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Abstract and Applied Analysis On principal eigenvalues for periodic parabolic Steklov problems
On principal eigenvalues for periodic parabolic Steklov problems
Godoy, T., Dozo, E. Lami, Paczka, S.How much do you like this book?
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2002
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Abstract and Applied Analysis
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10.1155/s1085337502204066
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ON PRINCIPAL EIGENVALUES FOR PERIODIC PARABOLIC STEKLOV PROBLEMS T. GODOY, E. LAMI DOZO, AND S. PACZKA Received 1 March 2002 Let Ω be a C 2+γ domain in RN , N ≥ 2, 0 < γ < 1. Let T >0 andlet L be a uniformly parabolic operator Lu = ∂u/∂t − i, j (∂/∂xi )(ai j (∂u/∂x j )) + j b j (∂u/∂xi ) + a0 u, a0 ≥ 0, whose coeﬃcients, depending on (x,t) ∈ Ω × R, are T periodic in t and satisfy some regularity assumptions. Let A be the N × N matrix whose i, j entry is ai j and let ν be the unit exterior normal to ∂Ω. Let m be a Tperiodic function (that may change sign) defined on ∂Ω whose restriction to ∂Ω × R belongs to 2−1/q,1−1/2q (∂Ω × (0,T)) for some large enough q. In this paper, we give necWq essary and suﬃcient conditions on m for the existence of principal eigenvalues for the periodic parabolic Steklov problem Lu = 0 on Ω × R, A∇u,ν = λmu on ∂Ω × R, u(x,t) = u(x,t + T), u > 0 on Ω × R. Uniqueness and simplicity of the positive principal eigenvalue is proved and a related maximum principle is given. 1. Introduction Let Ω be a C 2+γ and bounded domain in RN , N ≥ 2, 0 < γ < 1, let T > 0, let {ai j }1≤i, j ≤N , {b j }1≤ j ≤N be two families of real functions defined on Ω × R satisfying ai j ∈ C γ,γ/2 (Ω × R), b j ∈ C 1 (Ω × R), ai j = a j,i , and ∂ai j /∂xi ∈ C(Ω × R) for 1 ≤ i, j ≤ N. Assume also that ai j (x,t)ξi ξ j ≥ α0 ξ 2 (1.1) i, j for some positive constant α0 and all (x,t) ∈ Ω × R, ξ = (ξ1 ,...,ξN ) ∈ RN and that each ai j (x,t),b j (x,t) is T periodic in t. Let A be the N × N matrix whose i, j entry is ai j , let b = (b1 ,...,bN ), let a0 be a nonnegative and T periodic function Copyright © 2002 Hindawi Publishing Corporation Abstract and Applied Analysis 7:8 (2002) 401–421 2000 Mathematics Subject Classification: 35K20, 35P05, 35B10, 35B50 URL: http://dx.doi.org/10.1155/S1085337502204066 402 Principal eigenvalues for a Steklov problem belonging to C γ,γ/2 (Ω × R) and let L be the parabolic operator given by Lu = ut − div(A∇u) + b, ∇u + a0 u, (1.2) where , ; denotes the standard inner product on RN . For q ≥ 1, τ > 0, let Wq2,1 (Ω × (0,τ)) be the Sobolev space of the functions u ∈ Lq (Ω × (0,τ)), u = u(x,t), x = (x1 ,...,xN ) such that ∂u/∂t, ∂u/∂x j , and ∂2 u/∂xi ∂x j belong to Lq (Ω × (0,τ)) for 1 ≤ i, j ≤ N. We are interested in the periodic parabolic Steklov eigenvalue problem Lu = 0 in Ω × R, A∇u,ν = λmu on ∂Ω × R, (1.3) u(x,t) T periodic in t, where ν denotes the unit exterior normal to ∂Ω and the solution u is taken such that u Ω×(0,T) ∈ Wq2,1 (Ω × (0,T)) for a fixed and large enough q. The weight 2−1/q,1−1/2q function m is assumed T periodic such that m ∂Ω×(0,T) ∈ Wq (∂Ω × (0,T)) (the fractional Sobolev space defined, for example, as in [7, Chapter 2, paragraph 3]). Steklov introduced this eigenvalue problem in the elliptic case in connection with the study of the map, nowadays called Dirichlet to Neumann map (cf. [3, Part B, Chapter VI, pages 395–404]) which is also of interest in the inverse problem of reconstructing the coeﬃcients of L from this map. We say that λ∗ ∈ R is a principal eigenvalue for the weight m if (1.3) has a positive (i.e., a nonnegative and nontrivial) solution. In this paper, we give necessary and suﬃcient conditions, on a weight m as above, for existence of a positive principal eigenvalue. Uniqueness and simplicity of this positive principal eigenvalue is proved and a related form of the maximum principle is given. We remark that this weighted eigenvalue problem includes the corresponding elliptic case where the coeﬃcients are time independent. In Section 2, for given T periodic functions f and Φ defined on Ω × R and ∂Ω × R, respectively, and satisfying f ∂Ω×(0,T) ∈ Lq (Ω × (0,T)), Φ ∂Ω×(0,T) ∈ 2−1/q,1−1/2q Wq (∂Ω × (0,T)), we study existence of T periodic solutions u : Ω × R → R, such that u ∂Ω×(0,T) ∈ Wq2,1 (Ω × R) for the problem Lu = f on Ω × R, b0 u + A∇u,ν = Φ on ∂Ω × R, (1.4) u(x,t) T periodic in t. We prove that, under suitable hypothesis on a0 and b0 , this problem has a unique solution, and we state the boundedness (with respect to the natural topologies T. Godoy et al. 403 involved) of the corresponding solution operator u = Sb0 ( f ,Φ) (see Theorem 2.5). We also prove the compactness and the strong positivity of the operator Φ → Sb0 (0,Φ) (see Theorem 2.6). In Section 3, we study the following oneparameter family of principal eigenvalue problems: given λ ∈ R, we prove that there exists a unique principal eigenvalue µ = µm (λ) for the problem Lu = 0 on Ω × R, A∇u,ν − λmu = µu on ∂Ω × R, (1.5) u(x,t) T periodic in t, u>0 on Ω × R, we show that µm (λ) is concave and real analytic