• 検索結果がありません。

DIFFERENCE EQUATIONS

N/A
N/A
Protected

Academic year: 2022

シェア "DIFFERENCE EQUATIONS"

Copied!
20
0
0

全文

(1)

DIFFERENCE EQUATIONS

JES ´US RODRIGUEZ AND DEBRA LYNN ETHERIDGE Received 6 August 2004

We establish conditions for the existence of periodic solutions of nonlinear, second-order difference equations of the formy(t+ 2) +by(t+ 1) +cy(t)= f(y(t)), wherec=0 and f :RRis continuous. In our main result we assume that f exhibits sublinear growth and that there is a constant β >0 such thatu f(u)>0 whenever |u| ≥β. For such an equation we prove that ifNis an odd integer larger than one, then there exists at least one N-periodic solution unless all of the following conditions are simultaneously satisfied:

c=1,|b|<2, andNarccos1(b/2) is an even multiple ofπ.

1. Introduction

In this paper, we study the existence of periodic solutions of nonlinear, second-order, discrete time equations of the form

y(t+ 2) +by(t+ 1) +cy(t)= fy(t), t=0, 1, 2, 3,..., (1.1) where we assume thatbandcare real constants,cis different from zero, and f is a real- valued, continuous function defined onR.

In our main result we consider equations where the following hold.

(i) There are constantsa1,a2, ands, with 0s <1 such that

f(u)a1|u|s+a2 uinR. (1.2)

(ii) There is a constantβ >0 such that

u f(u)>0 whenever|u| ≥β. (1.3) We prove that ifNis odd and larger than one, then the difference equation will have a N-periodic solution unless all of the following conditions are satisfied:c=1,|b|<2, and Narccos1(b/2) is an even multiple ofπ.

Copyright©2005 Hindawi Publishing Corporation Advances in Dierence Equations 2005:2 (2005) 173–192 DOI:10.1155/ADE.2005.173

(2)

As a consequence of this result we prove that there is a countable subsetSof [2, 2]

such that ifb /S, then

y(t+ 2) +by(t+ 1) +cy(t)= fy(t) (1.4) will have periodic solutions of every odd period larger than one.

The results presented in this paper extend previous ones of Etheridge and Rodriguez [4] who studied the existence of periodic solutions of difference equations under signifi- cantly more restrictive conditions on the nonlinearities.

2. Preliminaries and linear theory

We rewrite our problem in system form, letting x1(t)=y(t),

x2(t)=y(t+ 1), (2.1)

wheretis inZ+≡ {0, 1, 2, 3,...}. Then (1.1) becomes x1(t+ 1)

x2(t+ 1)

=

0 1

c b

x1(t) x2(t)

+ 0

fx1(t)

(2.2) fortinZ+. For periodicity of periodN >1, we must require that

x1(0) x2(0)

= x1(N)

x2(N)

. (2.3)

We cast our problem (2.2) and (2.3) as an equation in a sequence space as follows.

LetXN be the vector space consisting of allN-periodic sequencesx:Z+R2, where we use the Euclidean norm| · |onR2. For suchx, ifx =supt∈Z+|x(t)|, then (XN, · ) is a finite-dimensional Banach space.

The “linear part” of (2.2) and (2.3) may be written as a linear operatorL:XNXN, where for eachtZ+,

Lx(t)=

x1(t+ 1) x2(t+ 1)

A x1(t)

x2(t)

, (2.4)

the matrixAbeing

0 1

c b

. (2.5)

The “nonlinear part” of (2.2) and (2.3) may be written as a continuous functionF: XNXN, where fortZ+,

F(x)(t)= 0

fx1(t)

. (2.6)

(3)

We have now expressed (2.2) and (2.3) in an equivalent operator equation form as

Lx=F(x). (2.7)

Following [4,5], we briefly discuss the purely linear problemsLx=0 andLx=h.

Notice thatLx=0 if and only if

x(t+ 1)=Ax(t) tinZ+,

x(0)=x(N), (2.8)

wherex(t) is inR2. But solutions of this system must be in the formx(t)=Atx(0), for t=1, 2, 3,..., where (IAN)x(0)=0. Accordingly, the kernel ofL (henceforth called ker(L)) consists of those sequences inXN for whichx(0)ker(IAN) and otherwise x(t)=Atx(0).

