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Existence of solutions of multi-point boundary value problems on time scales at resonance

Abstract

By applying the coincidence degree theorem due to Mawhin, we show the existence of at least one solution to the nonlinear second-order differential equation

u Δ (t)=f ( t , u ( t ) , u Δ ( t ) ) ,t [ 0 , 1 ] T ,

subject to one of the following multi-point boundary conditions:

u(0)= i = 1 m α i u( ξ i ), u Δ (1)=0,

and

u(0)= i = 1 m α i u( ξ i ),u(1)=0,

where T is a time scale such that 0T, 1 T k , ξ i (0,1)T, i=1,2,,m, f: [ 0 , 1 ] T × R 2 R is continuous and satisfies the Carathéodory-type growth conditions.

MSC:34B15, 39A10, 47G20.

1 Introduction

We assume that the reader is familiar with some notations and basic results for dynamic equations on time scales. Otherwise, the reader is referred to the introductory book on time scales by Bohner and Peterson [1, 2].

There is much current activity focused on dynamic equations on time scales, and a good deal of this activity is devoted to boundary value problems. We refer the readers to Agarwal [3], Morelli [4], Amster [5] and the references therein.

In [6], Anderson studied

u Δ (t)=ηa(t)f ( u ( t ) ) ,t ( t 1 , t n ) T ,
(1.1)

subject to one of the following boundary conditions:

u( t 1 )= i = 2 n 1 α i u( t i ), u Δ ( t n )=0,
(1.2)
u Δ ( t 1 )=0,u( t n )= i = 2 n 1 α i u( t i ).
(1.3)

By using a functional-type cone expansion-compression fixed point theorem, the authors get the existence of at least one positive solution to BVP (1.1), (1.2) and BVP (1.1), (1.3) when i = 2 n 1 α i 1.

When i = 2 n 1 α i =1, the operator L= u Δ is non-invertible, this is the so-called resonance case, and the theory used in [6] cannot be used. And to the best knowledge of the authors, the resonant case on time scales has rarely been considered. So, motivated by the papers mentioned above, in this paper, by making use of the coincidence degree theory due to Mawhin [7], we study

u Δ (t)=f ( t , u ( t ) , u Δ ( t ) ) ,t [ 0 , a ] T ,
(1.4)

subject to the following two sets of nonlocal boundary conditions:

u(0)= i = 1 m α i u( ξ i ), u Δ (1)=0,
(1.5)
u(0)= i = 1 m α i u( ξ i ),u(1)=0,
(1.6)

where T is a time scale such that 0T, 1 T k , ξ i (0,1)T, i=1,2,,m, i = 1 m α i =1 holds when (1.4), (1.5) are studied. While i = 1 m α i (1 ξ i )=1 when (1.4), (1.6) are studied. f: [ 0 , 1 ] T × R 2 R is continuous. We impose Carathéodory-type growth assumptions on f. It is possible, by other methods, to allow nonlinear growth on f, we refer to [8, 9] and the references therein when a time scale is R. A different m-point boundary value problem at resonance is studied in [10].

The main features in this paper are as follows. First, we study two new multi-point BVPs on time scales at resonance, which have rarely been considered, and thus we need to overcome some new difficulties. Second, we give reasons for every important step, which in turn makes this paper easier to be understood. Last but not the least, at the end of this paper, we give examples to illustrate our main results.

We will adopt the following notations throughout:

  1. (i)

    by [ a , b ] T we mean that [a,b]T, where a,bR, and ( a , b ) T is similarly defined.

  2. (ii)

    by u L 1 [a,b] we mean a b |u|t<.

2 Some definitions and some important theorems

For the convenience of the readers, we provide some background definitions and theorems.

Theorem 2.1 ([[1], p.137])

If f,g:TR are rd-continuous, then

a b f(t) g Δ (t)Δt+ a b f Δ (t)g ( σ ( t ) ) Δt=(fg)(b)(fg)(a).
(2.1)

Theorem 2.2 ([[2], p.332])

If f,g:TR are ld-continuous, then

a b f(t) g (t)t+ a b f (t)g ( ρ ( t ) ) t=(fg)(b)(fg)(a).
(2.2)

Theorem 2.3 ([[1], p.139])

The following formulas hold:

  1. (i)

    ( a t f ( t , s ) Δ s ) Δ =f(σ(t),t)+ a t f Δ (t,s)Δs;

  2. (ii)

    ( a t f ( t , s ) Δ s ) =f(ρ(t),ρ(t))+ a t f (t,s)Δs;

  3. (iii)

    ( a t f ( t , s ) s ) Δ =f(σ(t),σ(t))+ a t f Δ (t,s)s;

  4. (iv)

    ( a t f ( t , s ) s ) =f(ρ(t),t)+ a t f (t,s)s.

