HAL Id: hal-03172077
https://hal.archives-ouvertes.fr/hal-03172077
Preprint submitted on 17 Mar 2021
HAL is a multi-disciplinary open access
archive for the deposit and dissemination of
sci-entific research documents, whether they are
pub-lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d’enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.
Constructive exact control of semilinear 1D heat
equations
Jérôme Lemoine, Arnaud Munch
To cite this version:
Jérôme Lemoine, Arnaud Munch. Constructive exact control of semilinear 1D heat equations. 2021.
�hal-03172077�
Constructive exact control of semilinear 1D heat equations
J´
erˆ
ome Lemoine
∗Arnaud M¨
unch
†March 17, 2021
Abstract
The exact distributed controllability of the semilinear heat equation ∂ty − ∆y + g(y) = f 1ω
posed over multi-dimensional and bounded domains, assuming that g ∈ C1(R) satisfies the growth condition lim supr→∞g(r)/(|r| ln
3/2|r|) = 0 has been obtained by Fern´
andez-Cara and Zuazua in 2000. The proof based on a non constructive fixed point arguments makes use of precise estimates of the observability constant for a linearized heat equation. In the one dimensional setting, assuming that g0does not grow faster than β ln3/2|r| at infinity for β > 0 small enough and that g0
is uniformly H¨older continuous on R with exponent p ∈ [0, 1], we design a constructive proof yielding an explicit sequence converging to a controlled solution for the semilinear equation, at least with order 1 + p after a finite number of iterations.
AMS Classifications: 35K58, 93B05.
Keywords: Semilinear heat equation, Null controllability, Least-squares approach.
1
Introduction
Let Ω = (0, 1), ω ⊂⊂ Ω be any non-empty open set and let T > 0. We set QT = Ω×(0, T ), qT = ω ×(0, T )
and ΣT = ∂Ω × (0, T ). We are concerned with the null controllability problem for the following semilinear
heat equation
(
∂ty − ∂xxy + g(y) = f 1ω in QT,
y = 0 on ΣT, y(·, 0) = u0 in Ω,
(1)
where u0 ∈ H01(Ω) is the initial state of y and f ∈ L2(qT) is a control function. We assume moreover
that the nonlinear function g : R 7→ R is, at least, locally Lipschitz-continuous and, following [17], that g satisfies
|g0(r)| ≤ C(1 + |r|5) ∀r ∈ R. (2) Under this condition, (1) possesses exactly one local in time solution. Moreover, we recall (see [7]) that under the growth condition
|g(r)| ≤ C(1 + |r| ln(1 + |r|)) ∀r ∈ R, (3) the solutions to (1) are globally defined in [0, T ] and one has
y ∈ C0([0, T ]; H01(Ω)) ∩ L2(0, T ; H2(Ω)). (4) Without a growth condition of the kind (3), the solutions to (1) can blow up before t = T ; in general, the blow-up time depends on g and the size of ku0kL2(Ω).
∗Laboratoire de math´ematiques Blaise Pascal, Universit´e Clermont Auvergne, UMR CNRS 6620, Campus des C´ezeaux,
3, place Vasarely, 63178 Aubi`ere, France. e-mail: jerome.lemoine@uca.fr.
†Laboratoire de math´ematiques Blaise Pascal, Universit´e Clermont Auvergne, UMR CNRS 6620, Campus des C´ezeaux,
The system (1) is said to be controllable at time T if, for any u0 ∈ L2(Ω) and any globally defined
bounded trajectory y? ∈ C0([0, T ]; L2(Ω)) (corresponding to data u?
0 ∈ L2(Ω) and f? ∈ L2(qT)), there
exist controls f ∈ L2(q
T) and associated states y that are again globally defined in [0, T ] and satisfy (4)
and
y(x, T ) = y?(x, T ), x ∈ Ω. (5) The uniform controllability strongly depends on the nonlinearity g. Assuming a growth condition on the nonlinearity g at infinity, this problem has been solved by Fern´andez-Cara and Zuazua in [17] (which also covers the multi-dimensional case for which Ω ⊂ Rd is a bounded connected open set with Lipschitz
boundary).
Theorem 1. [17] Let T > 0 be given. Assume that (1) admits at least one solution y?, globally defined in [0, T ] and bounded in QT. Assume that g : R 7→ R is C1 and satisfies (2) and
(H1) lim sup|r|→∞ |g(r)| |r| ln3/2|r| = 0
then (1) is controllable in time T .
Therefore, if |g(r)| does not grow at infinity faster than |r| lnp(1 + |r|) for any p < 3/2, then (1) is controllable. We also mention [2] which gives the same result assuming additional sign condition on g, namely g(r)r ≥ −C(1 + r2
) for all r ∈ R and some C > 0. On the contrary, if g is too “super-linear” at infinity, precisely, if p > 2, then for some initial data, the control cannot compensate the blow-up phenomenon occurring in Ω\ω (see [17, Theorem 1.1]). The problem remains open when g behaves at infinity like |r| lnp(1 + |r|) with 3/2 ≤ p ≤ 2. We mention however the recent work of Le Balc’h [20] where uniform controllability results are obtained for p ≤ 2 assuming additional sign conditions on g, notably that g(r) > 0 for r > 0 or g(r) < 0 for r < 0, a condition not satisfied for g(r) = −r lnp(1 + |r|). Eventually, we also mention [9] where a positive boundary controllability result is proved for a specific class of initial and final data and T large enough.
In the sequel, for simplicity, we shall assume that g(0) = 0 and that f?≡ 0, u?
0≡ 0 so that y? is the
null trajectory. The proof given in [17] is based on a fixed point method, initially introduced in [30] for a one dimensional wave equation. Precisely, it is shown that the operator Λ : L∞(QT) → L∞(QT), where
y := Λ(z) is a null controlled solution of the linear boundary value problem ( ∂ty − ∂xxy + y ˜g(z) = f 1ω in QT y = 0 on ΣT, y(·, 0) = u0 in Ω , ˜g(r) := ( g(r)/r r 6= 0 g0(0) r = 0 (6) maps a closed ball B(0, M ) ⊂ L∞(QT) into itself, for some M > 0. The Kakutani’s theorem then
provides the existence of at least one fixed point for the operator Λ, which is also a controlled solution for (1). The control of minimal L∞(q
T) is considered in [17]. This allows, including in the multi-dimensional
case to obtain controlled solutions in L∞(QT).
The main goal of this work is to determine an approximation of the controllability problem associated with (1), that is to construct an explicit sequence (fk)k∈N converging strongly toward a null control
for (1). A natural strategy is to take advantage of the method used in [20, 17] and consider, for any element y0∈ L∞(QT), the Picard iterations defined by yk+1= Λ(yk), k ≥ 0 associated with the operator
Λ. The resulting sequence of controls (fk)k∈N is then so that fk+1 ∈ L2(qT) is a null control for yk+1
solution of
(
∂tyk+1− ∂xxyk+1+ yk+1eg(yk) = fk+11ω in QT, yk+1= 0 on ΣT, yk+1(·, 0) = u0 in Ω.
(7)
Numerical experiments reported in [13] exhibit the non convergence of the sequences (yk)k∈Nand (fk)k∈N
for some initial conditions large enough. This phenomenon is related to the fact that the operator Λ is in general not contracting, even ifeg is globally Lipschitz. We also refer to [4, 5] where this strategy is implemented. A least-squares type approach, based on the minimization over L2(Q
R : L2(Q
T) → R+ defined by R(z) := kz − Λ(z)k2L2(QT) has been introduced and analyzed in [13].
Assuming thateg ∈ C1(R) and g0 ∈ L∞
(R), it is proved that R ∈ C1(L2(QT); R+) and that, for some
constant C > 0
kR0(z)kL2(QT)≥ (1 − Ckg0k∞ku0k∞)
p
2R(z) ∀z ∈ L2(Q T)
implying that if kg0k∞ku0k∞is small enough, then any critical point for R is a fixed point for Λ (see [13,
Proposition 3.2]). Under this assumption on the data, numerical experiments reported in [13] display the convergence of gradient based minimizing sequences for R and a better behavior than the Picard iterates. The analysis of convergence is however not performed. As is usual for nonlinear problems and also considered in [13], we may employ a Newton type method to find a zero of the mapping eF : Y 7→ W defined by
e
F (y, f ) = (∂ty − ∂xxy + g(y) − f 1ω, y(· , 0) − u0) ∀(y, f ) ∈ Y (8)
where the Hilbert space Y and W are defined as follows
Y :=
(y, f ) : ρy ∈ L2(QT), ρ0(∂ty − ∂xxy) ∈ L2(QT), y = 0 on ΣT, ρ0f ∈ L2(qT)
and W := L2(ρ0; QT) × L2(Ω) for some appropriates weights ρi (defined in the next section). Here
L2(ρ
0; QT) stands for {z : ρ0z ∈ L2(QT)}. It is shown in [13] that, if g ∈ C1(R) and g0 ∈ L∞(R), then
e
F ∈ C1(Y ; W ) allowing to derive the Newton iterative sequence: given (y
0, f0) in Y , define the sequence
(yk, fk)k∈N in YN iteratively as follows (yk+1, fk+1) = (yk, fk) − (Yk, Fk) where Fk is a control for Yk
solution of (
∂tYk− ∂xxYk+ g0(yk) Yk= Fk1ω+ ∂tyk− ∂xxyk+ g(yk) − fk1ω in QT,
Yk = 0 on ΣT, Yk(·, 0) = u0− yk(·, 0) in Ω.
(9)
Numerical experiments in [13] exhibits the lack of convergence of the Newton method for large enough initial condition, for which the solution y is not close enough to the zero trajectory.
The controllability of nonlinear partial differential equations has attracted a large number of works in the last decades (see the monography [8] and references therein). However, as far as we know, few are concerned with the approximation of exact controls for nonlinear partial differential equations, and the construction of convergent control approximations for nonlinear equations remains a challenge.
In this article, given any initial data u0 ∈ H01(0, 1), we design an algorithm providing a sequence
(fk)k∈N converging to a controlled solution for (1), under assumptions on g that are slightly stronger
than (H1). Moreover, after a finite number of iterations, the convergence is super-linear. This is done
(following and improving [26] devoted to a linear case) by introducing a quadratic functional measuring how much a pair (y, f ) ∈ Y is close to a controlled solution for (1) and then by determining a particular minimizing sequence enjoying the announced property. A natural example of an error (or least-squares) functional is given by eE(y, f ) := 12k eF (y, f )k2
W to be minimized over Y . The controllability for (1) is
reflected by the fact that the global minimum of the nonnegative functional eE is zero, over all pairs (y, f ) ∈ Y solutions of (1).
