Internal controllability of the Kawahara equation on a bounded domain

Nonlinear Analysis 185 (2019) 356–373

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Nonlinear Analysis www.elsevier.com/locate/na

Internal controllability of the Kawahara equation on a bounded domain Mo Chen School of Mathematics and Statistics, Center for Mathematics and Interdisciplinary Sciences, Northeast Normal University, Changchun, 130024, PR China

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Article history: Received 9 May 2018 Accepted 22 March 2019 Communicated by Enzo Mitidieri MSC: 35Q53 93B05

abstract In this paper, we consider the Kawahara equation posed on a bounded interval with a distributed control. First, we establish a Carleman estimate for the Kawahara equation with internal observation. Then, applying this Carleman estimate, we can show that the Kawahara equation is null controllable. Furthermore, we prove the null controllability with constraints on the state for the Kawahara equation. © 2019 Elsevier Ltd. All rights reserved.

Keywords: Kawahara equation Null controllability Carleman estimate

1. Introduction The Kawahara equation yt + αyxxx + βy5x + γyyx = 0, where α, β and γ are real constants (β, γ ̸= 0), was first proposed by Kawahara [20] in 1972 as a model equation describing one-dimensional propagation of small-amplitude long waves in various problems of fluid dynamics and plasma physics. This equation can also describe the oscillatory tail of the wave and help us understand the effect of surface tension on solitary water waves [17]. On the other hand, the equation is known as a fifth-order KdV type equation as it becomes the well-known Korteweg–de Vries (KdV) equation yt + αyxxx + γyyx = 0, when β = 0. The KdV equation is a very useful approximation model in nonlinear studies whenever one wishes to include and balance a weak nonlinearity and weak dispersive effects. In particular, the equation is now commonly accepted as a mathematical model for the unidirectional propagation of small amplitude long waves in nonlinear dispersive systems. However, Kawahara pointed out that under certain conditions, E-mail address: [email protected]. https://doi.org/10.1016/j.na.2019.03.016 0362-546X/© 2019 Elsevier Ltd. All rights reserved.

M. Chen / Nonlinear Analysis 185 (2019) 356–373

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the coefficient α will become very small or even vanish. It is then necessary to take into account the higher order effect of dispersion in order to balance the nonlinear effect. The Kawahara equation was developed while the coefficient α of the third derivative in the KdV equation is close to zero. The Kawahara equation has been investigated widely on real line ([6,18]), half-line ([7,22]) and periodic domain [16,19]. However, if one is interested in implementing a numerical scheme to calculate solutions in these regions, there arises the issue of cutting off the spatial domain. In this occasion some boundary conditions are needed to specify the solution. Therefore, precise mathematical analysis of boundary value problems in bounded domains for the Kawahara-type equations is welcome and attracts attention of specialists in this area [2,7–10,13,21,22]. In this paper, we investigate the Kawahara equation from control point of view. Without loss of generality, we choose α = β = γ = 1 and consider the following system: ⎧ ⎪ in I × (0, T ), ⎨yt + yxxx + y5x + yyx = f χω (1.1) y(0, t) = y(L, t) = yx (0, t) = yx (L, t) = yxx (0, t) = 0 in (0, T ), ⎪ ⎩ y(x, 0) = y0 (x) in I, where T, L > 0, I = (0, L), Q = I × (0, T ) and ω ⊂ I is a nonempty open subset. y0 ∈ L2 (I), the control f belongs to L2 (Q) and χω represents the characteristic function of the control set ω. Remark 1.1. The results in this paper also holds if α, β and γ are real constants and β, γ ̸= 0. The only difference is that the role of the left and right boundaries should be converted for β < 0. The controllability of the KdV equation has drained much attention (see for instance [14,29] for boundary controllability and [1,23] for internal controllability). The strong physical background of the Kawahara equation and such similarities and differences between it and the KdV equation in both the form and the behavior of the solution render the controllability of the Kawahara equation particularly interesting. In [30,31], the authors considered the controllability and stabilization of the Kawahara equation on a periodic domain. [15] studied the boundary controllability of the fifth-order KdV equation on a bounded domain. As far as we know, there is no result considering internal controllability of the Kawahara equation on a bounded domain. The first result in this paper is the null controllability of (1.1). Theorem 1.1.

There exists a constant δ1 > 0 such that for any y0 ∈ L2 (I) satisfying ∥y0 ∥L2 (I) ≤ δ1 ,

one can find a control f ∈ L2 (Q) such that the solution y of (1.1) satisfies y(·, T ) = 0 in I.

(1.2)

Moreover, the control can be chosen such that ∥f ∥L2 (Q) ≤ C∥y0 ∥L2 (I) , where C = C(L, T, ω) is a positive constant. Based on the result in Theorem 1.1, we are interested in the controllability with some constraints for (1.1). The controllability problem with constraints on state or control was first proposed by Lions [24] while using sentinels method to identifying parameters in incomplete data problems. Nowadays, this kind of control problem has been extended to various kinds of partial differential equations:

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• • • • •

heat equation [25–28], 1-D Kuramoto–Sivashinsky equation [11], KdV equation [3,4], reaction–diffusion system [12], stochastic heat equation [5].

Motivated by these results, it is natural to ask Does the null controllability with constraints on the state hold for the Kawahara equation? The second result in this paper gives an affirmative answer to this question. Let E = Span(e1 , . . . , eM ) be the subspace of L2 (Q) generated by the functions ei ∈ L2 (Q), 1 ≤ i ≤ M . Assume that the functions ei , 1 ≤ i ≤ M , are such that ei χω , 1 ≤ i ≤ M, are linearly independent.

(1.3)

Theorem 1.2. For every ei ∈ L2 (Q), 1 ≤ i ≤ M , verifying (1.3), there exists a constant δ2 > 0 such that for any y0 ∈ L2 (I) satisfying ∥y0 ∥L2 (I) ≤ δ2 , one can find a control f ∈ L2 (Q) such that the solution y of (1.1) satisfies (1.2) and ∫ yei dxdt = 0, 1 ≤ i ≤ M.

