Polynomial bounds on the Sobolev norms of the solutions of the nonlinear wave equation with time dependent potential

Abstract

We consider the Cauchy problem for the nonlinear wave equation $u_{tt} - \Delta _x u +q(t, x) u + u^3 = 0$ with smooth potential $q(t, x) \ge 0$ having compact support with respect to x. The linear equation without the nonlinear term $u^3$ and potential periodic in t may have solutions with exponentially increasing $H^1(\mathbb {R}^{3}_{x})$ norm as $t\rightarrow \infty$. In Petkov and Tzvetkov (IMRN, https://doi.org/10.1093/imrn/rnz014), it was established that by adding the nonlinear term $u^3$, the $H^1(\mathbb {R}^{3}_{x})$ norm of the solution is polynomially bounded for every choice of q. In this paper, we show that the $H^k({{\mathbb {R}}}^3_x)$ norm of this global solution is also polynomially bounded. To prove this, we apply a different argument based on the analysis of a sequence $\{Y_k(n\tau _k)\}_{n = 0}^{\infty }$ with suitably defined energy norm $Y_k(t)$ and $0< \tau _k <1$.

1 Introduction

Consider for $t \in {{\mathbb {R}}},\, x \in {{\mathbb {R}}}^3$, the Cauchy problem

$$\begin{aligned} \partial _t^2 u - \Delta _x u + q(t, x) u + u^3 = 0, \, u(0, x) = f_1(x), \, \partial _t u(0, x) = f_2(x), \end{aligned}$$

(1.1)

where $0 \le q(t, x) \in C^{\infty }({{\mathbb {R}}}_t \times {{\mathbb {R}}}^3),\,q(t, x) = 0$ for $|x| \ge \rho > 0$ and

$$\begin{aligned} \sup _{t \in {{\mathbb {R}}}, |x|\le \rho }|\partial _t^k\partial _{x}^{\alpha } q(t,x)|\le C_{k, \alpha },\, \forall k, \forall \alpha . \end{aligned}$$

(1.2)

Set

$$\begin{aligned} \Vert u(t, x)\Vert _{{{\mathcal {H}}}} = \Vert u(t, x)\Vert _{H^1({{\mathbb {R}}}^3)} + \Vert u_t(t, x)\Vert _{L^2({{\mathbb {R}}}^3)}. \end{aligned}$$

For the Cauchy problem for the linear operator $\partial _t^2u - \Delta _x u+ q(t, x) u$, there exist potentials $q(t, x) \ge 0$ periodic in time with period $T > 0$ such that for suitable initial data $f = (f_1, f_2) \in {{\mathcal {H}}}({{\mathbb {R}}}^3) = H^1({{\mathbb {R}}}^3) \times L^2({{\mathbb {R}}}^3)$, we have

$$\begin{aligned} \Vert u(t, x) \Vert _{H^1({{\mathbb {R}}}^3)} \ge C e^{\alpha |t|} \end{aligned}$$

with $C> 0, \, \alpha > 0$ (see [1, 2]). This phenomenon is related to the so called parametric resonance. On the other hand, adding a nonlinear term $u^3$ for the Cauchy problem (1.1), there are no parametric resonances and for every potential q, the solution u(t, x) is defined globally for $t \in {{\mathbb {R}}}$ and satisfies a polynomial bound

$$\begin{aligned} \Vert u(t, x)\Vert _{H^1({{\mathbb {R}}}^3)} \le B_1 (1 + B_0 |t|)^2 \end{aligned}$$

with constants $B_0> 0, B_1 > 0$ depending on q and the initial data $f \in {{\mathcal {H}}}$. This result has been obtained in [2, Theorem 2], and the proof was based on the inequality

$$\begin{aligned} X'(t) \le C X(t)^{1/2}, \end{aligned}$$

where

$$\begin{aligned} X(t)=\frac{1}{2}\int \limits _{{{\mathbb {R}}}^3}\big (|\partial _t u|^2 + |\nabla _x u|^2 + q |u|^2 +\frac{1}{2} |u|^{4}\big )dx. \end{aligned}$$

In fact, the local Strichartz estimates and [2, Theorem 2] hold for every non-negative potential $q(t, x) \in C^{\infty }({{\mathbb {R}}}_t \times {{\mathbb {R}}}^3)$ with compact support with respect to x satisfying the estimates (1.2) since in the proofs of these results the periodicity of q with respect to t is not used.

In this paper, we study the problem (1.1) with initial data $f \in H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3),\, k \ge 2.$ Throughout the paper, we consider Cauchy problems with real-valued initial data f and real-valued solutions. First in Section 2, we establish a local result and we show the existence and uniqueness of the solution for $t \in [s, s+ \tau _k]$ with initial data $f \in H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3)$ on $t = s$ and

$$\begin{aligned} \tau _k = c_k(1 + \Vert (f_1, f_2)\Vert _{{{\mathcal {H}}}({{\mathbb {R}}}^3)})^{-\gamma }, \, \gamma > 0, \end{aligned}$$

where $c_k$ depends on q and k (see Proposition 2.1). It is important to notice that $\tau _k$ depends on the norm $\Vert f\Vert _{{{\mathcal {H}}}}$ and since we have a global bound for the ${{\mathcal {H}}}$ norm of $(u, u_t)(t, x)$, the interval of local existence depends on the ${{\mathcal {H}}}$ norm of the initial data. We prove this result without using local Strichartz estimates. Next we show that the global solution in ${{\mathbb {R}}}$ is in $H^k({{\mathbb {R}}}^3)$ for all $t \in {{\mathbb {R}}}$ and the problem is to examine if the norm $\Vert u(t, x)\Vert _{H^k({{\mathbb {R}}}^3)}, \,k \ge 2,$ is polynomially bounded. To do this, it is not possible to define a suitable energy $Y_k(t) \ge 0$ involving

$$\begin{aligned} \int \limits _{{{\mathbb {R}}}^3} (\Vert u(t, x)\Vert ^2_{H^k({{\mathbb {R}}}^3)} + \Vert u_t(t, x)\Vert _{H^{k-1}({{\mathbb {R}}}^3)}^2)dx \end{aligned}$$

for which $Y_k'(t) \le C_k Y_k^{\gamma _k}(t),\, 0<\gamma _k < 1.$ To overcome this difficulty, we follow another argument based on Lemma A.1 (see Appendix) which has an independent interest and apply local Strichartz estimates for the nonlinear equation. We first study the case $k = 2$ in Section 4 and by induction, one covers the case $k \ge 3$ in Section 5. Our principal result is the following

Theorem 1.1

For every potential q and every $k \ge 2$, the problem (1.1) with initial data $f \in H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3)$ has a global solution u(t, x) and there exist $A_k > 0$ and $m_k \ge 2$ depending on $q, \,k$, and $\Vert f\Vert _{{{\mathcal {H}}}}$ such that

$$\begin{aligned} \Vert u(t, x)\Vert _{H^k({{\mathbb {R}}}^3)} +\Vert \partial _t u(t, x)\Vert _{H^k({{\mathbb {R}}}^3)} \le A_k (1 + |t|)^{m_k}, \, t \in {{\mathbb {R}}}. \end{aligned}$$

(1.3)

We refer to [3] and the references therein for other results about polynomial bounds for the solutions of Hamiltonian partial differential equations. The method of the proof of Theorem 1.1 basically follows the approach in [3]. The main difficulty compared to [3] is that in our situation, we do not have a uniform bound on the $H^1({{\mathbb {R}}}^3)$ norm and for that purpose, we need to apply the estimate of Lemma A.1 in the Appendix.

