Proof. One can see that (2.3) implies (2.4) by taking $N = \widetilde {N}$. For the other direction, let $N = \widetilde {N} + \varepsilon P_M$ where $\varepsilon > 0$. The block matrices associated with $N$ and $\widetilde {N}$ are related by

\[ \Psi_{G,N,M} = \Psi_{G,\widetilde{N},M} + \left( \begin{array}{@{}cc@{}} -\varepsilon P_M & 0 \\ 0 & \varepsilon P_M \end{array}\right). \]

As before, (2.4) implies that the kernel of $L_1$ lies in the kernel of $M$ and as a consequence $\langle P_Mu,\,u\rangle = \langle P_MP_{L_1}u,\,P_{L_1}u\rangle$. By choosing $\varepsilon < \widetilde {\alpha } / \|P_M\|$, where $\|\cdot \|$ is the operator norm, we have $\widetilde {\alpha }|P_{L_1}u|^2 - \varepsilon \langle P_MP_{L_1}u,\,P_{L_1}u \rangle \geq (\widetilde {\alpha } - \varepsilon \|P_M\|)|P_{L_1} u|^2 > 0$. Then, we can see that (2.4) implies (2.3) with $\alpha = \widetilde {\alpha } - \varepsilon \|P_M\|$.

Proof. Given $(u,\,z)\in \mathbb {C}^n \times \mathbb {C}^n$ we have

(2.5)\begin{align} \langle \Psi_{I,M,M}(u,z),(u,z)\rangle = 2\langle L_1u,u\rangle - \langle MP_Mu,P_Mu\rangle + 2\text{Re}\langle Mu,z\rangle + \langle Mz,z\rangle. \end{align}

By the symmetry of the delay matrix $M$, we obtain $\langle Mz,\,z\rangle = \langle MP_Mz,\,P_Mz\rangle$ and $\langle Mu,\,z\rangle = \langle MP_Mu,\,P_Mz\rangle$, and therefore, by the Cauchy–Schwarz inequality we have

(2.6)\begin{equation} |\langle Mu,z\rangle| \leq \frac{1}{2}\langle MP_Mu,P_Mu\rangle + \frac{1}{2}\langle MP_Mz,P_Mz\rangle. \end{equation}

Using (2.6) in (2.5) yields the estimate

\[ \langle \Psi_{I,M,M}(u,z),(u,z)\rangle \geq 2\langle (L_1- P_MMP_M)u,u\rangle = 2\langle (L_1-M)u,u\rangle \geq \widetilde{\alpha}|P_{L_1}u|^2, \]

for some $\widetilde {\alpha } > 0$. The conclusion now follows from theorem 2.1.

Proof. Property (i) implies that the kernels of $MJ$ and $M$ coincide, and in particular, $\text {Ker}(MJ)^\perp = \text {Ker}(M)^\perp$ and $P_{MJ} = P_M$. Given $u \in \mathbb {C}^n$ there holds

\[ P_MJu = P_MJP_Mu + P_MJ(I-P_M)u = JP_Mu \]

since $JP_Mu \in \text {Ker}(M)^\perp$ and $J(I-P_M)u \in \text {Ker}(M)$. Hence, $P_M$ and $J$ commute, and consequently, $P_M$ and $J^T$ also commute by symmetry of $P_M$. If $(u,\,z)\in \mathbb {C}^n \times \mathbb {C}^n$, then we derive from (ii) and the preceding statement that

\begin{align*} & \langle \Psi_{G,N,M}(u,Jz),(u,Jz)\rangle\\ & \quad= 2\langle L_1u,u\rangle - \langle NP_Mu,P_Mu\rangle + 2\text{Re}\langle GMJz,u\rangle + \langle NJz,Jz\rangle\\ & \quad= 2\langle L_1u,u\rangle - \langle J^TNJP_{MJ}u,P_{MJ}u\rangle + 2\text{Re}\langle GMJz,u\rangle + \langle J^TNJz,z\rangle \\ & \quad= \langle \Psi_{G,J^TNJ,MJ}(u,z),(u,z)\rangle. \end{align*}

Using condition (M) and the fact that $P_M$ and $J$ commute, we can see from these equations that $\Psi _{G,J^TNJ,MJ} > 0$ on $\text {Ker}(L_1)^\perp \times \text {Ker}(M)^\perp$. Finally, since $J$ is bijective, it follows that $\langle GMJz,\, u \rangle = 0$ for every $z \in \mathbb {C}^n$ and $u \in \text {Ker}(L_1)$. These prove that $MJ$ satisfies condition (M).

Proof. Let $X = L^2((0,\,T)\times (-\tau,\,0)\times \mathbb {R}^d)$, $Y = L^2(\mathbb {R}^d; H^1((0,\,T)\times (-\tau,\,0)))$, and $Z = L^2(0,\,T;L^2) \times L_\theta ^2(L^2)$. Define the operators $\Lambda : Y \to X$, $\Psi : Y \to Z$, and $\Phi : Y \to Z$ as follows:

\[ \Lambda \psi := \mathscr{L}_1^* \psi, \qquad \Psi \psi := (\psi_{|\theta=0}, \psi_{|t=0}), \qquad \Phi\psi := (\psi_{|\theta={-}\tau}, \psi_{|t=T}) \]

for $\psi \in Y$. The variational equation (3.3) can now be written in form (3.5). From the a priori estimate (3.7), one can see that (3.4) is satisfied, and therefore by theorem 3.1, (3.1) has a weak solution.

To establish uniqueness, we proceed by a duality argument. Suppose that $z_1$ and $z_2$ are two weak solutions and let $z := z_1 - z_2$. Then, it follows that

(3.9)\begin{equation} \int_0^T\!\! \int_{-\tau}^0 (z,\mathscr{L}_1^* \psi)_{L^2} {\rm d} \theta {\rm d} t = 0 \end{equation}

for every $\psi \in \text {Ker}(\Phi )$. Let $(\phi _n)_{n=1}^\infty$be a sequence of infinitely differentiable functions with compact support in $(0,\,T)\times (-\tau,\,0)\times \mathbb {R}^d$ such that $\phi _n \to z$ in $L^2((0,\,T)\times (-\tau,\,0)\times \mathbb {R}^d)$. The backward-in-time transport equation

\[ - \partial_t\psi_{n} + \partial_\theta \psi_n + a \psi_n = \phi_n, \quad \psi_{n|t= T} = 0, \quad \psi_{n|\theta={-}\tau} = 0 \]

has a classical solution, so that $\psi _n \in Y$ for each $n$. Using this test function in (3.9) and then passing to the limit, we see that $z = 0$ almost everywhere. Therefore, the weak solution of (3.1) is unique.

Proof. Let $\rho _\varepsilon$ be a standard mollifier with respect to $x$, that is, $\rho _\varepsilon (x) := \varepsilon ^{-d}\rho ({x}/{\varepsilon })$ where $\rho \in \mathscr {D}(\mathbb {R}^d)$ satisfies $\int _{\mathbb {R}^d} \rho _\varepsilon (x) {\rm d} x = 1$, and let $R_\varepsilon v := \rho _\varepsilon \ast v$. Then, the regularized data $(R_\varepsilon z_0,\, R_\varepsilon v) \in H^k_\theta (H^\infty ) \times H^k(0,\,T;H^\infty )$ is still compatible up to order $k-1$ and

\[ (R_\varepsilon z_0, R_\varepsilon v) \to (z_0,v) \text{ in }H^k_\theta(H^s) \times H^k(0,T;H^s) \]

as $\varepsilon \to 0$. For a fix $\varepsilon > 0$, take a sequence $(z_{0n}^\varepsilon,\, v_{n1}^\varepsilon ) \in H^{k+1}_\theta (H^\infty ) \times H^{k+1}(0,\,T;H^\infty )$ such that as $n\to \infty$ there holds

\[ (z_{0n}^\varepsilon,v_{n1}^\varepsilon) \to (R_\varepsilon z_0, R_\varepsilon v) \text{ in }H^{k}_\theta(H^s) \times H^{k}(0,T;H^s). \]

For example, we first extend $R_\varepsilon z_0$ to a function in $H^k(\mathbb {R};H^\infty )$ by a standard reflection argument, see [Reference Adams1] for instance, and if $\widetilde {R}_\delta$ is the corresponding convolution operator with respect to $\theta$, then we may take $z_{0n}^\varepsilon$ to be the restriction of $\widetilde {R}_{1/n} (R_\varepsilon z_0)$ in $(-\tau,\,0)$.

