Lemma 2.1
Suppose that
\(E_{0}(x)\in L^{2}(\mathbf{R}^{2})\). Then, for the solution of problem (6)∼(9), we have
$$\Vert E\Vert ^{2}_{L^{2}(\mathbf{R}^{2})}= \bigl\Vert E_{0}(x) \bigr\Vert ^{2}_{L^{2}(\mathbf{R}^{2})}. $$
Proof
Taking the inner product of (6) and E, then taking the imaginary part, we have
$$\begin{aligned} &\operatorname {Im}(iE_{t},E )=\operatorname {Re}(E_{t},E )=\frac{1}{2} \frac {d}{dt}\Vert E\Vert ^{2}_{L^{2}}, \\ &\operatorname {Im}\bigl(\Delta E-H^{2}\Delta^{2}E-nE,E \bigr)=0. \end{aligned}$$
Hence, we get
$$\frac{d}{dt}\Vert E\Vert ^{2}_{L^{2}}=0. $$
We thus get Lemma 2.1. □
Lemma 2.2
Sobolev’s estimations
Assume that
\(u\in L^{q}(\mathbf{R}^{n})\), \(D^{m}u\in L^{r}(\mathbf{R}^{n})\), \(1\leq q,r\leq \infty, 0\leq j\leq m\), we have the estimations
$$\bigl\Vert D^{j}u \bigr\Vert _{L^{p}(\mathbf{R}^{n})}\leq C \bigl\Vert D^{m}u \bigr\Vert ^{\alpha}_{L^{r}(\mathbf{R}^{n})}\Vert u\Vert ^{1-\alpha}_{L^{q}(\mathbf{R}^{n})}, $$
where
C
is a positive constant, \(0\leq\frac{j}{m}\leq \alpha\leq1\),
$$\frac{1}{p}=\frac{j}{n}+\alpha \biggl(\frac{1}{r}- \frac{m}{n} \biggr)+(1-\alpha)\frac{1}{q}. $$
Lemma 2.3
Suppose that
\(E_{0}(x)\in H^{2}(\mathbf{R}^{2})\), \(n_{0}(x)\in H^{1}(\mathbf{R}^{2})\), \(\varphi_{0}(x)\in H^{1}(\mathbf{R}^{2})\). Then we have
$$\begin{aligned} \mathscr{F}(t)=\mathscr{F}(0), \end{aligned}$$
where
$$\begin{aligned} \mathscr{F}(t)=\Vert \nabla E\Vert ^{2}_{L^{2}}+H^{2} \Vert \Delta E\Vert ^{2}_{L^{2}}+ \int_{\mathbf{R}^{2}}n\vert E\vert ^{2}\,\mathrm{d} x+ \frac{1}{2}\Vert \nabla\varphi \Vert ^{2}_{L^{2}}+ \frac {1}{2}\Vert n\Vert ^{2}_{L^{2}}+ \frac{H^{2}}{2}\Vert \nabla n\Vert ^{2}_{L^{2}}. \end{aligned}$$
Proof
Take the inner products of (6) and \(-E_{t}\). Since
$$\begin{aligned} &\operatorname {Re}(iE_{t},-E_{t})=0, \qquad \operatorname {Re}(\Delta E,-E_{t})= \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla E\Vert ^{2}_{L^{2}}, \\ &\operatorname {Re}\bigl(-H^{2}\Delta^{2}E,-E_{t} \bigr)= \frac{H^{2}}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert \Delta E\Vert ^{2}_{L^{2}}, \\ &\operatorname {Re}(-n E,-E_{t})=\frac{1}{2} \int_{\mathbf{R}^{2}}n\vert E\vert ^{2}_{t} \,\mathrm{d} x \\ &\phantom{\operatorname {Re}(-n E,-E_{t})}=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbf{R}^{2}}n\vert E\vert ^{2}\,\mathrm{d} x- \frac{1}{2} \int_{\mathbf{R}^{2}}n_{t}\vert E\vert ^{2} \,\mathrm{d} x \\ &\phantom{\operatorname {Re}(-n E,-E_{t})}=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbf{R}^{2}}n\vert