Radial positive solutions of elliptic

systems with Neumann boundary

conditions

Denis Bonheure1, Enrico Serra∗,2, Paolo Tilli2

1Departement de Mathematique, Universite libre de Bruxelles

CP 214, Boulevard du Triomphe, B-1050 Bruxelles, Belgium2Dipartimento di Scienze Matematiche, Politecnico di Torino

Corso Duca degli Abruzzi, 24, 10129 Torino, Italy

Corresponding author: Enrico Serra. enrico.serra@polito.it

Tel. (+39) 011 090 7540. Fax. (+39) 011 090 7599

Abstract

We consider radial solutions of elliptic systems of the form8><>:−∆u + u = a(|x|)f(u, v) in BR

−∆v + v = b(|x|)g(u, v) in BR

∂νu = ∂νv = 0 on ∂BR,

where essentially a, b are assumed to be radially nondecreasing weightsand f, g are nondecreasing in each component. With few assumptionson the nonlinearities, we prove the existence of at least one couple ofnondecreasing nontrivial radial solutions. We emphasize that we do notassume any variational structure nor subcritical growth on the nonlineari-ties. Our result covers systems with supercritical as well as asymptoticallylinear nonlinearities.

1 Introduction

Let BR be the ball of radius R in RN , with N ≥ 2. We consider the Neumannproblem

−∆u+ u = a(|x|)f(u, v) in BR

−∆v + v = b(|x|)g(u, v) in BR

∂νu = ∂νv = 0 on ∂BR,

(1)

∗Author partially supported by the PRIN2009 grant “Critical Point Theory and Pertur-bative Methods for Nonlinear Differential Equations”.

1

where a, b and f , g satisfy the assumptions

(A) a, b ∈ L1(0, R) are nonnegative, nondecreasing and not identically zero;

(H1) f, g ∈ C(R+ ×R+) are nonnegative and nondecreasing in each variable;

(H2) lims+t→0+

f(s, t) + g(s, t)s+ t

= 0, lims+t→+∞

f(s, t) + g(s, t)s+ t

= +∞.

This class of systems includes for instance gradient type systems−∆u+ u = a(|x|)∂uG(u, v) in BR

−∆v + v = b(|x|)∂vG(u, v) in BR

∂νu = ∂νv = 0 on ∂BR,

(2)

or Hamiltonian type systems−∆u+ u = a(|x|)∂vH(u, v) in BR

−∆v + v = b(|x|)∂uH(u, v) in BR

∂νu = ∂νv = 0 on ∂BR,

(3)

as well as nonvariational systems, namely systems which are not the Euler–Lagrange equations of an energy functional.

When dealing with (2), it is usually assumed that G(s, t) grows at mostlike |s|p + |t|q where p, q < 2∗ := 2N/(N − 2). Such an assumption, whichis referred to as a subcriticality condition, gives the required compactness toallow a study of the system through critical point theory. In the case of (3),the subcriticality assumption, which brings the required compactness, takes arelaxed form. Assuming as a paradigm that H(s, t) = sp + tq, the subcriticalitycondition [11, 12] writes

1p

+1q> 1− 2

N. (4)

When dealing with non variational systems with topological methods, onealso requires some compactness. This compactness usually corresponds to apriori bounds on a class of auxiliary systems associated to the original one.Growth limitations on f and g then appear as a main assumption to derivethese a priori bounds. We refer to de Figueiredo [7] for further details.

Our main purpose in this work is to show that when looking for radial solu-tions of the Neumann problem (1), one can obtain existence results for broaderclasses of nonlinearities than when prescribing Dirichlet boundary conditions.Indeed, writing the Pohozaev identity, see e.g. [10], for the Dirichlet problem

−∆u+ u = a(|x|)vq−1 in BR

−∆v + v = b(|x|)up−1 in BR

u = v = 0 on ∂BR,

(5)

2

one realizes that there always exits a regime of parameters p, q for which thesystem has no positive solutions. This non existence regime could of coursedepend on the weights a and b. For instance, if a(|x|) = |x|β and b(|x|) = |x|α,with α, β ≥ 0, the non existence condition for (5) writes N+α

p + N+βq ≤ N − 2,

if N ≥ 3, see [4].In contrast with such non existence results for systems with Dirichlet bound-

ary conditions, we will show that when dealing with Neumann boundary condi-tions, there is no need at all to make growth assumptions to ensure existence,at least when looking for radial positive solutions under the set of assumptions(A), (H1-H2).

For a single equation, the authors of [15] proved that the Neumann problem −∆u+ u = a(|x|)f(u) in Bu > 0 in B∂νu = 0 on ∂B,

(6)

has at least one positive radially nondecreasing solution provided that a is anondecreasing nonnegative function and f ∈ C1([0,∞)) is such that

f(0) = f ′(0) = 0, f ′(t)t− f(t) > 0 and f(t)t ≥ µF (t) := µ

∫ t

0

f(s) ds, (7)

for t ∈ (0,∞), with some constant µ > 2. These assumptions hold for instancefor f(t) = tp, whatever p > 1. This result was further extended by the firstauthor et. al. [3] to a broader framework.

The key point in [15] consists in the observation that restricting the set oftrial functions to nonnegative and nondecreasing radial functions in H1(B) givesrise to boundedness and compactness properties even for supercritically growingnonlinearities. This idea is exploited in a variational flavor in [15] and combinedwith topological arguments in [3].

In this paper we develop these ideas for elliptic systems. Our main resultfor system (1) is

Theorem 1.1. Under assumptions (A), (H1), (H2), problem (1) admits at leastone solution (u, v) with u and v both nonnegative and nondecreasing.

The ambient set in this paper will be the cone of nonnegative and nonde-creasing functions, namely the set K ×K where

K = {u ∈ H1r (BR) | 0 ≤ u(r) ≤ u(s), ∀ 0 ≤ r < s ≤ R}.

Here H1r (BR) stands for the space of radially symmetric H1 functions on the

ball BR. The main advantage of working in the cone K is that it enjoys veryconvenient compactness properties, that rule out any concentration phenomena.

