# Impulsive Boundary Value Problems for Planar Hamiltonian Systems

Authors: Zeynep Kayar; Ağacık Zafer
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We give an existence and uniqueness theorem for solutions of inhomogeneous
impulsive boundary value problem (BVP) for planar Hamiltonian systems. Green's
function that is needed for representing the solutions is obtained and its properties are listed.
The uniqueness of solutions is connected to a Lyapunov type inequality for the corresponding
homogeneous BVP.


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## Body

## 1. Introduction

The planar Hamiltonian system of 2-linear first-order equations has the form(1)y′=JH(t)y,t∈ℝ,
where
(2)H(t)=[c(t)a(t)a(t)b(t)]
is a symmetric matrix with piecewise continuous real-valued entries, and
(3)J=[01-10]
is the so called symplectic identity. Setting
(4)y1(t)=x(t),y2(t)=u(t),
the Hamiltonian system can be rewritten in a more convenient way as
(5)x′=a(t)x+b(t)u,u′=-c(t)x-a(t)u.Our aim in this work is to prove an existence and uniqueness theorem for solutions of the related BVP for inhomogeneous Hamiltonian system under impulse effect of the form(6a)x′=a(t)x+b(t)u+f1(t),u′=-c(t)x-a(t)u+f2(t),00000000.00t∈(t1,t2)∖{τi},(6b)x(τi+)=αix(τi-)+ai1,u(τi+)=-βix(τi-)+αiu(τi-)+ai2,000000000000000000i=1,2,…,p,(6c)x(t1)=A,x(t2)=B,where(i)
{τi},{αi},{βi},{ai1}, and{ai2}are real sequences fori=1,2,…,pwith
(7)t1<τ1<τ2<⋯<τp<t2;(ii)
a,b,c,f1,f2∈PLC(J0), whereJ0=[t1,t2]and PLC(J0)={ω:J0→ℝis continuous on each interval(τi,τi+1), the limitsw(τi±)exist andw(τi-)=w(τi)fori=1,2,…,p};(iii)
b(t)>0fort∈(t1,t2)andαi≠0fori=1,2,…,p;AandBare given real numbers.We also setτ0=t1andτp+1=t2for convenience.By a solution of the impulsive BVP (6a)–(6c), we mean nontrivial functionsx,u∈PLC(J0)such that(x,u)satisfies system (6a)–(6c) for allt∈J0.The corresponding homogeneous BVP takes the form(8a)x′=a(t)x+b(t)u,u′=-c(t)x-a(t)u,0000000000000000000000000t∈(t1,t2)∖{τi},(8b)x(τi+)=αix(τi-),u(τi+)=-βix(τi-)+αiu(τi-),000000000000000000000000000000000i=1,2,…,p,(8c)x(t1)=0,x(t2)=0.Note that if we take(9)a(t)≡0,b(t)=1p(t),c(t)=q(t),f1(t)≡0,f2(t)=f(t),
then we obtain as a special case of (6a), (6b), and (6c) the impulsive BVP for second-order differential equations of the form (10a)(p(t)x′)′+q(t)x=f(t),t∈(t1,t2)∖{τi},(10b)x(τi+)=αix(τi-)+ai1,(10c)(px′)(τi+)=-βix(τi-)+αi(px′)(τi-)+ai2,i=1,2,…,p,(10d)x(t1)=A,x(t2)=B.To the best of our knowledge although many results have been obtained for linear impulsive boundary value problems by using different techniques, there is little known for the linear2×2Hamiltonian systems under impulse effect.The existence and uniqueness of linear impulsive boundary value problem for the first-order equations are considered in [1–4]. For the second-order case we refer to [5, 6] in which the integral representation of the solution of second order linear impulsive boundary value problems is given by using Green’s function and the existence and uniqueness of the solutions are obtained. Variational technique approach for the existence of the solutions of linear and nonlinear impulsive boundary value problems can be found in [7–10]. In [11], the method of upper and lower solutions is employed for the existence of solutions of nonlinear impulsive boundary value problems. For a detailed discussion on boundary value problems for higher-order linear impulsive equations we refer to [12]. Basic theory of impulsive differential equations is contained in the seminal book [13].Our method of proof is based on Green’s function formulation and Lyapunov type inequalities for linear Hamiltonian system under impulse effect. There are many studies on Lyapunov type inequalities and their applications for linear ordinary differential equations [14] and for systems [15–17] as well as for linear impulsive differential equations and systems [18, 19]. However, the use of a Lyapunov type inequality in connection with BVPs seems to be limited.


