# An Algebraic Criterion of Zero Solutions of Some Dynamic Systems

Authors: Ying Wang; Baodong Zheng; Chunrui Zhang
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We establish some algebraic results on the zeros of some exponential polynomials and a real coefficient polynomial. Based on the basic theorem, we develop a decomposition technique to investigate the stability of two coupled systems and their discrete versions, that is, to find conditions under which all zeros of the exponential polynomials have negative real parts and the moduli of all roots of a real coefficient polynomial are less than 1.


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## Body

## 1. Introduction

For an ordinary (delay) differential equation, the trivial solution is asymptotically stable if and only if all roots of the corresponding characteristic equation of the linearized system have negative real parts while the moduli of all roots of a real coefficient polynomial less than 1 mean the trivial solution is asymptotically stable for the difference equation. However, it is difficult to obtain the expression of the characteristic equation corresponding to the linearized systems. Special cases of the characteristic equation have been discussed by many authors. For example, Bellman and Cooke [1], Boese [2], Kuang [3], and Ruan and Wei [4–8] studied some exponential polynomials and used the results to investigate the stability and bifurcations for some systems. The well-known Jury criterion can be used to determine the moduli of the roots of a real coefficient polynomial less than one [9, 10], but the calculation is prolixly.The purpose of this paper is to provide a new algebraic criterion of zero for some exponential polynomials and a real coefficient polynomial.


## 2. Some Algebraic Results

LetV be a linear space over a number field F and W a subspace of V. A complement space of W in V is a linear space U of V such that V=W⊕U. A vector γ is said to be the projection of a vector α along W at U if α=β+γ, where β∈W and γ∈U.Let𝒜 be a linear transformation on V and W a subspace of V. A subspace W is said to be a 𝒜 invariant subspace if 𝒜α∈W for any α∈W. Let U be a k-complement space of 𝒜 for W, if(1)
Wis a 𝒜 invariant subspace;(2)
U is a complement space of W in V, that is, V=W⊕U;(3)
for anyα∈U, the projection of (𝒜-kIV)α along W at U is always 0.Let𝒜, ℬ be two linear transformations on V. We consider that 𝒜 and ℬ are (k,l) concordant if there exist a nontrivial invariant subspaces W of 𝒜 and ℬ, and a complement space U of W in V, such that(1)
at least one of constraints of𝒜,ℬ on W is a scalar transformation lIW;(2)
U is a k-complement space of 𝒜 or ℬ for W.LetA,B∈Rn×n. We say that A and B are (k,l) concordant if linear transformations 𝒜 and ℬ of Rn𝒜:Rn→Rn,X↦AX,  for all X∈Rn,ℬ:Rn→Rn,X↦BX,  for all X∈Rnare concordant.For example,In and a reducible matrix An×n are always in concordance.Theorem 2.1.
LetA,B∈Rn×n be (k,l) concordant. Then there exist an invertible matrix P, such that
(2.1)PAP-1=(A11A12OkIn2),PBP-1=(lIn1B12OB22),
or
(2.2)PAP-1=(lIn1A12OkIn2),PBP-1=(B11B12OB22),
or
(2.3)PAP-1=(A11A12OA22),PBP-1=(lIn1B12OkIn2),
or
(2.4)PAP-1=(lIn1A12OA22),PBP-1=(B11B12OkIn2),
where n1>0,n2>0,n1+n2=n.Proof.
Letα1,α2,…,αr be the base of W and αr+1,…,αn of U. If the constraints of 𝒜 on W are lIW and U is a k-complement space of 𝒜 for W, then Aαi=lαi,i=1,…,r and there exists a1j,…,arj making Aαj-kαj=a1jα1+⋯+arjαr, that is, Aαj=a1jα1+⋯+arjαr+kαj,j=r+1,…,n. So
(2.5)A(α1,…,αr,αr+1,…,αn)=(α1,…,αr,αr+1,…,αn)(l⋯0a1,r+1⋯a1n⋮⋱⋮⋮⋮0⋯lar,r+1⋯arn0k⋯0⋱⋮k).
Because a subspace W is a ℬ invariant subspace, we can write Bαi=b1iα1+⋯+briαr, i=1,…,r, Bαi=b1jα1+⋯+brjαr+⋯+bnjαn,j=r+1,…,n. So
(2.6)B(α1,…,αr,αr+1,…,αn)=(α1,…,αr,αr+1,…,αn)(b11⋯b1rb1,r+1⋯b1n⋮⋱⋮⋮⋮br1⋯brrbr,r+1⋯brn0br+1,r+1⋯br+1,n⋮⋱⋮bn,r+1⋯bnn).
If the constraint of 𝒜 on W is lIW and U is a k-complement space of ℬ for W, then Bαi=lαi,i=1,…,r. Aαj=a1jα1+⋯+arjαr+ar+1,jαr+1+⋯+anjαn,j=r+1,…,n. So
(2.7)A(α1,…,αr,αr+1,…,αn)=(α1,…,αr,αr+1,…,αn)(l⋯0a1,r+1⋯a1n⋮⋱⋮⋮⋮0⋯lar,r+1⋯arn0ar+1,r+1⋯ar+1,n⋮⋱⋮an,r+1⋯ann).
Because a subspace W is a ℬ invariant subspace, we can write Bαi=b1iα1+⋯+briαr, i=1,…,r. Because αj∈U (j=r+1,…,n), Bαj-kαj∈W, then there exists b1j,…,brj making Bαj=b1jα1+⋯+brjαr+kαj,j=r+1,…,n. So
(2.8)B(α1,…,αr,αr+1,…,αn)=(α1,…,αrαr+1,…,αn)(b11⋯b1rb1,r+1⋯b1,n⋮⋱⋮⋮⋮br1⋯brrbr,r+1⋯br,n0k⋯0⋱⋮k).
Other situations are similar.


