Principal eigenvalues for periodic parabolic Steklov problems with L&infin; weight function

Godoy, T.; Lami Dozo, E.; Paczka, S.

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Revista de la Unión Matemática Argentina

versão impressa ISSN 0041-6932versão On-line ISSN 1669-9637

Rev. Unión Mat. Argent. v.46 n.2 Bahía Blanca jul./dez. 2005

Principal eigenvalues for periodic parabolic Steklov problems with L^∞ weight function

T. Godoy, E. Lami Dozo and S. Paczka

Partially supported by Fundacion Antorchas, Agencia Cordoba Ciencia, Secyt-UNC, Secyt UBA and Conicet

Abstract: In this paper we give sufficient conditions for the existence of a positive principal eigenvalue for a periodic parabolic Steklov problem with a measurable and essentially bounded weight function. For this principal eigenvalue its uniqueness, simplicity and monotone dependence on the weight are stated. A related maximum principle with weight is also given.

2000 Mathematics Subject Classification. 35K20, 35P05, 35B10, 35B50

1. Introduction

Let be a C2+ θ and bounded domain in with N ≥ 2 and θ ∈ (0,1), let T > 0 and let {aij}1≤i,j≤N , {bj}1≤,j≤N be two families of real functions defined on -- Ω × ℝ and respectively, satisfying for 1 ≤ i,j ≤ N that a = a (x,t) ij ij and b = b (x, t) j j are periodic in aij = aji, ∂aij (-- ) ∂xi ∣[0,T] ∈ C Ω × ℝ and ∞ bj ∈ L (Ω × ℝ) . Let a0 : Ω × ℝ → ℝ be a nonnegative and periodic function belonging to Ls (Ω × ℝ) for some s > 1 + N. 2 Assume in addition that for some γ ∈ (1-,1) 2 and for all i,j

( ( -)) aij ∈ C γ ℝ,C Ω , bj ∈ Cγ (ℝ,L ∞ (Ω))

(1)

and that

(2)

where aij (t) (x) := aij (x,t), bj (t)(x) := bj (x,t) and a0(t)(x) := a0(x,t). Let b = (b1,...,bN) and let be the N × N matrix whose i,j entry is aij. Assume also that is uniformly elliptic on -- Ω × [0,T] , i.e., that there exists a positive constant such that

(3)

for all -- (x,t) ∈ Ω × ℝ , ξ = (ξ1,...,ξN ) ∈ ℝN . Let be the periodic parabolic operator defined by

(4)

where 〈,〉 denotes the standard inner product on N ℝ . Finally, let be a nonnegative and periodic function in ∞ L (∂Ω × ℝ) and let be the unit exterior normal to ∂ Ω. Under the above hypothesis and notations (that we assume from now on) we consider, for a periodic function (that may changes sign) m ∈ L∞ (∂Ω × ℝ) , the periodic parabolic Steklov principal eigenvalue problem with weight function

(5)

the solutions understood in the sense of the definition 2.1 below. In order to describe our results let us introduce, for m ∈ L ∞ (∂Ω × ℝ) , the quantities

∫ T ∫ T P (m) := ess sup m (x,t)dt, N (m) := ess inf m (x,t)dt 0 x∈∂Ω 0 x∈∂Ω

(6)

In this paper we prove (cf. Theorem 6.1) that if either a0 > 0 and b0 ≥ 0 or a0 = 0 and b > 0 0 and if P (m) > 0 (respectively N (m) < 0 ) then there exists a positive (resp. negative) principal eigenvalue for (5), that is, a whose associated eigenfunction satisfies (5). Under an additional assumption on a similar existence result is also given for the case a0 = 0, b0 = 0 .

Our approach, adapted from [4] and [8], reads as follows: If we change λmu in (5) by λmu + μu, we have the following one parameter eigenvalue problem: given λ ∈ ℝ find μ ∈ ℝ such that this modified (5) has a solution. We prove in section 4 that this problem has a unique solution μ = μm (λ) ∈ ℝ which satisfies that λ → μm (λ) is real analytic and concave. We also obtain an expression for ′ μm (0) which allows us to decide the sign of μ ′m (0) . In section 5 we prove that P (m) > 0 (respectively N (m) < 0 ) implie lim λ→∞ μm (λ) = - ∞ (resp. lim λ→- ∞ μm (λ) = - ∞ ). From these facts, and since the zeroes of the function are exactly the principal eigenvalues for (5), our results will follow.

Sections 2 and 3 have a preliminar character. In section 2 we collect some general facts about initial value parabolic problems and in section 3 we study existence and uniqueness of periodic solutions for parabolic problems and we prove some compactness and positivity properties of the corresponding solutions operators related.

2. Preliminaries

Let us start introducing the notations to be used along the paper. For a topological vector space we put for its topological dual and 〈,〉E*,E for the corresponding evaluation bilinear map 〈Λ,e〉 * = Λ (e). E ,E If E , 1 E 2 are normed spaces and if S : E → E 1 2 is a bounded linear map we denote by ∥S ∥E1,E2 (or simply by if no confusion arises) its corresponding operator norm. If is a real Banach, - ∞ ≤ t0 < t1 ≤ ∞ and 1 ≤ p < ∞ we put Lp(t0,t1;E) for the space of the measurable functions (in the Bochner sense) f : (t0,t1) → E such that (∫ t1 p )1p ∥f∥Lp(t0,t1;E) := t0 ∥f (t)∥E dt < ∞. We define also L ∞ (t0,t1;E) and, for 1 ≤ p ≤ ∞, the space Lp (t0,t1;E) loc , similarly (with the obvious changes) to the corresponding usual Lebesgue's spaces. For we put p L T (ℝ, E) for the space of the periodic functions f : ℝ →E satisfying that f∣(0,T) ∈ Lp (0,T ;E) . We write also (-- ) CT Ω × ℝ (respectively CT (∂ Ω × ℝ) ) for the space of the periodic functions belonging to (-- ) C Ω × ℝ (resp. to ). The spaces p L (t0,t1;E) , p L T (ℝ, E) , and CT (∂Ω × ℝ) , equipped with their respective norms ∥∥Lp(t0,t1;E), ∥∥Lp(0,T ;E), -- ∥∥C(Ω)×[0,T] and ∥∥C(∂Ω)×[0,T] are Banach spaces. For t0 < t1 we will identify (writing f (x, t) = f (t)(x) ) the spaces

and also the corresponding spaces of functions defined on ∂ Ω × (t0,t1)

Let X,V be the real Hilbert spaces X = L2 (Ω) , V = H1 (Ω) equipped with their usual norms. For t0 < t1 let D = C∞c (t0,t1;V) be the space of the indefinitely differentiable Frechet functions from (t0,t1) into , equipped with the topology of the uniform convergence on each compact subset of (t ,t) 0 1 of the function and all its derivatives. Let ′ D be its dual space. For 1 u ∈ L loc(t0,t1;V) , let ′ u be its distributional derivative defined by ′ ∫t1 〈u,φ 〉D′,D = - t0 〈u (t) ,φt(t)〉X dt for all φ ∈ D where 〈,〉X denotes the inner product in We will say that u′ ∈ L2 (t0,t1;V *) if there exists a function (denoted by t → u′(t) ) belonging to such that ′ ∫t1 ′ 〈u,φ 〉D′,D = t0 〈u (t),φ (t)〉V*,V dt for all φ ∈ D.

For t ∈ ℝ, let aL,b0 (t,.,.) : V × V → ℝ be the bilinear form defined by

∫ aL,b0 (t,g, h) = ∫ [〈A (.,t)∇g, ∇h 〉 + 〈b(.,t),∇g 〉h + a0 (.,t)gh] + b0(.,t) gh Ω ∂Ω

(7)

(the values on

and

understood in the trace sense) and let AL,b (t) : V → V * 0

be the bounded linear operator defined by

(8)

For t0 < t1, f ∈ L2 (Ω × (t0,t1)), Φ ∈ L2(∂ Ω × (t0,t1)) and t ∈ (t0,t1), let Λf,Φ (t) ∈ V * be defined by

∫ ∫ 〈Λf,Φ (t),h〉V*,V = f (.,t)h + Φ (.,t)h, h ∈ V. Ω ∂Ω

(9)

So Λf,Φ ∈ L2 (t0,t1;V *) and

( ) ∥Λf,Φ∥L2(t0,t1;V *) ≤ c ∥f ∥L2(Ω×(t0,t1)) + ∥Φ∥L2(,∂Ω ×(t0,t1))

(10)

for some positive constant depending only on t ,t , 0 1 and We set also

W := {u ∈ L2 (t ,t;V ) : u′ ∈ L2 (t ,t ;V*)} t0,t1 0 1 0 1

(11)

and ′ ∥u∥Wt0,t1 := ∥u∥L2(t0,t1;V) + ∥u∥L2(t0,t1;V*). So Wt0,t1, equipped with the norm ∥.∥W , t0,t1 is a Banach space. With these notations we can formulate the following definition

Definition 2.1. For - t0 < , 2 f ∈ L (Ω × (t0,t1)) and 2 Φ ∈ L (∂Ω × (t0,t1)) we say that u : Ω × (t0,t1) → ℝ is a solution of the problem

(12)

if u ∈ Wt0,t1 and u′(t) + AL,b0 (t)u (t) = Λf,Φ (t) a.e. t ∈ (t0,t1).

¿From now on, a solution of a boundary problem like (12) (except if otherwise is explicitely stated) will mean a solution in the above sense.

Remark 2.2. For t ∈ ℝ with k > 0, standard computations on the quadratic form g → aL+k,l(t,g,g) give, for all g ∈ V,

( ) ∥ ∣b∣∥2∞ ∫ aL+k,l(t,g,g) ≥ k - -----L--(Ω×-ℝ) ∥g∥2X + l g2 4α ∂Ω

and also

( ) ∥∣b∣∥2∞ ∫ aL+k,l(t,g,g) ≥ α - ----L--(Ω-×ℝ) ∥∇g ∥2X + l g2 4k ∂Ω

where is the ellipticity constant of So, for ∥∣b∣∥2∞ k > k0 := ---L4α(Ω×ℝ) and l ≥ 0, there exists a positive constant depending only on and ∥ ∣b∣∥L∞(Ω× ℝ) such that

(13)

for all t ∈ ℝ and g ∈ V. Moreover, for such and the assumptions on the coefficients of imply that there exists a positive constant such that

(14)

and that

∣aL+k,l(t,g, h) - aL+k,l(s,g,h) ∣ ≤ c ∣t - s∣γ ∥g∥ ∥h ∥ V V

(15)

for all s,t ∈ ℝ and g,h ∈ V.■

For as in Remark 2.2, k ≥ k0, - ∞ < τ < t < ∞ and u0 ∈ X consider the problem

(16)

Note that W τ,t ⊂ C ([τ,t],X) (cf. ([12], Lemma 5.5.1) and so the initial condition u (τ) = u0 makes sense. Taking into account the facts in Remark 2.2, ( [12], Theorem 5.5.1) applies to see that (16) has a unique solution Let UL+k,l(t,τ) : X → X be the linear operator defined by UL+k,l(t,τ)u0 = u(t).

