exponential stability of time-varying linear systems...articles gonz´alez & palencia (1998) and...

IMA Journal of Numerical Analysis(2011)31, 865−885doi:10.1093/imanum/drq002Advance Access publication on May 11, 2010

Exponential stability of time-varying linear systems

ADRIAN T. HILL∗

Departmentof Mathematical Sciences, University of Bath, Claverton Down,Bath BA2 7AY, UK

∗Correspondingauthor: [email protected]

AND

ACHIM ILCHMANN

Instituteof Mathematics, Technical University Ilmenau, Weimarer Straße 25,98693 Ilmenau, Germany

[email protected]

[Received on 16 May 2009; revised on 3 November 2009]

This paper considers the stability of both continuous and discrete time-varying linear systems. Stabilityestimates are obtained in either case in terms of the Lipschitz constant for the governing matrices and theassumed uniform decay rate of the corresponding frozen time linear systems. The main techniques usedin the analysis are comparison methods, scaling and the application of continuous stability estimates tothe discrete case. Counterexamples are presented to show the necessity of the stability hypotheses. Thediscrete results are applied to derive sufficient conditions for the stability of a backward Euler approxi-mation of a time-varying system and a one-leg linear multistep approximation of a scalar system.

Keywords: exponential stability; discrete time-varying linear systems; continuous time-varying linearsystems; one-leg multistep approximation.

1. Introduction

We study exponential growth (stability) of both continuous and discrete time-varying linear systems. Inthe continuous-time case we study solutions of the equation

u(t) = A(t)u(t), t > 0, (1.1)

whereA: R>0 → CN×N is a continuous function. Recall (see, for example,Rugh,1996) that (1.1) issaid to be (uniformly)exponentially stableif and only if

∃ (M, η) ∈ R>1 × R>0 ∀ slnsu of (1.1)∀ (t, s) ∈ D: ‖u(t)‖ 6 M e−η(t−s)‖u(s)‖, (1.2)

whereD := {(t, s) ∈ R>0 × R>0|t > s} and‘slns’ means solutions.To derive sufficient conditions for the exponential stability of (1.1) we assume that the frozen

systemsu(t) = A(τ )u(t) are exponentially stable andA is globally Lipschitz. More precisely, for(K , ω, L) ∈ R>1 × R>0 × R>0 weconsider the classes of generators

SK ,ω,L :=

{

A ∈ C(R>0,CN×N)

∣∣∣∣∣

∀ t, s ∈ R>0 : ‖eA(t)s‖ 6 K e−ωs

∀ t, s ∈ R>0 : ‖A(t) − A(s)‖ 6 L|t − s|

}

.

c© Theauthor 2010. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved.

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866 A. T. HILL AND A. ILCHMANN

In the discrete-time case we study solutions of

un+1 = Anun, n ∈ N0, (1.3)

where(An)n∈N0 is a sequence of matrices with elements inCN×N . System (1.3) is said to be (uniformly)exponentially stableif and only if

∃ (M, η) ∈ R>1 × R>0 ∀ slnsu of (1.3)∀ (n, m) ∈ Δ: ‖un‖ 6 M e−η(n−m)‖um‖, (1.4)

whereΔ := {(n, m) ∈ N0 × N0|n > m}. Sufficient conditions for the exponential stability of (1.3) arederived by considering the classes of generators

ΣK ,ω,L :=

{

(An) ∈ (CN×N)N

∣∣∣∣∣

∀ n, m ∈ N0 : ‖Amn ‖ 6 K e−ωm

∀ n, m ∈ N0 : ‖An − Am‖ 6 L|n − m|

}

,

where(CN×N)N is the set of all mappings fromN toCN×N .It may be worth knowing that, although we consider time-varying systems, due to the special system

classesSK ,ω,L andΣK ,ω,L , the decay rateη and the constantM prescribing the exponential stability in(1.2) and (1.4), respectively, hold for every initial value if and only if they hold for the initial value attime t = 0. To be more precise, see Lemmas5.1and5.10, respectively.

A nice textbook on time-varying systems, continuous as well as discrete time, isRugh(1996) (seealso the references therein). Bounds on the exponential growth of continuous/discrete time-varying sys-tems have been suggested by numerous authors, for example,Rosenbrock(1963),Desoer(1969,1970),Coppel(1978),Wu (1984),Kreisselmeier(1985),Krause & Kumar(1986), Ilchmannet al. (1987),Amatoet al. (1993) andSolo(1994).

A good description of the stability analysis available for numerical methods is given inHairer &Wanner(1991), which summarizes earlier work for linear multistep, Runge–Kutta and general linearmethods byNevanlinna(1977),Dahlquist(1978),Burrage & Butcher(1980) andButcher(1987). Theseresults relate to the propagation of errors made in the approximation of the problemu = f (t, u), wheref is assumed to satisfy a structural assumption ensuring the stability of solutionsu. More recently, thearticlesGonzalez & Palencia(1998) andOstermannet al.(2004) have investigated the stability of linearmultistep methods approximating time-varying systems of the formu(t) = A(t)u in a Banach spacesetting.

The paperSoderlind(1984) has a similar motivation to the current paper in that bounds for continu-ous and discrete linear time-varying equations are obtained in a unified way. There, bounds are derivedin terms of a scalar ordinary differential equation involving the logarithmic norm ofA(t), and this ideais also extended to the nonlinear case.

In this paper, for the continuous case, we use the variation of constants formula and the properties ofthe setSK ,ω,L to obtain an integral inequality for‖u(t)‖. The main idea revived in this paper is to usescaling to eliminate as many apparently independent parameters as possible in the integral inequality.Once the inequality is simplified in this way, it is then relatively simple to bound‖u(t)‖ sharply in termsof the solution of a comparison equation.

In the continuous case a natural timescale√

K L arises out of the scaling process. For the discreteproblem a similar timescaleβ :=

√K L eω battleswith the intrinsic unit timescale of the discrete

process. Whenβ is small, estimates from the continuous problem are also sharp for the correspondingdiscrete case. For largerβ a direct discrete approach yields sharper bounds.

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EXPONENTIAL STABILITY OF TIME-VARYING LINEAR SYSTEMS 867

The paper is organized as follows. In Section2 the two main results give sufficient conditions forthe uniform decay of solutions of continuous- and discrete-time systems. Counterexamples are also pre-sented in cases where the hypotheses do not hold. The proofs of the two main theorems are given inSection5. Applications of the main discrete result are given in Sections3 and4. In Section3 sufficientconditions are given for the exponential stability of a backward Euler approximation of a general con-tinuous time-varying system of the type considered in Section2. In Section4 the results of Section2are used to prove the stability of a one-leg multistep approximation of a time-varying scalar differentialequation. A list of notation is given at the end.

