# Spectrum of Discrete Second-Order Neumann Boundary Value Problems with Sign-Changing Weight

Authors: Ruyun Ma; Chenghua Gao; Yanqiong Lu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the spectrum structure of discrete second-order Neumann boundary value
problems (NBVPs) with sign-changing weight. We apply the properties of characteristic determinant of the
NBVPs to show that the spectrum consists of real and simple eigenvalues; the number of positive eigenvalues is
equal to the number of positive elements in the weight function, and the number of negative eigenvalues
is equal to the number of negative elements in the weight function. We also show that the eigenfunction
corresponding to thejth positive/negative eigenvalue changes its sign exactly j-1 times.


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## Body

## 1. Introduction

Letn>2 be an integer, 𝕋={1,2,…,n}. Let us consider the discrete second-order linear Neumann eigenvalue problem
(1)Δ[p(t-1)Δu(t-1)]+λm(t)u(t)=0,22222222222222222222222222222t∈𝕋,(2)u(0)=u(1),u(n)=u(n+1),
where 𝕋={1,2,…,n}, p:{0,1,…,n}→[0,∞) and m:𝕋→ℝ satsifies(A0)
p(j)>0, j∈{1,…,n-1}, p(0)=p(n)=0;(A1)
m(t)≠0 on 𝕋 and m changes its sign on 𝕋, that is, there exists a proper subset 𝕋+⊂𝕋, such that
(3)m(t)>0,t∈𝕋+;m(t)<0,t∈𝕋∖𝕋+.Let n+ be the number of elements in 𝕋+ and let n- be the number of elements in 𝕋∖𝕋+. Then
(4)n++n-=n.When the weight functionm(t) is of one sign, Atkinson [1] and Jirari [2] studied the eigenvalue of the second-order problem
(5)Δ[p(t-1)Δy(t-1)]-q(t)y(t)+λm(t)y(t)=0,22222222222222222222222222222222222222t∈𝕋,(6)y(0)-αy(1)=y(n+1)-βy(n)=0,
and obtained that (5), (6) have n real eigenvalues, which can be ordered as λ1<λ2<⋯<λn. Here q(t)≥0 and α,β∈[0,1] is constant. It can be seen that if we take q(t)=0, α=β=1, then (5), (6) will convert to (1), (2).However, these two results do not give any information on the sign-changing of the eigenfunction of (5), (6).In 1991, Kelley and Peterson [3] considered the linear eigenvalue problems (5), (6) with α=β=0, where p(t)>0 on {0,1,…n}, q(t) is defined and real valued on 𝕋 and m(t)>0 on 𝕋. They obtained that (5), (6) have exactly n real and simple eigenvalues λk, k∈𝕋 which satisfies
(7)λ1<λ2<⋯<λn
and the eigenfunction corresponding to λk changes its sign exactly k-1 times.Furthermore, whenm(t)≡1, Agarwal et al. [4] generalized the above results to the dynamic equations with Sturm-Liouville boundary condition. Moreover, under the assumption that the weight functions are of one sign, for further important results in linear Hamiltonian difference systems, including the oscillation properties of solutions, one can see Shi and Chen [5], Bohner [6], and the references therein. The spectrum results for the continuous case have been studied and used to deal with several nonlinear problems; see, for example, [7–13] and the references therein.However, there are few results on the spectrum of discrete second-order linear eigenvalue problems whenm(t) changes its sign on 𝕋. In 2007, Ji and Yang [14, 15] studied the structure of the eigenvalues of (5), (6) with m(t) changing its sign, and they obtained the number of positive eigenvalues equal to the number of positive elements in the weight function, and the number of negative eigenvalues equals to the number of negative elements in the weight function. It is worth remaking that they provided no information on the distribution of these eigenvalues of (1), (2) and no information on the sign-changing of the corresponding eigenfunctions.Naturally, there are two interesting questions: (a) how to distribute of the eigenvalues of (1), (2) and (b) how the sign-changing of the corresponding eigenfunctions occur.It is the purpose of this paper to establish the structure of eigenvalues and the oscillatory properties of the corresponding eigenfunctions of (1), (2).The main result of our paper is the following theorem.Theorem 1.
