JMATH Journal of Mathematics 2314-4785 2314-4629 Hindawi Publishing Corporation 578094 10.1155/2013/578094 578094 Research Article On the Support of Solutions to a Two-Dimensional Nonlinear Wave Equation Zhang Wenbin 1 0000-0002-6179-4786 Zhou Jiangbo 2 Tian Lixin 2 Kumar Sunil 3 Gao Ji 1 Taizhou Institute of Science and Technology NUST, Taizhou, Jiangsu 225300 China tzc.edu.cn 2 Nonlinear Scientific Research Center Faculty of Science Jiangsu University Zhenjiang, Jiangsu 212013 China ujs.edu.cn 3 Department of Mathematics National Institute of Technology Jamshedpur, Jharkhand 831014 India nitjsr.ac.in 2013 4 3 2013 2013 10 01 2013 21 01 2013 2013 Copyright © 2013 Wenbin Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

It is shown that if u is a sufficiently smooth solution to a two-dimensional nonlinear wave equation such that there exists L>0 with supp u(i)[L,L]×[L,L], for i=0,1, then u0.

1. Introduction

In this paper, we consider the following two-dimensional nonlinear wave equation: (1)ut+auux+buxxx+cuyyx+dux=0, where a, b, c, d 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 a=b=c=1 and d=0 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 b>c and L>0, if uC([0,1];H4(R2))C1([0,1];L2(R2)) is a solution of (1) such that (2)suppu(0),suppu(1)[-L,L]×[-L,L]; then, u0.

2. Preliminary Estimates Lemma 2 (see [<xref ref-type="bibr" rid="B15">13</xref>]).

Assume that s>0 and κ>0. (i) If fHs(R2)L2(e2κxdxdy), then (3)Jθs(e(1-θ)κxf)L2CJsfL2θeκxfL21-θ, (ii) if fHs(R2)L2(e2(κx+κy)dxdy), then (4)Jθs(e(1-θ)(κx+κy)f)L2CJsfL2θeκx+κyfL21-θ, where θ[0,1], C=C(s,κ), and [Jsf](ξ):=(1+|ξ|2)s/2[f](ξ).

Lemma 3.

Assume that κ>0, if uC([0,1];H4(R2))C1([0,1];L2(R2)) is a solution of (1) such that u(0)L2(e2κxdxdy); then, u(t) is bounded in H3(e2κxdxdy).

Proof.

Assume that φ(x)C(R) is a decreasing function with φ(x)=1 if x<1 and φ(x)=0 if x>10. Let θn(x)=0xφ(t/n)dt for nZ+ and ϕn(x)=e2kθn(x). It is easy to check that ϕn(x)ϕn+1(x) and (5)|ϕn(j)(x)|Cj,kϕn(x),jZ+,xR. Multiplying (1) by uϕn and integrating by parts in Rxy2, we obtain (6)12ddtu2ϕn=a3u3ϕ-3b2(ux)2ϕn+b2u2ϕn-c2(uy)2ϕ+d2u2ϕna3uL(R2)C1,κu2ϕn+b2C3,κu2ϕn+d2C1,κu2ϕn(a3CuC([0,1];H2)+b2C3,k+d2C1,k)u2ϕn=Ck,uu2ϕn. Applying Gronwall Lemma and the Monotone Convergence Theorem, we have (7)u(t)2e2κxdxdyCu(0)2e2κxdxdy,t[0,1]. This proves that u(t) is bounded in L2(e2κxdxdy).

Applying Lemma 2 with s=4 and θ=3/4, we have that u(t) is bounded in H3(e2κxdxdy). Here, we used the fact that uC([0,1];H4). This completes the proof of the lemma.

Lemma 4.

Assume that κ>0, b>c, and uC([0,1];H4(R2))C1([0,1];L2(R2)) is a solution of (1). (i) If u(0)L2(e2κxe2κ|y|dxdy), then u(t) is bounded in H3(e2κxe2κ|y|dxdy); (ii) if u(1)L2(e-2kxe2k|y|dxdy), then u(t) is bounded in H3(e-2κxe2κ|y|dxdy).

