The existence of solutions for a third order multi point boundary value problem at resonance

Mathematical Computation June 2013, Volume 2, Issue 2, PP.13-18

The Existence of Solutions for a Third-Order Multi-Point Boundary Value Problem at Resonance* Weihua Jiang#, Jiqing Qiu, Ruiyan Li College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018, Hebei, P. R. China #Email: weihuajiang@hebust.edu.cn

Abstract By means of the coincidence degree theory due to Mawhin, suitable Banach space has been constructed and appropriate operators have been defined, the solutions to a third-order multi-point boundary value problem at resonance have been obtained, in which dim KerL  3 .

Keywords: Resonance; Fredholm Operator; Multi-Point Boundary Value Problem; Coincidence Degree Theory

1

INTRODUCTION

In this paper, the solutions to the third-order multi-point boundary value problem have been studied:

u(t )  f  t , u(t ), u(t ), u(t )   e(t ), t  (0,1), m

n

l

i 1

j 1

k 1

(1.1)

u (0)   i u (i ) , u (0)   i u ( j ) , u (1)    k u ( k ) , where f :[0,1]  R3  R is a 0  1  2   n  1 , 0  1   2 

Caratheodory  l  1 .

function,

e(t )  L1[0,1],

0  1  2 

(1.2)  m  1

，

The boundary value problem (1.1)-(1.2) is a problem at resonance if Lu :  u(t )  0 has non-trivial solutions under the boundary condition (1.2), i.e. dim KerL  1 . The solutions to first-order, second-order and high-order multi-point boundary value problems at resonance have been studied previously (see, for example [1-5]), in which dim KerL  1 . In [6-8], the second-order multi-point boundary value problems at resonance have been discussed when dim KerL  2 .

Motivated by the above results, we will investigate the solutions to the problem (1.1)-(1.2) with dim KerL  3 . To the best of our knowledge, this is the first paper to study the resonance problems with dim KerL  3 .

In this section, the necessary background definitions and the key theorem duo to Mawhin are provided. In section 2, the main results of the problem (1.1)-(1.2) will be stated and proved.

Let Y and Z be real Banach spaces and let L :  domL  Y  Z be a Fredholm operator with index zero, P : Y  Y , Q : Z  Z be projectors such that

Im P  KerL , KerQ  Im L , Y  KerL  KerP , Z  Im L  Im Q .

It follows that

L

domL KerP

: domL

KerP  Im L

is invertible. The inverse is defined by K P . *

MSC: 34B10, 34B15 This work is supported by the Natural Science Foundation of China (11171088) and the Natural Science Foundation of Hebei Province (A2013208108) - 13 www.ivypub.org/mc

If  is an open bounded subset of Y , domL

   , the map N : Y  Z will be called L-compact on  if

 

QN  is bounded and K P ( I  Q) N :   Y is compact.

Theorem 2.1 [9]. Let L :  domL  Y  Z be a Fredholm operator of index zero and N : Y  Z L-compact on  .

It is assumed that the following conditions are satisfied: (1) Lu   Nu for every (u,  )   domL \ KerL  (2) Nu  Im L for every u  KerL (3) deg  QN

KerL

  (0,1) ;

 ;

KerL, 0   0 , where Q : Z  Z is a projection such that Im L  KerQ .

,

Then the equation Lu  Nu has at least one solution in domL

2

.

MAIN RESULTS

Let Y  C 2 [0,1] be a real Banach space with norm u  max  u  , u  , u   , where u 1 Z  L1[0,1] with norm u 1   u (t ) dt . L : domL  Y  Z is defined by 0



 max u(t ) and let t[0,1]

Lu  u, u  domL,

n l   where domL  u  Y : u (0)  i u (i ), u (0)    j u ( j ), u (1)    k u ( k ), u   Z  . i 1 j 1 k 1   m

A nonlinear operator N : Y  Z is defined by Nu(t )  f  t , u(t ), u(t ), u(t )   e(t ), t [0,1].

Then boundary value problem (1.1)-(1.2) can be written by Lu  Nu , u  domL .

In this paper, it is supposed that the following conditions hold: m

( C1 )

n

l

       i 1

i

j

j 1

k 1

k

 1,

m

m

     i i

i 1

Q11 Q21 Q31

( C2 ) △  Q1t Q2t Q31 Q1t

2

Q2t

2

Q3t

i i

i 1

n

2

   j j  0 ; j 1

a11

a12

a13

:  a21

a22

a23  0 .

a31

a32

a33

2

where Qi : Z  R, i  1, 2,3 are defined as follows: m

Q1 y  i i 1



i 0

n

j

l

(i  s)2 y(s)ds , Q2 y    j  (  j  s) y (s )ds , Q3 y    k  y (s)ds j 1

0

k 1

1

k

Lemma 2.1 Supposing that conditions ( C1 ) and ( C2 ) hold, then L :  domL  Y  Z is a Fredholm operator of index zero. Furthermore, the linear continuous operator Q : Z  Z can be defined by

Qy  T1 y  T2 y  t  T3 y  t 2 . where

T1 y 

1 △11Q1 y  △12Q2 y  △13Q3 y  , △

T2 y 

1 △21Q1 y  △22Q2 y  △23Q3 y  , △

1 △31Q1 y  △32Q2 y  △33Q3 y  . △ △ij is the algebraic cofactor of aij , i  1, 2,3, j  1, 2,3.

