Linear forced well-posedness by Duhamel formula

The closed annulus $K=\{x:R\le \mathrm{\mid }x\mathrm{\mid }\le R+1\}$ is disjoint from the obstacle and compact by [F7]. The cutoff and its classical derivatives through order two are continuous on $K$, so their images there are compact subsets of $\mathbb{R}$ and hence bounded by [F7]. Outside $K$, $\phi$ has the constant values zero or one and its positive-order derivatives vanish. Thus the function and all its classical derivatives through order two are bounded on $\mathrm{\Omega }$. Countable Choice follows from the assumed AC by [F4], so [F6] identifies these with the weak derivatives. Hence $\phi \in W^{2,\mathrm{\infty }}(\mathrm{\Omega })$. Apply the multiplier lemma in [F2] to obtain a bounded operator $M_{\phi }$ on $H_D^s$, and denote a bound for its norm by $C_s$. This uses no upper range restriction on $\phi$.

Put $I=(0,T)$ and $\rho (q)=T-q$. By [F8] and Countable Choice from [F4], $\rho$ is a measurable, measure-preserving involution of $I$ and takes null sets to null sets. For any strongly measurable $g:I\to X$, compose its almost-everywhere convergent simple approximation with $\rho$. The compositions are measurable simple functions, and they converge to $g\circ \rho$ outside the reflected exceptional null set. Hence $g\circ \rho$ is strongly measurable, independently of the representative of $g$. Applying [F9] to the scalar function $\mathrm{\parallel }g{\mathrm{\parallel }}_X$ gives ${\int }_I\mathrm{\parallel }g\circ \rho {\mathrm{\parallel }}_X={\int }_I\mathrm{\parallel }g{\mathrm{\parallel }}_X$. In particular reflection preserves $L^1(I;X)$ and its norm.

Put $X=H_D^s$ and $f(t)=-i\phi h(t)$. Bounded linear multiplication sends the simple approximants in [F3] to simple approximants, so $f$ is strongly measurable. By step 1.1, $\mathrm{\parallel }f(t){\mathrm{\parallel }}_X\le C_s\mathrm{\parallel }h(t){\mathrm{\parallel }}_X$ almost everywhere; hence $f$ is Bochner integrable. Restrict the group $W$ to nonnegative times and denote its generator by $B$. Its mild Cauchy problem is $z^{\mathrm{\prime }}=Bz+f$. This is the evolution interpretation of $i{\mathrm{\partial }}_{t}z+{\mathrm{\Delta }}_{D}z=\phi h$: on generator-domain data the free derivative is $i{\mathrm{\Delta }}_{D}W(t)v$ in the lower scale, and the forcing is $-i\phi h$.

Let $g\in L^1(I;X)$ and let $E\subset I$ be measurable. Choose defining integrable simple approximants $p_n$ with ${\int }_I\mathrm{\parallel }g-p_n{\mathrm{\parallel }}_X\to 0$, as in [F10]. Reflection sends each nonzero level set $A_j$ of $p_n$ to $\rho (A_j)$ and preserves its measure. Moreover $\rho (E\cap \rho (A_j))=\rho (E)\cap A_j$. The finite-sum formula in [F10] therefore gives ${\int }_E{p}_n\circ \rho ={\int }_{\rho (E)}{p}_n$. Step 1.2, also applied to $g-p_n$, shows that the reflected errors have the same $L^1$ norm. Restricting either error to the indicated subset can only decrease its norm integral, so [F5] and [F10] permit passage to the norm limits, yielding ${\int }_{E}g(\rho (q))dq={\int }_{\rho (E)}g(r)dr$. The same finite-sum limit argument gives ${\int }_{E}ig=i{\int }_{E}g$. These identities include null sets and hence apply to the interval integrals irrespective of their endpoints.

Apply variation of constants to this semigroup, initial state $z_0$ and forcing $f$. It gives the displayed forward formula, a continuous $X$-valued path, and uniqueness among mild solutions and separately among integral solutions. Unitarity and the Bochner norm inequality give $\mathrm{\parallel }z(t){\mathrm{\parallel }}_X\le \mathrm{\parallel }{z}_0{\mathrm{\parallel }}_X+C_s{\int }_0^t\mathrm{\parallel }h(r){\mathrm{\parallel }}_{X}dr$, proving the uniform estimate.

For the terminal problem set $y(\tau )=z(T-\tau )$, $0\le \tau \le T$. The reversed free group is $\tilde{W}(\tau )=W(-\tau )$, and its generator is $-B$ on $D(B)$: the group identity writes the reversed difference quotient as $-W(-\tau )(W(\tau )v-v)\mathrm{/}\tau$, and the converse follows by reversing the same identity. The reversed forcing is $\tilde{f}(\tau )=i\phi h(T-\tau )$, which belongs to $L^1(I;X)$ by step 2.1 and step 1.2. Variation of constants applied to $\tilde{W}$ gives a unique mild, equivalently unique integral, $y$ with $y(0)=z_T$. Fix $0\le t\le T$ and put $\tau =T-t$. Its Duhamel integrand $q\mapsto W(q-\tau )\tilde{f}(q)$ on $(0,\tau )$ is Bochner integrable by [F4]; extend it by zero to $I$. Apply step 2.2 with $E=(t,T)$ to this extended function. At $q=T-r$, its nonzero value is $iW(t-r)\phi h(r)$, so the result is $z(t)=W(t-T)z_T+i{\int }_t^{T}W(t-r)\phi h(r)dr$. Applying the same reflection twice gives the original mild formula. For the integral class, the reversed identity is $y(\tau )=z_T-B{\int }_0^{\tau }y+{\int }_0^{\tau }\tilde{f}$, with ${\int }_0^{\tau }y\in D(B)$; step 2.2 converts it to $z(t)=z_T-B{\int }_t^{T}z-{\int }_t^{T}f$, with ${\int }_t^{T}z\in D(B)$. Reflection converts these identities back as well. Thus reversal is bijective on both stated solution classes, proving backward uniqueness and continuity. The scale realization supplies the stated Dirichlet interpretation.