Rend. Sem. Mat. Univ. Pol. Torino
Vol. 1, 60 (2002)
T. Godoy - E. Lami Dozo - S. Paczka ∗
ON THE ANTIMAXIMUM PRINCIPLE FOR PARABOLIC
PERIODIC PROBLEMS WITH WEIGHT
Abstract. We prove that an antimaximum principle holds for the Neumann and Dirichlet periodic parabolic linear problems of second order
with a time periodic and essentially bounded weight function. We also
prove that an uniform antimaximum principle holds for the one dimensional Neumann problem which extends the corresponding elliptic case.
1. Introduction
inR N with C 2+γ boundary, 0 < γ < 1 and let τ > 0. Let
Let  be a bounded domain
ai, j (x, t) 1≤i, j ≤N b j (x, t) 1≤ j ≤N and a0 (x, t) be τ -periodic functions in t such
that ai, j , b j and a0 belong to C γ ,γ /2  × R , ai, j = a j,i for 1 ≤ i, j ≤ N and
P
P
ai, j (x, t) ξi ξ j ≥ c ξi2 for some c > 0 and all (x, t) ∈  × R, (ξ1 , ..., ξ N ) ∈ R N .
i, j
i
Let L be the periodic parabolic operator given by
(1)
Lu =
X ∂u
∂u X
∂ 2u
+
bj
+ a0 u
−
ai, j
∂t
∂ xi ∂ x j
∂x j
j
i, j
Let B (u) = 0 denote either the Dirichlet boundary condition u |∂×R = 0 or the
Neumann condition ∂u/∂ν = 0 along ∂ × R.
Let us consider the problem

 Lu = λmu + h in  × R,
u τ − periodic in t
Pλ,h

B (u) = 0 on ∂ × R
where the weight function m = m (x, t) is a τ - periodic and essentially bounded
function, h = h (x, t) is τ - periodic in t and h ∈ L p ( × (0, τ )) for some p > N + 2.
We say that λ∗ ∈ R is a principal eigenvalue for the weight m if Pλ∗ ,h has a
positive solution when h ≡ 0. The antimaximum principle can be stated as follows:
D EFINITION 1. We will say that the antimaximum principle (AMP) holds to the
right (respectively to the left) of a principal eigenvalue λ∗ if for each h ≥ 0, h 6= 0 (with
∗ Research partially supported by Agencia Córdoba Ciencia, Conicet and Secyt-UNC.
33
34
T. Godoy - E. Lami Dozo - S. Paczka
h ∈ L p ( × (0, τ )) for some p > N + 2) there exists a δ (h) > 0 such that Pλ,h has
a negative solution for each λ ∈ (λ∗ , λ∗ + δ (h)) (respectively λ ∈ (λ∗ − δ (h) , λ∗ )).
We prove that, depending on m, these two possibilities happen and that in some
cases the AMP holds left and right of λ∗ , similarly to the purely stationary case where
all data are independent of t (but in that case the period becomes artificial)
Our results are described by means of the real function µm (λ) , λ ∈ R, defined as
the unique µ ∈ R such that the homogeneous problem

 Lu − λmu = µu in  × R,
u τ − periodic in t
Pµ

B (u) = 0 on ∂ × R
has a positive solution.
This function was first studied by Beltramo - Hess in [2] for Hölder continuous
weight and Dirichlet boundary condition. They proved that µm is a concave and real
analytic function, for Neumann the same holds ([8], Lemmas 15.1 and 15.2). A given
λ ∈ R is a principal eigenvalue for the weight m iff µm has a zero at λ
We will prove that if µm is non constant and if λ∗ is a principal eigenvalue for the
weight m then the AMP holds to the left of λ∗ if µ0m (λ∗ ) > 0, holds to the right of
λ∗ if µ0m (λ∗ ) < 0 and holds right and left of λ∗ if µ0m (λ∗ ) = 0. As a consequence of
these results we will give (see section 3, Theorem 1), for the case a0 ≥ 0, conditions
on m that describe completely what happens respect to the AMP, near each principal
eigenvalue.
The notion of AMP is due to Ph. Clement and L. A. Peletier [3]. They proved an
AMP to the right of the first eigenvalue for m = 1, with all data independent of t and
a0 (x) ≥ 0,
i.e. the elliptic case. Hess [7] proves the same, in the Dirichlet case, for
m ∈ C  . Our aim is to extend these results to periodic parabolic problems covering
both cases, Neumann and Dirichlet. In section 2 we give a version of the AMP for a
compact family of positive operators adapted to our problem and in section 3 we state
the main results.