(1.4)

Q

Moreover, the control can be chosen such that ∥f ∥L2 (Q) ≤ C∥y0 ∥L2 (I) , where C = C(L, T, ω,

M ∑

∥ei ∥2L2 (Q) ) is a positive constant.

i=1

The rest of this paper has the following structure. In Section 2, we introduce some notations and present several preliminary results which will be used in the proof of the main theorems. Section 3 is devoted to a Carleman estimate for the Kawahara equation with internal observation. Theorems 1.1 and 1.2 are proved in Section 4. 2. Notations and preliminaries Throughout this paper, unless otherwise stated, C stands for a generic positive constant whose value can change from line to line. If it is essential, the dependence of a constant C on some parameters, say “ · ”, will be written by C(·). To prove Theorem 1.1, consideration is first given to the linearized system ⎧ ⎪ in I × (0, T ), ⎨yt + yxxx + y5x + (zy)x = g (2.1) y(0, t) = y(L, t) = yx (0, t) = yx (L, t) = yxx (0, t) = 0 in (0, T ), ⎪ ⎩ y(x, T ) = y0 (x) in I. The well-posedness of (2.1) is established in [15]. Proposition 2.1. Let y0 ∈ L2 (I), z ∈ C([0, T ]; L2 (I)) ∩ L2 (0, T ; H 2 (I)) and g ∈ L2 (0, T ; H −2 (I)) ∪ L1 (0, T ; L2 (I)). Then (2.1) admits a unique solution y ∈ C([0, T ]; L2 (I)) ∩ L2 (0, T ; H 2 (I)).

M. Chen / Nonlinear Analysis 185 (2019) 356–373

In the next section, we will use the regularity result for the following system: ⎧ ⎪ in I × (0, T ), ⎨ut + uxxx + u5x = g u(0, t) = u(L, t) = ux (0, t) = ux (L, t) = uxx (L, t) = 0 in (0, T ), ⎪ ⎩ u(x, T ) = 0 in I.

359

(2.2)

For reasons of simplicity, we define the spaces, for s ∈ [0, 5], { } Xs := u | u ∈ C([0, T ]; H s (I)) ∩ L2 (0, T ; H s+2 (I)) endowed with their natural norms. Proposition 2.2.

Let g ∈ L2 (0, T ; H0s−2 (I)) for s ∈ [0, 5], then the solution u of (2.2) satisfies ∥u∥Xs ≤ C∥g∥L2 (0,T ;H s−2 (I)) .

(2.3)

Proof . For s = 0. We multiply (2.2) by −(x + 1)u and integrate with respect to x over I. An integration by parts leads to ∫ ∫ 5 1 1 d 2 (x + 1)u dx + u2xx dx + u2xx (0, t) − 2 dt I 2 I 2 ∫ ∫ 3 2 = u dx − (x + 1)ugdx 2 I x I∫ ∫ 2 ≤2 uxx dx + C (x + 1)u2 dx + C∥g∥2H −2 (I) , I

I

where we used the inequality that ∫ ∫ ∫ u2x dx ≤ ε u2xx dx + C(ε) u2 dx, I

This implies −

d dt

∫

I

(x + 1)u2 dx +

for any ε > 0.

I

∫

I

u2xx dx ≤ C0

∫

I

I

(x + 1)u2 dx + C∥g∥2H −2 (I) .

Multiplying eC0 t on both sides of the above inequality and integrating with respect to t over (t, T ), we can deduce that ∫ ∫ T ∫ ∫ T C0 t 2 C0 s 2 e (x + 1)u dx + e uxx dxds ≤ C eC0 s ∥g∥2H −2 (I) ds ≤ C∥g∥2L2 (0,T ;H −2 (I)) . I

t

I

t

Take the supremum in t, we obtain (2.3) for s = 0. For s = 5, we apply the operator P = ∂x3 + ∂x5 to (2.2) to obtain that (P u)t + (P u)xxx + (P u)5x = P g. It follows from (2.2) that P u(0, t) = P u(L, t) = (P u)x (0, t) = (P u)x (L, t) = (P u)xx (L, t) = 0. Then applying the estimate in X0 , we have ∥P u∥X0 ≤ C∥P g∥L2 (0,T ;H −2 (I)) ≤ C∥g∥L2 (0,T ;H 3 (I)) . Combining the above inequality and estimate (2.3) for s = 0, it is not difficult to obtain (2.3) for s = 5. For s ∈ (0, 5), estimate (2.3) is a direct consequence of the estimates in X0 and X5 . □

360

M. Chen / Nonlinear Analysis 185 (2019) 356–373

In order to prove Theorem 1.2, we introduce some notations. Set E = C([0, T ]; L2 (I)) ∩ L2 (0, T ; H 2 (I)) ∩ H 1 (0, T ; H −3 (I)) endowed with its natural norm. We define the set B := {z ∈ E; ∥z∥E ≤ 1}. Considering the following system ⎧ ⎪ in Q, ⎨(pi )t + (pi )xxx + (pi )5x + z(pi )x = ei pi (0, t) = pi (L, t) = (pi )x (0, t) = (pi )x (L, t) = (pi )xx (L, t) = 0 in (0, T ), ⎪ ⎩ pi (x, T ) = 0, in I.

(2.4)

It is not difficult to prove that for z ∈ E, (2.4) admits a unique solution pi ∈ E. Set U = Span(p1 χQω , . . . , pM χQω ), the vector subspace of L2 (Qω ) generated by the M functions pi χQω , 1 ≤ i ≤ M , where Qω = ω × (0, T ). We denote by U ⊥ the orthogonal of U in L2 (Qω ) and Π the orthogonal projection operator from L2 (Qω ) into U. Then we set 1 1 1 Uθ = U = Span( p1 χQω , . . . , pM χQω ), θ θ θ the vector subspace of L2 (Qω ) generated by the M functions θ1 pi χQω , 1 ≤ i ≤ M , where θ is the function defined in Section 4. Following the methods developed in [26] with minor changes, we have the following lemmas. Lemma 2.1.

Assume that (1.3) holds. Let ξ ∈ L2 (0, T ; H 1 (I)) and let qi be the solution of ⎧ ⎪ ⎨(qi )t + (qi )xxx + (qi )5x + ξ(qi )x = ei qi (0, t) = qi (L, t) = (qi )x (0, t) = (qi )x (L, t) = (qi )xx (L, t) = 0 ⎪ ⎩ qi (x, T ) = 0,

in Q, in (0, T ), in I.

(2.5)

We set Uξ = Span(q1 χQω , . . . , qM χQω ). Then, any function w verifying wt + wxxx + w5x + ξwx = 0 in Qω and w|ω ∈ Uξ is identically zero in Qω . Lemma 2.2. Let (H, ∥ · ∥H ) be a Hilbert space. For n ∈ N∗ , let {pni , 1 ≤ i ≤ M } be a family of linearly independent functions and let hn ∈ Span(pn1 , . . . , pnM ). Assume that there exists a family of linear independent functions {qi , 1 ≤ i ≤ M } such that pni → qi ,

strongly in H, 1 ≤ i ≤ M.

