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Existence and asymptotic behavior of solutions for a class of nonlinear elliptic equations with Neumann condition
Nonlinear Analysis 61 (2005) 21 – 40 www.elsevier.com/locate/na
Existence and asymptotic behavior of solutions for a class of nonlinear elliptic equations with Neumann condition夡 Ilma Marques IMECC, Universidade Estadual de Campinas, Caixa Postal 6065, 13081-970 Campinas-SP, Brazil Received 24 January 2004; accepted 16 November 2004
Abstract In this paper, we study the existence of nonconstant solutions u and their asymptotic behavior (as
→ 0+ ) for the following class of nonlinear elliptic equations in radial form:
−2 (r |u | u ) = r f (u) u (0) = u (R) = 0,
in (0, R),
where , , are given real numbers and 0 < R < ∞. We use a version of the Mountain Pass Theorem and the main difficulty is to prove that the solutions so obtained are not constants. For that matter, we have to carryout a careful analysis of the solutions of some Dirichlet problems associated with the Neumann problem. 䉷 2004 Elsevier Ltd. All rights reserved. Keywords: Neumann problem; Nonconstant solutions; Existence; Asymptotic behaviour
1. Introduction In this paper, we consider the following class of nonlinear elliptic equations in radial form: 2 − (r |u | u ) = r f (u) in (0, R), (1) u (0) = u (R) = 0, 夡 These
results appear in the author’s UNICAMP doctoral dissertation. E-mail address: [email protected] (I. Marques).
0362-546X/$ - see front matter 䉷 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.11.006
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
where , , are given real numbers such that and 0, > 0 is a small parameter and 0 < R < ∞. We make the following assumptions on the nonlinearity f: (f1 ) f : [a, b] → R is a function of the class C 1 , with f (a) = f (b) = 0; (f2 ) f has exactly 2l + 1 zeros, a = a1 < 0 = a2 < a3 < · · · < a2l+1 = b with l = 1, 2, 3, . . . such that f (ai ) = 0 ∀i and f (ai ) < 0, if i is odd; 2l ) (f3 ) limt→a2l |t−af (t−a > 0. | (t−a ) 2l
2l
It has been emphasized in [5] that most of the phenomena occurring in the study of nonlinear elliptic equations can be more easily explained and understood when these equations are written in the radial form. Solutions of (1) will be absolutely continuous functions u : (0, R) → R and the equation will be satisfied a weak sense, defined below. Let us denote Lu := −(r |u | u ) . The motivation for our study is the fact that this operator includes the following class of operators, when considering acting in radial symmetric functions defined in a ball of RN : (i) Laplacian: = = N − 1, = 0, (ii) p-Laplacian (1 < p < N ): = = N − 1, = p − 2, (iii) k-Hessian (1 k < N/2): = N − k, = N − 1, = k − 1. R of absolutely continuous functions u : (0, R) → R We consider the Banach space X such that R R 2 +2 +2 r |u (r)| dr + r |u(r)|+2 dr < ∞. :=
u 0
0
R , a u(x) b for all x, is said to be a weak solution of (1) if A function u ∈ X R R R . 2 r |u | u dr − r f (u) dr = 0 ∀ ∈ X 0
0
q
Let us denote by L (0, R), q 1 and > − 1 the Banach space of Lebesgue measurable functions u : (0, R) → R such that R 1/q q q |u|L := r |u(r)| dr < ∞.
0
R and each weight , we define the critical exponent Associated with each space X q ∗ :=
( + 1)( + 2) , −−1
under the assumption that − − 1 > 0. R by Define a functional on X R R 2 +2 r |u | dr − r F (u) dr, I (u) = +2 0 0 u where F (u) = 0 f (t) dt.
(2)
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
23
The weak solutions of (1) are the critical points of the functional I . R is said to be a local minimum of function I if I (u0 ) I (u) for all A function u0 ∈ X R such that 0 < u − u0 0 for some 0 > 0. u∈X Using standard variational methods (local minimization and the Mountain Pass Theorem), we can prove that (1) has at least l nonconstant solutions and using monotone iteration methods we can prove that the nonconstant solutions approach either al or a2l+1 in (0, R) as → 0 almost everywhere. The hardest step of our proof is to show that the solutions given by the Mountain Pass Theorem are different from ai . For this matter, we use the existence of solutions for some Dirichlet problems associated with (1). A careful analysis of their asymptotic behavior is R . This fact is done in Section 2. If i is odd, we prove that ai is a local minimum in X not trivial either, since in this space we do not have any result saying that if ai is a local R . For the space W 1,p (), where is a minimum in C 1 , it is also a local minimum in X 0 smooth bounded domain in RN , an analogous result was obtained in [8] and for the space H01 (), an analogous result was obtained in [3]. Our main results are the following: Theorem 1.1. Suppose that f satisfies assumptions (f1 )–(f3 ). Then, there exists 0 > 0 such that, for all 0 < < 0 , (1) has at least l nonconstant solutions satisfying a1 < u1 (r) < a3 < u2 (r) < a5 < · · · < a2l−1 < ul (r) < a2l+1 , where l = 1, 2, 3, . . . . Remark 1.1. From a known regularity result (see [5, Proposition 2.2]), there exists ∈ (0, 1) such that ul ∈ C 1, (0, R). Remark 1.2. Theorem 1.1 is still true if we consider the ai s negative for all i 3, i.e., a = a2l+1 < · · · < a3 < a2 = 0 < a1 = b. The condition a2 = 0 is not essential, and it is only used to make the presentation easier. Theorem 1.2. Suppose that f satisfies assumptions (f1 )–(f3 ) and let u be a nonconstant solution of (1). Given any ∈ (a2l , a2l+1 ), let us consider the set
+ (, ) := {r ∈ (0, R) : a2l < u (r) < < a2l+1 }. Let B be the largest ball contained in + and let w ∗ (, ) be its radius. Then lim→0 w ∗ (, ) = 0. 2. Auxiliary results A fundamental step in the proof of Theorem 1.1 consists in showing the existence of solutions and their asymptotic behavior for the following Dirichlet problem: 2 − (r |u | u ) = r f (u) in (0, R), (3) u (0) = u(R) = 0,
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
where 0 < R < ∞, , , are given real numbers such that and 0. > 0 is a small parameter. The function f satisfies assumptions (f1 )–(f3 ) for l = 1, that is, f has only three zeros. We consider the Banach space XR of absolutely continuous functions u : (0, R) → R R +2 such that u(R) = 0 and u XR := 2 0 r |u (r)|+2 dr < ∞. We denote by 1 the first eigenvalue of the eigenvalue problem (see [5, p. 145])
Lu = r |u| u in (0, R), u (0) = u(R) = 0,
(4)
where and are given real numbers, 0 < R < ∞. 1 is characterized by
1 =
inf
u∈XR \{0}
R +2 dr 0 r |u (r)| . R +2 dr 0 r |u(r)|
The following theorem will be used in the proof of Theorem 1.1. Theorem 2.1. Let f be a function satisfying assumptions (f1 )–(f3 ) with l = 1. Then given any ball B(0, r1 ) ⊂ B(0, R), 0 < r1 < R, there exists 0 > 0 such that, for all 0 < 0 , the Dirichlet problem
−2 (r |u | u ) = r f (u) u (0) = u(r1 ) = 0,
in (0, r1 ),
(5)
has a positive solution 0 < u (r) < a3 in [0, r1 ] such that u → a3 as → 0 uniformly in every compact subset of (0, r1 ). Moreover, the Dirichlet problem
−2 (r |u | u ) = r f (u) u(r1 ) = u(R) = 0,
in (r1 , R),
(6)
has a negative solution a1 < v (r) < 0 in [r1 , R] such that v → a1 as → 0 uniformly in every compact subset of (r1 , R). The proof of Theorem 2.1 is done by combining an argument of truncation and the method of lower and upper-solution. A function u ∈ XR ∩ L∞ is said to be a lower solution of (3) if
2 0R r |u | u dr 0R r f (u) dr u (0) = 0, u(R) 0.
∀ ∈ XR , 0,
In the same way, a function u ∈ XR ∩ L∞ is said to be an upper-solution of (3) if the inequality signs are reversed.
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
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Lemma 2.1. Consider g : R → R, a continuous and increasing function, such that g(0) = 0. Let u1 , u2 ∈ XR ∩ L∞ be functions such that, for all ∈ XR , 0, we have R 2R 2 (r)) dr 0 r |u2 | u2 dr + 0 g(u R R 2 0 r |u1 | u1 dr + 0 g(u1 (r)) dr, u2 (R) u1 (R). Then u2 u1 a.e. (0, R). The proof of this result is similar to the proof of the Lemma 2.2 in [4]. So we omit it. Lemma 2.2. Let g : R → R be a continuous and increasing function, such that g(0) = 0. q Then, for every function h ∈ L (0, R), where 1/q + 1/q = 1, the problem
−2 (r |u | u ) + g(u(r)) = h(r), u (0) = u(R) = 0
r ∈ (0, R),
(7) q
admits a unique weak solution u ∈ XR . Moreover, the associated operator T : L → XR , h → u is continuous and nondecreasing. Proof. We split the proof into two parts. Part 1 (Existence): Consider the functional ∈ C 1 (XR , R) defined by R R R 2 +2 (u) := r |u | dr + G(u) dr − h(r)u dr, +2 0 0 0 u where G(u) = 0 g(t) dt. So the critical points u ∈ XR of are the weak solutions of (7), i.e.,
R
2 0
(r |u | u ) dr +
R
R
g(u) dr −
0
h(r) dr = 0
0
∀ ∈ XR .
(8)
We prove that the functional has a global minimum. For that matter, we observe that is coercive and weakly lower semicontinuous. The coerciveness follows from the estimate R 1 +2 (u) h(r)u dr.
u XR − +2 0 For the weakly lower semicontinuity, we use Proposition 1.1 [5] which states that the R functional (u) := 0 G(u) dr is weakly lower semicontinuous. Part 2 (Uniqueness): Let u1 , u2 ∈ XR be weak solutions of (7). Then
R
2 0
[(r |u1 | u1 ) − (r |u2 | u2 )] dr +
R 0
[g(u1 ) − g(u2 )] dr = 0.
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Choosing = (u1 − u2 )+ and using the fact that g is increasing, we obtain R [(r |u1 | u1 ) − (r |u2 | u2 )]((u1 − u2 )+ ) dr 0 = 2 0 R + [g(u1 ) − g(u2 )](u1 − u2 )+ dr 0
= 2
u1 u2
+
u1 u2
+
u1 u2
r
|u1 | + |u2 | |((u1 − u2 )+ ) |2 dr 2
r
|u1 | − |u2 | 2 [|u1 | − |u2 |2 ] dr 2
[g(u1 ) − g(u2 )](u1 − u2 )+ dr 0.
