Nonauton. Dyn. Syst. 2015; 2:63–76
Research Article
Open Access
Martin Bohner* and Nasrin Sultana*
Subexponential Solutions of Linear Volterra
Difference Equations
DOI 10.1515/msds-2015-0005
Received March 5, 2015; accepted September 17, 2015
Abstract: We study the asymptotic behavior of the solutions of a scalar convolution sum-difference equation.
The rate of convergence of the solution is found by determining the asymptotic behavior of the solution of
the transient renewal equation.
Keywords: Subexponential, transient renewal, convolutions, Banach space, linear operator
MSC: 39A10, 39A06, 39A22, 45D05, 45J05
1 Introduction
We consider the discrete equation
∆x(t) = −ax(t) +
t−1
X
k(t − 1 − s)x(s), t ∈ N0 ,
(1.1)
s=0
where N0 = {0, 1, 2, . . .} (as usual, we adopt the convention that a sum running from m to n with m > n is
equal to zero). We suppose that a ∈ (0, 1) and k : N0 → (0, ∞). We show that if
∞
X
k(s) < a,
s=0
then all solutions x of (1.1) satisfy x(t) → 0 as t → ∞ with the rate of convergence
lim
t→∞
x(t)
x(0)
P
=
2
k(t) (a − ∞
s=0 k(s))
provided k is in a class of subexponential sequences. The result is proved by determining the asymptotic
behavior of the solution of the transient renewal equation
r(t) = h(t) +
t−1
X
h(t − 1 − s)r(s),
where
s=0
∞
X
h(s) < 1.
(1.2)
s=0
If h is subexponential, then the solution r of (1.2) satisfies
lim
t→∞
r(t)
1
P
=
.
2
h(t) (1 − ∞
s=0 h(s))
The same problem has been studied in [2] for corresponding linear integro-differential equations. For basic
properties and formulas concerning difference equations, we refer the reader to [1, 4, 6]. We also refer to [3, 5]
for basic results on the existence of solutions of scalar linear Volterra equations and Lyapunov functionals in
the continuous case.
*Corresponding Author: Martin Bohner, Nasrin Sultana: Department of Mathematics, Missouri University of Science and
Technology, Rolla, MO 65409-0020 USA, E-mail: [email protected], [email protected]
© 2015 Martin Bohner and Nasrin Sultana, licensee De Gruyter Open.
This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 License.
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64 | Martin Bohner and Nasrin Sultana
2 Preliminaries
For f , g : N0 → R, we define the convolution of f and g (see [4, Section 3.10]) by
(f * g)(t) =
t−1
X
f (t − 1 − s)g(s),
t ∈ N0 .
s=0
(Note that this definition is slightly different from that one appearing elsewhere in the literature, e.g., in [6,
page 100].) The n-fold convolution f *n is given by f *1 = f and f *(n+1) = f * f *n for n ∈ N.
Definition 2.1. A subexponential sequence is a discrete mapping h : N0 → (0, ∞) with
P
*2
(S1 ) limt→∞ hh(t)(t) = 2 ∞
s=0 h(s);
(S2 ) limt→∞
h(t−s)
h(t)
P∞
s=0
h(s) < ∞ and
= 1 for each fixed s ∈ N0 .
The class of subexponential sequences is denoted by U.
Example 2.2. Consider h(t) =
1
,
(t+b)n
where t ∈ N0 , b ∈ N \ {1}, and n ∈ N \ {1}. Then h ∈ U.
Proof. We prove the statement in the case n = 2 and b = 2. Therefore h(t) =
P
(S2 ), (S2a ), and ∞
s=0 h(s) < 1. To see that h satisfies (S1 ), we calculate
1
,
(t+2)2
t ∈ N0 , clearly satisfies
t−1
h*2 (t) X h(s)h(t − 1 − s)
=
h(t)
h(t)
s=0
2
t−1
X
t+2
=
(s + 2)(t + 1 − s)
s=0
2 2
t−1 X
1
1
t+2
=
+
s+2 t+1−s
t+3
s=0
t+2
t+3
2
t+2
t+3
2
=
=
Now it is enough to show
Pt−1
t−1
X
s=0
2
s=0 (s+2)(t+1−s)
0≤
t−1
X
s=0
t−1
t−1
X
X
1
2
1
+
+
2
(s + 2)
(s + 2)(t + 1 − s)
(σ + 2)2
s=0
s=0
σ=0
!
t−1
t−1
X
X
1
1
2
+2
.
(s + 2)2
(s + 2)(t + 1 − s)
s=0
→ 0 as t → ∞:
t−1 1
1
1
1 X
=
+
s+2 t+1−s
(s + 2)(t + 1 − s) t + 3
s=0
t−1
2 X 1
=
t+3
s+2
s=0
≤
2
t+3
Zt+1
dx
x
1
=
2 ln(t + 1)
t+3
→ 0 as t → ∞.
