Publication Cover
Applicable Analysis
An International Journal
Volume 101, 2022 - Issue 1
1,095
Views
1
CrossRef citations to date
0
Altmetric
Articles

Inverse source problem related to one-dimensional Saint-Venant equation

ORCID Icon
Pages 35-47 | Received 22 Sep 2019, Accepted 04 Feb 2020, Published online: 13 Feb 2020

ABSTRACT

The one-dimensional Saint-Venant equation describes unsteady water flow in channels and is derived from the one-dimensional Euler equation by imposing several physical assumptions. In this paper, we consider the linearized and simplified equation in the one-dimensional case featuring a mixed derivative term and prove the global Lipschitz stability of the inverse source problem via the global Carleman estimate.

2010 MATHEMATICS SUBJECT CLASSIFICATIONS:

1. Introduction and main result

We consider (1) Au:=(t2x2+axt)u=H(x,t)in Q+:=(,)×(0,T),u(,0)=tu(,0)=0on (,),u(,)=u(,)=0on (0,T),(1) where a>0, >0 and T>0 are positive constants and H is the source term. The differential operator A has the form of a one-dimensional wave operator plus the mixed derivative term axt. This term appears when we linearize the one-dimensional Saint-Venant equation, which is the equation introduced by Saint-Venant in [Citation1] to describe unsteady water flow in channels. The one-dimensional Saint-Venant equation comprises continuity and momentum equations. Their formulations and physical meanings are written in Cunge et al. [Citation2]. Even though one-dimensional flow does not exist in nature, mathematically speaking, it is important to consider the simplified equation and observe its properties.

We consider the uniqueness and stability for the inverse source problem to determine H on (,) from the boundary observation data of the solution to (Equation1). The argument is based on the Carleman estimate and the Bukhgeim–Klibanov method in [Citation3]. The Carleman estimate was first introduced in Carleman [Citation4] to prove the unique continuation property for the elliptic operator whose coefficients are not necessarily analytic. Using the Carleman estimate, Bukhugeim and Klibanov proved global uniqueness results for multidimensional coefficient inverse problems. This methodology is widely applicable to various partial differential equations provided that we can prove the Carleman estimate for the considered equations. For a hyperbolic equation, Imanuvilov and Yamamoto [Citation5] considered the global Lipschitz stability for wave equations through interior observations. Baudouin et al. [Citation6] proved the global Carleman estimate for the wave equation and considered its applications to controllability, inverse problems, and reconstructions. Bellassoued and Yamamoto [Citation7] considered the inverse source and coefficient problems for the wave equation on a compact Riemannian manifold with a boundary.

In proving the Carleman estimate for the operator A, the main difficulties lie in the existence of the mixed derivative term axt. There are also difficulties when we apply the Carleman estimate to the extended solution to (Equation1). In the usual case of the wave equation, an evenly extended solution with respect to time t satisfies the wave equation as well. However, considering (Equation1), the evenly extended solution no longer satisfies the equation. We therefore need to consider a different extension.

To prove the global Lipschitz stability for inverse source problems of the hyperbolic partial differential equation, the observation time should be given for the distant wave to reach the boundary owing to the finite propagation speed. We define constants to describe this situation mathematically. Let x0[,]c be a given point and T0:=1ρmaxx|xx0|, where ρ:=((a2+4a)/2)2 is the square of the wave speed.

Theorem 1.1

Assume H(x,t)=f(x)R(x,t), where fL2(,),RH1(0,T;L(,)),f(x)=f(x),R(x,0)=R(x,0),a.e. x(,), and m0>0s.t. |R(x,0)|m0>0a.e.x(,). Let T>T0. We assume there exists a unique solution u to (Equation1) in the class uk=02H3k(0,T;Hk(,)). There then exists a constant C>0 that is independent of u and f such that fL2(,)Cxtu(,)L2(0,T)+xtu(,)L2(0,T).

2. Proof of Theorem 1.1

2.1. Preliminary

To prove Theorem 1.1 using the Bukhugeim–Klibanov method, we need to prove the Carleman estimate for the operator A in extended domain Q±. We consider A=t2x2+atx,Q±:=(,)×(T,T), where a>0 is a constant. The next proposition is the global Carleman estimate, whose proof is postponed to the Appendix section.

