Proof. We first consider the cases $\mathcal{X} = Y_0$ or $\mathcal{X} = L^2(\mathbb{R})$. Let $Z_m$ denote either the space $Y_m$ or $H^m(\mathbb{R})$ for $m \in \mathbb{N}_0$. We show that $\Lambda$ is a sectorial operator on $Z_0$ and $\mathcal{N}$ is a locally Lipschitz continuous nonlinearity on $Z_0$. Then, the existence of a maximal mild solution to (3.2) follows from standard semigroup theory for semilinear parabolic problems.

We observe that the elliptic operator $\partial_x^2$ acts on $Z_0$ with dense domain $Z_2$. Thus, the preconditioner $(1-\partial_x^2)^{-1}$ is a bounded linear operator from $Z_0$ into $Z_2$. We aim to show that $\Lambda$ is a sectorial operator on $Z_0$ by regarding $\Lambda$ as a bounded perturbation of the operator $\Lambda_0$ given by

\begin{align*} \Lambda_0 = \begin{pmatrix} \frac{\alpha}{2} \partial_x^2 & 1-\partial_x^2 \\ \frac{\alpha^2}{4} \partial_x^4 \left(1-\partial_x^2\right)^{-1} & \frac{\alpha}{2} \partial_x^2\end{pmatrix}. \end{align*}

The spectrum of the constant-coefficient operator $\Lambda_0$ is determined by the eigenvalues $\lambda_{0,\pm}(k)$ of its Fourier symbol

\begin{align*} \widehat{\Lambda}_0(k) = \begin{pmatrix} -\frac{\alpha}{2} k^2 & 1+k^2 \\ \frac{\alpha^2}{4} \frac{k^4}{1+k^2} &- \frac{\alpha}{2} k^2\end{pmatrix}, \end{align*}

which are given by

\begin{align*} \lambda_{0,-}(k) = -\alpha k^2, \qquad \lambda_{0,+}(k) = 0. \end{align*}

In particular, for $k \in \mathbb{R} \setminus \{0\}$, the matrix $\hat{\Lambda}_0(k)$ is diagonalizable. For later use, we note that the associated change of basis is represented by a matrix $S(k)$, whose columns are comprised of the eigenvectors of $\hat{\Lambda}_0(k)$, and its inverse, which are given by

\begin{align*} S(k) = \begin{pmatrix} \frac{2 \left(k^2+1\right)}{\alpha k^2} & -\frac{2 \left(k^2+1\right)}{\alpha k^2} \\ 1 & 1 \end{pmatrix}, \qquad S(k)^{-1} = \begin{pmatrix} \frac{\alpha k^2}{4 k^2+4} & \frac{1}{2} \\ -\frac{\alpha k^2}{4 k^2+4} & \frac{1}{2} \end{pmatrix}, \end{align*}

for $k \in \mathbb{R} \setminus \{0\}$. That is, we have $S(k)^{-1} \hat{\Lambda}_0(k) S(k) = \mathrm{diag}(0,-\alpha k^2)$ for $k \in \mathbb{R} \setminus \{0\}$. One readily observes that the coefficients of $S(\cdot)$ and $S(\cdot)^{-1}$ are bounded on $\mathbb{R} \setminus (-1,1)$.

Thus, we find $\sigma(\Lambda_0) = (-\infty,0]$. So, the resolvent set $\rho(\Lambda_0)$ contains the sector $\Sigma_0 = \{\lambda \in \mathbb{C} : \lambda \neq 1, |\mathrm{arg}(\lambda - 1)| \leq \frac{3\pi}{4}\}$. The resolvent $(\Lambda_0 - \lambda)^{-1}$ possesses the Fourier symbol

\begin{align*} \frac{1}{\lambda}\begin{pmatrix} -1 + \frac{\alpha k^2}{2(\alpha k^2 + \lambda)} & -\frac{1+k^2}{\alpha k^2 + \lambda} \\ -\frac{\alpha^2 k^4}{4\left(1+k^2\right)\left(\alpha k^2 + \lambda\right)} & -1 + \frac{\alpha k^2}{2(\alpha k^2 + \lambda)} \end{pmatrix}, \end{align*}

for $\lambda \in \Sigma_0$. For $\lambda \in \Sigma_0$ and $k \in \mathbb{R}$, we have the basic inequalities

\begin{align*} \left|1 + \frac{\lambda}{\alpha k^2}\right| \geq \frac{1}{\sqrt{2}}, \quad \ \ \left|\alpha k^2 + \lambda\right| \geq \frac{1}{\sqrt{2}}, \quad\ \ \frac{k^2}{1+k^2} \leq 1, \quad\ \ |\lambda - 1| \leq |\lambda| + 1 \leq |\lambda|(1+\sqrt{2}). \end{align*}

Hence, there exists a constant $M \gt 0$ such that

\begin{align*} \left\|(\Lambda_0 - \lambda)^{-1}\right\|_{Y_0} \leq \frac{M}{|\lambda - 1|}, \end{align*}

for all $\lambda \in \Sigma_0$. We conclude that $\Lambda_0$ is a sectorial operator on $Z_0$. By standard perturbation theory of sectorial operators [Reference Lunardi31, Proposition 2.4.1], it follows that $\Lambda$ is a sectorial operator on $Z_0$, since $\Lambda$ is a bounded perturbation of $\Lambda_0$. In addition, since the domain $D(\Lambda) = \{U \in Z_0 : \Lambda U \in Z_0\}$ obviously contains the dense subspace $C_c^\infty(\mathbb{R}) \subset Z_0$ of all test functions, $\Lambda$ is densely defined.

Furthermore, $\mathcal{N}$ is locally Lipschitz continuous on $Z_0$, because $N$ is smooth, the space $Z_1$ continuously embeds into $L^\infty(\mathbb{R})$ and $\smash{(1-\partial_x^2)^{-1}}$ is a bounded linear operator from $Z_0$ into $Z_2$. Combining the latter with the fact that $\Lambda$ is sectorial and densely defined on $Z_0$, it follows by standard local existence theory [Reference Lunardi31, Theorem 7.1.2 and Proposition 7.1.7] for semilinear parabolic equations that there exist $T_{\max} \in (0,\infty]$ and a unique, maximally defined, mild solution $U \in C\big([0,T_{\max}),Z_0\big)$ of (3.2) with initial condition $U(0) = U_0$. If $T_{\max} \lt \infty$, then we have (3.3). This establishes the result for the case $\mathcal{X} = Y_0$ or $\mathcal{X} = L^2(\mathbb{R})$.

Next, we consider the case $\mathcal{X} = X_0$. We introduce the tailor made variable $V = x U$ and observe that $(U,V)$ satisfies the four-component system

(3.4)\begin{equation} \begin{aligned} U_t &= \Lambda U + \mathcal{N}(U),\\ V_t &= \Lambda V + \widetilde{\Lambda} U + \widetilde{\mathcal{N}}(U,V) \end{aligned} \end{equation}

where $\widetilde{\Lambda}$ is the linear operator with Fourier symbol

\begin{align*} \mathrm{i} \hat{\Lambda}'(k) = \begin{pmatrix} -\alpha \mathrm{i} k & 2\mathrm{i} k \\ \frac{\alpha^2 k^3\left(2+k^2\right)}{2\left(1+k^2\right)^2} & -\alpha \mathrm{i} k \end{pmatrix}, \end{align*}

and we denote

\begin{align*} \widetilde{\mathcal{N}}(U,V) = \left(1-\partial_x^2\right)^{-1}\left(V_1\frac{N(U_1)}{U_1} \mathbf{e}_2\right) - 2 \partial_x \left(1-\partial_x^2\right)^{-2}\left(N(U_1) \mathbf{e}_2\right). \end{align*}

Since $\Lambda$ is sectorial and densely defined on $Y_0$, the operator $\mathcal{L}$ on the product space $Y_0 \times Y_0$ given by $(U,V) \mapsto (\Lambda U, \Lambda V)$ is also sectorial and densely defined. Next, we show that, for any $\epsilon \gt 0$, the operator $(U,V) \mapsto (0,\widetilde{\Lambda}U)$ is relatively bounded with respect to $\mathcal{L}$ with $\mathcal{L}$-bound $\epsilon \gt 0$. Firstly, we compute

\begin{align*} S(k)^{-1} \hat{\Lambda}'(k) S(k) = \begin{pmatrix} 0 & -\frac{\alpha k}{k^2+1} \\ \frac{\alpha k}{k^2+1} & -2 \alpha k \\ \end{pmatrix}, \end{align*}

for $k \in \mathbb{R} \setminus \{0\}$. Hence, abbreviating $S(k)^{-1} W = (w_1,w_2)$ and using that $S(\cdot)$, $S(\cdot)^{-1}$ and $k \mapsto k/(1+k^2)$ are bounded on $\mathbb{R} \setminus (-1,1)$, we establish

\begin{align*} \left|\hat{\Lambda}'(k) W\right| &\leq \left|S(k)\right| \left|S(k)^{-1} \hat{\Lambda}'(k) S(k) S(k)^{-1} W\right| \lesssim |k| \left|w_2\right| + \left|S(k)^{-1} W\right|\\ &\lesssim \left|S(k)^{-1} \hat{\Lambda}_0(k) S(k) S(k)^{-1} W\right|^{\frac12} \left|S(k)^{-1} W\right|^{\frac12} + \left|S(k)^{-1}\right| \left|W\right| \\ &\lesssim \left|S(k)^{-1}\right| \left|\hat{\Lambda}_0(k) W\right|^{\frac12} \left|W\right|^{\frac12} + \left|S(k)^{-1}\right| \left|W\right|\\ &\lesssim \left|\hat{\Lambda}_0(k) W\right|^{\frac12} \left|W\right|^{\frac12} + \left|W\right| \lesssim \left|\hat{\Lambda}(k) W\right|^{\frac12} \left|W\right|^{\frac12} + \left|W\right|, \end{align*}

for $W \in \mathbb{C}^2$ and $k \in \mathbb{R} \setminus (-1,1)$. Combining the latter with the fact that $\hat{\Lambda}'(\cdot)$ is bounded on $[-1,1]$, we obtain

(3.5)\begin{equation} \begin{aligned} \left|\hat{\Lambda}'(k) W\right|\lesssim \left|\hat{\Lambda}(k) W\right|^{\frac12} \left|W\right|^{\frac12} + \left|W\right|, \end{aligned} \end{equation}

for all $W \in \mathbb{C}^2$ and $k \in \mathbb{R}$. Thus, by employing Young’s inequality we infer that, for any $\epsilon \gt 0$, the operator $\widetilde{\Lambda}$ is relatively bounded with respect to $\Lambda$ with $\Lambda$-bound $\epsilon$. Consequently, for any $\epsilon \gt 0$, the operator $(U,V) \mapsto (0,\widetilde{\Lambda}U)$ is relatively bounded with respect to $\mathcal{L}$ with $\mathcal{L}$-bound $\epsilon \gt 0$. So, it follows by [Reference Lunardi31, Proposition 2.4.2] that

\begin{align*} \begin{pmatrix} U \\ V \end{pmatrix} \mapsto \begin{pmatrix} \Lambda U \\ \Lambda V + \widetilde{\Lambda} U \end{pmatrix}, \end{align*}

is a sectorial and densely defined operator on $Y_0 \times Y_0$ and, thus, generates an analytic semigroup on $Y_0 \times Y_0$.

