doi:10.1155/2010/623508
Research Article
Transformations of Difference Equations II
Sonja Currie and Anne D. Love
School of Mathematics, University of the Witwatersrand, Private Bag 3, Wits 2050, South Africa
Correspondence should be addressed to Sonja Currie,[email protected]
Received 13 April 2010; Revised 30 July 2010; Accepted 6 September 2010
Academic Editor: M. Cecchi
Copyrightq2010 S. Currie and A. D. Love. 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.
This is an extension of the work done by Currie and Love2010where we studied the effect of applying two Crum-type transformations to a weighted second-order difference equation with non-eigenparameter-dependent boundary conditions at the end points. In particular, we now consider boundary conditions which depend affinely on the eigenparameter together with various combinations of Dirichlet and non-Dirichlet boundary conditions. The spectra of the resulting transformed boundary value problems are then compared to the spectra of the original boundary value problems.
1. Introduction
This paper continues the work done in1, where we considered a weighted second-order difference equation of the following form:
cnyn1−bnyn cn−1yn−1 −cnλyn, 1.1
withcn>0 representing a weight function andbna potential function. This paper is structured as follows.
The relevant results from1, which will be used throughout the remainder of this paper, are briefly recapped inSection 2.
InSection 3, we show how non-Dirichlet boundary conditions transform to affineλ -dependent boundary conditions. In addition, we provide conditions which ensure that the linear function in λ in the affine λ-dependent boundary conditions is a Nevanlinna or Herglotz function.
that transforming the boundary value problem given by 2.2 with any one of the four combinations of Dirichlet and non-Dirichlet boundary conditions at the end points using3.1 results in a boundary value problem with one extra eigenvalue in each case. This is done by considering the degree of the characteristic polynomial for each boundary value problem.
It is shown, in Section 5, that we can transform affine λ-dependent boundary conditions back to non-Dirichlet type boundary conditions. In particular, we can transform back to the original boundary value problem.
To conclude, we outline briefly how the process given in the sections above can be reversed.
2. Preliminaries
Consider the second-order difference equation 1.1 for n 0, . . . , m −1 with boundary conditions
hy−1 y0 0, Hym−1 ym 0, 2.1
wherehandHare constants, see2. Without loss of generality, by a shift of the spectrum, we may assume that the least eigenvalue,λ0, of1.1,2.1isλ0 0.
We recall the following important results from1. The mappingy → ydefined for
n −1, . . . , m−1 byyn yn1−ynu0n1/u0n, whereu0nis the eigenfunction of1.1,2.1corresponding to the eigenvalueλ0 0, produces the following transformed equation:
cnyn1−bnyn cn−1yn−1 −cnλyn, n 0, . . . , m−2, 2.2
where
cn u0ncn
u0n1 >
0, n −1, . . . , m−1,
bn
u
0ncn
u0n1cn1−
cn−1u0n−1
cnu0n
bn
cn−λ0 u
0ncn
u0n1, n
0, . . . , m−2. 2.3
Moreover, y obeying the boundary conditions 2.1transforms to y obeying the Dirichlet boundary conditions as follows:
y−1 0, ym−1 0. 2.4
Applying the mappingy→ygiven byyn yn−yn−1zn/zn−1forn 0, . . . , m−1, whereznis a solution of2.2withλ λ0, whereλ0is less than the least eigenvalue of2.2, 2.4, such thatzn>0 for alln −1, . . . , m−1, results in the following transformed equation:
where, forn 0, . . . , m−1,
cn zn−z1ncn−1,
bn
zn−
1cn−1
zncn
zn
zn−1 zn−
1cn−1
zn .
2.6
Here, we takec−1 c−1, thuscnis defined forn −1, . . . , m−1.
In addition,yobeying the Dirichlet boundary conditions2.4transforms toyobeying the non-Dirichlet boundary conditions as follows:
hy−1 y0 0, Hym−1 ym 0, 2.7
where
h
c0
c−1
b0
c0 −
z1
z0 −
b0
c0
−1
,
H bcmm−−2
2−
bm−1
cm−1−
zm−2cm−2
zm−1cm−1.