Theorem 1.1 will be deduced in Section 3 from a fixed point argument inK × K. It will be clear from the proof that the result can be extended to a

3

larger class of systems of the formL1u = a(|x|)f(u, v) in BR

L2u = b(|x|)g(u, v) in BR

∂νu = ∂νv = 0 on ∂BR,

where the operators Li are uniformly elliptic, invariant under the action of theorthogonal groupO(N) and such that essentially Lemma 2.3 holds. For instance,one can include first order terms as in [3], namely consider the operators Li,i = 1, 2, defined by

Liu = −∆u+ c(|x|)x·∇u+ µiu,

where µi > 0 and

(c(r)r)r > −1− N − 1r2

,

for 0 < r < 1.

If both a and b are constant, we cannot expect that the solution is noncon-stant without further assumptions. Indeed, arguing as in [3, Proposition 4.1],one can provide examples of systems of the form (1) with a = b = 1 whoseunique positive solutions are constants. For the scalar problem (6) with a = 1,assuming there exists u0 > 0 such that f(u0) = u0 and f ′(u0) > λ2, one can usethe mountain pass Lemma in K to prove the existence of a nontrivial positivesolution, see [3]. As we do not want to restrict our attention to gradient systems,we cannot use variational arguments. Also, a variational approach seems quitedelicate in the case of a Hamiltonian system since the associated functional isstrongly indefinite. We therefore pursue with our topological treatment. Weexploit abstract results by Dancer, [5] and [6], on the local fixed point index fora map defined between cones, see Section 4, to derive a sufficient condition forthe existence of non trivial solutions of the system

−∆u+ u = f(u, v) in BR

−∆v + v = g(u, v) in BR

∂νu = ∂νv = 0 on ∂BR.

(8)

The method developed here is completely different from the one used in thescalar case [3] which by the way is also covered by our theorem. We denote by

1 = λ1 < λ2 < . . .

the eigenvalues of −∆+ I (as an operator in H1r (BR)) with Neumann boundary

conditions.

Theorem 1.2. Assume that f and g satisfy (H1) and (H2) and are differen-tiable. Assume also that the only constant nontrivial solution of (8) is (u0, v0).If

λ 6∈ {λ1, λ2} and λ > λ2, (9)

4

whereλ ≤ λ

are the (real) eigenvalues of the matrix

M =

fu(u0, v0) fv(u0, v0)

gu(u0, v0) gv(u0, v0)

, (10)

then problem (8) admits at least one nonnegative, non constant and nondecreas-ing solution.

The case of the Hamiltonian system−∆u+ u = f(v) in BR

−∆v + v = g(u) in BR

∂νu = ∂νv = 0 on ∂BR.

(11)

is particularly relevant in the applications. In this case the condition (9) inTheorem 1.2 takes the very simple form

f ′(v0)g′(u0) > λ22, (12)

see Corollary 4.9 in Section 4.

As previously mentioned, the existence theorems will be proved in Section3 and Section 4. We build the functional setting in Section 2 as well as somepreliminaries.

Notation. We denote by ‖u‖p, with 1 ≤ p ≤ ∞, the usual Lp norm of u onBR; norms on subsets of BR are always written down explicitly as, for example,‖u‖Lp(A). Throughout the paper, C denotes a positive constant that couldchange from line to line. If not specified, such positive constants always dependon R and the dimension N only.

2 Functional setting and preliminary properties

In this section we establish some properties that we will frequently use. Theserefer to single functions or equations, rather than to systems.

We let H1r (BR) be the Hilbert space of radially symmetric H1 functions on

the ball BR, endowed with the usual scalar product

〈u, v〉 =∫BR

∇u∇v dx+∫BR

uv dx.

We denote the associated norm by ‖·‖. With some abuse of notation, sometimeswe regard functions u ∈ H1

r (BR) as functions of one variable, thus writing u(|x|)for u(x), u′(|x|) for ∂νu(x) etc. We set

E = H1r (BR)×H1

r (BR).

5

The space E has a Hilbert structure with scalar product and norm given re-spectively by

〈(u1, v1), (u2, v2)〉 = 〈u1, u2〉+ 〈v1, v2〉

and‖(u, v)‖2 = 〈(u, v), (u, v)〉 = ‖u‖2 + ‖v‖2.

We also define

K = {u ∈ H1r (BR) | 0 ≤ u(r) ≤ u(s), ∀ 0 ≤ r < s ≤ R}. (13)

The functions in K can be thought of as continuous on the whole interval [0, R],by taking continuous representatives and by defining u(0) = limr→0+ u(r); thislimit of course exists by monotonicity. Note also that K ×K is a closed convexcone in E and that it has empty interior.

The following statement is taken from [15]; we report its elementary prooffor completeness.

Lemma 2.1. There exists a positive constant C > 0such that

‖u‖∞ ≤ C‖u‖ ∀u ∈ K. (14)

Proof. Fix ρ ∈ (0, R) and let u ∈ K. Since u is nonnegative and nondecreasing,we have

‖u‖∞ = ‖u‖L∞(BR\Bρ) ≤ C‖u‖H1(BR\Bρ) ≤ C‖u‖,

by the continuity of the embedding of H1rad(BR \Bρ) into L∞(BR \Bρ).

Lemma 2.2 (Harnack Inequality for nondecreasing supersolutions). Let u ∈H1r (BR) be a radially nondecreasing function such that

−∆u+ u ≥ 0 in BR

in the sense of distributions. Then

(i) u is nonnegative;

(ii) u is Lipschitz continuous, and ∇u admits a continuous extension at x = 0(namely, ∇u(0) = 0);

(iii) There holds‖∇u‖∞ + ‖u‖∞ ≤ Cu(0). (15)

Proof. For every nonnegative η ∈ C∞0 (BR) we have

−∫BR

∇u∇η dx ≤∫BR

uη dx

6

and, by a density argument, the same inequality holds true for every η ≥ 0which is Lipschitz continuous. Observe that u′(|x|) = |∇u(x)|, as u is radiallynondecreasing. Hence, for fixed r ∈ (0, R), choosing

ηε(x) =

1 if |x| < r,r+ε−|x|

ε if r ≤ |x| ≤ r + ε,0 if |x| > r + ε,

where ε ∈ (0, R− r), yields

1ε

∫Br+ε\Br

|∇u| dx ≤∫BR

uηε dx.

Letting ε→ 0+, we obtain that∫∂Br

|∇u| dHN−1 ≤∫Br

u dx for a.e. r ∈ (0, R).

