Exponential decay for the KdV equation on ℝ with new localized dampings

Abstract

In this paper, we prove several results on the exponential decay in $L^{2}$ norm of the KdV equation on the real line with localized dampings. First, for the linear KdV equation, the exponential decay holds if and only if the averages of the damping coefficient on all intervals of a fixed length have a positive lower bound. Moreover, under the same damping condition, the exponential decay holds for the (nonlinear) KdV equation with small initial data. Finally, with the aid of certain properties of propagation of regularity in Bourgain spaces for solutions of the associated linear system and the unique continuation property, the exponential decay for the KdV equation with large data holds if the damping coefficient has a positive lower bound on $E$, where $E$ is equidistributed over the real line and the complement $E^{c}$ has a finite Lebesgue measure.

1. Introduction

In this paper, we are interested in the exponential decay property of the KdV equation on the real line $\mathbb {R}$

(1.1)\begin{equation} \partial_tu+\partial_x^{3}u+u\partial_xu+a(x)u=0, \quad u(x,0)=u_0(x)\in L^{2}(\mathbb{R}). \end{equation}

Here, $u(x,\,t)$ is a real-valued function on $\mathbb {R}\times \mathbb {R}^{+}$, the function $a(x)$ satisfies the condition

(A1)\begin{equation} 0\leq a(x)\in L^{\infty}(\mathbb{R}). \end{equation}

In the case $a(x)=0$, (1.1) reduces to the classical KdV equation, which models the unidirectional propagation of small-amplitude long waves in nonlinear dispersive systems. If we multiply (1.1) by $2u$ and integrate over $\mathbb {R}$, then

(1.2)\begin{equation} \frac{{{\rm d}}}{{{\rm d}}t}\int_\mathbb{R} |u(x,t)|^{2}{{\rm d}}x + 2\int_\mathbb{R} a(x)|u(x,t)|^{2}{{\rm d}}x=0. \end{equation}

This, and condition $a(x)\geq 0$, clearly implies that $\|u(\cdot\,,t )\|_{L^{2}(\mathbb {R})}\leq \|u_0\|_{L^{2}(\mathbb {R})}$ for all $t\geq 0$. Moreover, if $a(x)\geq a_0>0$ for all $x\in \mathbb {R}$, then (1.2) gives the decay bound

(1.3)\begin{equation} \|u({\cdot},t)\|_{L^{2}(\mathbb{R})}\leq {\rm e}^{{-}a_0t}\|u_0\|_{L^{2}(\mathbb{R})}, \quad \text{for all } t\geq 0. \end{equation}

Thus, the term $au$ is called a damping in the literature. Now an interesting question arises naturally, whether the exponential decay as (1.3) holds if

(A2)\begin{equation} a(x)\geq a_0>0, \quad x\in E \end{equation}

is satisfied only on a subset $E\subset \mathbb {R}$? In this case, the term $au$ is referred to a localized damping.

Similar problems for KdV type equations on bounded domains have been studied extensively, we refer to e.g. [3, 14, 15, 17–19, 21, 23] and the survey [24]. But much less is known for the KdV equation on unbounded domains. For the KdV equation posed on $(x,\,t)\in \mathbb {R}^{2}_+$, the exponential decay of the $L^{2}(\mathbb {R}_+)$ norm was proved in [16] when (A2) holds on $E=(0,\,\delta )\bigcup (L,\,+\infty )$ for some $0<\delta < L$. The same result was obtained in [22] under a weaker localized damping, namely (A2) holds only on $E=(L,\,+\infty )$ for some $L>0$. For the KdV equation on $(x,\,t)\in \mathbb {R}\times \mathbb {R}_+$, namely (1.1), the exponential decay was established in [4] with damping on $E=(-\infty,\,-L)\bigcup (L,\,+\infty )$. If one considers the KdV equation with strong dissipation

\[ \partial_tu-\partial_x^{2}u+\partial_x^{3}u+u\partial_xu+a(x)u=0, \quad u(x,0)=u_0(x)\in L^{2}(\mathbb{R}), \]

called the Korteweg–de Vries–Burgers equation, then the exponential decay holds with an indefinite damping, namely $a(x)$ may change sign, see [5, 7, 9].

In this article, the main goal is to consider the following question: To what extent the set $E$ can be small so that the exponential decay holds for KdV equations with damping on $E$. First of all, we give a sufficient and necessary condition for the exponential decay of the linear KdV equation.

Theorem 1.1 Assume that (A1) holds. Then the following are equivalent:

1. (1) There exist constants $C,\,\lambda >0$ so that

   \[ \|u({\cdot},t)\|_{L^{2}(\mathbb{R})}\leq C{\rm e}^{-\lambda t}\|u_0\|_{L^{2}(\mathbb{R})}, \quad \forall t\geq 0 \]
   holds for all solutions of the initial value problem (IVP) $\partial _tu+\partial _x^{3}u+a(x)u=0,\, u(x,0)=u_0(x)\in L^{2}(\mathbb {R}).$

2. (2) There exists a constant $L>0$ so that

   (1.4)\begin{equation} \inf_{x\in \mathbb{R}}\int_{x-L}^{x+L}a(y){{\rm d}}y>0. \end{equation}

As noted in [8], under assumption (A1), condition (1.4) is equivalent to (A2) for some $a_0>0$ and a thick set $E$. Recall that (see e.g. [29]) a measurable set $E\subset \mathbb {R}$ is thick, if there exists $L>0$ so that

\[ \inf_{x\in \mathbb{R}} \Big|E\bigcap[x-L,x+L]\Big|>0. \]

Here and below, we use $|E|$ to denote the Lebesgue measure of $E$. Based on theorem 1.1 and the contraction mapping principle, we give the exponential decay for the KdV equation (1.1) with small data.

Theorem 1.2 (Decay for small data)

Assume that (A1) holds and (A2) holds on a thick set $E$. Then there exist constants $C>0,\, \lambda >0,\,\delta >0$ such that

\[ \|u(t)\|_{L^{2}(\mathbb{R})}\leq C{\rm e}^{-\lambda t}\|u_0\|_{L^{2}(\mathbb{R})}, \quad t\geq 0 \]

holds for all solutions of (1.1) with data $\|u_0\|_{L^{2}(\mathbb {R})}\leq \delta$.

To obtain the exponential decay for large data, we need to strength the damping effect. To state our result, we first introduce a set class. A set $E\subset \mathbb {R}$ is said to be satisfying the network control condition (NCC), named after [2], if there exist constants $r,\,L>0$ so that

\[ E\supset \bigcup_{n}(x_n-r,x_n+r), \quad \inf_{n}|x-x_n|\leq L, \text{ for all } x\in \mathbb{R}. \]

Clearly, a set satisfying NCC is a thick set, but a thick set could not satisfy NCC, see figure 1.

Remark 1.3 In figure 1, the set consisting of red intervals is a typical thick set. It has a fixed positive Lebesgue measure on every $[n,\,n+1]$, but it does not contain an interval with given length simultaneously on all $[n,\,n+1],\, n\in \mathbb {Z}$. The union of blue intervals is a typical set satisfying NCC.

FIG. 1. Comparison of thick set with NCC.

The set class satisfying NCC in higher dimensions (the definition is the same except minor modifications) was first introduced to study the observability of the Kolmogorov equation [10] (see also [6] for observability of heat equations).

Theorem 1.4 (Decay for general data)

Assume that (A1) holds and (A2) holds on $E,$ where $E$ satisfies NCC and the complement set $E^{c}$ has a finite Lebesgue measure. Then for every $R>0,$ there exist constants $C,\,\lambda >0$ depending only on $R$ and $a(x)$ so that

\[ \|u(t)\|_{L^{2}(\mathbb{R})}\leq C{\rm e}^{-\lambda t}\|u_0\|_{L^{2}(\mathbb{R})}, \quad t\geq 0 \]

holds for all solutions $u$ of IVP (1.1) with initial data satisfying $\|u_0\|_{L^{2}(\mathbb {R})}\leq R$.

Note that if $E$ is the complement of a compact set, then $E$ satisfies NCC and $m(E^{c})<\infty$. However, if

\[ E=\mathbb{R} \backslash \bigcup_{0\neq k\in \mathbb{Z}} \left[k,k+\frac{1}{k^{2}}\right], \]

then $E$ satisfies NCC and $m(E^{c})<\infty$, but $E^{c}$ cannot be contained in a compact set, see figure 2. Therefore, theorem 1.4 improves the results in [4] in the sense that the exponential decay holds with localized damping on more general sets. In fact, to obtain exponential decay results established in [4], it is assumed that the damping effect holds on the complement of a compact set (see the blue set in figure 2). On the other hand, conditions in theorem 1.4 allows no damping to occur on some small gaps at infinity (see the red set in figure 2).

FIG. 2. Damping domain considered in [4] vs damping domain considered in theorem 1.4.

Let us describe briefly the applications of conditions established on the set $E$ and the main arguments to prove theorem 1.4.

1. (1) $E$ satisfies NCC. This condition is necessary to show the propagation of regularity for the operator $\mathbb {L}=\partial _t+\partial _x^{3}$. Roughly speaking, let $u$ be a solution of the equation $\mathbb {L}u=f$ with a smooth $f$, then $u$ is smooth on $\mathbb {R}$ if $u$ is smooth on $E$. We refer to lemma 4.3 for precise statements.

