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Solutions to a boundary value problem of a fourth-order impulsive differential equation

Abstract

This paper is concerned with the existence of solutions to a boundary value problem of a fourth-order impulsive differential equation with a control parameter λ. By employing some existing critical point theorems, we find the range of the control parameter in which the boundary value problem admits at least one solution. It is also shown that under certain conditions there exists an interval of the control parameter in which the boundary value problem possesses infinitely many solutions. The main results are also demonstrated with examples.

MSC:34B15, 34B18, 34B37, 58E30.

1 Introduction

Fourth-order two-point boundary value problems of ordinary differential equations are widely employed by engineers to describe the beam deflection with two simply supported ends [13]. One example is the following fourth-order two-point boundary value problem:

{ u ( i v ) ( t ) + A u ( t ) + B u ( t ) = λ f ( t , u ( t ) ) , t [ 0 , 1 ] , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 ,
(1.1)

where u ( i v ) (t), u (t), u (t) are the fourth, third, and second derivatives of u(t) with respect to t, respectively, fC([0,1]×R,R), A and B are two real constants. System (1.1) has been studied in [47] and the references therein. For a beam, t=0 and t=1 in (1.1) refer to the two ends of the beam. At other locations of the beam, t(0,1), there may be some sudden changes in loads placed on the beam, or some unexpected forces working on the beam. These sudden changes may result in impulsive effects for the governing differential equation. This motivates us to consider the following boundary value problem for a fourth-order impulsive differential equation:

{ u ( i v ) ( t ) + A u ( t ) + B u ( t ) = λ f ( t , u ( t ) ) , t t j , t [ 0 , 1 ] , Δ u ( t j ) = I 1 j ( u ( t j ) ) , Δ u ( t j ) = I 2 j ( u ( t j ) ) , j = 1 , 2 , , m , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 ,
(1.2)

where I 1 j , I 2 j C(R,R), 0= t 0 < t 1 < t 2 << t m < t m + 1 =1, and the operator Δ is defined as ΔU( t j )=U( t j + )U( t j ), where U( t j + ) (U( t j )) denotes the right-hand (left-hand) limit of U at t j and λ>0 is referred to as a control parameter.

We are mainly concerned with the existence of solutions of system (1.2). A function u(t)C([0,1]) is said to be a (classical) solution of (1.2) if u(t) satisfies (1.2). In literature, tools employed to establish the existence of solutions of impulsive differential equations include fixed point theorems, the upper and lower solutions method, the degree theory, critical point theory and variational methods. See, for example, [820]. In this paper, we establish the existence of solutions of (1.2) by converting the problem to the existence of critical points of some variational structure. In this paper we regard λ as a parameter and find the ranges in which (1.2) admits at least one and infinitely many solutions, respectively. Note that when λ=1 system (1.2) reduces to the one studied in [21]. Our results extend those ones in [21].

The rest of this paper is organized as follows. In Section 2 we present some preliminary results. Our main results and their proofs are given in Section 3.

2 Preliminaries

Throughout we assume that A and B satisfy

A0B.
(2.1)

Let

H 0 1 ( [ 0 , 1 ] ) = { u L 2 ( [ 0 , 1 ] ) : u L 2 ( [ 0 , 1 ] ) , u ( 0 ) = u ( 1 ) = 0 } ,

and

H 2 ( [ 0 , 1 ] ) = { u L 2 ( [ 0 , 1 ] ) : u , u L 2 ( [ 0 , 1 ] ) } .

Take X:= H 2 ([0,1]) H 0 1 ([0,1]) and define

u X = ( 0 1 ( | u ( t ) | 2 A | u | 2 + B | u | 2 ) d t ) 1 2 ,uX.
(2.2)

Since A and B satisfy (2.1), it is straightforward to verify that (2.2) defines a norm for the Sobolev space X and this norm is equivalent to the usual norm defined as follows:

u= ( 0 1 u ( t ) 2 d t ) 1 2 .

It follows from (2.1) that

u u X .

For the norm in C 1 ([0,1]),

u =max ( max t [ 0 , 1 ] | u ( t ) | , max t [ 0 , 1 ] | u ( t ) | ) ,

we have the following relation.

