194

Nonlinear Analysis: Modelling and Control, 2012, Vol. 17, No. 2, 194–209

Existence of non-negative solutions for semilinear elliptic systems via variational methods Somayeh Khademlooa , Shapur Heidarkhanib a

Department of Basic Sciences, Babol University of Technology 47148-71167, Babol, Iran [email protected] b Department of Mathematics, Faculty of Sciences, Razi University Kermanshah 67149, Iran [email protected] Received: 19 June 2011 / Revised: 13 February 2012 / Published online: 21 May 2012

Abstract. In this paper we consider a semilinear elliptic system with nonlinearities, indefinite weight functions and critical growth terms in bounded domains. The existence result of nontrivial nonnegative solutions is obtained by variational methods. Keywords: variational methods, Robin boundary condition, first eigenvalue.

1

Introduction

In this paper we consider the existence results of the following two coupled semilinear equation
−∆u = λau − pav 2 u|u|p−2 + 2au|v|p , x ∈ Ω,
−∆v = λav − pau2 v|v|p−2 + 2av|u|p , x ∈ Ω, (1) ∂u ∂v (1 − α) + αu = (1 − α) + αv = 0, x ∈ ∂Ω, ∂n ∂n where α and λ are real parameters, p < 2∗ − 2, for 2∗ = N2N −2 , Ω is an open bounded domain in RN , N ≥ 3 with a smooth boundary ∂Ω, and a : Ω → R is a sign changing weight function. This work is motivated by the results in the literature for the single equation case, namely the equation of the form
−∆u = λg(x) 1 + |u|p u, x ∈ Ω, (2) ∂u (1 − α) + αu = 0, x ∈ ∂Ω, (3) ∂n with the sign changing weight function g. See [1–4] and references therein for the case where α 6= 0, and [5] for the case where α = 1. Recently in [6] some existence results

c Vilnius University, 2012

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195

were established for the case when a single equation is replace by a system of equations. Also we refer to [7] where ∆ is replaced by ∆p . In this work we extend this studies to classes of Robin boundary conditions. We prove our existence results via variational methods. The system (1) is posed in the framework of the Sobolev space H = H 1 (Ω)×H 1 (Ω) with the norm Z 1 Z



2 2 2 2 2

(u, v) = |∇u| + u + |∇v| + v . H Ω

Ω

Moreover a pair of functions (u, v) ∈ H is said to be a weak solution of the system (1) if Z Z Z Z α ∇u∇φ1 + ∇v∇φ2 − λ a(uφ1 + vφ2 ) + (uφ1 + vφ2 ) 1−α Ω Ω Ω ∂Ω Z Z

+p av 2 |u|p−2 uφ1 + au2 |v|p−2 vφ2 + 2 au|v|p φ1 + av|u|p φ2 = 0 Ω

Ω

for all (φ1 , φ2 ) ∈ H. Thus the corresponding energy functional to the system (1) is defined by Z  Z
1 α 2 2 2 Jλ (u, v) = |∇u| − λau + u 2 1−α Ω ∂Ω  Z Z Z

α 1 v 2 + a v 2 |u|p + u2 |v|p |∇v|2 − λav 2 + + 2 1−α Ω

Ω

∂Ω


1 = L(u) + L(v) + G1 (u, v) + G2 (u, v), 2 where

Z L(t) =


|∇t|2 − λat2 +

α 1−α

Ω

Z

t2

∂Ω

for t = u or v, and

Z G1 (u, v) =

av 2 |u|p

Ω

and

Z G2 (u, v) =

au2 |v|p .

Ω

It is well known that the weak solutions of the system (1) are the critical points of the Euler functional Jλ . Let I be the Euler functional associated with an elliptic problem on a Banach space X. If I is bounded below and has a minimizer on X, thus this minimizer is a critical point

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S. Khademloo, S. Heidarkhani

of I, and so is a solution of the corresponding elliptic problem. However, the Euler functional Jλ is not bounded below on the whole space H, but is bounded on an appropriate subset, and a minimizer on this set (if it exists) gives a solution to the system (1). Then we introduce the following notation: For any functional f : H → R we denote by f 0 (u, v)(h1 , h2 ) the Gâteaux derivative of f at (u, v) in the direction of (h1 , h2 ) ∈ H, and f (1) (u, v)h1 = f 0 (u + εh1 , v)|ε=0 , f (2) (u, v)h2 = f 0 (u, v + δh2 )|δ=0 . In fact we have f 0 (u, v)(h1 , h2 ) = f (1) (u, v)h1 + f (2) (u, v)h2 .

