- Email: [email protected]
On a p-Laplace Neumann problem with asymptotically asymmetric perturbations
Nonlinear Analysis 51 (2002) 369 – 389
www.elsevier.com/locate/na
On a p-Laplace Neumann problem with asymptotically asymmetric perturbations Mohssine Alif a , Pierpaolo Omarib;∗ a International
b Dipartimento
Centre for Theoretical Physics, Strada Costiera 11, I–34014 Trieste, Italy di Scienze Matematiche, Universit!a, Via A. Valerio 12=1, I–34127 Trieste, Italy Received 30 April 2001; accepted 16 May 2001
Keywords: Quasilinear elliptic equations; p-Laplacian; Neumann boundary conditions; Dancer– Fu0c12k spectrum; upper and lower solutions
1. Introduction and statements Let us consider the elliptic problem −4p u = f(u) − h(x) in ; @u=@ = 0
on @:
(1.1)
Throughout, we assume that (⊂ RN ) is a bounded domain, with a C 1; 1 boundary @, 4p u = div(|∇u|p−2 ∇u) is the p-Laplacian, with p ¿ 1, f : R → R is a continuous function and h ∈ L∞ (). By a solution of (1.1) we mean a function u ∈ W 1; p () ∩ L∞ (), such that |∇u|p−2 ∇u∇w = (f(u) − h)w
= for for every w ∈ W 1; p (). Known regularity results [15] then imply that u ∈ C 1; (), some ¿ 0. The aim of this paper is to improve and extend to the p-Laplacian, with p ¿ 1 arbitrary, a result obtained in [11] for the Laplacian, in the case where the function f exhibits an asymmetric behaviour at in?nity and may interact with the Dancer–Fu0c12k ∗
Corresponding author. E-mail addresses: [email protected] (M. Alif ), [email protected] (P. Omari).
c 2002 Elsevier Science Ltd. All rights reserved. 0362-546X/02/$ - see front matter PII: S 0 3 6 2 - 5 4 6 X ( 0 1 ) 0 0 8 3 5 - 5
370
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
spectrum of the involved diIerential operator. This is made possible also in the light of two recent contributions in this area given in [5,7] respectively. We recall that the Dancer–Fu0c12k spectrum of the Neumann p-Laplacian is de?ned as the set of all (+ ; − ) ∈ R2 such that −4p u = + |u|p−2 u+ − − |u|p−2 u−
in ;
@u=@ = 0
on @
(1.2)
admits a nontrivial solution. In [5] (see also [3], for more information about the Neumann problem) it was proved that contains, besides the trivial lines R × {1 } and {1 } × R, a ?rst nontrivial curve C which passes through (2 ; 2 ), where 1 = 0 is the principal eigenvalue and 2 ¿ 0 is the second eigenvalue. This curve is continuous, decreasing and symmetric with respect to the diagonal. Moreover, if p 6 N it is asymptotic to the lines R × {1 } and {1 } × R, while if p ¿ N is asymptotic to the = and {} = × R, where = ¿ 0 is de?ned by lines R × {} = = inf |∇u|p | u ∈ W 1; p (); |u|p = 1; min u = 0 : (1.3)
On the other hand, in [7, Theorem 8:2] the solvability of the Dirichlet problem for a quasilinear equation involving the p-Laplacian was proved in the presence of a pair of upper and lower solutions, for which no special ordering was required. In our main result, stated below as Theorem 1.1, the interaction of f with the principal eigenvalue 1 is controlled by means of a pair of upper and lower solutions, possibly without any ordering, and the interaction with the ?rst nontrivial branch C by an asymptotic condition similar to that considered in [11,16] and involving a primitive F of f. In this paper, the following de?nition of upper and lower solutions is adopted. We = is a lower solution of (1.1) if say that a function ∈ W 1; p () ∩ C 0 () |∇|p−2 ∇∇w 6 (f() − h)w (1.4)
for every w ∈ W 1; p (), with w(x) ¿ 0 a.e. in . An upper solution is de?ned similarly by reversing the inequality in (1.4). Theorem 1.1. Let (+ ; − ) ∈ C be given. Assume lim sup
f(s) 6 − |s|p−2 s
lim inf
pF(s) ¡ − |s|p
s→−∞
and lim sup s→+∞
f(s) 6 + |s|p−2 s
(1.5)
and s→−∞
or
lim inf s→+∞
pF(s) ¡ + : |s|p
(1.6)
Moreover; suppose that there exist a lower solution and an upper solution . Then; problem (1:1) has at least one solution.
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
371
The proof of Theorem 1.1 is divided in several intermediate steps and combines topological and variational methods. We point out in particular that one of the technical diLculties, which must be overcome for proving Theorem 1.1, consists in the extension to the Neumann problem of Lemma 5:3 in [5], which is given in Lemma 2.2 below. We feel that this result may have an independent interest, since the arguments developed in its proof could be used also in diIerent contexts. In order to apply Theorem 1.1, the simplest way is looking for constant upper and lower solutions. It is obvious that any ∈ R such that f() ¿ ess sup h is a lower solution and any ∈ R such that f() 6 ess inf h is an upper solution. Accordingly, we immediately get the following characterization of nonresonance, which generalizes [11, Theorem 1.1] to the quasilinear setting. Corollary 1.1. Assume (1:5) and (1:6). Then; problem (1:1) has at least one solution for any given h if and only if the range of f is R. Other assumptions, which are classical in the study of the semilinear problem (p = 2) around the principal eigenvalue and imply the existence of upper and lower solutions, are the following Landesman–Lazer conditions: there exists r ¿ 0 such that 1 f(s) sign(s) 6 h for every s; with |s| ¿ r (1.7) meas or 1 f(s) sign(s) ¿ meas
h
for every s; with |s| ¿ r:
(1.8)
Condition (1.7) yields the existence of a lower solution and an upper solution = whereas (1.8) yields the existence of a lower solution satisfying (x) ¡ 0 ¡ (x) in , = We stress that, when and an upper solution satisfying (x) ¿ 0 ¿ (x) in . compared with the semilinear case, extensively discussed for instance in [7,12,13], the construction of such upper and lower solutions appears in the quasilinear setting rather more delicate (cf. Lemma 2.3 below). Using conditions (1.7) and (1.8), we obtain the following corollary, which yields a partial extension of Theorem 1:3 in [2] to unbounded nonlinearities. Corollary 1.2. Assume (1:7); or (1:5); (1:6) and (1:8). Then; problem (1:1) has at least one solution. Now, we restrict ourselves to the case p ¿ N and brieOy consider the situation where the two-sided control imposed by (1.5) and (1.6) is replaced by the following one-sided condition lim sup s→−∞
f(s) ¡ = or |s|p−2 s
with = de?ned by (1.3).
lim sup s→+∞
f(s) = ¡ ; |s|p−2 s
(1.9)
372
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Proposition 1.1. Assume p ¿ N and (1:9). Moreover; suppose that there exist a lower solution and an upper solution . Then; problem (1:1) has at least one solution. This statement yields a partial extension of Theorem 7 in [18]. Note that in Proposition 1.1 no condition at all is assumed on f at +∞, or at −∞, respectively. It would be interesting to obtain analogous results if p 6 N . We are able to do this in the semilinear case (p = 2), or when is a ball and h is radially symmetric. In this frame, condition (1.9) must however be replaced by the stronger assumption f(s) f(s) lim sup p−2 6 0 or lim sup p−2 6 0 s s |s| |s| s→−∞ s→+∞ and a further restriction on the growth of f, involving the exponent (N (p−1))=(N −p), is required. More general results can be obtained in the one-dimensional case (N = 1) on the line of [10]. 2. Proofs We start with some preliminary results. The ?rst concerns the solvability of the problem −4p u = g(x; u) in ; @u=@ = 0
on @;
(2.1)
with g : × R → R a L∞ -Carath1eodory function, in the presence of upper and lower solutions without any special ordering. This is essentially an adaptation to the Neumann problem of [7, Theorem 8:2]. Lemma 2.1. Assume that; for a.e. x ∈ and every s ∈ R; g(x; s) = − q(x; s)|s|p−2 s− + r(x; s);
(2.2)
∞
where q; r : × R → R are L -Carath:eodory functions satisfying; for some constant M ¿ 0; 0 6 q(x; s) 6 M
for a:e: x ∈ and every s ∈ R
(2.3)
and r(x; s) =0 |s|→+∞ |s|p−1 lim
uniformly a:e: in :
(2.4)
Moreover; suppose that there exist a lower solution and an upper solution of (2:1) such that min( − ) ¡ 0: = satisfying Then; problem (2:1) has at least one solution u ∈ C 1; () min(u − ) 6 0 and
max(u − ) ¿ 0:
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
373
Proof of Lemma 2.1. We begin observing that, for each k ∈ L∞ (), there exists a unique solution u ∈ W 1; p () of the problem −4p u + |u|p−2 u = k(x) in ; @u=@ = 0
(2.5)
on @;
1=(p−1) 1=(p−1) = for which satis?es − k − ∞ 6 u(x) 6 k + ∞ in . Therefore, u ∈ C 1; () some ¿ 0 independent of k. = associated with = → C 0 () Hence, we can de?ne a solution operator T : C 0 () 0 = problem (2.1), which sends any function u ∈ C () onto the unique solution v ∈ = of C 1; ()
−4p v + |v|p−2 v = g(x; u) + |u|p−2 u
in ;
@v=@ = 0
on @:
= into C 1 () = imply that T The regularity theory and the compact embedding of C 1; () is completely continuous. Moreover, the ?xed points of T are precisely the solutions of (2.1). In the sequel, we say that a lower solution is strict if there is no solution u such that min(u − ) = 0. Similarly, an upper solution is said to be strict if there is no solution u such that max(u − ) = 0. Claim 1. Assume that ( and ) are strict lower and upper solutions of (2.1), satisfying min() − () ¿ 0, and introduce the open set = | min(u − () ¿ 0 and max(u − )) ¡ 0}: O(; ) = {u ∈ C 0 () Then, deg(I − T; O(; ) ) = 1 and, hence, problem (2.1) has at least one solution u ∈ = O(; ) ∩ C 1; (). The proof of this result follows standard lines; we refer e.g. to [7, Theorem 8:1] where the case of a quasilinear elliptic problem, involving the Dirichlet p-Laplacian, is discussed. Now, we are in position to prove Lemma 2.1. Let us de?ne, for each n, a L∞ Carath1eodory function gn : × R → R by setting, for a.e. x ∈ and every s ∈ R, 1 if s 6 − n − 1; (s + n + 1)g(x; s) − (s + n) if − n − 1 ¡ s ¡ − n; if |s| 6 n; gn (x; s) = f(x; s) (n + 1 − s)g(x; s) − (s − n) if n ¡ s ¡ n + 1; −1 if s ¿ n + 1:
374
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Testing against (u − n − 1)+ and (u + n + 1)− , it is easily seen that, for each n, any solution u of the problem −4p u = gn (x; u) in ; @u=@ = 0
on @
(2.6)
satis?es u ∞ 6 n + 1. Hence, it follows in particular that any constant A ¡ − n − 1 is a strict lower solution and any constant B ¿ n + 1 is a strict upper solution of (2.6). Claim 2. There exist an integer n= and a constant R such that; for every n ¿ n; = any solution u of (2:6); with min(u − ) 6 0 and max(u − ) ¿ 0; satis;es u ∞ ¡ R:
(2.7)
Proof of Claim 2. Assume, by contradiction, that for each k there are an integer nk and a solution uk of (2.6), with n = nk , such that min(uk − ) 6 0
and
max(uk − ) ¿ 0
(2.8)
and uk ∞ ¿ k: For each k, de?ne the function uk ; vk = uk ∞ which satis?es gnk (x; uk ) |∇vk |p−2 ∇vk ∇w = w p−1 uk ∞ for every w ∈ W 1; p (). We can write gnk (x; s) = − qk (x; s)(s− )p−1 + rk (x; s) for a.e. x ∈ and every s ∈ R, where qk and rk are L∞ -Carath1eodory functions satisfying 0 6 qk (x; s) 6 M for a.e. x ∈ and every s ∈ R and rk (x; s) =0 lim |s|→+∞ |s|p−1 uniformly with respect to k and to a.e. x ∈ . Since the sequence (gnk (·; uk )= uk p−1 ∞ )k is uniformly bounded in L∞ (), by the regularity theory and the compact embedding of = into C 1 (), = we have, possibly passing to a subsequence, that (vk )k converges C 1; () 1 = in C () to some function v, which satis?es v ∞ = 1 and |∇v|p−2 ∇v∇w = − q0 (v− )p−1 w (2.9)
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
375
for every w ∈ W 1; p (). Here, q0 ∈ L∞ () is the weak∗ limit, possibly of a subsequence, of (qk (·; uk ))k , with 0 6 q0 (x) 6 M
for a:e: x ∈ :
Now, we want to prove that v is constant in . Testing (2.9) against w = − 1, we get − p−1 q (v ) = 0, which implies that v− (x) = 0 on the set where q0 (x) ¿ 0. Testing 0 against w = v, we obtain |∇v|p =− q0 (v− )p = 0 and hence ∇v = 0 in . Therefore, we conclude that v = 1 or v = − 1 in . Accordingly, we have that uk (x) → +∞
or
uk (x) → −∞
uniformly in . This contradicts (2.8). Hence, Claim 2 is proved. == problem (2:6) has at Claim 3. There exists an integer n== such that; for every n ¿ n; least one solution un satisfying min(un − ) 6 0
and max(un − ) ¿ 0:
Proof of Claim 3. Let us de?ne the open set = | min(u − ) ¡ 0 and max(u − ) ¿ 0}: O = {u ∈ C 0 () Let n== be an integer satisfying n== ¿ max{ ∞ ; ∞ }. Fix any integer n ¿ n== and set B = n+2 and A =−B. As already observed, A and B are strict lower and upper solutions of (2.6). Assume also that there is no solution u of (2.6) with u ∈ @O, or equivalently that and are strict lower and upper solutions of (2.6); indeed, otherwise we are done. We have O ∩ OA; B = OA; B \(O= A; ∪ O= ; B ) and, by Claim 1, deg(I − Tn ; O ∩ OA; B ) = deg(I − Tn ; OA; B ) − deg(I − Tn ; OA; ) − deg(I − Tn ; O; B ) = − 1: = → C () = is the solution operator associated with problem (2.6), Here, Tn : C () = onto the unique solution v ∈ C 1; () = of which sends any function u ∈ C 0 () p−2 p−2 −4p v + |v| v = gn (x; u) + |u| u in ; 0
@v=@ = 0
0
on @:
== This Accordingly, there exists at least one solution un ∈ O of (2.6) for each n ¿ n. concludes the proof of Claim 3. By Claim 2, we have that, for each n ¿ max{n; = R}, every solution un ∈ O= of (2.6) is a solution of (2.1) too. Hence, Claim 3 yields the conclusion of Lemma 2.1. The second preliminary result extends to the Neumann problem Lemma 5:3 in [5], stated and proved there in the case of Dirichlet boundary conditions. We stress that, although the idea of the proof is substantially the same as in [5], there is an essential step (formulated below as a claim), which involves the consideration of a mixed problem on a nonsmooth domain and appears technically delicate.