Theorem 2.4 ([[1], p.89])

If f:TR is Δ-differentiable on  T k and if f Δ is continuous on  T k , then f is -differentiable on T k and

f (t)= f Δ ρ (t)for t T k .

If g:TR is -differentiable on T k and if g is continuous on T k , then g is Δ-differentiable on T k and

g Δ (t)= g σ (t)for t T k .

Theorem 2.5 [11]

If f is -integral on [a,b], then so is |f|, and

| 0 t f(t)t| 0 t | f ( t ) | t.
(2.3)

Definition 2.1 Let X and Y be normed spaces. A linear mapping L:domLXY is called a Fredholm operator with index 0, if the following two conditions hold:

  1. (i)

    ImL is closed in Y;

  2. (ii)

    dim KerL=codim ImL<.

Consider the supplementary subspaces X 1 and Y 1 such that X=KerLX1 and Y=ImLY1, and let P:XKerL and Q:YY1 be the natural projections. Clearly, KerL(domLX1)={0}; thus the restriction L P :=L | dom L X 1 is invertible. The inverse of L P :=L | dom L X 1 we denote by K P :ImLdomLKerP.

If L is a Fredholm operator with index zero, then, for every isomorphism J:ImQKerL, the mapping JQ+ K P (IQ):YdomL is an isomorphism and, for every udomL,

( J Q + K p ( I Q ) ) 1 u= ( L + J 1 P ) u.

Definition 2.2 Let L:domLXY be a Fredholm mapping, E be a metric space, and N:EY be a mapping. We say that N is L-compact on E if QN:EZ and K P (IQ)N:EX are compact on E.

Theorem 2.6 [7]

Suppose that X and Y are two Banach spaces, and L:domLXY is a Fredholm operator with index 0. Furthermore, ΩX is an open bounded set and N: Ω ¯ Y is L-compact on Ω ¯ . If:

  1. (i)

    LxλNx, xΩ(domLKerL), λ(0,1);

  2. (ii)

    NxImL, xΩKerL;

  3. (iii)

    deg{JQN,ΩKerL,0}0,

then Lx=Nx has a solution in Ω ¯ domL.

3 Related lemmas

Let

X= { u : [ 0 , 1 ] R : u Δ A C [ 0 , 1 ] , u Δ L 1 [ 0 , 1 ] }

with the norm u=sup{ u 0 , u Δ 0 }, where u 0 = sup t [ 0 , 1 ] T |u(t)|.

Let Y= L 1 [0,1] with the norm u 1 = 0 1 |u(t)|t.

Define the linear operator L 1 :dom L 1 XY by L 1 u= u Δ , with dom L 1 ={uX,u satisfies (1.5)}, and the linear operator L 2 :dom L 2 XY by L 2 u= u Δ , with dom L 2 ={uX,u satisfies (1.6)}.

For any open and bounded ΩX, we define N: Ω ¯ Y by

N=f ( t , u ( t ) , u Δ ( t ) ) ,t [ 0 , 1 ] T .
(3.1)

Then (1.4), (1.5) (respectively (1.4), (1.6)) can be written as

L 1 x=Nx(respectively  L 2 x=Nx).

Lemma 3.1 The mappings L 1 :dom L 1 XY and L 2 :dom L 2 XZ are Fredholm operators with index zero.

Proof We first show that L 1 is a Fredholm operator with index zero. We divide this process into two steps.

Step 1: Determine the image of L 1 .