2.8
3. Non-Dirichlet to Affine
In this section, we show howvobeying the non-Dirichlet boundary conditions3.2,3.13 transforms under the following mapping:
vn vn−vn−1 zn
zn−1, n 0, . . . , m−1, 3.1
to give v obeying boundary conditions which depend affinely on the eigenparamter λ. We provide constraints which ensure that the form of these affine λ-dependent boundary conditions is a Nevanlinna/Herglotz function.
Theorem 3.1. Under the transformation3.1,vobeying the boundary conditions
v−1−γv0 0, 3.2
forγ /0, transforms tovobeying the boundary conditions
v−1 v0aλb, 3.3
where a γk/c−1/c0−kγc−1/c0,b b0/c0−γkb0/c0−b0/c0
γc−1/c0 z1/z0/c−1/c0−γkc−1/c0,andk z0/z−1. Here,c−1:
Proof. The values of n for which v exists are n 0, . . . , m−1. So to impose a boundary condition at n −1, we need to extend the domain ofvto include n −1. We do this by forcing the boundary condition3.3and must now show that the equation is satisfied on the extended domain.
Substituting3.2into the above equation yields
This may be slightly rewritten as follows
Equating coefficients ofλon both sides gives the following:
a c− γk
1/c0−kγc−1/c0 3.11
and equating coefficients ofλ0on both sides gives the following:
b
b0/c0−γkb0/c0−b0/c0 γc−1/c0 z1/z0
c−1/c0−γkc−1/c0 , 3.12
wherek z0/z−1, and recallc−1 c−1.
Note that forγ 0, this corresponds to the results in1forb −1/h. Theorem 3.2. Under the transformation3.1,vsatisfying the boundary conditions
vm−2−δvm−1 0, 3.13
forδ /0, transforms tovobeying the boundary conditions
vm−2 vm−1pλq, 3.14
wherep δcm−2/{1−δK−Kcm−2 bm−2−λ0cm−2},q cm−21−δK−
δλ0/{1−δK−Kcm−2 bm−2−λ0cm−2}, andK zm−1/zm−2. Here,zn is a solution to2.2forλ λ0, whereλ0is less than the least eigenvalue of 2.2,3.2, and3.13 such thatzn>0in the given interval,−1, m−1.
Proof. Evaluating3.1atn m−1 andn m−2 gives the following:
vm−1 vm−1−vm−2zm−1
zm−2, 3.15
vm−2 vm−2−vm−3zm−2
zm−3. 3.16
By consideringvnsatisfying2.2atn m−2, we obtain that
vm−3
bm−2
cm−3 −λ
cm−2
cm−3
vm−2−cm−2
cm−3vm−1. 3.17
Substituting3.17into3.16gives the following:
vm−2 vm−2
1−
bm−2
cm−3−λ
cm−2
cm−3
zm−2
zm−3
vm−1zm−2cm−2
Now using3.13together with3.15yields
vm−1 vm−1
1−δzm−1/zm−2, 3.19
which in turn, by substituting into3.13, gives the following:
vm−2 δvm−1
1−δzm−1/zm−2. 3.20
Thus, by putting3.19and3.20into3.18, we obtain
vm−2 δvm−1 1−δzm−1/zm−2
1−
bm−2
cm−3 −λ
cm−2
cm−3
zm−2
zm−3
vm−1zm−2cm−2
1−δzm−1/zm−2zm−3cm−3.
3.21
The equation above may be rewritten as follows:
1−δzm−1
zm−2
vm−2
vm−1 ⎧ ⎨ ⎩
δcm−3zm−3−bm−2−λcm−2zm−2cm−2zm−2
cm−3zm−3
⎫ ⎬ ⎭.