The paper is organized as follows. In Section 2, we first derive a controllability result for a linearized wave equation with potential in L∞(QT) and source term in L2(QT). Then, in Section 3, we define the
least-squares functional E and the corresponding non convex optimization problem (29) over the Hilbert space A. We show that E is Gateaux-differentiable over A and that any critical point (y, f ) for E for which g0(y) belongs to L∞(QT) is also a zero of E (see Proposition 3). This is done by introducing a pair
(Y1, F1) for E(y, f ) for which E0(y, f ) · (Y1, F1) is proportional to E(y, f ). Then, in Section 4,assuming
that the nonlinear function g is such that g0 is uniformly Holder continuous with exponent p, for some p ∈ [0, 1], we determine a minimizing sequence based on (Y1, F1) which converges strongly to a controlled
pair for the semilinear heat equation (1). Moreover, we prove that after a finite number of iterates, the convergence enjoys a rate equal to 1 + p (see Theorem 3). Section 5 gathers several remarks on the
approach: we notably emphasize that this least-squares approach coincides with the damped Newton method one may use to find a zero of a mapping similar to eF mentioned above: this explains the super-linear convergence obtained. We also discuss some other super-linearizations of the system (1). We conclude in Section 6 with some perspectives.
As far as we know, the method introduced and analyzed in this work is the first one providing an explicit, algorithmic construction of exact controls for semilinear heat equations with non Lipschitz nonlinearity. It extends the study [21] assuming that g0 ∈ L∞
(R) which to obtain directly a uniform bound of the observability constant. The weaker assumption considered here required a refined analysis similar to the one recently developed by the author in [3, 28] for the wave equation. The parabolic case is however more much intricate (than the hyperbolic one) as it makes appear Carleman type weights depending on the controlled solution. These works devoted to controllability problems take their roots in the works [22, 23] concerned with the approximation of solution of Navier-Stokes type problem, through least-square methods: they refine the analysis performed in [24, 25] inspired from the seminal contribution [6].
Notations. Throughout, we denote by k · k∞ the usual norm in L∞(R), by (·, ·)X the scalar product
of X (if X is a Hilbert space) and by h·, ·iX,Y the duality product between X and Y .
Given any p ∈ [0, 1], we introduce for any g ∈ C1
(R) the following hypothesis : (Hp) [g0]p:= supa,b∈R
a6=b
|g0(a)−g0(b)|
|a−b|p < +∞
meaning, for p ∈ (0, 1], that g0 is uniformly H¨older continuous with exponent p. For p = 0, by extension, we set [g0]
0 := 2kg0k∞. In particular, g satisfies (H0) if and only if g ∈ C1(R) and g0 ∈ L∞(R), and g
satisfies (H1) if and only if g0 is Lipschitz continuous (in this case, g0 is almost everywhere differentiable
and g00∈ L∞
(R)), and we have [g0]1≤ kg00k∞.
We also denote by C a positive constant depending only on Ω and T that may vary from lines to lines.
2
A controllability result for a linearized heat equation with L
2right hand side
This section is devoted to a controllability result for a linear heat equation with potential in A ∈ L∞(QT)
and right hand side B ∈ L2(ρ0(s), QT) for a precise weight ρ0(s) parametrized by s ∈ R?+ defined in the
sequel. More precisely we are interested by the existence of a control v such that the solution z of ( ∂tz − ∂xxz + Az = v1ω+ B in QT, z = 0 on ΣT, z(·, 0) = z0 in Ω (10) satisfies z(·, T ) = 0 in Ω. (11) As this work concerns the null controllability of parabolic equation, we make use of Carleman weights introduced in this context in [18]). For any s ≥ 0, we consider the weight functions ρ(s) = ρ(x, t, s), ρ0(s) = ρ0(x, t, s) and ρ1(s) = ρ1(x, t, s) which are continuous, strictly positives and in L∞(QT −δ) for
any δ > 0. Precisely, we use the weights introduced in [1]: ρ0(s) = ξ−3/2ρ(s), ρ1(s) = ξ−1ρ(s) where ρ(s)
and ξ are defined, for all s ≥ 1 and λ ≥ 1, as follows
where θ ∈ C2([0, T )) is defined such that, noting µ = sλ2e2λ and 0 < T 1< min(14,3T8 ), θ(t) = 1 + 1 −4t T µ ∀t ∈ [0, T /4] 1 ∀t ∈ [T /4, T − 2T1] θ is increasing on [T − 2T1, T − T1], 1 T − t ∀t ∈ [T − T1, T ) (13) and ϕ ∈ C1([0, T )) is defined by ϕ(x, t) = θ(t) λ exp(12λ) − exp(λ bψ(x)) (14) with bψ = eψ + 6, where eψ ∈ C1(Ω) satisfies eψ ∈ (0, 1) in Ω, eψ = 0 on ∂Ω and |∂
xψ(x)| > 0 in Ω\ω. Wee emphasize that the weights blow up at t → T−.
Remark 1. We shall use in the sequel that 1 < ρ0≤ ρ1≤ ρ. Indeed, since ξ ≥ 1, ρ0≤ ρ1≤ ρ. Moreover,
for all (x, t) ∈ QT and λ ≥ 1, we check that ϕ(x, t) ≥ 32ξ(x, t) and thus, since s ≥ 1 and ξ(x, t) ≥ 1, we
get ρ0(x, t, s) = ξ−3/2(x, t)ρ(x, t, s) ≥ ξ−3/2(x, t) exp 3 2sξ(x, t) ≥ e3/2s ∀(x, t) ∈ QT.
The controllability property for the linear system (10) is based on the following Carleman estimate. Lemma 1. There exists λ0≥ 1 and s0≥ 1 such for all λ ≥ λ0 and for all s ≥ max(kAk
2/3 L∞(Q
T), s0) one
has the following Carleman estimate, for all p ∈ P0= {q ∈ C2(QT) : q = 0 on ΣT}:
Z Ω ρ−2(0, s)|∂xp(0)|2+ s3λ4e14λ Z Ω ρ−2(0, s)|p(0)|2+ sλ2 Z QT ρ−21 (s)|∂xp|2+ s3λ4 Z QT ρ−20 (s)|p|2 ≤ C Z QT ρ−2(s)| − ∂tp − ∂xxp + Ap|2+ Cs3λ4 Z qT ρ−20 (s)|p|2. (15)
Proof. This estimate is deduced from the one obtained in [1, Theorem 2.5] devoted to the case A ≡ 0: there exist λ1≥ 1 and s0≥ 1 such that for all smooth functions z on QT satisfying z = 0 on ΣT and for
all s ≥ s0 and λ ≥ λ1 Z Ω ρ−2(0, s)|∂xp(0)|2+ s3λ4e14λ Z Ω ρ−2(0, s)|p(0)|2+ sλ2 Z QT ρ−21 (s)|∂xp|2+ s3λ4 Z QT ρ−20 (s)|p|2 ≤ C Z QT ρ−2(s)|∂tp + ∂xxp|2+ Cs3λ4 Z qT ρ−20 (s)|p|2. Writing that Z QT ρ−2(s)|∂tp + ∂xxp|2≤ 2 Z QT ρ−2(s)| − ∂tp − ∂xxp + Ap|2+ 2 Z QT ρ−2(s)|Ap|2 ≤ 2 Z QT ρ−2(s)| − ∂tp − ∂xxp + Ap|2+ 2kAk2L∞(Q T) Z QT ρ−2(s)|p|2 we infer, since ρ0≤ ρ Z Ω ρ−2(0, s)|∂xp(0)|2+ s3λ4e14λ Z Ω ρ−2(0, s)|p(0)|2+ sλ2 Z QT ρ−21 (s)|∂xp|2+ s3λ4 Z QT ρ−20 (s)|p|2 ≤ C Z QT ρ−2(s)| − ∂tp − ∂xxp + Ap|2+ CkAk2L∞(Q T) Z QT ρ−20 (s)|p|2+ Cs3λ4 Z qT ρ−20 (s)|p|2.
Taking λ ≥ λ0= max(λ1, (2C)1/4) and s ≥ max(kAk 2/3 L∞(Q
In the sequel we assume that λ = λ0 and denote by C any constant depending only on Ω, ω, λ0 and
T .
Theorem 2. Assume A ∈ L∞(QT), s ≥ max(kAk 2/3 L∞(Q
T), s0), B ∈ L
2(ρ
0(s), QT) and z0∈ L2(Ω). Then
there exists a control v ∈ L2(ρ0(s), qT) such that the weak solution z of (10) satisfies (11).
Moreover, the unique control v which minimizes together with the corresponding solution z the func-tional J : L2(ρ(s), Q T) × L2(ρ0(s), qT) → R+ defined by J (z, v) := 12kρ(s) zk2L2(Q T)+ 1 2kρ0(s) vk 2 L2(q T)
satisfies the following estimates kρ(s) zkL2(Q T)+ kρ0(s) vkL2(qT)≤ Cs −3/2 kρ 0(s)BkL2(Q T)+ e cs kz0k2 (16) with c := kϕ(·, 0)k∞ and kρ1(s)zkL∞(0,T ;L2(Ω))+ kρ1(s)∂xzkL2(QT)≤ C1(s, A) kρ0(s)BkL2(QT)+ ecskz0k2 (17) where C1(s, A) := Cs−1/2(1 + kAk 1/2 L∞(Q T)).
Moreover, if z0∈ H01(Ω) then z ∈ L∞(QT) and
kzkL∞(Q T)≤ Ce −3 2s(1 + kAkL∞(Q T)) kρ0(s)BkL2(QT)+ e cskz 0kH1 0(Ω). (18)
We refer to [16] for an estimate of the null control of minimal L2(q
T)-norm (corresponding to ρ0≡ 1 and
ρ = 0) in the case B ≡ 0, refined later on in [12, 17]. Theorem 2 is based on several technical results. Remark first that the bilinear form
(p, q)P := Z QT ρ−2(s)L?Ap L?Aq + s3λ40 Z qT ρ−20 (s)p q where L?
Aq := −∂tq − ∂xxq + Aq for all q ∈ P0 is a scalar product on P0(see [14]). The completion P of
P0 for the norm k · kP associated with this scalar product is a Hilbert space. By density arguments, (15)
remains true for all p ∈ P , that is, for λ = λ0,
Z Ω ρ−2(0, s)|∂xp(0)|2+s3λ40e 14λ0 Z Ω ρ−2(0, s)|p(0)|2+sλ20 Z QT ρ−21 (s)|∂xp|2+s3λ40 Z QT ρ−20 (s)|p|2≤ Ckpk2P (19) for all s ≥ max(kAk2/3L∞(Q
T), s0).
