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Revista de la Unión Matemática Argentina
versión impresa ISSN 0041-6932versión On-line ISSN 1669-9637
Rev. Unión Mat. Argent. v.46 n.2 Bahía Blanca jul./dic. 2005
Principal eigenvalues for periodic parabolic Steklov problems with L∞ weight function
T. Godoy, E. Lami Dozo and S. Paczka
Partially supported by Fundacion Antorchas, Agencia Cordoba Ciencia, Secyt-UNC, Secyt UBA and Conicet
Abstract: In this paper we give sufficient conditions for the existence of a positive principal eigenvalue for a periodic parabolic Steklov problem with a measurable and essentially bounded weight function. For this principal eigenvalue its uniqueness, simplicity and monotone dependence on the weight are stated. A related maximum principle with weight is also given.
2000 Mathematics Subject Classification. 35K20, 35P05, 35B10, 35B50
Let be a and bounded domain in with and let and let , be two families of real functions defined on and respectively, satisfying for that and are periodic in and Let be a nonnegative and periodic function belonging to for some Assume in addition that for some and for all (1)
where and Let and let be the matrix whose entry is Assume also that is uniformly elliptic on , i.e., that there exists a positive constant such that (3)
for all , Let be the periodic parabolic operator defined by (4)
where denotes the standard inner product on . Finally, let be a nonnegative and periodic function in and let be the unit exterior normal to Under the above hypothesis and notations (that we assume from now on) we consider, for a periodic function (that may changes sign) the periodic parabolic Steklov principal eigenvalue problem with weight function (5)
the solutions understood in the sense of the definition 2.1 below. In order to describe our results let us introduce, for the quantities (6)
In this paper we prove (cf. Theorem 6.1) that if either and or and and if (respectively ) then there exists a positive (resp. negative) principal eigenvalue for (5), that is, a whose associated eigenfunction satisfies (5). Under an additional assumption on a similar existence result is also given for the case .
Our approach, adapted from [4] and [8], reads as follows: If we change in (5) by we have the following one parameter eigenvalue problem: given find such that this modified (5) has a solution. We prove in section 4 that this problem has a unique solution which satisfies that is real analytic and concave. We also obtain an expression for which allows us to decide the sign of In section 5 we prove that (respectively ) implie (resp. ). From these facts, and since the zeroes of the function are exactly the principal eigenvalues for (5), our results will follow.
Sections 2 and 3 have a preliminar character. In section 2 we collect some general facts about initial value parabolic problems and in section 3 we study existence and uniqueness of periodic solutions for parabolic problems and we prove some compactness and positivity properties of the corresponding solutions operators related.
Let us start introducing the notations to be used along the paper. For a topological vector space we put for its topological dual and for the corresponding evaluation bilinear map If are normed spaces and if is a bounded linear map we denote by (or simply by if no confusion arises) its corresponding operator norm. If is a real Banach, and we put for the space of the measurable functions (in the Bochner sense) such that We define also and, for the space , similarly (with the obvious changes) to the corresponding usual Lebesgue's spaces. For we put for the space of the periodic functions satisfying that We write also (respectively ) for the space of the periodic functions belonging to (resp. to ). The spaces and equipped with their respective norms and are Banach spaces. For we will identify (writing ) the spaces
and also the corresponding spaces of functions defined on
Let be the real Hilbert spaces equipped with their usual norms. For let be the space of the indefinitely differentiable Frechet functions from into , equipped with the topology of the uniform convergence on each compact subset of of the function and all its derivatives. Let be its dual space. For let be its distributional derivative defined by for all where denotes the inner product in We will say that if there exists a function (denoted by ) belonging to such that for all
For let be the bilinear form defined by
(the values on of and understood in the trace sense) and let be the bounded linear operator defined by
(7) (8)
for some positive constant depending only on and We set also (11)
and So equipped with the norm is a Banach space. With these notations we can formulate the following definition
Definition 2.1. For , and we say that is a solution of the problem (12)
if and a.e.
¿From now on, a solution of a boundary problem like (12) (except if otherwise is explicitely stated) will mean a solution in the above sense.