Define $v_n^\varepsilon := v_{n1}^\varepsilon - v_{n2}^\varepsilon$ where $v_{n2}^\varepsilon \in H^{k+1}(0,\,T;H^{s+1})$ is a function that will be constructed below that satisfies $v_{n2}^\varepsilon \to 0$ in $H^{k+1}(0,\,T;H^{s+1})$. For each $0 \leq j\leq k$, define

\[ \sigma_{nj} := \partial_t^j v^\varepsilon_{n1|t=0} - z^\varepsilon_{0nj|\theta = 0} \in H^{k - j + s + \frac{3}{2}}. \]

From the compatibility conditions for the data $(R_\varepsilon z_0,\, R_\varepsilon v)$, we have $\sigma _{nj} \to 0$ in $H^{k - j + s + {3}/{2}}$ as $n\to \infty$ for every $0\leq j < k$. According to trace theory, for each $n$ there exists $h_n \in H^{k+s+2}((0,\,T)\times \mathbb {R}^d) \subset H^{k+1}(0,\,T;H^{s+1})$ such that $\partial _t^jh_{n|t=0} = \sigma _{nj}$ for all $0 \leq j < k$, $\partial _t^k h_{n|t=0} = 0$, and $h_n \to 0$ in $H^{k+1}(0,\,T;H^{s+1})$.

Let $v_{n2}^\varepsilon := h_n + g_n$, where $g_n := \widetilde {g}_n \otimes \sigma _{n,k}$ and $\widetilde {g}_n \in H^{k+1}(0,\,T)$ satisfies

\[ \widetilde{g}_n^{(j)}(0) = 0, \text{ for } j=0,1,\ldots,k-1, \quad \widetilde{g}_n^{(k)}(0) = 1, \quad \|\widetilde{g}_n\|_{H^{k+1}(0,T)} \to 0. \]

For the construction of $\widetilde {g}_n$, we refer to [Reference Peralta and Propst22, Reference Rauch and Massey25]. If $j < k$, then

\[ \partial_t^j v^\varepsilon_{n|t=0} = \partial_t^j v^\varepsilon_{n1|t=0} - \partial_t^j h_{n|t=0} = z^\varepsilon_{0nj|\theta = 0}. \]

Also, $\partial _t^k v^\varepsilon _{n|t=0} = \partial _t^k v^\varepsilon _{n1|t=0} - \sigma _{nk} = z^\varepsilon _{0nk|\theta = 0}.$

We now construct the sequence $(z_{0n},\,v_n)$ as follows. Given a positive integer $n$, let $(z_{0n},\,v_n) := (z_{0N}^{1/n},\,v_N^{1/n})$ for a sufficiently large $N = N(n)$ be such that

\[ \|(z_{0n},v_n) - (R_{1/n}z_0,R_{1/n}v)\|_{H^{k}_\theta(H^{s}) \times H^{k}(0,T;H^{s})} < \frac{1}{n}. \]

From the above construction, we can see that the pair $(z_{0n},\,v_n)$ satisfies the desired properties.

Proof. By choosing $k$ and $s$ sufficiently large in corollary 3.4, one can construct a sequence $(z_{0n},\,v_n)$ of continuously differentiable data that are compatible up to order 1 and tends to $(z_0,\,v)$ in $L^2_\theta (L^2) \times L^2(0,\,T;L^2)$. For example one may take $k = 2$ and $s > {d}/{2} + 1$. It can be easily verified that the transport equation (3.1) with boundary data $v_n$ and initial data $z_{0n}$ has the classical solution

(3.11)\begin{equation} z_n(t,\theta,x) = \begin{cases} e^{a\theta}v_n(t+\theta,x) & \text{in } ( \{t>{-}\theta\} \cap (0,T)\times(-\tau,0)) \times \mathbb{R}^d,\\ e^{a\theta}z_{0n}(t+\theta,x) & \text{in }(\{t<{-}\theta\} \cap (0,T)\times(-\tau,0))\times \mathbb{R}^d. \end{cases} \end{equation}

Applying the a priori estimate (3.6) to $z_n - z_m$, one can see that $(z_n)_n$ is a Cauchy sequence in $C(0,\,T;L_\theta ^2(L^2))$, while $(z_{|\theta = -\tau })_n$ and $(z_{n|\theta = 0})_n$ are a Cauchy sequences in $L^2(0,\,T;L^2)$. By passing to the weak formulation of the equation for $z_n$, we can see that the limit in $C(0,\,T;L_\theta ^2(L^2))$ is a weak solution, and thus, the limit must be the weak solution of (3.1) by uniqueness. Note that the traces $z_{n|\theta = -\tau }$ and $z_{n|\theta = 0}$ tend to $z_{|\theta = -\tau }$ and $z_{|\theta = 0}$ in $L^2(\mathbb {R}^d;H^{-1/2}(0,\,T))$, respectively, and consequently in $L^2(0,\,T;L^2)$ by uniqueness of limits in the sense of distributions. Passing to the limit in (3.11), up to a subsequence we can see that the weak solution of (3.1) is given by (3.10). The energy estimate can be obtained by passing to the limit of the priori estimate (3.6) for $z_n$.

Proof. Taking $\phi = 0$ in (4.6) shows that $z$ is the weak solution of (4.9), and therefore, we have $z_\tau \in L^2(0,\,T;L^2)$. Given $\phi \in H^1((0,\,T)\times \mathbb {R}^d)$ such that $\phi _{|t= T} = 0$, the homogeneous backward-in-time Cauchy problem

\[ -\partial_t \psi + \partial_\theta \psi + \varepsilon \psi = 0, \quad \psi_{|\theta ={-}\tau} = e^{\varepsilon\tau}M^T \phi, \quad \psi_{|t=T} = 0 \]

has a compatible data, and thus, according to the previous section it has a solution satisfying

\[ \psi \in C(0,T; H^1_\theta(L^2) \cap L^2_\theta(H^1)) \cap C^1(0,T;L^2_\theta(L^2)), \]

and in particular, $\psi \in L^2(\mathbb {R}^d; H^1((0,\,T) \times (-\tau,\,0)))$. Choosing the pair $(\phi,\,\psi )$ in the variational formulation (4.6) and using (3.12), it follows that $u$ is the weak solution of (4.8). The energy estimate of the lemma follows from the energy estimates for solutions of (4.8) and (4.9), and by taking $\gamma$ sufficiently large.

Proof. Uniqueness follows immediately from the previous lemma. For existence, we apply theorem 3.1 and for this we introduce the function spaces $X := L^2(0,\,T;L^2) \times L^2(0,\,T;L^2_\theta (L^2))$, $Y := H^1((0,\,T)\times \mathbb {R}^d) \times L^2(\mathbb {R}^d,\, H^1((0,\,T)\times (-\tau,\,0)))$ any $Z := L^2(0,\,T;L^2) \times L^2 \times L^2_\theta (L^2)$. Define the operators $\Lambda : Y \to X$, $\Psi : Y \to Z$ and $\Phi : Y \to Z$ as follows:

\begin{align*} & \Lambda(\phi,\psi) := (\mathscr{L}_2^*\phi - P_M\psi_{|\theta =0}, \mathscr{L}_1^*\psi)\\ & \Psi(\phi,\psi) := (0, A^0\phi_{|t=0},e^{\varepsilon\theta}P_M\psi_{|t=0})\\ & \Phi(\phi,\psi) := (e^{\varepsilon\tau}M^T \phi - \psi_{|\theta ={-}\tau},\phi_{|t=T}, \psi_{|t=T}). \end{align*}

The variational equation (4.6) can now be expressed as

\[ ((u,z),\Lambda(\phi,\psi))_X = ((0,u_0,z_0), \Psi(\phi,\psi))_Z \]

for every $(\phi,\,\psi )\in \text {Ker}(\Phi )$. From the a priori estimates for the transport equation with parameter (3.7) and for hyperbolic systems, the dual version of (4.5), we obtain the priori estimate (3.4) with the help of an absorption argument. More precisely, the terms $\|\phi \|_{L^2(0,T;L^2)}$ and $\|\psi _{|\theta = 0}\|_{L^2(0,T;L^2)}$ arising on the right-hand side can be absorbed by the left-hand side by making $\gamma$ sufficiently large. Therefore, (4.6) is satisfied for some $(u,\,z) \in X$.