Remark 2. We denote by P (instead of PA) the completion of P0 for the norm k · kP since P does not
depend on A (see [13, Lemma 3.1]).
Lemma 2. Let s ≥ max(kAk2/3L∞(Q
T), s0). There exists p ∈ P unique solution of
(p, q)P = Z Ω z0q(0) + Z QT Bq, ∀q ∈ P. (20)
This solution satisfies the following estimate (with c := kϕ(·, 0)k∞)
kpkP ≤ Cs−3/2 kρ0(s) BkL2(QT)+ e
cskz
0k2. (21)
Proof. The linear map L1: P → R, q 7→RQ
T Bq is continuous. Indeed, for all q ∈ P
Z QT Bq ≤ Z QT |ρ0(s)B|2 1/2Z QT |ρ−10 (s)q|2 1/2
and since from the Carleman estimate (19) we have Z QT |ρ−10 (s)q|2 1/2 ≤ Cs−3/2kqk P, therefore |L1(q)| = Z QT Bq ≤ Cs −3/2kρ 0(s)BkL2(QT)kqkP. Thus L1is continuous.
From (19) we deduce that the linear map L2 : P → R, q 7→RΩz0q(0) is continuous. Indeed, noting
c := kϕ(·, 0)k∞and using s ≥ 1, we obtain for all q ∈ P that:
|L2(q)| = s−3/2ecskz0k2s3/2e−cskq(0)k2
≤ s−3/2ecskz0k2s3/2kq(0)e−sϕ(x,0)k2= s−3/2ecskz0k2s3/2kρ−1(0, s)q(0)k2
≤ Cs−3/2ecskz0k2kqkP.
Using Riesz’s theorem, we conclude that there exists exactly one solution p ∈ P of (20) and this solution satisfies (21).
Let us now introduce the convex set C(z0, T ) :=
(z, v) : ρ(s)z ∈ L2(QT), ρ0(s)v ∈ L2(qT), (z, v) solves (10)-(11) in the transposition sense
that is (z, v) is solution of Z QT zL?Aq = Z qT vq + Z Ω z0q(0) + Z QT Bq, ∀q ∈ P. (22)
Let us remark that if (z, v) ∈ C(z0, T ), then since v ∈ L2(qT) and B ∈ L2(QT), z coincides with the
unique weak solution of (10) associated with v. We can now claim that C(z0, T ) is non empty. Indeed we
have :
Lemma 3. Let s ≥ max(kAk2/3L∞(Q
T), s0), p ∈ P the unique solution of (20) given in Lemma 2 and (z, v)
defined by
z = ρ−2(s)L?Ap and v = −s3/2λ20ρ−20 (s)p|qT. (23)
Then (z, v) ∈ C(z0, T ) and satisfies the following estimate (with c := kϕ(0, ·)k∞)
kρ(s) zkL2(Q T)+ kρ0(s) vkL2(qT)≤ Cs −3/2 kρ 0(s)BkL2(Q T)+ e cskz 0k2. (24)
Proof. From the definition of P , ρ(s)z ∈ L2(QT) and ρ0(s)v ∈ L2(qT) and from the definition of ρ(s) and
ρ0(s) we have z ∈ L2(QT) and v ∈ L2(qT). In view of (20), (z, v) is solution of (22) and satisfies (24) that
is, z is the solution of (10)-(11) associated with v in the transposition sense. Thus (z, v) ∈ C(z0, T ).
Let us now consider the following extremal problem, introduced by Fursikov and Imanuvilov [18] Minimize J (z, v) = 1 2 Z QT ρ2(s)|z|2+1 2 Z qT ρ20(s)|v|2 Subject to (z, v) ∈ C(z0, T ). (25)
Then (z, v) 7→ J (z, v) is strictly convex and continuous on L2(ρ2(s); Q
T) × L2(ρ20(s); qT). Therefore (25)
possesses at most a solution in C(z0, T ). More precisely we have :
Proposition 1. Let s ≥ max(kAk2/3L∞(Q
T), s0). Then (z, v) ∈ C(z0, T ) defined in Lemma 3 is the unique
Proof. Let (y, w) ∈ C(z0, T ). Since J is convex and differentiable on L2(ρ2(s); QT) × L2(ρ20(s); qT) we have : J (y, w) ≥ J (z, v) + Z QT ρ2(s)z(y − z) + Z qT ρ20(s)v(w − v) = J (z, v) + Z QT L?p(y − z) − Z qT p(w − v) = J (z, v) y being the solution of (10) associated with w in the transposition sense.
Proof. of Theorem 2. Proposition 1 gives the existence of a control v ∈ L2(ρ0(s), qT) such that the
solution z of (10) satisfies (11). Moreover, this control is the unique control which minimizes together with the corresponding solution z the functional J and satisfies (16). To finish the proof of Theorem 2, it suffices to prove that (z, v) satisfies the estimate (17) and if z0∈ H01(Ω), then z ∈ L∞(QT) and satisfies
the estimate (18).
Multiplying (10) by ρ2
1(s)z and integrating by part we obtain
1 2 Z Ω (∂t|z|2) ρ21(s) + Z Ω ρ21(s)|∂xz|2+ 2 Z Ω ρ1(s)z∂xρ1(s) · ∂xz + Z Ω ρ21(s)Azz = Z ω vρ21(s)z + Z Ω Bρ21(s)z. (26) ButR Ω(∂t|z| 2)ρ2 1(s) = ∂tRΩ|z|2ρ21(s) − 2 R Ω|z| 2ρ 1(s)∂tρ1(s) and ∂tρ1(s) = −∂θtθρ1(s) + s∂tθθϕρ1(s). From
the definition of θ and ϕ we have:
∂tθ θ (t) ≤ Cs ∀t ∈ [0, T /4] 0 ∀t ∈ [T /4, T − 2T1] θ ∀t ∈ [T − T1, T ) and ∂tθ θ (t) ≤ C ∀t ∈ [T − 2T1, T − T1] since ∂tθ θ (T − 2T1) = 0, ∂tθ θ (T − T1) = 1 T1 and θ is C 2. Since θ ≤ ξ and s ≥ 1, on [0, T ): ∂tθ θ ≤ Csξ. From the definition of ϕ, ϕ ≤ Cθ ≤ Cξ and thus on [0, T ) :
∂tθ θ ϕ ≤ Csξ 2.
Thus, since s ≥ 1, ξ ≥ 1 and ρ(s) = ξρ1(s), on [0, T ) :
− Z Ω |z|2ρ 1(s)∂tρ1(s) = Z Ω ∂tθ θ |ρ1(s)z| 2− s Z Ω ∂tθ θ ϕ|ρ1(s)z| 2≤ Cs2 Z Ω ρ2(s)|z|2. On the other hand
∂xρ1(s) = ∂x(ξ−1ρ(s)) = ∂x(ξ−1)ρ(s) + ξ−1∂xρ(s) = −∂xψ λb 0ξ−1ρ(s) + sλ0ρ(s) = −∂xψλb 0ρ(s) ξ−1+ s
and thus, since ξ ≥ 1 and s ≥ 1, we write (ξ−1+ s) ≤ 2s and
Z Ω ρ1(s)z∂xρ1(s) · ∂xz ≤ 2k∂x ˆ ψk∞λ0 Z Ω s|ρ(s)z| |ρ1(s)∂xz| ≤ Cs2 Z Ω |ρ(s)z|2+1 2 Z Ω |ρ1(s)∂xz|2.
We also have, since ρ(s) = ξρ1(s) and ξ ≥ 1 the estimate
R Ωρ 2 1(s)Azz ≤ CkAk∞kρ(s)zk 2 2. Finally, since ρ2
1(s) = ξ−1/2ρ0(s)ρ(s) and ξ−1/2 ≤ 1, we infer that
Z ω vρ21(s)z ≤ Z ω ρ0(s)vξ−1/2ρ(s)z ≤ Z ω |ρ0(s)v|2 1/2 kρ(s)zk2
and |R
ΩBρ 2
1(s)z| ≤ kρ0(s)Bk2kρ(s)zk2. Thus (26) implies that
∂t Z Ω ρ21(s)|z|2+ Z Ω ρ21(s)|∂xz|2≤ C s2+kAkL∞(Q T)kρ(s)zk 2 2+ Z ω |ρ0(s)v|2 1/2 +kρ0(s)Bk2 kρ(s)zk2
and therefore for all t ∈ [0, T ), since kρ1(s, 0)z0k22≤ ecskz0k22 (with c := kϕ(·, 0)k∞), we get
Z Ω ρ21(s)|z|2 (t) + Z Qt ρ21(s)|∂xz|2≤ C s2+ kAkL∞(Q T)kρ(s)zk 2 L2(QT) + kρ0(s)vkL2(qT)+ kρ0(s)BkL2(QT)kρ(s)zkL2(QT)+ ecskz0k22.
Using (24) we obtain, since s ≥ 1, for all t ∈ [0, T ) : Z Ω ρ21(s)|z|2(t) + Z Qt ρ21(s)|∂xz|2≤ Cs−1 1 + kAkL∞(Q T) kρ0(s)Bk2L2(QT)+ e cs kz0k22 (27) which gives (17).
If z0 ∈ H01(Ω), since z is a weak solution of (10) associated with v, standard arguments give that
z ∈ L2(0, T ; H2(Ω)), ∂
tz ∈ L2(QT) and therefore z ∈ L∞(QT). Moreover, multiplying (10) by ∂xxz and
integrating by part we obtain 1 2∂t Z Ω |∂xz|2+ Z Ω |∂xxz|2≤ k − Az + v1ω+ Bk2k∂xxzk2
and thus, since ρ(s) ≥ ρ0(s) ≥ e
3 2s: ∂t Z Ω |∂xz|2≤ Z Ω | − Az + v1ω+ B|2≤ 3 Z Ω |Az|2+Z Ω |v1ω|2+ Z Ω |B|2 ≤ 3e−3skAk2 L∞(Q T)kρ(s)zk 2 2+ kρ0(s)v1ωk22+ kρ0(s)Bk22 which gives, a.e in t ∈ (0, T )
Z Ω |∂xz|2(t) ≤ 3e−3s kAk2 L∞(Q T)kρ(s)zk 2 L2(Q T)+ kρ0(s)vk 2 L2(q T)+ kρ0(s)Bk 2 L2(Q T) + kz0k 2 H1 0(Ω).