Remark 2.2. For with standard computations on the quadratic form give, for all
and also
where is the ellipticity constant of So, for and there exists a positive constant depending only on and such that (13)
for all and Moreover, for such and the assumptions on the coefficients of imply that there exists a positive constant such that (14)
for all and
For as in Remark 2.2, and consider the problem (16)
Note that (cf. ([12], Lemma 5.5.1) and so the initial condition makes sense. Taking into account the facts in Remark 2.2, ( [12], Theorem 5.5.1) applies to see that (16) has a unique solution Let be the linear operator defined by
Let us recall the following properties (cf. [12], Theorem 5.4.1) of the evolution operators
Remark 2.3. i) Given with there exists a positive constant such that, for (17)
ii) Since (the isomorphism given by duality) we can consider In this setting, it holds that for as above there exists a positive constant such that (18)
for and Since (and then also ) is dense in it follows that has a unique bounded extension to an operator (still denoted ) from into which satisfies, for as in (18), (19)
Finally, we recall also that for it holds that (20)
For , and consider the problem (21)
Taking into account (13), (14) and (15), ([12], Theorem 5.5.1) applies to see that (21) has a unique solution given by (22)
Remark 2.4. Observe that is a solution of the problem (23)
with defined by . Thus (23) has a unique solution given by (25)
Moreover, for we have (cf. [12], Lemma 5.5.2) (27)
¿From (27), standard computations show that there exists a positive constant independent of and such that (28)
Remark 2.5. The estimates (17), (18), (19) and (20) still hold (with another constants) for the operators given by (26) and satisfies (29)
for
Remark 2.6. For and the problem (30)
has a unique solution which satisfies in addition that
(31)
for some positive constant independent of and Indeed, the solutions of (30) are those of (23) taking there and Remark 2.4 applies.
Remark 2.7. It is easy to check that the constant in (28) and so also in Remark 2.5 and Remark 2.6 can be chosen depending only on and on an upper bound of
Lemma 2.8. Let and be as in Lemma 2.4 and let be a sequence of operators of the form
with and satisfying for each the conditions stated for at the introduction with the same and given there for Assume also that for each and and converge uniformly on to and respectively and that converges to in . Let and be sequences in and in respectively and assume that they converge to and in their respective spaces. Let be a sequence in that converges to in and let Thus the solution of the problem
converges in the norm to the solution of (30).
Proof. For as in Remark 2.2, , and , let be the solution of
and let be the solution of (24). We have
(32)
Our assumptions imply that and that uniformly on From Remarks 2.6 and 2.7 we have that is a bounded sequence. Then from (33) Thus from Remark 2.6 applied to (32) we obtain Since and the lemma follows.
Lemma 2.9. Assume that and are nonnegative. Then the solution of (30) is nonnegative.
Proof. We pick sequences and as in Lemma 2.8 satisfying in addition that and such that and belong to belongs to and Let be as in the proof of Lemma 2.8. Thus (cf. e.g., Theorem 5.3 in [9], p. 320)). The classical maximum principle gives and since by Lemma 2.8 in we get Since the solution of (30) is given by the lemma follows
Remark 2.10. Let us recall some well known facts concerning Sobolev spaces (see e.g. [9], Lemma 3.3, p 80 Lemma 3.4, p. 82)
i): For and with we have and the restriction map (in the trace sense) is continuous from into
ii) For with it holds that for and for such there exists a positive constant independent of such that
iii) For the following facts hold:
for some with continuous inclusion.
for some and with continuous inclusion.
iv) For let be defined by if and if Thus if and for all if in both cases with continuous inclusion
Remark 2.11. For it holds that continuously for some In this case, for , let be the space of the functions that satisfy (in the pointwise sense) where (34)
Let us recall that for such and for and there exists a unique satisfying almost everywhere
(for a proof, see [9], Theorem 9.1, p. 341, concerning the Dirichlet problem and its extension, to our boundary conditions, indicated there (at the end of chapter 4, paragraph 9, p. 351). Moreover, there exists a positive constant independent of and such that
Lemma 2.12. i) For is a compact and positive operator.
ii) Let with For and the restriction of to belongs to and there exists a positive constant such that for all
iii) for and and is a bounded operator from into
iv) For it hold that and is a bounded operator from into Moreover, if and then
v) For and is a compact and strongly positive operator
Proof. By Lemma 2.9 is a positive operator. It is also compact because is continuous (cf. Remark 2.5) and has compact inclusion in Thus (i) holds.