It remains to verify the constraint. For this purpose, let $u_\delta := R_\delta u \in L^2(0,\,T;H^\infty )$ and $z_{0\delta } := R_\delta z_0 \in L_\theta ^2(H^\infty )$. Let $z_\delta$ be the solution of the transport system with initial data $e^{\varepsilon \theta }P_Mz_{0\delta }$ and boundary data $P_M u_\delta$. Let $u^\delta$ be the solution of the hyperbolic system with source term $-e^{\varepsilon \tau }Mz_{\delta \tau }$ and initial data $u_{0\delta } := R_\delta u_0$. Then, we have $z_{\delta \tau } \in L^2(0,\,T;H^\infty )$, and consequently, $u \in H^1(0,\,T;H^{\infty })$. Moreover, $u^\delta \to u$ in $L^2$ by uniqueness of weak solutions. Therefore, for every $\varphi \in \mathscr {D}((0,\,T)\times \mathbb {R}^d)$ we obtain from the Parseval's identity that

\[ \int_0^T (\partial_t \mathscr{L}_3u^{\delta}, \varphi)_{L^2} {\rm d} t = \int_0^T \text{Re}((i|\xi|Q(\omega) + R)\widehat{u^{\delta}_t}, \widehat{\varphi})_{L^2}{\rm d} t \]

where $\, \widehat {\cdot }\,$ is the Fourier transform.

According to condition (Q) and the differential equation for $u^\delta$, we have

\begin{align*} & (i|\xi|Q(\omega) + R)\widehat{u^{\delta}_t} \\ & \quad =|\xi|^2Q(\omega)(A^0)^{{-}1}A(\omega)\widehat{u^\delta} - i|\xi| ( Q(\omega)(A^0)^{{-}1}L + R(A^0)^{{-}1}A(\omega)) \widehat{u^\delta}\\ & \qquad- R(A^0)^{{-}1}L \widehat{u^\delta} + ie^{\varepsilon\tau}|\xi|Q(\omega)(A^0)^{{-}1}M \widehat{z_{\delta\tau}} + e^{\varepsilon\tau}R(A^0)^{{-}1}M \widehat{z_{\delta\tau}} = 0. \end{align*}

Thus, $\mathscr {L}_3u^\delta$ is constant, and in particular, we have $\mathscr {L}_3u^\delta (t) = \mathscr {L}_3u_{0\delta } = R_\delta (\mathscr {L}_3u_0) = 0$ for every $t \geq 0$. Passing to the limit in $\int _0^T (u^\delta,\,\mathscr {L}_3^*\varphi )_{L^2} {\rm d} t = 0$ and using the density of $\mathscr {D}((0,\,T)\times \mathbb {R}^d)$ in $L^2(0,\,T;H^1)$, we infer that the weak solution satisfies the variational form (4.7) of the differential constraint.

Proof. Let $u^0 := u_0$ and $z^0 := e^{\varepsilon \theta }P_Mz_0$. Given $u^{n-1}$, let $z^n$ be the solution of the transport system (4.9) with boundary data $P_Mu^{n-1}$ and initial data $e^{\varepsilon \theta }P_Mz_0$. Then, we have $z^n_{\tau } \in L^2(0,\,T;H^s)$. Let $u^n$ be the solution of the hyperbolic system (4.8) with initial data $u_0$ and source term $-e^{\varepsilon \tau }Mz^{n}_\tau$. Hence, it follows that $u^n \in C(0,\,T;H^s)$. Using the energy estimates for the transport equation with parameter and hyperbolic systems, one can derive

\begin{align*} & \|u^n - u^{n-1}\|_{C(0,T;H^s)}^2 + \|z^n - z^{n-1}\|_{C(0,T;L^2_\theta(H^s))}^2 + \|z_\tau^n - z_\tau^{n-1}\|_{L^2(0,T;H^s)}^2 \\ & \quad\leq \frac{(Ce^{2\gamma T}T)^{n-1}}{(n-1)!}(\|u^{1} - u^{0}\|_{H^s}^2 + \|z^{1} - z^{0}\|_{L^2_\theta(H^s)}^2 ) \end{align*}

for some $C > 0$ and for every $n$. This implies that $(u^n)_n$, $(z^n)_n$, and $(z^{n}_\tau )_n$ are Cauchy sequences in $C(0,\,T;H^s)$, $C(0,\,T; L^2_\theta (H^s))$, and $L^2(0,\,T;H^s)$, respectively.

One can see that the limit of $(u^n,\,z^n)$ is the weak solution of system (4.1). In fact, this follows from

\begin{align*} & \int_0^T (u^n,\mathscr{L}_2^* \phi - P_M\psi_{|\theta = 0})_{L^2}{\rm d} t + \int_0^T (u^n - u^{n-1},P_M\psi_{|\theta = 0})_{L^2}{\rm d} t \\ & \quad+ \int_0^T\!\! \int_{-\tau}^0 (z^n,\mathscr{L}_2^*\psi)_{L^2}{\rm d} \theta {\rm d} t = (u_0,A^0\phi_{|t=0})_{L^2} + \int_{-\tau}^0 (z_0,e^{\varepsilon\theta}P_M\psi_{|t=0})_{L^2} {\rm d} \theta \end{align*}

which holds for every test function $(\phi,\,\psi ) \in H^1((0,\,T)\times \mathbb {R}^d) \times L^2(\mathbb {R}^d;H^1((0,\,T) \times (-\tau,\,0)))$ such that $\phi _{|t=T} = 0$, $\psi _{|t=T} = 0$, and $e^{\varepsilon \tau }M^T \phi = \psi _{|\theta = -\tau }$. Therefore, $u \in C(0,\,T;H^s)$, $z \in C(0,\,T;L^2_\theta (H^s))$ and $z_\tau \in L^2(0,\,T;H^s)$.

Proof. The case $m= 0$ has been already established from the previous theorem, and so, we only consider $m\geq 1$. Initially we have the regularity $u\in C(0,\,T;H^{m+k})$, $z \in C(0,\,T;L^2_\theta (H^{m+k}))$, and $z_\tau \in L^2(0,\,T;H^{m+k})$ according to theorem 4.3, and from the differential equation for $u$, we can see that $u\in H^1(0,\,T;H^{m+k-1})$. The compatibility of the data $(P_Mu,\,e^{\varepsilon \theta }P_Mz_0)$ implies that $z \in C(0,\,T;H^1_\theta (H^{m+k-1})) \cap C^1(0,\,T;L^2_\theta (H^{m+k-1}))$ and $z_\tau \in H^1(0,\,T;H^{m+k-1}) \subset C(0,\,T;H^{m+k-1})$. Consequently, $u\in C^1(0,\,T;H^{m+k-1}) \cap H^2(0,\,T;H^{m+k-2})$ according to the PDE for $u$ once more. Continuing the process, we obtain that $z \in Z_{m,k}$ and $z_\tau \in W_{m,k}$, and consequently, $u \in X_{m,k}$. The energy estimate (4.10) is a direct consequence of (3.14), (4.5), and by making $\gamma$ sufficiently large.

Proof. As mentioned in the preceding section, we can formally take the derivatives of the partial differential equations. We divide the derivation of the energy estimate in several steps.

