The stability result for the inverse problem is based on a global Carleman estimate for a plate equation

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

GLOBAL CARLEMAN ESTIMATE FOR THE PLATE EQUATION AND APPLICATIONS TO INVERSE PROBLEMS

PENG GAO

Abstract. In this article, we establish a Carleman estimate for the plate equation in order to solve an inverse problem retrieving the zeroth-order term for a plate equation from boundary measurements. We prove the local stability result for this inverse problem. Our proof relies on Carleman estimate.

1. Introduction

In this article, we discuss the local Lipschitz stability in determining a coefficient of the zeroth-order term for a plate equation from boundary measurements. Phys- ically speaking, we are required to determine a coefficientp(x) from measurements of boundary displacement. The stability result for the inverse problem is based on a global Carleman estimate for a plate equation.

Inverse Problem: Determinep(x) forx∈I such that u_tt+u_xxxx+pu= 0 in Q, u(0, t) = 0 =u(1, t) in (0, T), ux(0, t) = 0 =ux(1, t) in (0, T),

u(x,0) =a(x) in I, ut(x,0) =b(x) in I,

(1.1)

from the observed datau|_{1}×(0,T), where I= (0,1), T >0 andQ=I×(0, T).

The inverse problem for the wave equation has drawn the attention of many authors. In [22] it is discussed the global Lipschitz stability in determining a co- efficient of the zeroth-order term for a wave equation from data of the solution in a sub-domain over a time interval. In [2], uniqueness and Lipschitz stability are obtained for the inverse problem of retrieving a stationary potential for the wave equation with Dirichlet data and discontinuous principal coefficient from a single time dependent Neumann boundary measurement. Doubonva citeD1 solved an inverse problem of retrieving a stationary potential for the wave equation with Dirichlet data from a single time-dependent Neumann boundary measurement on a suitable part of the boundary. The uniqueness and the stability are also proved for this problem when a Neumann measurement is only located on a part of the boundary satisfying a rotated exit condition. Bellassoued [4] studied the global

2010Mathematics Subject Classification. 35R30, 35G15.

Key words and phrases. Carleman estimate; plate equation; inverse problem.

c

2016 Texas State University.

Submitted November 12, 2016. Published December 28, 2016.

1

logarithmic stability in determination of a coefficient of the zeroth-order term in a wave equation from data of the solution in a subboundary over a time interval. The similar inverse problems for parabolic equations can be found in the survey paper by Yamamoto [35] and the references therein. The inverse problem for dispersive equations is also interesting and was studied in [3, 5, 29] for Schr¨odinger equations.

Baudouin [1] studied the same problem for the Korteweg-de Vries equation. To our best knowledge, very little work is concerned with the inverse problem for plate equations, our work is motivated by [22]. The key ingredient we follow here to determine the coefficient relies on a Carleman estimate for the plate operator.

In this article we use the following assumptions:

(H1) x0<0,β∈(0,1), T >sup_x∈I|x−x0|.

We define the functions

ψ(x, t) = (x−x0)²−βt², ϕ(x, t) =e^µψ(x,t), l(x, t) =λϕ(x, t), θ=e^l(x,t).

Let

L^∞_M(I) ={p∈L^∞(I)|kpk_L^∞_(I)≤M}.

LetP be the operator P y:=ytt+yxxxx+py with domain U :=n

y ∈C¹(−T, T;H²(I))∩C²(−T, T;L²(I))∩C(−T, T;H⁴(I)) : y(t,0) =y(t,1) =yx(t,0) =yx(t,1) = 0, t∈(−T, T),

y(±T, x) =y_t(±T, x) = 0, x∈I, P y∈L²(−T, T;L²(I))o ,

wherep∈L^∞(I). Throughout this paper,C stands for a generic positive constant whose value can change from line to line.

(H2) a∈H₀²(I) satisfies|a(x)| ≥r₀>0 for allx∈I,b∈L²(I).

The main result in this paper is the following local stability for the inverse problem.

