Now we derive a second-order numerical method for (1.1). Since the employed approximations form the basis of our error analysis, we present some construction details. For an alternative derivation of this method, we refer to [3].
Let \(\tau >0\) be the time step size and \(t_{n}=n\tau \), \(n\ge 0\) the temporal grid points. First, by employing the twisted variable \(v=\mathrm{e}^{-it\Delta }u\) and Duhamel’s formula, we get
$$ v(t_{n}+\sigma )=v(t_{n})+i \int _{0}^{\sigma }\mathrm{e}^{-i(t_{n}+ \rho )\Delta } \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+\rho )\Delta }v(t_{n}+ \rho ) \bigr\vert ^{2} \mathrm{e}^{i(t_{n}+\rho )\Delta }v(t_{n}+\rho ) \bigr)\,d\rho . $$
(3.1)
Then, freezing the nonlinear interaction by approximating \(\mathrm{e}^{i(t_{n}+\rho )\Delta }\approx \mathrm{e}^{i(t_{n}+\sigma ) \Delta }\) and \(v(t_{n}+\rho )\approx v(t_{n})\), we get
$$\begin{aligned} v(t_{n}+\sigma )= v(t_{n})+i\sigma \mathrm{e}^{-i(t_{n}+\sigma ) \Delta } \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+\sigma )\Delta }v(t_{n}) \bigr\vert ^{2} \mathrm{e}^{i(t_{n}+\sigma )\Delta }v(t_{n}) \bigr) + \mathcal{R}_{3}^{n}(v, \sigma ). \end{aligned}$$
(3.2)
The remainder term \(\mathcal{R}_{3}^{n}(v,\sigma )\) satisfies the following estimate.
Lemma 3.1
Let \(\gamma >\frac{d}{2}\), \(\sigma \in [0,\tau ]\) and \(v\in L^{\infty }((0,T);H^{\gamma +2})\). Then,
$$ \bigl\Vert \mathcal{R}_{3}^{n}(v,\sigma ) \bigr\Vert _{H^{\gamma }} \lesssim \tau ^{2} \bigl( \Vert v \Vert _{L^{\infty }((0,T);H^{\gamma +2})}+ \Vert v \Vert _{L^{\infty }((0,T);H^{ \gamma +2})}^{3} \bigr). $$
We postpone the proof of the lemma to Sect. 3.1.
Next, we derive a second-order expansion of Duhamel’s formula
$$ v(t_{n}+\tau )=v(t_{n})+i \int _{0}^{\tau }\mathrm{e}^{-i(t_{n}+\sigma ) \Delta } \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+\sigma )\Delta }v(t_{n}+\sigma ) \bigr\vert ^{2} \mathrm{e}^{i(t_{n}+\sigma )\Delta }v(t_{n}+\sigma ) \bigr)\,d \sigma . $$
(3.3)
Replacing \(v(t_{n}+\sigma )\) by (3.2), we infer that
$$\begin{aligned} v(t_{n+1}) = v(t_{n})+I_{1}(t_{n})+I_{2}(t_{n})+ \mathcal{R}_{4}^{n}(v), \end{aligned}$$
(3.4)
where
$$\begin{aligned} &I_{1}(t_{n})=i \int _{0}^{\tau }\mathrm{e}^{-i(t_{n}+s)\Delta } \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}) \bigr\vert ^{2} \mathrm{e}^{i(t_{n}+s) \Delta }v(t_{n}) \bigr)\,ds, \\ &I_{2}(t_{n})=- \int _{0}^{\tau }s\mathrm{e}^{-i(t_{n}+s)\Delta } \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}) \bigr\vert ^{4} \mathrm{e}^{i(t_{n}+s) \Delta }v(t_{n}) \bigr)\,ds. \end{aligned}$$
(3.5)
The remainder term \(\mathcal{R}_{4}^{n}(v)\) can be bounded as stated in the following lemma. Again, the proof of this lemma is postponed to Sect. 3.1.
Lemma 3.2
Let \(\gamma >\frac{d}{2}\) and \(0<\tau \leq 1\). Then, for \(v\in L^{\infty }((0,T);H^{\gamma +2})\),
$$ \bigl\Vert \mathcal{R}_{4}^{n}(v) \bigr\Vert _{H^{\gamma }}\le C\tau ^{3}, $$
where the constant C only depends on \(\|v\|_{L^{\infty }((0,T);H^{\gamma +2})}\).
