Growth Rates of Delay–Differential
Equations and Uniform Euler Schemes
J. A. D. Appleby, M. McCarthy
A. Rodkina
School of Mathematical Sciences
Dept. Math. & Comp. Science
Dublin City University, Dublin 9, Ireland
UWI, Kingston, Jamaica
E-mail: john.appleby@dcu.ie
E-mail: alexandra.rodkina
E-mail: michael.mccarthy29@mail.dcu.ie
@uwimona.edu.jm
June 29, 2009
Abstract
In this note we compare the growth rates of scalar delay–differential
equations and related uniform Euler schemes. The solutions of the difference
and differential equations grow at the same rate if the feedback is sublinear,
but the Euler scheme grows more slowly if the instantaneous term is superlinear and dominates the delay term. In the sublinear case, the growth rate
of a continuous interpolant matches that of the differential equation; in the
superlinear case the interpolant underestimates the growth rate.
Keywords & Phrases: delay equation, Euler scheme, growth rate.
AMS Subject Classification: 39A11, 34K12, 34K28.
1
Introduction
In this note we analyse the growth rate of the solution of the differential equation
x0 (t) = f (x(t)) + g(x(t − τ )),
t > 0;
t ∈ [−τ, 0]
x(t) = ψ(t),
(1.1)
compared with that of the linear interpolant x̄h of the uniform mesh Euler scheme
xn+1 (h) = xn (h) + hf (xn (h)) + hg(xn−N (h)),
xn (h) = ψn (h) > 0,
n∈N
n = −N, −N + 1, . . . , 0.
1
(1.2a)
(1.2b)
John A. D. Appleby, Michael McCarthy, Alexandra Rodkina
We exclude equations (1.1) whose solutions explode in finite time (cf. e.g., [5]).
Under some asymptotic conditions, we show when f and g are sublinear (in
the sense that f (x)/x → 0, g(x)/x → 0 as x → ∞), x, xn and x̄h have the same
growth rate. However x̄h and xn grow more slowly than x when f is superlinear
(in the sense that f (x)/x → ∞ as x → ∞), so a uniform Euler method does not reproduce the correct growth rate. In a related paper, we analyse adaptive meshes
for superlinear equations which enable the asymptotic growth of global and exploding solutions to be mimicked. There is a huge literature on the asymptotic
behaviour of difference equations see e.g.,. [4, 7]. Many papers also consider the
preservation of asymptotic behaviour of DDEs, with [6, 8] particularly relevant.
The fact that decreasing h leads to a better estimate of the growth rate is seen
by discretising the linear ODE x0 (t) = αx(t), t > 0 with x(0) = 1 for α > 0 to get
xn+1 (h) = (1 + hα)xn (h), n ≥ 0 with x0 (h) = 1. Then limt→∞ log x(t)/t = α, while
the growth rate of xn depends on h as limn→∞ log xn (h)/(nh) = log(1 + αh)/h =:
αh . However, as h → 0, αh → α, justifying the increased computational effort.
2
Discussion of Results
First, we list some properties of regularly varying functions cf. e.g., [3].
Definition 2.1. A function k : [0, ∞) → (0, ∞) is regularly varying at infinity with
index α ∈ R if limx→∞ k(λx)/k(x) = λα for each λ > 0. We write k ∈ RV∞ (α).
Rx
(RV1). If k ∈ RV∞ (α), 1/k ∈ RV∞ (−α) and if K(x) := 1 k(u) du, x > 0, K ∈
RV∞ (α + 1) for α > −1, and K ∈ RV∞ (0) for α = −1 if K(x) → ∞ as x → ∞.
(RV2). If k ∈ RV∞ (α), ∃κ such that κ(x)/k(x) → 1, xκ0 (x)/κ(x) → α as x → ∞.
(RV3). If k ∈ RV∞ (α), and x and y are continuous with x(t)/y(t) → 1 and y(t) →
∞ as t → ∞, then limt→∞ k(x(t))/k(y(t)) = 1.