2 Existence and uniqueness of local solutions in $H^k({{\mathbb {R}}}^3),\, k\ge 3$

In this section, we study the existence and uniqueness of local solutions of the Cauchy problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{tt} - \Delta _ x u + q(t, x) u + u^3 = 0, t \in [s,s+ \tau ],\,x \in {{\mathbb {R}}}^3,\\ u(s, x) = f_1(x), u_t(s, x) = f_2(x),\end{array}\right. } \end{aligned}$$

(2.1)

where $f = (f_1, f_2) \in H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3),\, k\ge 1, 0< \tau < 1.$ We assume that $[s, s+ \tau ] \subset [0, a],$ where $a > 1$ is fixed. The cases $k = 1, 2$ have been investigated in [2, Section 3] by using the norms

$$\begin{aligned} \Vert u\Vert _{S_{k-1}}: = \Vert (u,u_t)\Vert _{C([s,s+ \tau ],H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3))}. \end{aligned}$$

For $k=1$, the space $S_0$ has been denoted as S. The number $\tau$ is given by

$$\begin{aligned} \tau = c_1(1 + \Vert (f_1, f_2)\Vert _{{{\mathcal {H}}}})^{-\gamma } < 1 \end{aligned}$$

(2.2)

with some positive constants $c_1> 0, \gamma > 0$ depending on q. The case $k \ge 3$ can be handled by a similar argument and we will show that with $\tau =\tau _k$ defined by (2.2) with the constant $c_1$ replaced by $0 < c_k \le c_1$ depending on k and q, one has a local existence and uniqueness in the interval $[s, s+ \tau _k].$ Consider the linear problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t^2 u_{n+1} - \Delta u_{n+1} + q(t, x) u_{n+1}+u_n^3= 0,\, n \ge 0,\\ u_{n+1}(s, x) = f_1(x), \, \partial _t u_{n+1}(s, x) = f_2(x)\end{array}\right. } \end{aligned}$$

(2.3)

for $t \in [s, s+ \tau _k]$ with $u_0 = 0.$ For the solution of the above problem with right hand part $- u_n^3$ and $f = (f_1, f_2)$, we have a representation

$$\begin{aligned} (u_{n+1}, (u_{n+1})_t)= & {} U_0(t-s)f - \int \limits _s^t \Bigl [U_0(t - \tau )Q(\tau ) u_{n+1}(\tau , x) \nonumber \\&+ U_0(t - \tau )Q_0 u_n^3(\tau , x)\Bigr ] d\tau . \end{aligned}$$

(2.4)

Here $U_0(t,s): {{\mathcal {H}}}\rightarrow {{\mathcal {H}}}$ is the propagator related to the free wave equation in ${{\mathbb {R}}}^3$ (see [2, Section 2]) and

$$\begin{aligned} Q(\tau ) = \begin{pmatrix} 0 &{}\quad 0\\ q(\tau , x) &{}\quad 0 \end{pmatrix},\, Q_0= \begin{pmatrix} 0 &{}\quad 0\\ 1 &{}\quad 0\end{pmatrix}. \end{aligned}$$

To estimate $\Vert u_{n+1}\Vert _{S_{k}},$ we apply the operator

$$\begin{aligned} L_k =\begin{pmatrix}(1- \Delta )^{k/2} &{}\quad 0\\ 0 &{}\quad ( 1- \Delta )^{(k- 1)/2}\end{pmatrix}. \end{aligned}$$

Notice that this operator commutes with $U_0(t- \tau )$ and $\Vert U_0(t-s)\Vert _{{{\mathcal {H}}}\rightarrow {{\mathcal {H}}}} \le A$ for $|t- s| \le 1$ with $A > 0$ independent of k. Therefore

$$\begin{aligned} \Vert U_0(t- s)L_k f\Vert _{{{\mathcal {H}}}} \le C \Vert f\Vert _{H^{k+1} \times H^k} \end{aligned}$$

and

$$\begin{aligned}&\Bigl \Vert \int \limits _s^{t} U_0(t- \tau ) L_k Q(\tau )u_{n+1}(\tau , x) d\tau \Bigr \Vert _{{{\mathcal {H}}}} \\&\quad \le \int \limits _s^{t}\Vert U_0(t-\tau )L_k Q(\tau ) u_{n+1}\Vert _{{{\mathcal {H}}}}d\tau \le A_k \tau \Vert u_{n+1}\Vert _{S_k}. \end{aligned}$$

For $A_k \tau \le 1/2$ with $A_k >0$, depending on k and q, the term involving $Q(\tau )u_{n+1}$ in (2.4) can be absorbed by $\Vert u_{n+1}\Vert _{S_k}$ and we deduce

$$\begin{aligned} \Vert u_{n+1}\Vert _{S_k} \le C\Vert (f_1, f_2)\Vert _{H^{k+1}({{\mathbb {R}}}^3) \times H^k({{\mathbb {R}}}^3)} + C \Vert u_n^3\Vert _{L^1([s, s+ \tau ], H^{k}({{\mathbb {R}}}^3))}. \end{aligned}$$

Here and below the constants C depend on k and q and they may change from line to line but we will omit this in the notations. Next we define the norm

$$\begin{aligned} \Vert f\Vert _{H^{s, p}({{\mathbb {R}}}^3)}: = \Vert (1- \Delta _x)^{s/2}f\Vert _{L^p({{\mathbb {R}}}^3)},\, 1 < p \le \infty . \end{aligned}$$

We will use the following product estimate

$$\begin{aligned} \Vert fg\Vert _{H^{s, p}} \le A_{s,p} \Vert f\Vert _{L^{q_1}} \Vert g\Vert _{H^{s,q_2}} + A_{s,p} \Vert g\Vert _{L^{r_1}} \Vert f\Vert _{H^{s, r_2}}, \end{aligned}$$

(2.5)

provided

$$\begin{aligned} \frac{1}{p} = \frac{1}{q_1} + \frac{1}{q_2} = \frac{1}{r_1}+\frac{1}{r_2},\, q_1, r_1\in (1, \infty ],\, q_2, r_2 \in (1, \infty ]. \end{aligned}$$

For the proof of the classical estimate (2.5), we refer to [4]. We apply (2.5) with $p = 2, q_1 = 3, q_2= 6, r_1 =6, r_2=3$ and get

$$\begin{aligned} \Vert u_n^3\Vert _{H^{k}({{\mathbb {R}}}^3)} \le C \Vert u_n\Vert _{H^{k, 6}({{\mathbb {R}}}^3)}\Vert u_n\Vert _{L^6({{\mathbb {R}}}^3)}^2 + C \Vert u_n^2\Vert _{H^{k, 3}({{\mathbb {R}}}^3)} \Vert u_n\Vert _{L^6({{\mathbb {R}}}^3)}. \end{aligned}$$

For the term involving $u_n^2$ in the above inequality, we apply the same estimate with $p = 3, q_1= q_2= r_1= r_2 = 6$ and deduce

$$\begin{aligned} \Vert u_n^2\Vert _{H^{k, 3}({{\mathbb {R}}}^3)} \le 2C \Vert u_n\Vert _{H^{k, 6}({{\mathbb {R}}}^3)} \Vert u_n\Vert _{L^6({{\mathbb {R}}}^3).} \end{aligned}$$

Consequently, by the Sobolev embedding theorem,

$$\begin{aligned} \Vert u_n^3\Vert _{H^{k}({{\mathbb {R}}}^3)} \le C_1 \Vert u_n\Vert _{H^{k+1}({{\mathbb {R}}}^3)} \Vert \nabla _x u_n\Vert _{L^2({{\mathbb {R}}}^3)}^2. \end{aligned}$$