2

Notations and preliminaries

First we consider the eigenvalue problem Z
µ(α, λ) = inf |∇u|2 − λau2 + Ω

α 1−α

Z

u2 ; u ∈ H 1 (Ω),

Z

 u2 = 1 .

Ω

∂Ω

Note that µ(α, 0) > 0 on α ∈ [α0 , 1] for some small negative α0 , and λ 7→ µ(α, λ) has + exactly two zeroes λ− α and λα and those are principal eigenvalues of the following linear problem −∆u = λau, x ∈ Ω, (4) ∂u (1 − α) + αu = 0, x ∈ ∂Ω ∂n (for more details, see [1]). Define


1

(u, v) = L(u) + L(v) 2 . λ + We prove that for λ ∈ (λ− α , λα ), k.kλ defines a norm in H which is equivalent to the usual norm for H. Our proof is motivated by that of [3]. Since k.kλ corresponds to the bilinear form Z Z

 α (u1 , v1 ), (u2 , v2 ) = (∇u1 ∇u2 − λau1 u2 ) + u1 u2 1−α Ω ∂Ω Z Z α + (∇v1 ∇v2 − λav1 v2 ) + v 1 v2 , 1−α Ω

∂Ω

in order to prove that k.kλ is a norm, it is sufficient to prove that h(u, v), (u, v)i ≥ 0 for all (u, v) ∈ H − {(0, 0)}. It follows from the variational characterization of µ(α, λ) that Z 


(u, v), (u, v) = L(u) + L(v) ≥ µ(α, λ) u2 + v 2 . (5) Ω

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197

+ Hence if λ ∈ (λ− α , λα ), then µ(α, λ) > 0 and so k(u, v)kλ > 0 whenever u, v 6= 0, thus k.kλ is a norm. Moreover for (u, v) ∈ H, there exists a constant k > 0 such that



(u, v) ≤ k (u, v) . H λ

In fact by using the embedding of H1 (Ω) into L2 (∂Ω), we have

(u, v) 2 Zλ Z Z Z

α α 2 2 2 2 2 2 |∇u| − λau + ≤ |∇v| − λav + u + v 1−α 1−α Ω Ω ∂Ω ∂Ω Z Z Z
α 2 2 + 2 u + ≤ |∇u| + |λ| a c1 |∇u| 1−α Ω Ω Ω Z Z Z
α 2 + 2 c1 |∇v|2 + |∇v| + |λ| a v + 1−α Ω

Ω

Ω

2 ≤ c2 (u, v) H , where c1 is the best Sobolev constant of the embedding of H1 (Ω) into L2 (∂Ω),   αc1 + + ,λ a c2 = max 1 + 1−α α and a+ = supx∈Ω |a(x)| > 0. Now suppose that there exists a sequence {(un , vn )} ⊆ H such that k(un , vn )kλ → 0 and k(un , vn )kH = 1. Since {(un , vn )} is bounded in H, there exists (u, v) ∈ H such that (un , vn ) * (u, v) in H. Applying compactly embedding of H in L2 (Ω) × L2 (Ω) and L2 (∂Ω) × L2 (∂Ω), we have (un , vn ) → (u, v) in L2 (Ω) × L2 (Ω) and L2 (∂Ω) × L2 (∂Ω), respectively. Taking k(un , vn )kλ → 0 into account, from (5), we have (un , vn ) → (0, 0) in L2 (Ω) × L2 (Ω), and so u = v = 0. This implies (un , vn ) → (0, 0) in L2 (∂Ω) × L2 (∂Ω), which concludes that Z 
2 2 lim |∇u| + |∇v| dx = 0. n→∞

Ω

This contradicts with the fact k(un , vn )kH = 1 for all n. Hence, k.kλ and k.kH are equivalent norms. Now we consider the Nehari minimizing problem  N (λ) = inf Jλ (u, v); (u, v) ∈ Mλ , where  

 (1) (2) Mλ = (u, v) ∈ H − (0, 0) ; Jλ0 (u, v), (u, v) = Jλ (u, v)u + Jλ (u, v)v = 0 . Nonlinear Anal. Model. Control, 2012, Vol. 17, No. 2, 194–209

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It is clear that all critical points of Jλ must lie on Mλ which is well-known as the Nehari manifold (see [2,8]). We will see below that local minimizers of Jλ on Mλ contains every + non-zero solution of the system (1). First we claim that for λ ∈ (λ− α , λα ) − {0} we have Mλ 6= ∅. In fact, since the function Ra changes sign, we can choose non-zero function R (u0 , v0 ) ∈ H such that G1 (u0 , v0 ) = av02 |u0 |p > 0 and G2 (u0 , v0 ) = au20 |v0 |p > 0. Let −L(u0 ) − L(v0 ) , tp = (2 + p)(G1 (u0 , v0 ) + G2 (u0 , v0 )) then

