First step : overriding the initial data

In order to get close to the steady stateh^ε, it is necessary to chooseH in such a way that a Rankine- Hugoniot type condition is satisfied. Once we get close enough to the steady state, the solution will very quickly converge to the steady state. Indeed, the eigenvalues of the linearized system around this steady state are real, negative, and of size at least 1/ε. This guarantees very quick convergence to the steady state. Such a study of the linearized problem around a steady state for the Burgers’ equation can be found in [114]. We give the following lemma describing the settling of the limit layer.

Lemma 4. LetT >0,H >0 andy0∈H₀¹(0,1)be given data. Then forε >0 small enough, there exists a boundary control ¯v∈H^3/4(0, T)such that v(·)¯ ≤H and such that the solution y¯∈X to system (2.5) with initial data y¯₀=εy₀ and controls ¯u= 0 and¯v satisfies :

ky(T,¯ ·)−hε(·)k_L2(0,1)=O_ε→0

ε^−1/2e^{−H4ε(HT−2)}

. (2.13)

h^ε(x)

x= 1

¯ y(t, x)

¯ y0(x) H

Figure2.3 – Example of evolution from an initial data towards a steady state.

Let us postpone the proof of Lemma4 for the moment. We start by giving a few remarks concerning this statement and its proof. The intuition is to choose a boundary control ¯v(t)≡H, just like we have done for the hyperbolic case. Moreover, we want to use the Cole-Hopf transform and Fourier series to compute explicitly ¯y(T,·). Let us introduce the Cole-Hopf transform :

Z(t, x) = exp

−1 2ε

Z x 0

¯ y(t, s)ds

.

This leads to the following heat system for the new unknownZ :











Zt−εZxx =− _4ε¹y¯²(t,0)−¹₂y¯x(t,0)

Z on (0, T)×(0,1),

Z(t,0) = 1 on (0, T),

Zx(t,1) = 0 on (0, T),

Z(0, x) =Z⁰(x) on (0,1),

(2.14)

where the initial dataZ⁰ is computed from the initial data ¯y0=εy0: Z⁰(x) = exp

−1 2

Z x 0

y0(s)ds

. (2.15)

Hence we see that it will not be possible to carry on explicit computations if we do choose ¯y(t,0)≡H.

Indeed, in that case, we would not know explicitly ¯yx(t,0) (which is needed to compute the solution to system (2.14)). However, we are confident that this term is very small. Hence, we are going to go the

other way around : we will choose our control explicitly in the Cole-Hopf domain and use it to compute our control ¯v(·). Therefore, we are interested in the following heat system :











Zt−εZxx =−^H_4ε²Z on (0, T)×(0,1), Z(t,0) = 1 on (0, T),

Zx(t,1) = 0 on (0, T), Z(0, x) =Z⁰(x) on (0,1).

(2.16)

If we go back to the Burgers’ domain, this means that we somehow use the following boundary condition atx= 0 :

¯

yx(t,0) = 1

2ε y¯²(t,0)−H²

. (2.17)

We expect that the solution Z will converge towards H^ε(·), where H^ε(·) is the Cole-Hopf transform of the steady stateh^ε:

H^ε(x) = cosh _2ε^H(1−x)

cosh_2ε^H . (2.18)

Indeed, we have the following lemma.

Lemma 5. Let T > 0 and Z⁰ ∈ H²(0,1) such that Z⁰(0) = 1 and Z_x⁰(1) = 0. Then system (2.16) has a unique solutionZ in the spaceL²((0, T);H³(0,1))∩H¹((0, T);H¹(0,1)). Moreover, there exists a constantC(Z⁰)>0 depending only on kZ⁰k_H1 such that :

kZ(T,·)−H^ε(·)k_H1(0,1)≤ε^−1/2C(Z⁰)e^−H

2T

4ε . (2.19)

Démonstration. Using standard techniques, it is classical to show that system (2.16) has a unique solution in the spaceL²((0, T);H³(0,1))∩H¹((0, T);H¹(0,1)). One can even get more smoothness if needed. An efficient method is the semi-group method that one can find for instance in [139]. To compute the dynamics of system (2.16), we introduce the adequate Fourier basis ofL² :

fn(x) =√ 2 sin

n+1

2

πx

for n≥0.

Hencefn(0) =f_n⁰(1) = 0. We will use the notationλn = (n+¹₂)π. Thus, f_n⁰⁰ =−λ²_nfn. Let us give the following scalar products, which can easily be computed using integration by parts :

h1|fni =

√2 λn

,

hH^ε|fni =

√2λn H²

4ε² +λ²_n, (2.20)

hZ⁰|fni ≤

√2 λn

1 + 1

2 Z⁰

_H1

. (2.21)

In these equationsh·|·idenotes the standard scalar product inL²(0,1). Let us writeZ = 1 +w. Hencew will satisfyw(t,0) =w_x(t,1) = 0. Easy computations lead to the following ordinary differential equations for the components ofw on our Fourier basis :

˙

w_n(t) =−ε

λ²_n+H² 4ε²

w_n(t)−H² 4εh1|fni.