in λ and its behavior at zero and at infinity is studied. In Section 4, using the properties of the function µm , we prove that, for the T case a0 > 0, the condition P(m) := 0 maxx∈Ω m(x,t)dt > 0 is a necessary and suﬃcient condition for the existence of a positive principal eigenvalue for the weighted problem (1.3) and that, for the case a0 = 0,there exists a positive principal eigenvalue for (1.3) if and only if P(m) > 0 and Ω×(0,T) Ψm < 0 where Ψ is a positive (unique up to a multiplicative constant and belonging to C 2+γ,1+γ/2 (Ω × R)) for the T periodic problem ∂Ψ + div(A∇Ψ) + b, ∇Ψ + div(b)Ψ = 0 on Ω × R, ∂t A∇Ψ,ν + b,ν Ψ = 0 on ∂Ω × R, (1.6) Ψ(x,t) T periodic in t. 2. Preliminaries We recall the following wellknown facts concerning Sobolev spaces (see, e.g., [7, Lemma 3.3, page 80 and Lemma 3.4, page 82]). (i) If u ∈ Wq2,1 (Ω × (0,τ)), q ≥ 1, τ > 0, then 2−1/q,1−1/2q u ∂Ω×(0,τ) ∈ Wq ∂Ω × (0,τ) (2.1) and the restriction map (understood in the trace sense) is continuous. (ii) For τ > 0 and q large enough, it holds that Wq2,1 Ω × (0,τ) ⊂ C 1+γ,(1+γ)/2 Ω × [0,τ] with continuous inclusion. (2.2) 404 Principal eigenvalues for a Steklov problem (iii) For τ > 0 and q large enough, it holds that 2−1/q,1−1/2q ∂Ω × (0,τ) ⊂ C 1+γ,(1+γ)/2 ∂Ω × [0,τ] Wq (2.3) with continuous inclusion. From now on, we fix, τ > T and a large enough q such that (ii) and (iii) hold. We recall also the following lemma. 2−1/q,1−1/2q (∂Ω × (0,τ)), b0 ≥ 0. Suppose also that f ∈ Lemma 2.1. Let b0 ∈ Wq 2−2/q 2−1/q,1−1/2q Lq (Ω × (0,τ)), ϕ ∈ Wq (Ω), and Φ ∈ Wq (∂Ω × (0,τ)), and that the compatibility condition b0 (·,0)ϕ + A∇ϕ,ν = Φ(·,0) on ∂Ω (2.4) is fulfilled, then the problem Lu = f on Ω × (0,τ), b0 u + A∇u,ν = Φ on ∂Ω × (0,τ), u(·,0) = ϕ (2.5) on Ω, has a unique solution u ∈ Wq2,1 (Ω × (0,τ)). Moreover, there exists a positive constant c independent of f , ϕ, and Φ, such that u Wq2,1 (Ω×(0,T)) ≤c f Lq (Ω×(0,T)) + Φ 2−1/q,1−1/2q Wq (∂Ω×(0,T)) + ϕ . (2.6) 2−2/q Wq (Ω) For a proof of Lemma 2.1, see [7, Theorem 9.1, page 341] concerning to the Dirichlet problem and its extension, to our boundary conditions, indicated there, at the end of Chapter 4, paragraph 9, page 351. For regular data, the following result holds (see, e.g., [7, Theorem 5.3, page 320]). Lemma 2.2. Suppose that b0 ∈ C 1+γ((1+γ)/2) (∂Ω × [0,τ]), b0 ≥ 0. Suppose also that f ∈ C γ,γ/2 (Ω × [0,τ]), ϕ ∈ C 2+γ (Ω), Φ ∈ C 1+γ((1+γ)/2) (∂Ω × [0,τ]) and that the compatibility condition (2.4) is fulfilled, then problem (2.5) has a unique solution u ∈ C 2+γ,1+γ/2 (Ω × [0,τ]). Moreover, there exists a positive constant c independent of f , ϕ, and Φ such that u C 2+γ,1+γ/2 (Ω×[0,τ]) ≤c f C γ,γ/2 (Ω×[0,τ]) + Φ C 1+γ((1+γ)/2) (∂Ω×[0,τ]) + ϕ . (2.7) C 2+γ (Ω) Remark 2.3. If, in addition to the hypothesis of Lemma 2.1, we have that f ∈ C γ,γ/2 (Ω × [0,τ]), then the solution u of (2.5) satisfies u ∈ C 2+γ,1+γ/2 (Ω × [δ,τ]) (2.8) T. Godoy et al. 405 for all δ > 0. Moreover, for such a δ, there exists a positive constant cδ independent of f and Φ such that u C 2+γ,1+γ/2 (Ω×[δ,τ]) ≤ cδ f Φ 2−1/q,1−1/2q Wq (∂Ω×(0,τ)) + ϕ . (2.9) Indeed, let h ∈ C ∞ (R) such that 0 ≤ h ≤ 1, h(t) = 0 for t < δ/4, h(t) = 1 for t ≥ 3δ/4, let u(x,t) = u(x,t)h(t), let f(x,t) = u(x,t)h (t) + f (x,t)h(t), and let Φ(x,t) = Φ(x,t)h(t). Then, C γ,γ/2 (Ω×[0,τ]) + 2−2/q Wq (Ω) Lu = f on Ω × (0,τ), on ∂Ω × (0,τ), b0 u + A∇ u,ν = Φ u(·,0) = 0 (2.10) on Ω. By Lemma 2.1, this problem has a unique solution in Wq2,1 (Ω × (0,τ)). Since ∈ C 1+γ,(1+γ)/2 (∂Ω × [0,τ]), Lemma 2.2 says that it f ∈ C γ,γ/2 (Ω × [0,τ]) and Φ has also a unique solution u ∈ C 2+γ,(1+γ)/2 (Ω × [0,τ]), and so, since h ≡ 1 on [δ,τ], we obtain (2.8). Also, Φ C 1+γ,(1+γ)/2 (∂Ω×[0,T]) ≤ cδ Φ 2−1/q,1−1/2q Wq (2.11) (∂Ω×(0,τ)) for some constant cδ independent of Φ, and so, using (2.6) and the definition of f, we get f C γ,γ/2 (Ω×[0,τ]) ≤ cδ f C γ,γ/2 (Ω×[0,τ]) + Φ 2−1/q,1−1/2q Wq (∂Ω×(0,τ)) + ϕ 2−2/q Wq (Ω) (2.12) for some positive constant cδ independent of f , Φ. Then (2.7), applied to problem (2.10), gives (2.9). 2−1/q,1−1/2q s (∂Ω × (0,τ)), b0 ≥ 0. For s > 1 + 1/q, let Wq,B (Ω) be the Let b0 ∈ Wq 0 s Banach space of the functions ϕ ∈ Wq (Ω) satisfying b0 (·,0)ϕ + A(·,0)∇ϕ,ν = 0 on ∂Ω. 2−2/q 2−1/q,1−1/2q (∂Ω × (0,τ)) provided with their natural orders Wq,B0 (Ω) and Wq are ordered Banach spaces. Enlarging q, if necessary, we can assume (from now on) that in both spaces the respective positive cones have nonempty interior. As usual, for f : Ω × R → R (resp., f : ∂Ω × R → R, f : Ω → R) we write f > 0 to mean f (x,t) ≥ 0 and f nonidentically zero. 2−2/q 2−2/q Let U : Wq,B0 (Ω) → Wq,B0 (Ω) be defined by Uϕ = u(·,T), where u ∈ Wq2,1 (Ω × (0,τ)) is the solution (given by Lemma 2.1) of Lu = 0 on Ω × (0,τ), b0 u + A∇u,ν = 0 u(·,0) = ϕ on ∂Ω × (0,τ), on Ω. (2.13) 406 Principal eigenvalues for a Steklov problem We have the following lemma. 