Theorem 1.1. Let assumptions(H1), (H2) be satisfied. Then there exists a con- stantC^?=C^?(T, x0, a, b, M)>0 such that for all p, q∈L^∞_M(I), we have

kp−qk_L2(I)≤C^? Z T

0

[(u(p)−u(q))²_xxt+ (u(p)−u(q))²_xxxt](1, t)dt (1.2) whereu(p), u(q)are the solutions of (1.1)depending onp, q.

Remark 1.2. Stability estimates play a special role in the theory of inverse prob- lems of mathematical physics that are ill posed in the classical sense. They de- termine the choice of regularization parameters and the rate at which solutions of regularized problems converge to an exact solution.

Remark 1.3. From Theorem 1.1, we can obtain the mapping p→u(p)|_{1}×(0,T)

is one to one, and the mappingu(p)|_{1}×(0,T)→pis continuous.

To obtain the stability of (1.2), the following Carleman estimate is essential.

Theorem 1.4. There exist three constants µ0 >1, λ0 >0 and C1 >0 such that forµ=µ0 and for everyλ≥λ0 andy∈ U, we have

Z T

−T

Z

I

(λµ²ϕθ²y_xxx² +λ³µ⁴ϕ³θ²y²_xx+λ⁵µ⁶ϕ⁵θ²y_x²+λ⁷µ⁸ϕ⁷θ²y² +λ³µ⁴ϕ³θ²y_t²+λµ²ϕθ²y_xt²)dx dt

≤C1[ Z T

−T

Z

I

θ²|P y|²dx dt+ Z T

0

(λ³µ³ϕ³θ²y_xx² +λµϕθ²y²_xxx)(1, t)dt].

(1.3)

Carleman estimate is anL²-weighted estimate with large parameter for a solution to a PDE and it is one of the major tools used in the study of unique continuation, observability, and controllability problems for various kinds of PDEs. Its history may date back to Carleman [6] for a two-dimensional elliptic equation. Then, many authors have considered this estimate such as Egorov [9], H¨ormander [19, 20], Isakov [24, 25, 26], Tataru [31], Taylor [32] and Tr`eves [33]. There have already been rich amounts of work for the Carleman estimates of second order parabolic equations, see [11, 23, 35]. Global Carleman estimate for fourth order parabolic equation we established in[12, 13, 14, 15, 18, 36]. A similar estimate for the hyperbolic equation can be found in [10, 21]. In [17, 30] there is a Carleman estimate for the KdV equation without the interior observation. Then a global Carleman estimate for the KdV equation was established in [7]. In [16] there are global Carleman estimates for forward stochastic fourth order parabolic equation and backward stochastic fourth order parabolic equation.

The remainder of this article is organized as follows. In Section 2, we establish the Carleman estimate (1.3), Section 3 we prove Theorem 1.1.

2. Proof of Theorem 1.4

As in [28], it is sufficient to prove (1.3) forP ye =ytt+yxxxx withy∈ U. In fact, assume that we have proved (1.3) forP y, we havee

Z

Q

θ²|P y|e ²dx dt≤ Z

Q

θ²|P y|²dx dt+kpk²_L∞(I)

Z

Q

θ²y²dx dt.

By choosing λ0 = λ0(µ, T) > 0 large, when λ > λ0, it is possible to absorb kpk²_L∞(I)

R

Qθ²y²dx dt with the left-hand side of (1.3), concluding that (1.3) also holds forP y.