Lemma 2.1 Let M 1 =1+ 1 π . Then u M 1 u X , uX.

Proof The proof follows easily from Wirtinger’s inequality [22], Lemma 2.3 of [23] and Hölder’s inequality. The detailed argument is similar to the proof of Lemma 2.2 in [21], and we thus omit it here. □

Define a functional φ λ as

φ λ (u)=Φ(u)λΨ(u),uX,
(2.3)

where

Φ(u)= 1 2 u X 2 + j = 1 m 0 u ( t j ) I 1 j (s)ds+ j = 1 m 0 u ( t j ) I 2 j (s)ds
(2.4)

and

Ψ(u)= 0 1 F(t,u)dt,
(2.5)

with

F(t,u)= 0 u ( t ) f(t,s)ds.

Note that φ λ is Fréchet differentiable at any uX, and for any vX we have

φ λ ( u ) ( v ) = lim h 0 φ λ ( u + h v ) φ λ ( u ) h = 0 1 ( u ( t ) v ( t ) A u ( t ) v ( t ) + B u ( t ) v ( t ) ) d t + j = 1 m I 2 j ( u ( t j ) ) v ( t j ) + j = 1 m I 1 j ( u ( t j ) ) v ( t j ) λ 0 1 f ( t , u ( t ) ) v ( t ) d t .
(2.6)

Next we show that a critical point of the functional φ λ is a solution of system (1.2).

Lemma 2.2 If uX is a critical point of φ λ , then u is a solution of system (1.2).

Proof Suppose that uX is a critical point of φ λ . Then for any vX one has

λ 0 1 f ( t , u ( t ) ) v ( t ) d t = 0 1 ( u ( t ) v ( t ) A u ( t ) v ( t ) + B u ( t ) v ( t ) ) d t + j = 1 m I 2 j ( u ( t j ) ) v ( t j ) + j = 1 m I 1 j ( u ( t j ) ) v ( t j ) .
(2.7)

For j{1,2,,m}, choose vX such that v(t)=0 for t[0, t j ][ t j + 1 ,1], then we have

t j t j + 1 ( u ( i v ) + A u ( t ) + B u ( t ) ) v(t)dt=λ t j t j + 1 f ( t , u ( t ) ) v(t)dt.

Thus

u ( i v ) +A u (t)+Bu(t)=λf ( t , u ( t ) ) a.e. t( t j , t j + 1 ).

Therefore, by (2.7) we have

j = 1 m ( Δ u ( t j ) + I 2 j ( u ( t j ) ) ) v( t j ) j = 1 m ( Δ u ( t j ) I 1 j ( u ( t j ) ) ) v ( t j )=0.

Next we show that u satisfies

Δ u ( t j )= I 2 j ( u ( t j ) ) ,j=1,2,,m.

Suppose on the contrary that there exists some j{1,2,,m} such that

Δ u ( t j )+ I 2 j ( u ( t j ) ) 0.

Pick

v(t)= ( t 3 3 t j 2 t ) i = 0 , i j m + 1 ( 1 3 t 3 1 2 ( t j + t i ) t 2 + t j t i t + 1 6 t i 3 1 2 t j t i 2 ) ,

then

v ( t ) = 3 ( t 2 t j 2 ) i = 0 , i j m + 1 ( 1 3 t 3 1 2 ( t j + t i ) t 2 + t j t i t + 1 6 t i 3 1 2 t j t i 2 ) + ( t 3 3 t j t ) k = 0 , k j m + 1 { ( t 2 ( t k + t j ) t + t k t j ) i = 0 , i j , k m + 1 ( 1 3 t 3 1 2 ( t j + t i ) t 2 + t j t i t + 1 6 t i 3 1 2 t j t i 2 ) } .