2. Exponential stability

2.1 Continuous-time systems

Below, we state our main result in the continuous case.

THEOREM2.1 Suppose thatA ∈ SK ,ω,L for some(K , ω, L) ∈ R>1×R>0×R>0. Then every solutionu of (1.1) satisfies

(i) ‖u(t)‖ 6 K exp{(

K L(t−s)4 − ω

)(t − s)

}‖u(s)‖ ∀ (t, s) ∈ D,

(ii) ‖u(t)‖ 6 K exp{(√

K L log(min{2, K }) − ω)(t − s)}‖u(s)‖ ∀ (t, s) ∈ D.

The above theorem is proved in Section5.1.

REMARK 2.2 Theorem2.1(ii) implies thatu(t) = A(t)u(t) is exponentially stable forM = K andη =

√K L log(min{2, K }) − ω if

K L log(min{2, K }) < ω2. (2.1)

2.2 Discrete-time systems

The main discrete-time result is stated below.

THEOREM 2.3 Suppose thatA ∈ ΣK ,ω,L for some(K , ω, L) ∈ R>1 × R>0 × R>0. Then, forβ :=√K L eω, every solutionu: N0 → CN of un+1 = Anun satisfies

(i) ‖un‖ 6 K n−m e−ω(n−m)‖um‖ ∀ (n, m) ∈ Δ,

(ii) ‖un‖ 6 K exp{(

β2(n−m)4 − ω

)(n − m)

}‖um‖ ∀ (n, m) ∈ Δ,

(iii) ‖un‖ 6 K exp{(β√

log2 − ω)(n − m)}

‖um‖ ∀ (n, m) ∈ Δ,

(iv) ‖un‖ 6 12{(1 + β)n−m + (1 − β)n−m}K e−ω(n−m)‖um‖ ∀ (n, m) ∈ Δ.

The above theorem is proved in Section5.2.

REMARK 2.4 (Bound comparison and timescales) Bounds (ii) and (iii) in Theorem2.3are the respectivesmall- and long-time analogues of the corresponding continuous results. Such bounds are particularlyuseful ifβ � 1, when it is harder to obtain sharp discrete bounds directly.

A direct discrete approach works better ifβ > 1. Comparing the growth rate of (iii) and (iv) forlargen − m, we see that the direct discrete bound (iv) is sharper than (iii) if

β > β0 ≈ 0.43such that exp(β0√

log2) = 1 + β0.

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For some problems the trivial bound (i) is best, particularly for smalln − m if a special norm is chosenso thatK is close to, or equal to, 1.

REMARK 2.5 (Exponential stability criteria) By inequalities (i), (iii) and (iv) in Theorem2.3, a systemA ∈ ΣK ,ω,L for some(K , ω, L) ∈ R>1 × R>0 × R>0 is exponentially stable if

ω > min{log(1 + β), β√

log2, log K }. (2.2)

2.3 Continuous and discrete generalizations of a counterexample of Hoppensteadt

The following examples generalize a well-known example ofHoppensteadt(1966).

EXAMPLE 2.6 Descriptionof system. DefineA: R>0 → C2×2 for a > θ > 0 by

A(t) := Q(t)A0QT(t), Q(t) :=

[cosθ t sinθ t

− sinθ t cosθ t

]

, A0 :=

[0 a

0 0

]

. (2.3)

Properties of the frozen systems. SinceQ(τ ) is orthogonal, we have

‖eA(τ )t‖2 = ‖Q(τ )eA0t QT(τ )‖2 = ‖eA0t‖2 =

∥∥∥∥∥

[1 at

0 1

]∥∥∥∥∥

2

6 1 + at ∀ τ, t > 0, (2.4)

where‖ ∙ ‖2 is the Euclidean norm.Lipschitzian properties of A. We observe for allt, s> 0 that

‖A(t) − A(s)‖2 = ‖Q(t)A0QT(t) − Q(s)A0QT(s)‖2 = ‖Q(t − s)A0QT(t − s) − A0‖2

andalso that

‖Q(t)A0QT(t) − A0‖2 = a

∥∥∥∥∥

[sinθ t cosθ t − sin2 θ t

− sin2 θ t − sinθ t cosθ t

]∥∥∥∥∥

2

= a| sinθ t | ∀ t > 0.

Hence

‖A(t) − A(s)‖2 = a| sin(θ(t − s))| 6 aθ |t − s| ∀ t, s> 0.

Properties of the time-varying system. Ifu: R>0 → C2 is a solution of (1.1), then we define

x: R>0 → C2, t 7→ x(t) := QT(t)u(t) and B :=

[0 a − θ

θ 0

]

.

SinceQ(t)TQ(t) =[

0 −θθ 0

]for all t > 0, we obtain that

x(t) = Q(t)Tu(t) + QT(t)u(t) = (Q(t)TQ(t) + A0)x(t) = Bx(t).

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SinceB2 = θ(a − θ)I , we have

eBt =∞∑

m=0

B2m t2m

(2m)!

(I + B

t

2m+ 1

)=

∞∑

m=0

[θ (a − θ)]mt2m

(2m)!

[1 (a−θ)t

2m+1θ t

2m+1 1

]

=

cosh[θ(a − θ)]1/2t

√a−θθ sinh[θ(a − θ)]1/2t

√θ

a−θ sinh[θ(a − θ)]1/2t cosh[θ(a − θ)]1/2t

.

For all t ∈ R>0 wehave‖u(t)‖2 = ‖Q(t)x(t)‖2 = ‖x(t)‖2 = ‖eBt x(0)‖2 andx(0) = u(0). Hence

maxu(0)∈C2\{0}

‖u(t)‖2

‖u(0)‖2= max

x(0)∈C2\{0}

‖eBt x(0)‖2

‖x(0)‖2= ‖eBt‖2 > exp([θ(a − θ)]1/2t). (2.5)

Comments.We observe that, while solutions of the frozen time systems only grow at most linearly withtime, those of the time-varying system may grow exponentially. Choosingε ∈ (0,

√θ(a − θ)) and

replacingA(t) by Aε(t) := A(t) − ε I hasthe effect of multiplying the right-hand side (RHS) of (2.4)and (2.5) by e−εt . This makes the frozen time systems exponentially stable, but leaves the time-varyingsystem exponentially unstable, and does not affect the Lipschitzian properties ofA(∙).

EXAMPLE 2.7 TheEuler method for Example2.6. The (forward) Euler method for (2.3) is

un+1 = (I + hA(tn))un =: Tn(h)un, n ∈ N0, (2.6)

whereh > 0 is a time step andtn = nh. Thus, forQ as in Example2.6, we have

Tn(h) = Q(tn)T0(h)Q(−tn), T0(h) =

[1 ah

0 1

]

.