2. (2) $m(E^{c})<\infty.$ This condition is necessary to establish the compactness of sequence $u_n\in L^{2}(0,\,T; L^{2}(E^{c}))$, where $u_n$ is the solution of IVP (1.1) with initial data $u_{0n}$ bounded in $L^{2}(\mathbb {R})$. Combining the compactness and the propagation of regularity, we show that the solution of (1.1) enjoys the observability

   \[ \int_0^{T}\int_\mathbb{R}|u(x,t)|^{2}{{\rm d}}x {{\rm d}}t\leq C\int_0^{T}\int_\mathbb{R} a(x)|u(x,t)|^{2}{{\rm d}}x {{\rm d}}t. \]
   Then the exponential decay follows from a standard argument.

The notation used in this paper is standard. We only point out that we use $A\lesssim B$ to denote $A\leq CB$ for some constant $C>0$, which may vary from place to place.

The paper is organized as follows. In § 2, we show theorem 1.1. The proof of theorem 1.2 is given in § 3. Finally, in § 4, the propagation of regularity and an observability inequality are established to prove theorem 1.4.

2. Exponential decay for linear KdV

Assume that $0\leq a(x)\in L^{\infty }(\mathbb {R})$. Consider the linear operator $A: H^{3}(\mathbb {R})\mapsto L^{2}(\mathbb {R})$

\[ Au=\partial_x^{3}u+a(x)u, \quad u\in H^{3}(\mathbb{R}). \]

Clearly, we have

\[ (Au,u)=(\partial_x^{3}u+a(x)u,u)=\int_\mathbb{R} a(x)|u(x)|^{2}{{\rm d}}x \geq 0. \]

This shows that $-A$ is dissipative. According to Lumer–Phillips theorem [20], $-A$ generates a $C_0$ semigroup of contractions in $L^{2}(\mathbb {R})$, namely

\[ \|{\rm e}^{{-}tA}\|_{L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})}\leq 1, \quad \forall t\geq 0. \]

This will be improved to an exponential decay upper bound if $a$ satisfies some further damping conditions.

Theorem 2.1 Assume that (A1) holds and (A2) holds on a thick set $E$. Then there exist constants $C,\,\lambda >0$ depending only on $a$ and $E$ so that

\[ \|{\rm e}^{{-}tA}\|_{L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})}\leq C{\rm e}^{-\lambda t}, \quad \forall t\geq 0. \]

Proof. The result has been stated in [31] without proof. We give a sketch here for the reader's convenience. Let $E$ be a thick set. Based on the uncertainty principle of the Fourier transform, one can show that (see [31, lemma 2.3]) there exists a constant $c_1>0$ so that for all $\tau \in \mathbb {R},\, u\in H^{3}(\mathbb {R})$,

(2.1)\begin{equation} c_1\|u\|_{L^{2}(\mathbb{R})}\leq \|(\partial_x^{3}+{{\rm i}}\tau)u\|_{L^{2}(\mathbb{R})}+\|u\|_{L^{2}(E)}. \end{equation}

Since $a\in L^{\infty }(\mathbb {R})$, by the triangular inequality

(2.2)\begin{equation} \|(\partial_x^{3}+{{\rm i}}\tau)u\|_{L^{2}(\mathbb{R})}\leq \|(A+{{\rm i}}\tau)u\|_{L^{2}(\mathbb{R})}+\|a\|_{L^{\infty}(\mathbb{R})}^{{1}/{2}}\|a^{{1}/{2}}u\|_{L^{2}(\mathbb{R})}. \end{equation}

By assumption (A2), $a(x)\geq a_0$ on $E$, we have

(2.3)\begin{equation} \|u\|_{L^{2}(E)}\leq a_0^{-({1}/{2})}\|a^{{1}/{2}}u\|_{L^{2}(\mathbb{R})}. \end{equation}

It follows from (2.1)–(2.3) that for some $c_2>0$

(2.4)\begin{equation} c_1\|u\|_{L^{2}(\mathbb{R})}\leq \|(A+{{\rm i}}\tau)u\|_{L^{2}(\mathbb{R})}+c_2\|a^{{1}/{2}}u\|_{L^{2}(\mathbb{R})}. \end{equation}

Moreover, taking the $L^{2}(\mathbb {R})$ inner product and using (A1), we find

(2.5)\begin{equation} \int_\mathbb{R} a(x)|u|^{2}{{\rm d}}x ={\bf Re}((A+{{\rm i}}\tau)u,u) \leq \|u\|_{L^{2}(\mathbb{R})}\|(A+{{\rm i}}\tau)u\|_{L^{2}(\mathbb{R})} .\end{equation}

Plugging (2.5) into (2.4) and using Cauchy–Schwartz, we infer that for all $\varepsilon >0$,

(2.6)\begin{equation} c_1\|u\|_{L^{2}(\mathbb{R})}\leq \|(A+{{\rm i}}\tau)u\|_{L^{2}(\mathbb{R})}+c_2\varepsilon\|u\|_{L^{2}(\mathbb{R})}+c_2\varepsilon^{{-}1}\|(A+{{\rm i}}\tau)u\|_{L^{2}(\mathbb{R})}. \end{equation}

Taking $\varepsilon =\varepsilon _0>0$ small enough such that $c_2\varepsilon _0\leq {c_1}/{2}$, then we deduce from (2.6) that

\[ \|u\|_{L^{2}(\mathbb{R})}\leq c_3\|(A+{{\rm i}}\tau)u\|_{L^{2}(\mathbb{R})}, \quad \forall \tau\in \mathbb{R}, \]

where $c_3>0$ is a constant independent of $\tau$. This implies that the resolvent set $\rho (A)\supset {{\rm i}}\mathbb {R}$ and $\sup _{\tau \in \mathbb {R}}\|(A+{{\rm i}} \tau )^{-1}\|_{L^{2}(\mathbb {R})\to L^{2}(\mathbb {R})}<\infty$. Thus according to Gearhart–Pruss–Huang criteria [11], the desired decay holds.

Proof of theorem 1.1. Assume that (A1) holds. We divide the proof into two steps.

(2)$\Longrightarrow$(1). First, following [8], we show that the condition $(2)$ implies that (A2) holds on a thick set $E$. In fact, if $(2)$ holds, then there exists $\gamma >0$, so that

(2.7)\begin{equation} \inf_{x\in \mathbb{R}} \int_{x-L}^{x+L}a(y){{\rm d}}y\geq \gamma>0. \end{equation}

Fix $x\in \mathbb {R}$. For every $\varepsilon >0$, we consider the set

\[ \Sigma_\varepsilon=\{y\in [x-L,x+L]: 0\leq a(y)< \varepsilon\}. \]

It follows from (2.7) that

(2.8)\begin{align} \gamma& \leq \int_{x-L}^{x+L}a(y){{\rm d}}y\leq \int_{[x-L,x+L]\bigcap \Sigma_\varepsilon}a(y){{\rm d}}y+\int_{[x-L,x+L]\bigcap \Sigma^{c}_\varepsilon}a(y){{\rm d}}y\nonumber\\ & \leq 2L\varepsilon +\|a\|_{L^{\infty}(\mathbb{R})}|\Sigma_\varepsilon^{c}|. \end{align}

Choose $\varepsilon =\varepsilon _0:={\gamma }/{4L}$. It follows from (2.8) and the definition of $\Sigma ^{c}_\varepsilon$ that

\[ |y\in [x-L,x+L]: a(y)\geq \varepsilon_0|\geq \frac{\gamma}{2\|a\|_{L^{\infty}(\mathbb{R})}}>0. \]

Since $x$ can be chosen arbitrarily, we conclude that the set $\{y\in \mathbb {R}: a(y)\geq \varepsilon _0\}$ is a thick set. Now according to theorem 2.1, $(1)$ holds.

(1)$\Longrightarrow$(2). Assume that $(1)$ holds for some constants $C,\,\lambda >0$, then

\[ \|{\rm e}^{{-}tA}\|_{L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})}\leq C{\rm e}^{-\lambda t}, \quad \forall t\geq 0, \]

where $A=\partial _x^{3}+a(x)$ with domain $D(A)=H^{3}(\mathbb {R})$. According to Gearhart–Pruss– Huang criteria [11], we have

\[ \sup_{\tau\in \mathbb{R}}\|(A+{{\rm i}}\tau)^{{-}1}\|_{L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})}<\infty. \]

In particular, letting $\tau =0$, this implies $\|A^{-1}\|_{L^{2}(\mathbb {R})\to L^{2}(\mathbb {R})}<\infty$. In other words, there exists a constant $c>0$ so that

(2.9)\begin{equation} c\|f\|_{L^{2}(\mathbb{R})}\leq \|(\partial_x^{3}+a(x))f\|_{L^{2}(\mathbb{R})}, \quad \forall f\in L^{2}(\mathbb{R}). \end{equation}

Now we are going to test (2.9) with a sequence of functions

\[ f_\varepsilon (x) = \varepsilon^{{1}/{4}}{\rm e}^{-\varepsilon x^{2}}, \quad \varepsilon>0, x\in \mathbb{R}. \]