Since z ∈ H01(Ω), a.e in (x, t) ∈ Ω × (0, T ) (recall that Ω = (0, 1)), z(x, t) =
Rx
0 ∂xz(r, t)dr ≤ k∂xzk2(t)
and thus, using (16), since s ≥ 1 : kzkL∞(Q T)≤ k∂xzkL∞(0,T ;L2(Ω)) ≤√3e−32s kAkL∞(QT)kρ(s)zkL2(QT)+ kρ0(s)vkL2(qT)+ kρ0(s)BkL2(QT) + kz0kH1 0(Ω) ≤ Ce−32s(1 + kAk L∞(Q T)) kρ0(s)BkL2(QT)+ e cskz 0kH1 0(Ω) that is (18).
Remark 3. Remark that c = kϕ(0, ·)k∞> 3/2 so that the previous bound of kzkL∞(Q
T) is not uniform
with respect to the parameter s ≥ 1.
3
The least-squares method
In this section, we assume that the nonlinear function g satisfies the hypothesis (Hp) for some p ∈ [0, 1]
and that
(H2) There exists α ≥ 0 and β > 0 such that |g0(r)| ≤ α + β ln3/2(1 + |r|) for every r in R.
We introduce the notation
ψ(r) := α + β ln3/2(1 + |r|), ∀r ∈ R. (28) We also assume that g(0) = 0 leading in particular to the estimate |g(r)| ≤ |r|(α + β ln3/2(1 + |r|)) for every r ∈ R. The case p = 0 corresponds to β = 0 and α = kg0kL∞(R) and thus ψ(r) ≤ kg0kL∞(R) for
3.1
The least-squares method
We introduce, for all s ≥ s0, the vector space A0(s)
A0(s) :=
(y, f ) : ρ(s) y ∈ L2(QT), ρ0(s)f ∈ L2(qT),
ρ0(s)(∂ty − ∂xxy) ∈ L2(QT), y(·, 0) = 0 in Ω, y = 0 on ΣT
where ρ(s), ρ1(s) and ρ0(s) are defined in (12). A0(s) endowed with the following scalar product
(y, f ), (y, f )A 0(s):= ρ(s)y, ρ(s)y 2,qT + ρ0(s)f, ρ0(s)f 2,qT + ρ0(s)(∂ty − ∂xxy), ρ0(s)(∂ty − ∂xxy)2
is a Hilbert space. The corresponding norm is k(y, f )kA0(s)=p((y, f ), (y, f ))A0(s). We also consider the
convex set A(s) := (y, f ) : ρ(s) y ∈ L2(QT), ρ0(s)f ∈ L2(qT), ρ0(s)(∂ty − ∂xxy) ∈ L2(QT), y(·, 0) = u0 in Ω, y = 0 on ΣT
so that we can write A(s) = (y, f ) + A0(s) for any element (y, f ) ∈ A(s). We endow A(s) with the same
norm. Clearly, if (y, f ) ∈ A(s), then y ∈ C([0, T ]; L2(Ω)) and since ρ(s) y ∈ L2(Q
T), then y(·, T ) = 0.
The null controllability requirement is therefore incorporated in the spaces A0(s) and A(s).
Remark 4. For any (y, f ) ∈ A(s), since ρ0(s) ≥ 1 (see Remark 1), we get that ∂ty − ∂xxy ∈ L2(QT);
since u0∈ H01(Ω), standard arguments imply that y ∈ L∞(QT) with
kykL∞(Q
T)≤ C ku0kH10(Ω)+ k∂ty − ∂xxykL2(QT)
In particular, for any (y, f ) ∈ A0(s), kykL∞(QT)≤ Ce−
3
2sk(y, f )kA
0(s) for some C independent of s.
For any fixed (y, f ) ∈ A(s) and s ≥ 0, we can now consider the following non convex extremal problem : min
(y,f )∈A0(s)
E(s, y + y, f + f ) (29)
where the least-squares functional E(s) : A(s) → R is defined as follows
E(s, y, f ) :=1 2 ρ0(s) ∂ty − ∂xxy + g(y) − f 1ω 2 L2(Q T) . (30)
We check that ρ0(s)g(y) ∈ L2(QT) for any (y, f ) ∈ A(s) so that E(s) is well-defined. Precisely, using
that |g(r)| ≤ |r| α + β ln3/2(1 + |r|) = |r|ψ(r) for every r and that ρ0≤ ρ, we write
kρ0(s)g(y)kL2(QT)≤ kρ0(s)|y|ψ(y)kL2(QT)
≤ ψ kykL∞(Q
T)kρ(s)ykL2(QT)≤ ψ kykL∞(QT)k(y, f )kA(s).
(31)
Any pair (y, f ) ∈ A for which E(y, f ) vanishes is a controlled pair of (1), and conversely. In this sense, the functional E is a so-called error functional which measures the deviation of (y, f ) from being a solution of the underlying nonlinear equation. Moreover, although the hypothesis (H2) is stronger (H1),
Theorem 1 proved in [17] does not imply the existence of zero of E in A(s), since controls of minimal L∞(qT) norm are considered in [17]. Nevertheless, our constructive approach will show that, for s large
We also emphasize that the L2(Q
T) norm in E indicates that we are looking for regular weak solutions
of the parabolic equation (1). We refer to [21] devoted to the case g0 ∈ L∞
(R) and the multidimensional case where the L2(0, T ; H−1(Ω)) is considered leading to weaker solutions.
A practical way of taking a functional to its minimum is through some use of its derivative. In doing so, the presence of local minima is always something that may dramatically spoil the whole scheme. The unique structural property that discards this possibility is the convexity of the functional E. However, for nonlinear equation like (1), one cannot expect this property to hold for the functional E. Nevertheless, we are going to construct a minimizing sequence which always convergence to a zero of E. To do so, we introduce the following definition.
Definition 1. For any s large enough and (y, f ) ∈ A(s), we define the unique pair (Y1, F1) ∈ A 0(s) solution of ( ∂tY1− ∂xxY1+ g0(y)Y1= F11ω+ ∂ty − ∂xxy + g(y) − f 1ω in QT, Y1= 0 on ΣT, Y1(·, 0) = 0 in Ω (32)
and which minimizes the functional J defined in Theorem 2. In the sequel, it is called the minimal controlled pair.
The next proposition shows that there do exists some (Y1, F1) in A
0(s). We emphasize that F1 is a null
control for the solution Y1. Preliminary, we prove the following result. Lemma 4. There exists (y, f ) ∈ L2(Q
T) × L2(qT) such (y, f ) ∈ A(s) for all s ≥ 0.
Proof. Let y? be the solution of (
∂ty?− ∂xxy?= 0 in QT,
y?= 0 on ΣT, y?(·, 0) = u0∈ H01(Ω) in Ω,
so that y? ∈ L2(0, T ; H2(0, 1)) and ∂
ty? ∈ L2(0, T ; L2(0, 1)). Let now any function φ ∈ C∞([0, T ]),
0 ≤ φ ≤ 1 such that φ(0) = 1 and φ ≡ 0 in [T /2, T ]. Then, we easily check that the pair (y, 0) with y := φ y? belongs to A(s) for any s ≥ 0.
Moreover, kykL∞(Q
T) ≤ ky
?k L∞(Q
T) ≤ Cku0kH01(Ω) so that any s ≥ max(kg
0k2/3 L∞(0,Cku
0kH1 0(Ω))
, s0)
satisfies s ≥ max(kg0(y)k2/3L∞(Q T), s0).
Proposition 2. Let (y, f ) ∈ A(s) with s ≥ max kg0(y)k2/3L∞(Q
T), s0. There exists a minimal controlled
pair (Y1, F1) ∈ A
0(s) solution of (32). It satisfies the estimate:
k(Y1, F1)k
A0(s)≤ C
p
E(s, y, f ) (33) for some C > 0.
Proof. For all (y, f ) ∈ A(s), ρ0(s)(∂ty − ∂xxy + g(y) − f 1ω) ∈ L2(QT). The existence of a null control
F1 is therefore given by Proposition 2. Choosing the control F1 which minimizes together with the corresponding solution Y1 the functional J defined in Theorem 2, we get from (16)-(17) the following
estimates (since Y1(·, 0) = 0) : kρ(s) Y1k L2(QT)+ kρ0(s)F 1k L2(qT)≤ Cs −3/2kρ 0(s)(∂ty − ∂xxy + g(y) − f 1ω)kL2(QT) ≤ Cs−3/2p E(s, y, f ). (34) Eventually, from the equation solved by Y1,
kρ0(s)(∂tY1− ∂xxY1)kL2(Q T) ≤ kρ0(s)F1kL2(qT)+ kρ0(s)g0(y)Y1kL2(QT)+ kρ0(s)(∂ty − ∂xxy + g(y) − f 1ω)kL2(QT) ≤ kρ0(s)F1kL2(q T)+ kρ0(s)g 0(y)Y1 kL2(Q T)+ p 2E(s, y, f ).
But, since ρ0(s) ≤ ρ(s), using (34), we have kρ0(s)g0(y)Y1k2L2(Q T)≤ kg 0(y)k2 L∞(Q T)kρ(s)Y 1k2 L2(Q T)≤ Cs −3kg0(y)k2 L∞(Q T)E(s, y, f ) (35) thus kρ0(s)(∂tY1− ∂xxY1− F11ω)kL2(Q T)≤ C 1 + s −3/2kg0(y)k L∞(Q T)pE(s, y, f ) (36)
which proves that (Y1, F1) belongs to A
0(s). Eventually, k(Y1, F1)k2 A0(s)= kρ(s)Y 1k2 2+ kρ0(s)F1k22+ kρ0(s)(∂tY1− ∂xxY1)k22 ≤ kρ(s)Y1k2 2+ 4kρ0(s)F1k22+ 3kρ0(s)g0(y)Y1k2L2(Q T)+ 3( p 2E(s, y, f ))2 ≤ 4Cs−3E(s, y, f ) + Cs−3kg0(y)k2L∞(Q T)E(s, y, f ) + 6E(s, y, f ) ≤ CE(s, y, f )(1 + s−3+ s−3kg0(y)k2L∞(Q T)).
Since s ≥ max(kg0(y)k2/3L∞(Q
T), s0) ≥ 1, we get s
−3 ≤ 1 and s−3kg0(y)k2 L∞(Q
T) ≤ 1 leading to the
result.
Remark 5. From (32), we observe that z := y − Y1∈ L2(ρ(s), Q
T) is a null controlled solution satisfying
(
∂tz − ∂xxz + g0(y)z = (f − F1)1ω+ g0(y)y − g(y) in QT,
z = 0 on ΣT, z(·, 0) = u0in Ω
(37)
by the control (f − F1) ∈ L2(ρ0(s), qT).
