# The Generalized Green’s Function for Boundary Value Problem of Second Order Difference Equation

Authors: Xiaoling Han; Juanjuan Huang
Journal: Journal of Function Spaces
 (2015)

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## Abstract

Letb>a+2 and [a+1,b+1]={a+1,a+2,…,b+1}. In this paper, by building the generalized Green’s function for the problems, we study the solvability of the S-L problem Lx=Δ[p(t-1)Δx(t-1)]+[q(t)+λr(t)]x(t)=-f(t), U1(x)=α1x(a)+α2Δx(a)=0, U2(x)=β1x(b+1)+β2Δx(b+1)=0, and the periodic S-L problem Lx=Δ[p(t-1)Δx(t-1)]+[q(t)+λr(t)]x(t)=-f(t), U3(x)=x(a)-x(b+1)=0, U4(x)=Δx(a)-Δx(b+1)=0, where the parameter λ is an eigenvalue of the linear problem Lx=0, U1(x)=0, U2(x)=0 or the problem Lx=0, U3(x)=0, U4(x)=0, and p:[a,b+1]→(0,+∞),r:[a+1,b+1]→(0,+∞), q(t) is defined and real valued on [a+1,b+1], α12+α22≠0,β12+β22≠0, and in the periodic S-L problem we have p(a)=p(b+1).


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## Body

## 1. Introduction

There are many results about the solvability of S-L problem(1)Lx=Δpt-1Δxt-1+qt+λrtxt=-ft,U1x=α1xa+α2Δxa=0,U2x=β1xb+1+β2Δxb+1=0
and the periodic S-L problem
(2)Lx=Δpt-1Δxt-1+qt+λrtxt=-ft,U3x=xa-xb+1=0,U4x=Δxa-Δxb+1=0,
where p:[a,b+1]→(0,+∞), r:[a+1,b+1]→(0,+∞), and q(t) is defined and real valued on [a+1,b+1], and [a+1,b+1]={a+1,a+2,…,b+1}. Consider α12+α22≠0, β12+β22≠0, in the periodic S-L problem we have p(a)=p(b+1).If the parameterλ is not an eigenvalue of the linear problem
(3)Lx=0,U1x=0,U2x=0
or the periodic S-L problem
(4)Lx=0,U3x=0,U4x=0;
then the linear operator is reversible; we can get an equivalent summation form for the problem (1) or problem (2) by using Green’s function; there are many results in this case; we refer readers to [1–4]. However, when λ is an eigenvalue, we cannot get the summation form in usual way. So, in this paper we introduce the concept of generalized Green’s function and construct the generalized Green function for the two problems. On this base we still can convert the originally irreversible difference operator into summation form. Moreover, we also obtain the necessary and sufficient condition for the solvability of problems (1) and (2). We refer readers to [5–10] for more details about discrete equation.


## 2. Preliminaries

We first wish to collect some basic lemmas that will be important to us in what follows.Definition 1 (see [5]).
Lety(t) and z(t) be solutions of Lx=0; then
(5)ωyt,zt=ytΔzt-ztΔyt.Lemma 2 (see [5]).
Ify(t) and z(t) are defined on [a,b+2], then
(6)ztLyt-ytLzt=Δpt-1ωzt-1,yt-1
for t∈[a+1,b+1].By summing both sides of (6) from a+1 to b+1, we get the following lemma.Lemma 3 (see [5]).
Assume thaty(t) and z(t) are defined on [a,b+2]. Then
(7)∑t=a+1b+1ztLyt-∑t=a+1b+1ytLzt=ptωzt,ytab+1.Lemma 4 (see [5]).
Ify(t) and z(t) are solutions of Lx=0, then ω[y(t),z(t)]=C/p(t), t∈[a,b+1], where C is a constant.Remark 5.
Ify(t) and z(t) are linearly dependent, then ω[y(t),z(t)]=0; if they are linearly independent, then ω[y(t),z(t)]≠0.