Assume also that there exists C > 0, independent of n, such that ∥hn ∥H ≤ C. Then, there exists a subsequence of {hn } still denoted by {hn } such that hn → h ∈ Span(q1 , . . . , qM ),

strongly in H.

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3. Carleman estimate In this section, we intend to obtain a Carleman estimate for the adjoint system (2.1). Note that ω ⊂ I is a nonempty open set, then there exists at least a small interval ω ˜ = (l1 , l2 ) ⊂ ω with 0 < l1 < l2 < L. Pick any function ψ ∈ C 8 ([0, L]) such that ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

ψ > 0 in [0, L]; ψ ′ (0) < 0 and ψ ′ (L) > 0; |ψ ′ | > 0, ψ ′′ < 0 in [0, L]\(l1 , l2 ); min ψ(x) = ψ(l3 ) < max ψ(x) = ψ(l1 ) = ψ(l2 );

x∈[l1 ,l2 ]

x∈[l1 ,l2 ]

max ψ(x) = ψ(0) = ψ(L) and ψ(L) < x∈[0,L]

4 ψ(l3 ) for some l3 ∈ (l1 , l2 ). 3

The existence of such ψ can be proved by a similar argument as in [1]. Set ψ(x) . φ(x, t) = t(T − T ) It is easy to show that C ≤ T φ, C1 φ ≤ |φkx | ≤ C2 φ, k = 1, 2, . . . , 7, 8, |φt | + |φxt | + |φxxt | + |φxxxt | ≤ CT φ2 .

(3.1)

Let Λ denote the operator Λu = ut + u5x with its domain D(Λ) = L2 (0, T ; H 5 (I) ∩ H02 (I)) ∩ H 1 (0, T ; L2 (I)). First, we establish a Carleman estimate for Λu. Proposition 3.1. There exists constant s0 = s0 (T ) > 0 such that for any u ∈ D(Λ) and all s ≥ s0 , one has ∫ (s9 φ9 u2 + s7 φ7 u2x + s5 φ5 u2xx + s3 φ3 u2xxx + sφu24x )e−2sφ dxdt Q (∫ ) ∫ 2 −2sφ 9 9 2 7 7 2 5 5 2 3 3 2 2 −2sφ ≤C |Λu| e dxdt + (s φ u + s φ ux + s φ uxx + s φ uxxx + sφu4x )e dxdt . ω Q Q˜ Proof . Let v = e−sφ u, after some elementary calculations, we have e−sφ Λu =e−sφ (ut + u5x ) =(s5 φ5x + 10s4 φ3x φxx + 10s3 φ2x φxxx + 15s3 φx φ2xx + 10s2 φxx φxxx + 5s2 φx φ4x + sφ5x + sφt )v + (5s4 φ4x + 30s3 φ2x φxx + 20s2 φx φxxx + 15s2 φ2xx + 5sφ4x )vx + (10s3 φ3x + 30s2 φx φxx + 10sφxxx )vxx + (10s2 φ2x + 10sφxx )vxxx + 5sφx v4x + v5x + vt :=I1 + I2 + I3 ,

M. Chen / Nonlinear Analysis 185 (2019) 356–373

362

where I1 =vt + v5x + 10s2 φ2x vxxx + 25s2 φx φxx vxx + 5s4 φ4x vx + 20s4 φ3x φxx v, I2 =5sφx v4x + 10sφxx vxxx + 10s3 φ3x vxx + (30s3 φ2x φxx − 5sφ4x )vx + s5 φ5x v, I3 =(−10s4 φ3x φxx + 10s3 φ2x φxxx + 15s3 φx φ2xx + 10s2 φxx φxxx + 5s2 φx φ4x + sφ5x + sφt )v + (20s2 φx φxxx + 15s2 φ2xx + 10sφ4x )vx + (5s2 φx φxx + 10sφxxx )vxx . It is not difficult to deduce that ∫ 2 I

I1 I2 dx ≤ ∥I1 + I2 ∥2L2 (I) = ∥e−sφ Λu − I3 ∥2L2 (I) ≤ 2∥I3 ∥2L2 (I) + 2∥e−sφ Λu∥2L2 (I) .

The rest of the proof is threefold. Step 1. We shall show that ∫

∫ I1 I2 dx =

2 I

2

({·}v +

{·}vx2

+

2 {·}vxx

+

2 {·}vxxx

+

2 {·}v4x )dx

∫ +

I

{·}t dx + {·}|x=L x=0 ,

(3.2)

I

where {·}v 2 ={−5s9 φ8x φxx + 155s7 (φ6x φxx )xx − 600s7 (φ5x φ2xx )x + 95s5 (φ4x φxx )4x − 200s5 (φ3x φ2xx )xxx − 5s5 φ4x φxt + 100s5 (φ3x φxx φ4x )x }v 2 , {·}vx2 ={−290s7 φ6x φxx − 150s5 (φ3x φ2xx )x + 50s5 φ4x φ4x − 150s5 (φ4x φxx )xx − 50s3 (φ2x φ4x )xx + 30s3 (φ2x φxx )4x + 125s3 (φx φxx φ4x )x + 30s3 φ2x φxt − 5sφ8x − 5sφxxxt }vx2 , 2 {·}vxx ={−150s5 φ4x φxx − 25s3 (φ2x φxx )xx − 250s3 (φx φ2xx )x + 100s3 φ2x φ4x 2 + 20sφ6x − 5sφxt }vxx , 2 2 {·}vxxx ={−50s3 φ2x φxx }vxxx , 2 2 {·}v4x ={−25sφxx }v4x , 2 {·}t ={5sφx vxx + 5sφxxx vx2 − 10s3 φ3x vx2 + s5 φ5x v 2 }t , 2 3 3 2 2 5 5 2 3 2 2 {·}|x=L x=0 ={5sφx v4x + 40s φx vxxx − 10sφxxx vxxx + 76s φx vxx + 250s φx φxx vxx 2 2 − 35s3 (φ2x φxx )x vxx − 10sφ5x vxx + 20sφxx vxxx v4x + 20s3 φ3x vxx v4x x=L + 130s3 φ2x φxx vxx vxxx + 10sφ4x vxx vxxx }|x=0 .

To prove (3.2), we need to calculate each term of ∫ I1 I2 dx.

2 I

Let (Ii )j (i = 1, 2, j = 1, 2, 3, 4, 5, 6) denote the jth term in the expression of Ii . Since u ∈ D(Λ), we have u(0, t) = u(L, t) = ux (0, t) = ux (L, t) = 0,

∀ t ∈ [0, T ].

