Theorem 1.5: cubic internal control near a trajectory

Put $E=H_D^s(\mathrm{\Omega })$, $z_0=v_0-w(0)$ and $z_T=v_T-w(T)$. These differences belong to $E$: (W) puts the trajectory endpoints in $H_D^2$, which embeds in $H_D^s$ for $s\le 2$ by $A\ge I$ and the spectral norm. Choose $\delta$ from F2. The two endpoint bounds give $\mathrm{\parallel }{z}_0{\mathrm{\parallel }}_E+\mathrm{\parallel }{z}_T{\mathrm{\parallel }}_E\le 2\delta$, so F2 gives the fixed point $z\in C([0,T];E)$ and $h\in C([0,T];E)$, with $z(0)=z_0$, $z(T)=z_T$ and the feedback integral identity. F3 gives $F(w,z)\in L^1(0,T;E)$ and $\phi h\in C([0,T];E)$.

Continuous $L^2$ paths have compatible jointly measurable representatives in this setting. Indeed, for such a path $q$, uniform continuity in F7 gives piecewise-constant sampled paths $q_j$ tending uniformly to $q$ in $L^2$. Choose measurable spatial representatives of their countably many values using F1. The resulting functions $Q_j(t,x)$ are measurable and Cauchy in product $L^2$, since F7 gives $\mathrm{\parallel }{Q}_j-Q_k{\mathrm{\parallel }}_{L^2((0,T)\times \mathrm{\Omega })}^2={\int }_0^T\mathrm{\parallel }{q}_j(t)-q_k(t){\mathrm{\parallel }}_2^{2}dt\le T{\sup }_t\mathrm{\parallel }{q}_j(t)-q_k(t){\mathrm{\parallel }}_2^2$. Completeness gives a measurable product limit $Q$. The section norms $d_j(t)=\mathrm{\parallel }{Q}_j(t,\cdot )-Q(t,\cdot ){\mathrm{\parallel }}_2$ tend to zero in scalar $L^2(0,T)$; select an almost-everywhere convergent subsequence by F7. For almost every $t$, both $q_j(t)\to q(t)$ and $Q_j(t,\cdot )\to Q(t,\cdot )$ in $L^2$, so $Q(t,\cdot )$ represents $q(t)$. Apply this to $z,h$ and use a compatible trajectory representative from (W), or the same construction for $w$. The pointwise polynomials then represent exactly the classes in F3, since changes of compatible representatives affect only product-null sets by F7.

Write $Z,H,\tilde{w}$ for these representatives and $M_w={\sup }_t\mathrm{\parallel }w(t){\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$, which follows from (W) and compact-time continuity. For the pointwise polynomial in F3, $\mathrm{\mid }Z{\mathrm{\mid }}^3\le \mathrm{\mid }\lambda {\mathrm{\mid }}^{-1}\mathrm{\mid }F(\tilde{w},Z)\mathrm{\mid }+3M_w\mathrm{\mid }Z{\mathrm{\mid }}^2$. For every compact spatial $K\subset \mathrm{\Omega }$, its right side has finite time-space integral: F8 bounds the first term by $\mathrm{\mid }\lambda {\mathrm{\mid }}^{-1}\mathrm{\mid }K{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\parallel }F(w,z){\mathrm{\parallel }}_{L_t^1{L}_x^2}$, and the second by $3M_{w}T{\sup }_t\mathrm{\parallel }z(t){\mathrm{\parallel }}_2^2$. Thus $Z\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^3((0,T)\times \mathrm{\Omega })$, and F7 also gives locally cubically integrable spatial sections almost everywhere in time. The control $H$ is locally integrable by F8. At $\theta =s\ge 0$, these representatives and both polynomial and control forcings have the regular $E$ realizations stipulated in F3; their strong measurability and Bochner integrability are provided by that fact. Hence every representative and forcing hypothesis of the nonlinear real-linear Duhamel equivalence is met.