Fix $\varepsilon >0$. Clearly,

(2.10)\begin{equation} \|f_\varepsilon\|_{L^{2}(\mathbb{R})}\geq c_0 \end{equation}

for some constant $c_0>0$ independent of $\varepsilon$. Moreover, we compute

\[ \partial_x^{3}({\rm e}^{\varphi}) = {\rm e}^{\varphi}\Big((\partial_x\varphi)^{3}+3\partial_x\varphi\partial_x^{2}\varphi+\partial_x^{3}\varphi\Big), \]

which implies that

\[ \partial_x^{3}f_\varepsilon=\varepsilon^{{1}/{4}}{\rm e}^{-\varepsilon x^{2}}({-}8\varepsilon^{3}x^{3}+12\varepsilon^{2}x). \]

This gives that for some constant $c_1>0$ independent of $\varepsilon$

(2.11)\begin{equation} \|\partial_x^{3}f_\varepsilon\|_{L^{2}(\mathbb{R})}\leq c_1\varepsilon^{{3}/{2}}. \end{equation}

It follows from (2.9)–(2.11) that

(2.12)\begin{equation} cc_0\leq c_1\varepsilon^{{3}/{2}}+\|af_\varepsilon\|_{L^{2}(\mathbb{R})}. \end{equation}

Choosing $\varepsilon =\varepsilon _0$ such that $c_1\varepsilon _0^{{3}/{2}}={cc_0}/{2}$ in (2.12), we find

(2.13)\begin{equation} \frac{cc_0}{2}\leq \|af_{\varepsilon_0}\|_{L^{2}(\mathbb{R})}\leq \|af_{\varepsilon_0}\|_{L^{2}(|x|\leq L)}+\|af_{\varepsilon_0}\|_{L^{2}(|x|>L)}, \end{equation}

for every $L>0$. Note that for some $c_2>0$ we have

(2.14)\begin{equation} \|af_{\varepsilon_0}\|_{L^{2}(|x|>L)}\leq \varepsilon_0^{{1}/{4}}\|a\|_{L^{\infty}(\mathbb{R})}\left(\int_{|x|\geq L}{\rm e}^{{-}2\varepsilon_0 x^{2}}{{\rm d}}x \right)^{{1}/{2}} \leq c_2\|a\|_{L^{\infty}(\mathbb{R})}{\rm e}^{-({\varepsilon_0}/{2})L^{2}}. \end{equation}

Choosing $L=L_0$ so that $c_2\|a\|_{L^{\infty }(\mathbb {R})}{\rm e}^{-({\varepsilon _0}/{2})L^{2}}\leq {cc_0}/{4}$, we deduce from (2.13)–(2.14) that

(2.15)\begin{equation} \frac{cc_0}{4}\leq \|af_{\varepsilon_0}\|_{L^{2}(|x|\leq L_0)}. \end{equation}

Squaring both sides of (2.15) and using Cauchy–Schwarz inequality, we infer that

(2.16)\begin{equation} \left(\frac{cc_0}{4}\right)^{2}\leq \|a\|_{L^{\infty}(\mathbb{R})} \int_{|x|\leq L_0}a(x)f^{2}_{\varepsilon_0}(x) {{\rm d}}x \leq \sqrt{\varepsilon_0}\|a\|_{L^{\infty}(\mathbb{R})} \int_{|x|\leq L_0}a(x) {{\rm d}}x. \end{equation}

It follows from (2.16) that

(2.17)\begin{equation} \int_{|x|\leq L_0}a(x) {{\rm d}}x\geq \gamma \end{equation}

with $\gamma = \left (\frac {cc_0}{4}\right )^{2}\frac {1}{\sqrt {\varepsilon _0} \|a\|_{L^{\infty }(\mathbb {R})}}>0$.

Now for every $x_0\in \mathbb {R}$, if we testing (2.9) with

\[ f_\varepsilon (x) = \varepsilon^{{1}/{4}}{\rm e}^{-\varepsilon (x-x_0)^{2}}, \]

then, similar to (2.17), we have

(2.18)\begin{equation} \int_{|x-x_0|\leq L_0}a(x) {{\rm d}}x\geq \gamma. \end{equation}

This shows that $(2)$ holds. Thus the proof is complete.

3. Exponential decay with small data

First we recall some estimates in Bourgain spaces. Let $s,\,b\in \mathbb {R}$, the Bourgain spaces $X^{s,b}$ are defined by the norm

\[ \|u\|_{X^{s,b}}:=\left( \int_{\mathbb{R}^{2}} (1+|\xi|)^{2s}(1+|\tau-\xi^{3}|)^{2b}|\widehat{u}(\xi,\tau)|^{2} {{\rm d}} \xi {{\rm d}} \tau\right)^{1/2}<\infty, \]

where $\widehat {u}(\xi,\,\tau )$ is the space–time Fourier transform of $u$, given by

\[ \widehat{u}(\xi,\tau) = \int_{\mathbb{R}^{2}} {\rm e}^{-{{\rm i}}(x\xi+t\tau)}u(x,t) {{\rm d}}x {{\rm d}}t. \]

For an open interval $I$ on $\mathbb {R}$, the restriction in time Bourgain spaces $X^{s,b}_{I}$ are endowed with the norm

\[ \|u\|_{X_{I}^{s, b}} :=\inf _{v \in X^{s, b}}\bigg\{\|v\|_{X^{s, b}}, v({\cdot})=u({\cdot}) \text{ on } I\bigg\}. \]

Let $\{W(t)\}_{t\in \mathbb {R}}$ be the Airy group, given by

(3.1)\begin{equation} (W(t)u_0) (x)= {\rm e}^{{-}t\partial_x^{3}}u_0 = c\int_{\mathbb{R}} {\rm e}^{{{\rm i}}x\xi} {\rm e}^{{{\rm i}}t\xi^{3}}\widehat{u_0}(\xi) {{\rm d}} \xi, \end{equation}

where $c$ is an absolute constant.

Lemma 3.1 Assume that $I=(0,\,\delta )$ with $0<\delta \leq 1$ and $s\geq 0$.

1. (1) If $b>{1}/{2},$ then

   (3.2)\begin{align} & \|W(t)u_0\|_{X^{s,b}_{I}}\lesssim_b \|u_0\|_{H^{s}(\mathbb{R})}, \end{align}
   (3.3)\begin{align} & \left\|\int_0^{t} W(t-\tau)f({\cdot},\tau){{\rm d}} \tau\right\|_{X^{s,b}_{I}}\lesssim_b \|f\|_{X^{s,b-1}_{I}}. \end{align}

2. (2) If $-({1}/{2})< b\leq b'<1/2$, then

   (3.4)\begin{equation} \|u\|_{X^{s, b}_{(-\delta,\delta)}} \lesssim_{s,b,b'} \delta^{b'-b}\|u\|_{X^{s, b'}_{I}}. \end{equation}

3. (3) If ${1}/{2}< b\leq b'\leq {3}/{4}$, then

   (3.5)\begin{equation} \|\partial_x(uv)\|_{X^{s,b'-1}_{I}}\lesssim_{s,b}\|u\|_{X^{s,b}_{I}}\|v\|_{X^{s,b}_{I}}. \end{equation}

Proof. See [30].

The item $(1)$ of lemma 3.1 can be understood as some estimates of the Airy group $W(t)$ in Bourgain spaces. Now we give some similar estimates of ${\rm e}^{-tA}$, $A=\partial _x^{3}+a(x)$, based on lemma 3.1.

Lemma 3.2 Assume that $a\in L^{\infty }(\mathbb {R}),$ $I=(0,\,\delta )$ and $b\in ({1}/{2},\,1]$. Then for some small $\delta =\delta (b,\,\|a\|_{L^{\infty }})>0,$ we have

(3.6)\begin{align} & \|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{I}}\lesssim_b \|u_0\|_{L^{2}(\mathbb{R})}, \end{align}

(3.7)\begin{align} & \left\|\int_0^{t} {\rm e}^{-(t-\tau)A}\partial_x(uv)(\tau){{\rm d}} \tau\right\|_{X^{0,b}_{I}}\lesssim_b \|u\|_{X^{0,b}_{I}}\|v\|_{X^{0,b}_{I}}. \end{align}

Proof. By Duhamel formula we have

(3.8)\begin{equation} {\rm e}^{{-}At}u_0= W(t)u_0-\int_0^{t}W(t-s)(a{\rm e}^{{-}As}u_0){{\rm d}}s. \end{equation}

Taking $X^{0,b}_{I}$ norm on both sides of (3.8), using $(1)$ of lemma 3.1, we find

(3.9)\begin{align} \|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{I}}& \leq \| W(t)u_0\|_{X^{0,b}_{I}}+ \left\|\int_0^{t}W(t-\tau)(a{\rm e}^{-\tau A}u_0){{\rm d}} \tau\right\|_{X^{0,b}_{I}}\nonumber\\ & \leq C\|u_0\|_{L^{2}(\mathbb{R})}+C \|a{\rm e}^{{-}tA}u_0\|_{X^{0,b-1}_{I}}, \end{align}

where $C=C(b)>0$. Since $b\leq 1$, noting $X^{0,0}=L^{2}_{t,x}$, we have

(3.10)\begin{align} \|a{\rm e}^{{-}tA}u_0\|_{X^{0,b-1}_{I}}\leq \|a{\rm e}^{{-}tA}u_0\|_{X^{0,0}_{I}}\leq \|a\|_{L^{\infty}}\|{\rm e}^{{-}tA}u_0\|_{X^{0,0}_{I}}\leq C'\delta^{b}\|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{I}}, \end{align}

where in the last step we used $(2)$ of lemma 3.1, and $C'>0$ is a constant depending on $b$ and $\|a\|_{L^{\infty }}$, but independent of $\delta$. It follows from (3.9)–(3.10) that

(3.11)\begin{equation} \|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{I}} \leq C\|u_0\|_{L^{2}(\mathbb{R})}+CC'\delta^{b}\|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{I}}. \end{equation}

If we take $\delta$ small such that $CC'\delta ^{b}\leq {1}/{2}$, then the last term of (3.11) can be absorbed by the left hand side, we have

\[ \|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{I}} \leq 2C\|u_0\|_{L^{2}(\mathbb{R})}. \]

This proves (3.6) for such $\delta$.