Clearly, vX. Simple calculations show that v( t i )=0, i=1,2,,j1,j+1,,m+1, v( t j )0 and v ( t i )=0, i=1,2,,m+1. Thus

1 3 t j 3 ( Δ u ( t j ) + I 2 j ( u ( t j ) ) ) i = 0 , i j m + 1 ( t i t j ) 3 =0,

which is a contradiction. Similarly, one can show that Δ u ( t j )= I 1 j ( u ( t j )), j=1,2,,m. Therefore, u is a solution of (1.2). □

For r 1 , r 2 R with r 1 < r 2 , we define

α( r 1 , r 2 )= sup v Φ 1 ( ( r 1 , r 2 ) ) Ψ ( v ) sup u Φ 1 ( ( , r 1 ) ) Ψ ( u ) Φ ( v ) r 1
(2.8)

and

β( r 1 , r 2 )= inf v Φ 1 ( ( r 1 , r 2 ) ) sup u Φ 1 ( ( r 1 , r 2 ) ) Ψ ( u ) Ψ ( v ) r 2 Φ ( v ) .
(2.9)

For rR, we define

ρ 1 (r)= inf v Φ 1 ( ( , r ) ) sup u Φ 1 ( ( , r ) ) Ψ ( u ) Ψ ( v ) r Φ ( v ) ,
(2.10)
ρ 2 (r)= sup v Φ 1 ( ( r , + ) ) Ψ ( v ) sup u Φ 1 ( ( , r ] ) Ψ ( u ) Φ ( v ) r .
(2.11)

3 Main results

3.1 Existence of at least one solution

In this section we derive conditions under which system (1.2) admits at least one solution. For this purpose, we introduce the following assumption.

(H1) Assume that there exist two positive constants k 1 and k 2 such that for each uX

0 j = 1 m 0 u ( t j ) I 1 j (s)ds k 1 max j { 1 , 2 , , m } | u ( t j ) | 2
(3.1)

and

0 j = 1 m 0 u ( t j ) I 2 j (s)ds k 2 max j { 1 , 2 , , m } | u ( t j ) | 2 .
(3.2)

Let k 0 =2 A 6 + B 60 and k 3 = k 0 + k 1 + 1 4 k 2 with k 1 given in (3.1) and k 2 given in (3.2). For constants c 1 , c 2 , and c satisfying

c 1 < 2 k 0 M 1 c< 2 k 3 M 1 c< c 2 ,
(3.3)

we define

a( c 2 ,c)= 0 1 max | u | c 2 F ( t , u ) d t 0 1 F ( t , u 1 ( t ) ) d t c 2 2 2 k 3 M 1 2 c 2
(3.4)

and

b( c 1 ,c)= 0 1 F ( t , u 1 ( t ) ) d t 0 1 max | u | c 1 F ( t , u ) d t 2 k 3 M 1 2 c 2 c 1 2 ,
(3.5)

where

u 1 (t)=ct(1t).
(3.6)

Note that for every c>0 and t[0,1] we have | u 1 (t)|=ct(1t) c 4 <c. Since A0B, then k 0 >2. Thus, if c and c 2 satisfy (3.3), then c 2 >c and 0 1 max | u | c 2 F(t,u)dt 0 1 F(t, u 1 (t))dt>0 and hence a( c 2 ,c)>0.

Theorem 3.1 Assume that (H1) is satisfied. If there exist constants c 1 , c 2 , and c satisfying (3.3) and

0<a( c 2 ,c)<b( c 1 ,c),
(3.7)

then, for each λ( λ 1 , λ 2 ), system (1.2) admits at least one solution u and u X < c 2 M 1 , where λ 1 = 1 2 M 1 2 b ( c 1 , c ) and λ 2 = 1 2 M 1 2 a ( c 2 , c ) .

Proof By Lemma 2.2, it suffices to show the functional φ λ defined in (2.3) has at least one critical point. We prove this by verifying the conditions given in [[10], Theorem 5.1]. Note that Φ defined in (2.4) is a nonnegative Gâteaux differentiable, coercive, and sequentially weakly lower semicontinuous functional, and its Gâteaux derivative admits a continuous inverse on X . Moreover, Ψ defined in (2.5) is a continuously Gâteaux differentiable functional whose Gâteaux derivative is compact. Set

r 1 = c 1 2 2 M 1 2 , r 2 = c 2 2 2 M 1 2 .