E\vert ^{2}\,\mathrm{d} x- \frac{1}{2} \int_{\mathbf{R}^{2}}n_{t} \bigl(\varphi_{t}-n+H^{2} \Delta n \bigr)\,\mathrm{d} x \\ &\phantom{\operatorname {Re}(-n E,-E_{t})}=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbf{R}^{2}}n\vert E\vert ^{2}\,\mathrm{d} x- \frac{1}{2} \int_{\mathbf{R}^{2}}n_{t}\varphi_{t}\,\mathrm{d} x+ \frac{1}{4}\frac{\mathrm{d}}{\mathrm{d} t}\Vert n\Vert ^{2}_{L^{2}}+ \frac{H^{2}}{4}\frac{\mathrm {d}}{\mathrm{d} t}\Vert \nabla n\Vert ^{2}_{L^{2}} \\ &\phantom{\operatorname {Re}(-n E,-E_{t})}=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbf{R}^{2}}n\vert E\vert ^{2}\,\mathrm{d} x- \frac{1}{2} \int_{\mathbf{R}^{2}}\Delta\varphi\varphi_{t}\,\mathrm{d} x+ \frac{1}{4}\frac{\mathrm{d}}{\mathrm{d} t}\Vert n\Vert ^{2}_{L^{2}}+ \frac{H^{2}}{4}\frac{\mathrm {d}}{\mathrm{d} t}\Vert \nabla n\Vert ^{2}_{L^{2}} \\ &\phantom{\operatorname {Re}(-n E,-E_{t})}=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int_{\mathbf{R}^{2}}n\vert E\vert ^{2}\,\mathrm{d} x+ \frac{1}{4}\frac{\mathrm{d}}{\mathrm{d} t}\Vert \nabla\varphi \Vert ^{2}_{L^{2}}+\frac{1}{4}\frac{\mathrm {d}}{\mathrm{d} t}\Vert n \Vert ^{2}_{L^{2}}+\frac{H^{2}}{4}\frac{\mathrm {d}}{\mathrm{d}t}\Vert \nabla n\Vert ^{2}_{L^{2}}, \end{aligned}$$
thus it follows that
$$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{d}t} \biggl[\Vert \nabla E\Vert ^{2}_{L^{2}}+H^{2}\Vert \Delta E\Vert ^{2}_{L^{2}}+ \int_{\mathbf{R}^{2}}n\vert E\vert ^{2}\,\mathrm{d} x+ \frac{1}{2}\Vert \nabla\varphi \Vert ^{2}_{L^{2}}+ \frac {1}{2}\Vert n\Vert ^{2}_{L^{2}}+ \frac{H^{2}}{2}\Vert \nabla n\Vert ^{2}_{L^{2}} \biggr] \\ &\quad=0. \end{aligned}$$
(10)
Letting
$$\mathscr{F}(t)=\Vert \nabla E\Vert ^{2}_{L^{2}}+H^{2} \Vert \Delta E\Vert ^{2}_{L^{2}}+ \int_{\mathbf{R}^{2}}n\vert E\vert ^{2}\,\mathrm{d} x+ \frac{1}{2}\Vert \nabla\varphi \Vert ^{2}_{L^{2}}+ \frac {1}{2}\Vert n\Vert ^{2}_{L^{2}}+ \frac{H^{2}}{2}\Vert \nabla n\Vert ^{2}_{L^{2}}, $$
and noticing (10), we obtain
$$\begin{aligned} \mathscr{F}(t)=\mathscr{F}(0). \end{aligned}$$
□
Lemma 2.4
Suppose that
\(E_{0}(x)\in H^{2}(\mathbf{R}^{2})\), \(n_{0}(x)\in H^{1}(\mathbf{R}^{2})\), \(\varphi_{0}(x)\in H^{1}(\mathbf{R}^{2})\). Then we have
$$\sup_{0\leq t\leq T} \bigl(\Vert E\Vert _{H^{2}}+\Vert n \Vert _{H^{1}}+\Vert \varphi \Vert _{H^{1}} \bigr)\leq C. $$
Proof
By Hölder’s inequality, Young’s inequality and Lemma 2.2, it follows that
