Hindawi Publishing Corporation
Discrete Dynamics in Nature and Society
Volume 2007, Article ID 57491, 25 pages
doi:10.1155/2007/57491
Research Article
Stability of a Second Order of Accuracy Difference Scheme for
Hyperbolic Equation in a Hilbert Space
Allaberen Ashyralyev and Mehmet Emir Koksal
Received 7 June 2007; Accepted 16 September 2007
The initial-value problem for hyperbolic equation d2 u(t)/dt 2 + A(t)u(t) = f (t) (0 ≤ t ≤
T), u(0) = ϕ,u (0) = ψ in a Hilbert space H with the self-adjoint positive definite operators A(t) is considered. The second order of accuracy difference scheme for the approximately solving this initial-value problem is presented. The stability estimates for the
solution of this difference scheme are established.
Copyright © 2007 A. Ashyralyev and M. E. Koksal. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use,
distribution, and reproduction in any medium, provided the original work is properly
cited.
1. Introduction
It is known (see, e.g., [1, 2]) that various mixed problems for the hyperbolic equations
can be reduced to the initial-value problem
d2 u(t)
+ A(t)u(t) = f (t) (0 ≤ t ≤ T),
dt 2
u(0) = ϕ,
u (0) = ψ
(1.1)
for differential equation in a Hilbert space H. Here, A(t) are the self-adjoint positive
definite operators in H with a t-independent domain D = D(A(t)).
A function u(t) is called a solution of the problem (1.1) if the following conditions are
satisfied:
(i) u(t) is twice continuously differentiable on the segment [0, T]; the derivatives
as the endpoints of the segment are understood as the appropriate unilateral
derivatives;
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Discrete Dynamics in Nature and Society
(ii) the element u(t) belongs to D for all t ∈ [0,T], and the function Au(t) is continuous on the segment [0,T];
(iii) u(t) satisfies the equation and the initial conditions (1.1).
A large cycle of works on difference schemes for hyperbolic partial differential equations (see, e.g., [3–6] and the references given therein), in which stability was established
under the assumption that the magnitudes of the grid steps τ and h with respect to the
time and space variables is connected. In abstract terms this means, in particular, that the
condition τ Aτ,h →0 when τ →0 is satisfied.
Of great interest is the study of absolute stable difference schemes of a high order of
accuracy for hyperbolic partial differential equations, in which stability was established
without any assumptions in respect of the grid steps τ and h. The stability inequalities for
solutions of the first order of accuracy difference scheme
Ak = A tk ,
τ
−1
τ −2 uk+1 − 2uk + uk−1 + Ak uk+1 = fk ,
fk = f tk ,
tk = kτ, 1 ≤ k ≤ N − 1, Nτ = T,
u1 − u0 + iA1/2
1 u1
= iA1/2
0 u0 + ψ,
(1.2)
u0 = ϕ
for approximately solving problem (1.1) were established without any assumtions for the
first time in the paper [7].
The study of the high order of accuracy of absolute stable difference schemes for approximately solving problem (1.1) in the case of A(t) = A has been studied in the papers
[3, 8–11]. The second order of accuracy difference schemes
τ2 2
A uk+1 = fk ,
4
tk = kτ, 1 ≤ k ≤ N − 1, Nτ = T,
τ −2 uk+1 − 2uk + uk−1 + Auk +
fk = f tk ,
τ −1 u1 − u0 + iA1/2 I +
iτ 1/2
u1 = z1 ,
A
2
τ
f0 + iA1/2 − τA u0 , f0 = f (0), u0 = ϕ,
2
1 τ −2 uk+1 − 2uk + uk−1 + A uk+1 + 2uk + uk−1 = fk ,
4
fk = f tk , tk = kτ, 1 ≤ k ≤ N − 1, Nτ = T,
z1 = I + iτA1/2 ψ +
(1.3)
i
τ −1 u1 − u0 + A1/2 u1 + u0 = z1 ,
2
iτ 1/2
τ
τA
ψ + f0 + iA1/2 −
u0 , f0 = f (0), u0 = ϕ
z1 = I + A
2
2
2
was presented in the paper [8]. The stability estimates for the solution of these difference
schemes; and its first and second order difference derivatives were established. Unfortunately, these difference schemes are generated by the A1/2 . In paper [9], the first order of
A. Ashyralyev and M. E. Koksal 3
accuracy difference scheme
τ −2 uk+1 − 2uk + uk−1 + Auk+1 = fk ,
fk = f tk ,
τ
−1
tk = kτ, 1 ≤ k ≤ N − 1, Nτ = T,
(1.4)
u0 = ϕ,
u1 − u0 = ψ,
and second order of accuracy difference scheme
τ2 2
A uk+1 = fk ,
4
fk = f tk , tk = kτ, 1 ≤ k ≤ N − 1, Nτ = T,
τ
I + τ 2 A τ −1 u 1 − u 0 =
f0 − Au0 + ψ, f0 = f (0), u0 = ϕ,
2
1
1 τ −2 uk+1 − 2uk + uk−1 + Auk + A uk+1 + uk−1 = fk ,
2
4
fk = f tk , tk = kτ, 1 ≤ k ≤ N − 1, Nτ = T,
τ
I + τ 2 A τ −1 u 1 − u 0 =
f0 − Au0 + ψ, f0 = f (0), u0 = ϕ
2
τ −2 uk+1 − 2uk + uk−1 + Auk +
(1.5)
(1.6)
for approximately solving this initial-value problem were presented. These difference
schemes were generated by the integer power of A. The stability estimates for the solution
of these difference schemes were established.
In papers [10, 11], the high order of accuracy two-step difference schemes generated
by an exact difference scheme or by the Taylor’s decomposition on three points for the numerical solutions of this problem was presented. The stability estimates for the solutions
of these difference schemes were established. In applications, the stability estimates for the
solutions of the high order of accuracy difference schemes of the mixed type boundary
value problems for hyperbolic equations were obtained.
We are interested in studying the high order of accuracy two-step difference schemes
for the approximate solutions of the problem (1.1) in a Hilbert space H with self-adjoint
positive definite operators A(t). In paper [12], second order of accuracy difference
scheme
1/2
−1
u k + u k −1
τ −2 uk+1 − 2uk + uk−1 + Ak+1/2 4−1 uk+1 + uk + A1/2
k+1/2 Ak−1/2 4
1/2
−1/2 −1
+ τ −1 A1/2
u k − u k −1
k−1/2 − Ak+1/2 Ak−1/2 τ
1/2
1/2 −1
+ 2−1 τ −1 A1/2
A−k+1/2
τ uk+1 − uk
k+1 − Ak
−1/2 −1 −1
1/2
−1/2 −1
+ A1/2
A1/2
u k − u k −1
k+1/2 Ak−1/2 2 τ
k − Ak−1 Ak−1/2 τ
1/2 −1/2
= 2−1 fk−1/2 + fk+1/2 + 2−1 A1/2
k+1/2 − Ak−1/2 Ak−1/2 fk−1/2 ,
1 ≤ k ≤ N − 1, u0 = u(0)
τ
τ
−1/2 −1 τ
−1/2 τ −1 u1 − u0 + A1/2 2−1 u1 + u0 + A1/2
A τ u1 − u0 = f1/2 + A1/2
1/2 A1/2 u0
2
2 1/2 1/2
2
(1.7)
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Discrete Dynamics in Nature and Society
generated by Crank-Nicholson difference scheme was presented. The following theorems
under the same smoothness assumption on A(t)A−1 (0) (see, e.g., [12]) on the stability estimates for the solution of this difference scheme and its first and second order difference
derivatives were established.