0  Jλ (tu0 , tv0 ), (tu0 , tv0 ) 
 = t2 L(u0 ) + L(v0 ) + tp (p + 2) G1 (u0 , v0 ) + G2 (u0 , v0 ) = 0, and so (u, v) = t(u0 , v0 ) ∈ Mλ . Define Fλ (u, v) = hJλ0 (u, v), (u, v)i. Then for (u, v) ∈ Mλ ,


Fλ0 (u, v), (u, v) = 2 L(u) + L(v) + (p + 2)2 (G1 + G2 )(u, v)
= p(p + 2)(G1 + G2 )(u, v) = −p L(u) + L(v) .

Set  Kλ+ = inf L(u) + L(v); (u, v) ∈ H, (G1 + G2 )(u, v) = 1 and  Kλ− = inf L(u) + L(v); (u, v) ∈ H, (G1 + G2 )(u, v) = −1 . By a similar way we can define K0+ and K0− . For α ∈ [0, 1], we have that λ → Kλ+ is a concave continuous curve on the interval − − [λα , λ+ α ]. By using a similar arguments we have these facts for Kλ . To state our main result, we now present some important properties of Kλ+ and Kλ− . Lemma 1. K0+ > 0 and K0− > 0 for α ∈ (0, 1]. Proof. Suppose otherwise, that is there exists sequence (un , vn ) ∈ H such that
lim L(un ) + L(vn ) = 0,

n→∞

(G1 + G2 )(un , vn ) = 1.

By the Sobolev embedding theorem and Hölder inequality, there exists ε0 < that Z Z

2 p 2 p + 1= avn |un | + aun |vn | ≤ a vn2 |un |p + u2n |vn |p Ω

2 N −2

such

Ω

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Existence of non-negative solutions for semilinear elliptic systems via variational methods

≤ a+

 Z


2∗ 2 2 −ε0

2  2∗ −2ε Z


2∗ −2ε0 |un |p 2∗ −2(ε0 +1)

0

vn Ω

0 +1)  2∗2−2(ε ∗ −2ε

Ω

Z +

u2n




2∗ 2

−ε0

2  2∗ −2ε Z 0

|vn |

p

199

∗ −2ε 0
2∗2−2(ε +1)

0

0 +1)   2∗2−2(ε ∗ −2ε 0

0

Ω

2∗


Ω 2∗ ≤ a+ 2 (un , vn ) H ≤ c0 a+ (un , vn ) λ → 0 as n → ∞, which is a contradiction. By the same argument we have K0− > 0. Lemma 2. Kλ++ = Kλ−− = 0. α

α

Proof. Suppose that ϕ+ is a positive eigenfunction of the linear problem (4) correspond+ ing to the principal eigenvalue λ+ α . Then L(ϕ ) = 0. On the other hand, if λ 6= 0 be the principal eigenvalue of (4) with corresponding positive principalReigenfunction ϕ, then R + p+1 λ Ω aϕp+1 > 0 (see [3]). So in this case, since λ+ > 0. We α > 0, we have Ω a(ϕ ) now let ϕ+ u= R 1 . (2 Ω a(ϕ+ )p+2 ) p+2 Then Z (G1 + G2 )(u, u) = 2 a|u|p+2 = 1, Ω

and L(u) + L(u) =

2L(ϕ+ ) = 0, R 2 2 2 p+2 ( Ω a(ϕ+ )p+2 ) p+2

i.e., Kλ++ = 0. By using a similar way we can prove that Kλ−− = 0. Since λ → Kλ+ α

α

is a concave continuous curve, we also have 0 < Kλ+ < K0+ for λ ∈ (0, λ+ α ), and , 0). 0 < Kλ− < K0− for all λ ∈ (λ− α + Lemma 3. For λ ∈ (λ− α , λα ), put   1 Y = (u, v) ∈ H; (G1 + G2 )(u, v) = − . p+2 2

−

2

Then k(u, v)kλp (u, v) ∈ Mλ if (u, v) ∈ Y , and k(u, v)kλ p+2 (u, v) ∈ Y if (u, v) ∈ Mλ . Proof. First suppose that (u, v) ∈ Y , then

2
Fλ (u, v) λp (u, v)

2

2


2

2


= Jλ0 (u, v) λp u, (u, v) λp v , (u, v) λp u, (u, v) λp v

4


2(2+p)
= (u, v) p L(u) + L(v) + (u, v) p (p + 2)(G1 + G2 )(u, v) λ

λ

4

2

2(2+p) = (u, v) λp (u, v) λ − (u, v) λ p = 0,

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S. Khademloo, S. Heidarkhani