Setu=θy,P ye =f. Direct computations show that

θ(y_tt+y_xxxx) =u_tt+A₀u_t+A₁u+A₂u_x+A₃u_xx+A₄u_xxx+u_xxxx where

A₀=−2l_t, A₁=l²_t−l_tt+l⁴_x+ 4l_xl_xxx−l_xxxx−6l²_xl_xx+ 3l²_xx, A₂= 12l_xl_xx−4l_x³−4l_xxx, A₃= 6l²_x−6l_xx, A₄=−4lx. Set

I₁=u_xxxx+B₁u+B₃u_xx+u_tt, I2=B2ux+B4uxxx+au+B0ut, Ru=θf−I₁−I₂=S₀u+S₁u_x+S₂u_xx,

where

a=−12l_x²l_xx, B₀=−2lt, B₁=l⁴_x, B₂=−4l³_x, B₃= 6l²_x, B₄=−4lx, S₀=l²_t−l_tt+ 4l_xl_xxx−l_xxxx+ 6l_x²l_xx+ 3l²_xx,

S1= 12lxlxx−4lxxx, S2=−6lxx. Step 1. We shall prove the equality

I1·I2=u²{· · · }+u²_x{· · · }+u²_xx{· · · }+u²_xxx{· · · }+u²_t{· · · } +u²_xt{· · · }+uxtuxxx{· · · }+utux{· · · }+utuxxx{· · · }

+{· · · }x+{· · · }xx+{· · · }xxx+{· · · }xxxx+{· · · }t+{· · · }tt,

(2.1)

where {· · · }x={3

2B2xxu²_x−3

2B2u²_xx+1

2B4u²_xxx+ 4axu²_x−2axxxu² +B0utuxxx+1

2B1B2u²+3

2(B1B4)xxu²−3

2B1B4u²_x+1

2B2B3u²_x +1

2B3B4u²_xx−(aB3)xu²+B0B3utux−1

2B2u²_t−3

2B4xxu²_t+3

2B4u²_xt}x, {· · · }xx={−3

2B2xu²_x+ 3axxu²−2au²_x−3

2(B1B4)xu²+1

2aB3u²+3

2B4xu²_t}xx, {· · · }xxx ={1

2B2u²_x−2axu²+1

2B1B4u²−1

2B4u²_t}xxx, {· · · }xxxx={1

2au²}xxxx, {· · · }t={1

2B₀B₁u²−1

2B₀B₃u²_x+B₂u_tu_x+1

2B₀u²_t−a_tu²+B₄u_tu_xxx}t, {· · · }tt={1

2au²}tt, u²{· · · }=u²{1

2a_xxxx−1

2(B₁B₂)_x−1

2(B₁B₄)_xxx+aB₁−1

2(B₀B₁)_t +1

2(aB₃)_xx+1 2a_tt}, u²_x{· · · }=u²_x{−1

2B_2xxx−2a_xx+3

2(B₁B₄)_x−1

2(B₂B₃)_x−aB₃+1

2(B₀B₃)_t}, u²_xx{· · · }=u²_xx{3

2B_2x+a−1

2(B₃B₄)_x}, u²_xxx{· · · }=u²_xxx{−1

2B_4x}, u²_t{· · · }=u²_t{1

2B_2x−1

2B_0t−a+1

2B_4xxx}, u²_xt{· · · }=u²_xt{−3

2B_4x}, uxtuxxx{· · · }=uxtuxxx{−B0}, utux{· · · }=utux{−(B0B3)x−B2t},

u_tu_xxx{· · · }=u_tu_xxx{−B_4t−B_0x}.

Indeed, (2.1) can be obtained from the following equations uxxxxB2ux= 1

2(B2u²_x)xxx−3

2(B2xu²_x)xx+3

2(B2xxu²_x−B2u²_xx)x

+3

2B2xu²_xx−1

2B2xxxu²_x, uxxxxB4uxxx =1

2[(B4u²_xxx)x−B4xu²_xxx], uxxxxau= 1

2(au²)xxxx−2(axu²)xxx+ (3axxu²−2au²_x)xx

+ (4axu²_x−2axxxu²)x+au²_xx−2axxu²_x+1

2axxxxu², uxxxxbuxx= 1

2(bu²_xx)xx−(bxu²_xx)x−bu²_xxx+1 2bxxu²_xx, u_xxxxB₀u_t= (B₀u_tu_xxx)_x−B_0xu_tu_xxx−B₀u_xtu_xxx,

