On the non-uniform null controllability of a linear KdV

equation∗

N. Carreno† S. Guerrero‡

Abstract

In this paper we consider a linear KdV equation with a transport term posed on afinite interval with the boundary conditions considered by Colin and Ghidaglia. Themain results concern the behavior of the cost of null controllability with respect tothe dispersion coefficient when the control acts on the left endpoint. In particular, forany final time we prove that this cost grows exponentially as the dispersion coefficientvanishes and the transport coefficient is negative.

AMS Subject Classification: 35B25, 93B05, 93C20

Keywords: Uniform controllability, dispersion limit, Carleman inequalities

1 Introduction

Let T > 0, L > 0 and Q := (0, T )×(0, L). We consider the following controlled Korteweg-deVries (KdV) equation posed in a finite domain:

yt + εyxxx −Myx = 0 in Q,y|x=0 = v, yx|x=L = 0, yxx|x=L = 0 in (0, T ),

y|t=0 = y0 in (0, L),(1.1)

where ε > 0 is the dispersion coefficient, M ∈ R is the transport coefficient, y0 ∈ L2(0, L)is the initial condition and v = v(t) stands for the control. Here and all along the paper weuse the notation, for a given function f = f(t, x),

f|x=x0:= f(·, x0) and f|t=t0 := f(t0, ·).

The boundary conditions in (1.1) were proposed by T. Colin and J.-M. Ghidaglia in [4](see also [3]) as a model for propagation of surface water waves in the situation where a wavemaker is putting energy in a finite-length channel from the left extremity and the right oneis free.

Most results for (1.1) are related to its well-posedness of its nonlinear version (see [13]and the references therein). However, let us observe that if M > 0, this system is notdissipative and therefore the existence of solutions needs to be justified. This will be donelater in Section 2.

∗This work has been partially supported by Fondecyt 3150089 (N. Carreno)†Departamento de Matematica, Universidad Tecnica Federico Santa Marıa, Casilla 110-V, Valparaıso,

Chile; email: nicolas.carrenog@usm.cl‡Universite Pierre et Marie Curie, UMR 7598, Laboratoire Jacques-Louis Lions, F-75005, Paris, France;

email: guerrero@ann.jussieu.fr

1

As for the controllability problem, in [1] the authors considered controls in all the bound-ary conditions and all the possible combinations. For controllability results of the KdV equa-tion in a finite interval with another boundary conditions, see [15, 16, 2] and the referencestherein.

In this paper, we are mainly interested in the study of how the size of the control behavesas a function of the dispersion coefficient ε. To this end, we define the quantity

Cε,0cost := supy0∈L2(0,L)

y0 6=0

minv∈L2(0,T )y|t=T=0

‖v‖2L2(0,T )

‖y0‖2L2(0,L)

, (1.2)

which stands for the cost of null controllability of (1.1). Notice that Cε,0cost is the best constantsuch that, for all y0 ∈ L2(0, L) and v ∈ L2(0, T ) such that y|t=T = 0, the estimate

‖v‖2L2(0,T ) ≤ Cε,0cost‖y0‖2L2(0,L)

is satisfied.Our first result states an improvement of the cost of the control with respect to the one

in [12]. From this work, one can deduce that there exists v ∈ L2(0, T ) such that

Cε,0cost ≤ C exp(Cε−1). (1.3)

The first main result of this paper is the following:

Theorem 1.1. Let T, L, ε > 0 and M ∈ R. Then, for any y0 ∈ L2(0, L), there exists a con-trol v ∈ L2(0, T ) such that the associated solution of (1.1) satisties y|t=T = 0. Furthermore,we have the estimate

Cε,0cost ≤ C exp(C(ε−1/2T−1/2 +M1/2ε−1/2 +MT )

), (1.4)

if M > 0, and

Cε,0cost ≤ C exp(C(ε−1/2T−1/2 + |M |1/2ε−1/2)

), (1.5)

if M < 0, where C > 0 is a constant independent of T , M and ε and C > 0 depends at mostpolynomially on ε−1, T−1, T and |M |−1.

We remark that (1.4)-(1.5) say that the cost of the control is at most of order exp(Cε−1/2),whereas in (1.3) is of order exp(Cε−1). This difference becomes of great importance whenstudying its behavior in the limit ε→ 0.

Before stating the second and third main results of this paper, let us consider the trans-port equation

yt −Myx = 0 in Q. (1.6)

Since (1.6) is controllable if and only if T ≥ L/|M | (see, for instance, [5, Theorem 2.6,page 29]), with a control y|x=0 = v1 if M < 0 and a control y|x=L = v2 if M > 0. Fur-thermore, the cost of null controllability is equal to zero. Indeed, the solution of (1.6) canbe brought to zero at time T just by taking v1 ≡ 0 when M < 0, and by taking v2 ≡ 0when M > 0. Thus, we should expect that the cost would decrease to zero in this case asε→ 0, or at least if the final time T is large enough. On the other hand, if T < L/|M | it isexpected that the cost of the control would explode as ε tends to zero.

In [10], the authors consider this problem for the classical boundary conditions

y|x=0 = v(t), y|x=L = 0, yx|x=L = 0 in (0, T ),

2

and in [9] with controls in all the boundary terms. We refer also to [6] and [11] for the caseof vanishing viscosity in one and arbitrary space dimension, respectively.

In these works, the strategy relies on the combination of a suitable Carleman inequality,which gives an observability constant that explodes with ε, with an exponential dissipationestimate for the adjoint equation such that for T large enough counteracts the previousconstant. It has been pointed out in [9] and [10] that such a result can only be expectedfor (1.1) when M > 0 due to the asymmetric effect of the dispersion term.

We will prove that the cost of null controllability explodes as ε → 0, even if T is large.Actually, we will prove it for the quantity

Cε,−1cost := supy0∈H3

n(0,L)y0 6=0

minv∈H−1(0,T )y|t=T=0

‖v‖2H−1(0,T )

‖y0‖2H3n(0,L)

(1.7)

where the space H3n(0, L) is defined as follows: for any a < b, let

H3n(a, b) := {h ∈ H3(a, b) : h′(b) = h′′(b) = 0},

endowed with the norm

‖h‖2H3n(a,b)

:= ‖h‖2L2(a,b) + ‖h′′′‖2L2(a,b).

The following theorem states a lower bound for this new cost of null controllability.

Theorem 1.2. Let T, L, ε,M > 0 and γ, δ ∈ (0, 1). Then,

Cε,−1cost >(1− γ)K(T, L, ε,M, δ)

C−1(T, L, ε,M, δ)C1(δL, L)− 1

where

K(T, L, ε,M, δ) :=sinh2

((1− δ)LM1/2ε−1/2

)εTM(1− δ)L

and, C−1(T, L, ε,M, δ) and C1(δL, L) are given by (4.2) and (4.25), respectively.

Remark 1.3. From the expression of C−1(T, L, ε,M, δ) in (4.2), one can see that it dependspolynomially on ε and ε−1.

From Theorem 1.2, we can deduce the following result that shows that the cost definedin (1.7) blows up as ε vanishes.

Corollary 1.4. Let T, L,M > 0 and κ ∈ (0, 1). Then, there exists ε0 > 0 such that forall ε ∈ (0, ε0) we have

Cε,−1cost ≥ exp

((2− κ)LM1/2

ε1/2

).

Notice that this implies that the cost defined in (1.2) goes to infinity as ε goes to zerosince Cε,−1cost ≤ Cε,0cost. What is interesting about this result is that it differs from the oneobtained in [10] (and [6] for that matter). Notice that for M > 0 and ε → 0 in (1.1), itseems very difficult (if not impossible) to prove convergence in some sense to a solutionof the transport equation (1.6) since we do not know the value at the right-end of theinterval (0, L). Actually, this convergence question has not been addressed even for theclassical boundary conditions (see [10]). Nevertheless, one could have expected to obtain anappropriate dissipation estimate as in [9], but Corollary 1.4 shows that this is not the case.

Finally, when M < 0 we are able to obtain an explosion result when T is smallerthan L/|M |.

3

Theorem 1.5. Let M < 0 and T < L/|M |. Then, there exist a constant C > 0 (independentof ε), ε0 > 0 and initial conditions y0 ∈ L2(0, L) such that, if v ∈ L2(0, T ) is a control suchthat the solution y of (1.1) satisfies y|t=T = 0, then, for every ε ∈ (0, ε0),

‖v‖2L2(0,T ) ≥ exp

(C

ε1/2

)‖y0‖2L2(0,L). (1.8)

In particular,

Cε,0cost ≥ exp

(C

ε1/2

).

The rest of the paper is organized as follows. In Section 2 we prove the existence ofsolutions of equation (1.1). In Section 3, we prove Theorem 1.1. We prove an observabilityinequality for the adjoint system (3.2) and then applying the Hilbert Uniqueness Method(HUM). The observability inequality is proved by means of a suitable Carleman estimate.In Section 4, we prove Theorem 1.2. In Section 5, we prove Theorem 1.5. Finally, we givethe proof of the Carleman estimate in Appendix A.

2 Existence of solutions when M > 0

In this section we will prove the existence (and uniqueness) of a solution of (1.1) whenM > 0, v ∈ L2(0, T ) and y0 ∈ L2(0, L). Let us first remark that if y solves (1.1), then

z(t, x) :=

∫ t

0

y(s, x) ds− (L− x)3

L3

∫ t

0

v(s) ds

would solve zt + εzxxx −Mzx = f in Q,z|x=0 = 0, zx|x=L = 0, zxx|x=L = 0 in (0, T ),z|t=0 = 0 in (0, L),

(2.1)

with

f(t, x) := − (L− x)3

L3v(t) +

(6ε

L3− 3M(L− x)2

L3

)∫ t

0

v(s) dt+ y0. (2.2)

Then, if we prove the existence (and uniqueness) of solution z of (2.1), we would haveproved the existence (and uniqueness) of a solution of (1.1) by simply defining

y(t, x) := zt(t, x) +(L− x)3

L3v(t). (2.3)

Lemma 2.1. Let ε, T,M > 0 and f ∈ L2(Q). Then, there exists a unique solution z of (2.1)which belongs to XT := L∞(0, T ;L2(0, L)) ∩ L2(0, T ;H1(0, L)).

