Long Nguyen
email: longnt@hcmc.netnam.vn

Alain Pham
email: alain.pham@univ-orleans.fr

Ngoc Dinh

Xuan Le

Truong
email: truong@math.net

Thanh Nguyen

Alain Long

Ngoc Pham

Dinh

Long

Ngoc Dinh Mapmo

EXISTENCE AND DECAY OF SOLUTIONS OF A NONLINEAR VISCOELASTIC PROBLEM WITH A MIXED NONHOMOGENEOUS CONDITION

Keywords: 2000 Mathematics Subject Classification. 35L05, 35L15, 35L70, 37B25 Faedo-Galerkin method, Global existence, contraction mapping theorem, nonlinear wave equation, Viscoelastic, Exponential decay. 1

d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Introduction

In this paper we will consider the following initial and boundary value problem:

u tt -u xx + t 0 k(t -s)u xx (s)ds + |u t | q-2 u t = f (x, t, u), 0 < x < 1; 0 < t < T, (1.1) u x (0, t) = u(0, t), u x (1, t) + ηu(1, t) = g(t), (1.2)
u(x, 0) = u 0 (x), u t (x, 0) = u 1 (x), (1.3) where η ≥ 0, q ≥ 2 are given constants and u 0 , u 1 , g, k, f are given functions satisfying conditions specified later.

In a recent paper [START_REF] Berrimia | Existence and decay of solutions of a viscoelastic equation with a nonlinear source[END_REF], Berrimia and Messaoudi considered the problem u tt -∆u + t 0 k(t -s)∆u(s)ds = |u| p-2 u, x ∈ Ω, t > 0, (1.4)

u = 0, on ∂Ω, (1.5)
u(x, 0) = u 0 (x), u t (x, 0) = u 1 (x), x ∈ Ω, (1.6) where p > 2 is a constant, k is a given positive function, and Ω is a bounded domain of R n (n ≥ 1), with a smooth boundary ∂Ω. This type of problems have been considered by many authors and several results concerning existence, nonexistence, and asymptotic behavior have been established. In this regard, Cavalcanti et al. [START_REF] Cavalcanti | Exponential decay for the solution of semilinear viscoelastic wave equations with localized damping[END_REF] studied the following equation

u tt -∆u + t 0 k(t -s)∆u(s)ds + |u| p-2 u + a(t)u t = 0, in Ω × (0, ∞), (1.7)
for a : Ω → R + , a function, which may be null on a part of the domain Ω. Under the conditions that a(x) ≥ a 0 > 0 on ω ⊂ Ω, with ω satisfying some geometry restrictions and -ζ 1 k(t) = k / (t) = -ζ 2 k(t), t ≥ 0, (1.8) the authors established an exponential rate of decay.

In [START_REF] Bergounioux | Mathematical model for a shock problem involving a linear viscoelastic bar[END_REF] Bergounioux, Long and Dinh studied problem (1.1), (1.3) with k = 0, q = 2, f (x, t, u) = -Ku + F (x, t), and the mixed boundary conditions (1.2) standing for u x (0, t) = g(t) + hu(0, t) -t 0 H(t -s)u(0, s)ds,

(1.9)

u x (1, t) + K 1 u(1, t) + λ 1 u t (1, t) = 0, (1.10)
where h ≥ 0, K, λ, K 1 , λ 1 are given constants and g, H are given functions. In [START_REF] Long | On a shock problem involving a nonlinear viscoelastic bar[END_REF], Long, Dinh and Diem obtained the unique existence, regularity and asymptotic expansion of the problem (1.1), (1.3), (1.9) and (1.10) in the case of k = 0, f (x, t, u) = -K|u| p-2 u + F (x, t), with p ≥ 2, q ≥ 2; K, λ are given constants.

In [START_REF] Long | On a shock problem involving a linear viscoelastic bar[END_REF], Long, Ut and Truc gave the unique existence, stability, regularity in time variable and asymptotic expansion for the solution of problem (1.1)-(1.3) when k = 0, q = 2, f (x, t, u) = -Ku + F (x, t) and ( u 0 , u 1 ) ∈ H 2 × H 1 . In this case, the problem (1.1)- (1.3) is the mathematical model describing a shock problem involving a linear viscoelastic bar.

In [START_REF] Long | A wave equation associated with mixed nonhomogeneous conditions: Global existence and asymptotic expansion of solutions[END_REF], Long and Giai obtained the unique existence and asymptotic expansion for the solution of problem (1.1), (1.3) when k = 0, q = 2, f (x, t, u) = -Ku + F (x, t) and ( u 0 , u 1 ) ∈ H 1 × L 2 , and the mixed boundary conditions (1.2) standing for u x (0, t) =g(t) + K 1 |u(0, t)| α-2 u(0, t) + λ 1 |u t (0, t)| β-2 u t (0, t) -t 0 H(t -s)u(0, s)ds, (1.11) u(1, t) = 0, (1.12)

where K, λ, K 1 , λ 1 , α, β are given constants and g, H are given functions. In this case, the problem (1.1), (1.3), (1.11), (1.12) is the mathematical model describing a shock problem involving a nonlinear viscoelastic bar. In [START_REF] Long | Existence and asymptotic expansion for a viscoelastic problem with a mixed nonhomogeneous condition[END_REF], Long and Truong obtained the unique existence and asymptotic expansion for the solution of problem (1.1) -(1.3) when f (x, t, u) = -K|u| p-2 u + F (x, t),

( u 0 , u 1 ) ∈ H 2 × H 1 ; F, F t ∈ L 2 (Q T ), k ∈ W 2,1 (0, T ), g ∈ H 2 (0, T ); K, η ≥ 0, η 0 > 0; p, q ≥ 2.
In this paper, we consider two main parts. In Part 1, under a certain local Lipschitzian condition on f with ( u 0 , u 1 ) ∈ H 1 × L 2 ; k, g ∈ H 1 (0, T ),λ > 0, η 0 > 0; η ≥ 0; q ≥ 2, a global existence and uniqueness theorem is proved. The proof is based on the paper [START_REF] Long | Existence and asymptotic expansion for a viscoelastic problem with a mixed nonhomogeneous condition[END_REF] associated to a contraction mapping theorem and standard arguments of density. In Part 2, the asymptotic behavior of the solution u as t → ∞ is studied, under more restrictive conditions, namely f

(x, t, u) = -|u| p-2 u+F (x, t), p ≥ 2, F ∈ L 1 (R + ; L 2 ) L 2 (R + ; L 2 ), +∞ 0 e σt F (t) 2 dt < +∞, with σ > 0, and ( u 0 , u 1 ) ∈ H 1 × L 2 , g = 0, k ∈ H 1 (R + ,
and some others ( • denotes the L 2 (0, 1) norm). It is proved that under these conditions, a unique solution u(t) exists on R + such that u / (t) + u x (t) decay exponentially to 0 as t → +∞. The results obtained here relatively are in part generalizations of those in [START_REF] Berrimia | Existence and decay of solutions of a viscoelastic equation with a nonlinear source[END_REF][START_REF] Bergounioux | Mathematical model for a shock problem involving a linear viscoelastic bar[END_REF][START_REF] Cavalcanti | Exponential decay for the solution of semilinear viscoelastic wave equations with localized damping[END_REF][START_REF] Long | Linear recursive schemes and asymptotic expansion associated with the Kirchhoff-Carrier operator[END_REF][START_REF] Long | On a shock problem involving a nonlinear viscoelastic bar[END_REF][START_REF] Long | On a shock problem involving a linear viscoelastic bar[END_REF][START_REF] Long | A wave equation associated with mixed nonhomogeneous conditions: Global existence and asymptotic expansion of solutions[END_REF][START_REF] Long | Existence and asymptotic expansion for a viscoelastic problem with a mixed nonhomogeneous condition[END_REF]. Finally, we present some numerical results.

Preliminary Results

Put Ω = (0, 1), Q T = Ω×(0, T ), T > 0. We omit the definitions of usual function spaces:

C m (Ω), L p (Ω), W m,p (Ω). We denote W m,p = W m,p (Ω), L p = W 0,p (Ω), H m = W m,2 (Ω), 1 ≤ p ≤ ∞, m = 0, 1, . . . The norm in L 2 is denoted by • .
We also denote by •, • the scalar product in L 2 or pair of dual scalar product of continuous linear functional with an element of a function space. We denote by

• X the norm of a Banach space X and by X ′ the dual space of X. We denote by L p (0, T ; X), 1 ≤ p ≤ ∞ for the Banach space of the real functions u : (0, T ) → X measurable, such that Without loss of generality, we can suppose that η 0 = λ = 1. For every η ≥ 0, we put

u L p (0,T ;X) = T 0 u(t) p X dt 1/p < ∞ for 1 ≤ p < ∞,
and u L ∞ (0,T ;X) = ess sup 0<t<T u(t) X for p = ∞. Let u(t), u ′ (t) = u t (t), u ′′ (t) = u tt (t), u x (t
a η (u, v) = 1 0 u x (x)v x (x)dx + u(0)v(0) + ηu(1)v(1), ∀u, v ∈ H 1 , (2.1)
v η = (a η (v, v)) 1/2 . (2.2)
On H 1 we shall use the following equivalent norm

v 1 = v 2 (0) + 1 0 |v x (x)| 2 dx 1/2 (2.3)
Then we have the following lemmas.

Lemma 2.1. The imbedding V ֒→ C 0 ([0, 1]) is compact and

v C 0 ([0,1]) ≤ v V , for all v ∈ V. ( 2
|a η (u, v)| = C η u 1 v 1 , for all u, v ∈ H 1 , (ii) a η (v, v) = v 2 1 , for all v ∈ H 1 , where C η = 1 + 2η.
The proofs of these lemmas are straightforward, and we omit the details. We also note that on

H 1 , v 1 , v H 1 = v 2 + v / 2 1/2 , v η = (a η (v, v)) are three equivalent norms. v 2 1 ≤ v 2 η ≤ C η v 2 1 , for all v ∈ H 1 , (2.5) 1 3 v 2 H 1 ≤ v 2 1 ≤ 3 v 2 H 1 , for all v ∈ H 1 , (2.6)

The Existence and uniqueness theorem of the solution

In this section we study the global existence of solutions for problem (1.1)-(1.3). For this purpose, we consider, first, a related nonlinear problem. Then, we use the well-known Banach's fixed point theorem to prove the existence of solutions to the nonlinear problem (1.1)- (1.3).

We make the following assumptions:

(H1) η ≥ 0, q ≥ 2, (H2) k, g ∈ H 1 (0, T ), (H3) u 0 ∈ H 1 and u 1 ∈ L 2 , (H4) f ∈ C 0 (Ω × R + × R) satisfies the conditions D 2 f, D 3 f ∈ C 0 (Ω × R + × R).
For each T > 0, we put

W (T ) = v ∈ L ∞ (0, T ; H 1 ) : v t ∈ L ∞ (0, T ; L 2 ) L q (Q T ). (3.1)
Then W (T ) is a Banach space with respect to the norm (see [START_REF] Lions | Quelques méthodes de résolution des problè mes aux limites nonlinéaires[END_REF]):

v W (T ) = v L ∞ (0,T ;H 1 ) + v t L ∞ (0,T ;L 2 ) + v t L q (QT ) , v ∈ W (T ). (3.2)
For each v ∈ W (T ), we associate with the problem (1.1)-(1.3) the following variational problem. Find u ∈ W (T ) which satisfies the variational problem

< u // (t), w > +a η (u(t), w) - t 0 k(t -s)a η (u(s), w)ds+ < ψ q (u / (t)), w > = g 1 (t)w(1)+ < f (•, t, v(•, t)), w > for all w ∈ H 1 , (3.3)
u(0) = u 0 , u t (0) = u 1 , (3.4)
where

ψ q (z) = |z| q-2 z, g 1 (t) = g(t) - t 0 k(t -s)g(s)ds. (3.5)
Then, we have the following theorem Theorem 3.1. Let (H1)-(H4) hold. Then, for every T > 0 and v ∈ W (T ), problem (3.3)-(3.5) has a unique solution u ∈ W (T ) and such that u // , u xx ∈ L q / (0, T ; (H 1 ) / ), where q / = q/(q -1).

(3.6)

Furthermore, we have

u / (t) 2 + u(t) 2 η + 2 t 0 u / (s) q L q ds ≤ C 1T exp(T C 2T ), ∀t ∈ [0, T ], (3.7)
where

C 1T = C 1T (v, u 0 , u 1 , k, g) = 2 u 1 2 + u 0 2 η + 2|g 1 (0) u 0 (1)| +6 g 1 2 L ∞ (0,T ) + 2 g / 1 2 L 2 (0,T ) + T 0 f (•, s, v(s)) 2 ds , (3.8)
C 2T = C 2T (k) = 2 3 + 2|k(0)| + 6 k 2 L 2 (0,T ) + T k / 2 L 2 (0,T ) , (3.9)
and

g 1 (t) = g(t) - t 0 k(t -s)g(s)ds. (3.10)
Proof of theorem 3.1. The proof consists of steps two steps a. The existence of solution. We approximate u 0 , u 1 , k, g by sequences

{u 0m } ⊂ C ∞ 0 Ω , u 1m ⊂ C ∞ 0 (Ω), k m , g m ⊂ C ∞ 0 ([0, T ]), respectively, such that u 0m → u 0 strongly in H 1 , u 1m → u 1 strongly in L 2 , k m → k strongly in H 1 (0, T ),
g m → g strongly in H 1 (0, T ).

(3.11)

Then we consider the following variational problem: Find u m ∈ W (T ) which satisfies the variational problem

< u // m (t), w > +a η (u m (t), w) - t 0 k m (t -s)a η (u m (s), w)ds + < ψ q (u / m (t)), w >= g 1m (t)w(1)+ < f (•, t, v(•, t)), w >, ∀w ∈ H 1 , (3.12
)

u(0) = u 0m , u / (0) = u 1m , (3.13)
and

u m ∈ L ∞ (0, T ; H 2 ), u / m ∈ L ∞ (0, T ; H 1 ), u // m ∈ L ∞ (0, T ; L 2 ), (3.14)
where

g 1m (t) = g m (t) - t 0 k m (t -s)g m (s)ds. (3.15)
The existence of a sequence of solutions u m satisfying (3.12)-(3.15) is a direct result of the theorem 2.1 in [START_REF] Long | Existence and asymptotic expansion for a viscoelastic problem with a mixed nonhomogeneous condition[END_REF]. We shall prove that u m is a Cauchy sequence in W (T ).

(i) A priori estimates.

We take w = u / m (t) in (3.12), afterwards integrating with respect to the time variable from 0 to t, we get after some rearrangements

σ m (t) =σ m (0) -2g 1m (0)u 0m (1) + 2g 1m (t)u m (1, t) -2 t 0 g / 1m (r)u m (1, r)dr -2k m (0) t 0 u m (r) 2 η dr + 2 t 0 k m (t -s)a η (u m (s), u m (t))ds -2 t 0 dr r 0 k / m (r -s)a η (u m (s), u m (r))ds + 2 t 0 < f (•, s, v(•, s)), u / m (s) > ds, (3.16)
where

σ m (t) = u / m (t) 2 + u m (t) 2 η + 2 t 0 ||u / m (s)|| q L q ds. (3.17)
Proving in the same manner as in [START_REF] Long | Existence and asymptotic expansion for a viscoelastic problem with a mixed nonhomogeneous condition[END_REF], we have the following results:

σ m (t) = C 1T (m) + C 2T (m) t 0 σ m (s)ds, ∀t ∈ [0, T ], (3.18)
where

C 1T (m) =2 u 1m 2 + u 0m 2 η + 2|g 1m (0)u 0m (1)| + 6 g 1m 2 L ∞ (0,T ) +2 g / 1m 2 L 2 (0,T ) + T 0 f (•, s, v(s)) 2 ds , (3.19)
C 2T (m) = 2 3 + 2|k m (0)| + 6 k m 2 L 2 (0,T ) + T k / m 2 L 2 (0,T ) . (3.20)
From the assumptions (H1)-(H4), afterwards using Gronwall's lemma, we deduce from (3.11), that

σ m (t) ≤ C T , for all m and t ∈ [0, T ], (3.21)
where C T is a constant independent of m.