## 2. Preliminaries

### 2.1. Lyapunov Type Inequality for Homogeneous Problem

In this section we provide a Lyapunov type inequality to be used for the uniqueness of the inhomogeneous BVP. The obtained inequality is sharper than the one given by the present authors in [20] in the sense that2|a(t)|is replaced by|a(t)|.Theorem 1.
If the homogeneous BVP (8a), (8b), and (8c) has a real solution(x(t),u(t))such thatx(t)≢0on(t1,t2), then one has the Lyapunov type inequality:
(11)exp(∫t1t2‍|a(t)|dt)[∫t1t2‍b(t)dt]0000×[∫t1t2‍c+(t)dt+∑t1≤τi<t2‍(βiαi)+]≥4,
wherec+(t)=max{c(t),0}and(βi/αi)+=max{βi/αi,0}.Proof.
Define(12)z(t)=1α1α2⋯αix(t),v(t)=1α1α2⋯αiu(t)
fort∈(τi,τi+1)andi=0,1,…,p, where we put againτ0=t1,τp+1=t2and make a convention thatα1α2⋯αi=1ifi=0.
It is not difficult to see from (8a), (8b), (8c), and (12) that
(13)z′=a(t)z+b(t)v,v′=-c(t)z-a(t)v,00000000000000000000000t∈(t1,t2)∖{τi},(14)z(τi+)=z(τi-),v(τi+)=-βiαiz(τi-)+v(τi-),000000000000i=1,2,…,p,(15)z(t1)=0,z(t2)=0.
Since we assumed thatz(τi)=z(τi-),z(t)is continuous on[t1,t2]. Moreover,z′∈PLC(J0),z(t1)=z(t2)=0, andz(t)≢0for allt∈(t1,t2). We may assume without loss of generality thatz(t)≥0on(t1,t2).
Using (13) and (14) we obtain
(16)(vz)′=-c(t)z2+b(t)v2,t≠τi,(17)(vz)(τi+)-(vz)(τi-)=-βiαiz2(τi).
Integrating (16) fromt1tot2and using (15) and (17) lead to
(18)∑t1≤τi<t2‍βiαiz2(τi)=∫t1t2‍[b(t)v2(t)-c(t)z2(t)]dt,
from which we have
(19)∫t1t2‍b(t)v2(t)dt≤∫t1t2‍c+(t)z2(t)dt+∑t1≤τi<t2‍(βiαi)+z2(τi).
On the other hand, from the first equation in (13), we have
(20)[z(t)exp(-∫t1t‍a(u)du)]′=b(t)v(t)exp(-∫t1t‍a(u)du),(21)[z(t)exp(∫tt2‍a(u)du)]′=b(t)v(t)exp(∫tt2‍a(u)du).
Let
(22)max{|z(t)|:t∈(t1,t2)}=z(t0)>0.
If we integrate (20) fromt1tot0, we see that
(23)z(t0)=∫t1t0‍b(t)v(t)exp(∫tt0‍a(u)du)dt
and so
(24)z(t0)≤∫t1t0‍b(t)|v(t)|exp(∫tt0|a(u)|du‍)dt.
Using the obvious estimate
(25)∫tt0‍|a(u)|du≤∫t1t0‍|a(u)|du
and then applying Cauchy-Schwarz inequality, we have
(26)z2(t0)≤exp(2∫t1t0‍|a(u)|du)[∫t1t0‍b(t)dt]×[∫t1t0‍b(t)v2(t)dt].
Similarly, by integrating (21) fromt0tot2and following the above procedure, we get
(27)z2(t0)≤exp(2∫t0t2‍|a(u)|du)[∫t0t2‍b(t)dt]×[∫t0t2‍b(t)v2(t)dt].
Now we recall the elementary inequality:(28)x2α+y2β≥4xy,α,β>0,α+β=1
forx≥0andy≥0. In view of (26) and (27) setting
(29)α=∫t1t0‍b(t)dt∫t1t2‍b(t)dt,β=∫t0t2‍b(t)dt∫t1t2‍b(t)dt,x=z(t0)exp(-∫t1t0‍|a(u)|du),y=z(t0)exp(-∫t0t2‍|a(u)|du)
we have
(30)4z2(t0)exp(∫t1t2‍|a(t)|dt)≤[∫t1t2‍b(t)dt][∫t1t2‍b(t)v2(t)dt].
Combining (19) and (30) results in
(31)4z2(t0)exp(∫t1t2‍|a(t)|dt)00≤[∫t1t2‍b(t)dt]0000×[∫t1t2‍c(t)z2(t)dt+∑t1≤τi<t2‍(βiαi)z2(τi)].
Finally, sincez(t0)≥z(t)fort∈[t1,t2], from (31) we obtain the desired inequality:
(32)exp(∫t1t2‍|a(t)|dt)[∫t1t2‍b(t)dt]0000×[∫t1t2‍c+(t)dt+∑t1≤τi<t2‍(βiαi)+]≥4.