Let yY and for t [ 0 , 1 ] T ,

u(t)= t 1 (st)y(s)s+c,

then by Theorem 2.3,

u Δ ( t ) = ( t 1 ( s t ) y ( s ) s + c ) Δ = ( 1 t ( t s ) y ( s ) s ) Δ = ( σ ( t ) σ ( t ) ) y ( σ ( t ) ) + 1 t ( t s ) Δ y ( s ) s = 1 t y ( s ) s ,

and consequently, u Δ (1)=0 and u Δ (t)=y(t). If, in addition, y(s) satisfies

0 1 sy(s)s= i = 1 m α i ξ i 1 (s ξ i )y(s)s,
(3.2)

then u(t) satisfies the multi-point boundary condition (1.5). That is, udom L 1 , and we conclude that

{ y Y , y  satisfies  ( 3.2 ) } Im L 1 .

Let uX, then, by Theorem 2.2,

t 1 ( s t ) u Δ ( s ) s = 1 t ( t s ) u Δ ( s ) s = ( t s ) u Δ ( s ) | 1 t + 1 t u Δ ( ρ ( s ) ) s = ( t 1 ) u Δ ( 1 ) + 1 t u ( s ) s = ( 1 t ) u Δ ( 1 ) + u ( t ) u ( 1 ) ,
(3.3)

that is, u(t)=u(1)(1t) u Δ (1)+ t 1 (st) u Δ (s)s. If yIm L 1 , there exists udom L 1 X such that

u Δ (t)=y(t),

and the boundary conditions (1.5) are satisfied. Then the expression above becomes

u(t)=u(1)+ t 1 (st)y(s)s.

Since i = 1 m α i =1 and u(0)= i = 1 m α i u( ξ i ), it follows that (3.2) holds. Hence,

Im L 1 = { y Y : y  satisfies  ( 3.2 ) } .

Step 2: Determine the index of L 1 .

Let a continuous linear operator Q 1 :YY be defined by

Q 1 y= 1 C 1 ( 0 1 s y ( s ) s i = 1 m α i ξ i 1 ( s ξ i ) y ( s ) s ) ,
(3.4)

where C 1 = 0 1 ss i = 1 m α i ξ i 1 (s ξ i )s0.

It is clear that Q 1 2 y= Q 1 y, that is, Q 1 :YY is a continuous linear projector. Furthermore, Im L 1 =Ker Q 1 . Let y=(y Q 1 y)+ Q 1 yY. It is easy to see that Q 1 (y Q 1 y)=0, thus y Q 1 yKer Q 1 =Im L 1 and Q 1 yIm Q 1 and so Y=Im L 1 +Im Q 1 . If yIm L 1 Im Q 1 , then y(t)0. Hence, we have Y=Im L 1 Im Q 1 .

It is clear that Ker L 1 ={u=a,aR}. Now, Ind L 1 =dim Ker L 1 codim Im L 1 =dim Ker L 1 dim Im Q 1 =0, and so L 1 is a Fredholm operator with index zero.

Next, we show that L 2 is also a Fredholm operator with index zero. We also divide it into two steps.

Step 1: Determine the image of L 2 .

Let yY and for t [ 0 , 1 ] T ,

u(t)= t 1 (st)y(s)s+c(1t),

it is obvious that u(1)=0, and then

u Δ ( t ) = ( t 1 ( s t ) y ( s ) s + c ( 1 t ) ) Δ = ( 1 t ( t s ) y ( s ) s c ) Δ = ( σ ( t ) σ ( t ) ) y ( σ ( t ) ) + 1 t ( t s ) Δ y ( s ) s c = 1 t y ( s ) s c ,

consequently, u Δ (t)=y(t). If, in addition, y(s) satisfies (3.2), then u(t) satisfies the multi-point boundary conditions (1.6). That is, udom L 2 , and we conclude that

{ y Y , y  satisfies  ( 3.2 ) } Im L 2 .

Let uX, by (3.3), we have u(t)=u(1)(1t) u Δ (1)+ t 1 (st) u Δ (s)s. If yIm L 2 , there exists udom L 2 X such that u Δ (t)=y(t) and the boundary conditions (1.6) are satisfied. The expression above becomes

u(t)= t 1 (st)y(s)s(1t) u Δ (1).

Since i = 1 m α i (1 ξ i )=1 and u(0)= i = 1 m α i u( ξ i ), it follows that (3.2) holds. Hence,

Im L 2 = { y Y : y  satisfies  ( 3.2 ) } .

Step 2: Determine the index of L 2 .

