On an elliptic equation of Kirchhoff-type with a potential asymptotically linear at infinity

Mathematical and Computer Modelling 49 (2009) 1089–1096

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On an elliptic equation of Kirchhoff-type with a potential asymptotically linear at infinity Ahmed Bensedik ∗ , Mohammed Bouchekif Université de Tlemcen, Faculté des Sciences, Département de mathématiques, BP 119 (13 000) Tlemcen, Algérie

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Article history: Received 19 February 2008 Received in revised form 26 June 2008 Accepted 24 July 2008

This paper examines the existence of positive solutions for a boundary value problem of Kirchhoff-type involving a positive potential function which is asymptotically linear at infinity. © 2008 Elsevier Ltd. All rights reserved.

Keywords: Variational methods Nonlocal problem Kirchhoff’s equation Asymptotically linear at infinity

1. Introduction We consider the existence of positive solutions for the following problem:

  

Z −M u=0

Ω



|∇ u (x)| dx ∆u = f (x, u) 2

in Ω

(P)

on ∂ Ω ,

where Ω is a bounded domain in RN with smooth boundary ∂ Ω , f (x, t ) is a continuous function on Ω × R, which is asymptotically linear at infinity with respect to t for any x in Ω , and M is continuous function on R+ . In 1883, Kirchhoff [8] proposed this type of problems as an extension of the classical d’Alembert’s wave equation for free vibrations of elastic strings. The model studied is

( Z l  2 ) 2 ∂ 2 u (x, t ) ∂u ∂ u (x, t ) − p0 + p1 dx = 0, 2 ∂t ∂x ∂ x2 0

0 < x < l, t > 0,

where p0 is connected with the initial tension, p1 is dependent on the characteristic of the material of the string and u (x, t ) denotes the vertical displacement of the point x of the string at time t. Such problems are often called nonlocal since the equation is no longer a pointwise identity because it contains an integral over Ω . After the famous article by Lions [9], this type of problems has been the subject of numerous studies; we quote, in particular, the papers by Chipot [4,5], Corrêa et al. [6], and the references therein. In [6], the existence of positive solutions was proved for the sublinear case f (u) = uα , 0 < α < 1; by using the Galerkin method.

∗

Corresponding author. E-mail addresses: [email protected] (A. Bensedik), [email protected] (M. Bouchekif).

0895-7177/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2008.07.032

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Several researchers have considered problem (P ) when f has a superlinear growth — see for example [1,2] and the references therein. In their works, they used the following Ambrosetti–Rabinowitz condition there exist µ > 2, and R0 > 0 such that 0 < µF (x, u) ≤ uf (x, u), for all u, |u| ≥ R0 , x ∈ Ω ,

(AR)

Ru

with F (x, u) = 0 f (x, s)ds. It is known that (AR) is an important technical condition to apply Mountain Pass theorem. This condition implies that lim F (x, u) /u2 = +∞,

u−→∞

that is f (x, u) must be superlinear with respect to u at infinity, which is not the case in our work. Note that the case M = 1, has been studied by Zhou [11]. He proved the existence of positive solutions by using a variant of the Mountain Pass theorem. A natural question follows: for M non-identically constant and f (x, t ) is asymptotically linear at infinity with respect to t, can one have existence results when (AR) is not satisfied? The main purpose of the present paper is to answer positively the above question. Thus we are interested in finding sufficient conditions for M and f which make problem (P ) solvable, in the weak sense, by mean of variational methods. We make the following assumptions. (M0 ) M is a continuous function on R+ such that for some m0 > 0 we have M (t ) ≥ m0 ,

for all t ∈ R+ .

(M1 ) there exists m1 > 0, such that M (t ) = m1 , ∀t ≥ t0 , for some t0 > 0. (f1 ) f (x, t ) is a continuous function on Ω × R such that, f (x, t ) ≥ 0 ∀t ≥ 0, x ∈ Ω (f2 ) (f3 )

and f (x, t ) = 0 ∀t ≤ 0, x ∈ Ω ;

f (x, t ) t t limt −→0 f

7→

and |p|∞

is a nondecreasing function for any fixed x ∈ Ω ; (x, t ) /t = p (x) ; limt −→+∞ f (x, t ) /t = q (x) 6= 0
uniformly in x ∈ Ω , where 0 ≤ p (x) , q (x) ∈ L∞ (Ω ) < m0 λ1 , λ1 is the first eigenvalue of −∆, H01 (Ω ) .

