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This article is part of the series Recent Advances in Operator Equations, Boundary Value Problems, Fixed Point Theory and Applications, and General Inequalities.

Open Access Research

Nonlinear biharmonic boundary value problem

Tacksun Jung1 and Q-Heung Choi2*

Author Affiliations

1 Department of Mathematics, Kunsan National University, Kunsan, 573-701, Korea

2 Department of Mathematics Education, Inha University, Incheon, 402-751, Korea

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Boundary Value Problems 2014, 2014:30  doi:10.1186/1687-2770-2014-30


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2014/1/30


Received:28 June 2013
Accepted:8 January 2014
Published:4 February 2014

© 2014 Jung and Choi; licensee Springer.

This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We consider the nonlinear biharmonic equation with variable coefficient and polynomial growth nonlinearity and Dirichlet boundary condition. We get two theorems. One theorem says that there exists at least one bounded solution under some condition. The other one says that there exist at least two solutions, one of which is a bounded solution and the other of which has a large norm under some condition. We obtain this result by the variational method, generalized mountain pass geometry and the critical point theory of the associated functional.

MSC: 35J20, 35J25, 35Q72.

Keywords:
biharmonic boundary value problem; polynomial growth; variational method; generalized mountain pass geometry; critical point theory; ( PS ) condition

1 Introduction

Let Ω be a bounded domain in R n with smooth boundary Ω and L 2 ( Ω ) be a square integrable function space defined on Ω. Let Δ be the elliptic operator and Δ 2 be the biharmonic operator. Let c R . In this paper we study the following nonlinear biharmonic equation with Dirichlet boundary condition:

Δ 2 u + c Δ u = a ( x ) g ( u ) in  Ω , u = 0 , Δ u = 0 on  Ω , (1.1)

where a : Ω ¯ R is a continuous function which changes sign in Ω.

We assume that g satisfies the following conditions:

(g1) g C ( R , R ) ,

(g2) there are constants a 1 , a 2 0 such that

| g ( u ) | a 1 + a 2 | u | μ 1 ,

where 2 < μ < 2 n n 2 if n 3 ,

(g3) there exists a constant r 0 0 such that

0 < μ G ( ξ ) = μ 0 ξ g ( t ) d t ξ g ( ξ ) for  | ξ | r 0 ,

(g4) g ( u ) = o ( | u | ) as u 0 .

We note that (g3) implies the existence of the positive constants a 3 , a 4 , a 5 such that

1 μ ( ξ g ( ξ ) + a 3 ) G ( ξ ) + a 4 a 5 | ξ | μ for  ξ R . (1.2)

Remark 1.1 The real number ξ in the definition (g3) is not automatically nonnegative. The reason is as follows.

Since 0 < μ G ( ξ ) < ξ g ( ξ ) and μ > 2 , G ( ξ ) > 0 and ξ g ( ξ ) > 0 . By ξ g ( ξ ) > 0 , we have two cases: one case is that ξ > 0 and g ( ξ ) > 0 . The other case is that ξ < 0 and g ( ξ ) < 0 . Thus ξ is not nonnegative.

Remark 1.2 We obtain the boundedness of 1 2 g ( u ) u G ( u ) as follows.

By the condition (g3), 1 2 g ( u ) u 1 2 μ G ( u ) for | u | r 0 . Since μ > 2 , 1 2 1 μ > 0 , and G ( u ) + a 4 a 5 | u | μ in (1.2),

1 2 g ( u ) u G ( u ) 1 2 μ G ( u ) G ( u ) = μ ( 1 2 1 μ ) G ( u ) μ ( 1 2 1 μ ) ( a 5 | u | μ a 4 ) .

Thus we obtain the boundedness of 1 2 g ( u ) u G ( u ) .

Remark 1.3 (i) Assumption (g4) implies that (1.1) has a trivial solution.

(ii) If n = 1 , (g2) can be dropped. If n = 2 , it suffices that

| g ( u ) | a 1 exp p ( ξ ) ,

where p ( ξ ) ξ 2 0 as | ξ | .

