About the controllability of y t εy xx + My x = 0 w.r.t. ε

About the controllability of y t εy xx + My x = w.r.t. ε YOUCEF AMIRAT and ARNAUD MÜNCH Nancy - IECN - November 7th 217

Introduction - The advection-diffusion equation Let T >, M R, ε > and Q T := (, 1) (, T ). >< L εy ε := yt ε εyxx ε + My x ε = Q T, y ε (, ) = v ε (t), y ε (1, ) = (, T ), >: y ε (, ) = y ε (, 1), (1) Well-poseddness: y ε H 1 (, 1), v ε L 2 (, T ),! y ε L 2 (Q T ) C([, T ]; H 1 (, 1)) Null control property: From (Russel 7), T >, y H 1 (, 1), v ε L 2 (, T ) such that y ε (, T ) = in H 1 (, 1). Main concern: Behavior of the controls v ε as ε Controllability of conservation law system; Toy model for fluids when Navier-Stokes Euler.

Cost of control We note the non empty set of null controls by C(y, T, ε, M) := {v L 2 (, T ); y = y(v) solves (5) and satisfies y(, T ) = } For any ε >, we define the cost of control by the following quantity : K (ε, T, M) := sup y L 2 (,1) =1 j ff min u L u C(y,T,ε,M) 2 (,T ). We denote by T M the minimal time for which the cost K (ε, T, M) is uniformly bounded with respect to ε. In other words, (5) is uniformly controllable with respect to ε if and only if T T M.

Reminds Remark Let v ε the control of minimal L 2 -norm for the initial data y ε : then v ε L 2 (,T ) K (ε, T, M) y ε L 2 (,1) Thus, K (ε, T, M) is the norm of the (linear) operator y ε v HUM where v HUM is the control of minimal L 2 -norm. Remark By duality, the controllability property is related to the existence of C > such that ϕ(, ) L 2 (,1) C εϕx (, ) L 2 (,T ), ϕ T H 1 (, 1) H2 (, 1) (2) The quantity C obs (ε, T, M) = sup ϕ T H 1 (,1) ϕ(, ) L2(,1). εϕ x (, ) L 2 (,T ) is the smallest constant for which (2) holds true and K (ε, T, M) = C obs (ε, T, M).

The case M = >< yt ε εyxx ε = Q T, y ε (, ) = v ε (t), y ε (1, ) = (, T ), >: y ε (, ) = y ε (, 1), Making the change of variable t = εt, we easily have Lemma T >, ε >, K (ε, T, ) = ε 1/2 K (1, εt, ) Theorem (Miller, Lissy, Tenebaum-Tucsnak,...) K (1, T, ) T + e κ T, κ (1/2, 3/4) Corollary K (ε, T, ) ε + ε 1/2 e κ εt, κ (1/2, 3/4)

M - Direct problem - Behavior of y ε as ε Theorem (Coron- Guerrero) Let T >, M R, y L 2 (, 1). Let (v ε ) (ε) be a sequence of functions in L 2 (, T ) such that for some v L 2 (, T ) v ε v in L 2 (, T ), ε +. For ε >, let us denote by y ε C([, T ]; H 1 (, 1)) the weak solution of >< yt ε εyxx ε + My x ε = Q T, y ε (, ) = v ε (t), y ε (1, ) = (, T ), >: y ε (, ) = y (, 1), (3) Let y C([, T ]; L 2 (, 1)) be the weak solution of y t + My x = Q T, >< y(, ) = v(t) if M > (, T ), y(1, ) = if M < (, T ), >: y(, ) = y (, 1), (4) Then, y ε y in L 2 (Q T ) as ε +.

