Contrôle de l’équation de Burgers en présence d’une couche limite
2.4 Passive stage and dissipation of the boundary layer
The goal of this section is to prove the following estimate concerning the dissipation of the boundary residue Φε created in the previous section. Indeed, although its L2-norm increases as ε goes to zero, regularization effects of the Burgers equation will dissipate it in any positive timeT.
Lemma 7. Let T >0 be a fixed positive time. For any ε >0, let us consider φ∈X the solution to the following system :
φt+φφx−φxx = 0 in (0, T)×(0,1), φ(t,0) = 0 in (0, T),
φ(t,1) = 0 in (0, T), φ(0, x) = Φε(x) in (0,1),
whereΦε(x)is the boundary residue defined by (2.34). Then for any δ >0, we have the estimate : kφ(T,·)kL2(0,1)=Oε→0 ε1−δ
. (2.35)
0
y(t, x)
Φε(x)
−3
x= 0.5 x= 1
Figure2.5 – Numerical simulation of the dissipation of the boundary residue Φε(·). At timet= 0, the boundary residue was of sizekΦε(·)k∞= 100.
2.4.1 Cole-Hopf transform
Once again, we are going to use the Cole-Hopf transform to derive precise estimates. Therefore, let us introduce the following change of unknown forx∈[0,1] andt∈[0, T] :
z(t, x) = exp
−1 2
Z x 0
φ(t, s)ds
. This leads to the following heat system for the new unknownz :
zt−zxx = 0 on (0, T)×(0,1), zx(t,0) = 0 on (0, T),
zx(t,1) = 0 on (0, T), z(0, x) =Zε(x) on (0,1),
(2.36)
where the initial dataZε is computed from the initial data Φε: Zε(x) = exp
−1 2
Z x 0
Φε(s)ds
= 1 +eHε(x−1)
1 +e−Hε . (2.37)
An important remark is that Φε ≤ 0. Thus, by the comparison principle from Lemma 1, φ ≤0 on [0, T]×[0,1] andz≥1 on [0, T]×[0,1]. The backwards Cole-Hopf transform will give us :
φ(T) =−2zx(T) z(T).
Hence, using the fact thatz≥1, we will have the following estimate :
|φ(T,·)| ≤2|zx(T,·)|. (2.38) All we have to do is to study theL2-norm ofzx(T). To ease computations, let us introduce :
w= (1 +e−Hε)zx, (2.39)
such thatwis the solution to :
wt−wxx = 0 on (0, T)×(0,1),
w(t,0) = 0 on (0, T),
w(t,1) = 0 on (0, T),
w(0, x) =HεeHε(x−1) on (0,1).
2.4.2 Fourier series decomposition
We use Fourier series to compute w(T,·). We will use the following Hilbert basis ofL2 made of the eigen-functions for the Laplace operator with Dirichlet boundary conditions on [0,1] :
en(x) =√
2 sin(nπx) for n≥1.
Let us compute the decomposition ofw(0,·) on this basis. We integrate by parts twice : hw(0,·)|eni = √
2H εe−Hε
Z 1 0
sin(nπx)eHεxdx
= √
2e−Hε h
sin(nπx)eHεxi1 0
−√
2nπe−Hε Z 1
0
cos(nπx)eHεx
= −ε√ 2
H nπe−Hε h
cos(nπx)eHεxi1 0
−εnπ H
2
hw(0,·)|eni
=
√2 H
εnπ 1 + ε2nH22π2
(−1)n+1+e−Hε .
Now we can estimate the size ofw(T,·) inL2(0,1) : kw(T,·)k2L2 = X
n≥1
hw(0,·)|eni ·e−n2π2T2
≤ 8X
n≥1
ε2n2π2H−2
(1 +ε2n2π2H−2)2e−2n2π2T. Forα∈R, the following easy inequality holds :
α2
(1 +α2)2 ≤min
α2,1 4
.
Hence we split the sum and cut at a levelN(ε) : kw(T,·)k2L2 ≤ 8
N−1
X
n=1
ε2n2π2
H2 + 2X
k≥0
e−2(N+k)2π2T
≤ 8ε2N3π2
3H2 + 2e−2N2π2TX
k≥0
e−4N kπ2T
≤ 8ε2N3π2
3H2 + 2 e−2N2π2T 1−e−4N π2T.
We want to chooseN(ε)→+∞ such thatε2N3 →0. For instance, we can take N =bε−ηc, where η >0 is small enough. Forεsmall enough, we have :
kw(T,·)k2L2 ≤ 8π2
3H2ε2−3η+ 4e−2ε−2ηπ2T =O ε2−3η
. (2.40)
Combining estimates (2.40) and (2.38), and the definition (2.39) we can easily deduce the estimate (2.35).
This concludes the proof of Lemma7.
2.4.3 Approximate controllability towards the null state
First, let us prove the following technical lemma. Indeed, we have proven that the particular boundary layer Φεdissipates, but all we also want to know what would happen if we were very close to it.