Note that u 1 (t)=ct(1t)X. It then follows from (H1) that

Φ ( u 1 ) = 1 2 u 1 X 2 + j = 1 m 0 u 1 ( t j ) I 1 j ( s ) d s + j = 1 m 0 u 1 ( t j ) I 2 j ( s ) d s = ( 2 A 6 + B 60 ) c 2 + j = 1 m 0 c 2 c t j I 1 j ( s ) d s + j = 1 m 0 c t j ( 1 t j ) I 2 j ( s ) d s = k 0 c 2 + j = 1 m 0 c 2 c t j I 1 j ( s ) d s + j = 1 m 0 c t j ( 1 t j ) I 2 j ( s ) d s k 0 c 2 + k 1 max j | c 2 c t j | 2 + k 2 max j | c t j ( 1 t j ) | 2 k 0 c 2 + k 1 c 2 + 1 4 k 2 c 2 = k 3 c 2
(3.8)

and

Φ( u 1 ) 1 2 u 1 X 2 = k 0 c 2 .
(3.9)

By (3.3) we have

r 1 = c 1 2 2 M 1 2 < k 0 c 2 Φ( u 1 ) k 3 c 2 < c 2 2 2 M 1 2 = r 2 .

For uX satisfying Φ(u)< r 2 , by Lemma 2.1, one has

| u | 2 u 2 M 1 2 u X 2 2 M 1 2 Φ(u)<2 M 1 2 r 2 = c 2 2 ,t[0,1],

which implies that

Ψ(u)= 0 1 F ( t , u ( t ) ) dt 0 1 max | u | c 2 F(t,u)dt.

Hence

sup u Φ 1 ( ( r 1 , r 2 ) ) Ψ(u) sup u Φ 1 ( ( , r 2 ) ) Ψ(u) 0 1 max | u | c 2 F(t,u)dt.
(3.10)

For uX with Φ(u)< r 1 , one can similarly obtain

sup u Φ 1 ( ( , r 1 ) ) Ψ(u) 0 1 max | u | c 1 F(t,u)dt.
(3.11)

It follows from the definition of β( r 1 , r 2 ) that

β( r 1 , r 2 ) sup u Φ 1 ( ( , r 2 ) ) Ψ ( u ) Ψ ( u 1 ) r 2 Φ ( u 1 ) .
(3.12)

Note that Φ( u 1 )< r 2 . By (3.10) one has

β( r 1 , r 2 ) 0 1 max | u | c 2 F ( t , u ) d t 0 1 F ( t , u 1 ( t ) ) d t r 2 Φ ( u 1 ) .

Making use of Φ( u 1 )< r 2 , 0 1 max | u | c 2 F(t,u)dt 0 1 F(t, u 1 (t))dt>0, and (3.8), we obtain

β ( r 1 , r 2 ) 0 1 max | u | c 2 F ( t , u ) d t 0 1 F ( t , u 1 ( t ) ) d t c 2 2 2 M 1 2 k 3 c 2 = 2 M 1 2 0 1 max | u | c 2 F ( t , u ) d t 0 1 F ( t , u 1 ( t ) ) d t c 2 2 2 k 3 M 1 2 c 2 = 2 M 1 2 a ( c 2 , c ) > 0 .

By (2.8), and note that Φ( u 1 )> r 1 , one has

α( r 1 , r 2 ) Ψ ( u 1 ) sup u Φ 1 ( ( , r 1 ) ) Ψ ( u ) Φ ( u 1 ) r 1 .

By (3.11) we have

α( r 1 , r 2 ) 0 1 F ( t , u 1 ( t ) ) d t 0 1 max | u | c 1 F ( t , u ) d t Φ ( u 1 ) r 1 .

Note that (3.7) implies that

0 1 F ( t , u 1 ( t ) ) dt 0 1 max | u | c 1 F(t,u)dt>0,

which, together with (3.8), gives

α ( r 1 , r 2 ) 0 1 F ( t , u 1 ( t ) ) d t 0 1 max | u | c 1 F ( t , u ) d t k 3 c 2 r 1 = 2 M 1 2 0 1 F ( t , u 1 ( t ) ) d t 0 1 max | u | c 1 F ( t , u ) d t 2 k 3 M 1 2 c 2 c 1 2 = 2 M 1 2 b ( c 1 , d ) .

