RACSAM DOI 10.1007/s13398-015-0217-7 ORIGINAL PAPER

On existence of weak solutions for a p-Laplacian system at resonance Bui Quoc Hung · Hoang Quoc Toan

Received: 10 September 2014 / Accepted: 23 January 2015 © Springer-Verlag Italia 2015

Abstract This article shows the existence of weak solutions of a resonance problem for uniformly p-Laplacian system in a bounded domain in R N . Our arguments are based on the Saddle Point Theorem (P.H.Rabinowitz) and rely on a generalization of the Landesman–Lazer type condition. Keywords condition

Semilinear elliptic equation · Saddle point theorem · Landesman–Lazer

Mathematics Subject Classification

35J20 · 35J60 · 58E05

1 Introduction and preliminaries Let  be a bounded domain in R N , (N ≥ 3), with smooth boundary ∂. In the present paper we consider the existence of weak solutions of the following Dirichlet problem at resonance for p-Laplacian system: 

− p u = λ1 |u|α−1 |v|β−1 v + f (x, u, v) − k1 (x) − p v = λ1 |u|α−1 |v|β−1 u + g(x, u, v) − k2 (x) in ,

(1.1)

Research supported by the National Foundation for Science and Technology Development of Viet Nam (NAFOSTED under Grant Number 101.02-2014.03). B. Q. Hung (B) Faculty of Information Technology, Le Quy Don Technical University, 236 Hoang Quoc Viet, Cau Giay, Hanoi, Vietnam e-mail: [email protected] H. Q. Toan Department of Mathematics, Hanoi University of Science, 334 Nguyen Trai, Thanh Xuan, Hanoi, Vietnam e-mail: [email protected]

B. Q. Hung, H. Q. Toan

where p ≥ 2, α ≥ 1, β ≥ 1, α + β = p

(1.2)

and f, g :  × R 2 → R are Carathéodory functions which will be specified later. 

ki (x) ∈ L p (), p  =

p ¯ i = 1, 2. , ki (x) > 0, for a.e x ∈ , p−1

λ1 denotes the first eigenvalue of the problem:  − p u = λ|u|α−1 |v|β−1 v − p v = λ|u|α−1 |v|β−1 u, 1, p

(1.3)

1, p

where (u, v) ∈ E = W0 () × W0 (), p ≥ 2, α ≥ 1, β ≥ 1, α + β = p. It’s well-known that the principle eigenvalue λ1 = λ1 ( p) of (1.3) is obtained using the Ljusternick–Schnirelmann theory by minimizing the functional   α β p J (u, v) = |∇u| d x + |∇v| p d x p  p  on the set:

  1, p 1, p S = (u, v) ∈ E = W0 () × W0 () : A(u, v) = 1 ,

where

 A(u, v) =



|u|α−1 |v|β−1 uvd x

that is λ1 = λ1 ( p) can be variational characterized as λ1 = λ1 ( p) =

J (u, v) . A(u, v) A(u,v)>0

(1.4)

in f

Moreover the eigenpair (ϕ1 , ϕ2 ) associated with λ1 is componentwise positive and unique (up to multiplication by nonzero scalar) (see Theorem 2.2 in [3] and Remark 5.4 in [5]). As 1, p usual W0 () denotes Sobolev space which can be defined as the completion of C0∞ () under the norm:  1 p ||u||W 1, p = |∇u| p d x 0



and 1  p p p for w = (u, v) ∈ E : ||w|| E = ||u|| 1, p + ||v|| 1, p . W0

W0

Observe that the existence of weak solutions of ( p, q)-Laplacian systems at resonance in bounded domains with Dirichlet boundary condition, was first considered by Zographopoulos in [9]. Later in [4] Kandilakis and Magiropoulos have studied a quasilinear elliptic system with resonance part and nonlinear boundary condition in an unbounded domain by assuming the nonlinearities f and g depending only one variable u or v. In [8] Zeng-Qi Ou and Chen Lei Tang have considered the same system as in [4] with Dirichlet condition in a bounded domain. In these the existence of weak solutions is obtained by critical point theory (the Minimum Principle or the Saddle Point Theorem ) under a Landesman–Lazer type condition.