3.22
Now, sinceznis a solution to2.2forλ λ0, we have that
cm−3zm−3 −cm−2zm−1 bm−2zm−2−λ0cm−2zm−2. 3.23
Substituting3.23into3.22gives the following:
1−δzm−1
zm−2
vm−2
vm−1
−δcm−2zm−1 δλ−λ0cm−2zm−2 cm−2zm−2 −cm−2zm−1 bm−2zm−2−λ0cm−2zm−2
.
3.24
Settingzm−1/zm−2 Kyields
1−δKvm−2 vm−1
−δcm−2Kδλ−λ0cm−2 cm−2 −cm−2Kbm−2−λ0cm−2
Hence,
vm−2 vm−1 ⎧ ⎨ ⎩
δcm−2λcm−2−δK−δλ01 1−δK−cm−2Kbm−2−λ0cm−2
⎫ ⎬
⎭, 3.26
which is of the form3.14, whereK zm−1/zm−2,p δcm−2/{1−δK−Kcm− 2 bm−2−λ0cm−2}, andq cm−21−δK−δλ0/{1−δK−Kcm−2 bm− 2−λ0cm−2}.
Note that if we require thataλbin3.3be a Nevanlinna or Herglotz function, then we must have thata≥0. This condition provides constraints on the allowable values ofk. Remark 3.3. In Theorems3.1and3.2, we have takenznto be a solution of2.2forλ λ0 withλ0less than the least eigenvalue of2.2,3.2, and3.13such thatzn>0 in−1, m− 1. We assume that zn does not obey the boundary conditions3.2and 3.13 which is sufficient for the results which we wish to obtain in this paper. However, this case will be dealt with in detail in a subsequent paper.
Theorem 3.4. Ifk z0/z−1whereznis a solution to2.2forλ λ0 withλ0less than the least eigenvalue of 2.2,3.2, and3.13andzn> 0in the given interval−1, m−1, then the values ofkwhich ensure thata≥0in3.3, that is, which ensure thataλbin3.3is a Nevanlinna function are
k∈
0,1
γ
, forγ >0. 3.27
Proof. FromTheorem 3.1, we have that
a γk
c−1/c0−kγc−1/c0. 3.28
Assume thatγ >0, then to ensure thata≥ 0 we require that eitherk ≥0 andc−1/c0−
kγc−1/c0 > 0 ork ≤ 0 andc−1/c0−kγc−1/c0 < 0. For the first case, since c−1/c0γ > 0, we getk ≥ 0 andk < 1/γ. For the second case, we obtain k ≤ 0 and
k >1/γ, which is not possible. Thus, allowable values ofkforγ >0 are
k∈
0,1
γ
. 3.29
Sincek z0/z−1/0.Ifγ < 0, then we must have that eitherk ≤ 0 and c−1/c0−
kγc−1/c0 > 0 ork ≥ 0 andc−1/c0−kγc−1/c0 <0. The first case ofk ≤ 0 is not possible sincecn zn−1/zncn−1andcn,cn−1 >0, which implies that
Also, if we require thatpλqfrom3.14be a Nevanlinna/Herglotz function, then we must havep≥0. This provides conditions on the allowable values ofK.
Corollary 3.5. IfK zm−1/zm−2whereznis a solution to2.2forλ λ0withλ0less than the least eigenvalue of 2.2,3.2, and3.13, andzn>0in the given interval−1, m−1, then
K∈
−∞,1δ
∪
bm−2
cm−2,∞ , forδ >
cm−2
bm−2,
K∈
−∞,bcmm−−2
2 ∪
1
δ,∞
, forδ < cm−2
bm−2.
3.30
Proof. Without loss of generality, we may shift the spectrum of2.2with boundary conditions 3.2,3.13, such that the least eigenvalue of2.2with boundary conditions3.2,3.13is strictly greater than 0, and thus we may assume thatλ0 0.
Sincecm−2>0, we consider the two cases,δ >0 andδ <0.
