T. Godoy - E. Lami Dozo - S. Paczka ON THE ANTIMAXIMUM PRINCIPLE FOR PARABOLIC PERIODIC PROBLEMS WITH WEIGHT

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 { Let be a bounded domain in R N with C 2+γ boundary, 0 < γ < 1 and let τ > 0. Let ai, j x, t } { 1 i, j N b j x, t } 1 j N and a 0 x, t be τ-periodic functions in t such that a i, j, b j and a 0 belong to C γ,γ/2 R, a i, j = a j,i for 1 i, j N and a i, j x, t ξ i ξ j c ξi 2 for some c > 0 and all x, t R, ξ 1,...,ξ N R N. i, j i Let L be the periodic parabolic operator given by 1 Lu = u t i, j a i, j 2 u x i x j + j b j u x j + a 0 u 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 = mu + h in R, P,h u τ periodic in t 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 R is a principal eigenvalue for the weight m if P,h has a positive solution when h 0. The antimaximum principle can be stated as follows: DEFINITION 1. We will say that the antimaximum principle AMP holds to the right respectively to the left of a principal eigenvalue if for each h 0, h = 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 P,h has a negative solution for each, + δ h respectively δ h,. We prove that, depending on m, these two possibilities happen and that in some cases the AMP holds left and right of, 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 µ m, R, defined as the unique µ R such that the homogeneous problem Lu mu = µu in R, Pµ u τ periodic in t 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 µ m is a concave and real analytic function, for Neumann the same holds [8], Lemmas 15.1 and 15.2. A given R is a principal eigenvalue for the weight m iff µ m has a zero at We will prove that if µ m is non constant and if is a principal eigenvalue for the weight m then the AMP holds to the left of if µ m > 0, holds to the right of if µ m < 0 and holds right and left of if µ m = 0. As a consequence of these results we will give see section 3, Theorem 1, for the case a 0 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 a 0 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 P Y with norm preserving order, i.e. u,v Y, 0 < u v implies u v. Let PY denote the interior of P Y in Y. We will assume, from now on, that PY =. Its dual Y is an ordered Banach space with positive cone P = { y Y : y, y 0 for all y P } For y Y we set y = { y Y : y, y = 0 } and for r > 0, Br Y 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 Y. 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: LEMMA 1. If T 0 is a bounded operator on Y and if r 0 is an algebraically simple eigenvalue for T 0, then there exists δ > 0 such that T T 0 < δ implies that there exists a unique r T R satisfying r T r 0 < 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]. LEMMA 2. Let {T } be a compact family of compact and strongly positive operators on Y. Denote by ρ the spectral radius of T and σ its spectrum. Then for all 0 < u v in Y there exists δ u,v > 0 in R such that ρ δu,v,ρ σ = and θ I T 1 h < 0 uniformly in h [u,v] Y and θ ρ δ u,v,ρ R. Proof. We have that 1 ρ is an algebraically simple eigenvalue for T with a positive eigenvector 2 T has an eigenvector associated to the eigenvalue ρ such that, x > 0 for all x P {0}. 3 normalized by = 1 and normalized by, = 1 imply that {,,ρ Y Y R } is compact. 4 There exists r > 0 such that B Y r 0, for all. 1 and 2 follow from Krein - Rutman Theorem. For 3 {ρ, } is compact in 0, because given ρ n, the sequence T n has a subsequence still denoted T n T {T } in B Y. Taking into account 1, Lemma 1 provides an r > 0 such that T B Y, T T < r imply that 0 / ρ r,ρ + r σ T = {ρ T}, so ρ n ρ > 0. This lemma also gives { : } and { : } compact in Y and Y respectively. 4 follows remarking that {, } has a lower bound v, v P Y. Indeed, 1 2 P Y so w 1 2 = 1 2 + w P Y for w B r with