To characterize the image ofL(henceforth called Im(L)), we observe that ifhis an element ofXN, andx(t) is inR2for alltinZ+, thenhis an element of Im(L) if and only if x(t+ 1)=Ax(t) +h(t) tinZ+, (2.9)

x(0)=x(N). (2.10)

It is well known [1,6,7] that solutions of (2.9) are of the form x(t)=Atx(0) +Att

1 l=0

Al+11h(l) (2.11)

fort=1, 2, 3,.... For such a solution also to satisfy theN-periodicity condition (2.10), it follows thatx(0) must satisfy

IANx(0)=ANN 1 l=0

Al+11h(l), (2.12)

which is to say thatANNl=01(Al+1)1h(l) must lie in Im(IAN). Because Im(IAN)= [ker(IAN)T], it follows that if we construct matrixWby letting its columns be a basis for ker(I AN)T, then for h in XN, h is an element of Im(L) if and only if WTANNl=01(Al+1)1h(l)=0. See [4].

Following [4], we let

Ψ(0)=

ANTW, Ψ(l+ 1)=

Al+1TANTW forlinZ+. (2.13) Thenhis in Im(L) if and only ifNl=01ΨT(l+ 1)h(l)=0.

As will become apparent inSection 3, in which we construct the projectionsU and IEfor specific cases, it is useful to know that the columns ofΨ(·) span the solution space of the homogeneous “adjoint” problem

Lx=0, (2.14)

(4)

whereL=XNXNis given by

Lx(t) =x(t+ 1)ATx(t) fortinZ+. (2.15) Further, this solution space and ker(L) are of the same dimension. See [4,5,9].

The proof appears in Etheridge and Rodriguez [4]. One observes that x(t+ 1)= (AT)x(t) if and only ifx(t)=(AT)tx(0) and next, by direct calculation, that

Ψ(t+ 1)=

ATΨ(t). (2.16) Furthermore,

IATNΨ(0)=0 (2.17)

so thatΨ(0)=Ψ(N), whence the columns ofΨ(·) lie inXN. One then observes that, just as the dimension of ker(L) is equal to that of ker(IAN), the dimension of ker(L) is equal to that of ker(I(AT)N). The two matrices have kernels of the same dimension.

Our eventual aim is to analyze (2.7) using the alternative method [2,3,8,9,10,11,12]

and degree-theoretic arguments [3,12,13]. To begin, we will “split”XNusing projections U:XNker(L) andE:XNIm(L). The projections are those of Rodriguez [9]. See also [4,5]. A sketch of their construction is given here.

Just as we let the columns ofWbe a basis for ker((IAN)T), we let the columns of the matrixVbe a basis for ker(IAN). Note that the dimensions of these two spaces are the same. LetCUbe the invertible matrixNl=01(AlV)T(AlV) andCIEthe invertible matrix N1

l=0 ΨT(l+ 1)Ψ(l+ 1). ForxinXN, define Ux(t)=AtVCU1N1

l=0

AlVTx(l), (2.18)

(IE)x(t)=Ψ(t+ 1)CI1E N1

l=0

ΨT(l+ 1)x(l) (2.19)

for eachtinZ+. Rodriguez [9] shows that these are projections which splitXN, so that XN=ker(L)Im(IU),

XN=Im(L)Im(IE), (2.20)

where

Im(E)=Im(L),

Im(U)=ker(L), (2.21)

the spaces Im(IE) and Im(U) having the same dimension.

Note that if we let ˜L be the restriction to Im(IU) of L, then ˜L is an invertible, bounded linear map from Im(IU) onto Im(E). If we denote byM the inverse of ˜L, then it follows that

(i)LMh=hfor allhin Im(L), (ii)MLx=(IU)xfor allxinXN, (iii)UM=0,EL=L, and (IE)L=0.

(5)

3. Main results

We haveXN=ker(L)Im(IU). Letting the norms on ker(L) and Im(IU) be the norms inherited from XN, we let the product space ker(L)×Im(IU) have the max norm, that is,(u,v) =max(u,v).