Therefore, β( r 1 , r 2 )2 M 1 2 a( c 2 ,c)<2 M 1 2 b( c 1 ,d)α( r 1 , r 2 ). Thus all the conditions in [[10], Theorem 5.1] are verified, and hence for each λ( λ 1 , λ 2 ) the functional φ λ =ΦλΨ admits at least one critical point u such that r 1 <Φ(u)< r 2 . Consequently, system (1.2) admits at least one solution u and u X < c 2 M 1 . □

In particular, if we take c 1 =0, then (3.4) and (3.5) become

a( c 2 ,c)= 0 1 max | u | c 2 F ( t , u ) d t 0 1 F ( t , u 1 ( t ) ) d t c 2 2 2 k 3 M 1 2 c 2

and

b( c 1 ,c)=b(0,c)= 0 1 F ( t , u 1 ( t ) ) d t 2 k 3 M 1 2 c 2 .

Correspondingly, conditions (3.3) and (3.7) reduce to

2 k 3 M 1 c< c 2
(3.13)

and

0 1 max | u | c 2 F(t,u)dt< c 2 2 2 k 3 M 1 2 c 2 0 1 F ( t , u 1 ( t ) ) dt.
(3.14)

If (3.13) and (3.14) hold, then

λ 1 = k 3 c 2 0 1 F ( t , u 1 ( t ) ) d t := λ ˆ 1

and

λ 2 = c 2 2 2 k 3 M 1 2 c 2 2 M 1 2 ( 0 1 max | u | c 2 F ( t , u ) d t 0 1 F ( t , u 1 ( t ) ) d t ) c 2 2 2 M 1 2 0 1 max | u | c 2 F ( t , u ) d t : = λ ˆ 2 > λ ˆ 1 .

As a consequence, we have the following result.

Corollary 3.2 Assume that (H1) is satisfied. If there exist two constants c and c 2 satisfying (3.13) and (3.14), then for each λ( λ ˆ 1 , λ ˆ 2 ) system (1.2) admits at least one nontrivial solution u.

Example 3.1 Consider the boundary value problem

{ u ( i v ) ( t ) = λ t , t t 1 , t [ 0 , 1 ] , Δ u ( t 1 ) = 1 4 u ( t 1 ) , t 1 = 1 2 , Δ u ( t 1 ) = 1 5 u ( t 1 ) , t 1 = 1 2 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 .
(3.15)

Here, f(t,u)=t, I 11 (s)= 1 4 s, I 21 (s)= 1 5 s, A=B=0 and m=1. It is easy to verify that (H1) is satisfied with k 1 = 1 8 and k 2 = 1 10 . Direct calculations give F(t,u)=tu, k 0 =2, k 3 = 43 20 and M 1 =1+ 1 π 1.318. Let c 1 =0.01, c=12, c 2 =1,500, then c 1 , c, c 2 satisfy (3.3) and a( c 2 ,c)3.33× 10 4 <b( c 1 ,c)9.25× 10 4 . Thus, it follows from Theorem 3.1 that system (3.15) has at least one solution for λ( λ 1 , λ 2 )=(311.2,864.4).

3.2 Existence of infinitely many solutions

In this section, we derive some conditions under which system (1.2) admits infinitely many distinct solutions. To this end, we need the following assumptions.

(H2) Assume that

{ t 1 , t 2 ,, t m } [ 1 4 , 3 4 ] .

(H3) Assume that

F(t,u)0,for (t,u) ( [ 0 , 1 4 ] [ 3 4 , 1 ] ) ×R.

Let k 4 =2,048( 3 8 9 10 4 4 A+ 79 14 4 8 B) and k 5 = k 4 + k 2 with k 2 given in (3.2). We define

γ 1 := lim inf r + ρ 1 (r), γ 2 := lim inf r ( inf X Φ ) + ρ 1 (r),
(3.16)

where ρ 1 (r) is given (2.10). Let

μ 1 =2 M 1 2 lim inf ξ + 0 1 max | u | ξ F ( t , u ) d t ξ 2 , μ 2 = 1 k 5 lim sup ξ + 1 4 3 4 F ( t , ξ ) d t ξ 2 .