### 2.2. Green’s Function

Here we derive Green’s function to be used for the representation of the solutions of the inhomogeneous BVP.Let(33)Φ(t)=[x1(t)x2(t)u1(t)u2(t)],Φ(0)=I
be a fundamental matrix for (8a), (8b) and set
(34)M=[1000],N=[0010].Define the rectangles(35)R11=[t1,τ1]×[t1,τ1],Ri1=(τi-1,τi]×[t1,τ1],i=2,3,…,p+1,R1j=[t1,τ1]×(τj-1,τj],j=2,3,…,p+1,Rij=(τi-1,τi]×(τj-1,τj],i,j=2,3,…,p+1,
and the triangles
(36)Tu={(t,s)∈[t1,t2]×[t1,t2]:s>t},Tl={(t,s)∈[t1,t2]×[t1,t2]:s<t},Tiiu={(t,s)∈Rii:s>t},i=1,2,3,…,p+1,Tiil={(t,s)∈Rii:s<t},i=1,2,3,…,p+1.Green’s function (pair) and its properties are given in the next theorem.Theorem 2.
Suppose that the homogeneous BVP (8a)–(8c) has only the trivial solution. Let
(37)K=-[MΦ(t1)+NΦ(t2)]-1NΦ(t2).
Note that the inverse of matrixMΦ(t1)+NΦ(t2)exists in view of the assumption (see also the proof of Theorem 4).
Then the pair of functions(38)G(t,s)={Φ(t)(I+K)Φ-1(s),s<t,Φ(t)KΦ-1(s),s≥t,G~(t,τi+)={Φ(t)(I+K)Φ-1(τi+),τi<t,Φ(t)KΦ-1(τi+),τi≥t,
constitutes Green’s function for (6a), (6b), and (6c). Moreover, we have the following properties: (G1)
G(t,s)is continuous and bounded onRij,(G2)
(∂G(t,s))/∂tis continuous and bounded on the rectanglesRijwithi≠jand on the trianglesTiiuandTiil,(G3)
G(t,s)satisfies the following jump conditions:
(a)
G(τi+,τi)-G(τi-,τi)=Bi+(Bi-I)G(τi-,τi)whereBi=[αi0-βiαi],(b)
G(s+,s)-G(s-,s)=I,s≠τi,(c)
(∂G(s+,s)/∂t)-(∂G(s-,s)/∂t)=JH(s),s≠τi,(G4)
G(t,s), considered as a function oft, is left continuous and satisfies