First consequence Theorem If T < 1 M, lim ε K (ε, T, M). Consequently, T M 1 M. PROOF. Assume that K (ε, T, M) +. There exists (ε n) (n N) positive tending to such that (K (ε n, T, M)) (n N) is bounded. Let v εn the optimal control driving y to at time T and y εn the corresponding solution. Let T (T, 1/ M ). We extend y εn and v εn by on (T, T ). From the inequality v εn L 2 (,T ) = v εn L 2 (,T ) K (εn, T, M) y L 2 (,1), we deduce that (v εn ) (n N) is bounded in L 2 (, T ), so we extract a subsequence (v εn ) (n N) such that v εn v in L 2 (, T ). We deduce that y εn y in L 2 (Q T ) solution of the transport equation. Necessarily, y on (, 1) (T, T ). Contradiction.

M - Direct problem - Behavior of y ε (, T ) L 2 (,1) as ε for T > 1/ M Lemma The uncontrolled solution (v ε = ) satisfies y ε (, t) L 2 (,1) y ε M2 (, ) L 2 (,1) e 4ε (t M 1 )2, t > 1 M PROOF. Let α >. We check z ε (x, t) = e Mαx 2ε y ε (x, t) solves >< >: Consequently zt ε εzxx ε + M(1 α)zε x M2 4ε (α2 2α)z ε = in Q T, z ε (, ) = z ε (1, ) = on (, T ), z ε (, ) = e Mαx 2ε y ε in (, L), (5) e Mαx 2ε y ε (, t) L 2 (,1) y ε (, t) L 2 (,1) Mαx e 2ε y ε (, ) L 2 (,1) e M2 4ε (α2 2α)t Mαx e+ 2ε L (,1) e Mαx 2ε y ε (, t) L 2 (,1) e + Mαx 2ε and the result with α = t 1 M >. L (,1) e Mαx 2ε y ε (, ) L 2 (,1) e Mα 2ε ` y ε (, ) L 2 (,1) e M2 4ε (α2 2α)t 1 Mt+ Mα 2

Lower bounds for T M We expect T M = L M and that lim ε K (ε, T, M) = + because the transport eq. is null controlled at time T 1 with v! But, M Theorem (Coron-Guerrero 26) If M >, then K (ε, T, M) Ce c/ε, c, C >, when ε for T < L M. If M <, then K (ε, T, M) Ce c/ε, c, C >, when ε for T < 2 L M. More precisely, the lower bound are obtained using specific initial condition: y (x) = K εe Mx 2ε sin(πx), (K ε = O(ε 3/2 ) s.t. y L 2 (,1) = 1) leading, for M >, to and for M <, to ε 3/2 T 1/2 M 2 «M K (ε, T, M) C 1 1 + M 3 ε 3 exp 2ε (1 TM) π2 εt ε 3/2 T 1/2 M 2 «M K (ε, T, M) C 1 1 + M 3 ε 3 exp 2ε (2 TM) π2 εt

Upper bounds for T M Theorem (Coron-Guerrero 26) If M >, then K (ε, T, M) Ce c/ε when ε for T 4.3 M. If M <, then K (ε, T, M) Ce c/ε when ε for T 57.2 M.

Estimates for T M Theorem (Coron-Guerrero 26) T M [1, 4.3] 1 M if M >, [2, 57.2] 1 M if M <. Theorem (Glass 29) T M [1, 4.2] 1 M if M >, [2, 6.1] 1 M if M <. Theorem (Lissy 215) T M [1, 2 3] 1 M if M >, [2 2, 2(1 + 1 3)] M if M <. (2 3 3.46) Theorem (Darde-Ervedoza 217) T M [1, K ] 1 M if M >, K 3.34

Objective Objective: estimate the uniform minimal control time T M!!?? We can try the following two approaches : Numerical estimation of K (ε, T, M) with respect to ε and T 1 M (for M > and M < ) Asymptotic analysis with respect to the parameter ε of the corresponding optimality system.