Analogously to the continuous-time case, we obtain that

‖Tmn (h)‖2 = ‖eA(tn)tm‖2 6 1 + atm ∀ (n, m) ∈ N2

0, (2.7)

‖Tn(h) − Tm(h)‖2 = ah| sin(θ(n − m)h)| 6 aθh2|n − m| ∀ (n, m) ∈ N20. (2.8)

Properties of the time-varying system. Ifun: N0 → C2 is a solution of (2.6), then we define

x: N0 → C2, n 7→ xn := Q(−tn)un.

Then

xn+1 =

[cosθh − sinθh + ah cosθh

sinθh cosθh + ahsinθh

]

xn =: B(h)xn ∀ n ∈ N0.

Sincex0 = u0 andQ(∙) is orthogonal, we have

maxu0∈C2\{0}

‖un‖2

‖u0‖2= max

x0∈C2\{0}

‖xn‖2

‖x0‖2= ‖(B(h))n‖2 > (μ(B(h)))n, n ∈ N0. (2.9)

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As a > θ > 0, there is a uniqueh0 ∈ (0,π/θ) such thatah0 = 2tan(θh0/2) andah > 2 tan(θh/2) forall h ∈ (0,h0). Elementary trigonometry now implies that

μ(B(h)) = σ(h) +√

σ 2(h) − 1 > 1, σ (h) := cosθh + (ah/2)sinθh > 1 ∀ h ∈ (0,h0).

Comments.We observe that, while solutions of the frozen systems only grow at most linearly with time,those of the time-varying system may grow exponentially. ReplacingA(t) by Aε(t) = A(t) − ε I forε ∈ (0,

√θ(a − θ)) multiplies the RHS of (2.7) by(1− εh)m for all sufficiently smallh > 0, rendering

the discrete frozen systems exponentially stable. This change has no effect on the Lipschitzian propertiesof the discrete system. A similar analysis to the above shows thatμ(Bε(h)) > 1 for all sufficiently smallh > 0, implying that the discrete time-varying system is exponentially unstable in that parameter range.

REMARK 2.8 The fact that the Euler method in Example2.7 inherits both the stability of the frozentime systems and the instability of the time-varying system for all sufficiently smallh is an expectedconsequence of consistency. This would be true for any consistent one-step method—even a very sta-ble one, such as backward Euler. Thus, to establish the stability of a discrete time-varying system forany consistent method, more is required than the stability of the frozen time systems combined witha Lipschitz condition of the form (2.8). In general, one also requires additional information about theunderlying system, such as (2.1), or a corresponding bound on the discretization, such as (2.2).

3. The stability of a backward Euler approximation

Here we consider the approximation of (1.1) by the backward Euler method. For a time steph > 0 theequation of the method is

un+1 = un + hA(tn+1)un+1, n ∈ N0. (3.1)

Undersuitable assumptions onA, Lemma3.1below shows that the sequence(Tn(h))n∈: N0 of matricesin CN×N given by

Tn(h) := (I − hA(tn+1))−1, n ∈ N0, (3.2)

is well defined. This allows us to rewrite (3.1) as the discrete time-varying system

un+1 = Tn(h)un, n ∈ N0, (3.3)

andto establish the stability properties of the backward Euler method using Theorem2.3.

LEMMA 3.1 Suppose thath > 0 and thatA ∈ SK ,ω,L for (K , ω, L) ∈ R>1 × R>0 × R>0. Then, foreachn ∈ N0, we have thatTn(h) given by (3.2) is well defined, and

‖Tmn (h)‖ 6 K (1 + hω)−m, m ∈ N0. (3.4)

Proof. For everym ∈ N we have

‖(I − h A(tn))−m‖ =

∥∥∥∥∥

∫ ∞

0

tm−1

(m − 1)!e−(I −hA(tn))t dt

∥∥∥∥∥

6∫ ∞

0

tm−1

(m − 1)!K e−(1+ωh)t dt = K (1 + ωh)−m,

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wherethe boundedness of the RHS ensures that the left-hand side is well defined. The casem = 0 isclear, asK > 1. �

LEMMA 3.2 Suppose thath > 0 and thatA ∈ SK ,ω,L for (K , ω, L) ∈ R>1 × R>0 × R>0. Then

‖Tn(h) − Tm(h)‖ 6 LK 2(1 + ωh)−2h2|n − m| ∀ (n, m) ∈ N20. (3.5)

Proof. As in the second resolvent identity, we have

(I − h A(tn))−1 − (I − hA(tm))−1 = h(I − hA(tn))

−1[ A(tn) − A(tm)]( I − hA(tm))−1.

Thus,by (3.4) and the Lipschitzian properties ofA, we have

‖Tn(h) − Tm(h)‖ 6 ‖(I − hA(tn))−1‖‖(I − hA(tm))−1‖h‖A(tn) − A(tm)‖

6 K 2(1 + ωh)−2hL|tn − tm| = LK 2(1 + ωh)−2h2|n − m|. �

LEMMA 3.3 Suppose thath > 0 and thatA ∈ SK ,ω,L for (K , ω, L) ∈ R>1 × R>0 × R>0. Then, for

δ :=√

K 3L, every solutionu: N0 → CN of un+1 = Tn(h)un satisfies

(i) ‖un‖ 6 K n−m(1 + ωh)−(n−m)‖um‖ ∀ (n, m) ∈ Δ,

(ii) ‖un‖ 6 K (1 + ωh)−(n−m) exp{δ2(n − m)2h2/4}‖um‖ ∀ (n, m) ∈ Δ,

(iii) ‖un‖ 6 K (1 + ωh)−(n−m) exp{(δ√

log2)(n − m)h}‖um‖ ∀ (n, m) ∈ Δ,

(iv) ‖un‖ 6 12{(1 + δh)n−m + (1 − δh)n−m}K (1 + ωh)−(n−m)‖um‖ ∀ (n, m) ∈ Δ.

Proof. By Lemmas3.1and3.2, we have

(Tn(h))n∈N0 ∈ ΣK ,ω,L for K := K , L := K 2Lh2(1 + ωh)−1, ω := log(1 + ωh)

(where L hasbeen increased by a factor of(1 + ωh) for convenience). Applying the conclusions ofTheorem2.3with

β :=√

K L eω =√

K 3Lh2 = δh and e−ω = (1 + ωh)−1,

weobtain (i)–(iv). �The following is clear from bounds (iii) and (iv) in Lemma3.3.