Let τ > 0, f, g ∈ C((0, ∞); (0, ∞)), ψ ∈ C([−τ, 0], (0, ∞)). If f is locally Lipschitz continuous, there is a unique continuous and strictly positive x obeying
(1.1) on [−τ, Tψ ). x is defined on [−τ, ∞) if F (∞−) = ∞, where we define F , G by
Z x
Z x
F (x) =
1/f (u) du, G(x) =
1/g(u) du, x > 0.
(2.1)
1
1
Theorem 2.2. Suppose f, g ∈ C((0, ∞); (0, ∞)), f is locally Lipschitz continuous,
lim f (x)/x = 0,
x→∞
lim g(x)/x = 0;
x→∞
and
there exists λ ∈ [0, ∞] such that λ := lim g(x)/f (x).
x→∞
2
(2.2)
(2.3)
Growth Rate of Euler Discretisation of DDEs
Let τ > 0, ψ ∈ C([−τ, 0], (0, ∞)) and x be the unique continuous solution of (1.1).
(i) If λ ∈ [0, ∞) and there is α ≤ 1 such that f ∈ RV∞ (α), then
lim F (x(t))/t = 1 + λ,
t→∞
(2.4)
where F is defined by (2.1). If α < 1, then limt→∞ x(t)/F −1 ((1 + λ)t) = 1.
(ii) If λ = ∞ and there is α ≤ 1 such that g ∈ RV∞ (α), then
lim G(x(t))/t = 1,
t→∞
(2.5)
where G is defined by (2.1). If α < 1, then limt→∞ x(t)/G−1 (t) = 1.
Theorem 2.2 applies when f and g are asymptotic to e.g., φ1 (x) = xα logβ (x)
(for α < 1 and β ∈ R) or to φ2 (x) = x logβ (x) or φ3 (x) = x(log log x)β (for β < 0) as
x → ∞. Neither f nor g need be monotone nor tend to infinity as x → ∞.
When f grows superlinearly and dominates g, the solution of (1.1) behaves
asymptotically as the non–exploding solution of the equation x0 (t) = f (x(t)).
Theorem 2.3. Suppose f, g ∈ C((0, ∞); (0, ∞)), f is locally Lipschitz continuous,
Z ∞
lim f (x)/x = ∞, f ∈ RV∞ (1);
1/f (u) du = +∞; and
(2.6)
x→∞
1
there exists Λ ∈ [0, ∞) such that Λ := lim sup g(x)/f (x).
(2.7)
x→∞
If ψ ∈ C([−τ, 0], (0, ∞)), x is the continuous solution of (1.1), limt→∞ F (x(t))/t = 1.
Functions obeying (2.6) include those asymptotic to φ5 (x) = x logβ x for β ∈
(0, 1] (but not β > 1) or to φ6 (x) = x(log log x)β for β > 0 (but not β ≤ 0).
Discretising (1.1) on a uniform mesh of length h > 0 yields (1.2). In (1.2),
N ≥ 1 is the maximal integer not exceeding τ /h and ψn = ψ(nh), n = −N, . . . , 0.
Theorem 2.4. Let f, g ∈ C((0, ∞); (0, ∞)) obey (2.2), (2.3), (xn (h)) obey (1.2).
(i) If λ ∈ [0, ∞) in (2.3) and there is α ≤ 1 such that f ∈ RV∞ (α), then
lim F (xn (h))/(nh) = 1 + λ,
n→∞
(2.8)
where F is defined by (2.1). If α < 1, then limn→∞ xn (h)/F −1 (nh) = 1 + λ.
(ii) If λ = ∞ in (2.3) and there is α ≤ 1 such that g ∈ RV∞ (α), then
lim G(xn (h))/(nh) = 1,
n→∞
where G is defined by (2.1). If α < 1, then limn→∞ xn (h)/G−1 (nh) = 1.
3
(2.9)
John A. D. Appleby, Michael McCarthy, Alexandra Rodkina
A similar result appears in [1] for first order difference equations.