Theorem 1.1. Let u(0) ∈ D(A1/2 (0)). Then for the solution of the difference scheme (1.7),
the stability estimate
u k − u k −1 N −1 τ
1
Cτ
+ uτ Cτ ≤ M A1/2 (0)u0 H + u0 H +
N
−1
s =0
fs+1/2 τ
H
(1.8)
holds, where M does not depend on u0 ,u0 , fs+1/2 (0 ≤ s ≤ N − 1) and τ.
Theorem 1.2. Let u(0) ∈ D(A(0)), u (0) ∈ D(A1/2 (0)). Then for the solution of the difference scheme (1.7), the stability estimat
N −1 1/2
A (0) uk − uk−1
τ
1
Cτ
1/2
−1
+ Ak+1/2 4−1 uk+1 + uk + A1/2
u k + u k −1
k+1/2 Ak−1/2 4
1/2 −1/2 −1 + τ −1 A1/2
u k − u k −1
k−1/2 − Ak+1/2 Ak−1/2 τ
1/2 −1/2 −1 + 2−1 τ −1 A1/2
Ak+1/2 τ uk+1 − uk
k+1 − Ak
1/2
−1/2 −1 −1/2 −1 −1
+ A1/2
Ak − A1/2
u k − u k −1 H
k+1/2 Ak−1/2 2 τ
k−1 Ak−1/2 τ
+ τ −2 uk+1 − 2uk + uk−1
N −1 1
Cτ
n
−2
fs+1/2 − fs−1/2 ≤ M A(0)u0 H + A1/2 (0)u0 H + max fs+1/2 H +
H
0≤s≤k
s =0
(1.9)
holds, where M does not depend on u0 ,u0 , fs+1/2 (0 ≤ s ≤ N − 1) and τ.
Note that the difference scheme (1.7) for approximately solving problem (1.1) in the
case of A(t) = A is (1.6). So, these stability estimates are generalization of the results of
paper [9] in the general case of A(t).
In the present paper, the difference scheme (1.5) for approximately solving problem
(1.1) in the general case of A(t) is presented. Unfortunately, the stability estimates for
−1
−1
{(uk − uk−1 )/τ }N
Cτ and {uk }N
Cτ cannot be obtained for the solution of this
1
1
difference scheme under the same conditions of Theorem 1.1. Nevertheless, the stabilN −1
−1
Cτ , {Ak uk }N
Cτ and {τ −2 (uk+1 − 2uk +
ity estimates for {A1/2
1
k−1/2 (uk − uk−1 /τ)}1
N −1
uk−1 )}1 Cτ are obtained for the solution of this difference scheme under the same conditions.
A. Ashyralyev and M. E. Koksal 5
2. The construction of one difference scheme of a second order of accuracy
By papers [13, 14], we have the equivalent initial-value problem for a system of the first
order linear differential equations
du(t)
= iA1/2 (t)v(t),
dt
u (0) = u0 ,
u(0) = u0 ,
0 < t < T,
dv(t)
= iA1/2 (t)u(t) − A−1/2 (t) A1/2 (t) v(t) − iA−1/2 (t) f (t).
dt
(2.1)
For construction of a two-step difference scheme, we consider the uniform grid space
[0,T]τ = tk = kτ, 0 ≤ k ≤ N, Nτ = T .
(2.2)
Using the central difference formula for the derivative and (2.1), we can write
2
1 ≤ k ≤ N,
= iA1/2
k v tk−1/2 + o τ ,
1/2 1/2 −1/2
−1
Ak v tk−1/2
τ v tk − v tk−1 = iAk u tk−1/2 − Ak
1/2
−iA−
fk + o τ 2 , 1 ≤ k ≤ N,
v0 = −iA−0 1/2 u0 ,
k
τ −1 u tk − u tk−1
(2.3)
where
1/2
A1/2
= (A )
tk−1/2 ,
k
τ
tk−1/2 = tk − ,
A0 = A(0).
2
1/2 t
A1/2
k−1/2 ,
k =A
fk = f tk−1/2 ,
(2.4)
Using the Taylor expansion, we can write
τ −1 u tk − u tk−1
τ = u tk − u tk + o τ 2 ,
1 ≤ k ≤ N,
2
w tk−1/2 =
1 w tk + w tk−1 + o τ 2 ,
2
(2.5)
τ w tk−1/2 = w tk − w tk + o τ 2 .
2
Applying (2.5), and the formulas
u tk = iA1/2 tk v tk ,
u tk = f tk − A tk u tk ,
(2.6)
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Discrete Dynamics in Nature and Society
we get
τ −1 u tk − u tk−1
τ 1/2 τ
τ
Ak
v tk − fk + o τ 2 , 1 ≤ k ≤ N,
= Ak u tk + i A1/2
k +
2
2
2
τ −1 v tk − v tk−1
τ
1/2 −1 −1/2
Ak
v tk + v tk−1
= iA1/2
k u tk + Ak v tk − 2 Ak
2
1/2
− iA−
fk + o τ 2 , 1 ≤ k ≤ N,
v0 = −iA−0 1/2 u0 ,
k
τ
τ
1/2
v tk = −iA−k+1/2
τ −1 u tk − u tk−1 − Ak u tk + fk + o τ 2 , 1 ≤ k ≤ N.
2
2
(2.7)
Neglecting the small terms o(τ 2 ), we obtain the following difference scheme:
τ
τ 1/2 τ
Ak
vk − fk , u0 = u(0), 1 ≤ k ≤ N
τ −1 uk − uk−1 = Ak uk + i A1/2
k +
2
2
2
τ
−1 −1/2
A1/2
vk +vk−1 − iA−k 1/2 fk , 1 ≤ k ≤ N,
τ −1 vk − vk−1 = iA1/2
k uk + Ak vk − 2 Ak
k
2
τ
τ
1/2
τ −1 uk − uk−1 − Ak uk + fk ,
vk = −iA−k+1/2
2
2
1 ≤ k ≤ N,
v0 = −iA−0 1/2 u0 .
(2.8)
for the approximate solution of the initial-value problem (1.1).
Using (2.8) and eliminating vk , collecting uk on the left side and tk on the right side of
the equation, and rearranging the terms in (2.8), we obtain two-step difference scheme
of a second order of accuracy
τ −2 uk+1 − 2uk + uk−1
τ 1/2 τ
1 −1/2 1/2 1/2
+
A
A
u
+
−
A
A
A
−
= A1/2
k+1
k+1
k+1
k+1
2 k+1
2
2 k+1 k+1
τ
τ
1/2
−1/2
−1
τ
u
−
u
u
+
+
A
f
A
f
× A−
−
k+1
k
k+1 k+1
k+1
k+3/2
k+1 k+1
2
2
τ 1/2 −1/2 1/2 −1/2
A
Ak+1 Ak+1 Ak+1/2
− A1/2
k+1 +
2 k+1
1/2 1/2 τ 1/2 1/2
− τ −1 A1/2
−
A
+
A
−
A
A−k+1/2
k+1
k
k+1
k
2
τ
τ
−1
× τ
uk − uk−1 − Ak uk + fk + 2−1 Ak+1 uk+1 − Ak uk
2
2
u0 = u(0),
− 2−1 fk+1 − fk , 1 ≤ k ≤ N − 1,
τ
τ
τ
1/2
−
−1/2 τ −1 u1 − u0 + A1/2 2−1 u1 + u0 + A1/2 A1/21/2 τ −1 u1 − u0 = f1 + A1/2
1/2 A1/2 u0
2
2
2
(2.9)
A. Ashyralyev and M. E. Koksal 7
for the approximate solution of the initial-value problem (1.1). Note that the difference
scheme (2.9) for approximately solving problem (1.1) in the case of A(t) = A is (1.5).
Let us establish the formula for the solution of this difference scheme (2.9).