THEOREM3.4 Suppose thatA ∈ SK ,ω,L for (K , ω, L) ∈ R>1×R>0×R>0. Then (3.1) is exponentiallystable if

ω > min

{√

K 3L,exp(h

√K 3L log2) − 1

h

}

. (3.6)

REMARK 3.5 (Stability criteria) Bounds (i) and (ii) in Lemma3.3may be the sharpest for smalln − m.Inequalities (ii)–(iv) all give rise to bounds of the form

∀ τ > 0 ∀ (h, n) ∈ R>0 × N0 suchthat tn ∈ [0, τ ]: ‖un‖ 6 M‖u0‖.

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The condition for exponential stability of the method, given by (3.6), is good in the sense that it isessentiallyh-independent (with a mild improvement ash → 0+). However,ω is required to be approx-imately a factor ofK larger than in (2.1). This factor arises in the Lipschitz analysis ofTn(h). In thenonstiff case, whereh‖A(∙)‖∞ � 1, approximation could be used to bound solutions of the methodindirectly.

4. Stability for a one-leg multistep approximation of a scalar problem

Here we show how the general observations made above apply to the approximation of a simple time-varying problem.

We consider the approximation of time-varying scalar equations of the form

u(t) = λ(t)u(t), t > 0, (4.1)

whereλ: R>0 → C is assumed to be Lipschitz continuous, i.e.,

∃ Lλ > 0 ∀ t, s> 0: |λ(t) − λ(s)| 6 Lλ|t − s|. (4.2)

A q-stepone-leg method approximating (4.1) can be taken to be of the form

q∑

j =0

α j yn+ j = hλ(tn)q∑

j =0

β j yn+ j , n ∈ N0, (4.3)

whereh > 0 is the time step, and the coefficientsα0, β0, . . . , αq, βq ∈ R arechosen so that the corre-sponding linear multistep method is irreducible and has an orderp ∈ N. It is assumed that numericalinitial datay0, y1, . . . , yq−1 is generated by some other method. It is known (see, e.g.,Hairer & Wanner,1991, Chapter V) that the stability properties of the linear multistep method may be studied in terms ofthose of the one-leg method.

LetC denoteC ∪ {∞}. We define the companion matrixC: C → Cq×q

by

C(z) :=

0 1

.... . .

0 1

c0(z) c1(z) ∙ ∙ ∙ cq−1(z)

, cj (z) := −α j − zβ j

αq − zβq, 06 j 6 q − 1, (4.4)

wherewe follow the convention thatcj (∞) = limz→∞ cj (z) for 06 j 6 q − 1.Settingzn := hλ(tn) for n ∈ N0, we define

An := C(zn), un := [yn, . . . , yn+q−1]T ∈ Cq, n ∈ N0, (4.5)

for a solution(yn) of (4.3). We now observe that(un) satisfiesthe system

un+1 = Anun, n ∈ N0, (4.6)

which is of the form (1.3).

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A matrix B ∈ CN×N is calledpower boundedif and only if

∃ K > 0 ∀ n ∈ N0: ‖Bn‖ 6 K .

The linear stability regionfor the linear multistep method corresponding to (4.3) is

S := {z ∈ C‖C(z) is power bounded}. (4.7)

EXAMPLE 4.1 (Numerical instability for rapidly varyingλ) Consider (4.1) for

λ: R>0 → C, t 7→ λ(t) = −81+ 79 cos(π t)

8. (4.8)

Supposethat the scalar system (4.1) is approximated by the fourth–order backwards differentiationformula (BDF4) method. In this case the method is given by (4.3) forq = 4 and the coefficients

[α0, α1, α2, α3, α4] =[

1

4, −

4

3, 3,−4,

25

12

], [β0, β1, β2, β3, β4] = [0, 0,0,0,1].

Thecompanion matrix corresponding to BDF4, namely,C: C → C4×4

, is defined by (4.4) withq = 4.If the time step ish = 1 then

zn = hλ(tn) =

{−20, n even,

−1/4, n odd.

As in (4.5), letAn := C(zn), wheren ∈ N0. Hence, forn ∈ N0, we have

A2n = C(−20)=

0 1 0 0

0 0 1 0

0 0 0 1

− 3265

16265 − 36

26548265

, A2n+1 = C(−1/4) =

0 1 0 0

0 0 1 0

0 0 0 1

− 328

1628 −36

284828

.

Thefollowing straightforward calculations show that{−1/4,−20} ⊂S:

μ(A2n) = μ(C(−20)) = 0.42(2 d.p.), μ(A2n+1) = μ(C(−1/4)) = 0.78(2 d.p.)

(wherex (2 d.p.) means thatx is correct to two decimal places). Nowu2n = A2n−1A2n−2, . . . , A1A0u0,wherethe product

A2n−1, . . . , A1A0 = (A1A0)n = (C(−1/4)C(−20))n, n ∈ N0.

Thestability of the system is therefore determined by

μ(C(−1/4)C(−20)) = 1.02(2 d.p.).

Sinceμ(C(−1/4)C(−20)) > 1, we deduce that there are initial conditions for the method such thatthe norm of the numerical solution grows exponentially withn, i.e., the BDF4 method withh = 1 isunstable for this differential equation.

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THEOREM 4.2 Consider the scalar time-varying equation (4.1) such thatλ: R>0 → C satisfies(4.2).Suppose that (4.1) is approximated by a one-leg method (4.3) with corresponding linear stability regionS ⊂ C, as in (4.7). Assume also thatD ⊆ S is closed and that

∀ h > 0 ∀ n ∈ N0: hλ(tn) ∈ D. (4.9)

Thenthere exist(K , ω, L) ∈ R>1×R>0×R>0 suchthat, for allh > 0 andβ :=√

K L eω, any solutionu of (4.6) satisfies

(a) ‖un‖ 6 K exp{[(log 2)1/2βh − ω]n}‖u0‖ ∀ n ∈ N0,

(b) ‖un‖ 6 12{(1 + βh)n + (1 − βh)n}K e−ωn‖u0‖ ∀ n ∈ N0.

REMARK 4.3 Under slightly stronger conditions than the hypotheses of Theorem4.2, the conclusion ofProposition4.4below may be strengthened to

∃ K , ω, h0 > 0 ∀ h ∈ (0,h0] ∀ (n, m) ∈ Δ: ‖Cm(hλ(tn))‖ 6 K e−ωtm.