On existence of weak solutions

In this paper by introducing a generalization of Landesman–Lazer type condition we shall prove an existence result for a p-Laplacian system on resonance in bounded domain with the nonlinearities f and g to be functions depending on both variables u and v. Our arguments are based on the Saddle Point Theorem (P.H.Rabinowitz) and generalization of the Landesman–Lazer type condition. We have the following definition. Definition 1.1 Function w = (u, v) ∈ E is called a weak solution of the problem (1.1) if and only if, for all w¯ = (u, ¯ v) ¯ ∈E   α |∇u| p−2 ∇u.∇ ud ¯ x +β |∇v| p−2 ∇v.∇ vd ¯ x    ¯ x −λ1 (α|u|α−1 |v|β−1 v u¯ + β|u|α−1 |v|β−1 u v)d    ¯ x + (αk1 (x)u¯ + βk2 (x)v)d ¯ x = 0. − (α f (x, u, v)u¯ + βg(x, u, v)v)d 



We will use the following conditions (H1 ) (i) For a.e x ∈  : f (x, .), g(x, .) ∈ C 1 (R 2 ) and f (x, 0, 0) = 0, g(x, 0, 0) = 0.  p (ii) There exists function τ ∈ L p (), p  = p−1 such that: | f (x, s, t)| ≤ τ (x), |g(x, s, t)| ≤ τ (x), for a.e x ∈ , ∀(s, t) ∈ R 2 . (iii) For (s, t) ∈ R 2 : α

∂g(x, s, t) ∂ f (x, s, t) =β ∂t ∂s

for a.e x ∈ .

(1.5)

For (u, v) ∈ R 2 , a.e x ∈ , define   β v α u ( f (x, s, v) + f (x, s, 0))]ds + (g(x, u, t) + g(x, 0, t))dt. H (x, u, v) = 2 0 2 0 (1.6) By hypotheses (1.5), from (1.6) with some simple computations we deduce that: ∂ H (x, s, t) ∂ H (x, s, t) = α f (x, s, t), = βg(x, s, t), for a.e x ∈ , ∀(s, t) ∈ R 2 . (1.7) ∂s ∂t Now, for i, j = 1, 2 we define 



1 τ Fi (x) = lim f x, (−1)1+i yϕ1 , (−1)1+i τ ϕ2 + f x, (−1)1+i yϕ1 , 0 dy τ →+∞ τ 0 



 1 τ g x, (−1)1+ j τ ϕ1 , (−1)1+ j yϕ2 + g x, 0, (−1)1+ j yϕ2 dy G j (x) = lim τ →+∞ τ 0 (1.8) and lim f (x, s, t) = f +∞ (x),

s→+∞ t→+∞

lim f (x, s, t) = f −∞ (x),

s→−∞ t→−∞

lim g(x, s, t) = g +∞ (x)

s→+∞ t→+∞

lim g(x, s, t) = g −∞ (x).

s→−∞ t→−∞

B. Q. Hung, H. Q. Toan

Assume that (H2 ) (i) f +∞ (x) < k1 (x) < f −∞ (x) g +∞ (x) < k2 (x) < g −∞ (x)

for a.e x ∈ 

(1.9)

(ii)  1 α β (α F2 (x)ϕ1 (x) + βG 2 (x)ϕ2 (x)) − f −∞ (x)ϕ1 (x) − g −∞ (x)ϕ2 (x) d x p p  2   1 (αk1 (x)ϕ1 (x) + βk2 (x)ϕ2 (x))]d x < 1− p     1 α β < (α F1 (x)ϕ1 (x)+βG 1 (x)ϕ2 (x)) − f +∞ (x)ϕ1 (x)− g +∞ (x)ϕ2 (x) d x. p p  2 (1.10)

 

The main result of this paper can be described in the following theorem: Theorem 1.1 Assuming conditions (H1 ), (H2 ) are fulfilled. Then the problem (1.1) has at least a nontrivial weak solution in E. Proof of Theorem 1.1 is based on variational techniques and the Saddle Point Theorem (P.H.Rabinowitz). Theorem 1.2 (Saddle Point Theorem, P.H.Rabinowitz in [6]) Let E = X ⊕ Y be a Banach space with Y closed in E and dim X < ∞. For > 0 define M := {u ∈ X : ||u|| ≤ }

M0 := {u ∈ X : ||u|| = }

Let F ∈ C 1 (E, R) be such that b := inf F(u) > a := max F(u) u∈M0

u∈Y

If F satisfies the (P S)c condition with c := inf max F(γ (u)) γ ∈ u∈M

where := {γ ∈ C(M, E) : γ | M0 = I },

then c is a critical value of F.

