Radial solutions for Neumann problems involving

Radial solutions for Neumann problems involving extrinsing curvature and periodic nonlinearities∗ Jean Mawhin Institut de Recherche en Mathématique et Physique Université Catholique de Louvain 2, Chemin du Cyclotron B-1348 Louvain-la-Neuve, Belgique

Abstract In this paper we survey some recent results on the existence and multiplicity of radial solutions for Neumann problems in a ball and in an annular domain, associated to pendulum-like perturbations of mean curvature operators in Euclidean and Minkowski spaces and of the p-Laplacian operator. Our approach relies on the Leray-Schauder degree, upper and lower solutions method, and critical point theory. AMS MSC 2010 : 35J65, 34B15 Key words : mean curvature and p-Laplacian operators, Neumann problem, pendulumlike nonlinearities, Leray-Schauder degree, upper and lower solutions

1

Introduction

In the flat Minkowski space LN +1 = {(x, t) : x ∈ RN , t ∈ R} with metric PN 2 2 j=1 (dxj ) − (dt) , the mean extrinsic curvature of hypersurfaces, i.e. of spacelike submanifolds of codimension one, is the trace of the second fundamental form. Maximal hypersurfaces have mean extrinsic curvature zero, and hypersurfaces with constant mean extrinsic curvature are also of interest for geometry and physics. More specifically, let Ω be a bounded domain in {(x, t) ∈ LN +1 : t = 0} ≃ N R , and let us restrict our attention to hypersurfaces which are the graph of functions v : Ω → R. An extension of the problem of maximal hypersurfaces consists, given f : Ω × R → R continuous and bounded, to maximize the functional I defined by # Z "p Z v(x) I(v) := 1 − k∇v(x)k2 − f (x, t) dt dx (1) Ω

∗ Milan

0

Journal of Mathematics 79 (2011), 95-112

1

over C(ϕ, Ω) := {v ∈ C 0,1 (Ω) : v = ϕ on ∂Ω, k∇vk ≤ 1 a.e. in Ω}.

(2)

It has been proved by Bartnik and Simon [2] that the maximum exists if and only if C(ϕ, Ω) 6= ∅, and is unique if, furthermore, f (x, ·) is nondecreasing for any x ∈ Ω. The existence of a maximum for I follows pfrom the fact that I is bounded on C(ϕ, Ω), that C(ϕ, Ω) is equicontinuous, 1 − kpk2 concave, and, using Serrin’s semicontinuity theorem [29], I is upper semi-continuous. If one introduces the differential operator ! ∇v M(v) := ∇ · p (3) 1 − k∇vk2 the problem is then to show that the Euler-Lagrange equation for a maximum of I over C(ϕ, Ω) is given by M(v) = f (x, v)

in Ω,

u = ϕ on

∂Ω

(4)

A long argument in [2], which essentially consists in showing that a maximizing function u for I is such that k∇uk ≤ η < 1, gives conditions which imply the existence of a solution u ∈ C 1 (Ω) ∩ W 2,2 (Ω) such that k∇uk < 1 for f and ϕ bounded, when ϕ has an extension ψ to Ω such that k∇ψk < 1. More generally, given f : Ω × R × RN → R continuous, one can raise the question of the solvability of the equation M(v) = f (x, v, ∇v)

in

Ω

(5)

submitted to Dirichlet or Neumann boundary conditions. When N = 1, more general results that the one mentioned above have been obtained in [11, 14]. In particular, it has been shown that the problem ′ v′ √ = f (x, v, v ′ ), v(0) = A, v(T ) = B (6) 1 − v ′2

is solvable for every f continuous on [0, T ] × R2 if and only if |B − A| < T . The results is based upon Leray-Schauder degree arguments, but some of the techniques appear to be purely one-dimensional. A natural question is then to see if corresponding results hold for the radial solutions of (4) when Ω is a ball or an annulus in RN . This problem has been considered, for homogeneous boundary conditions, in [4] and [8] and for non-homogeneous boundary conditions in [6]. In particular, if 0 ≤ ρ < R, B(ρ, R) = {x ∈ RN : ρ < kxk < R} for ρ > 0, B(0, R) = B(R), it has been shown that the problem M(v) = f (kxk, v, ∂ν v)

in B(ρ, R),

(7)

with Dirichlet boundary conditions v(x) = A

on

∂B(ρ),

v(x) = B 2

on

∂B(R)

(8)

when ρ > 0 and v(x) = B

on

∂B(R)

(9)

when ρ = 0 (case of the ball), is solvable for any continuous f on ]ρ, R[ ×R2 if and only if |B − A| < 1 when ρ > 0 and B arbitrary when ρ = 0. The Neumann problem ′ v′ √ = 1, v ′ (0) = 0 = v ′ (R) 1 − v ′2 has no solution, as immediately checked, and hence such general results cannot be expected for Neumann boundary conditions. In this paper we survey recent existence and multiplicity results for the Neumann problem T (v) + µ sin v = l(|x|) in

A,

∂v = 0 on ∂A, ∂ν

(10)

where T is in one of the following situations: ∇v √ (mean extrinsic curvature in Minkowski space), T (v) = div 1−|∇v|2 ∇v √ T (v) = div (mean curvature in Euclidean space), 2 1+|∇v|

T (v) = div(|∇v|p−2 ∇v) (p-Laplacian).