## 3. Algebraic Criterion of Zero for Some Exponential Polynomials

From Section2, we have the following theorem on the roots of exponential polynomial.Theorem 3.1.
LetA,B∈Rn be (k,l) concordant, τ a constant, τ>0, f(λ) a polynomial about λ. If λ0 is a root of
(3.1)|f(λ)In-A-Be-λτ|=0,
then
(3.2)u=f(λ0)-ae-λ0τ
is an eigenvalue of A, or
(3.3)v=(f(λ0)-a)eλ0τ
is an eigenvalue of B, where a=k or l.Proof.
Without loss of generality, assume that(3.4)PAP-1=(A11A12OkIn2),PBP-1=(lIn1B12OB22).
Because λ0 is a root of
(3.5)|f(λ)In-A-Be-λτ|=0,|f(λ0)In1-A11-lIn1e-λ0τ*Of(λ0)In2-kIn2-B22e-λ0τ|=0,
then
(3.6)|f(λ0)In1-A11-lIn1e-λ0τ|=0or|f(λ0)In2-kIn2-B22e-λ0τ|=0.If|f(λ0)In1-A11-lIn1e-λ0τ|=0, then u=f(λ0)-le-λ0τ is an eigenvalue of A.If|f(λ0)In2-kIn2-B22e-λ0τ|=0, then v=(f(λ0)-k)eλ0τ is an eigenvalue of B.
As a application of Theorem3.1, consider a BAM neural network model with delays:
(3.7)X˙=-aX+f(Y(t-τ1)),Y˙=-bY+g(X(t-τ2)),
where X∈Rn,Y∈Rn,a>0,b>0. Assume that f,g∈C1, and f(0)=0,g(0)=0.
Under the hypothesis, the originO is an equilibrium of (3.7), and the linearization of system (3.7) at the origin O is
(3.8)X˙=-aX+B1Y(t-τ1),Y˙=-bY+B2X(t-τ2),
where B1,B2 are Jacobi matrices.
The associated characteristic equation of (3.8) is
(3.9)|λIn+aIn-B1e-λτ1-B2e-λτ2λIn+aIn|=0.
Since λ=-a,λ=-b have no influence on the stability of system (3.7), then let λ≠-a,λ≠-b. We have
(3.10)|(λ+a)(λ+b)In-e-λ(τ1+τ2)B1B2|=0.
Letu1,u2,…,ur be eigenvalues of B1B2. From Theorem 3.1, we have
(3.11)(λ+a)(λ+b)-uke-λτ=0,(k=1,2,…,n),
where τ=τ1+τ2, that is,
(3.12)λ2+(a+b)λ+ab-uke-λτ=0.
Consider
(3.13)λ2+(a+b)λ+ab-uke-λτ=0,uk=ck+dki.Lemma 3.2.
Ifmax1≤k≤n|uk|>ab, then (3.12) has a root iw0(k)(w0>0), and
(3.14)w0(k)=12[-(a2+b2)+((a2-b2)2+4|uk|)]1/2,τ0(k)are determined by
(3.15)a+b[w0k]2=ckcosw0(k)τ-dksinw0(k)τ,[w0k]2a+b=cksinw0(k)τ-dkcosw0(k)τ.Proof.
Letiw(w>0) be a root of (3.13), then
(3.16)-w2+i(a+b)w+ab-(ck+dki)(coswτ+isinwτ)=0.
Separating the real and imaging parts, the roots can be obtained.Lemma 3.3.
Re(dλ/dτ)|τ=τj(k),w=w0(k)>0.Proof.
Differentiating both sides of (3.13) with respect to τ gives
(3.17)(dλdτ)-1=2λ+a+b-uke-λτukλe-λτ,
that is,
(3.18)Re(dλdτ)-1(dλdτ)-1|τ=τj(k),w=w0(k)=[w0(k)]2Δ[a2+b2+2[w0(k)]2]>0,
where Δ=(a+b)2[w0(k)]2+[w0(k)]2[[w0(k)]2-ab]2.Theorem 3.4.
Letτ0=min{τ0(1),τ0(2),…,τ0(n)}.(1)
Ifmax1≤k≤n|uk|<ab, then the zero solution of (3.7) is absolutely stable.(2)
Ifmin1≤k≤n|uk|≥ab,max1≤k≤nReuk<ab, then the zero solution of (3.7) is asymptotically stable when τ∈[0,τ0) and unstable when τ>τ0, and (3.7) undergoes a Hopf bifurcation at the origin 0 when τ=τj(k),j=0,1,…,k=1,2,…,n, where τ=τj(k) is defined in Lemma 3.2.(3)
Ifmin1≤k≤nReuk≥ab, then the zero solution of (3.7) is unstable for all τ≥0.Proof.
Forτ=0, (3.13) becomes
(3.19)λ2+(a+b)λ+ab-(ck+dki)=0.
Letλ=α+iβ be a root of (3.19). Separating the real and imaginary parts, we can obtain
(3.20)α2-β2+(a+b)α+ab-ck=0,2αβ+(a+b)β-dk=0.
Hence, α=(-(a+b)±(a+b)2-4(ab-ck))/2. If ck<0, then α<0. Using Lemmas 3.2 and 3.3 the conclusions follow.