To see (ii) we pick a strictly increasing sequence of positive numbers such that for all and we pick also a sequence of functions in satisfying for for Let and let and be the sequences of functions on inductively defined by and by respectively Then, for all
(35)
Let be defined by and by (with as in (iv) of Remark 2.10) and let For the rest of the proof will denote a positive constant independent of non necessarily the same at each occurrence (even in a same chain of inequalities). We claim that for (36)
with their respective norms bounded by .
If (36) holds, for Remark 2.10 (iv) gives Taking into account that on Remark 2.11 gives
and so (ii) holds.
To prove the claim we proceed inductively. Since satisfies 29, Remark 2.6 gives and so Then, by Remark 2.11, and so Since on and we get also that Thus (36) holds for Suppose that it holds for some Then
Since it follows that and so, from (35), a similar estimate holds for This complete the proof of the claim.
The imbedding theorems for Sobolev spaces and (ii) imply (iii). The first part of (iv) is again obtained applying (ii) with To see the second part of (iv), we observe that if and then and, by Lemma 2.9, Let and be as in the proof of (ii), Since the boundary condition for holds in the pointwise sense. Now, the Hopf parabolic maximum principle applied to
jointly with the fact that on gives (iv).
To see (v), let and let Since (with given by (34)), from (ii) we can consider the bounded operator defined by Since the operator is continuous from into and the inclusion map is compact, we obtain the compactness assertion of (v). Finally, the strong positivity in (v) follows from (iv).
Lemma 2.13. i) If and then for
ii) If and are nonnegative functions and if either or then
Proof. Let , be the positive cones in and respectively and let be the closure of in Observe that if then . Indeed, if not, the Hann Banach Theorem gives such that and For let be defined by Thus for all Since is reflexive there exists such that for all In particular we have for all This implies that and so which contradicts Thus
Let so and then there exists a sequence of nonnegative functions in that converges to in Since is continuous and, by Lemma 2.12 (i), it is a positive operator on we have and so (i) holds.
To see (ii), observe that and so (i) gives (38)
is the solution of the problem
Then, by (i), in and since (because either or ) we conclude that for some the set
has positive measure. Then, since , Lemma 2.12 (iv) gives for all Now (ii) follows from (38) and (39).
Remark 2.14. Let us recall the following version of the Krein Rutman Theorem for Banach lattices and one of its corollaries (for a proof, see e.g., [5], Theorem 12.3 and Corollary 12.4)
i) Let be a Banach lattice with cone positive and let be a bounded, compact, positive and irreducible linear operator. Then has a positive spectral radius which is an algebraically simple eigenvalue of and The associated eigenspaces are spanned by a quasi interior eigenvector and a strictly positive eigenfunctional respectively. Moreover, is the only eigenvalue of having a positive eigenvector.
ii) For and as above and for a positive the equation has a unique positive solution if no positive solution if and no solution at all if In particular this implies that if for some positive then
We recall also that a point is a quasi interior point if and only if and the order interval is total (i.e. the linear span of is dense in ) and that for a measure space equipped with a positive measure on and the quasi interior points in are the functions that are strictly positive almost everywhere. Moreover, for such a bounded and positive linear operator satisfying that for all is an irreducible operator (cf [13], Proposition 3, p. 409).
Lemma 2.15. For and is a positive irreducible operator and its spectral radius satisfies
Proof. By (i) and (iv) of Lemma 2.12, is a positive, irreducible and compact operator. Thus, by the Krein Rutman Theorem, is positive and that is the unique eigenvalue with positive eigenfunctions associated. Moreover, by Lemma 2.10 (iii), these eigenfunctions belong to for . Take By Lemma 2.12 (v), is a compact and strongly positive operator which, by the Krein Rutman Theorem, has a positive spectral radius Since the eigenfunctions of belong to we have Thus, to prove the lemma, it is enough to see that
We proceed by contradiction. Suppose , let be a positive eigenfunction with eigenvalue and let . Since By Lemma 2.12 (ii), and since the maximum principle gives that either is a constant or is achieved at some point If is a constant, since the boundary condition (which is satisfied in the pointwise sense because ) implies which is impossible and if the maximum is achieved at some point we would have in contradiction with the boundary condition.
with norm
Lemma 3.1. For and the problem
has a unique solution .