In this case the results of Theorem4.2can be improved to

(a′) ‖un‖ 6 K exp{[(log 2)1/2β − ω]tn}‖u0‖ ∀ n ∈ N0,

(b′) ‖un‖ 6 12{(1 + βh)n + (1 − βh)n}K e−ωtn‖u0‖ ∀ n ∈ N0,

whereβ :=√

K L eωh0. These bounds are similar to those already encountered in Sections 2 and 3.Considering (a′), for example, the condition for the exponent(log 2)1/2β − ω to be negative is qualita-tively similar to conditions found in Theorem2.1for the continuous problem to be exponentially stable.We observe thatL depends linearly on the Lipschitz constant forλ(∙) (see the proof of Theorem4.2below) andω is closely related to the exponential decay rate for the frozen time continuous systems.

Quantitatively, as observed for the backward Euler method in Remark3.5, the exponent(log 2)1/2β−ω alsodepends on the method. For (a′), L alsodepends on the Lipschitz constant for the companion ma-trix (see the proof of Theorem4.2). As for backward Euler, bounds (a′) and (b′) may be sharpened inthe nonstiff regime,h‖λ(∙)‖∞ � 1, by first bounding the continuous problem and then bounding themethod using an approximation argument.

In the remainder of this section we prove Theorem4.2. To this end, we quote some results from theliterature and prove two lemmas.

PROPOSITION4.4 (Dahlquistet al.,1983, Theorem 3 andHairer & Wanner, 1991, Lemmas V.7.3 andV.7.4) ConsiderC(∙) as defined in (4.4). If D ⊆ S is closedin C then

∃ K > 0 ∀ n ∈ N0 ∀ z ∈ D: ‖Cn(z)‖ 6 K . (4.10)

If D ⊂ int[S ] is closedin C then

∃ K , ω > 0 ∀ n ∈ N0 ∀ z ∈ D: ‖Cn(z)‖ 6 K e−ωn. (4.11)

REMARK 4.5 Without loss of generality, it may be assumed thatK > 1 in (4.10) and (4.11).

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LEMMA 4.6 (A uniformly bounded companion matrix is uniformly Lipschitz) ConsiderC as defined in(4.4). Suppose thatD is closedin C, and that

∃ K > 0 ∀ z ∈ D: ‖C(z)‖ 6 K . (4.12)

Then

∃ LC > 0 ∀ z, w ∈ D: ‖C(z) − C(w)‖ 6 LC|z − w|. (4.13)

Proof. From (4.12) we deduce that

∃ K0 > 0 ∀ z ∈ D ∀ j ∈ {0, . . . ,q − 1}: |cj (z)| 6 K0. (4.14)

Sincethe method is irreducible, there is noz ∈ C such thatα j = β j z for all j ∈ {0,1, . . . ,q}.Consequently, (4.14) implies that

∃ K1 > 0 ∀ z ∈ D:1

|αq − zβq|6 K1.

Hence

∃ K2 > 0 ∀ z ∈ D ∀ j ∈ {0, . . . ,q − 1}: |c′j (z)| =

|αqβ j − βqα j |

|αq − zβq|26 K2.

Thuswe obtain (4.13). �

LEMMA 4.7 Consider (4.3) and suppose thatλ: R>0 → C satisfies(4.2), D ⊆ S is closed and (4.9) issatisfied. Then, in the notation of (4.4) and (4.5), we have

∃ L > 0 ∀ h > 0 ∀ (n, m) ∈ D: ‖An − Am‖ 6 Lh2|n − m|. (4.15)

Proof. By (4.9), (4.13) and (4.2), we have

‖An − Am‖ = ‖C(hλ(tn)) − C(hλ(tm))‖ 6 LCh|λ(tn) − λ(tm)| 6 LC Lλh2|n − m|.

Proof of Theorem4.2. The hypotheses of Theorem2.3 are implied, firstly by the assumption onDand Proposition4.4, and secondly by the assumption on (4.9) together with Lemma4.7. Noting thatL = Lh2 andβ = βh, we obtain (a) and (b) from parts (iii) and (iv) of Theorem2.3. �

5. Proofs

5.1 Continuous-time systems

LEMMA 5.1 For (K , ω, L) ∈ R>1 × R>0 × R>0 and f : R>0 → R>0 the following statements areequivalent:

(i) ∀ A ∈ SK ,ω,L ∀ slnsu of (1.1)∀ (t, s) ∈ D: ‖u(t)‖ 6 f (t − s)‖u(s)‖,(ii) ∀ A ∈ SK ,ω,L ∀ slnsu of (1.1)∀ t > 0: ‖u(t)‖ 6 f (t)‖u(0)‖.

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Proof. It suffices to prove ‘(ii)⇒ (i)’. Let u be a solution ofu(t) = A(t)u(t) satisfying (ii). For arbitrarybut fixeds> 0 we have

d

dtu(t + s) = A(t + s)u(t + s) and A(∙ + s) ∈ SK ,ω,L .

Thus(ii) yields

∀ t > 0: ‖u(t + s)‖ 6 f (t)‖u(0 + s)‖.

Settingτ := t + s implies that

∀ τ > s: ‖u(τ )‖ 6 f (τ − s)‖u(s)‖,

and (i) follows sinces is arbitrary. �In the context of exponential stability, the above lemma is of particular interest when

f (t) := M e−ηt for some(M, η) ∈ R>1 × R>0.

LEMMA 5.2 (Primary integral inequality) Suppose thatA ∈ SK ,ω,L for some(K , ω, L) ∈ R>1 ×R>0 × R>0.

Thenevery solutionu of (1.1) satisfies

∀ t > 0 ∀ ρ > 0: ‖u(t)‖ 6 K e−ωt‖u(0)‖ + K L∫ t

0|s − ρ|e−ω(t−s)‖u(s)‖ds. (5.1)

Proof. Givenρ > 0, then every solutionu of (1.1) satisfies

u(t) = A(ρ)u(t) + [ A(t) − A(ρ)]u(t), t > 0.

By the variation of constants formula, we have

u(t) = eA(ρ)t u(0) +∫ t

0eA(ρ)(t−s)[ A(s) − A(ρ)]u(s)ds, t > 0.

Taking norms and invoking the properties of the classSK ,ω,L , we have

‖u(t)‖ 6 K e−ωt‖u(0)‖ +∫ t

0K e−ω(t−s)L|s − ρ|‖u(s)‖ds, t > 0,

andwe obtain (5.1). �The number of independent parameters is reduced by the following scaling lemma.