Making the transformation ηk = uk + vk and μk = uk − vk in (2.8), we obtain the following system of the difference equations:
τ
+
τ −1 ηk − ηk−1 = iA1/2
2 ≤ k ≤ N,
k + Ak ηk + ϕk ,
2
+
η1 = K B u0 + C + u0 + D+ f1 ,
τ
−
2 ≤ k ≤ N,
τ −1 μk − μk−1 = − iA1/2
k + Ak μk + ϕk ,
2
μ1 = K B − u0 + C − u0 + D− f1 ,
−1 τ
τ
vk − fk ∓ A−k 1/2 A1/2
2 vk + vk−1 ∓ iA−k 1/2 fk ,
ϕ±k = i A1/2
k
2 k 2
τ
τ
1/2
τ −1 uk − uk−1 − Ak uk + fk , 2 ≤ k ≤ N,
vk = −iA−k+1/2
2
2
(2.10)
where
−1
τ 4 2 τ −1/2 1/2 τ 3 1/2 1/2 A1
+ A1 A1
,
A + A
4 1 2 1
2
τ
τ2
± iτA1/2
B ± = 1 − A1 + A−1 1/2 A1/2
1
1 ,
2
2
τ 3 1/2 −1/2 1/2 −1/2
−1/2
−
A
A1
A1 A0 ∓ iA−0 1/2
C ± = τA1/2
1 A0
4 1
1/2 −1/2
τ 1/2 1/2 −1/2 τ 2
τ3
± i A−
A1 A0 ± i A1 A−0 1/2 ∓ i A1/2
A1 A0 ,
1
1
2
2
4
3 2 τ 3 1/2 −1/2
τ4
τ3
+ τ +
A
A1 ∓ iτA−1 1/2 .
D± = A1 − A−1 1/2 A1/2
1
4
4
2
2 1
K = 1+
(2.11)
From this, it follows the system of recursion formulas
−1
−1
τ2
τ2
η
+
I
−
ϕ+k , 2 ≤ k ≤ N,
Ak − iτA1/2
Ak − iτA1/2
k
k
k −1
2
2
−1
−1
τ2
τ2
1/2
μk = I − Ak + iτA1/2
μ
+
I
−
+
iτA
ϕ−k , 2 ≤ k ≤ N.
A
k
k
k
k −1
2
2
ηk = I −
(2.12)
Hence,
ηk = Pk− (k)η1 +
k
R−m (k)ϕ+m ,
m=2
μk = Pk+ (k)μ1 +
k
R+m (k)ϕ−m .
(2.13)
m=2
Here,
Pk± (k) = Xk± Xk±−1 · · · X2± ,
R±m (k) = Xk± Xk±−1 · · · Xm± ,
(2.14)
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Discrete Dynamics in Nature and Society
where
Xk± = I −
τ2
Ak ± iτA1/2
k
2
−1
.
(2.15)
Then, using the formula uk = (1/2)(ηk + μk ), we obtain
uk = 2−1
Pk+ (k)KB− + Pk− (k)KB+ u0 + Pk+ (k)KC − + Pk− (k)KC + u0
+ Pk+ (k)KD− + Pk− (k)KD+ f1 +
k
m=2
R+m (k)ϕ−m + R−m (k)ϕ+m
(2.16)
.
Furthermore, by making the transformation k − m = s, we obtain
u k = 2 −1
Pk+ (k)KB− + Pk− (k)KB+ u0 + Pk+ (k)KC − + Pk− (k)KC + u0
+ Pk+ (k)KD− + Pk− (k)KD+ f1 +
k
−2
s =0
Es+ (k)ϕ−k−s + Es− (k)ϕ+k−s
(2.17)
,
where
τ 1/2 1/2 τ 2 1/2 −1/2
= i ± A−
A
−i
A
Ak−s+1/2
2 k −s k −s
2 k −s
τ
τ
× τ −1 uk−s − uk−s−1 − Ak−s uk−s + fk−s
2
2
τ
τ −1/2 1/2 −1/2
τ
−1
± i Ak−s Ak−s Ak−s−1/2 τ (uk−s−1 − uk−s−2 − Ak−s−1 uk−s−1 + fk−s−1
2
2
2
τ2
+ − ∓ iτA−k−1/2
f k −s ,
s
2
E0± (k) = Xk± .
Es± (k) = Xk± Xk±−1 ···Xk±−s ,
(2.18)
ϕ±k−s
Finally, from the last formula, it follows that
−1
A1/2
u k − u k −1
k−1/2 τ
−1
= A1/2
k−1/2 (2τ)
Pk+ (k) − Pk+−1 (k − 1) KB − + Pk− (k) − Pk−−1 (k − 1) KB + u0
+ Pk+ (k) − Pk+−1 (k − 1) KC − + Pk− (k) − Pk−−1 (k − 1) KC + ]u0
+ Pk+ (k) − Pk+−1 (k − 1) KD− + Pk− (k) − Pk−−1 (k − 1)]KD+ f1
+ E0+ (k)ϕ−k + E0− (k)ϕ+k +
+
k
−2
s =1
k
−2
s =1
Es+ (k) − Es+−1 (k − 1) ϕ−k−s
Es− (k) − Es−−1 (k − 1) ϕ+k−s .
(2.19)
A. Ashyralyev and M. E. Koksal 9
In the following section, these formulas will be used to establish the stability inequality
of the difference scheme (2.9).
3. Stability of difference scheme (2.9)
First of all, let us give some subsidiary conditions for operators A(t) that will be needed
below. Let A(t) be self-adjoint positive definite operators in H with a t-independent domain D = D(A(t)) : A(t) ≥ δI > 0. Then, the following estimates hold:
−1 α α/2
τ4 2
τ A
≤ 1,
I
+
A
k
k
4
α = 0,1,2,
−1 √ α α/2
√
τ4 2
τ A
≤ 4− 2 α+4 2−3 ,
I
+
A
k
4 k
−1 α α/2
α2 − α
τ2
1/2
τ A
≤
I − Ak ± iτAk
+ 1,
k
2
2
(3.1)
α = 3,4,
α = 0,1,2.
(3.2)
(3.3)
Let the operator function Aρ (t)A−ρ (z), ρ ∈ [0,2] satisfies the conditions
ρ
A (t) − Aρ (s) A−ρ (z) ≤ Mρ t − s,
(3.4)
ρ ρ −ρ A (t) − A (s) A (z) ≤ Mρ t − s,
(3.5)
where Mρ is a positive constant independent of t, s, z for t,s,z ∈ [0,T]. From this, it
follows that the operator function Aρ (t)A−ρ (z) has a finite variation on [0,T], that is,
there exists a number Pρ such that
N
ρ A sk − Aρ sk−1 A−ρ (z) ≤ Pρ
(3.6)
k =1
for any 0 = s0 < s1 < · · · < sN = T. Here, Pρ is a positive constant independent of s0 ,
s1 ,...,sN , and z.
Furthermore, let the operator functions (Aρ (t)) A−ρ (z) and Aρ (p)(Ar (t)) A−ρ−r (z)
satisfy the conditions
ρ −ρ A (t) A (z) ≤ M3 ,
(3.7)
ρ
A (p) Ar (t) A−ρ−r (z) ≤ M4 ,
(3.8)
where M3 and M4 are positive constants independent of t, z for t,z ∈ [0,T] and t, z, p for
t,z, p ∈ [0,T], respectively.
10
Discrete Dynamics in Nature and Society
Finally, let Pk± (k) = Xk± Xk±−1 · · · X2± and Es± (k) = Xk± Xk±−1 · · · Xk±−s such that Xk± =
−1
(I − (τ 2 /2)Ak ± iτA1/2
k ) . We have
Ak P ± (k)A−1 ≤ eM1 ki=1 (A1i −A1i−1 )A0−1 ,
α2 − α
k
1
1
−1
α/2−1 α Ak Es± (k)A1,k
τ
eM1 i=1 (Ai −Ai−1 )A0 ,
≤
+
1
−s
2
1/2
A
(3.9)
1
k
α = 0,1,2,
(3.10)
−ρ 3M1/2 M k (Aρ −Aρ )A−ρ Pk (k) − Pk±−1 (k − 1) A1 ≤
e ρ i=1 i i−1 0 ,
4
−1 ±
k−1/2 (2τ)
(3.11)
ρ α/2−1 α 1/2
Ak−s−1/2 (2τ)−1 Es± (k) − Es±−1 (k − 1) Ak−s
τ ≤
3M1/2 Mρ ki=1 (Aρi −Aρi−1 )A0−ρ ,
e
4
(3.12)
α = 0,1,2.