M. Chen / Nonlinear Analysis 185 (2019) 356–373

Then performing integrations by parts with respect to x, we can obtain that ∫ ∫ 2 (I1 )1 (I2 )1 dx = 10sφx vt v4x dx I ∫I 2 = (5sφx vxx − 5sφxxx vx2 )t dx I ∫ 2 + (20sφxx vtx vxx − 10sφ4x vt vx − 5sφxt vxx + 5sφxxxt vx2 )dx, ∫ ∫ I 2 (I1 )1 (I2 )2 dx = 20sφxx vt vxxx dx I ∫I ∫ = (10sφxxx vx2 )t dx + (20sφ4x vt vx − 10sφxxxt vx2 − 20sφxx vtx vxx )dx, I ∫I ∫ 2 (I1 )1 (I2 )3 dx = 20s3 φ3x vt vxx dx I ∫I ∫ = (−10s3 φ3x vx2 )t dx + (−60s3 φ2x φxx vt vx + 30s3 φ2x φxt vx2 )dx, I ∫ ∫I 3 2 2 (I1 )1 (I2 )4 dx = (60s φx φxx vt vx − 10sφ4x vt vx )dx, ∫I ∫I 2 (I1 )1 (I2 )5 dx = 2s5 φ5x vt vdx I ∫I ∫ = (s5 φ5x v 2 )t dx − 5s5 φ4x φxt v 2 dx; I

∫

I

∫ (I1 )2 (I2 )1 dx =

10sφx v4x v5x dx ( )⏐⏐x=L ∫ 2 2 = 5sφx v4x ⏐⏐ − 5sφxx v4x dx, I x=0 ∫ ∫ 2 (I1 )2 (I2 )2 dx = 20sφxx vxxx v5x dx I I ( )⏐⏐x=L ∫ 2 2 2 ⏐ = 20sφxx vxxx v4x − 10sφxxx vxxx + (10sφ4x vxxx − 20sφxx v4x )dx, ⏐ I x=0 ∫ ∫ 2 (I1 )2 (I2 )3 dx = 20s3 φ3x vxx v5x dx I I [ ]⏐⏐x=L 2 2 ⏐ = 20s3 φ3x vxx v4x − 60s3 φ2x φxx vxx vxxx − 10s3 φ3x vxxx + 30s3 (φ2x φxx )x vxx ⏐ x=0 ∫ [ 3 2 ] 2 3 2 2 + 90s φx φxx vxxx − 30s (φx φxx )xx vxx dx, ∫ ∫ I 2 (I1 )2 (I2 )4 dx = (60s3 φ2x φxx − 10sφ4x )vx v5x dx I I [ ]⏐⏐x=L 2 = (60s3 φ2x φxx − 10sφ4x )x vxx − (60s3 φ2x φxx − 10sφ4x )vxx vxxx ⏐⏐ x=0 ∫ [ ] 3 2 2 3 2 2 + (60s φx φxx − 10sφ4x )vxxx − (120s φx φxx − 20sφ4x )xx vxx dx ∫I + (30s3 φ2x φxx − 5sφ4x )4x vx2 dx, ∫ ∫ I 2 (I1 )2 (I2 )5 dx = 2s5 φ5x vv5x dx I I ( )⏐⏐x=L ∫ [ ] 5 5 2 2 = s φx vxx ⏐⏐ + −25s5 φ4x φxx vxx + 5s5 (φ5x )xxx vx2 − s5 (φ5x )5x v 2 dx; 2

I

I

x=0

I

363

M. Chen / Nonlinear Analysis 185 (2019) 356–373

364

∫

∫ (I1 )3 (I2 )1 dx =

2 I

100s3 φ3x vxxx v4x dx

I

)⏐⏐x=L ∫ ( 2 2 ⏐ − 150s3 φ2x φxx vxxx = 50s3 φ3x vxxx dx, ⏐ I

x=0

∫

∫ (I1 )3 (I2 )2 dx =

2

200s

I

2 φ2x φxx vxxx dx,

I

∫

∫ (I1 )3 (I2 )3 dx =

2

3

I

200s5 φ5x vxx vxxx dx

I

(

5

= 100s

2 φ5x vxx

)⏐⏐x=L ∫ 2 ⏐ − 500s5 φ4x φxx vxx dx, ⏐ I

x=0

∫

∫ (I1 )3 (I2 )4 dx =

2 I

∫ = ∫I

∫ (I1 )3 (I2 )5 dx =

2

(600s5 φ4x φxx − 100s3 φ2x φ4x )vx vxxx dx

I

I

[ ] 2 (300s5 φ4x φxx − 50s3 φ2x φ4x )xx vx2 − (600s5 φ4x φxx − 100s3 φ2x φ4x )vxx dx, 20s7 φ7x vvxxx dx

I

∫ =

[ ] 210s7 φ6x φxx vx2 − 70s7 (φ6x φxx )xx v 2 dx;

I

∫

∫ (I1 )4 (I2 )1 dx =

2 I

250s3 φ2x φxx vxx v4x dx

I

[

3

φ2x φxx vxx vxxx

= 250s

3

− 125s

2 (φ2x φxx )x vxx

]⏐⏐x=L ⏐ ⏐ x=0

∫ + ∫

∫ (I1 )4 (I2 )2 dx =

2 I

[ ] 2 2 125s3 (φ2x φxx )xx vxx − 250s3 φ2x φxx vxxx dx,

I

500s3 φx φ2xx vxx vxxx dx

I

(

3

= 250s

2 φx φ2xx vxx

)⏐⏐x=L ∫ 2 ⏐ − 250s3 (φx φ2xx )x vxx dx, ⏐ I

x=0

∫

∫ (I1 )4 (I2 )3 dx =

2 I

∫

∫ (I1 )4 (I2 )4 dx =

2 I

(1500s5 φ3x φ2xx − 250s3 φx φxx φ4x )vx vxx dx

I

∫ = ∫I

∫ (I1 )4 (I2 )5 dx =

2

2 500s5 φ4x φxx vxx dx,

I

I

∫I =

(−750s5 φ3x φ2xx + 125s3 φx φxx φ4x )x vx2 dx, 50s7 φ6x φxx vvxx dx [ 7 6 ] 25s (φx φxx )xx v 2 − 50s7 φ6x φxx vx2 dx;