LEMMA 5.3 (Scaling and the functionr ) Suppose thatA ∈ SK ,ω,L for some(K , ω, L) ∈ R>1×R>0×R>0, andr : R>0 → R>0 is a bounded piecewise continuous function. For a solutionu of (1.1) suchthatu(0) 6= 0, the function

U : R>0 → R>0, U (t) := eωt/α ‖u(t/α)‖

K‖u(0)‖, whereα :=

√K L, (5.2)

satisfies the integral inequality

U (t) 6 1 +∫ t

0|s − r (t)|U (s)ds ∀ t > 0. (5.3)

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Proof. For the given functionr we may takeρ = r (αt)/α in (5.1) to obtain

eωt‖u(t)‖ 6 K‖u(0)‖ + α2∫ t

0|s − r (t)/α|eωs‖u(s)‖ds ∀ t > 0.

Dividing by‖u(0)‖ > 0 and scaling time by 1/α, we obtain (5.3). �

LEMMA 5.4 (A comparison inequality) Suppose for somet0 > 0 thatr : [0, t0] → R>0 andw: [0, t0] →R arebounded piecewise continuous functions satisfying

w(t) 6∫ t

0|s − r (t)|w(s)ds ∀ t ∈ [0, t0]. (5.4)

Then

w(t) 6 0 ∀ t ∈ [0, t0]. (5.5)

Proof. From (5.4) we have

w(t) 6∫ t

0|s − r (t)|w(s)ds6

∫ t

0|s − r (t)|w+(s)ds ∀ t ∈ [0, t0],

wherew+ := max{w, 0}. Since the RHS is non-negative and

∀ t ∈ [0, t0] ∀ s ∈ [0, t ]: |s − r (t)| 6 max{s, r (t)} 6 max{t, r (t)} =: R(t),

we conclude that

w+(t) 6 R(t)∫ t

0w+(s)ds ∀ t ∈ [0, t0]. (5.6)

Let {0 = τ0, τ1, . . . , τn = t0} be a partition of [0, t0] such that the restrictions ofR andw+ to eachsubinterval(τm, τm+1) arecontinuous. Then (5.6) implies that, forI (t) :=

∫ t0 w+(s)ds, wehave

∀ m ∈ {0,1, . . . ,n − 1} ∀ t ∈ (τm, τm+1):d

dtI (t) 6 R(t)I (t).

Thefundamental theorem of calculus now implies that

∀ m ∈ {0,1, . . . ,n − 1} ∀ t ∈ [τm, τm+1]: I (t) exp

(−∫ t

τm

R(s)ds

)6 I (τm). (5.7)

Taking t = τm+1 in (5.7), and observing thatI (0) = 0, it follows that I (τm) 6 0 for m = 0,1, . . . ,n,by induction. Inequality (5.7) then implies thatI (t) 6 0, wheret ∈ [0, t0]. Inequality (5.5) now followsfrom (5.6). �

LEMMA 5.5 (Integral supersolutions yield upper bounds) Suppose thatA ∈ SK ,ω,L for some(K ,ω, L) ∈ R>1 × R>0 × R>0. Suppose also thatv: R>0 → R>0 and r : R>0 → R>0 areboundedpiecewise continuous functions satisfying

v(t) > 1 +∫ t

0|s − r (t)|v(s)ds ∀ t ∈ [0, t0], (5.8)

for somet0 > 0. Then the functionU defined by (5.2) satisfies

U (t) 6 v(t) ∀ t ∈ [0, t0]. (5.9)

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Proof. By (5.3), we have

U (t) 6 1 +∫ t

0|s − r (t)|U (s)ds ∀ t ∈ [0, t0]. (5.10)

The function w: [0, t0] → R, defined byw(t) := U (t) − v(t), is bounded piecewise continuous.Furthermore, by subtracting (5.10) from (5.3) we deduce that

w(t) 6∫ t

0|s − r (t)|w(s)ds ∀ t ∈ [0, t0].

Inequality(5.9) now follows from Lemma5.4. �

LEMMA 5.6 (Sufficient conditions for a supersolution) Suppose for somet0 > 0 thatv: [0, t0] → R>0andr : [0, t0] → R>0 arecontinuous functions, withr differentiable on [0, t0] except at a finite numberof points. Suppose thatv(0) > 1 and that the following inequalities are satisfied for almost allt ∈[0, t0] :

t > r (t), (5.11)

r (t) > 0, (5.12)

v(t) > 2r (t)v(r (t)), (5.13)

v(t) > (t − r (t))v(t). (5.14)

Thenv andr satisfy (5.8).

Proof. Integrating (5.13), we obtain that

∫ t

0v(s)ds> 2

∫ t

0r (s)v(r (s))ds = 2

∫ r (t)

0v(s)ds ∀ t ∈ [0, t0].

Hence,applying (5.11), we obtain

0>∫ r (t)

0v(s)ds−

∫ t

r (t)v(s)ds =

∫ t

0sign[r (t) − s]v(s)ds ∀ t ∈ [0, t0]. (5.15)

For all but a finite number oft ∈ [0, t0], elementary calculus implies that

d

dt

∫ t

0|r (t) − s|v(s)ds = (t − r (t))v(t) + r (t)

∫ t

0sign[r (t) − s]v(s)ds.

Hence, applying (5.12) and (5.14), we conclude that

v(t) > (t − r (t))v(t) + r (t)∫ t

0sign[r (t) − s]v(s)ds =

d

dt

∫ t

0|r (t) − s|v(s)ds

for all but a finite set oft ∈ [0, t0]. Integration and the inequalityv(0) > 1 imply that vsatisfies (5.8). �

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LEMMA 5.7 (Supersolution suitable for smallt) The functionsv1, r1: R>0 → R>0, defined by

v1(t) := et2/4, r1(t) := t/2 ∀ t > 0, (5.16)

satisfy

v1(t) > 1 +∫ t

0|s − r1(t)|v1(s)ds ∀ t > 0. (5.17)

Proof. Verifying the hypotheses of Lemma5.6, we observe that the conditions (5.11) and (5.12) aresatisfied byr1(t) for all t > 0, whilev1(0) = 1,

v1(t) =t

2et2/4 = (t − r1(t))v1(t), v1(t) > v1(t/2) = 2r1(t)v1(r1(t)), t > 0.

Hence(5.8) is satisfied for allt0 > 0, and so we obtain (5.17) by Lemma5.6. �

LEMMA 5.8 (Long-time supersolution) Letv2, r2: R>0 → R bedefined by

r2(t) = max{t − c, 0},

v2(t) = exp(ct)

}

t > 0, c :=√

log2. (5.18)

Then

v2(t) > 1 +∫ t

0|s − r2(t)|v2(s)ds ∀ t > 0. (5.19)

Proof. Verifying the hypotheses of Lemma5.6, we see that conditions (5.11) and (5.12) onr2 aresatisfied,except att = c. Also,v2(0) = 1 and

v2(t) = cect = (t − (t − c))ect > (t − max{t − c, 0})ect = (t − r2(t))v2(t) ∀ t > 0.