Due to the complexity of the phase functions
$$ \phi _{3}= \vert \boldsymbol{\xi } \vert ^{2}+ \vert \boldsymbol{\xi }_{1} \vert ^{2}- \vert \boldsymbol{\xi }_{2} \vert ^{2}- \vert \boldsymbol{\xi }_{3} \vert ^{2},\qquad \phi _{5}= \vert \boldsymbol{\xi } \vert ^{2}+ \vert \boldsymbol{\xi }_{1} \vert ^{2}+ \vert \boldsymbol{\xi }_{2} \vert ^{2}- \vert \boldsymbol{\xi }_{3} \vert ^{2}- \vert \boldsymbol{\xi }_{4} \vert ^{2}- \vert \boldsymbol{\xi }_{5} \vert ^{2}, $$
we note that the terms in \(I_{1}\) and \(I_{2}\) cannot be easily expressed in physical space.
Therefore, we consider \(I_{1}\) first in Fourier space. Using
$$\begin{aligned} \mathrm{e}^{is\Delta } w(\boldsymbol{x})= \int \mathrm{e}^{is \boldsymbol{x}\cdot \boldsymbol{\eta }} \mathrm{e}^{-is|\boldsymbol{\eta }|^{2}} \hat{w} ( \boldsymbol{\eta }) (d\boldsymbol{\eta }), \end{aligned}$$
(3.6)
we get
$$\begin{aligned} \widehat{I_{1}}(t_{n},\boldsymbol{\xi })=i \int _{0}^{\tau } \int _{ \boldsymbol{\xi }=\boldsymbol{\xi }_{1}+\boldsymbol{\xi }_{2}+ \boldsymbol{\xi }_{3}}\mathrm{e}^{i(t_{n}+s)\phi _{3}} \hat{\bar{v}}(t_{n}, \boldsymbol{\xi }_{1})\hat{v}(t_{n},\boldsymbol{\xi }_{2})\hat{v}(t_{n}, \boldsymbol{\xi }_{3}) (d \boldsymbol{\xi }_{1}) (d\boldsymbol{\xi }_{2})\,ds. \end{aligned}$$
The main problem concerns the handling of the phase \(\mathrm{e}^{is\phi _{3}}\). Defining
$$ \alpha =2 \vert \boldsymbol{\xi }_{1} \vert ^{2}, \qquad \beta =2 \boldsymbol{\xi }_{1} \cdot \boldsymbol{\xi }_{2}+2 \boldsymbol{\xi }_{1}\cdot \boldsymbol{\xi }_{3}+2 \boldsymbol{\xi }_{2}\cdot \boldsymbol{\xi }_{3} $$
allows us to write
$$\begin{aligned} \mathrm{e}^{is\phi _{3}}=\mathrm{e}^{is\alpha +is\beta }. \end{aligned}$$
Applying the formulas presented in Lemma 2.2, we get
$$\begin{aligned} \widehat{I_{1}}(t_{n},\boldsymbol{\xi })={}&i\tau \int _{ \boldsymbol{\xi }=\boldsymbol{\xi }_{1}+\boldsymbol{\xi }_{2}+ \boldsymbol{\xi }_{3}} \varphi (i\tau \alpha ) \mathrm{e}^{it_{n}\phi _{3}} \hat{\bar{v}}(t_{n},\boldsymbol{\xi }_{1}) \hat{v}(t_{n},\boldsymbol{\xi }_{2}) \hat{v}(t_{n}, \boldsymbol{\xi }_{3}) (d\boldsymbol{\xi }_{1}) (d \boldsymbol{\xi }_{2}) \\ & {} -i\tau \int _{\boldsymbol{\xi }=\boldsymbol{\xi }_{1}+ \boldsymbol{\xi }_{2}+\boldsymbol{\xi }_{3}} \bigl(\mathrm{e}^{i\tau \beta }-1 \bigr)\psi (i\tau \alpha ) \mathrm{e}^{it_{n}\phi _{3}} \hat{\bar{v}}(t_{n},\boldsymbol{\xi }_{1})\hat{v}(t_{n},\boldsymbol{\xi }_{2}) \hat{v}(t_{n},\boldsymbol{\xi }_{3}) (d\boldsymbol{\xi }_{1}) (d \boldsymbol{\xi }_{2}) \\ & {} +\widehat{\mathcal{R}_{5}^{n}}(v) (\boldsymbol{\xi }), \end{aligned}$$
(3.7)
where the remainder term \(\mathcal{R}_{5}^{n}(v)\) obeys the bound given in the following lemma. Its proof will be postponed to Sect. 3.1.