If (xn ) obeys (1.2), define x̄h ∈ C([−τ, ∞), (0, ∞)) by x̄h (t) = ψ(t), t ∈ [−τ, 0],
x̄h (t) = xn + (xn+1 − xn )(t − nh)/h,
t ∈ [nh, (n + 1)h],
n ≥ 0,
(2.10)
so x̄h takes the value xn (h) at time nh for n ≥ 0 and interpolates linearly between
the values of (xn (h)) at the times {0, h, 2h, . . .}. As h → 0, x̄h approaches x on
any compact interval [0, T ] in the sense that limh→0 sup0≤t≤T |x(t) − x̄h (t)| = 0
(see e.g., [2]). In fact, Theorems 2.2 and 2.4 show that x̄h mimics the asymptotic
behaviour of x.
Theorem 2.5. Suppose the conditions of Theorem 2.2 hold and x̄h is given by (2.10).
Let α ≤ 1. If λ ∈ [0, ∞) in (2.3) and f ∈ RV∞ (α), then limt→∞ F (x̄h (t))/t = 1 + λ.
If α < 1, then limt→∞ x̄h (t)/x(t) = 1. If λ = ∞ in (2.3) and g ∈ RV∞ (α), then
limt→∞ G(x̄h (t))/t = 1. If α < 1, then limt→∞ x̄h (t)/x(t) = 1.
Finally, under the conditions of Theorem 2.3 the solution (xn (h)) of (1.2) and
the interpolant x̄h have different growth rates from the solution x of (1.1), irrespective of the mesh size h > 0. The rate of growth of xn (h) is independent of h,
suggesting we must use special meshes to deal with such superlinear equations.
Theorem 2.6. Suppose f, g ∈ C((0, ∞); (0, ∞)), f is locally Lipschitz continuous,
obeys (2.6), f and g obey (2.7), and there is a function f1 such that
f1 (x)
= 1. (2.11)
x→∞ f (x)
x 7→ f0 (x) = f1 (x)/x is positive, nondecreasing on (X1 , ∞), lim
Let h > 0, N ∈ N, and (xn (h))n≥−N obey (1.2). If H is defined by
Z x
H(x) =
1/{u log(1 + f (u)/u)} du, x > 1
(2.12)
1
and x̄h ∈ C([−τ, ∞), (0, ∞)) is defined by (2.10), then
lim H(xn (h))/n = 1,
n→∞
lim H(x̄h (t))/t = 1/h.
t→∞
(2.13)
The third part of (2.6) implies H(x) → ∞ as x → ∞. To see this, put y =
f (x)/x > 0 in the inequality log(1 + y) < y, y > 0. Thus 1/{x log(1 + f (x)/x)} >
1/f (x), and integration gives H(x) ≥ F (x). The last part of (2.6) implies F (∞−) =
∞, proving the claim. Indeed as H(x)/F (x) → ∞ as x → ∞, the Euler scheme
underestimates the growth rate of the solution of (1.1), as the second limit in (2.13)
implies F (x̄h (t))/t → 0 as t → ∞, but x obeys F (x(t))/t → 1 as t → ∞. Lastly, the
second limit in (2.13) is derived using the method of proof of Theorem 2.5.
The proof of Theorem 2.6 is facilitated by the following lemma.
4
Growth Rate of Euler Discretisation of DDEs
Lemma 2.7. Let k > 0. Suppose φ ∈ C((0, ∞), (0, ∞)), φ ∈ RV∞ (1), φ(y)/y → ∞
R∞
as y → ∞, 1 1/φ(u) du = +∞, and there is a function φ1 with φ1 (y)/φ(y) → 1 as
y → ∞ such that φ0 : (Y1 , ∞) → (0, ∞) : y 7→ φ0 (y) := φ1 (y)/y is non–decreasing. If
yn+1 = yn + kφ(yn ), n ≥ 0 and y0 = ξ > 0, then limn→∞ K(yn )/n = 1, where
Z y
1/{u log(1 + φ(u)/u)} du, y > 1.