I

∫

∫ (I1 )5 (I2 )1 dx =

2 I

50s5 φ5x vx v4x dx

I

(

5

= −25s

2 φ5x vxx

)⏐⏐x=L ∫ [ ] 2 ⏐ + 375s5 φ4x φxx vxx − 125s5 (φ4x φxx )xx vx2 dx, ⏐ x=0

I

M. Chen / Nonlinear Analysis 185 (2019) 356–373

∫

∫ (I1 )5 (I2 )2 dx =

2 I

∫I = ∫I

∫ (I1 )5 (I2 )3 dx =

2 I

365

100s5 φ4x φxx vx vxxx dx ] [ 5 4 2 dx, 50s (φx φxx )xx vx2 − 100s5 φ4x φxx vxx 100s7 φ7x vx vxx dx

I∫

= − 350s7 φ6x φxx vx2 dx, ∫ ∫ I 2 (I1 )5 (I2 )4 dx = (300s7 φ6x φxx − 50s5 φ4x φ4x )vx2 dx, ∫I ∫I 2 (I1 )5 (I2 )5 dx = 10s9 φ9x vx vdx I I∫ = − 45s9 φ8x φxx v 2 dx; I

∫

∫ (I1 )6 (I2 )1 dx =

2 I

∫I = ∫I

∫ (I1 )6 (I2 )2 dx =

2 I

∫I =

200s5 φ4x φxx vv4x dx [ ] 2 200s5 φ4x φxx vxx − 400s5 (φ4x φxx )xx vx2 + 100s5 (φ4x φxx )4x v 2 dx, 400s5 φ3x φ2xx vvxxx dx [ ] 600s5 (φ3x φ2xx )x vx2 − 200s5 (φ3x φ2xx )xxx v 2 dx,

I

∫

∫ (I1 )6 (I2 )3 dx =

2 I

∫I = ∫I

∫ (I1 )6 (I2 )4 dx =

2 I

∫I = ∫I

∫ (I1 )6 (I2 )5 dx =

2 I

400s7 φ6x φxx vvxx dx [ ] −400s7 φ6x φxx vx2 + 200s7 (φ6x φxx )xx v 2 dx, (1200s7 φ5x φ2xx − 200s5 φ3x φxx φ4x )vvx dx (−600s7 φ5x φ2xx + 100s5 φ3x φxx φ4x )x v 2 dx, 40s9 φ8x φxx v 2 dx.

I

Step 2. We claim that there exists sufficiently large constant s0 such that for s ≥ s0 , we have ∫ 2 2 2 (s9 φ9 v 2 + s7 φ7 vx2 + s5 φ5 vxx + s3 φ3 vxxx + sφv4x )dxdt Q (∫ 2 2 2 ≤C (s9 φ9 v 2 + s7 φ7 vx2 + s5 φ5 vxx + s3 φ3 vxxx + sφv4x )dxdt ω Q˜ ) ∫ 2 2 + (|Λu| e−2sφ + |I3 | )dxdt . Q

We shall estimate each term in (3.2). Consideration is first given to the internal terms. Choosing s0 > 1 large enough and applying the estimates in (3.1), we can deduce that {·}v 2 = {−5s9 φ8x φxx + R0 }v 2 , {·}vx2 = {−290s7 φ6x φxx + R1 }vx2 , 2 2 {·}vxx = {−150s5 φ4x φxx + R2 }vxx ,

(3.3)

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where R0 ≤ Cs8 φ8 , R1 ≤ Cs6 φ6 , R2 ≤ Cs4 φ4 ,

(x, t) ∈ Q.

According to the methods developed in [1] with minor changes, it follows immediately that ∫ 2 2 2 ({·}v 2 + {·}vx2 + {·}vxx + {·}vxxx + {·}v4x )dxt Q ∫ 2 2 2 ≥C (s9 φ9 v 2 + s7 φ7 vx2 + s5 φ5 vxx + s3 φ3 vxxx + sφv4x )dxdt Q ∫ 2 2 2 −C (s9 φ9 v 2 + s7 φ7 vx2 + s5 φ5 vxx + s3 φ3 vxxx + sφv4x )dxdt. ω Q˜ Next, we estimate the boundary terms in (3.2). Recalling the definition of φ, we have lim φ(·, t) = lim φ(·, t) = −∞, t→T −

t→0+

therefore, v(x, 0) = v(x, T ) = 0 for any x ∈ I. This implies ∫ {·}t dxdt = 0. Q

Set {·}|x=L x=0 := V (L) − V (0). Taking into account the fact that φx (L, ·) > 0, we can deduce that for s ≥ s0 with sufficiently large s0 , ( )⏐ ( )⏐ 2 2 ⏐ ⏐ , ≤ s3 φ3x vxxx −10sφxxx vxxx x=L x=L ( 3 )⏐ ( )⏐ 2 2 2 2 2 ⏐ ⏐ s (250φx φxx − 35(φx φxx )x )vxx − 10sφ5x vxx ≤ s5 φ5x vxx , x=L x=L ⏐ ⏐ ( 3 3 2 ) 2 (20sφxx vxxx v4x ) ⏐x=L ≤ s φx vxxx + sφx v4x ⏐x=L , ( 3 3 )⏐ ( )⏐ 2 2 ⏐ 20s φx vxx v4x ⏐x=L ≤ 50s5 φ5x vxx + 2sφx v4x , ( )⏐ ( 5 5 x=L )⏐ 3 2 2 2 ⏐ (130s φx φxx + 10sφ4x )vxx vxxx ⏐ ≤ s φx vxx + s3 φ3x vxxx . x=L

x=L

This leads to ( )⏐ 2 2 2 ⏐ V (L) ≥ 2sφx v4x + 37s3 φ3x vxxx + 24s5 φ5x vxx x=L > 0. Similarly, we can show that −V (0) > 0. Consequently, we can obtain (3.3). Step 3. Choosing s0 large enough, we have 2

2 |I3 | ≤C(s8 φ8 v 2 + s6 φ6 vx2 + s4 φ4 vxx ) 2 ≤ε(s9 φ9 v 2 + s7 φ7 vx2 + s5 φ5 vxx )

∫

2

|I3 | dxdt can be absorbed by the left hand of (3.3). Thus, we have

holds for any ε > 0. This means that Q

∫

2 2 2 (s9 φ9 v 2 + s7 φ7 vx2 + s5 φ5 vxx + s3 φ3 vxxx + sφv4x )dxdt (∫ ) ∫ 2 −2sφ 9 9 2 7 7 2 5 5 2 3 3 2 2 ≤C |Λu| e dxdt + (s φ v + s φ vx + s φ vxx + s φ vxxx + sφv4x )dxdt . ω Q Q˜ Q

Replacing v by ue−sφ , the proof is complete. □

M. Chen / Nonlinear Analysis 185 (2019) 356–373

Applying Proposition 3.1, we can get a Carleman estimate for the adjoint system of (2.1): ⎧ ⎪ in I × (0, T ), ⎨ut + uxxx + u5x + zux = 0 u(0, t) = u(L, t) = ux (0, t) = ux (L, t) = uxx (L, t) = 0 in (0, T ), ⎪ ⎩ u(x, T ) = uT (x) in I.