Let a continuous linear operator Q 2 :YY be defined by

Q 2 y= 1 C 2 ( 0 1 s y ( s ) s i = 1 m α i ξ i 1 ( s ξ i ) y ( s ) s ) (t1),
(3.5)

where C 2 =( 0 1 s(s1)s i = 1 m α i ξ i 1 (s ξ i )(s1)s)0.

It is clear that Q 2 2 y= Q 2 y, that is, Q 2 :YY is a continuous linear projector. Furthermore, Im L 2 =Ker Q 2 . The remainder of the argument is identical to that concerning L 1 and the proof is completed. □

Lemma 3.2 N is L 1 -compact and L 2 -compact.

Proof Let P 1 :XX and P 2 :XX be continuous linear operators defined by P 1 u(t)=u(1), t [ 0 , 1 ] T , and P 2 u(t)= u Δ (1)(1t), t [ 0 , 1 ] T , respectively.

By taking uX in the form u(t)=u(1)+(u(t)u(1)), it is clear that X=Ker L 1 Ker P 1 . Letting u(t)= u Δ (1)(1t)+(u(t)+ u Δ (1)(1t)), we derive X=Ker L 2 Ker P 2 . Note that the two pairs of projectors P 1 , Q 1 and P 2 , Q 2 are exact, that is, satisfy the relationships as desired.

Define K P 1 :Im L 1 dom L 1 Ker P 1 by

K P 1 y(t)= t 1 (st)y(s)s.
(3.6)

And K P 2 :Im L 2 dom L 1 Ker P 2 is defined by

K P 2 y(t)= t 1 (st)y(s)s.
(3.7)

Then, by Theorem 2.5,

sup t [ 0 , 1 ] T | K P 1 y ( t ) | = sup t [ 0 , 1 ] T | t 1 ( s t ) y ( s ) s | sup t [ 0 , 1 ] T t 1 | ( s t ) y ( s ) | s y 1 ,

so that

sup t [ 0 , 1 ] T | ( K P 1 y ( t ) ) Δ | = sup t [ 0 , 1 ] T | t 1 y ( s ) s | sup t [ 0 , 1 ] T t 1 | y ( s ) | s y 1 .

Therefore,

K P 1 u y 1 .
(3.8)

And similarly,

K P 2 u y 1 .
(3.9)

It is clear that K P 1 = ( L 1 | dom L 1 Ker P 1 ) 1 and K P 2 = ( L 2 | dom L 2 Ker P 2 ) 1 .

Now, by using (3.4) and (3.5), we have

Q 1 Nu= 1 C 1 ( 0 1 s f ( s , u ( s ) , u Δ ( s ) ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , u ( s ) , u Δ ( s ) ) s ) ,
(3.10)
Q 2 N u = 1 C 2 ( 0 1 s f ( s , u ( s ) , u Δ ( s ) ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , u ( s ) , u Δ ( s ) ) s ) ( t 1 ) .
(3.11)

And consequently,

K P 1 (I Q 1 )Nu(t)= t 1 (st)(N Q 1 N)u(s)s,
(3.12)
K P 2 (I Q 2 )Nu(t)= t 1 (st)(N Q 2 N)u(s)s.
(3.13)

Obviously, both QN and K P (IQ)N are compact, thus, N is L 1 -compact and L 2 -compact. The proof is complete. □

4 Existence of solution to BVP (1.4), (1.5)

For the existence result concerning (1.4), (1.5), we have the following assumptions.

(H1) There exists a constant A>0 such that for any udom L 1 Ker L 1 satisfying |u(t)|>A for all t [ 0 , 1 ] T , Q 1 Nu0 holds;

(H2) There exist functions p,q,r,δ L 1 [0,1] and a constant ε(0,1) such that for (u,v) R 2 and all t [ 0 , 1 ] T , we have

| f ( t , u , v ) | δ(t)+p(t)|u|+q(t)|v|+r(t) | v | ε ,
(4.1a)

or

| f ( t , u , v ) | δ(t)+p(t)|u|+q(t)|v|+r(t) | u | ε ;
(4.1b)

(H3) There exists a constant B>0 such that, for every bR with |b|>B, we have either

b ( 0 1 s f ( s , b , 0 ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , b , 0 ) s ) <0,
(4.2a)

or

b ( 0 1 s f ( s , b , 0 ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , b , 0 ) s ) >0.
(4.2b)

Theorem 4.1 If (H1)-(H3) hold, then the boundary value problem (1.4), (1.5) has at least one solution provided

p 1 + q 1 < 1 2 .
(4.3)

Proof Firstly, we define an open bounded subset Ω of X. It is based upon four steps to obtain Ω.

