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Solvability of boundary value problem with p-Laplacian at resonance

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Abstract

By generalizing the extension of the continuous theorem of Ge and Ren and constructing suitable Banach spaces and operators, we investigate the existence of solutions for p-Laplacian boundary value problems at resonance. An example is given to illustrate our results.

MSC:34B15.

1 Introduction

In this paper, we will study the boundary value problem

{ ( φ p ( u ) ) ( t ) = f ( t , u ( t ) , u ( t ) , u ( t ) ) , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = 0 1 k ( t ) u ( t ) d t ,
(1.1)

and

{ ( φ p ( u ) ) ( t ) = f ( t , u ( t ) , u ( t ) , u ( t ) ) , u ( 0 ) = 0 , u ( 0 ) = 0 1 g ( t ) u ( t ) d t , u ( 1 ) = 0 1 h ( t ) u ( t ) d t ,
(1.2)

where φ p (s)= | s | p 2 s, p>1, 0 1 k(t)dt=1, 0 1 g(t)dt=1, 0 1 h(t)dt=1.

A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a non-trivial solution. Mawhin’s continuous theorem [1] is an effective tool to solve this kind of problems when the differential operator is linear, see [210] and references cited therein. But it does not work for nonlinear cases such as boundary value problems with a p-Laplacian, which attracted the attention of mathematicians in recent years [1115]. Ge and Ren extended Mawhin’s continuous theorem [15] and many authors used their results to solve boundary value problems with a p-Laplacian, see [16, 17]. In this new theorem, two projectors P and Q must be constructed. But it is difficult to give the projector Q in many boundary value problems with a p-Laplacian. In this paper, we generalize the extension of the continuous theorem and show that the p-Laplacian problem is solvable when Q is not a projector. And we will use this new theorem to discuss problems (1.1) and (1.2), respectively.

In this paper, we will always suppose that

(H1) k(t),g(t),h(t) L 1 [0,1] are nonnegative and k 1 = g 1 = h 1 =1, where k 1 := 0 1 |k(t)|dt.

(H2) f(t,u,v,w) is continuous in [0,1]× R 3 .

2 Preliminaries

Definition 2.1 [15]

Let X and Y be two Banach spaces with norms X , Y , respectively. A continuous operator M:XdomMY is said to be quasi-linear if

  1. (i)

    ImM:=M(XdomM) is a closed subset of Y,

  2. (ii)

    KerM:={xXdomM:Mx=0} is linearly homeomorphic to R n , n<,

where domM denote the domain of the operator M.

Let X 1 =KerM and X 2 be the complement space of X 1 in X, then X= X 1 X 2 . Let P:X X 1 be a projector and ΩX an open and bounded set with the origin θΩ.

Definition 2.2 Suppose N λ : Ω ¯ Y, λ[0,1] is a continuous and bounded operator. Denote N 1 by N. Let Σ λ ={x Ω ¯ domM:Mx= N λ x}. N λ is said to be M-quasi-compact in Ω ¯ if there exists a vector subspace Y 1 of Y satisfying dim Y 1 =dim X 1 and two operators Q, R with Q:Y Y 1 , QY= Y 1 , being continuous, bounded, and satisfying Q(IQ)=0, R: Ω ¯ ×[0,1] X 2 domM continuous and compact such that for λ[0,1],

  1. (a)

    (IQ) N λ ( Ω ¯ )ImM(IQ)Y,

  2. (b)

    Q N λ x=θ, λ(0,1)QNx=θ,

  3. (c)

    R(,0) is the zero operator and R(,λ) | Σ λ =(IP) | Σ λ ,

  4. (d)

    M[P+R(,λ)]=(IQ) N λ .

Theorem 2.1 Let X and Y be two Banach spaces with the norms X , Y , respectively, and let ΩX be an open and bounded nonempty set. Suppose

M:XdomMY

is a quasi-linear operator and that N λ : Ω ¯ Y, λ[0,1] is M-quasi-compact. In addition, if the following conditions hold:

(C1) Mx N λ x, xΩdomM, λ(0,1),

(C2) deg{JQN,ΩKerM,0}0,

then the abstract equation Mx=Nx has at least one solution in domM Ω ¯ , where N= N 1 , J:ImQKerM is a homeomorphism with J(θ)=θ.