B₁uB₂u_x= 1

2[(B₁B₂u²)_x−(B₁B₂)_xu²], B1uB4uxxx =1

2(B1B4u²)xxx−3

2[(B1B4)xu²]xx

+3

2[(B1B4)xxu²−B1B4u²_x]x+3

2(B1B4)xu²_x−1

2(B1B4)xxxu², B1ubuxx= 1

2(bB1u²)xx−[(bB1)xu²]x−bB1u²_x+1

2(bB1)xxu², B1uB0ut= 1

2[(B0B1u²)t−(B0B1)tu²], B3uxxB2ux=1

2[(B2B3u²_x)x−(B2B3)xu²_x], B3uxxB4uxxx =1

2[(B3B4u²_xx)x−(B3B4)xu²_xx], B3uxxau= 1

2(aB3u²)xx−[(aB3)xu²]x−aB3u²_x+1

2(aB3)xxu², B3uxxB0ut=−(B0B3)xutux+1

2(B0B3)tu²_x−1

2(B0B3u²_x)t+ (B0B3uxut)x, uttB2ux= (B2utux)t−B2tutux−1

2[(B2u²_t)x−B2xu²_t], uttB0ut= 1

2[(B0u²_t)t−B0tu²_t], uttau= 1

2(au²)tt−(atu²)t−au²_t+1 2attu², uttB4uxxx= (B4utuxxx)t−B4tutuxxx−1

2(B4u²_t)xxx+3

2(B4xu²_t)xx

−3

2[B4xxu²_t−B4u²_xt]x−3

2B4xu²_xt+1

2B4xxxu²_t, uttbuxx= (butuxx)t+bxtutux−1

2bttu²_x+1 2(btu²_x)t

−(btutux)x−1

2(bu²_t)xx+ (bxu²_t)x+bu²_xt−1 2bxxu²_t.

Step 2. We shall prove the estimate Z T

−T

Z

I

(λµ²ϕθ²y_xxx² +λ³µ⁴ϕ³θ²y_xx² +λ⁵µ⁶ϕ⁵θ²y_x²+λ⁷µ⁸ϕ⁷θ²y² +λ³µ⁴ϕ³θ²y²_t+λµ²ϕθ²y²_xt)dx dt

≤C1

hZ T

−T

Z

I

θ²|P y|²dx dt+ Z T

−T

(λ³µ³ϕ³θ²y²_xx+λµϕθ²y_xxx² )(1, t)dti .

Indeed,

u²_xxx{· · · }=u²_xxx{2lxx}, u²_xx{· · · }=u²_xx{6l_x²lxx}, u²_x{· · · }=u²_x{102l⁴_xlxx+r1}, u²{· · · }=u²{2l⁶_xlxx+r2}, u²_t{· · · }=u²_t{6l²_xlxx+ltt−2lxxxx}, u²_xt{· · · }=u²_xt{6lxx}, uxtuxxx{· · · }=uxtuxxx{2lt}, utux{· · · }=utux{24lxtl²_x+ 24ltlxlxx},

u_tu_xxx{· · · }=u_tu_xxx{6l_xt}, where

r1= 60l_xx³ + 180lxlxxlxxx+ 30l²_xlxxxx−6lttl_x²−12ltlxlxt, r2=−6(l²_xlxx)xxxx+ 2(l⁵_x)xxx+ (ltl_x⁴)t−36(l⁴_xlxx)xx−6(l²_xlxx)tt. Now, we estimate the termRT

−T

R

I({· · · }_x+{· · · }_xx+{· · · }_xxx+{· · · }_xxxx+{· · · }_t+ {· · · }tt)dx dtin (2.1). Indeed, noting that

y(0, t) =y(1, t) =yx(0, t) =yx(1, t) = 0 ∀t∈(−T, T), y(x,±T) =yt(x,±T) = 0 ∀x∈I,

this implies

u(0, t) =u(1, t) =u_x(0, t) =u_x(1, t) = 0 ∀t∈(−T, T) u(x,±T) =ut(x,±T) = 0 ∀x∈I.