2. Preliminaries
Let Y be an ordered real Banach space with a total positive cone PY with norm preserving order, i.e. u, v ∈ Y, 0 < u ≤ v implies kuk ≤ kvk . Let PY◦ denote the interior of
PY in Y. We will assume, from now on, that PY◦ 6= ∅. Its dual Y 0 is an ordered Banach
space with positive cone
P 0 = y 0 ∈ Y 0 : y 0 , y ≥ 0 for all y ∈ P
For y 0 ∈ Y 0 we set y 0⊥ = y ∈ Y : y 0 , y = 0 and for r > 0, BrY (y) will denote
the open ball in Y centered at y with radius r. For v, w ∈ Y with v < w we put
(v, w) and [u, v] for the order intervals {y ∈ Y : v < y < w} and {y ∈ Y : v ≤ y ≤ w}
On the antimaximum principle
35
respectively. B (Y ) will denote the space of the bounded linear operators on Y and for
T ∈ B (Y ) , T ∗ will be denote its adjoint T ∗ : Y 0 → Y 0 .
Let us recall that if T is a compact and strongly positive operator on Y and if
ρ is its spectral radius, then, from Krein - Rutman Theorem, (as stated, e.g., in [1],
Theorem 3.1), ρ is a positive algebraically simple eigenvalue with positive eigenvectors
associated for T and for its adjoint T ∗ .
We will also need the following result due to Crandall - Rabinowitz ([4], Lemma
1.3) about perturbation of simple eigenvalues:
L EMMA 1. If T0 is a bounded operator on Y and if r0 is an algebraically simple
eigenvalue for T0 , then there exists δ > 0 such that kT − T0 k < δ implies that there
exists a unique r (T ) ∈ R satisfying |r (T ) − r0 | < δ for which r (T ) I − T is singular.
Moreover, the map T → r (T ) is analytic and r (T ) is an algebraically simple eigenvalue for T . Finally, an associated eigenvector v (T ) can be chosen such that the map
T → v (T ) is also analytic.
We start with an abstract formulation of the AMP for a compact family of operators.
The proof is an adaptation, to our setting, of those in [3] and [7].
L EMMA 2. Let {Tλ }λ∈3 be a compact family of compact and strongly positive operators on Y . Denote by ρ (λ) the spectral radius of Tλ and σ (λ) its spectrum. Then
for all 0 < u ≤ v in Y there exists δu,v > 0 in R such that
ρ (λ) − δu,v , ρ (λ) ∩ σ (λ) = ∅ and (θ I − Tλ )−1 h < 0
uniformly in h ∈ [u, v] ⊂ Y and θ ∈ ρ (λ) − δu,v , ρ (λ) ⊂ R.
Proof. We have that
(1) ρ (λ) is an algebraically simple eigenvalue for Tλ with a positive eigenvector 8λ
(2) Tλ∗ has an eigenvector 9λ associated to the eigenvalue ρ (λ) such that h9λ , xi >
0 for all x ∈ P − {0} .
1 and
(3) 8
λ normalized by k8λ k =
9λ normalized by h9λ , 8λ i = 1 imply that
(8λ , 9λ , ρ (λ)) ∈ Y × Y 0 × R is compact.
(4) There exists r > 0 such that BrY (0) ⊂ (−8λ , 8λ ) for all λ ∈ 3.
(1) and (2) follow from Krein - Rutman Theorem.
For (3) {ρ (λ) , λ ∈ 3} is compact in (0, ∞) because given ρ (λn ) , the sequence
Tλn has a subsequence (still denoted) Tλn → Tλ∞ ∈ {Tλ }λ∈3 in B (Y ) . Taking into
account (1), Lemma 1 provides an r > 0 such that T ∈ B (Y ) , kT − T∞ k < r imply
that 0 ∈
/ (ρ (λ∞ ) − r, ρ (λ∞ ) + r ) ∩ σ (T ) = {ρ (T )} , so ρ (λn ) → ρ (λ∞ ) > 0. This
lemma also gives {8λ : λ ∈ 3} and {9λ : λ ∈ 3} compact in Y and Y 0 respectively.
(4) follows remarking that {8λ , λ ∈ 3} has a lower bound v ≤ 8λ , v ∈ PY◦ .