Before stating our results we need to introduce some notations. Notation 1. Throughout this paper, we denote by |.|p the Lp -norm, 1 ≤ p ≤ ∞; and by k.k the norm of H01 (Ω ) induced by R the inner product (u, v) = Ω ∇ u.∇v dx, and we have to use the notation u± = max (±u, 0). The letter C will denote various positive constants whose value may change from line to line but are not essential to the analysis of the problem. Also if we take a subsequence of a sequence (un ) we shall denote it again (un ). We use o(1) to denote any quantity which tends to zero when n → +∞. We know from the article of de Figueiredo [7], that the first eigenvalue λ1 (q) of the problem



−∆u = λq (x) u u=0

in Ω on ∂ Ω ,

where q is a continuous positive function on Ω , is positive. Let φ1 ∈ H01 (Ω ) be its associated eigenfunction. Let also, µ1 and ψ1 ∈ H01 (Ω ) be the first eigenvalue and eigenfunction of the problem



−M (kuk2 )∆u = µq (x) u u=0

in Ω on ∂ Ω ,

from [1], µ1 and ψ1 are positive. The corresponding energy functional of (P ) is denoted by I ( u) =

1 2

b kuk2 − M


Z Ω

F (x, u) dx,

Rt

Ru

b (t ) = M (s) ds and F (x, u) = f (x, s)ds. where M 0 0 A function u ∈ H01 (Ω ) is said to be a weak solution of (P ) if M k uk 2




Z Ω

∇ u.∇φ dx −

Z Ω

f (x, u) φ dx = 0

∀φ ∈ H01 (Ω ) .

Now we give our main results. Theorem 1. Assume that (f1 ) to (f3 ) hold and M satisfies (M0 ) and (M1 ). Then 1. If µ1 > 1, (P ) has no solution. 2. If µ1 < 1, (P ) has a positive solution.

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3. If µ1 = 1 and m0 λ1 (q) ≥ 1 and if there is a positive solution u ∈ H01 (Ω ) of (P ) then f (x, u) = λ1 (q) q (x) M kuk2 u




a.e. in Ω .

Theorem 2. Suppose that M is a nondecreasing function and satisfying (M0 ). Assume further that (f1 ) to (f3 ) hold with |p|∞ > 0. Then if max (λ1 (q) |q|∞ , µ1 ) < 1,

(q0 )

problem (P ) admits a positive solution. This paper is organized as follows. Section 2 contains the proof of Theorem 1 and Section 3 is concerned with the proof of Theorem 2. 2. M is a bounded function We give some auxiliary preliminaries. We start by giving another version of Mountain Pass theorem. Lemma 1 ([3]). Let E be a real Banach space and suppose that I ∈ C 1 (E , R) satisfies the condition max {I (0) , I (u1 )} ≤ α < β ≤

inf I (u)

kukE =ρ

for some α < β , ρ > 0 and u1 ∈ E with ku1 kE > ρ . Let c ≥ β be characterized by c = inf max I (γ (τ )) where Γ = {γ ∈ C ([0, 1] , E ) ; γ (0) = 0, γ (1) = u1 } γ ∈Γ τ ∈[0,1]

is the set of continuous paths joining 0 and u1 . Then there exists a sequence (un ) ⊂ E such that



n

0



n

0

I (un ) → c ≥ β and (1 + kun kE ) I (un ) 0 → 0 (E dual of E ). E

Proposition 1. Under the assumptions of Theorem 1, if m1 λ1 (q) < 1,

(1)

then the energy functional I satisfies the following geometric properties: (i) there exist ρ, α > 0 such that I (u) ≥ α if kuk = ρ. (ii) there exists v ∈ H01 (Ω ) such that kvk > ρ and I (v) ≤ 0. Proof. It follows from (f1 ) and (f3 ); that for any ε > 0, there exists A = A (ε) ≥ 0 such that for all (x, s) ∈ Ω × R, F (x, s) ≤

1 2

(|p|∞ + ε) s2 + Asγ +1

(2)

2 where 1 < γ < ∞ if N = 1, 2 and 1 < γ < NN + if N ≥ 3. −2 Choosing 0 < ε < m0 λ1 − |p|∞ , then from (2) and Sobolev inequality, we obtain