$$\begin{aligned} \biggl\vert \int_{\mathbf{R}^{2}} n\vert E\vert ^{2}\,\mathrm{d}x \biggr\vert &\leq \Vert n\Vert _{L^{2}}\Vert E\Vert ^{2}_{L^{4}} \\ &\leq\frac{1}{4}\Vert n\Vert ^{2}_{L^{2}}+\Vert E \Vert ^{4}_{L^{4}} \\ &\leq\frac{1}{4}\Vert n\Vert ^{2}_{L^{2}}+C\Vert \Delta E\Vert _{L^{2}}\Vert E\Vert ^{3}_{L^{2}} \\ &\leq\frac{1}{4}\Vert n\Vert ^{2}_{L^{2}}+ \frac{H^{2}}{2}\Vert \Delta E\Vert ^{2}_{L^{2}}+C. \end{aligned}$$
From Lemma 2.3 we get
$$\Vert \nabla E\Vert ^{2}_{L^{2}}+\frac{H^{2}}{2}\Vert \Delta E\Vert ^{2}_{L^{2}}+\frac{1}{2}\Vert \nabla \varphi \Vert ^{2}_{L^{2}}+\frac{1}{4}\Vert n\Vert ^{2}_{L^{2}}+\frac{H^{2}}{2}\Vert \nabla n\Vert ^{2}_{L^{2}}\leq\mathscr{F}(0)+C. $$
Take the inner products of Eq. (8) and φ. It follows that
$$\begin{aligned} \bigl(\varphi_{t}-n+H^{2}\Delta n-\vert E \vert ^{2}, \varphi \bigr)=0 \end{aligned}$$
(11)
since
$$\begin{aligned} &(\varphi_{t}, \varphi )=\frac{1}{2}\frac{\mathrm{d}}{\mathrm {d}t}\Vert \varphi \Vert ^{2}_{L^{2}}, \\ &\bigl(n+\vert E\vert ^{2}, \varphi \bigr)\leq \bigl(\Vert n \Vert _{L^{2}}+\Vert E\Vert ^{2}_{L^{4}} \bigr) \Vert \varphi \Vert _{L^{2}}\leq\frac{1}{2}\Vert \varphi \Vert ^{2}_{L^{2}}+C, \end{aligned}$$
where
$$\begin{aligned} &\Vert E\Vert ^{2}_{L^{4}}\leq C \Vert \nabla E\Vert _{L^{2}}\Vert E\Vert _{L^{2}}\leq C. \\ &\bigl(H^{2}\Delta n, \varphi \bigr)=H^{2} (n, \Delta\varphi )=H^{2} (n, n_{t} )=\frac{H^{2}}{2}\frac{\mathrm{d}}{\mathrm {d}t} \Vert n\Vert ^{2}_{L^{2}}. \end{aligned}$$
Hence, from Eq. (11) we get
$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \bigl(\Vert \varphi \Vert ^{2}_{L^{2}}+H^{2} \Vert n\Vert ^{2}_{L^{2}} \bigr)\leq \Vert \varphi \Vert ^{2}_{L^{2}}+C. \end{aligned}$$
Using Gronwall’s inequality, we obtain that
$$\begin{aligned} \sup_{0\leq t\leq T} \bigl(\Vert \varphi \Vert ^{2}_{L^{2}}+H^{2} \Vert n\Vert ^{2}_{L^{2}} \bigr)\leq C. \end{aligned}$$
We thus get Lemma 2.4. □
Lemma 2.5
Suppose that
\(E_{0}(x)\in H^{4}(\mathbf{R}^{2})\), \(n_{0}(x)\in H^{2}(\mathbf{R}^{2})\), \(\varphi_{0}(x)\in H^{2}(\mathbf{R}^{2})\). Then we have
$$\sup_{0\leq t \leq T} \bigl(\Vert E\Vert _{H^{4}}+\Vert n \Vert _{H^{2}}+\Vert \varphi \Vert _{H^{2}}+\Vert E_{t}\Vert _{L^{2}}+\Vert n_{t}\Vert _{L^{2}}+\Vert \varphi_{t}\Vert _{L^{2}} \bigr)\leq C. $$
Proof
Differentiating (6) with respect to t, then taking the inner products of the resulting equation and \(E_{t}\), we have
$$\begin{aligned} \bigl(iE_{tt}+\Delta E_{t}-H^{2} \Delta^{2}E_{t}- (nE )_{t}, E_{t} \bigr)=0 \end{aligned}$$