Proposition 2.1

Choose x0[,],and β such that 0<β<ρ,

and set ψ(x,t):=|xx0|2βt2,ϕ(x,t):=eγψ(x,t),σ(x,t):=sγϕ(x,t),(x,t)Q±¯, where γ>0 and s>0 is some parameters. There then exists a constant γ>0 such that for all γγ, the following holds.

There exist constants s=s(γ) and C=C(s) such that Q±e2sϕ(σ|xu|2+σ|tu|2+σ3|u|2)dxdtCQ±e2sϕ|Au|2dxdt+CTTe2sϕ(,t)σ(,t)|xu(,t)|2+e2sϕ(,t)σ(,t)|xu(,t)|2dt, for all s>s and uk=02H2k(T,T;Hk(,)) such that u(,±T)=tu(,±T)=0 on (,) and u(±,)=0 on (T,T).

Proving our main theorem requires several energy estimates as follows.

Lemma 2.2

Assume fL2(,), RL2(T,T;L(,)) and u1L2(,). Let uk=02H2k(T,T;Hk(,)) be a solution to Au=(t2x2+axt)u=f(x)R(x,t)in Q±,u(,0)=0, tu(,0)=u1on (,),u(,)=u(,)=0on (T,T). There then exists a constant C such that tu(,t)L2(,)2+xu(,t)L2(,)2C(fL2(,)2+u1L2(,)2),a.e. t(T,T) holds.

Proof.

Set E(t):=tu(,t)L2(,)2+xu(,t)L2(,)2. Multiplying the first equation by tu and integrating over (,) yield ddtE(t)E(t)+fR(,t)L2(,)2 and so ddt(etE(t))etfR(,t)L2(,)2. Furthermore, integration over (0,t) yields E(t)C(RL2(T,T;L(,))2fL2(,)2+u1L2(,)2).

Lemma 2.3

Assume HL2(T,T;L2(,)). Let zk=02H2k(T,T;Hk(,)) be a solution to Az=H(x,t)in Q±,z(,0)=0,tz(,0)=z1on (,),z(,)=z(,)=0on (T,T), and z(,±T)=tz(,±T)=0on (,). There then exists a constant C>0 such that z1L2(,)2CHtzL1(Q±) holds.

Proof.

Multiplying the first equation by 2tz and integrating over Q+=(,)×(0,T), we have Q+t|tz|2+Q+t|xz|2+aQ+x|tz|2=2Q+Htz. Hence, we get z1L2(,)2CHtzL1(Q+). The proof for Q:=(,)×(T,0) is similar.