Theorem 3.3 Assume that (H1), (H2), and (H3) are satisfied. If

μ 1 < μ 2
(3.17)

holds, then for each λ( 1 μ 2 , 1 μ 1 ) system (1.2) has an unbounded sequence of solutions in X.

Proof We apply [[5], Theorem 2.1] to show that the functional φ λ defined in (2.3) has an unbounded sequence of critical points.

We first show that γ 1 <+. Let { ξ n } be a sequence of positive numbers such that ξ n + as n and

lim n + 0 1 max | u | ξ n F ( t , u ) d t ξ n 2 = lim inf ξ + 0 1 max | u | ξ F ( t , ξ ) d t ξ 2 .

For any positive integer n, we let r n = ξ n 2 2 M 1 2 . For uX satisfying Φ(u)< r n , similar to the proof of Theorem 3.1, one can show that

u 2 2 M 1 2 Φ(u)< ξ n 2 ,t[0,1],

which implies that

Ψ(u)= 0 1 F(t,u)dt 0 1 max | u | ξ n F(t,u)dt.

Note that Ψ(0)=Φ(0)=0, thus we have

ρ 1 ( r n ) = inf v Φ 1 ( ( , r n ) ) sup u Φ 1 ( ( , r n ) ) Ψ ( u ) Ψ ( v ) r n Φ ( v ) sup u Φ 1 ( ( , r n ) ) Ψ ( u ) Ψ ( 0 ) r n Φ ( 0 ) = sup u Φ 1 ( ( , r n ) ) Ψ ( u ) r n 2 M 1 2 0 1 max | u | ξ n F ( t , u ) d t ξ n 2 ,

which, together with (3.16), gives us

γ 1 2 M 1 2 lim inf ξ + 0 1 max | u | ξ F ( t , u ) d t ξ 2 = μ 1 <+.

This shows that ( 1 μ 2 , 1 μ 1 )(0, 1 γ 1 ). For any fixed λ( 1 μ 2 , 1 μ 1 ), it follows from [[5], Theorem 2.1] that either φ λ =ΦλΨ has a global minimum or there is a sequence { u n } of critical points (local minima) of φ λ such that lim n + u n X =+.

Next we show that the functional φ λ has no global minimum for λ( 1 μ 2 , 1 μ 1 ). Since λ> 1 μ 2 = k 5 / lim sup ξ + 1 4 3 4 F ( t , ξ ) d t ξ 2 , we can choose a constant M such that, for each nN={1,2,},

sup ξ n 1 4 3 4 F ( t , ξ ) d t ξ 2 >M> k 5 λ .

Thus, there exists ξ n n such that

1 4 3 4 F ( t , ξ n ) d t ξ n 2 >M.

Define u n (t) as follows:

u n (t)={ 64 ξ n ( t 3 3 4 t 2 + 3 16 t ) , t [ 0 , 1 4 ) , ξ n , t [ 1 4 , 3 4 ] , 64 ξ n ( t 3 + 9 4 t 2 27 16 t + 7 16 ) , t ( 3 4 , 1 ] .

This, together with (H2), yields

Φ ( u n ) = 2 , 048 ( 3 8 9 10 4 4 A + 79 14 4 8 B ) ξ n 2 + j = 1 m 0 ξ n I 2 j ( s ) d s k 4 ξ n 2 + k 2 ξ n 2 = k 5 ξ n 2 .

It then follows from (H3) that

φ λ ( u n ) = Φ ( u n ) λ Ψ ( u n ) k 5 ξ n 2 λ 1 4 3 4 F ( t , ξ n ) d t ξ n 2 ( k 5 λ M ) .

Note that k 5 λM<0. Thus the functional φ λ is unbounded from below and hence it has no global minimum and the proof is complete. □

Corollary 3.4 Assume that (H1), (H2), and (H3) are satisfied. If

lim inf ξ + 0 1 max | u | ξ F ( t , u ) d t ξ 2 < 1 2 M 1 2

and

lim sup ξ + 1 4 3 4 F ( t , ξ ) d t ξ 2 > k 5

hold, then (1.2) has an unbounded sequence of solutions in X.