376
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Lemma 2.2. Let (+ ; − ) ∈ C be given and let p0 ; q0 ∈ L∞ () satisfy 0 6 p0 (x) 6 +
and 0 6 q0 (x) 6 −
a:e: in :
(2.10)
Suppose that meas{x ∈ | p0 (x) ¿ 0} ¿ 0
and meas{x ∈ | q0 (x) ¿ 0} ¿ 0:
(2.11)
Then; any nontrivial solution u of −4p u = p0 (x)|u|p−2 u+ − q0 (x)|u|p−2 u−
in ;
@u=@ = 0
on @
(2.12)
changes sign in and p0 (x) = +
a:e: on {x ∈ | u(x) ¿ 0}
(2.13)
q0 (x) = −
a:e: on {x ∈ | u(x) ¡ 0}
(2.14)
and (i.e. u is an eigenfunction associated with (+ ; − )). = be a nontrivial solution of (2.12), which therefore Proof of Lemma 2.2. Let u ∈ C 1 () satis?es |∇u|p−2 ∇u∇w = p0 (u+ )p−1 w − q0 (u− )p−1 w (2.15)
for every w ∈ W 1; p (). First, we prove that u changes sign in . Indeed, suppose, by contradiction, that e.g. u(x) ¿ 0 in . Testing (2.15) against w = 1, we get p0 up−1 = 0;
which implies that p0 (x) = 0 a.e. on {x ∈ | u(x) ¿ 0}. Hence, testing (2.15) against w = u, we deduce that |∇u|p = 0 and therefore u is a positive constant. Accordingly, p0 (x) = 0 a.e. in , which is in contradiction with (2.11). Second, we show that (2.13) holds; condition (2.14) follows from a totally similar argument. Assume, by contradiction, that meas{x ∈ | u(x) ¿ 0 and p0 (x) ¡ + } ¿ 0:
(2.16)
We begin stating the following result, whose proof is postponed. Claim. There exists a function v ∈ W 1; p (), which changes sign on subsets of of positive measure and satis;es |∇v+ |p ¡ + |v+ |p
and
|∇v− |p ¡ − : |v− |p
(2.17)
Once this Claim is proved, the remainder of the argument proceeds exactly as in the case of Dirichlet boundary conditions (cf. [5, Lemma 5:3] or [6]). However, we give a
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
377
brief sketch for the sake of completeness. Let us consider, for each s ¿ 0, a functional Js : W 1; p () → R de?ned by p Js (u) = |∇u| − s (u+ )p
and let J˜ s be its restriction to S = {u ∈ W 1; p () | |u|p = 1}. It is proved in [5] (see also [3]) that C consists of the points (s + c(s); c(s)), for all s ¿ 0, and of their symmetric with respect to the diagonal, where c(s) = inf
max J˜ s (u):
(∈1 u∈([−1;1]
Here, we set 1 = {( ∈ C 0 ([−1; 1]; S) | ((−1) = −’1 and ((1) = ’1 }, where ’1 = (1=meas )1=p is the ?rst eigenfunction. Possibly replacing u with −u, one can suppose that + ¿ − . Then, putting s = + − − , one has − = c(s). Using the above Claim, it is possible, as in [5], to build a path ( ∈ 1 such that max J˜ s (u) ¡ − ;
(2.18)
u∈([−1;1]
thus getting a contradiction with the de?nition of c(s). Indeed, ( can be constructed as follows. One starts joining v= v p to v+ = v+ p by a path (1 , v+ = v+ p to v− = v− p by a path (2 and −v− = v− p to v= v p by a path (4 . These paths are de?ned by convex combinations on S, i.e. (1 (t) =
tv + (1 − t)v+ ; tv + (1 − t)v+ p
(4 (t) =
−tv− + (1 − t)v − tv− + (1 − t)v p
(2 (t) =
tv+ + (1 − t)v− ; tv+ + (1 − t)v− p
for t ∈ [0; 1]. Using (2.17), one veri?es that J˜ s ((i (t)) ¡ − , for each i = 1; 2; 4 and every t, and, moreover, that J˜ s (v− = v− p ) ¡ − − s. Then, one observes, as in [5], that any component of the set {u ∈ S | J˜ s (u) ¡ 1 −s} contains a critical point of J˜ s and ’1 and −’1 are the only critical points of J˜ s at levels ¡ − − s. Taking into account these facts, it is possible to join v− = v− p to ’1 , or to −’1 , by a continuous path (3 in such a way that J˜ s ((3 (t)) ¡ − − s, for every t. Assume, for instance, that ’1 is reached in this way. Then, one considers the path −(3 which joins −’1 to −v− = v− p . Again, one has J˜ s (−(3 (t)) ¡ − , for every t. Accordingly, it is proved that one can de?ne a continuous path ( on S from −’1 to ’1 , which satis?es (2.18). Proof of the Claim. If |∇u− |p ¡ − |u− |p , then we simply set v = u, which, by (2.16), satis?es (2.17). On the contrary, if |∇u− |p = − |u− |p , we must resort to the following construction. From (2.16), it follows that there exists 40 ¿ 0 such that meas{x ∈ | u(x) ¿ 40 and p0 (x) ¡ + } ¿ 0. Denote by 540 a connected component of {x ∈ | u(x) ¿ 40 } such that meas{x ∈ 540 | p0 (x) ¡ + } ¿ 0 and, for every 4 ∈ ]0; 40 [, let 54 be a connected component of {x ∈ | u(x) ¿ 4} such that 540 ⊂ 54 . Hence, we have, in particular,
378
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
meas{x ∈ 54 | p0 (x) ¡ + } ¿ 0. De?ne 50 = 4∈]0; 40 ] 54 . Clearly, 50 is open and connected, u(x) ¿ 0 in 50 and meas{x ∈ 50 | p0 (x) ¡ + } ¿ 0. De?ne, for each 4 ∈ [0; 40 ],
u(x) − 4 if x ∈ 54 ; u4 (x) = 0 if x ∈ \54 : Clearly, u4 → u0 uniformly, as 4 → 0+ . Since, for each 4 ∈ [0; 40 ], 54 is a nodal domain of u − 4, Lemma 5:2 in [9], which immediately carries over to W 1; p (), guarantees that u4 ∈ W 1; p (). Testing (2.15) against w = u0 , we get |∇u0 |p ¡ + : up 0 Since
|∇u4 |p = up 4
and
50
u0p
up 50 4
→1
50
|∇u4 |p 50
u4p
6
50
|∇u0 |p 50
u0p
50 50
u0p u4p
as 4 → 0+ ;
we can pick 4 ¿ 0 so small that u4 satis?es |∇u4 |p ¡ + : up 4 Accordingly, we ?x such an 4 and we set v1 = u4
and
1 = 54 :
Now, let 0 be a connected component of {x ∈ | u(x) ¡ 0}. It is clear that = 1 ∩ = 0 = ∅. By the regularity of @, we can choose, for each x0 ∈ @∩ = 0 , a N -dimensional open cube R(x0 ) such that R(x0 ) ∩ = 1 = ∅, and, possibly after an orthogonal change of coordinates, R(x0 ) = ]x10 − r; x10 + r[ × · · · × ]xN0 − r; xN0 + r[ for some r ¿ 0, @ ∩ R(x0 ) = {(x1 ; : : : ; xN ) ∈ R(x0 ) | xN = ’(x1 ; : : : ; xN −1 )} and ∩ R(x0 ) = {(x1 ; : : : ; xN ) ∈ R(x0 ) | xN ¡ ’(x1 ; : : : ; xN −1 )}; where (x10 ; : : : ; xN0 ) and (x1 ; : : : ; xN ) are the coordinates of x0 and x, respectively, and ’ : [x10 − r; x10 + r] × · · · × [xN0 −1 − r; xN0 −1 + r] → ]xN0 − r; xN0 + r[ is a C 1; 1 function. Note that these conditions imply the existence of a number 6(x0 ) ¿ 0 such that, for every x ∈ @ ∩ R(x0 ); N (x) ¿ 6(x0 ), where N (x) is the N th component
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
379
of the unit outer normal to at the point x. Since @ ∩ = 0 is compact, there exist points x1 ; : : : ; xp ∈ @ ∩ = 0 such that @ ∩ = 0 ⊂ R(x1 ) ∪ · · · ∪ R(xp ): Let us set A = [R(x1 ) ∪ · · · ∪ R(xp )] ∩
and
10 = @0 ∩ (\A):
Of course, if @ ∩ = 0 = ∅, then A = ∅. Observe that 1= 0 ∩ @ = ∅. Indeed, assume by contradiction that there exists a sequence (x n )n ⊂ 10 such that x n → x ∈ @. We have x ∈ @ ∩ @0 . Therefore, x ∈ R(xi ) for some i ∈ {1; : : : ; p} and hence eventually x n ∈ R(xi ) ∩ ⊂ A: a contradiction. This implies that 1= 0 = 10 . Moreover, it is obvious that 10 ∩ = 1 = ∅: Therefore, for every y ∈ 10 , there is an N -dimensional open cube Q(y) centered at y such that Q(y) ⊂ and Q(y) ∩ = 1 = ∅. Since 10 is compact, there exist points y1 ; : : : ; yq ∈ 10 such that 10 ⊂ Q(y1 ) ∪ · · · ∪ Q(yq ). Let us set B = Q(y1 ) ∪ · · · ∪ Q(yq ): Clearly, B= ∩ @ = ∅ and @0 ∩ ⊂ A ∪ B. Finally, let us de?ne 2 = 0 ∪ A ∪ B: This set 2 is open, connected and satis?es 2 \(A ∪ B) ⊂ 2 . Indeed, assume by contradiction that there exists a sequence (x n )n ⊂ 2 \(A ∪ B) such that x n → x ∈ @2 . We necessarily have (x n )n ⊂ 0 and x ∈ @0 . If x ∈ @0 ∩ , then x ∈ A ∪ B and hence eventually x n ∈ A ∪ B. Whereas, if x ∈ @0 ∩ @, then x ∈ R(xi ) for some i ∈ {1; : : : ; p} and hence eventually x n ∈ R(xi ) ∩ ⊂ A. Therefore, in both cases, we get a contradiction. Moreover, 2 satis?es the cone property, according to De?nition 4:2 in [1]. This assertion follows observing that there exists an open ?nite cone C such that if x ∈ A ∪ B, then, by construction of A and B, x is the vertex of an open ?nite cone Cx congruent to C, which is contained in A ∪ B and, therefore, in 2 . Once this property has been veri?ed for all points in A ∪ B, it is easy to extend its validity to the whole of 2 , just taking a ?nite covering, made by N -dimensional open cubes contained in = 2 , of the set 2 \(A ∪ B). Let us denote by V the closure in W 1; p (2 ) of the linear space {v ∈ W 1; p (2 ) | v(x) = 0 a:e: in a neighborhood of @2 ∩ }: V is a reOexive Banach space satisfying W01; p (2 ) ⊂ V ⊂ W 1; p (2 ). De?ne a func tional J : V → R by setting J (v) = 2 |∇v|p : Clearly, J is C 1 and weakly sequentially lower semicontinuous. Let us set = inf J (v)|v ∈ V; |v|p = 1 2