To prove (3.7), we apply the identity

\begin{align*} \int_0^{t}{\rm e}^{-(t-\tau)A}f(\tau){{\rm d}} \tau & =\int_0^{t}W(t-\tau)f(\tau){{\rm d}} \tau\\& \quad -\int_0^{t}W(t-\tau)a(x)\left( \int_0^{\tau} {\rm e}^{-(\tau-\tau')A}f(\tau'){{\rm d}} \tau' \right){{\rm d}} \tau \end{align*}

with $f=\partial _x(uv)$, similar to the above argument, we obtain

\begin{align*} & \left\|\int_0^{t} {\rm e}^{-(t-\tau)A}\partial_x(uv)(\tau){{\rm d}} \tau\right\|_{X^{0,b}_{I}}\\ & \quad\leq C\|u\|_{X^{0,b}_{I}}\|v\|_{X^{0,b}_{I}}+C\|a\|_{L^{\infty}}\delta^{b}\left\|\int_0^{t} {\rm e}^{-(t-\tau)A}\partial_x(uv)(\tau){{\rm d}} \tau\right\|_{X^{0,b}_{I}}. \end{align*}

This proves (3.7) for $\delta$ small enough.

Lemma 3.2 holds for the interval $I$ with small length. Now we extend it to general intervals.

Lemma 3.3 Assume that (A1) holds and (A2) holds on a thick set $E$. Let $T>0,$ $I=(0,\,T)$ and $b\in ({1}/{2},\,1]$. Then there exists $C=C(b,\,\|a\|_{L^{\infty }})>0$ we have

(3.12)\begin{align} & \|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{I}}\leq C \|u_0\|_{L^{2}(\mathbb{R})}, \end{align}

(3.13)\begin{align} & \left\|\int_0^{t} {\rm e}^{-(t-\tau)A}\partial_x(uv)(\tau){{\rm d}} \tau\right\|_{X^{0,b}_{I}}\leq C(T+1) \|u\|_{X^{0,b}_{I}}\|v\|_{X^{0,b}_{I}}. \end{align}

Proof. In the case $T\leq \delta$, (3.12) and (3.12) follows from (3.6) and (3.7), respectively. So we assume $T>\delta$ now. The proof then mainly relies on the following inequality: for all $s\geq 0,\, b\in ({1}/{2},\,1],\,\delta >0$ and $t_0\in \mathbb {R}$,

(3.14)\begin{equation} \|u\|_{X^{s,b}_{(t_0,t_0+2\delta)}}\lesssim \|u\|_{X^{s,b}_{(t_0,t_0+\delta)}}+\|u\|_{X^{s,b}_{(t_0+\delta,t_0+2\delta)}}. \end{equation}

See [32, lemma 6.2] for a proof.

To prove (3.12), choose an integer $k\geq 1$ so that $k\delta \geq T$. Then $(0,\,T)\subset (0,\,k\delta )$, using (3.14) repeatedly, we find

(3.15)\begin{equation} \|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{(0,T)}}\lesssim \sum_{j=1}^{k}\|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{((j-1)\delta, j\delta)}}. \end{equation}

But by (3.6) again, we have

\begin{align*} \|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{((j-1)\delta, j\delta)}}& =\|{\rm e}^{{-}tA}{\rm e}^{-(j-1)\delta A}u_0\|_{X^{0,b}_{(0, \delta)}}\\& \lesssim_b\|{\rm e}^{-(j-1)\delta A}u_0\|_{L^{2}(\mathbb{R})}\lesssim {\rm e}^{-(j-1)\delta \lambda}\|u_0\|_{L^{2}(\mathbb{R})}, \end{align*}

where in the last step we used theorem 2.1. This, together with (3.15), gives

\[ \|{\rm e}^{{-}tA}u_0\|_{X^{0,b}_{(0,T)}}\lesssim \sum_{j=1}^{k} {\rm e}^{-(j-1)\delta \lambda}\|u_0\|_{L^{2}(\mathbb{R})} \leq C\|u_0\|_{L^{2}(\mathbb{R})} \]

with some $C>0$ depending on $\delta$ and $b$. This proves (3.12).

To show (3.13), we choose $k$ so that $(k-1)\delta < T\leq k\delta$. Then by (3.14) and (3.7),

\begin{align*} \left\|\int_0^{t} {\rm e}^{-(t-\tau)A}\partial_x(uv)(\tau){{\rm d}} \tau\right\|_{X^{0,b}_{(0,T)}}& \lesssim \sum_{j=1}^{k} \left\|\int_0^{t} {\rm e}^{-(t-\tau)A}\partial_x(uv)(\tau){{\rm d}} \tau\right\|_{X^{0,b}_{((j-1)\delta,j\delta)}}\\ & \lesssim_b \sum_{j=1}^{k}\|u\|_{X^{0,b}_{((j-1)\delta,j\delta)}}\|v\|_{X^{0,b}_{((j-1)\delta,j\delta)}}\\ & \lesssim k \|u\|_{X^{0,b}_{(0,T)}}\|v\|_{X^{0,b}_{(0,T)}}\\ & \leq \left(\frac{T}{\delta}+1\right)\|u\|_{X^{0,b}_{(0,T)}}\|v\|_{X^{0,b}_{(0,T)}}. \end{align*}

This proves (3.13).

Now we can prove theorem 1.2.

Proof of theorem 1.2. By Duhamel principle, we can rewrite the KdV equation (1.1) into an integral form

(3.16)\begin{equation} u(t)={\rm e}^{{-}tA}u_0-\int_0^{t}{\rm e}^{-(t-s)A}(uu_x)(s){{\rm d}}s. \end{equation}

Taking $L^{2}(\mathbb {R})$ norm on both sides of (3.16), using theorem 2.1 and the embedding $X^{0,b}_{(0,T)}\hookrightarrow L^{\infty }(0,\,T;L^{2}(\mathbb {R}))$ when $b>{1}/{2}$, we find for all $T>0$

(3.17)\begin{align} \|u(T)\|_{L^{2}(\mathbb{R})}& \leq C_0{\rm e}^{-\lambda T}\|u_0\|_{L^{2}(\mathbb{R})}+C\left\|\int_0^{t}{\rm e}^{-(t-s)A}(uu_x)(s){{\rm d}}s\right\|_{X^{0,b}_{(0,T)}}\nonumber\\ & \leq C_0{\rm e}^{-\lambda T}\|u_0\|_{L^{2}(\mathbb{R})}+C_1(T+1)\|u\|^{2}_{X^{0,b}_{(0,T)}} \end{align}

for some $C_0>{1}/{2}$, where in the last step we used (3.13).

Now fix a large $T>0$ so that

(3.18)\begin{equation} C_0{\rm e}^{-\lambda T}= \frac{1}{2}{\rm e}^{-({\lambda}/{2}) T}. \end{equation}

Then (3.17) becomes

(3.19)\begin{equation} \|u(T)\|_{L^{2}(\mathbb{R})} \leq \frac{1}{2}{\rm e}^{-({\lambda}/{2}) T}\|u_0\|_{L^{2}(\mathbb{R})}+C_1(T+1)\|u\|^{2}_{X^{0,b}_{(0,T)}}. \end{equation}

Moreover, taking $X^{0,b}_{(0,T)}$ norm on both sides of (3.16), using lemma 3.3, we obtain

(3.20)\begin{equation} \|u\|_{X^{0,b}_{(0,T)}} \leq C_2\|u_0\|_{L^{2}(\mathbb{R})}+C_3(T+1)\|u\|^{2}_{X^{0,b}_{(0,T)}}. \end{equation}

Now consider the map $\Gamma$

(3.21)\begin{equation} \Gamma u ={\rm e}^{{-}tA}u_0-\int_0^{t}{\rm e}^{-(t-s)A}(uu_x)(s){{\rm d}}s \end{equation}

on the ball

\[ \mathcal {B}=\left\{ u: \|u\|_{X^{0,b}_{(0,T)}}\leq 2C_2\|u_0\|_{L^{2}(\mathbb{R})} \right\}. \]

The estimate (3.20) shows that if $u\in \mathcal {B}$ then

(3.22)\begin{equation} \|\Gamma u\|_{X^{0,b}_{(0,T)}} \leq C_2\|u_0\|_{L^{2}(\mathbb{R})}+4C_2^{2}C_3(T+1)\|u_0\|^{2}_{L^{2}(\mathbb{R})}. \end{equation}

Moreover, if $u,\,v\in \mathcal {B}$, then

(3.23)\begin{equation} \|\Gamma u-\Gamma v\|_{X^{0,b}_{(0,T)}} \leq 4C_2C_3(T+1)\|u_0\|_{L^{2}(\mathbb{R})}\|u-v\|_{X^{0,b}_{(0,T)}}. \end{equation}