Thus h satisfies (S1 ).
Remark 2.3. Condition (S2 ) is equivalent to
(S2a ) limt→∞
h(t+1)
h(t)
= 1,
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!
Subexponential Solutions |
65
and it is also equivalent to
(S2b ) limt→∞ sup0≤s≤T h(t−s)
−
1
= 0 for each T ∈ N0 .
h(t)
We show that (S2 ) and (S2a ) are equivalent: If h satisfies (S2 ), then (take s = 1)
lim
t→∞
h(t − 1)
h(t)
= 1, so lim
= 1,
t→∞
h(t)
h(t + 1)
and hence (S2a ) holds. If h satisfies (S2a ), then
lim
t→∞
and thus
h(t + v + 1)
= 1 for any v ∈ Z,
h(t + v)
−1
Y
h(t + v + 1)
h(t)
=
→ 1 as t → ∞ for all s ∈ N0 ,
h(t − s) v=−s h(t + v)
and hence (S2 ) holds.
Remark 2.4. The terminology “subexponential” is justified by the following: If k : N0 → (0, ∞) satisfies
(S2a ), then
(1 − a)t
lim
= 0 for all a ∈ (0, 1).
(2.1)
t→∞
k(t)
We prove (2.1): Define
K(t) :=
Then
k(t)
, t ∈ N0 .
(1 − a)t
K(t + 1)
k(t + 1)
1
=
→
as t → ∞.
1−a
K(t)
(1 − a)k(t)
(2.2)
Hence there exists T ∈ N0 so that
1
2−a
K(t + 1)
a
−
>
=
> 1 for all t ≥ T.
1 − a 2(1 − a) 2(1 − a)
K(t)
Thus
K(t + 1) > K(t) for all t ≥ T.
Hence K is eventually increasing, so either
lim K(t) = ∞
(2.3)
lim K(t) =: K * ∈ (0, ∞).
(2.4)
t→∞
or otherwise
t→∞
Since (2.4) implies limt→∞
K(t+1)
K(t)
= 1, a contradiction to (2.2), we must have (2.3). Thus (2.1) holds.
Remark 2.5. If h ∈ U, then there exists M h ∈ (0, ∞) such that
sup
t∈N0
h*2 (t)
= Mh .
h(t)
Lemma 2.6. Let h ∈ U, n ∈ N, and
µ=
∞
X
h(s).
(2.5)
(2.6)
s=0
Then
∞
X
h*n (s) = µ n .
s=0
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(2.7)
66 | Martin Bohner and Nasrin Sultana
Proof. For n = 2, we have
t−1
X
h*2 (s) =
s=0
t−1 X
s−1
X
h(s − 1 − u)h(u)
s=0 u=0
=
t−2 X
t−1
X
h(s − 1 − u)h(u)
u=0 s=u+1
=
t−2 t−2−u
X
X
h(v)h(u).
u=0 v=0
As t → ∞, we obtain
∞
X
h*2 (s) =
s=0
∞ X
∞
X
h(v)h(u) = µ2 .
u=0 v=0
Hence (2.7) holds for n = 2. Now assume (2.7) holds for some n ∈ N \ {1}. Then
t−1
X
h*(n+1) (s) =
s=0
t−1 X
s−1
X
h*n (s − 1 − u)h(u).
s=0 u=0
Following the same calculation as above, we get
∞
X
h*(n+1) (s) =
∞ X
∞
X
h*n (v)h(u) = µ n+1 .
u=0 v=0
s=0
By the principle of mathematical induction, the proof is complete.
Lemma 2.7. Let h ∈ U and assume µ < 1, where µ is defined as in (2.6). Let ε ∈ (0, 1) satisfy (1 + 4ε)µ < 1.
Then there exists C0 ∈ N and λ ≥ 1 such that
n−2
sup
t≥C0
X
h*n (t)
≤ max{λ, 1 + ε}µ n−1
(1 + 4ε)k + 2(1 + ε)µ n−1 (1 + 4ε)n−2
h(t)
(2.8)
k=0
for all n ∈ N \ {1}.
Proof. For a given ε > 0, due to h > 0, (2.6), and (S1 ) we can choose C0 ∈ N such that for all t ≥ C0 ,
Pt−1
(a)
s=0 h(s) > (1 − ε)µ,
(b)
h*2 (t)
h(t)
≤ 2µ(1 + ε),
and also due to h > 0 and (S2 ) we can choose an integer T0 > C0 such that for all t ≥ T0 ,
(c) h(t−s)
−
1
< ε for all 0 ≤ s ≤ C0 .
h(t)
Let
λ = λ(C0 , T0 ) := max
h(t − 1 − s)
: 0 ≤ s ≤ t − 1 and C0 ≤ t ≤ T0 − 1 .
h(t)
In order to prove (2.8), we will use the method of induction. Clearly, for n = 2, (2.8) holds by (b). Assume that
(2.8) is true for some n ∈ N \ {1}.