Thus

Z T

−T

Z

I

({· · · }_t+{· · · }_tt)dx dt= 0, Z T

−T

Z

I

({· · · }x+{· · · }xx+{· · · }xxx+{· · · }xxxx)dx dt

= Z T

−T

{u²_xx(−10l³_x) +u²_xxx(−2lx)}(·, t)|¹₀dt=:V(1)−V(0).

(2.2)

If we chooseλ≥λ0=λ0(µ, T) withλ0 large enough such thatλµ⁻¹ϕ≥1, then it holds

|r1| ≤C(λ³µ⁶ϕ³+λ³µ⁴ϕ³+λ³µ⁵ϕ³)≤Cλ⁵ϕ⁵(µ²+µ⁴+µ³)≤Cλ⁵µ⁵ϕ⁵,

|r2| ≤C(λ³µ⁶ϕ³+λ³µ⁸ϕ³+λ⁵µ⁸ϕ⁵)≤Cλ⁷ϕ⁷(µ²+µ⁴+µ⁶)≤Cλ⁷µ⁷ϕ⁷,

|S0|²≤C(λ²µ²ϕ²+λ²µ⁴ϕ²+λ⁴µ⁸ϕ⁴+λ⁶µ⁸ϕ⁶+λ⁴µ⁴ϕ⁴+λ²µ⁸ϕ²)

≤Cλ⁷ϕ⁷(µ⁻³+µ⁻¹+µ⁵+µ⁷+µ+µ³)

≤Cλ⁷µ⁷ϕ⁷,

|S1|²≤C(λ⁴µ⁶ϕ⁴+λ²µ⁶ϕ²)≤Cλ⁵µ⁵ϕ⁵,

|S2|²≤Cλ²µ⁴ϕ²≤Cλ³µ³ϕ³

(2.3)

and

u²_xt{· · · }+u²_xxx{· · · }+u_xtu_xxx{· · · } ≥C(λµ²ϕu²_xxx+λµ²ϕu²_xt), u²_x{· · · }+u²_t{· · · }+u_tu_x{· · · } ≥C(λ⁵µ⁶ϕ⁵u²_x+λ³µ⁴ϕ³u²_t), u²_xxx{· · · }+u²_t{· · · }+u_tu_xxx{· · · } ≥C(λµ²ϕu²_xxx+λ³µ⁴ϕ³u²_t).

(2.4)

SinceI1+I2=θf−S0u−S1ux−S2uxx, it is clear that 2

Z T

−T

Z

I

I₁I₂ dx dt≤ kI₁+I₂k²_L2(−T ,T;L²(I))

=kθf+S0u+S1ux+S2uxxk²_L2(Q)

≤C Z T

−T

Z

I

(θ²f²+S₀²u²+S²₁u²_x+S²₂u²_xx)dx dt.

(2.5)

From (2.2)-(2.5), we can obtain Z T

−T

Z

I

[λµ²ϕu²_xxx+λ³µ⁴ϕ³u²_xx+λ⁵µ⁶ϕ⁵u²_x+λ⁷µ⁸ϕ⁷u²+λ³µ⁴ϕ³u²_t +λµ²ϕu²_xt]dx dt+V(1)−V(0)

≤C Z T

−T

Z

I

(θ²f²+λ⁷µ⁷ϕ⁷u²+λ⁵µ⁵ϕ⁵u²_x+λ³µ³ϕ³u²_xx)dx dt.

Noting that−V(0)≥0 and that

|V(1)| ≤C Z T

0

(λ³µ³ϕ³u²_xx+λµϕu²_xxx)(1, t)dt, and choosingλ₀ large enough, whenλ≥λ₀, we conclude that

Z T

−T

Z

I

(λµ²ϕu²_xxx+λ³µ⁴ϕ³u²_xx+λ⁵µ⁶ϕ⁵u²_x+λ⁷µ⁸ϕ⁷u² +λ³µ⁴ϕ³u²_t+λµ²ϕu²_xt)dx dt

≤C[

Z T

−T

Z

I

θ²f²dx dt+ Z T

0

(λ³µ³ϕ³u²_xx+λµϕu²_xxx)(1, t)dt].