Thanks to (3.22)–(3.23), if $\|u_0\|_{L^{2}(\mathbb {R})}$ is small enough, say,

(3.24)\begin{equation} \|u_0\|_{L^{2}(\mathbb{R})}\leq \frac{1}{8C_2C_3(T+1)}:=\delta_1, \end{equation}

then $\Gamma \mathcal {B} \subset \mathcal {B}$ and $\|\Gamma u-\Gamma v\|_{X^{0,b}_{(0,T)}}\leq ({1}/{2})\|u-v\|_{X^{0,b}_{(0,T)}}$, thus $\Gamma$ is a contraction mapping on $\mathcal {B}$. So equation (3.16) has a unique solution $u\subset \mathcal {B}$. This, together with the bound (3.19), gives that

(3.25)\begin{equation} \|u(T)\|_{L^{2}(\mathbb{R})} \leq \frac{1}{2}{\rm e}^{-({\lambda}/{2}) T}\|u_0\|_{L^{2}(\mathbb{R})}+4C_2^{2}C_1(T+1)\|u_0\|^{2}_{L^{2}(\mathbb{R})}. \end{equation}

Assume further that

(3.26)\begin{equation} \|u_0\|_{L^{2}(\mathbb{R})}\leq \frac{1}{8C_2^{2}C_1(T+1)}{\rm e}^{-({\lambda}/{2}) T}:=\delta_2, \end{equation}

so that $4C_2^{2}C_1(T+1)\|u_0\|^{2}_{L^{2}(\mathbb {R})}\leq ({1}/{2}){\rm e}^{-({\lambda }/{2}) T}\|u_0\|_{L^{2}(\mathbb {R})}$, then (3.25) becomes

\[ \|u(T)\|_{L^{2}(\mathbb{R})} \leq {\rm e}^{-({\lambda}/{2}) T}\|u_0\|_{L^{2}(\mathbb{R})}. \]

By induction, we find that for all $n\geq 1$

\[ \|u(nT)\|_{L^{2}(\mathbb{R})} \leq {\rm e}^{-({\lambda}/{2}) nT}\|u_0\|_{L^{2}(\mathbb{R})} \]

if $\|u_0\|_{L^{2}(\mathbb {R})}\leq \delta = \min \{\delta _1,\,\delta _2\}$. Then by the semigroup property, we infer that

\[ \|u(t)\|_{L^{2}(\mathbb{R})} \leq C'{\rm e}^{-\lambda't}\|u_0\|_{L^{2}(\mathbb{R})}, \quad \forall t\geq 0 \]

for some constants $C',\,\lambda '>0$. This completes the proof of theorem 1.2.

4. Exponential decay for general data

4.1 A unique continuation

This subsection is devoted to the following unique continuation property (UCP) of the KdV equation, which is a key step to establish the exponential decay for general data.

Proposition 4.1 Let $T>0$ and $E$ be a set satisfying NCC. Assume that $u\in X^{0,({1}/{2})+}_{(0,T)}$ is a solution of the KdV equation

\[ \partial_tu+\partial_x^{3}u+u\partial_xu=0, \quad (x,t)\in\mathbb{R}\times (0,T) \]

and $u(x,\,t)=0$ on $E\times (0,\,T)$. Then

\[ u(x,t)=0, \quad \mbox{ for } x\in \mathbb{R}, \; t\in (0,T). \]

The proof of this result relies on the following unique continuation property, stated explicitly in [33], that follows from the results in [25].

Lemma 4.2 Let $T>0$. Assume that $u\in L^{\infty }(0,\,T;H^{3}(\mathbb {R}))$ is a solution of the KdV equation

\[ \partial_tu+\partial_x^{3}u+u\partial_xu=0, \quad (x,t)\in\mathbb{R}\times (0,T) \]

and $u=0$ on an open subset of $\mathbb {R}\times (0,\,T)$, then

\[ u(x,t)=0, \quad \text{ for } x\in \mathbb{R}, \ t\in (0,T). \]

Note that proposition 4.1 does not follow from lemma 4.2 directly, since the solution, $u\in X^{0,({1}/{2})+}_{(0,T)}$, has not higher enough regularity. To overcome this difficulty, we need to first establish the propagation of regularity for the operator $\partial _t+\partial _x^{3}$ on $\mathbb {R}$.

Lemma 4.3 Let $T>0,$ $r\geq 0$ and $f\in X^{r,-({1}/{2})}_{(0,T)}$. Let $u\in X^{r,{1}/{2}}_{(0,T)}$ be a solution of

\[ \partial_tu +\partial_x^{3}u=f. \]

If $E\subset \mathbb {R}$ satisfies NCC and $u\in L^{2}_{loc}(0,\,T;H^{r+{1}/{2}}(E)),$ then $u\in L^{2}_{loc}(0,\,T; H^{r+{1}/{2}}(\mathbb {R}))$.

Roughly speaking, lemma 4.3 means that the solution of the linear KdV equation has higher regularity on a set satisfying NCC, then the solution automatically has higher regularity on the whole space $\mathbb {R}$. This result is inspired by the work of [14], in which the propagation of regularity for the linear KdV equation on torus $\mathbb {T}$ is proved, when $E$ is an open subset of $\mathbb {T}$. In [14], the proof relies on a partition of unity

\[ 1=\sum_j \chi(x-x_j) \]

where $\chi$ is a smooth cutoff function supported on the open set $E$, and the sum is taken over for finite terms. However, the set $E$, satisfying NCC, may have complicated structure, so it is not clear whether the corresponding partition of unity exists or not. Even though, we shall use the following lemma instead, which is sufficient for our purpose.

Lemma 4.4 Assume that $E$ satisfy NCC. Then there exist constants $L_0>0,\,m_0\in \mathbb {N}$ and a smooth function $\chi \in C^{\infty }(\mathbb {R})$ such that the support $supp\, \chi \subset E,$ $|\partial _x^{k}\chi |\leq C_k$ for all $k\in \mathbb {N},$ and

(4.1)\begin{equation} \sum_{\ell={-}m_0}^{m_0}\chi^{2}(x+\ell L_0)\geq 1, \qquad \forall x\in \mathbb{R}. \end{equation}

Proof. By definition, for some constants $r,\,L>0$ we have

\[ E \supset \bigcup_{n}(x_n-r,x_n+r), \qquad \inf_{n\in \mathbb{Z}}|x-x_n|\leq L, \text{ for all } x\in \mathbb{R}. \]

Without loss of generality, we assume that $L> 2r$. Clearly, there exists $n_0\in \mathbb {Z}$ so that $|x_{n_0}|\leq L$. Then

\[ (x_{n_0}-r,x_{n_0}+r)\subset \left(-\frac{3L}{2},\frac{3L}{2}\right). \]

We set

\[ I_0:=(x_{n_0}-r,x_{n_0}+r). \]

Similarly, for every $j\in \mathbb {Z}$, there exists an interval

(4.2)\begin{equation} I_j:=(x_{n_j}-r,x_{n_j}+r)\subset \left(6jL-\frac{3L}{2},6jL+\frac{3L}{2}\right). \end{equation}

Clearly, we have

(4.3)\begin{align} & E\supset \bigcup_{j\in \mathbb{Z}}I_j, \end{align}

(4.4)\begin{align} & {\rm dist}(I_j,I_{j+1})\geq 3L, \text{ for all } j\in \mathbb{Z}. \end{align}

Now let $\chi _0\in C_c^{\infty }(I_0)$ (the set of smooth functions with compact support) so that

(4.5)\begin{equation} \chi_0(x)= \left\{\begin{array}{@{}ll} 1, & \displaystyle\text{ if }x\in \Big[x_{n_0}-\dfrac{r}{2}, x_{n_0}+\dfrac{r}{2}\Big]\\ 0, & \text{ if }|x-x_{n_0}|\geq r. \end{array} \right. \end{equation}

Moreover, we define for every $j\in \mathbb {Z}$

(4.6)\begin{equation} \chi_j(x)=\chi_0(x-x_{n_j}), \quad x\in \mathbb{R} \end{equation}

and

(4.7)\begin{equation} \chi(x)=\sum_{j\in \mathbb{Z}}\chi_j(x), \quad x\in \mathbb{R}. \end{equation}

Now we show that $\chi$ enjoys the desired property. First, ${\rm supp} \, \chi \subset E$ follows from (4.3) and (4.5)–(4.7). Moreover, for every $x\in \mathbb {R}$, by (4.2), all terms in the sum of (4.7) vanish except at most one term, so

\[ |\partial^{k}_x\chi(x)|\leq |\partial^{k}_x\chi_0(x)|\leq C_k, \quad \forall k\in \mathbb{N}, \]

and at the same time we have

(4.8)\begin{equation} \chi^{2}(x)=\sum_{j\in \mathbb{Z}}\chi^{2}_j(x), \quad x\in \mathbb{R}. \end{equation}

Since $\chi _0=1$ on a subinterval of $[-6L,\,6L]$ with length $r$, then for some $\mathbb {N}\ni m_0\geq 6L/r$ we have

(4.9)\begin{equation} \sum_{\ell={-}m_0}^{m_0}\chi^{2}_0(x+\ell r)\geq 1, \qquad \mbox{ for all } x\in [{-}6L,6L]. \end{equation}

Similarly,

(4.10)\begin{equation} \sum_{\ell={-}m_0}^{m_0}\chi^{2}_j(x+\ell r)\geq 1, \qquad \mbox{ for all } x\in [6(j-1)L,6(j+1)L]. \end{equation}

Let $L_0=r$. It follows from (4.8)–(4.10) that

\[ \sum_{\ell={-}m_0}^{m_0}\chi^{2}(x+\ell L_0)\geq 1, \qquad \forall x\in \mathbb{R}. \]

This completes the proof.