2

i.e., k(u, v)kλp (u, v) ∈ Mλ . Now if (u, v) ∈ Mλ , we get

 L(u) + L(v) + (p + 2)(G1 + G2 )(u, v) = Jλ0 (u, v)(u, v) = 0, thus

− 2


− 2 (G1 + G2 ) (u, v) λ p+2 u, (u, v) λ p+2 v

−2

−2 −L(u) − L(v) = (u, v) λ (G1 + G2 )(u, v) = (u, v) λ p+2 2 k(u, v)k−2 k(u, v)k 1 λ λ =− =− , p+2 p+2 −

2

i.e., k(u, v)kλ p+2 (u, v) ∈ Y . 2 + For λ ∈ (λ− α , λα ), we define Qλ : Y → R by Qλ (u, v) = k(u, v)kλ . Then for (u, v) ∈ Y we have

 Qλ (u, v) =

p  p+2 

p2 2(p + 2) . Jλ (u, v) λ (u, v) p

Indeed,

2
Jλ (u, v) λp (u, v)

4

2(p+2)
1 = (u, v) λp L(u) + L(v) + (u, v) λ p (G1 + G2 )(u, v). 2 1 Since (G1 + G2 )(u, v) = − p+2 for (u, v) ∈ Y , we conclude  

2

2(p+2)
1 1

(u, v) p , Jλ (u, v) λp (u, v) = − λ 2 p+2

which means that the claim is true. Using a similar argument, if (u, v) ∈ Mλ , then Jλ (u, v) =

− 2
p+2 p Qλ (u, v) λ p+2 (u, v) p . 2(p + 2)

Indeed,

− 2
p+2 p Qλ (u, v) λ p+2 (u, v) p 2(p + 2)


p+2

4

2p
p+2 p p

(u, v) − p+2 (u, v) 2 p =

(u, v) p+2 p = λ λ λ 2(p + 2) 2(p + 2)

p

(u, v) 2 = Jλ (u, v). = λ 2(p + 2)

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Existence of non-negative solutions for semilinear elliptic systems via variational methods

201

The latest equality follows from that L(u) + L(v) + (p + 2)(G1 + G2 )(u, v) = 0 for (u, v) ∈ Mλ , i.e., (G1 + G2 )(u, v) = −


1 1

(u, v) 2 , L(u) + L(v) = − λ p+2 p+2

and so
1 L(u) + L(v) + (G1 + G2 )(u, v) 2

2

1 1 p

(u, v) 2 =

(u, v) 2 . = (u, v) λ − λ λ 2 p+2 2(p + 2)

Jλ (u, v) =

In fact we proved the following result: Lemma 4. Define qλ = inf (u,v)∈Y Qλ (u, v) and jλ = inf (u,v)∈Mλ Jλ (u, v). Then, p jλ ) p+2 . qλ = ( 2(p+2) p

3

Main results

+ Lemma 5. (u, v) = (0, 0) is not a limit point of Mλ if λ ∈ (λ− α , λα ), α 6= 0.

Proof. Let {(un , vn )} in Mλ , so that k(un , vn )kλ → 0 as n → ∞. (un ,vn ) Now let (u0n , vn0 ) = k(u , then k(u0n , vn0 )kλ = 1, i.e., {(u0n , vn0 )} is a bounded n ,vn )kλ ∗

∗

sequence in L2 (Ω) × L2 (Ω) equipped with the norm

(u, v) 2∗ = L (Ω)×L2∗ (Ω)

Z |u|

2∗

2∗

+ |v|




 21∗ ,

Ω

therefore hJλ0 (un , vn ), (un , vn )i k(un , vn )k2λ L(un ) + L(vn )) + (p + 2)(G1 + G2 )(un , vn ) = k(un , vn )k2λ (G1 + G2 )(un , vn ) = 1 + (p + 2) k(un , vn )k2λ

p = 1 + (p + 2) (un , vn ) (G1 + G2 )(u0n , vn0 ).

0=

λ

{(u0n , vn0 )}

Moreover by the boundedness of in H, and applying the inequality mentioned in Lemma 1, we derive that the sequence {(G1 +G2 )(u0n , vn0 )} is bounded in H, and so the right hand side of the last equality tends to 1. This contradiction proves the lemma.