## 4. Algebraic Criterion of Zero for Some Real Polynomial

Theorem 4.1.
LetA,B be (k,l) concordant. If f(z) be a polynomial about z, z0 is a root of
(4.1)|zm[f(z)I-A]-B|=0,
then
(4.2)u=f(z0)-az0-m
is an eigenvalue of A, or
(4.3)v=z0m[f(z0)-a]
is an eigenvalue of B, where a=k or l.Proof.
Let us assume that(4.4)PAP-1=(A11A12OkIn2),PBP-1=(lIn1B12OB22).
Because z0 is a root of
(4.5)|zm[f(z)I-A]-B|=0,|z0m[f(z0)In1-A11]-lIn1*Oz0m[f(z0)In2-kIn2]-B22|=0,
then
(4.6)|z0m[f(z0)In1-A11]-lIn1|=0or|z0m[f(z0)In2-kIn2]-B22|=0.If|z0m[f(z0)In1-A11]-lIn1|=0, then u=f(z0)-az0-m is an eigenvalue of A.If|z0m[f(z0)In2-kIn2]-B22|=0, then v=z0m[f(z0)-a] is an eigenvalue of B.
For the application of Theorem4.1, consider the discretization of BAM neural network (3.7). Let U(t)=X(t-τ2),V(t)=Y(t),τ=(τ1+τ2), then (3.7) can be rewritten as
(4.7)U˙(t)=-aU(t)+f(V(t-τ)),V˙(t)=-bV(t)+g(U(t)).
Let M(t)=U(tτ),N(t)=V(tτ), then (4.7) can be rewritten as
(4.8)M˙(t)=-aτM(t)+τf(N(t-1)),N˙(t)=-bτN(t)+τg(M(t)).
Let h=1/m,m∈N+, using an Euler method to (4.8), we obtain
(4.9)Mn+1=Mn-haτMn+hτf(Nn-m),Nn+1=Nn-hbτNn+hτg(Mn).
Let Zn=(MnT,NnT,Nn-1T,…,Nn-mT). Using the notation, (4.9) can be expressed as
(4.10)Zn+1=F(Zn,τ),
where F=(F-1T,F0T,F1T,…,FmT) and
(4.11)Fk={(1-haτ)Mn+hτf(Nn-m),k=-1,(1-hbτ)Nn+hτg(Mn),k=0,Nn-k,1≤k≤m.
It is clear that 0 is also an equilibrium of (4.10). The linearization of (4.10) at the origin 0 is
(4.12)Zn+1=A~(τ)zn,(4.13)A~(τ)=((1-haτ)In00⋯0hτB1hτB2(1-hbτ)In0⋯000In0⋯00⋯⋯⋯⋯⋯⋯000⋯In0).
The characteristic equation of (4.12) is
(4.14)|zI-A~(τ)|=0.
Since |1-haτ|<1,|1-hbτ|<1, hence there is no influence on the stability of (4.10). Let z≠1-haτ,z≠1-hbτ, then (4.14) can be rewritten as
(4.15)|zm(z-1+haτ)zm(z-1+hbτ)In-(hτ)2B1B2|=0.
Let uk(1≤k≤n) be an eigenvalue of B1B2. Using Theorem 4.1, we have
(4.16)[zm(z-1+haτ)][zm(z-1+hbτ)]Ir-(hτ)2uk=0.
It is clear that the stability of system (4.10) is determined by the distribution of the roots of (4.16). Next, we consider a special case of system (3.7)
(4.17)X˙=-aX+f(X(t-τ)),