(41)
Proof. Let For the solution of
is given by
(42) (43)
By Lemma 2.15, has a bounded inverse. From (25), if and only if (44)
then there exists a unique solution of in on and . For such a and for let Thus in on and Then (i.e., ) for Thus can be extended to a solution of (41) which is unique by (44).
Let be the trace operator on and for let be the trace operator defined by
For we define the linear operators
by
where is the solution of (41) given by Lemma 3.1,
respectively.
Remark 3.2. Let and be Banach spaces, and reflexive. let be a compact and linear map and an injective bounded linear operator. For finite and
is a Banach space under the norm A variant of an Aubin-Lions ´s theorem (for a proof see [10], p. 57 or Lemma 3 in [6]) asserts that if is bounded then the set is precompact in
We will apply this result to and The map is the trace map, is defined by
and Hence above is a special case of in (11) for which is naturally isometric to the space of (40).
Lemma 3.3. i) For and are bounded linear operators and is also compact
ii) If and are nonnegative and if either or then and Moreover, if then
iii) is a bounded, positive, irreducible and compact operator on
Proof. For and the periodic solution of (42) is given by (43) with given by (44). Remark 2.6 gives
So, to see that is a bounded operator, it is enough to obtain see that (45)
(for the rest of the proof will denote a positive constant independent of and non necessarily the same at each occurrence, even in a same chain of inequalities). Let Thus solves in on and Since
(27) (applied to this problem and used with and ) gives
the last inequality by Remark 2.6. So
By Lemma 2.5, has a bounded inverse, and so (44) and (46) give (45). Then is bounded and this implies the boundedness, first of and then of
To see that and are compact, we consider a bounded sequence Then, from Remark 3.2 is bounded in so has a convergent subsequence in From we have that is also compact.
Suppose now that either or and let be given by (44). For Lemma 2.13 (iv) gives for and, by Lemma 2.12 (ii), we have Then has a positive minimum on Now,
for and so, by periodicity, Since and we get that and also that Then (ii) holds and is irreducible.
Lemma 3.4. .
Proof. For consider and let Let with Thus and
Along the proof will denote a positive constant independent of and (non necessarily the same even in a same chain of inequalities). Since in , and is periodic, Remark 2.6 gives So
and a similar estimate hold for and then also for Now, solves in on and is periodic. Then, from (27) used with and we get
(47)
Now
the last inequality by Remark 2.6. Lemma 3.3 (iii) and Remark 2.6 give also
(48)
Thus and the lemma holds.
We will use the multiplication operator given by (49)
For and let us observe that satisfies
(in the sense of the definition 2.1) if and only if for each it satisfies in on , i.e., we can "add" to both sides in the boundary condition of (50).
(50)
Lemma 3.5. i) For each there exists such that for and such that the problem (50) has a unique solution for all Moreover, it satisfies if
ii) For such and the solution operator is a bounded linear operator from into whose norm is uniformly bounded on for
Proof. Let such that By Lemma 3.4 there exists such that for For we have and so has a bounded inverse. If solves (50), it solves in on and so
| (51) |
Thus the solution of (50), if exists, is unique and given by (51).
To prove existence, consider the function defined by (51). It solves
and so
(52)
(53)
Then (52) can be rewritten as
and so solves (50).
Suppose now By (ii) and (iii) of Lemma 3.3, and are positive operators and also Thus (51) gives and so (i) holds. Finally, from (51) and since and are bounded and and we obtain (ii).
We will need to introduce two news operators. For let
(54)
be defined by where is the solution of (50) given by Lemma 3.5 and by respectively.
Corollary 3.6. For and as in Lemma 3.5, is a bounded, compact, positive and irreducible operator.
Proof. By (53) we have
and the corollary follows from Lemma 3.3 (iv).
4. A one parameter eigenvalue problem
Lemma 4.1. i) For and there exists a unique such that the problem
has a positive solution. Moreover, for positive and large enough let be the spectral radius of It holds that (where is the spectral radius of ).