This implies

$$\begin{aligned} \int \limits _s^{s + \tau } \Vert u_n^3(t, x)\Vert _{H^{k}({{\mathbb {R}}}^3)} dt \le C_1\tau \Vert u_n\Vert _{L^{\infty }([s, s+ \tau ]), H^1({{\mathbb {R}}}^3))}^2\Vert u_n\Vert _{S_{k}}. \end{aligned}$$

On the other hand, for the solution $u_n$, we have the estimate

$$\begin{aligned} \Vert u_n\Vert _{C([s,s+\tau ], H^1({{\mathbb {R}}}^3))} \le 2C_0 \Vert (f_1, f_2)\Vert _{{{\mathcal {H}}}}, \, \forall n \ge 1 \end{aligned}$$

with some constant $C_0 >0$ depending on q (see [2, Section 3]) and we deduce the bound

$$\begin{aligned} C\Vert u_n^3\Vert _{L^1([s, s+ \tau ], H^k({{\mathbb {R}}}^3))} \le C C_1\tau (2C_0)^2 \Vert (f_1, f_2)\Vert _{{{\mathcal {H}}}}^2\Vert u_n\Vert _{S_{k+1}}. \end{aligned}$$

Thus choosing $2CC_1\tau (2C_0)^2 \Vert (f_1, f_2)\Vert ^2_{{{\mathcal {H}}}} \le 1,$ we may prove by induction the estimate

$$\begin{aligned} \Vert u_n\Vert _{S_{k}}\le 2C \Vert (f_1,f_2)\Vert _{H^{k+1}({{\mathbb {R}}}^3) \times H^k({{\mathbb {R}}}^3)},\, \forall n \ge 1. \end{aligned}$$

(2.6)

Next, let $w_n =u_{n+1} - u_n$ be a solution of the problem

$$\begin{aligned} \partial _t^2 w_n - \Delta w_n + q(t, x) w_n = u_{n-1}^3 - u_{n}^3, \,\, w_n(0, x) = \partial _t w_n(0, x) = 0. \end{aligned}$$

By using the inequality

$$\begin{aligned} \Bigl |v^3- w^3 \Bigr |\le 2 |v-w|\Bigl (|v|^2 + |w|^2\Bigr ), \end{aligned}$$

we can similarly show that

$$\begin{aligned} \Vert u_{n+1}-u_n\Vert _{S_k}\le \frac{1}{2}\Vert u_{n}-u_{n-1}\Vert _{S_k} \end{aligned}$$

which implies the convergence of $(u_n)_{n \ge 0}$ with respect to the $\Vert \cdot \Vert _{S_k}$ norm. Repeating the argument of [2, Section 3], we obtain local existence and uniqueness. Thus we get the following

Proposition 2.1

For every $k \ge 1$, there exist $C_k> 0,\,c_k > 0$, and $\gamma > 0$ depending on q and k such that for every $(f_1,f_2) \in H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3)$, there is a unique solution $(u,u_t) \in C([s, s + \tau _k], H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3))$ of the problem (2.1) on $[s, s + \tau _k]$ with $\tau _k = c_k( 1 + \Vert (f_1, f_2)\Vert _{{{\mathcal {H}}}})^{-\gamma }$. Moreover, the solution satisfies

$$\begin{aligned} \Vert u\Vert _{S_{k}} \le C_k \Vert (f_1, f_2)\Vert _{H^{k}({{\mathbb {R}}}^3) \times H^{k_1}({{\mathbb {R}}}^3)}. \end{aligned}$$

(2.7)

It is important to note that for every k,  $\tau _k$ depends on the ${{\mathcal {H}}}$ norm of the initial data.

In [2], it was proved that one has a global solution $(u, u_t) \in C({{\mathbb {R}}}, {{\mathcal {H}}}({{\mathbb {R}}}^3))$ with initial data $(f_1, f_2) \in {{\mathcal {H}}}({{\mathbb {R}}}^3).$ It is natural to expect that for $(f_1, f_2) \in H^{k}({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3)$, we have a global solution $(u, u_t) \in C({{\mathbb {R}}},H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3)).$

Let $a >1$ be fixed and let $k \ge 1.$ We wish to prove that the global solution with initial data $f \in H^{k+1}({{\mathbb {R}}}^3) \times H^{k}({{\mathbb {R}}}^3)$ is such that

$$\begin{aligned} (u, u_t) (t, x) \in H^{k +1}({{\mathbb {R}}}^3)\times H^{k}({{\mathbb {R}}}^3),\, 0 \le t \le a. \end{aligned}$$

(2.8)

According to [2, Theorem 2], for $0 \le t \le a$, we have an estimate

$$\begin{aligned} \Vert (u, u_t)(t, x)\Vert _{{{\mathcal {H}}}} \le B_a = \Vert f\Vert _{{{\mathcal {H}}}}+ a(B_1 + B_2 a), \end{aligned}$$

where $B_1 > 0$ and $B_2 > 0$ depend only on $\Vert f\Vert _{{{\mathcal {H}}}}.$ Consider

$$\begin{aligned} \tau _k(a) = c_k(1 + B_a)^{-\gamma }. \end{aligned}$$

(2.9)

First for $0\le t \le \tau _k(a)$, we apply Proposition 2.1. Next we apply Proposition 2.1 for the problem with initial data on $t = \frac{2}{3}\tau _k(a)$ which is bounded by (2.7). Thus we obtain a solution in $[0, \frac{5}{3}\tau _k(a)]$ and we continue this procedure by step $\frac{2}{3} \tau _k(a)$. On every step, the norm $H^{k+1}({{\mathbb {R}}}^3) \times H^{k}({{\mathbb {R}}}^3)$ of $(u, u_t)$ will increase with a constant $C_k$. Finally, if

$$\begin{aligned} \frac{3}{2} a \le m \tau _k(a) \le \frac{3}{2} (a +1), \end{aligned}$$

we deduce

$$\begin{aligned} \Vert (u, u_t) (a, x)\Vert _{ H^{k +1} \times H^{k}} \le C_k^m \Vert (f_1, f_2)\Vert _{H^{k+1} \times H^{k}} \nonumber \\ \le e^{\frac{3 (a+1)}{2\tau _k(a)}\log C_k}\Vert (f_1, f_2)\Vert _{H^{k +1} \times H^{k}}. \end{aligned}$$

(2.10)

Hence, we established (2.8) and one has a bound of the $H^{k+1}\times H^k$ norm. Since a is arbitrary, we obtain (2.8) for $t \in {{\mathbb {R}}}$ and a global existence for $t \in {{\mathbb {R}}}.$ In Section 5, we will improve (2.10) to polynomial bounds of the Sobolev norms.

