Multiple solutions for some fourth-order nonlinear elliptic variational inequalities

Nonlinear Analysis 63 (2005) e23 – e31 www.elsevier.com/locate/na

Multiple solutions for some fourth-order nonlinear elliptic variational inequalities夡 Jihui Zhang School of Mathematics and Computer Sciences, Nanjing Normal University, Nanjing 210097, People’s Republic of China

Abstract We consider the solvability for fourth-order nonlinear elliptic variational inequalities. Some existence results are given by the variational methods. 䉷 2005 Elsevier Ltd. All rights reserved. Keywords: Variational inequalities; Existence results; Variational methods

1. Introduction Let
be a bounded smooth open set in Rn and K = {u ∈ H 1 (
) ∩ H01 (
) | u
0 a.e. in
}. We are concerned with a fourth-order elliptic variational inequality
(,   u) ∈ R × K, (1.1) − u) − c ∇u∇(v − u)} dx − 
u(v − u) dx
{
u
(v 

f (v − u) dx for all v ∈ K, where
is the Laplacian on R n , ∇ is the gradient on R n , c,  ∈ R and f ∈ L2 (
). The bilinear form  {
u
v + c∇u∇v} dx


夡

Supported by Foundation of Major Project of Science and Technology of Chinese Education Ministry. E-mail address: [email protected].

0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.02.102

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J. Zhang / Nonlinear Analysis 63 (2005) e23 – e31

corresponds to the fourth-order elliptic operator Lu =
2 u + c
u, where
2 denotes the biharmonic operator. In [13], Tarantello considered the fourth-order nonlinear elliptic problem under the Dirichlet boundary value condition  2
u + c
u = b[(u + 1)+ − 1] in
, (1.2) u|j
=
u|j
= 0. She showed by degree theory that if b
1 (1 − c), then (1.2) has a solution u such that u(x) < 0 in
. At the same time in [5] it has been pointed out that this kind of nonlinearity can furnish a model to study travelling waves in a suspension bridge. For problem (1.1) when f (x, u) = bg(x, u), Micheletti and Pistoia in [10,11] proved that there exist two or three solutions for a more general nonlinearity g by the variational method. In [14,15] by using variational methods, we also proved the existence of weak solutions of the problem (1.1) for a more general nonlinearity f. In this paper we investigate the existence of solutions for the fourth-order linear and semilinear elliptic variational inequalities.

2. Preliminaries Given a bounded smooth open set
⊂ R n and c < 1 . Let V denote the Hilbert space the norm on V is given by  (2.1) u2 = [(
u)2 − c |∇u|2 ] dx.

H 1 (
) ∩ H01 (
);


Let k (k = 1, 2, . . .) denote the eigenvalues and k (k = 1, 2, . . .) the corresponding eigenfunctions, suitably normalized with respect to L2 (
) inner product, of the eigenvalue problem 
u + u = 0 in
, (2.2) u|j
= 0, where each eigenvalue k is repeated as often as multiplicity. Recall that 0 < 1 < 2  3  · · · , k → ∞, and that 1 (x) > 0 for x ∈
. The eigenvalue problem  2
u + c
u = u in
, (2.3) u|j
=
u|j
= 0 has infinitely many eigenvalues k = k (k − c),

k = 1, 2, . . . ,

and corresponding eigenfunctions k (x).

J. Zhang / Nonlinear Analysis 63 (2005) e23 – e31

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The set of functions {k } is an orthogonal basis for V ; thus we may denote an element u of V as u=

∞ 

a k k ,

k=1

∞ 

ak2 < ∞,

(2.4)

k=1

and u2 =

∞ 

k (k − c)ak2 .

(2.5)

k=1

Let V  denote the dual of V , let , be the duality pairing between V  and V , and let , L2 be the inner product in L2 (
). Let V = V1 ⊕ V2 be a direct decomposition of V ; thus for u ∈ V , we have u =  + , where  ∈ V1 and  ∈ V2 . Let K = {u ∈ V | u
0 a.e. in
}; for a fixed  ∈ V1 , let K = { ∈ V2 |  +  ∈ K} and  = { ∈ V1 | K = ∅}. It is clear that  is convex; let 0 be the interior of  in V1 . Let mapping A : K → V  be continuous and satisfy the following two conditions: (i) A is of class C 1 and is the gradient of a weak lower semicontinuous functional E; (ii) there exists a positive constant  such that A (u), 
2

for all u ∈ K,  ∈ V2 ,

where V = V1 ⊕ V2 , 1 dimV1 < ∞ and A (u) denotes the Fréchet derivative of A at u. We have the following lemmas Lemma 2.1. Assume that (i) and (ii) hold. Then there exist the mapping  :  → V2 and the functional W :  → R such that W () = E( + ()) = min∈K E( + ),  and W are continuous in 0 . The proof is given by Lemma 1.2 and Proposition 1.4 in [12]. Lemma 2.2 (Szulkin [12]). Assume that (i) and (ii) hold. Let W be as in Lemma 2.1, suppose, in addition, that W has a local minimum or a local maximum. Then the variational inequality u ∈ K, has a solution.