(12)
since
$$\begin{aligned} &\operatorname {Im}(iE_{tt},E_{t})=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \Vert E_{t}\Vert ^{2}_{L^{2}}, \qquad \operatorname {Im}\bigl(\Delta E_{t}-H^{2}\Delta^{2}E_{t}-nE_{t},E_{t} \bigr)=0, \\ &\bigl\vert \operatorname {Im}(-n_{t}E,E_{t}) \bigr\vert \leq C\Vert E\Vert _{L^{\infty}} \Vert n_{t}\Vert _{L^{2}}\Vert E_{t}\Vert _{L^{2}}\leq C \bigl(\Vert E_{t} \Vert ^{2}_{L^{2}}+\Vert n_{t}\Vert ^{2}_{L^{2}} \bigr). \end{aligned}$$
By Lemma 2.2, it follows that
$$\begin{aligned} \Vert E\Vert _{L^{\infty}}\leq C\Vert \Delta E\Vert ^{\frac{1}{2}}_{L^{2}}\Vert E\Vert ^{\frac{1}{2}}_{L^{2}}; \end{aligned}$$
thus from Eq. (12) we get
$$ \frac{\mathrm{d}}{\mathrm{d}t}\Vert E_{t}\Vert ^{2}_{L^{2}}\leq C \bigl(\Vert E_{t}\Vert ^{2}_{L^{2}}+\Vert n_{t}\Vert ^{2}_{L^{2}} \bigr). $$
(13)
Differentiating Eq. (7) with respect to t, then taking the inner products of the resulting equation and \(n_{t}\), we have
$$\begin{aligned} (n_{tt}-\Delta\varphi_{t}, n_{t} )=0 \end{aligned}$$
(14)
since
$$\begin{aligned} &(n_{tt}, n_{t} )=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert n_{t}\Vert ^{2}_{L^{2}}, \\ &(-\Delta\varphi_{t}, n_{t} )= \bigl(-\Delta n+H^{2}\Delta^{2} n-\Delta \vert E\vert ^{2}, n_{t} \bigr) \\ & = \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla n\Vert ^{2}_{L^{2}}+\frac{H^{2}}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert \Delta n\Vert ^{2}_{L^{2}}- \bigl(\Delta \vert E\vert ^{2}, n_{t} \bigr). \end{aligned}$$
Noting that
$$\begin{aligned} \bigl\vert \bigl(\Delta \vert E\vert ^{2}, n_{t} \bigr) \bigr\vert \leq C \Vert E\Vert _{L^{\infty}} \Vert \Delta E\Vert _{L^{2}}\Vert n_{t}\Vert _{L^{2}}\leq C \bigl( \Vert n_{t}\Vert ^{2}_{L^{2}}+1 \bigr), \end{aligned}$$
from Eq. (14) we get
$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \bigl[\Vert n_{t}\Vert ^{2}_{L^{2}}+\Vert \nabla n\Vert ^{2}_{L^{2}}+H^{2} \Vert \Delta n\Vert ^{2}_{L^{2}} \bigr]\leq C \bigl(\Vert n_{t}\Vert ^{2}_{L^{2}}+1 \bigr). \end{aligned}$$
(15)
From Eq. (13) and (15) we get
$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \bigl[\Vert E_{t}\Vert ^{2}_{L^{2}}+ \Vert n_{t}\Vert ^{2}_{L^{2}}+\Vert \nabla n \Vert ^{2}_{L^{2}}+H^{2}\Vert \Delta n\Vert ^{2}_{L^{2}} \bigr]\leq C \bigl(\Vert E_{t}\Vert ^{2}_{L^{2}}+\Vert n_{t}\Vert ^{2}_{L^{2}}+1 \bigr). \end{aligned}$$
By Gronwall’s inequality, it follows that