Suppose that (A0), (A1) hold. Then one has the following.(i)
If∑t=1n‍m(t)>0, then (1), (2) have n real and simple eigenvalues, which can be ordered as follows:
(8)λn--<λn--1-<⋯<λ2-<λ1-<0=λ1+<λ2+<⋯<λn++.
Moreover, for k∈{1,…,n-}, the eigenfunction ψk- corresponding to the eigenvalue λk- has exactly k-1 simple generalized zeros; for k∈{1,…,n+}, the eigenfunction ψk+ corresponding to the eigenvalue λk+ has exactly k-1 simple generalized zeros.(ii)
If∑t=1n‍m(t)<0, then (1), (2) have n real and simple eigenvalues, which can be ordered as follows:
(9)λn--<λn--1-<⋯<λ2-<λ1-=0<λ1+<λ2+<⋯<λn++.
Moreover, for k∈{1,…,n-}, the eigenfunction ψk- corresponding to the eigenvalue λk- has exactly k-1 simple generalized zeros; for k∈{1,…,n+}, the eigenfunction ψk+ corresponding to the eigenvalue λk+ has exactly k-1 simple generalized zeros.(iii)
If∑t=1n‍m(t)=0, then λ=0 is an eigenvalue of (1), (2) and other n-2 eigenvalues are real and simple, which can be ordered as follows:
(10)λn--<λn--1-<⋯<λ2-<λ1-=0=λ1+<λ2+<⋯<λn++.
Moreover, for k∈{2,…,n-}, the eigenfunction ψk- corresponding to the eigenvalue λk- has exactly k-1 simple generalized zeros; for k∈{2,…,n+}, the eigenfunction ψk+ corresponding to the eigenvalue λk+ has exactly k-1 simple generalized zeros.Remark 2.
It is worth remarking that the number of sign changing of eigenfunction is given in Theorem1. Thus, this result is a generalization of the main results in [15].Remark 3.
Applying Theorem1 and the well-known Rabinowitz global bifurcation theorem, it is easy to obtain existence results of sign-changing solutions for the nonlinear analogue of (1)-(2); see Ma and Gao [12, 16] for some related results.The rest of the paper is devoted to proving Theorem1. To do this, we make use of the law of inertia for quadratic forms and some techniques form oscillation matrices [17].


## 2. Proof of the Main Result

Letc(t)=p(t-1)+p(t) for t=2,…,n-1, c(1)=p(1), c(n)=p(n-1). Then (1), (2) can be written as a linear pencil problem as follows:
(11)Ju=λDu,
where (12)J=(p(1)-p(1)0⋯000-p(1)p(1)+p(2)-p(2)⋯0000-p(2)p(2)+p(3)⋯000⋮⋮⋮⋱⋮⋮⋮000⋯-p(n-2)p(n-2)+p(n-1)-p(n-1)000⋯0-p(n-1)p(n-1)),D=(m10⋯000m2⋯00⋮⋮⋱⋮⋮00⋯mn-1000⋯0mn).LetJj denote the jth principal submatrix of J and Dj the jth principal submatrix of D. It is easy to verify that Ji is positive definite for i=1,…,n-1, J is positive semidefinite and
(13)detJ=0.In fact, for any real vectorx=(x1,x2,…,xn-1)∈ℝn-1, it follows that
(14)xJn-1xT=∑s=1n-2p(s)(xs+1-xs)2+p(n-1)xn-12≥0.
From (A0), p(s)>0, s∈{1,2,…,n-1}. If xJn-1xT=0, then xi=xi+1 for 1≤i≤n-1 and xn-1=0, leading to x=0. So, Jn-1 is positive definite. By the same method, with obvious changes, we can conclude that Ji is also positive definite for j=1,2,…,n-2.For any real vectory=(y1,y2,…,yn)∈ℝn, we have yJyT=∑s=1n-1p(s)(ys+1-ys)2≥0. Thus, J is positive semidefinite.Forj=1,2,…,n, let Qj(λ) denote the jth principal subdeterminant of J-λD and suppose that Q0(λ)=1; then Qn(λ)=det(J-λD), and
(15)Q0(λ)=1;Q1(λ)=c(1)-λm(1);Qj(λ)=(c(j)-λm(j))Qj-1(λ)-p2(j-1)Qj-2(λ),222222222222222222222222222222j=2,3,…,n-1;Qn(λ)=(c(n)-λm(n))Qn-1(λ)-p2(n-1)Qn-2(λ),
where c(j)=p(j-1)+p(j), j=2,…,n-1, c(0)=p(1) and c(n)=p(n-1).As we know, to find the eigenvalues of (1), (2) is equivalent to find the roots of Qn(λ). Thus, it is necessary to discuss some properties of the sequence (15).Forj∈{1,…,n}, let j+ be the number of the elements in {m(i)∣m(i)>0 for som i∈{1,…,j}}, and j- the number of the elements in {m(i)∣m(i)<0 for some  i∈{1,…,j}}.Lemma 4.