Hence, (u1 − u2 )+ = 0 a.e. in (0, R) or, equivalently, u1 u2 a.e. in (0, R). By choosing = (u2 − u1 )+ and using a similar argument, we obtain that u2 u1 a.e. in (0, R). Hence, problem (7) has a unique weak solution u ∈ XR . The fact that T is nondecreasing follows from Lemma 2.1. 3. Proof of auxiliary results Proposition 3.1. Suppose that f satisfies assumptions (f1 )–(f3 ) with l = 1. Then, given any ball B(0, r1 ) ⊂ B(0, R), 0 < r1 < R, there exists 0 > 0 such that, for all 0 < 0 , the Dirichlet problem (5) has a positive solution 0 < u (r) < a3 in [0, r1 ]. Moreover, the Dirichlet problem (6) has a negative solution a1 < v (r) < 0 in [r1 , R]. Proof. We start by proving the existence of a positive solution for problem (5). Let f1 : R → R be defined by f1 (u) = f (u) if 0 u a3 and f1 (u) = 0 elsewhere. By (f1 ), we have that there exists M > 0 such that r f1 (t) + Mt is nondecreasing in t for t ∈ [0, a3 ]. Let us consider the following auxiliary problem 2 − (r |u | u ) + Mu = r f1 (u) + Mu in (0, r1 ), (9) u (0) = u(r1 ) = 0. Next, we prove the existence of a positive solution of (9) by using the method of the lowerand upper-solution. First, we observe that the function u(r) ≡ a3 , for r ∈ [0, r1 ] is an upper-solution of (9). Construction of the lower-solution of (9): let a = limt→0 f1 (t)/|t| t. It follows from (f3 ) that given ∈ (0, a), there exists t0 > 0 such that for all |t| t0 , we have a −
f1 (t) . |t| t
(10)
Let 1 be the first eigenvalue of problem (4) and 1 > 0 be an eigenfunction corresponding to this first eigenvalue. Take 1 > 0 such that |1 1 (r)| t0 and 1 (max(0,r1 ) 1 ) < a3 . From
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
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(10) a − f1 (1 1 )/|1 1 | 1 1 . Choosing 0 > 0 such that 20 1 (r /r ) < a − , we have that 1 1 is a lower-solution of (9) for all 0 < 0 . Next we will show that there exists a minimal (and, respectively, a maximal) weak solution u∗ (resp. u∗ ) for (9) such that 1 1 = u u∗ u = a3 . Consider the set [u, u] := {u ∈ L∞ (0, r1 ) : u(r) u(r) u(r) a.e. in (0, r1 )} with the topology of a.e. convergence, and q
q
define the operator S : [u, u] → L by Sv=r f1 (v)+Mv ∈ L∞ (0, r1 ) ⊂ L (0, r1 ) ∀v ∈ [u, u],where q is such that 1/q + 1/q = 1. Clearly, S is nondecreasing and bounded. Applying the Lebesgue Dominated Convergence Theorem, we obtain that Sv n −Sv q → L
0, for vn , v ∈ [u, u] and vn → v a.e. in (0, r1 ). Hence S is continuous. q Recall that the operator T : L () → XR defined in Lemma 2.2 is continuous and nondecreasing. And now consider the continuous nondecreasing mapping F : [u, u] → XR defined by F := T oS, i.e., for a function v ∈ [u, u], u : F (v) is the unique solution in the weak form, namely r1 r1 r1 r |u | u + Mu = (r f1 (v) + Mv) for ∈ XR . 0
0
0
Writing u1 = F (u) and u1 = F (u), we obtain that u F (u) F (u) F (u) u, a.e. in (0, r1 ) ∀u ∈ [u, u]. Repeating the same reasoning, we can prove the existence of sequences (un ) and (un ) satisfying u0 = u, un+1 = F (un ), u0 = u, un+1 = F (un ) such that u = u0 u1 · · · un u un · · · u1 u0 = u,
a.e. in (0, r1 ).
Then, un → u∗ , un → u∗ and u∗ = F (u∗ ), u∗ = F (u∗ ). Then, u∗ is the minimal weak solution (respectively, u∗ maximal weak solution) of (9) such that u∗ , u∗ ∈ [u, u], for all 0 < 0 . In particular, every solution u ∈ [u, u] of (9) satisfies also u∗ u u∗ , a.e. in (0, r1 ). Since the solutions u∗ and u∗ are between 0 and a3 , u∗ and u∗ are solutions of (5). Therefore, there exists a solution u := u∗ ∈ [1 1 , a3 ] of (5), for all 0 < 0 . To prove the existence of the negative solution v (r), where a1 v 0, it follows similarly using the truncated function f2 : R → R defined by f2 (u)=f (u) if a1 u 0 and f2 (u)=0 elsewhere. And now consider the problem 2 − (r |u | u ) = r f2 (u) in (r1 , R), u(r1 ) = u(R) = 0. 3.1. Asymptotic behavior of the solutions of the Dirichlet problem We have the following proposition which shows the asymptotic behavior of the solutions of (5) as → 0. Proposition 3.2. Let 0 < u < a3 be a positive solution of (5) and let a1 < v < 0 be a negative solution of (6). Then (i) u → a3 as → 0 uniformly on every compact subset of (0, r1 ); (ii) v → a1 as → 0 uniformly on every compact subset of (r1 , R).
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Proof. (i) The proof follows by adapting some arguments from Theorem 4 in [7]. First, observe that there exists ∈ (0, 1) such that u ∈ C 1, (0, r1 ), by using Proposition 2.2 in [5]. Consider the function f1 : R → R defined in the proof of the Proposition 3.1 and let 1 > 0 be an eigenfunction corresponding to the first eigenvalue 1 of Lu := −(r |u | u ) in (0, r1 ) subject to boundary conditions u (0) = 0, u(r1 ) = 0 (see Eq. (4) above). Take 1 such that 1 XR = 1. Since f1 0, f1 ≡ / 0 and 1 > 0 from Lemma 3.2 in [5], we have u > 0 in (0, r1 ), u 0 in (0, r1 ). Consequently, there exists 1 > 0 such that for all 0 < 0 , we have u (r) 1 1 , and for a given > 0 there is C such that u (r) C > 0,
(11)
for all r ∈ := {r ∈ (0, r1 ) : dist(r, r1 ) > }. Since u is the solution of (9) it follows that r1 r1 2 r f1 (u ) dr ∀ 0, ∈ XR . (12) r |u | u dr = 0
0
In particular, for = 1 , we obtain r1 r1 2 r |u | u 1 dr = r f1 (u )1 dr. 0
(13)
0
Claim. The expression in the left-hand side of (13) goes to zero as → 0. Indeed, observe
Thus, applying Hölder inequality in (13) and taking = u that 0 < u a3 and f1 (u ) C. in (12), we obtain
r1
2 0
r |u | u 1 dr 2
r1 0
r1
×
C 2
r |u |+2 dr
0
1 2
r
2/(+2) C
0
(+1)/(+2)
|1 |+2 dr
r1
1/(+2)
r f1 (u )u dr
(+1)/+2
f or some constant C.
Define d := inf{1 (r) : r ∈ } > 0. Then, d
r f1 (u ) dr
r f1 (u )1 dr <
r1 0
r f1 (u )1 → 0,
→ 0.
(14)
Now, suppose by contradiction that there are a number C1 > 0 and a sequence j → 0 such that the Lebesgue measure of the sets
,j := {r ∈ : uj (r) < a3 − }
(15)
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
are bounded below by C1 . It follows from (14) that r f1 (uj ) dr → 0 as j → 0. Ij := ,j
29
(16)
Observe that in ,j , from (11) and (15), we have C uj a3 − . Since f1 is bounded below in the interval [C , a3 − ] by a number d > 0, from (15) it follows r f1 (uj ) dr d r dr = dC d | ,j | d C1 , Ij = ,j
,j
for any 0 < d dC/| ,j |, which contradicts (16). Therefore, | ,j | is not bounded below, i.e., u (r) → a3 , on every compact subset of (0, r1 ) as → 0. (ii) It follows similarly as in the previous case (i). Proof of Theorem 2.1. The proof follows directly from Propositions 3.1 and 3.2.
4. Proof of Theorem 1.1 The proof is done by using the Mountain Pass Theorem of Ambrosetti and Rabinowitz [1], in a version due to Hofer [9]. 4.1. A particular case First, we consider a particular case of Theorem 1.1, that is, the case when f has only three zeros. Theorem 4.1. Let f satisfy assumptions (f1 )–(f3 ) for l = 1, that is, f has only three zeros. Then, there exists 0 > 0 such that, for all 0 < < 0 , (1) has at least one nonconstant solution u verifying a1 < u (r) < a3 . In order to prove Theorem 4.1, we use the two lemmas below. Lemma 4.1. Let f be a function satisfying assumptions (f1 )–(f3 ) for l = 1. Then there exist functions of class C 1 , f1 : (−∞, a1 ] → R+ , f2 : [a3 , +∞) → R− and real numbers 1 and 1 such that (i) f1 (a1 ) = f (a1 ), f1 (a1 ) = f (a1 ) and f1 (t) > 0 for all t ∈ ( 1 , a1 ); (ii) f2 (a3 ) = f (a3 ), f2 (a3 ) = f (a3 ) and f2 (t) < 0 for all t ∈ (a3 , 1 ); (iii) 1 and 1 are such that 1 < a1 < a3 < 1 , 0 a1 1 a3 f1 (t) dt = f (t) dt , f2 (t) dt = f (t) dt a3
1 0 a1 and for all t ∈ [0, 1], we have 1 < t (1 − t)a3 + (a3 − a1 )t + a1 < 1 and 1 < t (1 − t)a1 + (a3 − a1 )t + a1 < 1 .
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Proof. We start by proving the existence of 1 . We take (t) := t (1 − t)a1 + (a3 − a1 )t + a1 . Thus, 0 = (d/dt)((t)) if, and only if, t = a3 /2a1 . Moreover, (a3 /2a1 ) = a32 /4a1 + a1 is a minimal value of (t). So, we define
1 := a32 /4a1 + 2a1 . To prove the existence of 1 , we take (t) := t (1 − t)a3 + (a3 − a1 )t + a1 . We note that 0 = (d/dt)((t)) if, and only if, t = 1 − a1 /2a3 . It follows that (1 − a1 /2a3 ) = a12 /4a3 + a3 is a maximal value of . Now, we take 1 := a12 /4a3 + 2a3 . Next, we prove the existence of the function f1 . We take g(t) = f (a1 )(t − a1 ) and 1 , 2 functions of class C 1 so that 1 ≡ 1 at a neighborhood of a1 , 2 ≡ 1 at a neighborhood of 1 , 1 (t) + 2 (t) = 1 for all t ∈ [ 1 , a1 ] and 0 a1 g(t)1 (t) dt < f (t) dt .
1
a1
Now, we choose r > 0 such that a1 [r 2 (t)g(t) + 1 (t)g(t)] dt =
1
0 a1
f (t) dt .
We define f1 : (−∞, a1 ] → R+ by r 2 (t)g(t) + 1 (t)g(t), 1 t a1 , f1 (t) := rf (a1 )t − ra 1 f (a1 ), t 1 . We note that for t 1 , the graph of f1 is the tangent line to f1 at the point ( 1 , f1 ( 1 )). Finally, we prove the existence of f2 . We take g(t) = f (a3 )(t − a3 ) and 1 , 2 functions of class C 1 so that 1 ≡ 1 at a neighborhood of a3 , 2 ≡ 1 at a neighborhood of 1 , 1 (t) + 2 (t) = 1 for all t ∈ [a3 , 1 ] and 1 a3 g(t)1 (t) dt < f (t) dt. a3 0 Now, we choose r > 0 such that 1 a3 [r 2 (t)g(t) + 1 (t)g(t)] dt = f (t) dt. a3 0 We define f2 : [a3 , +∞) → R− by r 2 (t)g(t) + 1 (t)g(t), a3 t 1 , f2 (t) := t 1 . rf (a1 )t − ra 1 f (a1 ), We observe that for t 1 , the graph of f2 is the tangent line to f2 at the point ( 1 , f2 ( 1 )). Thus the proof of the lemma is concluded. As a consequence of the Maximum Principle, we have the following lemma.