Theorem 3.1. Let u(0) ∈ D(A1/2 (0)) and f1 ∈ D((A1/2
1 ) ). Then, for the solution of the
difference scheme (2.9), the stability estimate
1/2 uk − uk−1 N A
+ Ak uk N k−1/2
1 Cτ
τ
1 Cτ
N
1/2
2 1/2 ≤ M A(0)u0 H + A (0)u0 H + max fk + τ (A1 ) f1 H +
fs − fs−1 H
1≤s≤N
s =1
(3.13)
holds, where M does not depend on u0 ,u0 , fs (1 ≤ s ≤ N), and τ.
N
Proof. Firstly, the estimate {A1/2
k−1/2 (uk − uk−1 /τ)}1 Cτ will be obtained. Applying formula (2.19), we can write
−1
A1/2
uk − uk−1 = J1k + J2k + J3k + J4k + J5k ,
k−1/2 τ
(3.14)
where
−1
Pk+ (k) − Pk+−1 (k − 1) KB − + Pk− (k) − Pk−−1 (k − 1) KB + u0 ,
J1k = A1/2
k−1/2 (2τ)
−1
Pk+ (k) − Pk+−1 (k − 1) KC − + Pk− (k) − Pk−−1 (k − 1) KC + u0 ,
J2k = A1/2
k−1/2 (2τ)
−1
J3k = A1/2
Pk+ (k) − Pk+−1 (k − 1) KD− + Pk− (k) − Pk−−1 (k − 1) KD+ f1 ,
k−1/2 (2τ)
−1
E0+ (k)ϕ−k + E0− (k)ϕ+k ,
J4k = A1/2
k−1/2 (2τ)
−1
J5k = A1/2
k−1/2 (2τ)
k −2
s =1
Es+ (k) − Es+−1 (k − 1) ϕ−k−s +
k
−2
s =1
Es− (k) − Es−−1 (k − 1) ϕ+k−s .
(3.15)
A. Ashyralyev and M. E. Koksal 11
Now, let us estimate the terms Jmk H , m = 1,5, separately. Let m = 1. Then applying
estimates (3.1), (3.2), (3.4), (3.8), and (3.11), we get
J1k ≤ 2A1/2 (2τ)−1 P + (k) − P + (k − 1) KB − u0 k−1/2
k
k −1
H
H
1/2
−1 −1 +
+
1 ≤ 2 Ak−1/2 (2τ) Pk (k) − Pk−1 (k − 1) A1 A1 kb− A−
A1 A−0 1 A0 u0 H
1
−1 +
≤ 2A1/2
Pk (k) − Pk+−1 (k − 1) A−1 1 k−1/2 (2τ)
−1
τ 4 2 τ −1/2 1/2 τ 3 1/2 1/2 ×
A
1
+
+
A
+
A
A
A
A
1
1
1
4 1 2 1
2 1
τ 1/2 1/2 τ2
1/2 × 1 − A1 + A−
A
−
iτA
A1 A−0 1 A0 u0 H
1
1
1
2
2
5τ
3
3
1
≤ M1/2 + 1 M1
1 + M4 + eM1 P1 A0 u0 H = c1 A0 u0 H .
4
1 − τM4
4
2
(3.16)
Let m = 2. Then applying estimates (3.1), (3.2), (3.4), (3.7), (3.8), and (3.11), we get
J2k ≤ 2A1/2 (2τ)−1 P + (k) − P + (k − 1) KC − u 0 H
k−1/2
k
k −1
H
−1 +
≤ 2A1/2
Pk (k) − Pk+−1 (k − 1) A−1 1 A1 KC − A−0 1/2 A1/2
0 u0 H
k−1/2 (2τ)
−1 ≤ 2A1/2 (2τ) P + (k) − P + (k − 1) A−1 k−1/2
k −1
k
1
−1
τ 4 2 τ −1/2 1/2 τ 3 1/2 1/2 ×
A
1
+
+
A
+
A
A
A
A
1
1
1
4 1 2 1
2 1
−1/2
τ
τ 3 1/2 −1/2 1/2 −1/2
−1/2
× τA1/2
−
A1 A1
A1 A0 − iA−0 1/2 +i A−1 1/2 A1/2
A0
1 A0
1
4
+i
τ2
2
A1 A−0 1/2 − i
τ3
4
1/2 A1
1/2 A1
A−0 1/2
2
1/2 A−0 1/2 A0 u0 H
1
M3
τ M2
+ √3 +
2
4 δ 1 − τM4
2
3τ 2
3M1 τ M3
√
+ M4
×
eM1 P1 A1/2
+ M4 1 + 3M1 ) +
M4
0 u0 H
2
4
8
δ
= C2 A1/2
0 u0 H .
(3.17)
≤
3
M1/2
2
M1 +
Let m = 3. Then applying estimates (3.1), (3.2), (3.4), (3.7), (3.8), and (3.11), we get
J3k ≤ 2A1/2 (2τ)−1 P + (k) − P + (k − 1) KD− f1 k−1/2
k
k −1
H
H
1/2
−1 −1 +
+
≤ 2 Ak−1/2 (2τ) Pk (k) − Pk−1 (k − 1) A1
−1
τ 4 2 τ −1/2 1/2 τ 3 1/2 1/2 ×
A1
+ A1 A1
A1 1 + A1 + A1
×
4
τ4
4
A1 −
2
τ3
4
−1/2
A1
1/2 A1
2
3τ 2 τ 3 1/2 −1/2
+
A1 A1 − iτA−1 1/2 +
f1 H
2
2
12
Discrete Dynamics in Nature and Society
≤
5τ M4
3τ 2 M4 M4 + M3
3
5 τ
M1/2 + 1 +
+
+ M 4 + M3 +
2
2 2
4 1 − τM4
4
1 − τM4
eM1 P1
× f1 H + τ 2 A1/2
f1 H = C3 f1 H + τ 2 A1/2
f 1 H .
1
1
(3.18)
Let m = 4. We have that
−1
J4k = A1/2
E0+ (k)ϕ−k + E0− (k)ϕ+k
k−1/2 (2τ)
−
−1/2 1/2 τ 1/2 −
1/2 1
+
+
X
−
X
A
A
+
X
+
X
A
A
= i A1/2
k
k
k
k
k−1/2
k
k
k−1/2
k
4
4
−1/2 uk−1 − uk−2
u
−
u
1
k
k
−
1
1/2
× A−
X − − Xk+ A−k 1/2 A1/2
Ak−1/2
+ i A1/2
k+1/2
k
τ
4 k−1/2 k
τ
2
τ 1/2 +
τ
1/2
X + + Xk− A1/2
A−k+1/2
Ak uk
+ i Ak−1/2 Xk − Xk− A−k 1/2 + A1/2
k
8
8 k−1/2 k
+
−1/2
τ
+ i A1/2
X − Xk− A−k 1/2 A1/2
Ak−1/2 Ak−1 uk−1
k
8 k−1/2 k
−1/2
τ
−1/2
+
+ A1/2
+ τ + Xk− iA−k 1/2 + τ A1/2
Ak+1/2 fk
k−1/2 Xk − iAk
k
8
τ2
τ2
−1
−1/2
−1/2
+
−
+ A1/2
(2τ)
X
+
X
−
iτA
fk
−
−
+
iτA
k−1/2
k
k
k
k
2
2
−1/2
τ −
+ A1/2
Xk − Xk+ A−k 1/2 A1/2
Ak−1/2 fk−1 .