By radial symmetry, this can be rewritten as

|∇u(r)| ≤ 1NωNrN−1

∫Br

u dx,

where ωN is the measure of the unit ball in RN . Hence, as u is radially nonde-creasing,

|∇u(r)| ≤ ru(r)N

for a.e. r ∈ (0, R),

which entails that u is nonnegative and Lipschitz continuous. In particular,limr→0+ ∇u(r) = 0, and

‖∇u‖∞ ≤ C‖u‖∞. (16)

Now let w ∈ H1r (BR) be the solution of the Dirichlet problem

−∆w + w = 0 in BR, w = 1 on ∂BR,

and observe that the function v(x) := u(R)w(x) coincides with u along ∂BRand solves the equation −∆v+v = 0 in BR. As u is a supersolution of the sameequation, we have from the maximum principle that u(x) ≥ u(R)w(x) in BRand, in particular,

u(0) ≥ u(R)w(0).

Since w(0) > 0, letting C = w(0)−1 we obtain that

‖u‖∞ = u(R) ≤ Cu(0).

On combining the last inequality with (16), one obtains (15).

7

Lemma 2.3. Given a nonnegative, radial and radially nondecreasing functionh ∈ L1(BR), there exists a unique function u ∈ H1(BR) which solves the Neu-mann problem

−∆u+ u = h in BR, ∂νu = 0 on ∂BR, (17)

that is, ∫BR

∇u∇ϕdx+∫BR

uϕdx =∫BR

hϕdx ∀ϕ ∈ H1(BR). (18)

Moreover, we have

(i) u is nonnegative, radial and radially nondecreasing, that is, u ∈ K;

(ii) there hold‖u‖ ≤ C‖h‖1 (19)

and‖u‖H2(BR/2) + ‖u‖W 2,1(BR\BR/4) ≤ C‖h‖1. (20)

Proof. We first observe that the integral in the right hand side of (18) makessense for every ϕ ∈ H1(BR). Indeed, from our assumptions on h it is easy tosee that

‖h‖L∞(B3R/4) ≤ C‖h‖L1(BR\B3R/4), (21)

and hence, in particular, for every ϕ ∈ H1(BR)∫BR/2

h|ϕ| dx ≤ ‖h‖L∞(BR/2)‖ϕ‖L1(BR/2) ≤ C‖h‖1‖ϕ‖.

On the other hand, from the trace inequalities

‖ϕ‖L1(∂Bρ) ≤ C‖ϕ‖, ρ ∈ (R/2, R),

we see that∫BR\BR/2

h|ϕ| dx =∫ R

R/2

h(ρ)

(∫∂Bρ

|ϕ| dHn−1

)dρ ≤ C‖h‖1‖ϕ‖

which, combined with the estimate on BR/2, reveals that∫BR

h|ϕ| dx ≤ C‖h‖1‖ϕ‖ ∀ϕ ∈ H1(BR). (22)

This shows that the strictly convex functional

J(u) =∫BR

(|∇u|2 + u2 − 2hu

)dx

is coercive in H1(BR), hence it admits a unique minimizer u ∈ H1(BR), whichsolves the Euler equation (18). Moreover, it is easily seen that the radial sym-metry of h entails that of u, and also the uniqueness of u is easy to establish.Finally, from J(u+) ≤ J(u), we infer that u ≥ 0.

8

Now, choosing ϕ = u in (18) and using (22), one obtains (19). In addition,from (17), (21) and standard elliptic regularity theory one obtains that

‖u‖H2(BR/2) ≤ C‖h‖1. (23)

Considering for a while radial functions as functions of one variable, we see thatu satisfies the ODE

−u′′ − N − 1r

u′ + u = h

in the open interval (0, R). Getting back to radial functions, e.g. in the annulusA = BR \BR/4, we immediately obtain that

‖D2u‖L1(A) ≤ C(‖u‖W 1,1(A) + ‖h‖L1(A)

)≤ C‖h‖1

having used (19), and hence the estimate

‖u‖W 2,1(BR\BR/4) ≤ C‖h‖1

is obtained which, combined with (23), gives (20).

Finally, to prove that u is radially nondecreasing, it suffices to prove that,for every r ∈ (0, R), one of the following two cases certainly occurs:

(a) u(t) ≤ u(r) for every t ∈ (0, r);

(b) u(t) ≥ u(r) for every t ∈ (r,R).

Indeed, if we had u(r1) > u(r2) for some r1 < r2, we could find some r ∈ (r1, r2)with u(r1) > u(r) > u(r2), which would violate both (a) and (b).

To prove that (a) or (b) occurs, consider an arbitrary r ∈ (0, R) and suppose,to fix the ideas, that

u(r) ≥ h(r). (24)

Define the radial test function

ϕ(x) =

{(u(|x|)− u(r)

)+ if |x| ≤ r,0 if r < |x| ≤ R.

Plugging this ϕ into (18), we see that∫Br

∇u∇ϕdx+∫Br

uϕdx =∫Br

hϕdx ≤ h(r)∫Br

ϕdx ≤ u(r)∫Br

ϕdx

having used (24), and hence∫Br

∇u∇ϕdx+∫Br

(u− u(r)

)ϕdx ≤ 0,

that is, ∫Br

|∇ϕ|2 dx+∫Br

ϕ2 dx ≤ 0,

9

which shows that ϕ ≡ 0, or equivalently that case (a) occurs. Much in the sameway as (a) follows from (24), testing with

ϕ(x) =

{0 if |x| ≤ r,(u(|x|)− u(r)

)− if r < |x| ≤ R,

one can see that (b) occurs, if (24) is replaced by the opposite inequality.

Remark 2.4. As a direct consequence of the previous lemma, if (hn)n areequibounded in L1(BR), from the corresponding sequence (un)n, where un solves(17) with h = hn, one can extract a subsequence strongly convergent in H1

r (BR).This compactness indeed follows from (20) and standard embedding theorems.

3 The main existence result

In this section we use an abstract fixed point result first due to Krasnosel’skiı([9]) in the framework of Ordered Banach Spaces; simpler proofs than the orig-inal one have been given in [2] and [13], but we refer the reader to the paper([1]) by Amann, which presents the material in a concise and ready–to–use way.Following [1] (see, in particular pp. 625–627 and p. 661), we introduce someterminology and notation needed to properly state the mentioned result.