367

(3.4)

Corollary 3.1. Let z ∈ X0 . Then there exists a constant s˜0 = s˜0 (T, ∥z∥X0 ) such that for all s ≥ s˜0 and all uT ∈ L2 (I), the solution u of (3.4) fulfills ∫ (s9 φ9 u2 + s7 φ7 u2x + s5 φ5 u2xx + s3 φ3 u2xxx + sφu24x )e−2sφ dxdt Q ∫ (3.5) 9 9 2 7 7 2 5 5 2 3 3 2 2 −2sφ ≤C (s φ u + s φ ux + s φ uxx + s φ uxxx + sφu4x )e dxdt. ω Q˜ Proof . Applying Proposition 2.2 with Λu = −uxxx − zux , the following holds ∫ (s9 φ9 u2 + s7 φ7 u2x + s5 φ5 u2xx + s3 φ3 u2xxx + sφu24x )e−2sφ dxdt Q (∫ ) 9 9 2 7 7 2 5 5 2 3 3 2 2 −2sφ ≤C (s φ u + s φ ux + s φ uxx + s φ uxxx + sφu4x )e dxdt ω Q˜ ∫ 2 +C |uxxx + zux | e−2sφ dxdt (∫ Q ) ≤C (s9 φ9 u2 + s7 φ7 u2x + s5 φ5 u2xx + s3 φ3 u2xxx + sφu24x )e−2sφ dxdt ω Q˜ ∫ + C(∥z∥X0 ) (u2xxx + u2xx + s2 φ2 u2x )e−2sφ dxdt. Q

Picking s˜0 sufficiently large, for s ≥ s˜0 , the last term in the above inequality can be absorbed by the left hand side terms. This proves Corollary 3.1. □ In order to remove the terms u4x , uxxx , uxx and ux from the right hand side of (3.5), we introduce the functions ψ(0) 1 max ψ(x) = , t(T − t) x∈I t(T − t) 1 ψ(l3 ) φ(t) ˇ = min ψ(x) = . t(T − t) x∈I t(T − t)

φ(t) ˆ =

Combining with the definition of ψ, we have φ(t) ˇ < φ(t) ˆ <

4 φ(t). ˇ 3

Lemma 3.1. Let z ∈ X0 and s˜0 be chosen as in Corollary 3.1. Then for any s ≥ s˜0 and any uT ∈ L2 (I), the solution u of (3.4) satisfies ∫ (s9 φˇ9 u2 + s7 φˇ7 u2x + s5 φˇ5 u2xx + s3 φˇ3 u2xxx + sφu ˇ 24x )e−2sφˆ dxdt Q

∫ ≤C 0

T ˆ φ) ˇ s10 φˇ19 es(6φ−8 ∥u(·, t)∥2L2 (˜ dt. ω)

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368

Proof . The schemes of the proof are similar to the ones in [1], but we have to fill to framework with a new content. It follows from Corollary 3.1 that ∫ (s9 φˇ9 u2 + s7 φˇ7 u2x + s5 φˇ5 u2xx + s3 φˇ3 u2xxx + sφu ˇ 24x )e−2sφˆ dxdt Q ∫ ≤C (s9 φˇ9 u2 + s7 φˇ7 u2x + s5 φˇ5 u2xx + s3 φˇ3 u2xxx + sφu ˇ 24x )e−2sφˇ dxdt ω Q˜ ∫ T ≤C (s9 φˇ9 ∥u(·, t)∥2L2 (˜ + s7 φˇ7 ∥u(·, t)∥2H 1 (˜ + s5 φˇ5 ∥u(·, t)∥2H 2 (˜ ω) ω) ω) 0

)e−2sφˇ dt + sφ∥u(·, ˇ t)∥2H 4 (˜ + s3 φˇ3 ∥u(·, t)∥2H 3 (˜ ω) ω) =:C

4 ∑

Jk .

k=0

According to interpolation in the Sobolev spaces, we have 3

13

∥u(·, t)∥H 1 (˜ ≤ C∥u(·, t)∥ 1616 ω)

H 3 (˜ ω)

∥u(·, t)∥ 162

L (˜ ω)

3

∥u(·, t)∥H 2 (˜ ≤ C∥u(·, t)∥ 8 ω)

16 H 3 (ω )

5

∥u(·, t)∥ 8 2

L (˜ ω)

˜

9

∥u(·, t)∥H 3 (˜ ≤ C∥u(·, t)∥ 1616 ω)

H 3 (˜ ω) 16 H 3 (ω )

L (˜ ω)

,

1

∥u(·, t)∥ 4 2

L (˜ ω)

˜

Then combining with Young inequality, we can deduce that ∫ T ∫ T −2 −5 −2sφ ˆ 2 J1 ≤ C s φˇ e ∥u(·, t)∥ 16 dt + C ω) H 3 (˜ 0 0 ∫ T ∫ T J2 ≤ C s−2 φˇ−5 e−2sφˆ ∥u(·, t)∥2 16 dt + C H 3 (˜ ω) 0 0 ∫ T ∫ T J3 ≤ C s−2 φˇ−5 e−2sφˆ ∥u(·, t)∥2 16 dt + C H 3 (˜ ω) 0 0 ∫ T ∫ T J4 ≤ C s−2 φˇ−5 e−2sφˆ ∥u(·, t)∥2 16 dt + C H 3 (˜ ω) 0 0

,

7

∥u(·, t)∥ 162

3

∥u(·, t)∥H 4 (˜ ≤ C∥u(·, t)∥ 4 ω)

,

s

118 13

s

46 5

ˆ φˇ8 es( 5 φ−

s

66 7

φˇ 7 es(

127

6

.