36 T. Godoy - E. Lami Dozo - S. Paczka r > 0. The open covering { B r } { of { : } admits a finite subcovering } B r j j, j = 1, 2,..., l and it is simple to obtain r j 0, 1 such that j > 1 r j 2 1 j = 1, 2,..., l, so v = r 1 2 1 j < for all and some j j depending on. We prove now the Lemma for each and h [u,v], i.e. we find δ u,v and we finish by a compactness argument thanks to 1-4. is a closed subspace of Y and then, endowed with the norm induced from Y, it is a Banach space. It is clear that is T invariant. and that T : is a compact operator. Now, ρ is a simple eigenvalue for T with eigenvector and /, but ρ > 0 and T is a compact operator, thus ρ / σ T. We have also Y = R, a direct sum decomposition with bounded projections P, P given by P y =, y and P y = y, y respectively. Let T : Y Y be defined by T = T P Thus T is a compact operator. Moreover, ρ does not belongs to its spectrum. Indeed, suppose that ρ is an eigenvalue for T, let v be an associated eigenvector. We write v = P v + P v. Then ρ v = T v = T P v and so v, but ρ / σ T. Contradic- tion. Thus, for each, ρ I T has a bounded inverse. Hence, from the compactness of the set {ρ : } it follows that there exists ε > 0 such that θ I T has a bounded inverse for θ, D where D = {θ, :, ρ ε θ ρ + ε} and that 1 BY θ I T remains bounded as,θ runs on D. 1 But θ I T : θ has a bounded inverse given by I T = θ 1 1 B I T and so θ I T remains bounded as,θ runs on D. For h [u,v] we set w,h = h, h. As θ / σ T we have 2 θ I T 1 h = [, h + θ ρ ] 1 θ I T θ ρ, h wh, and u,v w h, v, u that is wh, cu,v for some constant c u,v independent of h. Hence θ I T wh, remains bounded in Y, uni- 1 formly on θ, D and h [u,v]. Also,, h, u and since {, u } is compact in 0, it follows that, h c for some positive constant c and all and all h [u,v]. Thus the lemma follows from 4. REMARK 1. The conclusion of Lemma 2 holds if 1-4 are fulfilled.

On the antimaximum principle 37 We will use the following COROLLARY 1. Let T be continuous map from [a, b] R into B Y. If each T is a compact and strongly positive operator then the conclusion of Lemma 2 holds. 3. The AMP for periodic parabolic problems For 1 p, denote X = Lτ p 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 to 0,τ belong to L p 0,τ. We write also C 1+γ,γ τ,b R for the space of the τ - periodic Hölder continuous functions u on R satisfying the boundary condition B u = 0 and C τ R for the space of τ - periodic continuous functions on R. We set 3 Y = C 1+γ,γ τ,b R and Y = C τ 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 P Y of the positive elements in Y has non empty interior. Fix s 0 > a 0. If s s 0,, 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 τ R and δ 1 a + a 0 δ 2 for some positive constants δ 1 and δ 2, then L+a I : X Y has a bounded inverse L + a I 1 : X C 1+γ,γ B R Y, i.e. 4 L + a I 1 f C 1+γ,γ R c f Lτ p R for some positive constant c and all f [9], Lemma 3.1. So L + a I 1 : X X and its restriction L + a I 1 Y : Y Y are compact operators. Moreover, L + a I 1 Y : Y Y is a strongly positive operator [9], Lemma 3.7. If a i, j / x j C R for 1 i, j N, we recall that for f Lτ p R,