## 3. The Main Results for Problem (1)

Theorem 6.
The semihomogeneous boundary value problem (1) has a solution if and only if
(8)∑τ=a+1b+1fτxnτ=0,n=0,1,2,…,
where xn(t) is an eigenfunction corresponding to the eigenvalue λ.Proof.
Letx(t) be the solution of problem (1) and xn(t) the nontrivial solution for homogeneous problem (3); using
(9)xtLxnt-xntLxt=Δpt-1ωxt-1,xnt-1
and Lx=-f(t), Lxn=0, we have
(10)∑τ=a+1b+1fτxnτ=ptωxt,xntab+1=pb+1xb+1Δxnb+1-xnb+1Δxb+1-paxaΔxna-xnaΔxa.
Since x(t) and xn(t) satisfy the boundary condition of (1), then
(11)∑τ=a+1b+1‍fτxnτ=0.
On the other hand, choose a solutionx(t) for Lx=-f(t), satisfying x(b+1)=xn(b+1), Δx(b+1)=Δxn(b+1); then
(12)U2x=U2xn=β1xb+1+β2Δxb+1=0.
By conditions Lx=-f(t), Lxn=0, ∑τ=a+1b+1‍f(τ)xn(τ)=0, and (9), observe that
(13)∑τ=a+1b+1fτxnτ=ptωxt,xntab+1=0;
then
(14)pb+1xb+1Δxnb+1-xnb+1Δxb+1=paxaΔxna-xnaΔxa;
that is, U1(x)=α1x(a)+α2Δx(a)=0. Thus, x(t) is a solution of problem (9), and this completes the proof.In what follows, we begin to build the generalized Green function.Theorem 7.
If the S-L problem (3) has a solution x0(t), and ∑t=a+1b+1‍x02(t)=1, then there exists a unique function G(t,τ) which satisfies the following properties:(1)
LG(t,τ)=x0(t)x0(τ),U1(G)=0,t∈[a,τ];(2)
LG(t,τ)=x0(t)x0(τ),U2(G)=0,t∈[τ,b+1];(3)
ΔtG(t,τ)|t=τ+0-ΔtG(t,τ)|t=τ-0=-(1/p(τ));(4)
∑t=a+1b+1‍G(t,τ)x0(t)=0.The function having the above properties is called the generalized Green function of the problem (1).Proof.
Choose another solutionx1(t) of Lx=0 which is linearly independent with x0(t), such that
(15)ptx0tΔx1t-x1tΔx0t=1.
Letω(t) be any one particular solution of Lx=x0(t)x0(τ). Thus, by the structure of general solution for nonhomogeneous difference equations, we know that
(16)Gt,τ=ωt+c1x0t+c2x1t,t∈a,τ,ωt+c3x0t+c4x1t,t∈τ,b+1,
where ci(i=1,2,3,4) can be contained by the properties of G(t,τ).
Firstly, letG(t,τ) satisfy the boundary conditions
(17)U1G=U1ω+c1U1x0+c2U1x1=0,U2G=U2ω+c3U2x0+c4U2x1=0.
By the fact that U1(x0)=0 and U2(x0)=0, we get that
(18)U1(G)=U1ω+c2U1x1=0,U2G=U2ω+c4U2x1=0.
Since x0(t) and x1(t) are linearly independent, thus, U1(x1)≠0 and U2(x1)≠0, so we get that c2 and c4 are
(19)c2=-U1ωU1x1,c4=-U2ωU2x1;
then
(20)c2-c4=U2ωU2x1-U1(ω)U1(x1).
However, from (16) and ΔtG(t,τ)|t=τ+0-ΔtG(t,τ)|t=τ-0=-(1/p(τ)) we get
(21)c1-c3x0τ+c2-c4x1τ=0,c1-c3Δx0τ+c2-c4Δx1τ=1pτ.
By (21), we get
(22)c1-c3=-x1τ,c2-c4=x0τ.
In the following we will prove that (20) is consistent with (22).
By(23)pax0aΔx1a-x1aΔx0a=1,pb+1x0b+1Δx1b+1000000000-x1b+1Δx0b+1=1,α1x0a+α2Δx0a=0,β1x0b+1+β2Δx0b+1=0,
we get
(24)x0a=α2paU1x1,Δx0a=-α1paU1x1,x0b+1=β2pb+1U2x1,Δx0b+1=-β1pb+1U2x1.
From
(25)x0Lω-ωLx0=Δpx0Δω-ωΔx0(t-1)
and ∑t=a+1b+1‍x02(t)=1 we get