On the other hand, we deduce from (3.12), (3.21), that, for all w ∈ H 1 , we have

| < u // m (t), w > | ≤ u m (t) η w η + t 0 |k m (t -s)| u m (s) η w η ds + ψ q (u / m ) L q / (Ω) w L q (Ω) + |g 1m (t)| w η + f (•, t, v(•, t)) w ≤ C T (3C η ) 1 + ψ q (u / m ) L q / (Ω) w H 1 . (3.22)
This implies that

u // m (t) (H 1 ) / = sup 0 =w∈H 1 < u // m (t), w > w H 1 ≤ C T 3C η 1 + ψ q (u / m ) L q / (Ω) . (3.23) Hence u // m q / L q / (0,T ;(H 1 ) / ) = T 0 u // m (t) q / (H 1 ) / dt ≤ C T 3C η q / 2 q / -1 T 0 1 + u / m (t) q L q (Ω) dt ≤ C T , (3.24)
where C T always indicating a constant depending on T .

(ii) The convergence of sequence {u m } We shall prove that u m is a Cauchy sequence in W (T ). Let u = u m -u µ . Then u satisfies the variational problem

< u // (t), w > +a η ( u(t), w) - t 0 k m (t -s)a η ( u(s), w)ds - t 0 k(t -s)a η (u µ (s), w)ds+ < ψ q (u / m (t)) -ψ q (u / µ (t)), w > = g 1 (t)w(1) for all w ∈ H 1 , (3.25) u(0) = u 0 , u / (0) = u1, (3.26) where u 0 = u 0m -u 0µ , u 1 = u 1m -u 1µ , k = k m -k µ , g = g m -g µ , g 1 = g 1m -g 1µ , g 1 (t) = g(t) - t 0 k m (t -s) g(s)ds - t 0 k(t -s)g µ (s)ds. (3.27)
We take w = u / (t) in (3.25), after integrating with respect to the time variable from 0 to t, we get after some rearrangements

Z(t) =Z(0) -2 g 1 (0) u 0 (1) + 2 g 1 (t) u(1, t) -2 t 0 g1 / (r) u(1, r)dr -2k m (0) t 0 u(r) 2 η dr + 2 t 0 k m (t -s)a η ( u(s), u(t))ds -2 t 0 dr r 0 k / m (r -s)a η ( u(s), u(r))ds -2 k(0) t 0 a η (u µ (s), u(s))ds + 2 t 0 k(t -s)a η (u µ (s), u(t))ds -2 t 0 dr r 0 k / (r -s)a η (u µ (s), u(r))ds, (3.28)
where

Z(t) = u / (t) 2 + u(t) 2 η + 2 t 0 < ψ q (u / m (s)) -ψ q (u / µ (s)), u / m (s) -u / µ (s) > ds.
(3.29)

Using the following inequality

∀q ≥ 2, ∃C q > 0 : (|x| q-2 x -|y| q-2 y)(x -y) ≥ C q |x -y| q , ∀x, y ∈ R, (3.30)
it follows from (3.29) that

Z(t) ≥ u / (t) 2 + u(t) 2 η + 2C q t 0 u / (s) q L q ds. (3.31) Using the inequality 2ab ≤ ǫa 2 + 1 ǫ b 2 , ∀a, b ∈ R, ∀ǫ > 0, (3.32)
and the following inequalities

|a η (u, v)| ≤ u η v η , ∀u, v ∈ H 1 , (3.33) | u(1, t)| ≤ u(t) C 0 (Ω) ≤ √ 2 u(t) 1 ≤ √ 2 u(t) η ≤ 2Z(t), ( 3
.34) we shall estimate respectively the following terms on the right-hand side of (3.28) as follows

Z(0) -2 g 1 (0) u 0 (1) ≤ u 1m -u 1µ 2 + u 0m -u 0µ 2 η + 2|g 1m (0) -g 1µ (0)||u 0m (1) -u 0µ (1)|, (3.35) 2 g 1 (t) u(1, t) ≤ 8 g 1 2 L ∞ (0,T ) + 1 4 Z(t), with ǫ = 1 8 , (3.36) -2 t 0 g / 1 (r) u(1, r)dr ≤ 2 g / 1 2 L 2 (0,T ) + t 0 Z(r)dr, (3.37)
2 t 0 k m (t -s)a η ( u(s), u(t))ds ≤ 1 8 Z(t) + 8 k m 2 L 2 (0,T ) t 0 Z(s)ds, (3.38)
-2k m (0)

t 0 u(r) 2 η dr ≤ 2|k m (0)| t 0 Z(r)dr, (3.39) -2 t 0 dr r 0 k / m (r -s)a η ( u(s), u(r))ds ≤ 1 + T k / m 2 L 2 (0,T ) t 0 Z(s)ds, (3.40) 2 t 0 k(t -s)a η (u µ (s), u(t))ds ≤ 1 8 Z(t) + 8 C T k 2 L 1 (0,T ) , (3.41)
-2 k(0) t 0 a η (u µ (s), u(s))ds ≤ T C T | k(0)| 2 + t 0 Z(s)ds, (3.42) -2 t 0 dr r 0 k / (r -s)a η (u µ (s), u(r))ds ≤ T 2 C T k / 2 L 2 (0,T ) + t 0 Z (
Z(t) ≤ ρ 1T (m, µ) + ρ 2T (m) t 0 Z(s)ds, ∀t ∈ [0, T ], (3.44)
where

ρ 1T (m, µ) = 2 u 1 2 + u 0 2 η + 2| g 1 (0) u 0 (1)| + 8 g 1 2 L ∞ (0,T ) +2 g / 1 2 L 2 (0,T ) + 8 C T k 2 L 1 (0,T ) + T C T | k(0)| 2 + T 2 C T k / 2 L 2 (0,T ) , ρ 2T (m) = 2 4 + 2|k m (0)| + 8 k m 2 L 2 (0,T ) + T k / m 2 L 2 (0,T ) .
(3.45) By Gronwall's lemma, we deduce from (3.31), (3.44), (3.45), that On the other hand, by (3.47) and the continuity of ψ q , we obtain

u / (t) 2 + u(t) 2 η + 2C q t 0 u / (s) q L q ds ≤ Z(t) ≤ ρ 1T (m, µ)exp(T ρ 2T (m)), for all t ∈ [0, T ].
ψ q (u / m ) → ψ q (u / ) a.e. (x, t) ∈ Q T . (3.48)
By means of (3.21), it follows that

ψ q (u / m ) q / L q / (QT ) = u / m q L q (QT ) ≤ 1 2 C T , (3.49)
for all m. By Lions's lemma [5, Lemma 1.3, p. 12], it follows from (3.48) and (3.49) that ψ q (u / m ) → ψ q (u / ) in L q / (Q T ) weakly.

(3.50) Noticing (3.11) 3 and (3.47) we have

T 0 dt t 0 k m (t -s)a η (u m (s), w(t))ds - T 0 dt t 0 k(t -s)a η (u(s), w(t))ds ≤ T 0 dt t 0 k m (t -s)a η (u m (s) -u(s), w(t))ds + T 0 dt t 0 [k m (t -s) -k(t -s)]a η (u(s), w(t))ds ≤ 3C η w L 1 (0,T ;H 1 ) k m L 1 (0,T ) u m -u L ∞ (0,T ;H 1 ) + k m -k L 1 (0,T ) u L ∞ (0,T ;H 1 ) → 0 (3.51)
for all w ∈ L 1 (0, T ; H 1 ).

On the other hand, by (3.11) 3,4 and (3.15), we also obtain

g 1m → g 1 strongly in H 1 (0, T ). ( 3
u // (t), w + a η (u(t), w) - t 0 k(t -s)a η (u(s), w)ds + ψ q (u / (t)), w = g 1 (t)w(1)+ < f (•, t, v(•, t)), w >, ∀w ∈ H 1 , in L q / (0, T ) weak, (3.54)
and

u(0) = u 0 , u / (0) = u 1 . (3.55)
On the other hand, we deduce from (3.54), that

u xx (t) - t 0 k(t -s)u xx (s)ds = φ(t), (3.56)
where

φ(t) = u // (t) + |u / | q-2 u / -f (•, t, v(•, t)) ∈ L q / (0, T ; (H 1 ) / ). (3.57)
Hence, it follows from (3.56) and (3.57), that

u xx (t) q / (H 1 ) / ≤ φ(t) (H 1 ) / + t 0 |k(t -s)| u xx (s) (H 1 ) / ds q / ≤ 2 q / -1   φ(t) q / (H 1 ) / + t 0 |k(t -s)| u xx (s) (H 1 ) / ds q /   ≤ 2 q / -1 φ(t) q / (H 1 ) / + k q / L q (0,T ) t 0 u xx (s) q / (H 1 ) / ds . (3.58)
This implies that r 0 u xx (t) q / (H 1 ) / dt ≤2 q / -1 φ q / L q / (0,T ;(H 1 ) / ) + 2 q / -1 k q / L q (0,T )

r 0 dt t 0 u xx (s) q / (H 1 ) / ds (3.59)
Using Gronwall's lemma, we obtain

r 0 u xx (t) q / (H 1 ) / dt ≤ 2 q / -1 φ q / L q / (0,T ;(H 1 ) / ) exp 2 q / -1 k q / L q (0,T ) r ≤ C T , (3.60
) where C T always indicating a constant depending on T Thus b. Uniqueness of the solution. First, we shall now require the following lemma.

u xx ∈ L q / (0, T ; (H 1 ) / ). ( 3
Lemma 3.2. Let u be the weak solution of the following problem

u // -u xx + t 0 k(t -s)u xx (s)ds = Φ, 0 < x < 1, 0 < t < T, u x (0, t) = u(0, t), u x (1, t) + ηu(1, t) = 0, u(x, 0) = u 0 (x), u / (x, 0) = u 1 (x), u ∈ L ∞ (0, T ; H 1 ), u / ?L ∞ (0, T ; L 2 ), k ∈ H 1 (0, T ), Φ ∈ L 2 (Q T ).
(3.62)

Then we have

1 2 u / (t) 2 + 1 2 u(t) 2 η = 1 2 u 1 2 + 1 2 u 0 2 η -k(0) t 0 u(r) 2 η dr + t 0 k(t -s)a(u(s), u (
u i ∈ W (T ), u // i , u ixx ∈ L q / (0, T ; (H 1 ) / ), i = 1, 2. (3.64)
Then u = u 1 -u 2 is the weak solution of the following problem where

u // -u xx + t 0 k(t -s)u xx (s)ds + ψ q (u / 1 ) -ψ q (u / 2 ) = 0, u x (0, t) -u(0, t) = u x (1, t) + ηu(1, t) = 0, u(0) = u / (0) = 0, u ∈ W (T ), u // , u xx ∈ L q / (0, T ; (H 1 ) / ).
σ(t) = u / (t) 2 + u(t) 2 η + 2 t 0 ψ q (u / 1 (s)) -ψ q (u / 2 (s)), u / (s) ds. (3.67)
By using the same computations as in the above part we obtain from (3.66) that Proof. For each T 1 > 0, we put

σ(t) = 2 1 + 2 k 2 L 2 (0,T ) + 2|k(0)| + k / 2 L 1 (0,T ) t 0 σ(r)dr. ( 3
W 1 (T 1 ) = v ∈ L ∞ (0, T 1 ; H 1 ) : v t ∈ L ∞ (0, T 1 ; L 2 ) . (3.70)
Then W 1 (T 1 ) is a Banach space with respect to the norm (see [START_REF] Lions | Quelques méthodes de résolution des problè mes aux limites nonlinéaires[END_REF]):

v W1(T1) = v L ∞ (0,T1;H 1 ) + v t L ∞ (0,T1;L 2 ) , v ∈ W 1 (T 1 ). (3.71)
For M > 0 and T 1 > 0, we put

B(M, T 1 ) = v ∈ W 1 (T 1 ) : v W1(T1) ≤ M . (3.72)
We also define the operator ̥ from B(M, T 1 ) into W (T 1 ) by u = ̥(v), where u is the unique solution of problem (3.3)-(3.5). We would like to show that ̥ is a contraction operator from B(M, T 1 ) into itself. Applying the contraction mapping theorem, the operator ̥ has a fixed point in B(M, T 1 ) that is also a weak solution of the problem (1.1)-(1.3). First, by Theorem 3.1, we note that the unique solution of problem (3.3)-(3.5) satisfies (3.7), (3.8), (3.9). On the other hand, it follows from (H3), that

t 0 f (•, s, v(s)) 2 ds ≤ 2 t 0 f (•, s, v(s)) -f (•, s, 0) 2 ds + 2 t 0 f (•, s, 0) 2 ds ≤ 2T 1 K 2 1 M 2 + 2 T 0 f (•, s, 0) 2 ds, (3.73)
where

K 1 = K 1 (M, T, f ) = sup |D 3 f (x, t, u)| : 0 ≤ x ≤ 1, 0 ≤ t ≤ T, |u| ≤ √ 2M . ( 3
u / (t) 2 + u(t) 2 η + 2 t 0 u / (s) q L q ds ≤ C 1T + 2T 1 K 2 1 M 2 exp(T 1 C 2T ), ∀t ∈ [0, T 1 ], (3.75)
where

C 1T = C 1T ( u 0 , u 1 , k, g) = 2 u 1 2 + u 0 2 η + 2 |g 1 (0) u 0 (1)| +6 g 1 2 L ∞ (0,T ) + 2 g / 1 2 L 2 (0,T ) + 2 T 0 f (•, s, 0) 2 ds , C 2T = C 2T (k) = 2 3 + 2|k(0)| + 6 k 2 L 2 (0,T ) + T k / 2 L 2 (0,T ) .
(3.76)

By choosing M > 0 large enough so that C 1T = 1 4 M 2 , then T 1 sufficiently small so that 1 4

M 2 + 2T 1 K 2 1 M 2 exp(T 1 C 2T ) ≤ 1 2 M 2 , (3.77) and 2 2T 1 K 1 exp T 1 2 + 2|k(0)| + 2 k 2 L 2 (0,T ) + k / 2 L 1 (0,T ) < 1. (3.78)
From (3.75), (3.77) we have u W1(T1) ≤ M , hence u ∈ B(M, T 1 ). This shows that ̥ maps B(M, T 1 ) into itself. Next, we verify that ̥ is a contraction. Let

u 1 = ̥(v 1 ), u 2 = ̥(v 2 ), where v 1 , v 2 ∈ B(M, T 1 ). Put U = u 1 -u 2 and V = v 1 -v 2 .
Then U is the weak solution of the following problem

U // -U xx + t 0 k(t -s)U xx (s)ds + ψ q (u / 1 ) -ψ q (u / 2 ) = f (x, t, v 1 (t)) -f (x, t, v 2 (t)), 0 < x < 1, 0 < t < T 1 , U x (0, t) -U (0, t) = U x (1, t) + ηU (1, t) = 0, U (0) = U / (0) = 0, U ∈ W (T 1 ); U // , U xx ∈ L q / (0, T 1 ; (H 1 ) / ). (3.79) By using Lemma 3.2 with u 0 = u 1 = 0, Φ = -ψ q (u / 1 ) + ψ q (u / 2 ) + f (x, t, v 1 (t)) - f (x, t, v 2 (t)), we have δ(t) = -2k(0) t 0 U (r) 2 η dr + 2 t 0 k(t -s)a(U (s), U (t))ds -2 t 0 dr r 0 k / (r -s)a(U (s), U (r))ds + 2 t 0 < f (•, s, v 1 (s)) -f (•, s, v 2 (s)), U / (s) > ds, a.e. t ∈ [0, T 1], (3.80)
where

δ(t) = U / (t) 2 + U (t) 2 η + 2 t 0 ψ q (u / 1 ) -ψ q (u / 2 ), U / (s) ds ≥ U / (t) 2 + U (t) 2 η + 2C q t 0 U / (s) q L q ds.
(3.81)

By the assumption (H4), we have

2 t 0 f (•, s, v 1 (s)) -f (•, s, v 2 (s)), U / (s) ds ≤ t 0 U / (s) 2 ds + t 0 f (•, s, v 1 (s)) -f (•, s, v 2 (s)) 2 ds ≤ t 0 U / (s) 2 ds + 2T 1 K 2 1 V 2 W1(T1) , (3.82)
Therefore, we can prove in a similar manner as above that

δ(t) ≤2T 1 K 2 1 V 2 W1(T1) + 2 2 + 2|k(0)| + 2 k 2 L 2 (0,T ) + k / 2 L 1 (0,T ) t 0 δ(s)ds. (3.83)
By Gronwall's lemma, we obtain from (3.83) that

δ(t) = 2 ρ 1 (k, K 1 , T, T 1 ) V W1(T1) 2 , (3.84) where ρ 1 (k, K 1 , T, T 1 ) = 2T 1 K 1 exp T 1 2 + 2|k(0)| + 2 k 2 L 2 (0,T ) + k / 2 L 1 (0,T ) . ( 3
U W1(T1) ≤ 2ρ 1 (k, K 1 , T, T 1 ) V W1(T1) , (3.86)
Proof. By multiplying the equation (4.1) 1 by u t and integrate over (0, 1) × (0, t) we obtain

E(t) + 2 t 0 u / (s) q L q ds + t 0 k(s) u(s) 2 η ds - t 0 dr r 0 k / (r -s) u(s) -u(r) 2 η ds = E(0) + 2 t 0 < F (s), u / (s) > ds, (4.4)
where

E(t) = u / (t) 2 + 1 - t 0 k(s)ds u(t) 2 η + 2 p u(t) p L p + t 0 k(t -s) u(s) -u(t) 2 η ds. (4.5)
On the other hand, by ( H4) and the Cauchy's inequality, we obtain

2 t 0 F (s), u / (s) ds ≤ t 0 F (s) ds + t 0 F (s) u / (s) 2 ds ≤ +∞ 0 F (s) ds + t 0 F (s) E(s)ds. (4.6)
By Gronwall's lemma, we obtain from (4.4) and (4.6) that

E(t) ≤ E(0) + +∞ 0 F (s) ds exp t 0 F (s) ds ≤ E(0) + +∞ 0 F (s) ds exp +∞ 0 F (s) ds = C, ∀t ≥ 0. (4.7)
By ( H3, i), we have

E(t) ≥ u / (t) 2 + 1 - t 0 k(s)ds u(t) 2 η ≥ u / (t) 2 + k ∞ u(t) 2 η . (4.8)
Then we obtain (4.3) from (4.7) and (4.8). This completes the proof of Lemma 4.1.