Lemma 3.3
Let \(\gamma >\frac{d}{2}\) and \(v\in L^{\infty }((0,T);H^{\gamma +2})\). Then,
$$ \bigl\Vert \mathcal{R}_{5}^{n}(v) \bigr\Vert _{H^{\gamma }} \lesssim \tau ^{3} \Vert v \Vert _{L^{\infty }((0,T);H^{\gamma +2})}^{3}. $$
Using \(\beta =\phi _{3}-\alpha \) and (3.6), we transform (3.7) back to physical space to get
$$\begin{aligned} {I_{1}(t_{n})}={}&i\tau \mathrm{e}^{-it_{n}\Delta } \bigl\{ \bigl[ \varphi (-2i\tau \Delta )\mathrm{e}^{-it_{n}\Delta }\bar{v}(t_{n}) \bigr]\cdot \bigl(\mathrm{e}^{it_{n}\Delta }v(t_{n}) \bigr)^{2} \bigr\} \\ & {} -i\tau \mathrm{e}^{-it_{n+1}\Delta } \bigl\{ \bigl[ \psi (-2i\tau \Delta ) \mathrm{e}^{-it_{n-1}\Delta }\bar{v}(t_{n}) \bigr]\cdot \bigl( \mathrm{e}^{it_{n+1}\Delta }v(t_{n}) \bigr)^{2} \bigr\} \\ & {} +i\tau \mathrm{e}^{-it_{n}\Delta } \bigl\{ \bigl[ \psi (-2i\tau \Delta ) \mathrm{e}^{-it_{n}\Delta }\bar{v}(t_{n}) \bigr]\cdot \bigl( \mathrm{e}^{it_{n}\Delta }v(t_{n}) \bigr)^{2} \bigr\} + \mathcal{R}_{5}^{n}(v) . \end{aligned}$$
(3.8)
The term \(I_{2}\) is of higher order in τ. Therefore, it is sufficient to freeze the linear flow and approximate the term as
$$\begin{aligned} I_{2}(t_{n})&=- \int _{0}^{\tau }s\mathrm{e}^{-it_{n}\Delta } \bigl( \bigl\vert \mathrm{e}^{it_{n}\Delta }v(t_{n}) \bigr\vert ^{4} \mathrm{e}^{it_{n}\Delta }v(t_{n}) \bigr)\,ds+ \mathcal{R}_{6}^{n}(v) \end{aligned}$$
(3.9)
$$\begin{aligned} &= -\frac{1}{2} \tau ^{2}\mathrm{e}^{-it_{n}\Delta } \bigl( \bigl\vert \mathrm{e}^{it_{n} \Delta }v(t_{n}) \bigr\vert ^{4} \mathrm{e}^{it_{n}\Delta }v(t_{n}) \bigr)+ \mathcal{R}_{6}^{n}(v), \end{aligned}$$
(3.10)
where the remainder term \(\mathcal{R}_{6}^{n}(v)\) obeys the bound given in the following lemma. Again, its proof will be postponed to Sect. 3.1.