2−1/q,1−1/2q (∂Ω × (0,τ)), b0 ≥ 0. Then U is a Lemma 2.4. Suppose that b0 ∈ Wq compact operator. Moreover, if either a0 > 0 or b0 > 0 in their respective domains, then U is a strongly positive operator with positive spectral radius ρ < 1. Proof. From Lemma 2.1, the solution u of (2.13) satisfies u Wq2,1 (Ω×(0,T)) ≤c ϕ 2−2/q Wq (Ω) . (2.14) be as in Remark 2.3, taking there f = 0, Φ = 0. From (2.14) and Let h, u, f, and Φ (2.2), we have f Cγ,γ/2 (Ω×[0,T]) ≤ c ϕ Wq2−2/q (Ω) , and so, (2.9) applied to (2.13) implies u(·,T) C 2+γ (Ω) ≤c ϕ 2−2/q Wq (Ω) (2.15) for some positive constant c independent of ϕ. Now, (2.15) implies the compactness assertion of the lemma. 2−2/q Suppose now that for some ϕ > 0 in Wq,B0 (Ω), the minimum of Uϕ = u(·,T) is nonpositive. Then the minimum of u on Ω × (0,T) is nonpositive and it is achieved at some (x0 ,t0 ) ∈ Ω × (0,T]. If x0 ∈ Ω, the parabolic maximum principle (as stated, e.g., in [6, Proposition 13.3, page 33]) implies that u is a constant on Ω × [δ,T] for all δ > 0, so ϕ is a nonpositive constant which is a contradiction. If x0 ∈ ∂Ω, the same principle states that A∇u,ν < 0 at (x0 ,t0 ) contradicting b0 (x0 ,t0 )u(x0 ,t0 ) + A∇u,ν (x0 ,t0 ) = 0. So, U is a strongly posi2−2/q (Ω). Now, KreinRutman theorem (as stated, e.g., in [1, tive operator on Wq Theorem 3.1]) gives that its spectral radius ρ is a positive eigenvalue with positive 2−2/q (Ω) be such an eigenfunction. To see that ρ < 1, eigenfunctions. Let ϕρ ∈ Wq we proceed by contradiction. Suppose ρ ≥ 1. Then U(ϕρ ) = ρϕρ ≥ ϕρ , that is, the solution of (2.13) (assuming by taking ϕ = ϕρ ) would satisfy u(·,T) ≥ ϕρ , but the maximum principle states that u is a constant or maxΩ×[δ,T] u(x,t) is attained at some (x0 ,t0 ) ∈ ∂Ω × [0,T] and so a0 = 0 or b0 (x0 ,t0 ) < 0, respectively. 2−1/q,1−1/2q 2,1 Let Wq,T (Ω × R) (resp., Wq,T (∂Ω × R)) the Banach space of the T periodic functions v : Ω × R → R such that v Ω×(0,T) ∈ Wq2,1 (Ω × (0,T)) (resp., 2−1/q,1−1/2q v ∂Ω×(0,T) ∈ Wq (∂Ω × (0,T))), equipped with the norm v (resp., v Wq2,1 (∂Ω×(0,T)) ). Wq2,1 (Ω×(0,T)) 2−1/q,1−1/2q (∂Ω × R), b0 ≥ 0. If a0 > 0 or b0 > 0 and Theorem 2.5. Let b0 ,Φ ∈ Wq,T if f : Ω × R → R is T periodic and satisfies f Ω×(0,T) ∈ Lq (Ω × (0,T)), then the problem Lu = f on Ω × R, b0 u + A∇u,ν = Φ on ∂Ω × R, u(x,t) T periodic in t, (2.16) T. Godoy et al. 407 2,1 (Ω × R). Moreover, there exists a positive constant has a unique solution u ∈ Wq,T c independent of f and Φ such that u Wq2,1 (Ω×(0,T)) ≤c Φ 2−1/q,1−1/2q Wq (∂Ω×(0,T)) + f Lq (Ω×(0,T)) . (2.17) If in addition to the above hypothesis, f ∈ C γ,γ/2 (Ω × R), then u ∈ C 2+γ,1+γ/2 (Ω × R) and u C 2+γ,1+γ/2 (Ω×R) ≤c Φ 2−1/q,1−1/2q Wq (∂Ω×(0,T)) + f (2.18) C γ,γ/2 (Ω×R) for some constant c independent of f and Φ. Proof. We start constructing a function ϕ1 ∈ W 2−2/q (Ω) satisfying b0 (·,0)ϕ1 + A(·,0)∇ϕ1 ,ν = Φ(·,0) (2.19) and such that ϕ1 W 2−2/q (Ω) ≤c Φ 2−1/q,1−1/2q Wq (2.20) (∂Ω×(0,T)) for some constant c independent of Φ. To do so, consider F(x,s)=x − sA(x,0)ν(x) on ∂Ω × (−ε,ε). Since x(x) = x for x ∈ ∂Ω and A(x,0)ν(x) is nontangential to ∂Ω at x ∈ ∂Ω, F defines a diﬀeomorphism onto a neighborhood V of ∂Ω in RN for some ε > 0. So we have F −1 (x) = (x(x),s(x)) for x ∈ V . Then we solve the (noncharacteristic) Cauchy problem b0 x(x),0 w + A x(x),0 ∇w,ν x(x) = Φ x(x),0 , x ∈ V, w = 0 on ∂Ω, (2.21) the solution is, for x = x − sA(x,0)ν(x), w(x) = Φ(x,0) s 0 eb0 (x,0)(η−s) dη, x ∈ V. (2.22) Thanks to a cutoﬀ function h associated to V, we can extend w to Ω by ϕ1 = hw which satisfies (2.19) and (2.20). Let u1 ∈ W 2,1 (Ω × (0,τ)) be the solution, given by Lemma 2.1, of the problem Lu1 = f on Ω × (0,τ), b0 u1 + A∇u1 ,ν = Φ u1 (·,0) = ϕ1 on ∂Ω × (0,τ), (2.23) on Ω. Thus, taking into account (2.20) and the estimate given by Lemma 2.1, we obtain u1 Wq2,1 (Ω×(0,T)) ≤c Φ 2−1/q,1−1/2q Wq (∂Ω×(0,T)) + f Lq (Ω×(0,T)) . (2.24) 408 Principal eigenvalues for a Steklov problem 2−2/q 2−2/q Since b0 is T periodic, u1 (·,T) − u1 (·,0) ∈ Wq,B0 (Ω). Let ϕ2 ∈ Wq,B0 (Ω) be defined by ϕ2 = (I − U)−1 u1 (·,T) − u1 (·,0) . (2.25) From (2.24), we get ϕ2 2−2/q Wq (Ω) ≤c Φ 2−1/q,1−1/2q Wq (∂Ω×(0,T)) + f (2.26) Lq (Ω×(0,T)) with the constant c independent of Φ and f . Let u2 ∈ Wq2,1 (Ω × (0,T)) be the solution of the problem Lu2 = 0 on Ω × (0,τ), b0 u2 + A∇u2 ,ν = 0 on ∂Ω × (0,τ), (2.27) on Ω, u(·,0) = ϕ2 taking into account (2.26), Lemma 2.1 gives u2 ≤c Wq2,1 (Ω×(0,T)) Φ 2−1/q,1−1/2q Wq (∂Ω×(0,T)) + f Lq (Ω×(0,T)) . (2.28) Thus, u := u1 + u2 solves (2.16) on Ω × (0,τ). From (2.25), u satisfies u(·,0) = u(·,T). Also, (2.24) and (2.28) give (2.18). Moreover, it is easy to see that u(x,t) − u(x,t + T) is identically zero for 0 ≤ t ≤ τ − T. So, the T periodic extension of u (still denoted by u) solves (2.16) on Ω × R. The uniqueness assertion of the lemma follows easily from the maximum principle. Observe also that if f ∈ C γ,γ/2 (Ω × R), then, taking into account Remark 2.3, the periodicity of u implies that u ∈ C 2+γ,1+γ/2 (Ω × R). From (2.17), we have u(·,0) 2−1/q Wq (Ω) ≤c Φ 2−1/q,1−1/2q Wq (∂Ω×(0,T)) + f C γ,γ/2 (Ω×R) , (2.29) and so, Remark 2.3, applied to (2.16), gives u(·,T) C 2+γ (Ω) ≤c Φ 2−1/q,1−1/2q Wq (∂Ω×(0,T)) + f C γ,γ/2 (Ω×R) . (2.30) So, by the periodicity of u, the same estimate holds for u(·,0). Then, (2.18) fol lows from the estimate given in Lemma 2.2. Theorem 2.6. Let a0 ,b0 , and Φ be as in Theorem 2.5 and let 2−1/q,1−1/2q Sb0 : Wq,T 2−1/q,1−1/2q (∂Ω × R) −→ Wq,T (∂Ω × R) (2.31) T. Godoy et al. 409 be the operator defined by Sb0 Φ = u ∂Ω×R , where u is the T periodic solution of Lu = 0 on Ω × R, b0 u + A∇u,ν = Φ on ∂Ω × R, (2.32) u(x,t) T periodic in t, given by Theorem 2.5. Then Sb0 is a compact strongly positive operator. Proof. Theorem 2.5 gives u C 2+γ,1+γ/2 (Ω×[0,T]) ≤c Φ 2−1/q,1−1/2q Wq (∂Ω×(0,T)) . (2.33) From this estimate, the compactness of Sb0 follows and, taking into account the regularity of the solution of (2.32), the assertion about the strong positivity of Sb0 follows easily from the stated hypothesis on a0 and b0 and the maximum principle. Corollary 2.7. Let a0 ,b0 ,Sb0 be as in Theorem 2.6 and let ρ be the spectral radius of Sb0 . Then, ρ is positive and it is an algebraically simple eigenvalue of Sb0 with positive associated eigenfunctions. Moreover, no other eigenvalue of Sb0 has positive eigenfunctions associated. Proof. The proof follows from Theorem 2.6 and the KreinRutman theorem. 3. A oneparameter eigenvalue problem 2−1/q,1−1/2q (∂Ω × R) be fixed from now on. In order to study princiLet m ∈ Wq,T pal eigenvalues for the weighted problem (1.3), we can assume, without loss of generality, that m ∞ ≤ 1/2. For ε positive and small enough (i.e., such that 1 − ε(1 − m ∞ ) > 0) and λ > −ε, let 2−1/q,1−1/2q Sλ,m : Wq,T 2−1/q,1−1/2q (∂Ω × R) −→ Wq,T (∂Ω × R) (3.1) 2,1 be the operator defined by Sλ,m Φ = u ∂Ω×R , where u ∈ Wq,T (Ω × R) is the solution of the problem Lu = 0 on Ω × R, u + λ(1 − m)u + A∇u,ν = Φ on ∂Ω × R, (3.2) u(x,t) T periodic in t, and let µm (λ) be defined by 1 = ρλ,m , 1 + λ + µm (λ) where ρλ,m is the spectral radius of Sλ,m . (3.3) 410 Principal eigenvalues for a Steklov problem Remark 3.1. From Corollary 2.7, it follows that µm (λ) can be characterized as the unique real number µ, such that problem Luλ = 0 on Ω × R, on ∂Ω × R, A∇uλ ,ν = λmuλ + µuλ (3.4) uλ (x,t) T periodic in t, 2,1 (Ω × R). Since λm = (−λ)(−m), the above has a positive solution uλ ∈ Wq,T characterization of µm (λ) implies that µm (−λ) = µ−m (λ) for λ ∈ (−ε,ε). We extend µm to the whole real line setting µm (−λ) = µ−m (λ) and so the above characterization of µm (λ) holds for all λ ∈ R. In particular, this fact implies that µm+k (λ) = µm (λ) − kλ for all λ,k ∈ R. 2,1 (Ω × R) of the probNote also that, for fixed λ ∈ R, the solution space in Wq,T lem Lu = 0 on Ω × R, on ∂Ω × R, A∇u,ν = λmu + µm (λ)u (3.5) u(x,t) T periodic in t, is one dimensional and is contained in C 2+γ,1+γ/2 (Ω × R). Moreover, by Corollary 2.7, positive solutions have a positive minimum on Ω × R. From the above characterization of µm (λ), our problem (1.3) on principal eigenvalues is equivalent to find the zeroes of the function µm . Lemma 3.2. Suppose that v ∈ C 2+γ,1+γ/2 (Ω × R), such that Lv ≥ 0 on Ω × R, A∇v,ν ≥ λmv + µv on ∂Ω × R, v > 0 on Ω × R, v(x,t) (3.6) T periodic in t, for some λ,µ ∈ R. Then µm (λ) ≥ µ. If in addition either Lv > 0 or A∇v,ν > λmv + µv, then µm (λ) > µ. Proof. We proceed by contradiction. Suppose that µm (λ) < µ. From (3.6), we have, for r large enough, Lv ≥ 0 on Ω × R, r + λ(1 − m) v + A∇v,ν > r + λ + µm (λ) v > 0 on ∂Ω × R, (3.7) v(x,t) T periodic in t. Then, the maximum principle implies that v is bounded from below for some positive constant. Let uλ be a positive solution of (3.5). It follows that there exists a positive constant c, such that uλ ≤ cv on Ω × R. Take c minimal with respect to T. Godoy et al. 411 this property and let w = cv − uλ . Then Lw ≥ 0, (r + λ(1 − m))w + A∇w,ν > 0 on ∂Ω × R. Now, the maximum principle implies that minΩ×[0,T] w > 0 and this leads to a contradiction with the choice of c. Finally, note that the above argument gives also the last assertion of the lemma. Lemma 3.3. The function µm is a concave function. Proof. Let λ0 ,λ1 ∈ R and let uλ0 ,uλ1 be positive solutions of (3.5) for λ = λ0 ,λ1 , respectively. For θ ∈ (0,1), let uθ = uθλ0 u1λ−1 θ , so uθ ∈ C 2+γ,1+γ/2 (Ω × R), uθ is T periodic and uθ (x,t) > 0 for all (x,t) ∈ Ω × R. For w ∈ RN and (x,t) ∈ Ω × R, let w 2A(x,t) = A(x,t)w,w. We recall that for regular u,v ∈ C 2,1 (Ω × R) → R, such that u(x,t) > 0 and v(x,t) > 0 for all (x,t) ∈ Ω × R and for β ∈ R it holds that L(uβ ) = βuβ−1 Lu − β(β − 1)∇u2 uβ−2 and L(uv) = uLv + vLu − 2A∇u, ∇v. Using these formulas and the definition of · A(x,t) , a direct computation shows that Luθ (x,t) = θ(1 − θ) uλ1 uλ0 (1−θ)/2 ∇u λ 0 u1/2 λ0 − uλ0 uλ1 θ/2 ∇u λ 1 u1/2 λ1 2 (x,t) A(x,t) (3.8) for (x,t) ∈ Ω × R, and so, Luθ ≥ 0 on Ω × R. Another computation shows that A∇uθ ,ν = θλ0 + (1 − θ)λ1 muθ + θµm λ0 + (1 − θ)µm λ1 uθ on ∂Ω × R. Then, this lemma follows from Lemma 3.2. (3.9) Remark 3.4. Lemma 3.3 implies that µm is continuous. Moreover, taking into account Corollary 2.7, we can apply [4, Lemma 1.3] (proceeding, e.g., as in [5, Remark 3.9 and Lemma 3.10]) to obtain that µm (λ) is real analytic in λ for λ > −ε for some small enough positive ε, and since µm (−λ) = µ−m (λ) we get that µm is real analytic on the whole real line. Moreover, a positive solution uλ for (3.5) can be chosen such that λ → uλ∂Ω×(0,T) is a real analytic map from R into 2−1/q,1−1/2q (∂Ω × R). Wq,T Note also that if a0 = 0, then µm (0) = 0 and that, in this case, the eigenfunctions associated for (3.5) are the constant functions. Finally, for the case a0 > 0, applying Lemma 3.2 with v = 1, λ = 0 and µ = 0, we obtain that µm (0) > 0. 