Case 1. If t ≥ T0 , then using (c) and the induction hypothesis, we have
CX
C0 −1
t−1
0 −1
X
h*n (s) h(s)h(t − 1 − s)
h*(n+1) (t) X *n
h(t − 1 − s)
=
h (s)
−1 +
h*n (s) +
h(t)
h(t)
h(s)
h(t)
s=0
≤ (1 + ε)
s=0
CX
0 −1
s=C0
h*n (s)
(2.9)
s=0
+
max{λ, 1 + ε}µ
n−1
n−2
X
!
k
(1 + 4ε) + 2(1 + ε)µ
n−1
(1 + 4ε)
n−2
k=0
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t−1
X
h(s)h(t − 1 − s)
.
h(t)
s=C0
Subexponential Solutions |
67
Now, if t ≥ T0 , then using (c) and (a), we obtain
C0 −1
t−1
t−1
X
h(s)h(t − 1 − s) X h(s)h(t − 1 − s) X h(s)h(t − 1 − s)
=
−
h(t)
h(t)
h(t)
s=0
s=0
s=C0
!
CX
0 −1
h(t − 1 − s)
h(s)
−1 +
h(s)
h(t)
s=0
s=0
!
CX
CX
0 −1
0 −1
h(s)
h(s) +
≤ 2µ(1 + ε) − −ε
h*2 (t)
−
=
h(t)
CX
0 −1
s=0
s=0
= 2µ(1 + ε) − (1 − ε)
CX
0 −1
h(s)
s=0
≤ 2µ(1 + ε) − (1 − ε)2 µ
= (2 + 2ε − 1 + 2ε − ε2 )µ
≤ (1 + 4ε)µ.
(2.10)
Using (2.10) and Lemma 2.6 in (2.9), if t ≥ T0 , then we get
h*(n+1) (t)
≤ (1 + ε)µ n +
h(t)
max{λ, 1 + ε}µ
n−1
n−2
X
!
k
(1 + 4ε) + 2(1 + ε)µ
n−1
(1 + 4ε)
n−2
(1 + 4ε)µ
k=0
n−1
X
= (1 + ε)µ n + max{λ, 1 + ε}µ n
(1 + 4ε)k + 2(1 + ε)µ n (1 + 4ε)n−1 .
(2.11)
k=1
Case 2. Now suppose C0 ≤ t ≤ T0 − 1. Then
t−1
t−1
s=0
s=0
X
h*(n+1) (t) X *n h(t − 1 − s)
=
≤λ
h (s)
h*n (s) ≤ λµ n .
h(t)
h(t)
(2.12)
Therefore, if t ≥ C0 , then (2.11) and (2.12) imply
n−1
X
h*(n+1) (t)
≤ max{λ, 1 + ε}µ n + max{λ, 1 + ε}µ n
(1 + 4ε)k + 2(1 + ε)µ n (1 + 4ε)n−1
h(t)
k=1
= max{λ, 1 + ε}µ n
n−1
X
(1 + 4ε)k + 2(1 + ε)µ n (1 + 4ε)n−1 .
k=0
By the principle of mathematical induction, the proof is complete.
Lemma 2.8. Let f , g : N0 → (0, ∞). Suppose further that g is summable, satisfies (S2 ), and
lim
t→∞
f (t)
=: λ > 0,
g(t)
lim
t→∞
(f * g)(t)
=: ν.
g(t)
Then f is summable, satisfies (S2 ), and
∞
X
(f * f )(t)
=ν+
(f (s) − λg(s)).
t→∞
f (t)
lim
s=0
Proof. The fact that limt→∞
f (t)
g(t)
= λ > 0 implies that f is summable. Since g also satisfies (S2 ),
f (t − s) f (t − s) g(t − s) g(t)
=
·
·
f (t)
g(t − s)
g(t)
f (t)
implies that f must satisfy (S2 ). Now note that, for all t ∈ N, we have
t−1
t−1 X
f (t − 1 − s)
(f * f )(t)
(f * g)(t) X
−λ
−
(f (s) − λg(s)) =
− 1 (f (s) − λg(s)).
f (t)
f (t)
f (t)
s=0
s=0
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(2.13)
68 | Martin Bohner and Nasrin Sultana
To establish the remainder of the assertions, we show that the right-hand side of (2.13) tends to 0 as t → ∞.
Let ε > 0. Then there exists T ∈ N such that
f (t)
g(t) − λ < ε for all t ≥ T.