In this section we state and prove decay result.

Theorem 4.2. Suppose that ( H1) -( H5) hold. Then the solution u(t) of problem (4.1) decays exponentially to zero as t → +∞ in the following sense: there exist the positive constants N and γ such that

u / (t) + u(t) η ≤ N e -γt for all t ≥ 0. (4.9)
Proof. We use the following functional

Γ(t) = Γ(ε 1 , ε 2 , t) = E(t) + ε 1 E 1 (t) + ε 2 E 2 (t), (4.10)
where

E 1 (t) =< u(t), u / (t) >, (4.11)
E 2 (t) = - t 0 k(t -s) u / (t), u(t) -u(s) ds. (4.12)
Estimating Γ(t). By (2.3), (2.4), we obtain from ( H2, i) that

|E 1 (t)| = < u(t), u / (t) > ≤ 1 2 u / (t) 2 + u(t) 2 η , (4.13)
|E 2 (t)| = t 0 k(t -s) u / (t), u(t) -u(s) ds ≤ 1 2 u / (t) 2 + 1 2 t 0 k(t -s) u(t) -u(s) ds 2 ≤ 1 2 u / (t) 2 + (1 -k ∞ ) t 0 k(t -s) u(t) -u(s) 2 η ds. (4.14)
Hence, it follows from (4.10)-(4.14) that for ε 1 , ε 2 small enough, there exist two positive constants α 1 , α 2 , such that

α 1 E(t) ≤ Γ(t) ≤ α 2 E(t). (4.15)
Estimating Γ / (t). Now differentiating (4.4) with respect to t, we have

E / (t) = -2 u / (t) q L q + t 0 k / (t -s) u(s) -u(t) 2 η ds -k(t) u(t) 2 η + 2 F (t), u / (t) ≤ -2 u / (t) q L q + t 0 k / (t -s) u(s) -u(t) 2 η ds + 2 F (t), u / (t) , (4.16)
since k(t) ≥ 0. By multiplying the equation (4.1) 1 by u and integrate over (0, 1) we obtain

E / 1 (t) = u / (t) 2 -u(t) 2 η -u(t) p L p + F (t), u(t) + t 0 k(t -s)a(u(s), u(t))ds -|u / (t)| q-2 u / (t), u(t) = u / (t) 2 -u(t) 2 η -u(t) p L p + F (t), u(t) + I 1 (t) + I 2 (t).
(4.17)

We now estimate the last two terms in the right side of (4.17) as follows

Estimating I 1 (t).

Using the inequality

ab ≤ δ r a r + r -1 r δ -r r-1 b r r-1 , ∀a, b ≥ 0, ∀r > 1, ∀δ > 0, (4.18)
we have

I 1 (t) = t 0 k(t -s)a(u(s), u(t))ds = t 0 k(t -s)a (u(s) -u(t), u(t)) ds + t 0 k(t -s) u(t) 2 η ds ≤ δ 1 u(t) 2 η + 1 4δ 1 t 0 k(s)ds t 0 k(t -s) u(s) -u(t) 2 η ds + t 0 k(s)ds u(t) 2 η ≤ δ 1 u(t) 2 η + 1 -k ∞ 4δ 1 t 0 k(t -s) u(s) -u(t) 2 η ds + (1 -k ∞ ) u(t) 2 η ≤ (δ 1 + 1 -k ∞ ) u(t) 2 η + 1 -k ∞ 4δ 1 t 0 k(t -s) u(s) -u(t) 2 η ds, (4.19)
for all δ 1 > 0.

Estimating I 2 (t). We again use inequality (4.18) we obtain from (4.3) that

I 2 (t) = -|u / (t)| q-2 u / (t), u(t) ≤ u / (t) q-1 L q u(t) L q ≤ δ q 1 q u(t) q L q + q -1 q δ -q q-1 1 u / (t) q L q ≤ 2 δ q 1 q √ 2C q-2 u(t) 2 η + q -1 q δ -q q-1 1 u / (t) q L q , (4.20)
for all δ 1 > 0. By combining (4.17), (4.19) and (4.20), we obtain

E / 1 (t) ≤ -u(t) p L p + u / (t) 2 -k ∞ -δ 1 -2 δ q 1 q √ 2C q-2 u(t) 2 η + q -1 q δ -q q-1 1 u / (t) q L q + 1 -k ∞ 4δ 1 t 0 k(t -s) u(s) -u(t) 2 η ds + F (t), u(t) . (4.21)
Then, we can always choose the constant δ 1 > 0 such that

γ 1 = k ∞ -δ 1 -2 δ q 1 q √ 2C q-2 > 0. (4.22)
This implies that

E / 1 (t) ≤ -u(t) p L p + u / (t) 2 -γ 1 u(t) 2 η + γ 2 u / (t) q L q + γ 3 t 0 k(t -s) u(s) -u(t) 2 η ds + F (t), u(t) , (4.23)
where

γ 2 = q -1 q δ -q q-1 1 , γ 3 = 1 -k ∞ 4δ 1 . ( 4
+ t 0 k(t -s) |u / (t)| q-2 u / (t), u(t) -u(s) ds - t 0 k(t -s) F (t), u(t) -u(s) ds = 7 i=1 J i (t).
(4.25)

Similarly to (4.17), we estimate respectively the following terms on the right-hand side of (4.25) as follows.

Estimating J 1 (t). Since k is continuous and k(0) > 0 then there exists t 0 > 0, such that Estimating J 2 (t).

t 0 k(s)ds ≥ t0 0 k(s)ds = k 0 > 0 for all t ≥ t 0 . ( 4
J 2 (t) = - t 0 k / (t -s) u / (t), u(t) -u(s) ds ≤ δ 2 u / (t) 2 + 1 4δ 2 t 0 |k / (t -s)|ds t 0 |k / (t -s)| u(s) -u(t) 2 ds ≤ δ 2 u / (t) 2 + 1 2δ 2 t 0 |k / (t -s)|ds t 0 |k / (t -s)| u(s) -u(t) 2 η ds ≤ δ 2 u / (t) 2 - k(0) 2δ 2 t 0 k / (t -s) u(s) -u(t) 2 η ds.
(4.28)

Estimating J 3 (t).

J 3 (t) = t 0 k(t -s)a (u(t), u(t) -u(s)) ds ≤ δ 2 u(t) 2 η + 1 4δ 2 t 0 k(s)ds t 0 k(t -s) u(s) -u(t) 2 η ds ≤ δ 2 u(t) 2 η + 1 -k ∞ 4δ 2 t 0 k(t -s) u(s) -u(t) 2 η ds (4.29)
Estimating J 4 (t).

J 4 (t) = - t 0 k(t -s)a t 0 k(t -τ )u(τ )dτ, u(t) -u(s) ds ≤ t 0 k(t -τ ) u(τ ) η dτ t 0 k(t -s) u(s) -u(t) η ds ≤ δ 2 t 0 k(t -τ ) u(τ ) η dτ 2 + 1 4δ 2 t 0 k(t -s) u(s) -u(t) η ds 2 ≤ 2δ 2 t 0 k(t -τ ) u(τ ) η dτ 2 + 2δ 2 + 1 4δ 2 t 0 k(t -s) u(s) -u(t) η ds 2 ≤ 2δ 2 (1 -k ∞ ) 2 u(t) 2 η + 2δ 2 + 1 4δ 2 (1 -k ∞ ) t 0 k(t -s) u(s) -u(t) 2 η ds.
(4.30)

Estimating J 5 (t).

J 5 (t) = t 0 k(t -s) |u(t)| p-2 u(t), u(t) -u(s) ds ≤ 2 √ 2C p-2 t 0 k(t -s) u(t) η u(t) -u(s) η ds ≤ 2 √ 2C p-2 δ 2 u(t) 2 η + 1 4δ 2 t 0 k(t -s) u(t) -u(s) η ds 2 ≤ 2 √ 2C p-2 δ 2 u(t) 2 η + 1 4δ 2 (1 -k ∞ ) t 0 k(t -s) u(t) -u(s) 2 η ds .
(4.31)

Estimating J 6 (t). We again use inequality (4.18) with r = q, δ = δ 2 , we obtain from (4.3) that

|u / (t)| q-2 u / (t), u(t) -u(s) ≤ u / (t) q-1 L q u(t) -u(s) L q ≤ δ q 2 q u(t) -u(s) q L q + q -1 q δ -q q-1 2 u / (t) q L q ≤ 2 δ q 2 q 2 √ 2C q-2 u(t) -u(s) 2 η + q -1 q δ -q q-1 2 u / (t) q L q .
(4.32)

It follows from (4.32) that

J 6 (t) = t 0 k(t -s) |u / (t)| q-2 u / (t), u(t) -u(s) ds ≤ 2 δ q 2 q 2 √ 2C q-2 t 0 k(t -s) u(t) -u(s) 2 η ds + q -1 q δ -q q-1 2 u / (t) q L q t 0 k(t -s)ds ≤ 2 δ q 2 q 2 √ 2C q-2 t 0 k(t -s) u(t) -u(s) 2 η ds + q -1 q δ -q q-1 2 (1 -k ∞ ) u / (t) q L q (4.33) Estimating J 7 (t) J 7 (t) = - t 0 k(t -s) F (t), u(t) -u(s) ds ≤ t 0 k(t -s) F (t) u(t) -u(s) ds ≤ 1 4δ 2 F (t) 2 + δ 2 t 0 k(t -s)ds t 0 k(t -s) u(t) -u(s) 2 ds ≤ 1 4δ 2 F (t) 2 + 2δ 2 (1 -k ∞ ) t 0 k(t -s) u(t) -u(s)
E / 2 (t) = -(k 0 -δ 2 ) u / (t) 2 + δ 2 γ 1 u(t) 2 η + γ 2 u / (t) q L q + γ 3 t 0 k(t -s) u(t) -u(s) 2 η ds -γ 4 t 0 k / (t -s) u(s) -u(t) 2 η ds + 1 4δ 2 F (t) 2 , (4.35)
where

γ 1 = 1 + 2(1 -k ∞ ) 2 + 2 √ 2C p-2 , γ 2 = q q -1 δ -q q-1 2 (1 -k ∞ ), γ 3 = 2 δ q 2 q 2 √ 2C q-2 + (1 -k ∞ ) 1 2δ 2 √ 2C p-2 + 4δ 2 + 1 2δ 2 , γ 4 = k(0) 2δ 2 .
(4.36)

Combining of (4.10), (4.16), (4.23) and (4.35), we obtain

Γ / (t) + ε 1 u(t) p L p + ((k 0 -δ 2 )ε 2 -ε 1 ) u / (t) 2 + (ε 1 γ 1 -ε 2 δ 2 γ 1 ) u(t) 2 η + (2 -ε 1 γ 2 -ε 2 γ 2 ) u / (t) q L q -(ε 1 γ 3 + ε 2 γ 3 ) t 0 k(t -s) u(t) -u(s) 2 η ds -(1 -ε 2 γ 4 ) t 0 k / (t -s) u(s) -u(t) 2 η ds ≤ F (t), 2u / (t) + ε 1 u(t) + ε 2 4δ 2 F (t) 2 .
(4.37)

Whence δ 1 is fixed, choosing

δ 2 = 1 2 k(0)γ 1 γ 1 + γ 1 , ε 2 = 2 k 0 ε 1 , where ε 1 > 0 is arbitrary, (4.38)
we deduce from (4.37) and (4.38) that

Γ / (t) + ε 1 u(t) p L p + ε 1 γ 1 γ 1 + γ 1 u / (t) 2 + ε 1 γ 2 1 γ 1 + γ 1 u(t) 2 η + 2 -ε 1 1 + 2 k(0) γ 2 u / (t) q L q -ε 1 γ 3 + 2 k(0) γ 3 t 0 k(t -s) u(t) -u(s) 2 η ds -1 - 2 k(0) ε 1 γ 4 t 0 k / (t -s) u(s) -u(t) 2 η ds ≤ F (t), 2u / (t) + ε 1 u(t) + ε 1 k 2 0 1 + γ 1 γ 1 F (t) 2 . (4.39)
Next, we choose ε 1 > 0, with

ε 1 < min ζ γ 3 + 2 k0 γ 3 + 2 k0 γ 4 ζ , 2 1 + 2 k0 γ 2
and (4.15) is satisfied, then by the assumption ( H2, ii), we deduce that

Γ / (t) + ε 1 u(t) p L p + ε 1 γ 1 γ 1 + γ 1 u / (t) 2 + ε 1 γ 2 1 γ 1 + γ 1 u(t) 2 η + k 1 u / (t) q L q + k 2 t 0 k(t -s) u(s) -u(t) 2 η ds ≤ F (t), 2u / (t) + ε 1 u(t) + k 3 F (t) 2 , (4.40)
where This completes the proof of Theorem 4.2.

k 1 = 2 -ε 1 1 + 2 k 0 γ 2 > 0, k 2 = ζ 1 - 2 k 0 ε 1 γ 4 -ε 1 γ 3 + 2 k 0 γ 3 > 0, k 3 = ε 1 k 2 0 1 + γ 1 γ 1 . ( 4
Remark 4.3. The estimate (4.9) holds for any regular solution corresponding to ( u 0 , u 1 ) ∈ H 2 × H 1 . This remains holds for solutions corresponding to ( u 0 , u 1 ) ∈ H 1 × L 2 by simple density argument.

Numerical results

Consider the following problem:

u tt -u xx + t 0
k(t -s)u xx (s)ds + u 3 t = u 2 + F (x, t), 0 < x < 1, 0 < t < T, (5.1) with boundary conditions u x (0, t) = u(0, t), u x (1, t) + u(1, t) = 0, (

and initial conditions u(x, 0) = u 0 (x), u t (x, 0) = u 1 (x), (5.3) where u 0 (x) = -x 2 + x + 1, u 1 (x) = -u 0 (x), k(t) = 1 2 e -t , (5.4) F (x, t) = (2 -t)e -t + U ex (1 -U ex -U 2 ex ), (5.5) where U ex (x, t) = (-x 2 + x + 1)e -t .

(5.6) The exact solution of the problem (5.1)-(5.3) with u 0 (x), u 1 (x), k(t) and F (x, t) defined in (5.4) and (5.5) respectively, is the function U ex given in (5.6). To solve problem (5.1)-(5.3) numerically, we consider the differential system for the unknowns u j (t) = u(x j , t), v j (t) = duj dt (t), with x j = jh, h = 1 N , j = 0, 1, ..., N : u j (0) = u 0 (x j ), v j (0) = u 1 (x j ), j = 0, 1, ..., N.

du j dt (t) = v j (
(5.7)

In fig. 1 we have drawn the approximated solution of the problem (5.1)-(5.5) while fig. 2 represents his corresponding exact solution (5.6). The fig. 3 corresponds to the surface (x, t) → u(x, t) approximated solution in the case where F (x, t) = 0. So in both cases we notice the very good decay of these surfaces from T = 0 to T = 2.

  s)ds. (3.43) Combining (3.28), (3.29), (3.31) and (3.35)-(3.43), we obtain

( 3 .