Proposition3.1. The operator equationLx=F(x)is equivalent to vMEF(u+v)=0,

Q(IE)Fu+MEF(u+v)=0, (3.1)

whereuis inker(L)=Im(U),vIm(IU), andQmapsIm(IE)linearly and invertibly ontoker(L).

Proof.

Lx=F(x) (3.2)

⇐⇒

EL(x)F(x)=0,

(IE)LxF(x)=0 (3.3)

⇐⇒

L(x)EF(x)=0,

(IE)F(x)=0 (3.4)

⇐⇒

MLxMEF(x)=0,

Q(IE)F(x)=0 (3.5)

⇐⇒

x=Ux+MEF(x),

Q(IE)F(x)=0 (3.6)

⇐⇒

(IU)xMEF(x)=0,

Q(IE)FUx+MEF(x)=0. (3.7) Now, eachxinXNmay be uniquely decomposed asx=u+v, whereu=Uxker(L) andv=(IU)x. So (3.7) is equivalent to

vMEF(u+v)=0, (3.8)

Q(IE)Fu+MEF(u+v)=0. (3.9) By means of (2.18) and (2.19), we have split our operator equation (2.7);vMEF(u+ v) is in Im(IU), whileQ(IE)F(u+MEF(u+v)) is in Im(U).

Proposition 3.2. IfN is odd, c=0, and Narccos(b/2) is not an even multiple ofπ whenc=1and|b|<2, then eitherker(L)is trivial or bothker(L)andIm(IE)are one- dimensional. In the latter case, the projectionsUandIEand the bounded linear mapping

(6)

Q(IE)may be realized as follows. ForxinXN, and for alltZ+ Ux(t)=

1 1

1

2N

N1

l=0

x1(l)

+

N1

l=0

x2(l)

, (3.10)

(IE)x(t)= c

1

1

Nc2+ 1

(c)

N1

l=0

x1(l)

+

N1

l=0

x2(l)

. (3.11) Proof. The “homogeneous linear part” of our scalar problem (corresponding toLx=0) is

y(t+ 2) +by(t+ 1) +cy(t)=0, (3.12) where

y(0)=y(N), y(1)=y(N+ 1). (3.13) Calculations, detailed in the appendix of this paper, show that under the hypotheses of Proposition 3.2, the homogeneous linear part of our scalar problem has either only the trivial solutiony(t)=0 for alltinZ+or the constant solutiony(t)=1 for alltinZ+.

In the latter case, the constant function 1

1

(3.14) spans ker(L), so that for everytZ+,AtVof (2.18) may be taken to be

1 1

. (3.15)

Then

CU1= N

l=0

AlVTAlV 1

=

N1

l=0

[1, 1]

1 1

1

=(2N)1,

N1 l=0

AlVTx(l)=

N1 l=0

[1, 1]

x1(l) x2(l)

=

N1

l=0

x1(l)

+

N1

l=0

x2(l)

(3.16)

wheneverxis inXN. Therefore Ux(t)=

1 1

1

2N

N1

l=0

x1(l)

+

N1

l=0

x2(l)

(3.17)

(7)

fortZ+, a constant multiple of

1 1

. (3.18)

In the appendix, we also show that under the hypotheses ofProposition 3.2, the homo- geneous adjoint problemLx=0 has either only the trivial solution or a one-dimensional solution space spanned by the constant function

c 1

. (3.19)

Therefore in (2.19), we may take

Ψ(t)= c

1

(3.20) for alltinZ+, so that

CIE1

=

N1

l=0

[c 1]

c 1

1

=

Nc2+ 11,

N1 l=0

ΨT(l+ 1)x(l)=

N1 l=0

[c 1]

x1(l) x2(l)

=(c)

N1

l=0

x1(l)

+

N1

l=0

x2(l)

.

(3.21)

Therefore forxinXN, for alltZ+ (IE)x(t)=

c 1

1 Nc2+ 1

(c)

N1

l=0

x1(l)

+

N1

l=0

x2(l)

, (3.22) a constant multiple of

c 1

. (3.23)

Furthermore, since Q must map Im(IE) linearly and invertibly onto ker(L)= Im(U), our simplest choice forQis as follows.