Indeed, 21 8λ ∈ PY◦ so w − 21 8λ = 21 8λ + w − 8λ ∈ PY◦ for w ∈ Br(λ) (8λ ) with
36
T. Godoy - E. Lami Dozo - S. Paczka
rn (λ) > 0. The open covering oBr(λ) (8λ ) of {8λ : λ ∈ 3} admits a finite subcovering
Br (λ j ) 8λ j , j = 1, 2, ..., l and it is simple to obtain r j ∈ (0, 1) such that 8λ j >
r j 21 8λ1 j = 1, 2, ..., l, so v = r 8λ1 ≤ 21 8λ j < 8λ for all λ ∈ 3 and some j ( j
depending on λ).
We prove now the Lemma for each λ and h ∈ [u, v] , i.e. we find δu,v (λ) and we
finish by a compactness argument thanks to (1)-(4).
9λ⊥ is a closed subspace of Y and then, endowed with the norm induced from Y, it
is a Banach space. It is clear that 9λ⊥ is Tλ invariant. and that Tλ|9 ⊥ : 9λ⊥ → 9λ⊥ is
λ
a compact operator. Now, ρ (λ) is a simple eigenvalue for Tλ with eigenvector
8λ and
⊥
8λ ∈
/ 9λ , but ρ (λ) > 0 and Tλ is a compact operator, thus ρ (λ) ∈
/ σ Tλ|9 ⊥ .
λ
L ⊥
We have also Y = R8λ 9λ , a direct sum decomposition with bounded projections P8λ , P9 ⊥ given by P8λ y = h9λ , yi 8λ and P9 ⊥ y = y − h9λ , yi 8λ respecλ
λ
eλ is a compact operator.
eλ : Y → Y be defined by T
eλ = Tλ P ⊥ Thus T
tively. Let T
9λ
Moreover, ρ (λ) does not belongs to its spectrum. (Indeed, suppose that ρ (λ) is an
eλ , let v be an associated eigenvector. We write v = P8λ v + P ⊥ v.
eigenvalue for T
9λ
⊥
e
/ σ Tλ|9 ⊥ . ContradicThen ρ (λ) v = Tλ (v) = Tλ P9 ⊥ v and so v ∈ 9λ , but ρ (λ) ∈
λ
λ
eλ has a bounded inverse.
tion). Thus, for each λ, ρ (λ) I − T
Hence, from the compactness of the set {ρ (λ) : λ ∈ 3} it follows that there exists
eλ has a bounded inverse for (θ, λ) ∈ D where
ε > 0 such that θ I − T
D = {(θ, λ) : λ ∈ 3, ρ (λ) − ε ≤ θ ≤ ρ (λ) + ε}
eλ −1 remains bounded as (λ, θ ) runs on D.
and that θ I − T
B(Y )
−1
=
But θ I − Tλ|9 ⊥ : 9λ⊥ → 9λ⊥ has a bounded inverse given by θ I − Tλ|9 ⊥
λ
λ
−1
eλ −1
θI − T
and so θ I − Tλ|9 ⊥
remains bounded as (λ, θ ) runs
⊥
|9λ
λ
B 9λ⊥
on D.
For h ∈ [u, v] we set wλ,h = h − h9λ , hi 8λ . As θ ∈
/ σ (Tλ ) we have
−1
h9λ , hi
θ − ρ (λ) −1
wh,λ
8λ +
(2)
(θ I − Tλ )|9 ⊥
(θ I − Tλ ) h =
λ
h9λ , hi
θ − ρ (λ)
and u − h9λ , vi 8λ ≤ wh,λ ≤ v − h9λ , ui 8λ that is wh,λ ≤ cu,v for some con
−1
wh,λ remains bounded in Y, unistant cu,v independent of h. Hence (θ I − Tλ )|9 ⊥
λ
formly on (θ, λ) ∈ D and h ∈ [u, v] . Also, h9λ , hi ≥ h9λ , ui and since {h9λ , ui} is
compact in (0, ∞) it follows that h9λ , hi ≥ c for some positive constant c and all
λ ∈ 3 and all h ∈ [u, v] . Thus the lemma follows from (4).
R EMARK 1. The conclusion of Lemma 2 holds if (1)-(4) are fulfilled.
37
On the antimaximum principle
We will use the following
C OROLLARY 1. Let λ → Tλ be continuous map from [a, b] ⊂ R into B (Y ). If
each Tλ is a compact and strongly positive operator then the conclusion of Lemma 2
holds.