Proof The proof is similar to the one of Lemma 2.1 and Theorem 2.1 in [15]. □

We can easily get the following inequalities.

Lemma 2.1 For any u,v0, we have

  1. (1)

    φ p (u+v) φ p (u)+ φ p (v), 1<p2.

  2. (2)

    φ p (u+v) 2 p 2 ( φ p (u)+ φ p (v)), p2.

In the following, we will always suppose that q satisfies 1/p+1/q=1.

3 The existence of a solution for problem (1.1)

Let X= C 2 [0,1] with norm u=max{ u , u , u }, Y=C[0,1]×C[0,1] with norm ( y 1 , y 2 )=max{ y 1 , y 2 }, where y = max t [ 0 , 1 ] |y(t)|. We know that (X,) and (Y,) are Banach spaces.

Define operators M:XdomMY, N λ :XY as follows:

Mu=[ ( φ p ( u ) ) ( t ) T ( φ p ( u ) ) ( t ) ], N λ u=[ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) 0 ],

where Ty=c, yC[0,1], c satisfying

0 1 k ( t ) t 1 φ q ( 0 s y ( r ) c d r ) d s d t = 0 , dom M = { u X φ p ( u ) C 1 [ 0 , 1 ] , u ( 0 ) = u ( 0 ) = 0 } .
(3.1)

Lemma 3.1 For yC[0,1], there is only one constant cR such that Ty=c with |c| y and that T:C[0,1]R is continuous.

Proof For yC[0,1], let

F(c)= 0 1 k(t) t 1 φ q ( 0 s ( y ( r ) c ) d r ) dsdt.

Obviously, F(c) is continuous and strictly decreasing in . Take a= min t [ 0 , 1 ] y(t), b= max t [ 0 , 1 ] y(t). It is easy to see that F(a)0, F(b)0. Thus, there exists a unique constant c[a,b] such that F(c)=0, i.e. there is only one constant cR such that Ty=c with |c| y .

For y 1 , y 2 C[0,1], assume T y 1 = c 1 , T y 2 = c 2 . By k(t)0, 0 1 k(t)dt=1 and φ q being strictly increasing, we obtain, if c 2 c 1 > max t [ 0 , 1 ] ( y 2 (t) y 1 (t)), then

0 = 0 1 k ( t ) t 1 φ q ( 0 s ( y 2 ( r ) c 2 ) d r ) d s d t = 0 1 k ( t ) t 1 φ q ( 0 s [ ( y 1 ( r ) c 1 ) + ( y 2 ( r ) y 1 ( r ) ( c 2 c 1 ) ) d r ] ) d s d t < 0 1 k ( t ) t 1 φ q ( 0 s ( y 1 ( r ) c 1 ) d r ) d s d t = 0 .

This is a contradiction. On the other hand, if c 2 c 1 < min t [ 0 , 1 ] ( y 2 (t) y 1 (t)), then

0 = 0 1 k ( t ) t 1 φ q ( 0 s ( y 2 ( r ) c 2 ) d r ) d s d t = 0 1 k ( t ) t 1 φ q ( 0 s [ ( y 1 ( r ) c 1 ) + ( y 2 ( r ) y 1 ( r ) ( c 2 c 1 ) ) d r ] ) d s d t > 0 1 k ( t ) t 1 φ q ( 0 s ( y 1 ( r ) c 1 ) d r ) d s d t = 0 .

This is a contradiction, too. So, we have min t [ 0 , 1 ] ( y 2 (t) y 1 (t)) c 2 c 1 max t [ 0 , 1 ] ( y 2 (t) y 1 (t)), i.e. | c 2 c 1 | y 2 y 1 . So, T:C[0,1]R is continuous. The proof is completed. □

It is clear that udomM is a solution if and only if it satisfies Mu=Nu, where N= N 1 . For convenience, let ( a , b ) L := [ a b ] .

Lemma 3.2 M is a quasi-linear operator.

Proof It is easy to see that KerM={btbR}:= X 1 .

For uXdomM, if Mu= ( y , c ) L , then c satisfies (3.1). On the other hand, if yC[0,1], Ty=c, take

u(t)= 0 t (ts) φ q ( 0 s y ( r ) d r ) ds.