Proof. We will use the contracting map fixed-point theorem. We consider, for any g ∈L2(0, T ;H1(0, L)), the following problem

zt + εzxxx = Mgx + f in Q,z|x=0 = 0, zx|x=L = 0, zxx|x=L = 0 in (0, T ),z|t=0 = 0 in (0, L),

(2.4)

Notice that the operator operator defined by

A := −εzxxx

4

is dissipative with

D(A) = {h ∈ H3(0, L) : h(0) = h′(L) = h′′(L) = 0}.

It is easy to check that the adjoint of A is also dissipative, thus the existence of a unique zsolution of (2.4) is ensured.

Now, we multiply the equation in (2.4) by (x+L)z and integrate by parts in space. Weobtain

1

2

d

dt

∫ L

0

(x+ L)|z|2 dx+3ε

2

∫ L

0

|zx|2 dx ≤M∫ L

0

(x+ L)gxz dx+

∫ L

0

(x+ L)fz dx

Using Young’s and Poincare’s inequalities, we get

d

dt

∫ L

0

(x+ L)|z|2 dx+ ε

∫ L

0

|zx|2 dx

≤ 2M

(∫ L

0

(x+ L)|z|2 dx

)1/2(∫ L

0

(x+ L)|gx|2 dx

)1/2

+ Cε−1∫ L

0

|f |2 dx,

where C > 0 depends only on L. We integrate between 0 and t, and take the supremum:

supt∈(0,T )

∥∥(x+ L)1/2z∥∥2L2(0,L)

+ ε‖zx‖2L2(Q)

≤ 2M supt∈(0,T )

∥∥(x+ L)1/2z∥∥L2(0,L)

∫ T

0

∥∥(x+ L)1/2gx∥∥L2(0,L)

dt+ Cε−1‖f‖2L2(Q).

Using Young’s inequality once more and (2.2) we obtain∥∥z∥∥2L∞(0,T ;L2(0,L))

+ ε‖z‖2L2(0,T ;H1(0,L))

≤ CM2‖g‖2L1(0,T ;H1(0,L)) + Cε−1(

(1 + ε2T +M2T )‖v‖2L2(0,T ) + T‖y0‖2L2(0,L)

), (2.5)

where C depends only on L.We consider the norm

‖z‖2XT := ‖z‖2L∞(0,T ;L2(0,L)) + ε‖z‖2L2(0,T ;H1(0,L)).

Let us now prove the existence of β ∈ (0, T ] such that the map

A : Xβ → Xβg 7→ z

is a contraction. In view of (2.5), we have

‖A(g)‖2Xβ ≤ CM2ε−1β‖g‖2Xβ + Cε−1

((1 + ε2T +M2T )‖v‖2L2(0,T ) + T‖y0‖2L2(0,L)

).

Let now, for r > 0:Br := {h ∈ Xβ : ‖h‖Xβ ≤ r},

withr2 = 2Cε−1

((1 + ε2T +M2T )‖v‖2L2(0,T ) + T‖y0‖2L2(0,L)

).

Choosing β such that

CM2ε−1β ≤ 1

2

5

we have for any g ∈ Br:‖A(g)‖Xβ ≤ r.

Then, for any g1, g2 ∈ Br:

‖A(g1)−A(g2)‖Xβ ≤1√2‖g1 − g2‖Xβ .

Therefore, A is a contraction mapping on Br and admits a unique fixed point z = A(z)in Xβ which is easy to check that is the solution of (2.1) in (0, β)× (0, L). Since β does notdepend on v nor y0, we can repeat this argument in the time intervals (nβ, (n + 1)β), for

any n ∈ N, with initial condition znβ|t=nβ ∈ L2(0, L), where znβ ∈ Xnβ is the solution of (2.1)

in (0, nβ)× (0, L). Thus, z ∈ XT and is a solution of (2.1).

From (2.1)-(2.3), one sees that the solution of (1.1) belongs to L2(0, T ;H−2(0, L)) ∩C0([0, T ];H−5(0, L)). In the sequel, we will also need to establish the existence of solutionsof (1.1) when v ∈ H−1(0, T ).

Corollary 2.2. Let ε, T,M > 0, v ∈ H−1(0, T ) and y0 ∈ L2(0, L). Then, there exists aunique solution y ∈ L2(0, T ;H−5(0, L)) ∩ C0([0, T ];H−8(0, L)) of (1.1).

Proof. Let v ∈ H−1(0, T ). Then (see for instance [7, Theorem 1, page 283]), there existv0, v1 ∈ L2(0, T ) such that

v = v0 + v′1. (2.6)

Furthermore,

‖v‖H−1(0,T ) = infv0,v1∈L2(0,T )s.t.(2.6)holds

(‖v0‖2L2(0,T ) + ‖v1‖2L2(0,T )

)1/2. (2.7)

Then,y = p+ qt, (2.8)

where p and q are the solutions ofpt + εpxxx −Mpx = 0 in Q,p|x=0 = v0(t), px|x=L = 0, pxx|x=L = 0 in (0, T ),

p|t=0 = y0 in (0, L),(2.9)

and qt + εqxxx −Mqx = 0 in Q,q|x=0 = v1(t), qx|x=L = 0, qxx|x=L = 0 in (0, T ),

q|t=0 = 0 in (0, L),(2.10)

respectively.From Lemma 2.1 and the computations above, one deduces the desired result.

3 Proof of Theorem 1.1

In this section, we will prove Theorem 1.1 by applying the Hilbert Uniqueness Method(H.U.M.) (see, for instance, [14]), that is, we prove the following observability inequality:

‖ϕ|t=0‖2L2(0,L) ≤ Cobs‖ϕxx|x=0‖2L2(0,T ). (3.1)

6

Here, ϕ is the solution of the adjoint equation−ϕt − εϕxxx +Mϕx = 0 in Q,ϕ|x=0 = 0, ϕx|x=0 = 0, (εϕxx −Mϕ)|x=L = 0 in (0, T ),

ϕ|t=T = ϕT in (0, L),(3.2)

with ϕT ∈ L2(0, L) and Cobs > 0 is a constant independent of ϕ. Indeed, (3.1) is equivalentto prove that for every y0 ∈ L2(0, L), there exists a control v ∈ L2(0, T ) such that y|t=T = 0satisfying

‖v‖2L2(0,T ) ≤Cobsε2‖y0‖2L2(0,L)

where y is the solution of (1.1). Thus, once an inequality like (3.1) is established, the proofof Theorem 1.1 is finished since

Cε,0cost ≤Cobsε2

. (3.3)

The observability inequality (3.1) is proved by means of Carleman and energy estimates,which are the goals of the following sections.

3.1 A change of unknowns

A relevant system associated to (3.2) will be{−φt − εφxxx +Mφx = 0 in Q,φx|x=0 = 0, φxx|x=0 = 0, φ|x=L = 0 in (0, T ).

(3.4)

Notice that (3.4) comes from

φ := εϕxx −Mϕ. (3.5)

Furthermore, we notice that from (3.5) and the boundary conditions on x = 0 in (3.2),for every t ∈ (0, T ) we have the following initial value ordinary differential equation{

ϕxx −Mε−1ϕ = ε−1φ in (0, L),ϕ|x=0 = 0, ϕx|x=0 = 0.

We can actually find an explicit formula for ϕ in terms of φ, but we need to distinguishthe cases M > 0 and M < 0:

3.1.1 Case M > 0

It is not difficult to show that the solution is given by

ϕ(t, x) =1

ε1/2M1/2

∫ x

0

sinh(M1/2ε−1/2(x− s))φ(t, s) ds.

Thus,

|ϕ(t, x)| ≤ 1

ε1/2M1/2

(∫ L

0

sinh2(M1/2ε−1/2(1− s)) ds

)1/2(∫ L

0

|φ(t, s)|2 ds

)1/2

.

Taking the L2(0, L)-norm in this inequality, we get∫ L

0

|ϕ(t, x)|2 dx ≤ C

ε1/2M3/2exp

(CM1/2ε−1/2

) ∫ L

0

|φ(t, x)|2 dx. (3.6)

7

Moreover, since

ϕx(t, x) =1

ε

∫ x

0

cosh(M1/2ε−1/2(x− s))φ(t, s) ds,

same computations show that∫ L

0

|ϕx(t, x)|2 dx ≤ C

ε3/2M1/2exp

(CM1/2ε−1/2

) ∫ L

0

|φ(t, x)|2 dx. (3.7)

Using directly (3.5) and (3.6), we obtain∫ L

0

|ϕxx(t, x)|2 dx ≤( 2

ε2+CM1/2

ε5/2exp

(CM1/2ε−1/2

)) ∫ L

0

|φ(t, x)|2 dx. (3.8)

3.1.2 Case M < 0

In this case, ϕ is given by

ϕ(t, x) =1

ε1/2|M |1/2

∫ x

0

sin(|M |1/2ε−1/2(x− s))φ(s) ds.

The same computations as for the case M > 0 show that∫ L

0

|ϕ(t, x)|2 dx ≤ C

ε|M |

∫ L

0

|φ(t, x)|2 dx, (3.9)

∫ L

0

|ϕx(t, x)|2 dx ≤ C

ε2

∫ L

0

|φ(t, x)|2 dx, (3.10)

and ∫ L

0

|ϕxx(t, x)|2 dx ≤( 2

ε2+C|M |ε3

) ∫ L

0

|φ(t, x)|2 dx. (3.11)

3.2 Carleman estimates

To establish the Carleman estimate, we introduce some weight functions. Let

α(t, x) =− x2

L2 + 4xL + 1

tm(T − t)m,

where m ≥ 1/2. Notice that there exist positive constants C0 and C1 that do not dependon T such that

C0 ≤ T 2mα, C0α ≤ αx ≤ C1α, C0α ≤ −αxx ≤ C1α, (3.12)

and|αt|+ |αxt|+ |αxxt| ≤ C1Tα

1+1/m, |αtt| ≤ C1T2α1+2/m, (3.13)

for every (t, x) ∈ Q. This kind of weights has been introduced in [8] and widely used in theliterature (in particular, in [9, 10, 12]).