Next, we note that $\widetilde{\mathcal{N}} \colon Y_0 \times Y_0 \to Y_0$ is locally Lipschitz continuous, since the map $u \mapsto N(u)/u$ is smooth and because $\partial_x (1-\partial_x^2)^{-2}$ and $(1-\partial_x^2)^{-1}$ are bounded linear operators on $Y_0$. We conclude that

\begin{align*} \begin{pmatrix} U \\ V \end{pmatrix} \mapsto \begin{pmatrix} \mathcal{N}(U) \\ \widetilde{\mathcal{N}}(U,V) \end{pmatrix} \end{align*}

is locally Lipschitz continuous on $Y_0 \times Y_0$. It follows again by [Reference Lunardi31, Theorem 7.1.2 and Proposition 7.1.7] that there exist $T_{\max} \in (0,\infty]$ and a unique, maximally defined, mild solution

\begin{align*} \begin{pmatrix}U \\ V \end{pmatrix} \in C\big([0,T_{\max}),Y_0 \times Y_0\big), \end{align*}

of (3.4) with initial condition $U(0) = (U_0,V_0)^\top$, where $V_0(x) = x U_0(x)$. If $T_{\max} \lt \infty$, then it holds

\begin{align*} \limsup_{t \uparrow T_{\max}} \|(U(t),V(t))\|_{Y_0 \times Y_0} = \infty. \end{align*}

Upon recalling $V(x,t) = x U(x,t)$, we conclude that $U \in C\big([0,T_{\max}),X_0\big)$ is a solution of (3.2) with initial condition $U(0) = U_0$. Moreover, if $T_{\max} \lt \infty$, then we have (3.3).

Proof. Using again the variable $Z = (u,u_t)^\top$ we write (1.2) as the system

(3.6)\begin{equation} Z_t = \mathcal{A} Z + F(Z), \end{equation}

where $\mathcal{A}$ is the differential operator given by (1.18) defined on the space $H^2(\mathbb{R}) \times L^2(\mathbb{R})$ with dense domain $H^2(\mathbb{R}) \times H^2(\mathbb{R})$ and the nonlinear map $F \colon H^2(\mathbb{R}) \times L^2(\mathbb{R}) \to H^2(\mathbb{R}) \times L^2(\mathbb{R})$ is given by

\begin{align*} F(z_1,z_2) = \begin{pmatrix} 0 \\ N(z_1)\end{pmatrix}. \end{align*}

Thanks to the smoothness of $N$ and the fact that $H^1(\mathbb{R})$ continuously embeds into $L^\infty(\mathbb{R})$, we find that $F$ is well-defined and locally Lipschitz continuous. Moreover, since $\partial_x^2$ is a sectorial operator on $L^2(\mathbb{R})$ with dense domain $H^2(\mathbb{R})$, [Reference Webb39, Proposition 2.2] yields that $\mathcal{A}$ generates an analytic semigroup on $H^2(\mathbb{R}) \times L^2(\mathbb{R})$. Standard analytic semigroup theory [Reference Lunardi31, Proposition 7.1.8 and 7.1.10] now provides a time $\tau_{\max} \in (0,\infty]$ and a unique, maximally defined, classical solution

(3.7)\begin{equation} \begin{aligned} Z &\in C\big([0,\tau_{\max}),H^2(\mathbb{R}) \times L^2(\mathbb{R})\big) \cap C^1\big((0,\tau_{\max}),H^2(\mathbb{R}) \times L^2(\mathbb{R})\big)\\ &\qquad \cap C\big((0,\tau_{\max}),H^2(\mathbb{R}) \times H^2(\mathbb{R})\big), \end{aligned} \end{equation}

of (3.6) with initial condition $Z(0) = (u_0,w_0) \in H^2(\mathbb{R}) \times L^2(\mathbb{R})$ such that, if $\tau_{\max} \lt \infty$, then it must hold

(3.8)\begin{equation} \limsup_{t \uparrow \tau_{\max}} \|F(Z(t))\|_{H^2 \times L^2} = \infty. \end{equation}

Writing $Z(t) = (u(t),w(t))^\top$, one readily observes that $\partial_t u(t) = w(t)$ for all $t \gt 0$. Combining the latter with (3.7), we arrive at

\begin{equation*} \begin{aligned} u &\in C\big([0,\tau_{\max}),H^2(\mathbb{R})\big) \cap C^1\big([0,\tau_{\max}),L^2(\mathbb{R})\big) \cap C^1\big((0,\tau_{\max}),H^2(\mathbb{R})\big)\\ &\qquad \cap C^2\big((0,\tau_{\max}),L^2(\mathbb{R})\big). \end{aligned} \end{equation*}

Finally, due to the continuity of $N$ and the fact that $\|F(z_1,z_2)\|_{H^2 \times L^2} = \|N(z_1)\|_{L^2}$ for $(z_1,z_2) \in H^2(\mathbb{R}) \times L^2(\mathbb{R})$, identity (3.8) implies

(3.9)\begin{equation} \limsup_{t \uparrow \tau_{\max}} \|u(t)\|_{L^2} = \infty. \end{equation}

Define $\check{U}(t) = (u(t),v(t))$ with $v(t) = (1-\partial_x^2)^{-1}(w(t) - \frac{\alpha}{2} \partial_x^2 u(t))$. Then, by construction $\check{U} \in C\big([0,\tau_{\max}), L^2(\mathbb{R})\big)$ is a mild solution of (3.2) with initial condition $\check{U}(0) = (u_0,v_0)^\top$. On the other hand, $U \in C\big([0,T_{\max}), L^2(\mathbb{R})\big)$, established in Proposition 3.2, is also a mild solution of (3.2) with $U(0) = (u_0,v_0)^\top$. Recalling from the proof of Proposition 3.2 that $\Lambda$ generates a $C_0$-semigroup on $L^2(\mathbb{R})$ and the nonlinearity $\mathcal{N}$ is locally Lipschitz continuous on $L^2(\mathbb{R})$, it follows by uniqueness of mild solutions, cf. [Reference Cazenave and Haraux7, Lemma 4.3.2], that $\check{U}(t) = U(t)$ for all $t \in [0,\min\{T_{\max},\tau_{\max}\})$. Since (3.9) implies (3.3), we must necessarily have $\tau_{\max} \geq T_{\max}$.

Proof. Let

\begin{equation*}k_1 = \sqrt{\frac{2 + 2\sqrt{1 + \alpha^2}}{\alpha^2}}.\end{equation*}

The eigenvalues $\lambda_+(k)$ and $\lambda_-(k)$ are distinct for $k^2 \gt k_1^2$ and, thus, $\widehat{\Lambda}(k)$ can be diagonalized for such values. The associated change of basis is represented by a matrix $S(k)$, whose columns are comprised of eigenvectors of $\widehat{\Lambda}(k)$, and its inverse, which are given by

\begin{align*} S(k) = \begin{pmatrix} \frac{1+k^2}{\mu(k)} & -\frac{1+k^2}{\mu(k)}\\ 1 & 1 \end{pmatrix}, \qquad S(k)^{-1} = \frac12\begin{pmatrix} \frac{\mu(k)}{4 k^2+4} & 1 \\ -\frac{\mu(k)}{4 k^2+4} & 1 \end{pmatrix}. \end{align*}

One readily observes that the coefficients of $S(\cdot)$ and $S(\cdot)^{-1}$ are bounded on $\mathbb{R} \setminus (-2k_1,2k_1)$. Moreover, it holds

\begin{align*} \lambda_-(k) \leq -\dfrac{1}{2}\alpha k^2, \qquad \lambda_+(k) \leq -\dfrac{1}{\alpha} \end{align*}

for $|k| \gt k_1$. We conclude that there exists $\theta_2 \gt 0$ such that the matrix exponential and its temporal derivative

\begin{align*} \partial_t^j \mathrm{e}^{\widehat{\Lambda}(k) t} = S(k)^{-1} \mathrm{diag}\left(\lambda_+(k)^j \mathrm{e}^{\lambda_+(k) t}, \lambda_-(k)^j\mathrm{e}^{\lambda_-(k) t}\right) S(k), \end{align*}

obey the estimate

(4.9)\begin{equation} t^j \left|\partial_t^j \mathrm{e}^{\widehat{\Lambda}(k) t}\right| \lesssim \mathrm{e}^{-\theta_2 t}, \end{equation}

for $k \in \mathbb{R} \setminus (-2k_1,2k_1)$, $j = 0,1$ and $t \geq 0$. To bound the matrix exponential $\mathrm{e}^{\widehat{\Lambda}(k) t}$ and its temporal derivative on the compact set $\smash{J \overset{\scriptscriptstyle{\mathrm{def}}}{=} [-2k_1,-k_0/2] \cup [k_0/2,2k_1]}$, we collect some facts from [Reference Arendt, Grabosch, Greiner, Groh, Lotz, Moustakas, Nagel, Neubrander and Schlotterbeck1, Chapter A-III, §7]. Firstly, since $J$ is compact and $\widehat{\Lambda}$ is continuous on $J$, the multiplication operator $A \colon f \mapsto \widehat{\Lambda} f$ generates a strongly continuous semigroup $(T(t))_{t \geq 0}$ on $C(J,\mathbb{C}^2)$, which is given by

\begin{equation*}(T(t)f)(k) = \mathrm{e}^{\widehat{\Lambda}(k) t} f(k), \qquad k \in J.\end{equation*}

Secondly, the growth bound of the semigroup $(T(t))_{t \geq 0}$ coincides with the spectral bound of $A$. Thirdly, the spectrum of $A$ is given by

\begin{equation*}\sigma(A) = \bigcup_{k \in J} \sigma(\widehat{\Lambda}(k)).\end{equation*}

Combining the latter three observations with (4.6) yields that the growth bound of the semigroup $(T(t))_{t \geq 0}$ is smaller than $-\theta_1$, which implies

\begin{align*} \left|\mathrm{e}^{\widehat{\Lambda}(k) t}\right| \lesssim \mathrm{e}^{-\theta_1 t}, \qquad \left|\partial_t\mathrm{e}^{\widehat{\Lambda}(k) t}\right| = \left|\hat{\Lambda}(k) \mathrm{e}^{\widehat{\Lambda}(k) t}\right| \lesssim \mathrm{e}^{-\theta_1 t}, \end{align*}

for all $t \geq 0$ and $k \in J$. Combining the latter with (4.9) yields a constant $\theta_3 \gt 0$ such that

(4.10)\begin{equation} t^j \left|\partial_t^j \mathrm{e}^{\widehat{\Lambda}(k) t}\right| \lesssim \mathrm{e}^{-\theta_3 t}, \end{equation}

for $j = 0,1$, $t \geq 0$ and $k \in \mathbb{R} \setminus [-\frac{k_0}{2},\frac{k_0}{2}]$, proving the first estimate in (4.7).