Here, | · | denotes the Euclidean norm in RN , µ > 0 is a constant, A = {x ∈ RN : R1 < |x| < R2 } (0 ≤ R1 , < R2 ), l : [R1 , R2 ] → R is a given continuous ∂v stands for the outward normal derivative of v. function and ∂ν Setting r = |x| and v(x) = u(r), problem (10) reduces to (rN −1 φ(u′ ))′ + rN −1 µ sin u = rN −1 l(r),

u′ (R1 ) = 0 = u′ (R2 ),

(11)

v v in the Minkowski case, φ(v) = √1+v in the Euclidean case where φ(v) = √1−v 2 2 p−2 and φ(v) = |v| v (p > 1) in the p-Laplacian case. Actually, in what follows φ will be a general increasing homeomorphism, with φ(0) = 0, in one of the following situations:

φ : (−a, a) → R (singular ), φ : R → (−a, a) (bounded ), φ : R → R (classical ). Using degree arguments, we prove in Corollary 1 that the problem ! ∇v ∂v div p = 0 on ∂A, + µ sin v = l(|x|) in A, 2 ∂ν 1 − |∇v| 3

(12)

has at least two classical radial solutions not differing by a a multiple of 2π if 2(R2 − R1 ) < π and

Moreover, if

Z R2 N N −1 r l(r)dr N < µ cos (R2 − R1 ) . R2 − R1N R1 2(R2 − R1 ) = π,

problem (12) has at least one classical radial solution provided that Z R2 rN −1 l(r)dr = 0.

(13)

R1

For the p-Laplacian, we prove in Corollary 2 that problem div(|∇v|p−2 ∇v) + µ sin v = l(|x|) in

A,

∂v ∂ν

=0

on ∂A,

(14)

has at least two classical radial solutions not differing by a multiple of 2π if (13) holds and R2 is sufficiently small (or N sufficiently large). Moreover the same type of result holds true for the Neumann problem ! ∂v ∇v = 0 on ∂A. (15) + µ sin v = l(|x|) in A, div p ∂ν 1 + |∇v|2 In the case where

R1 > 0 (i.e., A is an annular domain) we show in Corollary 3, combining degree arguments with upper and lower solutions techniques, that (12) and (14) have at least two classical radial solutions, not differing by a multiple of 2π, if ||l||∞ < µ, and have at least one classical radial solution if ||l||∞ = µ. Moreover, if condition 2µR2 0 and a natural question is to know if the restriction 2(R2 − R1 ) ≤ π 4

(16)

can be dropped. In the analogous problem of the forced pendulum equation u′′ + µ sin u = h(t) with periodic or Neumann homogeneous boundary conditions on [0, T ], it has been shown that the corresponding necessary condition Z

T

h(t) dt = 0

(17)

0

is also sufficient for the existence of at least two solutions not differing by a multiple of 2π. But, in this case, all the known proofs are of variational or symplectic nature (see e.g. the survey [26]). Recently, it has been proved in [17] that the “relativistic forced pendulum equation” ′ u′ √ + µ sin u = h(t) 1 − u′2

has at least one T-periodic solution for any µ > 0 when the (necessary) condition (17) is satisfied. The approach is essentially variational, but combined with some topological arguments. The paper [9] has adapted the methodology of [17] to the radial Neumann problem for (11) and proved that, for the existence part, condition (16) can be dropped. We obtain here the same result using a purely variational approach introduced in [10]. The results are stated and proved for the more general class of equations of the form [rN −1 φ(u′ )]′ = rN −1 [g(r, u) + h(r)],

u′ (R1 ) = 0 = u′ (R2 )

(18)

where φ : (−a, a) → R is a suitable homeomorphism and g belongs to some class of functions 2π-periodic with respect to its second variable. The remaining of the paper is organized as follows. In Section 2 we introduce the function spaces and the operators which are needed in the sequel. Section 3 presents a fixed point operator and some degree computations in the singular case. Existence and multiplicity results for problem (11) are given in Sections 4 and 5, under conditions upon the radius and the mean value of the forcing term or upon the norm of the forcing term. The variational approach is developed in Section 6 and 7.

2

Notations, function spaces and operators

Let 0 ≤ R1 < R2 . We denote by C the Banach space of continuous functions defined on [R1 , R2 ] endowed with the usual norm || · ||∞ , by C 1 the Banach space of continuously differentiable functions defined on [R1 , R2 ] endowed with the norm ||u|| = ||u||∞ + ||u′ ||∞ , 5

and by C†1 the closed subspace of C 1 defined by C†1 := {u ∈ C 1 : u′ (R1 ) = 0 = u′ (R2 )}. The corresponding open ball with center in 0 and radius ρ is denoted by Bρ . For any continuous function w : [R1 , R2 ] → R, we write wL = min w,

wM = max w.

[R1 ,R2 ]

[R1 ,R2 ]

Let us introduce the continuous projectors Q : C → C,

u = Qu =

N R2N − R1N

P : C → C,

Z

R2

rN −1 u(r)dr,

R1

P u = u(R1 ),

the continuous function γ : (0, ∞) → R,

γ(r) =

1 , rN −1

and the linear operators Z r tN −1 u(t)dt (r ∈ (R1 , R2 ]), Lu(r) = γ(r) R Z r 1 1 u(t)dt (r ∈ [R1 , R2 ]). : C → C , Hu(r) =

L : C → C, H

Lu(0) = 0,

R1

It is not difficult to prove that L is compact and H is bounded. Finally, we b 1 the closed subspace of C 1 defined by denote by C † † and notice that

b1 := {u ∈ C 1 : u = 0}, C † † b1 , C†1 = R ⊕ C †

b1 . so that any u ∈ C†1 can be uniquely written as u = u + u b, with u ∈ R, u b∈C †

3

A fixed point operator and degree computations in the singular case

Throughout this section we assume that φ is singular. The case where N = 1 and R1 = 0 in the results of this section has been considered in [27]. Proposition 1 Assume that F : C†1 → C is continuous and takes bounded sets into bounded sets. The function u ∈ C†1 is a solution of the abstract Neumann problem (rN −1 φ(u′ ))′ = rN −1 F (u), 6

u′ (R1 ) = 0 = u′ (R2 )

(19)

if and only if it is a fixed point of the compact operator MF defined on C†1 by MF = P + QF + H ◦ φ−1 ◦ L ◦ (I − Q) ◦ F. Furthermore, one has k(MF (u))′ k∞ < a for all u ∈ C†1 . Proof. Let u ∈ C†1 and v = MF (u). One has that v ∈ C 1 and φ(v ′ ) = L ◦ (I − Q) ◦ F (u). So, φ(v ′ (R1 )) = 0 and because QF (u) is constant, φ(v ′ (R2 )) =

1 R2N −1

Z

R2

R1

tN −1 F (u)(t)dt −

1

QF (u) N −1

R2

Z

R2

tN −1 dt = 0.