3 Local Strichartz estimate for the nonlinear wave equation

Our purpose is to establish a local Strichartz estimate for the solution of the Cauchy problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{tt} - \Delta _ x u + q(t, x) u + u^3 = 0, t \in ]s, s + \tau ], x \in {{\mathbb {R}}}^3,\\ u(s, x) = f_1(x), u_t(s, x) = f_2(x),\end{array}\right. } \end{aligned}$$

(3.1)

where $f = (f_1, f_2) \in H^2({{\mathbb {R}}}^3) \times H^1({{\mathbb {R}}}^3),\, 0 < \tau \le 1.$ It is well known (see [2, Proposition 1]) that for the solution of the Cauchy problem

$$\begin{aligned} {\left\{ \begin{array}{ll} v_{tt} - \Delta _ x v + q(t, x) v= F, \, (t , x) \in ]s, s + \tau ] \times {{\mathbb {R}}}^3,\\ v(s, x) = h_1(x), v_t(s, x) = h_2(x),\end{array}\right. } \end{aligned}$$

(3.2)

we have an estimate

$$\begin{aligned}&\Vert v(t, x)\Vert _{L^p([s, s + \tau ], L^{r}_x({{\mathbb {R}}}^3))} \le C\Bigl ( \Vert (h_1, h_2)\Vert _{H^1({{\mathbb {R}}}^3) \times L^2({{\mathbb {R}}}^3)} \\&\quad + \Vert F\Vert _{L^1([s, s + \tau ],L^2({{\mathbb {R}}}^3))} \Bigr ), \end{aligned}$$

where $\frac{1}{p} + \frac{3}{r} = \frac{1}{2},\, 2 < p \le \infty .$ We will later choose $r = \frac{4 + 2 \epsilon }{\epsilon }$ with $0 < \epsilon \ll 1$ and this determines the choice of $p > 2.$ For the solution of (3.1), we get

$$\begin{aligned}&\Vert u(t, x) \Vert _ {L^p([s, s + \tau ], L^{r}_x({{\mathbb {R}}}^3))} \le C(p, r)\Bigl ( \Vert u(s, x), u_t(s, x)\Vert _{H^1({{\mathbb {R}}}^3) \times L^2( {{\mathbb {R}}}^3)} \nonumber \\&\quad + \tau \Vert u (t, x)\Vert ^3_{L^{\infty }([s, s + \tau ], H^1({{\mathbb {R}}}^3))}\Bigr ), \end{aligned}$$

(3.3)

where we have used the estimate

$$\begin{aligned} \Vert u^3(t, x)\Vert _{L^1([s, s + \tau ],L^2({{\mathbb {R}}}^3))} \le \tau \Vert u (t, x)\Vert ^3_{L^{\infty }([s, s + \tau ], H^1({{\mathbb {R}}}^3))}. \end{aligned}$$

Next, for the solution $u(t, x) \in H^1({{\mathbb {R}}}^3)$ of (3.1) in $]0, s+ \tau ]$ with initial data $f = (u, u_t)(0, x) \in {{\mathcal {H}}}({{\mathbb {R}}}^3)$, we have a polynomial bound (see [2, Section 3])

$$\begin{aligned} \sup _{t \in [0, s+ \tau ]}\Vert u(t, x)\Vert _{H^1({{\mathbb {R}}}^3)} \le \Vert f\Vert _{{{\mathcal {H}}}({{\mathbb {R}}}^3)} + s(B_1 +B_2 s), \end{aligned}$$

where $B_1> 0, B_2 >0$ depend only on $\Vert f\Vert _{{{\mathcal {H}}}},$ and this implies

$$\begin{aligned} \Vert u(t, x) \Vert _ {L^p([s, s + \tau ], L^{r}_x({{\mathbb {R}}}^3))} \le C_1(p, r, \Vert f\Vert _{{{\mathcal {H}}}})(1 + s)^6. \end{aligned}$$

(3.4)

Now we will examine the continuous dependence on the initial data of the local solution to (2.1) given in Section 2. Let $g_n = ((g_n)_1, (g_n)_2) \in H^{k+1}({{\mathbb {R}}}^3) \times H^k({{\mathbb {R}}}^3)$ be a sequence converging in $H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3)$ to $f = (f_1, f_2) \in H^k({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3).$ Let

$$\begin{aligned} w_n(t, x) \in C([s, s+ \tau ], H^{k+1}({{\mathbb {R}}}^3)) \cap C^1([s, s+ \tau ], H^{k}({{\mathbb {R}}}^3)) \end{aligned}$$

be the local solution of (3.1) with initial data $g_n$. Setting $v_n = w_n - u$, we obtain for $v_n$ the equation

$$\begin{aligned} \partial _t^2 v_n - \Delta _x v_n + q(t, x) v_n = u^3 - w^3_n. \end{aligned}$$

By the local Strichartz estimates for the linear equation with respect to $v_n$, we get

$$\begin{aligned}&\Vert (v_n, (v_n)_t)\Vert _{C([s, s+ \tau ], H^{k}({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3))} + \Vert v_n\Vert _{L_t^{\infty }([s, s+\tau ],H^{k-1, 6}_x({{\mathbb {R}}}^3))}\nonumber \\&\quad \le C_k(a) \Vert g_n - f\Vert _{H^{k}({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3)} + C_k(a) \Vert u^3 - w_n^3\Vert _{L^1_t([s, s+ \tau ], H^{k-1}_x({{\mathbb {R}}}^3))}.\nonumber \\ \end{aligned}$$

(3.5)

This estimate for $k = 1, 2$ has been proved in [2, Proposition 1]. The proof for $k \ge 3$ follows the same argument. The constant $C_k(a) > 0$ depends on k and on the interval [0, a], where $[s, s+ \tau ]\subset [0,a].$ In the notations below we will omit the dependence of the constants on k and a. Applying (2.5), we have

$$\begin{aligned}&\Vert u^3 {-} w_n^3\Vert _{H^{k-1}} {\le } C\Vert v_n\Vert _{H^{k-1,6}} \Vert u^2 {+} u w_n {+} w_n^2\Vert _{L^3}\\&\quad +C\Vert v_n\Vert _{L^6}\Vert u^2 {+} u w_n {+} w_n^2\Vert _{H^{k-1,3}} \\&\quad \le 2 C\Vert v_n\Vert _{H^{k-1, 6}}\Bigl (\Vert u\Vert _{L^6}^2 +\Vert w_n\Vert _{L^6}^2\Bigr ) + C\Vert v_n\Vert _{L^6}\Bigl (2\Vert u\Vert _{H^{k-1, 6}}\Vert u\Vert _{L^6} \\&\quad + 2\Vert w_n\Vert _{H^{k-1,6}}\Vert w_n\Vert _{L^6} {+} \Vert u\Vert _{H^{k-1, 6}} \Vert w_n\Vert _{L^6} + \Vert w_n\Vert _{H^{k-1,6}}\Vert u\Vert _{L^6}\Bigr ) = P_n + Q_n. \end{aligned}$$

To handle $P_n$, notice that the $L^{\infty }([s, s+ \tau ], L^6({{\mathbb {R}}}^3))$ norms of u and $w_n$, by local Strichartz estimates, can be estimated by $\Vert f\Vert _{{{\mathcal {H}}}}$ and $\Vert g_n\Vert _{{{\mathcal {H}}}}$. Therefore, for $n \ge n_0$, we have

$$\begin{aligned} \Bigl |\int \limits _s^{s + \tau } P_n dt \Bigr | \le A_k \tau \Vert v_n\Vert _{L^{\infty }([s, s+\tau ], H^{k-1, 6}({{\mathbb {R}}}^3))} \end{aligned}$$

with a constant $A_k$ depending on $C_k(a)$ and $\Vert f\Vert _{{{\mathcal {H}}}}$. Hence, we may absorb $P_n$ by the left hand side of (3.5) choosing $0 < \tau \le \frac{1}{2A_k}$ small. The analysis of $Q_n$ is easy since we proved in [2, Subsection 3.2] that for all $t \in [s, s+ \tau ]$, we have $\Vert \nabla _x v_n(t, x)\Vert _{L^2({{\mathbb {R}}}^3)} \rightarrow 0$ as $n \rightarrow \infty$ and the term in the braked $\Bigl (...\Bigr )$ for $t \in [0, a]$ is uniformly bounded with respect to n according to the analysis in Section 2 and estimate (2.10). Finally, we conclude that

$$\begin{aligned} \Vert (v_n, (v_n)_t)\Vert _{C([s, s+ \tau ], H^{k}({{\mathbb {R}}}^3) \times H^{k-1}({{\mathbb {R}}}^3))} \rightarrow _{n \rightarrow \infty } 0. \end{aligned}$$