Step 1. Applying $\partial _x^\ell$ to the PDE for $u$ and then taking the inner product with $G \partial _x^\ell u$ yields

(5.2)\begin{align} & \frac{1}{2}\frac{{\rm d}}{{\rm d}t}\langle GA^0\partial_x^\ell u,\partial_x^\ell u\rangle + \frac{1}{2}\sum_{j=1}^d \partial_{x_j}\langle GA^j\partial_x^\ell u,\partial_x^\ell u\rangle + \langle (GL)_1 \partial_x^\ell u,\partial_x^\ell u\rangle \nonumber\\ & \quad + e^{\varepsilon\tau}\langle GM\partial_x^\ell z_\tau, \partial_x^\ell u \rangle = 0 \end{align}

for $0 \leq \ell \leq s$. By taking the inner product of the transport equation for $z$ with $Nz$ and then integrating over $(-\tau,\,0)$, we obtain

(5.3)\begin{align} & \frac{1}{2}\frac{{\rm d}}{{\rm d}t}\int_{-\tau}^0 \langle N \partial_x^\ell z(\theta),\partial_x^\ell z(\theta)\rangle {\rm d} \theta + \varepsilon\int_{-\tau}^0 \langle N\partial_x^\ell z(\theta),\partial_x^\ell z(\theta)\rangle {\rm d} \theta\nonumber\\ & \quad- \frac{1}{2}\langle NP_M\partial_x^\ell u,P_M\partial_x^\ell u \rangle + \frac{1}{2}\langle N\partial_x^\ell z_\tau,\partial_x^\ell z_\tau\rangle = 0. \end{align}

Getting the sum of (5.2) and (5.3), integrating with respect to $x$, and using the fact that the range of $GM$ is orthogonal to the kernel of $L_1$, we have

(5.4)\begin{align} & \frac{1}{2} \frac{{\rm d}}{{\rm d}t} E_{1,\ell} + \varepsilon(N \partial_x^\ell z,\partial_x^\ell z)_{L^2_\theta(L^2)} \nonumber\\ & \quad+ \frac{1}{2} \int_{\mathbb{R}^d} \langle \Psi(\partial_x^\ell u, \partial_x^\ell z_\tau),(\partial_x^\ell u, \partial_x^\ell z_\tau)\rangle {\rm d} x + (e^{\varepsilon\tau} - 1)(GM\partial_x^\ell z_\tau, \partial_x^\ell v)_{L^2} = 0 \end{align}

where

\[ E_{1,\ell} := (GA^0\partial_x^\ell u,\partial_x^\ell u)_{L^2} + (N \partial_x^\ell z,\partial_x^\ell z)_{L^2_\theta(L^2)}. \]

Using condition (M) on (5.4) and then choosing $\varepsilon$ sufficiently small, we have

(5.5)\begin{equation} \frac{{\rm d}}{{\rm d}t}E_{1\ell} + C(\|\partial_x^\ell v\|^2_{L^2} + \|\partial_x^\ell z\|_{L^2_\theta(L^2)}^2 + \|\partial_x^\ell z_\tau\|_{L^2}^2) \leq 0. \end{equation}

Step 2. The next step is to derive dissipation terms involving $w$. For this purpose, we differentiate $\ell$ times the equation for $u$ and take the inner product with $S^T\partial _x^\ell u$ to obtain

(5.6)\begin{align} & \frac{1}{2}\frac{{\rm d}}{{\rm d}t}\langle SA^0\partial_x^\ell u,\partial_x^\ell u\rangle + \sum_{j=1}^d \biggl(\frac{1}{2}\partial_{x_j}\langle SA^j\partial_x^\ell u,\partial_x^\ell u\rangle + \langle (SA^j)_2 \partial_{x_j}\partial_x^\ell u, \partial_x^\ell u\rangle\biggl) \nonumber\\ & \quad+ \langle (SL)_1 \partial_x^\ell u,\partial_x^\ell u\rangle + e^{\varepsilon\tau}\langle SM\partial_x^\ell z_\tau, \partial_x^\ell w \rangle = 0. \end{align}

Here, we used the fact that the range of $SM$ is orthogonal to the kernel of $L$. The second term in the sum can be rewritten as

(5.7)\begin{equation} \sum_{j = 1}^d \langle (SA^j)_2 \partial_{x_j}\partial_x^\ell u, \partial_x^\ell u\rangle = \sum_{j=1}^d ( \langle Y_j\partial_{x_j}\partial_x^\ell u, \partial_x^\ell u\rangle + \langle (Q^{jT}\varPi_1WR)_2 \partial_{x_j}\partial_x^\ell u, \partial_x^\ell u\rangle)\end{equation}

where

\[ Y_j := (SA^j - Q^{jT}\varPi_1WR)_2, \qquad 1 \leq j \leq d. \]

Integrating both sides with respect to $x$ and then applying Parseval's identity to the first sum on the right-hand side of (5.7), we get

(5.8)\begin{equation} \sum_{j=1}^d ( Y_j \partial_{x_j}\partial_x^\ell u, \partial_x^\ell u)_{L^2} = (|\xi|^{2\ell+1}Y(\omega) \widehat{u},\widehat{u})_{L^2}, \end{equation}

where $\omega := \xi /|\xi |$ and

\[ Y(\omega) := i(SA(\omega)-Q(\omega)^T\varPi_1 WR)_2. \]

According to condition (S)$_s$, (5.8) is nonnegative. On the other hand, using the constraint $\mathscr {L}_3 u = 0$, we obtain

\[ \sum_{j=1}^d\langle (Q^{jT}\varPi_1WR)_2 \partial_{x_j}\partial_x^\ell u, \partial_x^\ell u\rangle = \langle W_1R\partial_x^\ell u, R\partial_x^\ell u\rangle \geq 0 \]

because $W_1$ is nonnegative on the range of $R$. Therefore, integrating (5.6) over $\mathbb {R}^d$, using the above information, and Young's inequality, we have the estimate

(5.9)\begin{equation} \frac{1}{2} \frac{{\rm d}}{{\rm d}t}E_{2,\ell} + ((SL)_1 \partial_x^\ell u,\partial_x^\ell u)_{L^2} - (\eta \|\partial_x^\ell w\|_{L^2}^2 + C_\eta \|\partial_x^\ell z_\tau\|_{L^2}^2)\leq 0 \end{equation}

where $\eta > 0$ and

\[ E_{2,\ell} := (SA^0\partial_x^\ell u,\partial_x^\ell u)_{L^2}. \]

From condition (S), there exist positive constants $C_1$ and $C_2$ such that

\begin{align*} ((SL)_1 \partial_x^\ell u,\partial_x^\ell u)_{L^2} & = ((SL + L)_1 \partial_x^\ell u,\partial_x^\ell u)_{L^2} - (L_1 \partial_x^\ell u,\partial_x^\ell u)_{L^2}\\ & \geq C_1\|\partial_x^\ell w\|_{L^2}^2 - C_2\|\partial_x^\ell v\|^2_{L^2}. \end{align*}

Plugging this estimate to (5.9) and choosing $\eta < C_1$, we obtain

(5.10)\begin{equation} \frac{1}{2} \frac{{\rm d}}{{\rm d}t}E_{2,\ell} + C(\|\partial_x^\ell w\|_{L^2}^2 - \|\partial_x^\ell z_\tau\|_{L^2}^2 - \|\partial_x^\ell v\|^2_{L^2})\leq 0. \end{equation}

Multiplying (5.10) by small enough $\alpha > 0$ and then adding with (5.5), we have

(5.11)\begin{equation} \frac{1}{2}\frac{{\rm d}}{{\rm d}t}(E_{1,\ell} + \alpha E_{2,\ell}) + C_\alpha (\|\partial_x^\ell v\|_{L^2} ^2 + \|\partial_x^\ell w\|_{L^2}^2 + \|\partial_x^\ell z\|_{L^2_\theta(L^2)}^2 + \|\partial_x^\ell z_\tau\|_{L^2}^2) \leq 0 \end{equation}

for some $C_\alpha > 0$ and for all $0 \leq \ell \leq s$.