and let (vn )n be a minimizing sequence. Since (J (vn ))n is bounded, it follows that (vn )n is bounded in V and hence, possibly passing to a subsequence, (vn )n is weakly convergent to some v0 ∈ V . Since 2 has the cone property, Theorem 6:2 in [1] guarantees that W 1; p (2 ) is compactly imbedded into Lp (2 ). Hence, vn → v0 in Lp (2 ) and therefore 2 |v0 |p = 1. Moreover, one has J (v0 ) 6 lim inf J (vn ) 6 : n→+∞
380
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
This implies that J (v0 ) = , i.e. v0 is a constrained minimizer. Then, Lagrange multipliers rule yields |v0 |p−2 v0 v |∇v0 |p−2 ∇v0 ∇v = 2
2
for every v ∈ V and, in particular, for every v ∈ W 1; p (2 ) with supp(v) ⊂ 2 . This means [8, p. 828] that v0 is a local weak solution of the equation −4p 9 = |9|p−2 9
in 2 :
Now, de?ne the function u(x) p 1=p z(x) = ( 0 |u| ) 0
(2.19)
if x ∈ 0 ;
if x ∈ 2 \0 : Obviously, z ∈ V ∩ L∞ (2 ) and 2 |z|p = 1. Testing (2.15) against the function w ∈ W 1; p () de?ned by
u(x) if x ∈ 0 ; w(x) = 0 if x ∈ \0 ; we get 0
|∇u|p =
0
q0 |u|p 6 −
0
|u|p
and therefore |∇z|p 6 − : 0
We are in position to prove that ¡ − :
(2.20)
Indeed, assuming by contradiction that ¿ − and therefore, necessarily, = − , we conclude that z is a constrained minimizer for J and hence is a bounded local weak solution of (2.19), with z ∈ C 1 (2 ) (see [8, Corollary]). The strong maximum principle [17, Theorem 5] ?nally implies that z(x) ¿ 0 everywhere in 2 ; which contradicts the de?nition of z, since 2 \0 has nonempty interior. Accordingly, (2.20) holds. Now, we de?ne the function
v0 (x) in 2 ; v2 (x) = 0 in \2 ; where we can suppose, possibly replacing v0 with |v0 |, that v0 (x) ¿ 0 a.e. in 2 . Finally, we set v = v1 − v2 : Clearly, v satis?es (2.17). Thus, the proof of the Claim is ?nished. Hence, also the conclusion of Lemma 2.2 follows.
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
381
Proof of Theorem 1.1. Without loss of generality, we can assume in the course of this proof that min( − ) ¡ 0;
supf(s) ¡ + ∞ and s60
inf f(s) ¿ − ∞:
s¿0
(2.21)
Indeed, otherwise we can ?nd a lower solution ˜ and an upper solution ˜ satisfying min(˜ − ) ˜ ¿ 0. Accordingly, standard results (cf. Claim 1 in the proof of Lemma 2.1) yield the solvability of problem (1.1). Suppose, for instance, that condition (1.6) holds at +∞ and observe that, by condition (1.5) too, there exist a constant ; ¿ 0 and a sequence (Sn )n , with Sn → +∞, such that f(Sn ) Snp−1
pF(Sn ) 6 + − ; Snp
6 + − ; and
for every n:
(2.22)
Indeed, two situations may occur: lim sup s→+∞
pF(s) ¡ + sp
or
lim sup s→+∞
pF(s) = + : sp
In the former case, by (1.5) and l’Hospital rule, we necessarily have lim inf s→+∞
f(s) ¡ + sp−1
and therefore (2.22) holds. In the latter case, by (1.6) there exist a constant ; ¿ 0 and a sequence (Sn )n , with Sn → +∞, such that, for each n, pF(Sn ) 6 + − ; Snp and Sn is a local minimum point of pF(s)=sp . Hence, for each n we get = 0, that is f(Sn ) Snp−1
=
d pF(s) ds ( sp )|s=Sn
pF(Sn ) 6 + − ;: Snp
Now, for each n we de?ne the truncations
f(s) if s 6 Sn ; fn (s) = f(Sn ) if s ¿ Sn : Let us observe that, for each n and for every s ¿ Sn , p−1 fn (s) f(Sn ) Sn = 6 + − ; sp−1 s Snp−1 and pFn (s) 6 + − ;; sp
(2.23)
382
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
where Fn (s) =
s 0
fn (<) d<. Indeed, if s ¿ Sn , we have
pFn (s) pF(Sn ) pf(Sn ) s − Sn = + p−1 p s s s sp p p−1 Sn Sn Sn 1− 6 (+ − ;) + p (+ − ;) s s s
p p−1 Sn Sn = (+ − ;) (1 − p) +p s s 6 + − ;; since (1 − p)t p + pt p−1 6 1 for every t ∈ [0; 1]. Without loss of generality, we can also assume that Sn ¿ max{ ∞ ; ∞ } for every n. Accordingly, and are, respectively, a lower and an upper solution of −4p u = fn (u) − h(x) in ; @u=@ = 0
(2.24)
on @;
where fn satis?es the assumptions of Lemma 2.1. Hence, we can conclude that, for each n, problem (2.24) has at least one solution un satisfying min(un − ) 6 0
and
max(un − ) ¿ 0:
(2.25)
The solvability of problem (1.1) will be obviously achieved if we succeed in proving that there exists an integer n˜ such that max un˜ ¡ Sn˜:
(2.26)
Assume by contradiction that for every n max un ¿ Sn : Hence, in particular, un ∞ → +∞. Observe that, for each n, we can write fn (s) = pn (s)(s+ )p−1 − qn (s)(s− )p−1 + rn (s)
for every s;
where pn ; qn and rn are continuous functions such that 0 6 pn (s) 6 + ;
0 6 qn (s) 6 −
for every s
and rn (s) =0 |s|→+∞ |s|p−1 lim
uniformly with respect to n. Let us set for every n un vn = : un ∞
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
383
Clearly, vn satis?es vn ∞ = 1 and |∇vn |p−2 ∇vn ∇w = pn (un )(vn+ )p−1 w
−
qn (un )(vn− )p−1 w +
rn (un ) − h un p−1 ∞
w
for every w ∈ W 1; p (). Hence, by the regularity theory, there is a constant C such that for every n vn C 1; 6 C:
(2.27)
= into C () = implies that, possibly passing to Then, the compact imbedding of C () = to some function v, with v ∞ = 1, which a subsequence, (vn )n converges in C 1 () satis?es |∇v|p−2 ∇v∇w = p0 (v+ )p−1 w − q0 (v− )p−1 w (2.28) 1;
1
for every w ∈ W 1; p (). Here, p0 ; q0 ∈ L∞ () are the weak∗ limits, possibly of subsequences, of (pn (un ))n and (qn (un ))n , respectively, and satisfy 0 6 p0 (x) 6 +
and
0 6 q0 (x) 6 −
a:e: in : = We now distinguish two cases: It is clear that, by (2.25), v vanishes somewhere in . • p0 (x) = 0 or q0 (x) = 0 a.e. in ; • p0 (x) ¿ 0 and q0 (x) ¿ 0 on subsets of of positive measure. In the ?rst case, if e.g. q0 (x) = 0 a.e. in , then testing (2.28) against w = v− , we v− = 0 and therefore v = v+ in get ∇v− = 0 in and, since v vanishes somewhere, + p−1 . Testing (2.28) against w = 1, we obtain p0 (v ) = 0, which implies v(x) = 0 where p0 (x) ¿ 0. Finally, testing (2.28) against w = v, we conclude that ∇v = 0 in and hence v = 0 in , because v vanishes somewhere. This yields a contradiction. In the second case, Lemma 2.2 implies that v changes sign in and, p0 (x) = + where v(x) ¿ 0 and q0 (x) = − where v(x) ¡ 0. Therefore, we have + p−1 |+ − pn (un )|(v ) = (+ − pn (un ))(v+ )p−1 → 0;
|− − qn (un )|(v− )p−1 =
(− − qn (un ))(v− )p−1 → 0
and hence 1−p | un ∞ fn (un ) − [+ (vn+ )p−1 − − (vn− )p−1 ]|
=
6
1−p |[pn (un ) − + ](vn+ )p−1 − [qn (un ) − − ](vn− )p−1 + un ∞ rn (un )|
|pn (un ) − + |(v+ )p−1 +
|qn (un ) − − |(v− )p−1
384
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
+
+
+
|pn (un ) − + (vn+ )p−1 − (v+ )p−1 | |qn (un ) − − (vn− )p−1 − (v− )p−1 | 1−p |rn (un )| → 0 un ∞
as n → + ∞. Moreover, since v changes sign in , there are an integer nˆ and constants ˆ ;1 ; ;2 , with 0 ¡ ;1 ¡ 1 ¡ ;2 , such that for every n ¿ n, max un 6 ;2 : (2.29) ;1 6 −min un In order to prove (2.26), we exploit condition (2.23) and the following technical result. = there exists a simple piecewise linear path (; joining x0 Claim. For every x0 ; y0 ∈ ; to y0 and having range contained in ; with the possible exception of the end points x0 and y0 ; such that; possibly for a subsequence; 1−p | un ∞ fn (un ) − [+ vn+ − − vn− ]| → 0; (
as n → +∞; where
(
denotes a line integral.