Assume thatδ > 0, then the numerator ofp is strictly positive. Thus, to ensure that
p > 0 the denominator must be strictly positive, that is,1−δK−Kcm−2 bm−2−
λ0cm−2>0. So either 1−δK >0 and−Kcm−2 bm−2−λ0cm−2>0 or 1−δK <0 and−Kcm−2 bm−2−λ0cm−2< 0. Sinceλ0 0, we have that eitherK <1/δand
K <bm−2/cm−2orK >1/δandK >bm−2/cm−2. Thus, if 1/δ <bm−2/cm−2, that is,δ >cm−2/bm−2, we get
K∈
−∞,δ1
∪
bm−2
cm−2,∞ , 3.31
and if 1/δ >bm−2/cm−2, that is,δ <cm−2/bm−2, we get
K∈
−∞,bm−2
cm−2 ∪
1
δ,∞
. 3.32
Now ifδ <0, then the numerator ofpis strictly negative. Thus, in order thatp >0, we require that the denominator is strictly negative, that is,1−δK−Kcm−2bm−2−λ0cm−2< 0. So either 1−δK > 0 and −Kcm−2 bm−2−λ0cm−2 < 0 or 1−δK < 0 and −Kcm−2 bm−2−λ0cm−2 > 0. As λ0 0, we obtain that either K > 1/δ and
K >bm−2/cm−2orK <1/δandK <bm−2/cm−2. These are the same conditions as we had onK forδ >0. Thus, the sign ofδ does not play a role in finding the allowable values ofKwhich ensure thatp≥0, and hence we have the required result.
4. Comparison of the Spectra
By combining the results of1, conclusionwith Theorems3.1and3.2, we have proved the following result.
Theorem 4.1. Assume that vn satisfies 2.2. Consider the following four sets of boundary
conditions:
v−1 0, vm−1 0, 4.1
v−1 0, vm−2 δvm−1, 4.2
v−1 γv0, vm−1 0, 4.3
v−1 γv0, vm−2 δvm−1. 4.4
The transformation3.1, whereznis a solution to2.2forλ λ0, whereλ0is less than the least eigenvalue of 2.2with one of the four sets of boundary conditions above, such thatzn>0in the given interval−1, m−1, takesvnobeying2.2tovnobeying2.5.
In addition,
ivobeying4.1transforms tovobeying
hv−1 v0 0, 4.5
whereh c0/c−1b0/c0−z1/z0−b0/c0−1and
Hvm−1 vm 0, 4.6
whereH bm−2/cm−2−bm−1/cm−1−zm−2cm−2/zm−1cm−1 withc−1 c−1.
iivobeying4.2transforms tovobeying4.5and3.14.
iiivobeying4.3transforms tovobeying3.3and4.6.
ivvobeying4.4transforms tovobeying3.3and3.14.
The next theorem, shows that the boundary value problem given by vnobeying 2.2together with any one of the four types of boundary conditions in the above theorem has m−1 eigenvalues as a result of the eigencondition being the solution of anm−1th order polynomial inλ. It should be noted that if the boundary value problem considered is self-adjoint, then the eigenvalues are real, otherwise the complex eigenvalues will occur as conjugate pairs.
Theorem 4.2. The boundary value problem given byvnobeying2.2together with any one of the four types of boundary conditions given by4.1to4.4hasm−1eigenvalues.
Proof. Sincevnobeys2.2, we have that, forn 0, . . . , m−2,
vn1 −cn−1vn−1
cn
bn
So settingn 0, in4.7, gives the following: Thus, by an easy induction, we have that
So our eigencondition is given by
Next,4.2givesvm−2 δvm−1, so
Pm−2
0 P1m−2λ· · ·Pmm−−22λm−2
v0 δP0m−1P1m−1λ· · ·Pmm−−11λm−1v0, 4.15
from which we obtain the following eigencondition:
Pm−2
0 P1m−2λ· · ·Pmm−−22λm−2
δPm−1
0 P1m−1λ· · ·Pmm−−11λm−1
. 4.16
This is again anm−1th order polynomial inλand therefore hasm−1 roots. Hence, the boundary value problem given byvnobeying2.2with4.2hasm−1 eigenvalues.