$$\begin{aligned} \sup_{0\leq t\leq T} \bigl[\Vert E_{t} \Vert ^{2}_{L^{2}}+\Vert n_{t}\Vert ^{2}_{L^{2}}+\Vert \nabla n\Vert ^{2}_{L^{2}}+H^{2} \Vert \Delta n\Vert ^{2}_{L^{2}} \bigr]\leq C. \end{aligned}$$
(16)
Take the inner products of Eq. (8) and \(\varphi_{t}\). It follows that
$$\begin{aligned} \bigl(\varphi_{t}-n+H^{2}\Delta n-\vert E \vert ^{2}, \varphi _{t} \bigr)=0 \end{aligned}$$
(17)
since
$$\begin{aligned} &(\varphi_{t}, \varphi_{t} )=\Vert \varphi_{t} \Vert ^{2}_{L^{2}}, \\ &\bigl(-n+H^{2}\Delta n-\vert E\vert ^{2}, \varphi_{t} \bigr)\leq \bigl(\Vert n\Vert _{L^{2}}+H^{2} \Vert \Delta n\Vert _{L^{2}}+\Vert E\Vert ^{2}_{L^{4}} \bigr)\Vert \varphi_{t}\Vert _{L^{2}} \\ & \phantom{\bigl(-n+H^{2}\Delta n-\vert E\vert ^{2}, \varphi_{t} \bigr)}\leq C+\frac{1}{2}\Vert \varphi_{t}\Vert ^{2}_{L^{2}}. \end{aligned}$$
From Eq. (17) we get
$$\begin{aligned} \Vert \varphi_{t}\Vert ^{2}_{L^{2}}\leq C. \end{aligned}$$
Take the inner products of Eq. (6) and ΔE. It follows that
$$\begin{aligned} \bigl(iE_{t}+\Delta E-H^{2} \Delta^{2}E-nE, \Delta E \bigr)=0 \end{aligned}$$
(18)
since
$$\begin{aligned} &(iE_{t}-nE, \Delta E )\leq \bigl(\Vert E_{t}\Vert _{L^{2}}+\Vert E\Vert _{L^{\infty}} \Vert n\Vert _{L^{2}} \bigr)\Vert \Delta E\Vert _{L^{2}}\leq C, \\ &\bigl(\Delta E-H^{2}\Delta^{2}E, \Delta E \bigr)=\Vert \Delta E\Vert ^{2}_{L^{2}}+H^{2} \bigl\Vert \nabla^{3} E \bigr\Vert ^{2}_{L^{2}}. \end{aligned}$$
From Eq. (18) we get
$$\begin{aligned} \bigl\Vert \nabla^{3} E \bigr\Vert ^{2}_{L^{2}} \leq C. \end{aligned}$$
From (6) we obtain
$$\begin{aligned} H^{2} \bigl\Vert \Delta^{2}E \bigr\Vert _{L^{2}} \leq \Vert E_{t}\Vert _{L^{2}}+\Vert \Delta E\Vert _{L^{2}}+\Vert nE\Vert _{L^{2}}\leq C, \end{aligned}$$
where
$$\begin{aligned} \Vert nE\Vert _{L^{2}}\leq \Vert n\Vert _{L^{4}}\Vert E \Vert _{L^{4}}\leq C \Vert \nabla n\Vert ^{\frac {1}{2}}_{L^{2}} \Vert n\Vert ^{\frac{1}{2}}_{L^{2}}\Vert \nabla E\Vert ^{\frac{1}{2}}_{L^{2}}\Vert E\Vert ^{\frac {1}{2}}_{L^{2}}\leq C. \end{aligned}$$
From (7) we obtain
$$\begin{aligned} \Vert \Delta\varphi \Vert _{L^{2}}=\Vert n_{t}\Vert _{L^{2}}\leq C. \end{aligned}$$
We thus get Lemma 2.5. □
Lemma 2.6
Suppose that
\(f_{1},f_{2}\in H^{s}(\Omega)\), \(\Omega\subseteq\mathbf{R}^{n}\). Then we have
$$\bigl\Vert D^{s}(f_{1}\cdot f_{2}) \bigr\Vert _{L^{2}}\leq C_{s} \bigl(\Vert f_{1}\Vert _{L^{2}} \bigl\Vert D^{s}f_{2} \bigr\Vert _{L^{2}}+ \bigl\Vert D^{s}f_{1} \bigr\Vert _{L^{2}}\Vert f_{2}\Vert _{L^{2}} \bigr), $$
where the constant
\(C_{s}\)
is independent of
\(f_{1}\)
and
\(f_{2}\).