Attempt 1 : Numerical estimation of K (ε, T, M)

Reformulation of the cost of control K 2 (ε, T, M) = sup y L 2 (,1) (A εy, y ) L 2 (,1) (y, y ) L 2 (,1) where A ε : L 2 (, 1) L 2 (, 1) is the control operator defined by A εy := ˆϕ() where ˆϕ solves the adjoint system < : ϕ t εϕ xx Mϕ x = in Q T, ϕ(, ) = ϕ(1, ) = on (, T ), ϕ(, T ) = ϕ T in (, 1), (6) associated to the initial condition ϕ T H 1 (, 1), solution of the extremal problem inf ϕ T H 1 (,L) J (ϕ T ) := 1 2 Z T (εϕ x (, )) 2 dt + (y, ϕ(, )) L 2 (,T ). REFORMULATION - K (ε, T, M) is solution of the generalized eigenvalue problem : ff λ supj R : y L 2 (, 1), y, s.t. A εy = λy in L 2 (, 1).

The generalized eigenvalue problem by the power iterated method In order to get the largest eigenvalue of the operator A ε, we may employ the power iterate method (Chatelain 9): y L2 (, 1) given such that y L 2 (,1) = 1, >< ỹ k+1 = A εy k, k, >: y k+1 ỹ k+1 = ỹ k+1, k. L 2 (,1) (7) The real sequence { ỹ k L 2 (,1) } k> converges to the eigenvalue with largest module of the operator A ε: q ỹ k L 2 (,1) K (ε, T, M) as k. () The L 2 sequence {y k } k then converges toward the corresponding eigenvector. The first step requires to compute the image of A ε: this is done by determining the control of minimal L 2 norm by minimizing J with y k as initial condition for (5).

Computation of the control of minimal L 2 -norm For a fixed initial data y L 2 (, 1) and ε small, the numerical approximation of controls of minimal L 2 -norm is a serious challenge : the minimization of J is ill-posed : the infimum ϕ T lives in a huge dual space!!! this implies that the minimizer ϕ T is highly oscillating at time T leading to highly oscillation of the control εϕ,x. Tychonoff like regularization inf ϕ T H 1 (,1) J β (ϕ T ) := J (ϕ T ) + β ϕ T H 1 (,1) y ε (, T ) H 1 (,1) β (9) is meaningless here for T > 1/ M because the uncontrolled solution y ε (, T ) goes to zero with ε. Boundary layers occurs for y ε and ϕ ε on the boundary and requires fine discretization. We use the variational approach developed in [Fernandez-Cara-Munch, 213], [De Souza-Munch, 215] leading to convergent approximation with respect to the discretization parameter (ε being fixed).

Picture of controls with respect to ε, y fixed y (x) = sin(πx); T = 1; M = 1.6.4.2.2.4.6 1 5 5 1 15 4 2 2 4 6 1..2.4.6. 1 t 2.2.4.6. 1 t 12.2.4.6. 1 t Control of minimal L 2 (, T )-norm v ε (t) [, T ] for ε = 1 1, 1 2 and 1 3

Picture of controls with respect to ε, y fixed y (x) = sin(πx); T = 1; M = 1 6 4 2 2 4 6 4 3 2 1 1 2 3 25 2 15 1 5 5 1 15 2.2.4.6. 1 t 4.2.4.6. 1 t 25.2.4.6. 1 t Control of minimal L 2 (, T )-norm v ε (t) [, T ] for ε = 1 1, 1 2 and 1 3

Cost of control K (ε, T, M) w.r.t. ε - M = 1 25 2 6 2 15 1 5 15 1 5 5 4 3 2 1.2.4.6..1.2.4.6..1.2.4.6..1 Cost of control w.r.t. ε for T =.95 1 M, T = 1 M and T = 1.5 1 M In agreement with Coron-Guerrero 26, ε 3/2 T 1/2 M 2 «M K (ε, T, M) C 1 1 + M 3 ε 3 exp 2ε (1 TM) π2 εt (1)