Proof of lemma 4.3. Fix $T>0$. We assume, without loss of generality, that $0\leq r<1$. Otherwise, we consider the equation

\[ \partial_t(\partial^{k}_xu) +\partial_x^{3}(\partial_x^{k}u)=\partial_x^{k}f \]

instead, where $k$ is a positive integer. We divide the proof into three steps.

Step 1. Let $\phi \in C_c^{\infty }(0,\,T)$ and $\varphi \in C^{\infty }(\mathbb {R})$ so that $|\partial _x^{k}\varphi |\leq C_k,\, \forall k\in \mathbb {N}$. We claim that

(4.11)\begin{equation} \Big| (\phi(t)J^{2r-1}(\partial_x\varphi)\partial_x^{2}u,u) \Big|\leq C. \end{equation}

Here and in the rest of the proof, $J^{s}$ is defined by the Fourier transform as $\widehat {J^{s}f}=(1+|\xi |)^{s}\widehat {f}(\xi )$, $(\cdot,\,\cdot )$ denotes the inner product in $L^{2}(0,\,T;L^{2}(\mathbb {R}))$.

In fact, let $\mathbb {L}=\partial _t+\partial _x^{3}$ and $\mathbb {A}=\phi (t)J^{2r-1}\varphi$. Then using Parseval's identity, we have

(4.12)\begin{equation} (\mathbb{L}u,\mathbb{A}^{*}u)+(\mathbb{A}u,\mathbb{L}u)=([\mathbb{A},\partial_x^{3}]u,u)-(\phi'(t)J^{2r-1}\varphi u, u), \end{equation}

where $\mathbb {A}^{*}=\varphi (x)J^{2r-1}\phi (t)$ is the dual operator of $\mathbb {A}$, the commutator $[A,\,B]=AB-BA$ as usual. Since $\mathbb {L}u=f\in X^{r,-({1}/{2})}_{(0,T)}$, we infer that

\begin{align*} |(\mathbb{L}u,\mathbb{A}^{*}u)+(\mathbb{A}u,\mathbb{L}u) |& \leq \|f\|_{X^{r,-({1}/{2})}_{(0,T)}}(\|\mathbb{A}u\|_{X^{{-}r,({1}/{2})}_{(0,T)}}+\|\mathbb{A}^{*}u\|_{X^{0,{1}/{2}}_{(0,T)}}) \\ (\mbox{by lemma A.5}) & \leq C\|f\|_{X^{r,-({1}/{2})}_{(0,T)}}\|u\|_{X^{r,{1}/{2}}_{(0,T)}}\leq C. \end{align*}

By (A.1) in the appendix, we also have $|(\phi '(t)J^{2r-1}\varphi u,\, u)|\leq C$. So by (4.12),

(4.13)\begin{equation} |([\mathbb{A},\partial_x^{3}]u,u)|\leq C. \end{equation}

A direct computation gives that

\[ [\mathbb{A},\partial_x^{3}]={-}3\phi(t)J^{2r-1}(\partial_x\varphi)\partial_x^{2}-3\phi(t)J^{2r-1}(\partial_x^{2}\varphi)\partial_x -\phi(t)J^{2r-1}\partial_x^{3}\varphi. \]

Similar to (A.1), we have

\[ \left|\Big(3\phi(t)J^{2r-1}(\partial_x^{2}\varphi)\partial_xu +\phi(t)J^{2r-1}\partial_x^{3}\varphi)u,u\Big)\right|\leq C\|u\|^{2}_{L_{loc}^{2}(0,T;H^{r}(\mathbb{R}))}\leq C. \]

Thus, the claim (4.11) follows from (4.13).

Step 2. For any $\phi \in C_c^{\infty }(0,\,T)$, $\chi \in C^{\infty }(\mathbb {R})$ with support ${\rm supp}\, \chi \subset E$ and $|\partial _x^{k} \chi |\leq C_k$, we claim that

(4.14)\begin{equation} \Big| (\phi(t)J^{2r-1}\chi^{2}\partial_x^{2}u,u) \Big|\leq C. \end{equation}

In fact, we rewrite

\[ (\phi(t)J^{2r-1}\chi^{2}\partial_x^{2}u,u)=I_1+I_2 \]

where

\begin{align*} I_1& = (\phi(t)J^{r-({3}/{2})}\chi\partial_x^{2}u,J^{r+{1}/{2}}\chi u),\\ I_2& =(\phi(t)J^{r-({3}/{2})}\chi\partial_x^{2}u,[\chi,J^{r+{1}/{2}}]u) +(\phi(t)[J^{r-({3}/{2})},\chi]\chi\partial_x^{2}u, J^{r+{1}/{2}}u). \end{align*}

From assumption $u\in L_{loc}^{2}(0,\,T;H^{r+{1}/{2}}(E))$, we infer $\chi \partial _x^{2}u\in L^{2}_{loc}(0,\,T; H^{r-({3}/{2})}(\mathbb {R}))$, thus

\[ |I_1|\leq C\|\phi(t)J^{r-({3}/{2})}\chi\partial_x^{2}u\|_{L^{2}(0,T;L^{2}(\mathbb{R}))}\|u\|_{L_{loc}^{2}(0,T;H^{r+{1}/{2}}(E))}\leq C. \]

Moreover, using the fact $u\in L^{2}(0,\,T;H^{r}(\mathbb {R}))$ and (A.2)–(A.3), one can show that $|I_2|\leq C$. This proves the claim (4.14).

Step 3. Complete the proof. Let $\chi \in C^{\infty }$ be the cutoff function constructed in lemma 4.4. For every $x_0\in \mathbb {R}$, define a function $\varphi$ by Fourier transform

\[ \widehat{\varphi}(\xi)=\frac{1-{\rm e}^{{{\rm i}}x_0\xi}}{{{\rm i}} \xi}\widehat{\chi^{2}}(\xi). \]

By the Fourier inversion, this implies that

(4.15)\begin{equation} \partial_x\varphi(x) = \chi^{2}(x)-\chi^{2}(x+x_0), \quad x\in \mathbb{R}. \end{equation}

We apply (4.11) with $\partial _x\varphi$ given by (4.15), and use (4.14) to find that

(4.16)\begin{equation} \Big| (\phi(t)J^{2r-1}\chi^{2}({\cdot}{+}x_0)\partial_x^{2}u,u) \Big|\leq C. \end{equation}

Since $0\leq r<1$ we infer

\[ |(\phi(t)J^{2r-1}\chi^{2}({\cdot}{+}x_0)u,u)|\leq C. \]

Noting $J^{2}=1-\partial _x^{2}$ we get from (4.16) that

(4.17)\begin{equation} \Big| (\phi(t)J^{2r-1}\chi^{2}({\cdot}{+}x_0)J^{2}u,u) \Big|\leq C. \end{equation}

We rewrite

\begin{align*} & (\phi(t)J^{2r-1}\chi^{2}({\cdot}{+}x_0)J^{2}u,u)\\ & \quad= \left(\phi(t)J^{r+{1}/{2}}u,\chi^{2}({\cdot}{+}x_0)J^{r+{1}/{2}}u\right) +\left(\phi(t)J^{r+{1}/{2}}u,[J^{r-({3}/{2})},\chi^{2}({\cdot}{+}x_0)]J^{2}u\right) \end{align*}

and use the bound $|(\phi (t)J^{r+{1}/{2}}u,\,[J^{r-({3}/{2})},\,\chi ^{2}(\cdot +x_0)]J^{2}u)|\leq C\|u\|^{2}_{L^{2}_{loc}(0,\,T;H^{r}(\mathbb {R}))}\leq C$ (follows from (A.4)), we deduce from (4.17) that

(4.18)\begin{equation} \int_0^{T}\int_\mathbb{R} \phi(t)|\chi(x+x_0)J^{r+{1}/{2}}u|^{2}{{\rm d}}x {{\rm d}}t \leq C. \end{equation}

Applying (4.18) with $x_0=\{lL_0\}_{l=-m_0}^{m_0}$, noting (4.1), we conclude that

\[ \int_0^{T}\phi(t)\int_\mathbb{R} |J^{r+{1}/{2}}u|^{2}{{\rm d}}x {{\rm d}}t\leq \sum_{l={-}m_0}^{m_0}\int_0^{T}\int_\mathbb{R} \phi(t) |\chi(x+lL_0)J^{r+{1}/{2}}u|^{2}{{\rm d}}x {{\rm d}}t \leq C. \]

This shows that $u\in L^{2}_{loc}(0,\,T;H^{r+{1}/{2}}(\mathbb {R}))$, and completes the proof.

Remark 4.5 Very recently, Panthee and Vielma Leal have established in [19] the propagation of regularity in Bourgain's spaces for the Benjamin equation on a periodic domain.

Corollary 4.6 Let $T>0$ and $u\in X^{0,({1}/{2})+}_{(0,T)}$ be a solution of the KdV equation

(4.19)\begin{equation} \partial_tu+\partial_x^{3}u+u\partial_xu=0, \quad (x,t)\in \mathbb{R}\times(0,T). \end{equation}

If $E$ satisfies NCC and $u(x,\,t)=0$ for $(x,\,t)\in E\times (0,\,T),$ then $u\in L^{2}_{loc}(0,\,T;H^{\infty }(\mathbb {R}))$.