Now for the boundary conditions4.3and 4.4, we have that v−1 γv0, thus 4.8becomes
v1
−c−1
c0
b0
c0 −λ 1
γ
v−1: Q1 0Q11λ
v−1, 4.17
whereQ1
0andQ11are real constants, that is, a first order polynomial inλ.
Usingv−1 γv0andv1from above, we can show thatv2can be written as the following:
v2: Q20Q21λQ22λ2v−1, 4.18
where againQ2
i,i 0,1,2 are real constants, that is, a quadratic polynomial inλ. Thus, by induction,
vm−1 Q0m−1Qm1−1λ· · ·Qmm−−11λm−1v−1,
vm−2 Qm−2
0 Qm1−2λ· · ·Qmm−−22λm−2
v−1,
4.19
whereQm−1
i , i 0,1, . . . , m−1 andQim−2,i 0,1, . . . , m−2 are real constants, thereby giving anm−1th and anm−2th order polynomial inλ, respectively.
Now,4.3givesvm−1 0, that is,
Qm−1
0 Qm1−1λ· · ·Qmm−−11λm−1
v−1 0. 4.20
So our eigencondition is given by
Qm−1
0 Qm1−1λ· · ·Qmm−−11λm−1
0, 4.21
Lastly,4.4givesvm−2 δvm−1, that is,
Qm−2
0 Qm1−2λ· · ·Qmm−−22λm−2
v−1 δQ0m−1Qm1−1λ· · ·Qmm−−11λm−1v−1, 4.22
from which we obtain the following eigencondition:
Qm−2
0 Qm1−2λ· · ·Qmm−−22λm−2
δQm−1
0 Qm1−1λ· · ·Qmm−−11λm−1
. 4.23
This is again anm−1th order polynomial inλand therefore hasm−1 roots. Hence, the boundary value problem given byvnobeying2.2with4.4hasm−1 eigenvalues.
In a similar manner, we now prove that the transformed boundary value problems given inTheorem 4.1havemeigenvalues, that is, the spectrum increases by one in each case.
Theorem 4.3. The boundary value problem given byvnobeying2.5,n 1, . . . , m−2, together with any one of the four types of transformed boundary conditions given in (i) to (iv) inTheorem 4.1 hasmeigenvalues. The additional eigenvalue is preciselyλ0 with corresponding eigenfunctionzn, as given inTheorem 4.1.
Proof. The proof is along the same lines as that of Theorem 4.2. ByTheorem 3.1, we have extendedyn, such thatynexists forn 1, . . . , m−1.
Sincevnobeys2.5, we have that, forn 0, . . . , m−2,
vn1 −cn−1vn−1
cn
bn
cn −λ vn. 4.24
For the transformed boundary conditions iniandiiofTheorem 4.1, we have that4.5is obeyed, and as inTheorem 4.2, we can inductively show that
vm−1 M0m−1M1m−1λ· · ·Mmm−−11λm−1v−1,
vm−2 M0m−2M1m−2λ· · ·Mmm−−22λm−2v−1,
4.25
and also by1, we can extend the domain ofvnto includen mif necessary by forcing 4.6and then
vm Mm
0 Mm1λ· · ·Mmmλm
v−1, 4.26
whereMmi −1, i 0,1, . . . , m−1,Mim−2, i 0,1, . . . , m−2, and Mim, i 0,1, . . . , mare real constants, that is, anm−1th,m−2th, andmth order polynomial inλ, respectively.
Now fori, the boundary condition4.6gives the following:
Mm−1
0 Mm1−1λ· · ·Mmm−−11λm−1
v−1 −1
H
Mm
0 Mm1λ· · ·Mmmλm
Therefore, the eigencondition is
which is anmth order polynomial in λand thus has mroots. Hence, the boundary value problem given byvnobeying2.5with transformed boundary conditionsi, that is,4.5
Thus, inductively we obtain
where all the coefficients ofλare real constants.