(26)x0τ=px0Δω-ωΔx0ab+1=pb+1x0b+1Δωb+100000000000-ωb+1Δx0b+1-pax0aΔωa-ωaΔx0a.
Now, we get
(27)x0τ=U2ωU2x1-U1ωU1x1=c2-c4;
that is, (20) and (22) are consistent.
In order to determinec1 and c3, by c1-c3=-x1(τ), we obtain c1=c3-x1(τ); putting c1 into ∑t=a+1b+1‍G(t,τ)x0(t)=0, we get
(28)∑t=a+1b+1‍ωtx0t+c3∑t=a+1τ-1‍x02(t)-x1τ∑t=a+1τ-1‍x02t-U1ωU1x1∑t=a+1τ-1‍x0tx1t+c3∑t=τb+1‍x02t-U2ωU2x1∑t=τb+1‍x0tx1t=0;
then
(29)c3=-∑t=a+1b+1‍ωtx0t+x1τ∑t=a+1τ-1‍x02t+U1ωU1x1∑t=a+1τ-1‍x0tx1t+U2ωU2x1∑t=τb+1‍x0tx1t.
All these show that we can getc1, c2, c3, and c4 uniquely one by one; then G(t,τ) is unique too; that is, the function G(t,τ) constructed by the above steps has all the properties of generalized Green’s function. This completes the proof.Example 8.
Consider the generalized Green function of the S-L problem(30)Δ2xt-1+2xt=0,x0=0,x2=0.Proof.
From the conditions∑t=12‍x02(t)=1 and p(t)[x0(t)Δx1(t)-x1(t)Δx0(t)]=1, we can get
(31)x0t=sin⁡π2t,x1t=-cos⁡⁡π2t.
It is easy to verify that
(32)ωt=14sin⁡π2τsin⁡π2t-t2sin⁡π2τcos⁡⁡π2t
is a particular solution of the following equation:
(33)Δ2ωt-1+2ωt=sin⁡π2τsin⁡π2t.
Note that
(34)U1x1=x10=-1,U1ω=ω0=0,U2x1=x12=1,U2ω=ω2=sinπ2τ.
So
(35)c2=0,c4=-sin⁡π2τ,c3=-12τcos⁡⁡π2τ+12sin⁡π2τ-14cos⁡⁡π2τ,c1=c3-x1τ=-12τcos⁡⁡π2τ+12sin⁡π2τ+34cos⁡⁡π2τ;
then we can get the generalized Green function
(36)Gt,τ=-t2sin⁡π2τcos⁡⁡π2t+3-2τ4sin⁡π2tcos⁡⁡π2τ,t∈0,τ,2-t2sin⁡π2τcos⁡⁡π2t-2τ+14sin⁡π2tcos⁡⁡π2τ,t∈τ,2.
Now, we prove that by the extended Green function it still can convert the originally irreversible difference operator into reversible form, and its summation form is unique.Theorem 9.
LetG(t,τ) be the generalized Green function of the S-L problem (2), and x0(t) satisfying ∑t=a+1b+1‍x02(t)=1 is a solution of the S-L problem, and ∑τ=a+1b+1‍f(τ)x0(τ)=0. Then the solution of the semihomogeneous boundary value problem
(37)Lx=-ft,U1x=0,U2x=0
can be expressed by
(38)xt=∑τ=a+1b+1‍Gt,τfτ.Proof.
By the fact that(39)Δxt-1=∑τ=a+1b+1‍ΔGt-1,τfτ,Δ2xt-1=Δ∑τ=a+1t-1‍ΔGt-1,τfτ00000+∑τ=tb+1‍ΔGt-1,τfτ=Δ∑τ=a+1t-1‍Gt,τfτ-Gt-1,τfτ000000+∑τ=tb+1‍Gt,τfτ-Gt-1,τfτ=ΔGt,t-1ft-100000-Gt-1,a+1fa+100000+Gt,b+1fb+1-Gt-1,tft=ΔGt,b+1fb+10000-ΔGt-1,a+1fa+1+ΔGt,t-1ft-1-ΔGt-1,tft=∑τ=a+1b+1‍Δ2Gt-1,τfτ+ΔtGτ,t|τ=t+0-ΔtGτ,t|τ=t-0ft=∑τ=a+1b+1‍Δ2G(t-1,τ)f(τ)-f(t)p(t),
then
(40)Lx=Δpt-1Δxt-1+ptΔ2xt-1+qtxt+λrtxt=∑τ=a+1b+1‍LGf(τ)-ft=∑τ=a+1b+1‍x0tx0τfτ-ft=-ft.
To prove the uniqueness, it needs to be proved that we cannot findx~(t) and f(t) which satisfy ∑τ=a+1b+1‍f(τ)x0(τ)=0 except x0(t) and f(t); if there exists x~(t), then by Theorem 6, we know that x~(t) is also an eigenfunction of the S-L problem; this shows that x~(t) and x0(t) are linearly dependent, which completes the proof.