Step 3. The final step is to derive dissipation terms for the derivatives. Applying $\partial _x^\ell$ to the equation for $u$, taking the $L^2$-inner product with $\sum _{k = 1}^d K^{kT} \partial _{x_k}\partial _{x}^\ell u$, and applying the anti-symmetry of $K^kA^0$, we obtain

(5.12)\begin{align} & \frac{1}{2}\frac{{\rm d}}{{\rm d}t} \sum_{k=1}^d (K^kA^0 \partial_{x}^\ell u, \partial_{x_k}\partial_{x}^\ell u )_{L^2} + \sum_{j,k=1}^d ( K^kA^j \partial_{x_j}\partial_x^\ell u, \partial_{x_k} \partial_x^\ell u )_{L^2}\nonumber\\ & \quad+ \sum_{k=1}^d ( ( K^k L\partial_x^\ell w, \partial_{x_k}\partial_x^\ell u)_{L^2} + e^{-\varepsilon\tau}( K^k M\partial_x^\ell z_\tau, \partial_{x_k}\partial_x^\ell u)_{L^2}) = 0 \end{align}

for every $0\leq \ell \leq s-1$. Let $I_1$ and $I_2$ denote the last two sums in this equation. The term $I_2$ can be estimated as

\[ |I_2| \leq \eta\|\partial_x^{\ell + 1} u\|_{L^2}^2 + C_\eta (\|\partial_x^\ell w\|^2_{L^2} + \|\partial_x^\ell z_\tau\|^2_{L^2}) \]

for every $\eta > 0$.

Now, applying (2.2), the fact that $\widehat {u}(t,\,\xi ) \in \text {Ker}(\varPi _2Q(\omega ))$ for every $t \geq 0$ and for every $\xi \in \mathbb {R}^d \setminus \{0\}$, and using Parseval's identity, we infer that

\begin{align*} I_1 & = \text{Re}(|\xi|^{2\ell+2} K(\omega)A(\omega) \widehat{u},\widehat{u})_{L^2}\\ & = (|\xi|^{2\ell+2}(K(\omega)A(\omega) + \vartheta (SL+L))_1\widehat{u},\widehat{u})_{L^2} - \vartheta (|\xi|^{2\ell+2}(SL+L)_1\widehat{w},\widehat{w})_{L^2}\\ & \geq C_1 \||\xi|^{2\ell+2}\widehat{u}\|_{L^2}^2 - C_2\||\xi|^{2\ell+2} \widehat{w}\|^2_{L^2} \end{align*}

for some constants $C_1,\,C_2 > 0$ and $\vartheta$ is the constant in (2.2). Here, $K(\omega ) := \sum _{k =1}^d K^k \omega _k$. Using Plancherel's identity to the latter terms and then combining the above estimates, we obtain from (5.12)

(5.13)\begin{equation} \frac{1}{2}\frac{{\rm d}}{{\rm d}t} E_{3,\ell} + C_\eta( \|\partial_x^{\ell +1}u\|_{L^2}^2 - \|\partial_x^\ell w\|^2_{H^1} - \|\partial_x^\ell z_\tau\|_{L^2}^2) \leq 0 \end{equation}

by choosing $\eta > 0$ small enough, where

\[ E_{3,\ell} := \sum_{k=1}^d (K^kA^0 \partial_{x}^\ell u, \partial_{x_k}\partial_{x}^\ell u )_{L^2}. \]

Multiplying (5.13) by $\beta > 0$ small enough and then adding the result to (5.11) yields, for $0 \leq \ell \leq s-1$, the estimate

(5.14)\begin{align} & \frac{1}{2}\frac{{\rm d}}{{\rm d}t} (E_{1,\ell} + E_{1,\ell + 1} + \alpha (E_{2,\ell} + E_{2,\ell + 1}) + \beta E_{3,\ell}) \nonumber\\ & \quad+ C(\|\partial_x^{\ell +1}u\|_{L^2}^2 + \|\partial_x^\ell v\|_{H^1} ^2+ \|\partial_x^\ell w\|_{H^1}^2 + \|\partial_x^\ell z\|_{L^2_\theta(H^1)}^2 + \|\partial_x^\ell z_\tau\|_{H^1}^2) \leq 0 \end{align}

for some constant $C = C_{\alpha,\beta,\eta } > 0$. By reducing $\alpha > 0$ and then $\beta > 0$ if necessary, we can see that there are constants $C_1,\,C_2 > 0$ such that

(5.15)\begin{align} C_1(\|\partial_x^\ell u\|_{H^1}^2 + \|\partial_x^\ell z\|_{L^2_\theta(H^1)}^2) & \leq E_{1,\ell} + E_{1,\ell + 1} + \alpha (E_{2,\ell} + E_{2,\ell + 1}) + \beta E_{3,\ell} \nonumber\\ & \leq C_2(\|\partial_x^\ell u\|_{H^1}^2 + \|\partial_x^\ell z\|_{L^2_\theta(H^1)}^2). \end{align}

Taking the sum of (5.14) for $0\leq \ell \leq s-1$, integrating with respect to time, and then using the equivalence (5.15), we acquire the estimate in the theorem.

Proof. First, since $Z_{m,k} \subset Z_{0,s} = L^2_\theta (H^s)$ it follows that (5.1) holds. The next step is to obtain an estimate for the time derivatives of $u$. Taking the $(\ell -1)$st derivative with respect to $t$ of the equation for $u$, for $1\leq \ell \leq m$, we have

(5.17)\begin{equation} \partial_t^\ell u ={-} (A^0)^{{-}1}\sum_{j = 1}^d A^j \partial_{x_j} \partial_t^{\ell-1} u - (A^0)^{{-}1} L\partial_t^{\ell-1}w - e^{\varepsilon\tau} (A^0)^{{-}1} M\partial_t^{\ell-1}z_\tau. \end{equation}

Using an induction argument, it can be easily seen that the estimate

(5.18)\begin{equation} \|\partial_t^\ell u(t)\|^2_{H^{s-\ell}} \leq C \biggl( \|\partial_x u(t)\|^2_{H^{s-1}} + \|w(t)\|^2_{H^{s-1}} +\sum_{j = 0}^{\ell - 1} \|\partial_t^j z_\tau(t)\|_{H^{s-j-1}}^2 \biggl) \end{equation}

holds for every $1 \leq \ell \leq m$ and $t\geq 0$.

On the other hand, by applying $\partial _t^\ell \partial _x^\nu$ to the transport equation for $z$ and multiplying with $\partial _t^\ell \partial _x^\nu z$, we have

(5.19)\begin{equation} \frac{1}{2}\partial_t (|\partial_t^\ell \partial_x^\nu z|^2) - \frac{1}{2}\partial_\theta(|\partial_t^\ell \partial_x^\nu z|^2) + \varepsilon|\partial_t^\ell \partial_x^\nu z|^2 = 0 \end{equation}

for every $0 \leq \ell \leq m$ and $0\leq \nu \leq m - \ell$. From the boundary condition at $\theta = 0$ and the fact that $\text {Ker}(L_1) \subset \text {Ker}(M)$, we have $z_{|\theta = 0} =P_Mu = P_M( v + (I - P_{L_1})u) = P_Mv$. Integrating (5.19) over $(0,\,t) \times (-\tau,\,0) \times \mathbb {R}^d$ and then taking the sum over all $0 \leq \nu \leq s - \ell$ produce the estimate

(5.20)\begin{align} & \|\partial_t^\ell z(t)\|^2_{L_\theta^2(H^{s-\ell})} + \varepsilon\int_0^t \|\partial_t^\ell z(\sigma)\|^2_{L_\theta^2(H^{s-\ell})} {\rm d} \sigma + \int_0^t \|\partial_t^\ell z_\tau(\sigma)\|_{H^{s-\ell}}^2 {\rm d} \sigma \nonumber\\ & \quad\leq C \biggl(\|\partial_t^\ell z(0)\|_{L_\theta^2(H^{s-\ell})}^2 + \int_0^t \|\partial_t^\ell v(\sigma)\|_{H^{s-\ell}}^2 {\rm d} \sigma \biggl) \end{align}

for every $0 \leq \ell \leq m$. We note that for every $0 \leq \ell \leq m$ it holds that $\|\partial _t^\ell z(0)\|_{L_\theta ^2(H^{s-\ell })} \leq \|z_0\|_{Z_{m,k}}$.