(39)y′=JH(t)y,t∈Js∖{τi},y(τi+)=Biy(τi-),i∈{i:τi∈Js},My(t1)+Ny(t2)=0,whereJsis any of the intervals[t1,s)or(s,t2],(G5)
Δ|t=τiG~(t,τi+)=G~(τi+,τi+)-G~(τi-,τi+)=(Bi-I)G~(τi-,τi+),(G6)
G~(t,s), considered as a function oft, is left continuous and satisfies (39).Proof.
(G1) and (G2) are trivial. Let us consider (G3)(a) follows from(40)G(τi+,τi)-G(τi-,τi)00=Φ(τi+)(I+K)Φ-1(τi)-Φ(τi-)KΦ-1(τi)00=Bi+(Bi-I)G(τi-,τi).
To see (b), we write fors≠τi,
(41)G(s+,s)-G(s-,s)=Φ(s+)(I+K)Φ-1(s)-Φ(s-)KΦ-1(s)=I.
For (c), lett≠τi; then
(42)∂G(t,s)∂t={Φ′(t)(I+K)Φ-1(s)=JH(t)Φ(t)(I+K)Φ-1(s),s<t,Φ′(t)KΦ-1(s)=JH(t)Φ(t)KΦ-1(s),s≥t,∂G(s+,s)∂t-∂G(s-,s)∂t=JH(s)Φ(s)(I+K)Φ-1(s)-JH(s)Φ(s)KΦ-1(s)=JH(s).
Next, we consider (G4). By definition, it is easy to see thatG(t,s)is left continuous function att=τi. Let us consider the interval[t1,s). The later is similar. The first equation in (39) is direct consequences of (c) and the definition ofG(t,s). Clearly,
(43)G(τi+,s)=Φ(τi+)KΦ-1(s)000000=BiΦ(τi-)KΦ-1(s)=BiG(τi-,s),MG(t1,s)+NG(t2,s)=MΦ(t1)KΦ-1(s)+NΦ(t2)(I+K)Φ-1(s)=[MΦ(t1)+NΦ(t2)]KΦ-1(s)+NΦ(t2)Φ-1(s)=0.
The proofs of (G5) and (G6) are similar to (a) and (G4), respectively.Remark 3.
One can easily rewrite Green’s function (pair) in terms of the solutions of system (8a), (8b). Indeed,
(44)K=-[MΦ(t1)+NΦ(t2)]-1NΦ(t2)=1x1(t1)x2(t2)-x1(t2)x2(t1)×[x1(t2)x2(t1)x2(t1)x2(t2)-x1(t1)x1(t2)-x1(t1)x2(t2)],
and since
(45)detΦ(t)=detΦ(0)exp(∫0t‍trace(JH(s))ds)×∏i=1p‍detBi=∏i=1p‍αi2,
we may write
(46)Φ-1(t)=1detΦ(t)[u2(t)-x2(t)-u1(t)x1(t)]=(∏i=1p‍αi-2)[u2(t)-x2(t)-u1(t)x1(t)].