 3 46)By(3.11),(3.27) and (3.45), we obtain ρ 1T (m, µ)exp(T ρ 2T (m)) → 0 as m, µ → +∞. Hence, it follows from (3.46) that {u m } is a Cauchy sequence in W (T ). Therefore there exists u ∈ W (T ) such that u m → u strongly in W (T ).(3.47)

  .52) From (3.24), we deduce the existence of a subsequence of {u m }, still denoted by {u m }, such that u // m → u // in L q / (0, T ; (H 1 ) / ) weak.(3.53)Passing to the limit in (3.12), (3.13) by (3.47) and (3.50)-(3.53) we have u satisfying the equation

  .61) On ther other hand, the estimate (3.7) hold by means of (3.11), (3.18), (3.19), (3.20), (3.47). The existence of the theorem is proved completely.

k 0 <

 0 / (r -s)a(u(s), u(r))ds+ t Φ(s), u / (s) > ds, a.e. t ∈ [0, T ]. (3.63) Furthermore, if u 0 = u 1 = 0 there is equality in (3.63).The idea of the proof is the same as in [4, Lemma 2.1, p. 79].We now return to the proof of the uniqueness of a solution of the problem (3.3)-(3.5). Let u 1 , u 2 be two weak solutions of problem (3.3)-(3.5), such that

(3. 65 )/ 1 ) + ψ q (u / 2 )k

 6512 By using Lemma 3.2 with u 0 = u 1 = 0, Φ = -ψ q (u -s)a(u(s), u(t))ds -2k(0) / (r -s)a(u(s), u(r))ds,(3.66)

Theorem 3 . 3 .

 33 .68) By Gronwall's lemma, we deduce that σ(t) = 0 and Theorem 3.1 is completely proved. Let T > 0 and (H1) -(H4) hold. Then there exists T 1 ∈ (0, T ) such that problem (1.1)-(1.3) has a unique weak solution u ∈ W (T 1 ) and such that u // , u xx ∈ L q / (0, T 1 ; (H 1 ) / ).(3.69)

  .74) It follows from (3.7)-(3.10) and (3.73) that

  .85) It follows from (3.81), (3.84) and (3.85) that

  Figure 2

Figure 3

 3 Figure 3

  F (t), 2u / (t) + ε 1 u(t) + k 3 F (t) 2 ≤ N F (t) 2 + γΓ(t),(4.43)for some constant N > 0. ThereforeΓ / (t) + γΓ(t) ≤ N F (t) 2 for all t ≥ t 0 . (4.44)Putting γ = 1 2 min{σ, γ}. A simple integration of (4.44) over (t 0 , t) givesΓ(t) ≤ Γ(t 0 )e σt0 + N +∞ t0 e σs F (s) 2 ds e -2γt = N 1 e -2γt ,(4.45)for all t ≥ t 0 . By the boundedness of Γ(t) on [0, t 0 ], we deduce from (4.45) that Γ(t) = Γ L ∞ (0,t0) e -2γ(t-t0) + N 1 e -2γt = N 2 e -2γt ,

	By combining (4.5), (4.15) and (4.40), we can always choose the constant γ > 0 is
	independent of t such that
	Γ / (t) + 2 γΓ(t) ≤ F (t), 2u / (t) + ε 1 u(t) + k 3 F (t) 2 ,	(4.42)
	for all t ≥ t 0 . On ther other hand,
						(4.46)
	for all t ≥ 0. By (4.15), it follows from (4.46) that
	E(t) ≤	1 α 1	Γ(t) ≤	1 α 1	N
						.41)

2 e -2γt , for all t ≥ 0. (4.47)

  t), j = 0, 1, ..., N,

						dv 0 dt	(t) =	1 h 2 [-(1 + h)u 0 (t) + u 1 (t)]
		-	1 h 2		0	t	k(t -s) [-(1 + h)u 0 (s) + u 1 (s)] ds -v 3 0 (t) + u 2 0 (t) + F (x 0 , t),
						dv j dt	(t) =	1 h 2 [u j-1 (t) -2u j (t) + u j+1 (t)]
						-	1 h 2	0	t	k(t -s) [u j-1 (s) -2u j (s) + u j+1 (s)] ds
						-v 3 j (t) + u 2 j (t) + F (x j , t), j = 1, 2, ..., N -1, dv N dt (t) = 1 h 2 [u N -1 (t) -(1 + h)u N (t)]
	-	1 h 2	0	t

k(t -s) [u N -1 (s) -(1 + h)u N (s)] ds -v 3 N (t) + u 2 N (t) + F (x N , t),

where 2ρ 1 (k, K 1 , T, T 1 ) < 1, (3.87) since (3.78) and (3.85). Hence, (3.86) shows that ̥ : B(M, T 1 ) → B(M, T 1 ) is a contraction. Applying the contraction mapping theorem, the operator ̥ has a fixed point in B(M, T 1 ) that is also a weak solution of the problem (1.1)- (1.3).

The solution of the problem (1.1)-(1.3) is unique, that can be showed using the same arguments as in the proof of Theorem 3.1. The proof of Theorem 3.3 is completed.

and the boundary condition u(0, t) = u(1, t) = 0 standing for (1.2), S. Berrimia, S. A. Messaoudi [START_REF] Berrimia | Existence and decay of solutions of a viscoelastic equation with a nonlinear source[END_REF] has obtained a global existence and uniqueness theorem.

Decay of solution

In this part, we will consider the problem of global existence and asymptotic behavior for t → +∞. We assume that g(t) = 0, f (x, t, u) = F (x, t) -|u| p-2 u, p ≥ 2 and consider the following problem

We make the following assumptions:

) There exists a constant σ > 0 such that ∞ 0 e st F (t) 2 dt < +∞. Under assumptions ( H1)-( H4) and let T > 0, by theorem 2.3, the problem (4.1) has a unique weak solution u(t) such that

Then, we have the following Lemma 4.1. Suppose that ( H1) -( H4) hold. Then there is a unique solution u(t) of problem (4.1) defined on R + . Moreover

where C is a positive constant depending only on u 0 , u 1 , F , k ∞ and p.

To solve the nonlinear differential system (5.7), we use the following linear recursive scheme generated by the nonlinear term

+ F (x j , t), j = 1, 2, ..., N -

and where u (n) j (i∆t), i = 1, ..., N 1 -1, j = 0, 1, ..., N , of the system (5.8) being calculated at the time t = N 1 ∆t.

The latter system is solved by a spectral method and since the matrix of this system is very ill-conditioned so we have to regularize it by adding to the diagonal terms a small parameter in order to have a good accuracy of the convergence.


Georgi Vodev
email: georgi.vodev@univ-nantes.fr

HIGH-FREQUENCY APPROXIMATION OF THE INTERIOR DIRICHLET-TO-NEUMANN MAP AND APPLICATIONS TO THE TRANSMISSION EIGENVALUES

We study the high-frequency behavior of the Dirichlet-to-Neumann map for an arbitrary compact Riemannian manifold with a non-empty smooth boundary. We show that far from the real axis it can be approximated by a simpler operator. We use this fact to get new results concerning the location of the transmission eigenvalues on the complex plane. In some cases we obtain optimal transmission eigenvalue-free regions.

Introduction and statement of results

Let (X, G) be a compact Riemannian manifold of dimension d = dim X ≥ 2 with a nonempty smooth boundary ∂X and let ∆ X denote the negative Laplace-Beltrami operator on (X, G). Denote also by ∆ ∂X the negative Laplace-Beltrami operator on (∂X, G 0 ), which is a Riemannian manifold without boundary of dimension d -1, where G 0 is the Riemannian metric on ∂X induced by the metric G. Given a function f ∈ H m+1 (∂X), let u solve the equation is defined by N (λ; n)f := ∂ ν u| ∂X where ν is the unit inner normal to ∂X. One of our goals in the present paper is to approximate the operator N (λ; n) when n(x) ≡ 1 in X by a simpler one of the form p(-∆ ∂X ) with a suitable complex-valued function p(σ), σ ≥ 0. More precisely, the function p is defined as follows p(σ) = σλ 2 , Re p < 0.

Our first result is the following Theorem 1.1. Let 0 < ǫ < 1 be arbitrary. Then, for every 0 < δ ≪ 1 there are constants C δ , C ǫ,δ > 1 such that we have

(1.2) N (λ; 1) -p(-∆ ∂X ) L 2 (∂X)→L 2 (∂X) ≤ δ|λ| for C δ ≤ |Im λ| ≤ (Re λ) 1-ǫ , Re λ ≥ C ǫ,δ .
Note that this result has been previously proved in [START_REF] Petkov | Localization of the interior transmission eigenvalues for a ball[END_REF] in the case when X is a ball in R d and the metric being the Euclidean one. In fact, in this case we have a better approximation of the operator N (λ; 1). In the general case when the function n is arbitrary the DN map can be approximated by h -ΨDOs, where 0 < h ≪ 1 is a semi-classical parameter such that Re (hλ) 2 = 1. To describe this more precisely let us introduce the class of symbols S k δ (∂X), 0 ≤ δ < 1/2, as being the set of all functions a(x ′ , ξ ′ ) ∈ C ∞ (T * ∂X) satisfying the bounds

∂ α x ′ ∂ β ξ ′ a(x ′ , ξ ′ ) ≤ C α,β h -δ(|α|+|β|) ξ ′ k-|β|
for all multi-indices α and β with constants C α,β independent of h. We let OPS k δ (∂X) denote the set of all h -ΨDOs, Op h (a), with symbol a ∈ S k δ (∂X), defined as follows (Op h (a)f ) (x ′ ) = (2πh) -d+1

T * ∂X e -i h x ′ -y ′ ,ξ ′ a(x ′ , ξ ′ )f (y ′ )dy ′ dξ ′ .
It is well-known that for this class of symbols we have a very nice pseudo-differential calculus (e.g. see [START_REF] Dimassi | Spectral asymptotics in semi-classical limit[END_REF]). It was proved in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF] that for |Im λ| ≥ |λ| 1/2+ǫ , 0 < ǫ ≪ 1, the operator hN (λ; n) is an h -ΨDO of class OPS 1 1/2-ǫ (∂X) with a principal symbol ρ(x ′ , ξ ′ ) = r 0 (x ′ , ξ ′ ) -(hλ) 2 n 0 (x ′ ), Re ρ < 0, n 0 := n| ∂X , r 0 ≥ 0 being the principal symbol of -∆ ∂X . Note that it is still possible to construct a semiclassical parametrix for the operator hN (λ; n) when |Im λ| ≥ |λ| ǫ , 0 < ǫ ≪ 1, if one supposes that the boundary ∂X is strictly concave (see [START_REF] Vodev | Transmission eigenvalues for strictly concave domains[END_REF]). This construction, however, is much more complex and one has to work with symbols belonging to much worse classes near the glancing region Σ = {(x ′ , ξ ′ ) ∈ T * ∂X : r ♯ (x ′ , ξ ′ ) = 1}, where r ♯ = n -1 0 r 0 . On the other hand, it seems that no parametrix construction near Σ is possible in the important region 1 ≪ Const ≤ |Im λ| ≤ |λ| ǫ . Therefore, in the present paper we follow a different approach which consists of showing that, for arbitrary manifold X, the norm of the operator hN (λ; n)Op h (χ 0 δ ) is O(δ) for every 0 < δ ≪ 1 independent of λ, provided |Im λ| and Re λ are taken big enough (see Proposition 3.3

below).

Here the function χ Section 3 for the precise definition of χ 0 δ ). Theorem 1.1 is an easy consequence of the following semi-classical version.

0 δ ∈ C ∞ 0 (T * ∂X) is supported in {(x ′ , ξ ′ ) ∈ T * ∂X : |r ♯ (x ′ , ξ ′ ) -1| ≤ 2δ 2 } and χ 0 δ = 1 in {(x ′ , ξ ′ ) ∈ T * ∂X : |r ♯ (x ′ , ξ ′ ) -1| ≤ δ 2 } (see
Theorem 1.2. Let 0 < ǫ < 1 be arbitrary. Then, for every 0 < δ ≪ 1 there are constants C δ , C ǫ,δ > 1 such that we have (1.3) hN (λ; n) -Op h (ρ(1χ 0 δ ) + hb) L 2 (∂X)→H 1 h (∂X) ≤ Cδ for C δ ≤ |Im λ| ≤ (Re λ) 1-ǫ , Re λ ≥ C ǫ,δ , where C > 0 is a constant independent of λ and δ, and b ∈ S 0 0 (∂X) is independent of λ and the function n. Here H 1 h (∂X) denotes the Sobolev space equipped with the semi-classical norm (see Section 3 for the precise definition). Thus, to prove (1.3) (resp. (1.2)) it suffices to construct semiclassical parametrix outside a δ 2 -neighbourhood of Σ, which turns out to be much easier and can be done for an arbitrary X. In the elliptic region {(x ′ , ξ ′ ) ∈ T * ∂X : r ♯ (x ′ , ξ ′ ) ≥ 1 + δ 2 } we use the same parametrix construction as in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF] with slight modifications. In the hyperbolic region {(x ′ , ξ ′ ) ∈ T * ∂X : r ♯ (x ′ , ξ ′ ) ≤ 1δ 2 }, however, we need to improve the parametrix construction of [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF]. We do this in Section 4 for 1 ≪ Const ≤ |Im λ| ≤ |λ| 1-ǫ . Then we show that the difference between the operator hN (λ; n) microlocalized in the hyperbolic region and its parametrix is O e -β|Im λ| + O ǫ,M |λ| -M , where β > 0 is some constant and M ≥ 1 is arbitrary. So, we can do it small by taking |Im λ| and |λ| big enough.

This kind of approximations of the DN map are important for the study of the location of the complex eigenvalues associated to boundary-value problems with dissipative boundary conditions (e.g. see [START_REF] Petkov | Location of eigenvalues for the wave equation with dissipative boundary conditions[END_REF]). In particular, Theorem 1.2 leads to significant improvements of the eigenvalue-free regions in [START_REF] Petkov | Location of eigenvalues for the wave equation with dissipative boundary conditions[END_REF]. In the present paper we use Theorem 1.2 to study the location of the interior transmission eigenvalues (see the next section). We improve most of the results in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF] as well as those in [START_REF] Petkov | Localization of the interior transmission eigenvalues for a ball[END_REF], [START_REF] Vodev | Transmission eigenvalues for strictly concave domains[END_REF], and provide a simpler proof. In some cases we get optimal transmission eigenvalue-free regions (see Theorem 2.1). Note that for the applications in the anisotropic case it suffices to have an weaker analogue of the estimate (1.3) with the space H 1 h replaced by L 2 , in which case the operator Op h (hb) becomes negligible. In the isotropic case, however, it is essential to have in (1.3) the space H 1 h and that the function b does not depend on the refraction index n.

Note finally that Theorem 1.2 can be also used to study the location of the resonances for the exterior transmission problems considered in [START_REF] Cardoso | Asymptotics of the number of resonances in the transmission problem[END_REF] and [START_REF] Galkowski | The quantum Sabine law for resonances in transmission problems[END_REF]. For example, it allows to simplify the proof of the resonance-free regions in [START_REF] Cardoso | Asymptotics of the number of resonances in the transmission problem[END_REF] and to extend it to more general boundary conditions.

