Well-posedness for measurable coercive parabolic forms

If $V=\{0\}$, density in $H$ forces $H=\{0\}$, and the only data and solution are zero, so the conclusion holds. Assume henceforth $V\mathrm{\ne }\{0\}$. By [F1], $W(0,T;V)$ has the stated graph norm. Set $Y=L^2(0,T;V)$ and $X=\{w\in W(0,T;V):w(0)=0\}$, with the inherited $W$ norm. By [F3], $W$ and $Y$ are Hilbert spaces; by the trace bound in [F2], $X$ is a closed Hilbert subspace. Define $b(w,y)={\int }_0^T(⟨w^{\mathrm{\prime }}(t),y(t){⟩}_{V^{\ast },V}+a(t;w(t),y(t)))dt\mathrm{.}$ The integrand is measurable by [F4], and Cauchy–Schwarz and the form bound give $\mathrm{\mid }b(w,y)\mathrm{\mid }\le (1+M)\mathrm{\parallel }w{\mathrm{\parallel }}_W\mathrm{\parallel }y{\mathrm{\parallel }}_Y$.

For $w\in X$, let $z_0(t)$ solve $a(t;z_0(t),v)=⟨w^{\mathrm{\prime }}(t),v⟩$ and set $z=z_0+\mu w$, where $\mu =M^4{\alpha }^{-4}$. By [F4], $z_0\in Y$ and $\mathrm{\parallel }{z}_0{\mathrm{\parallel }}_Y\le {\alpha }^{-1}\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(V^{\ast })}$, so $\mathrm{\parallel }z{\mathrm{\parallel }}_Y\le C_z\mathrm{\parallel }w{\mathrm{\parallel }}_W$ with $C_z={\alpha }^{-1}+\mu$. Since $A(t)z_0(t)=w^{\mathrm{\prime }}(t)$, coercivity and the operator bound give $⟨w^{\mathrm{\prime }},z_0⟩=a(t;z_0,z_0)\ge \alpha M^{-2}\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_{V^{\ast }}^2$. The energy identity [F2] and $w(0)=0$ give ${\int }_0^T⟨w^{\mathrm{\prime }},w⟩=\frac{1}{2}\mathrm{\parallel }w(T){\mathrm{\parallel }}_H^2$. The remaining mixed term obeys $\mathrm{\mid }a(t;w,z_0)\mathrm{\mid }\le (M\mathrm{/}\alpha )\mathrm{\parallel }w{\mathrm{\parallel }}_V\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_{V^{\ast }}$, and Young's inequality bounds it by $\frac{\alpha }{2M^2}\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_{V^{\ast }}^2+\frac{M^4}{2{\alpha }^3}\mathrm{\parallel }w{\mathrm{\parallel }}_V^2$. Therefore $b(w,z)\ge \frac{\alpha }{2M^2}\mathrm{\parallel }{w}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(V^{\ast })}^2+\frac{M^4}{2{\alpha }^3}\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2(V)}^2+\frac{\mu }{2}\mathrm{\parallel }w(T){\mathrm{\parallel }}_H^2\ge c_0\mathrm{\parallel }w{\mathrm{\parallel }}_W^2,$ where $c_0=\min \{\alpha \mathrm{/}(2M^2),M^4\mathrm{/}(2{\alpha }^3)\}>0$. For $w\mathrm{\ne }0$, positivity forces $z\mathrm{\ne }0$, and $b(w,z)\le \mathrm{\parallel }z{\mathrm{\parallel }}_Y{\sup }_{\mathrm{\parallel }y{\mathrm{\parallel }}_Y\le 1}\mathrm{\mid }b(w,y)\mathrm{\mid }$ together with $\mathrm{\parallel }z{\mathrm{\parallel }}_Y\le C_z\mathrm{\parallel }w{\mathrm{\parallel }}_W$ gives ${\sup }_{\mathrm{\parallel }y{\mathrm{\parallel }}_Y\le 1}\mathrm{\mid }b(w,y)\mathrm{\mid }\ge (c_0\mathrm{/}{C}_z)\mathrm{\parallel }w{\mathrm{\parallel }}_W$; the inequality is trivial for $w=0$.