3. The AMP for periodic parabolic problems
p
For 1 ≤ p ≤ ∞, denote X = L τ ( × R) the space of the τ - periodic functions
u :  × R → R (i.e. u (x, t) = u (x, t + τ ) a.e.(x, t) ∈  × R) such whose restrictions
1+γ ,γ
to  × (0, τ ) belong to L p ( × (0, τ )) . We write also Cτ,B
 × R for the space
of the τ - periodic Hölder continuous
functions u on  × R satisfying the boundary
condition B (u) = 0 and Cτ  × R for the space of τ - periodic continuous functions
on  × R. We set
1+γ ,γ
(3)
×R
Y = Cτ,B
and
Y = Cτ  × R
if B (u) = u |∂×R
if B (u) = ∂u/∂ν.
In each case, X and Y, equipped with their natural orders and norms are ordered Banach
spaces, and in the first one, Y has compact inclusion into X and the cone PY of the
positive elements in Y has non empty interior.
Fix s0 > ka0 k∞ . If s ∈ (s0 , ∞), the solution operator S of the problem
Lu + su = f on  × R, B (u) = 0, u τ − periodic,
f ∈Y
defined by S f = u, can be extended to an injective and bounded operator, that we
still denote by S, from X into Y (see [9], Lemma 3.1). This provides an extension of
the original differential operator L, which is a closed operator from a dense subspace
D ⊂ Y into X (see [9], p. 12). From now on L will denote this extension of the original
differential operator.
If a ∈ L ∞
δ2 ,
τ ( × R) and δ1 ≤ a + a0 ≤ δ2 for some positive constants δ1 and
1+γ ,γ
then L+a I : X → Y has a bounded inverse (L + a I )−1 : X → C B
 × R ⊂ Y,
i.e.
(4)
(L + a I )−1 f 1+γ ,γ
≤ c k f k L τp (×R)
C
×R
for some positive constant c and all f ([9], Lemma 3.1). So (L + a I )−1 : X → X and
−1
its restriction (L + a I )−1
|Y : Y → Y are compact operators. Moreover, (L + a I )|Y :
Y → Y is a strongly positive operator ([9], Lemma 3.7).
p
If ∂ai, j /∂ x j ∈ C  × R for 1 ≤ i, j ≤ N, we recall that for f ∈ L τ ( × R) ,
38
T. Godoy - E. Lami Dozo - S. Paczka
(L + a I )−1 f is a weak solution of the periodic problem
!
X ∂ai, j ∂u
X
∂u
bj +
− di v (A∇u) +
+ (a + a0 ) u = f on  × R,
∂t
∂ xi
∂x j
j
i
B (u) = 0 on ∂ × R
u (x, t) = u (x, t + τ )
where A is the N × N matrix whose i, j entry is ai, j (weak solutions defined as,
e.g., in [10], taking there, the test functions space adapted to the periodicity and to the
respective boundary condition) In fact, this is true for a Hölder continuous f (classical
solutions are weak solutions) and then an approximation process, using that L is closed
and (4), gives the assertion for a general f.
R EMARK 2. Let m ∈ L ∞
τ ( × R) and let M : X → X be the operator multiplication
by
m.
Then
for
each
λ
∈ R there exists a unique µ ∈ R such that the problem
Pµ from the introduction has a positive solution.
This is shown for λ > 0, a0 ≥ 0 in [9] (see Remark 3.9 and Lemma 3.10). A
slight modification of the argument used there shows that this is true for λ ∈ R, a0 ∈
L∞
τ ( × R) (start with L + r instead of L, with r ∈ R large enough). Thus µm (λ)
is well defined for all λ ∈ R, µm is a concave function, µm (λ) is real analytic in λ,
µm (λ) is an M simple eigenvalue for L and
µm (λ) = 0 if and only if λ is a principal eigenvalue for the weight m.
Moreover, the positive solution u λ of Pµm (λ) can be chosen real analytic in λ (as
a map from R into Y ). As in the case m Hölder continuous we have µm (−λ) =
µ−m (λ) , λ ∈ R. We recall also that for the Dirichlet problem with a0 ≥ 0 (and also
for the Neumann problem with a0 ≥ 0, a0 6= 0) we have µm (0) > 0 (see [8], also [5]
and [6]).