Step 1: Let

Ω 1 = { u dom L 1 Ker L 1 : L 1 u = λ N u , λ ( 0 , 1 ) } ,

then for u Ω 1 , L 1 u=λNu. Thus, we have NuIm L 1 =Ker Q 1 and

0 1 sf ( s , u ( s ) , u Δ ( s ) ) s i = 1 m α i ξ i 1 (s ξ i )f ( s , u ( s ) , u Δ ( s ) ) s=0.

It follows from (H2) that there exists t 0 [ 0 , 1 ] T such that |u( t 0 )|A. Hence, by Theorems 2.4 and 2.5, we have

| u ( 1 ) | = | u ( t 0 ) + t 0 1 u ( s ) s | = | u ( t 0 ) + t 0 1 u Δ ( ρ ( s ) ) s | A+ u Δ 0 .
(4.4)

Also, u Δ (t)= t 1 u Δ (s)s implies

u Δ 0 u Δ 1 = L 1 u 1 < N u 1 .
(4.5)

Combining (4.4), (4.5), one gets

| u ( 1 ) | A+ N u 1 .
(4.6)

Observe that (I P 1 )uIm K P 1 =dom L 1 Ker P 1 for u Ω 1 , then we obtain

( I P 1 ) u = K P 1 L 1 ( I P 1 ) u L 1 ( I P 1 ) u 1 = L 1 u 1 < N u 1 .
(4.7)

Using (4.6), (4.7), we get

u= P 1 u + ( I P 1 ) u P 1 u+ ( I P 1 ) u < | u ( 1 ) | + N u 1 <A+2 N u 1 ,
(4.8)

that is, for all u Ω 1 ,

u<A+2 N u 1 .
(4.9)

If (4.1a) holds, then

u 0 , u Δ 0 uA+2 ( δ 1 + p 1 u 0 + q 1 u Δ 0 + r 1 u Δ 0 ε ) ,
(4.10)

and consequently,

u 0 2 1 2 p 1 ( δ 1 + q 1 u Δ 0 + r 1 u Δ 0 ε + A 2 ) .
(4.11)

Further, by (4.10) and (4.11),

u Δ 0 2 p 1 u 0 + 2 ( δ 1 + q 1 u Δ 0 + r 1 u Δ 0 ε + A 2 ) 2 ( δ 1 + q 1 u Δ 0 + r 1 u Δ 0 ε + A 2 ) ( 2 p 1 1 2 p 1 + 1 ) = 2 q 1 1 2 p 1 u Δ 0 + 2 r 1 1 2 p 1 u Δ 0 ε + 2 δ 1 + A 1 2 p 1 ,
(4.12)

that is,

u Δ 0 2 r 1 1 2 ( p 1 + q 1 ) u Δ 0 ε + 2 δ 1 + A 1 2 ( p 1 + q 1 ) .
(4.13)

Since ε(0,1) and (4.13) holds, we know that there exists R 1 >0 such that u Δ 0 R 1 for all u Ω 1 . Inequality (4.11) then shows that there exists R 2 >0 such that u 0 R 2 for all u Ω 1 . Therefore, Ω 1 is bounded given (4.1a) holds. If, otherwise, (4.1b) holds, then with minor adjustments to the arguments above we derive the same conclusion.

Step 2: Define

Ω 2 ={uKer L 1 :NuIm L 1 }.

Then u=bR and NuIm L 1 =Ker Q 1 imply that

1 C 1 ( 0 1 s f ( s , b , 0 ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , b , 0 ) s ) =0.

Hence, by (H3), u=bB, that is, Ω 2 is bounded.

Step 3: Let

Ω 3 = { u Ker L 1 : H ( u , λ ) = 0 } ,

where

H(u,λ)={ λ I u + ( 1 λ ) J Q 1 N u if (4.2a) holds , λ I u + ( 1 λ ) J Q 1 N u if (4.2b) holds ,
(4.14)

and J:Im Q 1 Ker L 1 is a homomorphism such that J(b)=b for all bR.

Without loss of generality, we suppose that (4.2a) holds, then for every b Ω 3 ,

λb=(1λ) 1 C 1 ( 0 1 s f ( s , b , 0 ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , b , 0 ) s ) .