We now are in position to present our Carleman inequality whose proof is given at theend of the paper (Appendix A).

8

Proposition 3.1. Let T, ε > 0, M ∈ R and m = 1/2. There exists a positive constant Cindependent of T , ε and M such that, for any solution φ of (3.4), we have∫∫

Q

e−2sα(s5α5|φ|2 + s3α3|φx|2 + sα|φxx|2

)dx dt ≤ Cs5

∫ T

0

e−2sα|x=0α5|x=0|φ|x=0|2 dt,

(3.14)for all s ≥ C(T + ε−1/2T 1/2 + |M |1/2ε−1/2T ).

Furthermore, we can deduce from Proposition 3.1 and (3.6)-(3.11) a Carleman estimatefor the solutions of (3.2).

Proposition 3.2. Let T, ε > 0, M ∈ R \ {0} and m = 1/2. There exists a positive constantC independent of T , ε and M such that, for any solution ϕ of (3.2), we have∫∫

Q

e−2sα|x=L(s5α5|x=0|ϕ|

2 + s3α3|x=0|ϕx|

2 + sα|x=0|ϕxx|2)

dx dt

≤ C exp(C|M |1/2ε−1/2

)s5∫ T

0

e−2sα|x=0α5|x=0|ϕxx|x=0|2 dt, (3.15)

for all s ≥ C(T + ε−1/2T 1/2 + |M |1/2ε−1/2T ) and C depends at most polynomially on ε−1,ε, |M |−1 and |M |.

Remark 3.3. The lack of homogeneous Dirichlet condition on x = L plays an importantrole in the choice of the power m of the weight function to prove (3.15). Indeed, a similarinequality was proved in [12] where m ≥ 1 was needed to estimate a trace term on x = L.From this inequality one can deduce that the cost of the null controllability is bounded byexp(Cε−1).

Here, by means of the change of variable (3.5), which satisfies φ|x=L = 0, we manage totake the optimal power m = 1/2 as in [9, 10].

It would be interesting to know if a Carleman estimate can be obtained for the solutionsof (3.2) for m = 1/2 without using this change of variables.

3.3 Dissipation estimates

To prove (3.1), we will combine (3.14) with a dissipation estimate. For M < 0, it is easy tocheck that ∫ L

0

|φ|t=t1 |2 dx ≤

∫ L

0

|φ|t=t2 |2 dx, (3.16)

for every 0 ≤ t1 ≤ t2 ≤ T . For M > 0 we can prove the following result.

Proposition 3.4. Let ε,M > 0. Then, for every pair (t1, t2) such that 0 ≤ t1 < t2 ≤ Tand every solution φ of (3.4) the following inequality is satisfied

E(φ)(t1) ≤ exp(CM(t2 − t1))E(φ)(t2) (3.17)

where C > 0 is a constant independent of M , T and ε, and

E(φ)(t) :=

∫ L

0

(|φ(t, x)|2 + (L− x)3|φx(t, x)|2

)dx.

Proof. We proceed in two steps. First, we multiply (3.4) by (2L−x)φ and integrate in (0, L).We obtain after some integration by parts

− 1

2

d

dt

∫ L

0

(2L− x)|φ|2 dx+3ε

2

∫ L

0

|φx|2 dx+ε

2|φx|x=L|2 +

M

2

∫ L

0

|φ|2 dx = LM |φ|x=0|2.

(3.18)

9

Next, we take the derivative with respect to x of (3.4), multiply by (L−x)32 φx and proceed

as before. Straightforward computations lead to

− 1

2

d

dt

∫ L

0

(L− x)3

2|φx|2 dx+

9ε

4

∫ L

0

(L− x)2|φxx|2 dx

− 3ε

2

∫ L

0

|φx|2 dx+3M

4

∫ L

0

(L− x)2|φx|2 dx = 0. (3.19)

Adding (3.18) and (3.19) we obtain, after neglecting the positive terms on the left-handside,

−1

2

d

dt

∫ L

0

(2L− x)|φ|2 dx− 1

2

d

dt

∫ L

0

(L− x)3

2|φx|2 dx ≤ LM |φ|x=0|2.

Now, notice that|φ|x=0|2 ≤ C‖φ‖2H1(0,L2 )

,

and therefore we obtain

− 1

2

d

dt

∫ L

0

(2L− x)|φ|2 dx− 1

2

d

dt

∫ L

0

(L− x)3

2|φx|2 dx

≤ 16CM

(∫ L

0

(2L− x)|φ|2 dx+

∫ L

0

(L− x)3

2|φx|2 dx

).

Estimate (3.17) can be easily deduced by integrating in (t1, t2).

3.4 Observability inequality

In this section we combine the Carleman inequality (3.14) and the dissipation estimatesobtained in the previous paragraph to finish the proof of (3.1) and therefore the proof ofTheorem 1.1. Let us separate the cases M > 0 and M < 0.

3.4.1 Case M > 0

Notice that from (3.14) we have∫ 3T/4

T/4

∫ L

0

(|φ|2 + |φx|2

)dxdt ≤ Cs2(1 + T−2) exp

(sC/T

) ∫ T

0

|φ|x=0|2 dt.

From the dissipation estimate (3.17) with t1 = 0 and integrating between T/4 and 3T/4we obtain

T

2

∫ L

0

|φ|t=0|2 dx ≤ C exp(CMT

)∫ 3T/4

T/4

∫ L

0

(|φ|2 + |φx|2

)dx dt.

Combining these two inequalities and fixing s = C(T + ε−1/2T 1/2 + M1/2ε−1/2T ), weget∫ L

0

|φ|t=0|2 dx ≤ C(1+ε−1)(1+M) exp(C(MT+ε−1/2T−1/2+M1/2ε−1/2)

)∫ T

0

|φ|x=0|2 dt,

(3.20)where C depends at most polynomially on T−1 and T .

10

Let us now go back to ϕ. From (3.5), we obtain directly φ|x=0 = εϕxx|x=0. On the otherhand, from (3.6) with t = 0, we find∫ L

0

|ϕ|t=0|2 dx ≤ C

ε1/2M3/2exp

(CM1/2ε−1/2

) ∫ L

0

|φ|t=0|2 dx.

These two elements combined with (3.20) give (3.1) with

Cobs = C(ε1/2 + ε3/2) exp(C(ε−1/2T−1/2 +M1/2ε−1/2 +MT )

),

where C depends polynomially on T−1, T and M−1. Consequently, from (3.3), (1.4) isdeduced.

3.4.2 Case M < 0

This case is actually simpler. From (3.14) we have∫ 3T/4

T/4

∫ L

0

|φ|2 dx dt ≤ C exp(sC/T

) ∫ T

0

|φ|x=0|2 dt.

Using (3.16) and the same arguments as for the previous case yield∫ L

0

|φ|t=0|2 dx ≤ C

Texp

(C(ε−1/2T−1/2 + |M |1/2ε−1/2)

)∫ T

0

|φ|x=0|2 dt.

We recover ϕ as before from (3.9) instead of (3.6) and obtain (3.1) with

Cobs = εC exp(C(ε−1/2T−1/2 + |M |1/2ε−1/2)

),

where C depends polynomially on |M |−1 and T−1. From (3.3), this gives (1.5).

This completes the proof of Theorem 1.1.

4 Proof of Theorem 1.2

In this section we prove Theorem 1.2. For ease of comprehension, we have divided the proofin two parts presented in the following paragraphs.

4.1 Previous estimates

Here we will establish several estimates on y (solution of (1.1)) in terms of a constantdepending only on T , L, ε and M times the norm of v and the norm of y0. The principalresult of this part is:

Lemma 4.1. For any δ ∈ (0, 1), any T, L, ε,M > 0, any y0 ∈ H3n(0, L) and any v ∈

H−1(0, T ), the solution y of (1.1) satisfies

‖yxx|x=δL‖2L2(0,T ) ≤ C−1(T, L, ε,M, δ)(‖v‖2H−1(0,T ) + ‖y0‖2L2(0,L) + ‖y′′′0 ‖2L2(0,L)

), (4.1)

whereC−1(T, L, ε,M, δ) := C0(T, L, ε,M, δ) + C ′0(T, L, ε,M, δ) (4.2)

with C0(T, L, ε,M, δ) and C ′0(T, L, ε,M, δ) given by Lemma 4.2.

11

The proof of Lemma 4.1 relies on the following estimates for equation (1.1):

Lemma 4.2. There exists C > 0 such that for any δ ∈ (0, 1), any T, L, ε,M > 0, anyy0 ∈ H3

n(0, L) and any v ∈ L2(0, T ), the solution y of (1.1) satisfies

‖yxx|x=δL‖2L2(0,T ) ≤ C0(T, L, ε,M, δ)(‖v‖2L2(0,T ) + ‖y0‖2L2(0,L) + ‖y′′′0 ‖2L2(0,L)

)(4.3)

where

C0(T, L, ε,M, δ) :=CTeCMT/L

L4δ10

((εT

L3

)−1+

(εT

L3

)4

+

(MT

L

)4)(1

T+

1

L+ L5

).

Furthermore, if y0 ≡ 0, then

‖ytxx|x=δL‖2L2(0,T ) ≤ C′0(T, L, ε,M, δ)‖v‖2L2(0,T ) (4.4)

where

C ′0(T, L, ε,M, δ) :=CeCMT/L

T 2L4δ22

((εT

L3

)−1+

(εT

L3

)4

+

(MT

L

)4)×((

εT

L3

)−2+

(εT

L3

)4

+

(MT

L

)2

+

(MT

L

)8).