Remark 6. We emphasize that the presence of a right hand side term in (32), namely ∂ty − ∂xxy +
g(y) − f 1ω, forces us to introduce the non trivial weights ρ0(s), ρ1(s) and ρ(s) in the space A(s). This
can be seen in the equality (20): since ρ−10 (s)q belongs to L2(Q
T) for all q ∈ P , we need to impose
that ρ0(s)B ∈ L2(QT) with here B = ∂ty − ∂xxy + g(y) − f 1ω. Working with the linearized equation
(6) (introduced in [17]) which does not make appear any right hand side, we may avoid the introduction of Carleman type weights. Actually, [17] considers controls of minimal L∞(qT) norm. Introduction of
weights allows however the characterization (20), which is very convenient at the practical level. We refer to [14] where this is discussed at length.
We also emphasize that we have considered bounded weights at the initial time t = 0 because of the constraints “ρ(s)y ∈ L2(Q
T)” and “y(0) = u0 in Ω” appearing in the set A(s).
3.2
Main properties of the functional E
The interest of the minimal controlled pair (Y1, F1) ∈ A
0(s) lies in the following result.
Proposition 3. For any (y, f ) ∈ A(s) and s ≥ max kg0(y)k2/3L∞(Q
T), s0, let (Y
1, F1) ∈ A
0(s) defined in
Definition 1. Then the derivative of E(s) at the point (y, f ) ∈ A(s) along the direction (Y1, F1) given by
E0(s, y, f ) · (Y1, F1) := lim
η→0,η6=0E(s,(y,f )+η(Y
1,F1))−E(s,y,f )
η satisfies
E0(s, y, f ) · (Y1, F1) = 2E(s, y, f ). (38) Proof. We preliminary check that for all (Y, F ) ∈ A0(s), E(s) is differentiable at the point (y, f ) ∈ A(s)
along the direction (Y, F ) ∈ A0(s). For all λ ∈ R, simple computations lead to the equality
E(s, y + λY, f + λF ) = E(s, y, f ) + λE0(s, y, f ) · (Y, F ) + h s, (y, f ), λ(Y, F ) with E0(s, y, f ) · (Y, F ) = ρ0(s)(∂ty − ∂xxy + g(y) − f 1ω), ρ0(s)(∂tY − ∂xxY + g0(y)Y − F 1ω) L2(QT) (39)
and
h(s, (y, f ), λ(Y, F )) :=λ
ρ0(s)(∂tY − ∂xxY + g0(y)Y − F 1ω), ρ0(s)l(y, λY )
L2(Q T) +λ 2 2 kρ0(s)(∂tY − ∂xxY + g 0(y)Y − F 1 ω)k2L2(Q T) +
ρ0(s)(∂ty − ∂xxy + g(y) − f 1ω), ρ0(s)l(y, λY )
L2(QT) +1 2kρ0(s)l(y, λY )k 2 L2(Q T)
where l(y, λY ) := g(y + λY ) − g(y) − λg0(y)Y .
The application (Y, F ) → E0(s, y, f ) · (Y, F ) is linear and continuous from A0(s) to R as it satisfies
using (33), (34) and (35) : |E0(s, y, f ) · (Y, F )| ≤ kρ0(s)(∂ty − ∂xxy + g(y) − f 1ω)kL2(Q T)kρ0(s)(∂tY − ∂xxY + g 0(y)Y − F 1 ω)kL2(Q T) ≤p2E(s, y, f ) kρ0(s)(∂tY − ∂xxY )k2+ kρ0(s)F kL2(qT)+ kρ0(s)g0(y)Y kL2(QT) ≤√6 1 + kg0(y)kL∞(Q T)pE(s, y, f )k(Y, F )kA0(s). (40)
Similarly, for all λ ∈ R?
1 λh s, (y, f ), λ(Y, F ) ≤ |λ|kρ0(s)(∂tY − ∂xxY + g0(y)Y − F 1ω)kL2(QT)+ p 2E(s, y, f ) +1 2kρ0(s)l(y, λY )kL2(QT) 1 |λ|kρ0(s)l(y, λY )kL2(QT) +|λ| 2 kρ0(s)(∂tY − ∂xxY + g 0(y)Y − F 1 ω)k2L2(QT).
Since g0 ∈ C(R) we have, a.e in QT :
1 λl(y, λY = g(y+λY )−g(y) λ − g 0(y)Y → 0 as λ → 0 and, since Y ∈ L∞(QT) and y ∈ L∞(QT), a.e in QT 1 λl(y, λY =
g(y + λY ) − g(y) λ − g 0(y)Y ≤ sup θ∈[0,1] kg0(y + θY )k L∞(Q T)+ kg 0(y)k L∞(Q T)|Y |
and therefore (recalling that ρ0≤ ρ)
1
|λ|kρ0(s)l(y, λY )kL2(QT)≤ sup
θ∈[0,1] kg0(y + θY )kL∞(Q T)+ kg 0(y)k L∞(Q T)kρ0Y kL2(QT) ≤ sup θ∈[0,1] kg0(y + θY )kL∞(Q T)+ kg 0(y)k L∞(Q T)kρY kL2(QT).
It then follows from the Lebesgue dominated convergence theorem that λ1kρ0(s)l(y, λY )kL2(QT) → 0 as
λ → 0 and therefore that h(s, (y, f ), λ(Y, F )) = o(λ). Thus the functional E(s) is differentiable at the point (y, f ) ∈ A(s) along the direction (Y, F ) ∈ A0(s). Eventually, the equality (38) follows from the
definition of the pair (Y1, F1) given in (32).
Remark that from the equality (39), the derivative E0(s, y, f ) is independent of (Y, F ). We can then define the norm kE0(s, y, f )k(A0(s))0 := sup(Y,F )∈A0(s),(Y,F )6=(0,0)
E0(s,y,f )·(Y,F )
k(Y,F )kA0(s) associated to A00(s), the
set of the linear and continuous applications from A0(s) to R.
Combining the equality (38) and the inequality (33), we deduce the following estimates of E(s, y, f ) in term of the norm of E0(s, y, f ).
Proposition 4. For any (y, f ) ∈ A(s) and s ≥ max kg0(y)k2/3L∞(Q
T), s0, the inequalities hold true
1 √ 6 1 + kg0(y)k L∞(Q T) kE 0(s, y, f )k A0 0(s)≤ p E(s, y, f ) ≤ CkE0(s, y, f )kA0 0(s)
where C > 0 is the constant appearing in Proposition 2.
Proof. (38) rewrites E(s, y, f ) = 12E0(s, y, f ) · (Y1, F1) where (Y1, F1) ∈ A0(s) is solution of (32) and
therefore, with (33) E(s, y, f ) ≤ 1 2kE 0(s, y, f )k A0 0(s)k(Y 1, F1)k A0(s) ≤ CkE0(s, y, f )k A0 0(s) p E(s, y, f ). On the other hand, using (40), for all (Y, F ) ∈ A0(s) :
|E0(s, y, f ) · (Y, F )| ≤√6 1 + kg0(y)kL∞(Q
T)pE(s, y, f )k(Y, F )kA0(s)
leading to the left inequality.
In particular, any critical point (y, f ) ∈ A(s) for E(s) (i.e. for which E0(s, y, f ) vanishes) is a zero for E(s), a pair solution of the controllability problem. In other words, any sequence (yk, fk)k∈N of A(s)
satisfying kE0(s, yk, fk)kA0
0(s) → 0 as k → ∞ and for which (kg
0(y
k)k∞)k∈N is bounded is such that
E(s, yk, fk) → 0 as k → ∞. We insist that this property does not imply the convexity of the functional
E(s) (nor a fortiori the strict convexity of E(s), which actually does not hold here in view of the multiple zeros for E(s)) but show that a minimizing sequence for E(s) can not be stuck in a local minimum. Our least-squares algorithm, designed in the next section, is based on that property.
Eventually, the left inequality indicates that the functional E(s) is flat around its zero set. As a consequence, gradient based minimizing sequences for E(s) are inefficient as they usually achieve a low rate of convergence (we refer to [26] and also [24] devoted to the Navier-Stokes equation where this phenomenon is observed).
We end this section with the following crucial estimate.
Lemma 5. Assume that g satisfies (Hp) for some p ∈ [0, 1]. Let (y, f ) ∈ A(s), s ≥
max kg0(y)k2/3L∞(Q
T), s0, and (Y
1, F1) ∈ A
0(s) given in Definition 1 associated with (y, f ). For any
λ ∈ R+ the following estimate holds
q
E s, (y, f ) − λ(Y1, F1) ≤pE(s, y, f )
|1 − λ| + λp+1c 1(s) p E(s, y, f )p (41) with c1(s) := C1+p 1 + ps −3/2e−3p 2s[g0]p. (42)
Proof. For any (x, y) ∈ R2, y 6= 0, and λ ∈ R, we write g(x + λy) − g(x) =R0λyg
0(x + ξy)dξ leading to |g(x + λy) − g(x) − λg0(x)y| ≤ Z λ 0 |y||g0(x + ξy) − g0(x)|dξ ≤ Z λ 0 |y|1+p|ξ|p|g0(x + ξy) − g0(x)| |ξy|p dξ ≤ [g0] p|y|1+p |λ|1+p 1 + p. It follows that ρ0(s)| 1 λl(y, λY 1)| = ρ 0(s)
g(y + λY1) − g(y)
λ − g 0(y)Y1 ≤ [g 0] pρ0(s)|Y1|1+p |λ|p 1 + p
and thus, since Y1∈ L∞(Q
T) (see Remark 4) and ρ0(s) ≤ ρ(s) :
1 |λ|kρ0(s)l(y, λY 1)k L2(QT)≤ [g0]p |λ|p 1 + pkρ0(s)|Y 1|1+pk L2(QT) ≤ [g0]p |λ|p 1 + pkY 1kp L∞(Q T)kρ(s)Y 1k L2(QT) (43)
and obtain that
2E s, (y, f ) − λ(Y1, F1) =
ρ0(s) ∂ty − ∂xxy + g(y) − f 1ω − λρ0(s) ∂tY1− ∂xxY1+ g0(y)Y1− F 1ω + ρ0(s)l(y, −λY1)
2 L2(QT) =
ρ0(s)(1 − λ) ∂ty − ∂xxy + g(y) − f 1ω + ρ0(s)l(y, −λY1)
2 L2(Q T) ≤ ρ0(s)(1 − λ) ∂ty − ∂xxy + g(y) − f 1ω L2(QT)+ ρ0(s)l(y, −λY1) L2(QT) 2 ≤ 2 |1 − λ|pE(s, y, f ) + λ p+1 √ 2(p + 1)[g 0] pkY1kpL∞(Q T)kρ(s)Y 1k L2(QT) 2 . (44) Using Remark 4 and estimates (33) and (34), we obtain
2E s, (y, f ) − λ(Y1, F1) ≤ 2 |1 − λ|pE(s, y, f ) + Ce−3p2sλ p+1 p + 1[g 0] pk(Y1, F1)k p A0(s)s −3/2p E(s, y, f ) 2 ≤ 2 |1 − λ|pE(s, y, f ) + λp+1Cs −3/2e−3p 2sCp p + 1 [g 0] pE(s, y, f ) p+1 2 2
from which we get (41).