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Lemma 4.2. If u is a nonconstant solution of the Neumann problem 2 − (r |u | u ) = r f(u) in (0, R), u (0) = u (R) = 0,
31
(17)
where f : R → R is the function defined by f (t), a1 t a3 , f(t) := f1 (t), t a1 , f2 (t), t a3 , where f1 and f2 are defined in Lemma 4.1. Then a1 u (r) a3 . Proof. We start by proving that u (r) a3 . Indeed, let u (r) − a3 if u (r) a3 , v(r) := 0 if u (r) < a3 and + := {r ∈ (0, R) : u (r) a3 }. Note that R R 2 r |u | u v dr = r f(u )v dr = 0
0
+
r f2 (u )v dr 0.
The left-hand side of the above relation is 0. So this implies + r f2 (u )v = 0 ⇒ v ≡ 0. Therefore, u (r) a3 , for all r ∈ (0, R). Similarly, we have a1 u , if we take − := {r ∈ (0, R) : u (r) a1 } and u (r) − a1 , u (r) a1 , w(r) := 0, u (r) > a1 . This finishes the proof of the lemma.
Thus, from Lemma 4.2, it follows that a solution of (17) is a solution of (1), since f ≡ f in [a1 , a3 ]. Therefore, to prove Theorem 4.1 it suffices to prove the following theorem. Theorem 4.2. Let f be a function satisfying conditions (f1 )–(f3 ). Then there exists 0 > 0 such that, for all 0 < < 0 , (17) has at least one nonconstant solution u . R by Define a functional on X R R 2 +2 (u) dr, r |u | dr − r F J (u) := +2 0 0 (u) := u f(t) dt. The functional J is C 1 with derivative given by where F 0
J (u) =
R
2 0
R
r |u | dr − 0
r f(u) dr
R . ∀ ∈ X
The weak solutions of (17) are the critical points of the functional J .
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Lemma 4.3. The functional J satisfies the Palais–Smale condition. R be a sequence satisfying J (un ) → c and J (un ) → 0, as n → ∞. Proof. Let (un ) ⊂ X Hence, 2 R R +2 |J (un )| = r |u | dr − r F (un ) dr d +2 0 0 for some d > 0, (18) R R 2 r |un | un v dr − r f (un )v dr n v (19) |J (un )v| = 0
0
R where n → 0, n → ∞. for all v ∈ X Claim. (un ) is bounded. Indeed, define 1 := {r ∈ (0, R) : un (r) 1 } and un (r) − 1 if un (r) 1 , vn (r) = 0 if un (r) < 1 . Since f(un ) − C in 1 , it follows from (19) with v = vn that
R
C 0
r vn dr n vn .
(20)
Define wn := vn / vn . Taking subsequences if necessary we can assume that wn / w R , wn → w in Lp and wn (r) → w(r) a.e. in (0, R). Dividing (20) by vn weakly in X and taking limits at both sides and observing that wn 0, we have that w ≡ 0 a.e. in (0, R). From (19) with v = vn , taking the limit as n → ∞, we obtain
R
2
r
0
|vn |+2 → 0.
vn
(21)
On the other hand, wn = 1, then
R
2 0
r |wn |+2 dr → 1.
(22)
Multiplying and dividing (21) by vn +1 , from (22) we conclude that vn → 0 as n → ∞. Consequently, vn C. Next, we define 2 := {r ∈ (0, R) : 1 un (r) 1 } and zn (r) :=
un (r) if 1 un (r) 1 , 0 if 1 un (r), 0 if un (r) 1 .
By a similar argument, we prove that zn C.
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
33
Finally, we consider 3 := {r ∈ (0, R) : un (r) 1 < 0} and the sequence tn (r) :=
un (r) − 1 0
if un (r) 1 , if un (r) > 1 .
Similarly, as in the previous case, we prove that tn C. Let un := vn + zn + tn . Then, (un ) is bounded in . Thus, J satisfies the (PS) condition. Lemma 4.4. The functional J is bounded below. Proof. Let 1 : =3 ∪ 4 and 2 := 5 ∪ 6 , with
3 := {r ∈ (0, R) : 0 u(r) a3 }, 5 := {r ∈ (0, R) : a1 u(r) 0},
4 := {r ∈ (0, R) : a3 < u(r) < ∞}, 6 := {r ∈ (0, R) : −∞ < u(r) < a1 }.
Since u is bounded in 3 and in 5 , we have (a3 ) dr C,
r F r F (u) dr < C + 1
2
4
(u) dr < C + r F
6
(a1 ) dr < C. r F
Also,
R
r F (u) dr =
1
0
r F (u) dr +
2
(u) dr < C. r F
Hence, J (u)
2 +2
R
r |u |+2 dr − C − C.
0
Therefore, J is bounded below.
R . Lemma 4.5. ai , with i odd, is a strict local minimum of J in X Proof. First of all, we will prove that ai is a local minimum of J in the C 1 topology. Let (t), to |t − ai | < and let u ∈ C 1 such that u(r) − ai C 1 = (ai ) F > 0 such that F max {|u(r) − ai |, |u (r) − ai |} < . We claim that there exists > 0 such that := {r ∈ (0, R) : |u(r) − ai | > } has positive measure. Indeed, let u ∈ C 1 (0, R), u ≡ / ai . Then, there exists r0 ∈ (0, R) such that u(r0 ) = ai . Hence, there exists > 0 such that either u(r0 ) > ai + or u(r0 ) < ai − . Since u is continuous, there exists a ball B (r0 ) such that |u(r) − ai | > , for all r ∈ B (r0 ). Therefore, has positive measure.
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
(ai − ), F (ai + )}. Since ai is a strict local maximum of Now, we define c1 := max{F , we have F R (u) dr (ai ) dr r F r c1 dr + r F
0
(ai )
(0,R)\
r dr +
(0,R)\
r F (ai ) dr =
R 0
(ai ) dr. r F
Therefore, J (u) > J (ai ), i.e., ai is a strict local minimum of J in C 1 . Take ai =a3 . Assume that a3 is a local minimum of J in C 1 and a3 is not a local minimum R . Then, for all > 0 small enough, there exists v ∈ B ⊂ X R such that in X J (v ) = min J (v) < J (a3 ),
(23)
v∈B
R : v − a3 }, since B where B is a ball of radius centered at a3 , B := {v ∈ X is weakly close and J is bounded below in B . Therefore, v satisfies the Euler equation R , where G(v ) := v − a3 , i.e., J (v ) = G (v ) ∀ ∈ X
2
R
R
r f(v ) r |v | v − 0 0
R R 2 r |v | v + r |v − a3 | (v − a3 ) =
0
0
R and for some Lagrange multiplier . Since v is a minimum of J in B for all ∈ X then 0. Rewriting the equation above, we have R R 2 r |v | v = [r f(v ) + r |v − a3 | (v − a3 )] (24) (1 − ) 0
0
R . Thus, v is a weak solution of the problem for all ∈ X 2 −(1 − ) (r |v | v ) = r |v − a3 | (v − a3 ) +r f(v ) in (0, R), v (0) = v (R) = 0.
(25)
We will show that a1 v (r) a3 , for all r ∈ (0, R). R . Claim 1. v is nonconstant. Indeed, if v ≡ constant then v ≡ a3 , since v → a3 in X This contradicts (23). Claim 2. v a3 . Indeed, let v (r) − a3 if v(r) := 0 if
v (r) a3 , v (r) < a3
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
35
and + := {r ∈ (0, R) : v (r) a3 }. From (24), we have (1 − )
R
2 0
r |v | v v dr =
+
[r f2 (v )v + r |v − a3 |+2 ] dr 0,
a.e. since, f2 (t) 0, v 0, 0. The function f2 is a function defined in Lemma 4.1. Since, (1 − ) 0, it follows v ≡ constant. Also, if v (r) a3 , for all r ∈ (0, R) it follows that v ≡ constant. It is a contradiction. Therefore, there exists r ∈ (0, R) such that v (r) < a3 , i.e., v(r) = 0. Then v(r) ≡ 0, for all r ∈ (0, R), i.e., v a3 for all r ∈ (0, R). Claim 3. a1 v . Indeed, it follows similarly as in Claim 2 if we take v (r) − a1 , v (r) a1 , w(r) := 0, v (r) > a1 and − := {r ∈ (0, R) : v (r) a1 }. Therefore, from Claims 2 and 3, we have that a1 v (r) a3 , i.e., sup(0,R) |v | C ∗ . Then from Theorem 2 in [10], we have that there exists a positive constant ∈ (0, 1) such that v ∈ C 1, and, moreover, there exists a positive constant C = C((0, R), C ∗ , ) such that v C 1, C. Thus, from Ascoli–Arzelá Theorem we get that there exists a subsequence of (v j ) = (v ) such that v → a3 in C 1 . It contradicts the fact that a3 is a local minimum of J in C 1 . Indeed, if a3 is a local minimum of J in C 1 , it follows J (a3 ) J (v) for all v ∈ C 1 such that 0 < v − a3 C 1 < 0 for some 0 > 0. Since v → a3 in C 1 , it follows J (a3 ) J (v ). On the other hand, we have from (23) that J (v ) < J (a3 ), for all > 0 small enough. R . Hence, the contradiction. Therefore, a3 is a local minimum of J in X R . Similarly, it follows that a1 is a local minimum of J in X R . Without loss of generality, we suppose that ai is a strict local minimum of J in X R such that J (v ) = J (ai ). Hence, v is a Contrarily, for all > 0, there exists v ∈ X R . critical point of J in X We will use arguments from [6] to obtain the next lemma. Lemma 4.6. If a is a strict local minimum of J , i.e., J (a) < J (u)
(26)
R such that 0 < u − a < 0 for some 0 > 0. Then, for any 0 < < 0 , for all u ∈ X R and u − a = } > J (a). inf{J (u) : u ∈ X
(27)
Proof. Assume by contradiction that the infimum in (27) is equal to J (a) for some with R with un −a = and, say, J (un ) J (a)+ 0 < < 0 . So there exists a sequence un ∈ X 2 1/2n . Call R : − u − a + }, A := {u ∈ X
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
where > 0 is chosen so that 0 < − and + < 0 . In view of our contradiction hypothesis in (26), it follows that inf{J (u) : u ∈ A} = J (a). We now apply the Ekeland Variational Principle to the functional J on A to get the existence of vn ∈ A such that J (vn ) J (un ),
(28)
1
vn − un , n
(29)
J (vn ) J (u) +
1
u − vn n
∀u ∈ A.
(30)
R , i.e., J (vn ) C (it follows Our purpose is to show that vn is a (PS) sequence for J in X by (28)) and J (vn ) → 0, as n → ∞. Once this is proved, we get that vn has a convergent R . Note that subsequence. Denote this subsequence by vn and we have that vn → v in X v ∈ A, since A is complete. Hence, v ∈ XR and therefore it satisfies v − a = and J (v) = J (a), which contradicts (26). So in order to conclude the proof of Lemma 4.6 we will prove that J (vn ) → 0, as R and ut := vn + tw, for n > 1/. We observe that for |t| sufficiently n → ∞. Take w ∈ X small, ut = vn + tw ∈ A. Indeed, lim ut − a = vn − a vn − un + un − a
t→0
1 + < + . n
On the other hand,
vn − a a − un − un − vn −
1 > − . n
Also, we can take u = ut in (30), and then, for t > 0, J (vn ) − J (vn + tw) 1 1 1
vn − tw − vn tw . t nt nt
(31)
Taking the limit in (31) as t → 0, we obtain J (vn ), w (1/n) w . Consequently, R . So, J (vn ) → 0, n → ∞ and the proof of the |J (vn ), w | (1/n) w ∀w ∈ X lemma is concluded. Proof of Theorem 4.2. We define R ) : h(0) = a1 , h(1) = a3 } := {h ∈ C([0, 1], X and
:= inf max J (h(t)). h∈ t∈[0,1]
From Lemma 4.6 := inf h∈ maxt∈[0,1] J (h(t)) > c = max{J (a1 ), J (a3 )}. Since J satisfies the condition (PS) using the Mountain Pass Theorem from [9], it follows that there R , exists u critical point of J such that J (u) = . If the critical points are not isolated in X then there exists an infinite number of critical points of J . Otherwise, u is of the mountain pass type (see [9]). Since a1 and a3 are the strict local minimum, u = a1 and u = a3 .