Lemma 3.4
Let \(\gamma >\frac{d}{2}\) and \(v\in L^{\infty }((0,T);H^{\gamma +2})\). Then
$$ \bigl\Vert \mathcal{R}_{6}^{n}(v) \bigr\Vert _{H^{\gamma }} \lesssim \tau ^{3} \Vert v \Vert _{L^{\infty }((0,T);H^{\gamma +2})}^{5}. $$
Now combining (3.4), (3.8), and (3.10), we have that
$$\begin{aligned} v(t_{n+1})= \Phi ^{n} \bigl(v(t_{n}) \bigr)+\mathcal{R}_{4}^{n}(v)+ \mathcal{R}_{5}^{n}(v)+ \mathcal{R}_{6}^{n}(v), \end{aligned}$$
(3.11)
where the operator \(\Phi ^{n}\) is defined by
$$ \begin{aligned} \Phi ^{n} (f ) ={}&f+i\tau \mathrm{e}^{-it_{n}\Delta } \bigl\{ \bigl( \varphi (-2i\tau \Delta ) \mathrm{e}^{-it_{n}\Delta }\bar{f} \bigr)\cdot \bigl(\mathrm{e}^{it_{n}\Delta }f \bigr)^{2} \bigr\} \\ &{}-i\tau \mathrm{e}^{-it_{n+1}\Delta } \bigl\{ \bigl( \psi (-2i\tau \Delta ) \mathrm{e}^{-it_{n-1}\Delta }\bar{f} \bigr)\cdot \bigl(\mathrm{e}^{it_{n+1} \Delta }f \bigr)^{2} \bigr\} \\ &{}+i\tau \mathrm{e}^{-it_{n}\Delta } \bigl\{ \bigl( \psi (-2i\tau \Delta ) \mathrm{e}^{-it_{n}\Delta }\bar{f} \bigr)\cdot \bigl(\mathrm{e}^{it_{n} \Delta }f \bigr)^{2} \bigr\} \\ &{}-\frac{1}{2} \tau ^{2}\mathrm{e}^{-it_{n}\Delta } \bigl( \bigl\vert \mathrm{e}^{it_{n} \Delta }f \bigr\vert ^{4} \mathrm{e}^{it_{n}\Delta }f \bigr). \end{aligned} $$
(3.12)
Our second-order low-regularity integrator is obtained by dropping the remainder terms \(\mathcal{R}_{4}^{n}\), \(\mathcal{R}_{5}^{n}\), \(\mathcal{R}_{6}^{n}\) in (3.11). The method for the twisted variable is summarized as follows: let \(v^{0}=u_{0}\) and
$$\begin{aligned} v^{n+1}=& \Phi ^{n} \bigl(v^{n} \bigr) \quad \text{for } n\geq 0. \end{aligned}$$
(3.13)
Finally, setting \(u^{n}=\mathrm{e}^{it_{n}\Delta }v^{n}\), we obtain the announced numerical scheme (1.6) for the NLS equation (1.1).
3.1 Estimates of the remainder terms
Now we prove Lemmas 3.1 to 3.4.
Proof of Lemma 3.1
By (3.2), we have that
$$\begin{aligned} &\mathcal{R}_{3}^{n}(v,s) \\ &\quad =i \int _{0}^{s} \bigl(\mathrm{e}^{-i(t_{n}+ \sigma )\Delta }- \mathrm{e}^{-i(t_{n}+s)\Delta } \bigr) \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+\sigma )\Delta }v(t_{n}+ \sigma ) \bigr\vert ^{2} \mathrm{e}^{i(t_{n}+\sigma )\Delta }v(t_{n}+ \sigma ) \bigr)\,d\sigma \\ &\qquad {}+ i \int _{0}^{s} \mathrm{e}^{-i(t_{n}+s)\Delta } \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+\sigma )\Delta }v(t_{n}+\sigma ) \bigr\vert ^{2}- \bigl\vert \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}+s) \bigr\vert ^{2} \bigr) \mathrm{e}^{i(t_{n}+ \sigma )\Delta }v(t_{n}+\sigma ) \,d\sigma \\ &\qquad {}+i \int _{0}^{s} \mathrm{e}^{-i(t_{n}+s)\Delta } \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}+s) \bigr\vert ^{2} \bigl( \mathrm{e}^{i(t_{n}+ \sigma )\Delta }v(t_{n}+\sigma )- \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}) \bigr) \bigr)\,d\sigma . \end{aligned}$$
Note that from (3.1), Lemma 2.1(i), and the Sobolev embedding, we get
$$ \sup_{0\le \sigma \le \tau } \bigl\Vert v(t_{n}+\sigma )-v(t_{n}) \bigr\Vert _{H^{ \gamma }} \lesssim \tau \Vert v \Vert _{L^{\infty }((0,T);H^{\gamma })}^{3}. $$