Proof.

Letting v(t)=eκxu(t) and u a solution to (1), we have (8)ekxu=κ(au+bκ2+d)v-(au+3bκ2+d)vx+3bκvxx-bvxxx+cκvyy-cvyyx. Multiplying (8) by vϕn(y) and integrating by parts in Rxy2, we obtain (9)eκxuvϕn=aκuv2ϕn+a2v2uxϕn+(bκ3+dκ)v2ϕn-3bκ(vx)2ϕn-cκ(vy)2ϕn+c2κv2ϕn-cvxvyϕn. Note that |ϕn(y)|2κϕn(y) and (10)eκxuvϕn=12ddtv2ϕn,a.e  t[0,1]. It follows from (9) that (11)12ddtv2ϕnaκCuC([0,1];H2(R2))v2ϕn+aCuxC([0,1];H2(R2))v2ϕn+(bκ3+dκ)v2ϕn-cκ[(vx)2+2|vx||vy|+(vy)2]ϕn+c2κC2,κv2ϕn=Cκ,uv2ϕn-cκ(|vx|+|vy|)2ϕnCκ,uv2ϕn,a.e  t[0,1]. Since uC([0,1];H4) and u(t) is bounded in L2(e2κxe2κydxdy), applying Lemma 2 with s=4 and θ=3/4, we have that u(t) is also bounded in H3(e2κxe2κydxdy).

Similarly, we can prove that u(t) is bounded in H3(e2κxe-2κydxdy). Let u~(x,y,t)=u(x,-y,t); then, u~ is a solution to (1) and satisfies u~(0)L2(e2κxe2κ|y|dxdy), and therefore u~ is bounded in H3(e2κxe2κydxdy). This proves (i).

Now, we prove (ii). Let u-(x,y,t)=u(-x,y,1-t); then, u-(x,y,t) 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 u(0) and u(1) given in (i) and (ii), respectively, are satisfied, then u(t) is bounded in H3(e2κ|x|e2κ|y|dxdy).

Lemma 6 (see [<xref ref-type="bibr" rid="B15">13</xref>]).

Let κ>0, vC1([0,1];L2(R2)) is a function such that v(t) is bounded in L2(e2κ|x|e2κ|y|dxdy), and vL1([0,1];L2(e2κ|x|e2κ|y|dxdy)). Then, for all λR and all ξ=(ξ1,ξ2)R2, the functions t[eλxv(t)](ξ) and t[eλyv(t)](ξ) are absolutely continuous in [0,1] with derivatives [eλxv(t)](ξ) and [eλyv(t)](ξ) a.e. t[0,1], respectively.

Lemma 7.

Assume that κ>0, b>c, and λ0, if vC([0,1];H3(R2))C1([0,1];L2(R2)) is a function such that v(t) is bounded in H3(e2κ|x|e2κ|y|dxdy) and v(t)L1([0,1];L2(e2κ|x|e2κ|y|dxdy)). Then, (12)eλxvL2(R2×[0,1])eλxv(0)L2(R2)+eλxv(1)L2(R2)+eλx(vt+bvxxx+cvyyx+dvx)L2(R2×[0,1]),(13)eλyvL2(R2×[0,1])eλyv(0)L2(R2)+eλyv(1)L2(R2)+eλy(vt+bvxxx+cvyyx+dvx)L2(R2×[0,1]).

Proof.