Forj∈{1,…,n}, one has
(16)limλ→-∞(-1)j-Qj(λ)=+∞,limλ→+∞(-1)j+Qj(λ)=+∞.Proof.
Forj∈{1,…,n}, it is evident that Qj(λ) is a polynomial of degree precisely j, and
(17)Qj(λ)=m(1)⋯m(j)(-λ)j+O(λj-1).Lemma 5.
The roots ofQj(λ) are real, j=1,2,…,n. Moreover, Qj(λ)=0(j=1,2,…,n-1) has j+ positive roots and j- negative roots.Proof.
SinceJj is positive definite matrix for j=1,2,…,n-1 and J is positive semidefinite matrix, it follows that the roots of Qj(λ) are real, j=1,2,…,n.
For theJj, j=1,2,…,n-1, there exists a unique lower triangular real matrix L such that
(18)LLT=Jj
(this is the well-known Cholesky decomposition; see [18, Corollary 7.2.9]). It is easy to check that the matrix L-1Dj(LT)-1 is real and symmetric, and λ is a root of Qj(λ) if and only if 1/λ is an eigenvalue of L-1Dj(LT)-1.
The fact thatL-1Dj(LT)-1 is real and symmetric indicates that there exists an orthogonal matrix Q such that
(19)QTL-1Dj(LT)-1Q=diag(a1,…,aj),
where a1≥a2≥⋯≥aj are all eigenvalues of L-1Dj(LT)-1. Let x=(LT)-1Qz. It is seen from (19) that
(20)∑i=1j‍aizi2=∑i=1j‍m(i)xi2
are two representations of the real quadratic form xTDjx. In view of the law of inertia for quadratic forms [19, Theorem 1, p. 297], we immediately deduce that the number of positive and the number of negative elements in the set {a1,…,aj} are j+ and j-, respectively.Lemma 6.
Two consecutive polynomialsQj-1(λ), Qj(λ) have no common zeros for j=1,…,n.Proof.
Suppose on the contrary that there existsλ=λ0 such that Qj-1(λ0)=Qj(λ0)=0. Then by the recurrence relation (15), we get Qj-2(λ0)=0. Furthermore, we can get Qj-3(λ0)=⋯=Q1(λ0)=Q0(λ0)=0. However, this contradicts Q0(λ0)=1.Lemma 7.
Suppose thatλ=λ0 is a root of Qj(λ). Then Qj-1(λ0)Qj+1(λ0)<0 for j=1,…,n-1.Proof.
SinceQj(λ0)=0, by the Lemma 6, we have Qj-1(λ0)≠0. By the recurrence relation (15), Qj+1(λ0)=-p2(j)Qj-1(λ0), which implies that Qj+1(λ0)Qj-1(λ0)=-p2(i)Qj-12(λ0)<0. This completes the assertion.Lemma 8.
Assume that (A0),  (A1) hold. Then(1)
Qn′(0)=-p(1)⋯p(n-1)∑s=1nm(s);(2)
If∑s=1nm(s)=0, then Qn′′(0)<0.Proof.
(1) From (11), we have (21)Qn′(λ)=d|Ju-λDu|dλ=|-m(1)0⋯00-p(1)p(1)+p(2)-λm(2)⋯000-p(2)⋯00⋮⋮⋱⋮⋮00⋯p(n-2)+p(n-1)-λm(n-1)-p(n-1)00⋯-pn-1pn-1-λmn|+|p(1)-λm(1)-p(1)⋯000-m(2)⋯000-p(2)⋯00⋮⋮⋱⋮⋮00⋯p(n-2)+p(n-1)-λm(n-1)-p(n-1)00⋯-p(n-1)p(n-1)-λm(n)|+⋯+|p(1)-λm(1)⋯00-p(1)⋯000⋯00⋮⋱⋮⋮0⋯p(n-2)+p(n-1)-λm(n-1)-p(n-1)0⋯0-m(n)|.
