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Neumann problems of superlinear elliptic systems at resonance

Ruyun Ma

1, 2

, Zhongzi Zhao

B1

and Mantang Ma

2

1School of Mathematics and Statistics, Xidian University, Xi’an 710071, P. R. China

2Department of Mathematics, Northwest Normal University, Lanzhou 730070, P. R. China

Received 12 October 2021, appeared 2 February 2022 Communicated by Dimitri Mugnai

Abstract. We prove existence of weak solutions of Neumann problem of nonhomo- geneous elliptic system with asymmetric nonlinearities that may resonant atand superlinear at+∞. The proof is based on Mawhin’s coincidence theory and the product formula of Brouwer degree.

Keywords: elliptic equation, Neumann problem, weak solution, continuation methods.

2020 Mathematics Subject Classification: 35J25, 35J60, 47H11.

1 Introduction

Let Ω ⊂ RN (N ≥ 2) be a smooth bounded connected domain in real N-dimensional Eu- clidean space. We are concerned with the existence of weak solutions of the following Neu- mann problem of semilinear elliptic systems

∆u+ f(v) =h1(x), in Ω,

∆v+g(u) =h2(x), in Ω,

∂u

∂ν = ∂v

∂ν =0, on ∂Ω,

(1.1)

where f,g : RR are continuous functions, ∂ν denotes the outward normal derivative on

∂Ω, the boundary ofΩ, andh1,h2 ∈ L1().

The motivation for this work is the paper F. O. de Paiva, W. Rosa [12], in which the authors showed the following resonant Neumann problems

∆u= (v+)p+h1(x), inΩ,

v= (u+)q+h2(x), inΩ,

∂u

∂ν = ∂v

∂ν =0, on∂Ω

(1.2)

BCorresponding author. Email: [email protected]

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has at least one solution (u,v) in H1()×H1() under the assumptions h1,h2 ∈ Lr(), r> N2, 1< p,q< NN2 and

Z

hi(x)dx<0, i=1, 2. (1.3) We first define the bilinear form associated with the Laplacian operator. For u,v ∈ W1,1(),φ,ψ∈W1,(), we defineB1(u,φ)andB2(v,ψ)by

B1(u,φ) =−

N i=1

Z

∂u

∂xi

∂φ

∂xidx,

B2(v,ψ) =−

N i=1

Z

∂v

∂xi

∂ψ

∂xidx,

where the derivatives are taken in the distributional sense. By aweak solutionof (1.1), we mean a pair(u,v)∈W1,1()×W1,1(), such that f(v(·))∈L1(),g(u(·))∈ L1()and

B1(u,φ) +

Z

f(v)φdx =

Z

h1(x)φdx,φ∈W1,∞(), B2(v,ψ) +

Z

g(u)ψdx =

Z

h2(x)ψdx,ψ∈W1,∞(). Denote

f =lim sup

s→−

f(s), g =lim sup

s→−

g(s), f+ =lim inf

s→+ f(s), g+ =lim inf

s→+ g(s). We will make the following assumptions.

(C0) h1,h2 ∈ L1().

(Cl) There are the nonnegative constantsC1,C2 ∈(0,∞)such that f(t)≥ −C1, g(t)≥ −C2, t∈ R and for allt ≤0 we have also|f(t)| ≤C1,|g(t)| ≤C2.

(C2) There are the constants a,b ∈ R and p with 1 ≤ p < N/(N−2) for N ≥ 3 and 1≤ p<forN=2 such that for allt ≥0 the inequality

|f(t)|,|g(t)| ≤atp+b a.e. onΩ.

(C3) We assume f,gtends to be nondecreasing in that there is aγRand a numberM ≥0 such that the inequalities

f(t1)≤ f(t2) +γ, g(t1)≤g(t2) +γ hold a.e. onΩwhenevert2−t1≥ M.

(C4)

Z

f <

Z

h1(x)dx<

Z

f+, Z

g <

Z

h2(x)dx<

Z

g+.