38 T. Godoy - E. Lami Dozo - S. Paczka L + a I 1 f is a weak solution of the periodic problem u t div A u + b j + a i, j u + a + a 0 u = f on R, x i 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 a i, 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. REMARK 2. Let m L τ R and let M : X X be the operator multiplication by m. Then for each R there exists a unique µ R such that the problem Pµ from the introduction has a positive solution. This is shown for > 0, a 0 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 R, a 0 L τ R start with L + r instead of L, with r R large enough. Thus µ m is well defined for all R, µ m is a concave function, µ m is real analytic in, µ m is an M simple eigenvalue for L and µ m = 0 if and only if is a principal eigenvalue for the weight m. Moreover, the positive solution u of P µm can be chosen real analytic in as a map from R into Y. As in the case m Hölder continuous we have µ m = µ m, R. We recall also that for the Dirichlet problem with a 0 0 and also for the Neumann problem with a 0 0, a 0 = 0 we have µ m 0 > 0 see [8], also [5] and [6]. Given R, we will say that the maximum principle in brief MP holds for if is not an eigenvalue for the weight m and if h X with h 0, h = 0 implies that the solution u of the problem P,h belongs to P Y. The function µ m describes what happens, with respect to the MP, at a given R for the case m Hölder continuous see [8], Theorem 16.6: µ m > 0 if and only if is not an eigenvalue and M P holds for Indeed, for h X with h 0, h = 0, for r R large enough such that a 0 m + r > 0, problem P,h is equivalent to r 1 I S u = H with S = L + r M 1 and H = r 1 S h. Now H > 0. Also µ m > 0 if and only if ρ < r 1, where ρ is the spectral radius of S so Krein - Rutman Theorem ensures, for such a u, that µ m > 0 is equivalent to u P Y. Moreover, µ m > 0 implies also that is not an eigenvalue for the weight m, since, if would be an eigenvalue with an associated eigenfunction and if u is a positive solution of Lu =

On the antimaximum principle 39 mu + µ m u, B u = 0, then, for a suitable constant c, v = u + c would be a solution negative somewhere for the problem Lv = mv + µ m u, B v = 0. Next theorem shows that µ m also describes what happens, with respect to the AMP, near to a principal eigenvalue. THEOREM 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 be a principal eigenvalue for the weight m. Finally, let u,v L p τ R for some p > N + 2, with 0 < u v. Then a If µ m vanishes identically, then for all R and all h 0, h = 0 in L p τ R, problem P,h has no solution. b If µ m < 0 respectively µ m > 0, then the AMP holds to the right of respectivley to the left and its holds uniformly on h [u,v], i.e. there exists δ u,v > 0 such that for each, + δ u,v respectively δ u,v, and for each h [u,v], the solution u,h of P,h satisfies u,h P Y. c If µ m = 0 and if µ m does not vanishes identically, then the AMP holds uniformly on h for h [u,v] right and left of, i.e. there exists δ u,v > 0 such that for 0 < < δ u,v, h [u,v], the solution u,h of P,h is in P Y. Proof. Let M : X X be the operator multiplication by m. Given a closed interval I around we choose r 0, such that r > µ m and r m a 0 > 0 for all I. For a such r and for I, let T : Y Y defined by T = L + r I 1 M + r I so each T is a strongly positive and compact operator on Y with a positive spectral radius ρ that is an algebraically simple eigenvalue for T and T. Let, be the corresponding positive eigenvectors normalized by = 1 and, = 1. By Lemma 1, ρ is real analytic in and, are continuous in. As a consequence of Krein - Rutman, we have that ρ = 1 iff is a principal eigenvalue for the weight m. So ρ = 1. Since T is strongly positive we have PY, so there exists s > 0 such that Bs Y 0, for all I. Let H = L + r 1 h, U = L + r 1 u and V = L + r 1 v. The problem Lu = mu + h on R, B u = 0 on R is equivalent to 5 u = I T 1 H and u h v implies U H V. So, we are in the hypothesis of our Lemma 2 and 1 from its proof we get that ρ I T remains bounded for near to and from 2 with θ = ρ = 1 we obtain 6 u = [, H + 1 ρ ] 1 I T 1 ρ, H wh,