So, by simple computation, it follows that(22)Qn′(0)=d|Ju-λDu|dλ|λ=0=-m(1)p(1)p(2)⋯p(n-1)-m(2)p(1)p(2)⋯p(n-1)-⋯-m(n)p(1)p(2)⋯p(n-1)=-p(1)p(2)⋯p(n-1)∑s=1n‍m(s).
(2) If∑s=1nm(s)=0, then Qn(0)=Qn′(0)=0. Moreover, let j1,…,jn-2 be n-2 elements in {1,2,…,n-1} with
(23)j1<j2<⋯<jn-3<jn-2.
Let
(24)Λn={(j1,j2,…,jn-2)∣jk∈{1,2,…,n-1}2withj1<j2<⋯<jn-3<jn-2(j1,j2,…,jn-2)|}.
Let us arrange the elements of Λn in the increasing order,
(25)(1,2,…,n-3,n-2),(1,2,…,n-3,n-1),…,(2,3,…,n-2,n-1).
It is easy to check that the number of the elements on Λn is n-1. We denote the kth elements in (25) by βk, k=1,2,…,n-1. For given βk=(j1,j2,…,jn-2), define ak=Pj1Pj2⋯Pjn-2.
By computing and simplifying, we get that(26)Qn′′(λ)|λ=0=∑i,j=1n‍i<jm(i)m(j)Wn-2(…,i-1,i+1,…,j-1,j+1,…)=-2a1[m(1)+m(2)+⋯+m(n-1)]2-2a2[m(1)+m(2)+⋯+m(n-2)]2-⋯-2an-2[m(1)+m(2)]2-2an-1m(1)2=-2∑k=1n-1‍ak(∑s=1n-k‍m(s))2,
and here Wn-2(…,i-1,i+1,…,j-1,j+1,…) denote the principal minor determinant of order n-2 of J-λD, which is obtained by deleting the elements of i column and i row and deleting the elements of j column and j row. Thus, from (A0) and ∑s=1nm(s)=0, it follows that
(27)Qn′′(0)<0.Lemma 9.
Forj=1,…,n-1, the roots of Qj(λ)=0 are simple. Moreover, one has the following.(i)
The largest negative rootλj,1- and the smallest positive root λj,1+ of Qj(λ)=0 and the largest negative root λj+1,1- and the smallest positive root λj+1,1+ of  Qj+1(λ)=0 satisfy
(28)(λj,1-,λj,1+)⊃(λj+1,1-,λj+1,1+),j=2,…,n.(ii)
Forj=1,…,n-1, the positive roots of Qj(λ)=0 and Qj+1(λ)=0 separate one another; the negative roots of Qj(λ)=0 and Qj+1(λ)=0 separate one another.(iii)
If∑s=1n‍m(s)≠0, then the roots of Qn(λ)=0 are simple and 0 is a simple root; if ∑s=1n‍m(s)=0, then 0 is a double roots of Qn(λ)=0 and the other root are simple.Proof.
First, we deal with the casej=1.
Obviously,Q1(λ)=c(1)-λm(1). If m(1)>0, then
(29)j=1,j+=1,j-=0,λ1,1+=c(1)m(1)>0.
If m(1)<0, then
(30)j=1,j+=0,j-=1,λ1,1-=c(1)m(1)<0.
RecallQ2(λ)=(c(2)-λm(2))(c(1)-λm(1))-p2(1). If m(1)>0, m(2)>0, then
(31)j=2,j+=2,j-=0,
and Q2(λ)=0 has two different roots λ2,1+ and λ2,2+ as follows:
(32)λ2,1+=(c(1)m(2)+c(2)m(1)(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2)2-(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2))×(2m(1)m(2))-1>0,λ2,2+=(c(1)m(2)+c(2)m(1)(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2)2+(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2))×(2m(1)m(2))-1>0.
It is easy to see that 0<λ2,1+<λ1,1+<λ2,2+.