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R Theorem 1. Let α ∈ (0, 1] or Ω a 6= 0 if α = 0. Then there exist two positive constants δ1 , δ2 such that if {(un , vn )} is a minimizing sequence of Jλ on Mλ , then Z
2 2 lim inf a un + vn > 0, n→∞

Ω

where

λ− α

0, by using Lemma 4, if {(un , vn )} be a minimizing − 2 sequence of Jλ on Mλ , then k(un , vn )kλ p+2 {(un , vn )} is a minimizing sequence of Qλ on Y , and so we get

− 2
inf Qλ = lim Qλ (un , vn ) λ p+2 (un , vn ) n→∞ Y

− 4
= lim (un , vn ) λ p+2 L(un ) + L(vn ) n→∞

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Existence of non-negative solutions for semilinear elliptic systems via variational methods

203

− 2

− 2



= lim L (un , vn ) λ p+2 un ) + L (un , vn ) λ p+2 vn n→∞

< q < K0+ for some q > 0. Also we have

− 2

− 2


1 . (G1 + G2 ) (un , vn ) λ p+2 un , (un , vn ) λ p+2 vn = − p+2 Now by Lemma 5, we obtain k(un , vn )kλ 6→ 0, i.e., L(un )+L(vn ) 6→ 0, and if vn2 ) → 0, then we get Z  Z

4

2 2 2 2

(un , vn ) − p+2 |∇u | + |∇v | + u + v 6→ 0. n n n n λ Ω

R Ω

a(u2n +

∂Ω

So by taking λ = 0, we have

− 4


lim (un , vn ) λ p+2 L(un ) + L(vn ) = K0+ ≤ q < K0+ ,

n→∞

which is a contradiction. Therefore, Z
lim a u2n + vn2 > 0. n→∞ Ω

Now let R δ1 =

Ω

|∇ϕ− |2 + R

α 1−α Ω

(ϕ− )2 − K0−

R

∂Ω a(ϕ− )2

K0− . a(ϕ− )2 Ω

R − λ− α =−

− By a similar argument we get a same result for λ ∈ (λ− α , λα + δ1 ).

Lemma 6. The production of two Hilbert spaces, is a Hilbert space. Proof. For Hilbert spaces H1 and H2 , define

 (u1 , u2 ), (v1 , v2 ) H ×H = hu1 , v1 iH1 + hu2 , v2 iH2 . 1

2

It is easy to see that, the mentioned bilinear form defines an inner product on H1 ×H2 . To state our significant proposition form [9], first let B be a ball in the Hilbert space H, centered at 0 and of radial . Proposition 1. Let Φ be a C 1 -functional on a Hilbert space X = X1 × X2 , where X1 and X2 are Hilbert spaces, and let Γ be a closed subset in X such that for any (u, v) ∈ Γ with Φ0 (u, v) 6= 0 and  > 0 small arbitrary, there exists Frechet differentiable function

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0

(u,v) ) for 0 ≤ δ ≤ , we s(u,v) : B → R such that by setting t(u,v) (δ) = s(u,v) (δ kΦΦ0 (u,v)k X have t(u,v) (0) = 1, and   Φ(2) (u, v) Φ(1) (u, v) , v − δ (2) ∈ Γ. t(u,v) (δ) u − δ (1) kΦ (u, v)kX1 kΦ (u, v)kX2

If Φ is bounded below on Γ , then for any minimizing sequence {(un , vn )} for Φ in Γ , there exists another minimizing sequence {(u∗n , vn∗ )} for Φ in Γ such that

Φ(u∗n , vn∗ ) ≤ Φ(un , vn ), lim (u∗n , vn∗ ) − (un , vn ) = 0 n→∞

and

0 ∗ ∗

Φ (un , vn ) X






 1 √ ≤ 2 + (u∗n , vn∗ ) t0(u∗n ,vn∗ ) (0) + t0(u∗n ,vn∗ ) (0) Φ0 (u∗n , vn∗ ), (u∗n , vn∗ ) . n Proof. Let η = inf γ∈Γ Φ(γ). Using Ekland variational principle, we get a minimizing sequence {(u∗n , vn∗ )} in Γ , which (i) Φ(u∗n , vn∗ ) ≤ Φ(un , vn ) < η + n1 , (ii) limn→∞ k(u∗n , vn∗ ) − (un , vn )k = 0,

(iii) Φ(x, y) ≥ Φ(u∗n , vn∗ ) − n1 (x, y) − (u∗n , vn∗ )k for all (x, y) ∈ Γ . Let us assume kΦ0 (u∗n , vn∗ )kX > 0 for large n. Apply the hypothesis on the set Γ with (u, v) = (u∗n , vn∗ ) to find the function   Φ0 (u∗ , v ∗ ) tn (δ) = t(u∗n ,vn∗ ) (δ) = s(u∗n ,vn∗ ) δ 0 ∗n ∗n . kΦ (un , vn )kX Then,  (xδ , yδ ) = tn (δ) u∗n − δ