The proof of this result, when p = 2, is given in [11]. The extension to the present situation is immediate and therefore is omitted. We know that un changes sign in , provided that n is suLciently large. Let us denote by yn ∈ = a point where un vanishes and let x n ∈ = be such that un (x n ) = = where max un (¿ Sn ). Clearly, we can suppose that x n → x0 ∈ = and yn → y0 ∈ , v(x0 ) = max v and v(y0 ) = 0. Let (n denote the path joining x n to yn constructed in the = then join x0 to following way: ?rst join x n to x0 by a C 1 path n having range in , y0 by the piecewise linear path ( provided by the above claim, ?nally join y0 to yn = By the smoothness of @; n and =n can be taken by a C 1 path =n having range in . such that ‘(n ) → 0
and
‘(=n ) → 0
as n → + ∞;
(2.30)
where ‘(·) denotes the length of the corresponding path. Now, using (2.27) and the properties of fn , we get + (max un )p − Fn (max un ) p − − + + p p = (u (x n )) − (u (x n )) − Fn (un (x n )) p n p n + + − − p p − (u (yn )) − (u (yn )) − Fn (un (yn )) p n p n
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
=
0
+
1
+ (un+ (n ))p−1 − − (un− (n ))p−1 − fn (un (n )) ∇un (n ); n
1
0
+
385
1
0
[+ (un+ (())p−1 − − (un− (())p−1 − fn (un (())]∇un ((); ( [+ (un+ (=n ))p−1 − − (un− (=n ))p−1 − fn (un (=n ))]∇un (=n ); =n
6 (‘(n ) + ‘(=n )) fn (un ) − [+ (un+ )p−1 − − (un− )p−1 ] ∞ ∇un ∞ + |fn (un ) − [+ (un+ )p−1 − − (un− )p−1 ]| ∇un ∞
(
6 ‘(n ) + ‘(=n ) +
1−p un ∞
(
|fn (un ) −
[+ (un+ )p−1
−
− (un− )p−1 ]|
× C un p∞ : Therefore, by the above Claim and (2.30), we deduce + (max un )p − Fn (max un ) 6 4n un p∞ ; p
(2.31)
where 4n → 0 as n → +∞. On the other hand, using (2.23) and (2.29), we have ;;p ; + (max un )p − Fn (max un ) ¿ (max un )p ¿ 1 un p∞ : p p p
(2.32)
Comparing (2.31) and (2.32) yields a contradiction, letting n → +∞. This concludes the proof of Theorem 1.1. Proof of Corollary 1.1. The conclusion immediately follows combining the following lemma with Claim 1 in the proof of Lemma 2.1, if (1.7) holds, or with Theorem 1.1, if (1.5), (1.6) and (1.8) hold. Lemma 2.3. Set h= = (1=meas )
h: Assume that there is a constant r ¿ 0 such that
f(s) ¿ h= for every s ¿ r
(2.33)
(f(s) ¿ h= for every s 6 − r; respectively): Then; for any c ¿ 0 problem (1:1) has a lower solution ; with (x) ¿ c in ((x) 6 − c in ; respectively). Similarly; assume that there is a constant r ¿ 0 such that f(s) 6 h= for every s 6 − r (f(s) 6 h= for every s ¿ r; respectively): Then; for any c ¿ 0 problem (1:1) has an upper solution ; with (x) 6 − c in ((x) ¿ c in ; respectively).
386
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Proof of Lemma 2.3. Assume that condition (2.33) holds, the other cases being treated similarly. In order to built a lower solution of (1.1), we will prove that the problem −4p w = h= − h(x) in ; @w=@ = 0 w=0
on @;
(2.34)
= Then, for any constant has a solution w ∈ W 1; p ()∩L∞ (), and therefore w ∈ C 1; (). t ¿ w ∞ + r, we de?ne (x) = t + w(x) in . It is immediately seen, by condition (2.33), that is a lower solution. 1; p 1; p First, we prove that (2.34) has a solution w 1;∈pW (). Let us set W = {w ∈ W () | w = 0}: W is closed linear subspace of W () and therefore is a reOexive Banach space. De?ne a functional J : W → R by setting J (w) = (1=p) |∇w|p + (h= − h)w: Clearly, J is strictly convex and C 1 . Moreover, the Poincar1e–Wirtinger inequality implies that J is coercive. Hence, J has a unique global minimum point w ∈ W and, therefore, for every z ∈ W and every c ∈ R, we get |∇w|p−2 ∇w∇(z + c) = (h= − h)(z + c);
i.e. w is a solution of (2.34). Second, we show that w ∈ L∞ (). Of course, only the case where p 6 N requires a proof. Assume that w = 0 and observe that, as w = 0, w changes sign on subsets of of positive measure. Let us set B = {x ∈ | w(x) ¡ 0} and, for each n, An = {x ∈ | w(x) ¿ n} and wn = (w − n)+ . It is clear that wn (x) = 0 a.e. on B. We claim that there exists C ¿ 0 such that wn p 6 C ∇wn p
(2.35)
for every n. Indeed, otherwise, possibly passing to a subsequence, we should have ∇wn p = wn p → 0: Setting, for each n, zn = wn = wn p , we have, possibly passing to a further subsequence, that (zn )n converges weakly in W 1; p (), strongly in Lp () and pointwise a.e. in to some function z ∈ W 1; p (). Actually, (zn )n converges strongly in W 1; p () to z. Hence, z satis?es z p = 1 and ∇z = 0. The Poincar1e–Wirtinger inequality implies that z is constant in and, since z(x) = 0 a.e. in B, we conclude that z(x) = 0 a.e. in : a contradiction. To prove the boundedness from above of w, we apply Lemma 5:1 in [14]. Accordingly, we show that there exist constants C ¿ 0 and 4 ¿ 0 such that, for every n, wn 6 C(meas(An ))1+4 : (2.36) An
Assume that p ¡ N ; a similar argument applies if p = N . In the sequel, the same symbol C will denote constants which may change time by time, but are all independent of n. By HUolder inequality and the Sobolev imbedding theorem, we have, using (2.35)
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
too,
An
wn 6 (meas (An ))
1−(N −p)=Np
An
(wn )
Np=(N −p)
387
(N −p)=Np
= (meas (An ))1−(N −p)=Np wn Np=(N −p) 6 C(meas (An ))1−(N −p)=Np wn 1; p 6 C(meas (An ))1−(N −p)=Np ∇wn p : Testing the equation in (2.34) against wn ∈ W 1; p (), we get = ∞ |∇wn |p 6 h − h wn
and hence 1−(N −p)=Np wn 6 C(meas (An ))
An
An
that is An
1=p wn
;
wn 6 C(meas(An ))1+p=(N (p−1)) :
Similarly, one proves that w is bounded from below. Proof of Proposition 1.1. Assume, for instance, that condition (1.9) holds at −∞. As in the proof of Theorem 1.1, we can also suppose, without loss of generality, that (2.21) holds. Then, for each n, we de?ne the truncations
f(s) if s 6 n; fn (s) = f(n) if s ¿ n: = and C ¿ 0 such Conditions (1.9) and (2.21) imply the existence of constants ˆ ∈ ]0; [ that ˆ − )p−1 + 2C |fn (s) − h(x)| 6 fn (s) − h(x) + 2(s
(2.37)
for every n, every s and a.e. x. Consider problem (2.24) and observe that, for every n ¿ max{ ∞ ; ∞ }, and are a lower and an upper solution, respectively. Hence, the assumptions of Lemma 2.1 are satis?ed and therefore (2.24) has at least one solution un with min(un − ) 6 0
and
max(un − ) ¿ 0:
(2.38)
The solvability of problem (1.1) will be obviously achieved if we are able to prove the existence of an integer n˜ such that ˜ max un˜ 6 n:
(2.39)
388
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Assume, by contradiction, that max un ¿ n for every n. Hence, un ∞ → +∞ and un 1; p → +∞, as p ¿ N . Let us set vn =
un : un 1; p
Possibly passing to a subsequence, still denoted by (vn )n , we have that (vn )n converges = to some function v. Moreover, vn satis?es weakly in W 1; p () and strongly in C 0 () fn (un ) − h |∇vn |p−2 ∇vn ∇w = w (2.40) p−1 un 1; p for every w ∈ W 1; p (). Testing (2.40) against w = 1 and using (2.37), we get |fn (un ) − h| ˆ (v− )p−1 + 2C meas 6 2 (2.41) n un p−1 un p−1 1; p 1; p and hence there is a constant still denoted by C such that fn (un ) − h 1 un p−1 1; p
6C
(2.42)
for every n. Now, testing (2.40) against w = vn − v and using (2.42), we obtain |∇vn |p−2 ∇vn ∇(vn − v) 6 C vn − v ∞ → 0
as n → +∞. Hence, condition (S+) in [4, p. 18] implies that vn → v in W 1; p () and therefore, as vn 1; p = 1, also v 1; p = 1. Next, we show that v− = 0. Indeed, if v− = 0, then vn− → 0 in L∞ () and, by (2.41), fn (un ) − h 1 = un p−1 1; p → 0: Hence, testing (2.40) against w = v, we get |∇vn |p−2 ∇vn ∇v → 0: Since |∇vn |p−2 ∇vn → |∇v|p−2 ∇v in (Lp=(p−1) ())N , we conclude that |∇v|p = 0, i.e. ∇v = 0 a.e. in . As v is continuous and vanishes somewhere in = by (2.38), we get v = 0: a contradiction. Finally, testing (2.40) against −vn− and using (2.37) again, we easily have fn (un ) − h − (fn (un ) − h)− − − p |∇(vn )| = − v 6 vn n p−1 un p−1 un 1; p 1; p C ˆ 6 (vn− )p + un p−1 1; p and hence, passing to the limit, − p ˆ |∇(v )| 6 |v− |p :
= Since ˆ ¡ = and v− = 0, we get a contradiction with the de?nition of .
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
389
References [1] R.A. Adams, Sobolev Spaces, Academic Press, New York, 1975. [2] D. Arcoya, L. Orsina, Landesman-Lazer conditions and quasilinear elliptic equations, Nonlinear Analysis T.M.A. 28 (1997) 1623–1632. [3] M. Arias, J. Campos, J.P. Gossez, On the antimaximum principle and the Fu0c12k spectrum for the Neumann p-Laplacian, DiIerential Integral Equations 13 (2000) 217–226. [4] F.E. Browder, Existence theorem for nonlinear partial diIerential equations, Proceedings of the Symposium on Pure Mathematics, Vol. XVI, Berkeley, CA, Amer. Math. Soc., Providence, RI, 1968, pp. 1– 60. [5] M. Cuesta, D. de Figueiredo, J.P. Gossez, The beginning of the Fu0c12k spectrum for the p-Laplacian, J. DiIerential Equations 159 (1999) 212–238. [6] M. Cuesta, D. de Figueiredo, J.P. Gossez, A nodal domain property for the p-Laplacian, C. R. Acad. Sci. Paris 330 (2000) 669–673. [7] C. De Coster, M. Henrard, Existence and localization of solutions for second order elliptic BVP in presence of lower and upper solutions without any order, J. DiIerential Equations 145 (1998) 420–452. [8] E. Di Benedetto, C 1+ local regularity of weak solutions of degenerate elliptic equations, Nonlinear Analysis T.M.A. 7 (1983) 827–850. [9] D. de Figueiredo, J.P. Gossez, On the ?rst curve of the Fu0c12k spectrum of an elliptic operator, DiIerential Integral Equations 7 (1994) 1285–1302. [10] M. Garc12a-Huidobro, R. ManZasevich, F. Zanolin, Strongly nonlinear second-order ODEs with unilateral conditions, DiIerential Integral Equations 6 (1993) 1057–1078. [11] J.P. Gossez, P. Omari, On a semilinear elliptic Neumann problem with asymmetric nonlinearities, Trans. Amer. Math. Soc. 347 (1995) 2553–2562. [12] P. Habets, P. Omari, Existence and localization of solutions of second order elliptic problems using lower and upper solutions in the reversed order, Top. Meth. Nonlinear Analysis 8 (1996) 25–56. [13] J. Kazdan, F. Warner, Remarks on some quasilinear elliptic equations, Comm. Pure Appl. Math. 28 (1975) 567–597. [14] O.A. Ladyzhenskaya, N.N. Ural’tseva, Linear and Quasilinear Elliptic Equations, Academic Press, New York, 1968. [15] G.M. Lieberman, Boundary regularity for solutions of degenerate elliptic equations, Nonlinear Analysis T.M.A. 12 (1988) 1203–1219. [16] P. Omari, F. Zanolin, Nonresonance conditions on the potential for a second-order periodic boundary value problem, Proc. Amer. Math. Soc. 117 (1993) 125–135. [17] J.L. Vazquez, A strong maximum principle for some quasilinear elliptic equations, Appl. Math. Optim. 3 (1984) 191–202. [18] S. Villegas, A Neumann problem with asymmetric nonlinearity and a related minimizing problem, J. DiIerential Equations 145 (1998) 145–155.