K(y) =
(2.14)
1
3
Proofs
Proof of Theorem 2.2 Since x(t) > 0, t ≥ −τ and f , g are positive, x0 (t) > 0, t > 0.
Thus x(t) → L ∈ (0, ∞] as t → ∞. If L < ∞, integrate (1.1) over [0, t], divide by t
and let t → ∞ to get the contradiction f (L) + g(L) = 0, so limt→∞ x(t) = ∞. Since
x(t − τ ) < x(t), t ≥ τ , 0 < x0 (t)/x(t) ≤ f (x(t))/x(t) + g(x(t − τ ))/x(t − τ ). (2.2)
gives x0 (t)/x(t) → 0 as t → ∞, so x(t − τ )/x(t) → 1 as t → ∞.
In case (i), since f ∈ RV∞ (α), and x(t − τ )/x(t) → 1 as t → ∞, f (x(t −
τ ))/f (x(t)) → 1 as t → ∞. Hence, using this, (2.3), and dividing both sides of
(1.1) by f (x(t)), we get x0 (t)/f (x(t)) → 1 + λ as t → ∞. Thus for every ε ∈ (0, 1)
Rt
there is a T (ε) > 0 such that for all t > T we have t − T ≤ T x0 (s)/f (x(s)) ds ≤
(1 + ε)(t − T ). Using (2.1), letting t → ∞, and then ε → 0+ yields (2.4).
In case (ii), since g ∈ RV∞ (α), and x(t − τ )/x(t) → 1 as t → ∞, g(x(t −
τ ))/g(x(t)) → 1 as t → ∞. Hence using this, (2.3), and dividing both sides of (1.1)
by g(x(t)) we get x0 (t)/g(x(t)) → 1 as t → ∞. Proceeding as in (i) yields (2.5).
Proof of Theorem 2.3 (2.6) ensures (1.1) has a unique continuous solution. Since
x(t) > 0 for t ≥ −τ and f and g are positive, x0 (t) > 0 for t > 0, and so x(t) → ∞
as t → ∞. Thus x0 (t)/x(t) > f (x(t))/x(t) → ∞ as t → ∞, using (2.6). Hence for
every M > 0 there is a T (M ) > 0 such that x0 (t)/x(t) > M , t > T (M ). Thus for t >
Rt
τ +T (M ), log(x(t)/x(t−τ )) = t−τ x0 (s)/x(s) ds ≥ M τ , so limt→∞ x(t)/x(t−τ ) = ∞.
Hence for every ε > 0 there is a T1 (ε) > 0 such that x(t − τ ) < εx(t) for t > T1 (ε).
Since f ∈ RV∞ (1), there is a ϕ such that f (x)/ϕ(x) → 1 and xϕ0 (x)/ϕ(x) → 1 as
x → ∞, with ϕ increasing on [x2 , ∞). Since x(t) → ∞ as t → ∞ there is a T2 > 0
such that x(t − τ ) > x2 for t > T2 . Let T3 (ε) = max(T2 , T1 (ε)). Then for t > T3 (ε),
ϕ(x2 ) < ϕ(x(t − τ )) < ϕ(εx(t)), so ϕ(εx)/ϕ(x) → ε as x → ∞ gives
lim sup
t→∞
f (x(t − τ ))
f (x(t − τ )) ϕ(x(t − τ )) ϕ(εx(t)) ϕ(x(t))
= lim sup
·
·
·
≤ ε.
f (x(t))
ϕ(εx(t))
ϕ(x(t)) f (x(t))
t→∞ ϕ(x(t − τ ))
Thus f (x(t − τ ))/f (x(t)) → 0 as t → ∞. Using this, (2.7), and dividing both sides
of (1.1) by f (x(t)) gives limt→∞ x0 (t)/f (x(t)) = 1. Integration yields the result.