4
Convergence of the least-squares method
We now examine the convergence of an appropriate sequence (yk, fk) ∈ A(s). In this respect, we observe
from the equality (38) that, for any (y, f ) ∈ A(s), −(Y1, F1) given in Definition 1, is a descent direction
for the functional E(s) at the point (y, f ), as soon as s ≥ max(kg0(y)k2/3L∞(Q
T), s0). Therefore, we can
define at least formally, for any fixed m ≥ 1, a minimizing sequence (yk, fk)k∈N∈ A(s) as follows:
(y0, f0) ∈ A(s), (yk+1, fk+1) = (yk, fk) − λk(Yk1, F 1 k), k ≥ 0, λk = argminλ∈[0,m]E s, (yk, fk) − λ(Yk1, F 1 k) (45)
where (Yk1, Fk1) ∈ A0(s) is the minimal controlled pair solution of
( ∂tYk1− ∂xxYk1+ g 0(y k)Yk1= F 1 k1ω+ ∂tyk− ∂xxyk+ g(yk) − fk1ω in QT, Yk1= 0 on ΣT, Yk1(·, 0) = 0 in Ω (46)
associated with (yk, fk) ∈ A(s). In particular, the pair Yk1, F 1
k vanishes when E(s, yk, fk) vanishes. The
real number m ≥ 1 is arbitrarily fixed and is introduced in order to keep the sequence (λk)k∈N bounded.
We highlight that, in order to give a meaning to (45), we need to prove that we can choose the parameter s independent of k, that is s ≥ max kg0(yk)k
2/3
L∞(QT), s0 for all k ∈ N. In this respect, it
suffices to prove that there exists M > 0 such that kykkL∞(Q
T)≤ M for every k ∈ N. Under (H2), this
implies that kg0(yk)kL∞(Q
T) ≤ ψ(M ) for every k ∈ N, where ψ is defined in (28). We shall prove the
Proposition 5. Assume that g satisfies (H2) and (Hp) for some p ∈ [0, 1]. Let M > 0 large enough
and s ≥ max C(p)ψ(M )2/3, s
0 with C(p) = 1 if p ∈ (0, 1], and C(0) = (2C)3/2. Let (y0, f0) ∈ A(s)
such that M ≥ ky0kL∞(Q
T). Assume that, for some n ≥ 0, (yk, fk)0≤k≤n defined from (45) satisfies
kykkL∞(Q T)≤ M . Then kyn+1kL∞(Q T)≤ky0kL∞(QT)+ Cm max p + 1 p p E(s, y0, f0), (1 + p)1p+1 p c 1/p 1 (s)E(s, y0, f0) (47) if p ∈ (0, 1] and kyn+1kL∞(Q T)≤ ky0kL∞(QT)+ Cm pE(s, y0, f0) 1 − c1(s) (48) if p = 0.
We point out that the existence of (y0, f0) ∈ A(s) follows from Lemma 4.
Proof. The inequality kynkL∞(Q
T)≤ M implies that kg 0(y n)k 2/3 L∞(Q T)≤ ψ(M ) 2/3and then s ≥ max ψ(M )2/3, s 0 ≥ max kg0(yn)k 2/3 L∞(Q
T), s0. Proposition 2 allows to construct the pair sequence
(Y1
n, Fn1) ∈ A0(s) solution of (32). Then, (45) allows to define (yn+1, fn+1). Estimate (41) implies that
q E s, (yk, fk) − λ(Yk1, F 1 k) ≤ p E(s, yk, fk) |1 − λ| + λp+1c1(s) p E(s, yk, fk) p and then p E(s, yk+1, fk+1) ≤ p E(s, yk, fk) min λ∈[0,m]pk(s, λ) (49) where pk(s, λ) := |1 − λ| + λp+1c1(s)E(s, yk, fk)p/2, ∀λ ∈ R, ∀s > 0. (50)
Since E(s, yk, fk)0≤k≤n decreases, pk(s, λ)0≤k≤ndecreases for all λ (pk do not depend on k if p = 0)
and thus, defining pk(s, fλk) := minλ∈[0,m]pk(s, λ), pk(s, fλk)
0≤k≤ndecreases as well. (49) then implies,
for all 0 ≤ k ≤ n − 1, that p E(s, yk+1, fk+1) ≤ p E(s, yk, fk)pk(s, fλk) ≤ p E(s, yk, fk)p0(s, fλ0). (51)
First case : p ∈ (0, 1]. We prove that
n X k=0 p E(s, yk, fk) ≤ max 1 p p E(s, y0, f0), (1 + p)1p+1 p c2(s)E(s, y0, f0) (52) where c2(s) := c 1/p
1 (s) . Since p00(s, 0) = −1, p0(s, fλ0) < p0(s, 0) = 1 we deduce from (51) that : n X k=0 p E(s, yk, fk) ≤ p E(s, y0, f0) 1 − p0(s, fλ0)n+1 1 − p0(s, fλ0) ≤ pE(s, y0, f0) 1 − p0(s, fλ0) . (53)
If c2(s)pE(s, y0, f0) < (p+1)11/p, we check that p0(s, fλ0) ≤ p0(s, 1) = c1(s)pE(s, y0, f0)
p ≤ 1 p+1 and thus pE(s, y0, f0) 1 − p0(s, fλ0) ≤ p + 1 p p E(s, y0, f0).
If c2(s)pE(s, y0, f0) ≥ (p+1)11/p, then for all λ ∈ [0, 1], p
0
0(s, λ) = −1 + (p + 1)λpc1(s)E(s, y0, f0)p/2 and
thus p00(s, λ) = 0 if and only if λ = (p+1)1/pc 1 2(s) √ E(s,y0,f0) leading to p0(s, fλ0) = 1 − p (1 + p)1p+1 1 c2(s)pE(s, y0, f0)
and pE(s, y0, f0) 1 − p0(s, fλ0) ≤(1 + p) 1 p+1 p c2(s)E(s, y0, f0). (53) then leads to (52). Then (45) implies that yn+1= y0−P
n
k=0λkYk1and thus, using (33)
kyn+1kL∞(Q T)≤ ky0kL∞(QT)+ m n X k=0 kYk1kL∞(Q T)≤ ky0kL∞(QT)+ Cm n X k=0 k(Yk1, F 1 k)kA0(s) ≤ ky0kL∞(Q T)+ Cm n X k=0 p E(s, yk, fk)
which gives (47), using (52).
Second case : p = 0. Recall that for p = 0, ψ(r) = kg0k∞ for every r ∈ R. Then simply pk(s, λ) =
|1 − λ| + λc1(s) for all k and
p0(s, fλ0) = min
λ∈[0,m]p0(s, λ) = minλ∈[0,1]p0(s, λ) = p0(s, 1) = c1(s)
with c1(s) = Cs−3/2[g0]0 = 2Cs−3/2kg0k∞ = 2Cs−3/2ψ(M ). Taking s large enough, precisely s >
max((2C)1/3ψ(M )2/3, s0), we obtain that c1(s) < 1. We then have for all 0 ≤ k ≤ n − 1 that
p E(s, yk+1, fk+1) ≤ p E(s, yk, fk) c1(s) and thus n X k=0 p E(s, yk, fk) ≤ p E(s, y0, f0) 1 − c1(s)n+1 1 − c1(s) ≤pE(s, y0, f0) 1 − c1(s) . (54)
Proceeding as before, we get (48).
In view of estimates (47) and (48), we now intend to choose s such that kyn+1kL∞(Q
T)≤ M . To this
end, we need an estimate of E(s, y0, f0) = 12kρ0(s)(∂ty0− ∂xxy0+ g(y0) − f01ω)k2L2(QT) in terms of s.
Since ρ(s) /∈ L2(Q
T), such estimate is not straightforward for any (y0, f0) ∈ A(s). We select the pair
(y0, f0) ∈ A(s) solution of the linear problem, i.e. g ≡ 0 in (1).
Lemma 6. Assume that g satisfies (H2). For any s ≥ s0, let (y0, f0) ∈ A(s) be the solution of the
extremal problem (25) in the linear case for which g ≡ 0. Then, p E(s, y0, f0) ≤ α + β c3/2+ ln3/2(1 + Cku0kH1 0(Ω)) ecsku0kH1 0(Ω). (55) with c = kϕ(·, 0)k∞.
Proof. Estimate (16) of Proposition 2 with A = 0, B = 0 and z0= u0 leads to
kρ(s)y0kL2(Q
T)+ kρ0(s)f0kL2(qT)≤ Cs
−3/2ecsku
0k2 (56)
while (18) leads to, since s > 1 and ρ0≥ 1,
ky0kL∞(Q T)≤ Ce −3 2secsku0k H1 0(Ω)≤ Ce cs ku0kH1 0(Ω). (57)
It follows that, using (31), since ρ0≤ ρ and s ≥ 1 p E(s, y0, f0) = 1 √ 2kρ0(s)g(y0)kL2(QT)≤ 1 √ 2ψ(ky0kL∞(QT))kρ(s)y0kL2(QT) ≤ Cψ(Cecsku 0kH1 0(Ω))s −3/2ecsku 0kH1 0(Ω) ≤ C α + β ln3/2(1 + Cescku 0kH1 0(Ω))s −3/2ecsku 0kH1 0(Ω) ≤ C α + β ln3/2(esc(1 + Cku 0kH1 0(Ω)))s −3/2ecsku 0kH1 0(Ω) ≤ C α + β (sc)3/2+ ln3/2(1 + Cku0kH1 0(Ω)) s−3/2ecsku0kH1 0(Ω) ≤ C α + β c3/2+ ln3/2(1 + Cku0kH1 0(Ω)) ecsku0kH1 0(Ω).
We are now in position to prove to following result.