By a simple calculation, we get uXdomM and Mu= ( y , c ) L . Thus

ImM= { ( y , c ) L y C [ 0 , 1 ] , c  satisfies (3.1) } .

By the continuity of T, we find that ImMY is closed. So, M is quasi-linear. The proof is completed. □

Lemma 3.3 T(c)=c, T(y+c)=T(y)+c, T(cy)=cT(y), cR, yC[0,1].

Proof The proof is simple. Therefore, we omit it. □

Take a projector P:X X 1 and an operator Q:Y Y 1 as follows:

(Pu)(t)= u (0)t,Q ( y , y 1 ) L = ( 0 , T y 1 T y ) L ,

where Y 1 ={ ( 0 , c ) L cR}. Obviously, QY= Y 1 , and dim Y 1 =dim X 1 .

By the continuity and boundedness of T, we can easily see that Q is continuous and bounded in Y. It follows from Lemma 3.3 that Q(IQ) ( y , y 1 ) L = ( 0 , 0 ) L , y, y 1 C[0,1].

Define an operator R:X×[0,1] X 2 as

R(u,λ)(t)= 0 t (ts) φ q ( 0 s λ f ( r , u ( r ) , u ( r ) , u ( r ) ) d r ) ds,

where KerM X 2 =X. By (H2) and the Arzela-Asscoli theorem, we can easily see that R: Ω ¯ ×[0,1] X 2 domM is continuous and compact, where ΩX is an open bounded set.

Lemma 3.4 Assume that ΩX is an open bounded set. Then N λ is M-quasi-compact in  Ω ¯ .

Proof It is clear that ImP=KerM, Q N λ x=θ, λ(0,1)QNx=θ and R(,0)=0. For u Ω ¯ ,

( I Q ) N λ u = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) 0 ] [ 0 T [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] ] = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] ] Im M .

Since ImMKerQ and y=Qy+(IQ)y, we obtain ImM(IQ)Y. Thus, (IQ) N λ ( Ω ¯ )ImM(IQ)Y.

For u Σ λ ={u Ω ¯ domM:Mu= N λ u}, we get

R ( u , λ ) = 0 t ( t s ) φ q ( 0 s λ f ( r , u ( r ) , u ( r ) , u ( r ) ) d r ) d s = 0 t ( t s ) φ q ( 0 s ( φ p ( u ) ) ) d s = u ( t ) u ( 0 ) t = ( I P ) u ,

i.e. Definition 2.2(c) holds. For u Ω ¯ , we have

M [ P u + R ( u , λ ) ] =[ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] ]=(IQ) N λ u.

So, Definition 2.2(d) holds. Therefore, N λ is M-quasi-compact in Ω ¯ . The proof is completed. □

Theorem 3.1 Assume that the following conditions hold.

(H3) There exists a nonnegative constant K such that one of (1) and (2) holds:

  1. (1)

    Bf(t,A,B,C)>0, t[0,1], |B|>K, A,CR,

  2. (2)

    Bf(t,A,B,C)<0, t[0,1], |B|>K, A,CR.

(H4) There exist nonnegative functions a(t),b(t),c(t),e(t) L 1 [0,1] such that

| f ( t , x , y , z ) | a(t) φ p ( | x | ) +b(t) φ p ( | y | ) +c(t) φ p ( | z | ) +e(t),t[0,1],x,y,zR,

where φ q ( a 1 + b 1 + c 1 )< 2 2 q , if 1<p2; φ q ( 2 p 2 a 1 + 2 p 2 b 1 + c 1 )<1, if p2.

Then boundary value problem (1.1) has at least one solution.

In order to prove Theorem 3.1, we show two lemmas.

Lemma 3.5 Suppose (H3) and (H4) hold. Then the set

Ω 1 = { u dom M M u = N λ u , λ ( 0 , 1 ) }

is bounded in X.