We proceed with the second estimate in (4.7). Firstly, we express the derivative as

(4.11)\begin{equation} \partial_k \mathrm{e}^{\widehat{\Lambda}(k) t} = t\int_0^1 \mathrm{e}^{\widehat{\Lambda}(k) l t} \hat{\Lambda}'(k) \mathrm{e}^{\widehat{\Lambda}(k) (1-l) t} \mathrm{d} l. \end{equation}

Next, we apply the estimate (3.5) and obtain

\begin{align*} \left|\partial_k \mathrm{e}^{\widehat{\Lambda}(k) t} W\right| \leq t \int_0^1 \left|\mathrm{e}^{\widehat{\Lambda}(k) l t} \right| \left|\hat{\Lambda}(k) \mathrm{e}^{\widehat{\Lambda}(k) (1-l) t}W\right|^{\frac12} \left|\mathrm{e}^{\widehat{\Lambda}(k) (1-l) t}W\right|^{\frac12} \mathrm{d} l, \end{align*}

for $W \in \mathbb{C}^2$, $k \in \mathbb{R}$ and $t \geq 0$. So, using (4.10), we arrive at

\begin{align*} \left|\partial_k \mathrm{e}^{\widehat{\Lambda}(k) t} W\right| \lesssim t \mathrm{e}^{-\theta_0 t} \int_0^1 \frac{1}{\sqrt{(1-l)t}} \mathrm{d} l \left|W\right| \lesssim \sqrt{t}\, \mathrm{e}^{-\theta_3 t} \left|W\right| \lesssim \mathrm{e}^{-\frac{\theta_3}{2} t} \left|W\right|, \end{align*}

for $W \in \mathbb{C}^2$, $k \in \mathbb{R} \setminus [-\frac{k0}{2},\frac{k0}{2}]$ and $t \geq 0$, which proves the second estimate in (4.7).

Finally, we obtain bounds for $k\in [-k_0, k_0]$. For such values, the matrix $\smash{\widehat{\Lambda}(k)}$ is diagonalizable and both $S(\cdot)$ and $S^{-1}(\cdot)$ are bounded on $[-k_0,k_0]$. Moreover, we have $|\hat{\Lambda}'(k)| \lesssim |k|$ for $k \in [-k_0,k_0]$. Combining the latter two observations with (4.5) and (4.11) readily yields the last estimate (4.8), which finishes the proof.

Proof. Recall that $1-\chi$ vanishes on $[-k_0/2,k_0/2]$. So, letting $\upsilon \colon \mathbb{R} \to [0,1]$ be a smooth function supported on $[-k_0/2,k_0/2]$ with $\upsilon(0) = 1$, we obtain

\begin{align*} B_2\big({\widehat{V}},(1-\chi){\widehat{W}}\big)(k) = \int_\mathbb{R} \frac{1-\upsilon(l)}{l} (k + (l-k)) N_2(k,l)\big({\widehat{V}},(1-\chi){\widehat{W}}\big) \mathrm{d} l, \end{align*}

for $k \in \mathbb{R}$. Upon noticing that $l \mapsto \frac{1-\upsilon(l)}{l}$ is bounded on $\mathbb{R}$, we deduce

(6.4)\begin{equation} \left|\partial_k^j B_2\big({\widehat{V}},(1-\chi){\widehat{W}}\big)(k)\right| \lesssim |k| \big\|{\widehat{V}}\big\|_{W^{j,1}} \big\|{\widehat{W}}\big\|_{L^\infty} + \big\||\cdot| {\widehat{V}}\big\|_{W^{j,1}} \big\|{\widehat{W}}\big\|_{L^\infty}, \end{equation}

for $j = 0,1$, $k \in \mathbb{R}$, ${\widehat{V}} \in W^{j,1}(\mathbb{R})$ and ${\widehat{W}} \in L^\infty(\mathbb{R})$ with $\||\cdot|{\widehat{V}}\|_{W^{j,1}} \lt \infty$. On the other hand, using Young’s convolution inequality, we infer

(6.5)\begin{equation} \begin{aligned} \left\|\partial_k^j B_2\big({\widehat{V}},{\widehat{W}}\big) \right\|_{L^\infty} &\lesssim \big\|{\widehat{V}}\big\|_{W^{j,1}} \big\|{\widehat{W}}\big\|_{L^\infty},\\ \left\|\partial_k^j B_2\big({\widehat{V}},\widehat{Z}\big) \right\|_{L^1} &\lesssim \big\|{\widehat{V}}\big\|_{W^{j,1}} \big\|\widehat{Z}\big\|_{L^1},\\ \left\|\partial_k^j B_3\big({\widehat{V}},{\widehat{W}},\widehat{Z}\big)\right\|_{L^\infty} &\lesssim \big\|{\widehat{V}}\big\|_{W^{j,1}} \big\|{\widehat{W}}\big\|_{L^\infty} \big\|\widehat{Z}\big\|_{L^1}, \end{aligned} \end{equation}

for $j = 0,1$, ${\widehat{V}} \in W^{j,1}(\mathbb{R})$, ${\widehat{W}} \in L^\infty(\mathbb{R})$ and $\widehat{Z} \in L^1(\mathbb{R})$.

Next, we expand

(6.6)\begin{equation} \mathcal{F} \mathcal{N}\big(\mathcal{F}^{-1}{\widehat{V}}\big)\big) = B_2\big({\widehat{V}},{\widehat{V}}\big) + \mathcal{F} \mathcal{R}_2\left(\mathcal{F}^{-1}\big({\widehat{V}}\big)\right), \end{equation}

with $\mathcal{R}_2(V)$ defined by

(6.7)\begin{equation} \mathcal{R}_2(V) = (1-\partial_x^2)^{-1} R_0(V_1)\mathbf{e}_2, \qquad R_0(z) \overset{\scriptscriptstyle{\mathrm{def}}}{=} N(z) - 2\pi \kappa z^2. \end{equation}

The expansion (6.1) yields $|R_0(z)| \lesssim |z|^3$ and $|R(z)| \lesssim |z|^4$ for $z \in [-1,1]$. So, using the facts that we have $\|V\|_{L^2}^2 \leq \|V\|_{L^1} \|V\|_{L^\infty}$ for $V \in L^1(\mathbb{R}) \cap L^\infty(\mathbb{R})$ and the Fourier transform is an isomorphism on $L^2(\mathbb{R})$ and it maps $L^1(\mathbb{R})$ continuously into $L^\infty(\mathbb{R})$, we establish

(6.8)\begin{equation} \begin{aligned} \left\|\partial_k^j \mathcal{F} \mathcal{R}_2\left(\mathcal{F}^{-1}\big({\widehat{V}}\big)\right)\right\|_{L^\infty} &\lesssim \left\|\frac{1}{1+|\cdot|^{2}}\right\|_{W^{j,\infty}} \left\|(1+|\cdot|^j) R_0\left(\mathcal{F}^{-1}\big({\widehat{V}}\big)_1\right)\right\|_{L^1}\\ &\lesssim \left\|\mathcal{F}^{-1}\big({\widehat{V}}\big)\right\|_{L^2}^2 \left\|(1+|\cdot|^j)\mathcal{F}^{-1}\big({\widehat{V}}\big)\right\|_{L^\infty}\\ &\lesssim \big\|{\widehat{V}}\big\|_{L^2}^2 \big\|{\widehat{V}}\big\|_{W^{j,1}} \lesssim \big\|{\widehat{V}}\big\|_{L^1} \big\|{\widehat{V}}\big\|_{W^{j,1}},\\ \left\|\partial_k^j \mathcal{F} \mathcal{R}_2\left(\mathcal{F}^{-1}\big({\widehat{V}}\big)\right)\right\|_{L^1} &\lesssim \left\|\frac{1}{1+|\cdot|^{2}}\right\|_{H^j} \left\| (1+|\cdot|^j) R_0\left(\mathcal{F}^{-1}\big({\widehat{V}}\big)_1\right)\right\|_{L^2}\\ &\lesssim \big\|\mathcal{F}^{-1}\big({\widehat{V}}\big)\big\|_{L^2} \big\|\mathcal{F}^{-1}\big({\widehat{V}}\big)\big\|_{L^\infty} \big\|(1+|\cdot|^j) \mathcal{F}^{-1}\big({\widehat{V}}\big)\big\|_{L^\infty}\\ &\lesssim \big\|{\widehat{V}}\big\|_{L^2} \big\|{\widehat{V}}\big\|_{L^1} \big\|{\widehat{V}}\big\|_{W^{j,1}}\lesssim \big\|{\widehat{V}}\big\|_{L^1} \big\|{\widehat{V}}\big\|_{W^{j,1}}, \end{aligned} \end{equation}

for $j = 0,1$ and ${\widehat{V}} \in W^{j,1}(\mathbb{R}) \cap L^\infty(\mathbb{R})$ with $\|{\widehat{V}}\|_{L^1 \cap L^\infty} \leq 1$. Similarly, we obtain

(6.9)\begin{equation} \begin{aligned} \left\|\partial_k^j \mathcal{F} \mathcal{R}\left(\mathcal{F}^{-1}\big({\widehat{V}}\big)\right)\right\|_{L^\infty} &\lesssim \big\|{\widehat{V}}\big\|_{L^2}^2 \big\|{\widehat{V}}\big\|_{W^{j,1}} \big\|V\big\|_{L^1} \lesssim \big\|{\widehat{V}}\big\|_{L^1}^2 \big\|{\widehat{V}}\big\|_{W^{j,1}}, \end{aligned} \end{equation}

for $j = 0,1$ and ${\widehat{V}} \in W^{j,1}(\mathbb{R}) \cap L^\infty(\mathbb{R})$ with $\|{\widehat{V}}\|_{L^1 \cap L^\infty} \leq 1$. In summary, the desired bound on $\mathcal{F}\mathcal{N}(\mathcal{F}^{-1}({\widehat{V}}))$ follows by recalling (6.6) and combining the estimates (6.5) and (6.8), whereas the estimate on $\mathcal{E}({\widehat{V}},(1-\chi){\widehat{W}})$ follows by (6.4), (6.5) and (6.9).

Proof. We start by isolating the critical quadratic term in the nonlinearity in Equation (5.8) for $\widehat{U}_\mathrm{c}$. Thus, suppressing the $t$-dependency of $\widehat{U}_\mathrm{s}$ and $\widehat{U}_\mathrm{c}$, we write

(7.4)\begin{equation} \chi \mathcal{F} \mathcal{N}\big(\mathcal{F}^{-1}\big(\widehat{U}_\mathrm{c} + \widehat{U}_\mathrm{s}\big)\big) = \chi B_2\big(\widehat{U}_\mathrm{c},\widehat{U}_\mathrm{c}\big) + \chi R_2\big(\widehat{U}_\mathrm{c},\widehat{U}_\mathrm{s}\big), \end{equation}

where we denote

\begin{align*} R_2\big(\widehat{U}_\mathrm{c},\widehat{U}_\mathrm{s}\big) &= 2B_2\big(\widehat{U}_\mathrm{c},\widehat{U}_\mathrm{s}\big) + B_2\big(\widehat{U}_\mathrm{s},\widehat{U}_\mathrm{s}\big) + \mathcal{F} \mathcal{R}_2\left(\mathcal{F}^{-1}\big(\widehat{U}_\mathrm{c} + \widehat{U}_\mathrm{s}\big)\right), \end{align*}

with $\mathcal{R}_2(V)$ defined by (6.7). Combining (6.5) and (6.8), yields a $t$-independent constant $C \gt 0$ such that, if we have $t \in [0,T_{\max})$ with $\|\widehat{U}(t)\|_{L^1 \cap L^\infty} \leq 1$, then it holds

(7.5)\begin{equation} \begin{aligned} \big\|\partial_k^j R_2\big(\widehat{U}_\mathrm{c}(t),\widehat{U}_\mathrm{s}(t)\big)\big\|_{L^\infty} &\leq C\big\|\widehat{U}(t)\big\|_{W^{j,1}}\left(\big\|\widehat{U}_\mathrm{s}(t)\big\|_{L^\infty} + \big\|\widehat{U}(t)\big\|_{L^1} \right), \end{aligned} \end{equation}

for $j = 0,1$.