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
37
Therefore, in order to show the existence of a nonconstant critical point of J , we only need to prove that < 0, since 0 is the other constant solution and J (0) = 0. We claim that < 0. In fact, we consider B ⊂ (0, R) open ball and we define v (r), r ∈ B, u0 (r) := w (r), r ∈ [0, R]\B, where v is a positive solution of (5) in B and w is a negative solution of (6) in [0, R]\B. R . Now, given > 0, we Since (0, R) ∈ C 1 and v , w ∈ Lp (0, R), it follows that u0 ∈ X consider the special path h (t) := t (1 − t)u0 (r) + (a3 − a1 )t + a1
in .
Then we claim that there exists a small number 0 > 0, such that, for all 0 < < 0 and for all t ∈ [0, 1], max J (h (t)) < 0. In fact, suppose by contradiction that there is no such 0 > 0. Then, for any 0 > 0, there is < 0 such that J (h (t )) 0
(32)
for some t ∈ [0, 1]. Then choose a sequence (k ) such that limk→∞ k = 0, J (hk (tk )) 0, 0 limk→∞ tk = 1. For simplicity, we drop the indices and write k = and tk = t. So 2 2 r |h (t)|+2 dr r |h (t)|+2 dr + J (h (t)) = + 2 (0,R)\B +2 B R (h (t)) dr − r F 0 2 +2 = (1 − t)+2 r |v |+2 dr + r |w |+2 dr t +2 B (0,R)\B R (t (1 − t)u0 + (a3 − a1 )t + a1 ) dr. − r F 0
Since
2 and
B
r |v |+2 dr =
2
(0,R)\B
B
r f(v )v dr
r |w |+2 dr =
(0,R)\B
r f(w )w dr,
we get
t +2 (1 − t)+2 J (h (t)) = r f (v )v dr + r f (w )w dr +2 B (0,R)\B R (t (1 − t)u0 + (a3 − a1 )t + a1 ) dr. − r F 0
(33)
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Since v → a3 as → 0 uniformly on every compact subset of B and w → a1 as → 0 uniformly on every compact subset of \B, so lim r f(w )w dr = 0. r f(v )v dr = 0 and lim →0
→0
B
(0,R)\B
If we take the limit on both sides of (33), it follows from the Dominated Convergence Theorem and the above facts that
R (t (1 − t)u0 (r, ) + (a3 − a1 )t + a1 ) dr lim J (h (t)) = lim − r F →0
→0
0
= − r F ((1 − )a3 + (a3 − a1 ) + a1 )|B| ((1 − )a1 + (a3 − a1 ) + a1 )|(0, R)\B|, − r F where |A| is the Lebesgue measure of A ⊂ RN . It follows from Lemma 4.1 that ((1 − )a3 + (a3 − a1 ) + a1 ) 0, −F
(34)
((1 − )a1 + (a3 − a1 ) + a1 ) 0. −F
(35)
is zero only at 1 , 0 and In fact we have strict inequalities in either (34) or (35). Indeed, F
1 , so we conclude from Lemma 4.1(iii) that if (34) and (35) are zero, then
(1 − )a3 + (a3 − a1 ) + a1 = 0 = (1 − )a1 + (a3 − a1 ) + a1 , i.e., (1 − )a3 = (1 − )a1 . Since a3 = a1 , then (1 − ) = 0, i.e., either = 0 or 1. Hence, (a1 ) = 0 or F (a3 ) = 0. Then, either a1 = 0 or a3 = 0, which is impossible. Hence, either F lim→0 J (h (t)) < 0, which is in contradiction with (32). So we get finally that < 0. Corollary 4.1. Let f be satisfying (f1 )–(f3 ) for l = 1 and let u be a nonconstant solution of (1) such that J (u ) = . Then lim→0 sup J (u ) < 0. Proof. From the proof of Theorem 4.2, there exists a positive number 1 so that J (h1 (t)) < 0 for all t ∈ [0, 1], where h1 (t) = t (1 − t)u0 (r, 1 ) + (a3 − a1 )t + a1 , u0 as defined in the proof of Theorem 4.2. Hence, for all 0 < < 1 , J (h1 (t)) J (h1 (t)) < 0 for all t ∈ [0, 1]. It follows from the intermediate value theorem that there exists t ∈ [0, 1] such that h1 (t ) = u . Consequently, J (h1 (t )) = J (u ) for some t ∈ [0, 1], and so the assertion is true. Proof of Theorem 4.1. The proof follows directly from the Theorem 4.2 and Lemma 4.2. Proof of Theorem 1.1. For each l = 1, 2, . . . , we consider the function f : [a2l−1 − a2l , a2l+1 − a2l ] → R of class C 1 defined by f (t) := f (t + a2l ). Then from Theorem 4.1,
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
the problem 2 +2 + v ) = r f (v), − (r |v | v (0) = v (R) = 0,
r ∈ (0, R),
39
(36)
has a nonconstant solution vl (r) such that a2l−1 −a2l < vl (r) < a2l+1 −a2l , where v=u−a2l . So, ul = vl + a2l is a nonconstant solution of (1) with a2l−1 < ul < a2l+1 . Hence, there exist at least l nonconstant solutions of (1) satisfying a1 < u1 (r) < a3 < u2 (r) < a5 < · · · < a2l−1 < ul (r) < a2l+1 . 5. Asymptotic behavior of the solutions of the Neumann problem Proof of Theorem 1.2. Let a2l < < a2l+1 . Suppose by contradiction that lim→0 w ∗ (, ) = 0. Then, there is a convergent sequence {w∗ (k , )} so that limk→∞ w ∗ (k , ) = > 0. This means that, for each k > 0, there is rk =r(k , ) ∈ + (k , ) so that the ball B(rk , ), centered at the point rk with the radius , is contained in + (k , ). Consider f : [a2l−1 − a2l , a2l+1 − a2l ] → R of class C 1 defined by f (t) := f (t + a2l ) and 2 +2 −k (r |v | v ) = r f (v), r ∈ B(rk , ), (37) v (0) = v(r) = 0, r ∈ jB(rk , ), where v = u − a2l . Also, v = u − a2l with 0 v a2l+1 − a2l . Define f 1 : R → R by f 1 (v) = f (v) if 0 v a2l+1 − a2l and f 1 (v) = 0 elsewhere. By (f1 ), we have that there exists M > 0 such that r f 1 (t) + Mt is nondecreasing in t for t ∈ [0, a2l+1 − a2l ]. Consider the auxiliary problem 2 +2 −k (r |v | v ) + Mv = r f 1 (v) + Mv, r ∈ B(rk , ), (38) v (0) = v(r) = 0, r ∈ jB(rk , ). Note that v is in an upper solution of (38). Claim. There are k0 and 1 > 0 such that 1 1 is a lower solution of (38) for all 0 < k k0 , where 1 is an eigenfunction corresponding to the first eigenvalue 1 of problem (4). In fact, this follows similarly as in the proof of Proposition 3.1 step 2. Also, as in the proof of Proposition 3.1 step 3 with u = 1 and u = v , it follows that (38) has a minimal solution v∗ with 1 v∗ v a2l+1 − ra2l , and as in the proof of Proposition 3.2 with u = v∗ , f1 = f 1 , a3 = a2l+1 − a2l and 0 1 = B(rk , ) , it follows that v∗ → a2l+1 − a2l on every compact subset of B(rk , ) as k → 0. Therefore, there exists u∗ u with u∗ → a2l+1 so that v∗ = u∗ − a2l v = u − a2l . This leads to a contradiction for < a2l+1 . Remark 5.1. If we choose an open ball B = B(x, w ∗ (, )) centered at some point r = r(, ) whose radius w ∗ (, ) is the maximum of the radii of balls in
− (, ) = {r ∈ (0, R) : a2l−1 < − < u(r, ) < a2l } we can prove, by a similar method, that lim→0 w ∗ (, ) = 0.
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
5.1. Examples Example 5.1. Let f : R → R defined by f (u) :=
|u|p−2 u(p − p|u|p ) . (1 + |u|p )
Hence, f satisfies conditions (f1 )–(f3 ), for l = 1, where a1 = −1, a2 = 0 and a3 = 1. So, by Theorem 1.1, there is an 0 > 0 so that the Neumann problem (1) has a nonconstant solution a1 u a3 for all 0 < < 0 and p 2. 2 Example 5.2. Let f : R → R defined by f (u) = u(a R+ . Hence, f √ − u ), where a ∈ √ satisfies conditions (f1 )–(f3 ), for l = 1, where a1 = a, a2 = 0 and a3 = a. Then, by Theorem 1.1, 0 > 0 so that the Neumann problem (1) has at least l nonconstant solutions satisfying a1 u1 (x) a3 u2 (x) a5 < · · · < ul (x) < a2l+1 for all 0 < < 0 .
Example 5.3. Let f : [−, (2l − 1)] → R defined by f (u) = sen(u). Hence, f satisfies conditions (f1 )–(f3 ), for p=2, l=1, 2, 3, . . . , where a1 =− < a2 =0 < a3 = < · · · < a2l = (2l −2) < a2l+1 =(2l −1). Hence, from Theorem 1.1, there is 0 > 0 so that the Neumann problem (1) has at least l nonconstant solutions satisfying a1 u1 (x) a3 u2 (x) a5 < · · · < ul (x) < a2l+1 for all 0 < < 0 . References [1] A. Ambrosetti, P.H. Rabinowitz, Dual variational methods in critical point theory and applications, J. Funct. Anal. 14 (1973) 349–381. [3] H. Brézis, L. Nirenberg, H 1 versus local C 1 minimizers, C. R. Acad. Sci. Paris 317 (1993) 465–472. [4] A. Cañada, P. Drábek, J. Gamez, Existence of positive solutions for some problems with nonlinear diffusion, Trans. Amer. Math. Soc. 349 (1997) 4231–4249. [5] P. Clément, D.G. De Figueiredo, E. Mitidieri, Quasilinear elliptic equations with critical exponents, Top. Methods Nonlinear Anal. 7 (1996) 133–170. [6] M. Cuesta, D.G. De Figueiredo, J.P. Gossez, The beginning of the Fuˇcik spectrum for the p-Laplacian, J. Differential Equations 159 (1999) 212–238. [7] D.G. De Figueiredo, On the existence of multiple ordered solutions of nonlinear eigenvalue problems, Nonlinear Anal. TMA 11 (1987) 481–492. [8] J.P. García Azorero, I. Peral Alonso, J.J. Manfredi, Sobolev versus Hölder local minimizers and global multiplicity for some quasilinear elliptic equations, Comm. Contemp. Math. 3 (2000) 385–404. [9] H. Hofer, A geometric description of the neighborhood of a critical point given by the mountain-pass theorem, J. London Math. Soc. 31 (1985) 566–570. [10] G.M. Lieberman, Boundary regularity for solutions of degenerate elliptic equations, Nonlinear Anal. TMA 12 (1988) 1203–1219.