Let p(t)=eλxv(t) and q(t)=eλx(vt+bvxxx+cvyyx+dvx); then, (14)q(t)=eλxv(t)-(bλ3+dλ)p(t)+(3bλ2+d)px(t)-3bλpxx(t)+bpxxx(t)-cλpyy(t)+cpyyx(t). Taking the spatial Fourier transform in (14) and applying Lemma 6, we have (15)ddt[p(t)](ξ)+[-im(ξ)-n(ξ)][p(t)](ξ)=[q(t)](ξ),a.e  t[0,1], where (16)m(ξ)=-(3bλ2+d)ξ1+bξ13+cξ1ξ22,n(ξ)=(bλ3+dλ)-3bλξ12-cλξ22. According to (15), when n(ξ)0, we have (17)[p(t)](ξ)=eim(ξ)ten(ξ)t[p(0)](ξ)+0teim(ξ)(t-τ)en(ξ)(t-τ)[q(τ)](ξ)dτ, and when n(ξ)>0, we choose to write (18)[p(t)](ξ)=e-im(ξ)(1-t)e-n(ξ)(1-t)[p(1)](ξ)-t1e-im(ξ)(τ-t)e-n(ξ)(τ-t)[q(τ)](ξ)dτ. Therefore, we have (19)|[p(t)](ξ)||[p(0)](ξ)|+|[p(1)](ξ)|+01|[q(τ)](ξ)|dτ,t[0,1]. Appling Plancherel formula, we have inequality (12).

Similarly, letting p(t)=eλyw(t), we can also have (13). This completes the proof of the lemma.

Lemma 8 (see [<xref ref-type="bibr" rid="B12">12</xref>, <xref ref-type="bibr" rid="B15">13</xref>]).

Assume that L>0, and b>c, if uC([0,1];H4(R2)) is a solution to (1) such that (20)suppu(t)[-L,L]×[-L,L],t[0,1]; then, u0.

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 ϕ(x):=ϕ~(x-A) for A>L where ϕ~(x)C(R) is a nondecreasing function such that ϕ~(x)=0 for x<0 and ϕ~(x)=1 for x>1. Let g:=ϕ(x)u; then, g(0)=g(1)=0. According to Lemma 7, we obtain that (21)eλxweλx(gt+bgxxx+cgyyx+dgx)=eλx[ϕut+bϕuxxx+cϕuyyx+(bϕ+dϕ)u=eλx  +(3bϕ+dϕ)ux+3bϕuxx+cϕuyy]aeλxϕuux+eλxGϕ,u, where ·:=·L2(R2×[0,1]) and (22)Gϕ,u=(bϕ+dϕ)u+3bϕux+3bϕuxx+cϕuyy. Note that the derivatives of ϕ(x) are supported in the interval [A,A+1]; then, (23)eλxGϕ,u=Ceλ(A+1), where C is dependent on uC([0,1];  H2) and max{b,c,d}.

Combining (21) with (23), we obtain (24)eλxϕuaeλxϕuuxL([A,+]×R×[0,1])+Ceλ(A+1). Applying Lemma 4 with κ=1, we have that (25)|ux(t)(x,y)|C1e-x, then (26)eλxϕuaC1e-Aeλxϕu+Ceλ(A+1). Since eλxϕu<+, taking A>max{L,1} such that C1e-A<1/2a, we have (27)eλxϕuCeλ(A+1). Note that ϕ(x)=1 for x2A; we have (28)e2λA(01-2A|u(t)(x,y)|2)1/2eλxϕuCeλ(A+1). Letting λ+, we obtain (29)(01-2A|u(t)(x,y)|2)1/2=0, and this proves that u0 in [2A,)×R×[0,1].

Next, we will prove that u0 in R×[2A,)×[0,1]. Let h=ϕ(y)u; then, (30)eλyheλy(ht+bhxxx+chyyx+dhx)=eλy(ϕut+bϕuxxx+cϕuyyx+dϕux+2cϕuxy+cϕux)aeλyϕuux+eλyHϕ,u, where Hϕ,u:=2cϕuxy+cϕux.

In fact, Let u~(x,y,t)=u(-x,-y,1-t); it is easy to check that u~ also satisfies the hypotheses of this theorem, and then we find that u0 in (31)(-,-2A]×R×[0,1]R×(-,-2A]×[0,1]. Thus, there exists A>0 such that suppu(t)[-A,A]×[-A,A] for all t[0,1]. 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).

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