2 Proof of the main result We define the Euler–Lagrange functional associated to the problem (1.1) by    β α |∇u| p d x + |∇v| p d x − λ1 |u|α−1 |v|β−1 u.vd x I (w) = p  p     − H (x, u, v)d x + (αk1 (x)u + βk2 (x)v)d x 

= J (w) + T (w),



for w = (u, v) ∈ E,

(2.1)

On existence of weak solutions

where

 β |∇u| p d x + |∇v| p d x. (2.2) p      |u|α−1 |v|β−1 u.vd x − H (x, u, v)d x + (αk1 (x)u + βk2 (x)v)d x. T (w) = −λ1 J (w) =

α p









(2.3) We deduce that I ∈ C 1 (E). Remark 2.1 By similar arguments as those in the proof of Lemma 2.3 in [10] and Lemma 5 in [4], we infer that the functional A : E → R and the operator B : E → E ∗ given by, for any (u, v), (u, ¯ v) ¯ ∈E  A(u, v) = |u|α−1 |v|β−1 u.vd x 

and

 < B(u, v), (u, ¯ v) ¯ >=



|u|

α−1

|v|

β−1

 uvd ¯ x+



|u|α−1 |v|β−1 u vd ¯ x,

are compact. Remark 2.2 Applying Theorem 1.6 in [6, p9] we deduce that the functional J : E → R given by (2.2) is weakly lower semicontinuous on E. Hence the functional I = T + J is also weakly lower semicontinuous on E. Proposition 2.1 Assuming the hypotheses (H1 ) and (H2 ) are fulfilled. The functional I : E → R given by (2.1) satisfies the (P S) condition on E. Proof Let {wm = (u m , vm )} be a Palais–Smale sequence in E, i.e: |I (wm )| ≤ M, M is positive constant

(2.4)

I  (wm ) → 0 in E ∗ as m → +∞

(2.5)

First, we shall prove that {wm } is bounded in E. We suppose by contradiction that {wm } is not bounded in E. Without loss of generality we assume that ||wm || E → +∞ as m → +∞. u m , vm ) that is u m = ||wumm|| E and vm = ||wvmm|| E . Let w m = ||wwmm|| E = ( wm k = ( u m k , vm k )}k which Thus w m is bounded in E. Then there exists a subsequence { 1, p converges weakly to w = ( u , v ) in E. Since the embedding W0 () into L p () is compact, the sequences { u m k } and { vm k } converge strongly to u and v in L p () respectively. From (2.4) we have     α β |∇ umk | p d x + |∇ vm k | p d x − λ1 | u m k |α−1 | vm k |β−1 umk vm k d x lim sup k→+∞ p  p  

  H (x, wm k ) αk1 u m k + βk2 vm k − d x + d x ≤ 0. (2.6) p p−1  ||wm k || E  ||wm k || E

B. Q. Hung, H. Q. Toan

By hypotheses (H1 ), we deduce that   α umk β vm k H (x, wmk ) = ( f (x, s, vmk )+ f (x, s, 0))ds + (g(x, u mk , t)+g(x, 0, t))dt. 2 0 2 0 This implies that |H (x, wmk )| ≤ c.τ (x)(|u mk | + |vmk |), c is positive constant. Hence,      c H (x, wmk )   ≤ ||τ || L p () || vmk || L p () . u mk || L p () + ||   p−1 p  ||wmk || ||wmk || E vm k converge strongly in L p () then bounded in L p (), hence Since umk ,  H (x, wmk ) lim sup p =0 k→+∞  ||wmk || E and

 lim

αk1 u m k + βk2 vm k

k→+∞ 

p−1

||wm k || E

d x = 0.

From the compactness of operator A it follows that   lim λ1 | u m k |α−1 | vm k |β−1 umk vm k d x = λ1 | u |α−1 | v |β−1 u . v d x. k→+∞

(2.7)





(2.8)

Using the weak lower semicontinuity of the functional J and the variational characterization of λ1 from (2.6) we get    α β λ1 | u |α−1 | v |β−1 u . vd x ≤ |∇ u| p d x + |∇ v| p d x p p        α β p p |∇ umk | d x + |∇ vm k | d x ≤ lim inf k→+∞ p  p       α β p p ≤ lim sup |∇ umk | d x + |∇ vm k | d x ≤ λ1 | u |α−1 | v |β−1 u . v d x. k→+∞ p  p   (2.9) Thus, theses inequalities are indeed equalities and we have       α β α β |∇ umk | p d x + |∇ vm k | p d x = |∇ u| p d x + |∇ v| p d x lim k→+∞ p  p  p  p   | u |α−1 | v |β−1 u . v d x. (2.10) = λ1 