Sincer2(t) = 0 for t ∈ [0, c), (5.13) also holds fort ∈ [0, c). Sincer2(t) = 1 for t > c, we have

2r2(t)v2(r2(t)) = 2ec(t−c) = ect = v2(t) ∀ t > c,

andso (5.13) holds for allt ∈ R>0 \ {c}. Applying Lemma5.6, we deduce that (5.8) is satisfied for allt0 > 0. Hence (5.19) follows from Lemma5.6. �

LEMMA 5.9 (Short-time bound implies a long-time bound) Suppose for(K , ω, γ ) ∈ R>1×R>0×R>0thatu: R>0 → R>0 satisfies

u(t) 6 K exp{γ 2(t − s)2/4 − ω(t − s)}u(s) ∀ (t, s) ∈ D. (5.20)

Then

u(t) 6 K exp{(γ√

log K − ω)t }u(0) ∀ t > 0. (5.21)

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Proof. Sett1 := 2γ−1√

log K . Then (5.20) implies that

u((n + 1)t1) 6 exp((γ√

log K − ω)t1)u(nt1) ∀ n ∈ N0.

Hence

u(nt1) 6 exp((γ√

log K − ω)nt1)u(0) ∀ n ∈ N0. (5.22)

For n ∈ N0 andτ ∈ [0, t1), takingt = nt1 + τ ands = nt1 in (5.20) implies that

u(t) 6 K exp(γ 2τ2/4 − ωτ )u(nt1) 6 K exp((γ√

log K/2 − ω)τ )u(nt1).

Combiningwith (5.22), we obtain (5.21). �

Proof of Theorem2.1. If u(0) = 0 then both (i) and (ii) are clear. Assume now thatu(0) 6= 0.

Proof of (i). The functionsr1: R>0 → R>0 andv1: R>0 → R>0 definedin Lemma5.7 satisfy (5.8)and the other conditions of Lemma5.5for all t0 > 0. Consequently, by (5.2), we have

eωt/α ‖u(t/α)‖

K‖u(0)‖6 v1(t) = et2/4 ∀ t > 0,

whereα :=√

K L. Taking the scalingt = αt and multiplying byK‖u(0)‖e−ωt , we obtain that

‖u(t)‖ 6 K exp(α2t2/4 − ωt)‖u(0)‖ ∀ t > 0, (5.23)

which is inequality (i) in the statement of Theorem2.1for s = 0. By Lemma5.1, we deduce the generalcase for(t, s) ∈ D.

Proof of (ii) when K∈ [1, 2]. We observe that bound (i) implies that the hypotheses of Lemma5.9 aresatisfied foru = ‖u(∙)‖ andγ = α. Hence we obtain bound (ii) fors = 0. The general case(t, s) ∈ Dnow follows from Lemma5.1.

Proof of (ii) when K> 2. The functionsr2: R>0 → R>0 andv2: R>0 → R>0 given by Lemma5.8satisfy (5.8) and the other conditions of Lemma5.5for all t0 > 0. Consequently, by (5.2), we have

eωt/α ‖u(t/α)‖

K‖u(0)‖6 v2(t) = ect ∀ t > 0,

wherec :=√

log2. Taking the scalingt = αt and multiplying byK‖u(0)‖e−ωt , we obtain that

‖u(t)‖ 6 K‖u0‖ exp((√

K L log 2− ω)t) ∀ t > 0.

Applying Lemma5.1, we obtain statement (ii) of Theorem2.1for K > 2. �

5.2 Discrete-time systems

LEMMA 5.10 For (K , ω, L) ∈ R>1 × R>0 × R>0 and f : N0 → R>0 the following statements areequivalent:

(i) ∀ A ∈ ΔK ,ω,L ∀ slnsu of (1.3)∀ (n, m) ∈ Δ: ‖un‖ 6 f (n − m)‖um‖,(ii) ∀ A ∈ ΔK ,ω,L ∀ slnsu of (1.3)∀ n ∈ N0: ‖un‖ 6 f (n)‖u0‖.

We omit the proof since it is analogous to the proof of Lemma5.1.

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LEMMA 5.11 (Primary summation inequality) Suppose thatA ∈ ΣK ,ω,L for some(K , ω, L) ∈ R>1 ×R>0 × R>0. Then every solutionu of (1.3) satisfies

‖un‖ 6 K e−ωn‖u0‖ +n−1∑

i =0

K e−ω(n−1−i )L|i − ρ|‖ui ‖ ∀ n ∈ N, ∀ ρ > 0. (5.24)

Proof. Let u: N0 → CN bea solution of (1.3). We first consider the special case ofρ ∈ N0. Then

un+1 = Aρun + (An − Aρ)un ∀ n ∈ N0.

By the discrete variation of constants formula, we have

un = Anρu0 +

n−1∑

i =0

An−1−iρ (Ai − Aρ)ui ds ∀ n ∈ N0.

Taking norms and invoking the properties of the classΔK ,ω,L , (5.24) follows.Suppose now thatρ = θk + (1 − θ)(k + 1) for k ∈ N0 andθ ∈ (0,1). Since no integeri satisfies

k < i < k + 1, we have

∀ i ∈ N0: θ |i − k| + (1 − θ)|i − (k + 1)| = |θ(i − k) + (1 − θ)(i − (k + 1))| = |i − ρ|.

Hence (5.24) follows forρ by taking a weighted averageθ and(1−θ) of (5.24) forρ = k andρ = k+1,respectively. Thus we obtain (5.24). �

5.2.1 Connection to the continuous case

LEMMA 5.12 (The summation inequality implies a scaled integral inequality) Suppose thatA ∈ ΣK ,ω,L

for some(K , ω, L) ∈ R>1 × R>0 × R>0 andthatr : R>0 → R>0 is a bounded piecewise continuousfunction. Foru: N0 → CN asolution of (1.3) withu0 6= 0, the bounded piecewise continuous function

U : R>0 → R>0, t 7→ U (t) :=eωn‖un‖

K‖u0‖, n = bt/βc, t > 0, β =

√K L eω, (5.25)

satisfiesthe integral inequality (5.3), namely,

U (t) 6 1 +∫ t

0|s − r (t)|U (s)ds ∀ t > 0.

Proof. We first observe that

m+1∫

m

|s − ρ|ds =

{|m − (ρ − 1/2)|, ρ ∈ R>0 \ (m, m + 1),

(m−ρ)2

2 + (m+1−ρ)2

2 > |m − (ρ − 1/2)|, ρ ∈ (m, m + 1).(5.26)

We define the following bounded piecewise continuous function:

U : R>0 → R>0, t 7→ U (t) :=eωn‖un‖

K‖u0‖, n = btc, t > 0.