Proof For u Ω 1 , we have Q N λ u=0, i.e. Tf(t,u(t), u (t), u (t))=0. By (H3), there exists a constant t 0 [0,1] such that | u ( t 0 )|K. Since u(t)= 0 t u (s)ds, u (t)= u ( t 0 )+ t 0 t u (s)ds, we have

| u ( t ) | u , | u ( t ) | K+ u ,t[0,1].
(3.2)

It follows from Mu= N λ u, (H4), and (3.2) that

| u ( t ) | = | φ q ( 0 t λ f ( s , u ( s ) , u ( s ) , u ( s ) ) d s ) | φ q ( 0 1 a ( t ) φ p ( | u | ) + b ( t ) φ p ( | u | ) + c ( t ) φ p ( | u | ) + e ( t ) d t ) φ q [ ( a 1 + b 1 ) φ p ( K + u ) + c 1 φ p ( u ) + e 1 ] .

If 1<p2, by Lemma 2.1, we get

| u ( t ) | φ q ( B 1 + A 1 φ p ( u ) ) 2 q 2 [ φ q ( B 1 ) + φ q ( A 1 ) u ] ,

thus

u 2 q 2 φ q ( B 1 ) 1 2 q 2 φ q ( A 1 ) ,

where B 1 =( a 1 + b 1 ) φ p (K)+ e 1 , A 1 = a 1 + b 1 + c 1 .

If p>2, by Lemma 2.1, we get

| u ( t ) | φ q ( B 2 + A 2 φ p ( u ) ) [ φ q ( B 2 ) + φ q ( A 2 ) u ] ,

thus

u φ q ( B 2 ) 1 φ q ( A 2 ) ,

where B 2 = 2 p 2 ( a 1 + b 1 ) φ p (K)+ e 1 , A 2 = 2 p 2 ( a 1 + b 1 )+ c 1 .

These, together with (3.2), mean that Ω 1 is bounded in X. □

Lemma 3.6 Assume (H3) holds. Then

Ω 2 ={uKerMQNu=0}

is bounded in X, where N= N 1 .

Proof For u Ω 2 , we have u=bt and Tf(t,bt,b,0)=0. By (H3), we get |b|K. So, Ω 2 is bounded. The proof is completed. □

Proof of Theorem 3.1 Let Ω={uXu<r}, where r is large enough such that K<r<+ and Ω Ω 1 ¯ .

By Lemmas 3.5 and 3.6, we know Mu N λ u, udomMΩ and QNu0, uKerMΩ.

Let H(u,δ)=ρδu+(1δ)JQNu, δ[0,1], uKerM Ω ¯ , where J:ImQKerM is a homeomorphism with J ( 0 , b ) L =bt, ρ={ 1 , if (H 3 )(1) holds , 1 , if (H 3 )(2) holds .

Define a function Sgn(x)={ 1 , if  x > 0 , 1 , if  x < 0 .

For uKerMΩ, we have u=bt0. Thus

H(u,δ)=ρδbt+(1δ) ( T f ( t , b t , b , 0 ) ) t.

If δ=1, H(u,1)=ρbt0. If δ=0, by QNu0, we get H(u,0)=JQN(bt)0. For 0<δ<1, we now prove that H(u,δ)0. Otherwise, if H(u,δ)=0, then

Tf(t,bt,b,0)= ρ δ 1 δ b.
(3.3)

Since u=r>K, we have |b|>K. Thus, T[bf(t,bt,b,0)]=bTf(t,bt,b,0)= ρ δ 1 δ b 2 . So, we have Sgn(bf(t,bt,b,0))=Sgn{T[bf(t,bt,b,0)]}=Sgn( ρ δ 1 δ b 2 )=Sgn(ρ). A contradiction with the definition of ρ. So, H(u,δ)0, uKerMΩ, δ[0,1].

By the homotopy of degree, we get

deg ( J Q N , Ω Ker M , 0 ) = deg ( H ( , 0 ) , Ω Ker M , 0 ) = deg ( H ( , 1 ) , Ω Ker M , 0 ) = deg ( ρ I , Ω Ker M , 0 ) 0 .

By Theorem 2.1, we can see that Mu=Nu has at least one solution in Ω ¯ . The proof is completed. □

Example Let us consider the following boundary value problem at resonance:

{ ( φ p ( u ) ) ( t ) = 1 8 t sin x 3 + 1 16 y 3 + t 3 sin z 3 + cos t , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = 2 0 1 t u ( t ) d t ,
(3.4)

where p=4.