(55)
ii) The solution space for this problem is one dimensional and for positive and large enough is an algebraically simple eigenvalue of
iii) Each positive solution of (55) satisfies
Proof. Let let be as in Lemma 3.5 and for let be the spectral radius of . From Lemma 3.6 is a compact, positive and irreducible operator on Then, by the Krein Rutman theorem, is a positive eigenvalue of with a positive eigenfunction associated. Let Thus is a periodic solution of in on . It is also positive because, by Lemma 3.5, is a positive operator. Since it follows that solves (55) for
On the other hand, if is a positive solution of (55) then in and on . So, for From Corollary 3.6 and the Krein Rutman theorem it follows that and so Thus (55) has a positive solution if and only if In particular, this gives that does not depend on the choice of and . If is another positive solution of (55), for and as above, and since and is an eigenfunction of with eigenvalue the Krein Rutman theorem gives for some . Thus
then the solution space for (55) is one dimensional. Again by the Krein Rutmnan theorem, is an algebraically simple eigenvalue of .
Finally, each positive solution of (55) satisfies
and so Lemma 3.5 (iii) gives
The aim of the rest of this section is to given some properties of the function defined, for , by Lemma 4.1. Each zero of provides a principal eigenvalue with weight and the corresponding solutions in (55) are the respective positive eigenfunctions. We will prove that the map is strictly decreasing in (Lemma 4.6) and continuous for the convergence in (Lemma 4.7) hence continuous in is concave and analytic in (cf. Corollary 4.9 and Remark 4.11).
Remark 4.2. For let be the space of the periodic functions on whose restriction to belongs to and for let be the space of the periodic functions on belonging to .
We recall that if
, for
for such a , then (cf. Remark 3.1 in [8]) the solutions of (55) belong to and so for some Thus Theorem 2.5 in [8] gives
In order to make explicit the dependence on and we will write sometimes or for the function
Lemma 4.3. Let and suppose that satisfies
(56)
for some , and If and then If in addition either or then
Proof. If and then, by Lemma 4.1, Assume that either or Since for all , it suffices to prove the lemma in the case For let be as in Lemma 3.5 and let . Let and let Thus and, since So also Now,
If since we have If then (by Lemma 4.3) and so Then and thus, from (57), Then Also, from (57),
Let be the spectral radius of Remark 2.14 (ii) gives and so
Lemma 4.4. Suppose satisfies
(58)
for some , and If and then If in addition either or then
Proof. Consider first the case when and For let be as in Lemma 3.5 and let Let be the periodic solution of in , on and let be the periodic solution of in , on . Thus and, by Lemma 3.3 (iv), Then and so also Let
Since is periodic and
If then and so where is the spectral radius of Thus, Remark 2.14 (ii) gives and so Then, by Lemma 3.3 (iii), This implies in contradiction with the assumption Thus
Assume now that either or and that If then and so and This implies and if the same conclusion is obtained. So, in both cases, (59) gives now in contradiction with Remark 2.14, (ii).
Since for we have the case and arbitrary follows from the previous one and, finally, the case follows from the case by considering the identity
Let be the operator defined by We have
Corollary 4.5. i) Suppose Then for all
ii) Suppose Then for all
Proof. let be the solution of (55). Thus (60)
If since we have then Lemma 4.4 gives (i). If then Since
(ii) follows again from Lemma 4.4.
Lemma 4.6. For with imply for all and for all .
Proof. Suppose and Let be a positive and periodic solution of
Since on and Lemma 4.4 applies to give which contradicts our assumption The case follows from the case using that .
Lemma 4.7. Let be a bounded sequence in which converges to in . Then for each .
Proof. To prove the lemma it suffices to show that for each as in the statement of the lemma there exists a subsequence such that
Let be a positive number such that for all and let Thus, by Corollary 4.5, (61)
Let be the positive periodic solution of (62)
normalized by We observe that is a bounded sequence in and so, by Lemma 3.3 (i), is bounded in Thus is bounded in and is bounded in where and are the linear maps defined in Remark 3.2 Then there exists a subsequence that converges in to some From (61), after pass to a furthermore subsequence, we can assume also that for some Thus converges in to Since and is continuous we obtain that converges in to It follows that i.e., that is a periodic solution of in in . Since and converges in to and since we get Then
Corollary 4.8. For each the map is continuous from .