k−1/2 i
k
8
(3.19)
Then applying estimates (3.3), (3.4), (3.7), and (3.8), we get
J4k ≤ τ A1/2 A−1/2 X − + 1 X + A1/2 a1/2 A−1 k−1/2 k
k
k
k
k
k+1/2
H
2
u k − u k −1 −1/2 1/2
+ τ A1/2 A−1/2 X − X + × A1/2
A
A
k
k
k+1/2 k−1/2 k−1/2
k−1/2 k
τ
2
H
1/2 uk − uk−1 1/2 −1 1/2
× A1/2
Ak Ak−1/2 Ak−1/2 A−k−1/2
k
3/2 Ak−3/2
τ
H
τ
+ A1/2 A−1/2 X + + A1/2 τX + A1/2 A−1/2 Ak uk k
k
k−1/2 k
k
k
k+1/2
H
4
τ
1/2 + A1/2
A−1/2 Xk+ A1/2
A−k+1/2
Ak−1 uk−1 H
k
4 k−1/2 k
1/2 −1/2 τ2 −1/2 + + + A1/2
Xk + 2A1/2
Ak Ak+1/2 fk H
k−1/2 Ak
k τXk
2
−1/2 + 1/2
+ τ A1/2
2 Xk + Ak τXk+ fk H
k−1/2 Ak
τ2 A1/2 A−1/2 X + A1/2 A−1/2 fk−1 k
k−1/2 k
k
k+1/2
H
2
1/2 uk−1 − uk−2 1/2 uk − uk−1 ≤ τC4 Ak−1/2
+ Ak−3/2
τ
τ
H
H
+
3
+ Ak uk H + Ak−1 uk−1 H + fk H + fk−1 H + M1/2 fk H ,
2
(3.20)
A. Ashyralyev and M. E. Koksal 13
where
C4 = max
2
M1/2 + 1 M4 ,
1
M1/2 + 1 M3 .
2
(3.21)
Let m = 5. It is clear that
J5k = S1k + S2k + S3k + S4k + S5k + S6k + S7k ,
(3.22)
where
−1
S1k = A1/2
k−1/2 (2τ)
−2 k
s =1
Es+ (k) − Es+−1 (k − 1)
τ
2
1/2
− i A−
k −s +
k
−2
s =1
2
− Es+ (k) − Es+−1 (k − 1) + Es− (k) − Es−−1 (k − 1)
τ 1/2 1/2 −1/2
−1
u k −s −1 − u k −s −2 ,
× A−
k−s Ak−s Ak−s−1/2 τ
2
−1
S3k = A1/2
k−1/2 (2τ)
−2 k
s =1
Es+ (k) − Es+−1 (k − 1)
τ
2
1/2
− i A−
k −s +
τ2 −
Es (k) − Es−−1 (k − 1)
2
τ 1/2 τ 2 1/2 −1/2 τ
× i A−
+
Ak−s Ak−s+1/2 Ak−s uk−s ,
k −s
2
−1
S4k = A1/2
k−1/2 (2τ) i
τ2
+ Es− (k) − Es−−1 (k − 1)
2
τ 1/2 τ 2 1/2 −1/2
× i A−
+
Ak−s Ak−s+1/2 τ −1 uk−s − uk−s−1 ,
k −s
2
−1
S2k = A1/2
k−1/2 (2τ) i
k
−2
s =1
2
2
− Es+ (k) − Es+−1 (k − 1) + Es− (k) − Es−−1 (k − 1)
τ 1/2 1/2 −1/2 τ
A
A
Ak−s−1 uk−s−1 ,
× A−
2 k−s k−s k−s−1/2 2
−1
S5k = A1/2
k−1/2 (2τ)
−2 k
s =1
Es+ (k) − Es+−1 (k − 1)
+
−1
S6k = A1/2
k−1/2 (2τ)
k
−2 s =1
−
Es (k) − Es−−1 (k − 1)
Es+ (k) − Es+−1 (k − 1)
−1
S7k = A1/2
k−1/2 (2τ) i
τ −1/2 k
−2
s =1
× i Ak−s A1/2
k −s
2
τ
i
+ Es− (k) − Es−−1 (k − 1)
τ
2
1/2
− i A−
k −s +
−
2
A−k−1/2
s
τ2
2
τ2
+
2
τ2
1/2
− iτA−
k −s
2
−
τ2
+ iτA−k−1/2
s
2
−1/2 τ
A1/2
k−s Ak−s+1/2 fk−s ,
2
f k −s ,
− Es+ (k) − Es+−1 (k − 1) + Es− (k) − Es−−1 (k − 1)
τ
A−k−1/2
s−1/2 fk−s−1 .
2
(3.23)
14
Discrete Dynamics in Nature and Society
Now, let us estimate the terms Smk H , m = 1,7, separately. Let m = 1. Then applying
estimates (3.4), (3.8), and (3.12), we get
k−2
−1 +
S1k ≤ 2
A1/2
Es (k) − Es+−1 (k − 1) A−k−1/2
k−1/2 (2τ)
s
H
s =1
1/2 −1/2
τ τ2
× − i + A1/2
Ak−s Ak−s+1/2 τ −1 uk−s − uk−s−1 k −s
2 2
H
≤2
k
−2
1/2
A
−1 +
s =1
Es (k) − Es+−1 (k − 1) A−k−1s k−1/2 (2τ)
−1
+ A1/2
Es+ (k) − Es+−1 (k − 1) A−k−1/2
k−1/2 (2τ)
s τ
1/2
1/2
1/2 −1
−1/2
−1
A
A
× A1/2
u k −s − u k −s −1 H
k−s Ak−s Ak−s+1/2
k−s+1/2 Ak−s−1/2
k−s−1/2 τ
k
−2
1/2
A
≤ τC5
k−s−1/2 τ
s =1
−1
u k −s − u k −s −1 H ,
(3.24)
where
C5 =
2
3
M1/2 + 1 M4 eM1 P1 .
2
(3.25)
Let m = 2. Then applying estimates (3.4), (3.8), and (3.12), we get
k −2
1/2 −1 +
S2k ≤ 2
A1/2
Es (k) − Es+−1 (k − 1) τA−k−1/2
k−1/2 (2τ)
s Ak−s
H
s =1
−1/2
−1
× Ak−s−1/2 τ
u k −s −1 − u k −s −2 H
≤ 2τ
k
−2
1/2
A
s =1
k−1/2 (2τ)
−1 × A1/2 A1/2 A−1
k −s
k −s
Es+ (k) − Es+−1 (k − 1) A−k−1s k−s−1/2
1/2
A
−1/2
−1
A1/2
uk−s−1 − uk−s−2 H ,
k−s−1/2 Ak−s−3/2
k−s−3/2 τ
−2
k
−1
S2k ≤ τ C5 A1/2
u k −s −1 − u k −s −2 H .
k−s−3/2 τ
H
2
s =1
(3.26)
Let m = 3. Then applying estimates (3.7) and (3.12), we get
k −2
−1 +
S3k ≤ 2
A1/2
Es (k) − Es+−1 (k − 1) A−k−1/2
s
k−1/2 (2τ)
H
s =1
τ 1/2 1/2 τ 2 1/2 −1/2 τ
× i A−
Ak−s Ak−s+1/2 Ak−s uk−s k−s Ak−s −
2
2
2
H
A. Ashyralyev and M. E. Koksal 15
≤2
k
−2
1/2
A
k−1/2 (2τ)
s =1
−1 +
Es (k) − Es+−1 (k − 1) A−k−1/2
s τ
k
−2
−1
Ak−s uk−s ≤ τC6 Ak−s uk−s ,
× A1/2
k−s Ak−s+1/2
H
H
s=1
(3.27)
where
C6 =
3
M1/2 + 1 M4 eM1 P1 .
