ON THE STABILITY OF THE LINEAR DELAY
DIFFERENTIAL AND DIFFERENCE
EQUATIONS
A. ASHYRALYEV AND P. SOBOLEVSKII
Received 14 August 2001
We consider the initial-value problem for linear delay partial differential equations of the parabolic type. We give a sufficient condition for the stability of
the solution of this initial-value problem. We present the stability estimates
for the solutions of the first and second order accuracy difference schemes for
approximately solving this initial-value problem. We obtain the stability estimates in Hölder norms for the solutions of the initial-value problem of the delay
differential and difference equations of the parabolic type.
1. Introduction
Methods for numerical solutions of the delay ordinary differential equations
have been studied extensively by many researchers (cf. [1, 2, 3, 4, 6, 7, 8,
9, 10, 12, 13, 14, 15, 16] and the references therein) and developed over the
last two decades. In this literature, the linear delay equations was considering
under the condition, such that in the case of the constant coefficients follows
obviously from characteristic equations as the necessary condition. The stability
of the solution of this initial-value problem and of the difference schemes for
approximately solving the initial-value problem were obtained.
However, the theory and numerical solution of the delay partial differential
equations have received less attention than the delay ordinary differential equations. In the present paper, we consider the initial-value problem for linear delay
differential equations
du(t)
+ Au(t) = B(t)u(t − ω), t ≥ 0,
dt
u(t) = g(t), −ω ≤ t ≤ 0,
(1.1)
in an arbitrary Banach space E with unbounded linear operators A and B(t) in
E with dense domain D(A) ⊆ D(B(t)). Let A be a strongly positive operator,
Copyright © 2001 Hindawi Publishing Corporation
Abstract and Applied Analysis 6:5 (2001) 267–297
2000 Mathematics Subject Classification: 65J, 65M, 47D, 34K, 34G
URL: http://aaa.hindawi.com/volume-6/S1085337501000616.html
268
Delay differential and difference equations
that is, −A is the generator of the analytic semigroup exp{−tA} (t ≥ 0) of the
linear bounded operators with exponentially decreasing norm when t → ∞.
This means the following estimates hold:
exp{−tA}
E→E
≤ Me−δt ,
tA exp{−tA}
E→E
≤ Me−δt ,
t > 0 (1.2)
for some M > 0, δ > 0 (see [11]).
A function u(t) is called a solution of problem (1.1) if the following conditions are satisfied:
(i) The function u(t) is continuously differentiable on the interval [−ω, ∞).
The derivative at the endpoint t = −ω is understood as the appropriate unilateral
derivative.
(ii) The function u(t) belongs to D(A) for all t ∈ [−ω, ∞), and the functions Au(t) and B(t)u(t) are continuous on the interval [−ω, ∞).
(iii) The function u(t) satisfies the equation and the initial condition (1.1).
A solution u(t) of the initial-value problem (1.1) is said to be stable if
u(t) ≤ max g(t)
E
E
−ω≤t≤0
(1.3)
for every t, −ω ≤ t < ∞. We are interested to study the stability of solutions
of the initial-value problem under the assumption that
B(t)A−1 E→E
≤1
(1.4)
holds for every t ≥ 0. We have not been able to obtain the estimate (1.3) in the
arbitrary Banach space E. Nevertheless, we can establish the analog of estimates
(1.3) where the space E is replaced by the fractional spaces Eα (0 < α < 1)
under the more strong assumption than (1.4).
2. The delay differential equation
It is known that (cf. [17]) the strongly positive operator A defines the fractional
spaces Eα = Eα (A, E) (0 < α < 1) consisting of all v ∈ E for which the
following norms are finite:
vEα = sup λ1−α A exp{−λA}v E .
(2.1)
λ>0
First, we consider problem (1.1) when
A−1 B(t)x = B(t)A−1 x,
x ∈ D(A).
(2.2)
A. Ashyralyev and P. Sobolevskii
269
Theorem 2.1. Assume that the condition
B(t)A−1 E→E
≤
(1 − α)
M22−α
(2.3)
holds for every t ≥ 0. Then for every t ≥ 0,
u(t)
Eα
≤ max g(t)E .
(2.4)
exp − (t − s)A B(s)g(s − ω)ds,
(2.5)
−ω≤t≤0
α
Proof. Using the formula
t
u(t) = exp{−tA}g(0) +
0
the semigroup property, condition (2.3), and the estimates (1.2) we have
λ1−α A exp{−λA}u(t)E
≤ λ1−α A exp − (λ + t)A g(0)E
t
B(s)A−1 A exp − λ + t − s A + λ1−α
E→E
2
0
E→E
λ
+
t
−
s
A g(s − ω)
×
ds
A exp −
2
E
t Mλ1−α 22−α
λ1−α g(0) + 1 − α
ds max g(s − ω)E
≤
E
1−α
2−α
2−α
α
α
0≤s≤ω
(λ + t)
M2
0 (λ + t − s)
≤ max g(t)
−ω≤t≤0
Eα
(2.6)
for every t, 0 ≤ t ≤ ω and λ, λ > 0. This shows that
u(t)
Eα
≤ max g(t)E
−ω≤t≤0
α
(2.7)
for every t, 0 ≤ t ≤ ω. Applying the mathematical induction, one can easily
show that it is true for every t. Namely, assume that inequality (2.4) is true for
t, (n − 1)ω ≤ t ≤ nω, n = 1, 2, 3, . . . . Using the formula
u(t) = exp −(t −nω)A u(nω)+
t
nω
exp −(t −s)A B(s)u(s −ω)ds, (2.8)
270
Delay differential and difference equations
the semigroup property, condition (2.3), and the estimates (1.2) we have
λ1−α A exp{−λA}u(t)E
≤ λ1−α A exp − (λ + t − nω)A u(nω)E
t λ+t −s
1−α
B(s)A−1 A
+λ
A
exp
−
E→E
2
nω
E→E λ+t −s
A u(s − ω)
×
ds
A exp −
2
(2.9)
E
t
1−α
1−α
2−α
λ
Mλ 2
u(nω) + 1 − α
≤
ds
Eα
(λ + t − nω)1−α
M22−α nω (λ + t − s)2−α
u(s − ω)
×
max
E
nω≤s ≤(n+1)ω
≤
max
−(n−1)ω≤t≤nω
α
u(t)
Eα
for every t, nω ≤ t ≤ (n + 1)ω, n = 1, 2, 3, . . . and λ, λ > 0. This shows that
u(t) ≤ max g(t)
(2.10)
E
E
α
−ω≤t≤0
α
for every t, nω ≤ t ≤ (n + 1)ω, n = 1, 2, 3, . . . . This result completes the proof
of Theorem 2.1.
Now, we consider problem (1.1) when
A−1 B(t)x = B(t)A−1 x,
x ∈ D(A).