Next, we rewrite the critical quadratic term (6.2) using integration by parts with respect to time. Firstly, we recall that the spectral projections $P_\pm(k)$ of $\hat{\Lambda}(k)$ are well-defined on the interval $(-k_0,k_0)$ on which $\widehat{U}_\mathrm{c}(t)$ is supported. Thus, we can decompose (6.2) into

(7.6)\begin{equation} \begin{aligned} &\int_0^t \chi(k) \mathrm{e}^{\hat{\Lambda}(k) (t-s)} B_2\big(\widehat{U}_\mathrm{c}(s), \widehat{U}_\mathrm{c}(s)\big)(k) \, \mathrm{d} s \\ &\quad \ = \sum_{j \in \left\lbrace -1,1 \right\rbrace^3} \int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \chi(k) P_{j_0}(k) \int_\mathbb{R} N_2(k,l)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2}\widehat{U}_\mathrm{c}(s)\big) \mathrm{d} l \mathrm{d} s, \end{aligned} \end{equation}

with $k \in \mathbb{R}$ and $t \in [0,T_{\max})$. We will integrate by parts in each summand of (7.6) by integrating the exponential $\smash{\mathrm{e}^{\lambda_{j_0}(k)(t-s)}}$ and differentiating the quadratic contribution $B_2(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2}\widehat{U}_\mathrm{c}(s))(k)$ with respect to $s$. Thus, expressing the derivative $\partial_s \widehat{U}_\mathrm{c}(s)$ using Equation (5.8), recalling the expansion (7.4) and noting that $\hat{\Lambda}(k) P_{j_i}(k) = \lambda_{j_i}(k) P_{j_i}(k)$ for $i = 1,2$, we compute

(7.7)\begin{equation} \begin{aligned} &\partial_s N_2(k,l) \big(P_{j_1} \widehat{U}_\mathrm{c}, P_{j_2} \widehat{U}_\mathrm{c}\big)\\ &\quad\ = N_2(k,l) \big(P_{j_1} \partial_s \widehat{U}_\mathrm{c}, P_{j_2} \widehat{U}_\mathrm{c}\big) + N_2(k,l) \big(P_{j_1} \widehat{U}_\mathrm{c}, P_{j_2} \partial_s \widehat{U}_\mathrm{c}\big)\\ &\quad\ = \big(\lambda_{j_1}(k-l) + \lambda_{j_2}(l)\big) N_2(k,l)\big(P_{j_1}\widehat{U}_\mathrm{c}, P_{j_2}\widehat{U}_\mathrm{c}\big)\\ &\quad\ \qquad + \, N_2(k,l)\big(\chi P_{j_1} B_2\big(\widehat{U}_\mathrm{c}, \widehat{U}_\mathrm{c}\big), P_{j_2} \widehat{U}_\mathrm{c}\big) + N_2(k,l)\big(P_{j_1} \widehat{U}_\mathrm{c}, \chi P_{j_2} B_2\big(\widehat{U}_\mathrm{c}, \widehat{U}_\mathrm{c}\big)\big) \\ &\quad\ \qquad + \, N_2(k,l)\big(\chi P_{j_1} R_2\big(\widehat{U}_\mathrm{c}, \widehat{U}_\mathrm{s}\big), P_{j_2} \widehat{U}_\mathrm{c}\big) + N_2(k,l)\big(P_{j_1} \widehat{U}_\mathrm{c}, \chi P_{j_2} R_2\big(\widehat{U}_\mathrm{c},\widehat{U}_\mathrm{s}\big)\big), \end{aligned} \end{equation}

for $k, l \in \mathbb{R}$ and $j \in \left\lbrace -1,1 \right\rbrace^3$, where we have suppressed the $s$-dependency of $\widehat{U}_\mathrm{s}$ and $\widehat{U}_\mathrm{c}$. Since it holds $\phi_j^2(0,0) \in \{\pm \mathrm{i}\}$ for all $j\in \left\lbrace -1,1 \right\rbrace^3$, the phase function $\phi_j^2$ does not vanish on the square $[-k_0,k_0]^2$, upon taking $k_0 \gt 0$ smaller if necessary. Hence, using integration by parts, we obtain

\begin{align*} &\int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \chi(k) P_{j_0}(k) N_2(k, l)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2}\widehat{U}_\mathrm{c}(s)\big) \mathrm{d} l \mathrm{d} s\\ &\quad \ = \int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \chi(k) \frac{\lambda_{j_0}(k)}{\phi_j^2(k,l)} P_{j_0}(k) N_2(k, l)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2}\widehat{U}_\mathrm{c}(s)\big) \mathrm{d} l \mathrm{d} s\\ &\quad \ \qquad - \, \int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \chi(k)\frac{\lambda_{j_1}(k-l) + \lambda_{j_2}(l)}{\phi_j^2(k,l)} P_{j_0}(k) N_2(k, l)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2}\widehat{U}_\mathrm{c}(s)\big) \mathrm{d} l \mathrm{d} s \\ &\quad \ = -\left[\mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \frac{\chi(k) }{\phi_j^2(k,l)} P_{j_0}(k) N_2(k, l)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2}\widehat{U}_\mathrm{c}(s)\big) \mathrm{d} l \right]_{s = 0}^t\\ &\quad \ \qquad + \, \int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \frac{\chi(k)}{\phi_j^2(k,l)} P_{j_0}(k) \partial_s N_2(k, l)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2}\widehat{U}_\mathrm{c}(s)\big) \mathrm{d} l \mathrm{d} s\\ &\quad \ \qquad - \, \int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \chi(k) P_{j_0}(k) \frac{\lambda_{j_1}(k-l) + \lambda_{j_2}(l)}{\phi_j^2(k,l)} N_2(k, l)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2}\widehat{U}_\mathrm{c}(s)\big) \mathrm{d} l \mathrm{d} s \end{align*}

for $k \in \mathbb{R}$, $t \in [0,T_{\max})$ and $j\in \left\lbrace -1,1 \right\rbrace^3$. So, taking the sum of the latter over all $j \in \left\lbrace -1,1 \right\rbrace^3$, we obtain the desired identity (7.1) by (7.6) and (7.7) with error function

\begin{align*} \mathcal{E}_2(k,t) &= \sum_{j \in \left\lbrace -1,1 \right\rbrace^3} \int_\mathbb{R} \frac{1}{\phi_j^2(k,l)} P_{j_0}(k) \left( N_2(k,l)\big(\chi P_{j_1} R_2\big(\widehat{U}_\mathrm{c}(t), \widehat{U}_\mathrm{s}(t)\big), P_{j_2} \widehat{U}_\mathrm{c}\big)\right. \\ &\qquad \left. +\, N_2(k,l)\big(P_{j_1} \widehat{U}_\mathrm{c}(t), \chi P_{j_2} R_2\big(\widehat{U}_\mathrm{c}(t),\widehat{U}_\mathrm{s}(t)\big)\big)\right) \mathrm{d} l, \end{align*}

for $t \in [0,T_{\max})$. Applying Young’s convolution inequality, while using estimate (7.5) and recalling that the phase function $\phi_j^2$ is bounded away from $0$ on the square $[-k_0,k_0]^2$ and the cut-off function $\chi$ is supported on $(-k_0,k_0)$, we obtain $t$-independent constants $C_{1,2} \gt 0$ such that, if we have $t \in [0,T_{\max})$ with $\smash{\|\widehat{U}(t)\|_{L^1 \cap L^\infty}} \leq 1$, then it holds

\begin{align*} &\left\|\partial_k^j\left(\chi \mathcal{E}_2(\cdot,t)\right)\right\|_{L^\infty} \\ &\quad \ \leq C_1 \left(\big\|\widehat{U}_\mathrm{c}(t)\big\|_{L^1} \big\|R_2(\widehat{U}_\mathrm{c}(t),\widehat{U}_\mathrm{s}(t)\big\|_{W^{j,\infty}} + \big\|\widehat{U}_\mathrm{c}(t)\big\|_{W^{j,1}} \big\|R_2(\widehat{U}_\mathrm{c}(t),\widehat{U}_\mathrm{s}(t)\big\|_{L^\infty}\right) \\ &\quad \ \leq C_2 \big\|\widehat{U}(t)\big\|_{W^{j,1}} \big\|\widehat{U}(t)\big\|_{L^1} \left(\big\|\widehat{U}_\mathrm{s}(t)\big\|_{L^\infty} + \big\|\widehat{U}(t)\big\|_{L^1}\right), \end{align*}

for $j = 0,1$ (with $j = 0$ in case $\mathcal{X} = Y_0$). Similarly, we derive the bounds (7.2) and (7.3).

Next, let ${\widehat{V}} \in C_c(-k_0,k_0)$ with $\overline{{\widehat{V}}(k)} = {\widehat{V}}(-k)$ for $k \in \mathbb{R}$. We prove $Q_2({\widehat{V}},{\widehat{V}})(0) \in \mathbb{R}$. Firstly, we note that the maps $\tau, \xi \colon \left\lbrace -1,1 \right\rbrace^3 \to \left\lbrace -1,1 \right\rbrace^3$ given by $\tau(j_0,j_1,j_2) = (j_0,j_2,j_1)$ and $\xi(j) = -j$ are bijections. We have

\begin{align*} \phi_j^2(k,l) = \phi_{\tau(j)}^2(k,k-l), \end{align*}

for $k,l \in \mathbb{R}$ and $j \in \left\lbrace -1,1 \right\rbrace^3$. Hence, using (4.3), we obtain

\begin{align*} \overline{Q_2({\widehat{V}},{\widehat{V}})(0)} &= \sum_{j \in \left\lbrace -1,1 \right\rbrace^3} \int_\mathbb{R} \frac{P_{-j_0}(0)}{\phi_{-j}^2(0,l)} N_2(0,l)\left(P_{-j_1} \overline{{\widehat{V}}}, P_{-j_2} \overline{{\widehat{V}}}\right) \mathrm{d} l\\ &= \sum_{j \in \left\lbrace -1,1 \right\rbrace^3} \int_\mathbb{R} \frac{P_{j_0}(0)}{\phi_{j}^2(0,l)} N_2(0,l)\left(P_{j_2} {\widehat{V}}, P_{j_1} {\widehat{V}} \right) \mathrm{d} l\\ &= \sum_{j \in \left\lbrace -1,1 \right\rbrace^3} \int_\mathbb{R} \frac{P_{\tau(j)_0}(0)}{\phi_{\tau(j)}^2(0,l)} N_2(0,l)\left(P_{\tau(j)_1} {\widehat{V}}, P_{\tau(j)_2} {\widehat{V}} \right) \mathrm{d} l\\ &= Q_2({\widehat{V}},{\widehat{V}})(0), \end{align*}

which implies $Q_2({\widehat{V}},{\widehat{V}})(0) \in \mathbb{R}$.