Hence for t ≥ T
t−1 X
f (t − 1 − s)
− 1 (f (s) − λg(s)) =
f (t)
t−1 X
f (s)
f (t − 1 − s)
−1
− λ g(s)
f (t)
g(s)
s=T
t−1
X
f (t − 1 − s)g(s)
− g(s)
≤ε
f (t)
s=T
!
t−1
(f * g)(t) X
+
g(s)
≤ε
f (t)
s=0
!
∞
ν X
→ε
+
g(s) as t → ∞.
λ
s=T
s=0
Also, it follows from
f (t − 1 − s)
f (t − 1 − s)/g(t − 1 − s)
−1=
f (t)
f (t)/g(t)
g(t − 1 − s)
f (t − 1 − s)/g(t − 1 − s)
−1 +
−1
g(t)
f (t)/g(t)
that f satisfies (S2b ). Hence
T−1
T−1 X
f (t − s)
X
f (t − 1 − s)
− 1 (f (s) − λg(s)) ≤ sup − 1
(f (s) − λg(s)),
0≤s≤T f (t)
f (t)
s=0
s=0
which tends to 0 as t → ∞. Therefore it has been proved that
t−1 X
f (t − 1 − s)
lim sup − 1 (f (s) − λg(s)) ≤ ε
f
(t)
t→∞
s=0
!
∞
ν X
g(s) .
+
λ
s=0
Since ε > 0 is arbitrary small, the proof is complete.
Definition 2.9. Let h ∈ U. Then BCh (N0 ) is defined to be the space of sequences f on N0 such that f = ϕh for
some bounded sequence ϕ on N0 . We will use BCh in short in stead of BCh (N0 ). It is a Banach space equipped
with the norm
f (t) ,
||f ||h = M h sup t≥0 h(t)
where M h is defined in (2.5). We denote by BClh the closed subspace of sequences in BCh for which
L h f := lim
t→∞
f (t)
h(t)
exists. The operator L h : BClh → R is a bounded linear operator on BClh . BC0h is defined to be the closed
subspace of sequences in BClh for which L h f = 0.
Theorem 2.10. Suppose that h ∈ U. Then BCh is a commutative Banach algebra with the convolution as product, and BClh and BC0h are subalgebras. If f , g ∈ BClh , then
L h (f * g) = L h f
∞
X
s=0
g(s) + L h g
∞
X
f (s).
s=0
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(2.14)
Subexponential Solutions |
69
Proof. Let f , g ∈ BCh . Then
Mh
|(f * g)(t)|
h(t)
t−1 X
f (t − 1 − s) g(s) h(t − 1 − s)h(s)
1 h*2 (t)
·
≤ Mh
≤
||
f
||
||
g
||
.
·
h
h
h(t − 1 − s) h(s) M h h(t)
h(t)
s=0
Hence (2.5) implies that ||f * g ||h ≤ ||f ||h ||g ||h . Now let g ∈ BC0h . Then clearly g is summable. Let ε > 0. Then
there exists B > 0 such that
∞
X
g(t) < ε for all t ≥ B and
|g(s)| < ε.
h(t) s=B
Suppose that t ≥ B. Since
∞
B−1 X
(g * h)(t) X
−
g(s) =
h(t)
s=0
s=0
t−1
∞
X
h(t − 1 − s)
1 X g(s)
− 1 g(s) +
h(s)h(t − 1 − s) −
g(s),
h(t)
h(t)
h(s)
s=B
s=B
we have
B−1
∞
t−1
X
h(t − s)
g(s) 1 X
(g * h)(t) X
−
g(s) ≤ sup − 1
h(s)h(t − 1 − s) + ε
|g(s)| + sup h(t)
0≤s≤B h(t)
B≤s≤t−1 h(s) h(t) s=0
s=0
s=0
B−1
h(t − s)
X
εh*2 (t)
≤ sup − 1
|g(s)| +
+ ε.
h(t)
h(t)
0≤s≤B
s=0
By taking the limit superior on both sides as t → ∞ and then using (S1 ) and (S2b ), we get
∞
∞
X
(g * h)(t) X
h(s) + ε.
lim sup −
g(s) ≤ 2ε
h(t)
t→∞ s=0
s=0
Thus we have shown that
L h (g * h) =
∞
X
g(s) for all g ∈ BC0h .
(2.15)
s=0
Also we observe that if t ≥ 2B, then
B−1 t−1
X
g(t − 1 − s) h(t − 1 − s)
g(t − 1 − s)
1 X
−1 +
g(s)
g(t − 1 − s)g(s) ≤ h(t)
h(t − 1 − s)
h(t)
h(t − 1 − s)
s=0
s=0
t−1 1 X g(s) +
h(s) |g(t − 1 − s)|h(s)
h(t)
s=B
B−1
t−1
h(t − s)
X |g(t − 1 − s)|
ε X
≤ 1 + sup − 1
|g(s)| +
|g(t − 1 − s)|h(s)
h(t)
h(t − 1 − s)
h(t)
0≤s≤B
s=0
s=B
∞
h(t − s)
X
(|g | * h)(t)
≤ ε 1 + sup − 1
|g(s)| + ε
.
h(t)
h(t)
0≤s≤B
s=0
Taking limit superior on both sides as t → ∞ and then using (S2b ) and (2.15) implies
∞
t−1
X
1 X
|g(s)|.
lim sup g(t − 1 − s)g(s) ≤ 2ε
t→∞ h(t)
s=0
s=0
Thus we have shown
L h (g * g) = 0 for all g ∈ BC0h .
