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Infinitely many solutions for p-harmonic equation with singular term

Abstract

In this paper, we study the following p-harmonic problem involving the Hardy term:

Δ ( | Δ u | p 2 Δ u ) λ | u | p 2 u | x | 2 p =f(x,u),in Ω,u= u n =0on Ω,

where Ω is an open bounded domain containing the origin in R N , 1<p< N 2 and 0λ< [ N ( p 1 ) ( N 2 p ) / p 2 ] p . By using the variational method, we prove that the above problem has infinitely many solutions with positive energy levels.

AMS Subject Classification:35J60, 35J65.

1 Introduction

The main purpose of this paper is to show the existence of infinitely many solutions for the following p-harmonic equation:

{ Δ ( | Δ u | p 2 Δ u ) λ | u | p 2 u | x | 2 p = f ( x , u ) , x Ω , u = u n = 0 , x Ω ,
(1.1)

where Ω is an open bounded domain containing the origin in R N , the boundary Ω is smooth. 1<p< N 2 , 0λ< λ ¯ = [ N ( p 1 ) ( N 2 p ) / p 2 ] p . n is the outer normal derivative. Nonlinearity f(x,u) satisfies the following conditions:

(f1) f(x,u) is continuous on Ω×R and limits subcritical growing at infinity; that is,

lim u f ( x , u ) | u | p 1 =0,uniformly for xΩ,
(1.2)

where p =Np/(N2p) is the critical exponent of Sobolev’s embedding W 0 2 , p (Ω) L s (Ω).

(f2) For any xΩ, f(x,u) satisfies

lim u 0 f ( x , u ) u | u | p =0and lim u f ( x , u ) u | u | p =+.
(1.3)

(f3) Denote G(x,t)=f(x,t)tp 0 t f(x,s)ds. For all xΩ, there exists a constant M0 such that

G(x, t 1 )G(x, t 2 )for any M| t 1 || t 2 |.
(1.4)

(f4) f(x,u) is odd with respect to u.

By the Hardy-Rellich inequality (see [1, 2]), we know that

Ω | u | p | x | 2 p dx 1 λ ¯ Ω | Δ u | p dx.
(1.5)

Obviously, for any λ[0, λ ¯ ),

( 1 λ λ ¯ ) Ω | Δ u | p dx Ω ( | Δ u | p λ | u | p | x | 2 p ) dx Ω | Δ u | p dx.

In W 0 2 , p (Ω), forλ[0, λ ¯ ), we define

u= u W 0 2 , p ( Ω ) := ( Ω ( | Δ u | p λ | u | p | x | 2 p ) d x ) 1 p ,

this norm is equivalent to ( Ω | Δ u | p d x ) 1 / p .

A weak solution of the problem (1.1) is a critical point of the energy functional

I(u)= 1 p Ω ( | Δ u | p λ | u | p | x | 2 p ) dx Ω F(x,u)dx,u W 0 2 , p (Ω),
(1.6)

where F(x,u)= 0 u f(x,s)ds. It is easy to check that I(u) is a continuous even functional.

Biharmonic equations can describe the static form change of a beam or the sport of a rigid body. For example, this type of equation furnishes a model for studying traveling wave in suspension bridges (see [3]). By using variational arguments, many authors investigated nonlinear biharmonic equations under Dirichlet boundary conditions or Navier boundary conditions and got interesting results (see [410]).

Li and Squassina in [11] considered the superlinear p-harmonic equation with Navier boundary conditions

Δ ( | Δ u | p 2 Δ u ) =f(x,u),in Ω,u=Δu=0on Ω,
(1.7)

where Ω is an open bounded domain in R N with a smooth boundary Ω. p>1, N2p+1 and f:Ω×RR is a Carathéodory function such that for some positive constant C,

| f ( x , u ) | C ( 1 + | u | q 1 ) for all q[1, p ).
(1.8)