## 2.1. Lyapunov Type Inequality for Homogeneous Problem

In this section we provide a Lyapunov type inequality to be used for the uniqueness of the inhomogeneous BVP. The obtained inequality is sharper than the one given by the present authors in [20] in the sense that2|a(t)|is replaced by|a(t)|.Theorem 1.
If the homogeneous BVP (8a), (8b), and (8c) has a real solution(x(t),u(t))such thatx(t)≢0on(t1,t2), then one has the Lyapunov type inequality:
(11)exp(∫t1t2‍|a(t)|dt)[∫t1t2‍b(t)dt]0000×[∫t1t2‍c+(t)dt+∑t1≤τi<t2‍(βiαi)+]≥4,
wherec+(t)=max{c(t),0}and(βi/αi)+=max{βi/αi,0}.Proof.
Define(12)z(t)=1α1α2⋯αix(t),v(t)=1α1α2⋯αiu(t)
fort∈(τi,τi+1)andi=0,1,…,p, where we put againτ0=t1,τp+1=t2and make a convention thatα1α2⋯αi=1ifi=0.
It is not difficult to see from (8a), (8b), (8c), and (12) that
(13)z′=a(t)z+b(t)v,v′=-c(t)z-a(t)v,00000000000000000000000t∈(t1,t2)∖{τi},(14)z(τi+)=z(τi-),v(τi+)=-βiαiz(τi-)+v(τi-),000000000000i=1,2,…,p,(15)z(t1)=0,z(t2)=0.
Since we assumed thatz(τi)=z(τi-),z(t)is continuous on[t1,t2]. Moreover,z′∈PLC(J0),z(t1)=z(t2)=0, andz(t)≢0for allt∈(t1,t2). We may assume without loss of generality thatz(t)≥0on(t1,t2).
Using (13) and (14) we obtain
(16)(vz)′=-c(t)z2+b(t)v2,t≠τi,(17)(vz)(τi+)-(vz)(τi-)=-βiαiz2(τi).
Integrating (16) fromt1tot2and using (15) and (17) lead to
(18)∑t1≤τi<t2‍βiαiz2(τi)=∫t1t2‍[b(t)v2(t)-c(t)z2(t)]dt,
from which we have
(19)∫t1t2‍b(t)v2(t)dt≤∫t1t2‍c+(t)z2(t)dt+∑t1≤τi<t2‍(βiαi)+z2(τi).
On the other hand, from the first equation in (13), we have
(20)[z(t)exp(-∫t1t‍a(u)du)]′=b(t)v(t)exp(-∫t1t‍a(u)du),(21)[z(t)exp(∫tt2‍a(u)du)]′=b(t)v(t)exp(∫tt2‍a(u)du).
Let
(22)max{|z(t)|:t∈(t1,t2)}=z(t0)>0.
If we integrate (20) fromt1tot0, we see that
(23)z(t0)=∫t1t0‍b(t)v(t)exp(∫tt0‍a(u)du)dt
and so
(24)z(t0)≤∫t1t0‍b(t)|v(t)|exp(∫tt0|a(u)|du‍)dt.
Using the obvious estimate
(25)∫tt0‍|a(u)|du≤∫t1t0‍|a(u)|du
and then applying Cauchy-Schwarz inequality, we have
(26)z2(t0)≤exp(2∫t1t0‍|a(u)|du)[∫t1t0‍b(t)dt]×[∫t1t0‍b(t)v2(t)dt].
Similarly, by integrating (21) fromt0tot2and following the above procedure, we get
(27)z2(t0)≤exp(2∫t0t2‍|a(u)|du)[∫t0t2‍b(t)dt]×[∫t0t2‍b(t)v2(t)dt].
Now we recall the elementary inequality:(28)x2α+y2β≥4xy,α,β>0,α+β=1
forx≥0andy≥0. In view of (26) and (27) setting
(29)α=∫t1t0‍b(t)dt∫t1t2‍b(t)dt,β=∫t0t2‍b(t)dt∫t1t2‍b(t)dt,x=z(t0)exp(-∫t1t0‍|a(u)|du),y=z(t0)exp(-∫t0t2‍|a(u)|du)
we have
(30)4z2(t0)exp(∫t1t2‍|a(t)|dt)≤[∫t1t2‍b(t)dt][∫t1t2‍b(t)v2(t)dt].
Combining (19) and (30) results in
(31)4z2(t0)exp(∫t1t2‍|a(t)|dt)00≤[∫t1t2‍b(t)dt]0000×[∫t1t2‍c(t)z2(t)dt+∑t1≤τi<t2‍(βiαi)z2(τi)].
Finally, sincez(t0)≥z(t)fort∈[t1,t2], from (31) we obtain the desired inequality:
(32)exp(∫t1t2‍|a(t)|dt)[∫t1t2‍b(t)dt]0000×[∫t1t2‍c+(t)dt+∑t1≤τi<t2‍(βiαi)+]≥4.