We shall prove that u = 0 and v = 0. By contradiction suppose that u = 0, thus u m k → 0 in L p () as k → +∞. We have     α−1 β−1  |A( u m k , vm k )| =  | u m k | | vm k | umk vm k d x  

β

≤ || u m k ||αL p () .|| vm k || L p () . Since || u m k || L p () → 0, letting k → +∞ shows that lim A( u m k , vm k ) = 0.

k→+∞

(2.11)

On existence of weak solutions

From (2.6) taking lim sup with (2.7) and (2.10) we arrive at k→+∞



α lim sup k→+∞ p



β |∇ umk | d x + p 





p

|∇ vm k | d x = 0. p



(2.12)

On the other hand, since || wm k || E = 1 and       α β α β α β p p |∇ umk | d x + |∇ vm k | d x ≥ min , .|| wm k || E = min , >0 p  p  p p p p which contradicts (2.11). Thus u = 0. Similary we have v = 0. By again the definition of λ1 from (2.10) we deduce that = ( u , v ) = (−ϕ1 , −ϕ2 ), w = ( u , v ) = (ϕ1 , ϕ2 ) or w where (ϕ1 , ϕ2 ) is eigenpair associated with λ1 of the problem (1.3). Next, we shall consider following two cases: Firstly, assume that u m k → ϕ1 , vm k → ϕ2 in L p () as k → +∞. From (2.4) we have    α β −M ≤− |∇u m k | p d x − |∇vm k | p d x + λ1 |u m k |α−1 |vm k |β−1 u m k vm k d x p  p     H (x, wm k )d x − (αk1 u m k + βk2 vm k )d x ≤ M. (2.13) + 



Moreover, from (2.5) there exists the sequence k , k → 0+ , k → +∞ such that   u m k vm k 1 | < I  (wm k ), , > | ≤ k . ||wm || E . p p p This implies       1 umk vm k − k . ||wm k || E ≤ α |∇u m k | p−2 ∇u m k ∇ |∇vm k | p−2 ∇vm k ∇ dx + β dx p p p         umk vm k α|u m k |α−1 |vm k |β−1 vm k + β|u m k |α−1 |vm k |β−1 u m k dx −λ1 p p         umk  vm k   um vm α f x, wm k αk1 k + βk2 k d x + βg x, wm k dx + − p p p p   1 ≤ k . ||wm k || E . p

Remark that α + β = p, we get   1 α β − k . ||wm k || E ≤ |∇u m k | p d x + |∇vm k | p d x p p  p         umk  vm k   α|u m k |α−1 |vm k |β−1 u m k vm k d x − −λ1 + βg x, wm k dx α f x, wm k p p      α β 1 (2.14) k1 u m k + k2 vm k d x ≤ k . ||wm k || E . + p p  p Hence, summing (2.13), (2.14) we obtain

B. Q. Hung, H. Q. Toan

       

k β  α  H x, wm k − dx −M − ||wm k || E ≤ f x, wm k u m k + g x, wm k vm k p p p         1 1

k − (2.15) α 1− k1 u m k + β 1 − k2 vm k d x ≤ M + ||wm k || E . p p p  After dividing (2.15) by ||wm k || E , letting lim sup we deduce that k→+∞

 

 H (x, wm k ) α β lim sup − f (x, wm k ) u m k − g(x, wm k ) vm k d x k→+∞ ||wm k || E p p    1 (αk1 ϕ1 + βk2 ϕ2 )d x. = 1− p 

(2.16)

We remark that, from (1.6) by some standard computations we get   H (x, wm k ) 1 lim sup dx = (α F1 ϕ1 + βG 1 ϕ2 )d x, k→+∞ 2   ||wm k || E where F1 (x), G 1 (x) are given by (1.8). Letting lim sup (2.16) we obtain k→+∞

 

 1 α β (α F1 ϕ1 + βG 1 ϕ2 ) − f +∞ ϕ1 − g +∞ ϕ2 d x p p  2   1 (αk1 ϕ1 + βk2 ϕ2 )d x, = 1− p 