Now let f ∈ BClh . Then f̃ := f − (L h f )h ∈ BC0h . Therefore the linearity of L h , (S1 ), and (2.15) imply that
L h (f * h) =(L h f )L h (h * h) + L h (f̃ * h)
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(2.16)
70 | Martin Bohner and Nasrin Sultana
= 2L h f
= 2L h f
= Lh f
∞
X
s=0
∞
X
h(s) +
h(s) +
∞
X
s=0
∞
X
s=0
∞
X
s=0
∞
X
s=0
s=0
h(s) +
f̃ (s)
f (s) − L h f
∞
X
h(s)
s=0
f (s).
(2.17)
Now if f , g ∈ BC0h , then by (2.16), we get L h ((f + g) * (f + g)) = 0, which implies L h (f * g) = 0. Hence BC0h is a
subalgebra. This fact and (2.17) implies that BClh is also a subalgebra. For the remaining part of the proof, let
f , g ∈ BClh and put
f̃ := f − (L h f )h, g̃ := g − (L h g)h.
Using (2.16), (S1 ), and (2.15), we obtain
L h (f * g) =L h (((L h f )h + f̃ ) * ((L h g)h + g̃))
=L h fL h gL h (h * h) + L h fL h (g̃ * h) + L h gL h (f̃ * h) + L h (f̃ * g̃)
∞
∞
∞
X
X
X
=2L h fL h g
h(s) + L h f
g̃(s) + L h g
f̃ (s)
=2L h fL h g
s=0
s=0
∞
X
∞
X
h(s) + L h f
s=0
=L h f
∞
X
(g(s) − (L h g)h(s)) + L h g
s=0
g(s) + L h g
s=0
s=0
∞
X
∞
X
(f (s) − (L h f )h(s))
s=0
f (s).
s=0
Thus (2.14) holds and the proof is complete.
Corollary 2.11. Let h ∈ U. Then for every n ∈ N \ {1}, h*n ∈ BClh and L h (h*n ) = nµ n−1 , where µ is defined as in
(2.6).
Proof. An induction argument using Theorem 2.10 with Lemma 2.6 establishes the claim.
Lemma 2.12. If f ∈ BClh , f > 0, and L h f ≠ 0, then f ∈ U.
Proof. Since f ∈ BClh , by Theorem 2.10, we have
L h (f * f ) = 2L h f
∞
X
f (s).
s=0
Also since
(f * f )(t) (f * f )(t) h(t)
=
·
f (t)
h(t)
f (t)
and L h f ≠ 0, f satisfies (S1 ). Since
f (t − s) f (t − s) h(t − s) h(t)
=
·
·
,
f (t)
h(t − s)
h(t)
f (t)
L h f ≠ 0, and h satisfies (S2 ), f satisfies (S2 ). Hence f ∈ U.
3 Subexponential solutions of transient renewal equations
Consider the solution r of the linear scalar convolution equation
r(t) = h(t) +
t−1
X
h(t − 1 − s)r(s), t ∈ N0
s=0
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(3.1)
Subexponential Solutions
with h ∈ U and
µ=
∞
X
h(s) < 1.
|
71
(3.2)
s=0
Then r is positive on N. Summing (3.1) from s = 0 to t − 1, we get
t−1
X
r(s) =
s=0
t−1
X
t−1 X
s−1
X
h(s) +
s=0
=
t−1
X
t−2 X
t−1
X
h(s) +
s=0
=
h(s − 1 − u)r(u)
s=0 u=0
h(s − 1 − u)r(u)
u=0 s=u+1
t−1
X
t−2 t−u−2
X
X
h(s) +
s=0
h(v)r(u).
u=0 v=0
Now as t → ∞, we have
∞
X
r(s) =
s=0
∞
X
h(s) +
s=0
∞
X
∞
X
r(u)
u=0
Using (3.2), we obtain
∞
X
h(v) = µ + µ
v=0
∞
X
r(s).
s=0
µ
.
1−µ
r(s) =
s=0
(3.3)
It can also be represented, since r = h + (h * r) by (3.1), by the Neumann series
r=
∞
X
h*n .
