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 and
let 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 T-periodic 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 one-parameter 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 well-known 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, Krein-Rutman 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 cut-off 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 Krein-Rutman theorem.


3. A one-parameter 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)|∇u|2 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
Krein-Rutman theorem asserts that ρk = 1/k. Thus, also by Krein-Rutman 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 Krein-Rutman 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 well-defined 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 well-known 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 |∂Ω|µm (λ) ≥ µm (λ)

∂Ω×(0,T)

uλ

≥ −λP(m)|∂Ω| − T |∂Ω| b

∞ − div(b)

∞ |Ω|T,

(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 well-defined 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, Secyt-UNC,
Secyt-UBA, 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 periodic-parabolic 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, Periodic-Parabolic 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
E-mail address: godoy@mate.uncor.edu
E. Lami Dozo: Departement de Mathematique, Université Libre de Bruxelles,
Campus Plaine 214, 1050 Bruxelles, Belgique
Current address: IAM-Conicet and Universidad de Buenos Aires Saavedra 15, 3er
Piso 1083 Buenos Aires, Argentina
E-mail 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
E-mail address: paczka@mate.uncor.edu

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