## 2.2. Green’s Function

Here we derive Green’s function to be used for the representation of the solutions of the inhomogeneous BVP.Let(33)Φ(t)=[x1(t)x2(t)u1(t)u2(t)],Φ(0)=I
be a fundamental matrix for (8a), (8b) and set
(34)M=[1000],N=[0010].Define the rectangles(35)R11=[t1,τ1]×[t1,τ1],Ri1=(τi-1,τi]×[t1,τ1],i=2,3,…,p+1,R1j=[t1,τ1]×(τj-1,τj],j=2,3,…,p+1,Rij=(τi-1,τi]×(τj-1,τj],i,j=2,3,…,p+1,
and the triangles
(36)Tu={(t,s)∈[t1,t2]×[t1,t2]:s>t},Tl={(t,s)∈[t1,t2]×[t1,t2]:s<t},Tiiu={(t,s)∈Rii:s>t},i=1,2,3,…,p+1,Tiil={(t,s)∈Rii:s<t},i=1,2,3,…,p+1.Green’s function (pair) and its properties are given in the next theorem.Theorem 2.
Suppose that the homogeneous BVP (8a)–(8c) has only the trivial solution. Let
(37)K=-[MΦ(t1)+NΦ(t2)]-1NΦ(t2).
Note that the inverse of matrixMΦ(t1)+NΦ(t2)exists in view of the assumption (see also the proof of Theorem 4).
Then the pair of functions(38)G(t,s)={Φ(t)(I+K)Φ-1(s),s<t,Φ(t)KΦ-1(s),s≥t,G~(t,τi+)={Φ(t)(I+K)Φ-1(τi+),τi<t,Φ(t)KΦ-1(τi+),τi≥t,
constitutes Green’s function for (6a), (6b), and (6c). Moreover, we have the following properties: (G1)
G(t,s)is continuous and bounded onRij,(G2)
(∂G(t,s))/∂tis continuous and bounded on the rectanglesRijwithi≠jand on the trianglesTiiuandTiil,(G3)
G(t,s)satisfies the following jump conditions:
(a)
G(τi+,τi)-G(τi-,τi)=Bi+(Bi-I)G(τi-,τi)whereBi=[αi0-βiαi],(b)
G(s+,s)-G(s-,s)=I,s≠τi,(c)
(∂G(s+,s)/∂t)-(∂G(s-,s)/∂t)=JH(s),s≠τi,(G4)
G(t,s), considered as a function oft, is left continuous and satisfies
(39)y′=JH(t)y,t∈Js∖{τi},y(τi+)=Biy(τi-),i∈{i:τi∈Js},My(t1)+Ny(t2)=0,whereJsis any of the intervals[t1,s)or(s,t2],(G5)
Δ|t=τiG~(t,τi+)=G~(τi+,τi+)-G~(τi-,τi+)=(Bi-I)G~(τi-,τi+),(G6)
G~(t,s), considered as a function oft, is left continuous and satisfies (39).Proof.
(G1) and (G2) are trivial. Let us consider (G3)(a) follows from(40)G(τi+,τi)-G(τi-,τi)00=Φ(τi+)(I+K)Φ-1(τi)-Φ(τi-)KΦ-1(τi)00=Bi+(Bi-I)G(τi-,τi).
To see (b), we write fors≠τi,
(41)G(s+,s)-G(s-,s)=Φ(s+)(I+K)Φ-1(s)-Φ(s-)KΦ-1(s)=I.
For (c), lett≠τi; then
(42)∂G(t,s)∂t={Φ′(t)(I+K)Φ-1(s)=JH(t)Φ(t)(I+K)Φ-1(s),s<t,Φ′(t)KΦ-1(s)=JH(t)Φ(t)KΦ-1(s),s≥t,∂G(s+,s)∂t-∂G(s-,s)∂t=JH(s)Φ(s)(I+K)Φ-1(s)-JH(s)Φ(s)KΦ-1(s)=JH(s).
Next, we consider (G4). By definition, it is easy to see thatG(t,s)is left continuous function att=τi. Let us consider the interval[t1,s). The later is similar. The first equation in (39) is direct consequences of (c) and the definition ofG(t,s). Clearly,
(43)G(τi+,s)=Φ(τi+)KΦ-1(s)000000=BiΦ(τi-)KΦ-1(s)=BiG(τi-,s),MG(t1,s)+NG(t2,s)=MΦ(t1)KΦ-1(s)+NΦ(t2)(I+K)Φ-1(s)=[MΦ(t1)+NΦ(t2)]KΦ-1(s)+NΦ(t2)Φ-1(s)=0.
The proofs of (G5) and (G6) are similar to (a) and (G4), respectively.Remark 3.
One can easily rewrite Green’s function (pair) in terms of the solutions of system (8a), (8b). Indeed,
(44)K=-[MΦ(t1)+NΦ(t2)]-1NΦ(t2)=1x1(t1)x2(t2)-x1(t2)x2(t1)×[x1(t2)x2(t1)x2(t1)x2(t2)-x1(t1)x1(t2)-x1(t1)x2(t2)],
and since
(45)detΦ(t)=detΦ(0)exp(∫0t‍trace(JH(s))ds)×∏i=1p‍detBi=∏i=1p‍αi2,
we may write
(46)Φ-1(t)=1detΦ(t)[u2(t)-x2(t)-u1(t)x1(t)]=(∏i=1p‍αi-2)[u2(t)-x2(t)-u1(t)x1(t)].