(3.4)
n=1
r is called the resolvent of h, since every solution of
y(t) = f (t) +
t−1
X
y(t − 1 − s)h(s), t ∈ N0 ,
(3.5)
s=0
can be represented as
y(t) = f (t) +
t−1
X
r(t − 1 − s)f (s).
(3.6)
s=0
Theorem 3.1. Let h ∈ U satisfy (3.2). Then the resolvent r defined by (3.1) is in BClh and
Lh r =
1
.
(1 − µ)2
(3.7)
Also, r ∈ U.
Proof. By the representation (3.4) for r, Corollary 2.11, and the uniform convergence implied by Lemma 2.7,
L h r = lim
t→∞
∞
∞
∞
∞
n=1
n=1
n=1
n=1
X n−1
X h*n (t) X
r(t)
h*n (t) X
1
lim
L h (h*n (t)) =
nµ
=
= lim
=
=
.
t→∞ h(t)
h(t) t→∞
h(t)
(1 − µ)2
Since L h r > 0, it follows that r ∈
BClh ,
and also by Lemma 2.12, r ∈ U.
4 Linear sum-difference equations
Consider the asymptotic stability of the scalar linear Volterra sum-difference equation
∆x(t) = −ax(t) +
t−1
X
k(t − 1 − s)x(s) + f (t), x(0) = x0 ,
s=0
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72 | Martin Bohner and Nasrin Sultana
under the assumptions that f , k > 0 on N0 and
resolvent z, which is the solution of
∆z(t) = −az(t) +
P∞
s=0
t−1
X
k(s) < ∞. It is convenient to introduce the difference
k(t − 1 − s)z(s), z(0) = 1.
(4.2)
s=0
Theorem 4.1. If a ∈ (0, 1) and k > 0 on N0 , then all solutions z of (4.2) with z(0) > 0 are positive. If, moreover,
∞
X
k(s) < a,
s=0
then all solutions z of (4.2) satisfy
P∞
s=0
z(s) < ∞ and limt→∞ z(t) = 0.
Proof. To show z(t) > 0 for all t ∈ N0 , we use the method of induction. First, z(0) > 0 holds by assumption.
If for some T ∈ N0 , z(t) > 0 for all 0 ≤ t ≤ T, then from (4.2) and using the assumptions, we get
z(T + 1) = (1 − a)z(T) +
T−1
X
k(T − 1 − s)z(s) > 0.
s=0
Now we use a Lyapunov functional to show
of convergent series. Let
P∞
s=0
z(s) < ∞, and then limt→∞ z(t) = 0 follows from the property
A := a −
∞
X
k(s) > 0.
s=0
Since z, k > 0 on N0 , we have that
V(t) := z(t) +
t−1 X
∞
X
k(u − 1 − s)z(s) ≥ 0 for all t ∈ N0 .
s=0 u=t
Taking the difference of V(t) and substituting ∆z(t) from (4.2), we get
" ∞
#
t−1
t−1
∞
∞
X
X
X
X
X
∆V(t) = − az(t) +
k(t − 1 − s)z(s) +
k(u − 1 − s)z(s) −
k(u − 1 − s)z(s) +
k(u − 1 − t)z(t)
= − az(t) +
= −a +
= −a +
s=0
s=0
t−1
X
t−1
X
k(t − 1 − s)z(s) −
s=0
∞
X
u=t
u=t+1
∞
X
k(t − 1 − s)z(s) +
s=0
u=t+1
k(u − 1 − t)z(t)
u=t+1
!
k(u − 1 − t) z(t)
u=t+1
∞
X
!
k(s) z(t)
s=0
= − Az(t).
Now taking sums on both sides, we have
0 ≤ V(t) = V(0) − A
t−1
X
z(s).
s=0
Thus V(t) + A
Pt−1
s=0
z(s) = V(0). Hence
P∞
s=0
z(s) < ∞.
The following lemma shows the significance of the difference resolvent.
Lemma 4.2. If z is the difference resolvent defined by (4.2), then x defined by
x(t) = z(t)x0 +
t−1
X
z(t − 1 − s)f (s), t ∈ N0
s=0
solves (4.1).
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(4.3)
Subexponential Solutions
|
73
Proof. Taking differences in (4.3) and using (4.2), we get
∆x(t) =∆z(t)x0 +
t−1
X
∆z(t − 1 − s)f (s) + z(0)f (t)
s=0
=(−az(t) + (k * z)(t))x0 + ((−az + (k * z)) * f )(t) + f (t)
= − az(t)x0 + (k * z)(t)x0 − a(z * f )(t) + ((k * z * f )(t) + f (t)
= − ax(t) + (k * (zx0 + (z * f ))(t) + f (t)
= − ax(t) + (k * x)(t) + f (t)
and x(0) = z(0)x0 = x0 .
Note that in the following result, the hypotheses k > 0 and h ∈ U are not necessary.