If λ=1, then b=0. And in the case λ[0,1), if |b|>B, then by (4.2a),

0λ b 2 =b(1λ) 1 C 1 ( 0 1 s f ( s , b , 0 ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , b , 0 ) s ) <0,

which is a contradiction.

When (4.2b) holds, by a similar argument, again, we can obtain a contradiction. Thus, for any u Ω 3 , uB, that is, Ω 3 is bounded.

Step 4: In what follows, we shall prove that all the conditions of Theorem 2.6 are satisfied. Let Ω be an open bounded subset of X such that i = 1 3 Ω ¯ i Ω, clearly, we have

LuNx,λ(0,1),uΩ,

and

NuIm L 1 ,uΩKer L 1 .

It can be seen easily that

H(u,λ)0,λ[0,1],uΩKer L 1 .

Then assumptions (i) and (ii) of Theorem 2.6 are fulfilled. It only remains to verify that the third assumption of Theorem 2.6 applies.

We apply the degree property of invariance under homotopy. To this end, we define the homotopy

H(u,λ)=λIu+(1λ)J Q 1 Nu.

If uΩKer L 1 , then

deg { J Q 1 N , Ω Ker L 1 , 0 } = deg { H ( , 0 ) , Ω Ker L 1 , 0 } = deg { H ( , 1 ) , Ω Ker L 1 , 0 } = deg { I , Ω Ker L 1 , 0 } 0 .
(4.15)

So, the third assumption of Theorem 2.6 is fulfilled.

Therefore, Theorem 2.6 can be applied to obtain the existence of at least one solution to BVP (1.4) and (1.5). The proof is complete. □

5 Existence of solution to BVP (1.4), (1.6)

In this section, we give the existence result for BVP (1.4), (1.6). We first state the following assumptions:

(H4) There exists a constant C>0 such that for any udom L 2 Ker L 2 satisfying | u Δ (t)|>C for all t [ 0 , 1 ] T , Q 2 Nu0 holds;

(H5) There exists a constant D>0 such that, for every dR with |d|>D, we have either

d ( 0 1 s f ( s , d ( s 1 ) , d ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , d ( s 1 ) , d ) s ) <0,
(5.1a)

or

d ( 0 1 s f ( s , d ( s 1 ) , d ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , d ( s 1 ) , d ) s ) >0.
(5.1b)

Theorem 5.1 Assume that (H2), (H4) and (H5) hold, then BVP (1.4), (1.6) has at least one solution provided

p 1 + q 1 < 1 2 .
(5.2)

Proof Let

Ω 1 = { u dom L 2 Ker L 2 : L 2 u = λ N u , λ ( 0 , 1 ) } ,

then for u Ω 1 , L 2 u=λNu. Thus, we have NuIm L 2 =Ker Q 2 , and thus

0 1 sf ( s , u ( s ) , u Δ ( s ) ) s i = 1 m α i ξ i 1 (s ξ i )f ( s , u ( s ) , u Δ ( s ) ) s=0.

It follows from (H4) that there exists t 0 [ 0 , 1 ] T such that | u Δ ( t 0 )|C. Hence, by Theorem 2.5, we have

| u Δ ( 1 ) | = | u Δ ( t 0 ) + t 0 1 u Δ ( s ) s | A+ N u 1 .
(5.3)

Observe that (I P 2 )uIm K P 2 =dom L 2 Ker P 2 for u Ω 1 , then we obtain

( I P 2 ) u = K P 2 L 2 ( I P 2 ) u L 2 ( I P 2 ) u 1 = L 2 u 1 < N u 1 .
(5.4)

Using (5.3), (5.4), one gets

u = P 2 u + ( I P 2 ) u P 2 u + ( I P 2 ) u < | u Δ ( 1 ) | + N u 1 < C + 2 N u 1 ,
(5.5)

that is, for all u Ω 1 ,

u<C+2 N u 1 .
(5.6)

As in the proof of Theorem 4.1, by applying (H2) we can show that Ω 1 is bounded.

Step 2: Define

Ω 2 ={uKer L 2 :NuIm L 2 }.

Then u=d(t1), where dR and NuIm L 2 =Ker Q 2 imply that

1 C 2 ( 0 1 s f ( s , d ( s 1 ) , s ) Δ s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , d ( s 1 ) , s ) Δ s ) =0.