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Our main result is the following

Theorem 1.1. Under assumptions (C0)–(C4) the Neumann problem(1.1)has a weak solution(u,v)∈ W1,1()×W1,1(). Moreover the solution(u,v)∈W1,q()×W1,q()for all1≤q<N/(N−1). Remark 1.2. Obviously, (1.3) in F. O. de Paiva, W. Rosa [12] are the special case of (C0) and (C4).

Remark 1.3. Our proof is based upon ideas found in Ward Jr [16]. He used the well-known Mawhin’s continuation theorem to get a weak solution of the scale elliptic equation

∆u+ f(x,u) =k(x), inΩ,

∂u

∂ν =0, on∂Ω (1.4)

under the conditionsk∈ L1(),

|f(x,t)| ≤α(x)|t|p+β(x), x∈ Ω,

where p ∈[1,NN2),α∈ L(),β∈ L1(), and Landesman–Lazer condition Z

f <

Z

k(x)dx<

Z

f+.

Remark 1.4. Similar problems, under Dirichlet and Neumann boundary condition, can be found in D. Arcoya and S. Villegas [2], M. Cuesta and C. De Coster [3], F. M. Ferreira, F. O. de Paiva [4], R. Kannan and R. Ortega [6,7], S. Kyritsi and N. S. Papageorgiou [8], D. Motreanu, V. Motreanu, N. S. Papageorgiou [10], K. Perera [14], N. S. Papageorgiou and V. D. R˘adulescu [13], F. O. de Paiva and A. E. Presoto [11], L. Recova and A. Rumbos [15], J. R. Ward [16].

2 The preliminaries

Before proving Theorem 1.1 we will need a lemma. In the following we will write Lp for Lp() andW1,p forW1,p(). We denote the norm in Lp by| · |p, that of W1,p by| · |1,p. For h∈ L1. LetQhbe the projection

Qh =||1

Z

hdx.

Lemma 2.1([16]). For each h∈ L1()with Qh=0. There is a unique w∈W1,1()with Qw=0 such that

B(w,φ) =

Z

h(x)φdx, for all φ ∈ W1,∞, where B(w,φ) = −Ni=1R

∂w

∂xi

∂φ

∂xidx. Moreover w ∈ W1,q for each q satisfying 1≤q< N/(N−1)and there is a constant C(q)such that

|w|1,q≤C(q)|h|1.

By the Rellich–Kondrachov theorem W1,q is compactly imbedded in Lp for 1 ≤ p < NNqq since q < N/(N−1)≤ N for all N ≥ 2. (e.g. see [1, p. 144]). Assume that the number pin condition (C2) is fixed hereafter, satisfying 1 ≤ p < N/(N−2)if N ≥ 3 and 1 ≤ p < if N=2.

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Chooseqso that

p< Nq

N−q and 1<q< N N−1. We have thatW1,qis compactly imbedded in Lp.

LetX1 denote the closed subspace ofL1defined by h∈X1 if and only if Qh=0.

LetTdenotes the operator mappingX1intoW1,q∩X1given byh →wwherewis the unique weak solution to

∆w=h inΩ, Qu=0,

∂w

∂ν

=0 onΩ.

Note that W1,q = (W1,q∩X1)⊕R. T maps X1 intoW1,q and we see thatΨ◦T maps X1 compactly into LpifΨ is the imbedding ofW1,q intoLp. Let

K= Ψ◦T,

and define an operatorL: Lp→ L1. BecauseL1is not the dual space to L, we do not use the usual method of definingL. Instead, we let

D(L) =RangeK⊕R and

L(w1+α˜) =h,

whereh∈X1andKh= w1, forw1 ∈RangeKand ˜αR. It is easy to see thatLis a Fredholm operator: it has closed rangeX1and since ker(L) =Rand the dimension of L1\X1is clearly 1, the index ofLis 0,

index(L) =dim kerL−dim cokerL.

We now define the substitution operatorsN1,N2: Lp→ L1 by N1v(x) = f(v(x))−h1(x), v∈ Lp andx∈ Ω.

N2u(x) =g(u(x))−h2(x), u∈Lpandx∈ .