Corresponding worst initial condition 5 4 3 2 1.2.4.6. 1 x Figure: T = 1 - M = 1 - The optimal initial condition y in (, 1) for ε = 1 1 (full line), ε = 1 2 (dashed line) and ε = 1 3 (dashed-dotted line). = y is close to e Mx 2ε sin(πx)/ e Mx 2ε sin(πx) L 2 (,1)

Cost of control K (ε, T, M) w.r.t. ε - M = 1 x 14 7 6 5 4 3 2 1.2.4.6..1 2 1 1 2 3 x 14 3.95.96.97.9.99 1 t Left: Cost of control w.r.t. ε for T = 1 M ; Right: Corresponding control v ε in the neighborhood of T for ε = 1 3

Corresponding worst initial condition for M = 1 6 5 4 3 2 1.2.4.6. 1 x Figure: T = 1 - M = 1 - The optimal initial condition y in (, 1) for ε = 1 1 (full line), ε = 1 2 (dashed line) and ε = 1 3 (dashed-dotted line).

Part 2 Attempt 2 : Asymptotic analysis w.r.t. ε We take M >. Optimality system : L εy ε =, L ε ϕε =, (x, t) Q T, y ε (, ) = y ε, x (, 1), >< v ε (t) = y ε (, t) = εϕ ε x (, t), t (, T ), y ε (1, t) =,. t (, T ), ϕ >: ε (, t) = ϕ ε (1, t) =, t (, T ), βϕ ε xx (, T ) + y ε (, T ) =, x (, 1). (11) β > - Regularization parameter

Direct problem - Asymptotic expansion yt ε εyxx ε + Myx ε =, (x, t) (, 1) (, T ), >< y ε (, t) = v ε (t), t (, T ), y >: ε (1, t) =, t (, T ), y ε (x, ) = y (x), x (, 1), (12) y and v ε are given functions. We assume that mx v ε = ε k v k, k= the functions v, v 1,, v m being known. We construct an asymptotic approximation of the solution y ε of (12) by using the matched asymptotic expansion method.

Direct problem - Asymptotic expansion Let us consider two formal asymptotic expansions of y ε : the outer expansion mx ε k y k (x, t), (x, t) (, T ), the inner expansion k= mx ε k Y k (z, t), k= z = 1 x ε (, ε 1 ), t (, T ).

Direct problem - Outer expansion Putting mx ε k y k (x, t) into equation (12) 1, the identification of the powers of ε yields k= ε : y t + My x =, ε k : y k t + My k x = y k 1 xx, for any 1 k m. Taking the initial and boundary conditions into account we define y and y k (1 k m) as functions satisfying the transport equations, respectively, y >< t + Myx =, (x, t) Q T, y (, t) = v (t), t (, T ), >: y (x, ) = y (x), x (, 1), (13) and > y < t k + Myx k = y xx k 1, (x, t) Q T, y k (, t) = v k (t), t (, T ), >: y k (x, ) =, x (, 1). (14)

Direct problem - Outer expansion < y (x Mt) x > Mt, y (x, t) = : v t x, x < Mt. M Using the method of characteristics we find that, for any 1 k m, Z t >< yxx k 1 (x + (s t)m, s)ds, x > Mt, y k (x, t) = >: v k t x Z x/m + yxx k 1 (sm, t x + s)ds, x < Mt. M M Remark < t y y 1 (x Mt), x > Mt, (x, t) = : v 1 t x + x M M 3 (v ) t x, x < Mt, M >< y 2 (x, t) = v 2 t x M >: t 2 2 y (4) (x Mt), x > Mt, + x M 3 (v 1 ) t x M 2x M 5 (v ) (3) t x + x 2 M 2M 6 (v ) (4) t x, x < Mt. M