www.elsevier.com/locate/na
On a p-Laplace Neumann problem with asymptotically asymmetric perturbations Mohssine Alif a , Pierpaolo Omarib;∗ a International
b Dipartimento
Centre for Theoretical Physics, Strada Costiera 11, I–34014 Trieste, Italy di Scienze Matematiche, Universit!a, Via A. Valerio 12=1, I–34127 Trieste, Italy Received 30 April 2001; accepted 16 May 2001
Keywords: Quasilinear elliptic equations; p-Laplacian; Neumann boundary conditions; Dancer– Fu0c12k spectrum; upper and lower solutions
1. Introduction and statements Let us consider the elliptic problem −4p u = f(u) − h(x) in ; @u=@ = 0
on @:
(1.1)
Throughout, we assume that (⊂ RN ) is a bounded domain, with a C 1; 1 boundary @, 4p u = div(|∇u|p−2 ∇u) is the p-Laplacian, with p ¿ 1, f : R → R is a continuous function and h ∈ L∞ (). By a solution of (1.1) we mean a function u ∈ W 1; p () ∩ L∞ (), such that |∇u|p−2 ∇u∇w = (f(u) − h)w
= for for every w ∈ W 1; p (). Known regularity results [15] then imply that u ∈ C 1; (), some ¿ 0. The aim of this paper is to improve and extend to the p-Laplacian, with p ¿ 1 arbitrary, a result obtained in [11] for the Laplacian, in the case where the function f exhibits an asymmetric behaviour at in?nity and may interact with the Dancer–Fu0c12k ∗
Corresponding author. E-mail addresses: [email protected] (M. Alif ), [email protected] (P. Omari).
c 2002 Elsevier Science Ltd. All rights reserved. 0362-546X/02/$ - see front matter PII: S 0 3 6 2 - 5 4 6 X ( 0 1 ) 0 0 8 3 5 - 5
370
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
spectrum of the involved diIerential operator. This is made possible also in the light of two recent contributions in this area given in [5,7] respectively. We recall that the Dancer–Fu0c12k spectrum of the Neumann p-Laplacian is de?ned as the set of all (+ ; − ) ∈ R2 such that −4p u = + |u|p−2 u+ − − |u|p−2 u−
in ;
@u=@ = 0
on @
(1.2)
admits a nontrivial solution. In [5] (see also [3], for more information about the Neumann problem) it was proved that contains, besides the trivial lines R × {1 } and {1 } × R, a ?rst nontrivial curve C which passes through (2 ; 2 ), where 1 = 0 is the principal eigenvalue and 2 ¿ 0 is the second eigenvalue. This curve is continuous, decreasing and symmetric with respect to the diagonal. Moreover, if p 6 N it is asymptotic to the lines R × {1 } and {1 } × R, while if p ¿ N is asymptotic to the = and {} = × R, where = ¿ 0 is de?ned by lines R × {} = = inf |∇u|p | u ∈ W 1; p (); |u|p = 1; min u = 0 : (1.3)
On the other hand, in [7, Theorem 8:2] the solvability of the Dirichlet problem for a quasilinear equation involving the p-Laplacian was proved in the presence of a pair of upper and lower solutions, for which no special ordering was required. In our main result, stated below as Theorem 1.1, the interaction of f with the principal eigenvalue 1 is controlled by means of a pair of upper and lower solutions, possibly without any ordering, and the interaction with the ?rst nontrivial branch C by an asymptotic condition similar to that considered in [11,16] and involving a primitive F of f. In this paper, the following de?nition of upper and lower solutions is adopted. We = is a lower solution of (1.1) if say that a function ∈ W 1; p () ∩ C 0 () |∇|p−2 ∇∇w 6 (f() − h)w (1.4)
for every w ∈ W 1; p (), with w(x) ¿ 0 a.e. in . An upper solution is de?ned similarly by reversing the inequality in (1.4). Theorem 1.1. Let (+ ; − ) ∈ C be given. Assume lim sup
f(s) 6 − |s|p−2 s
lim inf
pF(s) ¡ − |s|p
s→−∞
and lim sup s→+∞
f(s) 6 + |s|p−2 s
(1.5)
and s→−∞
or
lim inf s→+∞
pF(s) ¡ + : |s|p
(1.6)
Moreover; suppose that there exist a lower solution and an upper solution . Then; problem (1:1) has at least one solution.
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
371
The proof of Theorem 1.1 is divided in several intermediate steps and combines topological and variational methods. We point out in particular that one of the technical diLculties, which must be overcome for proving Theorem 1.1, consists in the extension to the Neumann problem of Lemma 5:3 in [5], which is given in Lemma 2.2 below. We feel that this result may have an independent interest, since the arguments developed in its proof could be used also in diIerent contexts. In order to apply Theorem 1.1, the simplest way is looking for constant upper and lower solutions. It is obvious that any ∈ R such that f() ¿ ess sup h is a lower solution and any ∈ R such that f() 6 ess inf h is an upper solution. Accordingly, we immediately get the following characterization of nonresonance, which generalizes [11, Theorem 1.1] to the quasilinear setting. Corollary 1.1. Assume (1:5) and (1:6). Then; problem (1:1) has at least one solution for any given h if and only if the range of f is R. Other assumptions, which are classical in the study of the semilinear problem (p = 2) around the principal eigenvalue and imply the existence of upper and lower solutions, are the following Landesman–Lazer conditions: there exists r ¿ 0 such that 1 f(s) sign(s) 6 h for every s; with |s| ¿ r (1.7) meas or 1 f(s) sign(s) ¿ meas
h
for every s; with |s| ¿ r:
(1.8)
Condition (1.7) yields the existence of a lower solution and an upper solution = whereas (1.8) yields the existence of a lower solution satisfying (x) ¡ 0 ¡ (x) in , = We stress that, when and an upper solution satisfying (x) ¿ 0 ¿ (x) in . compared with the semilinear case, extensively discussed for instance in [7,12,13], the construction of such upper and lower solutions appears in the quasilinear setting rather more delicate (cf. Lemma 2.3 below). Using conditions (1.7) and (1.8), we obtain the following corollary, which yields a partial extension of Theorem 1:3 in [2] to unbounded nonlinearities. Corollary 1.2. Assume (1:7); or (1:5); (1:6) and (1:8). Then; problem (1:1) has at least one solution. Now, we restrict ourselves to the case p ¿ N and brieOy consider the situation where the two-sided control imposed by (1.5) and (1.6) is replaced by the following one-sided condition lim sup s→−∞
f(s) ¡ = or |s|p−2 s
with = de?ned by (1.3).
lim sup s→+∞
f(s) = ¡ ; |s|p−2 s
(1.9)
372
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Proposition 1.1. Assume p ¿ N and (1:9). Moreover; suppose that there exist a lower solution and an upper solution . Then; problem (1:1) has at least one solution. This statement yields a partial extension of Theorem 7 in [18]. Note that in Proposition 1.1 no condition at all is assumed on f at +∞, or at −∞, respectively. It would be interesting to obtain analogous results if p 6 N . We are able to do this in the semilinear case (p = 2), or when is a ball and h is radially symmetric. In this frame, condition (1.9) must however be replaced by the stronger assumption f(s) f(s) lim sup p−2 6 0 or lim sup p−2 6 0 s s |s| |s| s→−∞ s→+∞ and a further restriction on the growth of f, involving the exponent (N (p−1))=(N −p), is required. More general results can be obtained in the one-dimensional case (N = 1) on the line of [10]. 2. Proofs We start with some preliminary results. The ?rst concerns the solvability of the problem −4p u = g(x; u) in ; @u=@ = 0
on @;
(2.1)
with g : × R → R a L∞ -Carath1eodory function, in the presence of upper and lower solutions without any special ordering. This is essentially an adaptation to the Neumann problem of [7, Theorem 8:2]. Lemma 2.1. Assume that; for a.e. x ∈ and every s ∈ R; g(x; s) = − q(x; s)|s|p−2 s− + r(x; s);
(2.2)
∞
where q; r : × R → R are L -Carath:eodory functions satisfying; for some constant M ¿ 0; 0 6 q(x; s) 6 M
for a:e: x ∈ and every s ∈ R
(2.3)
and r(x; s) =0 |s|→+∞ |s|p−1 lim
uniformly a:e: in :
(2.4)
Moreover; suppose that there exist a lower solution and an upper solution of (2:1) such that min( − ) ¡ 0: = satisfying Then; problem (2:1) has at least one solution u ∈ C 1; () min(u − ) 6 0 and
max(u − ) ¿ 0:
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
373
Proof of Lemma 2.1. We begin observing that, for each k ∈ L∞ (), there exists a unique solution u ∈ W 1; p () of the problem −4p u + |u|p−2 u = k(x) in ; @u=@ = 0
(2.5)
on @;
1=(p−1) 1=(p−1) = for which satis?es − k − ∞ 6 u(x) 6 k + ∞ in . Therefore, u ∈ C 1; () some ¿ 0 independent of k. = associated with = → C 0 () Hence, we can de?ne a solution operator T : C 0 () 0 = problem (2.1), which sends any function u ∈ C () onto the unique solution v ∈ = of C 1; ()
−4p v + |v|p−2 v = g(x; u) + |u|p−2 u
in ;
@v=@ = 0
on @:
= into C 1 () = imply that T The regularity theory and the compact embedding of C 1; () is completely continuous. Moreover, the ?xed points of T are precisely the solutions of (2.1). In the sequel, we say that a lower solution is strict if there is no solution u such that min(u − ) = 0. Similarly, an upper solution is said to be strict if there is no solution u such that max(u − ) = 0. Claim 1. Assume that ( and ) are strict lower and upper solutions of (2.1), satisfying min() − () ¿ 0, and introduce the open set = | min(u − () ¿ 0 and max(u − )) ¡ 0}: O(; ) = {u ∈ C 0 () Then, deg(I − T; O(; ) ) = 1 and, hence, problem (2.1) has at least one solution u ∈ = O(; ) ∩ C 1; (). The proof of this result follows standard lines; we refer e.g. to [7, Theorem 8:1] where the case of a quasilinear elliptic problem, involving the Dirichlet p-Laplacian, is discussed. Now, we are in position to prove Lemma 2.1. Let us de?ne, for each n, a L∞ Carath1eodory function gn : × R → R by setting, for a.e. x ∈ and every s ∈ R, 1 if s 6 − n − 1; (s + n + 1)g(x; s) − (s + n) if − n − 1 ¡ s ¡ − n; if |s| 6 n; gn (x; s) = f(x; s) (n + 1 − s)g(x; s) − (s − n) if n ¡ s ¡ n + 1; −1 if s ¿ n + 1:
374
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Testing against (u − n − 1)+ and (u + n + 1)− , it is easily seen that, for each n, any solution u of the problem −4p u = gn (x; u) in ; @u=@ = 0
on @
(2.6)
satis?es u ∞ 6 n + 1. Hence, it follows in particular that any constant A ¡ − n − 1 is a strict lower solution and any constant B ¿ n + 1 is a strict upper solution of (2.6). Claim 2. There exist an integer n= and a constant R such that; for every n ¿ n; = any solution u of (2:6); with min(u − ) 6 0 and max(u − ) ¿ 0; satis;es u ∞ ¡ R:
(2.7)
Proof of Claim 2. Assume, by contradiction, that for each k there are an integer nk and a solution uk of (2.6), with n = nk , such that min(uk − ) 6 0
and
max(uk − ) ¿ 0
(2.8)
and uk ∞ ¿ k: For each k, de?ne the function uk ; vk = uk ∞ which satis?es gnk (x; uk ) |∇vk |p−2 ∇vk ∇w = w p−1 uk ∞ for every w ∈ W 1; p (). We can write gnk (x; s) = − qk (x; s)(s− )p−1 + rk (x; s) for a.e. x ∈ and every s ∈ R, where qk and rk are L∞ -Carath1eodory functions satisfying 0 6 qk (x; s) 6 M for a.e. x ∈ and every s ∈ R and rk (x; s) =0 lim |s|→+∞ |s|p−1 uniformly with respect to k and to a.e. x ∈ . Since the sequence (gnk (·; uk )= uk p−1 ∞ )k is uniformly bounded in L∞ (), by the regularity theory and the compact embedding of = into C 1 (), = we have, possibly passing to a subsequence, that (vk )k converges C 1; () 1 = in C () to some function v, which satis?es v ∞ = 1 and |∇v|p−2 ∇v∇w = − q0 (v− )p−1 w (2.9)
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
375
for every w ∈ W 1; p (). Here, q0 ∈ L∞ () is the weak∗ limit, possibly of a subsequence, of (qk (·; uk ))k , with 0 6 q0 (x) 6 M
for a:e: x ∈ :
Now, we want to prove that v is constant in . Testing (2.9) against w = − 1, we get − p−1 q (v ) = 0, which implies that v− (x) = 0 on the set where q0 (x) ¿ 0. Testing 0 against w = v, we obtain |∇v|p =− q0 (v− )p = 0 and hence ∇v = 0 in . Therefore, we conclude that v = 1 or v = − 1 in . Accordingly, we have that uk (x) → +∞
or
uk (x) → −∞
uniformly in . This contradicts (2.8). Hence, Claim 2 is proved. == problem (2:6) has at Claim 3. There exists an integer n== such that; for every n ¿ n; least one solution un satisfying min(un − ) 6 0
and max(un − ) ¿ 0:
Proof of Claim 3. Let us de?ne the open set = | min(u − ) ¡ 0 and max(u − ) ¿ 0}: O = {u ∈ C 0 () Let n== be an integer satisfying n== ¿ max{ ∞ ; ∞ }. Fix any integer n ¿ n== and set B = n+2 and A =−B. As already observed, A and B are strict lower and upper solutions of (2.6). Assume also that there is no solution u of (2.6) with u ∈ @O, or equivalently that and are strict lower and upper solutions of (2.6); indeed, otherwise we are done. We have O ∩ OA; B = OA; B \(O= A; ∪ O= ; B ) and, by Claim 1, deg(I − Tn ; O ∩ OA; B ) = deg(I − Tn ; OA; B ) − deg(I − Tn ; OA; ) − deg(I − Tn ; O; B ) = − 1: = → C () = is the solution operator associated with problem (2.6), Here, Tn : C () = onto the unique solution v ∈ C 1; () = of which sends any function u ∈ C 0 () p−2 p−2 −4p v + |v| v = gn (x; u) + |u| u in ; 0
@v=@ = 0
0
on @:
== This Accordingly, there exists at least one solution un ∈ O of (2.6) for each n ¿ n. concludes the proof of Claim 3. By Claim 2, we have that, for each n ¿ max{n; = R}, every solution un ∈ O= of (2.6) is a solution of (2.1) too. Hence, Claim 3 yields the conclusion of Lemma 2.1. The second preliminary result extends to the Neumann problem Lemma 5:3 in [5], stated and proved there in the case of Dirichlet boundary conditions. We stress that, although the idea of the proof is substantially the same as in [5], there is an essential step (formulated below as a claim), which involves the consideration of a mixed problem on a nonsmooth domain and appears technically delicate.