Let us recall the following properties (cf. [12], Theorem 5.4.1) of the evolution operators UL+k,l(t,τ)

Remark 2.3. i) Given t0,t1 ∈ ℝ with t0 < t1 there exists a positive constant such that, for t0 < τ < t ≤ t1,

(17)

ii) Since * * V ⊂ X ≃ X ⊂ V (the isomorphism * X ≃ X given by duality) we can consider * X ⊂ V . In this setting, it holds that for t0,t1 as above there exists a positive constant ′ c such that

1 ∥UL+k,l(t,τ)u0 ∥X ≤ c′(t - τ)-2 ∥u0∥V*

(18)

for t < τ < t ≤ t 0 1 and u ∈ X. 0 Since (and then also ) is dense in V *, it follows that UL+k,l(t,τ) : X → V has a unique bounded extension to an operator (still denoted ) from * V into which satisfies, for ′ c as in (18),

(19)

Finally, we recall also that for τ ≤ s ≤ t it holds that

UL+k,l (t,τ ) = UL+k,l (t,s) UL+k,l(s,τ).

(20)

For - ∞ < t0 < t1 < ∞ , Λ ∈ L2 (t0,t1;V *) and u0 ∈ X consider the problem

(21)

Taking into account (13), (14) and (15), ([12], Theorem 5.5.1) applies to see that (21) has a unique solution given by

∫ t vk (t) = UL+k,l (t,t0)u0 + UL+k,l(t,τ )Λ (τ)dτ.■ t0

(22)

Remark 2.4. Observe that u ∈ Wt0,t1 is a solution of the problem

(23)

if and only if vk (t) := e-k(t-t0)u (t) solves

(24)

with defined by Λk (t) := e- k(t-t0)Λ (t) . Thus (23) has a unique solution given by

∫ t u (t) = U (t,t )u + U (t,τ) Λ(τ )dτ L,l 0 0 t0 L,l

(25)

with UL,l(t,τ) defined by

(26)

Moreover, for t ∈ [t ,t ] 0 1 we have (cf. [12], Lemma 5.5.2)

(27)

¿From (27), standard computations show that there exists a positive constant independent of and such that

( ) ∥u∥Wt ,t ≤ c ∥Λ ∥L2(t ,t,V*) + ∥u0∥L2(Ω) .■ 0 1 0 1

(28)

Remark 2.5. The estimates (17), (18), (19) and (20) still hold (with another constants) for the operators UL,l(t,τ) given by (26) and u(t) := UL.l(t,τ )u0 satisfies

(29)

for 2 u0 ∈ L (Ω).■

Remark 2.6. For l ≥ 0, - ∞ < t0 < t1 < ∞, f ∈ L2 (Ω × (t0,t1)), Φ ∈ L2 (∂Ω × (t0,t1)) and u0 ∈ L2 (Ω) the problem

(30)

has a unique solution which satisfies in addition that

( ) ∥u ∥Wt0,t1 ≤ c ∥f ∥L2(Ω×(t0,t1)) + ∥Φ ∥L2(∂Ω ×(t0,t1)) + ∥u0∥L2(Ω) .

(31)

for some positive constant independent of and u0. Indeed, the solutions of (30) are those of (23) taking there Λ = Λ , f,Φ and Remark 2.4 applies.

Remark 2.7. It is easy to check that the constant in (28) and so also in Remark 2.5 and Remark 2.6 can be chosen depending only on and on an upper bound of Σi,j ∥aij∥L∞(Ω×(t0,t1)) + Σj ∥bj∥L ∞(Ω×(t0,t1)) + ∥a0∥Ls(Ω×(t0,t1)).■

Lemma 2.8. Let and be as in Lemma 2.4 and let be a sequence of operators of the form

with and satisfying for each the conditions stated for at the introduction with the same and given there for Assume also that for each and and converge uniformly on to and respectively and that converges to in . Let and be sequences in and in respectively and assume that they converge to and in their respective spaces. Let be a sequence in that converges to in and let Thus the solution of the problem

converges in the norm to the solution of (30).

Proof. For as in Remark 2.2, k ≥ k0 , l ≥ 0 and n ∈ ℕ , let v(n)∈ Wt ,t k 0 1 be the solution of pict

and let v k be the solution of (24). We have

(32)

where

(33)

Our assumptions imply that ∥ ∥ limn →∞ ∥∥ Λf(n),Φ(n)- Λfk,Φk∥∥ = 0 k k L2(t0,t1;V*) and that ∥ ∥ limn →∞ ∥AL+k,l (t) - AL(n)+k,l(t)∥V.V * = 0 uniformly on t ∈ [t0,t1]. From Remarks 2.6 and 2.7 we have that { } ∥∥ (n)∥∥ ∥vk ∥ 2 L (t0,t1;V) is a bounded sequence. Then from (33) ∥∥ ∥∥ limn →∞ ∥ ^Λ(n)∥ = 0. L2(t0,t1;V*) Thus from Remark 2.6 applied to (32) we obtain ∥ ∥ limn →∞ ∥∥v(nk) - vk∥∥ = 0. Wt0,t1 Since u(n)(t) = ek(t-t0)v(kn) and u (t) = ek(t- t0)vk the lemma follows.

Lemma 2.9. Assume that and are nonnegative. Then the solution of (30) is nonnegative.

Proof. We pick sequences {L }, n { f(n)} , { Φ(n)} and { } u(n) 0 as in Lemma 2.8 satisfying in addition that f (n) ≥ 0, Φ(n) ≥ 0, u(0n)≥ 0 and such that (n) aij , (n) bj , (n) a0 and (n) f belong to ∞ (-- ) C Ω × [t0,t1] , (n) Φ belongs to C ∞ (∂Ω × [t0,t1]) and u(0n)∈ C ∞c (Ω) . Let { } v(nk) be as in the proof of Lemma 2.8. Thus (n) σ vk ∈ C2+σ,1+2 (Ω × (t0,t1)) (cf. e.g., Theorem 5.3 in [9], p. 320)). The classical maximum principle gives v(n)≥ 0 k and since by Lemma 2.8 limn →∞ v(n)= vk k in L2(Ω × (t0,t1)) we get v ≥ 0. k Since the solution of (30) is given by u (t) = ektv (t) k the lemma follows

Remark 2.10. Let us recall some well known facts concerning Sobolev spaces (see e.g. [9], Lemma 3.3, p 80 Lemma 3.4, p. 82)

i): For - ∞ < t0 < t1 < ∞ and u ∈ W 2q,1(Ω × (t0,t1)) with 1 ≤ q < ∞ we have 2- 1q,1-21q u ∣∂Ω×(t0,t1) ∈ W q (∂ Ω × (t0,t1)) and the restriction map (in the trace sense) u → u∣∂Ω ×(t0,t1) is continuous from 2,1 W q (Ω × (t0,t1)) into 2- 1q,1- 12q W q (∂ Ω× (t0,t1)).

ii) For u ∈ W 2q,1(Ω × (t0,t1)) with 1 ≤ q < ∞ it holds that 2 u(.,t) ∈ W 2-q,q (Ω) for t ∈ [t ,t ] 0 1 and for such there exists a positive constant independent of such that ∥u(.,t)∥W2- 1q,q(Ω) ≤ c ∥u∥W 2q,1(Ω ×(t0,t1)).

iii) For q > N + 2 the following facts hold:

W 2,1 (Ω × (t,t )) ⊂ C1+ σ,1+2σ (Ω-× [t,t ]) q 0 1 0 1 for some σ ∈ (0,1) , with continuous inclusion.

2- 1,1- 1 1+σ W q q 2q(∂Ω × (t0,t1)) ⊂ C1+ σ, 2 (∂ Ω × [t0,t1]) for some σ ∈ (0,1) and with continuous inclusion.

iv) For 1 ≤ r ≤ ∞ let be defined by (r*)-1 = r-1 - (N + 1)-1 if r < N + 1 and * r = ∞ if r ≥ N + 1. Thus 2,1 r* W r (Ω × (t0,t1)) ⊂ L (Ω × (t0,t1)) if * r < ∞ and 2,1 q W r (Ω × (t0,t1)) ⊂ L (Ω × (t0,t1)) for all q ∈ [1,∞) if * r = ∞, in both cases with continuous inclusion

Remark 2.11. For q > N + 2 it holds that 2 (-) W 2-q,q (Ω) ⊂ C1+ σ Ω continuously for some σ ∈ (0, 1). In this case, for τ ∈ ℝ , let 2- 2q,q W Bl(τ) (Ω) be the space of the functions h ∈ W 2- 2q,q (Ω) that satisfy (in the pointwise sense) B (τ )h = 0 l where

(34)

Let us recall that for such and for - ∞ < t0 < t1 < ∞, f ∈ Lq (Ω × (t0,t1)), 2- 1q,1- 12q Φ ∈ W q (∂Ω × (t0,t1)) and 2- 2q,q u0 ∈ W Bl(t0) (Ω) there exists a unique 2,1 u ∈ W q (Ω × (t0,t1)) satisfying almost everywhere

(for a proof, see [9], Theorem 9.1, p. 341, concerning the Dirichlet problem and its extension, to our boundary conditions, indicated there (at the end of chapter 4, paragraph 9, p. 351). Moreover, there exists a positive constant independent of f,Φ and such that

Lemma 2.12. i) For is a compact and positive operator.

ii) Let with For and the restriction of to belongs to and there exists a positive constant such that for all

iii) for and and is a bounded operator from into

iv) For it hold that and is a bounded operator from into Moreover, if and then

v) For and is a compact and strongly positive operator

Proof. By Lemma 2.9 2 2 UL,l(t,τ) : L (Ω) → L (Ω) is a positive operator. It is also compact because 2 1 UL,l(t,τ) : L (Ω) → H (Ω) is continuous (cf. Remark 2.5) and 1 H (Ω) has compact inclusion in L2 (Ω) . Thus (i) holds.