It is well known that the conditions on f andgimply thatNjmapsLpintoL1continuously andNj obviously takes sets bounded inLp into sets bounded in L1for j=1, 2.

A function (u,v) ∈ W1,1×W1,1 is a weak solution of (1.1) if and only if (u,v) ∈ D(L)× D(L)and

Lu+N1v=0,

Lv+N2u=0. (2.1)

Recalling that for u ∈ L1 we have defined Qu to be the mean value of u, we have from our remarks above that K(I −Q)Nj : Lp → Lp is compact and continuous, clearly QNj is also compact and continuous for j = 1, 2. Thus Nj is L-compact (see [5]) on ¯G for any open bounded set ¯G in Lp forj= 1, 2. We will use a well known continuation theorem of Mawhin (see [5] and [9]).

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3 Proof of the main result

We are in the position to prove our main result.

Proof of Theorem1.1. By one of Mawhin’s continuation theorems (see [5, p. 40] or [9, The- orem 7.2]) and our remarks above, if we can show the existence of a bounded open set G := G¯ ×G¯ in Lp×Lp such that conditions (i) and (ii) below hold, then (2.1) has a solu- tion. The conditions are:

(i) For eachλ∈(0, 1)and each(u,v)∈(D(L)×D(L))∩∂G, Lu+λN1v̸=0,

Lv+λN2u̸=0. (3.1)

(ii) QNjw̸=0 for eachj=1, 2,w∈kerL∩G¯ and

d(Γ,G∩(kerL×kerL), 0)̸=0,

where Γ := (JQN1,JQN2), J : ImQ → kerL is an isomorphism, andd is the Brouwer topological degree.

We first verify (i). We consider

Lu+λN1v=0,

Lv+λN1u=0 (3.2)

for 0<λ<1. If((u,v),λ)is a solution of (3.2) then B1(u,φ) +λ

Z

f(v)φ=λ Z

h1φ,φ∈W1,, B2(v,ψ) +λ

Z

g(u)ψ=λ Z

h2ψ,ψ∈W1,∞. In particular by taking φ=ψ=1, then

Z

f(v) =

Z

h1, Z

g(u) =

Z

h2. It follows from (Cl) that for eacht ∈R

|f(t)| ≤ f(t) +2C1, |g(t)| ≤g(t) +2C2. Thus

|N1v|1=

Z

|f(v)−h1(x)|dx

Z

f(v) +2C1+|h1(x)|dx

Z

h1dx+2|C1| · ||+

Z

|h1(x)|dx=:d1,

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|N2u|1 =

Z

|g(u)−h2(x)|dx

Z

g(u) +2C2+|h2(x)|dx

Z

h2dx+2|C2| · ||+

Z

|h2(x)|dx =:d2.

Writingu=u1+α, v=v1+βwithu1,v1 ∈RangeKandα,βRby Lemma2.1we have

|u1|1,q≤C(q)d1 =:m1,

|v1|1,q≤C(q)d2=:m2,

wherem1andm2are independently ofλ∈ (0, 1). By the Sobolev imbedding theorem

|u1|p ≤m3, |v1|p ≤m4 for some constantsm3 andm4.

We now show that for solutions ((u,v),λ) = (u1+α,v1+β),λ

that α and β are also bounded independently ofλ∈ (0, 1).

Suppose this is not the case. Then there is a sequence ((un,vn),λn) of solutions to (3.2) with

un= u1n+αn, vn=v1n+βn

and

|αn|+|βn| →∞, asn→∞.

Suppose first that a subsequence of {αn}, relabeled as {αn}, tends to +∞. Then using

|u1n|1,q≤ m1is easy to show that

nlimun(x) = + a.e. (3.3) For otherwise there is a constantk1>0 and sets Ω(n)inΩfor infinitely manyn(without loss of generality we assume for alln) such that|(n)| ≥δ> 0 andun(x)≤k1 forx∈(n). We haveu1n+αn≤k1implies

k1|| ≥

Z

(n)k1dx≥

Z

(n)u1n+αndx

αn|(n)| −

Z

|u1n|

αnδ−C

forC, a constant, which contradictsαn→∞. Thus (3.3) holds and lim inf

n g(un(x)) = g+ a.e.