Lemma 2.7
Suppose that
\(E_{0}(x)\in H^{m+4}(\mathbf{R}^{2})\), \(n_{0}(x)\in H^{m+2}(\mathbf{R}^{2})\), \(\varphi_{0}(x)\in H^{m+2}(\mathbf{R}^{2})\), \(m\geq0\). Then we have
$$\begin{aligned} &\sup_{0\leq t\leq T} \bigl( \bigl\Vert E(x,t) \bigr\Vert _{H^{m+4}}+ \bigl\Vert n(x,t) \bigr\Vert _{H^{m+2}}+ \bigl\Vert \varphi(x,t) \bigr\Vert _{H^{m+2}} \bigr)\leq C \\ &\sup_{0\leq t\leq T} \bigl( \bigl\Vert E_{t}(x,t) \bigr\Vert _{H^{m}}+ \bigl\Vert n_{t}(x,t) \bigr\Vert _{H^{m}}+\Vert \varphi_{t}\Vert _{H^{m}} \bigr)\leq C. \end{aligned}$$
Proof
Lemma 2.7 is true when \(m=0\) (Lemma 2.5). Suppose that Lemma 2.7 is true when \(m=k, (k\geq0)\). Take the inner products of (8) and \((-1)^{k+1}\Delta ^{k+3}\varphi\). It follows that
$$\begin{aligned} \bigl(\varphi_{t}-n+H^{2}\Delta n-\vert E \vert ^{2}, (-1)^{k+1}\Delta^{k+3}\varphi \bigr)=0 \end{aligned}$$
(19)
since
$$\begin{aligned} &\bigl(\varphi_{t},(-1)^{k+1}\Delta^{k+3}\varphi \bigr)=\frac{1}{2}\frac {\mathrm{d}}{\mathrm{d}t} \bigl\Vert \nabla^{k+3} \varphi \bigr\Vert ^{2}_{L^{2}}, \\ &\bigl(-n,(-1)^{k+1}\Delta^{k+3}\varphi \bigr)= \bigl(-n,(-1)^{k+1}\Delta^{k+2} n_{t} \bigr)= \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \bigl\Vert \nabla^{k+2} n \bigr\Vert ^{2}_{L^{2}}, \\ &\bigl(H^{2}\Delta n,(-1)^{k+1}\Delta^{k+3}\varphi \bigr)=H^{2} \bigl(\Delta n,(-1)^{k+1}\Delta^{k+2} n_{t} \bigr)=\frac{H^{2}}{2}\frac{\mathrm{d}}{\mathrm{d}t} \bigl\Vert \nabla^{k+3} n \bigr\Vert ^{2}_{L^{2}}, \\ &\bigl\vert \bigl(-\vert E\vert ^{2},(-1)^{k+1}\Delta ^{k+3}\varphi \bigr) \bigr\vert = \bigl\vert \bigl( \nabla^{k+3} \vert E\vert ^{2},\nabla^{k+3} \varphi \bigr) \bigr\vert \leq \bigl\Vert \nabla^{k+3} \vert E \vert ^{2} \bigr\Vert _{L^{2}} \bigl\Vert \nabla^{k+3} \varphi \bigr\Vert _{L^{2}} \\ &\phantom{\bigl\vert \bigl(-\vert E\vert ^{2},(-1)^{k+1}\Delta ^{k+3}\varphi \bigr) \bigr\vert }\leq C\Vert E\Vert _{L^{2}} \bigl\Vert \nabla^{k+3} E \bigr\Vert _{L^{2}} \bigl\Vert \nabla^{k+3}\varphi \bigr\Vert _{L^{2}} \\ &\phantom{\bigl\vert \bigl(-\vert E\vert ^{2},(-1)^{k+1}\Delta ^{k+3}\varphi \bigr) \bigr\vert }\leq C \bigl( \bigl\Vert \nabla^{k+3}\varphi \bigr\Vert ^{2}_{L^{2}}+1 \bigr), \end{aligned}$$