4
(3.28)
Let m = 4. Then applying estimates (3.7) and (3.12), we get
k −2
−1 +
S4k ≤ 2
A1/2
Es (k) − Es+−1 (k − 1) A−k−1/2
s
k−1/2 (2τ)
H
s =1
τ 1/2 1/2 τ 2 1/2 −1/2 τ
× i A−
A
−
A
A
u
A
k
−
s
k
−
s
k−s+1/2
2 k −s k −s
2 k −s
2
H
≤ 2τ
k
−2
1/2
A
k−1/2 (2τ)
s =1
−1 +
Es (k) − Es+−1 (k − 1)
(3.29)
A−k−1/2
s τ
k
−2
−1
Ak−s−1 uk−s−1 ≤ τ C6 Ak−s−1 uk−s−1 .
× A1/2
k−s Ak−s−1/2
H
H
2
s =1
Let m = 5. Then applying estimates (3.7) and (3.12), we get
k
−2
1/2
−1 +
S5k ≤ τ
A
Es (k) − Es+−1 (k − 1) A−k−1/2
k−1/2 (2τ)
s τ
H
s =1
≤ τC6
−1
−1 +
f k −s + A1/2
Es (k) − Es+−1 (k − 1) τ 2 A1/2
k−1/2 (2τ)
k−s Ak−s−1/2
H
k
−2
f k −s .
H
s =1
(3.30)
+ −1
Let m = 6. It is easy to show that −(τ 2 /2)Ak−s + iτA1/2
k−s = −I + (Xk−s ) . Making s − 1 = m
for the first term in the parenthesis in S6 , we get
−1
A1/2
k−1/2 (2τ)
−2
k
s =1
1/2 +
+ −1 +
f k −s
− Es+ (k) − Es+−1 (k − 1) A−
k−s + Es (k) − Es−1 (k − 1) Xk−s
−1 +
= A1/2
k−1/2 (2τ)
E0 (k) − E−+ 1 (k − 1) fk−1
−1
+ A1/2
Ek+−2 (k) − Ek+−3 (k − 1) f1
k−1/2 (2τ)
−1
+ A1/2
k−1/2 (2τ)
k
−2
s =1
Es+ (k) − Es+−1 (k − 1) fk−s−1 − fk−s .
(3.31)
16
Discrete Dynamics in Nature and Society
Then applying estimates (3.3) and (3.4), we get
S6k ≤ 1 A1/2 A−1/2 τA1/2 X + + 2X + k−1/2 k
k
k
k
H
2
× fk−1 H + Xk+−1 ···X2+ f1 H
+
k
−2
1/2
A
s =1
−1/2 k−1/2 Ak
+ +
2−1 τA1/2
k Xk + Xk
(3.32)
× Xk+−1 · · · Xk+−s fk−s − fk−s−1 H
k
−2
3
≤ M1/2 + 1 fk−1 H + f1 H + fk−s − fk−s−1 H .
4
s =1
Let m = 7. Then applying estimates (3.7) and (3.12), we get
k
−2
S7k ≤ τ A1/2 (2τ)−1 E+ (k) − E+ (k − 1) A−1/2 τ s
s −1
k−1/2
k −s
H
s =1
k
−2
fk−s−1 ≤ τ C6 fk−s .
k−s−1/2
H
H
× A1/2 A−1
k −s
2
(3.33)
s =1
Using formula (3.22), the triangle inequality, and the last seven estimates, we obtain
−1
k
1/2 −1 −1
J5k ≤ τC7
A
us − us−1 H + A1/2
u s −1 − u s −2 H
s−1/2 τ
s−3/2 τ
H
s =2
+ As us H + As−1 us−1 H + fs H + fs−1 H
(3.34)
k
−1
+ 3 M1/2 + 1
f k −1 H + f 1 H +
f s − f s −1 H ,
s=2
where
3
C7 = max C5 , C6 .
2
(3.35)
A. Ashyralyev and M. E. Koksal 17
Using formula (3.14), the triangle inequality, and the estimates for Jmk H , m = 1,5 , we
obtain
1/2 uk − uk−1 A
k−1/2
τ
H
f1 + τ 2 A1/2 f1 ≤ C1 A0 u0 H + C2 A1/2
0 u0 H + C3
1
H
H
+ τC4 A1/2 τ −1 uk − uk−1 + Ak uk + A1/2 τ −1 uk−1 − uk−2 k−1/2
k−3/2
H
H
3
+ Ak−1 uk−1 H + fk H + fk−1 H + M1/2 fk H
H
2
+ τC7
k
−1
s =2
1/2 −1 −1
A
us − us−1 H + As us H + A1/2
u s −1 − u s −2 H
s−1/2 τ
s−3/2 τ
+ As−1 us−1 H + fs H + fs−1 H
k
−1
+ 3M1/2 fk−1 H + f1 H +
f s − f s −1 H .
s =2
(3.36)
From the above result, it follows that
1/2 uk − uk−1 A
k−1/2
τ
H
2 1/2 ≤ C8 A0 u0 H + A1/2
f1 H + max fs H
0 u0 H + τ A1
1≤s≤k
+τ
(3.37)
k
k
1/2 −1 A
us − us−1 H + As us H + fs − fs−1 H ,
s−1/2 τ
s =1
s =2
where
C8 = max {C1 ,C2 ,C3 + 3 M1/2 + 1 ,C4 + C7 }.
(3.38)
Second, let us estimate {Ak uk }N1 H . Applying formula (2.17), we can write
Ak uk = Y1k + Y2k + Y3k + Y4k ,
(3.39)
where
J1k = Ak 2−1 Pk+ (k)KB− + Pk− (k)KB+ u0 ,
J2k = Ak 2−1 Pk+ (k)KC − + Pk− (k)KC + u0 ,
J3k = Ak 2−1 Pk+ (k)KD− + Pk− (k)KD+ f1,
J4k = Ak 2−1
−2
k
s =0
Es+ (k)ϕ−k−s + Es− (k)ϕ+k−s .
(3.40)
18
Discrete Dynamics in Nature and Society
Now, let us estimate the terms Ymk H , m = 1,4, separately. Let m = 1. Then applying
estimates (3.1), (3.2), (3.4), (3.8), and (3.9), we get
Y1k ≤ A1/2 P + (k)KB − u0 k−1/2 k
H
H
1 ≤ Ak Pk+ (k)A−
A1 KB − A−1 1 A1 A−0 1 A0 u0 H
1
5τ
3
1
≤ M1
1 + M4 + eM1 P1 A0 u0 H = C9 A0 u0 H .
1 − τM4
4
2
(3.41)
Let m = 2. Then applying estimates (3.1), (3.2), (3.3), (3.4), (3.7), (3.8), and (3.9), we get
Y2k ≤ A1/2 P + (k)KC − u ≤ Ak P + (k)A−1 A1 KC − u 1
0 H
0 H
k−1/2 k
k
H
1
τ 15
1
3 2 2
5
≤
M 1 + M3 +
M 4 + M3 + τ M 4 + M4 M 3
4
2
4
8
2
1/2 × eM1 P1 A1/2
0 u0 H = C10 A0 u0 H .
(3.42)
4
Let m = 3. Then applying estimates (3.1), (3.2), (3.4), (3.7), (3.8), and (3.9), we get
Y3k ≤ Ak P + (k)KD− f1 ≤ Ak P + (k)A−1 A1 KD− f1 1
k
k
H
H
H
1
3
5
15
≤
+τ
M4 + M3 + τ 2 M42 + M4 M3 eM1 P1
4
8
2
(3.43)
4
× f1 H + τ 2 A1/2
f1 H = C11 f1 H + τ 2 A1/2
f 1 H .
