#### Abstract

It is shown that if is a sufficiently smooth solution to a two-dimensional nonlinear wave equation such that there exists with supp , for , then .

#### 1. Introduction

In this paper, we consider the following two-dimensional nonlinear wave equation: where , , , are arbitrary positive constants. Equation (1) was recently derived by Gottwald  for large scale motion from the barotropic quasigeostrophic equation as a two-dimensional model for Rossby waves. He  showed that (1) has traveling wave solutions via the homotopy perturbation method. Using a subequation method, the traveling wave solutions are also studied by Fu et al. . Aslan  constructed solitary wave solutions and periodic wave solutions to (1) by the Exp-function method.

For and in (1), one obtains the classical Zakharov-Kuznetsov (ZK) equation , which is a mathematical model to describe the propagation of nonlinear ion-acoustic waves in magnetized plasma. Solitary wave solutions and the Cauchy problem to ZK equation have extensively been studied in the literature (). Panthee  proved that if a sufficiently smooth solution to the initial value problem associated with the ZK equation is supported compactly in a nontrivial time interval, then it vanishes identically. Recently, Bustamante et al.  showed that sufficiently smooth solutions of the ZK equation that have compact support for two different times are identically zero.

The purpose of this paper is to investigate the support of solutions to (1). To solve the problem, we mainly use the ideas of . The main result is as follows.

Theorem 1. Assume that and , if is a solution of (1) such that then, .

#### 2. Preliminary Estimates

Lemma 2 (see ). Assume that and . (i) If , then (ii) if , then where , , and .

Lemma 3. Assume that , if is a solution of (1) such that ; then, is bounded in .

Proof. Assume that is a decreasing function with if and if . Let for and . It is easy to check that and Multiplying (1) by and integrating by parts in , we obtain Applying Gronwall Lemma and the Monotone Convergence Theorem, we have This proves that is bounded in .
Applying Lemma 2 with and , we have that is bounded in . Here, we used the fact that . This completes the proof of the lemma.

Lemma 4. Assume that , , and is a solution of (1). (i) If , then is bounded in ; (ii) if , then is bounded in .

Proof. Letting and a solution to (1), we have Multiplying (8) by and integrating by parts in , we obtain Note that and It follows from (9) that Since and is bounded in , applying Lemma 2 with and , we have that is also bounded in .
Similarly, we can prove that is bounded in . Let ; then, is a solution to (1) and satisfies , and therefore is bounded in . This proves (i).
Now, we prove (ii). Let ; then, is also a solution of (1) and satisfies the hypothesis of (i). This proves (ii) and completes the proof of the lemma.

Remark 5. In particular, if the conditions for and given in (i) and (ii), respectively, are satisfied, then is bounded in .

Lemma 6 (see ). Let , is a function such that is bounded in , and . Then, for all and all , the functions and are absolutely continuous in with derivatives and a.e. , respectively.

Lemma 7. Assume that , , and , if is a function such that is bounded in and . Then,

Proof. Let and ; then, Taking the spatial Fourier transform in (14) and applying Lemma 6, we have where According to (15), when , we have and when , we choose to write Therefore, we have Appling Plancherel formula, we have inequality (12).
Similarly, letting , we can also have (13). This completes the proof of the lemma.

Lemma 8 (see [12, 13]). Assume that , and , if is a solution to (1) such that then, .

Proof. The proof is similar to that of Theorem  1.1 in , and we omit the details.

#### 3. Proof of the Main Result

Assume that for where is a nondecreasing function such that for and for . Let ; then, . According to Lemma 7, we obtain that where and Note that the derivatives of are supported in the interval ; then, where is dependent on and .

Combining (21) with (23), we obtain Applying Lemma 4 with , we have that then Since , taking such that , we have Note that for ; we have Letting , we obtain and this proves that in .

Next, we will prove that in . Let ; then, where .

In fact, Let ; it is easy to check that also satisfies the hypotheses of this theorem, and then we find that in Thus, there exists such that for all . Applying Lemma 8, we complete the proof of Theorem 1.

#### Acknowledgments

This work was supported by the National Natural Science Foundation of China (no. 11171135), the Natural Science Foundation of Jiangsu (no. BK 2010329), the Project of Excellent Discipline Construction of Jiangsu Province of China, the Priority Academic Program Development of Jiangsu Higher Education Institutions, and the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (no. 09KJB110003), as well as the Taizhou Social Development Project (no. 2011213).