R1

MF is well defined and compact. From the above computation and since φ is singular, we get ||v ′ ||∞ < a. Now, let u ∈ C†1 be such that u = MF (u). It follows QF (u) = 0,

(20)

implying that u = P u + H ◦ φ−1 ◦ L ◦ F (u), u′ = φ−1 ◦ L ◦ F (u), Z r tN −1 F (u)(t)dt (r ∈ (R1 , R2 ]), φ(u′ (r)) = γ(r) R1

and u verifies the differential equation in (19). Conversely, let u be a solution of (19). Taking into account the fact that u verifies (20), after two integrations we deduce that u is a fixed point of MF . Leray-Schauder degree applied to operator MF provides the following existence result. Lemma 1 Let the continuous function h : [R1 , R2 ] × R2 → R be bounded on [R1 , R2 ] × R × (−a, a), µ 6= 0 and consider the Neumann problem (rN −1 φ(u′ ))′ + µrN −1 u = rN −1 h(r, u, u′ ),

u′ (R1 ) = 0 = u′ (R2 ).

(21)

If Mµ is the fixed point operator associated to (21), then there exists ρ > 0 such that any possible fixed point of Mµ is contained in Bρ and dLS [I − Mµ , Bρ , 0] = sign(µ). Proof. Let us consider say, the case where µ > 0, the other one being similar. We can find a constant R > 0 such that sign(u)[−µu + h(r, u, v)] < 0

7

(22)

for all r ∈ [R1 , R2 ], v ∈ (−a, a) and |u| ≥ R. Define the compact homotopy M : [0, 1] × C†1 → C†1 by M(λ, ·) = P + QFµ + H ◦ φ−1 ◦ λL ◦ (I − Q) ◦ Fµ , where Fµ : C†1 → C,

If (λ, u) ∈ [0, 1] × C†1 is such that

Fµ (u) = −µu + h(·, u, u′ ).

u = M(λ, u), then u′ = φ−1 ◦ λL ◦ (I − Q) ◦ Fµ (u) and ||u′ ||∞ < a.

(23)

QFµ (u) = 0.

(24)

Notice also that

Assume that uL ≥ R. Then, using (22) and (23) we have Fµ (u)(r) < 0

for all r ∈ [R1 , R2 ].

This implies that QFµ (u) < Q(0) = 0, a contradiction with (24). Hence uL < R, and analogously uM > −R. From uM ≤ uL +

Z

R2

R1

|u′ (r)|dr

and (23), we deduce that −R − a(R2 − R1 ) < uL ≤ uM < R + a(R2 − R1 ), which together with (23) gives ||u|| < R + a(R2 − R1 + 1) =: ρ0 . The homotopy invariance of the Leray-Schauder degree implies that dLS [I − Mµ , Bρ , 0] = dLS [I − (P + QFµ ), Bρ , 0], for any ρ ≥ ρ0 . As the range of P +QFµ is contained in the subspace of constant functions, the reduction property of the Leray Schauder degree implies that dLS [I − (P + QFµ ), Bρ , 0] = dB [I − (P + QFµ )|R , (−ρ, ρ), 0], 8

where dB denotes the Brouwer degree. But, I − (P + QFµ )|R = −QFµ |R , and we can take ρ sufficiently large such that QFµ (−ρ) > 0 > QFµ (ρ), implying that dB [−QFµ |R , (−ρ, ρ), 0] = 1 = sign(µ). Now, consider the Neumann boundary-value problem (BVP) (rN −1 φ(u′ ))′ = rN −1 f (r, u, u′ ),

u′ (R1 ) = 0 = u′ (R2 ),

(25)

where f : [R1 , R2 ] × R2 → R is continuous. Definition 1 A strict lower solution α (resp. strict upper solution β) of (25) is a function α ∈ C 1 such that kα′ k∞ < a, rN −1 φ(α′ ) ∈ C 1 , α′ (R1 ) ≥ 0 ≥ α′ (R2 ) (resp. β ∈ C 1 , kβ ′ k∞ < a, rN −1 φ(β ′ ) ∈ C 1 , β ′ (R1 ) ≤ 0 ≤ β ′ (R2 )) and (rN −1 φ(α′ (r)))′ (resp.

(r

N −1

′

> rN −1 f (r, α(r), α′ (r))

′

φ(β (r)))

< r

N −1

(26)

′

f (r, β(r), β (r)))

for all r ∈ [R1 , R2 ]. Lemma 2 Assume that (25) has a strict lower solution α and a strict upper solution β such that α(r) < β(r)

r ∈ [R1 , R2 ],

for all

and if N ≥ 2 assume also that R1 > 0. Then dLS [I − Mf , Ωα,β , 0] = −1, where Ωα,β = {u ∈ C†1 : α(r) < u(r) < β(r)

for all

r ∈ [R1 , R2 ],

||u′ ||∞ < a}

and Mf is the fixed point operator associated to (25). Proof. Let γ : [R1 , R2 ] × R → R be the   β(r) if u if γ(r, u) =  α(r) if

continuous function given by u > β(r) α(r) ≤ u ≤ β(r) u < α(r),

and define f1 : [R1 , R2 ] × R2 → R by f1 (r, u, v) = f (r, γ(r, u), v). We consider the modified problem (rN −1 φ(u′ ))′ = rN −1 [f1 (r, u, u′ ) + u − γ(r, u)], 9

u′ (R1 ) = 0 = u′ (R2 ),

(27)

and let Mf1 be the associated fixed point operator of (27). Then, arguing exactly as in the proof of Theorem 4.2 from [8], one has that if u is a solution of (27) then α(r) < u(r) < β(r) for all r ∈ [R1 , R2 ]. It follows that any fixed point of Mf1 is contained in Ωα,β , and using the excision property of the Leray-Schauder degree and Lemma 1 we infer that dLS [I − Mf1 , Ωα,β , 0] = dLS [I − Mf1 , Bρ , 0] = −1, for any ρ sufficiently large. On the other hand Mf (u) = Mf1 (u) for all u ∈ Ωα,β . Consequently, dLS [I − Mf , Ωα,β , 0] = −1.