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Givent > 1, letn = btc. Then (5.26) withρ = r (βt)/β and Lemma5.11imply that

U (t) :=eωn‖un‖

K‖u0‖6 1 + K L eω

n−1∑

m=0

|m − (r (βt)/β − 1/2)|eωm ‖um‖

K‖u0‖

6 1 + β2n−1∑

m=0

∫ m+1

m|s − r (βt)/β|U (s)ds

6 1 + β2∫ t

0|s − r (βt)/β|U (s)ds ∀ t > 1.

Givent ∈ [0, 1), we have 06 U (t) = 1/K 6 1.Combining these two estimates, we deduce that

U (t) 6 1 + β2∫ t

0|s − r (βt)/β|U (s)ds ∀ t > 0.

NotethatU (t) = U (t/β), wheret > 0, and we deduce thatU satisfies (5.3). �

LEMMA 5.13 (Comparison result for a continuous supersolution) Suppose thatA ∈ ΣK ,ω,L for some(K , ω, L) ∈ R>1 × R>0 × R>0 andthatv: R>0 → R>0 andr : R>0 → R>0 arebounded piecewisecontinuous functions satisfying

v(t) > 1 +∫ t

0|s − r (t)|v(s)ds ∀ t > 0.

Then, foru: N0 → CN asolution of (1.3), we have

‖un‖ 6 K e−ωnv(βn)‖u0‖, n ∈ N0, β =√

K L eω. (5.27)

Proof. If u0 = 0 then the result is trivial. Else,u0 6= 0, and by Lemma5.12the functionU defined by(5.25) satisfies (5.3). So, by Lemma5.5, we have

eωn‖un‖

K‖u0‖= U (βn) 6 v(βn), n ∈ N0.

Hencewe obtain (5.27). �

5.2.2 A direct discrete approach

LEMMA 5.14 (A scaled summation inequality) Suppose thatA ∈ ΣK ,ω,L for some(K , ω, L) ∈ R>1 ×R>0 × R>0, thatu is a solution of (1.3) withu0 6= 0 and thatr : N → R>0. Then the function

U : N0 → R>0, n 7→ Un :=eωn‖un‖

K‖u0‖(5.28)

satisfiesthe inequality

Un 6 1 + β2n−1∑

m=0

|m − r (n)|Um, n ∈ N, β =√

K L eω. (5.29)

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Proof. For n ∈ N we observe that we may chooseρ = r (n) in Lemma5.11to obtain

‖un‖ 6 K e−ωn‖u0‖ + K L eωn−1∑

m=0

e−ω(n−m)|m − r (n)|‖um‖, n ∈ N.

Dividing both sides byK e−ωn‖u0‖, we obtain (5.29). �

REMARK 5.15 Unlike the continuous case, the factorβ is not scaled out in (5.29).

LEMMA 5.16 (Discrete supersolution and comparison result) Suppose thatA ∈ ΣK ,ω,L for some(K , ω, L) ∈ R>1 × R>0 × R>0. Suppose also that the sequence(vn)n∈N0 with real elements satis-fiesv0 > 1/K and

vn > 1 + β2n−1∑

m=0

|m − r (n)|vm, n ∈ N, β =√

K L eω, (5.30)

for some functionr : N → R>0. Then every solutionu of (1.3) satisfies

‖un‖ 6 K e−ωnvn‖u0‖, n ∈ N0. (5.31)

Proof. If u0 = 0 then the result is trivial. Ifu0 6= 0 then consider the functionw: N0 → R>0,n 7→ wn := Un − vn, where the sequence(Un) is as in (5.28). Subtracting (5.30) from (5.29), we obtainthat

wn 6 β2n−1∑

m=0

|m − r (n)|wm, n ∈ N.

Sincew0 = U0 − v0 = 1/K − v0 6 0, the inequalitywn 6 0 for n ∈ N0 follows by induction. Hencewe deduce (5.31). �

LEMMA 5.17 (Discrete supersolution suitable for largen) Suppose thatβ > 0, and let the real sequence(vn)n∈N0 bedefined by

vn :=1

2{(1 + β)n + (1 − β)n}, n ∈ N0. (5.32)

Let r : N0 → R>0 bedefined by

r (n) := n − 1, n ∈ N. (5.33)

Thenv0 = 1 and

vn = 1 + β2n−1∑

m=0

|m − r (n)|vm, n ∈ N. (5.34)

Proof. Assume thatγ ∈ R \ {1} and defineSn(γ ) :=∑n−1

m=0(n − 1 − m)γ m, wheren ∈ N. Then

Sn+1(γ ) − Sn(γ ) =n−1∑

m=0

γ m =γ n − 1

γ − 1, n ∈ N.

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SinceS1(γ ) = 0, we have

Sn(γ ) =n−1∑

m=1

[Sm+1(γ ) − Sm(γ )] =n−1∑

m=1

γ m − 1

γ − 1=

γ n − 1 − n(γ − 1)

(γ − 1)2, n ∈ N.

Thus,for (vn) andr asin (5.32) and (5.33), respectively, we have

1 + β2n−1∑

m=0

|m − r (n)|vm = 1 + β2n−1∑

m=0

(n − 1 − m)vm

= 1 +β2

2{Sn(1 + β) + Sn(1 − β)}

= 1 +1

2{[(1 + β)n − 1 − nβ] + [(1 − β)n − 1 + nβ]} = vn, n ∈ N,

andv0 = 1. �

Proof of Theorem2.3. Bound (i) follows from the inequality

‖un+1‖ = ‖Anun‖ 6 ‖An‖‖un‖ 6 K e−ωn‖un‖, n ∈ N0.

Bounds(ii) and (iii) follow from the comparison result, Lemma5.13, and the properties of the continu-ous supersolutionsv1 andv2 shown in Lemmas5.7and5.8, respectively.

Bound (iv) follows from the discrete comparison result, Lemma5.16, and the properties of thesupersolution (5.32) shown in Lemma5.17. �

Notation

bxc := max{n ∈ Z|n 6 x}, x ∈ R,

sign[x] :=

{x/|x|, x ∈ R \ {0},

0, x = 0,

R>p := [ p, ∞), p ∈ R,

R>p := (p, ∞), p ∈ R,

D := {(t, s) ∈ R>0 × R>0|t > s},

Δ := {(n, m) ∈ N0 × N0|n > m},

μ(A) := max{|λ| |λ ∈ C and det(λI − A) = 0} for A ∈ CN×N .

Acknowledgement

The authors would like to express their thanks to the referees for their careful reading of the manuscriptand their suggestions for improving the presentation.

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