Au, v − u
0

for all v ∈ K

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J. Zhang / Nonlinear Analysis 63 (2005) e23 – e31

3. The main results Let
be a bounded smooth open subset of Rn . Let V = H 1 (
) ∩ H01 (
) be a subspace of H01 (
) and let K = {u ∈ V | u
0 a.e. in
} defined in Section 2. Let us define the mapping B : V → V  by  Bu, v = [
u
v − c ∇u∇v] dx (3.1)


for all u, v ∈ V , define the mapping I : V → V  by  I u, v = uv dx

(3.2)

for all u, v ∈ V , and define the mapping F : V → V  by  F u, v = f v dx

(3.3)







for all u, v ∈ V , where f ∈ L2 (
). Then the variational inequality (1.1) may be written as  (, u) ∈ R × K, Bu − I u, v − u
F u, v − u for all v ∈ K.

(3.4)

Theorem 3.1. Let c < 1 and  > 1 (1 − c). Suppose that  f 1 dx
0,

(3.5)




where f = 0. Then the variational inequality (1.1) has no solution. ∞ ∞ Proof. Let 1 1 + k=2 ak k ∈ K, then a1
0. Let v =(a1 +1)1 + k=2 ak k ∈ K u=a a  , if u is a solution of the variational inequality (1.1), then we have and  = ∞ k k k=2 Bu − I u, v − u
F u, v − u . Also,

  Bu − I u, v − u = B a1 1 +

∞  k=2

 =

(3.6)



ak k − I a1 1 +

∞ 





ak k , 1

k=2

[
(a1 1 + )
1 − c ∇(a1 1 + )∇1 ] dx  −  (a1 1 + )1 dx





= a1 1 + , 1 (1 − c)1 L2 −  a1 1 + , 1 L2  = a1 (1 (1 − c) − )21 dx


(3.7)

J. Zhang / Nonlinear Analysis 63 (2005) e23 – e31

and

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 F u, v − u =




f 1 dx.

(3.8)

By combining (3.5)–(3.8), it follows that a1 = 0 and  = 0. Then u = 0 is a solution of the variational inequality (1.1); thus we get  0
f v dx for all v ∈ K.


Hence f (x)0 a.e. in
; using (3.5), it implies that f (x) = 0 a.e. in
, a contradiction. This completes the proof of Theorem 3.1.  Theorem 3.2. Let c < 1 and 1 (1 − c) <  < 2 (2 − c). Assume that f (x) < 0 a.e. in
. Then the variational inequality (1.1) has at least two solutions. Proof. It is easy to see that u = 0 is a solution. Now we shall show the existence of the other solution. Let us define the mapping A : V → V  by Au = Bu − I u − F u.

(3.9)

Then the variational inequality (1.1) may be written as u ∈ K,

Au, v − u
0

for all v ∈ K.

Define the functional E : V → R by    1 2 1 2 2 Eu = [(
u) − c |∇u| ] dx − u − f u dx.
2
2

(3.10)

(3.11)

Then E  u = Au, and it is easy to check that E is weakly lower semicontinuous. Let V1 = span{1 } and V2 = V1⊥ . For u ∈ K write u = z1 +  (z ∈ R,  ∈ V2 ); we have   A (u),  = [(
)2 − c |∇|2 − 2 ] dx. (3.12)


Also,  2




 dx 

1



2 (2 − c)


[(
)2 − c |∇|2 ] dx.

By combining (3.12) and (3.13), we obtain

    A (u), 
1 − (
)2 − c |∇|2 dx 2 (2 − c)


  = 1− 2 2 (2 − c)

(3.13)

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J. Zhang / Nonlinear Analysis 63 (2005) e23 – e31

for all  ∈ V2 . Thus the mapping A satisfies conditions (i) and (ii) of Lemma 2.1. It is easy to see that  = {z1 | z ∈ [0, +∞)} and K0 = {0}; note that 

1  z2 1 2 1− E(z1 + ) = 1 (1 − c) 2 1 + B − I ,  − F , z1 +  . 2

(3.14)

Thus it implies that there exist z ∈ R and (z) ∈ V2 such that W (z) = E(z1 + (z)), = min E(z1 + ) ∈Kz1

E(z1 )

 1  = 1− z2 1 2 − z F 1 , 1 . 2 1 (1 − c)

(3.15)

Thus limz→+∞ W (z) = −∞. Then there exists  > 0 such that W (z) > 0 for all 0 < z < . In fact, suppose that there exists a sequence {zn } ⊂ (0, ) such that zn → 0 and W (zn )0 for all n. Let vn = (zn )/zn ; then W (zn ) = E(zn 1 + (zn ))

 1  = 1− z2  2 2 1 (1 − c) n 1 1 + zn2 Bv n − I v n , vn − zn F v n , 1 + vn . 2

(3.16)

Since f (x) < 0 a.e. in
, we have W (zn )


  1 1− z2  2 2 1 (1 − c) n 1 1 + zn2 Bv n − I v n , vn . 2

(3.17)

Using (3.17) and the fact that W (zn ) 0, it follows that {vn } is bounded; thus we may assume that vn → v0 weakly in V . Hence, we get

 1  1− zn 1 2 (1/zn )W (zn ) = 2 1 (1 − c) 1 + zn Bv n − I v n , vn − F v n , 1 + vn

2 → − F v 0 , 1 + v0 .