Direct problem - Inner expansion mx Now we turn back to the construction of the inner expansion. Putting ε k Y k (z, t) k= into equation (12) 1, the identification of the powers of ε yields ε 1 : Yzz (z, t) + MY z (z, t) =, ε k 1 : Yzz k (z, t) + MY z k k 1 (z, t) = Yt (z, t), for any 1 k m. We impose that Y k (, t) = for any k m and use the asymptotic matching conditions Y (z, t) y (1, t), as z +, Y 1 (z, t) y 1 (1, t) yx (1, t)z, as z +, Y 2 (z, t) y 2 (1, t) y 1 x (1, t)z + 1 2 y xx (1, t)z2, as z +, Y m (z, t) y m (1, t) y m 1 x as z +. (1, t)z + 1 2 y xx m 2 (1, t)z 2 + + 1 m! (y ) (m) x (1, t)( z) m,

Direct problem - Inner expansion Lemma Y (z, t) = y (1, t) 1 e Mz, (z, t) (, + ) (, T ). For any 1 k m, the solution of reads where Y k (z, t) = Q k (z, t) + e Mz P k (z, t), (z, t) (, + ) (, T ), (15) P k (z, t) = kx i= 1 i y k i i! x i (1, t)z i, Q k (z, t) = kx ( 1) i i y k i i! x i (1, t)z i. i=

Asymptotic approximation Let X : R [, 1] denote a C 2 cut-off function satisfying ( 1, s 2, X (s) =, s 1, 1 X ε(x) 1 2ε γ 1 ε γ Figure: The function X ε. 1 x We define, for γ (, 1), the function X ε(x) = X 1 x ε γ, then introduce the function mx mx «1 x wm(x, ε t) = X ε(x) ε k y k (x, t) + (1 X ε(x)) ε k Y k, t, (16) ε k= k= defined to be an asymptotic approximation at order m of the solution y ε of (12).

Compatibility conditions - Regularity Lemma (i) Assume that y C 2m+1 [, 1], v C 2m+1 [, T ] and the following C 2m+1 -matching conditions are satisfied M p (y ) (p) () + ( 1) p+1 (v ) (p) () =, p 2m + 1. (17) Then the function y belongs to C 2m+1 ([, 1] [, T ]). (ii) Additionally, assume that v k C 2(m k)+1 [, T ], and the following C 2(m k)+1 -matching conditions are satisfied, respectively, (v k ) (p) () = X ( 1) i M i p+1 y k 1 x i+2 t j (, ), p 2(m k) + 1. (1) i+j=p 1 Then the function y k belongs to C 2(m k)+1 ([, 1] [, T ]).

Compatibility conditions - example with m = 2 Lemma (i) Assume that y C 5 [, 1], v C 5 [, T ] and the following C 5 -matching conditions are satisfied M p (y ) (p) () + ( 1) p+1 (v ) (p) () =, p 5. (19) Then the function y belongs to C 5 ([, 1] [, T ]). (ii) Additionally, assume that v 1 C 3 [, T ], v 2 C 1 [, T ] and the following C 3 and C 1 -matching conditions are satisfied, respectively, >< >: v 1 () =, (v 1 ) (1) () = M 2 (v ) (2) () = y (2) (), (v 1 ) (2) () = 2M 2 (v ) (3) () = 2My (3) (), (v 1 ) (3) () = 3M 2 (v ) (4) () = 3M 2 y (4) (), (2) v 2 () =, (v 2 ) (1) () =. (21) Then the function y 1 belongs to C 3 ([, 1] [, T ]), and the function y 2 belongs to C 1 ([, 1] [, T ]).