To see (ii) we pick a strictly increasing sequence of positive numbers { ηj}j∈ℕ such that t < t + η < τ 0 0 j for all j ∈ ℕ and we pick also a sequence of functions {φ } j j∈ℕ in ∞ C (ℝ) satisfying φj (s) = 0 for s ≤ t0 + ηj, φj (s) = 1 for s ≥ t0 + ηj+1. Let u (t) := UL+k,l(t,t0)u0 and let {vj}j∈ℕ and {wj}j ∈ℕ be the sequences of functions on Ω × (t0,,t1) inductively defined by v1 := u φ1, vj+1 := φj+1vj and by w1 := φ′1u, wj+1 =: φ ′ vj + φj+1wj j+1 respectively Then, for all

(35)

Let {qj} j∈ℕ be defined by q1 = 2 and by qj+1 = q * j (with q* j as in (iv) of Remark 2.10) and let { * } j0 = min j : qj = ∞ . For the rest of the proof will denote a positive constant independent of non necessarily the same at each occurrence (even in a same chain of inequalities). We claim that for j ≤ j0

2,1 2,1 vj ∈ W qj (Ω × (t0 + ηj+1,t1)) and wj ∈ W qj (Ω × (t0 + ηj+1,t1))

(36)

with their respective norms bounded by c∥u0 ∥L2(Ω) .

If (36) holds, for 1 ≤ q < ∞ Remark 2.10 (iv) gives ∥w ∥ ≤ c∥u ∥ . j0 Lq(Ω×(t0+ηj0+1,t1)) 0 L2(Ω) Taking into account that u = vj0 on Ω × (τ,t1), Remark 2.11 gives

and so (ii) holds.

To prove the claim we proceed inductively. Since satisfies 29, Remark 2.6 gives ∥u∥L2(Ω×(t0+η1,t1)) ≤ ∥u ∥L2(Ω ×(t0,t1)) ≤ c∥u0 ∥L2(Ω) and so ∥w1∥L2(Ω×(t0+ η1,t1)) ≤ c∥u0∥L2(Ω). Then, by Remark 2.11, ∥v1∥W 2,1(Ω×(t+ η,t)) ≤ c ∥u0∥L2(Ω) 2 0 1 1 and so ∥v1∥W 2,1(Ω×(t+η ,t )) ≤ c ∥u0∥L2(Ω). 2 0 2 1 Since u = v 1 on Ω × (t + η ,t ) 0 2 1 and w = uφ 1 1 we get also that ∥w ∥ 2,1 ≤ c∥u ∥ . 1 W 2 (Ω ×(t0+η2,t1)) 0 L2(Ω) Thus (36) holds for j = 1. Suppose that it holds for some j < j0. Then

and so (since u = v j+1 on Ω × (t + η ,t ) 0 j+3 1 )

(37)

Since ∑ ∏ wj+1 = u φ′k φr 1≤k≤j+1 1≤r≤j+1 r⁄=j+1 it follows that ∥wj+1 ∥W 2,q1 (Ω ×(t0+ηj+2,t1)) ≤ c∥u0∥L2(Ω) j+1 and so, from (35), a similar estimate holds for vj+1. This complete the proof of the claim.

The imbedding theorems for Sobolev spaces and (ii) imply (iii). The first part of (iv) is again obtained applying (ii) with q > N + 2. To see the second part of (iv), we observe that if u0 > 0 and u := UL,l(t,τ)u0 then u ⁄= 0 and, by Lemma 2.9, u ≥ 0. Let and be as in the proof of (ii), Since 2,1 1+σ,1+σ v1 = φ1u ∈ W q (Ω × (t0,t1)) ⊂ C 2 (Ω × [t0,t1]) , the boundary condition for holds in the pointwise sense. Now, the Hopf parabolic maximum principle applied to

jointly with the fact that v1 = u on Ω × (τ,t1) gives (iv).

To see (v), let s ∈ (0,τ), q > N + 2 and let ^q > q. Since 2-2∕q,q 2 W Bl(τ) (Ω) ⊂ L (Ω) (with Bl (τ) given by (34)), from (ii) we can consider the bounded operator S : W 2-2∕q,q(Ω) → W 2,1(Ω × (τ,t)) Bl(τ) ^q defined by Su0 = (UL,l(.,s)u0) . ∣Ω×(τ,T) Since the operator u → u (t) is continuous from 2,1 W q^ (Ω × (τ, t)) into 2-2∕^q,^q W (Ω) and the inclusion map i : W 2- 2∕^q,^q (Ω) → W 2-2∕q,q (Ω) is compact, we obtain the compactness assertion of (v). Finally, the strong positivity in (v) follows from (iv).

Lemma 2.13. i) If and then for

ii) If and are nonnegative functions and if either or then

Proof. Let P 2 , L (Ω) P 1 H (Ω) , P 1 * H (Ω) be the positive cones in L2 (Ω) , H1 (Ω) and respectively and let -- P H1(Ω) be the closure of PH1(Ω) in 1 * H (Ω) . Observe that if Λ ∈ PH1( Ω)* ∪ {0} then -- Λ ∈ P H1(Ω) . Indeed, if not, the Hann Banach Theorem gives η ∈ H1 (Ω) ** such that η-- = 0 ∣P H1(Ω) and η(Λ) = 1. For 1 g ∈ H (Ω) let 1 * λg ∈ H (Ω) be defined by ∫ λg (f) = Ω fg. Thus λg ∈ PH1(Ω)* for all g ∈ PH1(Ω). Since H1 (Ω) is reflexive there exists φ ∈ H1 (Ω) such that η (λ) = λ (φ) for all λ ∈ H1 (Ω)* . In particular we have ∫ 0 = η (λg) = f g Ω for all g ∈ P 1 . H (Ω) This implies that φ = 0 and so η = 0 which contradicts η (Λ) = 1. Thus -- Λ ∈ PH1(Ω).

Let Λ ∈ PH1(Ω)*, so -- Λ ∈ P H1(Ω) and then there exists a sequence {u0,j}j∈N of nonnegative functions in H1 (Ω) that converges to in H1 (Ω)* . Since 1 * 2 UL,l(t,τ) : H (Ω) → L (Ω) is continuous and, by Lemma 2.12 (i), it is a positive operator on 2 L (Ω) , we have UL,l(t,τ) Λ = limj → ∞ UL,l(t,τ )u0,j ≥ 0 and so (i) holds.

To see (ii), observe that Λf,Φ ≥ 0 and so (i) gives

(38)

Moreover,

(39)

is the solution of the problem

Then, by (i), u ≥ 0 in Ω × (t0,t1) and since u ⁄= 0 (because either f ⁄= 0 or Φ ⁄= 0 ) we conclude that for some - t ∈ (t0,t1) the set

- { ( -) (- ) } Jt = τ ∈ 0,t : UL,l t,τ Λf,Φ(τ) ∈ PL2(Ω)

has positive measure. Then, since ( -) (- ) UL,l(T,τ) = UL,l T, t UL,l t,τ , Lemma 2.12 (iv) gives UL,l(T, τ)Λf,Φ (τ ) > 0 for all τ ∈ Jt. Now (ii) follows from (38) and (39).

Remark 2.14. Let us recall the following version of the Krein Rutman Theorem for Banach lattices and one of its corollaries (for a proof, see e.g., [5], Theorem 12.3 and Corollary 12.4)

i) Let be a Banach lattice with cone positive and let S : E → E be a bounded, compact, positive and irreducible linear operator. Then has a positive spectral radius ρ (S) which is an algebraically simple eigenvalue of and S *. The associated eigenspaces are spanned by a quasi interior eigenvector and a strictly positive eigenfunctional respectively. Moreover, ρ(S) is the only eigenvalue of having a positive eigenvector.

ii) For and as above and for a positive v ∈ E the equation ru - Su = v has a unique positive solution if r > ρ (S), no positive solution if r < ρ(S) and no solution at all if r = ρ (S) . In particular this implies that if Sv ≥ ρ (S) v for some positive then Sv = ρ(Sv) .

We recall also that a point a ∈ E is a quasi interior point if and only if a ∈ P and the order interval [0,a] is total (i.e. the linear span of [0,a] is dense in ) and that for a measure space equipped with a positive measure on and 1 ≤ p < ∞ the quasi interior points in Lp (Z, dσ) are the functions that are strictly positive almost everywhere. Moreover, for such a bounded and positive linear operator S : Lp (Z, dσ) → Lp (Z, dσ) satisfying that S (f) (x) > 0 a.e. x ∈ Z for all f > 0 is an irreducible operator (cf [13], Proposition 3, p. 409).

Lemma 2.15. For and is a positive irreducible operator and its spectral radius satisfies

Proof. By (i) and (iv) of Lemma 2.12, UL,l(t,τ) is a positive, irreducible and compact operator. Thus, by the Krein Rutman Theorem, is positive and that is the unique eigenvalue with positive eigenfunctions associated. Moreover, by Lemma 2.10 (iii), these eigenfunctions belong to 2- 2,q W q (Ω) for 1 ≤ q < ∞ . Take q > N + 2. By Lemma 2.12 (v), 2 2 U (t,τ) : W 2- q,q (Ω) → W 2-q,q(Ω) L,l Bl(τ) Bl(τ) is a compact and strongly positive operator which, by the Krein Rutman Theorem, has a positive spectral radius ρq. Since the eigenfunctions of UL,l(t,τ) belong to 2 W 2-q,q(Ω) Bl(τ) we have ρ = ρq. Thus, to prove the lemma, it is enough to see that ρq < 1.

We proceed by contradiction. Suppose ρq ≥ 1 , let be a positive eigenfunction with eigenvalue and let w = UL,l(.,τ)(φ) . Since UL,l(t,τ)(φ) = ρφ ≥ φ, By Lemma 2.12 (ii), w ∈ W 2q,1(Ω × (τ,t)) and since w (t) ≥ w (τ ) the maximum principle gives that either is a constant or max -- w (x, t) Ω ×[δ,T ] is achieved at some point (x*,t*) ∈ ∂Ω × (τ,t). If is a constant, since l > 0 the boundary condition (which is satisfied in the pointwise sense because q > N + 2 ) implies w = 0 which is impossible and if the maximum is achieved at some point (x*,t*) ∈ ∂ Ω × (τ,t) we would have 〈A ∇w, ν 〉(x*,t*) > 0 in contradiction with the boundary condition.

3. Periodic solutions

Let be the Banach space

{ ( ) ( )} W := u ∈ L2T ℝ, H1 (Ω) : u ′ ∈ L2T ℝ, H1 (Ω) *

(40)

with norm ∥u∥ = ∥u∥ 2 + ∥u ′∥ 2 * . W LT(ℝ,H1(Ω)) LT(ℝ,H1(Ω) )

Lemma 3.1. For and the problem

Lu = f in Ω × ℝ 〈A ∇u, ν〉 + lu = Φ on ∂ Ω × ℝ, u (x,t) T periodic in t

(41)

has a unique solution .