(3.6)

4 Polynomial bound of the $H^2({{\mathbb {R}}}^3)$ norm of the solution

Let $(u(t, x), u_t(t,x)) \in C([s, s + \tau ], H^2({{\mathbb {R}}}^3)) \times C([s, s+ \tau ], H^1({{\mathbb {R}}}^3)),$ where u(t, x) is the solution of the Cauchy problem (2.1) for $t \in [s, s + \tau ]$. Taking the derivative $\partial _{x_j} = \partial _j, \, j = 1, 2,3,$ and noting $u_ j = \partial _j u,\, u_{j t} = \partial _{j}\partial _t u,$ one gets in the sense of distributions

$$\begin{aligned} (u_{j t})_{t} - \Delta _ x u_j + (\partial _j q) u + q u_j + 3 u^2 u_ j = 0. \end{aligned}$$

(4.1)

It is easy to see that

$$\begin{aligned} (\partial _j q) u + q u_j + 3 u^2 u_ j \in C([s, s + \tau ], L^2({{\mathbb {R}}}^3)). \end{aligned}$$

In fact, our assumption implies that $u(t, x) \in C([s, s + \tau ], L_x^{\infty }({{\mathbb {R}}}^3))$ and this yields $u^2 u_j \in C([s, s + \tau ], L^2({{\mathbb {R}}}^3)).$ Therefore

$$\begin{aligned} (u_{j t})_{t} - \Delta _ x u_j \in C([s, s + \tau ], L^2({{\mathbb {R}}}^3)). \end{aligned}$$

Multiplying the equality (4.1) by $u_{j t}$, we have

$$\begin{aligned} \int \Bigl ((u_{j t})_{t} - \Delta _ x u_j \Bigr ) u_{j t} dx = - \int (\partial _j q) u u_{j t} dx - \int q u_j u_{j t} dx \nonumber \\ - 3 \int u^2 u_j u_{ j t} dx = I_1(t) + I_ 2(t) + I_ 3(t). \end{aligned}$$

(4.2)

Assuming $(u(t, x), u_t(t, x)) \in C([s, s + \tau ], H^3({{\mathbb {R}}}^3) \times H^2({{\mathbb {R}}}^3))$, we can write

$$\begin{aligned}&I_2(t) = -\frac{1}{2} \int q \partial _t( u_j^2) dx = - \frac{1}{2} \partial _t \Bigl ( \int q u_j^2 dx\Bigr ) + \frac{1}{2} \int q_t u_j^2 dx, \\&I_3(t) = -\frac{3}{2} \int u^2 \partial _t( u_j^2) dx = - \frac{3}{2} \partial _t \Bigl (\int u^2 u_j^2 dx \Bigr ) + 3 \int u u_t u_j^2 dx. \end{aligned}$$

After an integration by parts in the integral $\int \Delta _x (u_j) u_{j t} dx$ for solutions $(u(t, x), u_t(t, x)) \in C([s, s + \tau ], H^3({{\mathbb {R}}}^3) \times H^2({{\mathbb {R}}}^3))$, the equality (4.2) can be written as

$$\begin{aligned}&\frac{1}{2}\partial _t \sum _{j = 1}^3 \Bigl [\int \Bigl ( (u_{j t})^2 {+} |\nabla _x (u_j)|^2 + 3 u^2 u_j ^2 + q u_j ^2 \Bigr )(t, x) dx\Bigr ] {=} -\sum _{j= 1}^3\int (\partial _j q) u u_{j t} dx \nonumber \\&\quad + 3 \sum _{j= 1}^3\int u u_t u_j^2 dx + \frac{1}{2}\sum _{j = 1} ^3 \int q_t u_j ^2 dx= I_1(t) + J_1 (t)+ J_2(t), \end{aligned}$$

(4.3)

where the derivative with respect to t of the left hand side is taken in the sense of distributions.

4.1 Justification of (4.3) for $(u(t, x), u_t(t, x)) \in C([s,s+ \tau ], H^2({{\mathbb {R}}}^3) \times H^1({{\mathbb {R}}}^3))$

Introduce

$$\begin{aligned} X(t): = \frac{1}{2} \sum _{j = 1}^3\int \Bigl ( (u_{j t})^2 + |\nabla _x (u_j)|^2 + 3 u^2 u_j ^2 + q u_j ^2 \Bigr )(t, x) dx. \end{aligned}$$

Notice that the function X(t) is well defined. For the integral of $u^2 u_j^2$, we have

$$\begin{aligned} \int u^2 u_j^2 dx \le \Vert u\Vert ^2_{L^4({{\mathbb {R}}}^3)} \Vert u_j\Vert _{L^4({{\mathbb {R}}}^3)}^2\le \Vert u\Vert ^{1/2}_{L^2}\Vert \nabla _x u\Vert _{L^2}^{3/2}\Vert u_j\Vert _{L^2}^{1/2} \Vert \nabla _x u_j\Vert _{L^2}^{3/2}.\nonumber \\ \end{aligned}$$

(4.4)

Also a similar argument shows that the right hand side of (4.3) is well defined and it is a continuous function of t. For example,

$$\begin{aligned} \Bigl |\int u u_t u_j^2(t, x) dx \Bigr |\le \Vert u_j(t,x)\Vert _{L^6({{\mathbb {R}}}^3)}^2 \Vert u(t, x)\Vert _{L^6({{\mathbb {R}}}^3)}\Vert u_t(t,x)\Vert _{L^2({{\mathbb {R}}}^3)}. \end{aligned}$$

(4.5)

This implies that the derivative with respect to t is taken in the classical sense. Now let $(g_n, h_n) \in H^3({{\mathbb {R}}}^3) \times H^2({{\mathbb {R}}}^3)$ converge to $(u(s, x), u_t(s, x))$ in $H^2({{\mathbb {R}}}^3) \times H^1({{\mathbb {R}}}^3)$ as $n \rightarrow \infty .$ Denote, as in Section 3, by $w_n(t, x)$ the local solution of (3.1) with initial data $(g_n, h_n).$ Therefore, for $t \in [s, s+ \tau ]$, we have

$$\begin{aligned}&\int w_n^2 ((w_n)_j)^2(t, x) dx \rightarrow _{n \rightarrow \infty } \int u^2u_j^2(t, x) dx, \\&\int w_n (w_n)_t ((w_n)_j)^2(t, x) dx \rightarrow _{n \rightarrow \infty } \int u u_t u_j^2(t, x) dx. \end{aligned}$$

To justify these limits, we apply the estimates (4.4) and (4.5). For example,

$$\begin{aligned}&\Bigl |\int w_n (w_n)_t ((w_n)_j)^2(t, x) dx\Bigr | \le \Bigl |\int (w_n - u) (w_n)_t ((w_n)_j)^2 dx \Bigr | \\&\quad + \Bigl | \int u ((w_n)_t - u_t) ((w_n)_j)^2 dx\Bigr | + \Bigl |\int u u_t (((w_n)_j)^2 - u_j^2) dx\Bigr | \end{aligned}$$

and we use (3.6) for $k= 2$. Passing to limit in the equality (4.3) for $w_n$, we obtain it for u.