Proof of Lemma 4.2. To start the proof, we perform the change of variables

t :=t

Tand x :=

x

L.

Then, y(t, x) := y(t, x) satisfies system (1.1) for

(T, L, ε,M) = (1, 1, ε, M) and (v, y0) = (v, y0)

with

ε :=εT

L3, M :=

MT

L, v(t) := v(t) and y0(x) := y0(x).

For y we will prove the following estimates:

‖yxx|x=δ‖2L2(0,1) ≤C(1 + ε4 + M4)eCM

δ10(‖v‖2L2(0,1) + ‖y0‖2L2(0,1) + ‖y′′′0 ‖2L2(0,1)

)(4.5)

and, if y0 ≡ 0,

‖ytxx|x=δ‖2L2(0,1) ≤C(1 + ε4 + M4)eCM (ε−2 + ε4 + M2 + M8)

δ22‖v‖2L2(0,1) (4.6)

from where inequalities (4.3) and (4.4) are easily deduced by going back to the original vari-able. Here and in the following, C denotes a positive constant independent of all parameters.

For the sake of clearness we will denote y, t and x instead of y, t and x until the end ofthe proof of Lemma 4.2.

The proof of (4.5) and (4.6) is divided in several steps:

Step 0. A previous computation.

12

Let us first perform a previous computation which will be useful in the proof. Letθ ∈ C3([0, 1]) be a positive increasing function such that θ(0) = θ′(0) = θ′′(0) = 0. Aftersome integrations by parts we obtain, for all h ∈ H1(0, 1;L2(0, 1)) ∩ L2(0, 1;H3(0, 1)),

1

2

d

dt

∫ 1

0

θ|h|2 dx+3ε

2

∫ 1

0

θ′|hx|2 dx+M

2

∫ 1

0

θ′|h|2 dx

+ εθ(1)hxx|x=1h|x=1− εθ′(1)hx|x=1h|x=1−ε

2θ(1)|hx|x=1|2−

M

2θ(1)|h|x=1|2 +

ε

2θ′′(1)|h|x=1|2

=ε

2

∫ 1

0

θ′′′|h|2 dx+

∫ 1

0

θ(ht + εhxxx − Mhx)hdx. (4.7)

Step 1. Estimate of yx.

In this paragraph we lift the boundary and initial conditions. Let

z(t, x) :=

∫ t

0

y(s, x) ds− (1− x)6∫ t

0

v(s) ds (t, x) ∈ (0, 1)2. (4.8)

Then, z satisfies zt + εzxxx − Mzx = f in (0, 1)2,z|x=0 = 0, zx|x=1 = 0, zxx|x=1 = 0 in (0, 1),z|t=0 = 0 in (0, 1),

(4.9)

where

f(t, x) := −(1− x)6v(t) +(120(1− x)3ε− 6(1− x)5M

) ∫ t

0

v(s) dt+ y0. (4.10)

We multiply the equation in (4.9) by (1 + x)z and integrate in space. After integrationby parts we have

1

2

d

dt

∫ 1

0

(1 + x)|z|2 dx+3ε

2

∫ 1

0

|zx|2 dx ≤ M |z|x=1|2 +

∫ 1

0

(1 + x)fz dx.

From (4.7) with h = zx, θ = x3

4 we obtain

1

2

d

dt

∫ 1

0

x3

4|zx|2 dx+

9ε

8

∫ 1

0

x2|zxx|2 dx ≤ 3ε

4

∫ 1

0

|zx|2 dx+1

4

∫ 1

0

x3fxzx dx.

Adding these inequalities, we get

1

2

d

dt

∫ 1

0

(1 + x)|z|2 dx+1

2

d

dt

∫ 1

0

x3

4|zx|2 dx+

3ε

4

∫ 1

0

|zx|2 dx+9ε

8

∫ 1

0

x2|zxx|2 dx

≤ M |z|x=1|2 +

∫ 1

0

(1 + x)fz dx+1

4

∫ 1

0

x3fxzx dx. (4.11)

Now, notice that we can estimate the trace term in the following way:

M |z|x=1|2 = M

∫ 1

0

∂x(x3/2|z|2

)dx = 2M

∫ 1

0

x3/2zxz dx+3M

2

∫ 1

0

x1/2|z|2 dx

≤ CM∫ 1

0

((1 + x)|z|2 + x3|zx|2

)dx.

13

Then, we have:

1

2

d

dt

∫ 1

0

((1 + x)|z|2 +

x3

4|zx|2

)dx+

3ε

4

∫ 1

0

|zx|2 dx+9ε

4

∫ 1

0

x2|zxx|2 dx

≤ C(1 + M)

∫ 1

0

((1 + x)|z|2 +

x3

4|zx|2

)dx+ C

∫ 1

0

(|f |2 + |fx|2) dx.

From Gronwall’s Lemma, we obtain in particular

ε

∫ 1

0

∫ 1

0

x2|zxx|2 dx dt ≤ CeCM∫ 1

0

∫ 1

0

(|f |2 + |fx|2) dx dt. (4.12)

Now, we use (4.7) with h = zxx and θ = x5:

1

2

d

dt

∫ 1

0

x5|zxx|2 dx+15ε

2

∫ 1

0

x4|zxxx|2 dx

≤ 30ε

∫ 1

0

x2|zxx|2 dx+ε

2|zxxx|x=1|2 +

∫ 1

0

x5fxxzxx dx. (4.13)

To gain more derivatives of z and estimate the boundary term on zxxx, we use again (4.7)with h = zxxx and θ = ax7 (a := 1

28 ). Notice that from (4.9) and fx|x=1 = 0, we have thatz4x|x=1 = 0. Thus:

1

2

d

dt

∫ 1

0

ax7|zxxx|2 dx+21εa

2

∫ 1

0

x6|z4x|2 dx+ aεz5x|x=1zxxx|x=1 −aM

2|zxxx|x=1|2

+3ε

4|zxxx|x=1|2 ≤

15ε

4

∫ 1

0

x4|zxxx|2 dx+ a

∫ 1

0

x7fxxxzxxx dx.

Again from (4.9) and fxx|x=1 = 0, we have the relation εz5x|x=1 = Mzxxx|x=1. Using it inthis last inequality, we obtain

1

2

d

dt

∫ 1

0

ax7|zxxx|2 dx+21εa

2

∫ 1

0

x6|z4x|2 dx+3ε

4|zxxx|x=1|2

≤ 15ε

4

∫ 1

0

x4|zxxx|2 dx+ a

∫ 1

0

x7fxxxzxxx dx. (4.14)

Adding (4.13) and (4.14), we get

1

2

d

dt

∫ 1

0

(x5|zxx|2+ax7|zxxx|2

)dx+

15ε

4

∫ 1

0

x4|zxxx|2 dx+21εa

2

∫ 1

0

x6|z4x|2 dx+ε

4|zxxx|x=1|2

≤ 1

2

∫ 1

0

(x5|zxx|2 + ax7|zxxx|2

)dx+ C

∫ 1

0

|fxxx|2 dx+ 30ε

∫ 1

0

x2|zxx|2 dx.

Using Gronwall’s Lemma and (4.12), we find∫ 1

0

∫ 1

0

x5|zxx|2 dxdt+ ε

∫ 1

0

∫ 1

0

(x4|zxxx|2 + x6|z4x|2

)dxdt

≤ CeCM∫ 1

0

∫ 1

0

(|f |2 + |fxxx|2) dxdt. (4.15)

14

Finally, we go back to y. From (4.8), we get

yx = −εz4x + Mzxx + fx − 6(1− x)5v(t). (4.16)

Then, using (4.15), we have

ε

∫ 1

0

∫ 1

0

x6|yx|2 dx dt

≤ C(ε2 + εM2 + ε)eCM∫ 1

0

∫ 1

0

(|f |2 + |fxxx|2) dx dt+ Cε

∫ 1

0

|v|2 dt,

which, from the definition of f (see (4.10)) yields

ε

∫ 1

0

∫ 1

0

x6|yx|2 dxdt ≤ Cε(1 + ε4 + M4)eCM(∫ 1

0

|v|2 dt+

∫ 1

0

(|y0|2 + |y′′′0 |2

)dx

). (4.17)

For this last inequality we have used Young’s inequality.

Step 2. Estimate on yxx and yxxx, and conclusion of (4.5).

First, we use (4.7) with h = yx and θ = x9 . We have

1

2

d

dt

∫ 1

0

x9|yx|2 dx+27ε

2

∫ 1

0

x8|yxx|2 dx ≤ 252ε

∫ 1

0

x6|yx|2 dx.

Integrating in time and thanks to (4.17) we find

ε

∫ 1

0

∫ 1

0

x8|yxx|2 dxdt ≤ Cε(ε−1 + ε4 + M4)eCM(∫ 1

0

|v|2 dt+

∫ 1

0

(|y0|2 + |y′′′0 |2

)dx

).

(4.18)Next, integrating by parts, we get

ε

∫ 1

0

x10|yxxx|2 dx = −ε∫ 1

0

x10y4xyxx dx+ 45ε

∫ 1

0

x8|yxx|2 dx.

Then, by Young’s inequality and (4.18):

ε

∫ 1

0

x10|yxxx|2 dx = λε

∫ 1

0

x12|y4x|2 dx

+ Cλε(ε−1 + ε4 + M4)eCM

(∫ 1

0

|v|2 dt+

∫ 1

0

(|y0|2 + |y′′′0 |2

)dx

), (4.19)

for every ε > 0 and λ > 0 is to be chosen later on.Now, similar computations made to obtain (4.14) (taking h = yxxx and θ = x13 in (4.7))

yield

1

2

d

dt

∫ L

0

x13|yxxx|2 dx+39ε

2

∫ 1

0

x12|y4x|2 dx+ 78ε|yxxx|x=1|2 ≤ 858ε

∫ 1

0

x10|yxxx|2 dx.