Corollary 4.9. is a concave function.
Proof. Choose a sequence in that converges to in and such that for all By ([8], lemma 3.3), each is concave and the corollary follows from Lemma 3.8.
Let denote the space of the bounded linear operators on and for let be the open ball in with center and radius
Lemma 4.10. Let and let be as in Lemma 3.5. For the map is real analytic from into
Proof. Let and For the solution of (50) is periodic and solves in on , Then , i.e., we have (63)
Also, and then, from (63), An iteration of (63) gives, for ,
Since we have Thus
Since is real analytic the lemma follows.
Remark 4.11. Corollary 4.9 implies that is continuous. So, taking into account Corollary 3.3 and Lemma 4.10, ([3] lemma 1.3) applies to obtain that is real analytic in . Moreover, a positive solution for (55) can be chosen such that is a real analytic map from into
Observe also that if and then and that, in this case, the eigenfunctions associated for (55) are the constant functions. Finally, for the case when either or applying Lemma 4.3 with and we obtain that
Remark 4.12. Assume that and for large enough, consider the spectral radius of the operator Since is a positive eigenfunction associated to the eigenvalue the Krein Rutman Theorem asserts that and that there exists a positive eigenvector for the adjoint operator satisfying Moreover, such a is unique up a multiplicative constant.
Lemma 4.13. Suppose that and let and be as in remark 3.7. Then
Proof. For , let be a solution of (55) such that is real analytic and for Since
we get and so
Taking the derivative with respect to at and using that and that for the lemma follows.
We fix seen as compact Riemannian manifold of dimension For fixed in , we will find a closed curve of class and such that the tube (64)
To do let us introduce some additional notations to explain . For let denote the tangent space to at as a subspace of with the usual inner product of This Riemannian structure gives an exponential map and an area element For each where is the geodesic satisfying We have also the geodesic distance on and geodesic balls We denote the distance on given by (66)
and, for and we put for the corresponding open ball with center and radius So we have that is a cylinder. Concerning the measures on and on we denote indistinctly the measure of a Borel subset of or of
For let be an orthonormal basis of and let be the map defined by From well known properties of the exponential map there exists such that is a diffeomorphism for For such and let be the coordinate system defined by on let be the corresponding coordinate frame, let and let be the matrix whose entry is Finally, we put for the area of the unit sphere
Lemma 5.1. i) For it holds that uniformly in
ii) is doubling, that is for some independent of and
iii) Let be a Borel set. Then a.e. (the limit taken on balls in )
Proof. To obtain (i) we consider an orthonormal basis of and For small enough and we have
where Since is uniformly continuous on and we obtain (i) by taking limits.
As has finite diameter for we have (ii).
Finally, is also doubling in and so (iii) holds (cf. e.g. [11]).
Lemma 5.2. For each there exists a partition of and points in with such that is a family of disjoint sets and
Proof. Without lost of generality we can assume that For let and for let (67)
and let be the set of the density points (in the sense of Lemma 5.1, (iii)) in We fix For let be the set of the points such that
for all open ball containing and with radius Observe that for and that (from Lemma 3.16 (iii) Thus where
Given we fix such that For let and let be the partition of given by
Let and let Denote For let and let and, for let and let We also set and Since we have
Consider the case We have Also,
Since and we get ). So, the above inequalities give
For let From (68) and (69) we have
Hence
where and denote the cardinals of and respectively. Since and Lemma 3.11 gives that taking large enough and and small enough the lemma follows.
For a periodic curve and let defined by (64). We have
Lemma 5.3. Assume that is connected. Then for each there exist and such that
Proof. Let and let and be as in Lemma 5.2. For and let be a map satisfying and for and Let be defined by for for and by
for For small enough satisfies the conditions of the lemma.