Lemma 4.3. Let k be any sequence. The unique solution of (4.2) satisfies
z = e−a + e−a * r,
(4.4)
where e−a (t) = (1 − a)t for a ∈ (0, 1), t ≥ 0, h = e−a * k and r is the resolvent given by (3.1).
Proof. Taking differences in (4.4), we obtain
∆z(t) =∆(1 − a)t + ∆
t−1
X
(1 − a)t−1−s r(s)
s=0
= − a(1 − a)t +
t−1
X
∆(1 − a)t−s r(s) + r(t)
s=0
= − a(1 − a)t − a
t−1
X
(1 − a)t−1−s r(s) + r(t)
s=0
= − az(t) + r(t).
Now it remains to show that r = k * z. From (3.1), using h = e−a * k and (4.4), we get
r = h + (h * r) = (k * e−a ) + ((k * e−a ) * r) = (k * (e−a + e−a * r)) = (k * z).
Thus z given by (4.4) solves (4.2).
Theorem 4.4. Let k ∈ U. Suppose that
∞
X
k(s) < a.
s=0
Then the difference resolvent z given by (4.2) satisfies
lim
t→∞
z(t)
1
P
=
,
2
k(t) (a − ∞
s=0 k(s))
lim
t→∞
z(t + 1)
= 1,
z(t)
(4.5)
and z ∈ BClk . Moreover, z ∈ U.
Proof. Define h := e−a * k, where e−a is as in Lemma 4.3. First we prove that h ∈ U so that we can apply the
results for subexponential sequences that have already been established. Clearly h > 0 on N, and (3.2) holds
due to
µ :=
∞
X
h(s)
s=0
=
∞ X
s−1
X
(1 − a)s−1−u k(u)
s=0 u=0
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74 | Martin Bohner and Nasrin Sultana
=
∞ X
∞
X
(1 − a)s−1−u k(u)
u=0 s=u+1
=
∞
X
k(u)
u=0
∞
X
(1 − a)s
s=0
∞
1X
k(s) < 1.
=
a
(4.6)
s=0
Since k ∈ U, Remark 2.4 yields
e−a (t)
= 0.
k(t)
L k e−a = lim
t→∞
(4.7)
So e−a ∈ BC0k . Again by using the fact that k is subexponential (use (2.15)), h = e−a * k ∈ BClk . By Theorem
2.10, (4.7), and L k k = 1, we have
L k h = L k (e−a * k) = L k e−a
∞
X
k(s) + L k k
s=0
∞
X
e−a (s) =
s=0
Therefore by Lemma 2.12, h ∈ U. Now using (4.4), (4.7), (4.8), (2.14), (3.7),
(3.3), we obtain
1
(≠ 0).
a
P∞
s=0
e−a (s) =
(4.8)
1
a,
L h e−a = 0, and
L k z = L k e−a + L k (r * e−a )
= (L k h)L h (r * e−a )
1
=
a
=
Lh r
∞
X
e−a (s) + L h e−a
∞
X
s=0
!
r(s)
s=0
1
> 0.
a2 (1 − µ)2
(4.9)
Hence z ∈ BClk . Using (4.6) in (4.9), we get
z(t)
1
P
=
.
2
k(t) (a − ∞
s=0 k(s))
lim
t→∞
Now summing (4.2) from 0 to t − 1 and then using z(t) → 0 as t → ∞ from Theorem 4.1, by (4.6), we obtain
∞
X
z(s) =
s=0
1
.
a(1 − µ)
(4.10)
Using (2.14), (4.9), (4.6), and (4.10), we get
L k (k * z) = L k z
∞
X
k(s) + L k k
s=0
∞
X
z(s) =
s=0
1
.
a(1 − µ)2
Next, using (4.2), (4.11), and (4.9), we obtain
lim
t→∞
z(t + 1)
(1 − a)z(t) + (k * z)(t)
= lim
t→∞
z(t)
z(t)
(k * z)(t) k(t)
= (1 − a) + lim
·
t→∞
k(t)
z(t)
1
= (1 − a) + L k (k * z) ·
Lk z
= 1.
Finally, since z ∈ BClk and L k z > 0, Lemma 2.12 implies that z ∈ U.
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(4.11)
Subexponential Solutions
Corollary 4.5. Let k ∈ U. If
∞
X
|
75
k(s) < a,
s=0
then for every f ∈ BClk , the solution x of (4.1) satisfies x ∈ BClk and
!
∞
X
1
Lk f
P∞
Lk x =
f (s) +
.
2 x 0 +
P∞
(a
−
a − s=0 k(s)
s=0 k(s))
s=0
(4.12)
Proof. Since f ∈ BClk and z ∈ BClk (by Theorem 4.4), we get by Theorem 2.10 that z*f ∈ BClk . Hence, from (4.3),
we obtain x ∈ BClk . Applying (2.14) to (4.3) and then using (4.9) and (4.10), the proof of (4.12) is complete.