which contradicts (H2 (ii)). Similarly, in the case when u m k → −ϕ1 , vm k → −ϕ2 , in L p () as k → +∞, by similar computations, we also have    1 α β (α F2 ϕ1 + βG 2 ϕ2 ) − f −∞ ϕ1 − g −∞ ϕ2 d x p p  2   1 = 1− (αk1 ϕ1 + βk2 ϕ2 )d x, p  where F2 (x), G 2 (x) are given by (1.8), which contradicts (H2 (ii)). This implies that the (P S) sequence {wm } is bounded in E. Then there exists a subsequence wm k which converges weakly to w0 = (u 0 , v0 ) ∈ E. We shall prove that wm k converges strongly to w0 = (u 0 , v0 ) ∈ E. 1, p 1, p Indeed, since wm k  w0 = (u 0 , v0 ) in E and the embedding W0 × W0 → L p () × L p () is compact, the subsequences u m k , vm k converge strongly to u 0 , v0 in L p respectively. We have  α|u m k |α−1 |vm k |β |u m k − u 0 |d x |T  (wm k , (wm k − w0 ))| ≤ λ1      β|u m k |α |vm k |β−1 |vm k − v0 |d x + + α| f (x, wm k )||u m k − u 0 |       + β|g(x, wm k )||vm k − v0 | d x + αk1 (x)|u m k − u 0 | + βk2 (x)|vm k − v0 | d x 

On existence of weak solutions

  β β−1 α p + β||u m || p ||vm || p ||vm − v0 || L p ≤ λ1 α||u m k ||α−1 ||v || ||u − u || p p m m 0 L k L k k L k L k L + ||τ || L p (α||u m k − u 0 || L p + β||vm k − v0 || L p ) + α||k1 || L p ||u m k − u 0 || L p + β||k2 || L p ||u m k − u 0 || L p .

(2.17)

Letting k → +∞ and remark that ||u m k − u 0 || L p → 0, ||vm k − v0 || L p → 0. We obtain lim < T  (wm k ), (wm k − w0 ) > = 0.

k→+∞

Moreover, lim (J  (wm k ), (wm k − w0 )) = lim

k→+∞

k→+∞



 (I  (wm k ), (wm k − w0 )) − (T  (wm k ), (wm k − w0 )) .

We have lim (J  (wm k ), (wm k − w0 )) = 0

k→+∞

i.e (J  (wm k ), (wm k − w0 )) = α



|∇u m k | p−2 |∇u m k |∇(u m k − u 0 )d x  |∇vm k | p−2 |∇vm k |∇(vm k − v0 )d x → 0 +β 



as k → +∞. (2.18)

Since wm k  w0 in E and J  (w0 ) ∈ E ∗ ,(J  (w0 ), (wm − w0 )) → 0 as k → +∞. That is  (J  (w0 ), (wm k − w0 )) = α |∇u 0 | p−2 |∇u 0 |∇(u m k − u 0 )d x   +β |∇v0 | p−2 |∇v0 |∇(vm k − v0 )d x → 0, as k → +∞. 

(2.19) Using the well-know inequality: (|s|r −2 s − |¯s |r −2 )(s − s¯ ) ≥ cr |s − s¯ |r , for s, s¯ ∈ R N , r ≥ 2, we deduce that < J  (wm k ) − J  (w0 ), (wm k − w0 ) >  = α (|∇u m k | p−2 ∇u m k − |∇u 0 | p−2 ∇u 0 )∇(u m k − u 0 )d x   + β (|∇vm k | p−2 ∇vm k − |∇v0 | p−2 ∇v0 )∇(vm k − v0 )d x 

≥ c1 ||u m k − u 0 ||W 1, p + c2 ||vm k − v0 ||W 1, p . 0

0

From (2.18), (2.19) it follows that the left-hand side of this inequality converges to zero as 1, p k → +∞. Then we arrive at u m k → u 0 , vm k → v0 as k → +∞ in W0 (). Hence, we deduce that {wm k } converges strongly to w0 in E. Therefore, the functional I satisfies the Palais−Smale condition in E. The proof of the Proposition 2.1 is complete.

B. Q. Hung, H. Q. Toan

Splitting E as the direct sum of X, Y : E = X ⊕ Y where X = L(ϕ) = {tϕ = t (ϕ1 , ϕ2 ), t ∈ R}    α−1 β α β−1 Y = w = (u, v) ∈ E : (uϕ1 ϕ2 + vϕ1 ϕ2 )d x = 0 , 

where ϕ = (ϕ1 , ϕ2 ) is a nomarlized eigenpair associated with the eigenvalue λ1 of the problem (1.3)  ||(ϕ1 , ϕ2 )|| =

1





|∇ϕ1 | p d x +



p

|∇ϕ2 | p d x

= 1.

Since w = (u, v) ∈ E, w = t (ϕ1 , ϕ2 ) + w0 , w0 = (u 0 , v0 ) ∈ Y . u = tϕ1 + u 0

(2.20)

v = tϕ2 + v0

(2.21) β

β−1

Multiplying the equations in (2.20), (2.21) by ϕ1α−1 ϕ2 λ1 and ϕ1α ϕ2 have

λ1 respectively, we

λ1 uϕ1α−1 ϕ2 = λ1 tϕ1α ϕ2 + λ1 u 0 ϕ1α−1 ϕ2 .