2.2. Proof of the main theorem

Proof

Proof of Theorem 1.1

We use the weight function ψ(x,t)=|xx0|2βt2 for x0[,]c. We assume T>T0, i.e. ρT2>ρT02=maxx|xx0|2, and there thus exists 0<β<ρ such that maxx|xx0|2<βT2(<ρT2). There then exists 0<ϵ(<T2) such that maxx|xx0|2<β(T2ϵ)2. Thus, for all x[,] and t[T,T+2ϵ][T2ϵ,T], we have ϕ(x,t)=eγψ(x,t)<1. Let u be the solution in the class uk=02H3k(0,T;Hk(,)) and take the extension of u to (T,T): u(x,t)=u(x,t)in Q+:=(,)×(0,T),u(x,t)in Q:=(,)×(T,0). We also extend R: R(x,t)=R(x,t)in Q+,R(x,t)in Q. We can then prove uk=02H3k(T,T;Hk(,)) and RH1(T,T;L(,)). Indeed, we assume u(,0)=tu(,0)=0 and we thus have tu(x,t)=tu(x,t)in Q+,tu(x,t)in Q, and t2u(x,t)=t2u(x,t)in Q+,t2u(x,t)in Q. Furthermore, we assume the symmetry of the source term f and R(,0) and thus get t3u(x,t)=t3u(x,t)in Q+,t3u(x,t)in Q. Considering R, we have tR(x,t)=tR(x,t)in Q+,tR(x,t)in Q, owing to the symmetry of R(,0). This extended u satisfies Au=(t2x2+axt)u=f(x)R(x,t)in Q±,u(,0)=tu(,0)=0on (,),u(,)=u(,)=0on (T,T). From Lemma 2.2, we have (2) tu(,t)L2(,)2+xu(,t)L2(,)2CfL2(,)2,a.e. t(T,T).(2) Let v:=tu. v satisfies Av=(t2x2+axt)v=f(x)tR(x,t)in Q±,v(,0)=0, tv(,0)=f()R(,0)on (,),v(,)=v(,)=0in (T,T). Also from Lemma 2.2, we have (3) tv(,t)L2(,)2+xv(,t)L2(,)2CfL2(,)2,a.e. t(T,T).(3) We define a cut-off function η satisfying 0η(t)1: η(t):=1,|t|T2ϵ,0,|t|Tϵ. We set w:=ηtu=ηv. w satisfies Aw=ηftR+2tηtv+atηxv+t2ηvin Q±,w(,0)=0, tw(,0)=f()R(,0)on (,),w(,)=w(,)=0on (T,T). Moreover, w(,±T)=tw(,±T)=0,on (,) holds. We can therefore apply Proposition 2.1 to w to obtain Q±e2sϕ(s|xw|2+s|tw|2+s3|w|2)CQ±e2sϕ|ftR|2+CQ±e2sϕ(|tηtv|2+|tηxv|2+|t2ηv|2)+CseCsD2 for all ss, where D2:=TT|xw(,t)|2+|xw(,t)|2dtTT|xtu(,t)|2+|xtu(,t)|2dt. We here consider sufficiently large γ>γ as a fixed constant. Considering the second, third, and fourth terms on the right-hand side and using (Equation2) and (Equation3), we get Q±e2sϕ(|tηtv|2+|tηxv|2+|t2ηv|2)Ce2sfL2(,)2, because supp(tη), supp(t2η)[T+ϵ,T+2ϵ][T2ϵ,Tϵ], and for all x[,] and t[T+ϵ,T+2ϵ][T2ϵ,Tϵ], we have ϕ(x,t)=eγψ(x,t)<1. We therefore have (4) Q±e2sϕ(s|xw|2+s|tw|2+s3|w|2)CQ±e2sϕ|ftR|2+Ce2sfL2(,)2+CseCsD2,(4) for all ss.

We next set z:=esϕw and z satisfies Az=esϕ(Aw+G(x,t))in Q±,z(,0)=0,tz(,0)=esϕ(,0)f()R(,0)on (,),z(,)=z(,)=0on (T,T), where G:=(2stϕ+asxϕ)tw+(2sxϕ+astϕ)xw+(st2ϕ+s2|tϕ|2sx2ϕs2|xϕ|2+asxtϕ+as2xϕtϕ)w. From Lemma 2.3, we have (5) esϕ(x,0)fL2(,)2CQ±e2sϕ|tz|2+CQ±e2sϕ|ftR|2+Ce2sfL2(,)2+CesϕGtzL1(Q±)(5) and we also find esϕGtzCe2sϕ(s|tw|+s|xw|+s2|w|)(|tw|+s|w|) and thus obtain esϕGtzL1(Q±)CQ±e2sϕ(s|xw|2+s|tw|2+s3|w|2). Furthermore, we have Q±e2sϕ|ftR|2Ce2sϕ(x,0)|f(x)|2TTe2s(ϕ(x,0)ϕ(x,t))dtdxCe2sϕ(x,0)|f(x)|2TTe2seγ|xx0|2(1eγβt2)dtdxo(1)esϕ(x,0)fL2(,)2,as s, by the Lebesgue's dominated convergence theorem. We apply this estimate and (Equation4) to (Equation5) to get esϕ(x,0)fL2(,)2Ce2sfL2(,)2+CseCsD2. There exists κ>1 such that for all x[,], ϕ(x,0)=eγ|xx0|κ>1 holds. Therefore, taking sufficiently large s>s, we finally have fL2(,)CD, for a constant C>0 independent of f.

Acknowledgments

The author would like to thank Professor Masahiro Yamamoto (The University of Tokyo) for many valuable discussions and comments. The author also thank anonymous referees for invaluable comments and Glenn Pennycook, NSc, from Edanz Group (www.edanzediting.com/ac) for editing a draft for this manuscript.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

This work was supported by JSPS and RFBR under the Japan-Russia Research Cooperative Program (project No. J19-721).