376
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Lemma 2.2. Let (+ ; − ) ∈ C be given and let p0 ; q0 ∈ L∞ () satisfy 0 6 p0 (x) 6 +
and 0 6 q0 (x) 6 −
a:e: in :
(2.10)
Suppose that meas{x ∈ | p0 (x) ¿ 0} ¿ 0
and meas{x ∈ | q0 (x) ¿ 0} ¿ 0:
(2.11)
Then; any nontrivial solution u of −4p u = p0 (x)|u|p−2 u+ − q0 (x)|u|p−2 u−
in ;
@u=@ = 0
on @
(2.12)
changes sign in and p0 (x) = +
a:e: on {x ∈ | u(x) ¿ 0}
(2.13)
q0 (x) = −
a:e: on {x ∈ | u(x) ¡ 0}
(2.14)
and (i.e. u is an eigenfunction associated with (+ ; − )). = be a nontrivial solution of (2.12), which therefore Proof of Lemma 2.2. Let u ∈ C 1 () satis?es |∇u|p−2 ∇u∇w = p0 (u+ )p−1 w − q0 (u− )p−1 w (2.15)
for every w ∈ W 1; p (). First, we prove that u changes sign in . Indeed, suppose, by contradiction, that e.g. u(x) ¿ 0 in . Testing (2.15) against w = 1, we get p0 up−1 = 0;
which implies that p0 (x) = 0 a.e. on {x ∈ | u(x) ¿ 0}. Hence, testing (2.15) against w = u, we deduce that |∇u|p = 0 and therefore u is a positive constant. Accordingly, p0 (x) = 0 a.e. in , which is in contradiction with (2.11). Second, we show that (2.13) holds; condition (2.14) follows from a totally similar argument. Assume, by contradiction, that meas{x ∈ | u(x) ¿ 0 and p0 (x) ¡ + } ¿ 0:
(2.16)
We begin stating the following result, whose proof is postponed. Claim. There exists a function v ∈ W 1; p (), which changes sign on subsets of of positive measure and satis;es |∇v+ |p ¡ + |v+ |p
and
|∇v− |p ¡ − : |v− |p
(2.17)
Once this Claim is proved, the remainder of the argument proceeds exactly as in the case of Dirichlet boundary conditions (cf. [5, Lemma 5:3] or [6]). However, we give a
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
377
brief sketch for the sake of completeness. Let us consider, for each s ¿ 0, a functional Js : W 1; p () → R de?ned by p Js (u) = |∇u| − s (u+ )p
and let J˜ s be its restriction to S = {u ∈ W 1; p () | |u|p = 1}. It is proved in [5] (see also [3]) that C consists of the points (s + c(s); c(s)), for all s ¿ 0, and of their symmetric with respect to the diagonal, where c(s) = inf
max J˜ s (u):
(∈1 u∈([−1;1]
Here, we set 1 = {( ∈ C 0 ([−1; 1]; S) | ((−1) = −’1 and ((1) = ’1 }, where ’1 = (1=meas )1=p is the ?rst eigenfunction. Possibly replacing u with −u, one can suppose that + ¿ − . Then, putting s = + − − , one has − = c(s). Using the above Claim, it is possible, as in [5], to build a path ( ∈ 1 such that max J˜ s (u) ¡ − ;
(2.18)
u∈([−1;1]
thus getting a contradiction with the de?nition of c(s). Indeed, ( can be constructed as follows. One starts joining v= v p to v+ = v+ p by a path (1 , v+ = v+ p to v− = v− p by a path (2 and −v− = v− p to v= v p by a path (4 . These paths are de?ned by convex combinations on S, i.e. (1 (t) =
tv + (1 − t)v+ ; tv + (1 − t)v+ p
(4 (t) =
−tv− + (1 − t)v − tv− + (1 − t)v p
(2 (t) =
tv+ + (1 − t)v− ; tv+ + (1 − t)v− p
for t ∈ [0; 1]. Using (2.17), one veri?es that J˜ s ((i (t)) ¡ − , for each i = 1; 2; 4 and every t, and, moreover, that J˜ s (v− = v− p ) ¡ − − s. Then, one observes, as in [5], that any component of the set {u ∈ S | J˜ s (u) ¡ 1 −s} contains a critical point of J˜ s and ’1 and −’1 are the only critical points of J˜ s at levels ¡ − − s. Taking into account these facts, it is possible to join v− = v− p to ’1 , or to −’1 , by a continuous path (3 in such a way that J˜ s ((3 (t)) ¡ − − s, for every t. Assume, for instance, that ’1 is reached in this way. Then, one considers the path −(3 which joins −’1 to −v− = v− p . Again, one has J˜ s (−(3 (t)) ¡ − , for every t. Accordingly, it is proved that one can de?ne a continuous path ( on S from −’1 to ’1 , which satis?es (2.18). Proof of the Claim. If |∇u− |p ¡ − |u− |p , then we simply set v = u, which, by (2.16), satis?es (2.17). On the contrary, if |∇u− |p = − |u− |p , we must resort to the following construction. From (2.16), it follows that there exists 40 ¿ 0 such that meas{x ∈ | u(x) ¿ 40 and p0 (x) ¡ + } ¿ 0. Denote by 540 a connected component of {x ∈ | u(x) ¿ 40 } such that meas{x ∈ 540 | p0 (x) ¡ + } ¿ 0 and, for every 4 ∈ ]0; 40 [, let 54 be a connected component of {x ∈ | u(x) ¿ 4} such that 540 ⊂ 54 . Hence, we have, in particular,
378
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
meas{x ∈ 54 | p0 (x) ¡ + } ¿ 0. De?ne 50 = 4∈]0; 40 ] 54 . Clearly, 50 is open and connected, u(x) ¿ 0 in 50 and meas{x ∈ 50 | p0 (x) ¡ + } ¿ 0. De?ne, for each 4 ∈ [0; 40 ],
u(x) − 4 if x ∈ 54 ; u4 (x) = 0 if x ∈ \54 : Clearly, u4 → u0 uniformly, as 4 → 0+ . Since, for each 4 ∈ [0; 40 ], 54 is a nodal domain of u − 4, Lemma 5:2 in [9], which immediately carries over to W 1; p (), guarantees that u4 ∈ W 1; p (). Testing (2.15) against w = u0 , we get |∇u0 |p ¡ + : up 0 Since
|∇u4 |p = up 4
and
50
u0p
up 50 4
→1
50
|∇u4 |p 50
u4p
6
50
|∇u0 |p 50
u0p
50 50
u0p u4p
as 4 → 0+ ;
we can pick 4 ¿ 0 so small that u4 satis?es |∇u4 |p ¡ + : up 4 Accordingly, we ?x such an 4 and we set v1 = u4
and
1 = 54 :
Now, let 0 be a connected component of {x ∈ | u(x) ¡ 0}. It is clear that = 1 ∩ = 0 = ∅. By the regularity of @, we can choose, for each x0 ∈ @∩ = 0 , a N -dimensional open cube R(x0 ) such that R(x0 ) ∩ = 1 = ∅, and, possibly after an orthogonal change of coordinates, R(x0 ) = ]x10 − r; x10 + r[ × · · · × ]xN0 − r; xN0 + r[ for some r ¿ 0, @ ∩ R(x0 ) = {(x1 ; : : : ; xN ) ∈ R(x0 ) | xN = ’(x1 ; : : : ; xN −1 )} and ∩ R(x0 ) = {(x1 ; : : : ; xN ) ∈ R(x0 ) | xN ¡ ’(x1 ; : : : ; xN −1 )}; where (x10 ; : : : ; xN0 ) and (x1 ; : : : ; xN ) are the coordinates of x0 and x, respectively, and ’ : [x10 − r; x10 + r] × · · · × [xN0 −1 − r; xN0 −1 + r] → ]xN0 − r; xN0 + r[ is a C 1; 1 function. Note that these conditions imply the existence of a number 6(x0 ) ¿ 0 such that, for every x ∈ @ ∩ R(x0 ); N (x) ¿ 6(x0 ), where N (x) is the N th component
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
379
of the unit outer normal to at the point x. Since @ ∩ = 0 is compact, there exist points x1 ; : : : ; xp ∈ @ ∩ = 0 such that @ ∩ = 0 ⊂ R(x1 ) ∪ · · · ∪ R(xp ): Let us set A = [R(x1 ) ∪ · · · ∪ R(xp )] ∩
and
10 = @0 ∩ (\A):
Of course, if @ ∩ = 0 = ∅, then A = ∅. Observe that 1= 0 ∩ @ = ∅. Indeed, assume by contradiction that there exists a sequence (x n )n ⊂ 10 such that x n → x ∈ @. We have x ∈ @ ∩ @0 . Therefore, x ∈ R(xi ) for some i ∈ {1; : : : ; p} and hence eventually x n ∈ R(xi ) ∩ ⊂ A: a contradiction. This implies that 1= 0 = 10 . Moreover, it is obvious that 10 ∩ = 1 = ∅: Therefore, for every y ∈ 10 , there is an N -dimensional open cube Q(y) centered at y such that Q(y) ⊂ and Q(y) ∩ = 1 = ∅. Since 10 is compact, there exist points y1 ; : : : ; yq ∈ 10 such that 10 ⊂ Q(y1 ) ∪ · · · ∪ Q(yq ). Let us set B = Q(y1 ) ∪ · · · ∪ Q(yq ): Clearly, B= ∩ @ = ∅ and @0 ∩ ⊂ A ∪ B. Finally, let us de?ne 2 = 0 ∪ A ∪ B: This set 2 is open, connected and satis?es 2 \(A ∪ B) ⊂ 2 . Indeed, assume by contradiction that there exists a sequence (x n )n ⊂ 2 \(A ∪ B) such that x n → x ∈ @2 . We necessarily have (x n )n ⊂ 0 and x ∈ @0 . If x ∈ @0 ∩ , then x ∈ A ∪ B and hence eventually x n ∈ A ∪ B. Whereas, if x ∈ @0 ∩ @, then x ∈ R(xi ) for some i ∈ {1; : : : ; p} and hence eventually x n ∈ R(xi ) ∩ ⊂ A. Therefore, in both cases, we get a contradiction. Moreover, 2 satis?es the cone property, according to De?nition 4:2 in [1]. This assertion follows observing that there exists an open ?nite cone C such that if x ∈ A ∪ B, then, by construction of A and B, x is the vertex of an open ?nite cone Cx congruent to C, which is contained in A ∪ B and, therefore, in 2 . Once this property has been veri?ed for all points in A ∪ B, it is easy to extend its validity to the whole of 2 , just taking a ?nite covering, made by N -dimensional open cubes contained in = 2 , of the set 2 \(A ∪ B). Let us denote by V the closure in W 1; p (2 ) of the linear space {v ∈ W 1; p (2 ) | v(x) = 0 a:e: in a neighborhood of @2 ∩ }: V is a reOexive Banach space satisfying W01; p (2 ) ⊂ V ⊂ W 1; p (2 ). De?ne a func tional J : V → R by setting J (v) = 2 |∇v|p : Clearly, J is C 1 and weakly sequentially lower semicontinuous. Let us set = inf J (v)|v ∈ V; |v|p = 1 2