It is easy to see that the fixed points for these ODEs are the expected coefficientshH^ε−1|fni. We can solve these ODEs with our initial condition :

wn(t) =αne^{−ε λ2n+H}

2 4ε2

t

+hH^ε−1|fni, where :

αn=hZ⁰|fni − hH^ε|fni.

Now we can estimateZ(T,·)−H^ε(·) :

kZ(T,·)−H^ε(·)k²_H1(0,1)=X

n≥0

λ²_nα²_ne^{−2ε λ2n+H}

2 4ε2

T

.

From the expression ofαn, (2.20) and (2.21) we get the easy bound : λ²_nα²_n ≤16 +kZ⁰k²_H1(0,1), ∀n∈N. Thus, we get

kZ(T,·)−H^ε(·)k²_H1(0,1)≤

16 +kZ⁰k²_H1(0,1)

e^−H

2T

2ε X

n≥0

e^−2ελ2n.

Now we split the sum in two parts :n≤N =b1/εcand n≥N. We get : kZ(T,·)−H^ε(·)k²_H1(0,1) ≤

16 +kZ⁰k²_H1(0,1)

e^−H

2T 2ε



N+X

k≥0

e^{−2ε(N+k+12)2π2}





≤

16 +kZ⁰k²_H1(0,1)

e^−H

2T 2ε

N+ 1

1−e^{−4εN π2}e^−2εN2π2

. Hence, forεsmall enough, we have :

kZ(T,·)−H^ε(·)k²_H1(0,1)≤ 1

ε + 1

16 +kZ⁰k²_H1(0,1)

e^−H

2T 2ε . This concludes the proof of Lemma5.

Now we can prove Lemma 4.

Proof of Lemma 4. Definition of the control : Using Lemma5, we start by considering the solution Z∈L²((0, T);H³(0,1))∩H¹((0, T);H¹(0,1)) to system (2.16) with the initial data (2.15). SinceZ0(·)>

0, the usual strong maximum principle (see [149]) guarantees thatZ(t, x)>0. Thus, we can define :

¯

y(t, x) =−2εZ_x(t, x)

Z(t, x). (2.22)

Hence ¯y∈X is a solution to (2.5) with initial dataεy₀ and boundary control ¯v(t) =−2εZx(t,0). Since Z ∈L²((0, T);H³(0,1))∩H¹((0, T);H¹(0,1)), we can show that its boundary trace Z_x(t,0) belongs to H^3/4(0, T). Hence ¯v∈H^3/4(0, T).

Proof of an L^∞ bound on the solution : Ifεis small enough, thenεky₀k_∞≤H. Moreover, we know that ¯v ∈H^3/4(0, T). Hence, ¯v∈ C⁰[0, T]. Assume that sup_[0,T]v > H¯ . LetT₀ be a time such that

¯

v(T0) = sup_[0,T]v > H¯ . On the one hand, by the comparison principle from Lemma1, we know that :

¯

y≤¯v(T₀) on (0, T)×(0,1). (2.23)

On the other hand, we recall relation (2.17) : y¯_x(t,0) = 1

2ε y¯²(t,0)−H² .

Hence, since ¯v(T0) > 0, we get ¯yx(T0,0) > 0. Thus,there exists x > 0 such that ¯y(T0, x) > v(T¯ 0) = sup_[0,T]¯v. This is in contradiction with assertion (2.23). Hence, if ε is small enough, ¯v(·) ≤ H and

¯

y(T,·)≤H.

Derivation of theL²estimate at timeT : Now we want to prove estimate (2.13) from Lemma 4.

We want to use estimate (2.19) from Lemma5. We perform the following computation at timeT and for anyx∈[0,1] :

|¯y−h^ε|= 2ε

Z_x Z −H_x^ε

H^ε

= 2ε

Z(Z_x−H_x^ε) +Z_x(H^ε−Z) ZH^ε

≤ 2ε

Z_x−H_x^ε H^ε

+ 2ε

Z_x Z

·

Z−H^ε H^ε

.

Thus, we get :

k¯y(T,·)−h^ε(·)k_L2(0,1)≤(2ε+k¯y(T,·)k∞)×sup

[0,1]

1

H^ε × kZ(T,·)−H^ε(·)k_H1(0,1). Now we use thatk¯y(T,·)k_∞≤H and sup_[0,1]1/H^ε≤e^+H/2ε. Hence , using also (2.19),

k¯y(T,·)−h^ε(·)k_L2(0,1)≤ 1

√ε(2ε+H)C(Z⁰)e^{−4εH(HT−2)}. This estimate concludes the proof of Lemma4.

Remark 1. In Lemma 4, we take an initial data y0 ∈ H₀¹(0,1). This is a technical assumption that enables us to use stronger solutions. We will get rid of it later on, by letting the Burgers’ equation smooth our real initial data which is only inL²(0,1).

No documento DE L’UNIVERSITÉ PIERRE ET MARIE CURIE (páginas 47-50)