By means of the Morse theory, they proved the existence of two nontrivial solutions to (1.7). After that, Li and Tang [12] considered a more general problem than (1.7). They got three solutions by the three critical points theorem which was obtained in [13]. If f(x,u)=λg(x,u), Candito and Bisci [14] established a well-determined interval of values of the parameter λ for which the problem (1.7) admits at least two distinct weak solutions. The authors in [15] considered the p-harmonic equation with Dirichlet conditions and obtained the existence of a nontrivial solution. Under the condition of (1.8), [16] proved the existence and multiplicity of weak solutions for the nonuniformly nonlinear problem

Δ ( a ( x , Δ u ) ) =f(x,u),in Ω,u= u n =0on Ω,
(1.9)

where |a(x,t)| c 0 ( h 0 (x)+ h 1 (x) | t | p 1 ) with h 0 (x)0, h 1 (x)1. a(x,t) and f(x,t) are odd with respect to the second variable. Furthermore, f(x,t) is subcritical and satisfies the Ambrosetti-Rabinowitz condition, that is, there exists a constant θ>p such that

0<θ 0 t f(x,s)dsf(x,t)tfor any xΩ.
(1.10)

The aim of this paper is to obtain infinitely many solutions for the problem (1.1) when 2p<N. The main difficulty lies in the fact that the embeddings W 0 2 , p (Ω) L p (Ω) and W 0 2 , p (Ω) L p (Ω, | x | 2 p dx) are not compact. Furthermore, the assumption (f1) is not the usual subcritical growth (1.8). The condition (f3) is weaker than the A-R condition (1.10). We use the concentration compactness principle (see [17, 18]) to overcome those difficulties. The following theorem is our main result.

Theorem 1.1 Assume f(x,u) satisfies (f1)-(f4), then the problem (1.1) possesses infinitely many weak solutions and the corresponding critical values are positive.

This paper proceeds as follows. In the next section, we prove the energy functional I(u) satisfies the Palais-Smale condition. In Section 3, by using the symmetric mountain pass theorem (see [19]), we get the main result of this paper. Throughout the paper, denote

S p , λ := inf u W 0 2 , p ( Ω ) { 0 } u p u p p .

We use u q = ( Ω | u | q d x ) 1 / q to denote the norm of L q (Ω), C, C i stand for universal constants. We omit dx and Ω in the integrals if there is no other indication.

2 Palais-Smale condition

To show the (PS) sequence { u n } of the variational functional I(u) is compact in W 0 2 , p (Ω), we first prove the boundedness of { u n } by the analytic argument which has been used in [20]. Then, using the concentration compactness principle, we get the compactness of { u n }.

Lemma 2.1 The condition (f1) implies that for any ε>0, there exists a positive constant C ε such that

F(x,u)ε | u | p + C ε |u|,(x,u)Ω×R.
(2.1)

(f2) implies that F(x,u)=o( u p ) as u0. Furthermore, for any |u|M>0, there exists a small positive constant θ

F(x,u)C | u | p + θ andf(x,u)u>0.
(2.2)

Lemma 2.2 Under the conditions (f1), (f2) and (f3), if the sequence { u n } W 0 2 , p (Ω) satisfies

I( u n )c, I ( u n )0as n+,
(2.3)

then the sequence { u n } is bounded in W 0 2 , p (Ω).

Proof We prove this lemma by contradiction. Without loss of generality, we assume that

u n +as n+.
(2.4)

Set w n = u n / u n , then w n =1 for any nN. As n+, there exists w W 0 2 , p (Ω) such that

w n w weakly in  W 0 2 , p ( Ω ) , w n w strongly in  L s ( Ω )  for any  1 s < p , w n w a.e.  x Ω .

If w0, we denote Ω 0 ={xΩ:w=0}, set Ω Ω 0 is nonempty. (2.3) implies that

u n p Ω f(x, u n ) u n dx=o(1).

Therefore,

1o(1)= Ω f ( x , u n ) u n u n p dx= Ω 0 f ( x , u n ) u n u n p dx+ Ω Ω 0 f ( x , u n ) u n u n p dx.
(2.5)

As n+, Ω | u n | p dx u n 0. So, | u n |+ for xΩ Ω 0 . It follows from (f2) that

lim u n f ( x , u n ) u n u n p = lim u n f ( x , u n ) u n | u n | p | w n | p =+.