Lemma 8. Let us change the initial data from Lemma7toΦε(x) +1εδε. We assume : Φε(x) +1
εδε ≤ 0, (2.41)
kδε(·)kL2(0,1) = Oε→0(ε3). (2.42)
Then, the conclusion of Lemma7still holds.
Démonstration. We follow the same scheme than for the proof of Lemma 7. Hence, we start by taking the Cole-Hopf transform of the new initial data Φε(x) + 1εδε. Therefore, after the Cole-Hopf transform we have the following initial data :
Zε(x) +Zε(x)·
exp
−1 2ε
Z x 0
δε
−1
.
From our previous computation (2.37) of Zε, we know that |(Zε)x| = O(1/ε). Hence, using condition (2.42), we have :
Zε(x)·
exp
−1 2ε
Z x 0
δε
−1
H1(0,1)
=Oε→0(ε).
Let us use the fact that our heat system (2.36) is linear. Therefore, using the conclusion of Lemma7 we have :
kz(T,·)kH1(0,1)=Oε→0(ε1−δ) +Oε→0(ε).
Once again we apply the backwards Cole-Hopf transform. We use the fact thatz≥1 (this comes from the comparison principle and the hypothesis (2.41)). Hence,
kφ(T,·)kL2(0,1)≤2kz(T,·)kH1(0,1).
Thus, the conclusion (2.35) of Lemma 7 still holds with this new initial data.
Now everything is ready for us to show the following small time approximate controllability result for system (2.1). We have to combine the different estimates.
Theorem 4. LetT, r >0 andy0∈L2(0,1)be given data. Then there existsu, v∈L∞(0, T)×H1/4(0, T) such that the associated solutiony∈X to system (2.1) on [0, T] satisfies :
ky(T,·)kL2(0,1)≤r.
Démonstration. Take T, r >0 and y0 ∈L2(0,1) given data. Let us take a smallε >0 and break down our time interval into four parts. We introduceT1 =T /3,T2 =T1+ε andT3=T2+ε4. The first part [0, T1] of lengthT /3 is designed to smooth the initial data. The second part [T1, T2] of length ε is the part where the settling of the boundary layer takes place. The third part [T2, T3] of lengthε4is the quick push down to zero. The fourth part [T3, T] of length at leastT /3 (whenεis small enough) is the passive stage for the dissipation of the boundary layer. Let us give some details.
u(t) v(t)
T1 T2 T3 T
time Controls used
t= 0
smoothing
dissipation settling
push-down
Figure2.6 – Approximate null-controllability strategy.
Smoothing of the initial data : First, for t ∈ [0, T1], we choose u(t) = v(t) = 0. The system evolves freely. Regularization effects of the Burgers’ equation smooth our initial data y0 ∈L2(0,1). We have y(T1,·) ∈ H01(0,1). There are many ways to prove such a result. For instance, one can take the Cole-Hopf transform and use well-known regularization properties of the heat equation.
Settling of the boundary layer : Next, fort∈[T1, T2], we perform the scaling (2.4). We want to apply Lemma4for a duration 1. Hence, let us choose someHsuch thatH−2>0. We take ¯v∈H3/4(0, T) the control from Lemma4. Fort∈[T1, T2], we use :
u(t) = 0, v(t) = 1
εv¯
t−T1
ε
. From Lemma4, we know that :
y(T2,·)−1 εhε(·)
L2(0,1)
=Oε→0
ε−3/2e−4εH(H−2)
. (2.43)
Push-down towards zero : Then, still in the context of scaling (2.4), we want to apply Lemma6 during a very short durationε3. Hence, for t∈[T2, T3], we choose the controls found in Lemma6(with a total timeε3), and we scale them appropriately. That is to say :
u(t) = 1 ε2u¯
t−T2
ε
,
v(t) = 1 ε¯v
t−T2
ε
.
Combining (2.43) and Lemma 6, we get that, at the end of this hyperbolic stage : 0≥y(T3,·)≥Φε+1
εδ(ε3,·)−1
εkhε−kεk∞, where (using estimate (2.12)) :
δ(ε3,·)
L2+khε−kεk∞=Oε→0(ε3).
Dissipation of the boundary residue : Now we enter the passive stage. We choosev(t) =u(t) = 0 for t ∈ [T3, T]. Since ε goes to zero, T −T3 ≥ T /3. Hence we can apply Lemma 8 on a time interval independent ofε. By using the comparison principle from Lemma1we can conclude that :
ky(T,·)kL2 =Oε→0(ε1−η),
for any η > 0. For instance, one can choose η = 12. Then we choose ε small enough to ensure that ky(T,·)kL2 ≤r. This concludes the proof of Theorem4.
Remark 2. In the proof of Theorem4, we concatenate different controls found in different parts. This could be a problem for smoothness because we did not check compatibility conditions at the jointures.
However, the proof provides a controlv∈H1/4(0, T) and this doesn’t require compatibility conditions. If one wants smooth controls, it is also possible. One can choose a smooth control close to our control for the approximate controllability, then end with a smooth control for the exact controllability.