40 T. Godoy - E. Lami Dozo - S. Paczka where w H, = H, H. T = ρ is equivalent to L = ρ m + r > 0 this implies 1 7 µ m = r ρ ρ 1. 1 ρ 1 and, since If µ m vanishes identically then ρ = 1 for all and 5 has no solution for all h 0, h = 0. This gives assertion a of the theorem. For b suppose that µ m = 0. Taking the derivative in 7 at = and recalling that ρ = 1 we obtain 1 ρ = rρ µ m so ρ = µ m / µ m r. We have chosen r > µ m, thus µ m > 0 implies ρ < 0 and µ m < 0 implies ρ > 0 and then, proceeding as at the end of the proof of Lemma 2, assertion b of the theorem follows from 6. If µ m = 0, since µ m is concave and analytic we have µ m < 0 for =. Then 1/ρ has a local maximum at and c follows from 6 as above. To formulate conditions on m to fulfill the assumptions of Theorem 1 we recall the quantities P m = τ 0 ess sup x m x, t dt, N m = τ 0 ess in f x m x, t dt. The following two theorems describe completely the possibilities, with respect to the AMP, in Neumann and Dirichlet cases with a 0 0. THEOREM 2. Let L be given by 1. Assume that either B u = 0 is the Neumann condition and a 0 0, a 0 = 0 or that B u = 0 is the Dirichlet condition and a 0 0. Assume in addition that a i, 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 that is positive negative and the AMP holds to the right to the left of 2 If P m > 0, N m < 0 then there exist two principal eigenvalues 1 < 0 and 1 > 0 and the AMP holds to the right of 1 and to the left of 1. 3 If P m = then N m = 0 then there are no principal eigenvalues. Moreover if u,v L p 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 1 that is positive [9]. Since µ m 0 > 0, µ m 1 = 0 and µ is concave, we have µ m 1 < 0 and b of Theorem 1 applies. If P m > 0 and N m < 0 then there exist two eigenvalues 1 < 0 and 1 > 0 because µ m = µ m and in this case µ m is concave we have µ m 1 > 0 and µ m 1 < 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 µ m µ m 0 > 0 [9]. So 3 holds. Results in [9] for the Dirichlet problem remain valid for the Neumann condition with a 0 0, a 0 = 0, so the above proof holds. REMARK 3. If a 0 = 0 in the Neumann problem then 0 = 0 is a principal eigenvalue and µ m 0 = 0. To study this case we recall that L + 1 1 has a positive eigenvector X Y provided by the Krein - Rutman Theorem and 4. Then µ m 0 =, m /, 1 where, m = 0,τ ψm makes sense because L p 0,τ [9], remark 3.8. Indeed, let u be a positive τ - periodic solution of Lu = mu + µ m u on R, B u = 0 with u real analytic in and with u 0 = 1. Since vanishes on the range of L we have 0 =, mu + µ m, u. Taking the derivative at = 0 and using that u 0 = 1 we get the above expression for µ m 0. THEOREM 3. Let L be given by 1. Assume that B u = 0 is the Neumann condition and that a 0 = 0. Assume in addition that a i, j / x j C R, 1 i, j N. Let be as in Remark 3. Then, if m is not a function of t alone, we have 1 If, m < 0, m > 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, m < 0, m > 0, P m > 0 N m < 0, then there exists two principal eigenvalues, 0 and wich is positive negative and the AMP holds to the left to the right of 0 and to the right to the left of. 3 If, m = 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 1 If τ 0 m t dt = 0 then for all R the above problem Lu = mu + h has no solution. 2 If τ 0 m t dt = 0 and, m > 0, m < 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, m < 0, P m 0 then 0 is the unique principal eigenvalue and µ m 0 > 0. If, m < 0, P m > 0 then there exist two principal eigenvalues: 0 and some 1 > 0 and since µ m is concave we have µ m 0 > 0 and µ m 1 < 0. If, m = 0 and if m is not funtion of t alone, then µ m is not a constant and µ m 0 = 0 and 0 is the unique principal eigenvalue. In each case, the theorem follows from Theorem 1. The other cases are similar. If m is a function of t alone then µ m = Pm τ, this implies Pm τ =, m /, 1. If τ 0 m t dt = 0 then µ m = 0 and a of Theorem 1 applies. If τ 0 m t dt = 