(2.11)
Note that (see [11]) A is a strongly positive operator in a Banach space E if
and only if its spectrum σ (A) lies in the interior sector of angle ϕ, 0 < 2ϕ < π ,
symmetric with respect to the real axis, and if on the edges of this sector,
S1 = [z = ρ exp(iϕ) : 0 ≤ ρ < ∞] and S2 = [z = ρ exp(−iϕ) : 0 ≤ ρ < ∞],
and outside it the resolvent (z − A)−1 is subject to the bound
(z − A)−1 E→E
≤
M1
1 + |z|
(2.12)
for some M1 > 0. Throughout, the operator (zI −A)−1 will be written (z−A)−1 .
First of all we prove a lemma that will be needed in the sequel.
Lemma 2.2. For any z on the edges of the sector, S1 = [z = ρ exp(iϕ) : 0 ≤ ρ <
∞], and S2 = [z = ρ exp(−iϕ) : 0 ≤ ρ < ∞], and outside it the estimate
1−α (2−α)α
M1α M α 1 + M1
2
−1
A(z − A) ϕ ≤
α
ϕEα
E
α(1 − α) 1 + |z|
(2.13)
A. Ashyralyev and P. Sobolevskii
271
holds. Here and in the future M and M1 are some constants of the estimates
(1.2) and (2.12).
Proof. Applying the formula (see [11])
∞
ϕ=
A exp{−τ A}ϕ dτ,
(2.14)
0
and using the semigroup property we obtain
N
−1
A(z − A)−1 A exp{−τ A}ϕ dτ
A(z − A) ϕ =
0
∞
+
(z − A)
−1
N
(2.15)
τ
τ
A exp − A A exp − A ϕ dτ.
2
2
Using the estimates (1.2), (2.12), and the definition of the spaces Ea , we obtain
A(z − A)−1 ϕ ≤
E
N
A(z − A)−1 A exp{−τ A}ϕ d τ
E→E
E
0
A exp − τ A E→E 2 E→E
N
τ
A
ϕ
×
A
exp
−
dτ
2
E
N
∞
α−1
MM1 22−α τ α−2
dτ ϕEα
dτ +
1 + M1 τ
≤
|z| + 1
0
N
+
∞
(z − A)−1 = ψ(N)ϕEα .
(2.16)
Here
N α MM1 22−α N α−1
+ .
ψ(N) = 1 + M1
α
|z| + 1 (1 − α)
(2.17)
Taking the minimum of the function ψ(N) over all N, 0 < N < ∞, we obtain
Lemma 2.2.
Lemma 2.3. For all τ > 0 the following estimate holds:
−1 A A exp{−τ A}B(s) − B(s)A exp{−τ A} ϕ E
1−α (2−α)α
e(α + 1)M α M11+α 1 + 2M1 1 + M1
2
QE→E ϕEα
.
≤
1−α
2
τ
π α (1 − α)
(2.18)
Here Q = A−1 (AB(s) − B(s)A)A−1 .
272
Delay differential and difference equations
Proof. Applying the Cauchy-Riesz formula, we obtain the representation
A exp{−τ A}ϕ =
1
2π i
1
=−
2π i
+
1
2π i
S1 ∪S2
exp{−τ z}A(z − A)−1 dzϕ
S1 ∪S2
S1 ∪S2
exp{−τ z}dzϕ
(2.19)
exp{−τ z}z(z − A)−1 dzϕ
for every τ , τ > 0. From that it follows that
A−1 A exp{−τ A}B(s) − B(s)A exp{−τ A} ϕ
1
=
2πi
1
=
2πi
S1 ∪S2
S1 ∪S2
exp{−τ z}zA−1 (z − A)−1 B(s) − B(s)(z − A)−1 dzϕ
(2.20)
exp{−τ z}z(z − A)−1 QA(z − A)−1 dzϕ.
Making the substitution z = −1/τ +iy, −∞ < y < ∞ and using integration by
parts, we can write the formula
A−1 A exp{−τ A}B(s) − B(s)A exp{−τ A} ϕ
e
=
2π
−1
1
1
exp(−τ iy) − + iy − + iy − A
QA
τ
τ
−∞
∞
−1
1
dyϕ
× − + iy − A
τ
−1
∞
e
1
=
exp(−τ iy) − + iy − A
2π τ −∞
τ
−1
1
1
× Q − − + iy − + iy − A
Q
τ
τ
−1 1
1
− − + iy Q − + iy − A
τ
τ
−1
1
× A − + iy − A
dyϕ.
τ
(2.21)
A. Ashyralyev and P. Sobolevskii
273
By Lemma 2.2 and performing the change of variable τy = x, and using the
estimate (2.12), we obtain
−1 A A exp{−τ A}B(s) − B(s)A exp{−τ A} ϕ E
−1 ∞
1
e
exp(−τ iy) − + iy − A
≤
2πτ −∞
τ
E→E
−1
1
× 1 + 2M1 QE→E A − + iy − A
dyϕ τ
E
∞
e
dy
≤
2πτ −∞ 1/τ 2 + y 2 (1+α)/2
1−α (2−α)α
1 + 2M1 QE→E M11+α M α 1 + M1
2
ϕEα
×
α(1 − α)
∞
e
dx
=
(1+α)/2
πτ 1−α 0
1 + x2
1−α (2−α)α
1 + 2M1 QE→E M11+α M α 1 + M1
2
ϕEα
×
α(1 − α)
1−α (2−α)α
2
ϕEα
e(1 + α) 1 + 2M1 QE→E M11+α M α 1 + M1
.
≤
1−α
2
πτ
α (1 − α)
(2.22)
Lemma 2.3 is thus proved.
Suppose that
A−1 AB(t) − B(t)A A−1 E→E
≤
π(1 − α)2 α 2 1−α
2
22+α−α (1 + α)
1 + 2M1 1 + M1
1+α eM 1+α M1
(2.23)
holds for every t ≥ 0. Here and in the future is some constant, 0 ≤ ≤ 1.
Theorem 2.4. Assume that the condition
A−1 B(t)
E→E
≤
(1 − α)(1 − )
M22−α
holds for every t ≥ 0. Then for every t ≥ 0 the estimate (2.4) holds.
(2.24)
274
Delay differential and difference equations
Proof. Using formula (2.5) we have
λ1−α A exp{−λA}u(t)
= λ1−α A exp − (λ + t)A g(0)
λ+t −s
A B(s)
exp −
2
0
λ+t −s
A g(s − ω)ds
× A exp −
2
t
λ+t −s
A
exp −
+ λ1−α
2
0
λ+t −s
A B(s)
× A exp −
2
λ+t −s
A g(s − ω)ds
− B(s)A exp −
2
+ λ1−α
t
(2.25)
= I 1 + I2 + I3 .