An Ordered Banach Space (OBS) is a pair (E,P ) where E is a Banach space,and P ⊂ E is a closed cone in E, that is, a closed set P satisfying P + P ⊆ P ,αP ⊆ P for every α > 0, and P ∩ (−P ) = {0}. The linear ordering in E isinduced by P , according to

x ≤ y ⇐⇒ (y − x) ∈ P.

If (E,P ) is an OBS and ρ > 0, we let

Pρ = {x ∈ P such that ‖x‖ < ρ} (25)

and we denote by Pρ its closure. Finally, we let

S+ρ = {x ∈ P such that ‖x‖ = ρ} .

Theorem 3.1 (Krasnosel’skiı, [9]; Amann, [1]). Assume (E,P ) is an OBS. Forsome ρ > 0, let Φ : Pρ → P be a compact map, and let σ, τ ∈ (0, ρ] with σ 6= τ .Suppose that

(i) Φ(x) 6= λx for every x ∈ S+σ and every λ ≥ 1,

(ii) there exists an element p ∈ P , p 6= 0, such that x − Φ(x) 6= λp for everyx ∈ S+

τ and every λ ≥ 0.

Then Φ has at least one fixed point x with

min(σ, τ) < ‖x‖ < max(σ, τ).

10

Moreover, denoting by i(Φ, U) the fixed point index of the map Φ in some U ⊂ E,we have

i(Φ, Pσ \ P τ ) = 1 if τ < σ and i(Φ, Pτ \ Pσ) = −1 if σ < τ. (26)

We wish to apply this fixed point result in the concrete framework of theHilbert space

E = H1r (BR)×H1

r (BR)

and the coneP = K ×K (27)

where, as in (13),

K = {u ∈ H1r (BR) | 0 ≤ u(r) ≤ u(s), ∀ 0 ≤ r < s ≤ R}.

Indeed, it is easy to check that (E,P ) is an OBS.Now consider the (nonlinear) map Φ : P → E defined according to

Φ(ϕ,ψ) = (u, v) ⇐⇒

−∆u+ u = a(|x|)f(ϕ,ψ) in BR

−∆v + v = b(|x|)g(ϕ,ψ) in BR

∂νu = ∂νv = 0 on ∂BR.

(28)

Lemma 3.2. The map Φ is well defined and, in fact, Φ : P → P . Moreover, Φis continuous and, if U ⊂ P is a bounded set, then Φ(U) has compact closure.

Proof. Consider an arbitrary pair of functions (ϕ,ψ) ∈ P . Due to (A), (H1)and Lemma 2.1, the function

h(x) = a(|x|)f(ϕ(x), ψ(x)), x ∈ BR (29)

satisfies the assumptions of Lemma 2.3 (namely it is nonnegative, radially non-decreasing and L1). As a consequence, there exists a unique u ∈ H1

r (BR) whichsolves the first Neumann problem in (28), and from (i) of Lemma 2.3 we see thatu ∈ K. Similarly, the exists a unique v ∈ K which solves the second Neumannproblem in (28). Therefore the map Φ : P → E is well defined, with values inP .

Still from (A), (H1) and Lemma 2.1 we also see that

‖h‖1 ≤ ‖a‖1f(‖ϕ‖∞, ‖ψ‖∞) ≤ ‖a‖1f(C‖ϕ‖, C‖ψ‖),

(C being the constant that appears in Lemma 2.1), and likewise for the corre-sponding term in the second equation of (28).

Therefore whenever ϕ and ψ vary in a bounded subset U of P , the right–hand–sides of (28) are uniformly bounded in L1. Then from the last part of (ii)in Lemma 2.3, Φ(U) has compact closure in E. The same argument based onLemma 2.3 allows us to prove that Φ is continuous by passing to the limit inthe weak formulation of (28).

11

We are now in a position to prove the existence of a fixed point of Φ whichyields a positive solution of (1).

Proof of Theorem 1.1. We prove that, for suitable constants ρ = τ > σ > 0,one can apply Theorem 3.1 to the map Φ defined in (28) and the cone givenby (27). Note that our Φ is defined on the whole P , so we will not really beconcerned with the parameter ρ: we will prove that Φ satisfies assumptions (i)and (ii) of Theorem 3.1 for some τ > σ > 0, and then we will set ρ = τ .

Step 1 - proof of (i). Consider a number λ ≥ 1 and functions (u, v) ∈ P suchthat

Φ(u, v) = λ(u, v), (30)

that is, −∆u+ u = λ−1a(|x|)f(u, v) in BR

−∆v + v = λ−1b(|x|)g(u, v) in BR

∂νu = ∂νv = 0 on ∂BR,

(31)

and we show that we obtain a positive lower bound (depending only on N , R,‖a‖1 and ‖b‖1) for σ defined by

σ2 := ‖u‖2 + ‖v‖2. (32)

From (19) of Lemma 2.3, applied to u and then to v, we obtain

‖u‖ ≤ λ−1C‖a · f(u, v)‖1,‖v‖ ≤ λ−1C‖b · g(u, v)‖1.

But u, v are nonnegative and f, g are nondecreasing in each variable, hence

‖a · f(u, v)‖1 ≤ f(‖u‖∞, ‖v‖∞) · ‖a‖1,‖b · g(u, v)‖1 ≤ g(‖u‖∞, ‖v‖∞) · ‖b‖1

which, plugged into the previous two inequalitites, yield

‖u‖ ≤ Cf(‖u‖∞, ‖v‖∞), ‖v‖ ≤ Cg(‖u‖∞, ‖v‖∞)

where now C depends also on a, b but not on λ, because λ ≥ 1. Using Lemma2.1, we see that

‖u‖∞ ≤ Cf(‖u‖∞, ‖v‖∞), ‖v‖∞ ≤ Cg(‖u‖∞, ‖v‖∞)

and hencef(‖u‖∞, ‖v‖∞) + g(‖u‖∞, ‖v‖∞)

‖u‖∞ + ‖v‖∞≥ 1C. (33)

Using again Lemma 2.1 and (32), we see that

‖u‖∞ + ‖v‖∞ ≤ Cσ,

12

but this shows that σ cannot be arbitrarily small, otherwise (33) combined withthe first condition in (H2) would give a contradiction.