Remark 4.7 Here and below $X^{0,({1}/{2})+}=X^{0,({1}/{2})+\varepsilon }$ with an arbitrarily small ${\varepsilon >0}$.

Proof. Rewrite the KdV equation as

\[ \partial_tu+\partial_x^{3}u=f \]

with $f=-u\partial _xu$. Since $u\in X^{0,({1}/{2})+}_{(0,T)}$, by the bilinear estimate (3.5),

\[ \|f\|_{X^{0,-({1}/{2})+}_{(0,T)}}\leq C\|u\|^{2}_{X^{0,({1}/{2})+}_{(0,T)}}\leq C. \]

Since $u(x,\,t)=0$ for $(x,\,t)\in E\times (0,\,T)$, we have of course $u\in L^{2}_{loc}(0,\,T;H^{\infty }(E))$. By lemma 4.3, we obtain

\[ u\in L^{2}_{loc}(0,T;H^{{1}/{2}}(\mathbb{R})). \]

Let $t_0\in (0,\,T)$ so that $u(t_0)\in H^{{1}/{2}}(\mathbb {R})$. Then we find the solution $u$ of (4.19) satisfies

\[ u\in X^{{1}/{2},({1}/{2})+}_{(0,T)}. \]

Similarly, using lemma 4.3 repeatedly, we conclude that $u\in L^{2}_{loc}(0,\,T;H^{\infty }(\mathbb {R}))$.

Proof of proposition 4.1. According to corollary 4.6, we know that $u\in L^{2}_{loc}(0,\,T; H^{\infty }(\mathbb {R}))$. Since $u(x,\,t)=0$ for $(x,\,t)\in E\times (0,\,T)$ and $E$ contains an open set in $\mathbb {R}$, so $u=0$ on an open set in $\mathbb {R}\times (0,\,T)$, then by the UCP in lemma 4.2, we conclude that $u \equiv 0$.

4.2 Proof of theorem 1.4

Let $W(t)={\rm e}^{-t\partial _x^{3}}$ be the Airy group. Then we have the sharp Kato smoothing effect [12, theorem 4.1]

(4.20)\begin{equation} \|\partial_xW(t)u_0\|_{L_x^{\infty} L_t^{2}(\mathbb{R}^{2})}\lesssim \|u_0\|_{L^{2}(\mathbb{R})}. \end{equation}

The estimate (4.20) can be reformulated in Bourgain space as (see [13, p. 5])

(4.21)\begin{equation} \|\partial_xu\|_{L_x^{\infty} L_t^{2}(\mathbb{R}^{2})}\lesssim \|u\|_{X^{0,({1}/{2})+}}. \end{equation}

The bound (4.21) will be used to derive the compactness of some sequences later.

Proposition 4.8 Assume that $0\leq a(x)\in L^{\infty }(\mathbb {R})$. Then the IVP (1.1) has a unique global solution $u\in C([0,\,\infty );L^{2}(\mathbb {R}))$. Moreover, for every $T>0$

\[ \|u\|_{X^{0,({1}/{2})+}_{(0,T)}}\leq C(T,\|a\|_{L^{\infty}(\mathbb{R})},\|u_0\|_{L^{2}(\mathbb{R})})<\infty. \]

Proof. Thanks to lemma 3.1, using the contraction principle one can show that there exists a unique solution $u\in X^{0,({1}/{2})+}_{(0,\delta )}$ of (1.1) with the bound

\[ \|u\|_{X^{0,({1}/{2})+}_{(0,\delta)}}\leq 2\|u_0\|_{L^{2}(\mathbb{R})}, \]

where the life span

\[ \delta=\delta(\|a\|_{L^{\infty}(\mathbb{R})},\|u_0\|_{L^{2}(\mathbb{R})})>0 \]

is small enough. Multiplying (1.1) by $u$ and integrating over $\mathbb {R}$ we obtain

\[ \frac{1}{2}\frac{{{\rm d}}}{{{\rm d}}t}\int_\mathbb{R} |u(x,t)|^{2}{{\rm d}}x +\int_{\mathbb{R}}a(x)|u(x,t)|^{2}{{\rm d}}x=0, \]

which, together with the fact $a(x)\geq 0$, implies that

(4.22)\begin{equation} \|u(t,\cdot)\|_{L^{2}(\mathbb{R})}\leq \|u_0\|_{L^{2}(\mathbb{R})}, \quad \forall t\geq 0. \end{equation}

Thus we can take $u(\delta )$ as a new data, to find a solution on $(\delta,\,2\delta )$ so that $\|u\|_{X^{0,({1}/{2})+}_{(\delta,2\delta )}}\leq 2\|u_0\|_{L^{2}(\mathbb {R})}$. Repeat this process, we find that for every $T>0$

\[ \|u\|_{X^{0,({1}/{2})+}_{(0,T)}}\leq C(T,\|a\|_{L^{\infty}(\mathbb{R})},\|u_0\|_{L^{2}(\mathbb{R})}). \]

This completes the proof.

Corollary 4.9 Assume that $0\leq a(x)\in L^{\infty }(\mathbb {R})$. Let $u$ be the solution of (1.1) obtained in proposition 4.8. Then for every $T>0$ and for every measurable set $\Omega \subset \mathbb {R}$

(4.23)\begin{equation} \int_0^{T}\int_\Omega |\partial_xu(x,t)|^{2}{{\rm d}}x{{\rm d}}t \leq |\Omega|C(T,\|a\|_{L^{\infty}(\mathbb{R})},\|u_0\|_{L^{2}(\mathbb{R})}), \end{equation}

where $|\Omega |$ denotes the Lebesgue measure of $\Omega$.

Proof. Fix $T>0$. Combining (4.21) and proposition 4.8 we obtain

\[ \|\partial_xu\|_{L_x^{\infty} L_t^{2}(\mathbb{R}^{2})}\lesssim C(T,\|a\|_{L^{\infty}(\mathbb{R})},\|u_0\|_{L^{2}(\mathbb{R})}). \]

From this, we use Hölder inequality to find

\[ \|\partial_xu\|_{L_x^{2}(\Omega) L_t^{2}(0,T)}\leq |\Omega|^{{1}/{2}}\|\partial_xu\|_{L_x^{\infty}(\Omega) L_t^{2}(\mathbb{R})}\leq C. \]

Then we conclude (4.23) by Fubini theorem.

Now we prove an observability inequality, which means that we can recover the solution of the KdV equation if we observe the solution on $E\times (0,\,T)$ when $E$ satisfies NCC and $E^{c}$ has a finite Lebesgue measure.

Lemma 4.10 Let $R,\,T>0$ and let $E$ satisfy NCC, $|E^{c}|<\infty$. Assume that (A1) and (A2) hold. Then there exist a constant $C=C(R,\,T,\,a)>0$ such that for all $\|u_0\|_{L^{2}(\mathbb {R})}\leq R,$ the solution $u$ of the IVP (1.1) satisfies

(4.24)\begin{equation} \int_0^{T}\int_\mathbb{R} u^{2}(x,t){{\rm d}}x{{\rm d}}t \leq C \int_0^{T}\int_{\mathbb{R}}a(x)u^{2}(x,t){{\rm d}}x{{\rm d}}t. \end{equation}

Proof. Following [4], we argue by contradiction. Assume that there exists a sequence solutions $u_k$ of the KdV equation (1.1), with initial data $\|u_{0k}\|_{L^{2}(\mathbb {R})}\leq R$, such that

(4.25)\begin{equation} \lim_{k\rightarrow \infty} \frac{\int_0^{T}\int_\mathbb{R} a(x)u_k^{2}(x,t){{\rm d}}x{{\rm d}}t}{\int_0^{T}\int_\mathbb{R} u_k^{2}(x,t){{\rm d}}x{{\rm d}}t}=0. \end{equation}

Define

(4.26)\begin{equation} \alpha_k = \|u_k\|_{L^{2}(\mathbb{R}\times (0,T))}, \quad v_k(x,t) = \frac{u_k(x,t)}{\alpha_k}. \end{equation}

Then

(4.27)\begin{equation} \|v_k\|_{L^{2}( \mathbb{R}\times(0,T))} = 1, \quad k\in \mathbb{N} \end{equation}

and $v_k$ is a solution of

(4.28)\begin{equation} (v_k)_{t}+(v_k)_{xxx}+\alpha_kv_k(v_k)_{x}+a(x)v_k=0, \quad (x,t)\in \mathbb{R}\times \mathbb{R}^{+} \end{equation}

with initial data

\[ v_k(x,0) = u_{0k}(x)/\alpha_k. \]

It follows from (4.25)–(4.26) that

(4.29)\begin{equation} \lim_{k\rightarrow \infty}\int_0^{T}\int_\mathbb{R} a(x)v_k^{2}(x,t){{\rm d}}x{{\rm d}}t = 0. \end{equation}

This, since $a(x)\geq a_0>0$ for all $x\in E$, gives that

(4.30)\begin{equation} \lim_{k\rightarrow \infty}\int_0^{T}\int_{E}v_k^{2}(x,t){{\rm d}}x{{\rm d}}t = 0. \end{equation}