(iii) If n 3 and g ( ξ ) = ξ + 1 + ϵ , where ξ + = max { ξ , 0 } and ϵ > 0 is a small number, then (g1)-(g4) are satisfied.

The eigenvalue problem

Δ u + λ u = 0 in  Ω , u = 0 on  Ω

has infinitely many eigenvalues λ k , k 1 , and corresponding eigenfunctions ϕ k , k 1 , suitably normalized with respect to the L 2 ( Ω ) inner product, where each eigenvalue λ k is repeated as often as its multiplicity. The eigenvalue problem

Δ 2 u + c Δ u = Λ u in  Ω , u = 0 , Δ u = 0 on  Ω

has also infinitely many eigenvalues λ k ( λ k c ) , k 1 and corresponding eigenfunctions ϕ k , k 1 . We note that λ 1 ( λ 1 c ) λ 2 ( λ 2 c ) + , and that ϕ 1 ( x ) > 0 for x Ω .

Khanfir and Lassoued [1] showed the existence of at least one solution for the nonlinear elliptic boundary problem when g is locally Hölder continuous on R + . Choi and Jung [2] showed that the problem

Δ 2 u + c Δ u = b u + + s in  Ω , u = 0 , Δ u = 0 on  Ω (1.3)

has at least two nontrivial solutions when ( c < λ 1 , λ 1 ( λ 1 c ) < b < λ 2 ( λ 2 c ) and s < 0 ) or ( λ 1 < c < λ 2 , b < λ 1 ( λ 1 c ) and s > 0 ). The authors obtained these results by using the variational reduction method. The authors [3] also proved that when c < λ 1 , λ 1 ( λ 1 c ) < b < λ 2 ( λ 2 c ) and s < 0 , (1.3) has at least three nontrivial solutions by using degree theory. Tarantello [4] also studied

Δ 2 u + c Δ u = b ( ( u + 1 ) + 1 ) , u = 0 , Δ u = 0 on  Ω . (1.4)

She showed that if c < λ 1 and b λ 1 ( λ 1 c ) , then (1.4) has a negative solution. She obtained this result by degree theory. Micheletti and Pistoia [5] also proved that if c < λ 1 and b λ 2 ( λ 2 c ) then (1.4) has at least four solutions by the variational linking theorem and Leray-Schauder degree theory.

In this paper we are trying to find the weak solutions of (1.1), that is,

Ω ( Δ 2 u + c Δ u a ( x ) g ( u ) ) v d x = 0 for any  v H ,

where the space H is introduced in Section 2. Let us set

Ω + = { x Ω a ( x ) > 0 } , Ω = { x Ω a ( x ) < 0 }

and let

a + = a χ Ω + , a = a χ Ω .

Since a ( x ) changes sign, the open subsets Ω + and Ω are nonempty. Now we can write a = a + a . Our main results are as follows.

Theorem 1.1Assume that λ k < c < λ k + 1 , gsatisfies (g1)-(g4) and g ( u ) u μ G ( u ) is bounded. Then (1.1) has at least one bounded nontrivial solution.

Theorem 1.2Assume that λ k < c < λ k + 1 , gsatisfies (g1)-(g4), g ( u ) u μ G ( u ) is not bounded and there exists a small ϵ > 0 such that Ω a ( x ) d x < ϵ . Then (1.1) has at least two solutions, (i) one of which is nontrivial and bounded, and (ii) the other of which has a large norm such that

max x Ω | u ( x ) | > M for some  M .

The outline of Theorem 1.1 and Theorem 1.2 is as follows: In Section 2, we prove that the corresponding functional I ( u ) of (1.1), which is introduced in (2.1), is continuous and Fréchet differentiable and satisfies the ( PS ) condition. In Section 3, we prove Theorem 1.1. In Section 4, we prove Theorem 1.2 by the variational method, the generalized mountain pass geometry and the critical point theory.