4

Conditions on the radius and the mean value of the forcing term

We consider the Neumann boundary value problem (rN −1 φ(u′ ))′ + rN −1 g(u) =rN −1 l(r),

u′ (R1 ) = 0 = u′ (R2 ),

(28)

where g : R → R is continuous and l ∈ C. The idea of the following lemma comes from Theorem 2 in [28]. Lemma 3 Assume that φ is singular and that there exist t < s and A < B such that either

or

Qg(t + u b) ≤ A

and

Qg(s + u b) ≥ B

(29)

Qg(t + u b) ≥ B

and

Qg(s + u b) ≤ A

(30)

b1 satisfying ||b for any u b∈C u||∞ < a(R2 − R1 ). If † A < l < B,

(31)

then problem (28) has at least one solution u such that t < u < s. Proof. Let us assume that (29) holds true and let ε > 0 be fixed. For any λ ∈ [0, 1], consider the Neumann problem t+s = λrN −1 l(r) (rN −1 φ(u′ ))′ + λrN −1 g(u) + (1 − λ)εrN −1 u − 2 u′ (R1 ) = 0 = u′ (R2 ). (32) 10

Let also M(λ, ·) : C†1 → C†1

(λ ∈ [0, 1])

be the fixed point operator associated to (32) by Proposition 1. We will show that u − M(λ, u) 6= 0 for any (λ, u) ∈ (0, 1] × ∂Ω,

(33)

and u − M(0, u) = 0 implies

u ∈ Ω,

(34)

where Ω = {u ∈ C†1 :

t < u < s,

||b u||∞ < a(R2 − R1 ),

||u′ ||∞ < a}.

Then, using the invariance by homotopy, the excision property of the LeraySchauder degree and Lemma 1, one has that dLS [I − M(1, ·), Ω, 0] = dLS [I − M(0, ·), Ω, 0] = 1. Hence, the existence property of the Leray-Schauder degree implies the existence of some u ∈ Ω (in particular t < u < s) with u = M(1, u) which is also a solution of (28). So, let us consider (λ, u) ∈ (0, 1] × C†1 such that u = M(λ, u). It follows that b1 is a solution of (32). As Qb (23) holds true and u = u + u b∈R⊕C u = 0, there † exists r0 ∈ [R1 , R2 ] such that u b(r0 ) = 0, yielding ||b u||∞ ≤

Z

R2

r0

|b u′ (r)|dr < a(R2 − R1 ).

Integrating (32) over [R1 , R2 ] we obtain t+s b) − l) = 0. + λ(Qg(u + u (1 − λ)ε u − 2

(35)

(36)

On the other hand, from (29), (31) and (35) it follows that t+s t−s + λ(Qg(t + u b) − l) ≤ (1 − λ)ε + λ(A − l) < 0; (1 − λ)ε t − 2 2 (37) s−t t+s + λ(B − l) > 0. + λ(Qg(s + u b) − l) ≥ (1 − λ)ε (1 − λ)ε s − 2 2

Moreover, if u ∈ ∂Ω, from (23) and (35) one has u = t or u = s. But u verifies (36), contradiction with (37). Consequently, (33) is proved. Now, let u ∈ C†1 be such that u = M(0, u). We deduce that u verifies (23), (35) and (32) with t+s and u ∈ Ω. λ = 0. Hence, u = 2 If (30) holds true then one takes ε < 0. 11

Remark 1 From the proof above it can be seen that if the assumption “A < B” is replaced by “A ≤ B” then problem (28) has at least one solution u such that t ≤ u ≤ s, provided that A ≤ l ≤ B. Theorem 1 If φ is singular, l ∈ C, µ > 0 and 2a(R2 − R1 ) < π, then, the Neumann problem (11) has at least two solutions not differing by a multiple of 2π, provided that |l| < µ cos [a(R2 − R1 )] . Proof. We apply Lemma 3 with g(u) = µ sin(u) and A = −µ cos [a(R2 − R1 )] = −B. Taking t = −π/2, s = π/2, condition (29) is fulfilled and so, we get the existence of a solution u1 which satisfies −π/2 < u1 < π/2. Then, setting t = π/2, s = 3π/2, condition (30) is verified and we obtain a second solution u2 with π/2 < u2 < 3π/2. If we assume that there is some j ∈ Z such that u2 = u1 + 2jπ then necessarily one has 0 < 2jπ < 2π, a contradiction. Remark 2 Using Remark 1, if in Theorem 1 one has 2a(R2 − R1 ) = π, then problem (11) has at least one solution for any l ∈ C with l = 0. Corollary 1 Let µ > 0 and l ∈ C. If 2(R2 − R1 ) < π, then the Neumann problem (12) has at least two classical radial solutions not differing by a multiple of 2π, provided that |l| < µ cos (R2 − R1 ) . Moreover, if 2(R2 − R1 ) = π, then (12) has at least one classical radial solution for any l ∈ C with l = 0. We give now a second proof of Theorem 1, and consider also the classical case. The main idea of this proof comes from [1] and has been used for the classical forced pendulum in [20]. Let f : [R1 , R2 ] × R2 be a continuous function and Nf : C 1 → C be the Nemytskii operator associated to f. We first consider the modified problem of b 1 such that b) ∈ R × C finding (u, u † (rN −1 φ(b u′ ))′ = rN −1 [Nf (u + u b) − Q ◦ Nf (u + u b)].