Existence and asymptotic behavior of solutions for a class of nonlinear elliptic equations with Neumann condition夡 Ilma Marques IMECC, Universidade Estadual de Campinas, Caixa Postal 6065, 13081-970 Campinas-SP, Brazil Received 24 January 2004; accepted 16 November 2004
Abstract In this paper, we study the existence of nonconstant solutions u and their asymptotic behavior (as
→ 0+ ) for the following class of nonlinear elliptic equations in radial form:
−2 (r |u | u ) = r f (u) u (0) = u (R) = 0,
in (0, R),
where , , are given real numbers and 0 < R < ∞. We use a version of the Mountain Pass Theorem and the main difficulty is to prove that the solutions so obtained are not constants. For that matter, we have to carryout a careful analysis of the solutions of some Dirichlet problems associated with the Neumann problem. 䉷 2004 Elsevier Ltd. All rights reserved. Keywords: Neumann problem; Nonconstant solutions; Existence; Asymptotic behaviour
1. Introduction In this paper, we consider the following class of nonlinear elliptic equations in radial form: 2 − (r |u | u ) = r f (u) in (0, R), (1) u (0) = u (R) = 0, 夡 These
results appear in the author’s UNICAMP doctoral dissertation. E-mail address: [email protected] (I. Marques).
0362-546X/$ - see front matter 䉷 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2004.11.006
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
where , , are given real numbers such that and 0, > 0 is a small parameter and 0 < R < ∞. We make the following assumptions on the nonlinearity f: (f1 ) f : [a, b] → R is a function of the class C 1 , with f (a) = f (b) = 0; (f2 ) f has exactly 2l + 1 zeros, a = a1 < 0 = a2 < a3 < · · · < a2l+1 = b with l = 1, 2, 3, . . . such that f (ai ) = 0 ∀i and f (ai ) < 0, if i is odd; 2l ) (f3 ) limt→a2l |t−af (t−a > 0. | (t−a ) 2l
2l
It has been emphasized in [5] that most of the phenomena occurring in the study of nonlinear elliptic equations can be more easily explained and understood when these equations are written in the radial form. Solutions of (1) will be absolutely continuous functions u : (0, R) → R and the equation will be satisfied a weak sense, defined below. Let us denote Lu := −(r |u | u ) . The motivation for our study is the fact that this operator includes the following class of operators, when considering acting in radial symmetric functions defined in a ball of RN : (i) Laplacian: = = N − 1, = 0, (ii) p-Laplacian (1 < p < N ): = = N − 1, = p − 2, (iii) k-Hessian (1 k < N/2): = N − k, = N − 1, = k − 1. R of absolutely continuous functions u : (0, R) → R We consider the Banach space X such that R R 2 +2 +2 r |u (r)| dr + r |u(r)|+2 dr < ∞. :=
u 0
0
R , a u(x) b for all x, is said to be a weak solution of (1) if A function u ∈ X R R R . 2 r |u | u dr − r f (u) dr = 0 ∀ ∈ X 0
0
q
Let us denote by L (0, R), q 1 and > − 1 the Banach space of Lebesgue measurable functions u : (0, R) → R such that R 1/q q q |u|L := r |u(r)| dr < ∞.
0
R and each weight , we define the critical exponent Associated with each space X q ∗ :=
( + 1)( + 2) , −−1
under the assumption that − − 1 > 0. R by Define a functional on X R R 2 +2 r |u | dr − r F (u) dr, I (u) = +2 0 0 u where F (u) = 0 f (t) dt.
(2)
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
23
The weak solutions of (1) are the critical points of the functional I . R is said to be a local minimum of function I if I (u0 ) I (u) for all A function u0 ∈ X R such that 0 < u − u0 0 for some 0 > 0. u∈X Using standard variational methods (local minimization and the Mountain Pass Theorem), we can prove that (1) has at least l nonconstant solutions and using monotone iteration methods we can prove that the nonconstant solutions approach either al or a2l+1 in (0, R) as → 0 almost everywhere. The hardest step of our proof is to show that the solutions given by the Mountain Pass Theorem are different from ai . For this matter, we use the existence of solutions for some Dirichlet problems associated with (1). A careful analysis of their asymptotic behavior is R . This fact is done in Section 2. If i is odd, we prove that ai is a local minimum in X not trivial either, since in this space we do not have any result saying that if ai is a local R . For the space W 1,p (), where is a minimum in C 1 , it is also a local minimum in X 0 smooth bounded domain in RN , an analogous result was obtained in [8] and for the space H01 (), an analogous result was obtained in [3]. Our main results are the following: Theorem 1.1. Suppose that f satisfies assumptions (f1 )–(f3 ). Then, there exists 0 > 0 such that, for all 0 < < 0 , (1) has at least l nonconstant solutions satisfying a1 < u1 (r) < a3 < u2 (r) < a5 < · · · < a2l−1 < ul (r) < a2l+1 , where l = 1, 2, 3, . . . . Remark 1.1. From a known regularity result (see [5, Proposition 2.2]), there exists ∈ (0, 1) such that ul ∈ C 1, (0, R). Remark 1.2. Theorem 1.1 is still true if we consider the ai s negative for all i 3, i.e., a = a2l+1 < · · · < a3 < a2 = 0 < a1 = b. The condition a2 = 0 is not essential, and it is only used to make the presentation easier. Theorem 1.2. Suppose that f satisfies assumptions (f1 )–(f3 ) and let u be a nonconstant solution of (1). Given any ∈ (a2l , a2l+1 ), let us consider the set
+ (, ) := {r ∈ (0, R) : a2l < u (r) < < a2l+1 }. Let B be the largest ball contained in + and let w ∗ (, ) be its radius. Then lim→0 w ∗ (, ) = 0. 2. Auxiliary results A fundamental step in the proof of Theorem 1.1 consists in showing the existence of solutions and their asymptotic behavior for the following Dirichlet problem: 2 − (r |u | u ) = r f (u) in (0, R), (3) u (0) = u(R) = 0,
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
where 0 < R < ∞, , , are given real numbers such that and 0. > 0 is a small parameter. The function f satisfies assumptions (f1 )–(f3 ) for l = 1, that is, f has only three zeros. We consider the Banach space XR of absolutely continuous functions u : (0, R) → R R +2 such that u(R) = 0 and u XR := 2 0 r |u (r)|+2 dr < ∞. We denote by 1 the first eigenvalue of the eigenvalue problem (see [5, p. 145])
Lu = r |u| u in (0, R), u (0) = u(R) = 0,
(4)
where and are given real numbers, 0 < R < ∞. 1 is characterized by
1 =
inf
u∈XR \{0}
R +2 dr 0 r |u (r)| . R +2 dr 0 r |u(r)|
The following theorem will be used in the proof of Theorem 1.1. Theorem 2.1. Let f be a function satisfying assumptions (f1 )–(f3 ) with l = 1. Then given any ball B(0, r1 ) ⊂ B(0, R), 0 < r1 < R, there exists 0 > 0 such that, for all 0 < 0 , the Dirichlet problem
−2 (r |u | u ) = r f (u) u (0) = u(r1 ) = 0,
in (0, r1 ),
(5)
has a positive solution 0 < u (r) < a3 in [0, r1 ] such that u → a3 as → 0 uniformly in every compact subset of (0, r1 ). Moreover, the Dirichlet problem
−2 (r |u | u ) = r f (u) u(r1 ) = u(R) = 0,
in (r1 , R),
(6)
has a negative solution a1 < v (r) < 0 in [r1 , R] such that v → a1 as → 0 uniformly in every compact subset of (r1 , R). The proof of Theorem 2.1 is done by combining an argument of truncation and the method of lower and upper-solution. A function u ∈ XR ∩ L∞ is said to be a lower solution of (3) if
2 0R r |u | u dr 0R r f (u) dr u (0) = 0, u(R) 0.
∀ ∈ XR , 0,
In the same way, a function u ∈ XR ∩ L∞ is said to be an upper-solution of (3) if the inequality signs are reversed.
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
25
Lemma 2.1. Consider g : R → R, a continuous and increasing function, such that g(0) = 0. Let u1 , u2 ∈ XR ∩ L∞ be functions such that, for all ∈ XR , 0, we have R 2R 2 (r)) dr 0 r |u2 | u2 dr + 0 g(u R R 2 0 r |u1 | u1 dr + 0 g(u1 (r)) dr, u2 (R) u1 (R). Then u2 u1 a.e. (0, R). The proof of this result is similar to the proof of the Lemma 2.2 in [4]. So we omit it. Lemma 2.2. Let g : R → R be a continuous and increasing function, such that g(0) = 0. q Then, for every function h ∈ L (0, R), where 1/q + 1/q = 1, the problem
−2 (r |u | u ) + g(u(r)) = h(r), u (0) = u(R) = 0
r ∈ (0, R),
(7) q
admits a unique weak solution u ∈ XR . Moreover, the associated operator T : L → XR , h → u is continuous and nondecreasing. Proof. We split the proof into two parts. Part 1 (Existence): Consider the functional ∈ C 1 (XR , R) defined by R R R 2 +2 (u) := r |u | dr + G(u) dr − h(r)u dr, +2 0 0 0 u where G(u) = 0 g(t) dt. So the critical points u ∈ XR of are the weak solutions of (7), i.e.,
R
2 0
(r |u | u ) dr +
R
R
g(u) dr −
0
h(r) dr = 0
0
∀ ∈ XR .
(8)
We prove that the functional has a global minimum. For that matter, we observe that is coercive and weakly lower semicontinuous. The coerciveness follows from the estimate R 1 +2 (u) h(r)u dr.
u XR − +2 0 For the weakly lower semicontinuity, we use Proposition 1.1 [5] which states that the R functional (u) := 0 G(u) dr is weakly lower semicontinuous. Part 2 (Uniqueness): Let u1 , u2 ∈ XR be weak solutions of (7). Then
R
2 0
[(r |u1 | u1 ) − (r |u2 | u2 )] dr +
R 0
[g(u1 ) − g(u2 )] dr = 0.
26
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Choosing = (u1 − u2 )+ and using the fact that g is increasing, we obtain R [(r |u1 | u1 ) − (r |u2 | u2 )]((u1 − u2 )+ ) dr 0 = 2 0 R + [g(u1 ) − g(u2 )](u1 − u2 )+ dr 0
= 2
u1 u2
+
u1 u2
+
u1 u2
r
|u1 | + |u2 | |((u1 − u2 )+ ) |2 dr 2
r
|u1 | − |u2 | 2 [|u1 | − |u2 |2 ] dr 2
[g(u1 ) − g(u2 )](u1 − u2 )+ dr 0.