Proof. Let δ > 0. For 2 u0 ∈ L (Ω) the solution of

Lu = f in Ω × (0,T + δ) 〈A ∇u, ν〉 + lu = Φ on ∂Ω × (0,T + δ) , u (0) = u0

(42)

is given by

∫ t u(t) = UL,l(t,t0)u0 + UL,l(t,τ )Λf,Φ(τ )dτ. t0

(43)

By Lemma 2.15, I - UL,l(T,0) : L2 (Ω) → L2 (Ω) has a bounded inverse. From (25), u (0) = u(T ) if and only if

∫ -1 T u0 = (I - UL,l(T,0)) UL,l (T, τ)Λf,Φ (τ)dτ 0

(44)

then there exists a unique solution of Lu = f in Ω × (0,T + δ) , 〈A ∇u, ν〉 + lu = Φ on ∂ Ω × (0,T + δ) and u (0) = u(T ) . For such a and for t ∈ [0,T + δ], let v (t) = u (t + T ). Thus Lv = f in Ω × (0,δ), 〈A ∇v, ν〉 + lv = Φ on ∂Ω × (0, δ) and v (0) = u(0) . Then v (t) = u (t) (i.e., u(t + T) = u (t) ) for [0,T + δ]. Thus can be extended to a solution of (41) which is unique by (44).

Let tr : H1 (Ω) → L2(∂ Ω) be the trace operator on H1 (Ω) and for v ∈ W let T r(v) ∈ L2T (∂Ω × ℝ) be the trace operator defined by Tr (v)(t) = tr(v(t)).

For l > 0 we define the linear operators

l S1(f,Φ) = u where is the solution of (41) given by Lemma 3.1,

l ( l ) S2(f,Φ) = Tr S1 (f, Φ) ,

Sl(Φ) = Sl2(0,Φ)

respectively.

Remark 3.2. Let and be Banach spaces, and reflexive. let i : B → B 0 be a compact and linear map and j : B → B 1 an injective bounded linear operator. For finite and 1 < pi < ∞, i = 0,1

{ } p0 d- p1 W := v ∈ L (0,T;B0) :dt (j ∘ i ∘ v) ∈ L (0,T ;B1)

is a Banach space under the norm ∥ ∥ ∥v∥Lp0(0,T;B0) + ∥ ddt-(j ∘ i ∘ v)∥Lp1(0,T;B ). 1 A variant of an Aubin-Lions ´s theorem (for a proof see [10], p. 57 or Lemma 3 in [6]) asserts that if V ⊂ W is bounded then the set {i ∘ v : v ∈ V} is precompact in p L 0 (0,T;B) .

We will apply this result to B = L2 (∂ Ω) , B0 = H1 (Ω) and * B1 = H1 (Ω) . The map is the trace map, j : L2 (∂Ω) → H1 (Ω)* is defined by

∫ 〈j (g) ,h〉H1(Ω)*,H1(Ω) = tr (h)g, g ∈ L2 (∂ Ω) ∂Ω

and p = p = 2. 0 1 Hence above is a special case of in (11) for (t ,t ) = (0,T ) 0 1 which is naturally isometric to the space of (40).

Lemma 3.3. i) For and are bounded linear operators and is also compact

ii) If and are nonnegative and if either or then and Moreover, if then

iii) is a bounded, positive, irreducible and compact operator on

Proof. For 2 f ∈ L T (Ω × ℝ) and 2 Φ ∈ L T (∂ Ω × ℝ) the periodic solution of (42) is given by (43) with given by (44). Remark 2.6 gives

( ) ∥u∥W ≤ c ∥f∥L2T(Ω ×ℝ) + ∥ Φ∥L2T (∂Ω× ℝ) + ∥u0 ∥L2(Ω) .

So, to see that l S1 is a bounded operator, it is enough to obtain see that

( ) ∥u0 ∥L2(Ω) ≤ c ∥f ∥L2T(Ω×ℝ) + ∥Φ∥L2T(∂Ω× ℝ)

(45)

(for the rest of the proof will denote a positive constant independent of and non necessarily the same at each occurrence, even in a same chain of inequalities). Let v (t) := ∫t U (t,τ )Λ (τ ). 0 L+k,l f,Φ Thus solves (L + k)v = f in Ω × (0,T ), 〈A∇v, ν 〉 + lv = Φ on ∂Ω × (0,T) and v(0) = 0. Since

( ) ∥Λf,Φ∥ 2 1 * ≤ c ∥f∥ 2 + ∥Φ ∥ 2 L (0,T,H (Ω) ) LT(Ω× ℝ) LT(∂Ω×ℝ)

(27) (applied to this problem and used with t = 0 0 and t = T ) gives

the last inequality by Remark 2.6. So

( ) ∥v(T )∥ ≤ c ∥f∥ + ∥ Φ∥ . L2(Ω) L2T(Ω ×ℝ) L2T (∂Ω× ℝ)

Now,

and so

∥∥∫ T ∥∥ ( ) ∥∥ UL,l(T,τ )Λf,Φ(τ )dτ∥∥ ≤ c ∥f∥L2(Ω× ℝ) + ∥Φ ∥L2(∂Ω×ℝ) . 0 L2(Ω) T T

(46)

By Lemma 2.5, I - U (t,τ) : L2(Ω) → L2 (Ω) L,l has a bounded inverse, and so (44) and (46) give (45). Then l S1 is bounded and this implies the boundedness, first of l S2, and then of l S .

To see that Sl2 and are compact, we consider a bounded sequence {(fn,Φn)} ⊂ L2T (ℝ;L2 (Ω)) × L2T (ℝ; L2 (∂ Ω)) . Then, from Remark 3.2 { } Sl2(fn,Φn) is bounded in so {T r(Sl (f ,Φ ))} 1 n n has a convergent subsequence in L2 (ℝ; L2 (∂Ω)) . T From l l S (Φ) = S2 (0,Φ) we have that l S is also compact.

Suppose now that either f > 0 or Φ > 0 and let be given by (44). For δ > 0. Lemma 2.13 (iv) gives essinf UL,l(t,0)u0 > 0 for δ ≤ t ≤ T + δ and, by Lemma 2.12 (ii), we have (-- ) UL,l(.,0) u0 ∈ C Ω × (δ,T + δ) . Then UL,l(.,0)u0 has a positive minimum on -- Ω × [δ,T + δ]. Now,

∫ l t S1(f, Φ)(t) = UL,l(t,0)u0 + UL,l(t,τ) Λf,Φ (τ)d τ ≥ UL,l(t,0)u0 ≥ M 0

for t ∈ [δ,T + δ] and so, by periodicity, Sl1(f,Φ) ≥ M. Since Sl2(f,Φ) = Tr(Sl1(f, Φ)) and ( ) Sl(Φ) = T r Sl1(0,Φ) we get that Sl2(Φ) ≥ M and also that Sl(Φ) ≥ M. Then (ii) holds and is irreducible.

Lemma 3.4. ∥∥ l∥∥ liml→ ∞ S = 0 .

Proof. For l > 0 consider Φ ∈ L2T (∂ Ω × ℝ) and let u = Sl2(0,Φ) . Let u1 = Sl1 (0,Φ+) , u2 = Sl1(0,Φ - ) with Φ+ = max (Φ,0) , Φ - = max (- Φ,0) . Thus u1 ≥ 0, u2 ≥ 0 and u = u1 - u2.

Along the proof will denote a positive constant independent of and (non necessarily the same even in a same chain of inequalities). Since Lu1 = 0 in Ω × ℝ , + 〈A ∇u1, ν〉 + lu1 = Φ ≤ ∣Φ∣ and is periodic, Remark 2.6 gives 0 ≤ u1 ≤ Sl1(0,∣Φ ∣). So

and a similar estimate hold for u , 2 and then also for Now, solves Lu = 0 in Ω × ℝ, 〈A ∇u, ν 〉 + lu = Φ on ∂Ω × ℝ and is periodic. Then, from (27) used with t0 = 0 and t = T we get

(47)

Now

∫ - [〈A ∇u, ∇u 〉+ 〈b,∇u〉 u+ a u2] Ω×(0,T) 0 ∫ 〈 ( 1 ) 1 〉 ∫ [〈 1 〉 ] = - A ∇u + -A -1b ,∇u + -A- 1b + --A-1b,b - a0 u2 Ω×(0,T) ∥〈 2 〉∥ 2 Ω×(0,T) 4 ∥ 1 -1 ∥ ∫ 2 2 ≤ ∥∥ 4A b,b ∥∥ ∞ u ≤ c ∥Φ∥L2T (ℝ× ∂Ω) . L (Ω×(0,T)) Ω×(0,T )

(48)

the last inequality by Remark 2.6. Lemma 3.3 (iii) and Remark 2.6 give also

∫ uΦ ≤ ∥u∥ ∥Φ∥ ≤ c ∥Φ ∥2 . ∂Ω×(0,T) L2(∂Ω×(0,T ),) L2(∂Ω×(0,T ),) L2(∂Ω×(0,T ))

Thus ∥ ∥2 l∥Sl (Φ) ∥L2(∂Ω×(0,T)) 2 ≤ c∥Φ ∥L2(∂Ω ×(0,T),) and the lemma holds.

We will use the multiplication operator M ζ given by

M ζ (Φ) = ζΦ, ζ ∈ L∞ (∂ Ω × ℝ) , Φ ∈ L2 (∂Ω × ℝ) . T T

(49)

For ∞ ζ ∈ L T (∂Ω × ℝ) and 2 Φ ∈ L T (∂Ω × ℝ) let us observe that u ∈ W satisfies

Lu = 0 in Ω × ℝ, 〈A ∇u, ν〉 + lu = ζT r(u) + Φ on ∂Ω × ℝ

(50)

(in the sense of the definition 2.1) if and only if for each R ∈ ℝ

it satisfies Lu = 0

〈A ∇u, ν 〉 + (l + R) u = (ζ + R) Tr (u) + Φ

, i.e., we can "add"

to both sides in the boundary condition of (50).

Lemma 3.5. i) For each there exists such that for and such that the problem (50) has a unique solution for all Moreover, it satisfies if

ii) For such and the solution operator is a bounded linear operator from into whose norm is uniformly bounded on for

Proof. Let ∞ ζ ∈ L T (∂Ω × ℝ) such that ∥ζ ∥L∞T (∂Ω×ℝ) ≤ R. By Lemma 3.4 there exists l0 = l0 (R) > 0 such that ∥ ∥ ∥Sl+R ∥ ≤ 1-- 4R for l ≥ l0. For l ≥ l0 (R) we have ∥∥Sl+RM ∥∥ ≤ 1 ζ+R 2 and so I - Sl+RM ζ+R has a bounded inverse. If u ∈ W solves (50), it solves Lu = 0 in Ω × ℝ, 〈A∇u, ν 〉 + (l + R) u = (ζ + R) Tr (u) + Φ on ∂ Ω × ℝ and so

l+R ( l+R )-1 l+R T r(u) = S (M ζ+R (T r(u) + Φ)) ,i.e.,Tr (u) = I - S M ζ+R S (Φ).