We take λ = 39858×4 in (4.19) and we deduce

1

2

d

dt

∫ 1

0

x13|yxxx|2 dx+39ε

4

∫ 1

0

x12|y4x|2 dx+ 78ε|yxxx|x=1|2

≤ Cε(ε−1 + ε4 + M4)eCM(∫ 1

0

|v|2 dt+

∫ 1

0

(|y0|2 + |y′′′0 |2

)dx

).

15

Integrating in time between 0 and 1, together with (4.18) and (4.19), we obtain

ε

∫ 1

0

∫ 1

0

(x8|yxx|2 + x10|yxxx|2 + x12|y4x|2

)dx dt+ ε

∫ 1

0

|yxxx|x=1|2 dt

≤ Cε(ε−1 + ε4 + M4)eCM(∫ 1

0

|v|2 dt+

∫ 1

0

(|y0|2 + |y′′′0 |2

)dx

). (4.20)

To conclude, notice that for any δ > 0 we have∫ 1

0

|yxx|x=δ|2 dt = −2

∫ 1

0

∫ 1

δ

yxxyxxx dx dt ≤∫ 1

0

∫ 1

δ

(x8

δ8|yxx|2 +

x10

δ10|yxxx|2

)dxdt.

Combining this with (4.20), we obtain (4.5).

Step 3. Estimate on y5x and y6x, and conclusion of (4.6).

From here on, we consider y0 ≡ 0. We take h = y6x and θ = x22 in (4.7):

1

2

d

dt

∫ 1

0

x22|y6x|2 dx+ 33ε

∫ 1

0

x21|y7x|2 dx+ εy8x|x=1y6x|x=1 − 22εy7x|x=1y6x|x=1

− M

2|y6x|x=1|2 ≤ 4620ε

∫ 1

0

x19|y6x|2 dx+ε

2|y7x|x=1|2.

We recall that we already had found that y4x|x=1 = 0 and y5x|x=1 = M ε−1yxxx|x=1.With the help of these identities, the equation and the boundary conditions in (1.1), we findthat

y6x|x=1 = −ε−1ytxxx|x=1,

y7x|x=1 = M2ε−2yxxx|x=1,

andy8x|x=1 = M ε−1y6x|x=1 − ε−1yt5x|x=1 = −2M ε−2ytxxx|x=1.

Using these identities in the previous inequality we get

1

2

d

dt

∫ 1

0

x22|y6x|2 dx+ 33ε

∫ 1

0

x21|y7x|2 dx+3M ε−2

2|ytxxx|x=1|2

+ 11M2ε−2d

dt|yxxx|x=1|2 ≤ 4620ε

∫ 1

0

x19|y6x|2 dx+M4ε−3

2|yxxx|x=1|2. (4.21)

Let us estimate the first term in the right-hand side. First, notice that we have

ε

∫ 1

0

x17|y5x|2 dx = −ε∫ 1

0

x17y4xy6x dx+ 136ε

∫ 1

0

x15|y4x|2 dx

≤ C∫ 1

0

x22|y6x|2 dx+ C(ε+ ε2)

∫ 1

0

x12|y4x|2 dx.

(4.22)

Now, again by integration by parts we get

ε

∫ 1

0

x19|y6x|2 dx = εy5x|x=1y6x|x=1 − 171ε|y5x|x=1|2

+ 171ε

∫ 1

0

x17|y5x|2 dx− ε∫ 1

0

x19y5xy7x dx. (4.23)

16

Combining (4.22)-(4.23) with (4.21) we obtain

1

2

d

dt

∫ 1

0

x22|y6x|2 dx+33ε

2

∫ 1

0

x21|y7x|2 dx+

∫ 1

0

x17|y5x|2 dx

+ 2310M ε−1d

dt|yxxx|x=1|2 + 11M2ε−2

d

dt|yxxx|x=1|2 ≤ C

∫ 1

0

x22|y6x|2 dx

+ C(1 + ε2 + M4ε−3)

(∫ 1

0

x12|y4x|2 dx+ |yxxx|x=1|2).

Using (4.20) to estimate the last term and applying Gronwall’s lemma, we get∫ 1

0

∫ 1

0

(x17|y5x|2 + x22|y6x|2

)dx dt

≤ C(ε−1 + ε4 + M4)eCM(1 + ε2 + M4ε−3

) ∫ 1

0

|v|2 dt. (4.24)

Finally, using the equation in (1.1), we obtain from (4.20) and (4.24):∫ 1

0

∫ 1

0

(x17|ytxx|2 + x22|ytxxx|2

)dx dt

≤ C(ε−1 + ε4 + M4)eCM (ε2 + ε4 + M4ε−1 + M2)

∫ 1

0

|v|2 dt.

From here, we obtain (4.6) as in the end of Step 2.

Proof of Lemma 4.1. As in the proof of Corollary 2.2, we write v = v0 + v′1, for somev0, v1 ∈ L2(0, T ), and y = p + qt, where p and q are the solutions of (2.9) and (2.10),respectively. From Lemma 4.2, using (4.3) for p and (4.4) for q, we have from (2.8) that

‖yxx|x=δL‖2L2(0,T )

≤(C0(T, L, ε,M, δ)+C ′0(T, L, ε,M, δ)

)(‖v0‖2L2(0,T )+‖v1‖

2L2(0,T )+‖y0‖

2L2(0,L)+‖y

′′′0 ‖2L2(0,L)

),

for every v0, v1 ∈ L2(0, T ) such that (2.6) is satisfied. Therefore, from (2.6)-(2.7) we ob-tain (4.1).

Finally, let us state another technical result whose proof is given in Appendix B.

Lemma 4.3. Let 0 < L1 < L. For any h ∈ H3n(L1, L), there exists h ∈ H3

n(0, L) such that

h|(L1,L) = h and ‖h‖2H3n(0,L)

≤ C1(L1, L)‖h‖2H3n(L1,L)

whereC1(L1, L) := CL1

((L5

1 + (L− L1)5 + (L− L1)−1)

+ 1. (4.25)

4.2 Auxiliary problem and conclusion

Now, we introduce the following auxiliary control problem:wt + εwxxx −Mwx = 0 in (0, T )× (L0, L),wxx|x=L0

= u(t), wx|x=L = 0, wxx|x=L = 0 in (0, T ),w|t=0 = w0 in (L0, L).

(4.26)

17

For this problem, we define the cost associated to the null controllability as follows:

Kεcost := sup

w0∈H3n(L0,L)

w0 6=0

minu∈L2(0,T )w|t=T=0

‖u‖2L2(0,T )

‖w0‖2H3n(L0,L)

. (4.27)

We find now a lower bound of Kεcost.

Lemma 4.4. Let T, L, ε,M > 0 and L0 < L. Then

Kεcost ≥

sinh2((L− L0)M1/2ε−1/2

)εTM(L− L0)

. (4.28)

Proof. We remark that Kεobs is the smallest constant such that

suph∈H3

n(L0,L)

( ∫ LL0ϕ|t=0hdx

)2‖h‖2H3

n(L0,L)

≤ ε2Kεcost

∫ T

0

|ϕ|x=L0|2 dt, (4.29)

for any ϕT ∈ L2(L0, L), where ϕ is the solution of−ϕt − εϕxxx +Mϕx = 0 in (0, T )× (L0, L),ϕx|x=L0

= 0, (εϕxx −Mϕ)|x=L0= 0, (εϕxx −Mϕ)|x=L = 0 in (0, T ),

ϕ|t=T = ϕT in (L0, L).(4.30)

Now, letψ(x) := cosh

((x− L0)M1/2ε−1/2

).

It is straightforward to check that ψ satisfies the partial differential equation and theboundary conditions in (4.30). Furthermore, since h ≡ 1 belongs to H3

n(L0, L), we have

suph∈H3

n(L0,L)

( ∫ LL0ψ hdx

)2‖h‖2H3

n(0,L)

≥ 1

L− L0

(∫ L

L0

ψ dx

)2

=ε sinh2

((L− L0)M1/2ε−1/2

)M(L− L0)

.

On the other hand, ∫ T

0

|ψ|x=L0|2 dt = T.

Consequently, from (4.29) we deduce that (4.28) holds.

Conclusion of the proof of Theorem 1.2. We argue by contradiction, i.e., we suppose thatfor any y0 ∈ H3

n(0, L), there exists v ∈ H−1(0, T ) such that the solution of (1.1) satisfiesy|t=T = 0 and

‖v‖2H−1(0,T ) ≤(

(1− γ)K(T, L, ε,M, δ)

C−1(T, L, ε,M, δ)C1(δL, L)− 1

)‖y0‖2H3

n(0,L). (4.31)

Let w0 ∈ H3n(L0, L) where 0 < L0 < L. We choose y0 to be the extension of w0 given

by Lemma 4.3. In particular, we have

‖y0‖2H3n(0,L)

≤ C1(L0, L)‖w0‖2H3n(L0,L)

. (4.32)

Observe that y solves (4.26) with u := yxx|x=L0and w0 := y0|(L0,L). Let us take L0 := δL.

Then, using Lemma 4.1 we have

‖u‖2L2(0,T ) ≤ C−1(T, L, ε,M, δ)(‖v‖2H−1(0,T ) + ‖y0‖2H3

n(0,L)

).

18

From (4.31) and (4.32):

‖u‖2L2(0,T ) ≤ (1− γ)K(T, L, ε,M, δ)‖w0‖2H3n(δL,L)

which implies thatKεcost ≤ (1− γ)K(T, L, ε,M, δ).

This and (4.28) show that γ ≤ 0, which is a contradiction.

5 Proof of Theorem 1.5

The proof of Theorem 1.5 relies on finding a particular solution ϕ of (3.2) such that‖ϕxx|x=0‖L2(0,T ) decays exponentially as ε → 0+ and ‖ϕ|t=0‖L2(0,L) behaves like a con-stant in ε. The proof we perform here is inspired by [10, Theorem 1.4]. The main differencewith respect to [10] is that the boundary condition on x = L is not homogeneous.