Corresponding to problem (1.1), we have q= 4 3 , a(t)= 1 8 t, b(t)= 1 16 , c(t)= t 3 , e(t)=cost, k(t)=2t.

Take K=4. By a simple calculation, we find that the conditions (H1)-(H4) hold. By Theorem 3.1, we obtain the result that problem (3.4) has at least one solution.

4 The existence of a solution for problem (1.2)

Let X= C 2 [0,1] with norm u=max{ u , u , u }, Y=C[0,1]×C[0,1]×C[0,1] with norm ( y 1 , y 2 , y 3 )=max{ y 1 , y 2 , y 3 }, where y = max t [ 0 , 1 ] |y(t)|. We know that (X,) and (Y,) are Banach spaces.

Define operators M:XdomMY, N λ :XY as follows:

Mu=[ ( φ p ( u ) ) ( t ) T 1 ( φ p ( u ) ) ( t ) T 2 ( φ p ( u ) ) ( t ) ], N λ u=[ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) 0 0 ],

where T 1 y= c 1 , T 2 y= c 2 , yC[0,1], c 1 , c 2 satisfy

0 1 g ( t ) 0 t φ q ( 0 s y ( r ) c 1 d r ) d s d t = 0 , 0 1 h ( t ) t 1 φ q ( 0 s y ( r ) c 2 d r ) d s d t = 0 , dom M = { u X φ p ( u ) C 1 [ 0 , 1 ] , u ( 0 ) = 0 } .
(4.1)

Lemma 4.1 For yC[0,1], there is only one constant c i R such that T i y= c i with | c i | y . And T i :C[0,1]R are continuous, i=1,2.

The proof is similar to Lemma 3.1.

It is clear that udomM is a solution if and only if it satisfies Mu=Nu, where N= N 1 . For convenience, let ( a , b , c ) T := [ a b c ] .

Lemma 4.2 M is a quasi-linear operator.

Proof It is easy to get KerM={a+bta,bR}:= X 1 .

For uXdomM, if Mu= ( y , c 1 , c 2 ) T , then c 1 , c 2 satisfy (4.1). On the other hand, if yC[0,1], T 1 y= c 1 , T 2 y= c 2 , take

u(t)= 0 t (ts) φ q ( 0 s y ( r ) d r ) ds.

By simple calculation, we get uXdomM and Mu= ( y , c 1 , c 2 ) T . Thus

ImM= { ( y , c 1 , c 2 ) T y C [ 0 , 1 ] , c 1 , c 2  satisfy (4.1) } .

By the continuity of T i , i=1,2, we see that ImMY is closed. So, M is quasi-linear. The proof is completed. □

Take a projector P:X X 1 and an operator Q:Y Y 1 as follows:

(Pu)(t)=u(0)+ u (0)t,Q ( y , y 1 , y 2 ) T = ( 0 , T 1 y 1 T 1 y , T 2 y 2 T 2 y ) T ,

where Y 1 ={ ( 0 , c 1 , c 2 ) T c i R,i=1,2}. Obviously, QY= Y 1 , and dim Y 1 =dim X 1 .

By the continuity and boundedness of T i , i=1,2, we can easily see that Q is continuous and bounded in Y. It follows from Lemma 3.3 that Q(IQ) ( y , y 1 , y 2 ) T = ( 0 , 0 , 0 ) T , y, y 1 , y 2 C[0,1].

Define an operator R:X×[0,1] X 2 as

R(u,λ)(t)= 0 t (ts) φ q ( 0 s λ f ( r , u ( r ) , u ( r ) , u ( r ) ) d r ) ds,

where KerM X 2 =X. By (H2) and the Arzela-Asscoli theorem, we can easily see that R: Ω ¯ ×[0,1] X 2 domM is continuous and compact, where ΩX is an open bounded set.

Lemma 4.3 Assume that ΩX is an open bounded set. Then N λ is M-quasi-compact in  Ω ¯ .

Proof It is clear that ImP=KerM, Q N λ x=θ, λ(0,1)QNx=θ and R(,0)=0. For u Ω ¯ ,

( I Q ) N λ u = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) 0 0 ] [ 0 T 1 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 2 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 1 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 2 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] Im M .