(38)

Lemma 4 If φ is singular or classical, and if there exists α > 0 such that |f (r, u, v)| ≤ α

for all

(r, u, v) ∈ [R1 , R2 ] × R2 ,

then the set of solutions of problem (38) contains a continuum C whose projection b 1 is contained in on R is R and whose projection on C † b†1 : ||b Bφ = {b u∈C u′ ||∞ ≤ cφ , ||b u||∞ ≤ cφ (R2 − R1 )},

where cφ = max(|φ−1 (±2αR2 /N )|).

12

c:R×C b1 → C b 1 defined by Proof. Let us consider M † †

c(u, u M b) = (I − Q) ◦ H ◦ φ−1 ◦ L ◦ (I − Q) ◦ Nf (u + u b).

c is well defined and compact. Moreover, if It is not difficult to prove that M 1 b c b) ∈ R × C† satisfies u b = M (u, u b), then (u, u b) is a solution of (38). On the (u, u c is contained in Bφ other hand a simple computation shows that the range of M (in both of the two cases) and the proof follows now exactly like the proof of Lemma 2.1 in [8]. Remark 3 The assumption concerning the boundedness of f can be dropped in the singular case but then b 1 : ||b Bφ = {b u∈C u′ ||∞ < a, ||b u||∞ < a(R2 − R1 )}. †

Let ψ : (−b, b) → (−c, c) be a homeomorphism such that ψ(0) = 0 and 0 < b, c ≤ ∞. For l ∈ C and µ > 0 such that 2(||l||∞ + µ)R2 /N < c we introduce the notation ρ(ψ) = max{|ψ −1 (±2(||l||∞ + µ)R2 /N )|}. Theorem 2 Assume that φ is singular or classical, l ∈ C, µ > 0 and 2ρ(φ)(R2 − R1 ) < π,

(39)

then, the Neumann problem (11) has at least two solutions not differing by a multiple of 2π, provided that |l| < µ cos[ρ(φ)(R2 − R1 )]. Proof. Consider the continuous function Γ : R × C → R,

b) = Q ◦ Nf (u + u b). Γ(u, u

For any u b1 , u b2 such that (− π2 , u b1 ), ( π2 , u b2 ) ∈ C, one has that π π Γ(− , u b1 ) > 0 > Γ( , u b2 ). 2 2

Hence, using that C is a continuum and the continuity of Γ, we deduce the b) ∈ C such that − π2 < u < π2 and Γ(u, u b) = 0. Then, u = u + u b existence of (u, u is a solution of (11). Analogously, (11) has a solution w satisfying π2 < w < 3π 2 . Clearly, u − w is not a multiple of 2π. Remark 4 (i) If in (39) one has equality, then we have only existence in Theorem 2. (ii) In Theorem 2, if φ is singular, then ρ(φ) < a. Hence Theorem 1 follows from Theorem 2. The following result is a direct consequence of Lemma 2.5 from [8]. 13

Lemma 5 Let ψ : (−b, b) → (−c, c) be a homeomorphism such that ψ(0) = 0 and 0 < b, c ≤ ∞. Let µ > 0 and l ∈ C be such that (||l||∞ + µ)R2 /N < c. If u is a possible solution of the Neumann problem (rN −1 ψ(u′ ))′ + rN −1 µ sin u = rN −1 l(r),

u′ (R1 ) = 0 = u′ (R2 )

(40)

then ||u′ ||∞ ≤ max{|ψ −1 (±(||l||∞ + µ)R2 /N )|}. Theorem 3 Assume that ψ : R → (−c, c) is a homeomorphism such that ψ(0) = 0 and 0 < c ≤ ∞. If l ∈ C, µ > 0, 2(||l||∞ + µ)R2 /N < c and 2ρ(ψ)(R2 − R1 ) < π, then, the Neumann problem (40) has at least two solutions not differing by a multiple of 2π, provided that |l| < µ cos[ρ(ψ)(R2 − R1 )] is satisfied. Proof. Let d = ρ(ψ)+1 and b = ρ(ψ)+2. Consider φ : (−b, b) → R a singular homeomorphism which coincides with ψ on [−d, d]. Then ρ(ψ) = ρ(φ), and, using Lemma 5, we infer that the solutions of (11) coincide with the solutions of (40). Now the result follows from Theorem 2 (the singular case). Corollary 2 If (39) is satisfied with φ(u) = |u|p−2 u (p > 1), (resp. φ(u) = √ u ) then the Neumann problem (14) (resp. (15)) has at least two classical 1+u2 radial solutions not differing by a multiple of 2π for any l ∈ C with l = 0 (resp. l ∈ C with l = 0 and 2(||l||∞ + µ)R2 /N < 1).

5

Norm conditions on the forcing term

In the proof of the following theorem we adapt to our situation a strategy introduced in Theorem 3 from [28]. Theorem 4 Assume that φ is singular and let µ > 0, R1 > 0 in the case N ≥ 2, and assume that l ∈ C satisfies ||l||∞ < µ. Then problem (11) has at least two solutions not differing by a multiple of 2π. Moreover, if ||l||∞ = µ, then problem (11) has at least one solution.