Proof. We integrate by parts in each summand of

(8.9)\begin{equation} \begin{aligned} &\int_0^t \chi(k) \mathrm{e}^{\widehat{\Lambda}(k) (t-s)} Z_3^{\mathrm{c}}\big(\widehat{U}_\mathrm{c}(s),\widehat{U}_\mathrm{c}(s),\widehat{U}_\mathrm{c}(s)\big)(k) \, \mathrm{d} s\\ &\quad \ = \sum_{j \in \mathcal{T}^c} \int_0^t \chi(k) \mathrm{e}^{\lambda_0(k) (t-s)} P_{j_0}(k) Z_3\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2} \widehat{U}_\mathrm{c}(s), P_{j_3} \widehat{U}_\mathrm{c}(s)\big)(k) \, \mathrm{d} s\\ &\quad \ = \sum_{j \in \mathcal{T}^c} \int_0^t \chi(k) \mathrm{e}^{\lambda_0(k) (t-s)}P_{j_0}(k) \int_\mathbb{R}\int_\mathbb{R} \widetilde{K}_j(k,l_1,l_2,s) \mathrm{d} l_1 \mathrm{d} l_2 \mathrm{d} s, \end{aligned} \end{equation}

where we abbreviate

\begin{align*} \widetilde{K}_j(k,l_1,l_2,s) &= \widetilde{N}_3(k,l_1,l_2)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2}\widehat{U}_\mathrm{c}(s),P_{j_3}\widehat{U}_\mathrm{c}(s)\big). \end{align*}

We express the derivative $\partial_s \widehat{U}_\mathrm{c}(s)$ through Equation (5.8) and note that $\hat{\Lambda}(k) P_{j_i}(k) = \lambda_{j_i}(k) P_{j_i}(k)$ for $i = 1,2,3$, which leads to

(8.10)\begin{equation} \begin{aligned} &\partial_s \widetilde{K}_j(k,l_1,l_2,s)\\ &\quad \ = \widetilde{N}_3(k,l_1,l_2) \big(P_{j_1} \partial_s \widehat{U}_\mathrm{c}, P_{j_2} \widehat{U}_\mathrm{c},P_{j_3} \widehat{U}_\mathrm{c}\big) + \widetilde{N}_3(k,l_1,l_2) \big(P_{j_1} \widehat{U}_\mathrm{c}, P_{j_2} \partial_s \widehat{U}_\mathrm{c},P_{j_3} \widehat{U}_\mathrm{c}\big)\\ &\quad \ \qquad + \, \widetilde{N}_3(k,l_1,l_2) \big(P_{j_1} \widehat{U}_\mathrm{c}, P_{j_2} \widehat{U}_\mathrm{c},P_{j_3} \partial_s \widehat{U}_\mathrm{c}\big) \\ &\quad \ = \big(\lambda_{j_1}(k-l_1) + \lambda_{j_2}(l_1-l_2) + \lambda_{j_3}(l_2)\big) \widetilde{K}_j(k,l_1,l_2,s) + \check{K}_j(k,l_1,l_2,s), \end{aligned} \end{equation}

for $k, l_1,l_2 \in \mathbb{R}$ and $j \in \left\lbrace -1,1 \right\rbrace^4$, where we have suppressed the $s$-dependency of $\widehat{U}_\mathrm{s}$ and $\widehat{U}_\mathrm{c}$ and we denote

\begin{align*} \check{K}_j(k,l_1,l_2,s) &= \widetilde{N}_3(k,l_1,l_2)\big(\chi P_{j_1} \mathcal{F} \mathcal{N} \big(\mathcal{F}^{-1}\big(\widehat{U}_\mathrm{c}(s) + \widehat{U}_\mathrm{s}(s)\big)\big), P_{j_2} \widehat{U}_\mathrm{c}(s), P_{j_3} \widehat{U}_\mathrm{c}(s) \big)\\ &\qquad + \, \widetilde{N}_3(k,l_1,l_2)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), \chi P_{j_2} \mathcal{F} \mathcal{N} \big(\mathcal{F}^{-1}\big( \widehat{U}_\mathrm{c}(s) + \widehat{U}_\mathrm{s}(s)\big)\big), P_{j_3} \widehat{U}_\mathrm{c}(s) \big) \\ &\qquad + \, \widetilde{N}_3(k,l_1,l_2)\big(P_{j_1} \widehat{U}_\mathrm{c}(s), P_{j_2} \widehat{U}_\mathrm{c}(s), \chi P_{j_3} \mathcal{F} \mathcal{N} \big(\mathcal{F}^{-1}\big( \widehat{U}_\mathrm{c}(s) + \widehat{U}_\mathrm{s}(s)\big)\big) \big). \end{align*}

Since we have $\phi_j^3(0,0,0) \neq 0$ for all $j\in \mathcal{T}^c$, the phase function $\phi_j^3$ does not vanish on the cube $[-k_0,k_0]^3$, upon taking $k_0 \gt 0$ smaller if necessary. Hence, through integration by parts, we obtain

\begin{align*} &\int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \int_\mathbb{R} \chi(k) P_{j_0}(k) \widetilde{K}_j(k,l_1,l_2,s) \mathrm{d} l_1 \mathrm{d} l_2 \mathrm{d} s\\ &\quad \ = \int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \int_\mathbb{R} \chi(k) \frac{\lambda_{j_0}(k)}{\phi_j^3(k,l_1,l_2)} P_{j_0}(k) \widetilde{K}_j(k,l_1,l_2,s) \mathrm{d} l_1 \mathrm{d} l_2 \mathrm{d} s\\ &\quad \ \qquad + \, \int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \int_\mathbb{R} \chi(k)\frac{\phi_j^3(k,l_1,l_2) - \lambda_{j_0}(k)}{\phi_j^3(k,l_1,l_2)} P_{j_0}(k) \widetilde{K}_j(k,l_1,l_2,s) \mathrm{d} l_1 \mathrm{d} l_2 \mathrm{d} s \\ &\quad \ = -\left[\mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \int_\mathbb{R} \frac{\chi(k) }{\phi_j^3(k,l_1,l_2)} P_{j_0}(k) \widetilde{K}_j(k,l_1,l_2,s) \mathrm{d} l_1 \mathrm{d} l_2 \right]_{s = 0}^t\\ &\quad \ \qquad + \, \int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \int_\mathbb{R} \frac{\chi(k)}{\phi_j^3(k,l_1,l_2)} P_{j_0}(k) \partial_s \widetilde{K}_j(k,l_1,l_2,s) \mathrm{d} l \mathrm{d} s\\ &\quad \qquad - \, \int_0^t \mathrm{e}^{\lambda_{j_0}(k)(t-s)} \int_\mathbb{R} \int_\mathbb{R} \chi(k) P_{j_0}(k) \frac{\lambda_{j_1}(k-l_1) + \lambda_{j_2}(l_1-l_2) + \lambda_{j_3}(l_2)}{\phi_j^3(k,l_1,l_2)}\\ &\quad \ \qquad \cdot \, \widetilde{K}_j(k,l_1,l_2,s) \mathrm{d} l_1 \mathrm{d} l_2 \mathrm{d} s \end{align*}

for $k \in \mathbb{R}$, $t \in [0,T_{\max})$ and $j\in \mathcal{T}^c$. So, taking the sum of the latter over all $j \in \mathcal{T}^c$ while recalling (8.10), we arrive at (8.7) with error function

\begin{align*} \mathcal{E}_3(k,t) &= \sum_{j \in \mathcal{T}^c} \int_\mathbb{R} \int_\mathbb{R} \frac{1}{\phi_j^3(k,l_1,l_2)} P_{j_0}(k) \check{K}_j(k,l_1,l_2,s) \mathrm{d} l_1 \mathrm{d} l_2, \end{align*}

for $t \in [0,T_{\max})$. By Young’s convolution inequality, Lemma 6.1 and the facts that the phase function $\phi_j^3$ is bounded away from $0$ on the cube $[-k_0,k_0]^3$ for $j \in \mathcal{T}^c$ and the cut-off function $\chi$ is supported on $(-k_0,k_0)$, we find $t$-independent constants $C_{1,2} \gt 0$ such that, if we take $t \in [0,T_{\max})$ with $\smash{\|\widehat{U}(t)\|_{L^1}} \leq 1$, then we have

\begin{align*} \left\|\partial_k^j\left(\chi \mathcal{E}_3(\cdot,t)\right)\right\|_{L^\infty} &\leq C_1 \big\|\widehat{U}_\mathrm{c}(t)\big\|_{L^1} \left(\big\|\mathcal{F} \mathcal{N}\big(\mathcal{F}^{-1} \big(\widehat{U}_\mathrm{c}(t) + \widehat{U}_\mathrm{s}(t)\big)\big)\big\|_{W^{j,\infty}} \big\|\widehat{U}_\mathrm{c}(t)\big\|_{L^1} \right. \\ &\left.\qquad + \, \big\|\widehat{U}_\mathrm{c}(t)\big\|_{W^{j,1}} \big\|\mathcal{F} \mathcal{N}\big(\mathcal{F}^{-1} \big(\widehat{U}_\mathrm{c}(t) + \widehat{U}_\mathrm{s}(t)\big)\big)\big\|_{L^\infty}\right)\\ &\leq C_2 \big\|\widehat{U}(t)\big\|_{L^1}^2 \big\|\widehat{U}(t)\big\|_{W^{j,1}} \big\|\widehat{U}(t)\big\|_{L^\infty}, \end{align*}

for $j = 0,1$ (with $j = 0$ in case $\mathcal{X} = Y_0$). Similar considerations afford the bound (8.8).