Applications to the transmission eigenvalues

Let Ω ⊂ R d , d ≥ 2, be a bounded, connected domain with a C ∞ smooth boundary Γ = ∂Ω. A complex number λ ∈ C, Re λ ≥ 0, will be said to be a transmission eigenvalue if the following problem has a non-trivial solution:

(2.1)

     ∇c 1 (x)∇ + λ 2 n 1 (x) u 1 = 0 in Ω, ∇c 2 (x)∇ + λ 2 n 2 (x) u 2 = 0 in Ω, u 1 = u 2 , c 1 ∂ ν u 1 = c 2 ∂ ν u 2 on Γ,
where ν denotes the Euclidean unit inner normal to Γ, c j , n j ∈ C ∞ (Ω), j = 1, 2 are strictly positive real-valued functions. We will consider two cases:

(2.2)

c 1 (x) ≡ c 2 (x) ≡ 1 in Ω, n 1 (x) = n 2 (x) on Γ, (isotropic case) (2.3) (c 1 (x) -c 2 (x))(c 1 (x)n 1 (x) -c 2 (x)n 2 (x)) = 0 on Γ. (anisotropic case)
In Section 6 we will prove the following Theorem 2.1. Assume either the condition (2.2) or the condition

(2.4) (c 1 (x) -c 2 (x))(c 1 (x)n 1 (x) -c 2 (x)n 2 (x)) < 0 on Γ.
Then there exists a constant C > 0 such that there are no transmission eigenvalues in the region

(2.5) {λ ∈ C : Re λ > 1, |Im λ| ≥ C} .
Remark. It is proven in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF] that under the condition (2.2) (as well as the condition (2.6) below) there exists a constant C > 0 such that there are no transmission eigenvalues in the region

λ ∈ C : 0 ≤ Re λ ≤ 1, |Im λ| ≥ C .
This is no longer true under the condition (2.4) in which case there exist infinitely many transmission eigenvalues very close to the imaginary axis. Note that the eigenvalue-free region (2.5) is optimal and cannot be improved in general. Indeed, it follows from the analysis in [START_REF] Leung | Complex transmission eigenvalues for spherically stratified media[END_REF] (see Section 4) that in the isotropic case when the domain Ω is a ball and the refraction indices n 1 and n 2 constant, there may exist infinitely many transmission eigenvalues whose imaginary parts are bounded from below by a positive constant. Note also that the above result has been previously proved in [START_REF] Petkov | Localization of the interior transmission eigenvalues for a ball[END_REF] in the case when the domain Ω is a ball and the coefficients constant. In the isotropic case the eigenvalue-free region (2.5) has been also obtained in [START_REF] Sylvester | Transmission eigenvalues in one Inverse Problems[END_REF] when the dimension is one. In the general case of arbitrary domains transmission eigenvalue-free regions have been previously proved in [START_REF] Hitrik | The interior transmission problem and bounds of transmission eigenvalues[END_REF], [START_REF] Lakshtanov | Application of elliptic theory to the isotropic interior transmission eigenvalue problem[END_REF] and [START_REF] Robbiano | Spectral analysis of interior transmission eigenvalues[END_REF] (isotropic case), [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF] and [START_REF] Vodev | Transmission eigenvalues for strictly concave domains[END_REF] (both cases). For example, it has been proved in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF] that, under the conditions (2.2) and (2.4), there are no transmission eigenvalues in

λ ∈ C : Re λ > 1, |Im λ| ≥ C ε (Re λ) 1 2 +ε , C ε > 0,
for every 0 < ε ≪ 1. This eigenvalue-free region has been improved in [START_REF] Vodev | Transmission eigenvalues for strictly concave domains[END_REF] under an additional strict concavity condition on the boundary Γ to the following one

{λ ∈ C : Re λ > 1, |Im λ| ≥ C ε (Re λ) ε } , C ε > 0,
for every 0 < ε ≪ 1. When the function in the left-hand side of (2.3) is strictly positive, parabolic eigenvalue-free regions have been proved in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF] for arbitrary domains, which however are worse than the eigenvalue-free regions we have under the conditions (2.2) and (2.4). In Section 7 we will prove the following Theorem 2.2. Assume the conditions

(2.6) (c 1 (x) -c 2 (x))(c 1 (x)n 1 (x) -c 2 (x)n 2 (x)) > 0 on Γ and (2.7) n 1 (x) c 1 (x) = n 2 (x) c 2 (x) on Γ.
Then there exists a constant C > 0 such that there are no transmission eigenvalues in the region

(2.8) {λ ∈ C : Re λ > 1, |Im λ| ≥ C log(Re λ + 1)} .
Note that in the case when (2.6) is fulfilled but (2.7) is not, the method developed in the present paper does not work and it is not clear if improvements are possible compared with the results in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF]. To our best knowledge, no results exist in the degenerate case when the function in the left-hand side of (2.3) vanishes without being identically zero.

It has been proved in [START_REF] Petkov | Asymptotics of the number of the interior transmission eigenvalues[END_REF] that the counting function N (r) = #{λtrans. eig. : |λ| ≤ r}, r > 1, satisfies the asymptotics

N (r) = (τ 1 + τ 2 )r d + O ε (r d-κ+ε ), ∀ 0 < ε ≪ 1,
where 0 < κ ≤ 1 is such that there are no transmission eigenvalues in the region

λ ∈ C : Re λ > 1, |Im λ| ≥ C (Re λ) 1-κ , C > 0,
and

τ j = ω d (2π) d Ω n j (x) c j (x) d/2
dx, ω d being the volume of the unit ball in R d . Using this we obtain from the above theorems the following Corollary 2.3. Under the conditions of Theorems 2.1 and 2.2, the counting function of the transmission eigenvalues satisfies the asymptotics (2.9)

N (r) = (τ 1 + τ 2 )r d + O ε (r d-1+ε ), ∀ 0 < ε ≪ 1.
This result has been previously proved in [START_REF] Vodev | Transmission eigenvalues for strictly concave domains[END_REF] under an additional strict concavity condition on the boundary Γ. In the present paper we remove this additional condition to conclude that in fact the asymptotics (2.9) holds true for an arbitrary domain. We also expect that (2.9) holds with ε = 0, but this remains an interesting open problem. In the isotropic case asymptotics for the counting function N (r) with remainder o(r d ) have been previously obtained in [START_REF] Faierman | The interior transmission problem: spectral theory[END_REF], [START_REF] Pham | Weyl asymptotics of the transmission eigenvalues for a constant index of refraction[END_REF], [START_REF] Robbiano | Counting function for interior transmission eigenvalues[END_REF].

A priori estimates in the glancing region

Let λ ∈ C, Re λ > 1, 1 < |Im λ| ≤ θ 0 Re λ, where 0 < θ 0 < 1 is a fixed constant, and set h = µ -1 , where

µ = Re λ 1 - Im λ Re λ 2 ∼ Re λ ∼ |λ|.
Clearly, we have Re (hλ) 2 = 1 and

λ 2 = µ 2 (1 + izh), z = 2µ -1 Im λRe λ ∼ 2Im λ.
Given an integer m ≥ 0, denote by H m h (X) the Sobolev space equipped with the semi

-classical norm v H m h (X) = |α|≤m h |α| ∂ α x v L 2 (X) .
We define similarly the Sobolev space

H m h (∂X). It is well-known that v H m h (∂X) ∼ Op h ( ξ ′ m )v L 2 (∂X) ∼ v L 2 (∂X) + Op h ((1 -η)|ξ ′ | m )v L 2 (∂X) for any function η ∈ C ∞ 0 (T * ∂X) independent of h. Hereafter, ξ ′ = (1 + |ξ ′ | 2 ) 1/2
. Given functions V ∈ L 2 (X) and f ∈ L 2 (∂X), we let the function u solve the equation

(3.1) ∆ X + λ 2 n(x) u = λV in X, u = f on ∂X,
and set g = h∂ ν u| ∂X . We will first prove the following Lemma 3.1. There is a constant C > 0 such that the following estimate holds

(3.2) u H 1 h (X) ≤ C|Im λ| -1 V L 2 (X) + C|Im λ| -1/2 f 1/2 L 2 (∂X) g 1/2
L 2 (∂X) . Proof. By Green's formula we have

Im (λ 2 ) n 1/2 u 2 L 2 (X) = Im λV, u L 2 (X) + Im ∂ ν u| ∂X , f L 2 (∂X) which implies (3.3) |Im λ| u 2 L 2 (X) V L 2 (X) u L 2 (X) + f L 2 (∂X) g L 2 (∂X) .
On the other hand, we have

∇ X u 2 L 2 (X) -Re (λ 2 ) n 1/2 u 2 L 2 (X) = -Re λV, u L 2 (X) -Re ∂ ν u| ∂X , f L 2 (∂X) which yields (3.4) h∇ X u 2 L 2 (X) u 2 L 2 (X) + O(h 2 ) V 2 L 2 (X) + O(h) f L 2 (∂X) g L 2 (∂X) . Since h |Im λ| -1 , the estimate (3.
2) follows from (3.3) and (3.4).

2

We now equip X with the Riemannian metric nG. We will write the operator n -1 ∆ X in the normal coordinates (x 1 , x ′ ) with respect to the metric nG near the boundary ∂X, where 0 < x 1 ≪ 1 denotes the distance to the boundary and x ′ are coordinates on ∂X.

Set Γ(x 1 ) = {x ∈ X : dist(x, ∂X) = x 1 }, Γ(0) = ∂X. Then Γ(x 1
) is a Riemannian manifold without boundary of dimension d-1 with a Riemannian metric induced by the metric nG, which depends smoothly in x 1 . It is well-known that the operator n -1 ∆ X writes as follows

n -1 ∆ X = ∂ 2 x 1 + Q(x 1 ) + R where Q(x 1 ) = ∆ Γ(x 1 )
is the negative Laplace-Beltrami operator on Γ(x 1 ) and R is a firstorder differential operator. Clearly, Q(x 1 ) is a second-order differential operator with smooth coefficients and

Q(0) = ∆ (n)
∂X is the negative Laplace-Beltrami operator on ∂X equipped with the Riemannian metric induced by the metric nG.

Let

χ ∈ C ∞ 0 (R), 0 ≤ χ(t) ≤ 1, χ(t) = 1 for |t| ≤ 1, χ(t) = 0 for |t| ≥ 2. Given a parameter 0 < δ 1 ≪ 1 independent of λ and an integer k ≥ 0, set φ k (x 1 ) = χ(2 -k x 1 /δ 1 ). Given integers 0 ≤ s 1 ≤ s 2 we define the norm u s 1 ,s 2 ,k by u 2 s 1 ,s 2 ,k = u 2 H s 1 h (X) + s 1 ℓ 1 =0 s 2 -ℓ 1 ℓ 2 =0 ∞ 0 (h∂ x 1 ) ℓ 1 (φ k u)(x 1 , •) 2 H ℓ 2 h (∂X) dx 1 .
Clearly, we have

u H s 1 h (X) ≤ u s 1 ,s 2 ,k u H s 2 h (X) . Throughout this paper η ∈ C ∞ 0 (T * ∂X), 0 ≤ η ≤ 1, η = 1 in |ξ ′ | ≤ A, η = 0 in |ξ ′ | ≥ A + 1
, will be a function independent of λ, where A > 1 is a parameter we may take as large as we want. We will now prove the following Lemma 3.2. Let u solve the equation (3.1) with V ∈ H s-1 (X) and f ∈ H 2s (∂X) for some integer s ≥ 1. Then the following estimate holds

(3.5) u 1,s+1,k u H 1 h (X) + V 0,s-1,k+s-1 + Op h (1 -η)f 1/2 H 2s h (∂X) g 1/2 L 2 (∂X) . Proof. Note that u 1,s+1,k u H 1 h (X) + u s,k H 1 h (X)
where the function

u s,k = Op h ((1 -η)|ξ ′ | s )(φ k u) satisfies the equation h 2 ∂ 2 x 1 + h 2 Q(x 1 ) + 1 + ihz u s,k = U s,k with U s,k = h 2 Q(x 1 ), Op h ((1 -η)|ξ ′ | s ) (φ k u) + Op h ((1 -η)|ξ ′ | s ) h 2 ∂ 2 x 1 , φ k φ k+1 u -h 2 Op h ((1 -η)|ξ ′ | s )φ k Rφ k+1 u + h 2 λOp h ((1 -η)|ξ ′ | s )(φ k V ).
We also have

f s := u s,k | x 1 =0 = Op h ((1 -η)|ξ ′ | s )f, g s := h∂ x 1 u s,k | x 1 =0 = Op h ((1 -η)|ξ ′ | s )g ♭ , where g ♭ := h∂ x 1 u| x 1 =0
. Integrating by parts the above equation and taking the real part, we get

h∂ x 1 u s,k 2 L 2 (X) -(h 2 Q(x 1 ) + 1)u s,k , u s,k L 2 (X) ≤ U s,k , u s,k L 2 (X) + h f s , g s L 2 (∂X) u s,k H 1 h (X) ( V 0,s-1,k + u 1,s,k+1 ) (3.6) + Op h ((1 -η)|ξ ′ | s ) * Op h ((1 -η)|ξ ′ | s )f L 2 (∂X) g ♭ L 2 (∂X)
The principal symbol r of the operator

-Q(x 1 ) satisfies r(x, ξ ′ ) ≥ C ′ |ξ ′ | 2 , C ′ > 0, on suppφ k ,
provided δ 1 is taken small enough. Therefore, we can arrange by taking the parameter A big enough that r -1 ≥ C ξ ′ on supp (1-η)φ k , where C > 0 is some constant. Hence, by Gärding's inequality we have

(3.7) -(h 2 Q(x 1 ) + 1)u s,k , u s,k L 2 (X) ≥ C Op h ( ξ ′ )u s,k 2 L 2 (X)
with possibly a new constant C > 0. Since the norms of g and g ♭ are equivalent, by (3.6) and (3.7) we get

u s,k H 1 h (X) V 0,s-1,k + u H 1 h (X) + u s-1,k+1 H 1 h (X) (3.8) + Op h (1 -η)f 1/2 H 2s h (∂X) g 1/2
L 2 (∂X) . We may now apply the same argument to u s-1,k+1 . Thus, repeating this argument a finite number of times we can eliminate the term involving u s-1,k+1 in the RHS of (3.8) and obtain the estimate (3.5).

2

Let the functions

χ j ∈ C ∞ (R), 0 ≤ χ j (t) ≤ 1, j = 1, 2, 3, be such that χ 1 + χ 2 + χ 3 ≡ 1, χ 2 = χ, χ 1 (t) = 1 for t ≤ -2, χ 1 (t) = 0 for t ≥ -1, χ 3 (t) = 0 for t ≤ 1, χ 3 (t) = 1 for t ≥ 2. Given a parameter 0 < δ ≪ 1 independent of λ, set χ - δ (x ′ , ξ ′ ) = χ 1 ((r ♯ (x ′ , ξ ′ ) -1)/δ 2 ), χ 0 δ (x ′ , ξ ′ ) = χ 2 ((r ♯ (x ′ , ξ ′ ) -1)/δ 2 ), χ + δ (x ′ , ξ ′ ) = χ 3 ((r ♯ (x ′ , ξ ′ ) -1)/δ 2 ), where r ♯ = n -1 0 r 0 is the principal symbol of the operator -∆ (n) ∂X . Since (r ♯ -1) k χ 0 δ = O(δ 2k ), we have (3.9) (h 2 ∆ (n) ∂X + 1) k Op h (χ 0 δ ) = O(δ 2k ) : L 2 (∂X) → L 2 (∂X) for every integer k ≥ 0. Clearly, we also have Op h (χ 0 δ ) = O(1) : L 2 (∂X) → H m h (∂X)
, ∀ m ≥ 0, uniformly in δ. Using (3.9) we will prove the following Proposition 3.3. Let u solve (3.1) with f ≡ 0 and V ∈ H s (X) for some integer s ≥ 0. Then the function g = h∂ ν u| ∂X satisfies the estimate

(3.10) g H s h (∂X) ≤ C ′ |Im λ| -1/2 V 0,s,s with a constant C ′ > 0 independent of λ.
Let u solve (3.1) with f replaced by Op h (χ 0 δ )f and V ∈ H s+2 (X) for some integer s ≥ 0. Then the function g = h∂ ν u| ∂X satisfies the estimate

(3.11) g H s h (∂X) ≤ C δ + |Im λ| -1/4 f L 2 (∂X) + C δ 1/2 + |Im λ| -1/8 V 0,s+2,s+2 for 1 < |Im λ| ≤ δ 2 Re λ, Re λ ≥ C δ ≫ 1, with a constant C > 0 independent of λ and δ.
Proof. Set w = φ 0 (x 1 )u. We will first show that the estimates (3.10) and (3.11) with s ≥ 1 follow from (3.10) and (3.11) with s = 0, respectively. This follows from the estimate

(3.12) g H s h (∂X) g L 2 (∂X) + h∂ x 1 v s | x 1 =0 L 2 (∂X)
where the function

v s = Op h ((1 -η)|ξ ′ | s )w satisfies the equation (3.1) with V replaced by V s = nOp h ((1 -η)|ξ ′ | s )φ 0 n -1 V + λ -1 n n -1 ∆ X , Op h ((1 -η)|ξ ′ | s )φ 0 u.
We can write the commutator as

∂ 2 x 1 + R, φ 0 (x 1 ) Op h ((1 -η)|ξ ′ | s )φ 1 (x 1 ) + φ 0 Q(x 1 ) + R, Op h ((1 -η)|ξ ′ | s ) φ 1 (x 1
). Therefore, if f ≡ 0, in view of Lemmas 3.1 and 3.2, the function V s satisfies the bound

(3.13) V s 0,0,0 V 0,s,0 + u 1,s+1,1 u H 1 h (X) + V 0,s,s V 0,s,s .
Clearly, the assertion concerning (3.10) follows from (3.12) and (3.13). The estimate (3.11) can be treated similarly. Indeed, in view of Lemma 3.2, the function V s satisfies the bound

V s 0,2,2 V 0,s+2,0 + u 1,s+3,1 (3.14) u H 1 h (X) + V 0,s+2,s+2 + Op h (1 -η)Op h (χ 0 δ )f 1/2 H 2s+4 h (∂X) g 1/2 L 2 (∂X) .
Taking the parameter A big enough we can arrange that supp

χ 0 δ ∩ supp (1 -η) = ∅. Hence (3.15) Op h (1 -η)Op h (χ 0 δ ) = O(h ∞ ) : L 2 (∂X) → H m h (∂X)
, ∀m ≥ 0. By (3.14) and (3.15) together with Lemma 3.1 we conclude

V s 0,2,2 u H 1 h (X) + V 0,s+2,s+2 + O(h ∞ ) f 1/2 L 2 (∂X) g 1/2 L 2 (∂X) V 0,s+2,s+2 + O |Im λ| -1/2 + h ∞ f 1/2 L 2 (∂X) g 1/2
L 2 (∂X) . We now apply (3.11) with s = 0 to the function v s and note that

v s | x 1 =0 = Op h ((1 -η)|ξ ′ | s )Op h (χ 0 δ )f = O(h ∞ )f. Hence h∂ x 1 v s | x 1 =0 L 2 (∂X) ≤ O(h ∞ ) f L 2 (∂X) + O δ 1/2 + |Im λ| -1/8 V s 0,2,2 (3.16) ≤ O δ 1/2 + |Im λ| -1/8 V 0,s+2,s+2 + O |Im λ| -1/2 + h ∞ f 1/2 L 2 (∂X) g 1/2
L 2 (∂X) . Therefore, the assertion concerning (3.11) follows from (3.12) and (3.16).