Then

( (( ) ) ) u = Sl1+R 0,M ζ+R I - Sl+RM ζ+R -1Sl+R (Φ) + Φ .

(51)

Thus the solution of (50), if exists, is unique and given by (51).

To prove existence, consider the function defined by (51). It solves

Lu = 0 in Ω × ℝ, ( ) 〈A ∇u, ν〉 + (l + R) u = (ζ + R) I - Sl+RM ζ+R -1Sl+R (Φ) + Φ on ∂Ω × ℝ u(x, t) T periodic in T

(52)

and so

(53)

Then (52) can be rewritten as

Lu = 0 in Ω × ℝ, 〈A∇u, ν〉 + (l + R) u = (ζ + R) Tr (u) + Φ on ∂Ω × ℝ u(x, t) T periodic in T

and so

solves (50).

Suppose now Φ > 0. By (ii) and (iii) of Lemma 3.3, Sl+R 1 and Sl+R are positive operators and also l+R essinfΩ× ℝS 1 (Φ) > 0. Thus (51) gives ess inf Ω×ℝ u > 0 and so (i) holds. Finally, from (51) and since Sl+R and l+R S1 are bounded and ∥ ∥ ∥Sl+RM ζ+R∥ ≤ 12 and ∥M ζ+R ∥ ≤ 2R, we obtain (ii).

We will need to introduce two news operators. For R > 0, l ≥ l0 ((R)) , ∥ ζ∥ ∞ ≤ R LT (∂Ω×ℝ) let

(54)

be defined by Sl,1ζ(Φ) = u where is the solution of (50) given by Lemma 3.5 and by ( ) l,ζ l,ζ S (Φ) = T r S1 (Φ) respectively.

Corollary 3.6. For and as in Lemma 3.5, is a bounded, compact, positive and irreducible operator.

Proof. By (53) we have

( ) ( ) Sl,ζ (Φ) = Tr Sl,1ζ(Φ) = Sl I - Sl+RM ζ+R - 1Sl+R (Φ) + Sl (Φ )

and the corollary follows from Lemma 3.3 (iv).

4. A one parameter eigenvalue problem

Lemma 4.1. i) For and there exists a unique such that the problem

Lu = 0 in Ω × ℝ, 〈A ∇u, ν〉 + b0u = λmu + μu on ∂Ω × ℝ, u (x,t) T periodic in t

(55)

has a positive solution. Moreover, for positive and large enough let be the spectral radius of It holds that (where is the spectral radius of ).

ii) The solution space for this problem is one dimensional and for positive and large enough is an algebraically simple eigenvalue of

iii) Each positive solution of (55) satisfies

Proof. Let R > ∥λm - b0∥L∞(∂Ω× ℝ), let l0 = l0 (R) be as in Lemma 3.5 and for l ≥ l0, let be the spectral radius of Sl,λm-b0 . From Lemma 3.6 Sl,λm- b0 is a compact, positive and irreducible operator on L2T (∂ Ω × ℝ) . Then, by the Krein Rutman theorem, is a positive eigenvalue of Sl,λm -b0 with a positive eigenfunction associated. Let l,λm -b0 u = S1 (w) . Thus is a periodic solution of Lu = 0 in Ω × ℝ, 〈A ∇u, ν 〉 + lu = (λm - b0)u + w on . It is also positive because, by Lemma 3.5, l,λm-b S 1 0 is a positive operator. Since ( ) T r(u) = T r Sl,+ λm -b0(w) = Sl,+ λm-b0 (w) = ρw 1 it follows that solves (55) for μ = 1ρ - l.

On the other hand, if is a positive solution of (55) then Lv = 0 in Ω × ℝ and 〈A ∇u, ν〉 + (b + l)u = λmu + (μ + l) u 0 on ∂ Ω × ℝ . So, for l ≥ l (R) 0 l,λm-b0 -1- S (T r(u)) = μ+lT r(u) . From Corollary 3.6 and the Krein Rutman theorem it follows that μ1+l = ρ and so μ = 1ρ - l. Thus (55) has a positive solution if and only if μ = 1 - l. ρ In particular, this gives that does not depend on the choice of and . If is another positive solution of (55), for and as above, and since Tr (v) > 0 and Tr (v) is an eigenfunction of Sl,λm -b0 with eigenvalue the Krein Rutman theorem gives Tr (v) = ηTr (u) for some $η ∈ ℝ \{0}$ . Thus

then the solution space for (55) is one dimensional. Again by the Krein Rutmnan theorem, -1 (l + μm (λ)) is an algebraically simple eigenvalue of Sl+R,λmb0 .

Finally, each positive solution of (55) satisfies

l,-b0 u = S1 ((λmT r(u) + (μ + l)T r(u))),

and so Lemma 3.5 (iii) gives essinf u > 0.■ Ω×ℝ

The aim of the rest of this section is to given some properties of the function μm (λ), λ ∈ ℝ defined, for ∞ m ∈ LT (∂ Ω × ℝ) , by Lemma 4.1. Each zero of provides a principal eigenvalue with weight and the corresponding solutions in (55) are the respective positive eigenfunctions. We will prove that the map m → μm (λ) is strictly decreasing in (Lemma 4.6) and continuous for the a.e. convergence in (Lemma 4.7) hence continuous in L ∞ (∂Ω × ℝ) . T μ (λ) m is concave and analytic in (cf. Corollary 4.9 and Remark 4.11).

Remark 4.2. For q > N + 2 let 2,1 W q,T (Ω × ℝ) be the space of the periodic functions on Ω × R whose restriction to (0,T ) belongs to W q2,1(Ω × (0, T)) and for γ ∈ (0,1) let 1+γ1+2γ C T (∂ Ω × ℝ) be the space of the periodic functions on ∂ Ω × R belonging to C1+ γ1+2γ (∂ Ω × ℝ) .

We recall that if

γ,γ∕2(-- ) aij ∈ C Ω × ℝ , 1 (-- ) bj ∈ C Ω × ℝ for 1 ≤ i,j ≤ N ; γ,γ∕2( -- ) a0 ∈ C Ω × ℝ ,

1+γ b0 ∈ C1+T γ 2 (∂Ω × ℝ)

for such a , then (cf. Remark 3.1 in [8]) the solutions of (55) belong to 2,1 W q,T (Ω × ℝ) and so 1+η1+η λmu + μm (λ)u ∈ C T 2 (∂ Ω × ℝ) for some η ∈ (0,1). Thus Theorem 2.5 in [8] gives (-- ) u ∈ C2,1 Ω × ℝ .■

In order to make explicit the dependence on and b , 0 we will write sometimes μm,L,b0 or μ,m,L for the function μm.

Lemma 4.3. Let and suppose that satisfies

(56)

for some , and If and then If in addition either or then

Proof. If f = 0 and Φ = 0 then, by Lemma 4.1, μ = μm (λ) . Assume that either f > 0 or Φ > 0. Since μm,L,b0 (λ) = μm+ σ,L,b0+σλ (λ) for all λ, σ ∈ ℝ , it suffices to prove the lemma in the case m ≥ 0. For R > 0 let l0 (R) be as in Lemma 3.5 and let l ≥ l0(∥b0∥∞) + l0(∥ λm - b0∥ ∞) . Let w = Sl1,-b0(f,0) , and let z = Sl,-b0(0,(λm + μ + l)T r(v) + Φ) . 1 Thus w ≥ 0, z ≥ 0 and, since v = w + z, v ≥ z. So also T r(v) ≥ T r(z). Now,

then

l,λm-b0 l,λm- b z = S 1 (Φ + λmT r(v - z) + (μ + l) Tr (v)) ≥ S 0 ((μ + l)T r(z)).

(57)

If Φ > 0 since m ≥ 0 we have Φ + λmT r(v - z) + (μ + l)Tr (v) > 0. If f > 0 then (by Lemma 4.3) essinfΩ×ℝ w > 0 and so T r(w) > 0. Then T r(v - z) > 0 and thus, from (57), essinfΩ×ℝ z > 0. Then Tr (z) > 0. Also, from (57),

T r(z) ≥ Sl1,λm -b0((μ + l)Tr (v)) = (μ + l) Sl,λm- b0 (Tr (z)) .

Let ( ) ρ Sl,λm -b0 be the spectral radius of Sl,λm -b0. Remark 2.14 (ii) gives ( ) -1- ≥ ρ Sl,λm -b0 = ---1--- μ+l μm(λ)+l and so μm (λ) ≥ μ.■

Lemma 4.4. Suppose satisfies

(58)

for some , and If and then If in addition either or then

Proof. Consider first the case when λ ≥ 0 and m ≥ 0. For R > 0 let l0 (R) be as in Lemma 3.5 and let l ≥ l0(∥λm - b0∥∞) . Let be the periodic solution of Lw = f in Ω × ℝ , 〈A ∇w, ν〉 + (b0 + l)w = 0 on ∂Ω × ℝ and let be the periodic solution of Lz = 0 in , 〈A ∇z, ν〉 + (b0 + l)z = Φ + λmv + (μ + l)v on ∂ Ω × ℝ . Thus v = z + w and, by Lemma 3.3 (iv), w ≤ 0. Then 0 < ess inf Ω×ℝ v ≤ v ≤ z and so also 0 < T r(v) ≤ T r(z) . Let

^Φ := (λm + l + μ (λ))(T r(v) - T r(z)) + (μ - μ(λ)) Tr (v) + Φ.

Since is periodic and

we have ( ) T r(z) = Sl,λm -b0 (μ (λ) + l) Tr (z) + Φ^ . Thus

1 ( 1 ) ---------Tr (z) = Sl,λm -b0 T r (z) + ---------^Φ μ (λ) + l μ (λ) + l

(59)

If μ(λ) > μ then ^Φ ≤ 0 and so ( ) Sl,λm-b0 (T r(z)) ≥ ρ Sl,λm -b0 T r(z) where ( l,λm -b0) ρ S is the spectral radius of l,λm- b0 S . Thus, Remark 2.14 (ii) gives --1-- μ(λ)+l× T r(z) = Sl,λm-b0 (T r(z)) and so ( ) Sl,λm-b0 ^Φ = 0. Then, by Lemma 3.3 (iii), This implies μ = μ (λ) in contradiction with the assumption μ(λ) > μ. Thus μ (λ) ≤ μ.