Hence, by (H4), u=dD, which means Ω 2 is bounded.

Step 3: Let

Ω 3 = { u Ker L 2 : H ( u , λ ) } =0,

where

H(u,λ)={ λ I u + ( 1 λ ) J Q 2 N u if (5.1a) holds , λ I u + ( 1 λ ) J Q 2 N u if (5.1b) holds ,
(5.7)

and J:Im Q 2 Ker L 2 is a homomorphism such that J(d(t1))=d(t1) for all dR.

Without loss of generality, we suppose that (5.1a) holds, then for every d Ω 3 ,

λd=(1λ) 1 C 2 ( 0 1 s f ( s , d ( s 1 ) , d ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , d ( s 1 ) , d ) s ) .

If λ=1, then d=0. And in the case λ[0,1), if |d|>D, then by (5.1a),

0λ d 2 =d(1λ) 1 C 2 ( 0 1 s f ( s , d ( s 1 ) , d ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , d ( s 1 ) , d ) s ) <0,

which is a contradiction.

When (5.1b) holds, by a similar argument, again, we can obtain a contradiction. Thus, for any u Ω 3 , uD, that is, Ω 3 is bounded.

Step 4 is essentially the same as that of Theorem 4.1. Applying Theorem 2.6, we obtain the existence of at least one solution to BVP (1.4), (1.6). The proof is complete. □

6 Examples

In this section, we give an example to illustrate our main results.

Example 6.1 Let T=[ k 2 , 2 k + 1 4 ], where kZ. We consider the following BVP on T.

{ u Δ ( t ) = 1 6 ( 50 + 2 t 2 + u sin t + 20 t 2 ( u Δ ( t ) ) 1 / 3 + u Δ ( t ) ) , t [ 0 , 1 ] T , u ( 0 ) = 1 3 u ( 1 4 ) + 2 3 u ( 1 2 ) , u Δ ( 1 ) = 0 .
(6.1)

It is clear that f(t,u,v)= 1 6 (50+2 t 2 +usint+20 t 2 v 1 / 3 +v), 0T, 1 T k , ξ 1 = 1 4 , ξ 2 = 1 2 T, α 1 = 1 3 , α 2 = 2 3 , and α 1 + α 2 =1, thus BVP (6.1) is resonant.

In what follows, we try to show that all the conditions in Theorem 4.1 are satisfied.

Let δ(t)= t 2 + 25 3 , p(t)= | t | 6 , q(t)= 1 6 , r(t)= 10 t 2 3 , ε= 1 3 . We can see that

| f ( t , u , v ) | δ(t)+p(t)|u|+q(t)|v|+r(t) | v | ε

holds, which implies that (H2) is satisfied.

After a series of calculations, we obtain

C 1 = 0 1 s s i = 1 m α i ξ i 1 ( s ξ i ) s = ( 0 1 / 4 + 1 / 2 3 / 4 ) s d s + ( 1 / 4 1 / 2 + 3 / 4 1 ) s s 1 3 ( ( 1 / 4 1 / 2 + 3 / 4 1 ) ( s 1 4 ) s + 1 / 2 3 / 4 ( s 1 4 ) d s ) 2 3 ( 1 / 2 3 / 4 ( s 1 2 ) d s + 3 / 4 1 ( s 1 2 ) s ) = 11 32 0 .

For udom L 1 Ker L 1 , u(t)=at, u Δ (t)=a, then we have

6 C 1 Q 1 N u = 0 1 s f ( s , a s , a ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , a s , a ) s = 80 , 789 4 , 608 + 44 , 313 , 960 , 092 , 071 , 337 36 , 028 , 797 , 018 , 963 , 968 a + 23 , 915 2 , 304 a 1 / 3 17.5323 + 1.23 a + 10.3798 a 1 / 3 .

Let A=3, then when |u(t)|=|a|>3, Q 1 Nu0, which implies that (H1) holds while

b ( 0 1 s f ( s , b , 0 ) s i = 1 m α i ξ i 1 ( s ξ i ) f ( s , b , 0 ) s ) = 1 6 ( 47 , 545 2 , 304 b + 38 , 324 , 624 , 432 , 429 , 689 72 , 057 , 594 , 037 , 927 , 936 b 2 ) 1 6 ( 20.6359 b + 0.5319 b 2 ) .