Ifm(1)<0, m(2)>0, then
(33)j=2,j+=1,j-=1,
and Q2(λ)=0 has two different roots λ2,1+ and λ2,1- as follows:
(34)λ2,1+=((c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2)c(1)m(2)+c(2)m(1)2-(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2))×(2m(1)m(2))-1>0,λ2,1-=(c(1)m(2)+c(2)m(1)(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2)2+(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2))×(2m(1)m(2))-1<0.
It is easy to see that λ1,1-<λ2,1-<0<λ2,1+.
Ifm(1)>0, m(2)<0, then
(35)j=2,j+=1,j-=1,
and Q2(λ)=0 has two different roots λ2,1+ and λ2,1- as follows:
(36)λ2,1+=(c(1)m(2)+c(2)m(1)(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2)2-(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2))×(2m(1)m(2))-1>0,λ2,1-=(c(1)m(2)+c(2)m(1)(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2)2+(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2))×(2m(1)m(2))-1<0.
It is easy to see that λ1,1-<λ2,1-<0<λ2,1+.
Ifm(1)<0, m(2)<0, then
(37)j=2,j+=0,j-=2,
and Q2(λ)=0 has two different roots λ2,1- and λ2,2- as follows:
(38)λ2,1-=(c(1)m(2)+c(2)m(1)(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2)2+(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2))×(2m(1)m(2))-1<0,λ2,2-=(c(1)m(2)+c(2)m(1)(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2)2-(c(1)m(2)-c(2)m(1))2+4p2(1)m(1)m(2))×(2m(1)m(2))-1<0.
It is easy to see that λ2,2-<λ1,1-<λ2,1-<0. Thus, the assertion is true for j=1.
Second, suppose that forj=k(k≤n-3), and the relations of Qk(λ)=0 and Qk+1(λ)=0 are true, that is, the following two assertions hold.
Ifm(k+1)>0, then (k+1)+=k++1, (k+1)-=k- and accordingly,
(39)λk,k--<λk+1,k--<⋯<λk,1-<λk+1,1-<0,(40)0<λk+1,1+<λk,1+<⋯<λk+1,k++<λk,k++<λk+1,(k+1)++.
Ifm(k+1)<0, then (k+1)+=k+, (k+1)-=k-+1 and accordingly,
(41)λk+1,k-+1-<λk,k--<λk+1,k--<⋯<λk,1-<λk+1,1-<0,0<λk+1,1+<λk,1+<⋯<λk+1,k++<λk,k++.
Now, we consider the casej=k+1. It is enough to verify the following four cases.
Case
1. m(k+1)>0 and m(k+2)>0.
In this case,(k+2)+=k++2, (k+2)-=k-, we need to prove that
(42)λk+1,k--<λk+2,k--<⋯<λk+1,1-<λk+2,1-<0,(43)0<λk+2,1+<λk+1,1+<⋯<λk+2,k++1+<λk+1,k++1+<λk+2,k++2+.
Case
2. m(k+1)>0 and m(k+2)<0.
In this case,(k+2)+=k++1, (k+2)-=k-+1, we need to prove that
(44)λk+2,k-+1-<λk+1,k--<λk+2,k--<⋯<λk+1,1-<λk+2,1-<0,0<λk+2,1+<λk+1,1+<⋯<λk+2,k++1+<λk+1,k++1+.
Case
3.
m
(
k
+
1
)
<
0 and m(k+2)<0.
In this case,(k+2)+=k+, (k+2)-=k-+2, we need to prove that
(45)λk+2,k-+2-<λk+1,k-+1-<λk+2,k-+1-<⋯<λk+1,1-<λk+2,1-<0,0<λk+2,1+<λk+1,1+<⋯<λk+2,k++<λk+1,k++.
Case
4. m(k+1)<0 and m(k+2)>0.
In this case,(k+2)+=k++1, (k+2)-=k-+1, we need to prove that
(46)λk+2,k-+1-<λk+1,k--<λk+2,k--<⋯<λk+1,1-<λk+2,1-<0,0<λk+2,1+<λk+1,1+<⋯<λk+2,k++<λk+1,k++<λk+2,k++1+.
We only deal with Case 1. The other cases could be dealt via the same method.
First, we show that (42) holds.
Since(k+2)-=(k+1)-=k-, it follows from Lemma 4 that
(47)(-1)k-Qk(-∞)>0,(-1)(k+1)-Qk+1(-∞)=(-1)k-Qk+1(-∞)>0.
We only deal with the case thatk- is even. The case k- is odd could be treated by the same way.