Proposition 6. Assume that g satisfies (H2) and (Hp) for some p ∈ [0, 1]. Assume moreover that
2cC(p)β2/3 < 1 and let (y
0, f0) be the controlled pair given by Lemma 6. There exists M0 > 0 such
that, if we have constructed from (45) the pairs (yk, fk)0≤k≤n ∈ A(s) with s = max(C(p)ψ(M0)2/3, s0)
satisfying kykkL∞(Q
T) ≤ M0 for all 0 ≤ k ≤ n, then the pair (yn+1, fn+1) constructed from (45) also
belongs to A(s) and satisfies
kyn+1kL∞(Q
T)≤ M0.
Proof. Assume that for some M larhe enough, kykkL∞(QT)≤ M . The inequality (a + b)2/3≤ a2/3+ b2/3
for all a, b ≥ 0 allows to write
ψ(M )2/3= (α + β ln3/2(1 + M ))2/3≤ α2/3+ β2/3ln(1 + M ).
Assume that for some M large enough, kykkL∞(Q
T)≤ M . Estimate (57) with s = s = max(C(p)ψ(M0)
2/3, s 0) then leads to ky0kL∞(Q T)≤ Cku0kH01(Ω)e cs≤ Cku 0kH1 0(Ω)e c(s0+C(p)ψ(M )2/3)≤ Cecs0ku 0kH1 0(Ω)e cC(p)ψ(M )2/3 ≤ Cecs0ku 0kH1 0(Ω)e cC(p) α2/3+β2/3ln(1+M ) ≤ Cecs0ku 0kH1 0(Ω)e C(p)α2/3ecC(p)β2/3ln(1+M ) ≤ Cecs0ku 0kH1 0(Ω)e C(p)α2/3 (1 + M )cC(p)β2/3 ≤ c(α, u0)(1 + M )cC(p)β 2/3 . (58)
Similarly, this estimate of ecs and (55) leads to
p E(s, y0, f0) ≤ α + β c3/2+ ln3/2(1 + Cku0kH1 0(Ω)) ecs0eC(p)α2/3(1 + M )cC(p)β2/3ku 0kH1 0(Ω) ≤ c(α, β, u0)(1 + M )cC(p)β 2/3 . (59)
First case : p ∈ (0, 1]. Since s ≥ 1, the constant c1(s) defined in (42) satisfies c1(s) ≤ C
1+p
1+p [g 0]
p.Therefore,
by combining (47), (58) and (59), we get kyn+1kL∞(Q T)≤c(α, u0)(1 + M ) cC(p)β2/3 + Cm max 1 pc(α, β, u0)(1 + M ) cC(p)β2/3 , (1 + p) p [g 0]1/p p C (1+p)/pc2(α, β, u 0)(1 + M )2cC(p)β 2/3 ≤C(p, α, u0, [g0]p, β)(1 + M )2cC(p)β 2/3 .
Now, if β is small enough so that 2cC(p)β2/3< 1, the real M 0defined as follows M0:= infM > 0 | C(p, α, u0, [g0]p, β)(1 + M )2cC(p)β 2/3 ≤ M (60) exists and is independent of n. Moreover, for all M = M0 and s = max(C(p)ψ(M0)2/3, s0) :
kyn+1kL∞(Q
T)≤ M.
Second case : p = 0. In this case, c1(s) = Cs−3/2[g0]0 = 2Cs−3/2kg0k∞ < 1 for s large enough. By
combining (48), (58) and (59), we get kyn+1kL∞(Q T)≤ c(α, u0)(1 + M ) cC(0)β2/3+ 1 1 − c1(s) c(α, β, u0)(1 + M )cC(0)β 2/3 . ≤ C(α, u0, [g0]0, β)(1 + M )cC(0)β 2/3
and we conclude as in the previous case.
We are now in position to prove by induction the following decay result for the sequence (E(s, yk, fk))(k∈N).
Proposition 7. Assume that g satisfies (H2) and (Hp) for some p ∈ [0, 1]. Assume moreover that
2cC(p)β2/3 < 1. Let M
0 be given by (60) and s = max(C(p)ψ(M0)2/3, s0). Let (y0, f0) ∈ A(s) be
the solution of the extremal problem (25) in the linear situation for which g ≡ 0. Then the sequence (yk, fk)k∈N∈ A(s) defined by (45) satisfies
kykkL∞(QT)≤ M0, ∀k ∈ N.
Moreover, the sequence (E(s, yk, fk))k∈N → 0 tends to 0 as k → ∞. The convergence is at least linear,
and is at least of order 1 + p after a finite number of iterations.
Proof. The uniform boundedness of the sequence (yk)k∈N follows by induction from Proposition 6 and
implies the decay to 0 of E(s, yk, fk). Remark that, from the construction of M0, ky0kL∞(Q
T)< M0.
First case : p ∈ (0, 1]. From the definition of pk given in (50) we have pk(eλk) := minλ∈[0,m]pk(λ) ≤
pk(1) = c1(s)E(s, yk, fk)p/2 and thus
c2(s) p E(s, yk+1, fk+1) ≤ c2(s) p E(s, yk, fk) 1+p , c2(s) := c 1/p 1 (s). (61)
Thus, if c2(s)pE(s, y0, f0) < (p+1)11/p, then c2(s)pE(s, yk, fk) → 0 as k → ∞ with a rate 1 + p. On the
other hand, if c2(s)pE(s, y0, f0) ≥ (p+1)11/p, then I := {k ∈ N, c2(s)pE(s, yk, fk) ≥
1
(p+1)1/p} is a finite
subset of N. Indeed, for all k ∈ I and for all λ ∈ [0, 1] : p0k(s, λ) = −1 + (p + 1)λ pc 1(s)E(s, yk, fk)p/2 and thus p0 k(s, λ) = 0 if and only if λ = 1 (p+1)1/pc 2(s) √ E(s,yk,fk) , which gives pk(s, fλk) = min λ∈[0,m] pk(λ) = min λ∈[0,1] pk(λ) = pk 1 (1 + p)1/pc 2(s)pE(s, yk, fk) = 1 − p (1 + p)p1+1 1 c2(s)pE(s, yk, fk) and thus c2(s) p E(s, yk+1, fk+1) ≤ 1 − p (1 + p)1p+1 1 c2(s)pE(s, yk, fk) c2(s) p E(s, yk, fk) = c2(s) p E(s, yk, fk) − p (1 + p)1p+1 . (62)
This inequality implies that the sequence c2(s)pE(s, yk, fk)k∈I strictly decreases so that there exists
k0 ∈ I such that c2(s)pE(s, yk0+1, fk0+1) <
1
(p+1)1/p. Thus the sequence c2(s)pE(s, yk, fk)
k∈N
de-creases to 0 at least linearly and there exists k0∈ N such that for all k > k0, c2(s)pE(s, yk, fk) < (p+1)11/p,
that is I is a finite subset of N. Arguing as in the first case, it follows that c2(s)pE(s, yk, fk) → 0 as
k → ∞.
Second case : p = 0. Then for all k ∈ N, since c1(s) < 1, pk(s, fλk) = c1(s) (since fλk= 1) and therefore
p E(s, yk+1, fk+1) ≤ c1(s) p E(s, yk, fk) ≤ c1(s)k+1 p E(s, y0, f0) (63) ThuspE(s, yk, fk) → 0 as k → ∞.
We now prove the main result of this section.
Theorem 3. Assume that g satisfies (H2) and (Hp) for some p ∈ [0, 1]. Assume moreover that β is
small enough so that
2cC(p)β2/3< 1
with c = kϕ(·, 0)kL∞(Ω). Let M0be given by (60) and s = max(C(p)ψ(M0)2/3, s0). Let (y0, f0) ∈ A(s) be
the solution of the extremal problem (25) in the linear situation for which g ≡ 0 and let (yk, fk)k∈N be the
sequence defined by (45). Then, (yk, fk)k∈N → (y, f ) in A(s) where f is a null control for y solution of
(1). The convergence is at least linear, and is at least of order 1 + p after a finite number of iterations. Proof. For all k ∈ N, let Fk = −P
k
n=0λnFn1 and Yk = P k
n=0λnYn1. Let us prove that (Yk, Fk)k∈N
converges in A0(s), i.e. that the seriesP λn(Fn1, Yn1) converges in A0(s). Using that k(Yk1, F 1
k)kA0(s)≤
C(M0)pE(s, yk, fk) for all k ∈ N (see (33)), we write, using (52) and (53) : k X n=0 λnk(Yn1, F 1 n)kA0(s)≤ m k X n=0 k(Y1 n, F 1 n)kA0(s)≤ C(M0) k X n=0 p E(s, yn, fn) ≤ p E(s, y0, f0) C(M0) 1 − p0(fλ0) .
We deduce that the seriesP
nλn(Y 1
n, Fn1) is normally convergent and so convergent. Consequently, there
exists (Y, F ) ∈ A0(s) such that (Yk, Fk)k∈N converges to (Y, F ) in A0(s).
Denoting y = y0+ Y and f = f0+ F , we then have that (yk, fk)k∈N= (y0+ Yk, f0+ Fk)k∈Nconverges
to (y, f ) in A(s).
It suffices now to verify that the limit (y, f ) satisfies E(s, y, f ) = 0. Using that (Y1
k, Fk1) goes to zero in
A0(s) as k → ∞, we pass to the limit in (46) and get that (y, f ) ∈ A(s) solves (1), that is E(s, y, f ) = 0.
Moreover, we have
k(y, f ) − (yk, fk)kA0(s)≤ C(M0)
p
E(s, yk, fk), ∀k > 0 (64)
which implies, using Proposition 7, the announced order of convergence. Precisely,
k(y, f ) − (yk, fk)kA0(s)= ∞ X p=k+1 λp(Yp1, Fp1) A 0(s)≤ m ∞ X p=k+1 k(Y1 p, Fp1)kA0(s) ≤ m C(M0) ∞ X p=k+1 q E(s, yp, fp) ≤ m C(M0) ∞ X p=k+1 p0(eλ0)p−k p E(s, yk, fk) ≤ m C(M0) p0(eλ0) 1 − p0(eλ0) p E(s, yk, fk).
We emphasize, in view of the non uniqueness of the zeros of E, that an estimate (similar to (64)) of the form k(y, f ) − (y, f )kA0(s)≤ C(M0)
q
E(s, y, f ) does not hold for all (y, f ) ∈ A(s). We also mention the fact that the sequence (yk, fk)k∈N and its limits (y, f ) are uniquely determined from the initial guess
(y0, f0) and from our criterion of selection of the pair (Yk1, Fk1) for every k. In other words, the solution
(y, f ) is unique up to the element (y0, f0) and the functional J .