32

ˆ 13 φ) ˇ φˇ 13 es( 13 φ− ∥u(·, t)∥2L2 (˜ dt, ω) 6

93

16 φ) 5 ˇ

∥u(·, t)∥2L2 (˜ dt, ω)

18 φ− 32 φ) 7 ˆ 7 ˇ

∥u(·, t)∥2L2 (˜ dt, ω)

ˆ φ) ˇ s10 φˇ19 es(6φ−8 ∥u(·, t)∥2L2 (˜ dt. ω)

After some elementary calculation, we have ∫ T( ) 127 6 ˆ 32 φ) 46 8 s( 6 φ− 16 ˇ 66 93 18 ˆ 32 φ) 118 13 ˇ + s 5 φ 7 ˇ J0 + s 13 φˇ 13 es( 13 φ− ˇ e 5 ˆ 5 φ) + s 7 φˇ 7 es( 7 φ− ∥u(·, t)∥2L2 (˜ dt ω) 0 ∫ T ˆ φ) ˇ ≤C s10 φˇ19 es(6φ−8 ∥u(·, t)∥2L2 (˜ dt. ω) 0

Consequently, ∫

(s9 φˇ9 u2 + s7 φˇ7 u2x + s5 φˇ5 u2xx + s3 φˇ3 u2xxx + sφu ˇ 24x )e−2sφˆ dxdt

Q

∫ ≤C

T 10

19 s(6φ−8 ˆ φ) ˇ

s φˇ e 0

∥u(·, t)∥2L2 (˜ dt ω)

Now, our task is reduced to estimate ∫ 0

∫ +C

(3.6)

T −2

s

−5 −2sφ ˆ

φˇ

e

2

∥u(·, t)∥

16

dt.

H 3 (˜ ω)

0

T

s−2 φˇ−5 e−2sφˆ ∥u(·, t)∥2

16

H 3 (˜ ω)

dt.

(3.7)

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Let 1

ˆ u1 (x, t) = θ1 (t)u(x, t) with θ1 (t) = e−sφ(t) φ(t) ˇ −2 .

Then u1 is the solution of the following system: ⎧ ⎪ in I × (0, T ), ⎨(u1 )t + (u1 )3x + (u1 )5x = f1 := (θ1 )t u − zθ1 ux u1 (0, t) = u1 (L, t) = (u1 )x (0, t) = (u1 )x (L, t) = (u1 )xx (L, t) = 0 in (0, T ), ⎪ ⎩ u1 (x, T ) = 0 in I.

(3.8)

Arguing as in [1], we can obtain that ∥f1 ∥2L2 (0,T ;L2 (I)) ≤ C

∫

(s2 φˇ3 u2 + u2x + u2xx )e−2sφˆ dxdt.

Q

According to Proposition 2.2 and interpolation, it follows that 8

u1 ∈ X2 ↪→ L6 (0, T ; H 3 (I)). Let 5

ˆ u2 (x, t) = θ2 (t)u(x, t) with θ2 (t) = e−sφ(t) φ(t) ˇ −2 .

Then u2 satisfies (3.8) with f1 replaced by f2 = (θ2 )t u − zθ2 ux = (θ2 )t θ1−1 u1 − zθ2 θ1−1 (u1 )x . Obviously, |(θ2 )t θ1−1 | + |θ2 θ1−1 | ≤ Cs. Since 4

5

4

z ∈ X0 ↪→ L3 (0, T ; H 3 (I)) and (u1 )x ∈ L6 (0, T ; H 3 (I)) ↪→ L6 (0, T ; H 3 (I)), 4

4

It is clear that f2 belongs to L2 (0, T ; H 3 (I)), here we used the fact that H 3 (I) is an algebra. Then, we have ∥f2 ∥

4

L2 (0,T ;H 3 (I))

≤ Cs∥u1 ∥

8

L6 (0,T ;H 3 (I))

.

It follows from Proposition 2.2 again that u2 ∈ X 10 . 3

We conclude that ∫ 0

T

s−2 φˇ−5 e−2sφˆ ∥u(·, t)∥2

16

ω) H 3 (˜

dt ≤s−2 ∥u2 ∥2X 10 3

≤Cs−2 ∥f2 ∥2

4

L2 (0,T ;H 3 (I))

≤C∥u1 ∥2

8

L6 (0,T ;H 3 (I)) ≤C∥u1 ∥2X2 ≤C∥f1 ∥2L2 (0,T ;L2 (I))

∫ ≤C

(s2 φˇ3 u2 + u2x + u2xx )e−2sφˆ dxdt.

Q

This implies that (3.7) can be absorbed by the left hand side terms in (3.6). The proof of Lemma 3.1 is complete. □

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4. Proofs of main results This section is devoted to the proofs of Theorems 1.1 and 1.2. 4.1. Proof of Theorem 1.1 Since the analysis is similar as in [1], we only sketch it for the sake of completeness. Proceeding as in the proof of Proposition 2.2, we can obtain that max ∥u∥2L2 (I) + ∥u∥2L2 (0,T ;H 2 (I)) ≤ C(∥uT ∥2L2 (I) + ∥z∥2L2 (0,T ;H 1 (I)) ∥u∥2L2 (0,T ;H 2 (I)) ).

t∈[0,T ]

By choosing ∥z∥L2 (0,T ;H 1 (I)) sufficiently small, we have max ∥u∥2L2 (I) + ∥u∥2L2 (0,T ;H 2 (I)) ≤ C∥uT ∥2L2 (I) .

t∈[0,T ]

Similar as in [1], this leads to ∥u(·, 0)∥2L2 (I)

∫ ≤C

2T 3 T 3

∥u(·, t)∥2L2 (I) dt.

Combining with Lemma 3.1 and the properties of weight functions, we can obtain the observability estimate ∫ T ∫ T 2 2 ∥u(·, 0)∥L2 (I) ≤ C ∥u(·, t)∥L2 (˜ dt ≤ C ∥u(·, t)∥2L2 (ω) dt. ω) 0

0

Due to classical dual argument, this implies the null controllability of linearized system (2.1). Then applying Kakutani fixed point theorem as in [1], we can prove Theorem 1.1. 4.2. Proof of Theorem 1.2 To prove Theorem 1.2, we shall first show the null controllability with constraints on the state for linearized system (2.1), and then using Kakutani fixed point theorem with the same argument as in the proof of Theorem 1.1. Therefore, it remains only to prove the following proposition. Proposition 4.1. For every ei ∈ L2 (Q), 1 ≤ i ≤ M , verifying (1.3), z ∈ B, there exists a control f ∈ L2 (Q) such that the solution y of (2.1) satisfies (1.2) and (1.4). Moreover, the control can be chosen such that ∥f ∥L2 (Q) ≤ C∥y0 ∥L2 (I) , where C = C(L, T, ω,

M ∑

∥ei ∥2L2 (Q) ) is a positive constant.

i=1

Proof . Following the ideas as in [26], we can transform the null controllability problem with constraints on the state into a problem of null controllability with constraints on the control. More precisely, set ˆ θ(t) = φˇ−9/2 (t)esφ(t) ,

there exists a function h0 ∈ Uθ such that Proposition 4.1 is equivalent to the following problem: Given z ∈ B and y0 ∈ L2 (I), find h ∈ L2 (Q) such that h ∈ U⊥

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and if y = y(x, t, h) is the solution of ⎧ ⎪ in Q, ⎨yt + yxxx + y5x + (zy)x = (h0 + h)χω y(0, t) = y(L, t) = yx (0, t) = yx (L, t) = yxx (0, t) = 0 in (0, T ), ⎪ ⎩ y(x, 0) = y0 (x), in I, y satisfies y(·, T, h) = 0,

in I.