Z
1 1 γ +1 2 b I (u) = M kuk − F (x, u) dx ≥ m0 kuk2 − (|p|∞ + ε) |u|22 − A |u|γ +1 2 2 2 Ω   |p|∞ + ε 1 k uk 2 − C k uk γ + 1 . ≥ m0 − 2 λ1 1

Since γ > 1 the condition (i) follows. To prove (ii), we can write for t sufficiently large

b (t ) = M

t

Z

M (s) ds = 0

t0

Z 0

M (s) ds +

Z

t

m1 ds,

t0

b (t0 ) − m1 t0 + m1 t =M ≤ m2 + m1 t , with m2 = b M (t0 ) − m1 t0 .

for all t ≥ t0

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Using Fatou’s Lemma, we get lim

I (t φ1 ) t2

t →+∞

Z  F (x, t φ1 ) 2 kφ k − lim + m dx 1 1 t →+∞ Ω t →+∞ 2 t2 t2 Z F (x, t φ1 ) 2 1 lim ≤ m1 kφ1 k2 − φ1 dx t →+∞ 2 t 2 φ12 Ω 1  m2

≤ lim

1

=

2

m1 kφ1 k2 −

1 2λ1 (q)

kφ1 k2 ,

then lim

I (t φ1 ) t2

t →+∞

≤

1

 m1 −

2

1



λ1 (q)



ρ

kφ1 k2 < 0, since m1 λ1 (q) < 1.



Thus there exists t1 > max t0 , kφ k such that I (t1 φ1 ) ≤ 0. 1 By taking v = t1 φ1 , (ii) follows.  Now we are in position to prove Theorem 1. Proof of Theorem 1. We start by proving the first point. If µ1 > 1 then problem (P ) has no solution. Indeed, if u ∈ H01 (Ω ) is a solution of (P ), then from (f1 ) to (f3 ) we get M kuk2 kuk2 =




Z Ω

f (x, u) udx ≤

Z Ω

q (x) u2 dx,

then M kuk2 kuk2




µ1 =

inf

u∈H 1 (Ω ) 0 u6=0

R

Ω

≤ 1.

q (x) u2 dx



ρ

Claim 2: Suppose that µ1 < 1. From Proposition 1 there exists t1 > max t0 , kφ k 1 Define


Γ = γ ∈ C [0, 1] , H01 (Ω ) ; γ (0) = 0, γ (1) = t1 φ1 Then by Lemma 1 there exists a sequence (un ) ⊂ I ( un ) =

1 2

b k un k 2 − M


Z Ω

H01



such that I (t1 φ1 ) < 0.

and c = inf max I (γ (τ )) . γ ∈Γ τ ∈[0,1]

(Ω ) such that

F (x, un ) dx = c + o(1),

(3)

and

0

(1 + kun k) I (un )

n

H −1 (Ω )

→ 0,

(4)

which implies that

D

E

0

I (un ) , un = M kun k2 kun k2 −




Z Ω

f (x, un ) un dx = o(1).

(5) n

Let us show that the sequence (un ) is bounded. Suppose by contradiction that kun k → +∞, and let

√ wn =

t0

k un k

un .

(6)

Clearly (wn ) is bounded in H01 (Ω ), then there exists a subsequence (wn ) such that n

wn * w weakly in H01 (Ω ) , n

wn → w strongly in L2 (Ω ) , n

wn → w a.e. in Ω . We have

w 6= 0.

(7)

A. Bensedik, M. Bouchekif / Mathematical and Computer Modelling 49 (2009) 1089–1096

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f (x,t ) Indeed, supposing w = 0. By (f1 ) and (f3 ) there exists θ > 0 such that t ≤ θ for x ∈ Ω and t ≥ 0. Then from (5) and (6) we obtain t0 m0 ≤ t0 M kun k

2




f x , u+ n

Z =




u+ n

Ω

w

2 n dx

+ o(1) ≤ θ

Z Ω

wn2 dx + o(1) → 0,

which is impossible because t0 , m0 > 0, so w 6= 0. Now we show that w satisfies the identity

Z

m1 ∇w.∇φ − µ21 q (x) wφ dx = 0,

for all φ ∈ H01 (Ω ) .




Ω

Put for x ∈ Ω with un (x) > 0; for x ∈ Ω with un (x) ≤ 0.