5
John A. D. Appleby, Michael McCarthy, Alexandra Rodkina
Proof of Theorem 2.4 Let φ(x) = f (x) + g(x), x ≥ 0. In case (i), φ(x)/f (x) → 1 + λ
as x → ∞, and since f ∈ RV∞ (α), φ ∈ RV∞ (α). In case (ii), φ(x)/g(x) → 1 as
x → ∞, giving φ ∈ RV∞ (α). Thus there is ϕ ∈ C 1 (0, ∞) such that ϕ(x)/φ(x) → 1,
Rx
xϕ0 (x)/ϕ(x) → α as x → ∞. Let Φ(x) := 1 1/ϕ(u) du, x > 0, so Φ ∈ C 2 (0, ∞).
Since ψn > 0 and f , g are positive, xn > 0, n ≥ −N and xn+1 > xn , n ≥ 0,
limn→∞ xn = L ∈ (0, ∞]. If L < ∞, since f, g ∈ C(0, ∞), by taking limits as n → ∞
across (1.2), we get f (L) + g(L) = 0, a contradiction, so limn→∞ xn = ∞. By xn >
xn−N , xn+1 /xn ≤ 1 + f (xn )/xn + g(xn−N )/xn−N , so (2.2) gives limn→∞ xn+1 /xn = 1.
With hn := h(f (xn )/φ(xn ) + g(xn−N )/φ(xn )), xn+1 = xn + hn φ(xn ), n ≥ 0. In case
(i), φ(x)/f (x) → 1 + λ as x → ∞ so f (xn )/φ(xn ) → 1/(1 + λ) as n → ∞. Since
xn−N /xn → 1 as n → ∞ and φ ∈ RV∞ (α), φ(xn−N )/φ(xn ) → 1 as n → ∞. Since
g(x)/φ(x) → 1 as x → ∞, g(xn−N )/φ(xn ) → 1 as n → ∞. Thus hn → h as n → ∞.
By Taylor’s theorem, there is an ξn ∈ [xn , xn+1 ] such that
1
Φ(xn+1 ) − Φ(xn ) = Φ0 (xn )hn φ(xn ) + Φ00 (ξn )h2n φ2 (xn ).
2
(3.1)
By (2.2), φ(x)/x → 0 as x → ∞. Since 1 ≤ ξn /xn ≤ 1 + hn φ(xn )/xn , ξn /xn → 1 as
n → ∞. Consider the righthand side of (3.1) as n → ∞. First, Φ0 (xn )hn φ(xn ) =
hn φ(xn )/ϕ(xn ) → h as n → ∞. Rewrite the second term to get
1 ϕ0 (ξn ) 2
1 ξn ϕ0 (ξn ) ϕ(xn ) xn φ(xn ) φ(xn )
·
·
− h2n 2
φ (xn ) = − h2n
·
·
.
2 ϕ (ξn )
2
ϕ(ξn )
ϕ(ξn ) ξn
xn
ϕ(xn )
Since ξn → ∞, the first factor tends to α. Since ξn /xn → 1 and ϕ ∈ RV∞ (α),
ϕ(xn )/ϕ(ξn ) → 1 as n → ∞. The third factor has unit limit, and the fourth has
zero limit since φ(x)/x → 0 as x → ∞. Finally, the last factor has unit limit as
φ(x)/ϕ(x) → 1 as x → ∞. Since hn → h as n → ∞, limn→∞ Φ00 (ξn )h2n φ2 (xn ) = 0.
Thus by (3.1), Φ(xn+1 ) − Φ(xn ) → h as n → ∞, implying limn→∞ Φ(xn )/(nh) = 1.
In case (i), by L’Hôpital’s rule limx→∞ F (x)/Φ(x) = limx→∞ ϕ(x)/f (x) = 1 + λ,
so F (xn )/(nh) → 1 + λ as n → ∞. (ii) follows similarly as limx→∞ G(x)/Φ(x) = 1.