Each element of Im(IE) is of the form (3.11) for somexinXN. Now let Q(IE)x(t)=

1 1

1 Nc2+ 1

(c)

N1

l=0

x1(l)

+

N1

l=0

x2(l)

, (3.24) fortinZ+. We notice thatQis clearly linear, bounded, and maps onto Im(U), and that

Q((IE)x)=0 if and only if (IE)x=0.

(8)

Remark 3.3. In the case for which ker(L)= {0}, each ofUandIEis the zero projection onXN,Eis the identity onXN, andM isL1. Equation (3.9) then becomes trivial and (3.8) becomesL1F(v)=v, obviously equivalent to (2.7).

Theorem3.4. Suppose thatN3is odd,c=0, and f :RRis continuous. Assume also that

(i)there are nonnegative constants,a,˜ b, and˜ swiths <1such that|f(z)| ≤a˜|z|s+ ˜b for allzR,

(ii)there is a positive numberβsuch that for allz > β,f(z)>0andf(z)<0, (iii)whenc=1and|b|<2, thenNarccos(b/2)is not an even multiple ofπ.

Then there is at least oneN-periodic solution ofy(t+ 2) +by(t+ 1) +cy(t)=f(y(t)).

Proof. We have already seen that this scalar problem may be written equivalently as equa- tions of the form

0=Q(IE)Fu+MEF(u+v),

0=vMEF(u+v). (3.25)

Recall that our norm on ker(L)×Im(IU) is(u,v) =max{u,v}, whereuand vare, respectively, the norms onuandvas elements ofXN.

We defineH: ker(L)×Im(IU)ker(L)×Im(IU) by H(u,v)=

Q(IE)Fu+MEF(u+v) vMEF(u+v)

. (3.26)

We know that solving our scalar problem is equivalent to finding a zero of the continuous mapH.

We have shown that under the hypotheses of this theorem, ker(L) is either trivial or one-dimensional, and that when ker(L) is one-dimensional, it consists of the span of the constant function

1 1

. (3.27)

We will establish the existence of a zero ofH by constructing a bounded open subset, Ω, of ker(L)×Im(IU) and showing that the topological degree ofHwith respect toΩ and zero is different from zero. We will do this using a homotopy argument.

The reader may consult Rouche and Mawhin [13] and the references therein as a source of ideas and techniques in the application of degree-theoretic methods in the study of nonlinear differential equations.

We writeH(u,v)=(IG)(u,v), whereIis the identity and G(u,v)=

uQ(IE)Fu+MEF(u+v) MEF(u+v)

. (3.28)

It is obvious that ifΩcontains (0, 0), then the topological degree ofIwith respect toΩ and zero is one.

(9)

For 0τ1,

τH+ (1τ)I=τ(IG) + (1τ)I=IτG. (3.29) Therefore, if we can show that(IτG)(u,v)>0 for all (u,v) in the boundary ofΩ, then by the homotopy invariance of the Brouwer degree, it follows that the degree ofH with respect toΩand zero will be one, and consequentlyH(u,v)=(0, 0) for some (u,v) inΩ. Since, for 0τ1,

(IτG)(u,v)>(u,v)τG(u,v), (3.30) it suffices to show thatG(u,v)<(u,v)for all (u,v) in the boundary ofΩ.

We will let Ωbe the open ball in ker(L)×Im(IU) with center at the origin and radiusr, whereris chosen such thatr/2> β+ (2 ˜ars+ ˜b)(1 +ME). Observe that since 0< s <1, such a choice is always possible.

We will show that the second component function ofGmaps each boundary point of ΩintoΩitself and then, by breaking up the boundary ofΩinto separate pieces, consider the effect of the first component function ofGon those pieces.

The pieces will be, respectively, those boundary elements (u,v) for whichu[r,r]

and those for whichu[0,r), where r=

2(β+ME(2 ˜ars+ ˜b))< r.

Observation 3.5. For (u,v)Ω, (i)F(u+v)2 ˜ars+ ˜b,

(ii)MEF(u+v)ME(2 ˜ars+ ˜b).