## 4. The Main Results for Problem (2)

In what follows, we build the generalized Green function of the periodic S-L problem.Theorem 10.
If the periodic S-L problem (2) has a solution x0(t), and choosing a linearly independent solution x1(t) with x0(t) for Lx=0, then there exists a function G(t,τ) which satisfies the following properties:(1)
LG(t,τ)=x0(t)x0(τ),U3(G)=0,t∈[a,τ];(2)
LG(t,τ)=x0(t)x0(τ),U4(G)=0,t∈[τ,b+1];(3)
ΔtG(t,τ)|t=τ+0-ΔtG(t,τ)|t=τ-0=-(1/p(τ));(4)
∑t=a+1b+1‍G(t,τ)x0(t)=0 and ∑t=a+1b+1‍G(t,τ)x1(t)=0.The function having the above properties is called the generalized Green function to the problem (2).Proof.
Choosex1(t) so that
(41)ptx0tΔx1t-x1tΔx0t=1.
Letω(t) be any one particular solution to Lx=x0(t)x0(τ); then
(42)Gt,τ=ωt+c1x0t+c2x1t,t∈a,τ,ωt+c3x0t+c4x1t,t∈τ,b+1,
where ci(i=1,2,3,4) can be contained by the properties of G(t,τ).
Firstly, letG(t,τ) satisfy the boundary conditions
(43)U3G=U3ω+c1U3x0+c2U3x1=0,U4G=U4ω+c3U4x0+c4U4x1=0.
By the fact that U3(x0)=0,U4(x0)=0 and in the periodic problem we have U3x1=U4x1=0; then U3(ω)=U4(ω)=0.
Thus,ci can be contained; that is, by the conditions (3) and (4), we get
(44)c1-c3=-x1τ,c2-c4=x0τ.
In order to determine ci, by c1-c3=-x1(τ), we have c1=c3-x1(τ); by c2-c4=x0(τ), we have c2=c4+x0(τ); putting c1,c2 into ∑t=a+1b+1‍G(t,τ)x0(t)=0 and ∑t=a+1b+1‍G(t,τ)x1(t)=0 respectively, it shows that
(45)c3∑t=a+1b+1‍x02t+c4∑t=a+1b+1‍x0tx1t=-∑t=a+1b+1‍ωtx0t+x1τ∑t=a+1τ-1‍x02t-x0τ∑t=a+1τ-1‍x0tx1t,c3∑t=a+1b+1‍x0tx1t+c4∑t=a+1b+1‍x12t=-∑t=a+1b+1‍ωtx1t+x1τ∑t=a+1τ-1‍x0tx1t-x0(τ)∑t=a+1τ-1‍x12(t).
Combine the above equations; c3 and c4 can be determined; then c1 and c2 can be determined, so G(t,τ) also can be determined; that is, the function G(t,τ) set by the above steps has all the properties of generalized Green’s function. This completes the proof.Remark 11.