References

  • Saint-Venant. Théorie du mouvement non permanent des eaux, avec application aux crues des rivières et a l'introduction de marées dans leurs lits. C R Acad Sci. 1871;73:147–154 and 237–240.
  • Cunge JA, Holly FM, Verwey A. Practical aspects of computational river hydraulics. Boston: Pitman Advanced Publishing Program; 1980.
  • Bukhgeim AL, Klibanov MV. Global uniqueness of class of multidimensional inverse problems. Soviet Math Dokl. 1981;24:244–247.
  • Carleman T. Sur un problème d'unicité pur les systèmes d'équations aux dérivées partielles à deux variables indépendantes. Ark Mat Astr Fys. 1939;2B:1–9.
  • Imanuvilov OY, Yamamoto M. Global Lipschitz stability in an inverse hyperbolic problem by interior observations. Inverse Probl. 2001;17(4):717–728. doi: https://doi.org/10.1088/0266-5611/17/4/310
  • Baudouin L, de Buhan M, Ervedoza S. Global carleman estimates for waves and applications. Comm Partial Differ Equ. 2013;38(5):823–859. doi: https://doi.org/10.1080/03605302.2013.771659
  • Bellassoued M, Yamamoto. M. Carleman estimates and applications to inverse problems for hyperbolic systems. Tokyo: Springer Japan; 2017.