Hence, (u1 − u2 )+ = 0 a.e. in (0, R) or, equivalently, u1 u2 a.e. in (0, R). By choosing = (u2 − u1 )+ and using a similar argument, we obtain that u2 u1 a.e. in (0, R). Hence, problem (7) has a unique weak solution u ∈ XR . The fact that T is nondecreasing follows from Lemma 2.1. 3. Proof of auxiliary results Proposition 3.1. Suppose that f satisfies assumptions (f1 )–(f3 ) with l = 1. Then, given any ball B(0, r1 ) ⊂ B(0, R), 0 < r1 < R, there exists 0 > 0 such that, for all 0 < 0 , the Dirichlet problem (5) has a positive solution 0 < u (r) < a3 in [0, r1 ]. Moreover, the Dirichlet problem (6) has a negative solution a1 < v (r) < 0 in [r1 , R]. Proof. We start by proving the existence of a positive solution for problem (5). Let f1 : R → R be defined by f1 (u) = f (u) if 0 u a3 and f1 (u) = 0 elsewhere. By (f1 ), we have that there exists M > 0 such that r f1 (t) + Mt is nondecreasing in t for t ∈ [0, a3 ]. Let us consider the following auxiliary problem 2 − (r |u | u ) + Mu = r f1 (u) + Mu in (0, r1 ), (9) u (0) = u(r1 ) = 0. Next, we prove the existence of a positive solution of (9) by using the method of the lowerand upper-solution. First, we observe that the function u(r) ≡ a3 , for r ∈ [0, r1 ] is an upper-solution of (9). Construction of the lower-solution of (9): let a = limt→0 f1 (t)/|t| t. It follows from (f3 ) that given ∈ (0, a), there exists t0 > 0 such that for all |t| t0 , we have a −
f1 (t) . |t| t
(10)
Let 1 be the first eigenvalue of problem (4) and 1 > 0 be an eigenfunction corresponding to this first eigenvalue. Take 1 > 0 such that |1 1 (r)| t0 and 1 (max(0,r1 ) 1 ) < a3 . From
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
27
(10) a − f1 (1 1 )/|1 1 | 1 1 . Choosing 0 > 0 such that 20 1 (r /r ) < a − , we have that 1 1 is a lower-solution of (9) for all 0 < 0 . Next we will show that there exists a minimal (and, respectively, a maximal) weak solution u∗ (resp. u∗ ) for (9) such that 1 1 = u u∗ u = a3 . Consider the set [u, u] := {u ∈ L∞ (0, r1 ) : u(r) u(r) u(r) a.e. in (0, r1 )} with the topology of a.e. convergence, and q
q
define the operator S : [u, u] → L by Sv=r f1 (v)+Mv ∈ L∞ (0, r1 ) ⊂ L (0, r1 ) ∀v ∈ [u, u],where q is such that 1/q + 1/q = 1. Clearly, S is nondecreasing and bounded. Applying the Lebesgue Dominated Convergence Theorem, we obtain that Sv n −Sv q → L
0, for vn , v ∈ [u, u] and vn → v a.e. in (0, r1 ). Hence S is continuous. q Recall that the operator T : L () → XR defined in Lemma 2.2 is continuous and nondecreasing. And now consider the continuous nondecreasing mapping F : [u, u] → XR defined by F := T oS, i.e., for a function v ∈ [u, u], u : F (v) is the unique solution in the weak form, namely r1 r1 r1 r |u | u + Mu = (r f1 (v) + Mv) for ∈ XR . 0
0
0
Writing u1 = F (u) and u1 = F (u), we obtain that u F (u) F (u) F (u) u, a.e. in (0, r1 ) ∀u ∈ [u, u]. Repeating the same reasoning, we can prove the existence of sequences (un ) and (un ) satisfying u0 = u, un+1 = F (un ), u0 = u, un+1 = F (un ) such that u = u0 u1 · · · un u un · · · u1 u0 = u,
a.e. in (0, r1 ).
Then, un → u∗ , un → u∗ and u∗ = F (u∗ ), u∗ = F (u∗ ). Then, u∗ is the minimal weak solution (respectively, u∗ maximal weak solution) of (9) such that u∗ , u∗ ∈ [u, u], for all 0 < 0 . In particular, every solution u ∈ [u, u] of (9) satisfies also u∗ u u∗ , a.e. in (0, r1 ). Since the solutions u∗ and u∗ are between 0 and a3 , u∗ and u∗ are solutions of (5). Therefore, there exists a solution u := u∗ ∈ [1 1 , a3 ] of (5), for all 0 < 0 . To prove the existence of the negative solution v (r), where a1 v 0, it follows similarly using the truncated function f2 : R → R defined by f2 (u)=f (u) if a1 u 0 and f2 (u)=0 elsewhere. And now consider the problem 2 − (r |u | u ) = r f2 (u) in (r1 , R), u(r1 ) = u(R) = 0. 3.1. Asymptotic behavior of the solutions of the Dirichlet problem We have the following proposition which shows the asymptotic behavior of the solutions of (5) as → 0. Proposition 3.2. Let 0 < u < a3 be a positive solution of (5) and let a1 < v < 0 be a negative solution of (6). Then (i) u → a3 as → 0 uniformly on every compact subset of (0, r1 ); (ii) v → a1 as → 0 uniformly on every compact subset of (r1 , R).
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Proof. (i) The proof follows by adapting some arguments from Theorem 4 in [7]. First, observe that there exists ∈ (0, 1) such that u ∈ C 1, (0, r1 ), by using Proposition 2.2 in [5]. Consider the function f1 : R → R defined in the proof of the Proposition 3.1 and let 1 > 0 be an eigenfunction corresponding to the first eigenvalue 1 of Lu := −(r |u | u ) in (0, r1 ) subject to boundary conditions u (0) = 0, u(r1 ) = 0 (see Eq. (4) above). Take 1 such that 1 XR = 1. Since f1 0, f1 ≡ / 0 and 1 > 0 from Lemma 3.2 in [5], we have u > 0 in (0, r1 ), u 0 in (0, r1 ). Consequently, there exists 1 > 0 such that for all 0 < 0 , we have u (r) 1 1 , and for a given > 0 there is C such that u (r) C > 0,
(11)
for all r ∈ := {r ∈ (0, r1 ) : dist(r, r1 ) > }. Since u is the solution of (9) it follows that r1 r1 2 r f1 (u ) dr ∀ 0, ∈ XR . (12) r |u | u dr = 0
0
In particular, for = 1 , we obtain r1 r1 2 r |u | u 1 dr = r f1 (u )1 dr. 0
(13)
0
Claim. The expression in the left-hand side of (13) goes to zero as → 0. Indeed, observe
Thus, applying Hölder inequality in (13) and taking = u that 0 < u a3 and f1 (u ) C. in (12), we obtain
r1
2 0
r |u | u 1 dr 2
r1 0
r1
×
C 2
r |u |+2 dr
0
1 2
r
2/(+2) C
0
(+1)/(+2)
|1 |+2 dr
r1
1/(+2)
r f1 (u )u dr
(+1)/+2
f or some constant C.
Define d := inf{1 (r) : r ∈ } > 0. Then, d
r f1 (u ) dr
r f1 (u )1 dr <
r1 0
r f1 (u )1 → 0,
→ 0.
(14)
Now, suppose by contradiction that there are a number C1 > 0 and a sequence j → 0 such that the Lebesgue measure of the sets
,j := {r ∈ : uj (r) < a3 − }
(15)
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
are bounded below by C1 . It follows from (14) that r f1 (uj ) dr → 0 as j → 0. Ij := ,j
29
(16)
Observe that in ,j , from (11) and (15), we have C uj a3 − . Since f1 is bounded below in the interval [C , a3 − ] by a number d > 0, from (15) it follows r f1 (uj ) dr d r dr = dC d | ,j | d C1 , Ij = ,j
,j
for any 0 < d dC/| ,j |, which contradicts (16). Therefore, | ,j | is not bounded below, i.e., u (r) → a3 , on every compact subset of (0, r1 ) as → 0. (ii) It follows similarly as in the previous case (i). Proof of Theorem 2.1. The proof follows directly from Propositions 3.1 and 3.2.
4. Proof of Theorem 1.1 The proof is done by using the Mountain Pass Theorem of Ambrosetti and Rabinowitz [1], in a version due to Hofer [9]. 4.1. A particular case First, we consider a particular case of Theorem 1.1, that is, the case when f has only three zeros. Theorem 4.1. Let f satisfy assumptions (f1 )–(f3 ) for l = 1, that is, f has only three zeros. Then, there exists 0 > 0 such that, for all 0 < < 0 , (1) has at least one nonconstant solution u verifying a1 < u (r) < a3 . In order to prove Theorem 4.1, we use the two lemmas below. Lemma 4.1. Let f be a function satisfying assumptions (f1 )–(f3 ) for l = 1. Then there exist functions of class C 1 , f1 : (−∞, a1 ] → R+ , f2 : [a3 , +∞) → R− and real numbers 1 and 1 such that (i) f1 (a1 ) = f (a1 ), f1 (a1 ) = f (a1 ) and f1 (t) > 0 for all t ∈ ( 1 , a1 ); (ii) f2 (a3 ) = f (a3 ), f2 (a3 ) = f (a3 ) and f2 (t) < 0 for all t ∈ (a3 , 1 ); (iii) 1 and 1 are such that 1 < a1 < a3 < 1 , 0 a1 1 a3 f1 (t) dt = f (t) dt , f2 (t) dt = f (t) dt a3
1 0 a1 and for all t ∈ [0, 1], we have 1 < t (1 − t)a3 + (a3 − a1 )t + a1 < 1 and 1 < t (1 − t)a1 + (a3 − a1 )t + a1 < 1 .
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Proof. We start by proving the existence of 1 . We take (t) := t (1 − t)a1 + (a3 − a1 )t + a1 . Thus, 0 = (d/dt)((t)) if, and only if, t = a3 /2a1 . Moreover, (a3 /2a1 ) = a32 /4a1 + a1 is a minimal value of (t). So, we define
1 := a32 /4a1 + 2a1 . To prove the existence of 1 , we take (t) := t (1 − t)a3 + (a3 − a1 )t + a1 . We note that 0 = (d/dt)((t)) if, and only if, t = 1 − a1 /2a3 . It follows that (1 − a1 /2a3 ) = a12 /4a3 + a3 is a maximal value of . Now, we take 1 := a12 /4a3 + 2a3 . Next, we prove the existence of the function f1 . We take g(t) = f (a1 )(t − a1 ) and 1 , 2 functions of class C 1 so that 1 ≡ 1 at a neighborhood of a1 , 2 ≡ 1 at a neighborhood of 1 , 1 (t) + 2 (t) = 1 for all t ∈ [ 1 , a1 ] and 0 a1 g(t)1 (t) dt < f (t) dt .
1
a1
Now, we choose r > 0 such that a1 [r 2 (t)g(t) + 1 (t)g(t)] dt =
1
0 a1
f (t) dt .
We define f1 : (−∞, a1 ] → R+ by r 2 (t)g(t) + 1 (t)g(t), 1 t a1 , f1 (t) := rf (a1 )t − ra 1 f (a1 ), t 1 . We note that for t 1 , the graph of f1 is the tangent line to f1 at the point ( 1 , f1 ( 1 )). Finally, we prove the existence of f2 . We take g(t) = f (a3 )(t − a3 ) and 1 , 2 functions of class C 1 so that 1 ≡ 1 at a neighborhood of a3 , 2 ≡ 1 at a neighborhood of 1 , 1 (t) + 2 (t) = 1 for all t ∈ [a3 , 1 ] and 1 a3 g(t)1 (t) dt < f (t) dt. a3 0 Now, we choose r > 0 such that 1 a3 [r 2 (t)g(t) + 1 (t)g(t)] dt = f (t) dt. a3 0 We define f2 : [a3 , +∞) → R− by r 2 (t)g(t) + 1 (t)g(t), a3 t 1 , f2 (t) := t 1 . rf (a1 )t − ra 1 f (a1 ), We observe that for t 1 , the graph of f2 is the tangent line to f2 at the point ( 1 , f2 ( 1 )). Thus the proof of the lemma is concluded. As a consequence of the Maximum Principle, we have the following lemma.