Returninguto θy, we can obtain (1.3).

3. Proof of Theorem 1.1

In this section, we use the Bukhgeim-Klibanov method to study the inverse problem. We are now in a position to prove the stability result Theorem 1.1, the proof follows the ideas used in [22, 34]. The idea is to reduce the nonlinear inverse problem to some perturbed inverse problem which will be solved with the help of a global Carleman estimate. Firstly, we consider the system

ytt+yxxxx+py=G inQ, y(0, t) = 0 =y(1, t) in (0, T), y_x(0, t) = 0 =y_x(1, t) in (0, T),

y(x,0) = 0 inI, yt(x,0) = 0 inI .

(3.1)

Proposition 3.1. Let assumptions (H1), (H2) be satisfied. Assume that there exists a function g₀∈L²(0, T) such that

G∈H¹(0, T;L²(I)), kG_tk_L2(Q)≤M,

g0(t)|G(x,0)| ≥ |Gt(x, t)|, (x, t)∈Q. (3.2) Then there exists a constant C₁=C₁(M, T, r₀)such that

kGk_H1(0,T;L²(I))≤C₁ Z T

0

(y²_xxt+y_xxxt² )(1, t)dt. (3.3) Proof. Settingy1=yt, we have

y1tt+y1xxxx+py1=Gt in Q, y₁(0, t) = 0 =y₁(1, t) in (0, T), y1x(0, t) = 0 =y1x(1, t) in (0, T),

y1(x,0) = 0 inI, y_1t(x,0) =G(x,0) inI . Sincey(x,0) =yt(x,0) = 0, from [27] we have

y∈C([0, T], H⁴(I))∩C¹(0, T;H²(I))∩C²(0, T;L²(I)) and there exists a constantC=C(T, M)>0 such that

kykH²(I×(0,T))≤CkGkH¹(0,T;L²(I)).

We extend the functiony fromI×(0, T) by the formulay(x, t) =y(x,−t),(x, t)∈ I ×(−T,0) and denote the extension by the same symbol y. We know y ∈ C([−T, T], H⁴(I))∩C¹([−T, T];H²(I))∩C²([−T, T];L²(I)) and there exists a con- stantC=C(T, M)>0 such that

kyk_H2(I×(−T ,T))≤CkGk_H1(0,T;L²(I)).

We extend the functionGtonI×(−T, T) as the even function intand denote the extension by the same symbolGt. ThenGt∈L²(−T, T;L²(I)).

By assumption (H1), there exists aβ ∈(0,1) such thatβ >sup_x∈I|x−x0|/T². Therefore, by definition ofψ, ϕ, we have

ϕ(x,0)≥1, ϕ(x,−T) =ϕ(x, T)<1, x∈[0,1].

Therefore for givenε >0, we can choose a sufficiently smallδ=δ(ε)>0, such that ϕ(x, t)≥1−ε, (x, t)∈[0,1]×[−δ, δ],

ϕ(x, t)≤1−2ε, (x, t)∈[0,1]×([−T,−T + 2δ]∪[T−2δ, T]).

To apply Theorem 1.4, we introduce a cut-off function χ satisfying 0 ≤ χ ≤ 1, χ∈C^∞(R), and

χ(t) =

(0, [−T,−T +δ]∪[T−δ, T] 1, [−T+ 2δ, T−2δ].

We set

u=e^λϕχyt, w=χyt, D(y) = Z T

0

(y_xxt² +y²_xxxt)(1, t)dt.