Using the estimates (1.2), (2.12), and condition (2.24) we have
I1 = λ1−α A exp{−(λ + t)A}g(0)
E
E
≤
1−α
λ1−α g(0) ≤ λ
max g(t)E ,
E
1−α
1−α
α
α
−ω≤t≤0
(λ + t)
(λ + t)
I2 ≤ λ1−α
E
t
A exp − λ + t − s A 2
E→E
0
A−1 B(s)
E→E
λ+t −s
A g(s − ω)
× A exp −
ds
2
E
≤ max A−1 B(t)E→E
0≤t≤ω
≤ max g(t)E
−ω≤t≤0
t
0
Mλ1−α 22−α
ds max g(s − ω)E
2−α
α
0≤s≤ω
(λ + t − s)
α
λ1−α
1−
(1 − )
(λ + t)1−α
(2.26)
A. Ashyralyev and P. Sobolevskii
275
for every t, 0 ≤ t ≤ ω and λ, λ > 0. Now we estimate I3 . By Lemma 2.3 and
using the estimate (2.23), we obtain
t
A exp − λ + t − s A 2
I3 ≤ λ1−α
E
0
E→E
−1
λ+t −s
A B(s) − B(s)
× A
A exp −
2
λ+t −s
× A exp −
A g(s − ω)
ds
2
E
1−α (2−α)α
≤ λ1−α e(1 + α)M 1+α M11+α 1 + 2M1 1 + M1
2
t
×
A−1 AB(s) − B(s)A A−1 22−α g(s − ω)E ds
E→E
α
(λ + t − s)2−α πα 2 (1 − α)
0
≤ max A−1 AB(s) − B(s)A A−1 E→E
0≤s≤ω
1−α (2−α)α 2−α
e(1 + α)M 1+α M11+α 1 + 2M1 1 + M1
2
2
ds
×
2−α
2
(λ + t − s) πα (1 − α)
0
× max g(t)E
t
−ω≤t≤0
≤ max g(t)E
−ω≤t≤0
α
α
λ1−α
1−
(λ + t)1−α
(2.27)
for every t, 0 ≤ t ≤ ω and λ, λ > 0. Using the triangle inequality we obtain
λ1−α A exp{−λA}u(t)E ≤ max g(t)E
−ω≤t≤0
α
(2.28)
for every t, 0 ≤ t ≤ ω and λ, λ > 0. This shows that
u(t)
Eα
≤ max g(t)E
−ω≤t≤0
α
(2.29)
for every t, 0 ≤ t ≤ ω. Applying the mathematical induction, one can easily
show that it is true for every t. Namely, assume that inequality (2.4) is true for
every t, (n − 1)ω ≤ t ≤ nω, n = 1, 2, 3, . . . . Therefore, using formula (2.24)
276
Delay differential and difference equations
we have
λ1−α A exp{−λA}u(t)
= λ1−α A exp − (λ + t − nω)A u(nω)
+λ
λ+t −s
λ+t −s
A B(s)A exp −
A u(s − ω)ds
exp −
2
2
nω
λ+t −s
λ+t −s
A A exp −
A B(s)
exp −
2
2
nω
1−α
+ λ1−α
t
t
λ+t −s
A
− B(s)A exp −
2
× u(s − ω)ds
= I1n + I2n + I3n .
(2.30)
Using the estimates (1.2), (2.12), and condition (2.24), we have
I1n = λ1−α A exp − (λ + t − nω)A u(nω)
E
E
≤
λ1−α
λ1−α
u(nω) ≤
max g(t)E ,
E
1−α
1−α
α
α
−ω≤t≤0
(λ + t − nω)
(λ + t − nω)
I2n ≤ λ1−α
E
t
nω
A exp − λ + t − s A 2
A−1 B(s)
E→E
E→E
λ+t −s
ds
A
u(s
−
ω)
×
A
exp
−
2
E
≤
max
nω≤t≤(n+1)ω
×
max
nω≤s≤(n+1)ω
E→E
u(s − ω)
≤ max g(t)E
−ω≤t≤0
A−1 B(t)
Mλ1−α 22−α
ds
2−α
nω (λ + t − s)
t
Eα
α
λ1−α
1−
(1 − )
(λ + t)1−α
(2.31)
A. Ashyralyev and P. Sobolevskii
277
for every t, nω ≤ t ≤ (n + 1)ω and λ, λ > 0. Now we estimate I3n . By
Lemma 2.3 and using the estimate (2.23), we obtain
A exp − λ + t − s A 2
nω
E→E
−1
λ+t −s
A B(s)
× A
A exp −
2
λ+t −s
− B(s)A exp −
A u(s − ω)
ds
2
E
1−α (2−α)α
2
≤ λ1−α e(1 + α)M 1+α M11+α 1 + 2M1 1 + M1
I3n ≤ λ1−α
E
×
t
t
A−1 AB(s) − B(s)A A−1 22−α u(s − ω)E ds
E→E
α
nω
≤
max
nω≤s≤(n+1)ω
(λ + t − s)2−α πα 2 (1 − α)
A−1 AB(s) − B(s)A A−1 E→E
1−α (2−α)α 2−α
e(1 + α)M 1+α M11+α 1 + 2M1 1 + M1
2
2
ds
×
2−α
2
(λ + t − s) πα (1 − α)
nω
u(s − ω)
×
max
E
t
α
nω≤t≤(n+1)ω
≤ max g(t)E
−ω≤t≤0
α
1−
λ1−α
(λ + t)1−α
(2.32)
for every t, nω ≤ t ≤ (n + 1)ω and λ, λ > 0. Using the triangle inequality we
obtain
λ1−α A exp{−λA}u(t)E ≤ max g(t)E
(2.33)
−ω≤t≤0
α
for every t, nω ≤ t ≤ (n + 1)ω, n = 1, 2, 3, . . . , and λ, λ > 0. This shows that
u(t)
Eα
≤ max g(t)E
−ω≤t≤0
α
(2.34)
for every t, nω ≤ t ≤ (n + 1)ω, n = 1, 2, 3, . . . . This result completes the proof
of Theorem 2.4.
3. The delay difference equation
Using the first and second order accuracy implicit difference schemes for differential equations without delay (cf. [5]), we have the following approximate
278
Delay differential and difference equations
solution for the initial-value problem (1.1)
1
uk − uk−1 + Auk = Bk uk−N ,
τ
(3.1)
Nτ = ω,
Bk = B tk , tk = kτ, 1 ≤ k,
uk = g tk , tk = kτ, −N ≤ k ≤ 0;
1
τ
1
1 2
uk − uk−1 + A + τ A uk =
I + A Bk uk−N + uk−N−1 ,
τ
2
2
2
τ
(3.2)
, tk = kτ, 1 ≤ k,
Bk = B tk −
2
uk = g tk , tk = kτ, −N ≤ k ≤ 0.
Theorem 3.1. Assume that all the conditions of Theorem 2.1 are satisfied. Then
for the solution of difference scheme (3.1) the estimate
uk ≤ max g tk (3.3)
E
E
α
−N≤k≤0
α
holds for any k ≥ 1.
Proof. Consider 1 ≤ k ≤ N. In this case
uk = R k g(0) +
k
R k−j +1 g tj −N τ,
(3.4)
j =1
where R = (1 + τ A)−1 . Using the formula (see [5, 11])
∞
1
−k
t k−1 exp(−t) exp(−τ tA)dt,
(I + τ A) =
(k − 1)! 0
k ≥ 1,
(3.5)
and condition (2.3), we have
λ1−α A exp{−λA}uk E
∞
1
1−α
≤λ
t k−1 exp(−t)A exp{−(τ t + λ)A}g(0)E
(k − 1)! 0
+ λ1−α
j =1
τ
1
(k − j )!