Given λ ∈ R, we will say that the maximum principle (in brief MP) holds for λ if
λ is not an eigenvalue for the weight m and if h ∈ X with h ≥ 0, h 6= 0 implies that
the solution u λ of the problem Pλ,h belongs to PY◦ .
The function µm describes what happens, with respect to the MP, at a given λ ∈ R
(for the case m Hölder continuous see [8], Theorem 16.6):
µm (λ) > 0 if and only if λ is not an eigenvalue and M P holds for λ
Indeed, for h ∈ X with h ≥ 0, h 6= 0, for r ∈ R large enough such that − ka0 k∞ −
kλmk∞ + r > 0, problem Pλ,h is equivalent to r −1 I − Sλ u = Hλ with Sλ =
(L + r − λM)−1 and Hλ = r −1 Sλ h. Now Hλ > 0. Also µm (λ) > 0 if and only
if ρ
e(λ) < r −1 , where e
ρ (λ) is the spectral radius of Sλ so Krein - Rutman Theorem
ensures, for such a u, that µm (λ) > 0 is equivalent to u ∈ PY◦ . Moreover, µm (λ) > 0
implies also that λ is not an eigenvalue for the weight m, since, if λ would be an
eigenvalue with an associated eigenfunction 8 and if u λ is a positive solution of Lu =
39
On the antimaximum principle
λmu + µm (λ) u, B (u) = 0, then, for a suitable constant c, v = u λ + c8 would be a
solution negative somewhere for the problem Lv = λmv + µm (λ) u λ , B (v) = 0.
Next theorem shows that µm also describes what happens, with respect to the AMP,
near to a principal eigenvalue.
T HEOREM 1. Let L be the periodic parabolic operator given by (1) with coefficients satisfying the conditions stated there, let B (u) = 0 be either the Dirichlet
condition or the Neumann condition, consider Y given by (3), let m be a function
∗
in L ∞
τ ( × R) and let λ be a principal eigenvalue for the weight m. Finally, let
p
u, v ∈ L τ ( × R) for some p > N + 2, with 0 < u ≤ v. Then
(a) If λ → µm (λ) vanishes identically,
then for all λ ∈ R and all h ≥ 0, h 6= 0 in
p
L τ ( × R), problem Pλ,h has no solution.
(b) If µ0m (λ∗ ) < 0 (respectively µ0m (λ∗ ) > 0), then the AMP holds to the right
of λ∗ (respectivley to the left) and its holds uniformly on h ∈ [u, v] , i.e.
∗ ∗
there exists δu,v >
0 such that for each λ ∈ λ , λ + δu,v (respectively
∗
∗
λ ∈ λ − δu,v , λ ) and for each h ∈ [u, v] , the solution u λ,h of Pλ,h satisfies
u λ,h ∈ −PY◦ .
(c) If µ0m (λ∗ ) = 0 and if µm does not vanishes identically, then the AMP holds
uniformly on h for h ∈ [u, v] right and left of λ∗ , i.e. there exists δu,v > 0 such
that for 0 < |λ − λ∗ | < δu,v , h ∈ [u, v] , the solution u λ,h of Pλ,h is in −PY◦ .
Proof. Let M : X → X be the operator multiplication by m. Given a closed interval I
around λ∗ we choose r ∈ (0, ∞) such that r > λ∗ µ0m (λ∗ ) and r −kλmk∞ −ka0 k∞ > 0
for all λ ∈ I. For a such r and for λ ∈ I, let Tλ : Y → Y defined by
Tλ = (L + r I )−1 (λM + r I )
so each Tλ is a strongly positive and compact operator on Y with a positive spectral
radius ρ (λ) that is an algebraically simple eigenvalue for Tλ and Tλ∗ . Let 8λ , 9λ be the
corresponding positive eigenvectors normalized by k8λ k = 1 and h9λ , 8λ i = 1. By
Lemma 1, ρ (λ) is real analytic in λ and 8λ , 9λ are continuous in λ. As a consequence
of Krein - Rutman, we have that ρ (λ) = 1 iff λ is a principal eigenvalue for the weight
m. So ρ (λ∗ ) = 1. Since Tλ is strongly positive we have 8λ ∈ PY◦ , so there exists s > 0
such that BsY (0) ⊂ (−8λ , 8λ ) for all λ ∈ I. Let H = (L + r )−1 h, U = (L + r )−1 u
and V = (L + r )−1 v. The problem Lu λ = λmu λ + h on  × R, B (u λ ) = 0 on
∂ × R is equivalent to
(5)
u λ = (I − Tλ )−1 H
and u ≤ h ≤ v implies U ≤ H ≤ V. So, we are in the
hypothesis of our Lemma 2 and
−1 remains bounded for λ near to λ∗
from its proof we get that ρ (λ) I − Tλ|9λ⊥
∗
and from (2) with θ = ρ (λ ) = 1 we obtain
−1
h9λ , H i
1 − ρ (λ) w H,λ
8λ +
(6)
uλ =
(I − Tλ )|9 ⊥
λ
h9λ , H i
1 − ρ (λ)
40
T. Godoy - E. Lami Dozo - S. Paczka
where w H,λ = H − h9λ , H i 8λ .