β

β

(2.22)

β−1 λ1 vϕ1α ϕ2

β λ1 tϕ1α ϕ2

(2.23)

=

β

β−1 + λ1 v0 ϕ1α ϕ2 .

We remark that β

− p ϕ1 = −div(|∇ϕ1 | p−2 ∇ϕ1 ) = λ1 ϕ1α−1 ϕ2 . β

β

From (2.22) we have λ1 uϕ1α−1 ϕ2 = t (−div(|∇ϕ1 | p−2 ∇ϕ1 ))ϕ1 + λ1 u 0 ϕ1α−1 ϕ2 . By integrating both sides of (2.22), we obtain that      β β λ1 uϕ1α−1 ϕ2 d x = t u 0 ϕ1α−1 ϕ2 d x −div(|∇ϕ1 | p−2 ∇ϕ1 ) ϕ1 d x + λ1     α−1 β p =t |∇ϕ1 | d x + λ1 u 0 ϕ1 ϕ2 d x. (2.24) 



Similary, from (2.23) we also have    β−1 β−1 vϕ1α ϕ2 d x = t |∇ϕ2 | p d x + λ1 v0 ϕ1α ϕ2 d x. λ1 





Hence combining (2.24) and (2.25) we obtain   

β β−1 β uϕ1α−1 ϕ2 + vϕ1α ϕ2 dx = t |∇ϕ1 | p d x + λ1 u 0 ϕ1α−1 ϕ2 d x λ1      β−1 p |∇ϕ2 | d x + λ1 v0 ϕ1α ϕ2 d x. +t 

Since (u 0 , v0 ) ∈ Y , we have 

β β−1 u 0 ϕ1α−1 ϕ2 + v0 ϕ1α ϕ2 d x = 0. 



(2.25)

On existence of weak solutions

Thus, for any w ∈ E such that w = tϕ + w0 , w0 ∈ Y we get

 β β−1 

dx λ1  uϕ1α−1 ϕ2 + vϕ1α ϕ2 α−1 β α β−1   t= uϕ d x. (2.26) = λ ϕ + vϕ ϕ 1 1 1 2 2 p p   |∇ϕ1 | d x +  |∇ϕ2 | d x Moreover, if w = tϕ + w˜ where t is defined in (2.26) then w˜ ∈ Y. Therefore, E = X ⊕ Y. Lemma 2.1 Exists λ¯ > λ1 such that    α β |∇u| p d x + |∇v| p d x ≥ λ¯ |u|α−1 |v|β−1 uvd x, ∀w = (u, v) ∈ Y. p  p      Proof Let λ = inf{ αp  |∇u| p d x + βp  |∇v| p d x : (u, v) ∈ Y,  |u|α−1 |v|β−1 uvd x = 1}. We shall prove that this value is attained in Y . Let wm = (u m , vm ) ∈ Y be a minimizing sequence i.e  |u m |α−1 |vm |β−1 u m vm d x = 1, for m = 1, 2, ... 

and α m→+∞ p



lim



|∇u m | p d x +

β p

 

|∇vm | p d x = λ.

This implies that {wm } is bounded in E. Hence there exists a subsequence {wm k } of {wm } which weakly converges to w0 = (u 0 , v0 ) ∈ E and the compactness of the embedding 1, p W0 () into L p () implies that the subsequences {u m k } and {vm k } converge strongly to u 0 and v0 respectively in L p (). Observe further that with α + β = p 

β β−1 (u m k − u 0 )ϕ1α−1 ϕ2 + (vm k − v0 )ϕ1α ϕ2 dx 

β

β−1

α p ≤ ||u m k − u 0 || L p ||ϕ1 ||α−1 L p |ϕ2 || L p + ||vm k − v0 || L ||ϕ1 || L p |ϕ2 || L p .

Since ||u m k − u 0 || L p () → 0, ||vm k − v0 || L p () → 0 as k → +∞, we deduce that  



β β−1 β β−1 u m k ϕ1α−1 ϕ2 + vm k ϕ1α ϕ2 dx = u 0 ϕ1α−1 ϕ2 + v0 ϕ1α ϕ2 d x. lim k→+∞ 

From this it follows that





β β−1 u 0 ϕ1α−1 ϕ2 + v0 ϕ1α ϕ2 d x = 0, 

hence (u 0 , v0 ) ∈ Y . On the other hand, by the continuity of the operator A   lim |u m k |α−1 |vm k |β−1 u m k vm k d x = |u 0 |α−1 |v0 |β−1 u 0 v0 d x. k→+∞ 

This implies



 

So u 0 = 0 and v0 = 0.

|u 0 |α−1 |v0 |β−1 u 0 v0 d x = 1.