Moreover, multiplying (4.28) with $v_k$ and integrating over $\mathbb {R}\times (0,\,t)$ and changing the order of integration, we obtain

(4.31)\begin{equation} \int_\mathbb{R} v_k^{2}(x,t){{\rm d}}x + 2\int_0^{t}\int_\mathbb{R} a(x)v_k^{2}(x,\tau) {{\rm d}}x {{\rm d}} \tau = \int_\mathbb{R} v_{0k}^{2}{{\rm d}}x. \end{equation}

Integrating (4.31) with respect to $t$ over $[0,\,T]$, we obtain

\[ \int_\mathbb{R} v_{0k}^{2}{{\rm d}}x\leq \frac{1}{T}\int_0^{T}\int_\mathbb{R} v_k^{2}(x,t){{\rm d}}x {{\rm d}}t + 2\int_0^{T}\int_0^{t}\int_\mathbb{R} a(x)v_k^{2}{{\rm d}}x {{\rm d}}t. \]

This, together with (4.27) and (4.29), shows that

(4.32)\begin{equation} \int_\mathbb{R} v_{0k}^{2}(x){{\rm d}}x \leq C(T,\|a\|_{L^{\infty}}, R). \end{equation}

Furthermore, it follows from the bound (4.22) that

\[ \int_0^{T}\int_\mathbb{R} u_k^{2}(x,t){{\rm d}}x{{\rm d}}t \leq T\int_\mathbb{R} u_{0k}^{2}(x){{\rm d}}x. \]

This, together with (4.26), gives that

(4.33)\begin{equation} \alpha_k\leq \left(T \int u_{0k}^{2}(x){{\rm d}}x \right)^{1/2}\leq T^{1/2}R \end{equation}

since $\|u_{0k}\|_{L^{2}(\mathbb {R})}\leq R$ for all $k$. Thanks to (4.32) and (4.33), by the well posedness of (4.28), we have

(4.34)\begin{equation} \|v_k\|_{X^{0,({1}/{2})+}_{(0,T)}}\leq C(T,\|a\|_{L^{\infty}}, R). \end{equation}

By our assumption, the complement set $E^{c}$ has finite Lebesgue measure, ${|E^{c}|<\infty}$. Thanks to (4.32) and (4.33), we can apply corollary 4.9 to obtain that

(4.35)\begin{equation} \int_0^{T}\int_{E^{c}}|\partial_xv_k(x,t)|^{2}{{\rm d}}x {{\rm d}}t \leq C|E^{c}|<\infty, \quad \forall k\in \mathbb{N}. \end{equation}

Combining (4.27) and (4.35) we find that $v_k$ is uniformly bounded in $L^{2}(0,\,T;H^{1}(E^{c}))$. Also, using equation (4.28), we get that $(v_k)_t$ is uniformly bounded in $L^{2}(0,\,T;H^{-2}(E^{c}))$. Since $|E^{c}|<\infty$, it is easy to see that

\[ \lim_{x\to \infty}\left|E^{c}\bigcup \left(x-\frac{1}{2},x+\frac{1}{2}\right)\right|=0, \]

then according to [1, theorem 2.8], the embedding $H^{1}(E^{c})\hookrightarrow L^{2}(E^{c})$ is compact. Thus, by Aubin–Lions theorem, there exists a subsequence, still denoted by $v_k$, so that $v_k\to v$ in $L^{2}((0,\,T)\times E^{c})$. On the other hand, it follows from (4.30) that $v_k\to 0$ in $L^{2}((0,\,T)\times E)$. These show that

(4.36)\begin{equation} v_k \to v \text{ strongly in } L^{2}(0,T;L^{2}(\mathbb{R})) \end{equation}

with $\|v\|_{X^{0,({1}/{2})+}_{(0,T)}}\leq C$ (by (4.34)) and

(4.37)\begin{equation} v(x,t)=0, \text{ for } (x,t)\in E\times(0,T). \end{equation}

We assume that $\alpha _k\to \alpha \geq 0$. Let $\phi (x,\,t)$ be a function such that $\phi \in C([0,\,T]; H^{3}(\mathbb {R}))$, $\phi _t\in C([0,\,T];L^{2}(\mathbb {R}))$ with $\phi |_{t=0}=\phi |_{t=T}=0$. Testing (4.28) with $\phi$ we get

(4.38)\begin{equation} \int^{T}_0\int_{\mathbb{R}}\left(v_k\partial_t\phi+v_k\partial^{3}_x\phi-a(x)v_k\phi+\alpha_k\frac{v^{2}_k}{2}\partial_x\phi\right){{\rm d}}x {{\rm d}}t =0. \end{equation}

Now, taking the limit as $k\rightarrow \infty$ in (4.38), and applying (4.36)–(4.37) we arrive at

(4.39)\begin{equation} \int^{T}_0\int_{\mathbb{R}}\left(v\partial_t\phi+v\partial^{3}_x\phi+\alpha\frac{v^{2}}{2}\partial_x\phi\right){{\rm d}}x {{\rm d}}t =0, \end{equation}

where we used the fact that $\int _0^{T}\int _\mathbb {R} a(x)v_k(x,\,t)\phi (x,\,t){{\rm d}}x {{\rm d}}t\to 0$ as $k\to \infty$, which follows from (4.29) and the inequality

\[ \left| \int_0^{T}\int_\mathbb{R} a(x)v_k(x,t)\phi(x,t){{\rm d}}x {{\rm d}}t\right|\leq \left(\int_0^{T}\int_\mathbb{R} a(x)v^{2}_k(x,t){{\rm d}}x {{\rm d}}t \right)^{{1}/{2}}\|a^{{1}/{2}}\phi\|_{L^{2}(0,T;L^{2}(\mathbb{R}))}. \]

The identity (4.39) means that $v(x,\,t)\in X^{0,({1}/{2})+}_{(0,T)}$ is a weak solution of

(4.40)\begin{equation} \partial_tv+\partial^{3}_xv+\alpha v\partial_xv=0,\quad (x,t)\in \mathbb{R}\times (0,T). \end{equation}

Moreover, by (4.37) we have $v|_{E\times (0,T)}=0$.

If $\alpha =0$, then equation (4.40) becomes $\partial _tv+\partial ^{3}_xv=0$. Since $v=0$ on an open set in $(x,\,t)\in \mathbb {R}\times (0,\,T)$, according to [33, corollary 3.1], we have $v\equiv 0$.

If $\alpha >0$, set $V(x,\,t)=v(cx,\,c^{3}t)$ with $c=\alpha ^{-({1}/{2})}$, then $V\in X^{0,({1}/{2})+}_{(0,c^{-3}T)}$ is a solution of

\[ \partial_tV+\partial^{3}_xV+V\partial_xV=0,\quad (x,t)\in \mathbb{R}\times (0,c^{{-}3}T). \]

Moreover, we have $V|_{c^{-1}E\times (0,\,c^{-3}T)}=0$, where $c^{-1}E=\{c^{-1}x: x\in E\}$. It is easy to see that $c^{-1}E$ also satisfies NCC. Then by the unique continuation property in proposition 4.1 for the KdV equation, we get $V \equiv 0$ and thus $v \equiv 0$.

In both cases, we arrive at the conclusion $v \equiv 0$. However, this contradicts to (4.27). Therefore, (4.24) holds.

Proof of theorem 1.4. Let $T>0$ and $\|u_0\|_{L^{2}(\mathbb {R})}\leq R$. Multiplying (1.1) by $u$ and integrating over $\mathbb {R}\times (0,\,T)$, we get

(4.41)\begin{equation} \int_\mathbb{R} |u(x,T)|^{2}{{\rm d}}x + 2\int_0^{T}\int_\mathbb{R} a(x)|u(x,t)|^{2}{{\rm d}}x {{\rm d}}t =\int_\mathbb{R} |u_0(x)|^{2}{{\rm d}}x. \end{equation}

Since $E$ satisfies NCC, it follows from lemma 4.10 that

(4.42)\begin{equation} \int_0^{T}\int_\mathbb{R} a(x)|u(x,t)|^{2}{{\rm d}}x {{\rm d}}t \geq c\int_0^{T}\int_\mathbb{R} |u(x,t)|^{2}{{\rm d}}x {{\rm d}}t, \end{equation}

where $c>0$ is a constant depending only on $T,\,R$ and the damping coefficient $a(x)$. Moreover, since $a(x)\geq 0$, we have $\|u(\cdot,\,t)\|_{L^{2}(\mathbb {R})}\leq \|u(\cdot,\,t')\|_{L^{2}(\mathbb {R})}$ for all $t\geq t'$, which implies that

(4.43)\begin{equation} \int_0^{T}\int_\mathbb{R} |u(x,t)|^{2}{{\rm d}}x {{\rm d}}t \geq T\int_\mathbb{R} |u(x,T)|^{2}{{\rm d}}x. \end{equation}

Combining (4.41)–(4.43), we infer that

\[ \int_\mathbb{R} |u(x,T)|^{2}{{\rm d}}x \leq \alpha \int_\mathbb{R} |u_0(x)|^{2}{{\rm d}}x \]

with $\alpha ={1}/{(1+2cT)}\in (0,\,1)$. By iteration we have for all $n\geq 1$

\[ \int_\mathbb{R} |u(x,nT)|^{2}{{\rm d}}x \leq \alpha^{n} \int_\mathbb{R} |u_0(x)|^{2}{{\rm d}}x. \]

This gives the exponential decay clearly.