Sinceg(un(x))≥ −C2for all nandC2Rwe have by Fatou’s lemma Z

h2=lim inf

n Z

g(un(x))dx ≥

Z

g+dx which contradicts (C4). Thus the{αn}must be bounded above.

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Supposeαn→ −∞. It follows as for (3.3) that

nlimun(x) =− a.e.

Because g(t)is not everywhere bounded above by an L1 function, we cannot use the simple Fatou’s lemma argument as in the case ofαn → −∞.

We proceed as follows. Since|u1n|1,q≤ m1, we may without loss of generality assume the existence of ˜u1 ∈ Lp such thatu1n→u˜1 inLp.

Let 0 < ϵ < ||be given. Then ˜u1 ∈ Lp implies that there exists an integer n(ϵ) and a measurable set E⊆such that ifF=−Ethen |F|<ϵand

un(x)≤0, x ∈E, n≥n(ϵ), hence

g(un(x))≤ C2, x ∈E, n≥n(ϵ).

Moreover there exists another integermsuch that forn≥ mwe have αn≤ −M, where M is a positive constant.

Thus, forn≥max{n(ϵ),m}, Z

h2 =

Z

Eg(u1n+αn) +

Z

Fg(u1n+αn)

Z

Eg(u1n+αn) +

Z

Fg(u1n) +

Z

Fγ

and

Z

h2 ≤lim sup

n

Z

Eg(un) +

Z

Fg(u1n)

+

Z

F

γ

Z

Egdx+

Z

Fg(u˜1)dx+

Z

Fγ

(3.4)

by Fatou’s lemma for the integral overEand by convergence inL1for the integral over F.

Now chooseη>0 such that Z

gdx+η<

Z

h2dx. (3.5)

We may chooseϵ>0 so small that, since|F|< ϵ,

Z

Fgdx

< η 3,

Z

Fg(u˜1)dx

< η 3,

Z

Fγ

< η 3. For such asϵwe have from (3.4) and (3.5)

Z

h2

Z

gdx−

Z

F

gdx+

Z

F

g(u˜1)dx+

Z

F

γ

Z

gdx+η<

Z

h2. (3.6) Therefore we cannot have αn →+or αn→ − and this, combined with|u1|p ≤m3 shows that if ((u,v),λ) is a solution of (3.2) then |u|p = |u1+α|p ≤ m3+C3 for some constant C3. Similarly, We can obtain|v|p=|v1+α|p≤ m4+C4 for some constantC4.

This verifies condition (i) above for any ballG in L1×L1, centered at the origin and with radius larger thanρ1=max{m3+C3,m4+C4}.

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The verification of condition (ii) is now straightforward. Both the range of Q and the kernel ofL may be identified withR, so that the isomorphism J in (ii) we may take to be the identity onR. Now for α,βR,

QN1β= ||1

Z

h

f(β)−h1(x)idx, QN2α=||1

Z

h

g(α)−h2(x)idx.

We may now make two simple applications of Fatou’s lemma using (Cl) to show, using (C4), that there exists anr>0 such that

QN1(β)>0, QN1(−β)<0, forα>r, QN2(α)>0, QN2(−α)<0, for β> r.

Thus for ¯r ≥rmax{1,||},

d(QNj,[−r, ¯¯ r]∩kerL, 0)̸=0, j=1, 2.

By the product formula of Brouwer degree, we obtain

d(Γ,[−r, ¯¯ r]2∩(kerL×kerL), 0)̸=0.

Now let ρ := max{ρ1,r·max{1,||}}. Then we have that both (i) and (ii) are satisfied on [Bρ]2, whereBρ is the ball inLpwith radius ρcentered at the origin. Thus (2.1) has a solution (u,v)∈ D(L)×D(L)with

|u|pρ, |v|pρ,

and(u,v) ∈ W1,p×W1,p and is a weak solution of (1.1). This completes the proof of Theo- rem1.1.

Acknowledgements

The authors are very grateful to the anonymous referees for their valuable suggestions.

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