 2
1 2 µ1 f (x, un ) u− n / M k un k

pn (x) =

0

As above 0 ≤ pn (x) ≤ µ21 θ /m0 ∀x ∈ Ω , then there exists a subsequence (pn ) such that n

pn * h weakly in L2 (Ω ) , n

with 0 ≤ h ≤ µ21 θ /m0 .

n

Since kun k → +∞, and wn → w a.e. in Ω , it follows from (6) that un −→ +∞ a.e. in Ω ,

if w (x) > 0 a.e. in Ω .

Then by (f3 ) and (M1 ) we get h (x) =

µ21 m1

if w (x) > 0.

q ( x)

(8)

n

Since wn → w strongly in L2 (Ω ), one has

Z Ω

pn (x) wn (x) φ (x) dx =

Z

n

pn (x) wn (x) φ (x) dx → +

Ω

Z Ω

h (x) w + (x) φ (x) dx,

∀φ ∈ L2 (Ω ) ;

then n

pn wn * hw + weakly in L2 (Ω ) .



0

By (4) I (un )

(9)

n

H −1 (Ω )

n

→ 0 and the fact that kun k → +∞, we deduce that for any φ ∈ H01 (Ω ) ,

Z √ D 0 E t0 (∇wn .∇φ − pn (x) wn (x) φ) dx =
I ( un ) , φ 2 k k k k u M u Ω n n √

0

t0 n


≤ I u ( )

−1 kφk → 0. n 2 H Ω ( ) k un k M k un k n

So by using wn * w weakly in H01 (Ω ) and (9), we obtain

Z Ω


∇w.∇φ − h (x) w + φ dx = 0,

∀φ ∈ H01 (Ω ) .

Taking φ = w − , it follows that w − = 0 and so w = w + ≥ 0 on Ω . Then by the maximum principle we get w (x) > 0 on



µ2

Ω . Thus by (8) we get h (x) = m1 q (x) and 1 Z
m1 ∇w.∇φ − µ21 q (x) wφ dx = 0, Ω

∀φ ∈ H01 (Ω ) ,

thus

R |∇w|2 dx µ21 RΩ = , m1 q x w 2 dx Ω ( ) and this implies

|∇w|2 dx M (t0 ) 2 = µ1 = µ21 < µ1 < 1, 2 m1 q x w dx Ω ( )

M kwk2

R


R

Ω

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this is a contradiction with the fact that µ1 is the first eigenvalue. So (un ) is bounded in H01 (Ω ) , and since Ω and M are bounded and f (x, u) is subcritical with respect to u, by the compactness of Sobolev imbedding and standard arguments, we know that there exists a subsequence of (un ) which converges strongly to a nontrivial critical point of I , solution of the problem (P ). Recall that under the condition

|f (x, u)| ≤ C (1 + |u|γ ) , where C > 0, 1 < γ <

N +2 , N −2

∀x ∈ Ω , ∀u ∈ R, if N ≥ 3, and 1 < γ < ∞ if N = 1, 2; which is fulfilled in our case; using the compact 0

γ

imbedding (Ω ) ,→ L (Ω ) and the continuity of the Nemytskii map Nf from Lγ (Ω ) to L(γ +1) (Ω ) , Ma [10] proved that 2 if kun k → l and un * u weakly in H01 (Ω ), then kuk2 = l. Claim 3: Suppose that µ1 = 1. By the definition of µ1 , we have H01

M kψ1 k2

Z




Ω

Z

∇ψ1 .∇v dx =

Ω

q (x) ψ1 v dx, ∀v ∈ H01 (Ω ) .

(10)

If u is a positive solution of (P ), we have for ψ1 M k uk 2




Z Ω

∇ u.∇ψ1 dx =

Z Ω

f (x, u) ψ1 dx.

(11)

q (x) ψ1 udx.

(12)

Put u = v in (10) we obtain M kψ1 k2




Z Ω

∇ψ1 .∇ udx =

Z Ω

Then from (11) and (12) we deduce f (x, u)

Z Ω

M k uk

q (x) u

Z


ψ1 dx = 2

Ω

M kψ1 k2


ψ1 dx

that is f ( x , u)

Z Ω

M k uk 2


−

q (x) u M kψ1 k2

!
ψ1 dx = 0.

(13)

We have f (x, u) M k uk 2


−

q (x) u M kψ1 k2


≤ 0.