λ
m8λ + r
Tλ 8λ = ρ (λ) 8λ is equivalent to L8λ = ρ(λ)
8λ > 0 this implies
λ
1
(7)
µm
=r
−1 .
ρ (λ)
ρ (λ)
1
ρ(λ)
− 1 8λ and, since
If µm vanishes identically then ρ (λ) = 1 for all λ and (5) has no solution for all h ≥ 0,
h 6= 0. This gives assertion (a) of the theorem.
For (b) suppose that µ0m (λ∗ ) 6= 0. Taking the derivative in (7) at λ = λ∗ and
recalling that ρ (λ∗ ) = 1 we obtain
µ0m λ∗ 1 − λ∗ ρ 0 λ∗ = −rρ 0 λ∗
so ρ 0 (λ∗ ) = µ0m (λ∗ ) / λ∗ µ0m (λ∗ ) − r . We have chosen r > λ∗ µ0m (λ∗ ) , thus
µ0m (λ∗ ) > 0 implies ρ 0 (λ∗ ) < 0 and µ0m (λ∗ ) < 0 implies ρ 0 (λ∗ ) > 0 and then,
proceeding as at the end of the proof of Lemma 2, assertion (b) of the theorem follows
from (6).
If µ0m (λ∗ ) = 0, since µm is concave and analytic we have µm (λ) < 0 for λ 6= λ∗ .
Then 1/ρ has a local maximum at λ∗ and (c) follows from (6) as above.
To formulate conditions on m to fulfill the assumptions of Theorem 1 we recall the
quantities
Z τ
Z τ
P (m) =
ess supx∈m (x, t) dt,
N (m) =
ess i n f x∈ m (x, t) dt.
0
0
The following two theorems describe completely the possibilities, with respect to the
AMP, in Neumann and Dirichlet cases with a0 ≥ 0.
T HEOREM 2. Let L be given by (1). Assume that either B (u) = 0 is the Neumann
condition and a0 ≥ 0, a0 6= 0 or that B (u) = 0 is the Dirichlet condition and a0 ≥ 0.
Assume in addition that ∂ai, j /∂ x j ∈ C  × R , 1 ≤ i, j ≤ N. Then
(1) If P (m) > 0 (P (m) ≤ 0) , N (m) ≥ 0 (N (m) < 0) then there exists a unique
principal eigenvalue λ∗ that is positive (negative) and the AMP holds to the right
(to the left) of λ∗
(2) If P (m) > 0, N (m) < 0 then there exist two principal eigenvalues λ−1 < 0
and λ1 > 0 and the AMP holds to the right of λ1 and to the left of λ−1 .
(3) If P (m) = then N (m) = 0 then there are no principal eigenvalues.
p
Moreover if u, v ∈ L τ ( × R) satisfy 0 < u < v, then in (1) and (2) the AMP
holds uniformly on h for h ∈ [u, v] .
On the antimaximum principle
41
Proof. We consider first the Dirichlet problem. If P (m) > 0 and N (m) ≥ 0 then
there exists a unique principal eigenvalue λ1 that is positive ([9]). Since µm (0) > 0,
µm (λ1 ) = 0 and µ is concave, we have µ0m (λ1 ) < 0 and (b) of Theorem 1 applies.
If P (m) > 0 and N (m) < 0 then there exist two eigenvalues λ−1 < 0 and λ1 > 0
because µ−m (−λ) = µm (λ) and in this case (µm is concave) we have µ0m (λ−1 ) > 0
and µ0m (λ1 ) < 0, so Theorem 1 applies. In each case Theorem 1 gives the required
uniformity. The other cases are similar. If P (m) = N (m) = 0 then m = m (t) and
µm (λ) ≡ µm (0) > 0 ([9]). So (3) holds. Results in [9] for the Dirichlet problem
remain valid for the Neumann condition with a0 ≥ 0, a0 6= 0, so the above proof
holds.