2 Palais-Smale condition

Any element u in L 2 ( Ω ) can be written as

u = h k ϕ k with  h k 2 < .

We define a subspace H of L 2 ( Ω ) as follows:

H = { u L 2 ( Ω ) | | λ k ( λ k c ) | h k 2 < } .

Then this is a Banach space with a norm

u = [ | λ k ( λ k c ) | h k 2 ] 1 2 .

Since λ k + and c is fixed, we have

(i) Δ 2 u + c Δ u H implies u H ,

(ii) u C u L 2 ( Ω ) , for some C > 0 ,

(iii) u L 2 ( Ω ) = 0 if and only if u = 0 ,

which are proved in [6].

Let

H + = { u H h k = 0  if  λ k ( λ k c ) < 0 } , H = { u H h k = 0  if  λ k ( λ k c ) > 0 } .

Then H = H H + . Let P + be the orthogonal projection on H + and P be the orthogonal projection on H .

We are looking for the weak solutions of (1.1). The weak solutions of (1.1) coincide with the critical points of the associated functional

I ( u ) C 1 ( H , R ) , I ( u ) = Ω [ 1 2 | Δ u | 2 c 2 | u | 2 Ω a ( x ) G ( u ) ] d x I ( u ) = 1 2 ( P + u 2 P u 2 ) Ω a ( x ) G ( u ) d x . (2.1)

By (g1) and (g2), I is well defined. By Proposition 2.1, I C 1 ( H , R ) and I is Fréchet differentiable in H.

Proposition 2.1Assume that λ k < c < λ k + 1 , k 1 , and thatgsatisfies (g1)-(g4). Then I ( u ) is continuous and Fréchet differentiable inHwith Fréchet derivative

I ( u ) h = Ω [ Δ u Δ h c u h a ( x ) g ( u ) h ] d x . (2.2)

If we set

K ( u ) = Ω a ( x ) G ( u ) d x ,

then K ( u ) is continuous with respect to weak convergence, K ( u ) is compact, and

K ( u ) h = Ω a ( x ) g ( u ) h d x for all  h H .

This implies that I C 1 ( H , R ) and K ( u ) is weakly continuous.

The proof of Proposition 2.1 is the same as that of Appendix B in [7].

Proposition 2.2 (Palais-Smale condition)

Assume that λ k < c < λ k + 1 , k 1 , andgsatisfies (g1)-(g4). We also assume that g ( u ) u μ G ( u ) is bounded or that there exists an ϵ > 0 such that Ω a ( x ) d x < ϵ . Then I ( u ) satisfies the Palais-Smale condition.

Proof Suppose that ( u m ) is a sequence with I ( u m ) M and I ( u m ) 0 as m . Then by (g2), (g3), and the Hölder inequality and the Sobolev Embedding Theorem, for large m and μ > 2 with u = u m , we have

M + 1 2 u I ( u ) 1 2 I ( u ) u = Ω [ 1 2 a ( x ) g ( u ) u a ( x ) G ( u ) ] d x = Ω a + ( x ) [ 1 2 g ( u ) u G ( u ) ] d x Ω a ( x ) [ 1 2 g ( u ) u G ( u ) ] d x ( 1 2 1 μ ) μ Ω a + ( x ) G ( u ) d x max Ω | 1 2 g ( u ) u G ( u ) | Ω a ( x ) d x ( 1 2 1 μ ) μ Ω a + ( x ) ( a 5 | u | μ a 4 ) d x max Ω | 1 2 g ( u ) u G ( u ) | Ω a ( x ) d x .

Since 1 2 g ( u ) u G ( u ) is bounded or there exists an ϵ > 0 such that Ω a ( x ) < ϵ ; we have

1 + u M 1 Ω | u | μ d x M 2 ( Ω | u | 2 d x ) 1 2 μ . (2.3)

Moreover since

| I ( u m ) φ | φ (2.4)

for large m and all φ H , choosing φ = P + u m H + gives

P + u m 2 = Ω ( Δ 2 u m + c Δ u m ) P + u m d x = Ω a ( x ) g ( u m ) P + u m d x Ω | a ( x ) | | g ( u m ) | | u m | d x a Ω ( a 1 | u m | + a 2 | u m | μ ) d x C 1 Ω | u m | μ d x + C 2 u m L 2 ( Ω ) C 1 Ω | u m | μ d x + C 2 u m .