Assume now that either f < 0 or Φ < 0 and that μ(λ) < μ. If f < 0 then supw < 0 and so 0 < v < z and 0 < Tr (v) < Tr (z) This implies ^Φ < 0 and if Φ < 0 the same conclusion is obtained. So, in both cases, (59) gives now l,λm-b ( l,λm- b) S 0 (T r(z)) > ρ S 0 Tr (z) in contradiction with Remark 2.14, (ii).

Since for σ ∈ ℝ we have μL,m,b0 (λ) = μL,m+ σ,b0+σλ(λ) , the case λ ≥ 0 and arbitrary follows from the previous one and, finally, the case λ < 0 follows from the case λ > 0 by considering the identity μ (λ) = μ (- λ) .■ m -m

Let be the operator defined by ∂u L0u = ∂t - div(A ∇u) + 〈b,∇u 〉. We have

Corollary 4.5. i) Suppose Then for all

ii) Suppose Then for all

Proof. let be the solution of (55). Thus

(60)

If a0 > 0, since ess inf u > 0 we have - a0u < 0, then Lemma 4.4 gives (i). If b0 > 0 then - b T r(u) < 0. 0 Since

(ii) follows again from Lemma 4.4.

Lemma 4.6. For with imply for all and for all .

Proof. Suppose λ > 0 and μm1 (λ) ≤ μm2 (λ) . Let be a positive and periodic solution of

Since λm1u1 + μm1 (λ) u1 < λm2u1 + μm2 (λ) u1 on ∂Ω × (0,T) and essinfΩ× ℝu1 > 0, Lemma 4.4 applies to give μm2 (λ) < μm2 (λ) which contradicts our assumption μm1 (λ) ≤ μm2 (λ) . The case λ < 0 follows from the case λ > 0 using that μm (λ) = μ-m (- λ) .

Lemma 4.7. Let be a bounded sequence in which converges to in . Then for each .

Proof. To prove the lemma it suffices to show that for each {mn} as in the statement of the lemma there exists a subsequence {mnk } such that limk →∞ μmk (λ) = μm (λ).

Let be a positive number such that ∣mn ∣ ≤ M for all and let λ ∈ ℝ. Thus, by Corollary 4.5,

(61)

Let be the positive periodic solution of

(62)

normalized by ∥T r (un)∥L2T(∂Ω×ℝ) = 1. We observe that {λmnun + μmn (λ)un} is a bounded sequence in L2T (∂ Ω × ℝ) and so, by Lemma 3.3 (i), {un} is bounded in Thus {un} is bounded in L2 (ℝ,H1 (Ω)) T and {(j ∘ i ∘ un)′} is bounded in 2 1 * L T (ℝ, H (Ω) ) where 1 2 2 i : H (Ω) → L (∂Ω) × L (Ω) and 2 2 1 * j : L (∂ Ω) × L (Ω) → H (Ω) are the linear maps defined in Remark 3.2 Then there exists a subsequence {unk} that converges in L2T (∂Ω × ℝ) to some From (61), after pass to a furthermore subsequence, we can assume also that limk →∞ μmnk (λ) = μ for some μ ∈ ℝ. Thus { λm u + μ (λ)u } nk nk mnk nk converges in L2 (∂Ω × ℝ) T to λmu + μu. Since l,-b0 un = S2 (λmnun + μmn (λ)un) and l,-b0 S2 is continuous we obtain that {unk} converges in to Sl,2-b0(λmu + μu) . It follows that u = Sl,2-b0(λmu + μu) i.e., that is a periodic solution of Lu = 0 in Ω × ℝ, 〈A ∇u, ν〉 + b0u = λmu + μ in ∂ Ω × ℝ . Since u > 0 nk and {Tr (u )} nk converges in L2 (∂Ω × ℝ) T to and since ∥Tr (unk)∥L2T(∂Ω×ℝ) = 1 we get u > 0. Then μ = μm (λ) .■

Corollary 4.8. For each the map is continuous from .

Corollary 4.9. is a concave function.

Proof. Choose a sequence {mn} in ∞ CT (∂ Ω × ℝ) that converges a.e. to in and such that ∥mj ∥∞ ≤ 1 + ∥m ∥∞ for all By ([8], lemma 3.3), each μmn is concave and the corollary follows from Lemma 3.8.

Let B (L2T (∂Ω × ℝ)) denote the space of the bounded linear operators on L2T (∂ Ω × ℝ) and for ρ > 0, ζ ∈ L ∞T (∂Ω × ℝ) , let B ρ(ζ) be the open ball in L∞T (∂Ω × ℝ) with center and radius

Lemma 4.10. Let and let be as in Lemma 3.5. For the map is real analytic from into

Proof. Let l ≥ l0, ζ0 ∈ BR (0) and 2 Φ ∈ L T (∂Ω × ℝ) . For ζ ∈ BR -∥ζ0∥(ζ0), the solution uζ = Sl,ζ (Φ) of (50) is periodic and solves Lu ζ = 0 in Ω × ℝ, 〈A∇u ζ,ν 〉 + (b0 + l) uζ = Φ + ζ0T r(uζ) + (ζ - ζ0)Tr (uζ) on ∂ Ω × ℝ , Then T r(u ) = Sl,ζ0-b0Φ + Sl,ζ0- b0M T r(u ) ζ ζ-ζ0 ζ , i.e., we have

Sl,ζ-b0 = Sl,ζ0-b0 + Sl,ζ0-b0M ζ-ζ0Sl,ζ-b0

(63)

Also, ∥∥ l,ζ0- b0 ∥∥ ∥∥ l,ζ0-b0∥∥ S M ζ- ζ0 ≤ ∥ ζ - ζ0∥ S < 1 and then, from (63), ∥∥ l,ζ-b0∥∥ ∥∥ l,ζ0-b0∥∥ S ≤ 2 S . An iteration of (63) gives, for n ∈ ℕ ,

l,ζ-b l,ζ -b ∑n ( l,ζ- b )j l,ζ -b ( l,ζ -b )n+1 S 0 = S 0 0 S 0 0M ζ- ζ0 + S 0 0 M ζ-ζ0S 0 0 j=1

Since ∥ l,ζ -b ∥ ∥S 0 0M ζ- ζ0∥ < 1 we have ∥∥ l,ζ -b ( l,ζ -b )n+1∥∥ limn → ∞ ∥S 0 0 M ζ-ζ0S 0 0 ∥ = 0. Thus

∞ Sl,ζ-b0 = Sl,ζ0- b0 ∑ (Sl,ζ0- b0M )j = Sl,ζ0-b0 (I - Sl,ζ0-b0M )-1. ζ-ζ0 ζ-ζ0 j=1

Since ζ → M ζ-ζ 0 is real analytic the lemma follows.

Remark 4.11. Corollary 4.9 implies that μ m is continuous. So, taking into account Corollary 3.3 and Lemma 4.10, ([3] lemma 1.3) applies to obtain that μm (λ) is real analytic in . Moreover, a positive solution for (55) can be chosen such that λ → u λ∣∂Ω×R is a real analytic map from into L2T (∂Ω × ℝ).

Observe also that if a0 = 0 and b0 = 0 then μm (0) = 0 and that, in this case, the eigenfunctions associated for (55) are the constant functions. Finally, for the case when either a > 0 0 or b ⁄= 0, 0 applying Lemma 4.3 with v = 1, λ = 0 and μ = 0 we obtain that μm (0) > 0.■

Remark 4.12. Assume that a0 = 0, b0 = 0 and for large enough, consider the spectral radius of the operator Sl,λm- b0 : L2T (∂Ω × ℝ) → L2T (∂Ω × ℝ) . Since Φ = 1 is a positive eigenfunction associated to the eigenvalue 1, l the Krein Rutman Theorem asserts that ρ = 1 l l and that there exists a positive eigenvector Ψ ∈ L2 (∂Ω × ℝ) T for the adjoint operator ( l,λm -b0)* S satisfying ( l,λm -b0)* S Ψ = Ψ. Moreover, such a is unique up a multiplicative constant.

Lemma 4.13. Suppose that and let and be as in remark 3.7. Then

Proof. For λ ∈ ℝ , let be a solution of (55) such that λ → uλ is real analytic and u λ = 1 for λ = 0. Since

( { Lu λ = 0 on Ω × ℝ 〈A ∇u ,ν 〉 + (b + l) u = (λm + μ (λ) + l)u on ∂Ω × ℝ ( λ 0 λ m λ uλ(x,t) T periodic in t

we get l,λm- b0 l,λm -b0 T r(uλ) = λS (mT r(u λ)) + (μm (λ) + l)S (T r(uλ)) and so

λ〈Ψ, mT r (uλ)〉 + μm (λ)〈Ψ, T r(uλ)〉 = 0.

Taking the derivative with respect to at λ = 0 and using that μm (0) = 0 and that u λ = 1 for λ = 0, the lemma follows.

5. The behavior of at

We fix ∞ m ∈ L T (∂ Ω × ℝ) , seen as compact Riemannian 2 C manifold of dimension N - 1. For ρ > 0 fixed in , we will find a closed curve Γ ∈ CT (ℝ;∂ Ω) of class and δ = δ(ρ) such that the tube

{ } B Γ ,δ = (x,t) ∈ ∂Ω × [0,T ] : x ∈ exp Γ (t)D δ,Γ (t)

(64)

satisfies

∫ ∫ 1 b ω----δN--1 md σdt ≥ sup m (x, t) dt - 2ρ. N-1 BΓ,δ a x∈ ∂Ω

(65)

To do let us introduce some additional notations to explain ( ) exp Γ (t) D δ,Γ (t) . For x ∈ ∂Ω let T (∂Ω) x denote the tangent space to ∂ Ω at as a subspace of with the usual inner product of N ℝ . This Riemannian structure gives an exponential map expx : Tx (∂Ω) → ∂Ω and an area element dσ (x). For each X ∈ Tx (∂ Ω) , expxX = η (1) where η(t) is the geodesic satisfying η(0) = x, η′(0) = X. We have also the geodesic distance d ∂Ω on and geodesic balls Br (x) , x ∈ ∂Ω, r > 0. We denote the distance on ∂ Ω × (0,T ) given by

d ((x, t),(y,s)) = max (d (x,y),∣t - s∣) ∂Ω

(66)

and, for (x,t) ∈ ∂Ω × (0,T ) and r > 0 we put Br (x,t) for the corresponding open ball with center (x,t) and radius So we have that Br (x, t) = Br (x) × (t - r,t + r) is a cylinder. Concerning the measures on and dσdt on ∂Ω × (0, T) we denote indistinctly ∣E ∣ the measure of a Borel subset of or of ∂ Ω × (0,T ).