Since |Ω Ω 0 |>0, then

Ω Ω 0 f ( x , u n ) u n u n p dx+as n+.
(2.6)

Using (2.1) and (2.2), we derive that

Ω 0 f ( x , u n ) u n u n p d x = 1 u n p ( Ω 0 ( | u n | > M ) + Ω 0 ( | u n | M ) ) 1 u n p Ω 0 ( | u n | M ) | f ( x , u n ) u n | d x C u n p ( M + M p ) | Ω 0 | C 1 as  n + .
(2.7)

From (2.6) and (2.7), we get

Ω f ( x , u n ) u n u n p dx+as n+.
(2.8)

This contradicts (2.5).

If w0, then w n 0 strongly in L 1 (Ω). Since I( u n )c, we know that

c 1 p u n p F(x, u n )dx 1 p u n p ε | u n | p dx C ε | u n |dx.
(2.9)

In (2.9), choose ε=1, then

1 p u n p u n p p + C 2 u n 1 +c C 3 u n p p +c.

From (2.4), it follows that u n p + as n+. Denote

A n =M u n u n p M S p , λ 1 / p ,
(2.10)

where M is any fixed positive constant. It is clear that

A n u n = M u n p 0(as n+).

Therefore, for n large enough, A n / u n (0,1).

For every nN, we define a sequence of t n as follows:

I( t n u n )= max t [ 0 , 1 ] I(t u n ).

We claim that

I( t n u n )+(as n+).
(2.11)

In fact, by (2.1)-(2.10) and the definition of t n , we have

I ( t n u n ) I ( A n u n u n ) 1 p A n p ε | A n u n u n | p d x C ε A n | w n | d x = 1 p A n p ε M p C ε A n | w n | d x .

If A n + as n+, then (2.11) follows from the above estimate. If A n is bounded for any n, that is, there exists a constant K>0 such that A n K, then

I ( t n u n ) 1 p A n p ε M p C ε A n | w n | d x 1 p S p , λ M p ε M p C ε K | w n | d x 1 p S p , λ M p ε M p as  n + ,

where we use the fact that w n 1 0 as n+. The positive constant ε is arbitrary, thus

lim n + I( t n u n ) 1 p S p , λ M p for any positive constant M.

(2.11) follows from the above inequality.

From (2.3), we know that

pc+o(1)=pI( u n ) I ( u n ) , u n = ( f ( x , u n ) u n p F ( x , u n ) ) dx.
(2.12)

By the definition of t n , we obtain d d t I(t u n ) | t = t n =0 and t n [0,1], then | t n u n || u n |. The condition (f3) and (2.11) derive that

( f ( x , u n ) u n p F ( x , u n ) ) d x ( f ( x , t n u n ) t n u n p F ( x , t n u n ) ) d x = p I ( t n u n ) I ( t n u n ) , t n u n = p I ( t n u n ) t n d d t I ( t u n ) | t = t n + .

Which is a contradiction to (2.12). Therefore, the sequence { u n } is bounded in W 0 2 , p (Ω). □

Under the conditions of Lemma 2.2, we know u n C. Therefore, there exists a subsequence, still denoted by { u n }, and some u W 0 2 , p (Ω) such that

u n u weakly in  W 0 2 , p ( Ω ) , u n u weakly in  L p ( Ω , | x | 2 p d x ) , u n u strongly in  L s ( Ω )  for any  1 s < p , u n u a.e.  x Ω .

Obviously, u is a weak solution of (1.1). Now we prove that u0. According to the concentration compactness principle (see [17, 18]), there exists a subsequence, still denoted by { u n }, at most countable set J, a set of different points { x j } j J Ω ¯ {0} and two positive number sequences { μ j } j J { 0 } , { ν j } j J { 0 } such that

| Δ u n | p d μ | Δ u | p + j J μ j δ x j + μ 0 δ 0 weakly in the sense of measure , | u n | p d ν = | u | p + j J ν j δ x j + ν 0 δ 0 weakly in the sense of measure , | u n | p | x | 2 p d γ = | u | p | x | 2 p + γ 0 δ 0 weakly in the sense of measure , S p , λ ν 0 p p μ 0 λ γ 0 , and S p , 0 ν j p p μ j for  j J ,

where δ x j is the unit Dirac measure at x j .