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Lemma 4.2. If u is a nonconstant solution of the Neumann problem 2 − (r |u | u ) = r f(u) in (0, R), u (0) = u (R) = 0,
31
(17)
where f : R → R is the function defined by f (t), a1 t a3 , f(t) := f1 (t), t a1 , f2 (t), t a3 , where f1 and f2 are defined in Lemma 4.1. Then a1 u (r) a3 . Proof. We start by proving that u (r) a3 . Indeed, let u (r) − a3 if u (r) a3 , v(r) := 0 if u (r) < a3 and + := {r ∈ (0, R) : u (r) a3 }. Note that R R 2 r |u | u v dr = r f(u )v dr = 0
0
+
r f2 (u )v dr 0.
The left-hand side of the above relation is 0. So this implies + r f2 (u )v = 0 ⇒ v ≡ 0. Therefore, u (r) a3 , for all r ∈ (0, R). Similarly, we have a1 u , if we take − := {r ∈ (0, R) : u (r) a1 } and u (r) − a1 , u (r) a1 , w(r) := 0, u (r) > a1 . This finishes the proof of the lemma.
Thus, from Lemma 4.2, it follows that a solution of (17) is a solution of (1), since f ≡ f in [a1 , a3 ]. Therefore, to prove Theorem 4.1 it suffices to prove the following theorem. Theorem 4.2. Let f be a function satisfying conditions (f1 )–(f3 ). Then there exists 0 > 0 such that, for all 0 < < 0 , (17) has at least one nonconstant solution u . R by Define a functional on X R R 2 +2 (u) dr, r |u | dr − r F J (u) := +2 0 0 (u) := u f(t) dt. The functional J is C 1 with derivative given by where F 0
J (u) =
R
2 0
R
r |u | dr − 0
r f(u) dr
R . ∀ ∈ X
The weak solutions of (17) are the critical points of the functional J .
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Lemma 4.3. The functional J satisfies the Palais–Smale condition. R be a sequence satisfying J (un ) → c and J (un ) → 0, as n → ∞. Proof. Let (un ) ⊂ X Hence, 2 R R +2 |J (un )| = r |u | dr − r F (un ) dr d +2 0 0 for some d > 0, (18) R R 2 r |un | un v dr − r f (un )v dr n v (19) |J (un )v| = 0
0
R where n → 0, n → ∞. for all v ∈ X Claim. (un ) is bounded. Indeed, define 1 := {r ∈ (0, R) : un (r) 1 } and un (r) − 1 if un (r) 1 , vn (r) = 0 if un (r) < 1 . Since f(un ) − C in 1 , it follows from (19) with v = vn that
R
C 0
r vn dr n vn .
(20)
Define wn := vn / vn . Taking subsequences if necessary we can assume that wn / w R , wn → w in Lp and wn (r) → w(r) a.e. in (0, R). Dividing (20) by vn weakly in X and taking limits at both sides and observing that wn 0, we have that w ≡ 0 a.e. in (0, R). From (19) with v = vn , taking the limit as n → ∞, we obtain
R
2
r
0
|vn |+2 → 0.
vn
(21)
On the other hand, wn = 1, then
R
2 0
r |wn |+2 dr → 1.
(22)
Multiplying and dividing (21) by vn +1 , from (22) we conclude that vn → 0 as n → ∞. Consequently, vn C. Next, we define 2 := {r ∈ (0, R) : 1 un (r) 1 } and zn (r) :=
un (r) if 1 un (r) 1 , 0 if 1 un (r), 0 if un (r) 1 .
By a similar argument, we prove that zn C.
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
33
Finally, we consider 3 := {r ∈ (0, R) : un (r) 1 < 0} and the sequence tn (r) :=
un (r) − 1 0
if un (r) 1 , if un (r) > 1 .
Similarly, as in the previous case, we prove that tn C. Let un := vn + zn + tn . Then, (un ) is bounded in . Thus, J satisfies the (PS) condition. Lemma 4.4. The functional J is bounded below. Proof. Let 1 : =3 ∪ 4 and 2 := 5 ∪ 6 , with
3 := {r ∈ (0, R) : 0 u(r) a3 }, 5 := {r ∈ (0, R) : a1 u(r) 0},
4 := {r ∈ (0, R) : a3 < u(r) < ∞}, 6 := {r ∈ (0, R) : −∞ < u(r) < a1 }.
Since u is bounded in 3 and in 5 , we have (a3 ) dr C,
r F r F (u) dr < C + 1
2
4
(u) dr < C + r F
6
(a1 ) dr < C. r F
Also,
R
r F (u) dr =
1
0
r F (u) dr +
2
(u) dr < C. r F
Hence, J (u)
2 +2
R
r |u |+2 dr − C − C.
0
Therefore, J is bounded below.
R . Lemma 4.5. ai , with i odd, is a strict local minimum of J in X Proof. First of all, we will prove that ai is a local minimum of J in the C 1 topology. Let (t), to |t − ai | < and let u ∈ C 1 such that u(r) − ai C 1 = (ai ) F > 0 such that F max {|u(r) − ai |, |u (r) − ai |} < . We claim that there exists > 0 such that := {r ∈ (0, R) : |u(r) − ai | > } has positive measure. Indeed, let u ∈ C 1 (0, R), u ≡ / ai . Then, there exists r0 ∈ (0, R) such that u(r0 ) = ai . Hence, there exists > 0 such that either u(r0 ) > ai + or u(r0 ) < ai − . Since u is continuous, there exists a ball B (r0 ) such that |u(r) − ai | > , for all r ∈ B (r0 ). Therefore, has positive measure.
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
(ai − ), F (ai + )}. Since ai is a strict local maximum of Now, we define c1 := max{F , we have F R (u) dr (ai ) dr r F r c1 dr + r F
0
(ai )
(0,R)\
r dr +
(0,R)\
r F (ai ) dr =
R 0
(ai ) dr. r F
Therefore, J (u) > J (ai ), i.e., ai is a strict local minimum of J in C 1 . Take ai =a3 . Assume that a3 is a local minimum of J in C 1 and a3 is not a local minimum R . Then, for all > 0 small enough, there exists v ∈ B ⊂ X R such that in X J (v ) = min J (v) < J (a3 ),
(23)
v∈B
R : v − a3 }, since B where B is a ball of radius centered at a3 , B := {v ∈ X is weakly close and J is bounded below in B . Therefore, v satisfies the Euler equation R , where G(v ) := v − a3 , i.e., J (v ) = G (v ) ∀ ∈ X
2
R
R
r f(v ) r |v | v − 0 0
R R 2 r |v | v + r |v − a3 | (v − a3 ) =
0
0
R and for some Lagrange multiplier . Since v is a minimum of J in B for all ∈ X then 0. Rewriting the equation above, we have R R 2 r |v | v = [r f(v ) + r |v − a3 | (v − a3 )] (24) (1 − ) 0
0
R . Thus, v is a weak solution of the problem for all ∈ X 2 −(1 − ) (r |v | v ) = r |v − a3 | (v − a3 ) +r f(v ) in (0, R), v (0) = v (R) = 0.
(25)
We will show that a1 v (r) a3 , for all r ∈ (0, R). R . Claim 1. v is nonconstant. Indeed, if v ≡ constant then v ≡ a3 , since v → a3 in X This contradicts (23). Claim 2. v a3 . Indeed, let v (r) − a3 if v(r) := 0 if
v (r) a3 , v (r) < a3
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
35
and + := {r ∈ (0, R) : v (r) a3 }. From (24), we have (1 − )
R
2 0
r |v | v v dr =
+
[r f2 (v )v + r |v − a3 |+2 ] dr 0,
a.e. since, f2 (t) 0, v 0, 0. The function f2 is a function defined in Lemma 4.1. Since, (1 − ) 0, it follows v ≡ constant. Also, if v (r) a3 , for all r ∈ (0, R) it follows that v ≡ constant. It is a contradiction. Therefore, there exists r ∈ (0, R) such that v (r) < a3 , i.e., v(r) = 0. Then v(r) ≡ 0, for all r ∈ (0, R), i.e., v a3 for all r ∈ (0, R). Claim 3. a1 v . Indeed, it follows similarly as in Claim 2 if we take v (r) − a1 , v (r) a1 , w(r) := 0, v (r) > a1 and − := {r ∈ (0, R) : v (r) a1 }. Therefore, from Claims 2 and 3, we have that a1 v (r) a3 , i.e., sup(0,R) |v | C ∗ . Then from Theorem 2 in [10], we have that there exists a positive constant ∈ (0, 1) such that v ∈ C 1, and, moreover, there exists a positive constant C = C((0, R), C ∗ , ) such that v C 1, C. Thus, from Ascoli–Arzelá Theorem we get that there exists a subsequence of (v j ) = (v ) such that v → a3 in C 1 . It contradicts the fact that a3 is a local minimum of J in C 1 . Indeed, if a3 is a local minimum of J in C 1 , it follows J (a3 ) J (v) for all v ∈ C 1 such that 0 < v − a3 C 1 < 0 for some 0 > 0. Since v → a3 in C 1 , it follows J (a3 ) J (v ). On the other hand, we have from (23) that J (v ) < J (a3 ), for all > 0 small enough. R . Hence, the contradiction. Therefore, a3 is a local minimum of J in X R . Similarly, it follows that a1 is a local minimum of J in X R . Without loss of generality, we suppose that ai is a strict local minimum of J in X R such that J (v ) = J (ai ). Hence, v is a Contrarily, for all > 0, there exists v ∈ X R . critical point of J in X We will use arguments from [6] to obtain the next lemma. Lemma 4.6. If a is a strict local minimum of J , i.e., J (a) < J (u)
(26)
R such that 0 < u − a < 0 for some 0 > 0. Then, for any 0 < < 0 , for all u ∈ X R and u − a = } > J (a). inf{J (u) : u ∈ X
(27)
Proof. Assume by contradiction that the infimum in (27) is equal to J (a) for some with R with un −a = and, say, J (un ) J (a)+ 0 < < 0 . So there exists a sequence un ∈ X 2 1/2n . Call R : − u − a + }, A := {u ∈ X
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
where > 0 is chosen so that 0 < − and + < 0 . In view of our contradiction hypothesis in (26), it follows that inf{J (u) : u ∈ A} = J (a). We now apply the Ekeland Variational Principle to the functional J on A to get the existence of vn ∈ A such that J (vn ) J (un ),
(28)
1
vn − un , n
(29)
J (vn ) J (u) +
1
u − vn n
∀u ∈ A.
(30)
R , i.e., J (vn ) C (it follows Our purpose is to show that vn is a (PS) sequence for J in X by (28)) and J (vn ) → 0, as n → ∞. Once this is proved, we get that vn has a convergent R . Note that subsequence. Denote this subsequence by vn and we have that vn → v in X v ∈ A, since A is complete. Hence, v ∈ XR and therefore it satisfies v − a = and J (v) = J (a), which contradicts (26). So in order to conclude the proof of Lemma 4.6 we will prove that J (vn ) → 0, as R and ut := vn + tw, for n > 1/. We observe that for |t| sufficiently n → ∞. Take w ∈ X small, ut = vn + tw ∈ A. Indeed, lim ut − a = vn − a vn − un + un − a
t→0
1 + < + . n
On the other hand,
vn − a a − un − un − vn −
1 > − . n
Also, we can take u = ut in (30), and then, for t > 0, J (vn ) − J (vn + tw) 1 1 1
vn − tw − vn tw . t nt nt
(31)
Taking the limit in (31) as t → 0, we obtain J (vn ), w (1/n) w . Consequently, R . So, J (vn ) → 0, n → ∞ and the proof of the |J (vn ), w | (1/n) w ∀w ∈ X lemma is concluded. Proof of Theorem 4.2. We define R ) : h(0) = a1 , h(1) = a3 } := {h ∈ C([0, 1], X and
:= inf max J (h(t)). h∈ t∈[0,1]
From Lemma 4.6 := inf h∈ maxt∈[0,1] J (h(t)) > c = max{J (a1 ), J (a3 )}. Since J satisfies the condition (PS) using the Mountain Pass Theorem from [9], it follows that there R , exists u critical point of J such that J (u) = . If the critical points are not isolated in X then there exists an infinite number of critical points of J . Otherwise, u is of the mountain pass type (see [9]). Since a1 and a3 are the strict local minimum, u = a1 and u = a3 .