where X∈Rn, which is a Hopfiled neural network with delay. Its discrete system is
(4.18)Mn+1=(1-haτ)Mn+hτf(Mn-m).
The linear part is
(4.19)Mn+1=(1-haτ)Mn+hτBMn-m.
It is clear that we only need to discuss the distribution of the roots
(4.20)zm(z-1+haτ)=hτuk,
where uk(1≤k≤n) is an eigenvalue of B.Lemma 4.2.
Letmax1≤k≤nReuk<a. There exists a τ~>0 such that for 0<τ<τ~, all roots of (4.20) have moduli less than one.Proof.
Whenτ=0, (4.20) has an m-fold root z=0 and a simple root z=1 at τ=0. Consider the root z(t) such that z(0)=1. This root depends continuously on τ and is a differential function of τ:
(4.21)d|z|2dτ|z=1,τ=0=(zdz~dτ+z~dzdτ)|z=1,τ=0=2(-a+Reuk)h<0.
Hence, |z(t)|<1 for all sufficiently small τ>0. Thus, all roots of (4.20) lie in |z(t)|<1 for sufficiently small positive τ and existence of the maximal τ~ follows.Lemma 4.3.
Assume that the seep size h is sufficiently small. Letmax1≤k≤n|uk|<1, then there are no roots of (4.20) with moduli one for all τ>0.Proof.
Letz=eiw*/m be a root of (4.20) when τ=τ*, then
(4.22)cosm+1mw*-(1-haτk*)cosw*=hτk*ck,sinm+1mw*-(1-hbτk*)sinw*=hτk*dk,
where uk=ck+idk, hence
(4.23)cosw*m=1+(a2-|uk|2)h2(τ*)22(1-haτ*).
Let |uk|<a, for sufficiently small h>0, |cos(w*/m)|>1, which yields a contradiction. The proof is complete.Lemma 4.4.
If the seep sizeh is sufficiently small, then
(4.24)d|z|2dτ|τ=τk*,w=w*>0,
where τ*, w*satisfy (4.22).Proof.
Consider the following equation:(4.25)d|z|2dτ|τ=τk*,w=w*=(zdz~dτ+z~dzdτ)|τ=τk*,w=w*=2τ×(m+1)(1-cos(w*/m))+m(1-τk*ha)(1-cos(w*/m))(m+1)2+(1-haτk*)2+2m(m+1)(1-haτk*)cos(w*/m)>0.Using Lemmas 4.1–4.4 and Theorem 1 of [11], we have the following results.Theorem 4.5.
(1) Ifmax1≤k≤n|uk|<a, then the zero solution of (4.10) is unstable for τ≥0.
(2) Ifmin1≤k≤n|uk|≥a, max1≤k≤nuk<a, then the zero solution of (4.18) is asymptotically stable when τ∈(0,τ*), where τ*=min{τ1*,τ2*,…,τn*}, and (4.10) undergoes a Naimark-Sacker bifurcation.
(3) Ifmin1≤k≤nReuk≥a, then the zero solution of (4.10) is unstable for all τ≥0.