Lemma 2.3 Assume f(x,u) satisfies (f1), the sequence { u n } is bounded in W 0 2 , p (Ω) and satisfies (2.3). Then the set J={ x 1 , x 2 ,, x j } Ω ¯ is finite, μ 0 λ γ 0 =0.

Proof We first prove that μ j =0 for any x j J.

Let ε>0 be small enough such that 0 B ε ( x j ) and B ε ( x i ) B ε ( x j )= for ij. ϕ j (x)[0,1] is a cutting-off function in C 0 (Ω). ϕ j (x)1 for |x x j |<ε/2, ϕ j (x)0 for |x x j |ε, and |Δ ϕ j |4/ ε 2 . Since

I ( u n ) , u n ϕ j = | Δ u n | p ϕ j d x + 2 | Δ u n | p 2 Δ u n u n ϕ j d x + u n | Δ u n | p 2 Δ u n Δ ϕ j d x λ | u n | p | x | 2 p ϕ j d x f ( x , u n ) u n ϕ j d x ,
(2.13)

we deduce that

(2.14)
(2.15)

By Hölder’s inequality,

lim ε 0 lim n + | | Δ u n | p 2 Δ u n u n ϕ j d x | lim ε 0 lim n + Δ u n L p ( B ε ( x j ) ) p 1 u n L N p / ( N p ) ( B ε ( x j ) ) ϕ j L N ( B ε ( x j ) ) C lim ε 0 Δ u n L p ( B ε ( x j ) ) p = 0 , lim ε 0 lim n + | | Δ u n | p 2 Δ u n u n Δ ϕ j d x | lim ε 0 lim n + Δ u n L p ( B ε ( x j ) ) p 1 u n L p ( B ε ( x j ) ) Δ ϕ j L N / 2 ( B ε ( x j ) ) C lim ε 0 u n L p ( B ε ( x j ) ) = 0 .

Using (f1), (2.14) and (2.15), we have

lim ε 0 lim n + | f ( x , u n ) u n ϕ j | dx lim ε 0 lim n + ( ε | u n | p ϕ j d x + C ε | u n | ϕ j d x ) =0.

By the above estimates, (2.13) becomes

0= lim ε 0 lim n + I ( u n ) , u n ϕ j μ j .
(2.16)

Therefore, for any jJ, μ j =0, which implies that J is finite.

If the concentration is at the origin, let ϕ 0 (x)[0,1] be a cutting-off function in C 0 (Ω). ϕ 0 (x)1 for |x|<ε/2, ϕ 0 (x)0 for |x|ε, and |Δ ϕ 0 |4/ ε 2 . Choose ε>0 small enough such that x j B ε (0) for all jJ. Then

(2.17)
(2.18)

Similarly, we can get

(2.19)
(2.20)

From equalities (2.17)-(2.20), we get

0= lim ε 0 lim n + I ( u n ) , u n ϕ 0 μ 0 λ γ 0 .
(2.21)

On the other hand, μ 0 λ γ 0 S p , λ ν 0 p p 0, thus μ 0 λ γ 0 =0. The proof is complete. □

Lemma 2.4 Assume f(x,u) satisfies (f1), the sequence { u n } satisfying (2.3) is bounded in W 0 2 , p (Ω). Then there exist some u W 0 2 , p (Ω) and a subsequence, still denoted by { u n }, such that

u n u strongly in  L p ( Ω ϵ ) , u n u strongly in  L N p N p ( Ω ϵ ) ,

where Ω ϵ =Ω i = 0 j B ¯ ϵ ( x i ), x i J={ x 1 , x 2 ,, x j }{0}.

Proof Lemma 2.3 implies that J is finite. Choose φ C 0 (Ω) with φ( x i )=0 for any x i J{0}. Then

| φ u n | p d x = | φ u | p d x + i = 1 j ν i φ p ( x i ) + ν 0 φ p ( 0 ) = | φ u | p d x .
(2.22)

Since φ u n φu a.e. in Ω, we get that

u n ustrongly in  L p ( Ω ϵ ).