Let

μ 3 =2 M 1 2 lim inf ω 0 + 0 1 max | u | ω F ( t , u ) d t ω 2 , μ 4 = 1 k 5 lim sup ω 0 + 1 4 3 4 F ( t , ω ) d t ω 2 .

Theorem 3.5 Assume that (H1), (H2), and (H3) are satisfied. If

μ 3 < μ 4
(3.18)

holds, then for each λ( 1 μ 4 , 1 μ 3 ) system (1.2) has a sequence of non-zero solutions in X, which weakly converges to 0.

Proof The proof is similar to that of Theorem 3.3 by showing that γ 2 <+ and 0 is not a local minimum of the functional φ λ =ΦλΨ. □

Example 3.2 Consider

{ u ( i v ) ( t ) 2 u ( t ) + u = λ f ( t , u ) , t t 1 , t [ 0 , 1 ] , Δ u ( t 1 ) = 1 10 u ( t 1 ) , t 1 = 1 2 , Δ u ( t 1 ) = 1 20 u ( t 1 ) , t 1 = 1 2 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 ,
(3.19)

where f(t,u)=4tu(1+sinu)+2t u 2 cosu.

Here I 11 (s)= 1 10 s, I 21 (s)= 1 20 s, A=2, B=1 and m=1. Note that

0 u ( t 1 ) I 11 ( s ) d s = 0 u ( t 1 ) 1 10 s d s = 1 20 | u ( t 1 ) | 2 , 0 u ( t 1 ) I 21 ( s ) d s = 0 u ( t 1 ) 1 20 s d s = 1 40 | u ( t 1 ) | 2 , t 1 = 1 2 [ 1 4 , 3 4 ] , F ( t , u ) = 2 t ( 1 + sin u ) u 2 ,

so (H1), (H2), and (H3) are satisfied. Moreover, we have k 5 782.6, and

lim ξ + inf 0 1 max | u | ξ F ( t , u ) d t ξ 2 =0, lim ξ + sup 1 4 3 4 F ( t , ξ ) d t ξ 2 =1.

Therefore, condition (3.17) holds and Theorem 3.3 applies: For λ(782.6,+), (3.19) admits an unbounded sequence of solutions in X.

Example 3.3 Consider the boundary value problem

{ u ( i v ) ( t ) 2 u ( t ) + u = λ f ( t , u ) , t t 1 , t [ 0 , 1 ] , Δ u ( t 1 ) = 1 5 u ( t 1 ) , t 1 = 1 2 , Δ u ( t 1 ) = 1 8 u ( t 1 ) , t 1 = 1 2 , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 ,
(3.20)

where

f ( t , u ( t ) ) ={ 4 t u ( 0.5001 + 1 2 cos ( ln | u | ) 1 4 sin ( ln ( | u | ) ) ) if  u 0 , 0 if  u = 0 .

In this example, I 11 (s)= 1 5 s, I 21 (s)= 1 8 s, A=2 and B=1. The assumptions (H1), (H2), and (H3) clearly hold.

Direct calculations give

F ( t , u ( t ) ) ={ 2 t u 2 ( 0.5001 + 1 2 cos ( ln | u | ) ) if  u 0 , 0 if  u = 0 ,

k 5 762.64 and

lim w 0 + inf 0 1 max | u | w F ( t , u ) d t w 2 =0.0001, lim w 0 + sup 1 4 3 4 F ( t , w ) d t w 2 =0.50005.

Hence (3.18) holds. Therefore it follows from Theorem 3.5 that (3.20) admits a sequence of distinct solutions in X provided that λ(1525.1,2877.0).

Authors’ information

JX is with the College of Mathematics and Statistics, Jishou University, China and is a PhD candidate at the Department of Mathematics, Hunan Normal University, China. ZL is a professor at the Department of Mathematics, Hunan Normal University, China.

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Acknowledgements

The authors are very grateful to the referees for their valuable comments and suggestions, which greatly improved the presentation of this paper. The work is partially supported by Hunan Provincial Natural Science Foundation of China (No: 11JJ3012).

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Correspondence to Zhiguo Luo.

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The authors declare that they have no competing interests.

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Both authors made equal contribution. Both authors read and approved the final manuscript.

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Keywords

  • critical point theorem
  • impulsive differential equations
  • boundary value problem