(τ t + λ)A Bj A−1 t k−j exp(−t)
A
exp
−
E→E
2
0
E→E
(τ t + λ)A g tj −N × A exp −
2
E
×
k
∞
A. Ashyralyev and P. Sobolevskii
≤ λ1−α
1
(k − 1)!
+ λ1−α
∞
t k−1 exp(−t)
0
279
dt
g(0)
Eα
1−α
(tτ + λ)
k
1
1−α τ
2−α
(k − j )!
M2
j =1
∞
×
t k−j exp(−t)
0
M22−α dt g tj −N Eα
2−α
(tτ + λ)
≤ J max g tk E ,
−N ≤k≤0
(3.6)
α
where
J =λ
1−α
+λ
1
(k − 1)!
1−α
∞
t k−1 exp(−t)
0
(1 − α)
k
j =1
1
τ
(k − j )!
dt
(tτ + λ)1−α
∞
t
k−j
0
dt
exp(−t)
.
(tτ + λ)2−α
(3.7)
We will calculate J . Making the substitution k − j = m and using Taylor’s
formula and integration by parts, we obtain
∞
dt
1
J = λ1−α
t k−1 exp(−t)
(k − 1)! 0
(tτ + λ)1−α
+λ
1−α
(1 − α)
k−1
m=0
1
(k − 1)!
∞
1
τ
m!
∞
t m exp(−t)
0
dt
(tτ + λ)2−α
dt
(tτ + λ)1−α
0
∞
t k−1
dt
ξ
exp(−ξ )dξ
exp(−t)
+ λ1−α (1 − α)
exp(t) 1 −
(k − 1)!
(tτ + λ)2−α
0
0
∞
dt
1
= λ1−α
t k−1 exp(−t)
+1
(k − 1)! 0
(tτ + λ)1−α
∞
dt
1
t k−1 exp(−t)
− λ1−α
(k − 1)! 0
(tτ + λ)1−α
= λ1−α
t k−1 exp(−t)
= 1.
(3.8)
Therefore,
λ1−α A exp{−λA}uk E ≤ max g tk E
−N ≤k≤0
α
(3.9)
280
Delay differential and difference equations
for every k, 1 ≤ k ≤ N and λ, λ > 0. This shows that
uk Eα
≤ max g tk E
−N≤k≤0
(3.10)
α
for every k, 1 ≤ k ≤ N. Assume that the estimate (3.2) is true for k, where
(n − 1)N ≤ k ≤ nN, n = 1, 2, 3, . . . . Then for any k, with nN ≤ k ≤ (n + 1)N
using formula (3.5) and
k
uk = R k−nN unN +
τ R k−j +1 Bj uj −N
(3.11)
j =nN+1
and the estimates (1.2) and condition (2.3), we have
λ1−α A exp{−λA}uk E
≤λ
1−α
+λ
1
(k − nN − 1)!
1−α
k
j =nN+1
∞
0
t k−nN−1 exp(−t)A exp − (τ t + λ)A unN E
1
τ
(k − j )!
∞
t
k−j
0
(τ t + λ)A exp(−t)A exp −
2
E→E
(τ t + λ)A
u
× Bj A−1 E→E × A
exp
−
j −N 2
E
∞
dt
1
unN ≤ λ1−α
t k−nN−1 exp(−t)
Eα
(k − nN − 1)! 0
(tτ + λ)1−α
+λ
≤ Jn
1−α
1− α
M22−α
max
k
j =nN+1
(n−1)N≤k≤nN
1
τ
(k−j )!
uk Eα
∞
t k−j exp(−t)
0
M22−α dt uj −N Eα
2−α
(tτ + λ)
,
(3.12)
where
Jn = λ1−α
1
(k − nN − 1)!
+ λ1−α (1 − α)
∞
t k−nN −1 exp(−t)
0
k
j =nN+1
τ
1
(k − j )!
∞
0
dt
(tτ + λ)1−α
t k−j exp(−t)
dt
.
(tτ + λ)2−α
(3.13)
A. Ashyralyev and P. Sobolevskii
281
We will calculate Jn . Making the substitution k − j = m and using Taylor’s
formula and integration by parts, we obtain
Jn = λ
1−α
+λ
1
(k − nN − 1)!
1−α
(1 − α)
∞
t k−nN −1 exp(−t)
0
k−nN−1
m=0
1
τ
m!
∞
dt
(tτ + λ)1−α
t m exp(−t)
0
dt
(tτ + λ)2−α
∞
dt
1
t k−nN −1 exp(−t)
(k − nN − 1)! 0
(tτ + λ)1−α
∞
+ λ1−α (1 − α)
exp(t)
= λ1−α
0
× 1−
= λ1−α
1
(k − nN − 1)!
+ 1 − λ1−α
t
0
∞
dt
ξ k−nN −1 exp(−ξ )dξ
exp(−t)
(k − nN − 1)!
(tτ + λ)2−α
t k−nN −1 exp(−t)
0
1
(k − nN − 1)!
∞
dt
(tτ + λ)1−α
t k−nN −1 exp(−t)
0
dt
(tτ + λ)1−α
= 1.
(3.14)
Therefore
λ1−α A exp{−λA}uk E ≤ max g tk E
−N ≤k≤0
α
(3.15)
for every k, nN ≤ k ≤ (n + 1)N, n = 1, 2, 3, . . . , and λ, λ > 0. This shows that
u k Eα
≤ max g tk E
−ω≤k≤0
α
(3.16)
for every k, nN ≤ k ≤ (n+1)N, n = 1, 2, 3, . . . . This result completes the proof
of Theorem 3.1.
Theorem 3.2. Assume that all the conditions of Theorem 2.4 are satisfied. Then
for the solution of the difference scheme (3.1), the estimate (3.3) holds for any
k ≥ 1.
Proof. Consider 1 ≤ k ≤ N. Using formula (3.5) we have
282
Delay differential and difference equations
∞
1
t k−1 exp(−t)A exp − (τ t + λ)A g(0)
λ1−α A exp{−λA}uk = λ1−α
(k − 1)! 0
∞
k
1
τt +λ
1−α
k−j
+λ
A
τ
t
exp(−t) exp −
(k − j )! 0
2
j =1
τt +λ
× Bj A exp −
A g tj −N
2
+ λ1−α
k
τ
j =1
1
(k − j )!
τt +λ
×
A A−1
t
exp(−t)A exp −
2
0
τt +λ
τt +λ
A ABj − Bj A exp −
A g tj −N
× exp −
2
2
∞
k−j
= T1 + T 2 + T 3 .
(3.17)
Using the estimates (1.2), (2.12) and assumption (2.24), we have
T1 ≤ λ1−α
E
1
(k − 1)!