Corollary 5.4. Assume that is connected and let be defined by (6). If then for positive and small enough there exists such that
Remark 5.5. Let as in Lemma 5.3. Since the map belongs to there exists a and periodic map from into such that for . Let be an orthonormal basis of and let for . Thus each is a periodic map, and for each is an orthonormal basis of For and we set
and
(70) (71)
For let and . Thus, for positive and small enough is a diffeomorphism from onto an open neighborhood of the set satisfying
is periodic in
Moreover, and its inverse are of class on their respective domains. For as above, with we have and also (cf. (3.13) and (3.14) in [8])
for some satisfying for and for . Moreover, if denotes the Jacobian matrix of at from the definition of and taking into account that the differential of at the origin is the identity on we have that for .
Lemma 5.6. Assume that is connected and that Then
Proof. Let be a sequence in that converges to a.e in and satisfying for let be a sequence of operators as in Lemma 2.8 and let be the matrix whose entry , let be a sequence in for some and such in .
For positive and small enough let be as in Corollary 5.4 and let and be as in Remark 5.5.
For let
let with
and let be the symmetric and positive matrix whose entry is let be defined on by let be defined on by and For let be a positive and periodic solution of
normalized by Let be defined by Then, a computation shows that
Let (to be chosen latter), let such that for for and let be defined by for Finally, we set and, for a definite positive matrix and we put With these notations we have, as in the proof of Lemma 3.11 in [8],
Also
(72)
Thus, since and is continuous, we get for positive and small enough. Then (for a smaller if necessary) and some positive constant we have
for large enough. Since is continuous on and we can assume also (diminishing and if necessary) that, for large enough,
¿From these inequalities it is clear that we can pick small enough in the definition of such that for large enough
(73)
(74)
We have also
so, from (73), we get positive constants and independent of and such that for all large enough. Also, since
Lemma 4.3 gives Thus is bounded, and so, after pass to a subsequence we can assume that converges to some . Since is bounded in by Lemma 3.3 and after pass to a furthermore subsequence, we can assume that converges in to some By Lemma 2.8 satisfies in , on Thus and so .
6. Principal eigenvalues for periodic parabolic Steklov problems
Let and be defined by (6). We have
Theorem 6.1. Suppose one of the following assertions i), ii), iii), holds.
i) (respectively ) and either or
ii) (respectively ), (resp. ) with defined as in remark 3.7.
Then there exists a unique positive (resp. negative) principal eigenvalue for (55) and the associated eigenspace is one dimensional.
Proof. Suppose and Since and, by Lemma 3.14, the existence of a positive principal eigenvalue for (55) follows from Lemma 5.6. Since does not vanish identically, the concavity of gives the uniqueness of the positive principal eigenvalue.
Moreover, if are solutions in for (55), then, from Lemma 4.1, on for some constant Since, for , on and on . Thus, taking large enough, Lemma 2.9 gives on
If either or then (by Remark 3.12) and so the existence follows from Lemma 5.6. The other assertions of the theorem follow as in the case Since and the assertions concerning negative principal eigenvalues reduce to the above.
Theorem 6.2. Let such that Then for all the problem
(75)
has a unique solution. Moreover implies that
Proof. Since for large enough we have and so , is a well defined and positive operator. If is a solution of (75) then so the solution, if exists, is unique. To see that it exists, consider
and observe that solves (75). Finally, if then on and since
Lemma 2.18 (iii) gives
Let (respectively ) be the positive (resp. negative) principal eigenvalue for the weight with the convention that (respectively ) if there not exists such a principal eigenvalue. From the properties of Theorem 6.2 gives the following
Corollary 6.3. Assume that either or Then the interval does not contains eigenvalues for problem (55). If and the same is true for the intervals and
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T. Godoy
Facultad de Matemática, Astronomía y Física and CIEM - Conicet,
Universidad Nacional de Córdoba,
Ciudad Universitaria,
5000 Córdoba, Argentina
godoy@mate.uncor.edu
E. Lami Dozo
Département de Mathématique,
Université Libre de Bruxelles and
Universidad de Buenos Aires - Conicet,
Campus Plaine 214, 1050 Bruxelles
lamidozo@ulb.ac.be
S. Paczka
Facultad de Matemática, Astronomía y Física and CIEM - Conicet,
Universidad Nacional de Córdoba,
Ciudad Universitaria,
5000 Córdoba, Argentina
paczka@mate.uncor.edu
Recibido: 16 de diciembre de 2005
Aceptado: 7 de agosto de 2006