Báo cáo hóa học: "Research Article Antiperiodic Boundary Value Problems for Finite Dimensional Differential Systems" pptx

Nieto We study antiperiodic boundary value problems for semilinear differential and impulsive differential equations in finite dimensional spaces.. Pavel, “On a class of second-order anti-

Volume 2009, Article ID 541435, 11 pages

doi:10.1155/2009/541435

Research Article

Antiperiodic Boundary Value Problems for Finite Dimensional Differential Systems

1 Faculty of Applied Mathematics, Guangdong University of Technology, Guangzhou,

Guangdong 510006, China

2 Department of Mathematics, National University of Ireland, Galway, Ireland

Correspondence should be addressed to D O’Regan,donal.oregan@nuigalway.ie

Received 16 March 2009; Accepted 28 May 2009

Recommended by Juan J Nieto

We study antiperiodic boundary value problems for semilinear differential and impulsive differential equations in finite dimensional spaces Several new existence results are obtained Copyrightq 2009 Y Q Chen et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

The study of antiperiodic solutions for nonlinear evolution equations is closely related to the study of periodic solutions, and it was initiated by Okochi1 During the past twenty years, antiperiodic problems have been extensively studied by many authors, see1 31 and the references therein For example, antiperiodic trigonometric polynomials are important in the study of interpolation problems32,33, and antiperiodic wavelets are discussed in 34 Moreover, antiperiodic boundary conditions appear in physics in a variety of situations, see

35–40 InSection 2we consider the antiperiodic problem

ut  Aut  ft, ut, t ∈ R, ut  −ut  T, t ∈ R, E 1.1 where A is an n × n matrix, f : R × R n → R n is continuous, and f t  T, x  −ft, x for all

t, x ∈ R×R n Under certain conditions on the nondiagonal elements of A and f we prove an

existence result forE 1.1 InSection 3we consider the antiperiodic boundary value problem

ut  Gut  ft, ut, a.e t ∈ J  0, T, t / t k ,

u0  −uT, Δut k   I k ut k , k  1, 2, , p,

E 1.2

where G : R n → R n is a function satisfying G0  0, and f : J × R n → R nis a Caratheodory function,Δut k   ut

k  − ut−

k , and I k ∈ CR n , R n  Under certain conditions on G, f, and

I k u for k  1, 2, , p, we prove an existence result for E 1.2

Let| · | be the norm in R n In this section we study

ut  Aut  ft, ut, t ∈ R,

ut  −ut  T. E 2.1

First, we have the following result

Theorem 2.1 Let A  a ij  be an n × n matrix, where a ij is the element of A in the ith row and jth column, f : R → R n is continuous and f tT  −ft for t ∈ R Suppose T/2Σ1≤i<j≤n|a ij −a ji | <

1 Then the equation

ut  Aut  ft, t ∈ R, ut  −ut  T, t ∈ R E 2.2 has a unique solution.

Proof Put W a  {v· ∈ CR; R n  : vt  −vt  T} Then W ais a Banach space under the norm|v·|∞ maxt∈0,T |vt| For each v· ∈ W a, consider the following equation:

ut  Avt  ft, t ∈ R, ut  −ut  T, t ∈ R. E 2.3

It is easy to see that ut  −1/2T

0Avsfsdst

0Avsfsds is the unique solution

ofE 2.3

We define a mapping K : W a → W aas follows:

for any v· ∈ W a , Kv·  u·, u· is the solution of E 2.3. 2.1

First we prove that K is a continuous compact mapping Now assume v n · ∈ W a , n  1, 2, , and v n · → v· ∈ W a Then|Av n · − Av·|∞ → 0 as n → ∞ This immediately implies

thatT

0|Kv n t− Kvt|2dt → 0 as n → ∞.

We have Kv n t − Kvt  1/2{t

0Kv n s − Kvsds − T

t Kv n s −

Kvsds}, and so Kv n · → Kv· in W a

Now sinceKvt Avt  ft, t ∈ R, it is easy to see that

T

0Kvt2

dt

1/2

≤√T|Av·|∞

T

0

ft2

dt

1/2

. 2.2

Thus K maps a bounded subset of W a to a bounded equicontinuous subset in W a, therefore

K is completely continuous.