1
≤ λ1−α
(k − 1)!
k
T2 ≤ λ1−α
τ
E
j =1
≤
∞
0
∞
t k−1 exp(−t)A exp − (τ t + λ)A g(0)E
t k−1 exp(−t)
0
1
(k − j )!
dt
(tτ + λ)1−α
max g tk E ,
−N ≤k≤0
α
τt +λ A t
exp(−t)A exp −
2
0
E→E
τ t+λ
−1 A g tj −N × Bj A
× A exp −
E→E 2
E
∞
k−j
λ1−α (1 − α)(1 − )
M22−α
∞
k
1
M22−α dt g tj −N ×
τ
t k−j exp(−t)
Eα
2−α
(k − j )! 0
(tτ + λ)
j =1
≤ λ1−α (1 − α)(1 − )
k
j =1
× max g tk E
−N ≤k≤0
τ
1
(k − j )!
∞
0
t k−j exp(−t)
dt
(tτ + λ)2−α
α
(3.18)
A. Ashyralyev and P. Sobolevskii
283
for every k, 1 ≤ k ≤ N and λ, λ > 0. Now we estimate T3 . By Lemma 2.3 and
using the estimate (2.23), we obtain
k
T3 ≤ λ1−α
τ
E
j =1
1
(k − j )!
∞
0
τt +λ A
t k−j exp(−t)
A
exp
−
2
E→E
−1
τt +λ
τt +λ
A Bj −Bj A exp −
A g tj −N × A
A exp −
2
2
E
k
≤ λ1−α max A−1 ABj − Bj A A−1 E→E
τ
1≤k≤N
∞
×
t
k−j
0
×
≤λ
1−α
j =1
1
(k − j )!
1−α (2−α)α
e(1 + α)M α M11+α 1 + 2M1 1 + M1
2
exp(−t)
2
πα (1 − α)
M22−α dt g tj −N Eα
2−α
(tτ + λ)
(1 − α)
k
j =1
1
τ
(k − j )!
× max g tk E
−N ≤k≤0
∞
t k−j exp(−t)
0
dt
(tτ + λ)2−α
α
(3.19)
for every k, 1 ≤ k ≤ N and λ, λ > 0. Using the triangle inequality we obtain
λ1−α A exp{−λA}uk E ≤ max J g tk E
−N≤k≤0
α
(3.20)
for every k, 1 ≤ k ≤ N and λ, λ > 0. This shows that
u k Eα
≤ max g tk E
−N≤k≤0
α
(3.21)
for every k, 1 ≤ k ≤ N. Applying the mathematical induction, one can easily
show that it is true for every k. Namely, assume that inequality (3.2) is true for
every k, (n−1)N ≤ k ≤ nN , n = 1, 2, 3, . . . . Then for any k, nN ≤ k ≤ (n+1)N
using formula (2.8) we have
284
Delay differential and difference equations
λ1−α A exp{−λA}uk
= λ1−α
1
(k − nN − 1)!
k
+ λ1−α
τ
j =nN+1
∞
t k−nN−1 exp(−t)A exp − (τ t + λ)A unN
0
1
(k − j )!
∞
0
τt +λ
A Bj
t k−j exp(−t) exp −
2
τt +λ
A uj −N
× A exp −
2
∞
k
1
τt +λ
A
+ λ1−α
τ
t k−j exp(−t)A exp −
(k − j )! 0
2
j =nN+1
×A
−1
τt +λ
τt +λ
A ABj − Bj A exp −
A uj −N
exp −
2
2
= T1n + T2n + T3n .
(3.22)
Using the estimates (1.2), (2.12), and assumption (2.24) we have
∞
1
t k−nN −1 exp(−t)A exp −(τ t + λ)A unN E
(k−nN−1)! 0
∞
dt
1
1−α
g tk ,
≤λ
t k−nN−1 exp(−t)
max
Eα
(k−nN−1)! 0
(tτ +λ)1−α −N ≤k≤0
∞
k
1
k−j
T2n ≤ λ1−α
A exp − τ t + λ A τ
t
exp(−t)
E
(k − j )! 0
2
E→E
j =nN+1
τt +λ
× A−1 Bj E→E × A exp − 2 A uj −N E
T1n ≤ λ1−α
E
≤
λ1−α (1 − α)(1 − )
M22−α
×
k
τ
j =nN+1
≤λ
1−α
1
(k − j )!
∞
t k−j exp(−t)
0
k
(1 − α)(1 − )
j =nN+1
× max g tk E
−N≤k≤0
1
τ
(k − j )!
M22−α dt uj −N Eα
(tτ + λ)2−α
∞
0
t k−j exp(−t)
dt
(tτ + λ)2−α
α
(3.23)
A. Ashyralyev and P. Sobolevskii
285
for every k, nN ≤ k ≤ (n+1)N , n = 1, 2, 3, . . . , and λ, λ > 0. Now we estimate
T3n . By Lemma 2.3 and using the estimate (2.23), we obtain
τt +λ A
t
exp(−t)
A
exp
−
2
0
E→E
j =nN+1
−1
τt +λ
τt +λ
A
B
A
u
×
A
−
B
A
exp
−
A
exp
−
j
j
j −N 2
2
E
k
T3n ≤ λ1−α
τ
E
≤λ
1−α
1
(k − j )!
max
nN+1≤k≤(n+1)N
∞
k−j
A−1 ABj − Bj A A−1 t
k−j
e(1+α)M α M1
exp(−t)
0
×
≤λ
1−α
k
E→E
1+α ∞
×
j =1
1+2M1 1+M1
πα 2 (1−α)
τ
1
(k − j )!
1−α
2(2−α)α
M22−α dt uj −N Eα
2−α
(tτ + λ)
(1 − α)
k
j =nN+1
× max g tk E
−N≤k≤0
1
τ
(k − j )!
∞
t k−j exp(−t)
0
dt
(tτ + λ)2−α
α
(3.24)
for every k, nN ≤ k ≤ (n+1)N, n = 1, 2, 3, . . . , and λ, λ > 0. Using the triangle
inequality we obtain
λ1−α A exp{−λA}uk E ≤ max Jn g tk E
−N≤k≤0
α
(3.25)
for every k, nN ≤ k ≤ (n + 1)N, n = 1, 2, 3, . . . , and λ, λ > 0. This shows that
u k Eα
≤ max g tk E
−ω≤k≤0
α
(3.26)
for every k, nN ≤ (n + 1)N , n = 1, 2, 3, . . . . This result completes the proof of
Theorem 3.2.
Now we consider the difference scheme (3.2). We have not been able to
obtain the same result for solution of the difference scheme (3.2) in the spaces
Eα under the assumption for the difference scheme (3.1). But, we have the
asymptotical stability of the solution of the difference scheme under the supplementary restriction of the operator A that is considered in the following section.