Estimate of L ε w ε m Lemma Let wm ε be the function defined by (16). Assume the previous regularity and compatibility. Then there is a constant c m independent of ε such that L ε(w ε m)(x, t) = P 5 i=1 Ji ε(x, t), L ε(w ε m ) C([,T ];L 2 (,1)) Jε(x, 1 t) = ε m+1 yxx m (x, t)x ε(x), «1 x Jε 2 (x, t) = εm (1 X ε(x))yt m, t, «1 x Jε(x, 3 t) = MX ε γ ε γ «1 x Jε(x, 4 t) = X ε γ ε 1 2γ «1 x Jε 5 (x, t) = 2X ε γ ε m X k= m X cm ε (2m+1)γ 2. (22) «1 x ε k Y k, t ε «1 x ε k Y k, t ε! mx ε k y k (x, t), k=! mx ε k y k (x, t), k= k= X m «! 1 x mx ε 1 γ ε 1 ε k Yz k, t + ε k yx k ε (x, t). k= k=

Asymptotic approximation- Convergence Theorem Let y ε be the solution of problem (12) and let w ε m be the function defined by (16). Assume that the assumptions of Lemma 6 hold true. Then there is a constant c m independent of ε such that y ε w ε m C([,T ];L 2 (,1)) 2m+1 cmε 2 γ. (23) Introduce the function z ε = y ε wm ε. It satisfies >< L ε(z ε ) = L ε(wm ε ), (x, t) (, 1) (, t), z ε (, t) = z ε (1, t) =, t (, T ), (24) >: z ε (x, ) = z ε (x), x (, 1), Using the Gronwall Lemma it holds that z ε (, t) 2 L 2 (,1) L ε(wm) ε 2 L 2 (Q T ) + zε 2 L 2 e t, t (, T ]. (25) (,1) We have z ε (x) = (1 Xε(x)) y (x) mx k= 1 x ε k Y k, «!, ε

Convergence with respect to m Theorem Let, for any m N, y ε m denote the solution of problem (12). We assume that the initial condition y belongs to C [, 1] and there are c, b R such that y (m) C[,1] c b m, m N. (26) We assume that (v k ) k is a sequence of polynomials of degree p 1, p 1, uniformly bounded in C p 1 [, T ]. We assume in addition that, for any k N, for any m N, the functions v k and y satisfy the matching conditions. Let wm ε be the function defined by (16). Then, there is ε > such that, for any fixed < ε < ε, we have consequently y ε m w ε m in C([, T ], L2 (, 1)), as m +, lim w ε m + m (x, t) = Xε(x) X X «1 x ε k y k (x, t) + (1 X ε(x)) ε k Y k, t ε k= k= = y ε (x, t) a.e. in (, 1) (, T ), where y ε is the solution of problem (12) with y ε (, t) = P k= εk v k (t), t (, T ).

First approximate controllability result Proposition Let m N, T > 1 M and a ], T 1 M [. Assume regularity and matching conditions on the initial condition y and functions v k, k m. Assume moreover that T > L M t = L M v k (t) =, k m, t [a, T ]. Then, the solution y ε of problem (12) satisfies the following property y ε (, T ) L 2 (,1) cm ε (2m+1)γ 2, γ (, 1) for some constant c m > independent of ε. The function v ε C([, T ]) defined by v ε := P m k= εk v k is an approximate null control for (5). t = T L M x L

m? Remark For ε > small enough, we can not pass to the limit as m in order to get a null-controllability result. Let y (x) = 1. The functions v k, k defined as follows v (t) = X (t), v k (t) =, k > (27) with X = {f C [, T ], f () = 1, f (a) =, f (p) () = f (p) (a) =, p N }, a [, T ] satisfy the matching conditions. If a ], T 1/M[, then wm ε (, T ) L 2 (,T ) = m N. If lim m c mε (2m+1)γ 2 =, then the function v ε = P k= εk v k = v is a null control for y ε as time T. Contradiction.