Proof of Theorem 2.5 For every t > 0 there is an n = n(t) ∈ N such that t ∈
[n(t)h, (n(t) + 1)h), so xn (h) ≤ x̄h (t) < xn+1 (h). In case (i) as F is increasing,
n(t)h 1
1
(n(t) + 1)h
1
F (xn(t) (h)) ≤ F (x̄h (t)) ≤
F (xn(t)+1 (h)).
t n(t)h
t
t
(n(t) + 1)h
As n(t)h/t → 1 as t → ∞, (2.8) implies limt→∞ F (x̄h (t))/t = 1 + λ. To prove
x̄h (t)/x(t) → 1 as t → ∞, note that f ∈ RV∞ (α) implies F ∈ RV∞ (1−α). As α < 1,
F −1 ∈ RV∞ (1/(1 − α)). Since F (x̄h (t))/t → 1 as t → ∞, F −1 (F (x̄h (t)))/F −1 (t) → 1
as t → ∞, so x̄h (t)/x(t) → 1 as t → ∞. The proof is similar if λ = ∞.
6
Growth Rate of Euler Discretisation of DDEs
Proof of Lemma 2.7 Set c(u) = log(1 + kφ(eu )/eu ), u ∈ R. Since yn > 0 for n ≥ 0,
let un = log yn , so un+1 = un + c(un ), n ≥ 0. Since φ ∈ RV∞ (1), there is φ2 ∈ C 1
such that φ2 (y)/φ(y) → 1, yφ02 (y)/φ2 (y) → 1 as y → ∞. φ2 is positive on (Y2 , ∞)
for some Y2 := eU2 . Let c2 (u) = log(1 + φ2 (eu )/eu ), u > U2 , so c2 ∈ C 1 is positive.
Also, φ(y)/y → ∞, φ2 (y)/φ(y) → 1 as y → ∞ imply limu→∞ c2 (u)/c(u) = 1. Since
yφ02 (y)/φ2 (y) → 1, φ2 (y)/y → ∞ as y → ∞, c02 (u) = (φ2 (eu )/eu )/(1 + φ2 (eu )/eu ) ·
(eu φ02 (eu )/φ2 (eu ) − 1) → 0 as u → ∞. Let c1 (u) = log(1 + φ1 (eu )/eu ), u > U1 :=
log(Y1 ). c1 is positive and increasing, as y 7→ φ1 (y)/y is increasing on (Y1 , ∞), and
as φ(y)/y → ∞, φ1 (y)/φ(y) → 1 as y → ∞, we get c1 (u)/c(u) → 1 as u → ∞.
Ru
Set K2 (u) = U2 1/c2 (v) dv, u > U2 , so K2 ∈ C 2 (U2 , ∞), K20 (u) = 1/c2 (u) and
K200 (u) = −c02 (u)/c22 (u). Since un → ∞ as n → ∞, there is an N1 > 1 such that
un > max(U1 , U2 ), n ≥ N1 . By Taylor’s theorem, there is a ξn ∈ [un , un+1 ] such that
K2 (un+1 ) − K2 (un ) =
c2 (un ) c21 (un ) c21 (ξn )
c(un )
1
− c02 (ξn ) 2
·
·
.
c2 (un ) 2
c1 (un ) c21 (ξn ) c22 (ξn )
(3.2)
Since c(u) > 0, u ∈ R, ξn ≥ un . Consider the righthand side of (3.2) as n →
∞. Since c(u)/c2 (u) → 1 as u → ∞, the first term tends to unity as n → ∞.
In the second term, as ξn ≥ un and un → ∞ the third factor has unit limit as
c1 (u)/c2 (u) → 1 as u → ∞. For the second factor, as c21 is increasing on (U1 , ∞)
and ξn ≥ un > U1 , c21 (un )/c21 (ξn ) ≤ 1. The first factor has unit limit as n → ∞
because c(u)/c1 (u) → 1 as u → ∞. Finally, as ξn → ∞ as n → ∞, c02 (ξn ) → 0 as
n → ∞. Thus K2 (un+1 ) − K2 (un ) → 1 as n → ∞ and so K2 (un )/n → 1 as n → ∞.