286
Delay differential and difference equations
4. The difference scheme (3.2)
Theorem 4.1. Let condition (1.2) be satisfied. Suppose that the following estimate holds:
2 −1 (I + τ A) I + τ A + (τ A)
2
E→E
2 −1 I + τ A (I + τ A) I + τ A + (τ A)
2
2
≤ 1,
√
E→E
≤
(4.1)
1+ 2
,
2
and
B(t)A−1 E→E
≤
(1 − α)
√ 1+ 2
M21−α
(4.2)
holds for every t ≥ 0.
Then for the solution of difference scheme (3.2) the estimate
uk Eα
≤ max g tk E
−N≤k≤0
α
(4.3)
holds for any k ≥ 1.
Proof. Consider 1 ≤ k ≤ N. In this case
uk = R k g(0) +
k
j =1
τA g tj −N + g tj −N −1 τ,
R k−j +1 I +
2
(4.4)
where R = (I +τ A+(τ A)2 /2)−1 . Using formula (3.5), the estimates (1.2), and
assumption (4.1) we have
2 −1 k (τ
A)
A exp{−λA}uk E ≤ λ
(I + τ A) I + τ A + 2
E→E
∞
1
× λ1−α
t k−1 exp(−t)A exp − (τ t + λ)A g(0)E
(k − 1)! 0
k
(τ A)2 −1 k−j + λ1−α
τ
(I
+
τ
A)
I
+
τ
A
+
2
E→E
1−α j =1
(τ A)2 −1 τA
(I + τ A) I + τ A +
× I +
2
2
E→E
A. Ashyralyev and P. Sobolevskii 287
∞
τt +λ k−j
Bj A−1 A t
exp(−t)A exp −
E→E
2
1
(k − j )! 0
E→E
1 τt +λ
A
g
t
×
A
exp
−
+
g
t
j
−N
j
−N
−1
2
2
E
∞
dt
1
g(0)
t k−1 exp(−t)
≤ λ1−α
Eα
1−α
(k − 1)! 0
(tτ + λ)
√ ∞
k
1+ 2
1−α
1
M22−α dt
+
τ
t k−j exp(−t)
√ (k − j )! 0
2
(tτ + λ)2−α
M21−α 1 + 2
×
j =1
1 × g tj −N + g tj −N−1 2
Eα
≤ J max g tk E .
−N≤k≤0
(4.5)
α
Therefore
λ1−α A exp{−λA}uk E ≤ max g tk E
−N ≤k≤0
α
(4.6)
for every k, 1 ≤ k ≤ N and λ, λ > 0. This shows that
u k Eα
≤ max g tk E
−N≤k≤0
α
(4.7)
for every k, 1 ≤ k ≤ N. Assume that the estimate (4.3) is true for k, with
(n − 1)N ≤ k ≤ nN, n = 1, 2, 3, . . . . Then for any k, nN ≤ k ≤ (n + 1)N using
formula (3.5) and
uk = R
k−nN
unN +
k
j =nN+1
τR
k−j
τA
1
I+
Bj uj −N + uj −N −1 ,
2
2
(4.8)
the estimates (1.2), and assumption (4.1) we have that
λ1−α A exp{−λA}uk E
(τ A)2 −1 k−nN ≤ (I + τ A) I + τ A +
2
E→E
∞
1
× λ1−α
t k−nN−1 exp(−t)A exp −(τ t + λ)A unN E
(k − nN − 1)! 0
288
Delay differential and difference equations
+λ
1−α
k
j =nN+1
(τ A)2 −1 k−j τ (I + τ A) I + τ A +
2
E→E
(τ A)2 −1 τA
(I + τ A) I + τ A +
× I +
2
2
E→E
∞
1
(τ t + λ)A k−j
×
t
exp(−t)A exp −
(k − j )! 0
2
E→E
(τ t + λ)A 1 −1 A exp −
uj −N + uj −N −1 × Bj A
E→E 2
2
E
∞
dt
1
unN ≤ λ1−α
t k−nN−1 exp(−t)
Eα
1−α
(k − nN − 1)! 0
(tτ + λ)
√ k
1+ 2
1−α
1
+
τ
√ 1−α
2
(k
−
j )!
M2
1+ 2
j =nN+1
∞
×
0
≤ Jn
Therefore
M22−α dt 1 uj −N + uj −N −1 (tτ + λ)2−α 2
Eα
uk .
E
t k−j exp(−t)
max
(n−1)N≤k≤nN
α
λ1−α A exp{−λA}uk E ≤ max g tk E
−N ≤k≤0
α
(4.9)
(4.10)
for every k, nN ≤ k ≤ (n + 1)N , n = 1, 2, 3, . . . , and λ, λ > 0. This shows that
uk ≤ max g tk (4.11)
E
E
α
−N≤k≤0
α
for every k, nN ≤ (n + 1)N , n = 1, 2, 3, . . . . This result completes the proof of
Theorem 4.1.
Now, we consider the difference scheme (3.2) when
A−1 B(t)x = B(t)A−1 x,
Suppose that
A−1 AB(t) − B(t)A A−1 x ∈ D(A).
E→E
√ −1
1+ 2 ≤
1−α
2
eM 1+α M11+α 1 + 2M1 1 + M1
22+α−α
π(1 − α)2 α 2 (1 + α)−1
holds for every t ≥ 0.
(4.12)
(4.13)
A. Ashyralyev and P. Sobolevskii
289
Theorem 4.2. Let the assumptions (1.2) and (4.13) be satisfied. Suppose that
the following estimate holds:
A−1 B(t)
E→E
≤
(1 − α)(1 − )
√ M22−α 1 + 2
(4.14)
for every t ≥ 0. Then for the solution of difference scheme (3.2) the estimate
(4.3) holds for any k ≥ 1.
Proof. Consider 1 ≤ k ≤ N. Using formula (3.5) we obtain
λ1−α A exp{−λA}uk
(τ A)2 −1 k
= (I + τ A) I + τ A +
2
∞
1
× λ1−α
t k−1 exp(−t)A exp − (τ t + λ)A g(0)
(k − 1)! 0
k
(τ A)2 −1 k−j
τA
+ λ1−α
τ (I + τ A) I + τ A +
I+
2
2
j =1
(τ A)2 −1
× (I + τ A) I + τ A +
2
∞
1
(τ t + λ)A
k−j
×
t
exp(−t) exp −
(k − j )! 0
2
(τ t + λ)A 1 (4.15)
g tj −N + g tj −N −1
× Bj A exp −
2
2
k
(τ A)2 −1 k−j
τA
1−α
τ (I + τ A) I + τ A +
I+
+λ
2
2
j =1
(τ A)2 −1
× (I + τ A) I + τ A +
2
∞
1
(τ t + λ)A −1
A
×
t k−j exp(−t)A exp −
(k − j )! 0
2
(τ t + λ)A
(τ t + λ)A
ABj − Bj A exp −
× exp −
2
2
1
× g tj −N + g tj −N−1
2
= P1 + P 2 + P 3 .