2−1/q,1−1/2q Lemma 3.5. Let m1 ,m2 ∈ Wq,T µm1 (λ) > µm2 (λ) for all λ > 0. (∂Ω × R). Suppose that m1 < m2 . Then, Proof. Since for c ∈ R−{0} µcm j (λ) = µm j (λ/c), j = 1,2, we can assume, without loss of generality, that m j ∞ < 1/2, j = 1,2. For λ > 0, let Sλ,m j be defined as 2−1/q,1−1/2q before at the beginning of this section. Let Φ ∈ Wq,T (∂Ω × R) such that Φ > 0, let u j = Sλ,m j Φ, j = 1,2 and let v = u1 − u2 . A computation shows that v satisfies Lv = 0 on Ω × R and A∇v,ν + λ(1 − m1 )v = λ(m1 − m2 )v on ∂Ω × R; thus, Theorem 2.6 implies v < 0. Then, Sλ,m1 < Sλ,m2 , this gives ρλ,m1 < ρλ,m2 , and so, µm1 (λ) > µm2 (λ). 412 Principal eigenvalues for a Steklov problem In order to make explicit the dependence on L, we denote by SL,λ,m the operator Sλ,m as defined at the beginning of this section. We also denote by µm,L the function µm . Let L0 be the operator defined by L0 u = ∂u/∂t − div(A∇u) + b, ∇u. We have the following lemma. Lemma 3.6. Suppose that a0 = 0. Then µm,L (λ) > µm,L0 (λ) for all λ ∈ R. 2−1/q,1−1/2q Proof. Suppose that λ ≥ 0. Let Φ ∈ Wq,T (∂Ω × R), with Φ > 0, let k > m ∞ , let u = SL,λ,m Φ, let u0 = SL0 ,λ,m Φ and let v = u − u0 . Then L0 v = −a0 u < 0 on Ω × R, (λ(k − m) + 1)v + A∇v,ν = 0 on ∂Ω × (0,T), and v(x,t) T periodic in t. Thus the maximum principle gives v ≤ 0. So SL,λ,m < SL0 ,λ,m . This implies that µm,L (λ) > µm,L0 (λ). Since µm,L (λ) = µ−m,L (−λ) (and similarly for L0 ), the case λ < 0 reduces to the previous one. Remark 3.7. Suppose that a0 = 0. Let k ∈ R, k > 2−1/q,1−1/2q Sk : Wq,T bj 1≤ j ≤N 2−1/q,1−1/2q (∂Ω × R) −→ Wq,T ∞, let (∂Ω × R) (3.10) be defined by (2.17) and (2.18) taking b0 = k and let ρk be its spectral radius. Since Φ = 1 is a positive eigenfunction associated to the eigenvalue 1/k, the KreinRutman theorem asserts that ρk = 1/k. Thus, also by KreinRutman theorem, there exists a positive eigenvector Ψ for the adjoint operator S∗k satisfying S∗k Ψ = Ψ. Moreover, such a Ψ is unique up to a multiplicative constant. Lemma 3.8. Suppose that a0 = 0 and let Sk ,Ψ be as in Remark 3.7. Then µm (0) = −Ψ,m/ Ψ,1. Proof. For λ ∈ R, let uλ be a solution of (3.5) such that λ → uλ is real analytic and such that u0 = 1 Luλ = 0 on Ω × R, kuλ + A∇uλ ,ν = λm + µm (λ) + k uλ on ∂Ω × R, (3.11) uλ (x,t) T periodic in t, we get uλ = λSk (muλ ) + (µm (λ) + k)Sk uλ and so λ Ψ,muλ + µm (λ) Ψ,uλ = 0. (3.12) Taking the derivative with respect to λ at λ = 0 and using that µm (0) = 0 and that u0 = 1, the lemma follows. 2−1/q,1−1/2q For Φ, f ∈ Wq,T 2−1/q,1−1/2q (Wq,T (∂Ω × R)) (∂Ω × R), let i(Φ), f = ∂Ω×(0,T) Φ f . . We have the following lemma. So i(Φ) ∈ T. Godoy et al. 413 Lemma 3.9. Suppose that a0 = 0 and let k,Sk ,Ψ be as in Remark 3.7. Then, 2−1/q,1−1/2q (i) for f ∈ Wq,T (∂Ω × R), we have S∗k f = i(v ∂Ω×R ), where v is the T periodic solution of the problem ∂v + div(A∇v) + b, ∇v + div(b)v = 0 on Ω × R, ∂t A∇v,ν + k + b,ν v = f on ∂Ω × R, v(x,t) (3.13) T periodic in t; (ii) Ψ ∈ C 2+γ,1+γ/2 (Ω × R) and minΩ×R Ψ > 0. Moreover, Ψ can be characterized as the (unique up to a multiplicative constant) solution of the T periodic problem ∂Ψ + div(A∇Ψ) + b, ∇Ψ + div(b)Ψ = 0 on Ω × R, ∂t A∇Ψ,ν + b,ν Ψ = 0 on ∂Ω × R, (3.14) Ψ(x,t) T periodic in t. 2−1/q,1−1/2q Proof. Note that, for f ∈ Wq,T (∂Ω × R), (3.13) has a unique T periodic 2+γ,1+γ/2 (Ω × R). Indeed, the change of variable t = T − τ reduces solution v ∈ C (3.13) to the situation studied in Theorem 2.5. In order to prove part (i) of the lemma, we must show that i(v),Φ = ∂Ω×(0,T) S(Φ) f , that is, ∂Ω×(0,T) vΦ = ∂Ω×(0,T) f u, (3.15) where u is the T periodic solution of the problem ∂u − div(A∇u) + b, ∇u = 0 on Ω × R, ∂t ku + A∇u,ν = Φ on ∂Ω × R, (3.16) u(x,t) T periodic in t. Multiplying (3.13) by u, (3.16) by v, adding, and integrating on Ω × (0,T), we get 0= ∂(uv) + Ω×(0,T) ∂t Ω×(0,T) div(uA∇v) − div(vA∇u) (3.17) + vb, ∇u + ub, ∇v + uv div(b) . The first integral vanishes by the periodicity. Taking into account the boundary conditions of (3.13) and (3.16), an application of the divergence theorem gives (3.15). To prove (ii), consider the operator 2−1/q,1−1/2q S : Wq,T 2−1/q,1−1/2q (∂Ω × R) −→ Wq,T (∂Ω × R), (3.18) 414 Principal eigenvalues for a Steklov problem defined by S f = v ∂Ω×R , where v is the solution of (3.13). Note that, via the change of variable t = T − τ, Theorem 2.6 gives that S is a compact and strongly positive operator. Thus, S has a positive spectral radius which is an eigenvalue with an associated positive T periodic eigenfunction h, that, by Theorem 2.5, belongs to C 2+γ,1+γ/2 (Ω × R). Moreover, minΩ×R h > 0. Let Ψ be as in Remark 3.7. By Lemma 3.9, h is a positive eigenvector for S∗ and so, by KreinRutman theo rem, we get Ψ = ch for some positive constant c > 0. Thus (ii) holds. We set T P(m) = max m(x,t)dt, (3.19) min m(x,t)dt. (3.20) 0 x∈∂Ω T N(m) = 0 x∈∂Ω Proceeding as in [2], it can be shown that if P(m) > 0, then there exists a T periodic curve Γ ∈ C 2 (R,∂Ω), such that T 0 m Γ(t),t dt > 0 (3.21) we fix, from now on, such a Γ. For p ∈ ∂Ω, let T p (∂Ω) denotes the tangent space to ∂Ω at p and let exp p : T p (∂Ω)→T p (∂Ω) be the geodesic exponential map defined by exp p (X) = σ p,X (1), where σ p,X is the geodesic in ∂Ω (respect to the natural Riemannian structure on ∂Ω inherit from RN ) satisfying σ p,X (0) = p, (d/ds)(σ p,X (s)) = X. Since ∂Ω is of class C 2+γ , exp p is a welldefined map. Lemma 3.10. For δ positive and small enough, there exists Λ ∈ C 1 (−δ,δ)N × R, RN+1 , (3.22) such that Λ is a diﬀeomorphism from (−δ,δ)N × R onto an open neighborhood Wδ ⊂ RN × R of the set {(T(t),t) : t ∈ R} satisfying (1) Λ((−δ,δ)N −1 × (0,δ) × R) = Wδ ∩ (Ω × R), (2) Λ((−δ,δ)N −1 × {0} × R) = Wδ ∩ (∂Ω × R), (3) Λ(0,t) = (Γ(t),t), (4) Λ(·,t) is T periodic in t. Moreover, Λ : (−δ,δ)N × R → Wδ and its inverse Θ : Wδ → (−δ,δ)N × R are of class C 2,1 on their respective domains. Proof. The map t → ν(Γ(t)) is T periodic and belongs to the class C 1+γ (R, RN ). Then, there exists a C 1+γ and T periodic map t → A(t) from R into SO(N) such that A(t)ν(Γ(0)) = ν(Γ(t)), t ∈ R. Let {X1,0 ,...,XN −1,0 } be a basis of TΓ(0) (∂Ω) and let X j (t) = A(t)X j,0 , j = 1,2,...,N − 1. Thus, each X j is a T periodic map, T. Godoy et al. 415 X j ∈ C 1+γ (R, RN ) and for each t, {X1 (t),...,XN −1 (t)} is a basis of TΓ(t) (∂Ω). For δ positive and small enough, and for (s,t) ∈ (−δ,δ)N × R, let x(s,t) = expΓ(t) 1≤ j ≤N −1 s j X j (t) − sN ν expΓ(t) s j X j (t) , (3.23) 1≤ j ≤N −1 and let Λ(s,t) = x(s,t),t . (3.24) From the wellknown properties of the exponential map, it follows easily that, for δ small enough, (s,t) → Λ(s,t) is a C 2,1 map which satisfies the properties required by the lemma. Let δ,Λ,Θ,Wδ be as in Lemma 3.10, Θ(x,t) = (Θ1 (x,t),...,ΘN+1 (x,t)). Note that, since ΘN vanishes identically on Wδ ∩ (∂Ω × R), we have ∇ΘN = −gν on Wδ ∩ (∂Ω × R) (3.25) for some g ∈ C 1 (Wδ ∩ (∂Ω × R)) satisfying g(x,t) = 0 for all (x,t) ∈ Wδ ∩ (∂Ω × R). Moreover, Θ Γ(t),t Λ (0,t) = Id, (3.26) (where Λ and Θ denote the respective (N + 1) × (N + 1) Jacobian matrix of Λ and Θ, respectively). Thus, considering the (N,N) entries in this equality and using (3.25) and that (∂ΛN )/(∂sN (0,t) ) = −ν(Γ(t)), we get g Γ(t),t = 1, ∀t ∈ R. (3.27) Lemma 3.11. Suppose that P(m) > 0, then limλ→∞ µm (λ) = −∞. Proof. Let δ,Λ,Θ,Wδ be as in Lemma 3.10. Let QT,δ = (−δ,δ)N −1 × [0,δ) × (0,T) and let DT,δ = Λ(QT,δ ) ⊂ Ω × (0,T). If f : Dδ → R (resp., f : Dδ ∩ (∂Ω × R) → R) let f # : QT,δ → R (resp., f # : (−δ,δ)N −1 × {0} × (0,T) → R) be defined by f # = f ◦ Λ. For λ > 0, let uλ ∈ C 2+γ,1+γ/2 (Ω × R) be a positive T periodic solution of (3.5), since uλ = u#λ ◦ Θ on Dδ , the equation Luλ = 0 on Dδ gives ∂u#λ − div A# ∇u#λ + b# , ∇u#λ + a#0 u#λ = 0 on QT,δ , ∂t (3.28) where A# is the N × N symmetric and positive matrix whose (i, j) entry is a#i j (s,t) = 1≤l,r ≤N alr Λ(s,t) ∂Θi ∂Θ j ∂xl ∂xr Λ(s,t) Λ(s,t) (3.29) 416 Principal eigenvalues for a Steklov problem and where b# = (b1# ,...,bN# ) with each b#j belonging to C(QT,δ , R) and independent of λ. If ν(x,t) = (ν1 (x,t),...,νN (x,t)), the boundary condition A∇uλ ,ν = λmuλ + µm (λ)uλ on ∂Ω × (0,T) ∩ Dδ (3.30) transforms into ai j Λ(s,t) 1≤i, j,l≤N ∂u#λ ∂sl (s,t) = λm # ∂Θl Λ(s,t) νi Λ(s,t) ∂x j (3.31) (s,t)u#λ (s,t) + µm (λ)u#λ (s,t) for all (s,t) ∈ (−δ,δ)N −1 × {0} × (0,T). Let g be given by (3.25). Taking into account (3.29) and (3.25) from (3.31), we get on (−δ,δ)N −1 × {0} × (0,T), (3.32) A# ∇u#λ ,eN = −λm# g # u#λ − µm (λ)g # u#λ where ∇ denotes the gradient in the variables s1 ,...,sN and eN = (0,...,0,1). T T T Note that 0 m# (0,t)dt = 0 m(Λ(0,t))dt = 0 m(Γ(t),t)dt > 0 and thus, by (3.28), g # (0,t) = 1. Since m# and g # are continuous on (−δ,δ)N −1 × {0} × R, we have, for η small enough T 0 T {σ ∈RN −1 :σ <η} 0 {σ ∈RN −1 :σ <η} m# g # (σ, 0,t)dσ dt > 0, g # (σ,0,t)dσ dt > 0. (3.33) (3.34) Let β ∈ (0,η), let h ∈ C ∞ (R), such that 0 ≤ h ≤ 1, h(ζ) = 1 for ζ < η − β, h(ζ) = 0 for ζ ≥ η and let G : RN+1 → R be defined by G(s,t) = h(s); thus, G ∈ C ∞ (RN+1 ). From (3.33) and (3.34), it is easy to see that we can pick β small enough such that T 0 T {σ ∈RN −1 :σ <η} 0 {σ ∈RN −1 :σ <η} G2 m# g # (σ, 0,t)dσ dt > 0, G2 g # (σ, 0,t)dσ dt > 0. (3.35) (3.36) Let Bη,T = {(s,t) ∈ RN × (0,T) : s < η, sN ≥ 0}. We multiply (3.28) by G2 /u#λ and then, integrating on Bη,T and taking into account that u#λ (·,0) = u#λ (·,T) and that G does not depend on t, we get Bη,T − # G2 # G2 # # # 2 = 0. # div A ∇uλ + # b , ∇uλ + a0 G uλ uλ (3.37) T. Godoy et al. 417 Let vλ# = − logu#λ . Thus vλ# ∈ C 2,1 (Bη,T ). A computation gives that − # 2 # # G2 # # # # div A ∇uλ = div G A ∇vλ − 2 A G∇vλ , ∇G uλ # − A G∇vλ# ,G∇vλ# (3.38) on Bη,T . Also, 1 −1 G2 # b , ∇u#λ = −2 GA# ∇vλ# , G A# b# 2 u#λ on Bη,T , (3.39) so, from (3.40) the divergence theorem gives T 0 {σ ∈RN −1 :σ <η} G2 A# ∇vλ# ,ν = −2 + Bη,T 1 −1 GA# ∇vλ# , ∇G + G A# b# 2 A# G∇vλ# ,G∇vλ# + Bη,T Bη,T a#0 G2 . (3.40) For w ∈ RN and (s,t) ∈ Bη,T , let w A# (s,t) = A# (s,t)w,w. Taking into account the boundary condition (3.33) and that G(s) = 0 for s = η from (3.40), we get T µm (λ) 0 {σ ∈RN −1 :σ <η} = −λ − + ≤ −λ + T 0 Bη,T Bη,T {σ ∈RN −1 :σ <η} Bη,T G2 g # m# (σ, 0,t)dσ dt 1 −1 G∇vλ# + ∇G + G A# b# (s,t) 2 1 ∇G + A# Gb# (s,t) 2 T 0 G2 g # (σ,0,t)dσ dt {σ ∈RN −1 :σ <η} dsdt A# (s,t) 2 dsdt + A# (s,t) Bη,T (3.41) a#0 G2 # G2 g # m (σ, 0,t)dσ dt 1 ∇G + A# Gb# (s,t) 2 2 2 dsdt + A# (s,t) Bη,T a#0 G2 . From this inequality, (3.35), and (3.36), the lemma follows. 