*,V} \end{align*} for a.e. $t\in(0,T)$. Therefore the tested equation becomes \begin{align*} \frac{1}{2}\frac{d}{dt}\|w(t)\|_H^2+a(t;w(t),w(t))=\langle g(t),w(t)\rangle_{V^*,V}. \end{align*} Now we use the Gårding coercivity assumption with $v=w(t)$. For a.e. $t\in(0,T)$, \begin{align*} a(t;w(t),w(t))\geq \alpha\|w(t)\|_V^2-\beta\|w(t)\|_H^2. \end{align*} Substituting this lower bound into the previous identity yields \begin{align*} \frac{1}{2}\frac{d}{dt}\|w(t)\|_H^2+\alpha\|w(t)\|_V^2\leq \beta\|w(t)\|_H^2+\langle g(t),w(t)\rangle_{V^*,V}. \end{align*} This is the fundamental energy inequality: the coercive part gives the useful $V$-norm term, while the lower-order loss is the term $\beta\|w(t)\|_H^2$ that will be controlled by the scalar integral estimate below. [/guided] [/step] [step:Estimate the forcing term and absorb half of the coercive norm] For a.e. $t\in(0,T)$, the dual norm gives \begin{align*} \langle g(t),w(t)\rangle_{V^*,V}\leq |\langle g(t),w(t)\rangle_{V^*,V}|\leq \|g(t)\|_{V^*}\|w(t)\|_V. \end{align*} The needed algebraic absorption estimate with parameter $\alpha>0$ follows from the following displayed square estimate: \begin{align*} 0\leq \left(\frac{1}{\sqrt{2\alpha}}\|g(t)\|_{V^*}-\sqrt{\frac{\alpha}{2}}\|w(t)\|_V\right)^2. \end{align*} Expanding this non-negative square and rearranging gives \begin{align*} \|g(t)\|_{V^*}\|w(t)\|_V\leq \frac{1}{2\alpha}\|g(t)\|_{V^*}^2+\frac{\alpha}{2}\|w(t)\|_V^2. \end{align*} Hence, for a.e. $t\in(0,T)$, \begin{align*} \frac{1}{2}y'(t)+\frac{\alpha}{2}\|w(t)\|_V^2\leq \beta y(t)+\frac{1}{2\alpha}\|g(t)\|_{V^*}^2. \end{align*} Multiplying by $2$ gives \begin{align*} y'(t)+\alpha\|w(t)\|_V^2\leq 2\beta y(t)+\frac{1}{\alpha}\|g(t)\|_{V^*}^2. \end{align*} [/step] [step:Use a scalar comparison argument to control the $H$-norm uniformly in time] Integrating the preceding differential inequality over $(0,t)$ with respect to $\mathcal{L}^1$ gives, for each $t\in[0,T]$, \begin{align*} y(t)+\alpha\int_0^t\|w(s)\|_V^2\,d\mathcal{L}^1(s) \leq y(0)+2\beta\int_0^t y(s)\,d\mathcal{L}^1(s)+\frac{1}{\alpha}\int_0^t\|g(s)\|_{V^*}^2\,d\mathcal{L}^1(s). \end{align*} Discarding the non-negative integral term on the left gives \begin{align*} y(t)\leq y(0)+\frac{1}{\alpha}\|g\|_{L^2(0,T;V^*)}^2+2\beta\int_0^t y(s)\,d\mathcal{L}^1(s). \end{align*} Define the finite constant \begin{align*} A:=y(0)+\frac{1}{\alpha}\|g\|_{L^2(0,T;V^*)}^2. \end{align*} The finiteness follows from $w_0\in H$ and $g\in L^2(0,T;V^*)$. This is the scalar integral inequality handled by the [Gronwall Inequality](/theorems/872); for completeness we prove the comparison directly. Define $h:[0,T]\to[0,\infty)$ by \begin{align*} h(t):=A+2\beta\int_0^t y(s)\,d\mathcal{L}^1(s). \end{align*} Then $y(t)\leq h(t)$ and $h'(t)=2\beta y(t)\leq 2\beta h(t)$ for a.e. $t\in(0,T)$. Hence $t\mapsto e^{-2\beta t}h(t)$ is non-increasing on $[0,T]$, so $h(t)\leq e^{2\beta t}h(0)=e^{2\beta t}A$. Therefore \begin{align*} y(t)\leq e^{2\beta t}\left(y(0)+\frac{1}{\alpha}\|g\|_{L^2(0,T;V^*)}^2\right) \end{align*} for every $t\in[0,T]$. Therefore \begin{align*} \sup_{0\leq t\leq T}\|w(t)\|_H^2 \leq e^{2\beta T}\left(\|w_0\|_H^2+\frac{1}{\alpha}\|g\|_{L^2(0,T;V^*)}^2\right). \end{align*} [/step] [step:Recover the integral $V$-norm estimate and combine the bounds] Return to the integrated energy inequality with $t=T$: \begin{align*} y(T)+\alpha\int_0^{\mathsf T}\|w(s)\|_V^2\,d\mathcal{L}^1(s) \leq y(0)+2\beta\int_0^{\mathsf T} y(s)\,d\mathcal{L}^1(s)+\frac{1}{\alpha}\|g\|_{L^2(0,T;V^*)}^2. \end{align*} Since $y(T)\geq 0$ and \begin{align*} \int_0^{\mathsf T} y(s)\,d\mathcal{L}^1(s)\leq T\sup_{0\leq s\leq T}y(s), \end{align*} we obtain \begin{align*} \alpha\int_0^{\mathsf T}\|w(s)\|_V^2\,d\mathcal{L}^1(s) \leq \|w_0\|_H^2+2\beta T e^{2\beta T}\left(\|w_0\|_H^2+\frac{1}{\alpha}\|g\|_{L^2(0,T;V^*)}^2\right)+\frac{1}{\alpha}\|g\|_{L^2(0,T;V^*)}^2. \end{align*} Combining this estimate with the uniform $H$-norm estimate, there is a constant \begin{align*} C=C(\alpha,\beta,T)>0 \end{align*} such that \begin{align*} \sup_{0\leq t\leq T}\|w(t)\|_H^2+\int_0^{\mathsf T}\|w(t)\|_V^2\,d\mathcal{L}^1(t) \leq C\left(\|w_0\|_H^2+\|g\|_{L^2(0,T;V^*)}^2\right). \end{align*} Substituting $w=u_1-u_2$, $w_0=u_{1,0}-u_{2,0}$, and $g=f_1-f_2$ gives \begin{align*} \sup_{0\leq t\leq T}\|u_1(t)-u_2(t)\|_H^2+\int_0^{\mathsf T}\|u_1(t)-u_2(t)\|_V^2\,d\mathcal{L}^1(t) \leq C\left(\|u_{1,0}-u_{2,0}\|_H^2+\|f_1-f_2\|_{L^2(0,T;V^*)}^2\right). \end{align*} This is the desired continuous dependence estimate. [/step]