Ry
Let k(y) := Y2 1/(wc2 (log(w))) dw, so K2 (un ) = k(yn ), and c0 (u) := log(1 +
Ry
φ(eu )/eu ), u ∈ R, so limu→∞ c0 (u)/c(u) = 1, K(y) = 1 1/(wc0 (log(w))) dw. By
L’Hôpital’s rule and limu→∞ c2 (u)/c(u) = 1, limy→∞ k(y)/K(y) = 1, whence (2.14).
Proof of Theorem 2.6 Once again xn > 0, xn → ∞ as n → ∞. Since xn+1 >
xn + hf (xn ) and f (x)/x → ∞ as x → ∞, xn+1 /xn → ∞ as n → ∞, so xn−N /xn → 0
as n → ∞. Since f ∈ RV∞ (1), there is a y 0 > 1 and φ increasing on (y 0 , ∞) such that
f (x)/φ(x) → 1 as x → ∞. Also there is an N2 > 1 such that xn−N > y 0 , n ≥ N2 ,
and for every ε > 0 an N3 (ε) > 1 such that xn−N < εxn , n ≥ N3 . Then φ(xn−N ) <
φ(εxn ) for n ≥ N4 := max(N2 , N3 ), so by using φ(εx)/φ(x) → ε as x → ∞
lim sup
n→∞
f (xn−N )
f (xn−N ) φ(xn−N ) φ(xn )
= lim sup
·
·
≤ ε.
f (xn )
φ(xn )
f (xn )
n→∞ φ(xn−N )
Thus f (xn−N )/f (xn ) → 0 as n → ∞, and by (2.7), g(xn−N )/f (xn ) → 0 as n → ∞.
Set hn = (h + hg(xn−N )/f (xn ))f (xn )/φ(xn ). Then xn+1 = xn + hn φ(xn ) and hn → h
as n → ∞. Hence there is an N5 > 1 such that h/2 < hn < 2h, xn > 4y 0 , n ≥ N5 , so
7
John A. D. Appleby, Michael McCarthy, Alexandra Rodkina
xn+1 > xn + h/2φ(xn ), xn+1 < xn + 2hφ(xn ) for n ≥ N5 . Define (yn ), (ȳn ) by yN5 =
xN5 /2, ȳN5 = 2xN5 , yn+1 = yn +h/2·φ(yn ), n ≥ N5 and ȳn+1 = ȳn +2hφ(ȳn ), n ≥ N5 .
As φ is increasing on (y 0 , ∞), 1 < y 0 < yn < xn < ȳn , n ≥ N5 , and as K given by
(2.14) is increasing on (1, ∞), K(yn ) ≤ K(xn ) ≤ K(ȳn ), n ≥ N5 .
Since f (x)/φ(x) → 1 as x → ∞, φ ∈ RV∞ (1), φ(y)/y → ∞ as y → ∞,
R∞
1
1/φ(u) du = +∞ and φ1 := f1 is such that φ1 (y)/φ(y) → 1 as y → ∞ and
y 7→ φ0 (y) := φ1 (y)/y is increasing on (X1 , ∞). Hence by applying Lemma 2.7
to the sequences (yn ) and (ȳn ) we get limn→∞ K(yn )/n = 1, limn→∞ K(ȳn )/n = 1,
and so K(xn )/n → 1 as n → ∞. Finally, as f (x)/φ(x) → 1 and φ(x)/x → ∞ as
x → ∞, L’Hôpital’s rule gives H(x)/K(x) → 1 as x → ∞, so (2.13) holds.
Ackowledgement JA is supported by the SFI grant 07/MI/008 “Edgeworth
Centre for Financial Mathematics”, and MMcC by IRCSET under the Embark Fellowship project “Explosions of stochastic delay differential equations in finance”.
References
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Growth Rate of Euler Discretisation of DDEs
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