Direct computations show that

P u=u(λϕtt−λ²ϕ²_t−λ⁴ϕ⁴_x+ 6λ³ϕ²_xϕxx−3λ²ϕ²_xx+λϕxxxx

−4λ²ϕxϕxxx) + 2λϕtut+ux(4λ³ϕ³_x−12λ²ϕxϕxx+ 4λϕxxx)

+uxx(−6λ²ϕ²_x+ 6λϕxx) +uxxx·4λϕx+e^λϕχttyt+ 2e^λϕχtytt+e^λϕχGt. MultiplyingP u byutand integrating it overI×(−T,0), it follows that

Z 0

−T

Z

I

P u·u_tdx dt=1 2

Z

I

u_t(x,0)²dx+1 2

Z

I

u²_xx(0, x)dx

≥1 2

Z

I

u_t(x,0)²dx= 1 2 Z

I

G(x,0)²e^2λϕ(x,0)dx.

(3.4)

On the other hand, by the Cauchy inequality, we have Z 0

−T

Z

I

P u·utdx dt

≤C[

Z 0

−T

Z

I

(G²_tχ²θ²+θ²χ²_tty_t²+θ²χ²_ty_tt²)dx dt +

Z 0

−T

Z

I

(λ⁷µ⁸ϕ⁷u²+λ⁵µ⁶ϕ⁵u²_x+λ³µ⁴ϕ³u²_xx+λµ²ϕu²_xxx +λ³µ⁴ϕ³u²_t+λµ²ϕu²_xt)dx dt] =:J1+J2.

It is easy to see that J₁≤ChZ T

−T

Z

I

G²_tχ²θ²dx dt+Z −T+2δ

−T+δ

+ Z T−δ

T−2δ

(y²_t+y²_tt)θ²dx dti

≤ChZ T

−T

Z

I

G²_tχ²θ²dx dt+e^2λ(1−2ε)kyk²_H2(I×(−T ,T))

i

≤ChZ T

−T

Z

I

G²_tχ²θ²dx dt+e^2λ(1−2ε)kGk²_H1(0,T;L²(I))

i . NotingP w=χG_t+ 2y_ttχ_t+χ_tty_t, from Theorem 1.4 it follows that

Z T

−T

Z

I

(λµ²ϕθ²w²_xxx+λ³µ⁴ϕ³θ²w²_xx+λ⁵µ⁶ϕ⁵θ²w²_x+λ⁷µ⁸ϕ⁷θ²w² +λ³µ⁴ϕ³θ²w_t²+λµ²ϕθ²w²_xt)dx dt

≤C₁hZ T

−T

Z

I

θ²|P w|²dx dt+ Z T

0

(λ³µ³ϕ³θ²w²_xx+λµϕθ²w²_xxx)(1, t)dti

≤C[

Z T

−T

Z

I

θ²|χGt+ 2yttχt+χttyt|²dx dt +

Z T

0

(λ³µ³ϕ³θ²y_xxt² +λµϕθ²y_xxxt² )(1, t)dt]

≤ChZ T

−T

Z

I

G²_tθ²dx dt+e^2λ(1−2ε)kGk²_H1(0,T;L²(I))+D(y)i

; thus

J2≤ Z 0

−T

Z

I

(λ⁷µ⁸ϕ⁷u²+λ⁵µ⁶ϕ⁵u²_x+λ³µ⁴ϕ³u²_xx+λµ²ϕu²_xxx +λ³µ⁴ϕ³u²_t+λµ²ϕu²_xt)dx dt

≤ Z T

−T

Z

I

(λ⁷µ⁸ϕ⁷u²+λ⁵µ⁶ϕ⁵u²_x+λ³µ⁴ϕ³u²_xx+λµ²ϕu²_xxx +λ³µ⁴ϕ³u²_t+λµ²ϕu²_xt)dx dt

≤C Z T

−T

Z

I

(λµ²ϕθ²w_xxx² +λ³µ⁴ϕ³θ²w²_xx+λ⁵µ⁶ϕ⁵θ²w²_x+λ⁷µ⁸ϕ⁷θ²w² +λ³µ⁴ϕ³θ²w_t²+λµ²ϕθ²w²_xt)dx dt

≤ChZ T

−T

Z

I

G²_tθ²dx dt+e^2λ(1−2ε)kGk²_H1(0,T;L²(I))+D(y)i . From the above estimates, we know that

Z 0

−T

Z

I

P u·utdx dt

≤ChZ T

−T

Z

I

G²_tθ²dx dt+e^2λ(1−2ε)kGk²_H1(0,T;L²(I))+D(y)i .