Finally, let ${\widehat{V}}, \widehat{Y}_{1,2},{\widehat{W}}_{1,2},\widehat{Z}_{1,2} \in C_c(-k_0,k_0)$ satisfy

\begin{align*} \overline{{\widehat{V}}(k)} = {\widehat{V}}(-k), \qquad \overline{\widehat{Y}_1(k)} = \widehat{Y}_2(-k), \qquad \overline{{\widehat{W}}_1(k)} = {\widehat{W}}_2(-k), \qquad, \overline{\widehat{Z}_1(k)} = \widehat{Z}_2(-k), \end{align*}

for $k \in (-k_0,k_0)$. We prove $K_3({\widehat{V}},{\widehat{V}},{\widehat{V}})(0) \in \mathbb{R}$. Firstly, we note that $\xi \colon \{-1,1\}^3 \to \{-1,1\}^3$ and $\zeta \colon \mathcal{T}^c \to \mathcal{T}^c$ given by $\xi(j) = -j$ and $\zeta(j) = -j$ are bijections. So, using (4.3), we arrive at

\begin{align*} &\overline{\widetilde{N}_3(0,l_1,l_2)\big(\widehat{Y}_1,{\widehat{W}}_1,\widehat{Z}_1\big)}\\ &\quad \ = N_3(0,-l_1,-l_2)\big(\widehat{Y}_2,{\widehat{W}}_2,\widehat{Z}_2\big)\\ &\quad \ \qquad + \sum_{j \in \left\lbrace -1,1 \right\rbrace^3} \frac{P_{-j_0}(0)}{\phi^2_{-j}(0,-l_2)} N_2(0,-l_2)\big(\chi P_{-j_1} N_2(\cdot,-l_1+l_2) \big(\widehat{Y}_2, \widehat{W}_2\big), P_{-j_2}\widehat{Z}_2\big)\\ &\quad \ \qquad + \sum_{j \in \left\lbrace -1,1 \right\rbrace^3} \frac{P_{-j_0}(0)}{\phi^2_{-j}(0,-l_1)} N_2(0,-l_1)\big(P_{-j_1} \widehat{Y}_2, \chi P_{-j_2} N_2(\cdot,-l_2)\big(\widehat{W}_2, \widehat{Z}_2\big)\big)\\ &\quad \ = \widetilde{N}_3(0,-l_1,-l_2)\big(\widehat{Y}_2,{\widehat{W}}_2,\widehat{Z}_2\big) \end{align*}

for $l_1,l_2 \in \mathbb{R}$. Therefore, using (4.3) again and applying the substitution rule, we infer

\begin{align*} \overline{K_3\big(\widehat{V}, \widehat{V}, \widehat{V}\big)(0)} &= \sum_{j \in \mathcal{T}^c} \int_\mathbb{R} \int_\mathbb{R} \frac{P_{-j_0}(0)}{\phi^3_{-j}(0,-l_1,-l_2)} \widetilde{N}_3(0,-l_1,-l_2)\big(P_{-j_1} \widehat{V}, P_{-j_2} \widehat{V}, P_{-j_3} \widehat{V}\big) \mathrm{d} l_1 \mathrm{d} l_2\\ &= \sum_{j \in \mathcal{T}^c} \int_\mathbb{R} \int_\mathbb{R} \frac{P_{-j_0}(0)}{\phi^3_{-j}(0,l_1,l_2)} \widetilde{N}_3(0,l_1,l_2)\big(P_{-j_1} \widehat{V}, P_{-j_2} \widehat{V}, P_{-j_3} \widehat{V}\big) \mathrm{d} l_1 \mathrm{d} l_2\\ &= K_3\big(\widehat{V}, \widehat{V}, \widehat{V}\big)(0) \end{align*}

implying $K_3\big(\widehat{V}, \widehat{V}, \widehat{V}\big)(0) \in \mathbb{R}$.

Proof. Young’s convolution inequality yields

\begin{align*} \big\|\partial_k^j B_3\big({\widehat{V}},{\widehat{W}},\widehat{Z}\big)\big\|_{L^\infty} \lesssim \big\|{\widehat{V}}\big\|_{W^{j,1}}\big\|{\widehat{W}}\big\|_{L^\infty}\big\|\widehat{Z}\big\|_{L^1}, \end{align*}

for $j = 0,1$ and ${\widehat{V}} \in W^{j,1}(\mathbb{R}), {\widehat{W}} \in L^\infty(\mathbb{R})$ and $\widehat{Z} \in L^1(\mathbb{R})$. Combining the latter with the estimates on $Q_3^1$ and $Q_3^2$ established in Proposition 7.1 we infer (9.6).

Proof. We recall that $Z_3^{\mathrm{res}}$ is a trilinear form. So, the result follows by combining Lemma 9.1 with the estimates on $\mathcal{E}, Q_2, \mathcal{E}_2, K_3$ and $\mathcal{E}_3$ established in Lemma 6.1 and Propositions 7.1 and 8.1.

Proof. With the aid of Lemma 4.1, we infer

(10.8)\begin{equation} \left\|\partial_k^j \left(\chi \mathrm{e}^{\hat{\Lambda}(\cdot) (t+1)}\right)\right\|_{L^p} \lesssim \left\|(1 + j|\cdot|(t+1) ) \mathrm{e}^{-\frac12 \alpha |\cdot|^2 (t+1)}\right\|_{L^p} \lesssim (1+t)^{-\frac{1}{2p} + \frac{j}{2}}, \end{equation}

for $t \geq 0$, $p = 1,\infty$ and $j = 0,1$. Therefore, we find a $t$-independent constant $C_1 \gt 0$ such that

(10.9)\begin{equation} \big\|\partial_k^j \widehat{Z}_{\mathrm{c}}(t)\big\|_{L^p} \leq C_1 \left(\big\|\partial_k^j \widehat{\rho}_{\mathrm{c}}(t)\big\|_{L^p} + (1+t)^{-\frac{1}{2p} + \frac{j}{2}} |A(t)| \right), \end{equation}

for $j = 0,1$, $p = 1,\infty$ and $t \in [0,T_{\max})$. On the other hand, combining (10.8) with identity (10.3) and Lemma 9.1, we obtain a $t$-independent constant $C_2 \gt 0$ such that

\begin{align*} \left\|\partial_k^j \mathcal{E}_5(\cdot,t)\right\|_{L^\infty} &\leq C_2\big\|\widehat{Z}_{\mathrm{c}}(t)\big\|_{L^1} \big\|\widehat{Z}_{\mathrm{c}}(t)\big\|_{L^\infty} \left(\big\|\widehat{Z}_{\mathrm{c}}(t)\big\|_{W^{j,1}} + (1+t)^{\frac{j}2} \big\|\widehat{Z}_{\mathrm{c}}(t)\big\|_{L^1}\right), \end{align*}

for $j = 0,1$ and $t \in [0,T_{\max})$, which proves the result by inserting (10.9) into the latter.

Proof. Firstly, we note that by Lemma 4.1, there exists a $t$-independent constant $C_1 \gt 0$ such that

(10.13)\begin{equation} \|\varsigma_{\mathrm{c}}(t)\|_{L^1} \leq |A(t)| \left\|\mathrm{e}^{\hat{\Lambda}(\cdot) (t+1)}\right\|_{L^1} \leq C_1 \frac{|A(t)|}{\sqrt{1+t}}, \end{equation}

for $t \in [0,T_{\max})$. Next, we expand

\begin{align*} \chi(k) Z_3^{\mathrm{res}}(\widehat{Z}_{\mathrm{c}}(t),\widehat{Z}_{\mathrm{c}}(t),\widehat{Z}_{\mathrm{c}}(t))(k) &= \chi(k) Z_3^{\mathrm{res}}\big(\varsigma_{\mathrm{c}}(t),\varsigma_{\mathrm{c}}(t),\varsigma_{\mathrm{c}}(t)\big)(k) + E_1(k,t), \end{align*}

with remainder

\begin{align*} E_1(k,t) = \chi(k) \left(Z_3^{\mathrm{res}}\big(\widehat{Z}_{\mathrm{c}}(t),\widehat{Z}_{\mathrm{c}}(t),\widehat{Z}_{\mathrm{c}}(t)\big)(k) - Z_3^{\mathrm{res}}\big(\varsigma_{\mathrm{c}}(t),\varsigma_{\mathrm{c}}(t),\varsigma_{\mathrm{c}}(t)\big)(k)\right), \end{align*}

for $k \in \mathbb{R}$ and $t \in [0,T_{\max})$. The fact that $Z^3_{\mathrm{res}}$ is trilinear in combination with Lemma 9.1, identity (10.2) and estimate (10.13) yield $t$-independent constants $C_1,C_2 \gt 0$ such that

(10.14)\begin{equation} \begin{aligned} \left\|E_1(\cdot,t)\right\|_{L^\infty} &\leq C_1\big\|\widehat{\rho}_{\mathrm{c}}(t)\big\|_{L^\infty}\left( \big\|\widehat{\rho}_{\mathrm{c}}(t)\big\|_{L^1} + \big\|\varsigma_{\mathrm{c}}(t)\big\|_{L^1} \right)^2\\ &\leq C_2 \big\|\widehat{\rho}_{\mathrm{c}}(t)\big\|_{L^\infty}\left( \big\|\widehat{\rho}_{\mathrm{c}}(t)\big\|_{L^1} + \frac{|A(t)|}{\sqrt{1+t}} \right)^2, \end{aligned} \end{equation}

for $t \in [0,T_{\max})$.

The expansion (4.2) yields

(10.15)\begin{equation} \left|\lambda_\pm(k) - \tilde{\lambda}_\pm(k)\right| \lesssim |k|^4, \end{equation}

for $k \in (-k_0,k_0)$, where $\tilde{\lambda}_\pm(k)$ are the quadratic approximants of the eigenvalues $\lambda_\pm(k)$, given by (1.21). Clearly, it holds

(10.16)\begin{equation} \operatorname{Re}(\tilde{\lambda}_\pm(k)) \leq -\frac14\alpha\left(k_0^2 + k^2\right), \end{equation}

for $k \in \mathbb{R} \setminus (-k_0,k_0)$. Moreover, the mean value theorem implies

(10.17)\begin{equation} \left|\mathrm{e}^z - 1\right| \leq \mathrm{e}^{|z|} - 1 \leq |z| \mathrm{e}^{|z|}, \end{equation}

for $z \in \mathbb{C}$. Thus, combining estimates (10.15), (10.16), (10.17) and taking $k_0 \gt 0$ smaller if necessary, we find $t$- and $k$-independent constants $C_{1,2} \gt 0$ such that

\begin{align*} &\left|P_{\pm}(k) \varsigma_{\mathrm{c}}(k,t) - P_{\pm}(0) \mathrm{e}^{\tilde{\lambda}_\pm(k) (t+1)} A(t)\right|\\ &\quad \ \leq \left(\left|P_\pm(k)\chi(k)\right| \left|\mathrm{e}^{(\lambda_\pm(k) - \tilde{\lambda}_\pm(k))(t+1)} - 1 \right| + \left|P_\pm(k)\chi(k) - P_\pm(0)\right|\right) \mathrm{e}^{\operatorname{Re}(\tilde{\lambda}_\pm(k)) (t+1)} |A(t)| \\ &\quad \ \leq C_1\left(|k|^4 (t+1) + |k|\right) \mathrm{e}^{-\frac38 \alpha k^2 (t+1)} |A(t)| \leq C_2 |k| \mathrm{e}^{-\frac14 \alpha k^2 (t+1)} |A(t)|, \end{align*}

for $k \in (-k_0,k_0)$ and $t \in [0,T_{\max})$, and

\begin{align*} \left|P_{\pm}(k) \varsigma_{\mathrm{c}}(k,t) - P_{\pm}(0) \mathrm{e}^{\tilde{\lambda}_\pm(k) (t+1)} A(t)\right| &= \left|P_{\pm}(0) \mathrm{e}^{\tilde{\lambda}_\pm(k) (t+1)} A(t)\right|\\ &\leq C_1 \mathrm{e}^{-\frac14 \alpha k^2(t+1)} \mathrm{e}^{-\frac14 \alpha k_0^2(t+1)} |A(t)|, \end{align*}

for $k \in \mathbb{R} \setminus (-k_0,k_0)$ and $t \in [0,T_{\max})$. Therefore, we obtain a $t$-independent constant $C \gt 0$ such that

(10.18)\begin{equation} \begin{aligned} \left\|P_{\pm} \varsigma_{\mathrm{c}}(t) - P_{\pm}(0) \mathrm{e}^{\tilde{\lambda}_\pm(\cdot) (t+1)} A(t)\right\|_{L^p} &\leq \frac{C}{(1+t)^{\frac{1}{2} + \frac{1}{2p}}} |A(t)|, \\ \left\|P_\pm(0) \mathrm{e}^{\tilde{\lambda}_\pm(\cdot) (t+1)} A(t)\right\|_{L^p} &\leq \frac{C}{(1+t)^{\frac{1}{2p}}} |A(t)|, \end{aligned} \end{equation}

for $t \in [0,T_{\max})$ and $p = 1,\infty$.