Let M < 0. In this case, we can look at (3.2) as−ϕt − εϕxxx − |M |ϕx = 0 in Q,ϕ|x=0 = 0, ϕx|x=0 = 0, (εϕxx + |M |ϕ)|x=1 = 0 in (0, T ),

ϕ|t=T = ϕT in (0, L).(5.1)

First, notice that we have the dissipation estimate:

‖ϕ|t=t1‖L2(0,L) ≤ ‖ϕ|t=t2‖L2(0,L) for every 0 ≤ t1 ≤ t2 ≤ T. (5.2)

Now, we choose R > 0 such that

0 < 7R < L− |M |T, (5.3)

and a non-negative function ϕT ∈ C∞0 (0, L) such that

Supp(ϕT ) ⊂ (L− 2R,L−R) and ‖ϕT ‖L2(0,L) = 1. (5.4)

Let ϕ be the solution of (5.1) associated to ϕT as initial condition. We will prove that

‖ϕ|t=0‖L2(0,L) ≥ c > 0 (5.5)

and

‖ϕxx|x=0‖L2(0,T ) ≤ C(ε) exp

[− R3/2

33/2ε1/2T 1/2

]‖ϕT ‖L2(0,L), (5.6)

where C(ε) > 0 depends on ε−1 at most polynomially.Let us explain how (5.5) and (5.6) allow us to conclude Theorem 1.5. Let v ∈ L2(0, T )

be a control which drives the solution y of (1.1) from y0 to 0 (we know such a v exists byTheorem 1.1 and [12]). We multiply (1.1) by ϕ and integrate by parts to get

−∫ L

0

y0 ϕ|t=0 dx = ε

∫ T

0

v ϕxx|x=0 dt ≤ ε‖v‖L2(0,T )‖ϕxx|x=0‖L2(0,T ).

Setting y0 := −ϕ|t=0 and using (5.4), (5.5) and (5.6) in this last inequality we obtain (1.8).

Proof of (5.5). Let us define θ(t, x) := ϕT (x + |M |(T − t)) in Q. It follows from (5.3)and (5.4) that θ(t, ·) ∈ C∞0 (0, L) for all t ∈ [0, T ].

19

We multiply (5.1) by θ and after integration by parts we obtain∫ L

0

ϕT θ|t=T dx =

∫ L

0

ϕ|t=0 θ|t=0 dx+ ε

∫∫Q

ϕ θxxx dxdt. (5.7)

From (5.2) with t2 = T , we find

‖ϕ‖L2(Q) ≤ T 1/2‖ϕT ‖L2(0,L).

On the other hand, from the definition of θ, (5.3) and (5.4), it is easy to see that

‖θ|t=0‖L2(0,L) = ‖ϕT ‖L2(0,L)

and‖θxxx‖L2(Q) ≤ T 1/2R1/2‖ϕ′′′T ‖L∞(0,L).

Using these elements in (5.7), together with Young’s inequality, we obtain (5.5) for εsmall enough depending on T and R.

Proof of (5.6). The objective is to prove that

‖ϕ‖L2(0,T ;H3(0,R)) ≤ C(ε) exp

[− R3/2

33/2ε1/2T 1/2

]‖ϕT ‖L2(0,L), (5.8)

from where (5.6) will readily follow. This is done by proving the estimate

‖ϕ(t)‖L2(0,3R) ≤ C exp

[− R3/2

33/2ε1/2T 1/2

]‖ϕT ‖L2(0,L), (5.9)

and then applying an internal regularity result proved in [10] to conclude.We consider a cut-off function γ ∈ C∞(R) such that

γ ≥ 0, γ′ ≤ 0, γ = 1 in (−∞, L− 3R), γ = 0 in (L− 2R,+∞). (5.10)

We set β(t, x) := −|M |(T − t) − x and multiply (5.1) by γ(−β)erβϕ, where r > 0 is tobe chosen later on. We perform several integrations by parts, but observe that from (5.3)and (5.10), we have that γ(−β(t, x)) = 0 for all (t, x) ∈ [0, T ]× [L− 2R,L], so there are noboundary terms. We get, after neglecting the positive terms,

− 1

2

d

dt

∫ L

0

γ(−β)erβ |ϕ|2 dx− εr3

2

∫ L

0

γ(−β)erβ |ϕ|2 dx

≤ εC(r)(‖γ′‖∞ + ‖γ′′‖∞ + ‖γ′′′‖∞)

∫ L−2R−|M |(T−t)

L−3R−|M |(T−t)erβ |ϕ|2dx,

where C(r) is a polynomial function of degree 2 in r and we have used (5.10) to restrict thelimits in the integral. Multiplying by exp

(− εr3(T − t)

)and using that β is decreasing, we

have

− 1

2

d

dt

(exp

(− εr3(T − t)

) ∫ L

0

γ(−β)erβ |ϕ|2 dx

)≤ εC(r) exp

(− εr3(T − t) + r(3R− L)

) ∫ L

0

|ϕ|2 dx.

20

By (5.2), we obtain

−1

2

d

dt

(exp

(− εr3(T − t)

) ∫ L

0

γ(−β)erβ |ϕ|2 dx

)≤ εC(r) exp

(r(3R− L)

) ∫ L

0

|ϕT |2 dx.

Integrating in (t, T ), we get:∫ L

0

γ(−β)erβ |ϕ|2 dx ≤ εC(r)T exp(εr3(T − t) + r(3R− L)

) ∫ L

0

|ϕT |2 dx,

where we have used the fact that γ(s)ϕT (s) = 0 for all s ∈ R.Now, notice that γ(−β(t, x)) = 1 for all (t, x) ∈ [0, T ] × [0, 3R] thanks to (5.3), so we

have

exp(− r(|M |(T − t)

) ∫ 3R

0

|ϕ|2 dx ≤ εC(r) exp(εr3(T − t)− r(L− 6R))

) ∫ L

0

|ϕT |2 dx,

and thus ∫ 3R

0

|ϕ|2 dx ≤ εC(r) exp(εr3T − r(L− |M |T − 6R))

) ∫ L

0

|ϕT |2 dx.

Again from (5.3), we obtain∫ 3R

0

|ϕ|2 dx ≤ εC(r) exp(εr3T −Rr)

) ∫ L

0

|ϕT |2 dx.

We finish the proof of (5.9) by choosing r > 0 such that it minimises the expressioninside the exponential, that is.

r :=R1/2

31/2ε1/2T 1/2.

To prove (5.8), we will use the following lemma, which corresponds to Proposition 3.3in [10]

Lemma 5.1. Let ε ∈ (0, 1] and M ∈ R. Consider a solution w of−wt − εwxxx +Mwx = 0 in Q,w|x=0 = 0, wx|x=0 = 0, w|x=L = u(t) in (0, T ),w|t=T = wT in (0, L),

(5.11)

for some u ∈ L2(0, T ) and wT ∈ H3(0, L) ∩H20 (0, L). Then, w ∈ L2(0, T ;H4(0, L/2)), with

the estimate‖w‖L2(0,T ;H4(0,L/2)) ≤ C(ε)

(‖wT ‖H3(0,L) + ‖u‖L2(0,T )

), (5.12)

for some constant C(ε) depending at most polynomially in ε−1 and |M |.

Let w := ϕ|[0,2R] and apply Lemma 5.1 with (0, 2R) and (0, R) instead of (0, L) and(0, L/2), respectively. Notice that with this setting we have wT = 0 and u = ϕ|x=2R ∈L2(0, T ). Thus, from (5.12) we have

‖ϕ‖L2(0,T ;H4(0,R)) ≤ C(ε)‖ϕx‖L2(0,T ;L2(0,2R)). (5.13)

Now, we estimate the term in the right-hand side in a slightly larger interval. To do this,we multiply (5.1) by (3R− x)3ϕ and integrate in (0, 3R). We obtain

− 1

2

d

dt

∫ 3R

0

(3R− x)3|ϕ|2 dx+9ε

2

∫ 3R

0

(3R− x)2|ϕx|2 dx

= 3ε

∫ 3R

0

|ϕ|2 dx+3|M |

2

∫ 3R

0

(3R− x)2|ϕ|2 dx.

21

Since ϕT = 0 in (0, 3R), we get by integrating between 0 and T

‖(3R− ·)ϕx‖L2(0,T ;L2(0,3R)) ≤C

ε1/2‖ϕ‖L2(0,T ;L2(0,3R)). (5.14)

Combining this with (5.13) and (5.9), we obtain (5.8).

The proof of Theorem 1.5 is complete.

A Proof of Proposition 3.1

We now follow the steps of [9] and [10]. Let ψ := e−sαφ. Using equation (3.4), we get

L1ψ + L2ψ = L3ψ,

where we have denoted

L1ψ := εψxxx + ψt + 3εs2α2xψx −Mψx,

L2ψ := (εs3α3x + sαt −Msαx)ψ + 3εsαxxψx + 3εsαxψxx

andL3ψ := −3εs2αxαxxψ.

Notice that we have the following boundary values for ψ:

ψ|x=L = 0 (A.1)

ψx|x=0 = −sαx|x=0ψ|x=0 (A.2)

ψxx|x=0 = (s2α2x − sαxx)|x=0ψ|x=0 (A.3)

Taking the L2-norm we have

‖L1ψ‖2L2(Q) + ‖L2ψ‖2L2(Q) + 2(L1ψ, L2ψ)L2(Q) = ‖L3ψ‖2L2(Q). (A.4)

In the following, our efforts will be devoted to computing the double product in theprevious equation. Let us denote by (Liψ)j the j-th term of Liψ.