Picture y (x) = 1, v (t) = X (t), v k (t) =, k > (2) X (t) = e kt2m+2, with k > so that X p ()) = for all p 2m + 1. M = 1, k = 1 2, T = 1 and m = 2. Y (z, t) = (1 e z ) independent of t while Y 1 and Y 2 are identically zero. 1 1..6 5.4.2 1.5 x.2.4 t.6. 1 5 1.5 x.2.4 t.6. 1 2 x 1 4 1 1 2 1.5 x.2.4 t.6. 1

The case T = 1/M 1 x w ε (x, T ) = Xε(x)y (x, T ) + (1 X ε(x)) Y ε = X ε(x)v (T (1 x)) + (1 X ε(x)) y () «, T, (x, t) Q T, 1 e M(1 x) ε «If supp(v ) [, T ε γ ], T = L M Lε γ w ε (, T ) L 2 (,1) y () ε γ/2 and y ε (, T ) L 2 (,1) w ε (, T ) L 2 (,1) + (y ε w ε )(, T ) L 2 (,1) y () ε γ/2 + ε v L 2 (,ε γ T ) y () (cε τ(γ) + ε γ/2 ), t = Tε γ x L

The case of initial condition y ε of the form y ε Mx (x) = e 2ε f (x) Let us consider the case of the C ([, 1]) initial data (with L = 1): y ε Mx (x) := Kε sin(πx)e 2ɛ, K ε = O(ε 3/2 ). (29) Taking m =, the function L ε(w ε) involves the term εy xx (x, t)x ε(x). For points in C := {(x, t) Q T, x > Mt}, we obtain y (x, t) = y ε (x Mt); this leads, to (writing that X ε = 1 on (, 1 2ε γ )) Z 1 2ε γz x/m «2 «1/2 ε yxx Xε L 2 (C ) εkε (sin(π(x Mt))e M(x Mt) 2ε ) xx dtdx = εk εo(1) = O(ε 1/2 ). (3)

The case of initial condition y ε of the form y ε Mx (x) = e 2ε >< >: ` y ε (x, t) = e Mα 2ε L εy ε (x, t) = e Mα 2ε x (2 α)mt 2 ` x (2 α)mt 2 z ε (x, t), zt ε εz ε xx + M(1 α)z ε x «(31) We then define the approximations >< m zm ε (x, t) = Xε(x) X mx ε k z k (x, t) + (1 X ε(x)) >: ` ym(x, ε t) = e Mα 2ε k= x (2 α)mt 2 z ε m(x, t) k= «1 x ε k Z k, t, ε (32) The main issue is now to find control functions v k satisfying the matching conditions such that L εym ε C([,T ],L 2 (,1)) goes to zero with ε.

L ε y ε m C([,T ],L 2 (,1)) T 1 M L εy ε m ` Mα (x, t) = e 2ε x (2 α)mt 2 L ε,αzm ε (x, t) 2 M(2 α) In D β C+ α, L ε,αz ε m(x, t) = ε m+1 z m xx (x, t) ` e Mα 2ε (v ) (2) t t L εy ε (x, t) = ε x (2 α)mt x M 2 2 α M α «C 1 = ε M(1 α) M 2 e Mα2 α(2 α)m 2 x x 4ε Mα t 4ε(1 α) e (v ) (2) t x «α M α β >< (v ) (2) (t) = (C 1 + C 2 t)e η+α(2 α)m2 4ε t, t [, β], x 1 v () = z ε Figure: (, 1) (, T ) = >: (), v (β) =, (v ) (1) ()) = M α(z ε D + β (D β C α ) (D β C+ α ). ) (), (v ) (1) (β) =, (33) for some constants C 1 and C 2 and η >. «, L ε(y ε ) L 1 (L 2 (D β C+ α)) (v )()O(ε 1/2 )+(v ) (1) ()O(ε 3/2 ).

L ε y ε m C([,T ],L 2 (,1)) T 1 M t 2 M(2 α) Thus, the corresponding control is given >< v (t) = e γm2 t 4ε v (t) 1 [,β] (t), γ = α(2 α), >: v (t) = kc 1 2C 2 + kc 2 t k 3 e kt η + α(2 α)m + C 3 t + C 4, k := 4ε (34) C β 1 M(1 α) x 1 Figure: (, 1) (, T ) = D + β (D β C α ) (D β C+ α ). Figure: Control v (t) for ε = 1 2 and ε = 1 3 associated to y ε Mx (x) = e 2ε sin(πx).