Recalling (8.5) and (8.6), we further expand

\begin{align*} &Z_3^{\mathrm{res}}\big({\widehat{V}},{\widehat{W}},\widehat{Z}\big)(k)\\ &\quad \ = \sum_{j \in \mathcal{T}} P_{j_0}(k) Z_3\big(P_{j_1} {\widehat{V}}, P_{j_2} {\widehat{W}}, P_{j_3} \widehat{Z}\big)(k) \\ &\quad \ = \sum_{j \in \mathcal{T}} P_{j_0}(k) \int_\mathbb{R} \int_\mathbb{R} \widetilde{N}_3(k,l_1,l_2)\big(P_{j_1} {\widehat{V}}, P_{j_2} {\widehat{W}}, P_{j_3} \widehat{Z}\big) \mathrm{d} l_1 \mathrm{d} l_2\\ &\quad \ = \sum_{j \in \mathcal{T}} \int_\mathbb{R} \int_\mathbb{R} \left[P_{j_0}(k)N_3(k,l_1,l_2)\big(P_{j_1} {\widehat{V}}, P_{j_2} {\widehat{W}}, P_{j_3} \widehat{Z}\big) \phantom{\sum_{m \in \{\pm 1\}} \frac{P_{j_0}(k)}{\phi^2_{(j_0,m,j_3)}}} \right.\\ &\qquad \ \quad + \sum_{m \in \{\pm 1\}} \frac{P_{j_0}(k)}{\phi^2_{(j_0,m,j_3)}(k,l_2)} N_2(k,l_2)\big(\chi P_m N_2(\cdot,l_1-l_2) \big(P_{j_1} {\widehat{V}}, P_{j_2} {\widehat{W}} \big), P_{j_3} \widehat{Z} \big) \\ &\left. \quad \qquad\ + \sum_{m \in \{\pm 1\}} \frac{P_{j_0}(k)}{\phi^2_{(j_0,j_1,m)}(k,l_1)} N_2(k,l_1)\big(P_{j_1} {\widehat{V}} , \chi P_{m} N_2(\cdot,l_2)\big(P_{j_2} {\widehat{W}}, P_{j_3} \widehat{Z}\big)\big) \right] \mathrm{d} l_1 \mathrm{d} l_2, \end{align*}

for ${\widehat{V}},{\widehat{W}},\widehat{Z} \in C_c(-k_0,k_0) \cap X_0$. So, we readily observe that there exist smooth functions $\mathcal{K}_j \colon (-k_0,k_0)^2 \to \mathbb{C}^2$ such that

\begin{align*} Z_3^{\mathrm{res}}\big({\widehat{V}},{\widehat{W}},\widehat{Z}\big)(0) &= \sum_{j \in \mathcal{T}} \int_\mathbb{R} \int_\mathbb{R} \mathcal{K}_j(l_1,l_2) \left(P_{j_1}(-l_1) {\widehat{V}}(-l_1)\right)_1 \left(P_{j_2}(l_1-l_2) {\widehat{W}}(l_1-l_2)\right)_1\\ &\qquad \cdot \left(P_{j_3}(l_2) \widehat{Z}(l_2)\right)_1 \mathrm{d} l_1 \mathrm{d} l_2, \end{align*}

for ${\widehat{V}},{\widehat{W}},\widehat{Z} \in C_c(-k_0,k_0) \cap X_0$ and $j \in \mathcal{T}$. With the aid of (4.4) we compute

\begin{align*} \mathcal{K}_j(0,0) &= \left(\beta + \sum_{m \in \{\pm 1\}} \frac{\kappa^2 (P_m(0) \mathbf{e}_2)_1}{\phi^2_{j_0,m,j_3}(0,0)} + \sum_{m \in \{\pm 1\}} \frac{\kappa^2 (P_m(0) \mathbf{e}_2)_1}{\phi^2_{j_0,j_1,m}(0,0)}\right)P_{j_0}(0)\mathbf{e}_2\\ &= \left(\beta + \sum_{m \in \{\pm 1\}} \frac{\kappa^2 m}{2\left(m + j_3 - j_0\right)} + \sum_{m \in \{\pm 1\}} \frac{\kappa^2 m}{2\left(m + j_1 - j_0\right)}\right)P_{j_0}(0)\mathbf{e}_2, \end{align*}

for $j \in \mathcal{T}$, which implies

(10.19)\begin{equation} P_\pm(0) \sum_{j \in \mathcal{T}} \mathcal{K}_j(0,0) = \left(3\beta + \frac{10}{3} \kappa^2\right) P_\pm(0) \mathbf{e}_2. \end{equation}

In addition, by the smoothness of $\mathcal{K}_j$, we have

\begin{align*} \left|\mathcal{K}_j(l_1,l_2) - \mathcal{K}_j(0,0)\right| \lesssim |l_1| + |l_2|, \end{align*}

for $l_1,l_2 \in (-k_0,k_0)$ and $j \in \mathcal{T}$. Thus, combining the latter with Lemma 4.1, (10.13) and (10.18) and setting

\begin{align*} \mathcal{Z}(t) &= \sum_{j \in \mathcal{T}} \int_\mathbb{R} \int_\mathbb{R} \mathcal{K}_j(0,0) \left(P_{j_1}(0) \mathrm{e}^{\tilde{\lambda}_{j_1}(l_1)(t+1)} A(t) \right)_1 \left(P_{j_2}(0) \mathrm{e}^{\tilde{\lambda}_{j_2}(l_1-l_2)(t+1)} A(t) \right)_1 \\ &\qquad \cdot \left(P_{j_3}(0) \mathrm{e}^{\tilde{\lambda}_{j_3}(l_2)(t+1)} A(t) \right)_1 \mathrm{d} l_1 \mathrm{d} l_2, \end{align*}

we obtain $t$-independent constants $C_{1,2} \gt 0$ such that

(10.20)\begin{equation} \begin{aligned} &\left|Z_3^{\mathrm{res}}\big(\varsigma_{\mathrm{c}}(t),\varsigma_{\mathrm{c}}(t),\varsigma_{\mathrm{c}}(t)\big)(0) - \mathcal{Z}(t)\right|\\ &\quad \ \leq C_1 \left(\int_\mathbb{R} \int_\mathbb{R} \left(|l_1| + |l_2|\right) \left|\mathrm{e}^{\hat{\Lambda}(l_1)(t+1)}\right|\left|\mathrm{e}^{\hat{\Lambda}(l_1-l_2)(t+1)}\right|\left|\mathrm{e}^{\hat{\Lambda}(l_2)(t+1)}\right| |A(t)|^3 \mathrm{d} l_1 \mathrm{d} l_2\right.\\ &\quad \ \qquad \left.+ \, \frac{1}{(1+t)^{\frac32}} |A(t)|^3\right) \\ &\quad \ \leq \frac{C_2}{(1+t)^{\frac{3}{2}}} |A(t)|^3, \end{aligned} \end{equation}

for $t \in [0,T_{\max})$. Using the standard integral

(10.21)\begin{equation} \int_\mathbb{R} \mathrm{e}^{-a^2 z^2 + bz} \mathrm{d} z = \frac{\sqrt{\pi}}{a}\mathrm{e}^{\frac{b^2}{4a^2}}, \end{equation}

for $a, b \in \mathbb{C}$ with $\operatorname{Re}(a^2) \gt 0$, cf. [Reference Gradshteyn and Ryzhik19, Integral 3.323.2], and recalling $j_0 - j_1 - j_2 - j_3 = 0$, we compute

(10.22)\begin{equation} \begin{aligned} &\int_\mathbb{R} \int_\mathbb{R} \mathrm{e}^{\left(\tilde{\lambda}_{j_1}(l_1) + \tilde{\lambda}_{j_2}(l_1-l_2) + \tilde{\lambda}_{j_3}(l_2)\right)(t+1)} \mathrm{d} l_1 \mathrm{d} l_2\\ &\quad \ = \frac{2 \pi \mathrm{e}^{(j_1+j_2+j_3) \mathrm{i} (t+1)}}{(t+1)\sqrt{3 \alpha^2 - j_1 j_2 - j_1 j_3 - j_2 j_3 - 2 \mathrm{i} \alpha \left(j_1 + j_2 + j_3\right) }}\\ &\quad \ = \frac{2\pi \mathrm{e}^{j_0 \mathrm{i} (t+1)}}{(t+1)\sqrt{3\alpha^2 + 1 - 2 \mathrm{i} \alpha j_0}}, \end{aligned} \end{equation}

for $j \in \mathcal{T}$ and $t \geq 0$. Thus, invoking (4.3), (10.19) and (10.22), we calculate

(10.23)\begin{equation} P_\pm(0) \mathcal{Z}(t) = \frac{2\pi\left(9\beta + 10 \kappa^2\right)\mathrm{e}^{\pm \mathrm{i}(t+1)}}{3(t+1) \sqrt{3\alpha^2 + 1 \mp 2 \mathrm{i} \alpha}} \left|\left(P_{\pm}(0) A(t)\right)_1\right|^2 \left(P_{\pm}(0) A(t)\right)_1 P_{\pm}(0) \mathbf{e}_2, \end{equation}

for $t \in [0,T_{\max})$. The desired estimate now follows from (4.4), (10.14), (10.20) and (10.23), which concludes the proof.

Proof of Theorem 1.1

We close a nonlinear iteration argument, controlling the variables $\widehat{U}_\mathrm{c}(k,t) = \chi(k) \widehat{U}(k,t)$, $\widehat{U}_\mathrm{s}(k,t) = (1-\chi(k))\widehat{U}(k,t)$, $\widehat{U}(k,t)$, and the remainder

\begin{equation*} \widehat{\rho}_{\mathrm{c}}(t) = \smash{\mathrm{e}^{\hat{\Lambda}(k)(t+1)} \mathrm{e}^{-\hat{\Lambda}(0)(t+1)}} \chi(k) \widehat{Z}_{\mathrm{c}}(0,t) - \widehat{Z}_{\mathrm{c}}(k,t), \end{equation*}

where we recall that $\widehat{Z}_{\mathrm{c}}(k,t)$ can be expressed in terms of $\widehat{U}_\mathrm{c}(k,t)$ through (9.2).