Computing ((L1ψ)1, L2ψ)L2(Q).For the first term, we integrate by parts twice in space:

((L1ψ)1, (L2ψ)1)L2(Q) = −1

2ε

∫∫Q

(εs3α3x + sαt −Msαx)∂x|ψx|2 dx dt

−ε∫∫

Q

(3εs3α2xαxx + sαxt −Msαxx)ψψxx dx dt

−ε∫ T

0

(εs3α3x + sαt −Msαx)|x=0ψ|x=0ψxx|x=0 dt

=3

2ε

∫∫Q

(3εs3α2xαxx + sαxt −Msαxx)|ψx|2 dxdt

−1

2ε

∫ T

0

(εs3α3x + sαt −Msαx)|x=L|ψx|x=L|2dt

22

−3ε2s3∫∫

Q

α3xx|ψ|2 dxdt− 1

2ε

∫ T

0

(6εs3αxα2xx + sαxxt)|x=0|ψ|x=0|2 dt

+1

2ε

∫ T

0

(εs3α3x + sαt −Msαx)|x=0|ψx|x=0|2 dt

+ε

∫ T

0

(3εs3α2xαxx + sαxt −Msαxx)|x=0ψ|x=0ψx|x=0 dt

−ε∫ T

0

(εs3α3x + sαt −Msαx)|x=0ψ|x=0ψxx|x=0 dt

Using the properties (3.12), (3.13), (A.2) and (A.3), we obtain

((L1ψ)1, (L2ψ)1)L2(Q) ≥9

2ε2s3

∫∫Q

α2xαxx|ψx|2 dxdt− Cεs(T + |M |T 2)

∫∫Q

α3|ψx|2 dx dt

−ε2 s3

2

∫ T

0

α3x|x=L|ψx|x=L|

2 dt−Cεs(T+|M |T 2)

∫ T

0

α3x|x=L|ψx|x=L|

2 dt−Cε2s3T 2

∫∫Q

α5|ψ|2 dxdt

−Cε(εs5 + εs4T + s3(T + εT 2 + |M |T 2) + s2(T 2 + |M |T 3) + sT 3

)∫ T

0

α5|x=0|ψ|x=0|2 dt.

We integrate by parts again in space in the second term, and using the boundary valuesfor ψx|x=0 and ψxx|x=0:

((L1ψ)1, (L2ψ)2)L2(Q) = −3ε2s

∫∫Q

αxx|ψxx|2 dxdt+ 3ε2s

∫ T

0

αxx|x=Lψx|x=Lψxx|x=L dt

− 3ε2s

∫ T

0

αxx|x=0ψx|x=0ψxx|x=0 dt ≥ −3ε2s

∫∫Q

αxx|ψxx|2 dx dt− ε2s

2

∫ T

0

αx|x=L|ψxx|x=L|2 dt

− Cε2sT 2

∫ T

0

α3x|x=L|ψx|x=L|

2 dt− Cε2(s4T + s3T 2)

∫ T

0

α5|x=0|ψ|x=0|2 dt.

Here, we have used the boundary values (A.2), (A.3), the properties in (3.12) and Young’sinequality.

For the third term, we proceed in a similar manner:

((L1ψ)1, (L2ψ)3)L2(Q) = −3

2ε2s

∫∫Q

αxx|ψxx|2 dx dt+3

2ε2s

∫ T

0

αx|x=L|ψxx|x=L|2 dt

− 3

2ε2s

∫ T

0

αx|x=0|ψxx|x=0|2 dt ≥ −3

2ε2s

∫∫Q

αxx|ψxx|2 dx dt

+3

2ε2s

∫ T

0

αx|x=L|ψxx|x=L|2 dt− Cε2(s5 + s3T 2)

∫ T

0

α5|x=0|ψ|x=0|2 dt.

23

Putting together these computations, we obtain

((L1ψ)1, L2ψ)L2(Q) ≥9

2ε2s3

∫∫Q

α2xαxx|ψx|2 dx dt− 9

2ε2s

∫∫Q

αxx|ψxx|2 dx dt

+ ε2s

∫ T

0

αx|x=L|ψxx|x=L|2 dt− ε2 s3

2

∫ T

0

α3x|x=L|ψx|x=L|

2 dt

− Cεs(T + |M |T 2)

∫∫Q

α3|ψx|2 dxdt− Cε2s3T 2

∫∫Q

α5|ψ|2 dxdt

− Cεs(T + εT 2 + |M |T 2)

∫ T

0

α3x|x=L|ψx|x=L|

2 dt

− Cε(εs5 + εs4T + s3(T + εT 2 + |M |T 2) + s2T (T + |M |T 2) + sT 3

)∫ T

0

α5|x=0|ψ|x=0|2 dt.

(A.5)

Computing ((L1ψ)2, L2ψ)L2(Q).For the first term, we integrate by parts in time:

((L1ψ)2, (L2ψ)1)L2(Q) = −1

2

∫∫Q

(3εs3α2xαxt + sαtt −Msαxt)|ψ|2 dxdt

≥ −C(εs3 + s(T + |M |T 2))T

∫∫Q

α5|ψ|2 dx dt.

The second terms gives:

((L1ψ)2, (L2ψ)2)L2(Q) = 3εs

∫∫Q

αxxψxψt dxdt.

In the third term, we integrate by parts first in space and then in time. We obtain

((L1ψ)2, (L2ψ)3)L2(Q) = −3

2εs

∫∫Q

αx∂t|ψx|2 dxdt− 3εs

∫∫Q

αxxψxψt dxdt

− 3εs

∫ T

0

αx|x=0ψx|x=0ψt|x=0 dt =3

2εs

∫∫Q

αxt|ψx|2 dx dt

− 3εs

∫∫Q

αxxψxψt dxdt− 3εs2∫ T

0

αx|x=0αxt|x=0|ψ|x=0|2 dt

≥ −3εs

∫∫Q

αxxψxψt dxdt− CεsT∫∫

Q

α3|ψx|2 − Cεs2T 2

∫ T

0

α5|x=0|ψ|x=0|2 dt,

where we have used the boundary value (A.2).Putting together this inequalities, we have

((L1ψ)2, L2ψ)L2(Q) ≥ −C(εs3 + s(T + |M |T 2))T

∫∫Q

α5|ψ|2 dxdt

− CεsT∫∫

Q

α3|ψx|2 − Cεs2T 2

∫ T

0

α5|x=0|ψ|x=0|2 dt.

(A.6)

Computing ((L1ψ)3, L2ψ)L2(Q).

24

We integrate by parts in space and the properties (3.12)-(3.13) to treat the first term:

((L1ψ)3, (L2ψ)1)L2(Q)

= −1

2ε

∫∫Q

(15εs5α4xαxx + 6s3αxαxxαt + 3s3α2

xαxt − 9Ms3α2xαxx)|ψ|2 dxdt

− 1

2ε

∫ T

0

(3εs5α5x + 3s3α2

xαt + 3Ms3α3x)|x=0|ψ|x=0|2 dt

≥ 15

2C5

0ε2s5∫∫

Q

α5|ψ|2 dxdt− Cεs3(T + |M |T 2)

∫∫Q

α5|ψ|2 dxdt

− Cε(εs5 + s3(T + |M |T 2))

∫ T

0

α5|x=0|ψ|x=0|2 dt.

The second term simply gives:

((L1ψ)3, (L2ψ)2)L2(Q) = 9ε2s3∫∫

Q

α2xαxx|ψx|2 dxdt.

As for the third term, we integrate by parts in space once and use the boundaryvalue (A.2). We obtain:

((L1ψ)3, (L2ψ)3)L2(Q) = −27

2ε2s3

∫∫Q

α2xαxx|ψx|2 dx dt+

9

2ε2s3

∫ T

0

α3x|x=L|ψx|x=L|

2 dt

− 9

2ε2s3

∫ T

0

α3x|x=0|ψx|x=0|2 dt = −27

2ε2s3

∫∫Q

α2xαxx|ψx|2 dx dt

+9

2ε2s3

∫ T

0

α3x|x=L|ψx|x=L|

2 dt− 9

2ε2s5

∫ T

0

α5x|x=0|ψ|x=0|2 dt.

Putting together these estimates, we get:

((L1ψ)3, L2ψ)L2(Q) ≥ −9

2ε2s3

∫∫Q

α2xαxx|ψx|2 dxdt+

15

2C5

0ε2s5∫∫

Q

α5|ψ|2 dxdt

+9

2ε2s3

∫ T

0

α3x|x=L|ψx|x=L|

2 dt− Cεs3(T + |M |T 2)

∫∫Q

α5|ψ|2 dxdt

− Cε(εs5 + s3(T + |M |T 2))

∫ T

0

α5|x=0|ψ|x=0|2 dt.

(A.7)

Computing ((L1ψ)4, L2ψ)L2(Q).For the first term, we integrate by parts in space:

((L1ψ)4, (L2ψ)1)L2(Q) =M

2

∫∫Q

(3εs3α2xαxx + sαxt −Msαxx)|ψ|2 dxdt

+M

2

∫ T

0

(εs3α3x + sαt −Msαx)|x=0|ψ|x=0|2 dt

≥ −C(εs3 + s(T + |M |T 2))|M |T 2

∫∫Q

α5|ψ|2 dx dt

− C(εs3 + s(T + |M |T 2))|M |T 2

∫ T

0

α5|x=0|ψ|x=0|2 dt.

25

The second term gives directly:

((L1ψ)4, (L2ψ)2)L2(Q) ≥ −C|M |εsT 2

∫∫Q

α3|ψx|2 dx dt.

The third and final term gives, after integration by parts:

((L1ψ)4, (L2ψ)3)L2(Q) =3

2Mεs

∫∫Q

αxx|ψx|2 dxdt− 3

2Mεs

∫ T

0

αx|x=L|ψx|x=L|2 dt

+3

2Mεs

∫ T

0

αx|x=0|ψx|x=0|2 dt ≥ −C|M |εsT 2

∫∫Q

α3|ψx|2 dxdt

− C|M |εsT 2

∫ T

0

α3x|x=L|ψx|x=L|

2 dt− C|M |εs3T 2

∫ T

0

α5|x=0|ψ|x=0|2 dt.