Apply that equivalence to the feedback identity. It gives the free-flow identity for the perturbation equation. Set $V_1=2\lambda \mathrm{\mid }w{\mathrm{\mid }}^2$, $V_2=\lambda w^2$ and $f=i(V_{1}z+V_2\bar{z}+F(w,z)-\phi h)$. Since $w$ is continuous and bounded in $L^{\mathrm{\infty }}$, the potential terms are continuous $L^2$ paths, using $\mathrm{\parallel }au{\mathrm{\parallel }}_2\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }u{\mathrm{\parallel }}_2$ and continuity of $z$ in $L^2$. The other terms are Bochner integrable in $L^2$ by F3 and F5. Thus $f\in L^1(0,T;L^2)$. The consistent free group and commutation of the bounded inclusion with the Bochner integrals in F8 transport the free-flow identity to $L^2$ as $z(t)=W(t)z_0+{\int }_0^{t}W(t-r)f(r)dr$. All nonlinear and forcing terms in this equation are locally integrable by steps 2.1–3.1 and F8.

Apply F6 on $L^2(\mathrm{\Omega })$ with this fixed forcing $f$, the free semigroup, and the Dependent Choice supplied by F1. It gives $Q(t)={\int }_0^{t}z(r)dr\in D(B)$ and $z(t)=z_0+BQ(t)+P(t)$, where $P(t)={\int }_0^{t}f(r)dr$ and $B$ is the right-quotient generator on $L^2$. For $x\in D(B)$, the negative difference quotient satisfies $(W(-a)x-x)\mathrm{/}(-a)=W(-a)(W(a)x-x)\mathrm{/}a$ for $a>0$. Unitarity and strong continuity make its limit $Bx$ as well. The two-sided characterization in F5 therefore identifies the full domain $D(B)=D({\mathrm{\Delta }}_D)$ and $Bx=i{\mathrm{\Delta }}_{D}x$. Consequently $z(t)=z_0+i{\mathrm{\Delta }}_{x}Q(t)+P(t)$ as spatial distributions at every time. This does not put $z(t)$ itself in $D({\mathrm{\Delta }}_D)$.

For $a=z$ or $a=f$, the primitive $P_a(t)={\int }_0^{t}a(r)dr$ satisfies ${\mathrm{\partial }}_t{P}_a=a$ in space-time distributions. For an arbitrary $\eta \in C_c^{\mathrm{\infty }}((0,T)\times \mathrm{\Omega };\mathbb{C})$, F7–F8 give ${\int }_0^T{\int }_{\mathrm{\Omega }}{P}_a(t,x){\mathrm{\partial }}_t\eta (t,x)dxdt={\int }_0^T{\int }_{\mathrm{\Omega }}a(r,x)({\int }_r^T{\mathrm{\partial }}_t\eta (t,x)dt)dxdr=-{\int }_0^T{\int }_{\mathrm{\Omega }}a(r,x)\eta (r,x)dxdr$. The integrals are absolutely integrable because the test and its derivative are bounded on a compact spatial set, $a\in L_t^1{L}_x^2$, and F8 gives the local $L^1$ bound; the extra time integration contributes at most a factor $T$. The identity of step 5.1 consequently holds in space-time distributions. Differentiate it in time and commute the spatial and temporal distributional derivatives, using their test-function definitions in F1. This gives ${\mathrm{\partial }}_{t}z=i{\mathrm{\Delta }}_{x}z+f$, hence $i{z}_t+{\mathrm{\Delta }}_{x}z+V_{1}z+V_2\bar{z}+F(w,z)=\phi h$ in distributions, with the bilinear test pairing throughout.

Set $v=w+z$. The bounded trajectory is locally cubically integrable, as is $Z$ by step 3.1; $\mathrm{\mid }w+Z{\mathrm{\mid }}^3\le 4(\mathrm{\mid }w{\mathrm{\mid }}^3+\mathrm{\mid }Z{\mathrm{\mid }}^3)$ gives the required local cubic integrability of $v$. Apply the expansion and distributional equivalence in F4 to step 6.1 and the equation for $w$. They give $i{\mathrm{\partial }}_{t}v+{\mathrm{\Delta }}_{x}v+\lambda \mathrm{\mid }v{\mathrm{\mid }}^{2}v=\phi h$. Both $w$ and $z$ take values in $H_0^1$, which is a linear space, so $v$ has the specified homogeneous Dirichlet condition. Their continuity in $E$ and the endpoint differences give $v\in C([0,T];E)$, $v(0)=v_0$ and $v(T)=v_T$. The control is continuous in $E$ by step 1.1, and its force is supported in the closure of $\omega$ because $\phi$ vanishes on its complement. These are all the claimed properties.