Indeed, µ1 = 1 implies that 1 M kψ1 k

kψ1 k2 ≥ q x ψ 2 dx Ω ( )


= R 2

inf

u∈H 1 (Ω ) 0 u6=0

R

kuk2 = λ1 (q) . q x u2 dx Ω ( )

Then M kψ1 k2 λ1 (q) ≤ 1, but M kψ1 k2 λ1 (q) ≥ m0 λ1 (q) ≥ 1,



2




λ1 (q) = 1. Using this equality we can write " # f (x, u) q (x) u f (x, u)
−
− q (x) u ,
= λ1 (q) M k uk 2 M kψ1 k2 λ1 (q) M kuk2
and since λ1 (q) M kuk2 ≥ λ1 (q) m0 ≥ 1, we get so M kψ1 k

f (x, u)
− q (x) u ≤ f (x, u) − q (x) u ≤ 0. λ1 (q) M kuk2 Now since ψ1 > 0 a.e. in Ω , we deduce from (13) that f (x, u) =

M k uk 2




M kψ1 k2



q (x) u = λ1 (q) q (x) M kuk2 u,

a.e. in Ω . 

A. Bensedik, M. Bouchekif / Mathematical and Computer Modelling 49 (2009) 1089–1096

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3. M is a nondecreasing function In order to prove Theorem 2, we establish a lemma which shows that any solution of (P ) satisfies a priori estimate. Lemma 2. Assume that (M1 ) and (f1 ) to (f3 ) are satisfied with |p|∞ > 0. Then if |q|∞ < m1 , every positive solution u of (P ) satisfies the estimate

k uk ≤ C , where C is a positive constant depending on m0 , m1 , |p|∞ , |q|∞ and t0 . Proof. Set p0 := infx∈Ω p (x) . Then we have F (x, u) ≥ 12 p0 |u|2 , and thus c0 ≥ c, where c and c0 are the Mountain Pass levels related to the functionals I and I0 respectively, where I0 (u) =

1 2

b k uk 2 − M


Z

p0 2

Ω

|u|2 dx.

Let u be a positive solution of (P ). If kuk2 < t0 , we take C = So suppose that kuk2 ≥ t0 . From (f3 ), we have

√

t0 .

|p|∞ < m0 λ1 , then using the definition of λ1 , one has

|u|22 ≤

m0

|p|∞

k uk 2 .

(14)

Furthermore c0 ≥ c =

≥

1 2

1 2

b k uk 2 − M


m 1 k uk 2 −

1 2

Z

F (x, u) dx

Ω

m1 t0 −

1 2

|q|∞ |u|22 .

(15)

Using (14), we get from (15)

(2c0 + m1 t0 ) |p|∞ ≥ (m1 |p|∞ − m0 |q|∞ ) kuk2 . The constant m0 can be chosen such that m0 ≤ |p|∞ and since |q|∞ < m1 , we obtain

k uk 2 ≤

(2c0 + m1 t0 ) |p|∞ m1 |p|∞ − m0 |q|∞

,

and then

k uk ≤ C q (2c +m t )|p| with C = m |0p| −1 m0 |q|∞ . 1 0 ∞



∞

Now let us prove Theorem 2. Proof of Theorem 2. The function M is supposed nondecreasing, let us define the truncated problem (Pk )




−Mk kuk2 ∆u = f (x, u) u=0

in Ω on ∂ Ω ,

(Pk )

where

 M (t )   at + b Mk (t ) =  δ  λ1 (q)

if t ≤ k if k < t ≤ k + 1 if t > k + 1,

with λ1 (q) |q|∞ < δ < 1, k > 0 sufficiently large and a=

δ − M (k) , λ1 (q)

b = (1 + k) M (k) −

δk . λ1 (q)

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A. Bensedik, M. Bouchekif / Mathematical and Computer Modelling 49 (2009) 1089–1096

The existence of δ follows from the condition (q0 ) of Theorem 2. We see that Mk satisfies (M1 ) with m1 = λ δ(q) and the 1 condition (1) of Proposition 1 is obviously fulfilled. Then since Mk kuk2 kuk2




µ1k =

inf

u∈H 1 (Ω ) 0 u6=0

R

Ω

q (x) u2 dx

M kuk2 kuk2




≤

inf

u∈H 1 (Ω ) 0 u6=0

R

Ω

q (x) u2 dx

= µ1 < 1,

we conclude, from Theorem 1, that problem (Pk ) admits a positive solution uk . We have |q|∞ < m1 because λ1 (q) |q|∞ < δ = m1 λ1 (q), then from Lemma 2 we deduce that 2

k uk k 2 ≤ C . If kuk k2 > k we obtain 2

k


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