Since ImMKerQ and y=Qy+(IQ)y, we obtain ImM(IQ)Y. Thus, (IQ) N λ ( Ω ¯ )ImM(IQ)Y.

For u Σ λ ={u Ω ¯ domM:Mu= N λ u}, we get

R ( u , λ ) = 0 t ( t s ) φ q ( 0 s λ f ( r , u ( r ) , u ( r ) , u ( r ) ) d r ) d s = 0 t ( t s ) φ q ( 0 s ( φ p ( u ) ) ) d s = u ( t ) u ( 0 ) u ( 0 ) t = ( I P ) u ,

i.e. Definition 2.2(c) holds. For u Ω ¯ , we have

M [ P u + R ( u , λ ) ] =[ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 1 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 2 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ]=(IQ) N λ u.

Thus, Definition 2.2(d) holds. Therefore, N λ is M-quasi-compact in Ω ¯ . The proof is completed. □

Theorem 4.1 Assume that the following conditions hold:

(H5) There exists a nonnegative constant L such that if |u(t)|>L, t[0,1] then either

T 1 f ( t , u ( t ) , u ( t ) , u ( t ) ) 0

or

T 2 f ( t , u ( t ) , u ( t ) , u ( t ) ) 0.

(H6) There exist nonnegative constants K 1 , K 2 such that one of (1) and (2) holds:

  1. (1)
    Bf(t,A,B,C)>0,t[0,1],|B|> K 1 ,A,CR,

and

Af(t,A,B,C)>0,t[0,1],|B| K 1 ,|A|> K 2 ,CR.
  1. (2)
    Bf(t,A,B,C)<0,t[0,1],|B|> K 1 ,A,CR,

and

Af(t,A,B,C)<0,t[0,1],|A|> K 2 ,|B| K 1 ,CR.

(H7) There exist nonnegative functions a(t),b(t),c(t),e(t) L 1 [0,1] such that

| f ( t , x , y , z ) | a(t) φ p ( | x | ) +b(t) φ p ( | y | ) +c(t) φ p ( | z | ) +e(t),t[0,1],x,y,zR,

where φ q ( a 1 + b 1 + c 1 )< 2 2 q , if 1<p2; φ q ( 2 p 2 a 1 + 2 p 2 b 1 + c 1 )<1, if p2.

Then boundary value problem (1.2) has at least one solution.

In order to prove Theorem 4.1, we show two lemmas.

Lemma 4.4 Suppose (H5)-(H7) hold. Then the set

Ω 1 = { u dom M M u = N λ u , λ ( 0 , 1 ) }

is bounded in X.

Proof For u Ω 1 , we have Q N λ u=0, i.e. T i f(t,u(t), u (t), u (t))=0, i=1,2. By (H5) and (H6), there exist constants t 0 , t 1 [0,1] such that |u( t 0 )|L, | u ( t 1 )| K 1 . Since u(t)=u( t 0 )+ t 0 t u (s)ds, u (t)= u ( t 1 )+ t 1 t u (s)ds, then

| u ( t ) | L+ u , | u ( t ) | K 1 + u ,t[0,1].
(4.2)

It follows from Mu= N λ u, (H7), and (4.2) that

| u ( t ) | = | φ q ( 0 t λ f ( s , u ( s ) , u ( s ) u ( s ) ) d s ) | φ q ( 0 1 a ( t ) φ p ( | u | ) + b ( t ) φ p ( | u | ) + c ( t ) φ p ( | u | ) + e ( t ) d t ) φ q ( a 1 φ p ( K 1 + L + u ) + b 1 φ p ( K 1 + u ) + c 1 φ p ( u ) + e 1 ) .

If 1<p2, by Lemma 2.1, we get

| u ( t ) | φ q ( B 1 + A 1 φ p ( u ) ) 2 q 2 [ φ q ( B 1 ) + φ q ( A 1 ) u ] ,

thus

u 2 q 2 φ q ( B 1 ) 1 2 q 2 φ q ( A 1 ) ,

where B 1 = a 1 φ p ( K 1 +L)+ b 1 φ p ( K 1 )+ e 1 , A 1 = a 1 + b 1 + c 1 .