T3 y 

Projection P : Y  Y is defined by - 14 www.ivypub.org/mc

1 Pu (t )  u (0)  u (0)t  u (0)t 2 , t [0,1].

2 The linear operator K P :  Im L  domL be written by

KerP , which is the inverse of L

KP y 

1 t (t  s)2 y(s)ds, 2 0

domL KerP

: domL

KerP  Im L , can

y  Im L.

It is clear that

KP y  y 1 ,

y  Z.

(2.1)

Proof Obviously, KerL  a  bt  ct 2 | a, b, c  R . Now, it is shown that Im L   y | y  Z , Q1 y  Q2 y  Q3 y  0.

In fact, if Lu(t )  y(t ), u  domL , then

1 1 t u (t )  u (0)  u (0)t  u (0)t 2   (t  s)2 y(s)ds .

2 2 0 By (1.2), Q1 y  Q2 y  Q3 y  0 can be obtained.

On the other hand, if Q1 y  Q2 y  Q3 y  0 , take

u (t ) 

1 t (t  s)2 y(s)ds.

2 0

Then u(t )  y(t ) and u satisfies condition (1.2).

For y  Z , considering the definition of △ and △ij , we have 1 T1 T1 y   △11Q1 T1 y   △12Q2 T1 y   △13Q3 T1 y  △ 1  △11Q11  △12Q21  △13Q31 T1 y △  T1 y .

1 △11Q1 T2 y  t   △12Q2 T2 y  t   △13Q3 T2 y  t  △ 1  △11Q1 t  △12Q2 t  △13Q3 t  T2 y △  0.

T1 T2 y  t  

T1 T3 y  t 2  

1 △11Q1 T3 y  t 2   △12Q2 T3 y  t 2   △13Q3 T3 y  t 2   △ 1  △11Q1 t 2  △12Q2 t 2  △13Q3 t 2  T3 y △  0.

Similarly,

it

can

be

noted

that

T2 T1 y   T2 T3 y  t 2   T3 T1 y   T3 T2 y  t   0 ， T2 T2 y  t   T2  y  ，

T3 T3 y  t 2   T3  y  . Thus, Q2 y  Qy is obtained, meaning that Q : Z  Z is a projector operator.

Now, it will be proved that KerQ  Im L .

Obviously, Im L  KerQ . If y  KerQ , then Qy  0 , thus T1 y  T2 y  T3 y  0 , i.e. △11Q1 y  △12Q2 y  △13Q3 y  0 , △21Q1 y  △22Q2 y  △23Q3 y  0 , △31Q1 y  △32Q2 y  △33Q3 y  0 .

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△11 △12 △13 Since △21 △22 △23 

△31 △32 △33 For y  Z ,

1  0 , we get Q1 y  Q2 y  Q3 y  0 , i.e. y  Im L . So, KerQ  Im L . △2

y  Qy  ( y  Qy) , by

y  Qy  KerQ  Im L

and Qy  Im Q , we get Z  Im L 

ImQ .

Take y  Im L Im Q . By y  Im L , we have Q1 y  Q2 y  Q3 y  0 . Considering y  Im Q , we get y  Qy  0 .

So, Z  Im L  Im Q .

Since dim KerL  co dimIm L  dimIm Q  3 , it can be said that L is a Fredholm operator of index zero.

Now, it is shown that the generalized inverse K P : Im L  domL

KerP can be written by

1 t (t  s)2 y (s)ds .

2 0 In fact, for y  Im L , we have LK P y   K P y   y . For u  domL KerP , we get KP y 

1 t 1 (t  s)2 u (s)ds  u (0)  u (0)t  u (0)t 2  u (t ).  0 2 2 Considering u  KerP , we get u(0)  u(0)  u(0)  0. Thus, K P Lu  u. So, we get K P Lu 



KP  L

domL KerP



1

.

The following theorem is our main results.

Theorem 2.1 Suppose ( C1 ), ( C2 ) and the following conditions hold: ( H1 ) functions  ,  ,  ,  ,   L1[0,1] and a constant  [0,1) exist such that, for all x1 , x2 , x3  R, t [0,1] , one of the following conditions holds:

f  t , x1 , x2 , x3    (t )   (t ) x1   (t ) x2   (t ) x3   (t ) xi , i  1, 2,3. 

Where

1 4

 1  1  1 .

( H 2 ) a constant A  0 exists such that if u (t )  A or u(t )  A or u(t )  A , for all t  [0,1] , one of the following inequality holds:

Qi Nu (t )

0 , t [0,1], i  1, 2,3.

( H 3 ) a constant B  0 exists such that, for any a, b, c  R satisfying a2  b2  c2  B , either (1) aT1 N  a  bt  ct 2   bT2 N  a  bt  ct 2   cT3 N  a  bt  ct 2   0, t  [0,1] or (2) aT1 N  a  bt  ct 2   bT2 N  a  bt  ct 2   cT3 N  a  bt  ct 2   0, t  [0,1] Then the problem (1.1)-(1.2) has at least one solution.