Theorem 4.6. Let k : N0 → (0, ∞) and z the difference resolvent defined in (4.2). If
∞
X
k(s) < a,
s=0
then the two conditions
(i) k ∈ U,
(ii) z satisfies the conditions (4.5)
are equivalent.
Proof. If k ∈ U, then by Theorem 4.4, z satisfies the conditions (4.5). Suppose now that z satisfies the conditions (4.5). Applying the second expression of (4.5) to (4.2), we get
lim
t→∞
z(t + 1) − (1 − a)z(t)
(k * z)(t)
= lim
= a.
t→∞
z(t)
z(t)
(4.13)
Since z is summable (by Theorem 4.1) and satisfies (by the second condition of (4.5)) (S2a ), z satisfies (by
Remark 2.3) (S2 ), and we may apply Lemma 2.8 with f = k and g = z. Using (4.13) and (4.10), we have
∞
X
(k * k)(t)
a2 (1 − µ)2
= a + aµ −
= 2aµ = 2
k(s).
t→∞
k(t)
a(1 − µ)
lim
s=0
Thus k satisfies the subexponential property (S1 ). From the subexponential property (S2 ) of z and L k z > 0
from (4.9), it follows that
lim
t→∞
k(t − s)
= lim
t→∞
k(t)
k(t−s) z(t−s)
z(t−s) z(t)
k(t)
z(t)
=1
holds, i.e., k satisfies the subexponential property (S2 ). Hence k ∈ U.
Example 4.7. In Example 2.2, we have shown that
k(t) =
Note now that
1
,
(t + 2)2
∞
X
s=0
t ∈ N0 ,
k(s) =
satisfies
k ∈ U.
π2
3
− 1 < =: a.
6
4
Thus all assumptions of Theorem 4.4 are satisfied. Therefore, by Theorem 4.4, the solution z of (4.2), i.e., of
t−1
z(t + 1) =
X
1
z(s)
z(t) +
,
4
(t − s + 1)2
z(0) = 1
s=0
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76 | Martin Bohner and Nasrin Sultana
Table 4.1: All numbers are rounded to three decimal places
n
z(n)
z(n+1)
z(n)
0
1
2
3
4
5
6
7
8
9
1.000
0.250
0.313
0.252
0.231
0.211
0.196
0.183
0.172
0.163
0.250
1.250
0.806
0.919
0.913
0.928
0.934
0.940
0.945
0.949
z(n)
k(n)
n
4.000
2.250
5.008
6.293
8.328
10.345
12.537
14.826
17.213
19.678
10
20
30
40
50
100
1000
10000
100000
1000000
z(n)
z(n+1)
z(n)
0.154
0.102
0.075
0.058
0.046
0.018
0.000
0.000
0.000
0.000
0.952
0.966
0.972
0.976
0.978
0.984
0.998
1.000
1.000
1.000
z(n)
k(n)
22.212
49.452
76.717
101.616
123.330
186.956
118.331
93.669
90.968
90.635
satisfies z ∈ U,
lim
t→∞
z(t + 1)
= 1,
z(t)
and
n
o
(t + 2)2 z(t) = t→∞
1
lim
7
4
−
π2
6
2 ≈ 90.589.
Table 4.1 shows some relevant numerical values. All numbers in Table 4.1 are rounded to three decimal places.
Acknowledgement: The authors wish to thank Professor David Grow for technical help with the statement
and proof of Lemma 2.7. The authors are grateful to an anonymous reviewer for his/her valuable comments
and suggestions to improve the quality of the paper.
References
[1] Ravi P. Agarwal. Difference equations and inequalities, volume 228 of Monographs and Textbooks in Pure and Applied Mathematics. Marcel Dekker Inc., New York, second edition, 2000. Theory, methods, and applications.
[2] John A. D. Appleby and David W. Reynolds. Subexponential solutions of linear integro-differential equations and transient
renewal equations. Proc. Roy. Soc. Edinburgh Sect. A, 132(3):521–543, 2002.
[3] Cezar Avramescu and Cristian Vladimirescu. On the existence of asymptotically stable solutions of certain integral equations.
Nonlinear Anal., 66(2):472–483, 2007.
[4] M. Bohner and A. Peterson. Dynamic equations on time scales. Birkhäuser Boston Inc., Boston, MA, 2001. An introduction
with applications.
[5] Theodore Allen Burton. Volterra integral and differential equations, volume 167 of Mathematics in Science and Engineering.
Academic Press Inc., Orlando, FL, 1983.
[6] Walter G. Kelley and Allan C. Peterson. Difference equations. Harcourt/Academic Press, San Diego, CA, second edition, 2001.
An introduction with applications.
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