B. Q. Hung, H. Q. Toan

Moreover, since the functional J given by (2.2) is lower weakly semicontinuous, we obtain   α β λ ≤ J (u 0 , v0 ) = |∇u m k | p d x + |∇vm k | p d x p  p      α β p p |∇u m k | d x + |∇vm k | d x = λ, ≤ lim in f m→+∞ p  p  hence λ = J (u 0 , v0 ) =

α p

 

|∇u 0 | p d x +

β p

 

|∇v0 | p d x.

It means that λ is attained at w0 . Our goal is to show that λ > λ1 . By the variational characterization of λ1 , it is clear that: λ ≥ λ1 . If λ = λ1 , by simplicity of λ1 there exists t ∈ R such that w0 = (u 0 , v0 ) = t (ϕ1 , ϕ2 ). Since w0 = (u 0 , v0 ) ∈ Y  

β β−1 β 0= tϕ1 ϕ1α−1 ϕ2 + tϕ2 ϕ1α ϕ2 dx = t ϕ1α ϕ2 d x. 



This contradicts the fact that   β α−1 β−1 1= |u 0 | |v0 | u 0 v0 d x = t ϕ1α ϕ2 d x. 



Thus, there exists λ¯ such that: λ¯ > λ1 and the proof of proposition is complete.



Proposition 2.2 The functional I given by (2.1) is coercive on Y provided hypotheses (H1 ) and (H2 ) hold. Proof Observe that by Holder inequality, Lemma 2.1, hypotheses (H1 ), (H2 ), we have    β α p p |∇u| d x + |∇v| d x − λ1 |u|α−1 |v|β−1 uvd x |I (w)| = | p  p     H (x, u, v)d x + (αk1 u + βk2 v)d x| −         α β λ1 α β p ≥ |min ; ||w|| E − |∇u| p d x + |∇v| p d x p p p  p  λ¯  τ (x)(|u| + |v|)d x − α||k1 || L p ||u|| L p − β||k2 || L p ||v|| L p | −     α β λ1 p ||w|| E − (||τ || L p min ≥| 1− ; p p λ¯ + α||k1 || L p )||u|| L p − (||τ || L p + β||k2 || L p )||v|| L p |       α β λ1 p ≥ | 1− ; ||w|| E −max (||τ || L p +α||k1 || L p ), (||τ || L p +β||k2 || L p ) . min p p λ¯ .c(||u||W 1, p + ||v||W 1, p )|. 0

0

Since ||w E || → +∞ and 1 − λλ¯1 > 0, p ≥ 2, we obtain I (w) → +∞. Thus the functional I given by (2.1) is coercive on Y and Proposition 2.2 is proved.



On existence of weak solutions

From Proposition 2.1 the functional I is coercive on Y , so that BY = min I (w) > −∞. w∈Y

On the other hand, for every t ∈ R we have    α β |∇(tϕ1 )| p d x + |∇(tϕ2 )| p d x − λ1 |tϕ1 |α−1 |tϕ2 |β−1 (tϕ1 )(tϕ2 )d x = 0 p  p   as follows from the definition of λ1 and ϕ. Thus,   H (x, tϕ)d x I (tϕ) = t (αk1 ϕ1 + βk2 ϕ2 )d x −    H (x, tϕ)

=t (αk1 ϕ1 + βk2 ϕ2 ) − d x. t  Remark that H (x, tϕ) 1 = t t



α 2



tϕ1

( f (x, s, tϕ2 ) + f (x, s, 0))ds

0

  β tϕ2 (g(x, tϕ1 , τ ) + g(x, 0, τ ))dτ 2 0   1 α t = (( f (x, yϕ1 , tϕ2 ) + f (x, yϕ1 , 0))dy)ϕ1 t 2 0   t β + ((g(x, tϕ1 , yϕ2 ) + g(x, 0, yϕ2 ))dy)ϕ2 . 2 0 +

Hence, lim

t→+∞

Therefore,

1 H (x, tϕ) = (α F1 (x)ϕ1 + βG 1 (x)ϕ2 ). t 2

 H (x, tϕ)