Thus (47) reduces to
(48)Qk(-∞)>0,Qk+1(-∞)>0.
Recall (39) as follows:
(49)λk,k--<λk+1,k--<λk,k--1-<λk+1,k--1-<⋯<λk,1-<λk+1,1-<0
and the fact that
(50)Qk+1(λk+1,k--)=Qk+1(λk+1,k--1-)=⋯=Qk+1(λk+1,1-)=0.
It follows from (47), (48), and (49) that
(51)(-1)jQk(λk+1,k--j-)<0,j=0,…,k--1.
Combining this with (50) and using Lemma 7, it concludes that
(52)(-1)jQk+2(λk+1,k--j-)>0,j=0,…,k--1.
In particular,
(53)Qk+2(λk+1,1-)<0.
By Lemma 5, Qk+2(λ)=0 has exactly k- zeros in (-∞,0). This together with (52), (53), and the fact that
(54)Qk(0)>0,Qk+1(0)>0,Qk+2(0)>0
implies that there exist λk+2,k--j-∈(λk+1,k--j-,λk+1,k--j-1-), j=0,…,k--2, and λk+2,1-∈(λk+1,1-,0), such that
(55)Qk+2(λk+2,k--j-)=0,j=0,…,k--1.
Therefore, (42) is valid.
Next, we show that (43) is true.
Obviously,m(k+1)>0, m(k+2)>0 yields
(56)(k+2)+=(k+1)++1=k++2.
In the following, we only deal with the case that k+ is even. The case k+ is odd could be treated by the same way.
From Lemma4, we have that
(57)Qk(+∞)>0,Qk+1(+∞)<0,Qk+2(+∞)>0.
Combining this with (40) as
(58)0<λk+1,1+<λk,1+<⋯<λk+1,k++<λk,k++<λk+1,(k+1)++,
and using the fact that
(59)Q(λk+1,(k+1)+-j+)=0,j=0,…,(k+1)+-1,
it concludes that
(60)(-1)jQk(λk+1,(k+1)+-j+)>0,j=0,…,(k+1)+-1.
This together with Lemma 7 implies that
(61)(-1)jQk+2(λk+1,(k+1)+-j+)<0,j=0,…,(k+1)+-1.
In particular, for j=0,
(62)Qk+2(λk+1,(k+1)++)<0.
This together with the third inequality in (57) implies that
(63)Qk+2(λk+2,(k+1)++1)=0
for some λk+2,(k+1)++1∈(λk+1,(k+1)++,∞).
Using (61) with j=(k+1)+-1, we get
(64)Qk+2(λk+1,1+)<0,
which together with the fact
(65)Qk+2(0)>0
implies that
(66)Qk+2(λk+2,1+)=0
for some λk+2,1+∈(0,λk+1,1+).
Now, forj=1,…,(k+1)+-1, there exist λk+2,(k+2)+-j+∈(λk+1,(k+1)+-j+,λk+1,(k+1)+-j+1+), such that
(67)Qk+2(λk+2,(k+2)+-j+)=0,j=0,…,(k+1)+.
Therefore, (43) is valid.
Finally, forj=n, the relation Qn(λ)=0 and Qn-1(λ)=0 are also true. From above conclusions, we have Qn-1(λ)=0 has (n-1)- negative roots and (n-1)+ positive roots satisfying
(68)λn-1,(n-1)--<λn-1,(n-1)--1-<⋯<λn-1,1-<0<λn-1,1+<λn-1,2+<⋯<λn-1,(n-1)++.
Ifm(n)>0, we have that n+=(n-1)++1,n-=(n-1)-. By a similar argument and together with the fact Qn(0)=0 and Lemma 8, it follows that(i)
if∑s=0nm(s)<0, then
(69)λn-1,(n-1)--<λn,n--<⋯<λn-1,1-<λn,1-=0<λn,1+<λn-1,1+<⋯<λn-1,(n-1)++<λn,n++;(ii)
if∑s=0nm(s)>0, then
(70)λn-1,(n-1)--<λn,n--<⋯<λn-1,1-<λn,1-<0=λn,1+<λn-1,1+<λn,2+<⋯<λn-1,(n-1)++<λn,n++;(iii)
if∑s=0nm(s)=0, then
(71)λn-1,(n-1)--<λn,n--<⋯<λn-1,1-<λn,1-=0=λn,1-<λn-1,1+<λn,2+<⋯<λn-1,(n-1)++<λn,n++.