Proof This theorem will be verified by the following three steps.

Step 1 Set 1  u  domL \ KerL : Lu   Nu, for some  [0,1] 

It is proved that 1 is bounded.

In fact, u 1 means Nu  Im L . Therefore, we get Q1 Nu  Q2 Nu  Q3 Nu  0. By ( H 2 ), ti [0,1], i  1, 2,3 exists such that u(t0 )  A , u(t1 )  A , u(t2 )  A .

It follows from t0

u (0)  u (t0 )   u (s)ds  A  u  0

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

,

t1

u(t )  u(t1 )   u (s)ds  A  u  t



, t [0,1]

and t2

u(t )  u (t2 )   u (s)ds  A  u  1 , t  [0,1] t

that  Pu  max  Pu  , 

 Pu 

, 

 Pu 



   6 A  3 Nu 1 . 

In addition, ( I  P)u  K P L( I  P)u  L( I  P)u 1  Lu 1  Nu 1 .

Therefore, u  Pu  ( I  P)u  Pu  (I  P)u  6 A  4 Nu 1 .

This, together with ( H1 ), implies



u  6A  4  1  

1

u  

1

u 

1

u  

1

u



 e1



So, we have u 

1 1  4  1   1  

1

6A  4  

1

4 e14 

1

u



.

Since  [0,1) , we get that there exists a constant M  0 such that u  M , i.e. 1 is bounded.

Step 2 Let 2  u  KerL : Nu  Im L.

u 2 means that constants a, b, c  R exist such that u(t )  a  bt  ct 2 and Qi N  a  bt  ct 2   0, i  1, 2,3.

It follows from ( H 3 ) that u  3 B , i.e.  2 is bounded.

Step 3 A linear isomorphism J : KerL  Im Q is defined by

J  a  bt  ct 2  

1 △11a  △12b  △13c  △21a  △22b  △23c  t  △31a  △32b  △33c  t 2  △

It is assumed that ( H 3 )(1) holds. Take

3  u  KerL :   Ju  (1   )QNu  0, for some  [0,1]  .

It will be verified that 3 is bounded.

For u 3 , we have u(t )  a  bt  ct 2 and

 J  a  bt  ct 2   QN  a  bt  ct 2  .

Thus

 △11a  △12b  △13c   (1   )T1 N  a  bt  ct 2  ,

 △21a  △22b  △23c   (1   )T2 N  a  bt  ct 2  ,  △31a  △32b  △33c   (1   )T3 N  a  bt  ct 2  .

If   1 , then a  b  c  0 . If   0 , then Qi N  a  bt  ct 2   0 , by ( H 3 ), u  3 B . If   (0,1) , it is assumed that a2  b2  c2  B , by ( H 3 ) (1), we get

  a 2  b2  c2   (1   ) aT1 N  a  bt  ct 2   bT2 N  a  bt  ct 2   cT3 N  a  bt  ct 2   0 a contradiction. So, a2  b2  c2  B , i.e. 3 is bounded.

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Remark 2.1 If ( H 3 ) (2) holds, then take

3  u  KerL :  Ju  (1   )QNu  0, for some  [0,1]  .

Similarly, we can get that 3 is bounded.

Then, it is shown that all the conditions of Theorem 1.1 are satisfied.

Take an open bounded subset Y   

3

i

 

{0} . By ( H1 ), it can be inferred that QN 

i 1

is bounded. By

Arzela- Ascoli theorem, K p ( I  Q) N :   Y that is compact is acquired. Thus, N is L-compact on  .

It follows from Step 1 and Step 2 that (1) Lu   Nu for every (u,  )   domL \ KerL  (2) Nu  Im L , for every u  KerL It will be further shown that (3) deg  QN

KerL

, 

  (0,1);

. .

KerL, 0   0.

Let

H (u,  )   Ju  (1   )QNu,  [0,1].

It follows from Step 3 that H (u,  )  0 , u  KerL  ,  [0,1] . Therefore, by the homotopy degree, we get deg  QN KerL ,  KerL, 0   deg  H  , 0  ,  KerL, 0 

 deg  H  ,1 , 

KerL, 0   deg   J , 

property of

KerL, 0   1  0.

By Theorem 1.1, it can be obtained that Lu  Nu has at least one solution in domL one solution in Y . The proof is completed.

 , i.e. (1.1)-(1.2) has at least

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AUTHORS 1Weihua

Jiang was born in Hebei, China, in 1964. She received

the B.S. degree in mathematics from Xi'an Jiaotong University, Xi'an, China, in 1991 and the Ph.D. degree in mathematics from Hebei Normal University, Shijiazhuang, China, in 2009. Her research interests are in boundary value problems and functional

2Jiqing

Qiu (1956-): Male, Han nationality, Ph.D., professor.

His research interests are in Rrobust control.

3Ruiyan

Li (1983-): Female, Han nationality, master. Research

direction: Boundary value problems.

analysis.

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