Φ(1) (u∗n , vn∗ ) Φ(2) (u∗n , vn∗ ) ∗ , v − δ kΦ(1) (u∗n , vn∗ )kX1 n kΦ(2) (u∗n , vn∗ )kX2

 ∈Γ

for all small enough δ ≥ 0. By the mean value theorem we have

1

(xδ , yδ ) − (u∗n , vn∗ ) n

 ≥ Φ(u∗n , vn∗ ) − Φ(xδ , yδ ) = Φ0 (xδ , yδ ), (u∗n , vn∗ ) − (xδ , yδ ) + o(δ)

  = Φ0 (xδ , yδ ), (u∗n , vn∗ ) − hΦ0 (xδ , yδ ), (xδ , yδ ) + o(δ)

 = hΦ0 (xδ , yδ ), (u∗n , vn∗ )i − tn (δ) Φ0 (xδ , yδ ), (u∗n , vn∗ )    Φ(1) (u∗n , vn∗ ) Φ(2) (u∗n , vn∗ ) + δtn (δ) Φ0 (xδ , yδ ), , + o(δ) kΦ(1) (u∗n , vn∗ )kX1 kΦ(2) (u∗n , vn∗ )kX2


 = 1 − tn (δ) Φ0 (xδ , yδ ), (u∗n , vn∗ )    Φ(1) (u∗n , vn∗ ) Φ(2) (u∗n , vn∗ ) + δtn (δ) Φ0 (xδ , yδ ), , + o(δ), kΦ(1) (u∗n , vn∗ )kX1 kΦ(2) (u∗n , vn∗ )kX2

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Existence of non-negative solutions for semilinear elliptic systems via variational methods

where

o(δ) δ

205

→ 0 as δ → 0. So

 (1) ∗ ∗ 1

tn (δ)u∗n − δtn (δ) Φ (un , vn ) − u∗n ,

(1) ∗ ∗ nδ kΦ (un , vn )kX1 

Φ(2) (u∗ , v ∗ ) tn (δ)vn∗ − δtn (δ) (2) ∗ n ∗ n − vn∗

kΦ (un , vn )kX2  1 − tn (δ) 0 ≥ Φ (xδ , yδ ), (u∗n , vn∗ ) δ    Φ(2) (u∗n , vn∗ ) Φ(1) (u∗n , vn∗ ) o(δ) + tn (δ) Φ0 (xδ , yδ ), , + . δ kΦ(1) (u∗n , vn∗ )kX1 kΦ(2) (u∗n , vn∗ )kX2 Now we pass to the limit as δ → 0, and we derive

     1 tn (δ) − 1 ∗ tn (δ) − 1 ∗ lim u vn , n

δ→0 n δ δ

 

Φ(1) (u∗n , vn∗ ) 1 Φ(2) (u∗n , vn∗ )

+ lim tn (δ) ,

(1) ∗ ∗ (2) ∗ ∗ δ→0 n kΦ (un , vn )kX1 kΦ (un , vn )kX2

1  ∗ ∗ 1 Φ(1) (u∗n , vn∗ ) 2 Φ(2) (u∗n , vn∗ ) 2 2 1 0





= tn (0) (un , vn ) + + (2) ∗ ∗ n n kΦ(1) (u∗n , vn∗ )kX1 kΦ (un , vn )kX2


1 √ 2 + t0n (0) (u∗n , vn∗ ) = n   ≥ −t0n (0) Φ0 (u∗n , vn∗ ), (u∗n , vn∗ )    Φ(1) (u∗n , vn∗ ) Φ(2) (u∗n , vn∗ ) + Φ0 (u∗n , vn∗ ), , kΦ(1) (u∗n , vn∗ )kX1 kΦ(2) (u∗n , vn∗ )kX2

 = −t0n (0) Φ0 (u∗n , vn∗ ), (u∗n , vn∗ )     Φ(1) (u∗n , vn∗ ) Φ(2) (u∗n , vn∗ ) (2) ∗ ∗ + Φ(1) (u∗n , vn∗ ), + Φ (u , v ), n n kΦ(1) (u∗n , vn∗ )kX1 kΦ(2) (u∗n , vn∗ )kX2



 = −t0n (0) Φ0 (u∗n , vn∗ ), (u∗n , vn∗ ) + Φ(1) (u∗n , vn∗ ) X1 + Φ(2) (u∗n , vn∗ ) X2 . Thus, we have