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
37
Therefore, in order to show the existence of a nonconstant critical point of J , we only need to prove that < 0, since 0 is the other constant solution and J (0) = 0. We claim that < 0. In fact, we consider B ⊂ (0, R) open ball and we define v (r), r ∈ B, u0 (r) := w (r), r ∈ [0, R]\B, where v is a positive solution of (5) in B and w is a negative solution of (6) in [0, R]\B. R . Now, given > 0, we Since (0, R) ∈ C 1 and v , w ∈ Lp (0, R), it follows that u0 ∈ X consider the special path h (t) := t (1 − t)u0 (r) + (a3 − a1 )t + a1
in .
Then we claim that there exists a small number 0 > 0, such that, for all 0 < < 0 and for all t ∈ [0, 1], max J (h (t)) < 0. In fact, suppose by contradiction that there is no such 0 > 0. Then, for any 0 > 0, there is < 0 such that J (h (t )) 0
(32)
for some t ∈ [0, 1]. Then choose a sequence (k ) such that limk→∞ k = 0, J (hk (tk )) 0, 0 limk→∞ tk = 1. For simplicity, we drop the indices and write k = and tk = t. So 2 2 r |h (t)|+2 dr r |h (t)|+2 dr + J (h (t)) = + 2 (0,R)\B +2 B R (h (t)) dr − r F 0 2 +2 = (1 − t)+2 r |v |+2 dr + r |w |+2 dr t +2 B (0,R)\B R (t (1 − t)u0 + (a3 − a1 )t + a1 ) dr. − r F 0
Since
2 and
B
r |v |+2 dr =
2
(0,R)\B
B
r f(v )v dr
r |w |+2 dr =
(0,R)\B
r f(w )w dr,
we get
t +2 (1 − t)+2 J (h (t)) = r f (v )v dr + r f (w )w dr +2 B (0,R)\B R (t (1 − t)u0 + (a3 − a1 )t + a1 ) dr. − r F 0
(33)
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
Since v → a3 as → 0 uniformly on every compact subset of B and w → a1 as → 0 uniformly on every compact subset of \B, so lim r f(w )w dr = 0. r f(v )v dr = 0 and lim →0
→0
B
(0,R)\B
If we take the limit on both sides of (33), it follows from the Dominated Convergence Theorem and the above facts that
R (t (1 − t)u0 (r, ) + (a3 − a1 )t + a1 ) dr lim J (h (t)) = lim − r F →0
→0
0
= − r F ((1 − )a3 + (a3 − a1 ) + a1 )|B| ((1 − )a1 + (a3 − a1 ) + a1 )|(0, R)\B|, − r F where |A| is the Lebesgue measure of A ⊂ RN . It follows from Lemma 4.1 that ((1 − )a3 + (a3 − a1 ) + a1 ) 0, −F
(34)
((1 − )a1 + (a3 − a1 ) + a1 ) 0. −F
(35)
is zero only at 1 , 0 and In fact we have strict inequalities in either (34) or (35). Indeed, F
1 , so we conclude from Lemma 4.1(iii) that if (34) and (35) are zero, then
(1 − )a3 + (a3 − a1 ) + a1 = 0 = (1 − )a1 + (a3 − a1 ) + a1 , i.e., (1 − )a3 = (1 − )a1 . Since a3 = a1 , then (1 − ) = 0, i.e., either = 0 or 1. Hence, (a1 ) = 0 or F (a3 ) = 0. Then, either a1 = 0 or a3 = 0, which is impossible. Hence, either F lim→0 J (h (t)) < 0, which is in contradiction with (32). So we get finally that < 0. Corollary 4.1. Let f be satisfying (f1 )–(f3 ) for l = 1 and let u be a nonconstant solution of (1) such that J (u ) = . Then lim→0 sup J (u ) < 0. Proof. From the proof of Theorem 4.2, there exists a positive number 1 so that J (h1 (t)) < 0 for all t ∈ [0, 1], where h1 (t) = t (1 − t)u0 (r, 1 ) + (a3 − a1 )t + a1 , u0 as defined in the proof of Theorem 4.2. Hence, for all 0 < < 1 , J (h1 (t)) J (h1 (t)) < 0 for all t ∈ [0, 1]. It follows from the intermediate value theorem that there exists t ∈ [0, 1] such that h1 (t ) = u . Consequently, J (h1 (t )) = J (u ) for some t ∈ [0, 1], and so the assertion is true. Proof of Theorem 4.1. The proof follows directly from the Theorem 4.2 and Lemma 4.2. Proof of Theorem 1.1. For each l = 1, 2, . . . , we consider the function f : [a2l−1 − a2l , a2l+1 − a2l ] → R of class C 1 defined by f (t) := f (t + a2l ). Then from Theorem 4.1,
I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
the problem 2 +2 + v ) = r f (v), − (r |v | v (0) = v (R) = 0,
r ∈ (0, R),
39
(36)
has a nonconstant solution vl (r) such that a2l−1 −a2l < vl (r) < a2l+1 −a2l , where v=u−a2l . So, ul = vl + a2l is a nonconstant solution of (1) with a2l−1 < ul < a2l+1 . Hence, there exist at least l nonconstant solutions of (1) satisfying a1 < u1 (r) < a3 < u2 (r) < a5 < · · · < a2l−1 < ul (r) < a2l+1 . 5. Asymptotic behavior of the solutions of the Neumann problem Proof of Theorem 1.2. Let a2l < < a2l+1 . Suppose by contradiction that lim→0 w ∗ (, ) = 0. Then, there is a convergent sequence {w∗ (k , )} so that limk→∞ w ∗ (k , ) = > 0. This means that, for each k > 0, there is rk =r(k , ) ∈ + (k , ) so that the ball B(rk , ), centered at the point rk with the radius , is contained in + (k , ). Consider f : [a2l−1 − a2l , a2l+1 − a2l ] → R of class C 1 defined by f (t) := f (t + a2l ) and 2 +2 −k (r |v | v ) = r f (v), r ∈ B(rk , ), (37) v (0) = v(r) = 0, r ∈ jB(rk , ), where v = u − a2l . Also, v = u − a2l with 0 v a2l+1 − a2l . Define f 1 : R → R by f 1 (v) = f (v) if 0 v a2l+1 − a2l and f 1 (v) = 0 elsewhere. By (f1 ), we have that there exists M > 0 such that r f 1 (t) + Mt is nondecreasing in t for t ∈ [0, a2l+1 − a2l ]. Consider the auxiliary problem 2 +2 −k (r |v | v ) + Mv = r f 1 (v) + Mv, r ∈ B(rk , ), (38) v (0) = v(r) = 0, r ∈ jB(rk , ). Note that v is in an upper solution of (38). Claim. There are k0 and 1 > 0 such that 1 1 is a lower solution of (38) for all 0 < k k0 , where 1 is an eigenfunction corresponding to the first eigenvalue 1 of problem (4). In fact, this follows similarly as in the proof of Proposition 3.1 step 2. Also, as in the proof of Proposition 3.1 step 3 with u = 1 and u = v , it follows that (38) has a minimal solution v∗ with 1 v∗ v a2l+1 − ra2l , and as in the proof of Proposition 3.2 with u = v∗ , f1 = f 1 , a3 = a2l+1 − a2l and 0 1 = B(rk , ) , it follows that v∗ → a2l+1 − a2l on every compact subset of B(rk , ) as k → 0. Therefore, there exists u∗ u with u∗ → a2l+1 so that v∗ = u∗ − a2l v = u − a2l . This leads to a contradiction for < a2l+1 . Remark 5.1. If we choose an open ball B = B(x, w ∗ (, )) centered at some point r = r(, ) whose radius w ∗ (, ) is the maximum of the radii of balls in
− (, ) = {r ∈ (0, R) : a2l−1 < − < u(r, ) < a2l } we can prove, by a similar method, that lim→0 w ∗ (, ) = 0.
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I. Marques / Nonlinear Analysis 61 (2005) 21 – 40
5.1. Examples Example 5.1. Let f : R → R defined by f (u) :=
|u|p−2 u(p − p|u|p ) . (1 + |u|p )
Hence, f satisfies conditions (f1 )–(f3 ), for l = 1, where a1 = −1, a2 = 0 and a3 = 1. So, by Theorem 1.1, there is an 0 > 0 so that the Neumann problem (1) has a nonconstant solution a1 u a3 for all 0 < < 0 and p 2. 2 Example 5.2. Let f : R → R defined by f (u) = u(a R+ . Hence, f √ − u ), where a ∈ √ satisfies conditions (f1 )–(f3 ), for l = 1, where a1 = a, a2 = 0 and a3 = a. Then, by Theorem 1.1, 0 > 0 so that the Neumann problem (1) has at least l nonconstant solutions satisfying a1 u1 (x) a3 u2 (x) a5 < · · · < ul (x) < a2l+1 for all 0 < < 0 .
Example 5.3. Let f : [−, (2l − 1)] → R defined by f (u) = sen(u). Hence, f satisfies conditions (f1 )–(f3 ), for p=2, l=1, 2, 3, . . . , where a1 =− < a2 =0 < a3 = < · · · < a2l = (2l −2) < a2l+1 =(2l −1). Hence, from Theorem 1.1, there is 0 > 0 so that the Neumann problem (1) has at least l nonconstant solutions satisfying a1 u1 (x) a3 u2 (x) a5 < · · · < ul (x) < a2l+1 for all 0 < < 0 . References [1] A. Ambrosetti, P.H. Rabinowitz, Dual variational methods in critical point theory and applications, J. Funct. Anal. 14 (1973) 349–381. [3] H. Brézis, L. Nirenberg, H 1 versus local C 1 minimizers, C. R. Acad. Sci. Paris 317 (1993) 465–472. [4] A. Cañada, P. Drábek, J. Gamez, Existence of positive solutions for some problems with nonlinear diffusion, Trans. Amer. Math. Soc. 349 (1997) 4231–4249. [5] P. Clément, D.G. De Figueiredo, E. Mitidieri, Quasilinear elliptic equations with critical exponents, Top. Methods Nonlinear Anal. 7 (1996) 133–170. [6] M. Cuesta, D.G. De Figueiredo, J.P. Gossez, The beginning of the Fuˇcik spectrum for the p-Laplacian, J. Differential Equations 159 (1999) 212–238. [7] D.G. De Figueiredo, On the existence of multiple ordered solutions of nonlinear eigenvalue problems, Nonlinear Anal. TMA 11 (1987) 481–492. [8] J.P. García Azorero, I. Peral Alonso, J.J. Manfredi, Sobolev versus Hölder local minimizers and global multiplicity for some quasilinear elliptic equations, Comm. Contemp. Math. 3 (2000) 385–404. [9] H. Hofer, A geometric description of the neighborhood of a critical point given by the mountain-pass theorem, J. London Math. Soc. 31 (1985) 566–570. [10] G.M. Lieberman, Boundary regularity for solutions of degenerate elliptic equations, Nonlinear Anal. TMA 12 (1988) 1203–1219.