Next take r0 > 1 − T/2Σ1≤i<j≤n|a ij − a ji|−1√T/2T

0|ft|2dt 1/2 We show that

Kv · / λv· for all λ ≥ 1, and |v·|∞ r0 If this is not true, there exist λ0≥ 1, w· ∈ W awith

|w·|∞ r0such that Kw·  λ0w·, that is, wt  −wt  T, t ∈ R and

λ0wt  Awt  ft, t ∈ R. 2.3 Multiply 2.3 by wt i.e., take inner product and integrate over 0, T, and notice that

T

0w i tw

j tdt  −T

0w itw j tdt to get

λ0

T

0

|wt|2dt≤ Σ1≤i<j≤na ij − a jiT

0



w i tw

j tdt T

0

ft2

dt

1/2T

0

wt2

dt

1/2

,

2.4

where wt  w i t, i  1, 2, , n Notice that wt  1/2t

0wsds −T

t wsds, so we

have

|w·|∞≤

√

T

2

T

0

wt2

dt

1/2

From2.4, 2.5, we have

λ0

T

0

wt2

dt

1/2

≤√TΣ1≤i<j≤na ij − a ji |w·|∞

T

0

ft2

dt

1/2

. 2.6

This with2.5 gives

λ0|w·|∞≤ T

2Σ1≤i<j≤na ij − a ji |w·|∞

√

T

2

T

0

f t2

dt

1/2

. 2.7

As a result

|w·|∞≤



1−T

2Σ1≤i<j≤na ij − a ji−1√

T

2

T

0

ft2

dt

1/2

, 2.8

which contradicts|w·|∞ r0

Thus the Leray-Schauder degree degI − K, B0, r0, 0  1, where B0, r0 is the open

ball centered at 0 with radius r0 in C a Consequently, K has a fixed point in B0, r0, that is,

E 2.2 has a solution For the uniqueness, if u·, v· are two solutions of E 2.2, set wt  ut − vt, then wt  Awt, and wt  −wt  T, for t ∈ R Following the obvious

strategy abovesee the clear adjustment of 2.8 gives |w·|∞  0 Thus the solution of

E 2.2 is unique

FromTheorem 2.1we have immediately the following result

Corollary 2.2 Let A  a ij  be an n × n symmetric matrix, f : R → R n is continuous and ft  T  −ft for t ∈ R Then

ut  Aut  ft, t ∈ R, ut  −ut  T, t ∈ R, E 2.4 has a unique solution.

Using a proof similar toTheorem 2.1, we have the following result

Theorem 2.3 Let A  a ij  be an n × n matrix, G : R n → R n is an even continuously differentiable function, and f t, u : R × R n → R n is continuous and ft  T, u  −ft, u for t, u ∈ R × R n Suppose the following conditions are satisfied:

1 |ft, x| ≤ M|x|  gt, for a.e t, x ∈ R × R n , where M > 0 is a constant, and g· ∈

L20, T;

2 T/2Σ1≤i<j≤n|a ij − a ji |  M < 1.

Then

ut  Aut  ∂Gut  ft, ut, t ∈ R,

ut  −ut  T, t ∈ R E 2.5 has a solution.

Remark 2.4 Equation E 2.5 was studied by Haraux 18 and Chen et al 14 in the case

A 0, and also by Chen 12 with different assumptions on f and A.

3 Antiperiodic Boundary Value Problem for Impulsive ODE

In this section, we prove an existence result for the equation

ut  Gut  ft, ut, a.e t ∈ J  0, T, t / t k ,

u0  −uT, Δut k   I k ut k , k  1, 2, , p,

E 3.1

where G : R n → R n is a Lipschitz function We first introduce some notations Let

J  0, T, and 0  t0 < t1 < · · · < t p < t p1  T PCJ  {u : J → R n , u t k ,t k1 ∈

Ct k , t k1 , R n , k  0, 1, , p, ut−

k  exist for k  1, 2, , p, and u0  u0}, and

P W 1,2 J  {u ∈ PCJ : u t ,t  ∈ W 1,2 t k , t k1 , R n , k  1, , p} It is clear that PCJ

and P W 1,2 J are Banach spaces with the respective norm u P CJ  sup{|ut|, t ∈ J}, and

u P W 1,2 J  p

k0 u k W 1,2 t k ,t k1 , where u k : t k , t k1  → R is defined by u k t  ut for

t ∈ t k , t k1 , k  0, 1, , p.