For x ∈ ∂Ω let {X ,...,X } 1,x N -1,x be an orthonormal basis of T (∂ Ω) x and let { } φx : z ∈ ℝN -1 : ∣z∣ < r → ∂ Ω be the map defined by (∑N -1 ) φx (z1,...zN -1) = expx j=1 zjXj,x . From well known properties of the exponential map there exists ɛ > 0 such that φ : {z ∈ ℝN -1 : ∣z∣ < r} → B (x) x r is a diffeomorphism for 0 < r < ɛ, x ∈ ∂Ω. For such and let y → (z1(y) ,...,zN-1 (y)) be the coordinate system defined by on Br (x), let { } ∂∂z-,..., ∂z∂-- 1 N -1 be the corresponding coordinate frame, let 〈 〉 gij (y) := ∂-- , ∂- , ∂zi∣y ∂zj∣y 1 ≤ i,j ≤ N - 1, y ∈ Br (x) and let (gij (y)) be the (N - 1) × (N - 1) matrix whose i,j entry is gij.(y). Finally, we put ωN- 1 for the area of the unit sphere SN - 1 ⊂ ℝN .

Lemma 5.1. i) For it holds that uniformly in

ii) is doubling, that is for some independent of and

iii) Let be a Borel set. Then a.e. (the limit taken on balls in )

Proof. To obtain (i) we consider an orthonormal basis {X1.x,...XN - 1.x} of Tx (∂ Ω) and z ∈ ℝN - 1. For small enough and 0 < r < ɛ we have

∣B (x)∣ 1 ∫ ----r-N--1 - 1 = ------N--1 (f (x, z) - 1) dz1...dzN- 1 ωN -1r ωN -1r ∣z∣<r

where 12 ( ( (∑N - 1 ))) f (x,z) := det gij expx j=1 zjXj,x . Since (x,z) → f (x,z) - 1 is uniformly continuous on ∂Ω × D1 and f (x, 0) = 1, x ∈ ∂Ω we obtain (i) by taking limits.

As has finite diameter for d ∂Ω we have (ii).

Finally, dσdt is also doubling in ∂Ω × ℝ and so (iii) holds (cf. e.g. [11]).

Lemma 5.2. For each there exists a partition of and points in with such that is a family of disjoint sets and

∫ ∫ -----1---- T ω δN -1 n m (x,t)d σ(x) dt ≥ ess sup m (x,t)dt - ρ N- 1 ∪i=1B δ(xi)×(ti-1,ti) 0 x∈∂Ω

Proof. Without lost of generality we can assume that ∥m ∥ ≤ 1. ∞ For t ∈ [0,T ] let m^ (t) = ess supx∈∂Ω m (x,t) and for η > 0 let

$E (η) = {(x,t) ∈ ∂ Ω × ℝ : m (x,t) > ^m (t) - η}.$

(67)

and let E (η)d be the set of the density points (in the sense of Lemma 5.1, (iii)) in E (η). We fix ( 1) α ∈ 0,2 . For k ∈ ℕ, let (k) E (η) be the set of the points d (x,t) ∈ E (η) such that

∣B ρ(y,s) ∩ E (η)∣ -----------------> 1 - α ∣B ρ(y,s)∣

for all open ball B ρ(y,s) ⊂ ∂ Ω × ℝ containing (x,t) and with radius ρ < 1. k Observe that (k) (s) E (η) ⊂ E (η) for k < s and that (from Lemma 3.16 (iii) (k) E (η) = ∪k∈ ℕE (η) . Thus ∣ ( ) ∣ limk→ ∞ ∣∣π E (η)(k) ∣∣ = ∣π (E (η))∣ = T where π(x,t) := t.

Given ɛ > 0 we fix k ∈ ℕ such that ∣ ( ) ∣ ∣∣π E (η)(k) ∣∣ ≥ T - ɛ. For n ∈ ℕ let l = T2n and let {t0,....tn} be the partition of [0,T] given by ti = 2il.

Let { } I = i ∈ {1,2,...n}: (∂Ω × (ti-1,ti)) ∩ E (η)(k) ⁄= ∅ and let $Ic = {1,2,...n} \I.$ Denote -T δ = 4n. For $i ∈ I\{n}$ let * (k) (xi,ti) ∈ (∂Ω × (ti-1,ti)) ∩ E (η) and let Qi = Bδ (xi) × (ti- 1,ti) and, for $j ∈ Ic\ {n}$ let xj ∈ ∂Ω and let Qj = Bδ (xj) × (tj-1,tj). We also set x = x n 1 and Q = B (x ) × (t ,t) . n δ n n- 1 n Since ∣ ( )∣ ∣π E (η)(k) ∣ ≥ T - ɛ ∣ ∣ we have ∑ i∈Ic (ti - ti-1) ≤ ɛ.

Consider the case i ∈ I. We have ∫ ∫ ∫ m (x,t)dσ (x) dt = m (x,t)dσ (x)dt + c m (x,t)dσ (x)dt. Qi Qi∩E(η) Qi∩E(η) Also,

∫ ∫ Q ∩E(η)m (x, t) dσ (x) dt ≥ Q ∩E(η) ^m (t)dσ (x) dt - η ∣Qi ∩ E (η)∣ ∫ iti i ∫ ti ≥ ^m (t)(∣(Q ∩ E (η))∣ - ∣(Q )∣)dt + ^m (t)∣(Q )∣dt - η ∣Q ∣ ti-1 i t i t ti-1 i t i ∫ ti ≥ ∣Q ∩ E (η)∣ - ∣Q ∣ + ∣B (x )∣ ^m (t)dt - 2lη∣B (x )∣. i i δ i ti-1 δ i

Since

and

( ) (xi,t*i) ∈ B δ (xi) × ti+ti--1- l, ti+ti-1 + l 2 2

we get

). So, the above inequalities give

∫ ( ∫ t ) i Q ∩E(η) m (x,t)dσ (x)dt ≥ - 2l(α + η) + t ^m (t)dt ∣Bδ (xi)∣. i i-1

Moreover, ∫ Qi∩E(η)c m (x,t)dσ (x)dt ≤ ∣(Qi ∩ E (η))c∣ = ∣Qi∣ - ∣Qi ∩ E (η)∣ ≤ 2lα ∣B δ (xi)∣. Thus

∫ ( ∫ t ) i Q m (x,t)dσ (x)dt ≥ - 2l(2α + η) + t ^m ∣B δ (xi)∣. i i-1

(68)

Also, for c j ∈ I ,

∫ m (x, t) dσ(x) dt ≥ - ∣Qj∣ = - 2l∣Bδ (xj)∣ Qj

(69)

For i ∈ I let ɛi (δ) = -∣Bδ(xNi)∣-1 - 1. ωN-1δ From (68) and (69) we have

$∫ m (x,t)dσ (x)dt ∪ni=1Qi ∑ ∫ ∑ ∫ ∫ = m (x,t)dσ (x)dt+ m (x,t)dσ (x)dt+ m (x,t)dσ (x)dt i∈I\{n} Qi i∈Ic\{n} Qi Qn ( ∫ t ) ≥ ∑ i m^ (t)dt - 2l(2α+ η) ∣B (x)∣- ∑ 2αl∣B (x)∣- T-∣B (x )∣ ti-1 δ i c δ i n δ n i∈I\{n} i∈I$

$( ) ∫ T ∑ ∫ ti = ωN- 1δN -1( ^m (t)dt- ^m (t) dt- 2l# (I)(2α + η)- 2l# (Ic)α) 0 i∈Ic\{n} ti-1 T - ωN -1δN -1- ( ( n ) ) N-1 ∑ ∫ ti ∑ + ωN- 1δ ( ɛi(δ) - 2l(2α + η)+ ^m (t) dt - 2αlɛi(δ)) i∈I\{n} ti-1 i∈Ic\{n} T - ωN -1δN -1-ɛn(δ). n$

Hence

where # (I) and # (Ic) denote the cardinals of and respectively. Since δ = 4Tn and Lemma 3.11 gives that lim max ∣∣ɛ (T-)∣∣ = 0, n→ ∞ 1≤i≤n i 4n taking large enough and α, η and small enough the lemma follows.

For a periodic curve Γ ∈ C2 (ℝ, ∂Ω) and δ > 0, let B Γ ,δ defined by (64). We have

Lemma 5.3. Assume that is connected. Then for each there exist and such that

∫ ∫ T ----1----- ωN -1δN- 1 B m (x, t)d(x) σdt ≥ 0 ess sx∈up∂Ω m (x, t) dt - 2ρ Γ,δ

Proof. Let ρ > 0 and let x1,...,xn, t0,...,tn and be as in Lemma 5.2. For θ < T2n and i = 1,...,n - 1, let γ : [t - θ,t + θ] → ∂ Ω i i i be a map satisfying γ (t - θ) = x , i i i- 1 γi(ti + θ) = xi and (j) γi (t) = 0 for j = 1,2 and t = ti ± θ. Let 2 Γ ∈ C T (ℝ, ∂Ω) be defined by Γ (t) = x1 for t ∈ [t0,t1 - θ], Γ (t) = xn for t ∈ [tn + θ,tn] and by

for i = 1,...,n - 1. For small enough satisfies the conditions of the lemma.

Corollary 5.4. Assume that is connected and let be defined by (6). If then for positive and small enough there exists such that

Remark 5.5. Let ( ) Γ ∈ C2T ℝ, ℝN as in Lemma 5.3. Since the map t → ν (Γ (t)) belongs to ( ) C1+ θ ℝ, ℝN there exists a C1+ θ and periodic map t → A (t) from into SO (N ) such that A (t)ν (Γ (0)) = ν (Γ (t)) for t ∈ ℝ . Let {X ,...,X } 1,0 N-1,0 be an orthonormal basis of TΓ (0)(∂Ω) and let Xj (t) = A (t)Xj,0, for j = 1, 2,...N - 1, . Thus each is a periodic map, 1+ γ ( N ) Xj ∈ C ℝ, ℝ and for each {X1 (t),...,XN -1 (t)} is an orthonormal basis of TΓ (t)(∂Ω) . For z ∈ ℝN and t ∈ ℝ we set

x (z, t) ( ) ( ( )) ∑ ∑ := expΓ (t) zjXj (t) - zN+1ν expΓ (t) zjXj (t) , 1≤j≤N -1 1≤j≤N -1

(70)

and

(71)

For δ > 0 let { N -1 } D δ = z ∈ ℝ : ∣z∣ < δ and Q δ := Dδ × (0,δ) × ℝ . Thus, for positive and small enough is a diffeomorphism from Q δ onto an open neighborhood W δ ⊂ ℝN × ℝ of the set {(T (t),t) : t ∈ ℝ} satisfying

Λ(Q δ) = W δ ∩ (Ω × ℝ) ,

Λ(Q δ) = W δ ∩ (∂ Ω × ℝ) ,

Λ(D × {0} × {t}) = B (Γ (t)) × {t}, δ δ

Λ(0,t) = (Γ (t) ,t),

Λ(.,t) is periodic in

Moreover, Λ : Q δ → W δ and its inverse Θ : W δ → Q δ are of class C2,1 on their respective domains. For δ,Λ, Θ,W δ as above, with Θ (x,t) = (Θ1 (x,t),...,ΘN+1(x, t)) we have ΘN+1 (x,t) = t and also (cf. (3.13) and (3.14) in [8])

for some 1 g ∈ C (W δ ∩ (∂Ω × ℝ)) satisfying g (x,t) ⁄= 0 for (x,t) ∈ W δ ∩ (∂Ω × ℝ) and g (Γ (t),t) = 1 for t ∈ ℝ . Moreover, if Λ ′(Γ (t),t) denotes the Jacobian matrix of at (Γ (t),t), from the definition of and taking into account that the differential of expx at the origin is the identity on Tx (∂Ω) , we have that detΛ ′(Γ (t),t) = 1 for .

