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 $0< \tau _k <1$.

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). $\square$