R EMARK 3. If a0 = 0 in the Neumann problem then λ0 = 0 is a principal
eigenvalue and µm (0) = 0. To study this case we recall that (L + 1)−1∗ has a
positive eigenvector 9 ∈ X 0 ⊂ Y 0 provided by the KreinR - Rutman Theorem and
(4). Then µ0m (0) = − h9, mi / h9, 1i where h9, mi = ×(0,τ ) ψm makes sense
0
because 9 ∈ L p ( × (0, τ )) ([9], remark 3.8). Indeed, let u λ be a positive τ periodic solution of Lu λ = λmu λ + µm (λ) u λ on  × R, B (u λ ) = 0 with u λ
real analytic in λ and with u 0 = 1. Since 9 vanishes on the range of L we have
0 = λ h9, mu λ i + µm (λ) h9, u λ i . Taking the derivative at λ = 0 and using that
u 0 = 1 we get the above expression for µ0m (0) .
T HEOREM 3. Let L be given by (1). Assume that B (u) = 0 is the Neumann
condition and that a0 = 0. Assume in addition that ∂ai, j /∂ x j ∈ C  × R , 1 ≤
i, j ≤ N. Let 9 be as in Remark 3.
Then, if m is not a function of t alone, we have
(1) If h9, mi < 0 (h9, mi > 0), P (m) ≤ 0 (N (m) ≥ 0) , then 0 is the unique
principal eigenvalue and the AMP holds to the left (to the right) of 0.
(2) If h9, mi < 0 (h9, mi > 0) , P (m) > 0 (N (m) < 0) , then there exists two
principal eigenvalues, 0 and λ∗ wich is positive (negative) and the AMP holds to
the left (to the right) of 0 and to the right (to the left) of λ∗ .
(3) If h9, mi = 0,hen 0 is the unique principal eigenvalue and the AMP holds left
and right of 0.
If m = m (t) is a function of t alone, then we have
Rτ
(1’) If 0 m (t) dt = 0 then for all λ ∈ R the above problem Lu = λmu + h has no
solution.
Rτ
(2’) If 0 m (t) dt 6= 0 and h9, mi > 0 (h9, mi < 0) then 0 is the unique principal
eigenvalue and the AMP holds to the right (to the left) of 0.
Moreover, if u, v ∈ X satisfy 0 < u < v, then in each case (except (1’)) the AMP
holds uniformly on h for h ∈ [u, v] .
42
T. Godoy - E. Lami Dozo - S. Paczka
Proof. Suppose that m is not a function of t alone. If h9, mi < 0, P (m) ≤ 0 then 0 is
the unique principal eigenvalue and µ0m (0) > 0. If h9, mi < 0, P (m) > 0 then there
exist two principal eigenvalues: 0 and some λ1 > 0 and since µm is concave we have
µ0m (0) > 0 and µ0m (λ1 ) < 0. If h9, mi = 0 and if m is not funtion of t alone, then
µm is not a constant and µ0m (0) = 0 and 0 is the unique principal eigenvalue. In each
case, the theorem follows from Theorem 1. The other cases are similar.
P(m)
If m is a function of t alone then µm (λ) = − P(m)
=
τ λ, this implies
τ
Rτ
h9, mi / h9, 1i . If 0 m (t) dt = 0 then µm = 0 and (a) of Theorem 1 applies. If
Rτ
0 m (t) dt 6 = 0 and h9, mi > 0 then P (m) = N (m) > 0 so 0 is the unique principal
eigenvalue and µ0m (0) < 0, in Rthis case Theorem 1 applies also. The case h9, mi < 0
τ
is similar. The remaining case 0 m (t) dt 6= 0 and h9, mi = 0 is impossible because
P(m)
= h9, mi / h9, 1i.
τ
For one dimensional Neumann problems, similarly to the elliptic case, a uniform
AMP holds.
T HEOREM 4. Suppose N = 1,  = (α, β) and the Neumann condition. Let L be
γ ,γ /2
given by Lu = u t − au x x + bu x + a0 u, where a0 , b ∈ Cτ
 × R , a0 ≥ 0 and
with a ∈ Cτ1  × R , min a (x, t) > 0. Then the AMP holds uniformly in h (i.e.
x∈×R
holds on an interval independent of h) in each situation considered in Theorem 3.