We now turn to the proof of (3.10) and (3.11) with s = 0. In view of Lemma 3.1, the function

U := h(n -1 ∆ X + λ 2 )w = h[n -1 ∆ X , φ 0 (x 1 )]u + hλn -1 φ 0 V satisfies the bound U L 2 (X) u H 1 h (X) + V L 2 (X) (3.17) V L 2 (X) + O |Im λ| -1/2 f 1/2 L 2 (∂X) g 1/2
L 2 (∂X) . Observe now that the derivative of the function

E(x 1 ) = h∂ x 1 w 2 + h 2 Q(x 1 ) + 1 w, w ,
• and •, • being the norm and the scalar product in L 2 (∂X), satisfies

E ′ (x 1 ) = 2Re h 2 ∂ 2 x 1 + h 2 Q(x 1 ) + 1 w, ∂ x 1 w + h 2 Q ′ (x 1 )w, w = 2Re (U -izw -hRw) , h∂ x 1 w + h 2 Q ′ (x 1 )w, w . If we put g ♭ := h∂ x 1 u| x 1 =0 , we have g ♭ 2 + h 2 ∆ (n) ∂X + 1 Op h (χ 0 δ )f, Op h (χ 0 δ )f = E(0) = - ∞ 0 E ′ (x 1 )dx 1 U L 2 (X) + |z| w L 2 (X) + hRw L 2 (X) h∂ x 1 w L 2 (X) + w 2 H 1 h (X) (3.18) ≤ O(|z|) h∂ x 1 w L 2 (X) w L 2 (X) + O |Im λ| -1 F 2
where we have used Lemma 3.1 together with (3.17) and we have put

F = f 1/2 g 1/2 + V L 2 (X) .
Clearly, (3.10) with s = 0 follows from (3.18) applied with f ≡ 0 and Lemma 3.1. To prove (3.11) with s = 0, observe that (3.9) and (3.18) lead to

(3.19) g ≤ O(δ) f + O |Im λ| -1/2 F + O(|Im λ| 1/2 ) h∂ x 1 w 1/2 L 2 (X) w 1/2
L 2 (X) . We need now to bound the norm h∂ x 1 w L 2 (X) in the RHS of (3.19) better than what the estimate (3.2) gives. To this end, observe that integrating by parts yields

h∂ x 1 w 2 L 2 (X) -h 2 Q(x 1 ) + 1 w, w L 2 (X) = -hRe (U -hRw), w L 2 (X) -hRe f, g ♭ (3.20) ≤ O(h) w 2 H 1 h (X) + O(h) U 2 L 2 (X) + O(h) f g ≤ O(h)
g ≤ O(δ) f + O(|Im λ| 1/2 ) w 1 1/4 L 2 (X) w 3/4 L 2 (X) +O(h 1/4 |Im λ| 1/2 )F 1/2 w 1/2 L 2 (X) + O |Im λ| -1/2 F (3.21) ≤ O(δ) f + O(|Im λ| 1/8 ) w 1 1/4 L 2 (X) F 3/4 + O |Im λ| -1/2 + h 1/4 |Im λ| 1/4
F where we have put w 1 := h 2 Q(x 1 ) + 1 w. We need now the following Lemma 3.4. The function w 1 satisfies the estimate

|Im λ| 1/2 w 1 L 2 (X) ≤ O δ 2 + |Im λ| -1 + h ∞ f 1/2 g 1/2 (3.22) +O h 1/2 f + O |Im λ| -1/2 V 0,2,2 .
Let us see that this lemma implies the estimate (3.11) with s = 0. Set

F = f 1/2 g 1/2 + V 0,2,2 ≥ F.
By (3.21) and (3.22),

g ≤ O (δ) f + O δ 1/2 + |Im λ| -1/8 + h ∞ F +O h 1/8 ( f + F ) + O |Im λ| -1/2 + h 1/4 |Im λ| 1/4 F (3.23) ≤ O δ + h 1/8 f + O δ 1/2 + |Im λ| -1/8 + h 1/8 + h 1/4 |Im λ| 1/4 F .
Since by assumption h 1/4 |Im λ| 1/4 = O δ 1/2 , one can easily see that (3.11) with s = 0 follows from (3.23). 2

Proof of Lemma 3.4. Observe that the function w 1 satisfies the equation

h 2 ∂ 2 x 1 + h 2 Q(x 1 ) + 1 + ihz w 1 = hU 1 where U 1 := h 2 Q(x 1 ) + 1 (U -hRw) + 2h 3 Q ′ (x 1 )∂ x 1 w + h 3 Q ′′ (x 1 )
w. We also have

f 1 := w 1 | x 1 =0 = (h 2 Q(0) + 1)Op h (χ 0 δ )f, g 1 := h∂ x 1 w 1 | x 1 =0 = (h 2 Q(0) + 1)g ♭ + h 2 Q ′ (0)Op h (χ 0 δ )f.
Integrating by parts above equation and taking the imaginary part, we get

|z| w 1 2 L 2 (X) ≤ U 1 , w 1 L 2 (X) + | f 1 , g 1 | ≤ U 1 L 2 (X) w 1 L 2 (X) + O(1) (h 2 Q(0) + 1) 2 Op h (χ 0 δ )f g +O(h) Op h (χ 0 δ )f H 2 h (∂X) (h 2 Q(0) + 1)Op h (χ 0 δ )f ≤ U 1 L 2 (X) w 1 L 2 (X) + O(δ 4 ) f g + O(h) f 2
where we have used (3.9). Hence (3.24) |z| w 1 2

L 2 (X) ≤ O |z| -1 U 1 2 L 2 (X) + O(δ 4 ) f g + O(h) f 2 .
Recall that the function U is of the form (2h∂ x 1 + a(x))φ 1 (x 1 )u + hλn -1 φ 0 V , where a is some smooth function. Hence the function U 1 satisfies the estimate

U 1 L 2 (X) u 1,3,1 + V 0,2,0 (3.25) u H 1 h (X) + V 0,2,2 + O(h ∞ ) f 1/2 L 2 (∂X) g 1/2 L 2 (∂X)
where we have used Lemma 3.2 together with (3.15). By (3.24) and (3.25),

|z| w 1 2 L 2 (X) ≤ O |z| -1 u 2 H 1 h (X) + O |z| -1 V 2 0,2,2 (3.26) +O(δ 4 + h ∞ ) f g + O(h) f 2 .
Clearly, (3.22) follows from (3.26) and Lemma 3.1. 2

Parametrix construction in the hyperbolic region

Let λ be as in Theorems 1.1 and 1.2, and let h, z, δ, r 0 , n 0 , r ♯ , χ and χ - δ be as in the previous sections. Set θ = Im (hλ

) 2 = hz = O(h ǫ ), |θ| ≫ h, and ρ(x ′ , ξ ′ ) = r 0 (x ′ , ξ ′ ) -(1 + iθ)n 0 (x ′ ), Re ρ < 0.
It is easy to see that ρχ - δ ∈ S 0 0 (∂X). In this section we will prove the following Proposition 4.1. There are constants C, C 1 > 0 depending on δ but independent of λ such that

(4.1) hN (λ; n)Op h (χ - δ ) -Op h (ρχ - δ ) L 2 (∂X)→H 1 h (∂X) ≤ C 1 h + e -C|Im λ| . Proof.
To prove (4.1) we will build a parametrix near the boundary of the solution to the equation (1.1) with f replaced by Op h (χ - δ )f . Let x = (x 1 , x ′ ), x 1 > 0, be the normal coordinates with respect to the metric G, which of course are different from those introduced in the previous section. In these coordinates the operator ∆ X writes as follows

∆ X = ∂ 2 x 1 + Q + R
where Q ≤ 0 is a second-order differential operator with respect to the variables x ′ and R is a first-order differential operator with respect to the variables x, both with coefficients depending smoothly on x. Let (x 0 , ξ 0 ) ∈ supp χ - δ and let U ⊂ T * ∂X be a small open neighbourhood of (x 0 , ξ 0 ) contained in {r ♯ ≤ 1δ 2 /2}. Take a function ψ ∈ C ∞ 0 (U ). We will construct a parametrix u - ψ of the solution of (1.1) with u

- ψ | x 1 =0 = Op h (ψ)f in the form u - ψ = φ(x 1 )K -f
, where φ(x 1 ) = χ(x 1 /δ 1 ), 0 < δ 1 ≪ 1 being a parameter independent of λ to be fixed later on depending on δ, and

(K -f )(x) = (2πh) -d+1 e i h ( y ′ ,ξ ′ +ϕ(x,ξ ′ ,θ)) a(x, ξ ′ , λ)f (y ′ )dξ ′ dy ′ .
The phase ϕ is complex-valued such that ϕ| x 1 =0 =x ′ , ξ ′ and satisfies the eikonal equation mod O(θ M ):

(4.2) (∂ x 1 ϕ) 2 + B(x)∇ x ′ ϕ, ∇ x ′ ϕ = (1 + iθ)n(x) + θ M R M
where M ≫ 1 is an arbitrary integer, the function R M is bounded uniformly in θ, and B is a matrix-valued function such that r(x, ξ ′ ) = B(x)ξ ′ , ξ ′ , r(x, ξ ′ ) ≥ 0 being the principal symbol of the operator -Q. We clearly have r 0 (x ′ , ξ ′ ) = r(0, x ′ , ξ ′ ). Let us see that for (x ′ , ξ ′ ) ∈ U , 0 ≤ x 1 ≤ 3δ 1 , the equation (4.2) has a smooth solution safisfying

(4.3) ∂ x 1 ϕ| x 1 =0 = -iρ + O(θ M/2 )
provided δ 1 and U are small enough. We will be looking for ϕ in the form

ϕ = M -1 j=0 (iθ) j ϕ j (x, ξ ′ )
where ϕ j are real-valued functions depending only on the sign of θ and satisfying the equations

(4.4) (∂ x 1 ϕ 0 ) 2 + B(x)∇ x ′ ϕ 0 , ∇ x ′ ϕ 0 = n(x), (4.5)
k j=0 ∂ x 1 ϕ j ∂ x 1 ϕ k-j + k j=0 B(x)∇ x ′ ϕ j , ∇ x ′ ϕ k-j = ε k n(x), 1 ≤ k ≤ M -1, ϕ 0 | x 1 =0 = -x ′ , ξ ′ , ϕ j | x 1 =0 = 0 for j ≥ 1, where ε 1 = 1, ε k = 0 for k ≥ 2.
It is easy to check that with this choice the function ϕ satisfies (4.2) with R M being polynomial in θ.

Clearly, if ϕ 0 is a solution to (4.4), then we have (∂

x 1 ϕ 0 | x 1 =0 ) 2 = n 0 (x ′ ) -r 0 (x ′ , ξ ′ ) ≥ C ′ with some constant C ′ > 0 depending on δ.
It is well-known that the equation (4.4) has a local (that is, for δ 1 and U small enough) real-valued solution ϕ ± 0 such that

∂ x 1 ϕ ± 0 | x 1 =0 = ± √ n 0 -r 0 . We now define the function ϕ 0 by ϕ 0 = ϕ + 0 if θ > 0, ϕ 0 = ϕ - 0 if θ < 0. Hence |∂ x 1 ϕ 0 (x, ξ ′ )| ≥ Const > 0 for x 1 small
enough. Therefore, the equations (4.5) can be solved locally. Taking x 1 = 0 in the equation (4.5) with k = 1 we find (4.6)

θ∂ x 1 ϕ 1 | x 1 =0 = θn 0 (2∂ x 1 ϕ 0 | x 1 =0 ) -1 = |θ| 2 n 0 (n 0 -r 0 ) -1/2 ≥ C|θ| 2
on U , where C = min n 0 (x ′ ). Hence

(4.7) Im ∂ x 1 ϕ| x 1 =0 = θ∂ x 1 ϕ 1 | x 1 =0 + O(θ 2 ) ≥ C|θ| 3 if |θ| is
taken small enough. On the other hand, taking x 1 = 0 in the equation (4.2) we find

(4.8) (∂ x 1 ϕ| x 1 =0 ) 2 = (iρ) 2 + O(θ M ) = (iρ) 2 (1 + O(θ M ))
where we have used that |ρ| ≥ Const > 0 on U . Since Re ρ < 0, we get (4.3) from (4.7) and (4.8). By (4.6) we also get

θϕ 1 (x 1 , x ′ , ξ ′ ) = θx 1 ∂ x 1 ϕ 1 (0, x ′ , ξ ′ ) + O(θx 2 1 ) ≥ Cx 1 |θ| 2 -O(|θ|x 2 1 ) ≥ Cx 1 |θ| 3 provided x 1 is taken small enough. This implies (4.9) Im ϕ(x, ξ ′ , θ) = θϕ 1 (x 1 , x ′ , ξ ′ ) + O(θ 2 x 1 ) ≥ Cx 1 |θ| 4 .
The amplitude a of the form

a = m k=0 h k a k (x, ξ ′ , θ)
where m ≫ 1 is an arbitrary integer and the functions a k satisfy the transport equations mod O(θ M ):

(4.10) 2i∂ x 1 ϕ∂ x 1 a k + 2i B(x)∇ x ′ ϕ, ∇ x ′ a k + i (∆ X ϕ) a k + ∆ X a k-1 = θ M Q (k) M , 0 ≤ k ≤ m, a 0 | x 1 =0 = ψ, a k | x 1 =0 = 0 for k ≥ 1
, where a -1 = 0. Let us see that the transport equations have smooth solutions for (x ′ , ξ ′ ) ∈ U , 0 ≤ x 1 ≤ 3δ 1 , provided δ 1 and U are taken small enough. As above, we will be looking for a k in the form

a k = M -1 j=0 (iθ) j a k,j (x, ξ ′ ).
We let a k,j satisfy the equations (4.11) 2i

j ν=0 ∂ x 1 ϕ ν ∂ x 1 a k,j-ν + 2i j ν=0 B(x)∇ x ′ ϕ ν , ∇ x ′ a k,j-ν + i (∆ X ϕ j ) a k,j + ∆ X a k-1,j = 0, 0 ≤ j ≤ M -1, a 0,0 | x 1 =0 = ψ, a k,j | x 1 =0 = 0 for k + j ≥ 1. Then the functions a k satisfy (4.10) with Q (k)
M being polynomial in θ. As in the case of the equations (4.5) one can solve (4.11) locally. Then we can write

