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This article is part of the series Recent Trends on Boundary Value Problems and Related Topics.

Open Access Research

On the numerical solution of hyperbolic IBVP with high-order stable finite difference schemes

Allaberen Ashyralyev1 and Ozgur Yildirim2*

Author Affiliations

1 Department of Mathematics, Fatih University, Istanbul, 34500, Turkey

2 Department of Mathematics, Yildiz Technical University, Istanbul, 34210, Turkey

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Boundary Value Problems 2013, 2013:29  doi:10.1186/1687-2770-2013-29

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2013/1/29


Received:30 September 2012
Accepted:29 January 2013
Published:15 February 2013

© 2013 Ashyralyev and Yildirim; licensee Springer

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The abstract Cauchy problem for the hyperbolic equation in a Hilbert space H with self-adjoint positive definite operator A is considered. The third and fourth orders of accuracy difference schemes for the approximate solution of this problem are presented. The stability estimates for the solutions of these difference schemes are established. A finite difference method and some results of numerical experiments are presented in order to support theoretical statements.

MSC: 65J10, 65M12, 65N12, 35L30.

Keywords:
abstract hyperbolic equation; stability; initial boundary value problem

1 Introduction

Partial differential equations of the hyperbolic type play an important role in many branches of science and engineering. For example, acoustics, electromagnetics, hydrodynamics, elasticity, fluid mechanics, and other areas of physics lead to partial differential equations of the hyperbolic type (see, e.g., [1-5] and the references given therein). The stability has been an important research topic in the development of numerical techniques for solving these equations (see [6-27]). Particularly, a convenient model for analyzing the stability is provided by a proper unconditionally absolutely stable difference scheme with an unbounded operator.

A large cycle of works on difference schemes for hyperbolic partial differential equations, in which stability was established under the assumption that the magnitude of the grid steps τ and h with respect to time and space variables are connected (see, e.g., [5-7] and the references therein). 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. Such type stability inequalities for the solutions of the first order of accuracy difference scheme for the differential equations of hyperbolic type were established for the first time in [20].

It is known (see [25,26]) that various initial boundary value problems for a hyperbolic equation can be reduced to the initial value problem

{ d 2 u ( t ) d t 2 + A u ( t ) = f ( t ) , 0 < t < T , u ( 0 ) = φ , u ( 0 ) = ψ , (1)

where A is a self-adjoint positive definite linear operator with the domain D ( A ) in a Hilbert space H.

A function u ( t ) is called a solution of problem (1) if the following conditions are satisfied:

(i) u ( t ) is twice continuously differentiable on the segment [ 0 , 1 ] . The derivatives at the endpoints of the segment are understood as the appropriate unilateral derivatives.

(ii) The element u ( t ) belongs to D ( A ) for all t [ 0 , 1 ] and the function A u ( t ) is continuous on the segment [ 0 , 1 ] .

(iii) u ( t ) satisfies the equations and the initial conditions (1).

In recent decades, many scientists have worked in the field of a finite difference method for the numerical solutions of hyperbolic PDEs and have published many scientific papers. For problem (1), the first and two types of second order difference schemes were presented and the stability estimates for the solution of these difference schemes and for the first- and second-order difference derivatives were obtained in [11]. The high order of accuracy two-step difference schemes generated by an exact difference scheme or by the Taylor decomposition on the three points for the numerical solution of the same problem were presented and the stability estimates for approximate solution of these difference schemes were obtained in [10]. However, the difference methods developed in these references are generated by square roots of A. This action is very difficult for the realization. Therefore, in spite of theoretical results, the role of their application to a numerical solution for an initial value problem is not great. In this paper, the third order of accuracy difference scheme

{ τ 2 ( u k + 1 2 u k + u k 1 ) + 2 3 A u k + 1 6 A ( u k + 1 + u k 1 ) + 1 12 τ 2 A 2 u k + 1 = f k , f k = 2 3 f ( t k ) + 1 6 ( f ( t k + 1 ) + f ( t k 1 ) ) f k = 1 12 τ 2 ( A f ( t k + 1 ) + f ( t k + 1 ) ) , 1 k N 1 , u 0 = φ , ( I + τ 2 12 A + τ 4 144 A 2 ) τ 1 ( u 1 u 0 ) = τ 2 A φ + ( I τ 2 12 A ) ψ + f 1 , 1 τ , (2)

and the fourth order of accuracy difference scheme

{ τ 2 ( u k + 1 2 u k + u k 1 ) + 5 6 A u k + 1 12 A ( u k + 1 + u k 1 ) τ 2 72 A 2 u k + τ 2 144 A 2 ( u k + 1 + u k 1 ) = f k , f k = 5 6 f ( t k ) + 1 12 ( f ( t k + 1 ) + f ( t k 1 ) ) + τ 2 72 ( A f ( t k ) + f ( t k ) ) f k = 1 144 τ 2 ( A ( f ( t k + 1 ) + f ( t k 1 ) ) + f ( t k + 1 ) + f ( t k 1 ) ) , 1 k N 1 , t k = k τ , N τ = 1 , u 0 = φ , ( I + τ 2 A 12 + τ 4 A 2 144 ) τ 1 ( u 1 u 0 ) = τ 2 A φ + ( I τ 2 A 12 ) ψ + f 2 , 2 τ (3)