(3.5)

Consequently, (3.4) and (3.5) imply 1

2 Z

I

G(x,0)²e^2λϕ(x,0)dx

≤ChZ T

−T

Z

I

G²_tθ²dx dt+e^2λ(1−2ε)kGk²_H1(0,T;L²(I))+D(y)i .

(3.6)

Next we estimate the first term of the right-hand side of (3.6).

Z T

−T

Z

I

G²_tθ²dx dt

= Z

I

Z T

−T

|G_t(x, t)|²e^{2λeµψ(x,t)}dt dx

≤ Z

I

Z T

−T

|g0(t)G(x,0)|²e^{2λeµ[(x−x0 )}

2−βt2 ]

dt dx

= Z

I

Z T

−T

|g0(t)|²e^{2λeµ[(x−x0 )}

2−βt2 ]

e^{−2λϕ(x,0)}dt|G(x,0)|²e^2λϕ(x,0) dx

= Z

I

Z T

−T

|g0(t)|²e^{2λeµ[(x−x0 )}

2−βt2 ]

e^{−2λeµ(x−x0 )}

2

dt|G(x,0)|²e^2λϕ(x,0) dx

= Z

I

Z T

−T

|g0(t)|²e^{2λeµ(x−x0 )}

2(e^−βt2−1)dt|G(x,0)|²e^2λϕ(x,0) dx

≤ Z

I

Z T

−T

|g0(t)|²e^2λ(e−βt

2−1)dt|G(x,0)|²e^2λϕ(x,0) dx

=:

Z

I

( Z T

−T

h_λ(t)dt|G(x,0)|²e^2λϕ(x,0))dx.

We have thathλ(t) is in L¹(−T, T), and lim_λ→∞hλ(t) = 0 fort6= 0 and|hλ(t)| ≤

|g0|²∈L¹(−T, T). Hence the Lebesgue theorem implies Z T

−T

|g0(t)|²e^2λ(e−βt

2−1)dt=o(1) asλ→ ∞, so that

Z T

−T

Z

I

G²_tθ²dx dt=o(1) Z

I

G(x,0)²e^2λϕ(x,0)dx. (3.7) By the assumption in (3.2), we have

kGk²_H1(0,T;L²(I))≤C Z

I

G(x,0)²dx. (3.8)

Therefore (3.2), (3.6) and (3.8) yieldkGk²_H1(0,T;L²(I)) ≤CD(y).

Now we can prove Theorem 1.1. In Proposition 3.1, we sety=u(p)−u(q) and G(x, t) = (p−q)u(q). Recalling that H¹(0, T;L²(I))⊂C([0, T];L²(I)), it is easy to see that assumption (3.2) holds, and that (3.3) implies

kp−qk_L2(I)≤Ck(p−q)(x)u(q)(0, x)k_L2(I)

≤Ck(p−q)u(q)k_C([0,T];L2(I))

≤Ck(p−q)u(q)kH¹(0,T;L²(I))

=CkGk_H1(0,T;L²(I))

≤C Z T

0

(y²_xxt+y_xxxt² )(1, t)dt

≤C Z T

0

[(u(p)−u(q))²_xxt+ (u(p)−u(q))²_xxxt](1, t)dt, this completes the proof.

Acknowledgements. This research is supported by NSFC Grant (11601073). I am grateful to the anonymous referee and editor for their very helpful comments and suggestions. I sincerely thank Professor Yong Li for many useful suggestions and help.

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Peng Gao

School of Mathematics and Statistics, and Center for Mathematics and Interdisci- plinary Sciences, Northeast Normal University, Changchun 130024, China

E-mail address:[email protected]