Putting together these expressions, we obtain:

((L1ψ)4, L2ψ)L2(Q) ≥ −C(εs3 + s(T + |M |T 2))|M |T 2

∫∫Q

α5|ψ|2 dxdt

− C|M |εsT 2

∫∫Q

α3|ψx|2 dx dt− C|M |εsT 2

∫ T

0

α3x|x=L|ψx|x=L|

2 dt

− C(εs3 + s(T + |M |T 2))|M |T 2

∫ T

0

α5|x=0|ψ|x=0|2 dt.

(A.8)

The entire product (L1ψ, L2ψ)L2(Q).Adding inequalities (A.5)-(A.8), we find four positive terms, namely:

A1 :=15

2C5

0ε2s5∫∫

Q

α5|ψ|2 dxdt, A2 :=9

2C0ε

2s

∫∫Q

α|ψxx|2 dx dt,

A3 := 4ε2s3∫ T

0

α3x|x=L|ψx|x=L|

2 dt, A4 := ε2s

∫ T

0

αx|x=L|ψxx|x=L|2 dt.

In the following, we explain how to estimate the nonpositive integrals coming from theaddition of (A.5)-(A.8) in terms of Ai.

Let us start with the terms concerning |ψ|2 in Q. We can easily check that they can allbe bounded by

C(s3(εT + ε2T 2 + |M |εT 2) + s(T 2 + |M |T 3 + |M |2T 4)

) ∫∫Q

α5|ψ|2 dx dt,

which by taking s ≥ C(T + T 1/2ε−1/2 + |M |1/2ε−1/2T ) can be absorbed by A1.The integral of |ψx|x=L|2, can be bounded by

Cεs(T + εT 2 + |M |T 2)

∫ T

0

α3x|x=L|ψx|x=L|

2 dt.

Taking s ≥ C(T + T 1/2ε−1/2 + |M |1/2ε−1/2T ), this term can be absorbed by A3.Furthermore, taking s ≥ C(T + T 1/2ε−1/2 + |M |1/2ε−1/2T ) shows that all the integrals

concerning |ψ|x=0|2 can be estimated by

Cε2s5∫ T

0

α5|x=0|ψ|x=0|2 dt. (A.9)

26

Finally, let us treat the terms containing |ψx|2 in Q. They can be estimated by

Cεs(T + |M |T 2)

∫∫Q

α3|ψx|2 dxdt.

Now, similarly as the previous steps, integration by parts in space shows that:

Cεs(T + |M |T 2)

∫∫Q

α3|ψx|2 dxdt =3

2Cεs(T + |M |T 2)

∫∫Q

(2αα2x + α2αxx)|ψ|2 dxdt

+3

2Cεs(T + |M |T 2)

∫ T

0

α2|x=0αx|x=0|ψ|x=0|2 dt− Cεs(T + |M |T 2)

∫∫Q

α3ψψxx dx dt

− Cεs(T + |M |T 2)

∫ T

0

α3|x=0ψ|x=0ψx|x=0 dt

≥ −C(εsT 2(T + |M |T 2) + ε1/2s2(T 3/2 + |M |3/2T 3)

) ∫∫Q

α5|ψ|2 dxdt

− Cε3/2(T 1/2 + |M |1/2T )

∫∫Q

α|ψxx|2dxdt

− Cε(sT 2(T + |M |T 2) + s2T (T + |M |T 2)

) ∫ T

0

α5|x=0|ψ|x=0|2 dt. (A.10)

Notice that here we have used (A.2) and Young’s inequality. By taking s ≥ C(T +ε−1/2T 1/2 + |M |1/2ε−1/2T ), the first two integrals can be absorbed by A1 and A2, respec-tively, and the last one can be estimated by (A.9).

Finally, all these estimations give

(L1ψ, L2ψ)L2(Q) ≥ Cε2s5∫∫

Q

α5|ψ|2 dxdt+ Cε2s

∫∫Q

α|ψxx|2 dxdt

+ ε2s3∫ T

0

α3x|x=L|ψx|x=L|

2 dt+ Cε2s

∫ T

0

αx|x=L|ψxx|x=L|2 dt− Cε2s5∫ T

0

α5|x=0|ψ|x=0|2 dt,

for every s ≥ C(T + ε−1/2T 1/2 + |M |1/2ε−1/2T ).Coming back to (A.4), and together with the fact that

‖L3ψ‖2L2(Q) ≤ Cε2s4T

∫∫Q

α5|ψ|2 dx dt,

we obtain

ε2s5∫∫

Q

α5|ψ|2 dxdt+ ε2s

∫∫Q

α|ψxx|2 dxdt+ ε2s3∫ T

0

α3x|x=L|ψx|x=L|

2 dt

+ ε2s

∫ T

0

αx|x=L|ψxx|x=L|2 dt ≤ Cε2s5∫ T

0

α5|x=0|ψ|x=0|2 dt, (A.11)

for every s ≥ C(T + ε−1/2T 1/2 + |M |1/2ε−1/2T ).

Coming back to the original variable.Let us now go back to the original variable φ. First, we point out that the same compu-

tations made in (A.10), show that

ε2s3∫∫

Q

α3|ψx|2 dxdt ≤Cε2s5∫∫

Q

α5|ψ|2 dxdt+ Cε2s

∫∫Q

α|ψxx|2 dxdt

+ Cε2s5∫ T

0

α5|x=0|ψ|x=0|2 dt,

27

as long as s ≥ CT . This means that we can add this term to the left-hand side of (A.11),and together with ψ = e−sαφ, we have directly from (A.11) that

ε2s5∫∫

Q

e−2sαα5|φ|2 dxdt+ ε2s3∫∫

Q

α3|ψx|2 dxdt+ ε2s

∫∫Q

α|ψxx|2 dx dt

+ ε2s3∫ T

0

α3x|x=L|ψx|x=L|

2 dt+ ε2s

∫ T

0

αx|x=L|ψxx|x=L|2 dt

≤ Cε2s5∫ T

0

e−2sα|x=0α5|x=0|φ|x=0|2 dt, (A.12)

for every s ≥ C(T + ε−1/2T 1/2 + |M |1/2ε−1/2T ).Now, from ψ = e−sαφ, we find that

s3/2e−sαα3/2φx = s3/2α3/2ψx + s5/2e−sαα3/2αxφ,

and taking the L2(Q)-norm, we see that we can add

ε2s3∫∫

Q

e−2sαα3|φx|2 dxdt

to the left-hand side of (A.12). Similarly, from

s1/2e−sαα1/2φxx = s1/2α1/2ψxx + s3/2e−sαα1/2αxxφ+ 2s3/2α1/2αxψx + s5/2e−sαα1/2α2xφ,

we can add

ε2s

∫∫Q

e−2sαα|φxx|2 dx dt

to the left-hand side of (A.12) if s ≥ CT . Finally, using (A.1), we can add to the left-handside of (A.12) the respective boundary integrals and obtain

ε2s5∫∫

Q

e−2sαα5|φ|2 dxdt+ ε2s3∫∫

Q

e−2sαα3|φx|2 dx dt+ ε2s

∫∫Q

e−2sαα|φxx|2 dx dt

+ ε2s3∫ T

0

e−2sα|x=Lα3x|x=L|φx|x=L|

2 dt+ ε2s

∫ T

0

e−2sα|x=Lαx|x=L|φxx|x=L|2 dt

≤ Cε2s5∫ T

0

e−2sα|x=0α5|x=0|φ|x=0|2 dt, (A.13)

for every s ≥ C(T + ε−1/2T 1/2 + |M |1/2ε−1/2T ). The proof of Proposition 3.1 is complete.

B Proof of Lemma 4.3

We define h by:

h(x) := p(x), x ∈ [0, L1], h(x) := h(x), x ∈ (L1, L],

where p(x) = ax2 + bx+ c is given by

a :=h′′(L1)

2, b := h′(L1)− L1h

′′(L1), and c := h(L1)− L1h′(L1) + L2

1

h′′(L1)

2.

It is easy to check that h ∈ H3n(0, L) and

‖h‖2H3n(0,L)

= ‖p‖2L2(0,L1)+ ‖h‖2H3

n(L1,L). (B.1)

28

Let us estimate the term concerning p. Notice that

h(L1) = − 1

L− L1

∫ L

L1

((L− x)h)′ dx =1

L− L1

∫ L

L1

hdx− 1

L− L1

∫ L

L1

(L− x)h′ dx.

Taking the square and using Cauchy-Schwarz inequality we get

|h(L1)|2 ≤ 2

L− L1

∫ L

L1

|h|2 dx+2(L− L1)

3

∫ L

L1

|h′|2 dx

≤ 2

L− L1

∫ L

L1

|h|2 dx+(L− L1)5

6

∫ L

L1

|h′′′|2 dx

(B.2)

where we have used that h′(L) = h′′(L) = 0 in the last inequality.Now, using that h′(L) = 0, we have

h′(L1) = −∫ L

L1

h′′ dx ≤ (L− L1)1/2(∫ L

L1

|h′′|2 dx

)1/2

and therefore

|h′(L1)|2 ≤ (L− L1)

∫ L

L1

|h′′|2 dx ≤ (L− L1)3

2

∫ L

L1

|h′′′|2 dx. (B.3)

Similarly, we get

|h′′(L1)|2 ≤ (L− L1)

∫ L

L1

|h′′′|2 dx. (B.4)

From (B.2)-(B.4) we obtain

‖p‖2L2(0,L1)≤ CL1

(|h′′(L1)|2L4

1 + |h′(L1)|2L21 + |h(L1)|2

)≤ CL1

((L− L1)L4

1 + (L− L1)3L21 + (L− L1)5 + (L− L1)−1

)‖h‖2H3

n(L1,L).

Combining this with (B.1) and using Young’s inequality, we obtain the desired result.

Acknowledgements

Great part of this work has been accomplished while the first author was a PhD student atLaboratoire Jacques-Louis Lions, Universite Pierre et Marie Curie, Paris, France. He wouldlike to thank them for their support and hospitality during those years.

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