V -:= h -1 (h 2 ∆ X + (1 + iθ)n(x)) u - ψ = K - 1 f + K - 2 f where K - 1 f = h[∆ X , φ]K -f = h(2φ ′ (x 1 )∂ x 1 + c(x)φ ′′ (x 1 ))K -f = (2πh) -d+1 e i h ( y ′ ,ξ ′ +ϕ(x,ξ ′ ,θ)) A - 1 (x, ξ ′ , λ)f (y ′ )dξ ′ dy ′
c being some smooth function,

A - 1 = 2iφ ′ a∂ x 1 ϕ + hcφ ′′ ∂ x 1 a and (K - 2 f )(x) = (2πh) -d+1 e i h ( y ′ ,ξ ′ +ϕ(x,ξ ′ ,θ)) A - 2 (x, ξ ′ , λ)f (y ′ )dξ ′ dy ′ where A - 2 = φ(x 1 ) h -1 θ M R M a + θ M m k=0 h k Q (k) M + h m+1 ∆ X a m .
Let us see that Proposition 4.1 follows from the following Lemma 4.2. The function V -satisfies the estimate

(4.12) V -H 1 h (X) e -C|Im λ| f + O m h m-d f + O M h ǫM -d f with some constant C > 0.
Indeed, if u - ψ denotes the solution to the equation (1.1) with f replaced by Op h (ψ)f and u - ψ is the parametrix built above, then the function v = u - ψu - ψ satisfies the equation (3.1) with f ≡ 0. Therefore, by the estimates (3.10) and (4.12) we have (4.13) hN

(λ; n)Op h (ψ) -T - ψ L 2 (∂X)→H 1 h (∂X) e -C|Im λ| + O m h m-d + O M h ǫM -d
where the operator T - ψ is defined by

T - ψ f = h∂ x 1 K -f | x 1 =0 . Hence, in view of (4.3), T - ψ f (x ′ ) = (2πh) -d+1 e i h y ′ -x ′ ,ξ ′ (iψ∂ x 1 ϕ(0, x ′ , ξ ′ , θ) + h∂ x 1 a(0, x ′ , ξ ′ , λ))f (y ′ )dξ ′ dy ′ = Op h (ρψ + O(θ M/2 ))f + m k=0 h k+1 Op h (∂ x 1 a k (0, x ′ , ξ ′ , θ))f. Since Op h (∂ x 1 a k (0, x ′ , ξ ′ , θ)) = O(1) : L 2 (∂X) → H 1 h (∂X) uniformly in θ, it follows from (4.13) that (4.14) hN (λ; n)Op h (ψ) -Op h (ρψ) L 2 (∂X)→H 1 h (∂X) e -C|Im λ| + O(h).
On the other hand, using a suitable partition of the unity we can write the function χ - δ as J j=1 ψ j , where each function ψ j has the same properties as the function ψ above. In other words, we have (4.14) with ψ replaced by each ψ j , which after summing up leads to (4.1). 2

Proof of Lemma 4.2. Let α be a multi-index such that |α| ≤ 1. Since

i|α|A - 2 ∂ α x ϕ + (h∂ x ) α A - 2 = O m h m+1 + O M h ǫM -1 and Im ϕ ≥ 0, the kernel of the operator (h∂ x ) α K - 2 : L 2 (∂X) → L 2 (X) is O m h m-d + O M h ǫM -d
, and hence so is its norm. Since the function A - 1 is supported in the interval [δ 1 /2, 3δ 1 ] with respect to the variable x 1 , to bound the norm of the operator

K - 1,α := (h∂ x ) α K - 1 : L 2 (∂X) → L 2 (X) it suffices to show that (4.15) K - 1,α L 2 (∂X)→L 2 (∂X) e -C|θ|/h + O(h ∞ ) uniformly in x 1 ∈ [δ 1 /2, 3δ 1 ].
Since |θ|/h ∼ |Im λ|, (4.15) will imply (4.12). We would like to consider K - 1,α as an h-FIO with phase Re ϕ and amplitude

A α = e -Im ϕ/h i|α|A - 1 ∂ α x ϕ + (h∂ x ) α A - 1 .
To do so, we need to have that the phase satisfies the condition

(4.16) det ∂ 2 Re ϕ ∂x ′ ∂ξ ′ ≥ C > 0
for |θ| small enough, where C is a constant independent of θ. Since Re ϕ = ϕ 0 +O(|θ|), it suffices to show (4.16) for the phase ϕ 0 . This, however, is easy to arrange by taking x 1 small enough because ϕ 0 =x ′ , ξ ′ + O(x 1 ) and (4.16) is trivially fulfilled for the phasex ′ , ξ ′ . On the other hand, using Im ϕ = O(|θ|) together with (4.9) we get the following bounds for the amplitude:

(4.17)

∂ β 1 x ′ ∂ β 2 ξ ′ A α ≤ C β 1 ,β 2 0≤k≤|β 1 |+|β 2 | |θ| h k e -Cδ 1 |θ| 8h ≤ C β 1 ,β 2 e -Cδ 1 |θ| 9h
for all multi-indices β 1 and β 2 . It follows from (4.16) and (4.17) that, mod O(h ∞ ), the operator (K - 1,α ) * K - 1,α is an h -ΨDO in the class OPS 0 0 (∂X) uniformly in θ with a symbol which is O e -2C|θ|/h together with all derivatives, where C > 0 is a new constant. Therefore, its norm is also O e -2C|θ|/h , which clearly implies (4.15). 2

Parametrix construction in the elliptic region

We keep the notations from the previous sections and note that ρχ + δ ∈ S 1 0 (∂X). It is easy also to see that 0

< C 1 ξ ′ ≤ |ρ| ≤ C 2 ξ ′ on supp χ + δ ,
where C 1 and C 2 are constants depending on δ. In this section we will prove the following Proposition 5.1. There is a constant C > 0 depending on δ but independent of λ such that

(5.1) hN (λ; n)Op h (χ + δ ) -Op h (ρχ + δ + hb) L 2 (∂X)→H 1 h (∂X)
≤ Ch where b ∈ S 0 0 (∂X) does not depend on λ and the function n. Proof. The estimate (5.1) is a consequence of the parametrix built in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF]. In what follows we will recall this construction. We will first proceed locally and then we will use partition of the unity to get the global parametrix. Fix a point x 0 ∈ ∂X and let U 0 ⊂ ∂X be a small open neighbourhood of x 0 . Let (x 1 , x ′ ), x 1 > 0, x ′ ∈ U 0 , be the normal coordinates used in the previous section. Take a function ψ 0 ∈ C ∞ 0 (U 0 ) and set ψ = ψ 0 χ + δ . As in the previous section, we will construct a parametrix u + ψ of the solution of (1.1) with u

+ ψ | x 1 =0 = Op h (ψ)f in the form u + ψ = φ(x 1 )K + f
, where φ(x 1 ) = χ(x 1 /δ 1 ), 0 < δ 1 ≪ 1 being a parameter independent of λ to be fixed later on, and

(K + f )(x) = (2πh) -d+1 e i h ( y ′ ,ξ ′ +ϕ(x,ξ ′ ,θ)) a(x, ξ ′ , λ)f (y ′ )dξ ′ dy ′ .
The phase ϕ is complex-valued such that ϕ| x 1 =0 =x ′ , ξ ′ and satisfies the eikonal equation mod O(x M 1 ):

(5.2) (∂ x 1 ϕ) 2 + B(x)∇ x ′ ϕ, ∇ x ′ ϕ -(1 + iθ)n(x) = x M 1 R M where M ≫ 1 is
an arbitrary integer, the function R M is smooth up to the boundary x 1 = 0. It is shown in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF], Section 4, that for (x ′ , ξ ′ ) ∈ supp ψ, the equation (5.2) has a smooth solution of the form

ϕ = M -1 k=0 x k 1 ϕ k (x ′ , ξ ′ , θ), ϕ 0 = -x ′ , ξ ′ , safisfying (5.3) ∂ x 1 ϕ| x 1 =0 = ϕ 1 = -iρ.
Moreover, taking δ 1 small enough we can arrange that

(5.4) Im ϕ ≥ - x 1 2 Re ρ ≥ Cx 1 ξ ′ , C > 0, for 0 ≤ x 1 ≤ 3δ 1 , (x ′ , ξ ′ ) ∈ supp ψ. The amplitude a is of the form a = m j=0 h j a j (x, ξ ′ , θ)
where m ≫ 1 is an arbitrary integer and the functions a j satisfy the transport equations mod O(x M 1 ): (5.5) 2i∂

x 1 ϕ∂ x 1 a j + 2i B(x)∇ x ′ ϕ, ∇ x ′ a j + i (∆ X ϕ) a j + ∆ X a j-1 = x M 1 Q (j) M , 0 ≤ j ≤ m, a 0 | x 1 =0 = ψ, a j | x 1 =0 = 0 for j ≥ 1
, where a -1 = 0 and the functions Q

(j)
M are smooth up to the boundary x 1 = 0. It is shown in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF], Section 4, that the equations (5.5) have unique smooth solutions of the form

a j = M -1 k=0 x k 1 a k,j (x ′ , ξ ′ , θ)
with functions a k,j ∈ S -j 0 (∂X) uniformly in θ. We can write

V + := h -1 (h 2 ∆ X + (1 + iθ)n(x)) u + ψ = K + 1 f + K + 2 f where K + 1 f = h[∆ X , φ]K + f = h(2φ ′ (x 1 )∂ x 1 + c(x)φ ′′ (x 1 ))K + f = (2πh) -d+1 e i h ( y ′ ,ξ ′ +ϕ(x,ξ ′ ,θ)) A + 1 (x, ξ ′ , λ)f (y ′ )dξ ′ dy ′ , A + 1 = 2iφ ′ a∂ x 1 ϕ + hcφ ′′ ∂ x 1 a and (K + 2 f )(x) = (2πh) -d+1 e i h ( y ′ ,ξ ′ +ϕ(x,ξ ′ ,θ)) A + 2 (x, ξ ′ , λ)f (y ′ )dξ ′ dy ′ where A + 2 = φ(x 1 )   h -1 x M 1 R M a + x M 1 m j=0 h j Q (j) M + h m+1 ∆ X a m   .
As in the previous section, we will derive Proposition 5.1 from (5.3) and the following Lemma 5.2. The function V + satisfies the estimate

(5.6) V + H 1 h (X) ≤ O m h m-d f + O M h M -d f . Proof.
Let α be a multi-index such that |α| ≤ 1. In view of (5.4) we have

e iϕ/h i|α|A + 1 ∂ α x ϕ + (h∂ x ) α A + 1 sup δ 1 /2≤x 1 ≤3δ 1 e -Im ϕ/h e -C ξ ′ /h = O M (h/ ξ ′ ) M
for every integer M ≫ 1. Therefore, the kernel of the operator (h∂

x ) α K + 1 : L 2 (∂X) → L 2 (X) is O M h M -d+1
, and hence so is its norm. By (5.4) we also have

x M 1 e -Im ϕ/h ≤ x M 1 e -Cx 1 ξ ′ /h = O M (h/ ξ ′ ) M . This implies that e iϕ/h i|α|A + 2 ∂ α x ϕ + (h∂ x ) α A + 2 = O M (h/ ξ ′ ) M -1 + O m (h/ ξ ′
) m which again implies the desired bound for the norm of the operator (h∂ x ) α K + 2 .

2

By the estimates and (5.6) we have

(5.7) hN (λ; n)Op h (ψ) -T + ψ L 2 (∂X)→H 1 h (∂X) ≤ O m h m-d + O M h M -d
where the operator T + ψ is defined by

T + ψ f = h∂ x 1 K + f | x 1 =0
. In view of (5.3), we have

T + ψ f (x ′ ) = (2πh) -d+1 e i h y ′ -x ′ ,ξ ′ (iψ∂ x 1 ϕ(0, x ′ , ξ ′ , θ) + h∂ x 1 a(0, x ′ , ξ ′ , λ))f (y ′ )dξ ′ dy ′ = Op h (ρψ)f + m j=0 h j+1 Op h (a 1,j (x ′ , ξ ′ , θ))f
where a 1,j ∈ S -j 0 (∂X). Hence Op h (a 1,j ) = O(1) : L 2 (∂X) → H j h (∂X). Therefore it follows from (5.7) that

(5.8) hN (λ; n)Op h (ψ) -Op h (ρψ + ha 1,0 ) L 2 (∂X)→H 1 h (∂X) ≤ O(h).
We need now the following Lemma 5.3. There exists a function b 0 ∈ S 0 0 (∂X) independent of λ and n such that (5.9) a 1,0b 0 ∈ S -1 0 (∂X). Proof. We will calculate the function a 1,0 explicitly. Note that this lemma (resp. Proposition 5.1) is also used in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF], but the proof therein is not correct since a 1,0 is calculated incorrectly. Therefore we will give here a new proof. Clearly, it suffices to prove (5.9) with a 1,0 replaced by

(1 -η)a 1,0 with some function η ∈ C ∞ 0 (T * ∂X) independent of h. Since ρ = - √ r 0 1 + O(r -1 0 ) as r 0 → ∞, it is easy to see that (5.10) (1 -η)ρ -k -(1 -η)(- √ r 0 ) -k ∈ S -k-1 0 (∂X)
for every integer k ≥ 0, provided η is taken such that η = 1 for |ξ ′ | ≤ A with some A > 1 big enough. We will now calculate the function ϕ 2 from the eikonal equation. To this end, write

B(x) = B 0 (x ′ ) + x 1 B 1 (x ′ ) + O(x 2 1 ), n(x) = n 0 (x ′ ) + x 1 n 1 (x ′ ) + O(x 2
1 ) and observe that the LHS of (5.2) is equal to

x 1 (4ϕ 1 ϕ 2 + 2 B 0 ∇ x ′ ϕ 0 , ∇ x ′ ϕ 1 + B 1 ∇ x ′ ϕ 0 , ∇ x ′ ϕ 0 -(1 + iθ)n 1 ) + O(x 2 1
). Hence, taking into account that ϕ 0 =x ′ , ξ ′ and ϕ 1 = -iρ, we get

ϕ 2 = (2ρ) -1 B 0 ξ ′ , ∇ x ′ ρ + (4iρ) -1 B 1 ξ ′ , ξ ′ -(1 + iθ)(4iρ) -1 n 1 .

Using the identity 2ρ∇

x ′ ρ = ∇ x ′ r 0 -(1 + iθ)∇ x ′ n 0 we can write ϕ 2 in the form ϕ 2 = (2ρ) -2 B 0 ξ ′ , ∇ x ′ r 0 + (4iρ) -1 B 1 ξ ′ , ξ ′ -(1 + iθ)(2ρ) -2 B 0 ξ ′ , ∇ x ′ n 0 -(1 + iθ)(4iρ) -1 n 1 .
By (5.10) we conclude that, mod S -1 0 (∂X), (5.11) (1η)

ϕ 2 ϕ 1 = -i4 -1 (1 -η)r -3/2 0 B 0 ξ ′ , ∇ x ′ r 0 + (1 -η)(4r 0 ) -1 B 1 ξ ′ , ξ ′ .
Write now the operator ∆ X in the form

∆ X = ∂ 2 x 1 + B 0 ∇ x ′ , ∇ x ′ + q 1 (x ′ )∂ x 1 + q 2 (x ′ ), ∇ x ′ + O(x 1
) and observe that ∆ X ϕ = 2ϕ 2 + q 1 ϕ 1q 2 (x ′ ), ξ ′ + O(x 1 ).