14

Proof. Assume that ||l||∞ ≤ µ. Then α = − 3π 2 is a constant lower solution for (11) and β = − π2 is a constant upper solution for (11) such that α < β. Hence, using Theorem 4.2 from [8], it follows that (11) has a solution u1 such that α ≤ u1 ≤ β. Notice that if ||l||∞ < µ, then α, β are strict and α < u1 < β. Now, let us assume that ||l||∞ < µ, let Mµ be the fixed point operator associated to (11), and let 3π \ (Ω 3π ∪ Ω π2 , 3π ). Ω = Ω− 3π − 2 ,− π 2 , 2 2 2

(see Lemma 2)

Then using the additivity property of the Leray-Schauder degree and Lemma 2, we deduce that dLS [I − Mµ , Ω, 0] = 1. Hence, the existence property of the Leray-Schauder degree yields the existence of a solution u2 ∈ Ω of (11). If we assume that u2 = u1 + 2jπ for some j ∈ Z then, as −3π/2 < u1 < −π/2, one has −

3π π + 2jπ < u2 = u1 + 2jπ < − + 2jπ. 2 2

This leads to one of the contradictions: u2 ∈ Ω π2 , 3π if j = 1 or u2 = u1 ∈ 2 π for j = 0. Ω− 3π 2 ,− 2 Using Lemma 5, Theorem 4 and arguing exactly as in the proof of Theorem 3 with ρ(ψ) replaced by max{|ψ −1 (±2µR2 /N )|} we obtain the following result. Theorem 5 Let ψ : R → (−c, c) be an increasing homeomorphism such that ψ(0) = 0 and 0 < c ≤ ∞. Let also µ > 0, R1 > 0 in the case N ≥ 2 and l ∈ C 2 be such that 2µR < c. If ||l||∞ < µ, then (40) has at least two solutions not N differing by a multiple of 2π. If ||l||∞ = µ, then (40) has at least one solution. 2 Corollary 3 Let µ > 0, R1 > 0 and l ∈ C be such that 2µR < 1. If ||l||∞ < µ, N then the Neumann problem (15) has at least two classical radial solutions not differing by a multiple of 2π. Moreover, if ||l||∞ = µ, then (15) has at least one classical radial solution. The same conclusion holds also for (12) and (14) 2 < 1. without the assumption 2µR N

6

The variational functional framework

In what follows, we assume that Φ : [−a, a] → R satisfies the following hypothesis : (HΦ ) Φ(0) = 0, Φ is continuous, of class C 1 on (−a, a), with φ := Φ′ : (−a, a) → R an increasing homeomorphism such that φ(0) = 0. Clearly, Φ is strictly convex and Φ(x) ≥ 0 for all x ∈ [−a, a] .

15

Given 0 ≤ R1 < R2 and g : [R1 , R2 ] × R → R a continuous function, we denote by G : [R1 , R2 ] × R → R the indefinite integral of g, i.e., Z x G(r, x) := g(r, ξ) dξ, (r, x) ∈ [R1 , R2 ] × R. 0

We set C := C[R1 , R2 ], L1 := L1 (R1 , R2 ), L∞ := L∞ (R1 , R2 ) and W 1,∞ := W (R1 , R2 ). The usual norm k · k∞ is considered on C and L∞ . The space 1,∞ W is endowed with the norm 1,∞

kvk = kvk∞ + kv ′ k∞ ,

v ∈ W 1,∞ .

Denoting L1N −1

:= {v : (R1 , R2 ) → R measurable :

Z

R2

R1

rN −1 |v(r)| dr < +∞},

each v ∈ L1N −1 can be written v(r) = v + v˜(r), with v :=

N N R2 − R1N

Z

R2

v(r) rN −1 dr,

Z

R2

v˜(r) rN −1 dr = 0.

R1

R1

If v ∈ W 1,∞ then v˜ vanishes at some r0 ∈ (R1 , R2 ) and Z R2 |v ′ (t)| dt ≤ (R2 − R1 )kv ′ k∞ , |˜ v (r)| = |˜ v (r) − v˜(r0 )| ≤ R1

so, one has that ||e v ||∞ ≤ (R2 − R1 )kv ′ k∞ .

(41)

Putting K := {v ∈ W 1,∞ : kv ′ k∞ ≤ a},

it is clear that K is a convex subset of W 1,∞ .

Let Ψ : C → (−∞, +∞] be defined by  if v ∈ K,  ϕ(v), Ψ(v) =  +∞, otherwise,

where ϕ : K → R is given by ϕ(v) =

Z

R2

rN −1 Φ(v ′ ) dr,

R1

v ∈ K.

Obviously, Ψ is proper and convex. Lemma 6 If {un } ⊂ K and u ∈ C are such that un (r) → u(r) for all r ∈ [R1 , R2 ], then 16

(i) u ∈ K; (ii) u′n → u′ in the w∗ –topology σ(L∞ , L1 ). Proof. From the relation Z |un (r1 ) − un (r2 )| =

r1

r2

letting n → ∞, we get

u′n (r) dr

≤ a|r1 − r2 |,

|u(r1 ) − u(r2 )| ≤ a|r1 − r2 | (r1 , r2 ∈ [R1 , R2 ]), which yields u ∈ K.

Next, we show that that if {u′k } is a subsequence of {u′n } with u′k → v ∈ L∞ in the w∗ –topology σ(L∞ , L1 ) then v = u′

a.e. on [R1 , R2 ].

(42)

Indeed, as Z

R2

R1

u′k (r)f (r) dr →

Z

R2

v(r)f (r) dr

R1

for all f ∈ L1 ,

taking f ≡ χr1 ,r2 , the characteristic function of the interval having the endpoints r1 , r2 ∈ [R1 , R2 ], it follows Z r2 Z r2 ′ v(r) dr (r1 , r2 ∈ [R1 , R2 ]). uk (r) dr → r1

r1

Then, letting k → ∞ in uk (r2 ) − uk (r1 ) = we obtain u(r2 ) − u(r1 ) = which, clearly implies (42).