Consequently, after an integration with respect to t in (4.3), one deduces

$$\begin{aligned} X(s + \tau ) = X(s) +\int \limits _{s}^{s+ \tau } \Bigl (J_1(t) + J_2(t) + I_1(t)\Bigr ) dt. \end{aligned}$$

4.2 Estimation of $\int _{s}^{s +\tau }J_1(t)dt$

Let $0 < \epsilon \ll 1$ be a small number. First by the generalized Hölder inequality, one estimates

$$\begin{aligned}&|J_1(t)| \le 3\sum _{j=1}^3\Vert u(t, x)\Vert _{L^{r}({{\mathbb {R}}}^3)} \Vert u_t(t, x)\Vert _{L^{2 + \epsilon }({{\mathbb {R}}}^3)} \Vert u_j (t, x)\Vert ^2_{L^4({{\mathbb {R}}}^3)} \\&\quad \le 3\sum _{j=1}^3\Vert u(t, x)\Vert _{L^{r}({{\mathbb {R}}}^3)} \Vert u_t(t, x)\Vert _{L^{2 + \epsilon }({{\mathbb {R}}}^3)} \Vert u_j(t, x)\Vert ^{1/2}_{L^2({{\mathbb {R}}}^3)}\Vert u_j(t, x)\Vert ^{3/2}_{L^6({{\mathbb {R}}}^3)}, \end{aligned}$$

where

$$\begin{aligned} \frac{1}{r} = \frac{\epsilon }{4 + 2 \epsilon }. \end{aligned}$$

According to the estimate (2.7), for $s \le t \le s + \tau ,$ by the local existence of a solution of (3.1) with initial data $(u(s, x), u_t(s, x)) \in H^2({{\mathbb {R}}}^3) \times H^1({{\mathbb {R}}}^3)$ on $t = s$, we obtain

$$\begin{aligned} \Vert u_j(t, x)\Vert ^{3/2}_{L^6({{\mathbb {R}}}^3)} {\le } \Vert \nabla _x u_j(t, x)\Vert ^{3/2}_{L^2({{\mathbb {R}}}^3)} {\le } C_2 \Bigl ( \Vert u(s, x)\Vert _{H^2({{\mathbb {R}}}^3)} {+} \Vert u_t(s, x)\Vert _{H^1({{\mathbb {R}}}^3)}\Bigl )^{3/2} \end{aligned}$$

with constant $C_2 > 0$ depending on q. Next

$$\begin{aligned}&\Vert u(s, x)\Vert ^2 _{H^2({{\mathbb {R}}}^3)} \le C \Bigl (\sum _{i, j = 1}^3 \Vert \partial _{x_i} \partial _{x_j}u(s, x)\Vert ^2_{L^2({{\mathbb {R}}}^3)} + \Vert u(s, x)\Vert ^2_{H^1({{\mathbb {R}}}^3)}\Bigr ), \\&\quad \Vert u_t(s, x)\Vert ^2_{H^1({{\mathbb {R}}}^3)} \le C\Bigl ( \sum _{j= 1}^3\Vert u_{j t}(s, x)\Vert ^2_{L^2({{\mathbb {R}}}^3)} + \Vert u_t(s, x)\Vert ^2_{L^2({{\mathbb {R}}}^3)}\Bigr ). \end{aligned}$$

Notice that we have a polynomial bound with respect to s for the norms $\Vert u(s, x)\Vert _{H^1({{\mathbb {R}}}^3)}$ and $\Vert u_t(s, x)\Vert _{L^2({{\mathbb {R}}}^3)}$ of the solution u(s, x) (see [2, Theorem 2]). Consequently, we obtain

$$\begin{aligned}&\sup _{t \in [s, s + \tau ]} \Vert u_j(t, x)\Vert ^{3/2}_{L^6({{\mathbb {R}}}^3)} \\&\quad \le C_1\Bigl ( X(s)^{3/4} +(1 + s)^3\Bigr ),\,\sup _{t \in [s, s + \tau ]}\Vert u_j(t, x)\Vert _{L^2({{\mathbb {R}}}^3)} \le C_0( 1+ s), \end{aligned}$$

where $C_0> 0, C_1 > 0$ depend on $\Vert (u,u_t)(0,x)\Vert _{H^1({{\mathbb {R}}}^3)}.$

Now we pass to the estimate of $\Vert u_t(t, x)\Vert _{L^{2+\epsilon }({{\mathbb {R}}}^3)}.$ By the Hölder inequality, we obtain

$$\begin{aligned}&\Bigl | \int u_t^{2 +\epsilon } dx \Bigr | = \Bigl | \int u_t^{2(1 - \frac{\epsilon }{4})} u_t^{\frac{3\epsilon }{2}} dx \Bigr | \le \Vert u_t\Vert _{L^2({{\mathbb {R}}}^3)}^{2(1 - \epsilon /4)} \Vert u_t\Vert _{L^6({{\mathbb {R}}}^3)}^{\frac{3\epsilon }{2}} \\&\quad \le C_3(1 + t)^2 \Vert \nabla _x u_t\Vert _{L^2({{\mathbb {R}}}^3)}^{\frac{3\epsilon }{2}} \le C_4(1 + s)^2\Bigl ( X(s)^{\frac{3\epsilon }{4}} + (1 +s)^{3 \epsilon }\Bigr ). \end{aligned}$$

Hence, one deduces

$$\begin{aligned} \sup _{t \in [s, s+\tau ]} \Bigl | \int u_t ^{2 + \epsilon } dx \Bigr | ^{\frac{1}{2 + \epsilon }} \le C_5(1+ s)^{3/2}\Bigl (X(s)^{\frac{3\epsilon }{8 + 4\epsilon }} + 1\Bigr ). \end{aligned}$$

Taking into account the above estimates for the integral with respect to t, one applies the Hölder inequality and for small $\epsilon$, we have

$$\begin{aligned} \Bigl |\int \limits _{s}^{s + \tau } J_1(t)dt \Bigr | \le C_6 \tau ^{1/p'} (1 + s ) ^6\Vert u(t, x)\Vert _{L^p([s, s+ \tau ]; L_x^{r}({{\mathbb {R}}}^3))} \Bigl (X(s)^{\frac{3}{4} +\frac{3\epsilon }{8}} + 1\Bigr ), \end{aligned}$$

where

$$\begin{aligned} \frac{1}{p} + \frac{3\epsilon }{4 + 2 \epsilon } = \frac{1}{2},\, \frac{1}{p'} +\frac{1}{p} = 1. \end{aligned}$$

To complete the analysis, we apply the Strichartz estimate (3.4) and deduce

$$\begin{aligned} \Vert u(t, x)\Vert _{L^p([s, s+ \tau ]; L_x^{r}({{\mathbb {R}}}^3))} \le C(\epsilon ) (1 + s)^6. \end{aligned}$$

Finally, for $0 < \tau \le 1$ with $y = 12$, we have

$$\begin{aligned} \Bigl |\int \limits _{s}^{s + \tau } J_1(t)dt \Bigr | \le C'(\epsilon ) \Bigl (X(s)^{\frac{3}{4} + \frac{3 \epsilon }{8}}+ 1 \Bigr )(1 + s)^y. \end{aligned}$$

(4.6)

4.3 Estimation of $\int _{s}^{s +\tau }I_1(t)dt$

We apply a similar argument.

$$\begin{aligned} |I_1(t)| {\le } C\sum _{j= 1}^3 \Vert u(t, x)\Vert _{L^2({{\mathbb {R}}}^3)} \Vert u_{jt}(t, x)\Vert _{L^2({{\mathbb {R}}}^3)} {\le } C_7 (1 + |t|)^2 \sum _{j= 1}^3 \Vert u_{jt}(t, x)\Vert _{L^2({{\mathbb {R}}}^3)}. \end{aligned}$$