Step 2 - proof of (ii). Choose p ∈ P as the pair of constant functions p = (1, 1).With this choice of p, we will prove that, if τ > 0 is large enough, then theequation

(u, v)− Φ(u, v) = λp

has no solution (u, v) ∈ S+τ , for every scalar λ ≥ 0. In concrete terms, we

assume that some (u, v) ∈ K ×K solve the system−∆u+ u = λ+ a(|x|)f(u, v) in BR

−∆v + v = λ+ b(|x|)g(u, v) in BR

∂νu = ∂νv = 0 on ∂BR

(34)

for some λ ≥ 0, and we obtain an upper bound for τ given by

τ2 := ‖u‖2 + ‖v‖2. (35)

Testing with η ≡ 1 the weak formulation of the two equations in (34) weobtain ∫

BR

u dx ≥∫BR

a f(u, v) dx,∫BR

v dx ≥∫BR

b g(u, v) dx,

since λ ≥ 0. Moreover, since u, v, are nondecreasing, from assumption (H1) wehave ∫

BR

u dx ≥ f(u(0), v(0)

)‖a‖1 and

∫BR

v dx ≥ g(u(0), v(0)

)‖b‖1.

As the two right hand sides in (34) are nonnegative, we may apply Lemma 2.2to both u and v, and in particular (15) gives∫

Br

u dx ≤ Cu(0),∫Br

u dx ≤ Cv(0). (36)

Plugging into the previous inequalities and adding, we obtain

C(u(0) + v(0)

)≥(f(u(0), v(0)) + g(u(0), v(0))

)min {‖a‖1, ‖b‖1} ,

which provides an upper bound for u(0) + v(0) (otherwise the second conditionin (H2) would give a contradiction). Due to (36) and (35), this gives an upperbound for τ as well.

Conclusion. We now take σ so small and τ so large that (i) and (ii) of Theorem3.1 are satisfied, and we set ρ = τ . Then Theorem 3.1 applied to Φ : P ρ → Pyields a fixed point (u, v) of the map Φ such that ‖u‖2 + ‖v‖2 = σ2 > 0, namelya solution of (1). The functions u and v are of course nonnegative and radiallynondecreasing. Finally, we note for further reference that

i(Φ, Pτ \ Pσ) = −1.

13

A few comments are in order. The assumptions used to prove Theorem 1.1,namely (A), (H1) and (H2), are fairly general, and contain many particularcases. For example, the functions a and b may be constant, the nonlinearitiesf and g may vanish on some “large” subset of R2, and so on. In view of theseconsiderations, one cannot expect to obtain all the time a solution (u, v) withu and v both not constant or both not identically vanishing. These cases areallowed provided the data of the problem have some degree of degeneracy. Wenow discuss some of these issues. The autonomous case (a and b identically con-stant) is particularly relevant and deserves a separate treatment. We postponeits discussion to Section 4.

Theorem 1.1 yields a solution (u, v) with u and v not both identically zero,since ‖u‖2 + ‖v‖2 > 0. We describe some particular cases where more can besaid.

Suppose that a and b are both nonconstant. Then at least one among u andv is not constant. Indeed, if u ≡ c1 and v ≡ c2, then{

c1 = a(|x|)f(c1, c2)c2 = b(|x|)g(c1, c2).

By (A) and (H1) we must have f(c1, c2) = g(c1, c2) = 0, and then c1 = c2 = 0,which is ruled out by Theorem 1.1.

Suppose again that a and b are both nonconstant. Then u and v are bothnonconstant unless f (or g) vanishes on a segment. Indeed, assume for instancethat u ≡ c1; then the system reads{

c1 = a(|x|)f(c1, v(x))−∆v + v = b(|x|)g(c1, v(x)).

(37)

This implies (by (A) and (H1)) that f(c1, v(x)) ≡ 0, and then c1 = 0. Since nowv cannot vanish identically too, by (H1) it must be f = 0 on {0} × [0, ‖v‖∞].

Suppose now that a and b are not both constant. Then, as in the previouscase, solutions with at least one constant component are allowed only if f (or g)is constant on a segment. For example, assume that (c1, v(x)) solves the system,namely (37). If a is not constant, then as above, c1 = 0, v 6≡ 0 and f vanisheson the segment {0}× [0, ‖v‖∞]. On the other hand, if a is constant, then eitherv is not constant, in which case f is constant on {c} × [min v, ‖v‖∞], or v ≡ c2.Since necessarily b is not constant, then, as above, g(c1, c2) = 0, and therefore,by (H1), g vanishes on the segment {c1} × [0, c2].

Summing up, if f and g are for example strictly increasing in each variable,then if a or b are not constant, we deduce that u and v are both non constant.Similar and in fact simpler phenomena, which we do not describe, occur in theHamiltonian case f(u, v) = f(v), g(u, v) = g(u).

Let us also discuss the superlinearity assumption at infinity. One can in factweaken assumption (H2) at infinity in several ways. We will only discuss the

14

special case where f(u, v) = f(v), g(u, v) = g(u). Other variants can be de-rived. The following theorem extends [3, Theorem 1.1] to separate Hamiltoniansystems.

Theorem 3.3. Assume that f and g are nonnegative, nondecreasing and satisfy

lims→0+

f(s)s

= lims→0+

g(s)s

= 0.

Assume also that (A) holds with a0 := a(0) > 0 and b0 := b(0) > 0 and that

lim infs→+∞

f(s)s

>1a0

and lim infs→+∞

g(s)s

>1b0. (38)

Then the problem −∆u+ u = a(|x|)f(v) in BR

−∆v + v = b(|x|)g(u) in BR

∂νu = ∂νv = 0 on ∂BR.

(39)

admits at least one solution with u and v both nonnegative, non constant andnondecreasing.

Proof. The argument is the same as for the proof of Theorem 1.1 except in theway to deduce that the solutions of (34) for λ ≥ 0 are a priori bounded fromabove.