The transformed boundary conditions iii mean that 4.6 is obeyed, thus, our eigencondition is
which is anmth order polynomial in λand thus has mroots. Hence, the boundary value problem given by vn obeying 2.5with transformed boundary conditions iii, that is, 3.3and4.6, hasmeigenvalues.
Also, the transformed boundary conditions in iv give 3.14 which produces the following eigencondition:
which is anmth order polynomial in λand thus has mroots. Hence, the boundary value problem given byvnobeying2.5with transformed boundary conditionsiv, that is,3.3 and3.14, hasmeigenvalues.
Lastly, we have that 3.1 transforms eigenfunctions of any of the boundary value problems in Theorem 4.2 to eigenfunctions of the corresponding transformed boundary value problem, seeTheorem 4.2. In particular, ifλ1 < · · · < λm−1are the eigenvalues of the original boundary value problem with corresponding eigenfunctionsu1n, . . . ,um−1n, then
problem with eigenvalues λ0, λ1, . . . , λm−1. Since we know that the transformed boundary
value problem hasmeigenvalues, it follows thatλ0, λ1, . . . , λm−1constitute all the eigenvalues of the transformed boundary value problem, see1.
5. Affine to Non-Dirichlet
In this section, we now show that the process inSection 3may be reversed. In particular, by applying the following mapping:
vn vn1−vnu0n1
u0n ,
5.1
we can transform v obeying affine λ-dependent boundary conditions to v obeying non-Dirichlet boundary conditions.
Theorem 5.1. Consider the boundary value problem given byvnsatisfying2.5with the following boundary conditions:
v−1 v0αλβ, 5.2
vm−2 vm−1ζλη. 5.3
The transformation5.1, forn −1, . . . , m−1, whereu0nis an eigenfunction of 2.5,5.2, and 5.3corresponding to the eigenvalueλ0 0, yields the following equation:
cnvn1−bnvn cn−1vn−1 −cnλvn, n 0, . . . , m−3, 5.4
where, forc−1 c−1,
cn u0ncn
u0n1 >
0, n −1, . . . , m−2,
bn
u0ncn
u0n1cn1−
cn−1u0n−1
cnu0n
bn
cn −λ0
u0ncn
u0n1, n
0, . . . , m−2. 5.5
In addition,vobeying5.2and5.3transforms tovobeying the non-Dirichlet boundary conditions
v−1 Bv0, 5.6
vm−2 Avm−1, 5.7
whereB αc0/{αλ0βc0 αc−1}andA ηcm−2/cm−1 1/ζ−1.
5.2. Thus,vis defined forn −1, . . . , m−2 giving that5.4is valid forn 0, . . . , m−3. For
n 0 andn −1,5.1gives the following:
v0 v1−v0u01
u00,
5.8
v−1 v0−v−1 u00
u0−1.
5.9
Settingn 0 in2.5gives the following:
v1 −λv0 b0
c0v0−
c−1
c0 v−1, 5.10
which by using5.2becomes
v1
−λc0 b0−c−1αλβ
c0αλβ
v−1. 5.11
Sinceu0nis an eigenfunction of2.5,5.2, and5.3corresponding to the eigenvalueλ
λ0 0, we have that
u0−1
u00 αλ0β, 5.12
and hence
u01
−λ0c0 b0−c−1
αλ0β
c0αλ0β
u0−1. 5.13
Substituting5.11and5.13into5.8and using5.2, we obtain
v0
−λc0 b0−c−1αλβ
c0αλβ
v−1
−αλv−1β
−λ0c0 b0−c−1
αλ0β
c0αλ0β
u0−1
u00 .