Proof. For (u,v)Ω,

F(u+v)=sup

t∈Z+

fu1(t) +v1(t)

sup

t∈Z+

a˜u1(t) +v1(t)s+ ˜b

2sa˜(u,v)s+ ˜b2 ˜ars+ ˜b.

(3.31)

This establishes (i), from which (ii) follows immediately.

Observation 3.6. If (u,v) is in the boundary ofΩ, thenMEF(u+v)< r.

Proof.

MEF(u+v)ME

2 ˜ars+ ˜b<1 +ME

2 ˜ars+ ˜b+β <r

2< r. (3.32) For convenience’s sake, we will letg(u,v)(t)=[MEF(u+v)]1(t) for eachtZ+. The functiong maps ker(L)×Im(IU) continuously intoR. Keep in mind that for eachu in ker(L), there is a uniquely determinedαfor whichuis the constant function

α 1

1

. (3.33)

(10)

Observation 3.7. For (u,v) in the boundary of Ω, and for every lZ+, if α > β+ ME(2 ˜ars+ ˜b), then f(α+g(u,v)(l))>0, while if α <(β+ME(2 ˜ars+ ˜b)), then

f(α+g(u,v)(l))<0.

Proof. When (u,v) lies in the boundary ofΩandαβ+ME(2 ˜ars+ ˜b), we have for eachlZ+,

0< β=

β+ME

2 ˜ars+ ˜bME

2 ˜ars+ ˜b

αME

2 ˜ars+ ˜bαMEF(u+v)

αMEF(u+v)(l)αMEF(u+v)1(l)

=αg(u,v)(l)α+g(u,v)(l)

(3.34)

so that for eachl,f(α+g(u,v)(l))>0.

Similarly, if (u,v) lies in the boundary ofΩandα≤ −βME(2 ˜ars+ ˜b), then for eachlinZ+,

0>β= −

β+ME

2 ˜ars+ ˜b+ME

2 ˜ars+ ˜b

α+ME

2 ˜ars+ ˜bα+MEF(u+v)

α+MEF(u+v)(l)α+MEF(u+v)1(l)

=α+g(u,v)(l)α+g(u,v)(l)

(3.35)

so that for eachl,f(α+g(u,v)(l))<0.

Observation 3.8. If (u,v) is in the boundary ofΩ, uQ(IE)Fu+MEF(u+v)=

2α 1 Nc2+ 1

N1 l=0

fα+g(u,v)(l), (3.36) where

u=α 1

1

. (3.37)

Proof. Since for alltinZ+,

F(x)(t)= 0

fx1(t)

, Fu+MEF(u+v)(t)=

0

fu1(t) +MEF(u+v)1(t)

=

0 fα+g(u,v)(t)

(3.38)

so that

u(t)Q(IE)F(u+v)(t)

=

α 1

Nc2+ 1

(c)(0) +

N1

l=0

fα+g(u,v)(l)

1

1

, (3.39)

(11)

a constant function oft, hence

uQ(IE)Fu+MEF(u+v)=

2α 1 Nc2+ 1

N1 l=0

fα+g(u,v)(l). (3.40) Observation 3.9. For (u,v) in the boundary ofΩ,uQ(IE)F(u+MEF(u+v))< r.

Proof. For (u,v) in the boundary ofΩ,(u,v) =max{u,v} =r. We consider first those elements of the boundary ofΩfor whichu[r,r], and then those for which u[0,r).

For (u,v) in the boundary ofΩ, u is in [r,r]=

2β+ME

2 ˜ars+ ˜b,r (3.41) if and only if

|α| is in β+ME

2 ˜ars+ ˜b,r 2

!

= r 2,r

2

!

. (3.42)

We consider the subcases (i)α >0 and (ii)α <0.

(i)

α is in β+ME

2ars+ ˜b,r 2

!

= r 2,r

2

!