If p>2, by Lemma 2.1, we get

| u ( t ) | φ q ( B 2 + A 2 φ p ( u ) ) [ φ q ( B 2 ) + φ q ( A 2 ) u ] ,

thus

u φ q ( B 2 ) 1 φ q ( A 2 ) ,

where B 2 = 2 p 2 a 1 φ p ( K 1 +L)+ 2 p 2 b 1 φ p ( K 1 )+ e 1 , A 2 = 2 p 2 a 1 + 2 p 2 b 1 + c 1 .

These, together with (4.2), mean that Ω 1 is bounded in X. □

Lemma 4.5 Assume (H6) holds. Then

Ω 2 ={uKerMQNu=0}

is bounded in X, where N= N 1 .

Proof For u Ω 2 , we have u=a+bt and Q(Nu)=0. By (H6), we see that there exists a constant t 0 [0,1] such that |u( t 0 )|=|a+b t 0 | K 2 , | u (t)|=|b| K 1 . So, Ω 2 is bounded. The proof is completed. □

Proof of Theorem 4.1 Let Ω={uXu<r}, where r is large enough such that K 1 + K 2 <r<+ and Ω Ω 1 ¯ Ω 2 ¯ .

By Lemmas 4.4 and 4.5, we know Mu N λ u, udomMΩ and QNu0, uKerMΩ.

Let H(u,δ)=ρδu+(1δ)JQNu, δ[0,1], uKerM Ω ¯ , where J:ImQKerM is a homeomorphism with J ( 0 , a , b ) T =a+bt, ρ={ 1 , if (H 6 )(1) holds , 1 , if (H 6 )(2) holds .

Take the function Sgn(x) is the same as the one in Proof of Theorem 3.1.

For uKerMΩ, we have u=a+bt0. Thus

H(u,δ)=ρδ(a+bt)+(1δ) ( T 1 f ( t , a + b t , b , 0 ) T 2 f ( t , a + b t , b , 0 ) t ) .

If δ=1, H(u,1)=ρ(a+bt)0. If δ=0, by QNu0, we get H(u,0)=JQN(a+bt)0. For 0<δ<1, we now prove that H(u,δ)0. Otherwise, if H(u,δ)=0, then

T 1 f(t,a+bt,b,0)= ρ δ 1 δ a, T 2 f(t,a+bt,b,0)= ρ δ 1 δ b.
(4.3)

Since u=max{ a + b t ,|b|}=r> K 1 + K 2 , we have either |b|> K 1 or a + b t > K 1 + K 2 . If |b|> K 1 , then T 2 bf(t,a+bt,b,0)=b T 2 f(t,a+bt,b,0)= ρ δ 1 δ b 2 . So, we have Sgn(bf(t,a+bt,b,0))=Sgn( T 2 bf(t,a+bt,b,0))=Sgn( ρ δ 1 δ b 2 )=Sgn(ρ). This is a contradiction with the definition of ρ. If |b| K 1 , then a + b t > K 1 + K 2 . Thus min t [ 0 , 1 ] |a+bt|> K 2 and Sgn(a)=Sgn(a+bt). By T 1 af(t,a+bt,b,0)=a T 1 f(t,a+bt,b,0)= ρ δ 1 δ a 2 , we get Sgn( T 1 (a+bt)f(t,a+bt,b,0))=Sgn( T 1 af(t,a+bt,b,0))=Sgn( ρ δ 1 δ a 2 )=Sgn(ρ). This is a contradiction with the definition of ρ, too. So, H(u,δ)0, uKerMΩ, δ[0,1].

By the homotopy of degree, we get

deg ( J Q N , Ω Ker M , 0 ) = deg ( H ( , 0 ) , Ω Ker M , 0 ) = deg ( H ( , 1 ) , Ω Ker M , 0 ) = deg ( ρ I , Ω Ker M , 0 ) 0 .

By Theorem 2.1, we find that (1.2) has at least one solution in Ω ¯ . The proof is completed. □

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Acknowledgements

This work is supported by the National Science Foundation of China (11171088) and the Natural Science Foundation of Hebei Province (A2013208108).

The author is grateful to anonymous referees for their constructive comments and suggestions, which led to improvement of the original manuscript.

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Correspondence to Weihua Jiang.

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Keywords

  • continuous theorem
  • resonance
  • p-Laplacian
  • boundary value problem