We establish by strong induction the following estimate for $0 \leq \ell \leq m$ and $t \geq 0$

(5.21)\begin{equation} \|\partial_t^\ell z(t)\|^2_{L_\theta^2(H^{s-\ell})} + \int_0^t \|\partial_t^\ell z(\sigma)\|^2_{L_\theta^2(H^{s-\ell})} {\rm d} \sigma + \int_0^t \|\partial_t^\ell z_\tau(\sigma)\|_{H^{s-\ell}}^2 {\rm d} \sigma \leq CI_0^2. \end{equation}

The case $\ell = 0$ has been already established in theorem 5.1. Suppose that (5.21) holds for every $0,\, 1,\,\ldots,\,\ell -1$. Applying (5.1), (5.18), and the induction hypothesis, one has

\begin{align*} & \int_0^t \|\partial_t^\ell v(\sigma)\|_{H^{s-\ell}}^2 {\rm d} \sigma \\ & \quad\leq C \int_0^t \biggl( \|\partial_x u(\sigma)\|^2_{H^{s-1}} +\|w(\sigma)\|^2_{H^{s-1}} + \sum_{j = 0}^{\ell - 1} \|\partial_t^j z_\tau(\sigma)\|_{H^{s-j-1}}^2 \biggl) {\rm d} \sigma \leq CI_0^2.\qquad \end{align*}

Plugging this in inequality (5.20) proves (5.21).

Similarly, using a strong induction argument and the equation $\partial _\theta z = \partial _t z + \varepsilon z$, we can obtain estimates involving derivatives with respect to $\theta$. More precisely, we have

(5.22)\begin{equation} \|\partial_t^\ell \partial_\theta^\mu z(t)\|^2_{L_\theta^2(H^{s-\ell- \mu})} + \int_0^t \|\partial_t^\ell\partial_\theta^\mu z(\sigma)\|^2_{L_\theta^2(H^{s-\ell-\mu})} {\rm d} \sigma \leq CI_0^2 \end{equation}

for every $0 \leq \ell + \mu \leq m$. Given $0 \leq \ell \leq m$, taking the sum over all $0 \leq \mu \leq m-\ell$ in (5.22) results to

(5.23)\begin{equation} \sum_{\mu = 0}^{m-\ell} \biggl( \|\partial_t^\ell z(t)\|^2_{H_\theta^\mu(H^{s-\ell - \mu})} + \int_0^t \|\partial_t^\ell z(\sigma)\|^2_{H_\theta^\mu(H^{s-\ell - \mu})} {\rm d} \sigma\biggl) \, \leq CI_0^2. \end{equation}

Combining (5.21) and (5.23), and using the definition of $Z_{m,k}$, we obtain

\[ \sum_{\ell = 0}^m \, \biggl[\|\partial_t^\ell z(t)\|_{Z_{m-\ell,k}}^2 + \int_0^t \left( \|\partial_t^\ell z_\tau(\sigma)\|_{H^{s-\ell}}^2 + \|\partial_t^\ell z(\sigma)\|_{Z_{m-\ell,k}}^2 \right){\rm d} \sigma\biggl] \ \leq\, CI_0^2. \]

By the Poincaré inequality we have, for every $0 \leq j \leq m -1$,

\[ \|\partial_t^j z_\tau(t)\|_{H^{s-j-1}} \leq C\|\partial_t^j z(t)\|_{H_\theta^1(H^{s-j-1})}. \]

Utilizing this estimate in (5.18) together with (5.1) and (5.23), we have the other part of the desired estimate

\[ \sum_{\ell=1}^m \, \biggl[ \|\partial_t^\ell u(t)\|_{H^{s-\ell}}^2 + \int_0^t \|\partial_t^\ell u(\sigma)\|_{H^{s-\ell}}^2{\rm d} \sigma \biggl] \leq CI_0^2. \]

This completes the proof of the theorem.

Proof. First, let us prove the uniform decay (5.24). To do this, we introduce the functional $\Phi _1(t) := \|\partial _x\partial _t^\ell u(t)\|^2_{H^{s-\ell -2}}$. According to (5.1) and (5.16), we have $\Phi _1 \in W^{1,1}(0,\,\infty )$ and therefore $\Phi _1(t) \to 0$ as $t \to \infty$. Let $r = d/(2s_0)$. Then, $r = d/(d+2)$ if $d$ is even and $r = d/(d+1)$ if $d$ is odd, and in any case, we have $r \in [{1}/{2},\,1)$. In virtue of the Gagliardo–Nirenberg inequality [Reference Nirenberg20], we get

\begin{align*} \|\partial_t^\ell u(t)\|_{W^{s-s_0-\ell-1,\infty}} & \leq C\|\partial_x \partial_t^\ell u(t)\|_{H^{s-\ell-2}}^r\|\partial_t^\ell u(t)\|_{H^{s-s_0 - \ell-1}}^{1-r} \\ & \leq CI_0^{1-r}\|\partial_x \partial_t^\ell u(t)\|_{H^{s-\ell-2}}^r \to 0 \end{align*}

as $t \to \infty$. To prove (5.25) and (5.26), we consider the functional

\[ \Phi_2(t) := \|\partial_t^\ell (v,w)(t)\|_{H^{s-\ell-1}}^2 + \sum_{j = 0}^{m-\ell} \|\partial_t^\ell z(t)\|^2_{H_\theta^j(H^{s-\ell-j-1})}. \]

From the energy estimates (5.1) and (5.16) we can see that $\Phi _2 \in W^{1,1}(0,\,\infty )$, and hence, $\Phi _2(t) \to 0$ as $t \to \infty$. The Gagliardo–Nirenberg inequality once more imply (5.25), and similarly

(5.28)\begin{align} & \|\partial_t^\ell \partial_\theta^\mu z(t,\theta)\|_{L_\theta^2(W^{s-s_0-\ell - j,\infty})}^2 \nonumber\\ & \quad\leq C\int_{-\tau}^0 \|\partial_t^\ell \partial_\theta^\mu \partial_x z(t,\theta)\|_{H^{s-\ell-j-1}}^{2r}\|\partial_t^\ell \partial_\theta^\mu z(t,\theta)\|_{H^{s-s_0-\ell - j}}^{2(1-r)}{\rm d} \theta \end{align}

for every $0\leq \mu \leq j$ and $0\leq j \leq m-\ell$. Applying Hölder's inequality and using the fact that $r < 1$, we get

\begin{align*} \|\partial_t^\ell \partial_\theta^\mu z(t)\|_{L_\theta^2(W^{s-s_0-\ell - j,\infty})} & \leq C\|\partial_t^\ell \partial_\theta^\mu z(t)\|_{L_\theta^2(H^{s-\ell-j})}^{r} \|\partial_t^\ell \partial_\theta^\mu z(t)\|_{L_\theta^2(H^{s-s_0-\ell - j})}^{1-r}\\ & \leq CI_0^{r}\|\partial_t^\ell \partial_\theta^\mu z(t)\|_{L_\theta^{2}(H^{s-\ell-j-1})}^{1-r}\\ & \leq CI_0^{r} \|\partial_t^\ell z(t)\|_{H_\theta^j(H^{s-\ell-j-1})}^{1-r} \end{align*}

for every $0\leq \mu \leq j$. Passing to the limit $t \to \infty$, this estimate imply (5.26). Finally, (5.27) is a consequence of (5.26) and the Sobolev embedding.