. (3.43)

Then byObservation 3.7, we have f(α+g(u,v)(l))>0 for eachlinZ+, so that α

1 Nc2+ 1

N1

l=0

fα+g(u,v)(l)< αr

2. (3.44)

To show that

α

1 Nc2+ 1

N1

l=0

fα+g(u,v)(l)<r

2, (3.45)

it remains to show that 1 Nc2+ 1

N1 l=0

fα+g(u,v)(l)< α+r

2. (3.46)

Now sinceβ+ME(2 ˜ars+ ˜b)α, and each f(α+g(u,v)(l)) in the sum just above is positive, it suffices to show that

1 c2+ 1

2 ˜ars+ ˜b<r

2+β+ME

2 ˜ars+ ˜b (3.47)

(12)

or equivalently, that

2 ˜ars+ ˜b 1

c2+ 1ME

β < r

2. (3.48)

This follows, of course, from our having chosenrso that

r

2>2 ˜ars+ ˜b1 +ME

+β. (3.49)

(ii)

α is in r

2,βME

2 ˜ars+ ˜b!= r 2,r

2

!

. (3.50)

Then byObservation 3.7, we have f(α+g(u,v)(l))<0 for eachlZ+, so that α

1 Nc2+ 1

N1

l=0

fα+g(u,v)(l)> α≥ −r

2. (3.51)

To show that

α

1 Nc2+ 1

N1

l=0

fα+g(u,v)(l)<r

2, (3.52)

it remains to show that α

1 Nc2+ 1

N1 l=0

fα+g(u,v)(l)<r

2, (3.53)

or equivalently, to show that

1

Nc2+ 1 N1

l=0

fα+g(u,v)(l)<r

2α. (3.54)

Now sinceβME(2 ˜ars+ ˜b)α, so thatβ+ME(2 ˜ars+ ˜b)≤ −α, it suffices to show that

1

Nc2+ 1 N1

l=0

fα+g(u,v)(l)< β+ME

2 ˜ars+ ˜b+r

2. (3.55)

(13)

Further, for eachlin the sum just above, f(α+g(u,v)(l)) is negative and

fα+g(u,v)(l)a˜α+g(u,v)(l)s+ ˜bar˜ s+ ˜b (3.56) so that

N1 l=0

fα+g(u,v)(l)=

N1 l=0

fα+g(u,v)(l)

N1 l=0

2 ˜ars+ ˜b, (3.57) hence,

1

Nc2+ 1 N1

l=0

fα+g(u,v)(l)≤ − 1

Nc2+ 1 N1

l=0

2 ˜ars+ ˜b

= 1

c2+ 1

2 ˜ars+ ˜b,

(3.58)

so that it suffices to show that 1

c2+ 1

2 ˜ars+ ˜b< β+ME

2 ˜ars+ ˜b+r

2. (3.59)

This, as we have seen in the proof of (i), follows from our choice ofr.

Finally, we consider those elements (u,v) of the boundary ofΩfor which

|α|< β+ME

2 ˜ars+ ˜b=r. (3.60) Clearly

α 1 Nc2+ 1

N1 l=0

fα+g(u,v)(l)≤ |α|+ N Nc2+ 1

2 ˜ars+ ˜b

β+1 +ME

2 ˜ars+ ˜b

<r 2

(3.61)

so thatuQ(IE)F(u+MEF(u+v))< r.

For (u,v) in the boundary ofΩ,(u,v) =r. For each such (u,v), we have shown by means ofObservation 3.6, thatMEF(u+v)< r, and, by means ofObservation 3.9, that uQ(IE)F(u+MEF(u+v))< r. Therefore for such (u,v),G(u,v)<(u,v), so that no element of the boundary ofΩis a zero ofH; hence the degree ofHwith respect toΩand zero is 1, so that at least one solution of (2.7) exists insideΩ. Remark 3.10. If inTheorem 3.4, we change (ii) so that we require f(z)<0 andf(z)>0 for allz > β, the conclusions of the theorem still hold.

We let

=

"

2kπ

j :kandjare integers, 02k < jand jis odd

#

. (3.62)

(14)

It is easy to see that if arccos(b/2)/∆, then for any odd integerN,Narccos(b/2) cannot be an even multiple ofπ. It is also obvious thatS≡ {b: arccos(b/2)}is a countable subset of [2, 2]. The following result is now evident.

Corollary 3.11. Suppose f :RR is continuous,c=0, and the following conditions hold:

(i)there are constantsa,˜ b, and˜ s, with0s <1such that

f(u)a˜|u|s+ ˜b uinR, (3.63)

(ii)there is a constantβ >0, such thatu f(u)>0whenever|u| ≥β.