(αk1 ϕ1 + βk2 ϕ2 ) − dx t→+∞  t    1 = lim t (αk1 ϕ1 + βk2 ϕ2 ) − (α F1 (x)ϕ1 + βG 1 (x)ϕ2 ) d x. t→+∞  2 lim t

On the other hand, from (H2 (i)) we obtain   1 1 (α f +∞ ϕ1 + βg +∞ ϕ2 )d x < (αk1 ϕ1 + βk2 ϕ2 ) d x. p  p  It follows from H2 (ii) that    1 α β (α F1 (x)ϕ1 + βG 1 (x)ϕ2 ) − f +∞ (x)ϕ1 − g +∞ (x)ϕ2 d x p p  2   1 (αk1 ϕ1 + βk2 ϕ2 )d x. > 1− p  Thus,

  

 1 (α F1 (x)ϕ1 + βG 1 (x)ϕ2 ) − (αk1 ϕ1 + βk2 ϕ2 ) d x > 0. 2

B. Q. Hung, H. Q. Toan

This shows that lim I (tϕ) = −∞.

t→+∞

Next, with t < 0 we also have   1 α tϕ1 H (x, tϕ) = ( f (x, s, tϕ2 ) + f (x, s, 0))ds t t 2 0   β tϕ2 (g(x, tϕ1 , τ ) + g(x, 0, τ ))dτ + 2 0   1 α −|t|ϕ1 =− ( f (x, s, −|t|ϕ2 ) + f (x, s, 0))ds |t| 2 0   β −|t|ϕ2 (g(x, −|t|ϕ1 , τ ) + g(x, 0, τ ))dτ . + 2 0 Set s = −yϕ1 → ds = −ϕ1 dy and s = −|t|ϕ1 = −yϕ1 ⇒ y = |t|   H (x, tϕ) 1 α −|t| =− (( f (x, −yϕ1 , −|t|ϕ2 ) + f (x, −yϕ1 , 0))dy)(−ϕ1 ) t |t| 2 0   −|t| β + ((g(x, −|t|ϕ1 , −yϕ2 ) + g(x, 0, −yϕ2 ))dy)(−ϕ2 ) . 2 0 Now, letting t → −∞, we get H (x, tϕ) 1 = t→−∞ t 2



lim

We deduce that

 

lim I (tϕ) = lim t

t→−∞

t→−∞





(α F2 (x)ϕ1 + βG 2 (x)ϕ2 )d x.

 1 (αk1 ϕ1 + βk2 ϕ2 ) − (α F2 (x)ϕ1 + βG 2 (x)ϕ2 ) d x. 2

Similarly above from (H2 (ii)) we obtain   1 (α F2 (x)ϕ1 + βG 2 (x)ϕ2 )d x < (αk1 ϕ1 + βk2 ϕ2 )d x. 2   This implies that lim I (tϕ) = −∞.

t→−∞

Thus, there exists t0 such that |t0 | large enough, we have I (t0 ϕ) < 0. Set w0 (x) = (t0 ϕ1 , t0 ϕ2 ) we get I (w0 ) = I (t0 ϕ) < BY ≤ I (tϕ). Proof of theorem 1.1 By Propositions 2.1 and 2.2, applying the Saddle Point Theorem (P.H.Rabinowitz) (see Theorem 2.1), we deduce that the functional I attains its proper infimum at some w0 = (u 0 , v0 ) ∈ E, so that the problem (1.1) has at least a weak solution w0 ∈ E. Moreover w0 is nontrivial weak solution of the Problem (1.1). The Theorem 1.1 is completely proved. 

On existence of weak solutions

Remark 2.3 We will get the same result as above if the hypotheses (H2 ) is replaced by reverse inequalities as follows. We assume that (H2 )∗    1 α β (α F2 (x)ϕ1 (x) + βG 2 (x)ϕ2 (x)) − f −∞ (x)ϕ1 (x) − g −∞ (x)ϕ2 (x) d x p p  2   1 (αk1 (x)ϕ1 (x) + βk2 (x)ϕ2 (x))d x > > 1− p     1 α β > (α F1 (x)ϕ1 (x) + βG 1 (x)ϕ2 (x)) − f +∞ (x)ϕ1 (x) − g +∞ (x)ϕ2 (x) d x. p p  2 (2.27) This means that, if the conditions (H1 ), (H2 )∗ holds, then the problem (1.1) has at least a nontrivial weak solution in E. This assertion is proved by using variational techniques, the Minimum Principle and generalization of the Landesman–Lazer type condition. Acknowledgments The authors would like to thank the referees for their suggestions and helpful comments which improved the presentation of the paper.

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