We say a function u is a solution ofE 3.1 if u ∈ PW 1,2 J and u satisfies E 3.1

We first prove the following result

Lemma 3.1 Let I i : R n → R n be continuous functions for i  1, 2, , p, and Σ p

k1 |I k x k| ≤

α{max1≤k≤p|x k |}  δ for all x k ∈ R n , k  1, 2, , p, where α, δ > 0 are constants, and α < 2 Suppose u ∈ PW 1,2 J with u0  −uT, and Δut i   I i ut i , for i  1, 2, , p Then

u P CJ≤



1−1

2α

−1⎡

⎣1

2δ

√

T

2

T

0

us2ds

1/2⎤

⎦. 3.1

Proof By assumption, we have ut  u0 t

0usds for t ∈ 0, t1, and

ut  u0  Σ k

i1 I i ut i 

t

0

usds 3.2

for t ∈ t k , t k1 , k  1, 2, , p Since u0  −uT, it follows that ut  −1/2Σ p

i1 I i ut i 

T

0usds t

0usds for t ∈ 0, t1, and

ut  −1

2



Σp i1 I i ut i 

T

0

usds



 Σk i1 I i ut i 

t

0

usds 3.3

for t ∈ t k , t k1 , k  1, 2, , p Hence we have

u P CJ≤ 1

2



α u P CJ  δ

√

T

2

T

0

us2

ds

1/2

. 3.4

Thus

u P CJ≤



1−1

2α

−1⎡

⎣1

2δ

√

T

2

T

0

us2

ds

1/2⎤

⎦. 3.5

Theorem 3.2 Let G : R n → R n be a function satisfying G0  0, and f : 0, T → R n such that f· ∈ L20, T, and let I k : R n → R n be continuous functions for k  1, 2, , p Suppose the following conditions are satisfied:

1 |Gu − Gv| ≤ L|u − v| for all u, v ∈ R n , and L > 0 is a constant;

2 Σp

k1 |I k x k | ≤ γ{max1≤k≤p|x k |}  δ for all x k ∈ R n , k  1, 2, , p, where γ, δ > 0 are constants;

3 γ  TL < 2.

Then the problem

ut  Gut  ft, a.e t ∈ J  0, T, t / t k ,

u0  −uT, Δut k   I k ut k , k  1, 2, , p

E 3.2

has a solution.

Proof For each v ∈ PCJ, consider the problem

ut  Gvt  ft a.e t ∈ J  0, T, t / t k ,

u0  −uT, Δut k   I k vt k , k  1, 2, , p.

E 3.3

One can easily show that the solution u ofE 3.3 is given by the following:

ut  −1

2



Σp i1 I i vt i 

T

0



Gvs  fsds





t

0



Gvs  fsds, for t ∈ 0, t1,

ut  −1

2



Σp i1 I i vt i 

T

0



Gvs  fsds



 Σk i1 I i vt i



t

0



Gvs  fsds,

3.6

for t ∈ t k , t k1 , k  1, , p.

Obviously, the solution ofE 3.3 is unique Now we define K : PCJ → PW 1,2 J ⊂

P CJ by u  Kv We prove that K is continuous Let v n ∈ PCJ and v n → v in PCJ It is

easy to see that

T

0

Kv n t − Kvt2

dt

T

0

|Gv n t − Gvt|2dt ≤ L2

T

0

|v n t − vt|2dt. 3.7

ThereforeT

0|Kv n t − Kvt|2dt 1/2≤√TL v n − v P CJ → 0 as n → ∞.

Note thatΔKv n − Kvt k   I k v n t k  − I k vt k, and we have

Kv n t − Kvt  −1

2



Σp i1 I i v n t i  − I i vt i 

T

0

Kv n − Kvsds





t

0

Kv n − Kvsds, for t ∈ 0, t1,

Kv n t − Kvt  −1

2



Σp i1 I i v n t i  − I i vt i 

T

0

Kv n − Kvsds



 Σk i1 I i v n t i  − I i vt i 

t

0

Kv n − Kvsds

3.8

for t ∈ t k , t k1 , k  1, 2, , p From the continuity of I i , i  1, 2, , p, and T

0|Kv n t − Kvt|2dt → 0 as n → ∞, we deduce that K is continuous.

For each v ∈ PCJ, notice that 0  G0, so we have

T

0

|Kv|2dt

1/2

≤√TL v P CJ

T

0

f s2

ds

1/2

. 3.9

From3.9 and Lemma 3.1, we know that K maps bounded subsets of P CJ to relatively compact subsets of P CJ.