0 ∗ ∗

Φ (un , vn ) X

= Φ(1) (u∗n , vn∗ ) + Φ(2) (u∗n , vn∗ )



≤ Φ(1) (u∗n , vn∗ ) X1 + Φ(2) (u∗n , vn∗ ) X2




 1 √ ≤ 2 + t0n (0) (u∗n , vn∗ ) + t0n (0) Φ0 (u∗n , vn∗ ), (u∗n , vn∗ ) . n X

This completes the proof. Let α ∈ (0, 1] or that following results:

R Ω

+ a 6= 0 for α = 0. Then for λ ∈ (λ− α , λα ) we have the

Lemma 7. Jλ is bounded below on Mλ .

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Proof. Firstly we note that for (u, v) ∈ Mλ , (G1 + G2 )(u, v) = − + Now for λ ∈ (λ− α , λα ), let w =

u R 1 ( Ω u2 ) 2

L(u) µ(α, λ) ≤ L(w) = R 2 = u Ω


1 L(u) + L(v) . p+2

(6)

. Then R Ω

(|∇u|2 − λau2 ) + R u2 Ω

α 1−α

R ∂Ω

u2

,

+ and so L(u) > 0 for λ ∈ (λ− α , λα ). A similar argument shows that L(v) > 0 for − + λ ∈ (λα , λα ), and (6) implies (G1 + G2 )(u, v) < 0 for (u, v) ∈ Mλ . Hence, Jλ (u, v) = − p2 (G1 + G2 )(u, v) > 0, and Jλ is bounded below on Mλ .

Theorem 2. There exists a minimizing sequence {(u∗n , vn∗ )} of Jλ on Mλ such that

lim Jλ0 (u∗n , vn∗ ) λ = 0. n→∞

Proof. For (u, v) ∈ Mλ , let Ψ : R × H → R such that
Ψ s, (w1 , w2 ) = Fλ (su − w1 , sv − w2 ). Since (u, v) ∈ Mλ , we have Ψ(1, (0, 0)) = 0. Also,

∂ Ψ 1, (0, 0) = 2 L(u) + L(v) + (p + 2)2 (G1 + G2 )(u, v) ∂s
= 2 −(p + 2)(G1 + G2 )(u, v) + (p + 2)2 (G1 + G2 )(u, v) = p(p + 2)(G1 + G2 )(u, v). Now we want to apply the Implicit function theorem at (1, (0, 0)) to get for any δ > 0 small enough, a differentiable function s(u,v) : Bδ → R such that
s(u,v) (0, 0) = 1, s(u,v) (w1 , w2 ) (u, v) − (w1 , w2 ) ∈ Mλ ,   hFλ0 (u, v), (w1 , w2 )i s0(u,v) (0, 0), (w1 , w2 ) = hFλ0 (u, v), (u, v)i for all (w1 , w2 ) ∈ Bδ , where Bδ is defined before. Now for every (u, v) ∈ H, let Jλ0 (u, v) , kJλ0 (u, v)kλ
ρ(x, y)(u,v) for 0 ≤ ρ ≤ δ.

(x, y)(u,v) = t(u,v) (ρ) = s(u,v)

Then we have the following results: (i) t(u,v) (0) = 1, (ii) t0(u,v) (0) = hs0(u,v) (0, 0), (x, y)(u,v) i,

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207

(iii) Fλ (s(u,v) (ρ(x, y)(u,v) )((u, v) − ρ(x, y)(u,v) )) = 0, (iv) t(u,v) (ρ)((u, v) − ρ(x, y)(u,v) ) = s(u,v) (ρ(x, y)(u,v) )((u, v) − ρ(x, y)(u,v) ) ∈ Mλ . From Proposition 1, there exists a minimizing sequence {(u∗n , vn∗ )} for Jλ such that 1 Jλ (u∗n , vn∗ ) ≤ Jλ (un , vn ) < inf Jλ + , Mλ n

lim (u∗n , vn∗ ) − (un , vn ) = 0 n→∞

and √

0 ∗ ∗




Jλ (un , vn ) ≤ 1 2 + (u∗n , vn∗ ) t0 ∗ ∗ (0) + t0 ∗ ∗ (0)Fλ (u∗n , vn∗ ) . (un ,vn ) λ λ (un ,vn ) n Since Jλ (u∗n , vn∗ ) =

∗ ∗ 2 p

(un , vn ) < inf Jλ + 1 , λ Mλ 2(p + 2) n

so that {(u∗n , vn∗ )} is a bounded sequence in H, i.e., there exits c1 > 0 such that k(u∗n , vn∗ )kλ < c1 for all n. Then, √