By the local existence result for $t \in [s, s + \tau ]$, one has

$$\begin{aligned} \Vert u_{jt}(t, x)\Vert _{L^2({{\mathbb {R}}}^3)} \le C ( \Vert u(s, x)\Vert _{H^2({{\mathbb {R}}}^3)} + \Vert u_t(s, x)\Vert _{H^1({{\mathbb {R}}}^3)}) \end{aligned}$$

and repeating the above argument, we deduce

$$\begin{aligned} \Bigl |\int \limits _{s}^{s + \tau } I_1(t)dt \Bigr | \le C_8(X(s)^{1/2} + 1)(1 + s)^2. \end{aligned}$$

(4.7)

4.4 Estimation of $\int _{s}^{s +\tau }J_2(t)dt$

It is easy to find a bound for this term since we have a polynomial estimate

$$\begin{aligned} \int u_j^2(t, x) dx \le C_0( 1 + |t|)^2 \end{aligned}$$

and this yields

$$\begin{aligned} \Bigl |\int \limits _{s}^{s + \tau } J_2(t)dt \Bigr | \le C_9(1 + s)^2. \end{aligned}$$

(4.8)

Combining (4.6), (4.7), (4.8), finally we get

$$\begin{aligned} X(s + \tau ) \le X(s) + C_{10}\Bigl (X(s)^{\frac{3}{4} + \frac{3\epsilon }{8}} + 1\Bigr )(1 + s)^y. \end{aligned}$$

(4.9)

4.5 Growth of $H^2({{\mathbb {R}}}^3)$ norm

Let $a > 1$ be a fixed number. According to [2] and Proposition 2.1, there exists a solution in $[s , s + \tau (a)] \subset [0, a]$ with initial data $g \in H^2({{\mathbb {R}}}^3) \times H^1({{\mathbb {R}}}^3)$ on $t = s$. Here

$$\begin{aligned} \tau (a) = c \Bigl (( 1 + \Vert f\Vert _{H^1({{\mathbb {R}}}^3) \times L^2({{\mathbb {R}}}^3)} + a (B_1 + B_2 a)\Bigr )^{-\gamma }< 1, \end{aligned}$$

where $c> 0, \gamma> 0, B_1> 0, B_2 > 0$ are independent of a and f. We choose $N(a) \in {{\mathbb {N}}}$ so that $a- \tau (a) < N(a)\tau (a) \le a.$ Setting $X(n\tau (a)) = \alpha _n,\,n \le N(a),$ and exploiting (4.9), one deduces

$$\begin{aligned} \alpha _{n} \le \alpha _{n-1} + C_{10}(\alpha _{n-1}^{7/8} + 1)(1 + n)^{12}. \end{aligned}$$

We are in the position to apply Lemma A.1 in the Appendix and obtain

$$\begin{aligned}&X(N(a)\tau (a)) \le {\tilde{C}} (N(a))^{104} \\&\quad \le {\tilde{C}} \Bigl (\frac{a}{c}\Bigr )^{104} \Bigl ( 1 + \Vert f\Vert _{H^1({{\mathbb {R}}}^3) \times L^2({{\mathbb {R}}}^3)} + a (B_1 + B_2 a)\Bigr )^{104\gamma }. \end{aligned}$$

This estimate and the bound of the $H^1({{\mathbb {R}}}^3)$ norm of the solution u(a, x) established in [2] imply a polynomial with respect to a bound of $\Vert u(a, x)\Vert _{H^2({{\mathbb {R}}}^3)} + \Vert \partial _t u(a, x)\Vert _{H^1({{\mathbb {R}}}^3)}$. This implies the statement of Theorem 1.1 for $k = 2.$

5 Polynomial growth of the $H^k({{\mathbb {R}}}^3)$ norm of the solution.

To examine the growth of the $H^k({{\mathbb {R}}}^3)$ norm of the solution, we will proceed by induction. Assume that for $1 \le s \le k-1, k \ge 3,$ we have polynomial bounds

$$\begin{aligned} \Vert u(t, x)\Vert _{H^s_x({{\mathbb {R}}}^3)} + \Vert u_t(t, x)\Vert _{H^{s-1}_x({{\mathbb {R}}}^3)} \le A_k (1 + |t|)^{m_s},\, t \in {{\mathbb {R}}}\end{aligned}$$

for the global solution of the Cauchy problem of $u_{tt} - \Delta _x u +qu+ u^3 = 0$ with initial data $(f_1, f_2) \in H^s({{\mathbb {R}}}^3) \times H^{s-1}({{\mathbb {R}}}^3).$ Consider the equality

$$\begin{aligned} \partial _t^2\partial _x^{\alpha }u - \Delta _ x (\partial _x^{\alpha }u) + \partial _x^{\alpha }(qu) + \partial _x^{\alpha } (u^3) = 0 \end{aligned}$$

with $|\alpha | = k - 1.$ After an integration by parts which we can justify as in Section 4, we write

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int \Bigl (|\nabla _x \partial _x^{\alpha } u|^2 + |\partial _t \partial _x^{\alpha } u|^2\Bigr ) dx \nonumber \\&\quad = - \int \partial _x^{\alpha }(q u) \partial _x^{\alpha } \partial _t u dx - \int \partial _x^{\alpha } (u^3) \partial _x^{\alpha } \partial _t u dx = K_1(t) + K_2(t). \end{aligned}$$

(5.1)

Clearly,

$$\begin{aligned} \Bigr |\int \Bigr (\partial _x^{\alpha } (u^3) \partial _x^{\alpha } \partial _t u\Bigr ) dx\Bigr | \le \Vert \partial _x^{\alpha } (u^3)\Vert _{L^2({{\mathbb {R}}}^3)}\Vert \partial _x^{\alpha } \partial _t u\Vert _{L^2({{\mathbb {R}}}^3)}. \end{aligned}$$

Applying (2.5) two times, one gets

$$\begin{aligned} \Vert \partial _x^{\alpha } (u^3)\Vert _{L^2({{\mathbb {R}}}^3)} \le C\Vert \partial _x^{\alpha } u\Vert _{L^2({{\mathbb {R}}}^3)} \Vert u\Vert _{L^{\infty }({{\mathbb {R}}}^3)}^2 \end{aligned}$$

and by the Sobolev theorem, $\Vert u\Vert _{L^{\infty }({{\mathbb {R}}}^3)} \le C \Vert u\Vert _{H^2({{\mathbb {R}}}^3)}.$ Thus by our assumption,

$$\begin{aligned} \Vert \partial _x^{\alpha } (u^3(t, x))\Vert _{L^2({{\mathbb {R}}}^3)} \le C A_{k-1} A_2^2 (1 + |t|)^{m_{k-1} + 2m_2}. \end{aligned}$$

Therefore, using the notation of Subsection 4.5 for $n \tau (a) \le t \le (n+1) \tau (a),$ one deduces

$$\begin{aligned} \Vert \partial _x^{\alpha } (u^3(t, x))\Vert _{L^2({{\mathbb {R}}}^3)} \le C A_{k-1} A_2^2(1 + n)^{m_{k-1} + 2m_2}. \end{aligned}$$

On the other hand, applying (2.7), one obtains

$$\begin{aligned} \Vert \partial _x^{\alpha } \partial _t u(t, x)\Vert _{L^2({{\mathbb {R}}}^3)} \le C_{k}\Bigl ((\Vert u(n\tau (a),x)\Vert _{H^{k}({{\mathbb {R}}}^3)} + \Vert u_t(n\tau (a),x)\Vert _{H^{k-1}({{\mathbb {R}}}^3)}\Bigr ). \end{aligned}$$