Then, if eitherbR\S, orc=1, then

y(t+ 2) +by(t+ 1) +cy(t)= fy(t) (3.64) will haveN-periodic solutions for every odd periodN >1.

Appendix

We demonstrate here that, providedNis odd,c=0, andNarccos(b/2) is not an even multiple ofπwhenc=1 and|b|<2, the kernel ofLis either trivial or is a one-dimensional space spanned by the constant function

1 1

. (A.1)

We show also that the solution space of the homogeneous adjoint problem (2.14) is, in the latter case, the span of the constant function

c 1

. (A.2)

Of course, if ker(L) is trivial, so is ker(L).

As we have seen,Lis invertible if and only if the matrixIANis invertible. That matrix is invertible if and only if no eigenvalue ofAis anNth root of unity. Those eigenvalues may be complex conjugates, real and repeated, or real and distinct. We will consider each of those three cases after we examine the kernels ofLand ofLin more detail than before.

(i) The kernel ofLconsists of all functionsxinXNfor whichx(t)=Atx(0) fortinZ+, where

A=

0 1

c b

(A.3)

参照

関連したドキュメント

of ISE

Polubarinova{Kochina, Application of the theory of linear dierential equations to the problems of motion of ground water..

About the optimal control, Saint Jean Paulin &amp; Zoubairi [12] studied the problem of a mixture of two fluids periodically distributed one in the other... Throughout this proof,

There is a stable limit cycle between the borders of the stability domain but the fix points are stable only along the continuous line between the bifurcation points indicated

Under suitable assumptions on the weight of the damping and the weight of the delay, we prove the existence and the uniqueness of the solution using the semigroup theory.. Also we

The angular velocity decreases with increasing the material parameter, the slip parameter, the buoyancy parameter, and the heat generation parameter, while it increases with

By using the averaging theory of the first and second orders, we show that under any small cubic homogeneous perturbation, at most two limit cycles bifurcate from the period annulus

We study existence of solutions with singular limits for a two-dimensional semilinear elliptic problem with exponential dominated nonlinearity and a quadratic convection non

A new method is suggested for obtaining the exact and numerical solutions of the initial-boundary value problem for a nonlinear parabolic type equation in the domain with the

Kilbas; Conditions of the existence of a classical solution of a Cauchy type problem for the diffusion equation with the Riemann-Liouville partial derivative, Differential Equations,

Then it follows immediately from a suitable version of “Hensel’s Lemma” [cf., e.g., the argument of [4], Lemma 2.1] that S may be obtained, as the notation suggests, as the m A

We construct a sequence of a Newton-linearized problems and we show that the sequence of weak solutions converges towards the solution of the nonlinear one in a quadratic way.. In

It is known that if the Dirichlet problem for the Laplace equation is considered in a 2D domain bounded by sufficiently smooth closed curves, and if the function specified in the

Indeed, when using the method of integral representations, the two prob- lems; exterior problem (which has a unique solution) and the interior one (which has no unique solution for

A Darboux type problem for a model hyperbolic equation of the third order with multiple characteristics is considered in the case of two independent variables.. In the class

There arises a question whether the following alternative holds: Given function f from W ( R 2 ), can the differentiation properties of the integral R f after changing the sign of

[2])) and will not be repeated here. As had been mentioned there, the only feasible way in which the problem of a system of charged particles and, in particular, of ionic solutions

Under small data assumption, we prove the existence and uniqueness of the weak solution to the corresponding Navier-Stokes system with pressure boundary condition.. The proof is

(Non periodic and nonzero mean breather solutions of mKdV were already known, see [3, 5].) By periodic breather we refer to the object in Definition 1.1, that is, any solution that

Rostamian, “Approximate solutions of K 2,2 , KdV and modified KdV equations by variational iteration method, homotopy perturbation method and homotopy analysis method,”

In the present paper, we show that, under the same hypothesis on the diameter of the tree, the group is an HNN extension with finitely presented base group, and hence that the

[r]

Amount of Remuneration, etc. The Company does not pay to Directors who concurrently serve as Executive Officer the remuneration paid to Directors. Therefore, “Number of Persons”