0 ∗ ∗


Jλ (un , vn ) ≤ 1 2 + t0 ∗ ∗ (0) c1 . (un ,vn ) λ n Moreover, 0 t

 |hFλ0 (u∗n , vn∗ ), (x, y)(u∗n ,vn∗ ) i| = |hFλ0 (u∗n , vn∗ ), (u∗n , vn∗ )i| |hFλ0 (u∗n , vn∗ ), (x, y)(u∗n ,vn∗ ) i| |hFλ0 (u∗n , vn∗ ), (x, y)(u∗n ,vn∗ ) i| = = |p(p + 2)(G1 + G2 )(u∗n , vn∗ )| pk(u∗n , vn∗ )k2λ

0 = s

∗ (0) (u∗ n ,vn )

∗ (0, 0), (x, y)(u∗ ,v ∗ ) (u∗ n n n ,vn )

and

lim inf (u∗n , vn∗ ) λ > 0, n→∞

since (0, 0) is not a limit point of Mλ . So, if we show that |t0(u∗ ,v∗ ) (0)| is uniformly bounded on n, we are done. n n By using the Hölder inequality, Sobolev embedding theorem, boundedness of the sequence {(u∗n , vn∗ )} and k(x, y)(u∗n ,vn∗ ) k = 1, we have 0 ∗ ∗

 Fλ (un , vn ), (x, y)(u∗ ,v∗ ) ≤ c2 (u∗n , vn∗ ) + c3 . n n λ This proves the theorem. We can now prove the main result of the paper: R − Theorem 3. Let α ∈ (0, 1] or that Ω a 6= 0 for α = 0. For any λ ∈ (λ− α , λα + δ1 ) ∪ + + (λα − δ2 , λα ), λ 6= 0, The system (1) has a nontrivial nonnegative solution.

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S. Khademloo, S. Heidarkhani

Proof. Let c = inf Jλ (Mλ ) and {(un , vn )} be a sequence in Mλ such that lim Jλ (un , vn ) = c.

n→∞

By using Theorem 2, we have limn→∞ kJλ0 (un , vn )kλ = 0. Then {(un , vn )} is bounded and we can find a weak limit point of the sequence in H, i.e., un * u and vn * v both in H 1 (Ω), for some u, v ∈ H 1 (Ω), and so un → u and vn → v both in Lq (Ω) for q < N2N −2 . In particular for any (h1 , h2 ) ∈ H,

 Jλ0 (un , vn ), (h1 , h2 ) Z Z Z = ∇un ∇h1 + ∇vn ∇h2 − λ a(un h1 + vn h2 ) + Ω

Ω

Z +p

a

α 1−α

Ω

vn2 |un |p−2 un h1

+

Z (un h1 + vn h2 ) ∂Ω

u2n |vn |p−2 vn h2




Z +2

Ω

p


a un |vn | h1 + vn |un |p h2 ,

Ω

which converges to

0  Jλ (u, v), (h1 , h2 ) Z Z Z = ∇u∇h1 + ∇v∇h2 − λ a(uh1 + vh2 ) + Ω

Ω

Z

2

p−2

a v |u|

+p

α 1−α

Ω 2

(uh1 + vh2 ) ∂Ω

p−2

uh1 + u |v|

Z




Z

vh2 + 2

Ω

p

a u|v| h1 + v|u|p h2




Ω

as n → ∞. So, we derive 0  Jλ (u, v), (h1 , h2 )



 = lim Jλ0 (un , vn ), (h1 , h2 ) ≤ lim Jλ0 (un , vn ) λ (h1 , h2 ) = 0, n→∞

n→∞

that means hJλ0 (u, v), (h1 , h2 )i = 0 for all (h1 , h2 ) ∈ Y . Therefore, (u, v) is a weak solution for the system (1). R In particular, hJλ0 (u, v), (u, v)i = 0. Since lim inf | Ω (au2n + avn2 )| > 0, we have (u, v) 6= (0, 0). Hence, (u, v) ∈ Mλ . On the other hand, Jλ is weakly lower semicontinuous, and so we have c ≤ Jλ (u, v) ≤ lim Jλ (un , vn ) = c, n→∞

which follows that Jλ (u, v) = c and that k(un , vn )kλ → k(u, v)kλ which implies that un → u and vn → v both in H 1 (Ω). Since Jλ0 is continuous at (u, v), we get Jλ0 (u, v) = 0. One can check that Jλ (u, v) = Jλ (|u|, |v|), so (u, v) is a nontrivial nonnegative solution for the system (1).

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