The analysis of $K_1(t)$ is easy and

$$\begin{aligned}&|K_1(t)| \le C \Vert u(t, x)\Vert _{H^{k-1}({{\mathbb {R}}}^3)} \Vert \partial _x^{\alpha }\partial _t u(t, x)\Vert _{L^2({{\mathbb {R}}}^3)} \\&\le C_k A_{k-1}(1 + n)^{m_{k-1}}(\Vert u(n\tau (a),x)\Vert _{H^{k}({{\mathbb {R}}}^3)} + \Vert u_t(n\tau (a),x)\Vert _{H^{k-1}({{\mathbb {R}}}^3)}). \end{aligned}$$

Now define $Y_k(t): = \Vert u(t, x)\Vert _{H^k({{\mathbb {R}}}^3)}^2 + \Vert \partial _t u(t, x)\Vert _{H^{k-1}({{\mathbb {R}}}^3)}^2$ and integrate the equality (5.1) from $n\tau _k(a)$ to $(n+1)\tau _k(a)$ with respect to t, where $0< \tau _k(a) < 1$ is defined by (2.9). Taking into account the above estimates, we have

$$\begin{aligned}&Y_k((n+1)\tau _k(a)) \le Y_k(n \tau _k(a)) + C_q A_{k-1}(1 + n)^{m_{k-1}} Y_k^{\frac{1}{2}}(n\tau _k(a)) \\&\quad + C A_{k-1} A_2^2 (1+ n)^{m_{k-1} + 2m_2} Y_k^{1/2}(n\tau _k(a)). \end{aligned}$$

Applying Lemma A.1 and repeating the argument of Subsection 4.5, we obtain a polynomial bound for $Y_k(t)$ and this completes the proof of Theorem 1.1.

Appendix

The aim of this Appendix is to prove the following

Lemma A.1

Let $\{\alpha _n\}_{n=0}^{\infty }$ be a sequence of nonnegative numbers such that with some constants $0< \gamma <1$, $C > 0$, and $y \ge 0$, we have

$$\begin{aligned} \alpha _n \le \alpha _{n-1} + C ((\alpha _{n-1})^{1 - \gamma }+ 1)(1 + n)^y,\, \forall n \ge 1. \end{aligned}$$

Then there exists a constant ${\tilde{C}} > 0$ such that

$$\begin{aligned} \alpha _n \le {\tilde{C}}(1 + n)^{\frac{1 + y}{\gamma }},\,\forall n \ge 1. \end{aligned}$$

(A.1)

Remark A.1

A similar estimate has been established in [3] for sequences $\{\alpha _n\}$ satisfying the inequality

$$\begin{aligned} \alpha _n \le \alpha _{n-1} + C \alpha _{n-1}^{1 - \gamma }. \end{aligned}$$

Proof

We can choose a large constant $C_1 > 0$ such that

$$\begin{aligned} (\alpha _{n-1})^{1 - \gamma } + 1 \le C_1 (\alpha _{n-1} + 1)^{1-\gamma }, \, \forall n\ge 1. \end{aligned}$$

This implies with a new constant $C_2 > 0$ the inequality

$$\begin{aligned} \alpha _n + 1\le \alpha _{n-1} + 1+ C_2(\alpha _{n-1} + 1)^{1 - \gamma } (1 + n)^y,\, \forall n \ge 1. \end{aligned}$$

Setting $\beta _n = \alpha _n + 1,$ we reduce the proof to a sequence $\alpha _n$ satisfying the inequality

$$\begin{aligned} \alpha _n \le \alpha _{n-1} + C_2 (\alpha _{n-1})^{1 - \gamma }(1 + n)^y, \,n \ge 1. \end{aligned}$$

We will prove (A.1) by recurrence. Assume that (A.1) holds for $n-1$. Therefore

$$\begin{aligned}&\alpha _n \le {\tilde{C}} n^{\frac{1 + y}{\gamma }} + C_2 \Bigl ( {\tilde{C}} n^{\frac{1 + y}{\gamma }}\Bigr )^{1-\gamma } (1+n)^y \\&\quad = {\tilde{C}} n^{\frac{1 + y}{\gamma }} \Bigl [ 1 + C_2 {\tilde{C}}^{-\gamma }n^{-1 - y}(1+ n)^y\Bigr ] \\&\quad = {\tilde{C}} (1+n)^{\frac{1 + y}{\gamma }}\Bigl ( 1 -\frac{1}{n + 1}\Bigr )^{\frac{1 + y}{\gamma }}\Bigl [ 1 + C_2 {\tilde{C}}^{-\gamma }n^{-1}\Bigl (\frac{n}{n+1}\Bigr )^{- y}\Bigr ]. \end{aligned}$$

To establish (A.1) for n, it is sufficient to show that for large ${\tilde{C}}$, one has

$$\begin{aligned} f(n): = \Bigl ( 1 -\frac{1}{n + 1}\Bigr )^{\frac{1 + y}{\gamma }}\Bigl [ 1 + C_2 {\tilde{C}}^{-\gamma }n^{-1}\Bigl (\frac{n}{n+1}\Bigr )^{- y}\Bigr ] \le 1,\, n\ge 1. \end{aligned}$$

(A.2)

Setting $C_2 {\tilde{C}}^{-\gamma } = \epsilon$, a simple calculus yields

$$\begin{aligned} f'(n)= & {} \frac{1 + y}{\gamma } \Bigl (1 - \frac{1}{n+1}\Bigr )^{\frac{1 + y}{\gamma }- 1}\frac{1}{(n+1)^2}\Bigl [ 1 + \frac{\epsilon }{n}\Bigl (\frac{n}{n+ 1}\Bigr )^{- y}\Bigr ] \\&+ \epsilon \Bigl (1 {-} \frac{1}{n+1}\Bigr )^{\frac{1 + y}{\gamma }}\Bigl [ -\frac{1}{n^2}\Bigl (\frac{n}{n+ 1}\Bigr )^{-y}{-}y n^{-1} \frac{1}{(n+1)^2} \Bigl (1 - \frac{1}{n+1}\Bigr )^{-y -1}\Bigr ] \\&= \Bigl (1 - \frac{1}{n+1}\Bigr )^{\frac{1 + y}{\gamma }-1}\frac{1}{(n+ 1)^2}\Bigl [ \frac{1 + y}{\gamma }+ \frac{\epsilon }{n}\frac{1 + y}{\gamma }\Bigl ( 1 - \frac{1}{n+1}\Bigr )^{-\gamma } \\&\quad - [\epsilon \frac{n+1}{n}+ \frac{\epsilon y}{n}] \Bigl ( 1 - \frac{1}{n+1}\Bigr )^{-y}\Bigr ]. \end{aligned}$$

Notice that since $\frac{1}{2} \le 1- \frac{1}{n+ 1}$, we have

$$\begin{aligned} \Bigl ( 1 - \frac{1}{n+1}\Bigr )^{-\gamma } \le \Bigl (\frac{1}{2}\Bigr )^{-\gamma } \end{aligned}$$

which implies

$$\begin{aligned} \frac{1 + y}{\gamma }- \epsilon [\frac{n+1 + y}{n}] \Bigl ( 1 - \frac{1}{n+1}\Bigr )^{-y} \ge \frac{1 + y}{\gamma }- \epsilon [\frac{n+1 + y}{n}] \Bigl (\frac{1}{2}\Bigr )^{-y}. \end{aligned}$$

For small $\epsilon > 0$, the right hand side of the above inequality is positive. Consequently, for the derivative, we have $f'(n) > 0$ and one deduces

$$\begin{aligned} f(n) < \lim _{n \rightarrow +\infty } f(n)= 1 \end{aligned}$$

This completes the proof of (A.2). $\square$