Ifm(n)<0, we have that n+=(n-1)+, n-=(n-1)-+1. By a similar method and together with Qn(0)=0 and Lemma 8, we get that(i)
if∑s=0nm(s)<0, then
(72)λn,n--<λn-1,(n-1)--<⋯<λn-1,1-<λn,1-=0<λn,1+<λn-1,1+<⋯<λn,n++<λn-1,(n-1)++;(ii)
if∑s=0nm(s)>0, then
(73)λn,n--<λn-1,(n-1)--<⋯<λn-1,1-<λn,1-<0=λn,1+<λn-1,1+<λn,2+<⋯<λn,n++<λn-1,(n-1)++;(iii)
if∑s=0nm(s)=0, then
(74)λn,n--<λn-1,(n-1)--<⋯<λn,2-<λn-1,1-<λn,1-=0=λn,1+<λn-1,1+<⋯<λn,n++<λn-1,(n-1)++.Thus the proof is complete.Lemma 10.
Letw(λ) be the number of sign changes in the sequence (15). Then for i∈{1,…,n+},
(75)limλ→(λn,i+)-w(λ)=i-1,limλ→(λn,i+)+w(λ)=i,
where λ→(a)- means that λ→a in (-∞,a) and λ→(a)+ means that λ→a in (a,∞).Proof.
It is motivated by the proof of Strums Theorem; see [20, Theorem 1.4.3] and its proof.
The idea of the proof is to follow the changes inw as λ passes through the interval [0,n]. In particular, we will show that w is a monotonically increasing function and that each root of Qn and only a root of Qn make w jump by 1.
IfQj(λ^)=0 for some j∈{1,…,n-1}, then for Qj-1,Qj,Qj+1 we have from Lemma 7 that Qj-1 and Qj+1 have opposite, but constant signs, since Qj-1 and Qj+1 cannot be zero in a sufficiently small neighborhood U(λ^) and thus cannot change sign. Hence, whatever the sign of Qj is in U(λ^), it does not change the overall sign change count (to see this, note that Qj-1 and Qj+1 have opposite signs, and hence if the sign sequence before is +--, it is after ++- and the number of sign changes remains the same. The same for the other cases). In other word, w(λ) stays constant when λ passing through a root of Qj from some j∈{1,…,n-1}.
It is easy to see from Lemma9 that
(76)sgnQn-1(λn,i+)=(-1)i-1,i∈{1,…,n+}.
Next, we show that each root ofQn and only a root of Qn make w jump by 1.
In fact, fori=1, Qn-1(λn,1+)>0, which implies that there exists a neighborhood U(λn,1+) of λn,1+, such that
(77)Qn-1(λ)>0,λ∈U(λn,1+).
From the definition of λn,1+,
(78)Qn(λ)>0,λ∈[0,λn,1+).
The chain of signs switches from “⋯++” to “⋯+-” when passing through λn,1+, so w increases by 1.
Fori=2,
(79)Qn-1(λn,2+)<0,Qn(λ)<0,λ∈(λn,1+,λn,2+).
The chain of signs switches from “⋯--” to “⋯-+” when passing through λn,2+, so w increases by 1.
Repeating the above argument, we may deduce that(80)limλ→(λn,i+)-w(λ)=i-1,limλ→(λn,i+)+w(λ)=i.
This completes the proof.Lemma 11.
Ifu(·,λ) satisfies (1), (2) with u(1,λ)=1, then
(81)u(k+1,λ)=p(1)⋯p(k)Qk(λ),k=1,…,n-1.Proof.
Let(82)u=(u(0),u(1),u(2),…,u(n),u(n+1))T
be a solution of (1), (2). Then
(83)(c(1)-λm(1))u(1)-p(1)u(2)=0,-p(1)u(1)+(c(2)-λm(2))u(2)-p(2)u(3)=0,⋯-p(n-2)u(n-2)+(c(n-1)-λm(n-1))u(n-1)-p(n-1)u(n)=0,-p(n-1)u(n-1)+(c(n)-λm(n))u(n)=0.