Lemma 5.6. Assume that is connected and that Then

Proof. Let {mn} be a sequence in C∞T (∂ Ω × ℝ) that converges to a.e in ∂Ω × ℝ and satisfying ∥mn ∥∞ ≤ 1 + ∥m ∥∞ for n ∈ ℕ, let { } L(n) be a sequence of operators as in Lemma 2.8 and let (n) A be the N × N matrix whose i,j entry (n) aij , let { (n)} b0 be a sequence in 1 1- W 2- q,1- 2q q,T for some q > N + 2 and such limn →∞ b0(n) = b0 a.e. in ∂ Ω × ℝ .

For positive and small enough let be as in Corollary 5.4 and let Q δ, W δ, and be as in Remark 5.5.

For (s,t) ∈ Q δ let

(n) ∑ ∂Θi ∂Θj ^aij (s,t) = alr (Λ (s,t))----(Λ (s,t)) ----(Λ (s,t)), 1≤l.r≤N ∂xl ∂xr

let (n) ( (n) (n) ) ^b (s,t) = ^b1 (s,t) ,...,^bN (s,t) with

and let (n) A^ (s,t) be the N × N symmetric and positive matrix whose (i,j) entry is ^a(inj)(s,t), let ^a(0n) be defined on Q δ by ^a0 = a0 ∘ Λ, let ^mn, ^b0 be defined on D δ × {0} × [0,T] by m^n = mn ∘ Λ and ^ b0 = b0 ∘ Λ. For λ > 0 let un,λ be a positive and periodic solution of

normalized by ∥un,λ∥W = 1. Let ^un,λ ∈ C2,1(Q δ) be defined by ^un,λ = un,λ ∘ Λ. Then, a computation shows that

Let β ∈ (0,δ) (to be chosen latter), let ∞ h ∈ C (ℝ) such that 0 ≤ h ≤ 1, h (ζ) = 1 for ζ < δ - β, h (ζ) = 0 for ζ ≥ δ and let ( ) G ∈ C ∞ ℝN+! be defined by G (z,s,t) = h (∣(z,s)∣) for (z,s,t) ∈ ℝN - 1 × ℝ × ℝ. Finally, we set ^g = g ∘ Λ and, for a definite positive matrix P ∈ M (ℝ) N and w ∈ ℝN we put ∥w ∥ := 〈Pw, w 〉. P With these notations we have, as in the proof of Lemma 3.11 in [8],

$∫ ( 2 ) μmn,L(n),b(n0)(λ) G ^g (ξ, 0,t) dξdt D δ×(0,T) ∫ ( ) ≤ - λ G2 ^gm^n (ξ,0,t)dξdt [ Dδ×(0,T) ] ∫ ∥∥ ( G ) ∥∥2 + ∥∥ ∇G + --A^(n)^b(n) ∥∥ + ^a(n0)(s,t)G2 (s,t) dsdt. {s∈ℝN:∣s∣< δ:}×(0,T) 2 A^(n)(s,t)$

(72)

Also

┌│ ----(---(--------(------------)))--- ∫ │ N∑- 1 ∫ m^ (z,0,t)∘ det gij exp Γ (t) zjXj (t) dzdt = m > 0. Dδ×(0,T) j=1 BΓ,δ

Thus, since ∘ --------------- det (gij (Γ (t))) = 1 and ∘ ---(----(-------(--------------)))-- ∑N -1 z → det gij expΓ (t) j=1 zjXj (t) is continuous, we get ∫ ^m (z,0,t) dzdt > 0 D δ×(0,T) for positive and small enough. Then (for a smaller if necessary) and some positive constant we have

for large enough. Since is continuous on D δ × {0} × ℝ and ^g (0,t) = 1 we can assume also (diminishing and if necessary) that, for large enough,

∫ ∫ (^mn ^g)(z,0,t) dzdt > c and ^g (z, 0,t)dzdt > c Dδ×(0,T) Dδ×(0,T)

¿From these inequalities it is clear that we can pick small enough in the definition of such that for large enough

(73)

(74)

We have also

so, from (73), we get positive constants and independent of and such that μmn,L(n),b(n0)(λ) ≤ - c1 - c2λ for all large enough. Also, since

Lemma 4.3 gives μ (λ) ≥ - (1+ ∥m ∥ )λ - (1+ ∥b ∥ ). mn,L(n) ∞ 0 ∞ Thus { } μ (λ) mn,L(n) is bounded, and so, after pass to a subsequence we can assume that { } μmn,L(n) (λ) converges to some μ ≤ - c1 - c2λ . Since { } λmnT r(un,λ) + μ (n) (λ) T r(un,λ) mn,L is bounded in 2 L T (∂ Ω × ℝ) , by Lemma 3.3 and after pass to a furthermore subsequence, we can assume that {un,λ} converges in to some u λ ≥ 0. By Lemma 2.8 satisfies Lu = 0 in , 〈A ∇u, ν 〉 + b0u = λmu + μu on ∂Ω × ℝ. Thus μm,L,b0 (λ) = μ and so μm,L,b0 (λ) ≤ - c1 - c2λ .

6. Principal eigenvalues for periodic parabolic Steklov problems

Let P (m) and N (m) be defined by (6). We have

Theorem 6.1. Suppose one of the following assertions i), ii), iii), holds.

i) (respectively ) and either or

ii) (respectively ), (resp. ) with defined as in remark 3.7.

Then there exists a unique positive (resp. negative) principal eigenvalue for (55) and the associated eigenspace is one dimensional.

Proof. Suppose a0 = 0, b0 = 0, P (m) > 0 and 〈Ψ, m 〉 < 0. Since μm (0) = 0 and, by Lemma 3.14, μ′m (0) > 0 the existence of a positive principal eigenvalue λ = λ1 (m) for (55) follows from Lemma 5.6. Since does not vanish identically, the concavity of gives the uniqueness of the positive principal eigenvalue.

Moreover, if u,v are solutions in for (55), then, from Lemma 4.1, u = cv on ∂ Ω × R for some constant Since, for l ∈ R , L (u - cv) = 0 on Ω × R, Bb0+l (u - cv) = λm (u - cv) + μm (λ) (u - cv) and u - cv = 0 on ∂Ω × R . Thus, taking large enough, Lemma 2.9 gives u = cv on Ω × R.

If either a0 > 0 or b0 > 0 then (by Remark 3.12) μm (0) > 0 and so the existence follows from Lemma 5.6. The other assertions of the theorem follow as in the case a0 = 0. Since μm (- λ) = μ -m (λ) and N (m) = - P (- m) , the assertions concerning negative principal eigenvalues reduce to the above.

Theorem 6.2. Let such that Then for all the problem

(75)

has a unique solution. Moreover implies that

Proof. Since μm (λ) > 0 for large enough we have ( ) 1 ρ Sl,λm-b0 < l and so , (1I - Sl,λm -b0)-1 l is a well defined and positive operator. If is a solution of (75) then l,-b0 u = S λm+lΦ so the solution, if exists, is unique. To see that it exists, consider

1 (1 ) -1 w := -Sl,λm -b0 -I - Sl,λm-b0 Φ. l l

and observe that u = Sl,-b0((λm + l)w + Φ) 1 solves (75). Finally, if Φ > 0, then w > 0 on ∂ Ω × ℝ and since

u = Sl+R,-b0((λmT r(u) + (μ + l + R) Tr (u))), 1

Lemma 2.18 (iii) gives essinfΩ×ℝ u > 0.■

Let λ1(m) (respectively λ- 1(m) ) be the positive (resp. negative) principal eigenvalue for the weight with the convention that λ1(m) = + ∞ (respectively λ- 1(m) = - ∞ ) if there not exists such a principal eigenvalue. From the properties of μm, Theorem 6.2 gives the following

Corollary 6.3. Assume that either or Then the interval does not contains eigenvalues for problem (55). If and the same is true for the intervals and

References

[1] Amann, H., Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SIAM Review, Vol. 18, No 4, (1976), 620-709. [ Links ]

[2] Beltramo, A. and Hess, P. On the principal eigenvalues of a periodic -parabolic operator , Comm. in Partial Differential Equations 9, (1984), 914-941. [ Links ]

[3] Crandall, M.G. and Rabinowitz, P. H., Bifurcation, perturbation of simple eigenvalues and stability, Arch. Rat. Mech. Anal. V. 52, No2, (1973) 161-180. [ Links ]

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[5] D. Daners and P. Koch-Medina, Abstract evolution equations, periodic problems and applications, Pitman Research Notes in Mathematics Series 279. Harlow, Longman Scientific & Technical, New York Wiley (1992). [ Links ]

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T. Godoy
Facultad de Matemática, Astronomía y Física and CIEM - Conicet,
Universidad Nacional de Córdoba,
Ciudad Universitaria,
5000 Córdoba, Argentina
godoy@mate.uncor.edu

E. Lami Dozo
Département de Mathématique,
Université Libre de Bruxelles and
Universidad de Buenos Aires - Conicet,
Campus Plaine 214, 1050 Bruxelles
lamidozo@ulb.ac.be

S. Paczka
Facultad de Matemática, Astronomía y Física and CIEM - Conicet,
Universidad Nacional de Córdoba,
Ciudad Universitaria,
5000 Córdoba, Argentina
paczka@mate.uncor.edu

Recibido: 16 de diciembre de 2005
Aceptado: 7 de agosto de 2006

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Revista de la Unión Matemática Argentina

versão impressa ISSN 0041-6932versão On-line ISSN 1669-9637

Rev. Unión Mat. Argent. v.46 n.2 Bahía Blanca jul./dez. 2005