## 5. Computer Simulation

To illustrate the analytical results found, let us consider the following particular case of (3.7):
(5.1)x˙1=-x1+2tanh(y2(t-τ1)),x˙2=-x2+2tanh(y3(t-τ1)),x˙3=-x3+2tanh(y1(t-τ1)),y˙1=-2y1-1.5tanh(x2(t-τ2)),y˙2=-2y2-1.5tanh(x3(t-τ2)),y˙3=-2y3-1.5tanh(x1(t-τ2)),
where τ1,τ2 are parameters.Using the conclusions of Theorem3.4, the phase-locked periodic solutions appear with τ1+τ2=1.731. See Figure 1.Two phase-locked oscillations:x1(t)=x2(t+PT/3)=x3(t+2PT/3), y1(t)=y2(t+PT/3)=y3(t+2PT/3).
(a)
(b)For the special case of (4.8), we have the following equations:
(5.2)x1,n+1=(1-τh)x1,n+2τhtanh(y2,n-m),x2,n+1=(1-τh)x2,n+2τhtanh(y3,n-m),x3,n+1=(1-τh)x3,n+2τhtanh(y1,n-m),y1,n+1=(1-2τh)y1,n-1.5τhtanh(x2,n-m),y2,n+1=(1-2τh)y2,n-1.5τhtanh(x3,n-m),y3,n+1=(1-2τh)y3,n-1.5τhtanh(x1,n-m),
where τ=1.731,m=3,h=1/m. The phase-locked periodic solutions appear. See Figure 2.Two phase-locked oscillations:x1(n)=x2(n+PT/3)=x3(n+2PT/3), y1(n)=y2(n+PT/3)=y3(n+PT/3).
(a)
(b)



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