Proof. Multiplying (5.11) by $(1+t)^j$, integrating with respect to $t$, and then taking the sum of the corresponding inequalities for $j \leq \ell \leq s-j$, we obtain

\begin{align*} & (1+t)^j (\|\partial_x^j u(t)\|_{H^{s-j}}^2 + \|\partial_x^j z(t)\|^2_{L_\theta^2(H^{s-j})}) \\ & \quad+ \int_0^t (1+ \sigma)^j (\|\partial_x^j (v,w,z_\tau)(\sigma)\|_{H^{s-j}}^2 + \|\partial_x^j z(\sigma)\|^2_{L_\theta^2(H^{s-j})}) {\rm d} \sigma \\ & \leq CI_0^2 + Cj \int_0^t (1+\sigma)^{j-1}(\|\partial_x^j u(\sigma)\|_{H^{s-j}}^2 + \|\partial_x^j z(\sigma)\|^2_{L_\theta^2(H^{s-j})}) {\rm d} \sigma \end{align*}

for every $0 \leq j \leq s$. On the other hand, if we multiply (5.14) by $(1+ t)^{j}$, integrate from $0$ to $t$, and then take the sum for every $j \leq \ell \leq s-j-1$, we get

\begin{align*} & \int_0^t (1+ \sigma)^j(\|\partial_x^{j+1}u(\sigma) \|_{H^{s-j-1}}^2 + \|\partial_x^{j+1} z(\sigma)\|^2_{L_\theta^2(H^{s-j-1})}){\rm d} \sigma \nonumber\\ & \quad\leq CI_0^2 + Cj \int_0^t (1+\sigma)^{j-1} (\|\partial_x^{j+1} u(\sigma)\|_{H^{s-j-1}}^2 + \|\partial_x^{j+1} z(\sigma)\|^2_{L_\theta^2(H^{s-j-1})}){\rm d} \sigma. \end{align*}

Induction argument yields that these estimates imply (5.33) and (5.34).

Proof. Taking the inner product of the differential equation for $u$ with $G\partial _x^\ell w = GP_L\partial _x^\ell u$ and then using the fact that $P_L$ and $GA^0$ commute, we have

(5.36)\begin{equation} \frac{1}{2} \frac{{\rm d}}{{\rm d}t} \langle GA^0\partial_x^\ell w,\partial_x^\ell w \rangle + \langle (GL)_1\partial_x^\ell w, \partial_x^\ell w\rangle + e^{\varepsilon\tau}\langle GM\partial_x^\ell z_\tau, \partial_x^\ell w \rangle = R_{1\ell}, \end{equation}

where $R_{1\ell } := - \sum _{k= 1}^d \langle GA^k\partial _{x_k}\partial _x^\ell u,\, \partial _x^\ell w \rangle$. We claim that $P_{L_1}w = v$. Indeed, since $(I-P_L)u$ is in the kernel of $L$, and hence in the kernel of $L_1$, we have $P_{L_1} (u - P_Lu) = 0$. Using the definition of $w$ and $v$, this implies our claim. Since $P_{(GL)_1} = P_{L_1}$ and the range of $GM$ is orthogonal to the kernel of $L_1$, equation (5.36) can be written as

(5.37)\begin{equation} \frac{1}{2} \frac{{\rm d}}{{\rm d}t} \langle GA^0\partial_x^\ell w,\partial_x^\ell w \rangle + \langle (GL)_1\partial_x^\ell v, \partial_x^\ell v\rangle + e^{\varepsilon\tau}\langle GM \partial_x^\ell z_\tau, \partial_x^\ell v \rangle = R_{1\ell}. \end{equation}

Similarly, from the fact that $P_L$ and $SA^0$ commute, we obtain by multiplying the equation for $u$ by $S^T \partial _x^\ell w$

(5.38)\begin{equation} \frac{1}{2} \frac{{\rm d}}{{\rm d}t} \langle SA^0\partial_x^\ell w,\partial_x^\ell w \rangle + \langle (SL)_1\partial_x^\ell w,\partial_x^\ell w\rangle = R_{2\ell}, \end{equation}

where $R_{2\ell } := - e^{\varepsilon \tau }\langle SM \partial _x^\ell z_\tau,\, \partial _x^\ell w \rangle - \sum _{l = 1}^d \langle SA^k\partial _{x_k}\partial _x^\ell u,\, \partial _x^\ell w \rangle$. Here, we use the fact that $SLu = SLw$. Multiplying (5.38) by $\alpha$, taking the sum with (5.37) and (5.3), and then the sum for all $j \leq \ell \leq s-j-1$, we obtain

\begin{align*} \frac{1}{2}\frac{{\rm d}}{{\rm d}t} \widetilde{E}_j & + \varepsilon (N\partial_x^{j}z,\partial_x^j z)_{L_\theta^2(H^{s-j-1})} + \frac{1}{2}\sum_{\ell = j}^{s-j-1} \int_{\mathbb{R}^d} \langle \Psi(\partial_x^\ell v,\partial_x^\ell z_\tau), (\partial_x^\ell v,\partial_x^\ell z_\tau) \rangle {\rm d} x\\ & \quad+ (e^{\varepsilon\tau}-1)(GM\partial_x^j z_\tau, \partial_x^j v)_{H^{s-j-1}} + \alpha ((SL)_1 \partial_x^{j}w,\partial_x^j w)_{H^{s-j-1}} = R_j, \end{align*}

where $\widetilde {E}_j := ((GA^0 + \alpha SA^0) \partial _x^\ell w,\,\partial _x^\ell w)_{H^{s-j-1}} + (N \partial _x^\ell z,\, \partial _x^\ell z)_{L_\theta ^2(H^{s-j-1})}$ and the right-hand side is given by $R_j := \sum _{\ell = j}^{s-j-1} (R_{1\ell } + \alpha R_{2\ell })$.

Applying the Cauchy–Schwarz inequality, using condition (M), and then making $\alpha$ and $\varepsilon$ small enough, we can see that there exist positive constants $c$ and $C$ such that

\[ \frac{{\rm d}}{{\rm d}t}\widetilde{E}_j + c\widetilde{E}_j \leq C \|\partial_x^{j+1} u(t)\|_{H^{s-j-1}}^2. \]

Multiplying both sides by $e^{ct}$ and then integrating from $0$ to $t$ yields

\begin{align*} \widetilde{E}_j(t) & \leq \widetilde{E}_j(0)e^{{-}ct} + C\int_0^t e^{{-}c(t-\sigma)} \|\partial_x^{j+1} u(\sigma)\|_{H^{s-j-1}}^2 {\rm d} \sigma\\ & \leq \widetilde{E}_j(0)e^{{-}ct} + C\int_0^t e^{{-}c(t-\sigma)}(1+ \sigma)^{{-}j-1}{\rm d} \sigma\\ & \leq C(1+ t)^{{-}j-1}. \end{align*}

This estimate implies (5.35) and this completes the proof of the theorem.

Proof. Applying $\partial _x^\nu$ to (5.17) and then taking the sum over $j \leq \nu \leq s - \ell -j$ yields

(5.43)\begin{align} \|\partial_t^\ell \partial_x^j u(t)\|^2_{H^{s-\ell - j}} & \leq C\|\partial_t^{\ell-1}\partial_x^{j+1} u(t)\|^2_{H^{s-\ell -j-1}} + C\|\partial_t^{\ell-1}\partial_x^j w(t)\|^2_{H^{s-\ell-j}} \nonumber\\ & \quad + C\|\partial_t^{\ell-1} \partial_x^j z_\tau(t)\|^2_{H^{s-\ell - j}} . \end{align}

Multiplying equation (5.19) by $(1 + t)^j$, integrating with respect to $(t,\,\theta,\,x)$, and then getting the sum for $j \leq \nu \leq s-\ell - j$, we have

(5.44)\begin{align} & (1+t)^j \|\partial_t^\ell \partial_x^jz(t)\|_{L_\theta^2(H^{s-\ell - j})}^2 + \int_0^t (1+\sigma)^j \|\partial_t^\ell \partial_x^jz(\sigma)\|_{L_\theta^2(H^{s-\ell - j})}^2 {\rm d} \sigma \nonumber\\ & \quad+ \int_0^t (1+\sigma)^j \|\partial_t^\ell \partial_x^jz_\tau(\sigma)\|_{H^{s-\ell - j}}^2 {\rm d} \sigma \leq C I_0^2 + C\int_0^t (1+\sigma)^j \|\partial_t^\ell \partial_x^jv(\sigma)\|_{H^{s-\ell - j}}^2 {\rm d} \sigma \nonumber\\ & \quad+ Cj \int_0^t (1+\sigma)^{j-1} \|\partial_t^\ell \partial_x^jz(\sigma)\|_{L_\theta^2(H^{s-\ell - j})}^2 {\rm d} \sigma . \end{align}

The estimate involving $z$ in (5.39) when $\nu = 0$ can be obtained from (5.43) and (5.44), and when $\nu > 0$ we use the equation $\partial _\theta z = \partial _t z + \varepsilon z$ together with strong induction. The constraint and (5.39) immediately imply (5.40). Finally, (5.41) and (5.42) can be shown using the same argument as in the preceding theorem.