Proof. Let λ∗ be a principal eigenvalue for Lu = λmu. Without loss of generality
we can assume that khk p = 1 and that the AMP holds to the right of λ∗ . Denote M
the operator multiplication by m. Let Iλ∗ be a finite closed interval around λ∗ and, for
λ ∈ Iλ∗ , let Tλ , 8λ and 9λ be as in the proof of Theorem 1. Each 8λ belongs to the
interior of the positive cone in C  and λ → 8λ is a continuous map from Iλ∗ into
C  , thus there exist positive constants c1 , c2 such that
c1 ≤ 8λ (x) ≤ c2
(8)
for all λ ∈ Iλ∗ and x ∈ . As in the proof
of Lemma 2 we obtain (6). Taking into
−1 ∗
account (8) and that (I − Tλ )|9 ⊥
remains bounded for λ near λ , in order to
λ
prove our theorem,
to see that there exist a positive constant c independent
it is enough
−1
w H,λ of h, such that / h9λ , H i < c for λ ∈ Iλ∗ . Since
I − Tλ|9λ⊥
∞
I − Tλ|9 ⊥
λ
−1
h9λ , H i
w H,λ
−1 = I − Tλ|9 ⊥
λ
H
− 8λ ,
h9λ , H i
it suffices to prove that there exists a positive constant c such that
(9)
kH k∞ ≤ c h9λ , H i
43
On the antimaximum principle
for all h ≥ 0 with khk p = 1. To show (9) we proceed by contradiction. If (9)
does not
p
holds, we would have for all j ∈ N, h j ∈ L τ ( × R) with h j ≥ 0, h j p = 1 and
λ j ∈ Iλ∗ such that
E 1
9λ j , (L + r I )−1 h j < ( L + r I )−1 h j .
∞
j
Thus lim j →∞ 9λ j , (L + r I )−1 h j = 0. We claim that
(11)
(L + r I )−1 h / min (L + r I )−1 h ≤ c
(10)
D
∞
×[0,τ ]
p
for some positive constant c and all nonnegative and non zero h ∈ L τ ( × R) . If (11)
holds, then
E D
9λ j , (L + r I )−1 h j ≥ 9λ j , 1 min (L + r I )−1 h j ≥
×[0,τ ]
≥ c0 c ( L + r I )−1 h j 9λ j , 8λ j = cc0 ( L + r I )−1 h j ∞
∞
for some positive constant c0 independent of j, contradicting (10).
It remains to prove (11) which looks like an elliptic Harnack inequality. We may
suppose α = 0. Extending u := (L + r I )−1 h by parity to [−β, β] we obtain a function
e
u with e
u (−β, t) = e
u (β, t) , so we can assume that e
u is 2β - periodic in x and τ
periodic in t. e
u solves weakly the equation e
u t −e
ae
uxx +e
be
u x + (a0 + r )e
u =e
h in R × R
e
where e
a, e
a0 , h are extensions to R × R like e
u , but b is extended to an odd function e
b
e
in (−β, β) then 2β periodically.
In
spite
of
discontinuities
of
b,
a
parabolic
Harnack
inequality holds for e
u =e
u e
h ([10], Th. 1.1) and using the periodicity of e
u in t, we
obtain (11).
References
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[3] C LEMENT P H . AND P ELETIER L. A., An anti maximum principle for second
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44
T. Godoy - E. Lami Dozo - S. Paczka
[6] DANERS D., Existence and perturbation of principal eigenvalues for a periodic parabolic problem, Electron. J. Differ. Equ. Conf. 5 (1999), 51–67.
[7] H ESS P., An antimaximum principle for linear elliptic equations with an indefinite
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AMS Subject Classification: 35K20, 35P05, 47N20.
Tomas GODOY, Sofia PACZKA
Facultad de Matemática, Astronomı́a y Fı́sica
and CIEM-Conicet
Medina Allende y Haya de la Torre, Ciudad Universitaria
5000 Córdoba, ARGENTINA
e-mail: godoy@mate.uncor.edu
e-mail: paczka@mate.uncor.edu
Enrique LAMI DOZO
IAM, Conicet - Universidad de Buenos Aires
and Université libre de Bruxelles
Département de Mathématique
Campus Plaine CP 214
1050 Bruxelles, BELGIQUE
e-mail: lamidozo@ulb.ac.be
Lavoro pervenuto in redazione il 03.05.2001 e, in forma definitiva, il 25.02.2002.