Z

Z

r2

r1

u′k (r) dr

r2

(r1 , r2 ∈ [R1 , R2 ])

v(r) dr

r1

Now, to prove (ii) it suffices to show that if {u′j } is an arbitrary subsequence of {u′n }, then it contains itself a subsequence {u′k } such that u′k → u′ in the w∗ – topology σ(L∞ , L1 ). Since L1 is separable and {u′j } is bounded in L∞ = (L1 )∗ , we know that it has a subsequence {u′k } convergent to some v ∈ L∞ in the w∗ –topology σ(L∞ , L1 ). Then, as shown before (see (42)), we have v = u′ . Lemma 7 If {un } ⊂ K is a sequence such that uk (r) → u(r) for every r ∈ [R1 , R2 ] with u ∈ K, then Z R2 Z R2 Φ[u′ (r)] rN −1 dr. (43) Φ[u′n (r)] rN −1 dr ≥ lim inf n→∞

R1

R1

17

Proof. By virtue of (HΦ ) the function Φ is convex, hence for all y ∈ [−a, a] and z ∈ (−a, a) one has Φ(y) − Φ(z) ≥ φ(z)(y − z).

(44)

This implies that for any λ ∈ [0, 1) it holds Z

R2

R1

Φ[u′n (r)] rN −1 dr

Z

≥

R2

R1 Z R2

+

R1

Φ[λu′ (r)] rN −1 dr

(45)

φ[λu′ (r)][u′n (r) − λu′ (r)] rN −1 dr.

From Lemma 6 we have that u′n → u′ in the w∗ –topology σ(L∞ , L1 ). Since the map r 7→ rN −1 φ[λu′ (r)] belongs to L∞ ⊂ L1 , using (45) we infer that lim inf n→∞

Z

R2

R1

Φ[u′n (r)] rN −1 dr

Z

≥

R2

Φ[λu′ (r)] rN −1 dr

R1

+ (1 − λ) As φ(t)t ≥ 0, for all t ∈ (−a, a), we get lim inf n→∞

Z

R2

R1

Φ[u′n (r)] rN −1 dr ≥

Z

Z

R2

φ[λu′ (r)]u′ (r) rN −1 dr.

R1

R2

Φ[λu′ (r)] rN −1 dr,

R1

which, using Lebesgue’s dominated convergence theorem, gives (43) by letting λ → 1. Consequently, if {un } ⊂ K and u ∈ C are such that un (r) → u(r) for all r ∈ [R1 , R2 ], then u ∈ K and ϕ(u) ≤ lim inf ϕ(un ).

(46)

n→∞

This implies that Ψ is lower semicontinuous on C. Also, note that K is closed in C. Next, let G : C → R be defined by G(u) =

Z

R2

rN −1 G(r, u) dr,

R1

u ∈ C.

A standard reasoning (also see [23, Remark 2.7]) shows that G is of class C 1 on C and its derivative is given by hG ′ (u), vi =

Z

R2

rN −1 g(r, u)v dr,

R1

18

u, v ∈ C.

The functional I : C → (−∞, +∞] defined by I =Ψ+G

(47)

has the structure required by Szulkin’s critical point theory [30]. Accordingly, a function u ∈ C is a critical point of I if u ∈ K and it satisfies the inequality Ψ(v) − Ψ(u) + hG ′ (u), v − ui ≥ 0

for all v ∈ C,

or, equivalently Z

R2

R1

7

rN −1 [Φ(v ′ ) − Φ(u′ )] dr +

Z

R2

R1

rN −1 g(r, u)(v − u) dr ≥ 0 for all v ∈ K.

Critical points and solutions

Now, we consider the Neumann boundary value problem [rN −1 φ(u′ )]′ = rN −1 g(r, u),

u′ (R1 ) = 0 = u′ (R2 ).

(48)

under the basic hypothesis (HΦ ). Recall that by a solution of (48) we mean a function u ∈ C 1 [R1 , R2 ], such that ku′ k∞ < a, φ(u′ ) is differentiable and (48) is satisfied. We need an auxiliary result. Lemma 8 For every f ∈ C, problem [rN −1 φ(u′ )]′ = rN −1 [u + f ],

u′ (R1 ) = 0 = u′ (R2 )

(49)

has a unique solution uf , which is also the unique solution of the variational inequality Z

R2

R1

rN −1 [Φ(v ′ ) − Φ(u′ ) + u(v − u) + f (v − u)] dr ≥ 0

for all v ∈ K,

(50)

and the unique minimum over K of the strictly convex functional J defined by Z R2 u2 ′ N −1 Φ(u ) + r + f u dr. (51) J(u) = 2 R1 Proof. Problem (49) is equivalent to finding u = u + u e with u and u e solutions of  N −1 φ(e u′ )]′ = rN −1 fe, u e′ (R1 ) = 0 = u e′ (R2 ),  [r Z R2 (52)  u = −f , rN −1 u e(r) dr = 0. R1

19

Now the first equation gives, using the first boundary condition, Z r sN −1 fe(s) ds . u e′ (r) = φ−1 r1−N

(53)

R1

From (53) we get ′

ke u k∞ < a,

′

−1

u e (R2 ) = φ

"

R21−N

Z

R2

s

N −1

R1

# e f (s) ds = φ−1 (0) = 0.

Then the unique solution of (53) is given by Z t Z r sN −1 fe(s) ds dt, φ−1 t1−N u e(r) = c +

(54)

R1

R1

where

N c=− N R2 − R1N

Z

R2

R1

r

N −1

Z

r

R1

−1

φ

t

1−N

Z

t

s

N −1

R1

e f (s) ds dt dr.