Similarly,

u n ustrongly in  L N p N p ( Ω ϵ ).

 □

By using the Lebesgue decomposition theorem, we have

dμλdγ= | Δ u | p λ | u | p | x | 2 p +dσ,
(2.23)

where ( | Δ u | p λ | u | p | x | 2 p )dσ. Then dσ j J μ j δ x j +( μ 0 λ γ 0 ) δ 0 .

Lemma 2.5 Assume f(x,u) satisfies (f1)-(f3), the sequence { u n } W 0 2 , p (Ω) satisfies (2.3). Then there exist u W 0 2 , p (Ω) and a subsequence, still denoted by { u n }, such that u n converges to u strongly in W 0 2 , p (Ω).

Proof We first prove that

( σ j J μ j δ x j ( μ 0 λ γ 0 ) δ 0 ) ( Ω ¯ )=0.
(2.24)

Claim for any closed set F Ω ¯ J ¯ , J ¯ =J{0}, σ(F)=0. In fact, denote r=dist(F, J ¯ )>0. Then, by using a finite covering theorem, there exist finite open balls B r / 4 ( z i ), with z i F, i=1,,m, such that F i = 1 m B r / 4 ( z i ) Ω ¯ J ¯ . Let ϱ(t) be a smooth cutting-off function. 0ϱ(t)1 for any t[0,). ϱ(x)1 for 0t1/2, and ϱ0 for t1. Denote η ε , y =ϱ(|xy|/ε), then

(2.25)
(2.26)

The sequence { η ε , y u n } is still bounded in W 0 2 , p (Ω). Define

η(x)= i = 1 m η r / 4 , z i (x),
(2.27)

then η(x) C 0 ( Ω ¯ J ¯ ) and 0η(x)m. Furthermore,

F i = 1 m B r 4 ( z i )sptη(x) Ω r 4 ,
(2.28)

where sptη(x) denotes the support of η(x). As n+, (2.3) implies

0 I ( u n ) I ( u ) , ( u n u ) η = Ω η ( | Δ u n | p | Δ u n | p 2 Δ u n Δ u | Δ u | p 2 Δ u Δ u n + | Δ u | p ) d x λ Ω η | x | 2 p ( | u n | p | u n | p 2 u n u | u | p 2 u u n + | u | p ) d x + 2 Ω η ( u n u ) ( | Δ u n | p 2 Δ u n | Δ u | p 2 Δ u ) d x + Ω Δ η ( u n u ) ( | Δ u n | p 2 Δ u n | Δ u | p 2 Δ u ) d x Ω η ( u n u ) ( f ( x , u n ) f ( x , u ) ) d x = I + II + III + IV + V .

As n+,

I + II = Ω η ( | Δ u n | p | Δ u n | p 2 Δ u n Δ u | Δ u | p 2 Δ u Δ u n + | Δ u | p ) d x λ Ω η | x | 2 p ( | u n | p | u n | p 2 u n u | u | p 2 u u n + | u | p ) d x = Ω η ( | Δ u n | p | Δ u | p λ | u n | p | u | p | x | 2 p ) d x + o ( 1 ) = Ω η d σ + o ( 1 ) ,

where we have used (2.23). Lemma 2.2 implies that u n M. Hölder’s inequality, (2.25)-(2.28) and Lemma 2.4 deduce that

By (2.1), for any ε>0, we have

| V | = | Ω ( f ( x , u n ) f ( x , u ) ) ( u n u ) η d x | Ω ( | f ( x , u n ) | + | f ( x , u ) | ) | u n u | η d x Ω ( C ε + ε ( | u n | p 1 + | u | p 1 ) ) | u n u | η d x C ε Ω | u n u | η d x + ε ( u n p p 1 + u p p 1 ) ( spt ( η ) | u n u | p d x ) 1 p C ε Ω | u n u | d x + ε C ( M ) 0 as  n + , ε 0 .