Template function. By (5.2), (5.6), (9.3) and (10.4), the template function $\eta \colon [0,T_{\max}) \to \mathbb{R}$ given by

\begin{align*} \eta(t) &= \sup_{0 \leq s \leq t} \left[(\log(2+s))^{\frac23} \left(\big\|\widehat{\rho}_{\mathrm{c}}(s)\big\|_{L^\infty} + \big\|\partial_k \widehat{\rho}_{\mathrm{c}}(s)\big\|_{L^1} + \sqrt{1+s} \big\|\widehat{\rho}_{\mathrm{c}}(s)\big\|_{L^1} \right) + \big\|\widehat{U}(s)\big\|_{L^\infty} \right. \\ &\left.\left.\phantom{\frac{\big\|\widehat{U}_\mathrm{c}\big\|_{L^\infty}}{(t)^{\frac13}}} + \big\|\partial_k \widehat{U}(s)\big\|_{L^1} + \sqrt{1+s} \left(\frac{\big\||\cdot|\widehat{U}_\mathrm{c}(s)\big\|_{L^\infty} + \big\||\cdot|\partial_k \widehat{U}_\mathrm{c}(s)\big\|_{L^1}}{(\log(2+s))^{\frac13}} + \big\|\widehat{U}(s)\big\|_{L^1} \right. \right.\right. \\ &\left.\left.\phantom{\frac{\big\|\widehat{U}_\mathrm{c}\big\|_{L^\infty}}{(t)^{\frac13}}} + \big\|\widehat{U}_\mathrm{s}(s)\big\|_{L^\infty} + \big\|\partial_k \widehat{U}_\mathrm{s}(s)\big\|_{L^1} \right) + (1+s)\left(\big\|\widehat{U}_\mathrm{s}(s)\big\|_{L^1} + \frac{\big\||\cdot|\widehat{U}_\mathrm{c}(s)\big\|_{L^1}}{(\log(2+s))^{\frac13}}\right)\right], \end{align*}

is well-defined, continuous, monotonically increasing and, if $T_{\max} \lt \infty$, then it satisfies

(11.1)\begin{equation} \lim_{t \uparrow T_{\max}} \eta(t) = \infty, \end{equation}

by (3.3).

Clearly, (5.4) implies

(11.2)\begin{equation} \big\|\partial_k^j |\cdot|^m \widehat{U}_\mathrm{c}(t)\big\|_{L^p} + \big\|\partial_k^j \widehat{U}_\mathrm{s}(t)\big\|_{L^p} \lesssim \big\|\widehat{U}(t)\big\|_{W^{j,p}}, \end{equation}

for $j = 0,1$, $m = 0,1$, $p = 1,\infty$ and $t \in [0,T_{\max})$, where we use that $\chi$ has compact support. Moreover, (9.2), (7.2) and (8.8) and the fact that $\chi$ has compact support, afford the estimate

(11.3)\begin{equation} \big\||\cdot|^m \partial_k^j \big(\widehat{Z}_{\mathrm{c}}(t) - \widehat{U}_\mathrm{c}(t)\big)\big\|_{L^p} \lesssim \big\|\widehat{U}_\mathrm{c}(t)\big\|_{W^{j,1}} \big\|\widehat{U}_\mathrm{c}(t)\big\|_{L^p}, \end{equation}

implying

(11.4)\begin{equation} \big\|\partial_k^j \widehat{Z}_{\mathrm{c}}(t) \big\|_{L^p} \lesssim \big\|\widehat{U}_\mathrm{c}(t)\big\|_{W^{j,p}} \lesssim \big\|\widehat{U}(t)\big\|_{W^{j,p}}, \end{equation}

for $m = 0,1$, $j = 0,1$, $p = 1,\infty$ and $t \in [0,T_{\max})$ with $\|\widehat{U}(t)\|_{W^{j,1} \cap L^\infty} \leq 1$. Furthermore, thanks to (4.4), (10.3) and (11.4) it holds

(11.5)\begin{equation} r_\pm(t) = \frac12 |A(t)| = \frac12 \big|\widehat{Z}_{\mathrm{c}}(0,t)\big| \lesssim \big\|\widehat{U}(t)\big\|_{L^\infty}, \end{equation}

for all $t \in [0,T_{\max})$ with $\|\widehat{U}(t)\|_{W^{j,1} \cap L^\infty} \leq 1$. Finally, from (10.8) we have

(11.6)\begin{equation} \begin{aligned} \big\|\partial_k^j \widehat{\rho}_{\mathrm{c}}(t)\big\|_{L^p} &\leq \big\|\partial_k^j \varsigma_{\mathrm{c}}(t)\big\|_{L^p} + \big\|\partial_k^j \widehat{Z}_{\mathrm{c}}(t)\big\|_{L^p} \lesssim (1+t)^{-\frac{1}{2p} + \frac{j}{2}} |A(t)| + \big\|\partial_k^j \widehat{Z}_{\mathrm{c}}(t)\big\|_{L^p}\\ &\lesssim (1+t)^{-\frac{1}{2p} + \frac{j}{2}} \big\|\widehat{U}(t)\big\|_{L^\infty} + \big\|\widehat{U}(t)\big\|_{W^{j,p}}, \end{aligned} \end{equation}

for $j = 0,1$, $p = 1,\infty$ and $t \in [0,T_{\max})$ with $\|\widehat{U}(t)\|_{W^{j,1} \cap L^\infty} \leq 1$. Thus, using (5.1), (11.2) and (11.6), we conclude that there exists a constant $K_0 \geq 1$, independent of $E_0$, such that

(11.7)\begin{equation} \eta(0) \leq K_0E_0. \end{equation}

Approach. Our goal is to prove that there exist constants $C \geq K_0$ and $\eta_0 \in (0,1)$ such that for all $t \in [0,T_{\max})$ with $\eta(t) \leq \eta_0$ the key inequality

(11.8)\begin{equation} \eta(t) \leq C\left(E_0 + \eta(t)^2\right), \end{equation}

is fulfilled. Then, taking

\begin{equation*}{\varepsilon} = \min\left\{\frac{1}{4C^2},\frac{\eta_0}{2C}\right\},\end{equation*}

it follows by the continuity of $\eta$ that, provided $E_0 \in (0,{\varepsilon})$, we have $\eta(t) \leq 2CE_0 \leq \eta_0$ for all $t \in [0,T_{\max})$. Indeed, given $t \in [0,T_{\max})$ with $\eta(t) \gt 2CE_0$, there must, by continuity of $\eta$ and the fact that $\eta(0) \lt 2CE_0$ by (11.7), exist $s \in (0,t]$ with $\eta(s) = 2CE_0 \leq \eta_0$. Now (11.8) yields

\begin{equation*}\eta(s) \leq C\left(E_0 + 4C^2E_0^2\right) \lt 2CE_0,\end{equation*}

contradicting $\eta(s) = 2CE_0$. We conclude that, if the key inequality (11.8) holds, then we have $\eta(t) \leq 2CE_0$ for all $t \in [0,T_{\max})$. In this case, we must have $T_{\max} = \infty$ by (11.1). We then infer $\eta(t) \leq 2CE_0$ for all $t \geq 0$, which leads to the estimates (1.10) and (1.11). Hence, all that remains is to prove the key inequality (11.8) and derive the estimates (1.10) and (1.11).

Controlling $r_\pm(t)$. First of all, by (11.5), there exists an $E_0$-independent constant $K_1 \gt 0$ such that

(11.9)\begin{equation} r_\pm(0) \leq K_1 E_0. \end{equation}

Define the set

\begin{align*} \mathcal{S} = \left\{t \in [0,T_{\max}) : r_\pm(t) \sqrt{\left(K_1E_0 + \eta(t)^{\frac32}\right)^{-2} - 2 \operatorname{Re}(\omega_\pm) \log(1+t)} \leq 2\right\}. \end{align*}

By (11.9), it holds

(11.10)\begin{equation} r_\pm(0) \sqrt{\left(K_1E_0 + \eta(0)^{\frac32}\right)^{-2}} \leq 1, \end{equation}

implying $0 \in \mathcal{S}$. Our aim is to show that, we have, $\mathcal{S} = \{t \in [0,T_{\max}) : \eta(t) \leq \eta_0\}$, provided $\eta_0 \gt 0$ is sufficiently small (but independent of $E_0$). We argue by contradiction and assume that there exists $t \in [0,T_{\max}) \setminus \mathcal{S}$ with $\eta(t) \leq \eta_0$. So, by identity (11.10) and continuity of $r_\pm$ and $\eta$, there must exist $t_1 \in (0,t)$ such that

\begin{align*} r_\pm(s)\sqrt{\left(K_1E_0 + \eta(s)^{\frac32}\right)^{-2} - 2 \operatorname{Re}(\omega_\pm) \log(1+s)} \geq 1,\\ r_\pm(t_1)\sqrt{\left(K_1E_0 + \eta(t_1)^{\frac32}\right)^{-2} - 2 \operatorname{Re}(\omega_\pm) \log(1+t_1)} = 1, \end{align*}

for $s \in [t_1,t]$, implying

(11.11)\begin{equation} \begin{aligned} r_\pm(s)^{-2} &\leq \left(K_1E_0 + \eta(s)^{\frac32}\right)^{-2} - 2 \operatorname{Re}(\omega_\pm) \log(1+s),\\ r_\pm(t_1)^{-2} &= \left(K_1E_0 + \eta(t_1)^{\frac32}\right)^{-2} - 2 \operatorname{Re}(\omega_\pm) \log(1+t_1), \end{aligned} \end{equation}

for $s \in [t_1,t]$. Integrating (10.29) from $t_1$ to $t$ and employing (11.11), we arrive at

(11.12)\begin{equation} \begin{aligned} r_\pm(t)^{-2} &= \left(K_1E_0 + \eta(t_1)^{\frac32}\right)^{-2} - 2\operatorname{Re}(\omega_\pm) \log(t+1)\\ &\qquad -\, 2\int_{t_1}^t \operatorname{Re}\left(\frac{\mathrm{e}^{-\mathrm{i} \psi_\pm(s)\mp \mathrm{i} (s+1)}}{r_\pm(s)^3} \left(P_\pm(0) \mathcal{E}_4(0,s) + \mathcal{E}_6(s)\right)_1\right) \mathrm{d} s. \end{aligned} \end{equation}

The next steps are devoted to estimating the integral term in (11.12). Lemmas 9.2 and 10.2, Young’s inequality and estimate (11.5) yield a $t$- and $E_0$-independent constant $C_1 \gt 0$ such that

(11.13)\begin{equation} \begin{aligned} |\mathcal{E}_4(0,s)| &\leq C_1\frac{\eta(s)^2 (\log(2+s))^{\frac13}}{(1+s)^{\frac32}},\\ |\mathcal{E}_6(s)| &\leq \frac{C_1}{1+s} \left(\frac{\eta(s)^3}{(\log(2+s))^{2}} + \frac{\eta(s) r_\pm(s)^2}{(\log(2+s))^{\frac23}}\right), \end{aligned} \end{equation}