Taking φ = P u m in (2.4) yields

P u m 2 = Ω ( Δ 2 u m + c Δ u m ) ( P u m ) d x = Ω a ( x ) g ( u m ) ( P u m ) d x Ω | a ( x ) | | g ( u m ) | | u m | d x C 3 Ω | u m | μ d x + C 4 u m .

Thus, by (2.3), we have

u m 2 = P + u m 2 + P u m 2 M 3 Ω | u m | μ d x + M 4 u m M 5 ( 1 + u m ) + M 4 u m M 6 ( 1 + u m ) ,

from which the boundedness of ( u m ) follows. Thus ( u m ) converges weakly in H. Since P ± I ( u m ) = ± P ± u m + P ± P ˜ ( u m ) with P ˜ compact and the weak convergence of P ± u m imply the strong convergence of P ± u m and hence ( PS ) condition holds. □

3 Proof of Theorem 1.1

We shall show that I ( u ) satisfies the generalized mountain pass geometrical assumptions.

We recall the generalized mountain pass geometry.

Let H = V X , where V { 0 } and is finite dimensional. Suppose that I C 1 ( H , R ) , satisfies the Palais-Smale condition, and

(i) there are constants ρ , α > 0 and a bounded neighborhood B ρ of 0 such that I | B ρ X α , and

(ii) there is an e B 1 X and R > ρ such that if Q = ( B ¯ R V ) { r e 0 < r < R } , then I | Q 0 .

Then I possesses a critical value b α . Moreover b can be characterized as

b = inf γ Γ max u Q I ( γ ( u ) ) ,

where

Γ = { γ C ( Q ¯ , H ) γ = id  on  Q } .

Let H k = span { ϕ 1 , , ϕ k } . Then H k is a subspace of H such that

H = k N H k and H = H k H k .

Let

B r = { u H u r } .

We have the following generalized mountain pass geometrical assumptions.

Lemma 3.1Assume that λ k < c < λ k + 1 andgsatisfies (g1)-(g4). Then

(i) there are constants ρ > 0 , α > 0 and a bounded neighborhood B ρ of 0 such that I | B ρ H k α , and

(ii) there is an e B 1 H k and R > ρ such that if Q = ( B ¯ R H k ) { r e 0 < r < R } , then I | Q 0 , and

(iii) there exists u 0 H such that u 0 > ρ and I ( u 0 ) 0 .

Proof (i) Let u H k . Then

Ω ( Δ 2 u + c Δ u ) u d x λ k + 1 ( λ k + 1 c ) u L 2 ( Ω ) 2 > 0 .

Thus by (g2), (g4), and the Hölder inequality, we have

I ( u ) = 1 2 P + u 2 1 2 P u 2 Ω a ( x ) G ( u ) d x 1 2 P + u 2 a Ω C 1 | u | μ d x 1 2 P + u 2 a C 1 u μ

for C 1 , C 1 > 0 . Since μ > 2 , there exist ρ > 0 and α > 0 such that if u B ρ , then I ( u ) α .

(ii) Let B r be a ball with radius r > 0 , e be a fixed element in B 1 H k and u ( B ¯ r H k ) { r e 0 < r } . Then u = v + w , v B r H k , w = r e . We note that

since  v H k , Ω ( Δ 2 v + c Δ v ) v d x λ k ( λ k c ) v L 2 ( Ω ) 2 < 0 .

Thus we have

I ( u ) = 1 2 r 2 1 2 P v 2 Ω a ( x ) G ( v + r e ) d x 1 2 r 2 + 1 2 ( λ k ( λ k c ) ) v L 2 ( Ω ) 2 Ω + a ( x ) ( a 5 | v + r e | μ a 4 ) d x .