4. Principal eigenvalues for periodic parabolic Steklov problems Let P(m) and N(m) be defined by (3.19) and (3.20), respectively. We have the following theorem. Theorem 4.1. Suppose either a0 > 0 and P(m) > 0 (resp., a0 > 0 and N(m) < 0) or a0 = 0, P(m) > 0, and Ψ,m < 0 (resp., a0 = 0, N(m) < 0, and Ψ,m > 0) with 418 Principal eigenvalues for a Steklov problem Ψ defined as in Remark 3.7. Then, there exists a unique positive (resp., negative) principal eigenvalue for (1.3) and the associated eigenspace is one dimensional. Proof. Suppose a0 = 0 and P(m) > 0, Ψ,m < 0. Since µm (0) = 0 and, by Lemma 3.8, µm (0) > 0 the existence of a positive principal eigenvalue λ = λ1 (m) for (1.3) follows from Lemma 3.11. Since µm does not vanish identically, the concavity of µm gives the uniqueness of the positive principal eigenvalue. Moreover, if u,v are solutions in C 2+γ,1+γ/2 (Ω × R) for (1.3), then, by the facts stated in Remark 3.1, u = cv on ∂Ω × R for some constant c. Since L(u − cv) = 0 on Ω × R, u − cv = 0 on ∂Ω × R, and u − cv is T periodic, it follows easily from the maximum principle that u = cv on Ω × R. If a0 > 0, then (by Remark 3.4) µm (0) > 0; thus, the existence follows from Lemma 3.11. The other assertions of the theorem follows as in the case a0 = 0. Taking into account that µm (−λ) = µ−m (λ) and that N(m) = −P(−m), the assertions about negative principal eigenvalues follow from the previous cases. Lemma 4.2. Suppose that a0 = 0. Then for all λ > 0, µm (λ) ≥ − P(m) λ− b T ∞ − Ω  div(b) ∂Ω ∞. (4.1) Proof. We consider first the case m ≥ 0. Let λ > 0 and let uλ be a positive solution of (3.5) normalized by uλ ∞ = 1. From ∂uλ − div A∇uλ + b, ∇uλ = 0 on Ω × (0,T), ∂t A∇uλ ,ν = λmuλ + µm (λ)uλ on ∂Ω × (0,T), (4.2) uλ (·,0) = uλ (·,T), and since b, ∇uλ = div(uλ b) − uλ div(b), integrating (4.2) on Ω × (0,T) and taking into account the periodicity of uλ and the boundary conditions, the divergence theorem gives µm (λ) ∂Ω×(0,T) uλ = −λ ∂Ω×(0,T) muλ + ∂Ω×(0,T) uλ b,ν − Ω×(0,T) uλ div(b). (4.3) Since m ≥ 0 and uλ  ≤ 1, we have ∂Ω×(0,T) muλ ≤ P(m)∂Ω, also uλ div(b) ≤ div(b) ∞ and uλ b,ν  ≤ b ∞ . Thus T ∂Ωµm (λ) ≥ µm (λ) ∂Ω×(0,T) uλ ≥ −λP(m)∂Ω − T ∂Ω b ∞ − div(b) ∞ ΩT, (4.4) so the lemma holds for m ≥ 0. For the general case, pick k ∈ R, k > m ∞ taking into account that P(m + k) = P(m) + kT and that µm+k (λ) = µm (λ) − kλ the lemma follows from the previous case applied to m + k instead of m. T. Godoy et al. 419 Corollary 4.3. Suppose that a0 = 0. Then, limλ→∞ µm (λ) ≥ −P(m)/T. Proof. Suppose that P(m)=0, then Lemmas 3.11 and 4.2 imply that limλ→∞ µm (λ) = ±∞. Also, µm is concave; thus, there exists limλ→∞ µm (λ). Then, the L’Hopital rule gives limλ→∞ µm (λ) = limλ→∞ µm (λ)/λ ≥ −P(m)/T, the last inequality by Lemma 4.2. If P(m) = 0 and if µm (λ) < 0 for some λ > 0 then, since µm (0) = 0, the concavity of µm implies that limλ→∞ µm (λ) = −∞ and the above argument applies. If µm (λ) ≥ 0 for all λ > 0, the concavity implies that µm (λ) ≥ 0 for all λ > 0 and so the corollary is also true in this case. Lemma 4.4. Suppose that a0 = 0 and let Ψ be as in Remark 3.7. Then, P(m) < 0 implies that Ψ,m < 0. Proof. Suppose P(m) < 0. By Corollary 4.3, we have limλ→∞ µm (λ) > 0. Then, since µm is concave, we have µm (0) > 0 and so Ψ,m < 0. Lemma 4.5. Suppose that a0 = 0. Then, µm vanishes identically if and only if P(m) = Ψ,m = 0. Proof. Suppose that µn vanishes identically. Lemma 3.8 gives that Ψ,m = 0. = Also, by Lemma 3.11, we have P(m) ≤ 0. Suppose that P(m) < 0 and let m(t) maxx∈∂Ω m(x,t). Since m ∈ C(∂Ω × R), it follows easily that m ∈ C[0,T]. Take ε such that 0 < εT < −P(m) and take a T periodic function m∗ ∈ C 1 (R) such < m∗ (t) < m(t) + ε, t ∈ [0,T]. Thus, −εT > P(m) = P(m) > P(m∗ ) − that m(t) ∗ ∗ εT. Thus, P(m ) < 0 and so, by Lemma 2.4, Ψ,m < 0. Thus µm∗ (0) > 0 and then, since m < m∗ , for λ positive and small enough, we have µm (λ) ≥ µm∗ (λ) > 0 contradicting our original assumption. Suppose now that P(m) = Ψ,m = 0. Then µm (0) = 0 and also, by Corollary 4.3, limλ→∞ µm (λ) ≥ 0. Then, the concavity of µm implies that µm vanishes identically on the positive axis, and so, since µm (0) = 0 the same is true for µm and since µm is analytic, vanishes on the whole line. Theorem 4.6. Suppose that a0 = 0 and that µm does not vanish identically. Then, the conditions P(m) > 0 and Ψ,m < 0 (resp., N(m) < 0 and Ψ,m > 0) are necessary for the existence of a positive (resp., negative) principal eigenvalue for (1.3). Proof. Suppose that µm (λ1 ) = 0 for some λ1 > 0. Since µm (0) = 0 and µm is concave, we must have µm (0) > 0, and so, Ψ,m < 0. To see that P(m) > 0, we proceed by contradiction. Suppose that P(m) ≤ 0. Corollary 4.3 implies that limλ→∞ µm (λ) ≥ 0 and so, since µm is concave, we have µm (λ) ≥ 0 for all λ > 0, and then, since µm (0) > 0, µm cannot vanish on the positive