1
1
Let m = 4. We have that
Y4 = Q1k + Q2k + Q3k + Q4k + Q5k + Q6k + Q7k ,
(3.44)
where
Q1k = Ak 2
−1
k
−2 s =1
Es+ (k)
τ 1/2 τ 2
+
− i A−
2 k −s
2
τ
τ2
+ Es− (k) i A−k−1/2
s +
2
2
Q2k = Ak 2−1
−2
k
s =1
Q3k = −Ak 2−1
i − Es+ (k) + Es− (k)
−2 k
s =1
τ
2
−1/2
−1
A1/2
k−s Ak−s+1/2 τ (uk−s − uk−s−1 ),
1/2
−1/2
−1
u k −s −1 − u k −s −2 ,
A−k−1/2
s Ak−s Ak−s−1/2 τ
τ
τ2
Es+ (k) − i A−k−1/2
+
s
2
2
τ
τ2
+ Es− (k) i A−k−1/2
s +
2
2
−1/2 τ
A1/2
k−s Ak−s+1/2 Ak−s uk−s ,
2
A. Ashyralyev and M. E. Koksal 19
Q4k = −Ak 2−1
k
−2 s =1
τ
τ2
Es+ (k) − i A−k−1/2
s +
2
2
τ
τ2
+ Es (k) i A−k−1/2
s +
2
2
−
Q5k = Ak 2−1
k
−2 s =1
Q6k = Ak 2−1
−1/2 τ
A1/2
k−s Ak−s−1/2 Ak−s−1 uk−s−1 ,
2
τ
τ2
Es+ (k) − i A−k−1/2
s +
2
2
k
−2 s =1
τ
τ2
τ
τ2
Es+ (k) − i A−k−1/2
+ Es− (k) i A−k−1/2
s +
s +
2
2
2
2
k
−2 s =1
Q7k = Ak 2−1
A1/2
k −s
τ
2
f k −s ,
τ
τ2
+ Es− (k) i A−k−1/2
+
s
2
2
Es+ (k) iτA−k−1/2
s −
−1/2 τ
A1/2
k−s Ak−s−1/2 fk−s−1 ,
2
τ2
τ2
+ Es− (k) − iτA−k−1/2
s −
2
2
f k −s .
(3.45)
Now, let us estimate the terms Qmk H , m = 1,7, separately. Let m = 1. Then applying
estimates (3.4), (3.8), and (3.10), we get
k
−2
τ Q1k ≤
Ak E+ (k)A−1 + Ak E+ (k)A−1/2 τ s
s
k −s
k −s
H
s =1
2
× A1/2 A1/2 A−1
k −s
k −s
k−s+1/2
1/2
A
−1/2
−1
A1/2
u k −s − u k −s −1 H ,
k−s+1/2 Ak−s−1/2
k−s−1/2 τ
k
−2
−1
Q1k ≤ 2τ C6 A1/2
u k −s − u k −s −1 H .
k−s−1/2 τ
H
3
s =1
(3.46)
Let m = 2. Then applying estimates (3.4), (3.8), and (3.10), we get
k
−2
τ
Q2k ≤
Ak E+ (k)A−1 A1/2 A1/2 A−1
s
k −s
k −s
k −s
k−s+1/2
H
s =1
2
× A1/2
−1/2
−1
A1/2
u k −s − u k −s −1 H .
k−s+1/2 Ak−s−1/2
k−s−1/2 τ
(3.47)
k
−2
−1
Q2k ≤ 2τ C6 A1/2
u k −s −1 − u k −s −2 H .
k−s−3/2 τ
H
3
s =1
Let m = 3. Then applying estimates (3.7) and (3.10), we get
k
−2
τ Q3k ≤
Ak E+ (k)A−1/2 τ + Ak E+ (k)τ 2 s
s
k −s
H
s =1
8
−2
k
Ak−s uk−s ≤ τC12 Ak−s uk−s ,
k−s+1/2
H
× A1/2 A−1
k −s
s =1
(3.48)
20
Discrete Dynamics in Nature and Society
where
3
C12 = M3 eM1 P1 .
2
(3.49)
Let m = 4. Then applying estimates (3.7) and (3.10), we get
k
−2
τ
Q4k ≤
Ak E+ (k)A−1/2 τ A1/2 A−1
Ak−s−1 uk−s−1 s
k−s−1/2
k −s
k −s
H
H
s =1
4
(3.50)
k −2
τ
≤ C12 Ak−s−1 uk−s−1 H .
3
s =1
Let m = 5. Then applying estimates (3.7) and (3.10), we get
k
−2
τ Q5k ≤
fk−s Ak E+ (k)A−1/2 τ + Ak E+ (k)τ 2 A1/2 A−1
s
s
k−s+1/2
k −s
k −s
H
H
s =1
8
≤ τC12
k
−2
s =1
f k −s .
H
(3.51)
Let m = 6. Then applying estimates (3.7) and (3.10), we get
k
−2
τ
Q6k ≤
Ak E+ (k)A−1/2 τ A1/2 A−1/2 fk−s−1 s
k −s
k −s
k−s−1/2
H
H
s =1
4
(3.52)
k −2
τ
≤ C12 fk−s−1 H .
3
s =1
Let m = 7. We have
Q7k = Ak 2−1
−2 k
s =1
Es+ (k) −
τ2
τ2
1/2
−
+
E
(k)
− iτA−
+ iτA−k−1/2
−
s
s
k −s
2
2
f k −s .
(3.53)
Using similar manner in Q6k , we get
k
−2
τ
Q7k ≤
Ak E+ (k)A−1/2 τ A1/2 A−1/2 fk−s−1 s
k −s
k −s
k−s−1/2
H
H
s =1
4
k
1 fk + eM1 P1 f1 + eM1 P1 fs − fs−1 .
≤
H
H
H
2
s =2
(3.54)
A. Ashyralyev and M. E. Koksal 21
Using formula (3.22), the triangle inequality, and the last seven estimates, we obtain
k
−1
2 −1
−1
Y4k ≤ τ
us − us−1 H + A1/2
u s −1 − u s −2 H
C6 A1/2
s−1/2 τ
s−3/2 τ
H
s =2
3
1 + C12 3As us H + As−1 us−1 H + 3 fs H + fs−1 H
3
(3.55)
k
1 + fk H + eM1 P1 f1 H + eM1 P1 fs − fs−1 H .
2
s=2
Using formula (3.14), the triangle inequality, and the estimates Ymk H , m = 1,4, we
obtain
Ak uk ≤ C9 A0 u0 + C10 A1/2 u + C11 f1 + τ 2 A1/2 f1 0
1
0 H
H
H
H
H
k
−1
2 +τ
C6 A1/2 τ −1 us − us−1 + A1/2 τ −1 us−1 − us−2 s =2
s−1/2
3
H
s−3/2
H
1
2
+ C12 As us H + As−1 us−1 H + fs H + fs−1 H
3
3
k
1 M1 P1 M1 P1
+
fk H + e
f1 H + e
f s − f s −1 H .
2
s =2
(3.56)
From the above result, it follows that
Ak uk ≤ C13 A0 u0 + A1/2 u + τ 2 A1/2 f1 + max fs 0
1
0 H
H
H
H
H
1≤s≤k
+τ
k
s =1
k
1/2 −1 A
us − us−1 H + As us H +
f s − f s −1 H ,
s−1/2 τ
s =2
(3.57)
where
4
C13 = max C9 ,C10 ,C11 + eM1 P1 , C6 ,2C12 .
3
(3.58)
Combining estimates (3.37) and (3.57), we get
1/2 uk − uk−1 A
+ Ak uk k−1/2
H
τ
H
2 1/2 ≤ c14 A0 u0 H + A1/2
f 1 H
0 u0 H + τ A1
+ max fs H +
1≤s≤k
k
k
1/2 −1 f s − f s −1 A
τ
u
−
u
+
A
u
+
s
s −1 H
s s H
s−1/2
H
s =1
s =2
(3.59)
22
Discrete Dynamics in Nature and Society
for any k,1 ≤ k ≤ N. Here,
C14 =
1
.