Let B=39, then when |b|>B, (4.2a) or (4.2b) holds, which implies that (H3) is satisfied.

Finally, it is obvious that p 1 + q 1 < 1 2 . Thus, all the conditions in Theorem 4.1 are satisfied, then BVP (6.1) has at least one solution.

Example 6.2 Let T=[0, 2 3 ]{1}. We consider the following BVP on T.

{ u Δ ( t ) = 100 + t + t 2 u 4 + t 2 u 1 / 5 + t u Δ ( t ) 8 , t [ 0 , 1 ] T , u ( 0 ) = u ( 1 2 ) , u ( 1 ) = 0 .
(6.2)

It is clear that f(t,u,v)=100+t+ t 2 u 4 + t 2 u 1 / 5 + t u Δ ( t ) 8 , 0T, 1 T k , ξ 1 = 1 2 T, α 1 =1, thus BVP (6.2) is resonant.

In what follows, we try to show that all the conditions in Theorem 5.1 are satisfied.

Let δ(t)=100+t, p(t)= t 2 4 , q(t)= t 8 , r(t)= t 2 , ε= 1 5 . We can see that

| f ( t , u , v ) | δ(t)+p(t)|u|+q(t)|v|+r(t) | u | ε

holds, which implies that (H2) is satisfied.

After a series of calculations, we obtain

C 2 = 0 1 s ( s 1 ) s 1 / 2 1 ( s 1 2 ) ( s 1 ) s = 0 2 / 3 s ( s 1 ) d s + 2 / 3 1 s ( s 1 ) s 1 / 2 2 / 3 ( s 1 2 ) ( s 1 ) d s 2 / 3 1 ( s 1 2 ) ( s 1 ) s = 0.1181 0 .

For udom L 2 Ker L 2 , u(t)=a, u Δ (t)=0, then we have

C 2 Q 2 N u = 0 1 s f ( s , a , 0 ) s 1 / 2 1 ( s 1 2 ) f ( s , a , 0 ) s = 2 , 875 20 , 736 a + 2 , 875 5 , 184 a 1 / 5 + 5 , 437 144 0.1386 a + 0.5546 a 1 / 5 + 37.7569 .

Let C=300, then when | u Δ (t)|=|a|>300, Q 2 Nu0, which implies that (H4) holds, while

d ( 0 1 s f ( s , d ( s 1 ) , d ) s 1 / 2 1 ( s 1 2 ) f ( s , d ( s 1 ) , d ) s ) = 4 , 937 , 627 , 091 , 458 , 903 72 , 057 , 594 , 037 , 927 , 936 d 2 4 , 954 , 847 , 807 , 426 , 245 1 , 152 , 921 , 504 , 606 , 846 , 976 d 6 / 5 + 7 , 490 , 236 , 786 , 645 , 797 140 , 737 , 488 , 355 , 328 d 0.0685 d 2 0.0043 d 6 / 5 + 53.2213 d .

Let D=780, then when |d|>D, (5.1a) or (5.1b) holds, which implies that (H5) is satisfied.

Finally, it is obvious that p 1 + q 1 < 1 2 . Thus, all the conditions in Theorem 5.1 are satisfied, then BVP (6.2) has at least one solution.

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Acknowledgements

The authors were very grateful to the anonymous referee whose careful reading of the manuscript and valuable comments enhanced presentation of the manuscript. The study was supported by the National Natural Science Foundation (No. 11226133), the Fundamental Research Funds for the Central Universities (No. 2652012141), the Young Talents Programme of Beijing, Beijing higher education’s reform project of 2013 ‘Teaching research with mathematical thinking to promote the innovative ability of modern Geosciences talents’ and Beijing support to the central authorities in Beijing University Construction Projects ‘The reform of mathematics teaching content and methods of service in the modern geoscience of personnel training’.

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BZ and HL conceived of the study and participated in its coordination. JZ drafted the manuscript. All authors read and approved the final manuscript.

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Zhao, J., Chu, B. & Lian, H. Existence of solutions of multi-point boundary value problems on time scales at resonance. Adv Differ Equ 2013, 351 (2013). https://doi.org/10.1186/1687-1847-2013-351

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Keywords

  • multi-point BVP
  • time scale
  • resonance
  • coincidence degree