Proof. Note that inequality (5.5) is still satisfied. The next step is to revise the estimation of Step 2 in the proof of theorem 5.1. Since $Y(\omega )$ is nonnegative only on the kernel of $L_1$, we need to proceed in a different way to treat the first term on the right-hand side of (5.7). Recall that $Y_j := (SA^j - Q^{jT}\varPi _1WR)_2$. We rewrite the said term as follows:

(6.2)\begin{align} (Y_j\partial_{x_j}\partial_x^\ell u, \partial_x^\ell u)_{L^2} & = (Y_j\partial_{x_j}\partial_x^\ell (u-v), \partial_x^\ell (u-v))_{L^2} - (Y_j\partial_x^\ell v,\partial_{x_j} \partial_x^\ell u)_{L^2} \nonumber\\ & \quad+ (Y_j\partial_{x_j}\partial_x^\ell (u-v), \partial_x^\ell v)_{L^2}. \end{align}

Applying Parseval's identity and the fact that $\widehat {u}(\xi ) - \widehat {v}(\xi ) \in \text {Ker}(L_1)$ for every $\xi$, we obtain from condition (S)$_r$ that

\[ \sum_{j=1}^d ( Y_j\partial_{x_j}\partial_x^\ell (u-v), \partial_x^\ell (u-v))_{L^2} = (|\xi|^{2\ell + 1}Y(\omega)(\widehat{u} - \widehat{v}), \widehat{u} - \widehat{v})_{L^2} \geq 0. \]

We apply Young's inequality to estimate the last two terms on the right-hand side of (6.2) as follows:

\[ \sum_{j = 1}^d |( Y_j\partial_x^\ell v, \partial_{x_j}\partial_x^\ell u)_{L^2}| + |( Y_j\partial_{x_j}\partial_x^\ell (u-v), \partial_x^\ell v)_{L^2}| \leq \frac{\varrho}{2}\|\partial_x^{\ell+1}u\|_{L^2}^2 + C_{\varrho}\|\partial_x^{\ell} v\|_{H^1}^2 \]

for every $0 \leq \ell \leq s-1$ and $\varrho > 0$. Plugging these estimates to (5.6) and (5.7), we obtain

(6.3)\begin{equation} \frac{1}{2}\frac{{\rm d}}{{\rm d}t}E_{2,\ell} + C(\|\partial_x^\ell w\|_{L^2}^2 - \|\partial_x^\ell z_\tau\|_{L^2}^2) - C_{\varrho} \|\partial_x^{\ell} v\|_{H^1}^2 - \frac{\varrho}{2}\|\partial_x^{\ell +1}u\|_{L^2}^2 \leq 0 \end{equation}

for every $0 \leq \ell \leq s-1$. We perform a similar procedure and integrate by parts to pass the derivatives on $v$ to get

(6.4)\begin{align} \frac{1}{2}\frac{{\rm d}}{{\rm d}t}E_{2,\ell+1} + C(\|\partial_x^{\ell+1} w\|_{L^2}^2 - \|\partial_x^{\ell+1} z_\tau\|_{L^2}^2) - C_{\varrho} \|\partial_x^{\ell + 1} v\|_{H^1}^2 - \frac{\varrho}{2}\|\partial_x^{\ell +1}u\|_{L^2}^2 \leq 0 \end{align}

for every $0 \leq \ell \leq s-2$.

Combining estimates (5.5), (5.14), (6.3), and (6.4), we obtain the following:

(6.5)\begin{align} & \frac{1}{2}\frac{{\rm d}}{{\rm d}t}\widetilde{E}_\ell + (C- \alpha C_\varrho)\|\partial_x^{\ell}v\|^2_{H^2} + (C- \alpha C - \alpha \beta C_\eta)\|\partial_x^\ell z_\tau\|^2_{H^2}\nonumber\\ & \quad+ \alpha( C - \beta C_\eta)\|\partial_x^\ell w\|_{H^1}^2 + \alpha(\beta C_\eta - \varrho)\|\partial_x^{\ell+1} u\|_{L^2}^2 + C\|\partial_x^\ell z\|^2_{L_\theta^2(H^2)}\leq 0 \end{align}

for every $0 \leq \ell \leq s-2$, where we used the abbreviation

\[ \widetilde{E}_\ell := E_{1,\ell} + E_{1,\ell+1} + E_{1,\ell+2} + \alpha(E_{2,\ell} + E_{2,\ell+1}) + \alpha\beta E_{3,\ell}. \]

Choosing the positive constants $\varrho,$ $\beta$, and $\alpha$ in such a way that $\beta < C/C_\eta$, $\varrho < \beta C_{\eta }$, and $\alpha < \max (C/C_{\varrho },\, C/(C + \beta C_\eta ))$, we can see that every constants appearing on the left-hand side of inequality (6.5) are positive. Integrating this inequality with respect to $t$ and taking the sum for all $0\leq \ell \leq s-2$, we get the desired inequality stated in the theorem after reducing $\alpha$ and then $\beta$ if necessary.

Proof. Given $0\leq j \leq [{s}/{2}]$, we multiply (5.5) by $(1+t)^j$, take the sum over all $j \leq \ell \leq s-j$, and integrate with respect to time to obtain the time-weighted inequality

\begin{align*} & (1+t)^j(\|\partial_x^j u(t)\|^2_{H^{s-2j}} + \|\partial_x^j z(t)\|^2_{L_\theta^2(H^{s-2j})}) \\ & \qquad+ \int_0^t (1+\sigma)^j(\|\partial_x^j(v,z_\tau)(\sigma)\|_{H^{s-2j}}^2 + \|\partial_x^j z(\sigma)\|^2_{L_\theta^2(H^{s-2j})}){\rm d} \sigma \\ & \quad\leq CI_0^2 + Cj\int_0^t (1+\sigma)^{j-1}(\|\partial_x^j u(\sigma)\|^2_{H^{s-2j}} + \|\partial_x^j z(\sigma)\|^2_{L_\theta^2(H^{s-2j})}) {\rm d} \sigma. \end{align*}

Now, given $0 \leq j < [{s}/{2}]$ we multiply (6.5) by $(1+t)^j$ and then add the corresponding terms for $j \leq \ell \leq s-j-2$ to have the estimate

\begin{align*} & \int_0^t (1+\sigma)^j (\|\partial_x^{j}w(\sigma)\|_{H^{s-2j-1}}^2 + \|\partial_x^{j+1}u(t)\|_{H^{s-2j-2}}^2) {\rm d} \sigma \\ & \quad\leq CI_0^2 + Cj\int_0^t (1+\sigma)^{j-1}(\|\partial_x^j u(\sigma)\|^2_{H^{s-2j}} + \|\partial_x^j z(\sigma)\|^2_{L_\theta^2(H^{s-2j})}) {\rm d} \sigma. \end{align*}

A straightforward induction argument shows that these two estimates imply $N_r(t)^2 + D_r(t)^2 \leq CI_0^2$ for every $t \geq 0$ and for some constant $C > 0$. The rest of the theorem follows from the latter estimate and the constraint.