5.14
Since u0−1/u00 αλ0 β, everything can be written over the common denominator
c0αλβ. Taking outv−1and simplifying, we get
v0 v−1λ0−λc
0 αc−1
Thus,
v−1 v0 c0
αλβ
λ0−λc0 αc−1.
5.16
Substituting5.2into5.9gives the following:
v−1 v−1 αλ0−λ
αλβαλ0β
. 5.17
Hence, by putting5.16into5.17, we get
v−1 v0
αc0
αλ0β
c0 αc−1
. 5.18
So to impose the boundary condition 5.7, it is necessary to extend the domain of v by forcing the boundary condition5.7. We must then check thatv satisfies the equation on the extended domain.
Evaluating5.4atn m−2 and using5.7give the following:
1
A−
bm−2
cm−2λ
vm−2 cm−3
cm−2vm−3 0. 5.19
Using5.1withn m−2 andn m−3 together with5.3, we obtain
vm−2 vm−1−vm−2u0m−1
u0m−2 vm− 1
1−ζληu0m−1
u0m−2
,
vm−3 vm−2−vm−3u0m−2
u0m−3 vm−
1ζλη−vm−3u0m−2
u0m−3. 5.20
Substituting the above two equations into5.19yields
vm−1
1
A−
bm−2
cm−2λ
1−ζληu0m−1
u0m−2
ccmm−−3 2
ζλη
−vm−3cm−3
cm−2
u0m−2
u0m−3 0.
Sinceu0nis an eigenfunction of2.5,5.2, and5.3corresponding to the eigenvalueλ
Thus, equating coefficients ofλgives the following:
Sincecm−2/cm−3/0, we can divide and solve forAto obtain
A
η
cm− 2
cm−1 1
ζ
−1
. 5.28
Note that the case ofζ 0, that is, a non-Dirichlet boundary condition, givesA 0, that is,vm−2 0 which corresponds to the results obtained in1.
If we setu0n 1/zn−1cn−1, withzn a solution of2.2forλ λ0 0 whereλ0 less than the least eigenvalue of 2.2,3.2, and3.13andzn > 0 in the given interval−1, m−1, thenu0nis an eigenfunction of2.5,5.2, and5.3corresponding to the eigenvalueλ0 0. To see thatu0nsatisfies2.5, see1, Lemma 4.1with, as previously,
lu01 0, and nowu0−1 αλ0β β. Then, by construction,u0nobeys5.2. We now show thatu0obeys5.3. LetK zm−1/zm−2,
ζ δcm−2
1−δK−Kcm−2 bm−2−λ0cm−2 ,
η cm−21−δK−δλ0
1−δK−Kcm−2 bm−2−λ0cm−2 .
5.29
Nowznis a solution of2.2forλ λ0, thus,
u0m−2
u0m−1
zm−2cm−2
zm−3cm−3
−zzmm−−1 2
bm−2
cm−2−λ0 −1
ζλ0η η. 5.30
Remark 5.2. For u0n, α, β, ζ, and η as above, the transformation 5.1, in Theorem 5.1, results in the original given boundary value problem. In particular, we obtain that in Theorem 5.1cn cnandbn bn, see1, Theorem 4.2. In addition,
B αc0
αλ0β
c0 αc−1 γ,
A
η
cm− 2
cm−1 1
ζ
−1
δ.
5.31
We now verify thatB γ. Let
β
b0/c0−γz0/z−1b0/c0−b0/c0 γc−1/c0 z1/z0
c−1/c0−γc−1/c0z0/z−1 ,
α γz0/z−1
c−1/c0−c−1z0/c0z−1
γ
c−1/c0z−1/z0−c−1/c0.