1 − τ C13 + C8
(3.60)
Applying difference analogy of the integral inequality, we obtain
1/2 uk − uk−1 N A
+ Ak uk N k−1/2
k=1 Cτ
τ
k=1 Cτ
n
1/2 2 1/2 ≤ C15 A0 u0 H + A0 u0 H + τ A1
f1 H + max fs H +
f s − f s −1 H ,
1≤s≤N
s =1
(3.61)
where
C15 = C14 ekτC14 .
(3.62)
Theorem 3.1 is proved.
Theorem 3.2. Let u(0) ∈ D(A(0)), u (0) ∈
tion of the difference scheme (2.9), the stability estimate
−2 N −1 τ
uk+1 − 2uk + uk−1
1
Cτ
D(A1/2 (0)),
and fk+1 ∈ D. Then for the solu-
≤ M A(0)u0 H + A1/2 (0)u0 H + max fk H
1≤k≤N
+ max τ 2 Ak+1 fk+1 H +
1≤k≤N
N
s =1
fs − fs−1 H
(3.63)
holds, where M does not depend on u0 ,u0 , fs (1 ≤ s ≤ N), and τ.
Proof. Using (2.9), we get
uk+1 − 2uk + uk−1 τ2
H
−1/2 3/2 −3/2 1 τ
A A
Ak+3/2 A−1 τ 2 A2 uk+1 ≤
A
+ A1/2
k+1
k+1
k+1
k+1
k+1 k+3/2
H
4
+
8
−1/2 τ 1/2 −1/2 1 τ
A
+ A1/2
k+1 Ak+3/2 +
k+1 Ak+1
2 4
2
+
+
τ2 A1/2 A−1/2 A1/2 A−1/2 Ak+1 uk+1 k+1
k+1
k+1
k+3/2
H
8
−1/2 τ 2 1/2 −1/2 1/2 −1/2 1 τ
A
A
+ A1/2
k+1 Ak+3/2 +
k+1 Ak+1
k+1 Ak+1/2
2 2
4
1/2 −1/2 1 1/2 −1/2 τ 1/2 Ak uk + A1/2
−
A
A
+
A
−
A
A
k+1
k
k+1
k+1
k
k+1/2
H
2
2
A. Ashyralyev and M. E. Koksal 23
+
1
A3/2 A−3/2 Ak+3/2 A−1 + τ A1/2 A−1/2 k+1
k+1 k+3/2
k+1
k+1
2
4
uk+1 − uk −3/2 −1 A
A
A
τA
× A3/2
k+3/2 k+1
k+1
k+1 k+3/2
τ
H
−1/2 τ 1/2 −1/2 1/2 −1/2 1 A
A
+ A1/2
k+1 Ak+3/2 +
k+1 Ak+1
k+1 Ak+3/2
2
4
1/2 1/2 uk+1 − uk A
× A−
k+1/2 k+1/2
τ
H
−1/2 τ 1/2 −1/2 1/2 −1/2 A
A
+ A1/2
k+1 Ak+1/2 +
k+1 Ak+1
k+1 Ak+1/2
2
−1/2 1 1/2 −1/2 1 1/2 + A1/2
Ak+1/2
Ak+1 − A1/2
Ak+1/2 k+1 − Ak
k
τ
2
−1/2 1/2 uk − uk−1 × Ak−1/2 Ak−1/2
τ
H
1
−3/2 1 + A3/2
Ak+3/2 A−k+1
k+1 Ak+3/2
4
1/2 −1/2 3/2 −3/2 τ
−1 2
+
Ak+1 Ak+1 Ak+1 Ak+3/2 Ak+3/2 Ak+1
τ Ak+1 fk+1 H
8
−1/2 τ 2 1/2 −1/2 1/2 −1/2 τ A
A
+ 1 + A1/2
k+1 Ak+3/2 +
k+1 Ak+1
k+1 Ak+3/2
4
4
+
1/2 −1/2 τ
+
Ak+1 Ak+1 fk+1 H
2
2 τ
A1/2 A−1/2 + τ A1/2 A−1/2 A1/2 A−1/2 k+1
k+1/2
k+1
k+1
k+1
k+1/2
2
4
1/2
1/2 −1/2 1
1/2 −1/2 τ 1/2 +
Ak+1 − Ak Ak+1/2 +
Ak+1 − Ak
Ak+1/2 fk H
2
4
(3.64)
for any k,1 ≤ k ≤ N. In a similar manner as the proof of estimates (3.37) and (3.57), we
get
τAk+1 uk − uk−1 τ
H
2 1/2 ≤ 2C8 A0 u0 H + A1/2
f1 H + max fs H
0 u0 H + τ A1
1≤s≤k
+τ
k
k
1/2 −1 A
τ
u
−
u
+
A
u
+
f
−
f
s
s −1 H
s s H
s
s −1 H ,
s−1/2
s =1
s =2
24
Discrete Dynamics in Nature and Society
2 2
τ A u k k+1
H
2 1/2 ≤ 2C13 A0 u0 H + A1/2
f1 H + max fs H
0 u0 H + τ A1
1≤s≤k
+τ
k
k
1/2 −1 A
τ
u
−
u
+
A
u
+
f
−
f
s
s−1 H
s s H
s
s −1 H ,
s−1/2
s=1
s =2
(3.65)
respectively. Now, putting them in (3.64) and applying estimates (3.4), (3.5), and (3.7),
we get
uk+1 − 2uk + uk−1 ≤ C16 τ 2 A2 uk+1 + τAk+1 uk+1 − uk + Ak+1 uk+1 k+1
H
H
2
τ
τ
H
H
1/2 uk+1 − uk + Ak uk + A1/2 uk − uk−1 +
Ak+1/2
k−1/2
H
τ
τ
H
H
2
+ τ Ak+1 fk+1 + fk+1 + fk ,
H
H
H
(3.66)
where
C16 = max
1 3τ
τ2
1 τ
τ2
1 τ
+ M3 , + M3 + M32 , + M3 + M32 + τM1/2 ,
4 2
2 4
8
2 2
4
τ 1
τ
3
1
M3/2 + 1 M1 + M3 M3/2 + 1 M1 , √ M3 + M32 + M1/2 + 1 ,
2
4
2
2
δ
2
τ
3τ
τ
τ
τ2 2
1 1
√
M3 + M1/2 .
M3 + M32 , 1 + M3 + M32 , M3 + M1/2 +
4
4
4
2
4
δ 2
(3.67)
Using estimate (3.37), we get
uk+1 − 2uk + uk−1 max τ2
1≤k≤N −1
H
N
1/2 2
≤ C17 A0 u0 H + A0 u0 H + max τ Ak+1 fk+1 H + max fs H +
f s − f s −1 H ,
1≤k≤N
1≤s≤N
s =1
(3.68)
where
C17 = 5C15 C16 .
Theorem 3.2 is proved.
(3.69)
Acknowledgment
The authors would like to thank Professor Dr. Pavel Sobolevskii (Jerusalem, Israel) for his
helpful suggestions to the improvement of this paper.
A. Ashyralyev and M. E. Koksal 25
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Allaberen Ashyralyev: Department of Mathematics, Fatih University, 34900 Buyukcekmece,
Istanbul, Turkey
Email address: aashyr@fatih.edu.tr
Mehmet Emir Koksal: Graduate Institute of Sciences and Engineering, Fatih University,
34900 Buyukcekmece, Istanbul, Turkey; Department of Mathematics, Gebze Institute of Technology,
41400 Gebze/Kocaeli, Turkey
Email address: mekoksal@fatih.edu.tr