First, observe that (34) does not admit any solution in P for λ > λ forsome λ > 0. Moreover, there exists a constant M > 0 such that every solution(u, v) of (34) with 0 ≤ λ ≤ λ satisfies max(‖u‖L1(B), ‖v‖L1(B)) ≤ M . Fromthese L1 bounds, we then deduce L1 bounds on the derivatives of u and v andthereafter L∞ bounds from the fact that u, v ∈ K. The H1 bounds follow fromthe equations.

4 The autonomous case

In the autonomous case a(|x|) ≡ b(|x|) ≡ 1, a further difficulty arises due to thepresence of constant solutions. In other words, it is not clear that Theorem 1.1gives a noncontant solution (see [3] for a discussion of this aspect in the case ofa single equation).

In this section we indicate which assumptions on f and g guarantee theexistence of nonconstant solutions in the autonomous case. It turns out thatthese assumptions are a quite natural generalization of those found in [3] in thescalar case.

For simplicity we assume that the system has only one constant nontrivialsolution. Extensions to the case of a finite number of constant solutions can beeasily obtained by straightforward modifications of our argument.

The main point consists in calculating the local fixed point index of theconstant solution to derive a contradiction with Theorem 1.1.

15

4.1 Abstract setting

The computation of the local index of a fixed point of a map between cones hasbeen carried out by Dancer in [5] and [6] in an abstract setting. Since we aregoing to use the results from [5] and [6], we first recall them here in the sameabstract setting and then we show how our concrete case fits into the generalscheme.

Whenever possible we use in this abstract introduction the same notation asin Dancer’s papers.

Let E be a real Banach space and let W be a wedge in E, namely a closed,convex subset of E such that αW ⊂ W for every α ≥ 0. Recall that a wedgeW is called a cone if it also satisfies W ∩ −W = {0}.

To apply the abstract results it is necessary to assume that

W −W is dense in E. (40)

Definition 4.1. Let W be a wedge satisfying (40), and let y ∈W . We define

Wy = {x ∈ E | ∃γ > 0 such that y + γx ∈W}

andSy = {x ∈W y | − x ∈W y}.

Note (for all details we refer the reader to [5]) that the set Wy is convex,contains W and ±y, and αWy ⊂ Wy for every α ≥ 0. Thus W y is a wedgecontaining W and ±y.

Concerning Sy, it can be easily proved that it is a closed subspace of Econtaining y.

Still following [5] we introduce the following notion.

Definition 4.2. We say that a compact operator L : E → E mapping W y intoitself has property α if

there exist t ∈ (0, 1) and w ∈W y \ Sy such that w − tLw ∈ Sy.

In [5] it is shown that the validity of property α is strongly related to thespectral radius of L.

We can now turn to the statement of the main abstract result.Let Φ : W → E be a (nonlinear) map satisfying

(A1) Φ is completely continuous,(A2) Φ(W ) ⊂W ,(A3) Φ(y) = y,(A4) Φ is differentiable at y “in W” (see [5]),(A5) Φ′(y) =: L is compact from E to E.

Under these assumptions it can be proved that L maps W y into W y.Denoting by

iW (Φ, y) (41)

16

the local fixed point index of y in W , see for example [8], the results by Dancerthat we need, precisely Theorem 1 in [5] and Proposition 1 in [6] can be collectedin a single statement as follows.

Theorem 4.3 (Dancer, [5]–[6]). Let E be a Banach space and let W ⊂ E be awedge satisfying (40). Let Φ : W → E satisfy (A1)–(A5). Then the followingstatements hold.

(i) If I − L is invertible and L has property α, then iW (Φ, y) = 0.

(ii) If I −L is not invertible but ker (I −L)∩W y = {0}, then iW (Φ, y) = 0.

We are going to apply this result to the map Φ associated to the autonomoussystem (8) as in (28). Here of course a(|x|) ≡ b(|x|) ≡ 1.

4.2 Local index of the constant solution

We will denote by1 = λ1 < λ2 < λ3 ≤ . . .

the radial eigenvalues of −∆ + I with Neumann boundary conditions. It is wellknown that the corresponding eigenfunctions ϕk are obtained by scaling

ϕk(x) = J((λk − 1)|x|/R),

where the function J is the solution of the Bessel–type differential equation

t2J ′′(t) + (N − 1)tJ ′(t) + t2J(t) = 0, J(0) = 1, J ′(0) = 0.

The numbers λk − 1 are the nonnegative zeros of J ′(t), which is oscillating. Inparticular we point out that ϕk is radially monotone if and only if k ∈ {1, 2}.

To keep notation coherent with the first part of this section, we let, as inthe preceding section,

E = H1r (BR)×H1

r (BR),

and we define the wedgeW = K ×K (42)

where K is the cone of nonnegative radially nondecreasing functions defined in(13). Of course in this case W is not just a wedge, but a true cone and indeedit coincides with the cone P used in the preceding section.

To compute the local index of (u0, v0) we are going to use Theorem 4.3,which amounts to check that its assumptions hold true in our setting.

First of all, it is easy to see that W satisfies (40); this is a consequence ofthe fact that K −K is dense in H1

r (BR), as we now show.Let u ∈ H1

r (BR); for every ε > 0 there exists uε ∈ C1(BR) such that‖u− uε‖ < ε. Define

ϕε(r) = uε(0) +∫ r

0

(u′ε)+(t) dt, ψε(r) =

∫ r

0

(u′ε)−(t) dt

17

if uε(0) ≥ 0 and

ϕε(r) =∫ r

0

(u′ε)+(t) dt, ψε(r) =

∫ r

0

(u′ε)−(t) dt− uε(0)

if uε(0) < 0. Then ϕε, ψε ∈ K and uε(r) = ϕε(r)− ψε(r) for every r. Finally,

‖u− (ϕε − ψε)‖ = ‖u− uε‖ < ε.

To have lighter notation from now on we denote by y the (unique) nontrivialconstant solution (u0, v0) of Problem (8). As prescribed by Defintion 4.1 weobserve that

Wy = {(u, v) ∈ E | u, v are both bounded and nondecreasing}

andSy = {(u, v) ∈ E | u, v are both constant}. (43)

Notice that Wy is not closed in H1r (BR), and that

W y = {(u, v) ∈ E | u, v are both nondecreasing}. (44)

We begin by checking that properties (A1)–(A5) hold for Φ. The first twoproperties have already been proved in Section 3; the third is the assumptionthat y = (u0, v0) is a solution of Problem (8). Finally, (A4) is standard and(A5) has also been proved in Section 3, as well as the fact that Φ′(u0, v0) mapsW y into itself.