0 and, m > 0 then P m = N m > 0 so 0 is the unique principal eigenvalue and µ m 0 < 0, in this case Theorem 1 applies also. The case, m < 0 is similar. The remaining case τ 0 m t dt = 0 and, m = 0 is impossible because Pm τ =, m /, 1. For one dimensional Neumann problems, similarly to the elliptic case, a uniform AMP holds. THEOREM 4. Suppose N = 1, = α,β and the Neumann condition. Let L be given by Lu = u t au x x + bu x + a 0 u, where a 0, b Cτ γ,γ/2 R, a0 0 and with a Cτ 1 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 be a principal eigenvalue for Lu = mu. Without loss of generality we can assume that h p = 1 and that the AMP holds to the right of. Denote M the operator multiplication by m. Let I be a finite closed interval around and, for I, let T, and be as in the proof of Theorem 1. Each belongs to the interior of the positive cone in C and is a continuous map from I into C, thus there exist positive constants c 1, c 2 such that 8 c 1 x c 2 for all I and x. As in the proof of Lemma 2 we obtain 6. Taking into 1 account 8 and that I T remains bounded for near, in order to prove our theorem, it is enough to see that there exist a positive constant c independent 1 of h, such that I T wh, /, H < c for I. Since I T 1 wh,, H 1 = I T H, H it suffices to prove that there exists a positive constant c such that 9 H c, H,

On the antimaximum principle 43 for all h 0 with h p = 1. To show 9 we proceed by contradiction. If 9 does not holds, we would have for all j N, h j L p τ R with h j 0, h j p = 1 and j I such that 10 j,l + r I 1 h j < 1 L + r I 1 h j. j Thus lim j j,l + r I 1 h j = 0. We claim that 11 L + r I 1 h / min L + r I 1 h c [0,τ] for some positive constant c and all nonnegative and non zero h Lτ p R. If 11 holds, then j,l + r I 1 h j j, 1 min L + r I 1 h j [0,τ] c c L + r I 1 h j j, j = cc L + r I 1 h j for some positive constant c 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 ũ with ũ β, t = ũ β, t, so we can assume that ũ is 2β - periodic in x and τ periodic in t. ũ solves weakly the equation ũ t ãũ x x + bũ x +a 0 + r ũ = h in R R where ã, ã 0, h are extensions to R R like ũ, but b is extended to an odd function b in β, β then 2β periodically. In spite of discontinuities of b, a parabolic Harnack inequality holds for ũ = ũ h [10], Th. 1.1 and using the periodicity of ũ in t, we obtain 11. References [1] AMANN H., Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SIAM Review 18 4 1976, 620 709. [2] BELTRAMO A. AND HESS P. Uber den Haupteigenwert von peridiosh parabolischen Differenialoperatoren, Ph. D. Thesis, Univ of Zurich 1984 [3] CLEMENT PH. AND PELETIER L. A., An anti maximum principle for second order elliptic problems, J. Diff. Equat. 34 2 1979, 218 229. [4] CRANDALL M.G. AND RABINOWITZ P. H., Bifurcation, perturbation of simple eigenvalues and stability, Arch. Rat. Mech. Anal. 52 2 1973, 161 180. [5] DANERS D., Periodic parabolic eigenvalue problems with an indefinite weight function, Archiv der Matematik Basel 68 1997, 388 397.

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] HESS P., An antimaximum principle for linear elliptic equations with an indefinite weight function, J. Diff. Equat. 41 1981, 369 374. [8] HESS P., Periodic Parabolic Boundary Problems and Positivity, Pitman Res. Notes Math. Ser. 247, Harlow, Essex 1991. [9] GODOY T., LAMI DOZO E. AND PACZKA S., The periodic parabolic eigenvalue problem with L weight, Math. Scand. 81 1997, 20 34. [10] TRUDINGER N. S., Pointwise estimates and quasilinear parabolic equations, Comm Pure Appl. Math. 21 1968, 205 226. AMS Subject Classification: 35K20, 35P05, 47N20. Tomas GODOY, Sofia PACZKA Facultad de Matemática, Astronomía y Física 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.