We now calculate the LHS of the equation (5.5) with j = 0 modulo O(x 1 ). Recall that a 0,0 = ψ. We obtain 2iϕ 1 a 1,0 + 2i

B 0 ∇ x ′ ϕ 0 , ∇ x ′ a 0,0 + i(∆ X ϕ)a 0,0 = 2iϕ 1 a 1,0 + 2i B 0 ξ ′ , ∇ x ′ ψ + i(2ϕ 2 + q 1 ϕ 1 -q 2 (x ′ ), ξ ′ )ψ.
Since the RHS is O(x M 1 ), the above function must be identically zero. Thus we get the following expression for the function a 1,0 :

(5.12)

a 1,0 = -ϕ -1 1 B 0 ξ ′ , ∇ x ′ ψ -(ϕ -1 1 ϕ 2 + 2 -1 q 1 -(2ϕ 1 ) -1 q 2 (x ′ ), ξ ′ )ψ.
Taking into account that ψ = ψ 0 on supp (1η), we find from (5.10), (5.11) and (5.12) that (5.9) holds with

b 0 = i(1 -η)r -1/2 0 B 0 ξ ′ , ∇ x ′ ψ 0 (5.13) -4 -1 (1 -η)ψ 0 -ir -3/2 0 B 0 ξ ′ , ∇ x ′ r 0 + r -1 0 B 1 ξ ′ , ξ ′ + 2q 1 + 2r -1/2 0 q 2 (x ′ ), ξ ′ .
Clearly, b 0 ∈ S 0 0 (∂X) is independent of λ and n, as desired. 2

Lemma 5.3 implies that

(5.14) Op h (a 1,0 -b 0 ) = O(1) : L 2 (∂X) → H 1 h (∂X)
. Now, using a suitable partition of the unity on ∂X we can write 1 = J j=1 ψ 0 j . Hence, we can write the function χ + δ as J j=1 ψ j , where ψ j = ψ 0 j χ + δ . Since we have (5.8) and (5.

(-∆ ∂X )χ 2 ((-h 2 ∆ ∂X -1)δ -2 ) = -h 2 ∆ ∂X -1 -iθχ((-h 2 ∆ ∂X -1)δ -2 ) ≤ sup σ≥0 √ σ -1 -iθχ((σ -1)δ -2 ) ≤ sup δ 2 ≤|σ-1|≤2δ 2 |σ -1| + |θ| (5.16) ≤ O(δ + |θ| 1/2 ) = O(δ + h ǫ/2 ).
On the other hand, it is well-known that the operator hp(-

∆ ∂X )(1 -χ 2 )((-h 2 ∆ ∂X -1)δ -2
) is an h -ΨDO in the class OPS 1 0 (∂X) with principal symbol ρ(1χ 0 δ ). This implies the bound (5.17

) hp(-∆ ∂X )(1 -χ 2 )((-h 2 ∆ ∂X -1)δ -2 ) -Op h (ρ(1 -χ 0 δ )) = O(h) : L 2 (∂X) → L 2 (∂X).
It is easy to see that Theorem 1.1 follows from (1.3) together with (5.16) and (5.17). 2

6. Proof of 2.1

Define the DN maps N j (λ), j = 1, 2, by

N j (λ)f = ∂ ν u j | Γ
where ν is the Euclidean unit normal to Γ and u j is the solution to the equation (6.1) ∇c j (x)∇ + λ 2 n j (x) u j = 0 in Ω,

u j = f on Γ,
and consider the operator

T (λ) = c 1 N 1 (λ) -c 2 N 2 (λ).
Clearly, λ is a transmission eigenvalue if there exists a non-trivial function f such that T (λ)f = 0. Therefore Theorem 2.1 is a consequence of the following Theorem 6.1. Under the conditions of Theorem 2.1, the operator T (λ) sends

H 1+k 2 (Γ) into H 1-k 2 (Γ), where k = -1 if (2.
2) holds and k = 1 if (2.4) holds. Moreover, there exists a constant C > 0 such that T (λ) is invertible for Re λ ≥ 1 and |Im λ| ≥ C with an inverse satisfying in this region the bound

(6.2) T (λ) -1 H 1-k 2 (Γ)→H 1+k 2 (Γ) |λ| k-1 2
where the Sobolev spaces are equipped with the classical norms.

Proof. We may suppose that λ

∈ Λ ǫ = {λ ∈ C : Re λ ≥ C ǫ ≫ 1, |Im λ| ≤ |λ| ǫ }, 0 < ǫ ≪ 1
, since the case when λ ∈ {Re λ ≥ 1} \ Λ ǫ follows from the analysis in [START_REF] Vodev | Transmission eigenvalue-free regions[END_REF]. We will equip the boundary Γ with the Riemannian metric induced by the Euclidean metric g E in Ω and will denote by r 0 the principal symbol of the Laplace-Beltrami operator -∆ Γ . We would like to apply Theorem 1.2 to the operators N j (λ). However, some modifications must be done comming from the presence of the function c j in the equation (6.1). Indeed, in the definition of the operator N (λ; n) in Section 1 the normal derivative is taken with respect to the Riemannian metric g j = c -1 j g E , while in the definition of the operator N j (λ) it is taken with respect to the metric g E . The first observation to be done is that the glancing region corresponding to the problem (6.1) is defined by Σ j := {(x ′ , ξ ′ ) ∈ T * Γ : r j (x ′ , ξ ′ ) = 1}, where r j := m -1 j r 0 , m j := n j c j | Γ . We define now the cut-off functions χ 0 δ,j by replacing in the definition of χ 0 δ the function r ♯ by r j . Secondly, the function ρ must be replaced by

ρ j (x ′ , ξ ′ ) = r 0 (x ′ , ξ ′ ) -(1 + iθ)m j (x ′ ), Re ρ j < 0.
With these changes the operator N j (λ) satisfies the estimate (1.3). Set

τ δ = c 1 ρ 1 (1 -χ 0 δ,1 ) -c 2 ρ 2 (1 -χ 0 δ,2 ) = τ -c 1 ρ 1 χ 0 δ,1 + c 2 ρ 2 χ 0 δ,2 where
(6.3) τ = c 1 ρ 1 -c 2 ρ 2 = c(x ′ )(c 0 (x ′ )r 0 (x ′ , ξ ′ ) -1 -iθ) c 1 ρ 1 + c 2 ρ 2
where c and c 0 are the restrictions on Γ of the functions

c 1 n 1 -c 2 n 2 and c 2 1 -c 2 2 c 1 n 1 -c 2 n 2
respectively. Clearly, under the conditions of Theorem 2.1, we have c(x ′ ) = 0, ∀x ′ ∈ Γ. Moreover, (2.2) implies c 0 ≡ 0, while (2.4) implies c 0 (x ′ ) < 0, ∀x ′ ∈ Γ. Hence,

0 < C 1 ≤ |c 0 r 0 -1 -iθ| ≤ C 2 , if (2.2) holds, and 0 < C 1 r 0 ≤ |c 0 r 0 -1 -iθ| ≤ C 2 r 0 , if (2
.4) holds. Using this together with (6.3) and the fact that ρ j ∼ -√ r 0 as r 0 → ∞, we get

(6.4) 0 < C ′ 1 ξ ′ k ≤ C 1 r 0 k/2 ≤ |τ | ≤ C 2 r 0 k/2 ≤ C ′ 2 ξ ′ k where k = -1 if (2.2) holds, k = 1 if (2.4) holds. Let η ∈ C ∞ 0 (T * Γ) be such that η = 1 on |ξ ′ | ≤ A, η = 0 on |ξ ′ | ≥ A + 1,
where A ≫ 1 is a big parameter independent of λ and δ.

Taking A big enough we can arrange that (1η)τ δ = (1η)τ . On the other hand, we have ητ δ = ητ + O(δ + |θ| 1/2 ). Therefore, δ and |θ| small enough we get from (6.4) that the function τ δ satisfies the bounds (6.5)

C 1 ξ ′ k ≤ |τ δ | ≤ C 2 ξ ′ k
with positive constants C 1 and C 2 independent of δ and θ. Furthermore, one can easily check that (1η)τ ∈ S k 0 (Γ) and ητ δ ∈ S -2 0 (Γ). Hence, τ δ ∈ S k 0 (Γ), which in turn implies that the operator Op h (τ δ ) sends H 1+k 2 (Γ) into H 1-k 2 (Γ). Moreover, it follows from (6.5) that the operator Op h (τ δ ) :

H 1+k 2 h (Γ) → H 1-k 2 h
(Γ) is invertible with an inverse satisfying the bound

(6.6) Op h (τ δ ) -1 H 1-k 2 h (Γ)→H 1+k 2 h (Γ) ≤ C
with a constant C > 0 independent of λ and δ. We now apply Theorem 2.1 to the operators N j (λ). We get, for λ

∈ Λ ǫ , |Im λ| ≥ C δ ≫ 1, Re λ ≥ C ǫ,δ ≫ 1, that (6.7) hT (λ) -Op h (τ δ ) L 2 (Γ)→L 2 (Γ) ≤ Cδ
in the anisotropic case, and (6.8) hT (λ) -Op h (τ δ ) L 2 (Γ)→H 1 h (Γ) ≤ Cδ in the isotropic case, where C > 0 is a constant independent of λ and δ. Introduce the operators

A 1 (λ) = (hT (λ) -Op h (τ δ )) Op h (τ δ ) -1 , A 2 (λ) = Op h (τ δ ) -1 (hT (λ) -Op h (τ δ )) .
It follows from (6.6), (6.7) and (6.8) that in the anisotropic case we have the bound (6.9)

A 1 (λ) L 2 (Γ)→L 2 (Γ) ≤ C ′ δ
while in the isotropic case we have the bound (6.10)

A 2 (λ) L 2 (Γ)→L 2 (Γ) ≤ C ′ δ
where C ′ > 0 is a constant independent of λ and δ. Hence, taking δ small enough we can arrange that the operators 1 + A j (λ) are invertible on L 2 (Γ) with inverses whose norms are bounded by 2. We now write the operator hT (λ) as

hT (λ) = (1 + A 1 (λ))Op h (τ δ )
in the anisotropic case, and as

hT (λ) = Op h (τ δ )(1 + A 2 (λ))
in the isotropic Therefore, the operator hT (λ) is invertible in the desired region and by (6.6) we get the bound (6.11) (hT (λ)) -1

H 1-k 2 h (Γ)→H 1+k 2 h (Γ) ≤ 2 C.
Passing from semi-classical to classical Sobolev norms one can easily see that (6.11) implies (6.2). 2

7. Proof of Theorem 2.2

We keep the notations from the previous section. Theorem 2.2 is a consequence of the following Proof. As in the previous section we may suppose that λ ∈ Λ ǫ . We will again make use of the identity (6.3) with the difference that under the condition (2.6) we have c 0 (x ′ ) > 0, ∀x ′ ∈ Γ. This means that |τ | can get small near the characteristic variety Σ = {(x ′ , ξ ′ ) ∈ T * Γ : r(x ′ , ξ ′ ) = 1}, where r := c 0 r 0 . Clearly, the assumption (2.7) implies that Σ 1 ∩ Σ 2 = ∅. This in turn implies that Σ ∩ Σ j = ∅, j = 1, 2. Indeed, if we suppose that there is a ζ 0 ∈ Σ ∩ Σ j for j = 1 or j = 2, then it is easy to see that ζ 0 ∈ Σ 1 ∩ Σ 2 , which however is impossible in view of (2.7). Therefore, we can choose a cut-off function χ 0 ∈ C ∞ (T * Γ) such that χ 0 = 1 in a small neighbourhood of Σ, χ 0 = 0 outside another small neighbourhood of Σ, and supp χ 0 ∩ Σ j = ∅, j = 1, 2. This means that supp χ 0 belongs either to the hyperbolic region {r j ≤ 1δ 2 } or to the elliptic region {r j ≥ 1 + δ 2 }, provided δ > 0 is taken small enough. Therefore, we can use Propositions 4.1 and 5.1 to get the estimate hN j (λ)Op h (χ 0 ) -Op h (ρ j χ 0 ) L 2 (Γ)→L 2 (Γ) h + e -C|Im λ| which implies (7.2) hT (λ)Op h (χ 0 ) -Op h (τ χ 0 ) L 2 (Γ)→L 2 (Γ) h + e -C|Im λ| .

It follows from (6.3) that near Σ the function τ is of the form τ = τ 0 (r -1iθ) with some smooth function τ 0 = 0. We now extend τ 0 globally on T * Γ to a function τ 0 ∈ S 0 0 (Γ) such that τ 0 = τ 0 on supp χ 0 and | τ 0 | ≥ Const > 0 on T * Γ. Hence, we can write the operator Op h (τ χ 0 ) as follows Op h (τ χ 0 ) = Op h (χ 0 )Op h ( τ 0 )(Biθ) + O(h) where B = 1 2 Op h (r -1) + 1 2 Op h (r -1) * is a self-adjoint operator. Hence (Biθ) -1 = O(|θ| -1 ) : L 2 (Γ) → L 2 (Γ).

Since τ 0 is globally elliptic, we also have As in the proof of (6.5) can see that the function τ δ satisfies (7.5)

C 1 ξ ′ ≤ |τ δ | ≤ C 2 ξ ′ on supp (1 -χ 0 )
with positive constants C 1 and C 2 independent of δ and θ. Moreover, τ δ ∈ S 1 0 (Γ). We extend the function τ δ on the whole T * Γ to a function τ δ ∈ S 1 0 (Γ) such that τ δ (1χ 0 ) = τ δ (1χ 0 ) and (7. It follows from (7.9) that if δ is taken small enough, the operator hT (λ) is invertible with an inverse satisfying the bound (7.10) (hT (λ)

) -1 L 2 (Γ)→L 2 (Γ) ≤ 2 K 1 L 2 (Γ)→L 2 (Γ) + 2 K 2 L 2 (Γ)→L 2 (Γ) |θ| -1 + 1.
It is easy to see that (7.10) implies (7.1).

2

( 1

 1 .1) ∆ X + λ 2 n(x) u = 0 in X, u = f on ∂X, where λ ∈ C, 1 ≪ |Im λ| ≪ Re λ and n ∈ C ∞ (X)is a strictly positve function. Then the Dirichlet-to-Neumann (DN) map N (λ; n) : H m+1 (∂X) → H m (∂X)

F 2 .

 2 By (3.19) and (3.20) together with Lemma 3.1 we get

2

 2 14) with ψ replaced by each ψ j , we get (5.1) by summing up all the estimates. It follows from the estimate (3.11) applied with V ≡ 0 that (5.15) hN (λ; n)Op h (χ 0 δ ) = O(δ) : L 2 (∂X) → H 1 h (∂X) provided |Im λ| ≥ δ -4 and Re λ ≥ C δ ≫ 1. Now Theorem 1.2 follows from (5.15) and Propositions 4.1 and 5.1. Let us now see that Theorem 1.1 follows from Theorem 1.2. Since the operator -h 2 ∆ ∂X ≥ 0 is self-adjoint, we have the bound hp

Theorem 7 . 1 . 1 L 2 (

 7112 Under the conditions of Theorem 2.2, there exists a constant C > 0 such that the operator T (λ) : H 1 (Γ) → L 2 (Γ) is invertible for Re ≥ 1 and |Im λ| ≥ C log(Re λ + 1) with an inverse satisfying in this region the bound(7.1) T (λ) -Γ)→L 2 (Γ)1.

Op h ( τ 0 ) - 1 =

 1 O(1) : L 2 (Γ) → L 2 (Γ).This impliesK 1 := Op h (χ 0 )(Biθ) -1 Op h ( τ 0 ) -1 = O(|θ| -1 ) : L 2 (Γ) → L 2 (Γ)and (7.2) leads to the estimatehT (λ)K 1 -Op h (χ 0 ) L 2 (Γ)→L 2 (Γ) |θ| -1 h + e -C|Im λ| (7.3) |Im λ| -1 + Re λ e -C|Im λ| ≤ δ for any 0 < δ ≪ 1, provided |Im λ| ≥ C δ log(Re λ), Re λ ≥ C δ with some constants C δ , C δ > 0.On the other hand, by Theorem 1.2 we have, forλ ∈ Λ ǫ , |Im λ| ≥ C δ ≫ 1, Re λ ≥ C ǫ,δ ≫ 1, (7.4) hT (λ)Op h (1χ 0 ) -Op h (τ δ (1χ 0 )) L 2 (Γ)→L 2 (Γ) ≤ Cδ.

6 ) C ′ 1 ξ

 61 ′ ≤ | τ δ | ≤ C ′ 2 ξ ′ on T * Γ. Hence (7.7) Op h ( τ δ ) -1 L 2 (Γ)→L 2 (Γ)≤ C with a constant C > 0 independent of λ and δ. By (7.4) and (7.7) we obtain(7.8) hT (λ)K 2 -Op h (1χ 0 ) L 2 (Γ)→L 2 (Γ) ≤ Cδwith a new constant C > 0 independent of λ and δ, whereK 2 := Op h (1χ 0 )Op h ( τ δ ) -1 = O(1) : L 2 (Γ) → L 2 (Γ).By(7.3) and (7.8), (7.9) hT (λ)(K 1 + K 2 ) -1 L 2 (Γ)→L 2 (Γ) ≤ (C + 1)δ.