Moreover, for any \(f\in H^{\gamma }\),
$$\begin{aligned} \bigl\Vert \bigl(\mathrm{e}^{-i(t_{n}+\sigma )\Delta }- \mathrm{e}^{-i(t_{n}+s) \Delta } \bigr)f \bigr\Vert _{H^{\gamma }} \lesssim \vert \sigma -s \vert \Vert f \Vert _{H^{ \gamma +2}}. \end{aligned}$$
(3.14)
Applying these two estimates, we obtain
$$ \bigl\Vert \mathrm{e}^{i(t_{n}+\sigma )\Delta }v(t_{n}+\sigma )- \mathrm{e}^{i(t_{n}+s) \Delta }v(t_{n}) \bigr\Vert _{H^{\gamma }} \lesssim \tau \bigl( \Vert v \Vert _{L^{\infty }((0,T);H^{ \gamma +2})}+ \Vert v \Vert _{L^{\infty }((0,T);H^{\gamma +2})}^{3} \bigr) $$
and thus
$$\begin{aligned} \bigl\Vert \mathcal{R}_{3}^{n}(v,s) \bigr\Vert _{H^{\gamma }} \lesssim \tau ^{2} \bigl( \Vert v \Vert _{L^{\infty }((0,T);H^{\gamma +2})}+ \Vert v \Vert _{L^{\infty }((0,T);H^{ \gamma +2})}^{3} \bigr). \end{aligned}$$
This is the desired result. □
Proof of Lemma 3.2
Inserting (3.2) with \(\sigma =\rho \) in (3.1) and using (3.4), we find that the remainder \(\mathcal{R}_{4}^{n}(v)\) consists of terms of the form
$$ i \int _{0}^{\tau }\mathrm{e}^{-i(t_{n}+s)\Delta } \bigl( \mathrm{e}^{i(t_{n}+s) \Delta }\mathcal{W}_{j}\cdot \mathrm{e}^{-i(t_{n}+s)\Delta } \overline{\mathcal{W}_{k}}\cdot \mathrm{e}^{i(t_{n}+s)\Delta } \mathcal{W}_{\ell } \bigr)\,ds,\quad j+k+\ell \ge 5, $$
where
$$\begin{aligned} &\mathcal{W}_{1} = v(t_{n}), \\ & \mathcal{W}_{2} = is\mathrm{e}^{-i(t_{n}+s) \Delta } \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}) \bigr\vert ^{2} \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}) \bigr), \\ & \mathcal{W}_{3} = \mathcal{R}_{3}^{n}(v,s). \end{aligned}$$
By Lemma 3.1 and Lemma 2.1(i), we thus get
$$\begin{aligned} \bigl\Vert \mathcal{R}_{4}^{n}(v) \bigr\Vert _{H^{\gamma }} \lesssim C \bigl( \bigl\Vert v(t_{n}) \bigr\Vert _{L^{\infty }((0,t);H^{\gamma +2})} \bigr)\tau ^{3}. \end{aligned}$$
This finishes the proof of the lemma. □
Proof of Lemma 3.3
Without loss of generality, we may assume that \(\hat{v}(t_{n})\) and \(\hat{\bar{v}}(t_{n})\) are positive (otherwise, one may replace them with their absolute values).
From Lemma 2.2, we have
$$\begin{aligned} \widehat{\mathcal{R}_{5}^{n}}(v) (\boldsymbol{\xi }) = \int _{ \boldsymbol{\xi }=\boldsymbol{\xi }_{1}+\boldsymbol{\xi }_{2}+ \boldsymbol{\xi }_{3}}\mathcal{R}_{2}(\alpha ,\beta ,\tau ) \mathrm{e}^{it_{n}\phi _{3}} \hat{\bar{v}}(t_{n},\boldsymbol{\xi }_{1}) \hat{v}(t_{n},\boldsymbol{\xi }_{2}) \hat{v}(t_{n},\boldsymbol{\xi }_{3}) (d\boldsymbol{\xi }_{1}) (d\boldsymbol{\xi }_{2}) \end{aligned}$$
and further
$$\begin{aligned} \bigl\vert \widehat{\mathcal{R}_{5}^{n}}(v) ( \boldsymbol{\xi }) \bigr\vert \lesssim \tau ^{3} \int _{\boldsymbol{\xi }=\boldsymbol{\xi }_{1}+ \boldsymbol{\xi }_{2}+\boldsymbol{\xi }_{3}} \beta ^{2} \hat{\bar{v}}(t_{n}, \boldsymbol{\xi }_{1})\hat{v}(t_{n},\boldsymbol{\xi }_{2})\hat{v}(t_{n}, \boldsymbol{\xi }_{3}) (d \boldsymbol{\xi }_{1}) (d\boldsymbol{\xi }_{2}). \end{aligned}$$