Because the choice ofx0(t) in the above structure of generalized Green’s function is not unique, so the generalized Green function of periodic S-L problem (2) is changed with x0(t), to have different form.Example 12.
Consider the generalized Green function of the periodic S-L problem(46)Δ2xt-1+2xt=0,x0=x4,Δx0=Δx4.Proof.
Letx0(t)=sin⁡(π/2)t, and from
(47)ptx0tΔx1t-x1tΔx0t=1,
we can let x1(t)=-cos⁡⁡(π/2)t, (where p(t)=1). It is easy to verify that
(48)ωt=14sin⁡π2τsin⁡π2t-t2sin⁡π2τcos⁡⁡π2t
is a particular solution of the following equation:
(49)Δ2ωt-1+2ωt=sin⁡π2τsin⁡π2t.
By Theorem 10, we have
(50)c3=-14sin⁡π2τ-12cos⁡⁡π2ττ2+14cos⁡πτ-14-18sin⁡π2τsin⁡πτ,c4=-18cos⁡⁡π2τsin⁡πτ-12sin⁡π2τ-14cos⁡⁡πτ+τ2+94;
then
(51)c1=-14sin⁡π2τ-12cos⁡⁡π2ττ2-94+14cos⁡⁡πτ-18sin⁡π2τsin⁡πτ,c2=-18cos⁡⁡π2τsin⁡πτ-12sin⁡π2τ-14cos⁡⁡πτ+τ2+14.
So we can get the generalized Green function(52)G(t,τ)=-12sin⁡π2tcos⁡⁡π2ττ2+14cos⁡⁡πτ-94-18sin⁡π2τsin⁡πτsin⁡π2t+18cos⁡⁡π2tcos⁡⁡π2τsin⁡πτ+12sin⁡π2τcos⁡⁡π2t-14cos⁡⁡πτ+τ2+14-t,t∈[0,τ],-12sin⁡π2tcos⁡⁡π2ττ2-14+14cos⁡⁡πτ-18sin⁡π2τsin⁡πτsin⁡π2t+18cos⁡⁡π2tcos⁡⁡π2τsin⁡πτ+12sin⁡π2τcos⁡⁡π2t-14cos⁡⁡πτ+τ2+94-t,t∈[τ,4].
However, if we putx0(t)=-cos⁡⁡π/2t, then the generalized Green function is(53)Gt,τ=12sinπ2tcos⁡π2τ14cos⁡πτ-17-t-2τ4-18cos⁡π2τsinπτcos⁡π2t+18sinπ2tsinπ2τsinπτ+14cos⁡π2τcos⁡π2t-14cos⁡πτ+τ2-154,t∈0,τ,12sinπ2tcos⁡π2τ2τ-9+4t4+14cos⁡πτ-18cos⁡π2τsinπτcos⁡π2t+18sinπ2tsinπ2τsinπτ-12sinπ2τcos⁡π2t-14cos⁡πτ+τ2-34,t∈τ,4.Be similar to Theorem9, we can get the following result; however, because the generalized Green function of the periodic S-L problem (2) is not unique, summation form of difference operator is also not unique.Theorem 13.
LetG(t,τ) be the generalized Green function of the periodic S-L problem (2); then the solution of the semihomogeneous boundary value problem
(54)Lx=-ft,U3x=0,U4x=0
can be expressed by
(55)xt=∑τ=a+1b+1‍Gt,τfτ.



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