We let L = Φ′(u0, v0) be the differential of Φ at (u0, v0) in W , namely themap from E to E defined by

L(ϕ,ψ) = (u, v) ⇐⇒

−∆u+ u = fu(u0, v0)ϕ+ fv(u0, v0)ψ in BR

−∆v + v = gu(u0, v0)ϕ+ gv(u0, v0)ψ in BR

∂νu = ∂νv = 0 on ∂BR,

or

L(ϕ,ψ) = (u, v) ⇐⇒

(−∆ + I)

(u

v

)= M

(ϕ

ψ

)in BR

∂νu = ∂νv = 0 on ∂BR,

where M is the matrix defined in (10).

Remark 4.4. With some abuse of notation, we denote by the same symbolboth the 2 × 2 matrix M and the operator induced by M in the product spaceE = H1

r (BR)×H1r (BR) in the natural way. This convention will be adopted for

every 2× 2 matrices.

In order to apply Theorem 4.3 it remains to check that the assumptions ofTheorem 1.2 imply assertion (i) or (ii) of Theorem 4.3.

Recall that we denote by λ and λ the two eigenvalues of M , and by λ1 <λ2 ≤ λ3 < . . . the Neumann eigenvalues of −∆ + I on H1

r (BR).

18

Lemma 4.5. The spectrum of L is given by

σ (L) ={λλ−1

j

}∞j=1∪{λλ−1

j

}∞j=1

. (45)

In particular, the operator I − L : E → E is invertible if and only if

λ, λ 6∈ {λ1, λ2, λ3, . . . }.

Proof. Let U be an orthogonal matrix such that U tMU is upper triangular,namely

U tMU =(λ a12

0 λ

), a12 ∈ R.

Recalling Remark 4.4, one can see that

U tLU = U t(

(−∆ + I)−1 00 (−∆ + I)−1

)MU

=(

(−∆ + I)−1 00 (−∆ + I)−1

)U tMU =

(λ(−∆ + I)−1 a12(−∆ + I)−1

0 λ(−∆ + I)−1

),

and (45) follows.

Lemma 4.6. Assume that I − L is invertible. Then L has property α if andonly if

λ > λ2. (46)

Proof. Rephrasing Definition 4.2 in the light of (43) and (44), we see that prop-erty alpha for L reduces to the following statement: there exist t ∈ (0, 1) andtwo functions (w1, w2) ∈ E, both nondecreasing but not both constant, such that(−∆ + I)

(w1

w2

)= tM

(w1

w2

)+

(c1

c2

)in BR

∂νw2 = ∂νw2 = 0 on ∂BR,

(47)

where c1, c2 are constant functions.

Part I - the sufficient condition. Assuming (46) holds, we let ϕ2 denote thesecond eigenfunction for the Neumann problem in H1

r (BR), that is

−∆ϕ2 + ϕ2 = λ2ϕ2 in BR, ∂νϕ2 = 0 on ∂BR. (48)

As already emphasized, we may assume that ϕ2 is nondecreasing. Now consideran eigenvector of the matrix M relative to the largest eigenvalue λ, namely

M

(x1

x2

)= λ

(x1

x2

). (49)

From the Perron–Frobenius Theorem, x1, x2 ≥ 0 and hence the two functionsw1 = x1ϕ2, w2 = x2ϕ2 are nondecreasing (and not both constant). Now using(48) we see that

(−∆ + I)(w1

w2

)= (−∆ + I)

(x1ϕ2

x2ϕ2

)= λ2

(x1ϕ2

x2ϕ2

)

19

From (46), we may write λ2 = tλ for some t ∈ (0, 1), and recalling (49) fromthe last equation we obtain that

(−∆ + I)(w1

w2

)= tM

(w1

w2

). (50)

Since w1, w2 inherit from ϕ2 the Neumann condition along ∂BR, they solve asystem of the kind (47) (with c1 = c2 = 0), and hence L has property alpha.

Part II - the necessary condition. Here we show that (46) follows from (47).Note that, if we add any two constants to w1 and w2, the resulting functions(which remain nondecreasing and not both constant) solve a system that hasexactly the same form of (47), thus we may assume that∫

BR

w1 dx =∫BR

w2 dx = 0. (51)

With this condition, integrating over BR both equations in (47) wee see thatc1 = c2 = 0.

Now suppose that w1 ≡ 0. In this case, labelling the entries of M as in (10),we see that (47) reduces to

0 = tm12w2 in BR

−∆w2 + w2 = tm22w2 in BR

∂νw2 = 0 on ∂BR.

(52)

As w2 6≡ 0 (otherwise w1, w2 would be two constant functions), the first equationreveals that m12 = 0, hence M is a triangular matrix and therefore m22 is aneigenvalue of M . Thus, from the second equation, letting v = w2 and recalling(51), we have found a function v 6≡ 0 satisfying∫

BR

v dx = 0, −∆v + v = λv in BR, ∂νv = 0 on ∂BR (53)

for some λ ∈ {tλ, tλ}. Therefore, λ is a Neumann eigenvalue in BR, and thefirst condition on v reveals that λ 6= λ1. In other words, we have that

λ2 ≤ λ ≤ max{tλ, tλ} = tλ,

and λ > λ2 follows since t ∈ (0, 1).The case where w2 ≡ 0 (and w1 6= 0) can be treated in exactly the same

way.Finally, consider the (generic) case where neither w1 nor w2 is identically

zero. Let (x1, x2) be a left eigenvector of the matrix M corresponding to λ, thatis,

(x1, x2)M = λ(x1, x2).

If we multiply the first equation in (47) by x1, the second by x2 and take thesum, recalling that c1 = c2 = 0 we see that the function v = x1w1 +x2w2 solvesthe Neumann problem (53), with λ = tλ.

20

Then the condition that λ > λ2 is obtained arguing as in the previous cases,observing that v 6≡ 0. Indeed, from the Perron–Frobenius Theorem we have thatx1, x2 ≥ 0 (and of course x2

1 + x22 > 0). If, say, x1 > 0, then v ≡ 0 means that

w1 = −x2/x1w2 is nonincreasing, hence constant. Then also x2w2 is constantand, since w2 cannot be constant too, we must have that x2 = 0. But then alsow1 ≡ 0, contrary to our initial assumptions.