To solve this problem, the key point is to establish an adapted Carleman estimate for the adjoint system (3.4): ∫ ∫ ∫ 1 2 2 2 u dxdt ≤ C |u − Π u| dxdt, (4.1) |u(x, 0)| dx + 2 Qω I Q θ where u is the solution of (3.4) and Π is defined in Section 2. (4.1) is proved by contradiction. Suppose that (4.1) does not hold. Then, for any n ∈ N∗ , we can find unT ∈ L2 (I) and zn ∈ B such that ∫ ∫ ∫ 1 1 2 2 2 un dxdt = 1 and |un − Πn un | dxdt ≤ , (4.2) |un (x, 0)| dx + 2 n Qω I Q θ where un is the solution of (3.4) corresponding to uT = unT and z = zn , Πn = Πn (zn ) is the orthogonal projection operator from L2 (Qω ) into U(zn ) = Span(p1 (zn )χQω , . . . , p1 (zn )χQω ), and pi (zn ) is the solution of (2.4) corresponding to z = zn . It is easy to deduce that ∥pi (zn )∥E ≤ C∥ei ∥L2 (Q) . It follows from Aubin–Lions lemma that we can extract a subsequence of {pi (zn )} (still called {pi (zn )}) such that pi (zn ) → qi strongly in L2 (0, T ; H 1 (I)). Similarly, we can extract a subsequence of {zn } (still called {zn }) such that zn → ξ strongly in L2 (0, T ; H 1 (I)). Therefore, we have zn (pi (zn ))x ⇀ ξ(qi )x weakly in D′ (Q). This means that qi is the solution of (2.5). Due to the fact that U(zn ) is finite dimensional, it is standard to obtain that ∫ 2 |Πn un | dxdt ≤ C. Qω 2

ω

Applying Lemma 2.2 with H = L (Q ),

pni

= pi (zn ) and hn = Πn un , there exists w ∈ Uξ such that

Πn un → w,

strongly in L2 (Qω ).

Combining with (4.2), we have un → w,

strongly in L2 (Qω ).

We claim that w satisfies wt + wxxx + w5x + ξwx = 0

in Qω .

In fact, it is sufficient to check the convergence zn (un )x → ξwx

in D′ (Qω ).

(4.3)

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For any ϕ ∈ D(Qω ), ⏐ ⏐∫ ∫ ⏐ ⏐ ⏐ ⏐ z (u ) ϕdxdt − ξw ϕdxdt n n x x ⏐ ⏐ ω Qω ⏐ ⏐ ⏐∫ ⏐∫Q ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ (zn − ξ)vx ϕdxdt⏐⏐ zn (un − w)x ϕdxdt⏐ + ⏐ ≤⏐ ω ω ⏐ ⏐ ⏐Q ⏐∫Q ∫ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ zn (un − w)ϕx dxdt⏐⏐ (zn )x (un − w)ϕdxdt⏐ + ⏐ ≤⏐ ω Qω ⏐ ⏐ ⏐∫ Q ⏐∫ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ ⏐ (zn − ξ)wϕx dxdt⏐⏐ (zn − ξ)x wϕdxdt⏐ + ⏐ + ⏐ Qω

Qω

≤C∥zn ∥L2 (0,T ;H 1 (ω)) ∥(un − w)∥L2 (Qω ) + C∥zn − ξ∥L2 (0,T ;H 1 (ω)) ∥w∥L2 (Qω ) → 0. Then, according to Lemma 2.1 with the fact that w ∈ Uξ and (4.3), we have w = 0 in L2 (Qω ). Thus un → 0, It follows from the proof of Theorem 1.1 that ∫ ∫ 2 |un (x, 0)| dx + I

Q

strongly in L2 (Qω ).

1 2 u dxdt ≤ C θ2 n

∫ Qω

u2n dxdt → 0,

which contradicts to (4.2). Proposition 4.1 can be proved by the adapted Carleman estimate (4.1) with the same approach as in [26]. □ Acknowledgment We sincerely thank Professor Yong Li for many useful suggestions and help. This work is supported by NSFC Grant (11701078). References [1] R.A. Capistrano-Filho, A.F. Pazoto, L. Rosier, Internal controllability of the Korteweg–de Vries equation on a bounded domain, ESAIM Control Optim. Calc. Var. 21 (2015) 1076–1107. [2] J.C. Ceballos, M. Sep´ ulveda, V. Villagr´ an, P. Octavio, The Korteweg–de Vries-Kawahara equation in a bounded domain and some numerical results, Appl. Math. Comput. 190 (2007) 912–936. [3] M. Chen, Null controllability with constraints on the state for the linear Korteweg–de Vries equation, Arch. Math 104 (2015) 189–199. [4] M. Chen, Null controllability with constraints on the state for the Korteweg–de Vries equation, Acta Appl. Math. 146 (2016) 17–28. [5] M. Chen, Null controllability with constraints on the state for stochastic heat equation, J. Dyn. Control Syst. 24 (2018) 39–50. [6] S. Cui, S. Tao, Strichartz estimates for dispersive equations and solvability of the Kawahara equation, J. Math. Anal. Appl. 304 (2005) 683–702. [7] G.G. Doronin, N.A. Larkin, Kawahara equation in a quarter-plane and in a finite domain, Bol. Soc. Parana. Mat. 25 (2007) 9–16. [8] G.G. Doronin, N.A. Larkin, Boundary value problems for the stationary Kawahara equation, Nonlinear Anal. TMA 69 (2008) 1655–1665. [9] G.G. Doronin, N.A. Larkin, Kawahara equation in a bounded domain, Discrete Contin. Dyn. Syst. Ser. B 10 (2008) 783–799. [10] A.V. Faminskii, R.V. Kuvshinov, Initial–boundary value problems for the generalized Kawahara equation, Russian Math. Surveys 66 (2011) 187–188. [11] P. Gao, Null controllability with constraints on the state for the 1-D Kuramoto-Sivashinsky equation, Evol. Equ. Control Theory 4 (2015) 281–296. [12] P. Gao, Null controllability with constraints on the state for the reaction–diffusion system, Comput. Math. Appl. 70 (2015) 776–788.

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