Since μ > 2 , there exists R > 0 such that if u Q = ( B ¯ R H k ) { r e 0 < r < R } , then I ( u ) < 0 .

(iii) If we choose ψ H such that ψ = 1 , ψ 0 in Ω and supp ( ψ ) Ω + , then we have

I ( t ψ ) 1 2 P + ( t ψ ) 2 1 2 P ( t ψ ) 2 Ω + a ( x ) ( a 5 t μ ψ μ a 4 ) d x 1 2 t ψ 2 Ω + a ( x ) ( a 5 t μ ψ μ a 4 ) d x = 1 2 t 2 Ω + a ( x ) ( a 5 t μ ψ μ a 4 ) d x

for all t > 0 . Since μ > 2 , for t 0 great enough, u 0 = t 0 ψ is such that u 0 > ρ and I ( u 0 ) 0 . □

Proof of Theorem 1.1 By Proposition 2.1 and Proposition 2.2, I ( u ) C 1 ( H , R ) and satisfies the Palais-Smale condition. By Lemma 3.1, there are constants ρ > 0 , α > 0 and a bounded neighborhood B ρ of 0 such that I | B ρ H m α , and there is an e B 1 H k and R > ρ such that if Q = ( B ¯ R H k ) { r e 0 < r < R } , then I | Q 0 , and there exists u 0 H such that u 0 > ρ and I ( u 0 ) 0 . By the generalized mountain pass theorem, I ( u ) has a critical value b α . Moreover, b can be characterized as

b = inf γ Γ max u Q I ( γ ( u ) ) ,

where

Γ = { γ C ( Q ¯ , H ) γ = id  on  Q } .

We denote by u ˜ a critical point of I such that I ( u ˜ ) = b . We claim that there exists a constant C > 0 such that

a + ( x ) 1 μ u ˜ L 2 ( Ω ) C ( 1 + L Ω a ( x ) d x ) 1 μ ,

where L = max Ω | 1 2 g ( u ˜ ) u ˜ G ( u ˜ ) | .

In fact, we have

b max 0 t 1 I ( t u 0 )

and

I ( t u 0 ) = t 2 ( 1 2 P + u 0 2 1 2 P u 0 2 ) Ω a ( x ) G ( t u 0 ) d x t 2 u 0 2 Ω a + ( x ) G ( t u 0 ) d x + Ω a ( x ) G ( t u 0 ) d x t 2 u 0 2 a 5 t μ Ω a + ( x ) u 0 μ d x + a 4 Ω a + ( x ) d x + a 5 t μ Ω a ( x ) u 0 μ d x = C t 2 C t μ + C + C t μ .

Since 0 t 1 , b is bounded: b < C ˜ .

We can write

b = I ( u ˜ ) 1 2 I ( u ˜ ) u ˜ = Ω a ( x ) ( 1 2 g ( u ˜ ) u ˜ G ( u ˜ ) ) d x = Ω a + ( x ) ( 1 2 g ( u ˜ ) u ˜ G ( u ˜ ) ) d x Ω a ( x ) ( 1 2 g ( u ˜ ) u ˜ G ( u ˜ ) ) d x ( 1 2 1 μ ) Ω a + ( x ) g ( u ˜ ) u ˜ d x max Ω | 1 2 g ( u ˜ ) u ˜ G ( u ˜ ) | Ω a ( x ) d x ( 1 2 1 μ ) μ Ω a + ( x ) ( a 3 | u ˜ | μ a 4 ) d x L Ω a ( x ) d x ,

where L = max Ω | 1 2 g ( u ˜ ) u ˜ G ( u ˜ ) | . Thus we have

C ( 1 + L Ω a ( x ) d x ) Ω a + ( x ) | u ˜ | μ d x C [ Ω ( a + ( x ) 1 μ | u ˜ | ) 2 d x ] μ 2 (3.1)

for some constants C , C > 0 , from which we conclude that u ˜ is bounded and the proof of Theorem 1.1 is complete. □