Finally, for∀λ ∈ 0, 1, we prove that the set of solutions of u  λKu is bounded If

u  λKu for some λ ∈ 0, 1, then

ut  λGut  λft a.e t ∈ J  0, T, t / t k ,

u0  −uT, Δut k   λI k ut k , k  1, 2, , p.

3.10

Therefore we have

ut  −1

2λ



Σp i1 I i u i t i 

T

0



Gus  fsds



 λ

t

0



Gus  fsds 3.11

for t ∈ 0, t1, and

ut  −1

2λ



Σp i1 I i u i t i 

T

0



Gus  fsds



 λ Σ k i1 I i u i t i

 λ

t

0



Gus  fsds

3.12

for t ∈ t k , t k1 , k  1, , p This implies that

u P CJ≤ 1

2



γ u P CJ  δ 

T

0



|Gus| fsds. 3.13

Since 0 G0, and |Gu| ≤ L|u|, so we have

u P CJ≤ 1

2



1−1 2



γ  TL−1



δ

T

0

fsds. 3.14

The Leray-Schauder principle guarantees a fixed point of K, which is easily seen to be a

solution ofE 3.2

By using a similar method toTheorem 3.2, one can deduce the following result

Theorem 3.3 Let G : R n → R n be a function satisfying G0  0, and ft, x : 0, T × R n → R n

a Caratheodory function, that is, f is measurable in t for each x ∈ R n , and f is continuous in x for each t ∈ 0, T, such that |ft, x| ≤ gt for t, x ∈ 0, T × R n , where g· ∈ L20, T, and let I k : R n → R n be continuous functions for k  1, 2, , p Suppose the following conditions are satisfied:

1 |Gu − Gv| ≤ L|u − v| for all u, v ∈ R n , and L > 0 is a constant;

2 Σp

k1 |I k x k | ≤ γ{max1≤k≤p|x k |}  δ for all x k ∈ R n , k  1, 2, , p, where γ, δ > 0 are constants;

3 γ  TL < 2.

Then the equation

ut  Gut  ft, ut, a.e t ∈ J  0, T, t / t k ,

u0  −uT, Δut k   I k ut k , k  1, 2, , p

E 3.4

has a solution.

4 Examples

In this section, we give examples to show the application of our results to differential and impulsive differential equations

Example 4.1 Consider the antiperiodic problem

u1t  λ1u1t  5u2t  sin πt, t ∈ R,

u2t  7

2u1t  λ2u2t  cos πt, t ∈ R,

u1t  −u1t  1, u2t  −u2t  1, t ∈ R.

E 4.1

Set

u



u1

u2



, ft 



sin πt cos πt



, A

⎛

⎝λ71 5

2 λ2

⎞

⎠. 4.1

NowE 4.1 is equivalent to

ut  Aut  ft, t ∈ R, ut  −ut  1, t ∈ R. E 4.2

Also ft  −ft1, for t ∈ R and 1/2|a12−a21|  3/4 ByTheorem 2.1,E 4.2 has a unique solution, soE 4.1 has a unique solution

Example 4.2 Consider the antiperiodic boundary value problem

u1t  1

2 u2

1t  u2

2t 3u1t − 2u2t  sin πt, t ∈ 0, 1, t /1

4,

u2t  1

2 u2

1t  u2

2t 2u1t  3u2t − cos πt, t ∈ 0, 1, t /1

4,

Δu1

 1 4



 51  |u1

21/4| , Δu2

 1 4



 81  |u1

11/4| ,

u10  −u11, u20  −u21.

E 4.3

Set

u



u1

u2



, ft 



sin πt

− cos πt



, Gu

⎛

⎜

⎝

3u1− 2u2

2 u2

1 u2 2

2u1 3u2

2 u2

1 u2 2

⎞

⎟

⎠, Iu

⎛

⎜

⎝

1 51  |u2| 1 81  |u1|

⎞

⎟

⎠.

4.2

It is easy to check that|Gu − Gv| ≤ √13/2|u − v| for u, v ∈ R2,|Iu| < 2/5 for u ∈ R2, and

√

13/2 < 2 NowE 4.3 is equivalent to the equation

ut  Gut  ft, t ∈ 0, 1, t /1

4,

Δu

 1 4



 I



u

 1 4



, u0  −u1.

E 4.4

ByTheorem 3.2, we know thatE 4.4 has a solution, so E 4.3 has a solution

Acknowledgment

The first author is supported by an NSFC Grant, Grant no 10871052

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