Appendix

Proof

Proof of Proposition 2.1

For simplicity, we write p(x):=xψ(x,t),q(t):=tψ(x,t). Simple calculation yields xϕ=γpϕ,tϕ=γqϕ,x2ϕ=(γ2p2+2γ)ϕ,t2ϕ=(γ2q22βγ)ϕ,xtϕ=γ2pqϕ. We introduce w:=esϕu,Pw:=esϕAesϕw. Then, Pw=t2wx2w+2s(xϕxwtϕtw)s2(|xϕ|2axϕtϕ|tϕ|2)w+s(x2ϕaxtϕt2ϕ)w+axtwas(tϕxw+xϕtw). We decompose Pw into P+w and Pw as follows. P+w:=t2wx2w+axtws2(|xϕ|2|tϕ|2axϕtϕ)w and Pw:=s(2xϕatϕ)xws(2tϕ+axϕ)tw+s(x2ϕaxtϕt2ϕ)w. Note that Pw=P+w+Pw. We wish to make the lower bound of PwL2(Q±)2, and we therefore try to estimate (P+w,Pw)L2(Q±). (P+w,Pw)L2(Q±)=Q±t2ws(2xϕatϕ)xwQ±t2ws(2tϕ+axϕ)tw+Q±t2ws(x2ϕaxtϕt2ϕ)wQ±x2ws(2xϕatϕ)xw+Q±x2ws(2tϕ+axϕ)twQ±x2ws(x2ϕaxtϕt2ϕ)w+Q±axtws(2xϕatϕ)xwQ±axtws(2tϕ+axϕ)tw+Q±axtws(x2ϕaxtϕt2ϕ)wQ±s2(|xϕ|2|tϕ|2axϕtϕ)ws(2xϕatϕ)xw+Q±s2(|xϕ|2|tϕ|2axϕtϕ)ws(2tϕ+axϕ)twQ±s2(|xϕ|2|tϕ|2axϕtϕ)ws(x2ϕaxtϕt2ϕ)w=:k=112Jk. Through integration by parts, we get J1=sQ±(2xϕatϕ)12x(|tw|2)sQ±t(2xϕatϕ)twxw=s2Q±x(2xϕatϕ)|tw|2sQ±t(2xϕatϕ)twxw,J2=sQ±(2tϕ+axϕ)12t(|tw|2)=s2Q±t(2tϕ+axϕ)|tw|2,J3=sQ±(x2ϕaxtϕt2ϕ)|tw|2sQ±t(x2ϕaxtϕt2ϕ)12t(|w|2)=sQ±(x2ϕaxtϕt2ϕ)|tw|2+s2Q±t2(x2ϕaxtϕt2ϕ)|w|2,J4=s2Q±x(2xϕatϕ)|xw|2s2TT(2xϕatϕ)|xw|2x=dt,J5=sQ±x(2tϕ+axϕ)xwtwsQ±(2tϕ+axϕ)12t(|xw|2)=sQ±x(2tϕ+axϕ)xwtw+s2Q±t(2tϕ+axϕ)|xw|2,J6=sQ±(x2ϕaxtϕt2ϕ)|xw|2+sQ±x(x2ϕaxtϕt2ϕ)12x(|w|2)=sQ±(x2ϕaxtϕt2ϕ)|xw|2s2Q±x2(x2ϕaxtϕt2ϕ)|w|2,J7=s2Q±at(2xϕatϕ)|xw|2,J8=s2Q±ax(2tϕ+axϕ)|tw|2,J9=sQ±a(x2ϕaxtϕt2ϕ)xwtwsQ±ax(x2ϕaxtϕt2ϕ)12t(|w|2)=sQ±a(x2ϕaxtϕt2ϕ)xwtw+s2Q±atx(x2ϕaxtϕt2ϕ)|w|2,J10=s32Q±x((|xϕ|2|tϕ|2axϕtϕ)(2xϕatϕ))|w|2,J11=s32Q±t((|xϕ|2|tϕ|2axϕtϕ)(2tϕ+axϕ))|w|2,J12=s3Q±(|xϕ|2|tϕ|2axϕtϕ)(x2ϕaxtϕt2ϕ)|w|2. First, we consider a part of the sum k=19Jk. Set σ(x,t):=sγϕ(x,t) and B:=s2TT(2xϕatϕ)|xw|2x=dt. (A1) k=19Jk=Q±σγ2q2+2apq+a22p2+σ(4β+a2)|tw|2+Q±σγ2p22apq+a22q2+σ(4βa2)|xw|2+Q±[σγ((a24)pq+2a(q2p2))+σ(4βa4a)]twxw+Q±O(sγ4ϕ)|w|2B=2Q±σγq+ap2twpaq2xw2+Q±σ[(4β+a2)|tw|2+(4βa2)|xw|24a(β+1)twxw]+Q±O(sγ4ϕ)|w|2BQ±σ[(4β+a2)|tw|2+(4βa2)|xw|24a(β+1)twxw]+Q±O(sγ4ϕ)|w|2B.(A1) To estimate the terms σ× (first-order terms), we use the next energy inequality. (P+w,σw)L2(Q±)=Q±P+w(σw)=Q±σt2ww+Q±σx2ww+Q±σs2(|xϕ|2axϕtϕ|tϕ|2)|w|2Q±aσxtww=:k=14Ik. Integration by parts yields I1=Q±sγtϕ12t(|w|2)+Q±sγϕ|tw|2=Q±sγϕ|tw|212Q±sγt2ϕ|w|2,I2=Q±sγxϕ12x(|w|2)Q±sγϕ|xw|2=Q±sγϕ|xw|2+12Q±sγx2ϕ|w|2,I3=Q±s3γϕ(|xϕ|2atϕxϕ|tϕ|2)|w|2,I4=Q±asγxϕ12t(|w|2)+Q±asγϕtwxw=12Q±asγtxϕ|w|2+Q±asγϕtwxw. Hence, we have (A2) Q±P+w(σw)=k=14Ik=Q±σ|tw|2Q±σ|xw|2+Q±aσtwxw+Q±12sγt2ϕ+12sγx2ϕ+s3γϕ(|xϕ|2axϕtϕ|tϕ|2)a2sγtxϕ12|w|2=Q±σ|tw|2Q±σ|xw|2+Q±aσtwxw+Q±[σ3(p2q2apq)+O(sγ3ϕ)]|w|2.