axis. Theorem 4.7. Suppose that a0 > 0. Then, the condition P(m) > 0 (resp., N(m) < 0) is necessary for the existence of a positive (resp., negative) principal eigenvalue for (1.3). Proof. For λ > 0, by Lemmas 3.5 and 3.6, we have µm,L (λ) ≥ µm,L (λ) ≥ µm,L 0 (λ). Suppose that P(m) ≤ 0. Corollary 4.3 gives limλ→∞ µm,L 0 (λ) ≥ 0, and so, 420 Principal eigenvalues for a Steklov problem µm,L 0 (λ) ≥ 0 for all λ > 0. Since µm (0) > 0, the concavity of µm implies that µm cannot vanish on the positive axis. 2−1/q,1−1/2q Theorem 4.8. Let λ ∈ R such that µm (λ) > 0. Then, for all h ∈ Wq,T R), the problem (∂Ω × Lu = 0 in Ω × R, A∇u,ν = λmu + h on ∂Ω × R, (4.5) u(x,t) T periodic in t, has a unique solution. Moreover, h > 0 implies that minΩ×(0,T) u > 0. Proof. Let k, Sλ,k,m , and ρλ,k,m be as in Remark 3.1. Since µm (λ) > 0, we have ρλ,k,m < 1/(λk+1), and so, since Sλ,k,m is a strongly positive operator, ((1/(λk+1))I − Sλ,k,m )−1 is a welldefined and positive operator. Equation (4.5) is equivalent to u = (λk + 1)Sλ,k,m u + Sλ,k,m h, that is, to u= So the theorem follows. 1 1 Sλ,k,m I − Sλ,k,m λk + 1 λk + 1 −1 h. (4.6) Let λ1 (m) (resp., λ−1 (m)) be the positive (resp., negative) principal eigenvalue for the weight m with the convention that λ1 (m) = +∞ (resp., λ−1 (m) = −∞) if there does not exist such a principal eigenvalue. From the properties of µm we obtain the following corollary as an immediate consequence of Theorem 4.8. Corollary 4.9. Assume that a0 > 0. Then, the interval (λ−1 (m),λ1 (m)) does not contain eigenvalues for problem (1.3). If a0 = 0, the same is true for the intervals (λ−1 (m),0) and (0,λ1 (m)). Acknowledgment This work was partially supported by Agencia Cordoba Ciencia, SecytUNC, SecytUBA, and Conicet. References [1] [2] [3] [4] [5] H. Amann, Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SIAM Rev. 18 (1976), no. 4, 620–709. A. Beltramo and P. Hess, On the principal eigenvalue of a periodicparabolic operator, Comm. Partial Diﬀerential Equations 9 (1984), no. 9, 919–941. S. Bergman and M. Schiﬀer, Kernel Functions and Elliptic Diﬀerential Equations in Mathematical Physics, Academic Press, New York, 1953. M. G. Crandall and P. H. Rabinowitz, Bifurcation, perturbation of simple eigenvalues and linearized stability, Arch. Rational Mech. Anal. 52 (1973), no. 2, 161–180. T. Godoy, E. Lami Dozo, and S. Paczka, The periodic parabolic eigenvalue problem with L∞ weight, Math. Scand. 81 (1997), no. 1, 20–34. T. Godoy et al. 421 [6] [7] P. Hess, PeriodicParabolic Boundary Value Problems and Positivity, Pitman Research Notes in Mathematics Series, vol. 247, Longman Scientific & Technical, Harlow, 1991. O. A. Ladyženskaja, V. A. Solonnikov, and N. N. Ural’ceva, Linear and Quasilinear Equations of Parabolic Type, Translations of Mathematical Monographs, vol. 23, American Mathematical Society, Rhode Island, 1967. T. Godoy: Facultad de Matemática, Astronomı́a y Fı́sica and CIEM  Conicet Universidad Nacional de Córdoba, Ciudad Universitaria, 5000 Córdoba, Argentina Email address: godoy@mate.uncor.edu E. Lami Dozo: Departement de Mathematique, Université Libre de Bruxelles, Campus Plaine 214, 1050 Bruxelles, Belgique Current address: IAMConicet and Universidad de Buenos Aires Saavedra 15, 3er Piso 1083 Buenos Aires, Argentina Email address: lamidozo@ulb.ac.be S. Paczka: Facultad de Matemática, Astronomı́a y Fı́sica and CIEM  Conicet Universidad Nacional de Córdoba, Ciudad Universitaria, 5000 Córdoba, Argentina Email address: paczka@mate.uncor.edu Advances in Operations Research Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Advances in Decision Sciences Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Journal of Applied Mathematics Algebra Hindawi Publishing Corporation http://www.hindawi.com Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Journal of Probability and Statistics Volume 2014 The Scientific World Journal Hindawi Publishing Corporation http://www.hindawi.com Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 International Journal of Differential Equations Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Volume 2014 Submit your manuscripts at http://www.hindawi.com International Journal of Advances in Combinatorics Hindawi Publishing Corporation http://www.hindawi.com Mathematical Physics Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Journal of Complex Analysis Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 International Journal of Mathematics and Mathematical Sciences Mathematical Problems in Engineering Journal of Mathematics Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Volume 2014 Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Discrete Mathematics Journal of Volume 2014 Hindawi Publishing Corporation http://www.hindawi.com Discrete Dynamics in Nature and Society Journal of Function Spaces Hindawi Publishing Corporation http://www.hindawi.com Abstract and Applied Analysis Volume 2014 Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 International Journal of Journal of Stochastic Analysis Optimization Hindawi Publishing Corporation http://www.hindawi.com Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Volume 2014