(55)

The unique solution uf = u + u e of (49) follows from (52), (54) and (55). Now, if u is a solution of (49), then, taking v ∈ K, multiplying each member of the differential equation by v − u, integrating over [R1 , R2 ], and using integration by parts and the boundary conditions, we get Z R2 rN −1 [φ(u′ )(v ′ − u′ ) + u(v − u) + f (v − u)] dr = 0, R1

which gives (50) if we use the convexity inequality for Φ Φ(v ′ ) − Φ(u′ ) ≥ φ(u′ )(v ′ − u′ ). v2 u2 − ≥ u(v − u) introduced in (50) implies that 2 2 Z R2 v2 u2 + fv − − f u dr ≥ 0 for all v ∈ K, rN −1 Φ(v ′ ) − Φ(u′ ) + 2 2 R1

The inequality

which shows that J has a minimum on K at u. Conversely if it is the case, then, for all λ ∈ (0, 1] and all v ∈ K, we get Z R2 [(1 − λ)u + λv]2 rN −1 {Φ[(1 − λ)u′ + λv ′ ] + + f [(1 − λ)u + λv]} dr 2 R1 Z R2 u2 rN −1 [Φ(u′ ) + ≥ + f u] dr, 2 R1 which, using the convexity of Φ, simplifying, dividing both members by λ and letting λ → 0+ , gives the variational inequality (50). Thus solving (50) is equivalent to minimizing (51) over K. Now, it is straightforward to check that J is strictly convex over K and therefore has a unique minimum there, which gives the required uniqueness conclusions of Lemma 8. 20

Proposition 2 If u is a critical point of I, then u is a solution of problem (48). Proof. We set fu := g(·, u) − u ∈ C and consider the problem [rN −1 φ(w′ )]′ = rN −1 [w + fu (r)],

w′ (R1 ) = 0 = w′ (R2 ).

(56)

By virtue of Lemma 8, problem (56) has an unique solution u b and it is also the unique solution of Z

R2

R1

b(v − u b) + fu (r)(v − u b)] dr ≥ 0 rN −1 [Φ(v ′ ) − Φ(b u′ ) + u for all v ∈ K.

(57)

Since u is a critical point of I, we infer that Z

R2

R1

rN −1 [Φ(v ′ ) − Φ(u′ ) + u(v − u) + fu (r)(v − u)] dr ≥ 0 for all v ∈ K.

(58)

It follows by uniqueness that u = u b. Hence, u solves problem (48).

The following lemma is useful in minimization problems. For any ρ > 0, set b ρ := {u ∈ K : |u| ≤ ρ}. K

Lemma 9 Assume that there is some ρ > 0 such that inf I = inf I. K

bρ K

(59)

bρ, Then I is bounded from below on C and attains its infimum at some u ∈ K which solves problem (48).

Proof. By virtue of (59) and inf I = inf I, it suffices to prove that there is some C

b ρ such that u∈K

K

I(u) = inf I. bρ K

(60)

Then, we get that u is a minimum point of I on C and, on account of [30, Proposition 1.1], is a critical point of I. The proof will be accomplished by virtue of Proposition 2. b ρ then, using (41) we obtain If v ∈ K v (r)| ≤ ρ + (R2 − R1 )a. |v(r)| ≤ |v| + |˜ 21

b ρ is bounded in W 1,∞ and, by the This, together with kv ′ k∞ ≤ a show that K b ρ is relatively compact in compactness of the embedding W 1,∞ ⊂ C, the set K b C. Let {un } ⊂ Kρ be a minimizing sequence for I. Passing to a subsequence if necessary and using Lemma 6, we may assume that {un } converges uniformly b ρ . From (46) and the to some u ∈ K. It is easily seen that actually u ∈ K continuity of G on C, we obtain I(u) ≤ lim inf I(un ) = lim I(un ) = inf I, n→∞

n→∞

bρ K

showing that (60) holds true. Theorem 6 Let f : [R1 , R2 ] × R → R be continuous and F : [R1 , R2 ] × R → R be defined by Z x F (r, x) := f (r, ξ) dξ, (r, x) ∈ [R1 , R2 ] × R. 0

If there is some ω > 0 such that F (r, x) = F (r, x+ω) for all (r, x) ∈ [R1 , R2 ]×R, then, for any h ∈ C with h = 0, the problem [rN −1 φ(u′ )]′ = rN −1 [f (r, u) + h(r)],

u′ (R1 ) = 0 = u′ (R2 ).

b ω which is a minimizer of the corresponding has at least one solution u ∈ K energy functional I on C. Proof. We have

G(r, x) = F (r, x) + h(r)x,

(r, x) ∈ [R1 , R2 ] × R.

Due to the ω–periodicity of F (r, ·) and because of h = 0, it holds I(v + jω) = I(v) for all v ∈ K and j ∈ Z. Then, the conclusion follows from the equality

and Lemma 9.

bω} {I(v) : v ∈ K} = {I(v) : v ∈ K

Corollary 4 For any µ ∈ R and h ∈ C with h = 0 the problem !′ u′ N −1 p + rN −1 µ sin u = rN −1 h(r), u′ (R1 ) = 0 = u′ (R2 ) r 2 ′ 1−u has at least one solution.

22

Corollary 5 For any µ ∈ R and h ∈ C such that Z h(|x|) dx = 0, A

the problem ∇·

∇v p 1 − |∇v|2

!

+ µ sin v = h(|x|)

in

A,

∂ν v = 0

on

∂A

has at least one classical radial solution. Proof. Indeed, going to spherical coordinates, we have Z

h(|x|) dx =

A

2π n/2 Γ(n/2)

Z

R2

h(r) rN −1 dr. R1

Remark 5 If D is a bounded domain with sufficiently smooth boundary, a necessary condition for the existence of at least one solution to the Neumann problem ! ∇v + µ sin v = h(x) in D, ∂ν v = 0 on ∂D (61) ∇· p 1 − |∇v|2 for any µ > 0 is that condition Z

h(x) dx = 0

(62)

D

holds, as it is easily seen by integrating both members of (61) over D and using divergence theorem and the boundary conditions. It is an open problem to know if condition (62) is sufficient. A proof of the existence of a minimum for the functional Z h p i G(u) = − 1 − |∇v(x)|2 + µ cos v(x) + h(x)v(x) dx D

on the closed convex set

K := {v ∈ W 1,∞ (D) : |∇v(x)| ≤ 1 a.e. on D} can be done following the lines of the proof of Corollary 6, but our way to go from the variational inequality to the differential equation seems to be specific to a one-dimensional situation, i.e. to the radial case.

23

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