We also have the following convergence of the optimal sequence (λk)k∈N.
Lemma 7. Under hypotheses of Theorem (3) with p ∈ (0, 1], the sequence (λk)k∈N defined in (45)
converges to 1 as k → ∞.
Proof. If p ∈ (0, 1], in view of (44) we have, as long as E(yk, fk) > 0, since λk∈ [0, m]
(1 − λk)2=
E(s, yk+1, fk+1)
E(s, yk, fk)
− (1 − λk)
ρ0(s) ∂tyk+ ∆yk+ g(yk) − fk1ω, ρ0(s)l(yk, −λkYk1)
L2(Q T) E(s, yk, fk) − ρ0(s)l(yk, −λkYk1) 2 L2(QT) 2E(s, yk, fk) ≤ E(s, yk+1, fk+1) E(s, yk, fk) − (1 − λk)
ρ0(s) ∂tyk+ ∆yk+ g(yk) − fk1ω, ρ0(s)l(yk, −λkYk1)
L2(QT) E(s, yk, fk) ≤ E(s, yk+1, fk+1) E(s, yk, fk) +√2mpE(s, yk, fk)kρ0(s)l(yk, −λkY 1 k)kL2(QT) E(s, yk, fk) ≤ E(s, yk+1, fk+1) E(s, yk, fk) +√2mkρ0(s)l(yk, −λkY 1 k)kL2(QT) pE(s, yk, fk) .
But, from (43), (33) and Remark 4, we infer that
kρ0(s)l(yk, −λkYk1)kL2(Q T)≤ [g 0] p λp+1k p + 1kY k p L∞(Q T)kρ(s)Y kL2(QT) ≤ λp+1k Cs −3/2e−3p 2s p + 1 [g 0] pE(s, yk, fk) p+1 2 ≤ C(s)mp+1[g0] pE(s, yk, fk) p+1 2 and thus (1 − λk)2≤ E(s, yk+1, fk+1) E(s, yk, fk) + mp+2C(s)[g0]p(E(s, yk, fk))p/2.
Consequently, since E(s, yk, fk) → 0 and E(s,yE(s,yk+1,fk+1)
k,fk) → 0, we deduce that (1 − λk)
2→ 0 as k → ∞.
If p = 0 and if s is large enough, then c1(s) < 1 and fλk = 1 for every k ∈ N leading to the decay of
(E(s, yk, fk))k∈N to 0 (see (63)). Moreover, estimate (41) implies that the sequence (λk)k∈N with λk = 1
for every k also leads to the decay (E(s, yk, fk))k∈N with an order at least linear. Whether or not this
Remark 7. In Theorem 3, the sequence (yk, fk)k∈N is initialized with the solution of minimal norm
corresponding to g ≡ 0. This natural choice in practice leads to a precise estimate ofpE(s, y0, f0) with
respect to the parameter s. Many other pairs are available such as for instance the pair (y0, f0) = (y, 0) =
(φ y?, 0) constructed in Lemma 4 since it leads to the following estimate in term of s:
p E(s, φy?, 0) = √1 2kρ0 ∂t(φy ?) − ∆(φy?) + g(φy?)k 2= 1 √ 2kρ0(s) φty ?+ g(φy?)k 2 ≤T 2kρ0(s)kL∞(QT /2)k∂tφkL∞(QT /2)ky ?k L∞(Q T /2)+ kρ0(s)g(φy ?))k 2 ≤ C(T ) kρ0(s)kL∞(Q T /2)ku0kH1(Ω)+ ψ(kφy ?k L∞(Q T /2))kρ(s)φy ?k L∞(Q T /2) ≤ C(T )kρ(s)kL∞(Q T /2)ku0kH1(Ω) 1 + ψ(ku0kL∞(Ω)) ≤ C(T )ku0kH1(Ω) 1 + α + β ln3/2(1 + ku0kH1(Ω)) eskϕkL∞ (QT /2).
Remark 8. As stated in Theorem 3, the convergence is at least of order 1 + p after a number k0 of
iterations. Using (62), k0 is given by
k0= 1 + p p (1 + p)1/pc2(s) p E(s, y0, f0) − 1 + 1, (65)
(where b·c is the integer part) if (1 + p)1/pc
2(s)pE(s, y0, f0) − 1 > 0, and k0= 1 otherwise.
5
Comments
Several comments are in order.
Asymptotic condition. The asymptotic condition (H2) on g0is slightly stronger than the asymptotic
condition (H1) made in [17]: this is due to our linearization of (1) which involves r → g0(r) while the
linearization (6) in [17] involves r → g(r)/r. There exist cases covered by Theorem 1 in which exact controllability for (1) is true but that are not covered by Theorem 3. Note however that the example g(r) = a + br + cr ln3/2(1 + |r|), for any a, b ∈ R and for any c > 0 small enough (which is somehow the limit case in Theorem 1) satisfies (H2) as well as (Hp) for any p ∈ [0, 1].
While Theorem 1 was established in [17] by a nonconstructive fixed point argument, we obtain here, in turn, a new proof of the exact controllability of semilinear multi-dimensional wave equations, which is moreover constructive, with an algorithm that converges unconditionally, at least with order 1 + p.
Minimization functional. The estimate (33) is a key point in the convergence analysis and is inde-pendent of the choice of the functional J defined by J (y, f ) = 12kρ0(s)f k2L2(q
T)+ 1 2kρ(s)yk 2 L2(Q T) (see
Proposition 2) in order to select a pair (Y1, F1) in A
0(s). Thus, we may consider other weighted
func-tionals, for instance J (y, f ) = 12kρ0(s)f k2L2(qT)as discussed in [27].
Link with Newton method. If we introduce F : A(s) → L2(QT) by F (y, f ) := ρ−10 (s)(∂ty − ∂xxy +
g(y) − f 1ω), we get that E(s, y, f ) = 12kF (y, f )k2L2(Q
T) and check that, for λk = 1, the algorithm (45)
coincides with the Newton algorithm associated with the mapping F . This explains the super-linear convergence in Theorem 3. The optimization of the parameter λk is crucial here as it allows to get a
global convergence result. Its leads to so-called damped Newton method (for F ) (we refer to [10, Chapter 8]). As far as we know, the analysis of damped type Newton methods for partial differential equations has deserved very few attention in the literature. We mention [22, 29] in the context of fluid mechanics.
A variant. To simplify, let us take λk = 1, as in the standard Newton method. Then, for each k ∈ N,
the optimal pair (Yk1, Fk1) ∈ A0 is such that the element (yk+1, fk+1) minimizes over A(s) the functional
(z, v) → J (z − yk, v − fk). Alternatively, we may select the pair (Yk1, F 1
k) so that the element (yk+1, fk+1)
minimizes the functional (z, v) → J (z, v). This leads to the sequence (yk, fk)k∈Ndefined by
(
∂tyk+1− ∂xxyk+1+ g0(yk)yk+1= fk+11ω+ g0(yk)yk− g(yk) in QT,
yk = 0 on ΣT, yk+1(·, 0) = u0 in Ω.
(66)
In this case, for every k ∈ N, (yk, fk) is a controlled pair for a linearized heat equation, while, in the case
of the algorithm (45), (yk, fk) is a sum of controlled pairs (Yj1, F 1
j) for 0 ≤ j ≤ k. This analysis of this
variant used in [13] is apparently less straightforward.
Local controllability when removing the growth condition (H2). As in [3, 28] devoted to the
wave equations, we may expect to remove the growth condition (H2) on g0 if the initial value E(s, y0, f0)
is small enough. For s fixed, in view of Lemma 6, this is notably true if g(0) = 0 and if the norm ku0kH1 0(Ω)
of the initial data to be controlled is small enough. This would allow to recover the local controllability of the heat equation (usually obtained by an inverse mapping theorem, see [18, chapter 1]) and would be in agreement with the usual convergence of the standard Newton method. In the parabolic case considered here, the proof is however open, since in order to prove the convergence of (E(s, yk, yk))k∈N to zero, for
some s large enough independent of k, we need to prove that the sequence (kykkL∞(Q
T))k∈N is bounded.
This is in contrast with the wave equation where the parameter s does not appear.
Weakening of the condition (Hp). Given any p ∈ [0, 1], we introduce for any g ∈ C1(R) the following
hypothesis :
(H0p) There exist α, β, γ ∈ R+ such that |g0(a) − g0(b)| ≤ |a − b|p α + β(|a|γ+ |b|γ), ∀a, b ∈ R
which coincides with (Hp) if γ = 0 for α + 2β = [g0]p. If γ ∈ (0, 1) is small enough and related to the
constant β appearing in the growth condition (H2), Theorem 3 still holds if (Hp) is replaced by the
weaker hypothesis (H0p).
Influence of the parameter s and a simpler linearization. Taking s large enough in the case p = 0 (corresponding to g0 ∈ L∞
(R)) allows to ensure that the coefficient c1(s) (see (42)) is strictly
less than one, and then to prove the strong convergence of the sequence (yk, fk)k∈N. This highlights the
influence of the parameter s appearing in the Carleman weights ρ, ρ0 and ρ1. Actually, in this case, a
similar convergence can be obtained by considering a simpler linearization of the system (1). For any s ≥ s0and z ∈ L2(ρ(s), QT), we define the controlled pair (y, f ) ∈ A(s) solution of
(
∂ty − ∂xxy = f 1ω− g(z) in QT,
y = 0 on ΣT, y(·, 0) = u0 in Ω,
and which minimizes the weighted cost J . If g(0) = 0 and g is globally Lipschitz, then ρ0(s)g(z) ∈ L2(QT)
and Theorem 2 implies kρ(s) ykL2(Q
T) ≤ Cs
−3/2 kρ
0(s)g(z)kL2(Q T)+ e
csku
0k2. This allows to define
the operator K : L2(ρ(s), Q
T) → L2(ρ(s), QT) by y := K(z). From Lemma 3, for any zi∈ L2(ρ(s), QT),
i = 1, 2, yi:= K(zi) is given by yi= ρ−2(s)L?0pi where pi∈ P0solves
(pi, q)P = Z Ω u0q(0) − Z QT g(zi)q, ∀q ∈ P0.
Taking q := p1− p2, we then get
kp1− p2k2P ≤ Z QT |g(z1) − g(z2)||p1− p2| ≤ kρ0(s)(g(z1) − g(z2))k2kρ−10 (s)(p1− p2)k2 ≤ kg0kL∞(R)kρ0(s)(z1− z2)k2kρ−10 (s)(p1− p2)k2.