Suppose $y\in Y$ and $b(w,y)=0$ for every $w\in X$. By [F4], $g(t):=A^{\ast }(t)y(t)$ belongs to $L^2(0,T;V^{\ast })$. Taking $w(t)=\varphi (t)v$ for $v\in V$ and $\varphi \in C_c^{\mathrm{\infty }}(0,T)$ in $b(w,y)=0$ gives $\int {\varphi }^{\mathrm{\prime }}(t)(iv(t),iy(t){)}_H+\varphi (t)a(t;v,y(t))dt=0$. By [F1], this means $y^{\mathrm{\prime }}=g$ weakly, so $y\in W(0,T;V)$. Testing the weak derivative identity against $t\varphi$ and expanding $(t\varphi {)}^{\mathrm{\prime }}$ shows that $w(t)=ty(t)$ has weak derivative $y+t{y}^{\mathrm{\prime }}$. Its $H$ trace at zero is zero, so $w\in X$. Substituting this $w$ and using $y^{\mathrm{\prime }}=A^{\ast }y$ gives $0=b(ty,y)={\int }_0^T(\mathrm{\parallel }y(t){\mathrm{\parallel }}_H^2+2ta(t;y(t),y(t)))dt\ge {\int }_0^T\mathrm{\parallel }y(t){\mathrm{\parallel }}_H^2+2t\alpha \mathrm{\parallel }y(t){\mathrm{\parallel }}_V^{2}dt\mathrm{.}$ Hence $y=0$ in $Y$, proving the trivial right-kernel condition.

The preceding steps verify the boundedness, inf-sup and right-kernel hypotheses of [F5]. Consequently, for every $g\in L^2(0,T;V^{\ast })$ the functional $y\mapsto {\int }_0^T⟨g(t),y(t)⟩dt$ has a unique solution $w\in X$ to $b(w,y)=\int ⟨g,y⟩$ for all $y\in Y$. Its boundedness follows from Cauchy–Schwarz.

By density of $V$ in $H$ and [A1], choose $v_n\in V$ with $\mathrm{\parallel }i{v}_n-u_0{\mathrm{\parallel }}_H<1\mathrm{/}(n+1)$. The measurable-action result [F4] gives $g_n=f-A(t)v_n\in L^2(0,T;V^{\ast })$. Let $w_n\in X$ be the solution from step 3.1 for $g_n$, and let ${\mathrm{\ell }}_n(t)=v_n$ be the constant path. Then $u_n=w_n+{\mathrm{\ell }}_n\in W(0,T;V)$ has $u_n(0)=i{v}_n$ and $b(u_n,y)={\int }_0^T⟨f(t),y(t)⟩dt$ for every $y\in Y$.

For $0\le \tau \le T$, test the equation for $u_n$ with $y={\mathbf{1}}_{(0,\tau )}{u}_n\in Y$. By [F2], coercivity, and $2ab\le \alpha a^2+{\alpha }^{-1}{b}^2$, $\mathrm{\parallel }{u}_n(\tau ){\mathrm{\parallel }}_H^2+\alpha {\int }_0^{\tau }\mathrm{\parallel }{u}_n(t){\mathrm{\parallel }}_V^{2}dt\le \mathrm{\parallel }i{v}_n{\mathrm{\parallel }}_H^2+{\alpha }^{-1}{\int }_0^{\tau }\mathrm{\parallel }f(t){\mathrm{\parallel }}_{V^{\ast }}^{2}dt\mathrm{.}$ The right side is bounded uniformly in $n$. Also $G_n=u_n^{\mathrm{\prime }}+A(t)u_n-f\in L^2(0,T;V^{\ast })$ by [F4]. Testing with $y=\varphi v$ shows $\int \varphi ⟨G_n,v⟩=0$ for every fixed $v\in V$ and $\varphi \in C_c^{\mathrm{\infty }}(0,T)$. Choose a countable dense subset of $V$, apply [F6] to its scalar pairings, and take the union of their null sets. Outside that null set continuity of $G_n(t)\in V^{\ast }$ gives $G_n(t)=0$. Thus $u_n^{\mathrm{\prime }}=f-A(t)u_n$ a.e. and $\mathrm{\parallel }{u}_n^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(V^{\ast })}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(V^{\ast })}+M\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_{L^2(V)}$. Together these estimates bound $(u_n)$ in $W(0,T;V)$.

By [F3], a subsequence converges weakly in $W(0,T;V)$ to some $u$. The trace map $w\mapsto w(0)$ is bounded linear by [F2], so weak trace convergence and $i{v}_n\to u_0$ in $H$ imply $u(0)=u_0$. For every $y\in Y$, $b(u_n,y)=\int ⟨f,y⟩$ and boundedness of $b$ from step 1.1 let us pass to the weak limit, giving $b(u,y)=\int ⟨f,y⟩$. With $G=u^{\mathrm{\prime }}+A(t)u-f\in L^2(V^{\ast })$, testing by $\varphi v$ and applying [F6] on a countable dense subset of $V$ as in step 5.1, with [F7] to unite the null sets, proves the pointwise variational equation for a.e. $t$ and every $v\in V$. Testing the limiting identity with ${\mathbf{1}}_{(0,\tau )}u$ and using [F2] gives the stated energy estimate for every $\tau \in [0,T]$.

If $u_1,u_2$ are two solutions with the same data, $z=u_1-u_2$ lies in $X$ and satisfies $b(z,y)=0$ for every $y\in Y$. The uniqueness in [F5] applied to the zero functional gives $z=0$. This proves uniqueness.