Optimality condition For any ε > and β >, the optimality system associated to the extremal problem min v ε 2 v ε L 2 L (,T ) 2 (,T ) + β 1 y ε (, T ) 2 L 2 (,1) (35) where the pair (v ε, y ε ) solves (5) is given by L εy ε =, L εϕ ε =, (x, t) Q T, >< y ε (, ) = y ε, x (, 1), v ε (t) = y ε (, t) = εϕ ε x (, t), t (, T ), y ε (1, t) =, ϕ ε (, t) = ϕ ε (1, t) =, t (, T ), >: βϕ ε xx (, T ) + y ε (, T ) =, x (, 1). (36)

Optimality condition v (t) + εv 1 (t) + = Φ z(, t) + εφ 1 z(, t) +, t (, T ). At the zero order, we get therefore v (t) = Φ z (, t) leading simultaneously, to v (t) = Mϕ (, t) = ( Mϕ T (M(T t)), t ]T 1/M, T ],, t [, T 1/M] The last equality contradicts the matching conditions (17). Jε (ϕ ε T ) = 1 Z T (εϕ ε x 2 (, ))2 dt + (y, ϕ ε (, )) L 2 (,1), = 1 Z T (Φ 2 z(, t)) 2 dt + (y, ϕ (, )) + ε..., Z = M2 T «2 ϕ T 2 (M(T t)) dt + ε... T 1/M

Optimality condition v (t) + εv 1 (t) + = Φ z(, t) + εφ 1 z(, t) +, t (, T ). At the zero order, we get therefore v (t) = Φ z (, t) leading simultaneously, to v (t) = Mϕ (, t) = ( Mϕ T (M(T t)), t ]T 1/M, T ],, t [, T 1/M] The last equality contradicts the matching conditions (17). Jε (ϕ ε T ) = 1 Z T (εϕ ε x (, )) 2 dt + (y, ϕ ε (, )) 2 L 2 (,1), = 1 Z T (Φ 2 z(, t)) 2 dt + (y, ϕ (, )) + ε..., Z = M2 T «2 ϕ T 2 (M(T t)) dt + ε... T 1/M

Final remarks Instead of imposing regularity assumptions and matching conditions, we may introduce an additional C 2 cut-off X function to take into account the discontinuity of the solutions y k on the characteristic line. This allows to deal with the initial optimality system. The negative case is very similar except that the control v ε lives in the boundary layer. (still in progress! )

Easier case >< ytt ε + ε 2 y ε y ε =, in Q T, y ε =, νy ε = v ε 1 ΓT, on Σ T, >: (y ε (, ), yt ε (, )) = (y, y 1 ), on Ω. Theorem (Lions) Assume (y, y 1 ) L 2 (Ω) H 1 (Ω). Assume that (Ω, Γ T, T ) satisfies a geometric control condition. For any ε >, let v ε be the control of minimal L 2 (Γ T ) for y ε. Then, ( εv ε, y ε ) (v, y) in L 2 (Γ T ) L (, T ; L 2 (Ω)), as ε where v is the control of minimal L 2 (Γ T )-norm for y, solution in C ([, T ]; L 2 (Ω)) C 1 ([, T ]; H 1 (Ω)) of : >< y tt y =, in Q T, y = v1 ΓT, on Σ T, >: (y(, ), y t (, )) = (y, y 1 ), in Ω.

Fin A. Münch : Numerical estimations of the cost of boundary controls for the equation y t εy xx + My x = with respect to ε. Y. Amirat, A. Münch : Asymptotic analysis of the equation y t εy xx + My x = and controllability results. MERCI DE VOTRE ATTENTION