thus from Eq. (19) it follows that
$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \bigl( \bigl\Vert \nabla^{k+3} \varphi \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \nabla^{k+2} n \bigr\Vert ^{2}_{L^{2}}+H^{2} \bigl\Vert \nabla^{k+3} n \bigr\Vert ^{2}_{L^{2}} \bigr)\leq C \bigl( \bigl\Vert \nabla^{k+3}\varphi \bigr\Vert ^{2}_{L^{2}}+1 \bigr). \end{aligned}$$
(20)
By using Gronwall’s inequality, we have
$$\sup_{0\leq t\leq T} \bigl( \bigl\Vert \nabla^{k+3}\varphi \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \nabla^{k+2} n \bigr\Vert ^{2}_{L^{2}}+ \bigl\Vert \nabla^{k+3} n \bigr\Vert ^{2}_{L^{2}} \bigr)\leq C. $$
From (7) and (8) we get
$$\begin{aligned} & \bigl\Vert \nabla^{k+1}n_{t} \bigr\Vert _{L^{2}}= \bigl\Vert \nabla ^{k+3}\varphi \bigr\Vert _{L^{2}}\leq C, \\ & \bigl\Vert \nabla^{k+1}\varphi_{t} \bigr\Vert _{L^{2}}\leq C \bigl( \bigl\Vert \nabla^{k+1}n \bigr\Vert _{L^{2}}+ \bigl\Vert \nabla^{k+3} n \bigr\Vert _{L^{2}}+ \Vert E\Vert _{L^{2}} \bigl\Vert \nabla ^{k+1}E \bigr\Vert _{L^{2}} \bigr)\leq C. \end{aligned}$$
Differentiating (6) with respect to t, then taking the inner products of the resulting equation and \((-1)^{k+1}\Delta ^{k+1}E_{t}\), we obtain
$$\begin{aligned} \bigl(iE_{tt}+\Delta E_{t}-H^{2} \Delta^{2}E_{t}- (nE )_{t}, (-1)^{k+1} \Delta^{k+1}E_{t} \bigr)=0. \end{aligned}$$
(21)
Since
$$\begin{aligned} &\operatorname {Im}\bigl(iE_{tt},(-1)^{k+1}\Delta^{k+1}E_{t} \bigr)=\frac{1}{2}\frac {\mathrm{d}}{\mathrm{d}t} \bigl\Vert \nabla^{k+1}E_{t} \bigr\Vert ^{2}_{L^{2}}, \\ &\operatorname {Im}\bigl(\Delta E_{t}-H^{2}\Delta^{2}E_{t},(-1)^{k+1} \Delta^{k+1}E_{t} \bigr)=0, \\ &\bigl\vert \operatorname {Im}\bigl(-n_{t}E,(-1)^{k+1} \Delta^{k+1}E_{t} \bigr) \bigr\vert \leq \bigl\vert \bigl( \nabla^{k+1}(n_{t}E),\nabla^{k+1}E_{t} \bigr) \bigr\vert \\ &\phantom{\bigl\vert \operatorname {Im}\bigl(-n_{t}E,(-1)^{k+1} \Delta^{k+1}E_{t} \bigr) \bigr\vert }\leq C \bigl( \bigl\Vert \nabla^{k+1}n_{t} \bigr\Vert _{L^{2}}\Vert E\Vert _{L^{2}}+ \bigl\Vert \nabla^{k+1}E \bigr\Vert _{L^{2}}\Vert n_{t}\Vert _{L^{2}} \bigr) \bigl\Vert \nabla^{k+1}E_{t} \bigr\Vert _{L^{2}} \\ &\phantom{\bigl\vert \operatorname {Im}\bigl(-n_{t}E,(-1)^{k+1} \Delta^{k+1}E_{t} \bigr) \bigr\vert }\leq C \bigl( \bigl\Vert \nabla^{k+1}E_{t} \bigr\Vert ^{2}_{L^{2}}+1 \bigr), \\ &\bigl\vert \operatorname {Im}\bigl(-nE_{t},(-1)^{k+1} \Delta^{k+1}E_{t} \bigr) \bigr\vert \leq \bigl\vert \bigl( \nabla^{k+1}(nE_{t}),\nabla^{k+1}E_{t} \bigr) \bigr\vert \\ &\phantom{\bigl\vert \operatorname {Im}\bigl(-nE_{t},(-1)^{k+1} \Delta^{k+1}E_{t} \bigr) \bigr\vert }\leq C \bigl( \bigl\Vert \nabla^{k+1}n \bigr\Vert _{L^{2}} \Vert E_{t}\Vert _{L^{2}}+ \bigl\Vert \nabla^{k+1}E_{t} \bigr\Vert _{L^{2}}\Vert n\Vert _{L^{2}} \bigr) \bigl\Vert \nabla^{k+1}E_{t} \bigr\Vert _{L^{2}} \\ &\phantom{\bigl\vert \operatorname {Im}\bigl(-nE_{t},(-1)^{k+1} \Delta^{k+1}E_{t} \bigr) \bigr\vert }\leq C \bigl( \bigl\Vert \nabla^{k+1}E_{t} \bigr\Vert ^{2}_{L^{2}}+1 \bigr), \end{aligned}$$
thus from Eq. (21) we get
$$ \frac{\mathrm{d}}{\mathrm{d}t} \bigl\Vert \nabla^{k+1}E_{t} \bigr\Vert ^{2}_{L^{2}}\leq C \bigl( \bigl\Vert \nabla^{k+1}E_{t} \bigr\Vert ^{2}_{L^{2}}+1 \bigr). $$
(22)
By using Gronwall’s inequality, we get
$$\sup_{0\leq t \leq T} \bigl\Vert \nabla^{k+1}E_{t} \bigr\Vert ^{2}_{L^{2}}\leq C. $$
From (6) we obtain
$$\begin{aligned} \bigl\Vert \nabla^{k+5}E \bigr\Vert _{L^{2}}\leq C \bigl( \bigl\Vert \nabla ^{k+1}E_{t} \bigr\Vert _{L^{2}}+ \bigl\Vert \nabla^{k+3} E \bigr\Vert _{L^{2}}+ \bigl\Vert \nabla^{k+1}n \bigr\Vert _{L^{2}}\Vert E\Vert _{L^{2}}+\Vert n\Vert _{L^{2}} \bigl\Vert \nabla^{k+1}E \bigr\Vert _{L^{2}} \bigr)\leq C. \end{aligned}$$
Hence
$$\begin{aligned} &\sup_{0\leq t\leq T} \bigl( \bigl\Vert E(x,t) \bigr\Vert _{H^{k+5}}+ \bigl\Vert n(x,t) \bigr\Vert _{H^{k+3}}+ \bigl\Vert \varphi(x,t) \bigr\Vert _{H^{k+3}} \bigr)\leq C, \\ &\sup_{0\leq t\leq T} \bigl( \bigl\Vert E_{t}(x,t) \bigr\Vert _{H^{k+1}}+ \bigl\Vert n_{t}(x,t) \bigr\Vert _{H^{k+1}}+\Vert \varphi_{t}\Vert _{H^{k+1}} \bigr)\leq C. \end{aligned}$$
This means Lemma 2.7 is true when \(m=k+1\). Thus Lemma 2.7 is proved completely. □