and let (vn )n be a minimizing sequence. Since (J (vn ))n is bounded, it follows that (vn )n is bounded in V and hence, possibly passing to a subsequence, (vn )n is weakly convergent to some v0 ∈ V . Since 2 has the cone property, Theorem 6:2 in [1] guarantees that W 1; p (2 ) is compactly imbedded into Lp (2 ). Hence, vn → v0 in Lp (2 ) and therefore 2 |v0 |p = 1. Moreover, one has J (v0 ) 6 lim inf J (vn ) 6 : n→+∞
380
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
This implies that J (v0 ) = , i.e. v0 is a constrained minimizer. Then, Lagrange multipliers rule yields |v0 |p−2 v0 v |∇v0 |p−2 ∇v0 ∇v = 2
2
for every v ∈ V and, in particular, for every v ∈ W 1; p (2 ) with supp(v) ⊂ 2 . This means [8, p. 828] that v0 is a local weak solution of the equation −4p 9 = |9|p−2 9
in 2 :
Now, de?ne the function u(x) p 1=p z(x) = ( 0 |u| ) 0
(2.19)
if x ∈ 0 ;
if x ∈ 2 \0 : Obviously, z ∈ V ∩ L∞ (2 ) and 2 |z|p = 1. Testing (2.15) against the function w ∈ W 1; p () de?ned by
u(x) if x ∈ 0 ; w(x) = 0 if x ∈ \0 ; we get 0
|∇u|p =
0
q0 |u|p 6 −
0
|u|p
and therefore |∇z|p 6 − : 0
We are in position to prove that ¡ − :
(2.20)
Indeed, assuming by contradiction that ¿ − and therefore, necessarily, = − , we conclude that z is a constrained minimizer for J and hence is a bounded local weak solution of (2.19), with z ∈ C 1 (2 ) (see [8, Corollary]). The strong maximum principle [17, Theorem 5] ?nally implies that z(x) ¿ 0 everywhere in 2 ; which contradicts the de?nition of z, since 2 \0 has nonempty interior. Accordingly, (2.20) holds. Now, we de?ne the function
v0 (x) in 2 ; v2 (x) = 0 in \2 ; where we can suppose, possibly replacing v0 with |v0 |, that v0 (x) ¿ 0 a.e. in 2 . Finally, we set v = v1 − v2 : Clearly, v satis?es (2.17). Thus, the proof of the Claim is ?nished. Hence, also the conclusion of Lemma 2.2 follows.
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
381
Proof of Theorem 1.1. Without loss of generality, we can assume in the course of this proof that min( − ) ¡ 0;
supf(s) ¡ + ∞ and s60
inf f(s) ¿ − ∞:
s¿0
(2.21)
Indeed, otherwise we can ?nd a lower solution ˜ and an upper solution ˜ satisfying min(˜ − ) ˜ ¿ 0. Accordingly, standard results (cf. Claim 1 in the proof of Lemma 2.1) yield the solvability of problem (1.1). Suppose, for instance, that condition (1.6) holds at +∞ and observe that, by condition (1.5) too, there exist a constant ; ¿ 0 and a sequence (Sn )n , with Sn → +∞, such that f(Sn ) Snp−1
pF(Sn ) 6 + − ; Snp
6 + − ; and
for every n:
(2.22)
Indeed, two situations may occur: lim sup s→+∞
pF(s) ¡ + sp
or
lim sup s→+∞
pF(s) = + : sp
In the former case, by (1.5) and l’Hospital rule, we necessarily have lim inf s→+∞
f(s) ¡ + sp−1
and therefore (2.22) holds. In the latter case, by (1.6) there exist a constant ; ¿ 0 and a sequence (Sn )n , with Sn → +∞, such that, for each n, pF(Sn ) 6 + − ; Snp and Sn is a local minimum point of pF(s)=sp . Hence, for each n we get = 0, that is f(Sn ) Snp−1
=
d pF(s) ds ( sp )|s=Sn
pF(Sn ) 6 + − ;: Snp
Now, for each n we de?ne the truncations
f(s) if s 6 Sn ; fn (s) = f(Sn ) if s ¿ Sn : Let us observe that, for each n and for every s ¿ Sn , p−1 fn (s) f(Sn ) Sn = 6 + − ; sp−1 s Snp−1 and pFn (s) 6 + − ;; sp
(2.23)
382
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
where Fn (s) =
s 0
fn (<) d<. Indeed, if s ¿ Sn , we have
pFn (s) pF(Sn ) pf(Sn ) s − Sn = + p−1 p s s s sp p p−1 Sn Sn Sn 1− 6 (+ − ;) + p (+ − ;) s s s
p p−1 Sn Sn = (+ − ;) (1 − p) +p s s 6 + − ;; since (1 − p)t p + pt p−1 6 1 for every t ∈ [0; 1]. Without loss of generality, we can also assume that Sn ¿ max{ ∞ ; ∞ } for every n. Accordingly, and are, respectively, a lower and an upper solution of −4p u = fn (u) − h(x) in ; @u=@ = 0
(2.24)
on @;
where fn satis?es the assumptions of Lemma 2.1. Hence, we can conclude that, for each n, problem (2.24) has at least one solution un satisfying min(un − ) 6 0
and
max(un − ) ¿ 0:
(2.25)
The solvability of problem (1.1) will be obviously achieved if we succeed in proving that there exists an integer n˜ such that max un˜ ¡ Sn˜:
(2.26)
Assume by contradiction that for every n max un ¿ Sn : Hence, in particular, un ∞ → +∞. Observe that, for each n, we can write fn (s) = pn (s)(s+ )p−1 − qn (s)(s− )p−1 + rn (s)
for every s;
where pn ; qn and rn are continuous functions such that 0 6 pn (s) 6 + ;
0 6 qn (s) 6 −
for every s
and rn (s) =0 |s|→+∞ |s|p−1 lim
uniformly with respect to n. Let us set for every n un vn = : un ∞
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
383
Clearly, vn satis?es vn ∞ = 1 and |∇vn |p−2 ∇vn ∇w = pn (un )(vn+ )p−1 w
−
qn (un )(vn− )p−1 w +
rn (un ) − h un p−1 ∞
w
for every w ∈ W 1; p (). Hence, by the regularity theory, there is a constant C such that for every n vn C 1; 6 C:
(2.27)
= into C () = implies that, possibly passing to Then, the compact imbedding of C () = to some function v, with v ∞ = 1, which a subsequence, (vn )n converges in C 1 () satis?es |∇v|p−2 ∇v∇w = p0 (v+ )p−1 w − q0 (v− )p−1 w (2.28) 1;
1
for every w ∈ W 1; p (). Here, p0 ; q0 ∈ L∞ () are the weak∗ limits, possibly of subsequences, of (pn (un ))n and (qn (un ))n , respectively, and satisfy 0 6 p0 (x) 6 +
and
0 6 q0 (x) 6 −
a:e: in : = We now distinguish two cases: It is clear that, by (2.25), v vanishes somewhere in . • p0 (x) = 0 or q0 (x) = 0 a.e. in ; • p0 (x) ¿ 0 and q0 (x) ¿ 0 on subsets of of positive measure. In the ?rst case, if e.g. q0 (x) = 0 a.e. in , then testing (2.28) against w = v− , we v− = 0 and therefore v = v+ in get ∇v− = 0 in and, since v vanishes somewhere, + p−1 . Testing (2.28) against w = 1, we obtain p0 (v ) = 0, which implies v(x) = 0 where p0 (x) ¿ 0. Finally, testing (2.28) against w = v, we conclude that ∇v = 0 in and hence v = 0 in , because v vanishes somewhere. This yields a contradiction. In the second case, Lemma 2.2 implies that v changes sign in and, p0 (x) = + where v(x) ¿ 0 and q0 (x) = − where v(x) ¡ 0. Therefore, we have + p−1 |+ − pn (un )|(v ) = (+ − pn (un ))(v+ )p−1 → 0;
|− − qn (un )|(v− )p−1 =
(− − qn (un ))(v− )p−1 → 0
and hence 1−p | un ∞ fn (un ) − [+ (vn+ )p−1 − − (vn− )p−1 ]|
=
6
1−p |[pn (un ) − + ](vn+ )p−1 − [qn (un ) − − ](vn− )p−1 + un ∞ rn (un )|
|pn (un ) − + |(v+ )p−1 +
|qn (un ) − − |(v− )p−1
384
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
+
+
+
|pn (un ) − + (vn+ )p−1 − (v+ )p−1 | |qn (un ) − − (vn− )p−1 − (v− )p−1 | 1−p |rn (un )| → 0 un ∞
as n → + ∞. Moreover, since v changes sign in , there are an integer nˆ and constants ˆ ;1 ; ;2 , with 0 ¡ ;1 ¡ 1 ¡ ;2 , such that for every n ¿ n, max un 6 ;2 : (2.29) ;1 6 −min un In order to prove (2.26), we exploit condition (2.23) and the following technical result. = there exists a simple piecewise linear path (; joining x0 Claim. For every x0 ; y0 ∈ ; to y0 and having range contained in ; with the possible exception of the end points x0 and y0 ; such that; possibly for a subsequence; 1−p | un ∞ fn (un ) − [+ vn+ − − vn− ]| → 0; (
as n → +∞; where
(
denotes a line integral.