Using the estimates (1.2), (2.12), and assumption (2.24), we have
290
Delay differential and difference equations
2 −1 k P1 ≤ (I + τ A) I + τ A + (τ A)
E
2
E→E
∞
1
× λ1−α
t k−1 exp(−t)A exp − (τ t + λ)A g(0)E
(k − 1)! 0
∞
dt
1
1−α
≤λ
t k−1 exp(−t)
max g tk E ,
1−α
α
−N ≤k≤0
(k − 1)! 0
(tτ + λ)
k
(τ A)2 −1 k−j P2 ≤ λ1−α
τ (I + τ A) I + τ A +
E
2
E→E
j =1
(τ A)2 −1 τA
(I + τ A) I + τ A +
× I +
2
2
E→E
∞
1
(τ t + λ)A ×
t k−j exp(−t)
A
exp
−
(k − j )! 0
2
E→E
(τ t + λ)A 1 −1
g tj −N + g tj −N −1 × A Bj E→E A exp −
2
2
E
√ 1 + 2 λ1−α (1 − α)(1 − )
≤
√ 2
M21−α 1 + 2
×
k
j =1
τ
1
(k − j )!
∞
t k−j exp(−t)
0
M22−α dt
(tτ + λ)2−α
1 × g tj −N + g tj −N−1 2
Eα
≤ (1 − α)(1 − )
k
j =1
1
τ
(k − j )!
× max g tk E
−N≤k≤0
∞
0
t k−j exp(−t)
λ1−α dt
(tτ + λ)2−α
α
(4.16)
for every k, 1 ≤ k ≤ N and λ, λ > 0. Now we estimate P3 . By Lemma 2.3 and
using the estimate (2.23), we obtain
k
2 −1 k−j P3 ≤ λ1−α
(I + τ A) I + τ A + (τ A)
τ
E
2
E→E
j =1
(τ A)2 −1 τA
(I + τ A) I + τ A +
× I +
2
2
E→E
A. Ashyralyev and P. Sobolevskii 291
∞
(τ t + λ)A 1
k−j
t
exp(−t)A exp −
×
(k − j )! 0
2
E→E
−1
(τ t + λ)A
(τ t + λ)A
B
−
B
A
exp
−
A
exp
−
×
A
j
j
2
2
1 × g tj −N + g tj −N−1 2
E
√
k
λ1−α 1 + 2
1
−1
−1
max A ABj − Bj A A
≤
τ
E→E
1≤j ≤N
2
(k − j )!
j =1
1+α ∞
×
t
k−j
exp(−t)
e(1 + α)M α M1
0
1 + 2M1 1 + M1
πα 2 (1 − α)
1−α
2(2−α)α
1
g tj −N + g tj −N −1 Eα
2
∞
k
1
dt
1−α
≤ λ (1 − α)
τ
t k−j exp(−t)
(k − j )! 0
(tτ + λ)2−α
×
M22−α dt
(tτ + λ)2−α
j =1
× max g tk E
−N ≤k≤0
(4.17)
α
for every k, 1 ≤ k ≤ N and λ, λ > 0. Using the triangle inequality we obtain
λ1−α A exp{−λA}uk E ≤ J max g tk E
−N ≤k≤0
α
(4.18)
for every k, 1 ≤ k ≤ N and λ, λ > 0. This shows that
u k Eα
≤ max g tk E
−N≤k≤0
α
(4.19)
for every k, 1 ≤ k ≤ N. Assume that inequality (3.3) is true for k, with (n −
1)N ≤ k ≤ nN , n = 1, 2, 3, . . . . Then for any k, nN ≤ k ≤ (n + 1)N using
formula (2.8) we have that
λ1−α A exp{−λA}uk
(τ A)2 −1 k−nN
= (I + τ A) I + τ A +
2
∞
1
× λ1−α
t k−nN−1 exp(−t)A exp − (τ t + λ)A unN
(k − nN − 1)! 0
292
Delay differential and difference equations
+λ
(τ A)2 −1 k−j
τ (I + τ A) I + τ A +
2
k
1−α
j =nN+1
(τ A)2 −1
τA
(I + τ A) I + τ A +
× I+
2
2
∞
1
(τ t + λ)A
k−j
Bj A
×
t
exp(−t) exp −
2
(k − j )! 0
(τ t + λ)A 1 uj −N + uj −N −1
× exp −
2
2
+λ
(τ A)2 −1 k−j
τ (I + τ A) I + τ A +
2
k
1−α
j =nN+1
(τ A)2 −1
τA
(I + τ A) I + τ A +
× I+
2
2
∞
1
(τ t + λ)A
×
t k−j exp(−t)A exp −
(k − j )! 0
2
(τ t + λ)A
(τ t + λ)A
−1
×A
exp −
ABj − Bj A exp −
2
2
×
1
uj −N + uj −N−1
2
= P1n + P2n + P3n .
(4.20)
Using the estimates (1.2), (2.12), and assumption (2.24) we have
2 −1 k−nN (τ
A)
P1n ≤ (I + τ A) I + τ A +
E
2
E→E
∞
t k−1 exp(−t)A exp − (τ t + λ)A unN E
1
(k − nN − 1)! 0
∞
dt
1
1−α
≤λ
t k−1 exp(−t)
max g tk E ,
1−α
α
−N ≤k≤0
(k − nN − 1)! 0
(tτ + λ)
× λ1−α
P2n ≤ λ1−α
E
k
j =nN+1
(τ A)2 −1 k−j τ
(I
+
τ
A)
I
+
τ
A
+
2
E→E
(τ A)2 −1 τA
(I + τ A) I + τ A +
× I +
2
2
E→E
A. Ashyralyev and P. Sobolevskii 293
∞
(τ t + λ)A 1
k−j
t
exp(−t)A exp −
×
(k − j )! 0
2
E→E
(τ t + λ)A 1 u
× A−1 Bj E→E +
u
A
exp
−
j −N
j −N −1 2
2
E
√ 1 + 2 λ1−α (1 − α)(1 − )
≤
√ 2
M21−α 1 + 2
×
k
j =nN+1
1
τ
(k − j )!
∞
t k−j exp(−t)
0
M22−α dt
(tτ + λ)2−α
1
u
×
+
u
j −N−1 2 j −N
Eα
≤ (1 − α)(1 − )
k
j =nN+1
× max g tk E
−N≤k≤0
τ
1
(k − j )!
∞
t k−j exp(−t)
0
λ1−α dt
(tτ + λ)2−α
(4.21)
α
for every k, nN ≤ k ≤ (n+1)N, n = 1, 2, 3, . . . , and λ, λ > 0. Now we estimate
P3n . By Lemma 2.3 and using the estimate (2.23), we obtain
P3n ≤ λ1−α
E
k
j =nN+1
(τ A)2 −1 k−j τ
(I
+
τ
A)
I
+
τ
A
+
2
E→E
(τ A)2 −1 τA
(I
+
τ
A)
I
+
τ
A
+
×
I
+
2
2
E→E
∞
(τ t + λ)A t k−j exp(−t)
A
exp
−
2
1
(k − j )! 0
E→E
−1
(τ t + λ)A
A exp −
Bj
×
A
2
(τ t + λ)A 1 uj −N + uj −N −1 − Bj A exp −
2
2
E
√
1−α
λ
1+ 2
A−1 ABj − Bj A A−1 max
≤
E→E
nN+1≤j ≤(n+1)N
2
×
×
k
j =nN +1
τ
1
(k − j )!