Clearly, (83) is equivalent to
(84)-p(k-1)u(k-1)+(p(k-1)+p(k)-λm(k))u(k)-p(k)u(k+1)=0,k=1,…,n,
with u(0) and u(n+1) determined by (2).
Let(85)v0=u(0),v1=u(1),vk=p(1)⋯p(k-1)u(k),2222222222222222222222222222222222k=2,…n.
Then
(86)v1=u(1,λ)=1=Q0(λ),v2=Q1(λ),vk+1=(c(k)-λm(k))vk-p2(k-1)vk-1,222222222222222k=2,3,…n-1.
Obviously, vk+1 and Qk(λ) satisfy the same recurrence formula (15), and it follows that
(87)vk+1=Qk(λ),k=1,…,n-1,
and accordingly,
(88)Qk(λ)=p(1)p(2)⋯p(k)u(k+1,λ),k=1,…,n-1.Proof of Theorem1.
From Lemmas5 and 9, we can obtain the following consequence.(i)
If∑t=1n‍m(t)>0, then (1), (2) has n real and simple eigenvalues, which can be ordered as follows:
(89)λn--<λn--1-<⋯<λ2-<λ1-<0=λ1+<λ2+<⋯<λn++.(ii)
If∑t=1n‍m(t)<0, then (1), (2) has n real and simple eigenvalues, which can be ordered as follows:
(90)λn--<λn--1-<⋯<λ2-<λ1-=0<λ1+<λ2+<⋯<λn++.(iii)
If∑t=1n‍m(t)=0, then λ=0 is an eigenvalue of (1), (2) and other n-2 eigenvalues are real and simple, which can be ordered as follows:
(91)λn--<λn--1-<⋯<λ2-<λ1-=0=λ1+<λ2+<⋯<λn++.
Now, we consider the numbers of sign changing of eigenfunction. From Lemma11, we may determine that the number of sign changing of
(92){u(0),u(1),…,u(n),u(n+1)}
via that of
(93){Q1(λn,i+),…,Qn-1(λn,i+)}
since (2) implies that u(0)u(1)≥0 and u(n)u(n+1)≥0.
Letw^(λ) be the number of sign changes in the sequence
(94)Q0(λ)=1;Q1(λ)=c(1)-λm(1);Qj(λ)=(c(j)-λm(j))Qj-1(λ)-p2(j-1)Qj-2(λ),2222222222222222222222222222222j=2,3,…,n-1;Qn-1(λ)=(c(n-1)-λm(n-1))Qn-1(λ)=×Qn-2(λ)-p2(n-2)Qn-3(λ).
Using the same method to prove Lemma 10, with obvious changes, we may obtain that for i∈{2,…,(n-1)+},
(95)limλ→λn-1,i-1++0w^(λ)=i-1,limλ→λn-1,i+-0w^(λ)=i-1.
Thus, for i∈{2,…,(n-1)+}, Lemma 9 yields
(96)λn-1,i-1+<λn,i+<λn-1,i+.
This together with (95) and the fact that w^(λ) is nondecreasing in (0,∞) implies
(97)w^(λn,i+)=w^(λn-1,i-1++λn-1,i+2)=i-1,
and accordingly
(98)w(λn,i+)=w^(λn,i+)=i-1.
For the case i=1, since Qj(0)>0 for j∈{0,1,…,n-1}, it follows that
(99)limλ→0+w^(λ)=0.
This together with the facts 0≤λn,1+<λn-1,1+ and
(100)limλ→(λn-1,1+)-w^(λ)=0
implies that
(101)w(λn,1+)=w^(λn,1+)=0.
Ifn+=(n-1)+, then it has been done! If n+=(n-1)++1, then by the same method, with obvious changes, we may get
(102)w(λn,n++)=w^(λn,n++)=n+-1.
Finally, by using the above method, with obvious changes, we may prove that the number of sign changesψn,i- is i-1.Remark 12.
If one extra termq(t)y(t) is present in (1), then λ=0 is generally not an eigenvalue of
(103)Δ[p(t-1)Δu(t-1)]-q(t)u(t)+λm(t)u(t)=0,22222222222222222222222222222222222222t∈𝕋,u(0)=u(1),u(n)=u(n+1),
anymore. Recall the fact that λ=0 is essential in Lemma 8 and its proof. Therefore, the spectrum structure of the more general problem (103) is still open.



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*Source: 280508-2013-08-29.xml*