(3.18)

J. Zhang / Nonlinear Analysis 63 (2005) e23 – e31

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Note that 1 + vn ∈ K, 1 , vn L2 = 0 and 1 + v0 ∈ K\{0}; we have − F v 0 , 1 + v0 > 0. Hence W (zn ) > 0 for all sufficiently n, a contradiction. By Lemma 2.1 we know that W is continuous. Using W (0) = 0 we can conclude that supz∈Z W (z) is attained at some z0 ∈ (0, +∞); thus by Lemma 2.2 the variational inequality (1.1) has the second solution. This proves Theorem 3.2.  Now let g :
× R → R be a Carathéodory function such that g(·, 0) = 0

(3.19) jg

for a.e. x ∈
. Moreover, assume that js is a Carathéodory function and there exist nonnegative constants c1 , c2 and 1  < (n + 2)/(n − 2)(n
3) such that 0

j g(x, s) c1 + c2 |s| −1 js

(3.20)

for a.e. x ∈
, for all s ∈ R and lim

|s|→+∞

j j g(x, s) = +∞, lim g(x, s) = 0 s→0 js js

uniformly for a.e. x ∈
. Let us define the mapping P : V → V  by  P u, v = g(x, u)v dx


(3.21)

(3.22)

for all u, v ∈ V . From Lemma 2.3 in [2] and the compact embedding theorem, we know the mapping P is completely continuous and of class C 1 . Now let we consider a fourth-order semilinear elliptic variational inequality
(,  u) ∈ R × K,  (3.23) − u) − c ∇u∇(v − u)} dx − 
u(v − u) dx
{
u
(v 

(f − g(x, u))(v − u) dx for all v ∈ K. Theorem 3.3. Let c < 1 and 1 (1 −c) <  < 2 (2 −c). Assume g satisfies (3.19)–(3.21). Suppose also that f (x) < 0 a.e. in
and f L∞ is sufficiently small. Then the variational inequality (3.23) has at least three solutions. Proof. Let us define the mapping A : V → V  by Au = Bu − I u + P u − F u.

(3.24)

Then the variational inequality (3.23) may be written as u ∈ K,

Au, v − u
0

for all v ∈ K.

(3.25)

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J. Zhang / Nonlinear Analysis 63 (2005) e23 – e31

Define the functional E : V → R by    1 2 1 (3.26) u + G(x, u) − f u dx, [(
u)2 − c |∇u|2 ] dx − Eu =
2
2 u where G(x, u) = 0 g(x, s) ds; then E  u = Au. It is easy to check that E is weakly lower semicontinuous. Let V1 = span{1 } and V2 = V1⊥ . For u ∈ K write u = z1 +  (z ∈ R,  ∈ V2 ); we have   j 2 2 2  g(x, u)2 dx. (3.27) A (u),  = [(
) − c |∇| −  ] dx +

ju By combining (3.13) and (3.27), we obtain 

 j  2 2  g(x, u)2 dx [(
) − c |∇| ] dx + A (u), 
1 − ju 2 (2 − c)



 
1− [(
)2 − c |∇|2 ] dx 2 (2 − c)


  = 1− 2 2 (2 − c) for all  ∈ V2 . Thus the mapping A satisfies conditions (i) and (ii) of Lemma 2.1. It is easy to see that u = 0 is the first solution. By using (3.20) and (3.21), there exist constants b and c such that G(x, u)
21 u2 for all |u|
b a.e. in
and    1 1 G(x, u) − u2 dx − f u dx E(u)
u2 + 2 2
|u|
z→+∞

z→+∞

On the other hand, an easy computation show that W (z)E(z1 )

  1  2 2 G(x, z1 ) dx − z F 1 , 1 . = 1− z 1  + 2 1 (1 − c)


(3.28)

Using (3.19) and (3.21) one sees that W (z1 ) < 0 for some z1 if f L∞ is sufficiently small. Since G(x, u)
0 a.e. in
, we know as in the proof of Theorem 3.2 that W (z) > 0 for all small 0 < z < z1 . Thus W has a local maximum and a local minimum in (0, +∞); therefore the variational inequality (3.23) has the second and the third solution from Lemma 2.2. This completes the proof of Theorem 3.3. 

J. Zhang / Nonlinear Analysis 63 (2005) e23 – e31

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