4 Proof of Theorem 1.2

Assume that 1 2 g ( u ) u G ( u ) is not bounded and there exists an ϵ > 0 such that Ω a ( x , t ) d x < ϵ . By Proposition 2.1 and Proposition 2.2, I C 1 ( H , R ) and satisfies the Palais-Smale condition. By Lemma 3.1 and the generalized mountain pass theorem, I ( u ) has a critical value b with critical point u ˜ such that I ( u ˜ ) = b . If Ω a ( x ) d x is sufficiently small, by (3.1), we have

Ω a + ( x ) | u ˜ | μ d x C

for C > 0 , from which we can conclude that u ˜ is bounded and the proof of Theorem 1.2(i) is complete.

Next we shall prove Theorem 1.2(ii). We may assume that R n < R n + 1 for all n N . Let us set D n = B R n H n , D n = B R n H n .

Lemma 4.1Assume thatgsatisfies (g1)-(g4), 1 2 g ( u ) u G ( u ) is not bounded and there exists an ϵ > 0 such that Ω a ( x , t ) < ϵ . Then there exists an R n > 0 such that

I ( u ) 0 for  u H n B R n . (4.1)

Proof Let us choose ψ H such that ψ = 1 , ψ 0 in Ω and supp ( ψ ) Ω + . Then, by (g2), (g4), and the Hölder inequality, we have

I ( t ψ ) = 1 2 P + t ψ 2 1 2 P t ψ 2 Ω a ( x ) G ( t ψ ) d x = 1 2 P + t ψ 2 1 2 P t ψ 2 Ω + a + ( x ) G ( t ψ ) d x + Ω a ( x ) G ( t ψ ) d x 1 2 P + t ψ 2 1 2 P t ψ 2 Ω + a + ( x ) ( a 5 t μ ψ μ a 4 ) d x + G ( t ψ ) Ω a ( x ) d x 1 2 t 2 Ω + a + ( x ) ( a 5 t μ ψ μ a 4 ) d x + ϵ

for small ϵ > 0 . Since μ > 2 , there exist t n great enough for each n and an R n > 0 such that u n = t n ψ and I ( u n ) < 0 if u n H n B R n and u n > R n , so the lemma is proved. □

Let us set

Γ n = { γ C ( [ 0 , 1 ] , H ) γ ( 0 ) = 0  and  γ ( 1 ) = u n }

and

b n = inf γ Γ n max [ 0 , 1 ] I ( γ ( u ) ) , n N .

Proof of Theorem 1.2(ii) We assume that g ( u ) u μ G ( u ) is not bounded and there exists an ϵ > 0 such that Ω a ( x ) d x < ϵ . By Proposition 2.1 and Proposition 2.2, I C 1 ( H , R ) and satisfies the Palais-Smale condition. By Lemma 4.1, there exists an R n > 0 such that I ( u n ) 0 for u n H n B R n . We note that I ( 0 ) = 0 . By Lemma 4.1 and the generalized mountain pass theorem, for n large enough b n > 0 is a critical value of I and lim n b n = + . Let u ˜ n be a critical point of I such that I ( u ˜ n ) = b n . Then for each real number M, max Ω | u ˜ n ( x ) | M . In fact, by contradiction, Δ 2 u + c Δ u = a ( x ) g ( u ) and max Ω | u ˜ n ( x ) | K imply that

I ( u ˜ n ) max | u ˜ n | K ( 1 2 g ( u ˜ n ) u ˜ n G ( u ˜ n ) ) Ω | a ( x ) | d x ,

which means that b n is bounded. This is absurd because of the fact that lim n b n = + . Thus we complete the proof. □

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

TJ and Q-HC participated in the sequence alignment and drafted the manuscripted. Both authors read and approved the final manuscript.

Acknowledgements

This work (Choi) was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (KRF-2013010343).

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