Lemma 4.7. Assume that I − L is not invertible. If

λ > λ2 and λ 6∈ {λ1, λ2}, (54)

then ker (I − L) ∩W y = {0}.Proof. Consider w ∈ ker (I −L)∩W y. By (44), this means that w = (w1, w2)with w1, w2 nondecreasing functions such that(−∆ + I)

(w1

w2

)= M

(w1

w2

)in BR

∂νw2 = ∂νw2 = 0 on ∂BR.

(55)

Now the argument is the same as in part I of the previous proof.Consider first the case where w1 ≡ 0. Then (55) reduces to (52), now

rewritten with t = 1. If also w2 ≡ 0 then the proof is complete, otherwise asbefore, we infer from (52) with t = 1 that m12 = 0, hence M is lower triangularand m22 is an eigenvalue of M , as well as of the Neumann Laplacian with w2

as coresponding eigenfunction. Since w2 is nondecreasing, this can only happenif m22 is either λ1 or λ2. But this is in contrast with (54), hence also w2 ≡ 0.

The case where w2 ≡ 0 is similar.Finally, if w1 6≡ 0 and w2 6≡ 0, we multiply the first equation in (55) by x1

and the second by x2, where (x1, x2) is a left eigenvector of M correspondingto the eigenvalue λ, and we sum the two equations. As before, letting

v = x1w1 + x2w2, (56)

we see that v satisfies

−∆v + v = λv in BR, ∂νv = 0 on ∂BR.

As x1, x2 ≥ 0, we must have that v ≡ 0, otherwise v would be a nondecreasingeigenfunction and hence, as before, λ would be either λ1 or λ2, a contradiction.Now if x1 = 0 (and hence x2 > 0), we see using (56) and v ≡ 0, that w2 ≡ 0, acontradiction. Similarly, we can exclude that x2 = 0, hence x1 > 0 and x2 > 0.But then

w1 = −x2

x1w2, w2 = −x1

x2w1,

and each of w1, w2 (being both nondecreasing and nonincreasing) must beconstant. From (55), we see that the (non zero) vector w = (w1, w2) solvesMw = w, hence 1 = λ1 is an eigenvalue of M , which is in contrast with (54).Summing up, the assumption that w1 6≡ 0 and w2 6≡ 0 leads to a contradiction.

21

4.3 Proof of Theorem 1.2

Wa have now all the ingredients to prove Theorem 1.2.Theorem 1.1 provides a solution of (8) as a fixed point of the map Φ in the

set Pτ \ Pσ, for some suitable 0 < σ < τ , through the application of Theorem3.1. The same theorem shows that in our assumptions

i(Φ, Pτ \ Pσ) = −1.

Now if the constant solution (u0, v0) is the only (nontrivial) solution, it is wellknown ([8]) that the above index equals the local index of (u0, v0) (introducedin (41)):

i(Φ, Pτ \ Pσ) = iW (Φ, (u0, v0)). (57)

The proof of Theorem 1.2 reduces then to showing that the local index of theconstant solution is zero. If this is proved, then there must be another solutionin P , which is necessarily non constant.

Proof of Theorem 1.2. Assume that Problem (8) has no nonconstant solutions.Then, as pointed out in (57), and by Theorem 3.1,

−1 = i(Φ, Pτ \ Pσ) = iW (Φ, (u0, v0)).

Now if I − L = I − Φ′(u0, v0) is invertible, then the assumption λ > λ2 andLemma 4.6 show that I − L has property α. Therefore, by (i) of Theorem 4.3,

iW (Φ, (u0, v0)) = 0,

a contradiction.If I−L is not invertible, then (9) and Lemma 4.7 show that (ii) of Theorem

4.3 holds true. TheniW (Φ, (u0, v0)) = 0,

again a contradiction. Thus there must be another solution of the system in P ,which is necessarily nonconstant, since (u0, v0) is the only constant solution.

Remark 4.8. In the statement of the Theorem 1.2, “non constant” means ofcourse that at least one component of the solution (u, v) is not constant. It iseasy to see that one constant component is possible only if f or g are in somesense degenerate (e.g. constant on a segment). It is just as easy to find simpleconditions that prevent this phenomenon. Since these conditions have alreadybeen discussed at the end of Section 3, we do not describe them again here.

As mentioned in the Introduction, for the case of the Hamiltonian system−∆u+ u = f(v) in BR

−∆v + v = g(u) in BR

∂νu = ∂νv = 0 on ∂BR,

(58)

the statement of Theorem 1.2 is simpler.

22

Corollary 4.9. Assume that f and g are nonnegative, differentiable, nonde-creasing and satisfy the superlinearity conditions

lims→0+

f(s)s

= lims→0+

g(s)s

= 0.

and

lims→+∞

f(s)s

= lims→+∞

g(s)s

= +∞.

Assume also that (u0, v0) is the only nontrivial constant solution. If

f ′(v0)g′(u0) > λ22, (59)

then problem (58) admits at least one solution with u and v both nonnegative,non constant and nondecreasing.

Proof. As f(u, v) = f(v) and g(u, v) = g(u), the Jacobian in the statement ofTheorem 1.2 is

M =(

0 f ′(v0)g′(u0) 0

).

Therefore, the eigenvalues of M are ±√f ′(v0)g′(u0). It only remains to show

that if (u, v) is the solution found with Theorem 1.2, then u and v are both nonconstant. To see this assume for example that u ≡ c. Then the second equationreads

−∆v + v = g(c), ∂νv = 0.

Since this equation has only one solution, v must be constant too, which con-tradicts Theorem 1.2.

Remark 4.10. It can also be of interest to compare our results to those con-cerning a single equation. This case is of course contained in Theorem 1.2and corresponds to the choice f(u, v) = f(u), g(u, v) = f(v). In this settingthe eigenvalues are λ = λ = f ′(u0), so that the main condition (9) reduces tof ′(u0) > λ2. This is exactly the condition used in [3].

Remark 4.11. As a final remark and as discussed in Theorem 3.3, we canweaken the super linearity condition at infinity in Corollary 4.9 by assuming(59).

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