By symmetry, we may assume that \(|\boldsymbol{\xi }_{1}|\ge |\boldsymbol{\xi }_{2}|\ge | \boldsymbol{\xi }_{3}|\). This yields
$$\begin{aligned} \langle \boldsymbol{\xi }\rangle ^{\gamma }\beta ^{2} &\lesssim \langle \boldsymbol{\xi }\rangle ^{\gamma } \bigl( \vert \boldsymbol{\xi }_{1} \vert ^{2} \vert \boldsymbol{\xi }_{2} \vert ^{2}+ \vert \boldsymbol{\xi }_{1} \vert ^{2} \vert \boldsymbol{\xi }_{3} \vert ^{2}+ \vert \boldsymbol{\xi }_{2} \vert ^{2} \vert \boldsymbol{\xi }_{3} \vert ^{2} \bigr) \\ &\lesssim \vert \boldsymbol{\xi }_{1} \vert ^{2+\gamma } \vert \boldsymbol{\xi }_{2} \vert ^{2}. \end{aligned}$$
Using this estimate, we get
$$\begin{aligned} &\langle \boldsymbol{\xi }\rangle ^{\gamma } \bigl\vert \widehat{ \mathcal{R}_{5}^{n}}(v) (\boldsymbol{\xi }) \bigr\vert \\ &\quad \lesssim \tau ^{3} \int _{\boldsymbol{\xi }=\boldsymbol{\xi }_{1}+ \boldsymbol{\xi }_{2}+\boldsymbol{\xi }_{3}, \vert \boldsymbol{\xi }_{1} \vert \ge \vert \boldsymbol{\xi }_{2} \vert \ge \vert \boldsymbol{\xi }_{3} \vert } \vert \boldsymbol{\xi }_{1} \vert ^{2+ \gamma } \vert \boldsymbol{\xi }_{2} \vert ^{2} \hat{\bar{v}}(t_{n}, \boldsymbol{\xi }_{1}) \hat{v}(t_{n},\boldsymbol{\xi }_{2})\hat{v}(t_{n}, \boldsymbol{\xi }_{3}) (d \boldsymbol{\xi }_{1}) (d \boldsymbol{\xi }_{2}) \\ &\quad \lesssim \tau ^{3}\mathcal{F} \bigl( ({-}\Delta )^{1+\gamma /2} \bar{v}\cdot ({-}\Delta )v\cdot v \bigr) (t_{n}, \boldsymbol{\xi }). \end{aligned}$$
Therefore, by Plancherel’s identity and Lemma 2.1(ii) with \(\delta =0\), we obtain that for any \(\gamma _{1}>\frac{d}{2}\),
$$\begin{aligned} \bigl\Vert \mathcal{R}_{5}^{n}(v) \bigr\Vert _{H^{\gamma }} &\lesssim \tau ^{3} \bigl\Vert ({-}\Delta )^{1+\gamma /2}\bar{v}\cdot ( {-}\Delta )v\cdot v \bigr\Vert _{L^{\infty }((0,T);L^{2})} \\ &\lesssim \tau ^{3} \Vert v \Vert _{L^{\infty }((0,T);H^{\gamma +2})} \Vert v \Vert _{L^{\infty }((0,T);H^{\gamma _{1}+2})} \Vert v \Vert _{L^{\infty }((0,T);H^{\gamma _{1}})}. \end{aligned}$$
Since \(\gamma >\frac{d}{2}\), choosing \(\gamma _{1}=\gamma \), we get the desired result. □
Proof of Lemma 3.4
By (3.5) and (3.9), we have that
$$\begin{aligned} \mathcal{R}_{6}^{n}= {}&- \int _{0}^{\tau }s \bigl(\mathrm{e}^{-i(t_{n}+s) \Delta }- \mathrm{e}^{-it_{n}\Delta } \bigr) \bigl( \bigl\vert \mathrm{e}^{i(t_{n}+s) \Delta }v(t_{n}) \bigr\vert ^{4} \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}) \bigr) \,ds \\ &{}- \int _{0}^{\tau }s \mathrm{e}^{-it_{n}\Delta } \bigl( \bigl[ \bigl\vert \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}) \bigr\vert ^{4} - \bigl\vert \mathrm{e}^{it_{n} \Delta }v(t_{n}) \bigr\vert ^{4} \bigr] \mathrm{e}^{i(t_{n}+s)\Delta }v(t_{n}) \bigr) \,ds \\ &{}- \int _{0}^{\tau }s \mathrm{e}^{-it_{n}\Delta } \bigl( \bigl\vert \mathrm{e}^{it_{n} \Delta }v(t_{n}) \bigr\vert ^{4}\cdot \bigl( \mathrm{e}^{-i(t_{n}+s)\Delta }- \mathrm{e}^{-it_{n}\Delta } \bigr)v(t_{n}) \bigr)\,ds. \end{aligned}$$
Then, the claimed result follows directly from (3.14) and Lemma 2.1(i). □