5.32
Sincec−1 c−1, we obtainc−1/c0 z0/z−1, and thus
α γ
1−γz0/z−1. 5.33
Also,B αc0/αλ0βc0αc−1. Dividing through byαc0and usingλ0 0 together withc−1/c0 z0/z−1gives the following:
1
B β
1
α z0
z−1
. 5.34
Now,
b0
c0
z−1c−1
z0c0
z0
z−1, 5.35
and sincezsatisfies2.2atn 0 forλ λ0 0, we get
b0
c0
z−1c−1
z0c0
z1
z0. 5.36
Thus, using5.35and5.36, the numerator ofβis simplified to
z0
z1
1−γz0
z1
. 5.37
The denominator ofβcan be simplified usingc−1/c0 z0/z−1to
z0
z1
1−γz0
z1
, 5.38
Finally, substituting in forα, we obtain
1
α z0
z−1
1
γ. 5.39
Thus, 1/B 1/γ, that is,B γ.
Next, we show thatA δ. Recall thatλ0 0 and
1
A η
cm− 2
cm−1 1
ζ
. 5.40
Let
ζ δcm−2
1−δzm−1/zm−2−zm−1/zm−2cm−2 bm−2,
η cm−2
−zm−1/zm−2cm−2 bm−2.
5.41
Note that
cm−2
cm−1
zm−3cm−3zm−1
zm−2zm−2cm−2, 5.42
and sincezsatisfies2.2atn m−2 forλ λ0 0, we get
bm−2
cm−2
zm−3cm−3
zm−2cm−2
zm−1
zm−2. 5.43
We now substitute in forζandηinto the equation for 1/Aand use5.42and5.43to obtain that
1
A
1
δ, 5.44
that is,A δ.
To summarise, we have the following.
Considervnobeying2.5with one of the following 4 types of boundary conditions: anon-Dirichlet and non-Dirichlet, that is,4.5and4.6;
bnon-Dirichlet and affine, that is,4.5and3.14; caffine and non-Dirichlet, that is,3.3and4.6; daffine and affine, that is,3.3and3.14.
Now, the transformation5.1, withu0n 1/zn−1cn−1an eigenfunction of2.5 with boundary conditionsa b,c,d, resp.corresponding to the eigenvalueλ λ0 0, transformsvnobeying2.5tovnobeying2.2and transforms the boundary conditions as follows:
1boundary conditionsatransform tov−1 0 andvm−1 0; 2boundary conditionsbtransform tov−1 0 and3.13; 3boundary conditionsctransform to3.2andvm−1 0; 4boundary conditionsdtransform to3.2and3.13.
ByTheorem 4.2, we know that the above transformed boundary value problems invneach havem−1 eigenvalues. In particular, if 0 λ0 < λ1 < · · · < λm−1 are the eigenvalues of 2.5,a b,c,d, resp.with eigenfunctions u0n,v1n, . . . ,vm−1n, thenu0n ≡ 0 andv1n, . . . ,vm−1nare eigenfunctions of2.2,1 2,3,4, resp.with eigenvalues
λ1, . . . , λm−1. Since we know that these boundary value problems havem−1 eigenvalues, it follows thatλ1, . . . , λm−1constitute all the eigenvalues.
6. Conclusion
To conclude, we outlinethe details are left to the reader to verifyhow the entire process could also be carried out the other way around. That is, we start with a second order difference equation of the usual form, given in the previous sections, together with boundary conditions of one of the following forms:
inon-Dirichlet at the initial point and affine at the terminal point; iiaffine at the initial point and non-Dirichlet at the terminal point; iiiaffine at the initial point and at the terminal point.
We can then transform the above boundary value problemby extending the domain where necessary, as done previouslyto an equation of the same type with, respectively, transformed boundary conditions as follows:
ADirichlet at the initial point and non-Dirichlet at the terminal point; Bnon-Dirichlet at the initial point and Dirichlet at the terminal point; Cnon-Dirichlet at the initial point and at the terminal point.
It is then possible to return to the original boundary value problem by applying a suitable transformation to the transformed boundary value problem above.
Acknowledgments
References
1 S. Currie and A. Love, “Transformations of difference equations I,”Advances in Difference Equations, vol. 2010, Article ID 947058, 22 pages, 2010.