(A2) Let μR be a constant to be determined later. Multiplying (EquationA2) by 2μ, and adding it to (EquationA1) yields k=19Jk+2μk=14Ik+BQ±σ(4β+a2+2μ)|tw|2+Q±σ(4βa22μ)|xw|2+Q±σ(4βa4a+2aμ)twxw+Q±[σ32μ(p2apqq2)+O(sγ4ϕ)+O(sγ3ϕ)]|w|2Q±σ(4β+a2+2μϵ|2βa2a+aμ|)|tw|2+Q±σ4βa22μ1ϵ|2βa2a+aμ||xw|2+Q±[σ32μ(p2apqq2)+O(sγ4ϕ)+O(sγ3ϕ)]|w|2, for all ϵ>0 to be determined later. We wish to choose ϵ>0 such that both coefficients are positive, i.e. (A3) 4β+a2+2μϵ|2βa2a+aμ|>0,4βa22μ1ϵ|2βa2a+aμ|>0(A3) holds. This claim follows only if μ satisfies μ22(β+1)μ+β(a2+4)<0(μβ1)2<β2(a2+2)β+1. Therefore, if we choose μ:=β+1, by our assumption of β>0, this inequality holds and (EquationA3) also holds. Hence, there exists a positive constant C>0 such that (A4) k=19Jk+2μk=14Ik+BCQ±σ(|tw|2+|xw|2)+Q±[2μσ3(p2apqq2)+O(sγ4ϕ)+O(sγ3ϕ)]|w|2.(A4) Finally, we estimate zeroth-order terms. We have J10=Q±σ3γ3p4+3(a22)2p2q29a2p3q+3a2pq3+(6p2+(a22)q26apq)|w|2,J11=Q±σ3γ3q4+3(a22)2p2q23a2p3q+9a2pq3((a22)βp2+6βq2+6βapq)|w|2,J12=Q±σ3[γ(p2apqq2)22(β+1)(p2apqq2)]|w|2. Hence, we have k=1012Jk=Q±σ3[γ2(p2apqq2)2+(4a2β)p2+(a24β)q24(β+1)apq]|w|2. Adding this equality to (EquationA4) yields (A5) k=112Jk+2μk=14Ik+B=(P+w,Pw)L2(Q±)+2(P+w,μσw)L2(Q±)+BCQ±σ(|tw|2+|xw|2)+Q±[σ3γ2(p2apqq2)2+σ3{(4a2β)p2+(a24β)q24(β+1)apq+2μ(p2apqq2)}+O(sγ4ϕ)+O(sγ3ϕ)]|w|2.(A5) We consider the coefficient of σ3. Noting μ=β+1, we have (4a2β)p2+(a24β)q24(β+1)apq+2μ(p2apqq2)=(4a2β)p2+(a24β)q24(β+1)apq+(2μ2ν)(p2apqq2)+2ν(p2apqq2)2(μν)(p2apqq2)+(2ν+4a2βδa|ν+2+2β|)p2+2ν+a24βaδ|ν+2+2β|q2, for all δ>0 to be determined later. We wish to choose δ>0 such that (A6) 2ν+4a2βδa|ν+2+2β|>0,2ν+a24βaδ|ν+2+2β|>0.(A6) Such δ exists only if ν satisfies ν2+2(β+1)ν+β(a2+4)<0(ν+β+1)2<β2(a2+2)β+1. Therefore, if we take ν=μ=β1, both the above inequality and (EquationA6) hold. Hence, there exists constant a C>0 such that (4a2β)p2+(a24β)q24(β+1)apq+2μ(p2apqq2)4(1+β)(p2apqq2)+C(p2+q2). We apply this estimate to (EquationA5) to obtain k=112Jk+2μk=14Ik+BCQ±σ(|tw|2+|xw|2)+Q±[σ3γ2(p2apqq2)2+σ3{4(β+1)(p2apqq2)+C(p2+q2)}+O(sγ4ϕ)+O(sγ3ϕ)]|w|2CQ±σ(|tw|2+|xw|2)+Q±σ32γ(p2apqq2)+β+1γ2+σ3C(p2+q2)2(β+1)2γ(p2apqq2)+β+1γ2+O(sγ4ϕ)+O(sγ3ϕ)|w|2CQ±σ(|tw|2+|xw|2)+Cminxp(x)22(β+1)2γQ±[σ3+O(sγ4ϕ)+O(sγ3ϕ)]|w|2.CQ±σ(|tw|2+|xw|2)+CQ±[σ3+O(sγ4ϕ)+O(sγ3ϕ)]|w|2, for sufficiently large γ>0.

The Cauchy–Schwarz inequality then yields CQ±(σ|tw|2+σ|xw|2+σ3|w|2)+Q±[O(sγ4ϕ)+O(sγ3ϕ)]|w|212PwL2(Q±)2+Q±O(σ2)|w|2+B. We can choose s>0 large enough to ensure Q±(σ|tw|2+σ|xw|2+σ3|w|2)CPwL2(Q±)2+CB. We define w=esϕu and thus obtain tw=esϕ(tu+stϕu) and xw=esϕ(xu+sxϕu). Again by choosing s>0 large, we can rewrite the inequality in terms of u=wesϕ: Q±e2sϕ(σ|xu|2+σ|tu|2+σ3|u|2)CQ±e2sϕ|Au|2+CB. Here, we get B=TTσpa2q|xw|2dtCTTe2sϕ(,t)σ(,t)|xu(,t)|2+e2sϕ(,t)σ(,t)|xu(,t)|2dt.