294
Delay differential and difference equations
∞
×
t
k−j
0
×
1−α (2−α)α
e(1 + α)M α M11+α 1 + 2M1 1 + M1
2
exp(−t)
2
πα (1 − α)
M22−α dt 1 uj −N + uj −N −1 Eα
2−α
2
(tτ + λ)
≤ λ1−α (1 − α)
k
τ
j =nN+1
∞
×
t k−j exp(−t)
0
1
(k − j )!
dt
(tτ + λ)2−α
max g tk E
−N ≤k≤0
α
(4.22)
for every k, nN ≤ k ≤ (n + 1)N and λ, λ > 0. Using the triangle inequality we
obtain
(4.23)
λ1−α A exp{−λA}uk ≤ Jn max g tk E
−N ≤k≤0
Eα
for every k, nN ≤ k ≤ (n + 1)N, n = 1, 2, 3, . . . , and λ, λ > 0. This shows that
uk ≤ max g tk (4.24)
E
E
α
−ω≤k≤0
α
for every k, nN ≤ (n + 1)N , n = 1, 2, 3, . . . . This result completes the proof of
Theorem 4.2.
5. Applications
We consider the initial-value problem on the range {0 ≤ t ≤ 1, x ∈ Rn } for
2m-order multidimensional delay differential equations of parabolic type
∂ |τ | u(t, x)
∂u(t, x)
+
aτ (x) τ1
+ δu(t, x)
∂t
∂x1 · · · ∂xnτn
|τ |=2m

= b(t)
|τ |=2m

∂ |τ | u(t − ω, x)
aτ (x)
+ δu(t − ω, x) ,
∂x1τ1 · · · ∂xnτn
0 < t < ∞, x ∈ Rn ,
(5.1)
u(t, x) = g(t, x),
|τ | = τ1 + · · · + τn ,
−ω ≤ t ≤ 0, x ∈ R ,
n
(5.2)
where ar (x), b(t), g(t, x) are given sufficiently smooth functions and δ > 0 is
the sufficiently large number. We will assume that the symbol
ar (x)(iξ )r1 · · · (iξ )rn
(5.3)
Ax1 (ξ ) =
|τ |=2m
of the differential operator of the form
Ax1 =
|r=2m|
ar (x)
∂ |r|
∂x1r1 · · · ∂xnrn
(5.4)
A. Ashyralyev and P. Sobolevskii
acting on functions defined on the space Rn , satisfies the inequalities
0 < M1 |ξ |2m ≤ Ax1 (ξ ) ≤ M2 |ξ |2m < ∞ for ξ = 0.
295
(5.5)
Problem (5.1) has a unique smooth solution. This allows us to reduce the
initial-value problem (5.1) to the initial-value problem (1.1) in the Banach space
E with a strongly positive operator Ax = Ax1 + δI defined by (5.4). We give a
number of corollaries of the abstract Theorems 2.1 and 3.1.
Theorem 5.1. Assume that
1−α
sup b(t) ≤
.
M22−α
0≤t<∞
(5.6)
Then the solutions of the initial-value problem (5.1) satisfy the following stability estimates:
sup v(t, ·)C 2mα (Rn ) ≤ M2 (α) max g(t, ·)C 2mα (Rn ) ,
1
,
−ω≤t≤0
2m
0≤t<∞
(5.7)
where M2 (α) does not depend on g(t, x). Here C ε (Rn ) is the space of functions
satisfying a Hölder condition with the indicator ε ∈ (0, 1).
The proof of Theorem 5.1 is based on the estimate
exp − tAx n
≤ M, t ≥ 0,
C(R )→C(Rn )
0<α<
(5.8)
and on the abstract Theorem 2.1, on the strong positivity of the operator Ax in
C ε (Rn ), and on the fact of the equivalence of the norms in the spaces Eα =
Eα (A, C(Rn )) and C 2mα (Rn ) when 0 < α < 1/2m (see [5]).
We define the grid space Rnh (0 < h ≤ h0 ) as the set of all points of the
Euclidean space Rn whose coordinates are given by
xi = si h,
si = 0, ±1, ±2, . . . ,
i = 1, . . . , n.
(5.9)
To the differential operator Ax = Ax1 + δI defined by (5.4) we assign the difference operator Axh by the formula
Axh uh (x) =
brh (x)Dhr uh (x)
(5.10)
2m≤|r|≤S
of an arbitrary high order of accuracy that approximates this multidimensional
elliptic operator Ax which acts on functions uhx defined on the entire space Rnh .
We will assume that for |ξk h| ≤ π , the symbol Ax (ξ h, h) of the operator Axh
satisfies the inequalities
x
A (ξ h, h) ≥ M3 |ξ |2m , arg Ax (ξ h, h) ≤ φ < π.
(5.11)
296
Delay differential and difference equations
With the help of Axh we arrive at the initial-value problem
h
u (t, x) + Axh uh (t, x) = b(t)Axh uh (t − ω, x), t ≥ 0, x ∈ Rn ,
uh (t, x) = g h (t, x) = g(t, x)
(−ω ≤ t ≤ 0), x ∈ Rn ,
(5.12)
for an infinite system of ordinary differential equations. Now, we replace problem (5.12) by the first order accuracy in t difference scheme
1 h
uk (x) − uhk−1 (x) + Axh uhk (x) = b tk Axh uhk−N (x),
τ
tk = kτ, 1 ≤ k, N τ = ω, x ∈ Rnh ,
uhk (x) = g h tk , x , tk = kτ, −N ≤ k ≤ 0, x ∈ Rnh .
(5.13)
Theorem 5.2. Assume that all the conditions of Theorem 5.1 are satisfied. Then
for the solution of difference scheme (5.13) the estimate
sup uhk C 2mα (Rn ) ≤ M2 (α) max gkh C 2mα (Rn ) ,
1≤k<∞
h
−N≤t≤0
0<α<
h
1
, (5.14)
2m
holds.
The proof of Theorem 5.2 is based on the estimate
exp − tk Ax n
h C(R )→C(Rn ) ≤ M, k ≥ 0,
h
(5.15)
h
and on the abstract Theorem 3.1, the positivity of the operator Axh in C ε (Rnh ),
and on the fact that the Eα = Eα (Axh , C(Rnh ))-norms are equivalent to the norms
C 2mα (Rnh ) uniformly in h for 0 < α < 1/2m (see [5]).
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A. Ashyralyev: Department of Mathematics, Fatih University, 34900,
Buyukcekmece, Istanbul, Turkey
Current address: Department of Mathematics, International Turkmen-Turkish
University, 84, Gerogly, 744012, Ashgabat, Turkmenistan
E-mail address: aashyr@fatih.edu.tr
P. Sobolevskii: Institute of Mathematics, Hebrew University, Jerusalem,
Israel
E-mail address: pavels@math.hiji.ae.il