The proof of this result, when p = 2, is given in [11]. The extension to the present situation is immediate and therefore is omitted. We know that un changes sign in , provided that n is suLciently large. Let us denote by yn ∈ = a point where un vanishes and let x n ∈ = be such that un (x n ) = = where max un (¿ Sn ). Clearly, we can suppose that x n → x0 ∈ = and yn → y0 ∈ , v(x0 ) = max v and v(y0 ) = 0. Let (n denote the path joining x n to yn constructed in the = then join x0 to following way: ?rst join x n to x0 by a C 1 path n having range in , y0 by the piecewise linear path ( provided by the above claim, ?nally join y0 to yn = By the smoothness of @; n and =n can be taken by a C 1 path =n having range in . such that ‘(n ) → 0
and
‘(=n ) → 0
as n → + ∞;
(2.30)
where ‘(·) denotes the length of the corresponding path. Now, using (2.27) and the properties of fn , we get + (max un )p − Fn (max un ) p − − + + p p = (u (x n )) − (u (x n )) − Fn (un (x n )) p n p n + + − − p p − (u (yn )) − (u (yn )) − Fn (un (yn )) p n p n
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
=
0
+
1
+ (un+ (n ))p−1 − − (un− (n ))p−1 − fn (un (n )) ∇un (n ); n
1
0
+
385
1
0
[+ (un+ (())p−1 − − (un− (())p−1 − fn (un (())]∇un ((); ( [+ (un+ (=n ))p−1 − − (un− (=n ))p−1 − fn (un (=n ))]∇un (=n ); =n
6 (‘(n ) + ‘(=n )) fn (un ) − [+ (un+ )p−1 − − (un− )p−1 ] ∞ ∇un ∞ + |fn (un ) − [+ (un+ )p−1 − − (un− )p−1 ]| ∇un ∞
(
6 ‘(n ) + ‘(=n ) +
1−p un ∞
(
|fn (un ) −
[+ (un+ )p−1
−
− (un− )p−1 ]|
× C un p∞ : Therefore, by the above Claim and (2.30), we deduce + (max un )p − Fn (max un ) 6 4n un p∞ ; p
(2.31)
where 4n → 0 as n → +∞. On the other hand, using (2.23) and (2.29), we have ;;p ; + (max un )p − Fn (max un ) ¿ (max un )p ¿ 1 un p∞ : p p p
(2.32)
Comparing (2.31) and (2.32) yields a contradiction, letting n → +∞. This concludes the proof of Theorem 1.1. Proof of Corollary 1.1. The conclusion immediately follows combining the following lemma with Claim 1 in the proof of Lemma 2.1, if (1.7) holds, or with Theorem 1.1, if (1.5), (1.6) and (1.8) hold. Lemma 2.3. Set h= = (1=meas )
h: Assume that there is a constant r ¿ 0 such that
f(s) ¿ h= for every s ¿ r
(2.33)
(f(s) ¿ h= for every s 6 − r; respectively): Then; for any c ¿ 0 problem (1:1) has a lower solution ; with (x) ¿ c in ((x) 6 − c in ; respectively). Similarly; assume that there is a constant r ¿ 0 such that f(s) 6 h= for every s 6 − r (f(s) 6 h= for every s ¿ r; respectively): Then; for any c ¿ 0 problem (1:1) has an upper solution ; with (x) 6 − c in ((x) ¿ c in ; respectively).
386
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Proof of Lemma 2.3. Assume that condition (2.33) holds, the other cases being treated similarly. In order to built a lower solution of (1.1), we will prove that the problem −4p w = h= − h(x) in ; @w=@ = 0 w=0
on @;
(2.34)
= Then, for any constant has a solution w ∈ W 1; p ()∩L∞ (), and therefore w ∈ C 1; (). t ¿ w ∞ + r, we de?ne (x) = t + w(x) in . It is immediately seen, by condition (2.33), that is a lower solution. 1; p 1; p First, we prove that (2.34) has a solution w 1;∈pW (). Let us set W = {w ∈ W () | w = 0}: W is closed linear subspace of W () and therefore is a reOexive Banach space. De?ne a functional J : W → R by setting J (w) = (1=p) |∇w|p + (h= − h)w: Clearly, J is strictly convex and C 1 . Moreover, the Poincar1e–Wirtinger inequality implies that J is coercive. Hence, J has a unique global minimum point w ∈ W and, therefore, for every z ∈ W and every c ∈ R, we get |∇w|p−2 ∇w∇(z + c) = (h= − h)(z + c);
i.e. w is a solution of (2.34). Second, we show that w ∈ L∞ (). Of course, only the case where p 6 N requires a proof. Assume that w = 0 and observe that, as w = 0, w changes sign on subsets of of positive measure. Let us set B = {x ∈ | w(x) ¡ 0} and, for each n, An = {x ∈ | w(x) ¿ n} and wn = (w − n)+ . It is clear that wn (x) = 0 a.e. on B. We claim that there exists C ¿ 0 such that wn p 6 C ∇wn p
(2.35)
for every n. Indeed, otherwise, possibly passing to a subsequence, we should have ∇wn p = wn p → 0: Setting, for each n, zn = wn = wn p , we have, possibly passing to a further subsequence, that (zn )n converges weakly in W 1; p (), strongly in Lp () and pointwise a.e. in to some function z ∈ W 1; p (). Actually, (zn )n converges strongly in W 1; p () to z. Hence, z satis?es z p = 1 and ∇z = 0. The Poincar1e–Wirtinger inequality implies that z is constant in and, since z(x) = 0 a.e. in B, we conclude that z(x) = 0 a.e. in : a contradiction. To prove the boundedness from above of w, we apply Lemma 5:1 in [14]. Accordingly, we show that there exist constants C ¿ 0 and 4 ¿ 0 such that, for every n, wn 6 C(meas(An ))1+4 : (2.36) An
Assume that p ¡ N ; a similar argument applies if p = N . In the sequel, the same symbol C will denote constants which may change time by time, but are all independent of n. By HUolder inequality and the Sobolev imbedding theorem, we have, using (2.35)
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
too,
An
wn 6 (meas (An ))
1−(N −p)=Np
An
(wn )
Np=(N −p)
387
(N −p)=Np
= (meas (An ))1−(N −p)=Np wn Np=(N −p) 6 C(meas (An ))1−(N −p)=Np wn 1; p 6 C(meas (An ))1−(N −p)=Np ∇wn p : Testing the equation in (2.34) against wn ∈ W 1; p (), we get = ∞ |∇wn |p 6 h − h wn
and hence 1−(N −p)=Np wn 6 C(meas (An ))
An
An
that is An
1=p wn
;
wn 6 C(meas(An ))1+p=(N (p−1)) :
Similarly, one proves that w is bounded from below. Proof of Proposition 1.1. Assume, for instance, that condition (1.9) holds at −∞. As in the proof of Theorem 1.1, we can also suppose, without loss of generality, that (2.21) holds. Then, for each n, we de?ne the truncations
f(s) if s 6 n; fn (s) = f(n) if s ¿ n: = and C ¿ 0 such Conditions (1.9) and (2.21) imply the existence of constants ˆ ∈ ]0; [ that ˆ − )p−1 + 2C |fn (s) − h(x)| 6 fn (s) − h(x) + 2(s
(2.37)
for every n, every s and a.e. x. Consider problem (2.24) and observe that, for every n ¿ max{ ∞ ; ∞ }, and are a lower and an upper solution, respectively. Hence, the assumptions of Lemma 2.1 are satis?ed and therefore (2.24) has at least one solution un with min(un − ) 6 0
and
max(un − ) ¿ 0:
(2.38)
The solvability of problem (1.1) will be obviously achieved if we are able to prove the existence of an integer n˜ such that ˜ max un˜ 6 n:
(2.39)
388
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
Assume, by contradiction, that max un ¿ n for every n. Hence, un ∞ → +∞ and un 1; p → +∞, as p ¿ N . Let us set vn =
un : un 1; p
Possibly passing to a subsequence, still denoted by (vn )n , we have that (vn )n converges = to some function v. Moreover, vn satis?es weakly in W 1; p () and strongly in C 0 () fn (un ) − h |∇vn |p−2 ∇vn ∇w = w (2.40) p−1 un 1; p for every w ∈ W 1; p (). Testing (2.40) against w = 1 and using (2.37), we get |fn (un ) − h| ˆ (v− )p−1 + 2C meas 6 2 (2.41) n un p−1 un p−1 1; p 1; p and hence there is a constant still denoted by C such that fn (un ) − h 1 un p−1 1; p
6C
(2.42)
for every n. Now, testing (2.40) against w = vn − v and using (2.42), we obtain |∇vn |p−2 ∇vn ∇(vn − v) 6 C vn − v ∞ → 0
as n → +∞. Hence, condition (S+) in [4, p. 18] implies that vn → v in W 1; p () and therefore, as vn 1; p = 1, also v 1; p = 1. Next, we show that v− = 0. Indeed, if v− = 0, then vn− → 0 in L∞ () and, by (2.41), fn (un ) − h 1 = un p−1 1; p → 0: Hence, testing (2.40) against w = v, we get |∇vn |p−2 ∇vn ∇v → 0: Since |∇vn |p−2 ∇vn → |∇v|p−2 ∇v in (Lp=(p−1) ())N , we conclude that |∇v|p = 0, i.e. ∇v = 0 a.e. in . As v is continuous and vanishes somewhere in = by (2.38), we get v = 0: a contradiction. Finally, testing (2.40) against −vn− and using (2.37) again, we easily have fn (un ) − h − (fn (un ) − h)− − − p |∇(vn )| = − v 6 vn n p−1 un p−1 un 1; p 1; p C ˆ 6 (vn− )p + un p−1 1; p and hence, passing to the limit, − p ˆ |∇(v )| 6 |v− |p :
= Since ˆ ¡ = and v− = 0, we get a contradiction with the de?nition of .
M. Alif, P. Omari / Nonlinear Analysis 51 (2002) 369 – 389
389
References [1] R.A. Adams, Sobolev Spaces, Academic Press, New York, 1975. [2] D. Arcoya, L. Orsina, Landesman-Lazer conditions and quasilinear elliptic equations, Nonlinear Analysis T.M.A. 28 (1997) 1623–1632. [3] M. Arias, J. Campos, J.P. Gossez, On the antimaximum principle and the Fu0c12k spectrum for the Neumann p-Laplacian, DiIerential Integral Equations 13 (2000) 217–226. [4] F.E. Browder, Existence theorem for nonlinear partial diIerential equations, Proceedings of the Symposium on Pure Mathematics, Vol. XVI, Berkeley, CA, Amer. Math. Soc., Providence, RI, 1968, pp. 1– 60. [5] M. Cuesta, D. de Figueiredo, J.P. Gossez, The beginning of the Fu0c12k spectrum for the p-Laplacian, J. DiIerential Equations 159 (1999) 212–238. [6] M. Cuesta, D. de Figueiredo, J.P. Gossez, A nodal domain property for the p-Laplacian, C. R. Acad. Sci. Paris 330 (2000) 669–673. [7] C. De Coster, M. Henrard, Existence and localization of solutions for second order elliptic BVP in presence of lower and upper solutions without any order, J. DiIerential Equations 145 (1998) 420–452. [8] E. Di Benedetto, C 1+ local regularity of weak solutions of degenerate elliptic equations, Nonlinear Analysis T.M.A. 7 (1983) 827–850. [9] D. de Figueiredo, J.P. Gossez, On the ?rst curve of the Fu0c12k spectrum of an elliptic operator, DiIerential Integral Equations 7 (1994) 1285–1302. [10] M. Garc12a-Huidobro, R. ManZasevich, F. Zanolin, Strongly nonlinear second-order ODEs with unilateral conditions, DiIerential Integral Equations 6 (1993) 1057–1078. [11] J.P. Gossez, P. Omari, On a semilinear elliptic Neumann problem with asymmetric nonlinearities, Trans. Amer. Math. Soc. 347 (1995) 2553–2562. [12] P. Habets, P. Omari, Existence and localization of solutions of second order elliptic problems using lower and upper solutions in the reversed order, Top. Meth. Nonlinear Analysis 8 (1996) 25–56. [13] J. Kazdan, F. Warner, Remarks on some quasilinear elliptic equations, Comm. Pure Appl. Math. 28 (1975) 567–597. [14] O.A. Ladyzhenskaya, N.N. Ural’tseva, Linear and Quasilinear Elliptic Equations, Academic Press, New York, 1968. [15] G.M. Lieberman, Boundary regularity for solutions of degenerate elliptic equations, Nonlinear Analysis T.M.A. 12 (1988) 1203–1219. [16] P. Omari, F. Zanolin, Nonresonance conditions on the potential for a second-order periodic boundary value problem, Proc. Amer. Math. Soc. 117 (1993) 125–135. [17] J.L. Vazquez, A strong maximum principle for some quasilinear elliptic equations, Appl. Math. Optim. 3 (1984) 191–202. [18] S. Villegas, A Neumann problem with asymmetric nonlinearity and a related minimizing problem, J. DiIerential Equations 145 (1998) 145–155.