## 3. The Main Result

Our main result is the following theorem.Theorem 4.
Let (i)–(iii) hold. If(47)exp(∫t1t2‍|a(t)|dt)[∫t1t2‍b(t)dt]0000×[∫t1t2‍c+(t)dt+∑t1≤τi<t2‍(βiαi)+]00<4,
then BVP (6a), (6b), and (6c) has a unique solution(x(t),u(t)). Moreover,y=(x(t),u(t))is expressible as
(48)y(t)=w(t)+∫t1t2‍G(t,s)f(s)ds+∑t1≤τi<t2‍G~(t,τi+)ai,
where
(49)w(t)=Φ(t)[MΦ(t1)+NΦ(t2)]-1η,η=(A,B)T,
and Green’s function pair(G,G~)is given by (38).Proof.
We first prove the uniqueness. It suffices to show that the homogeneous BVP (8a)–(8c) has only the trivial solution. Letx(t)≢0on(t1,t2). By Theorem 1, we see that Lyapunov type inequality (11) holds contradicting the inequality (47). Thusx(t)=0for allt∈[t1,t2]. Moreover, by (6a), (6b), and (6c) we have
(50)b(t)u=0,t≠τi,
which results inu(t)=0fort≠τi. Taking limit we see thatu(τi±)=0. As a result we obtain(x(t),u(t))=(0,0)for allt∈[t1,t2]. This completes the uniqueness of the solutions.
For the existence, we start with the variation of parameters formula and write the general solution of system (6a), (6b) as
(51)y(t)=Φ(t)c+∫t1t‍Φ(t)Φ-1(s)f(s)ds+∑t1≤τi<t‍Φ(t)Φ-1(τi+)ai.
Clearly, the boundary condition is satisfied if
(52)[MΦ(t1)+NΦ(t2)]c=η-NΦ(t2)×[∫t1t2‍Φ-1(s)f(s)ds+∑t1≤τi<t2‍Φ-1(τi+)ai],
whereη=(A,B)T.
Since we have the uniqueness of solutions, the matrixMΦ(t1)+NΦ(t2)must have an inverse. Setting
(53)K=-[MΦ(t1)+NΦ(t2)]-1NΦ(t2),
we may solvecfrom (52) uniquely:
(54)c=[MΦ(t1)+NΦ(t2)]-1η+K[∫t1t2‍Φ-1(s)f(s)ds+∑t1≤τi<t2‍Φ-1(τi+)ai].
Hence,
(55)y(t)=Φ(t)[MΦ(t1)+NΦ(t2)]-1η+Φ(t)(I+K)×[∫t1t‍Φ-1(s)f(s)ds+∑t1≤τi<t‍Φ-1(τi+)ai]+Φ(t)K[∫tt2‍Φ-1(s)f(s)ds+∑t≤τi<t2‍Φ-1(τi+)ai].
Therefore the unique solution of the BVP (6a)–(6c) can be expressed as
(56)y(t)=w(t)+∫t1t2‍G(t,s)f(s)ds+∑t1≤τi<t2‍G~(t,τi+)ai.Let us now consider the BVP (10a), (10b), (10c), and (10d). In this case it is not difficult to see that the corresponding Green’s function (pair) becomes
(57)G(t,s)={ψ(t)(I+K)Ψ-1(s)1p(s)e2,s<t,ψ(t)KΨ-1(s)1p(s)e2,s≥t,G~(t,τi+)={ψ(t)(I+K)Ψ-1(τi+),τi<t,ψ(t)KΨ-1(τi+),τi≥t,
whereψ(t)=[ψ1,ψ2]is the first row of the (Wronskian) matrix:
(58)Ψ(t)=[ψ1(t)ψ2(t)ψ1′(t)ψ2′(t)],K=-[MΨ(t1)+NΨ(t2)]-1NΨ(t2),e2=[0,1]T.Corollary 5.
Suppose thatpandcare piece-wise continuous on[t1,t2],p(t)>0, andαi≠0fori=1,2,…,p.If
(59)[∫t1t2‍1p(t)dt][∫t1t2‍c+(t)dt+∑t1≤τi<t2‍(βiαi)+]<4,
then the BVP (10a), (10b), (10c), and (10d) has a unique solutionx(t)which is expressible as
(60)x(t)=w(t)+∫t1t2‍G(t,s)f(s)ds+∑t1≤τi<t2‍G~(t,τi+)ai,
where
(61)w(t)=ψ(t)[MΨ(t1)+NΨ(t2)]-1η,
and Green’s function pair(G,G~)is given by (57).Remark 6.
The results in this work are new even if the impulses are absent. The statements of the corresponding theorems are left to the reader.



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