Therefore, as n+,

0σ(F)σ ( spt ( η ) ) Ω ηdσ|III|+|IV|+|V|=0.
(2.29)

Since the set F is arbitrary, (2.24) is obtained.

Lemma 2.3 means that μ j =0 for any jJ and μ 0 λ γ 0 =0, so σ( Ω ¯ )=0, which implies that as n+,

Ω ( | Δ u n | p λ | u n | p | x | 2 p ) dx Ω ( | Δ u | p λ | u | p | x | 2 p ) dx.
(2.30)

Together with u n u a.e. in Ω, we complete the proof. □

3 The proof of the main result

In this section, by using the following symmetric mountain pass theorem (see [19]), we give the proof of Theorem 1.1.

Lemma 3.1 (Symmetric mountain pass theorem)

Assume functional I satisfies the following conditions:

  1. (1)

    I C 1 ( W 0 2 , p (Ω),R) is even and satisfies the Palais-Smale condition.

  2. (2)

    There exists a finite dimensional subspace X W 0 2 , p (Ω) such that I | X B r a>0.

  3. (3)

    There exists a sequence of the finite dimensional subspace { X j }, dim( X j )=j and r j >0 such that

    I(u)0for any u X j B r j ,j=1,2,.

Then I has infinitely many different critical points, and the corresponding energy values are positive.

Proof of Theorem 1.1 We know the energy functional I(u) is continuous and even. Lemma 2.5 implies that I(u) satisfies the PS condition in W 0 2 , p (Ω). In any finite dimensional subspace X W 0 2 , p (Ω), all norms are equivalent. For any uX, since F(x,u)=o( u p ) as u0, there exists a constant q>p such that

I ( u ) = 1 p Ω ( | Δ u | p λ | u | p | x | 2 p ) d x Ω F ( x , u ) d x 1 p u p C u q > 0 for  u  small enough .
(3.1)

Thus, there exists r>0 such that I | X B r a>0. On the other hand, by using (2.2), we have

I ( u ) = 1 p Ω ( | Δ u | p λ | u | p | x | 2 p ) d x Ω F ( x , u ) d x 1 p u p C { x Ω : | u | > M } | u | p + θ d x C 1 p u p C u p + θ C as  u .

There exists a sequence of the finite dimensional subspace { X j }, dim( X j )=j and r j >0 such that

I(u)0for any u X j B r j ,j=1,2,.

Finally, all the assumptions of Lemma 3.1 are satisfied. Hence, the problem (1.1) possesses infinitely many weak solutions, and the corresponding critical values are positive. □

Remark 3.1 Under the same conditions of Theorem 1.1, we can also prove that the following p-harmonic type equation with Navier boundary conditions:

{ Δ ( | Δ u | p 2 Δ u ) λ | u | p 2 u | x | 2 p = f ( x , u ) , x Ω , u = Δ u = 0 , x Ω ,
(3.2)

has infinitely many solutions { u n } W 2 , p (Ω) W 0 1 , p (Ω), where Ω containing the origin is an open bounded domain in R N , Ω is smooth. 1<p< N 2 , 0λ< λ ¯ = [ N ( p 1 ) ( N 2 p ) / p 2 ] p .

Remark 3.2 All the results obtained above obviously hold if we choose f(x,u)= | u | q 2 u+ | u | r 2 u with p<q<r< p .

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Acknowledgements

The authors are thankful to the referees for their helpful suggestions and necessary corrections in the completion of this paper. This research is supported by the Education Department of Henan Province (12B110002) and NSF of China (61143002).

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Correspondence to Huazhao Xie.

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The authors declare that they have no competing interests.

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HX carried out the existence of infinitely many solutions for the p-harmonic equation studies and drafted the manuscript. JW participated in its coordination. All authors read and approved the final manuscript.

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Xie, H., Wang, J. Infinitely many solutions for p-harmonic equation with singular term. J Inequal Appl 2013, 9 (2013). https://doi.org/10.1186/1029-242X-2013-9

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Keywords

  • infinitely many solutions
  • p-harmonic
  • Hardy term