for the approximate solution of initial value problem (1) are constructed using the integer powers of the operator A, and stability estimates for the solution of these difference schemes are obtained. Here,

f 1 , 1 = ( I + τ 2 12 A + τ 4 144 A 2 ) τ 1 f 1 , 1 3 = { f ( 0 ) + ( f ( 0 ) + τ f ( 0 ) ) 1 2 2 f ( 0 ) τ 6 } (4)

and

f 2 , 2 = ( I + τ 2 12 A + τ 4 144 A 2 ) τ 1 f 1 , 1 4 = { ( I τ 2 A 12 ) f ( 0 ) + ( ( I 5 τ 2 A 12 ) f ( 0 ) + τ f ( 0 ) ) 1 2 + ( A τ f ( 0 ) 2 f ( 0 ) + τ f ( 0 ) ) τ 6 + ( A f ( 0 ) 3 f ( 0 ) ) τ 2 24 } . (5)

Some results of this paper without proof are accepted and will be published in 2013 (see [14]).

Note that boundary value problems for parabolic equations, elliptic equations, and equations of mixed type have been studied extensively by many scientists (see, e.g., [17-27] and the references therein).

2 The stability estimates

In this section, construction of difference schemes (2), (3) and stability estimates for the solutions of these difference schemes are presented.

Let us obtain the third and fourth orders of approximation formulas for the solution of problem (1). If the function f ( t ) is not only continuous, but also continuously differentiable on [ 0 , T ] , φ D ( A ) and ψ D ( A 1 2 ) , it is easy to show that (see [25]) the formula

u ( t ) = c ( t ) φ + s ( t ) ψ + 0 t s ( t λ ) f ( λ ) d λ (6)

gives a solution of problem (1). Throughout this paper, { c ( t ) , t 0 } is a strongly continuous cosine operator-function defined by the formula

c ( t ) = e i t A 1 / 2 + e i t A 1 / 2 2 . (7)

Then, from the definition of the sine operator-function

s ( t ) u = 0 t c ( s ) u d s , (8)

it follows that

s ( t ) = A 1 / 2 e i t A 1 / 2 e i t A 1 / 2 2 i . (9)

For the theory of cosine operator-function, we refer to [25] and [27].

A uniform grid is considered on the segment [ 0 , T ]

[ 0 , T ] τ = { t k = k τ , k = 0 , 1 , , N , N τ = T } . (10)

In the construction of two-step difference schemes for the solution of initial value problem (1), it is necessary to approximate differential equation (1) and the derivative u ( 0 ) .

In the first step, the approximation of differential equation (1) is considered. Using Taylor’s decomposition on three points, the following formulas for the third order of approximation and the fourth order of approximation of (1) are obtained respectively:

(11)

(12)

Applying equation (1), one can write

u ( t k ) = A u ( t ) + f ( t ) , u ( 4 ) ( t ) = A u ( t ) + f ( t ) = A 2 u ( t ) A f ( t ) + f ( t ) . (13)

Using (11) and (13), the following formula:

(14)

for the third order of approximation of (1), and using (12), (13), the following formula

(15)

for the fourth order of approximation of (1) are obtained.

Neglecting the last small term, we get

τ 2 ( u k + 1 2 u k + u k 1 ) + 2 3 A u k + 1 6 A ( u k + 1 + u k 1 ) + 1 12 τ 2 A 2 u k + 1 = f k , f k = 2 3 f ( t k ) + 1 6 ( f ( t k + 1 ) + f ( t k 1 ) ) f k = 1 12 τ 2 ( A f ( t k + 1 ) + f ( t k + 1 ) ) (16)

and

(17)

for the third and fourth orders of approximations of (1), respectively.

In the second step, the approximation of u ( 0 ) is considered. Applying (6), we can write

u ( τ ) u ( 0 ) τ = c ( τ ) I τ φ + s ( τ ) τ ψ + 1 τ 0 τ s ( τ λ ) f ( λ ) d λ . (18)

From (18) it is obvious that for the approximation of u ( 0 ) , it is necessary to approximate the expressions

s ( τ ) , c ( τ ) , and 1 τ 0 τ s ( τ λ ) f ( λ ) d λ . (19)

Using the definitions of s ( τ ) , c ( τ ) , and Padé fractions for the function e z (see [9]), the following approximation formulas are obtained:

c ( τ ) = R ( i τ B ) + R ( i τ B ) 2 + o ( τ 3 ) , s ( τ ) = B 1 R ( i τ B ) R ( i τ B ) 2 i + o ( τ 3 ) (20)

for the third order of approximation of (1), where

R ( i τ B ) = R = D E , R ( i τ B ) = R ˜ = D ˜ E , (21)

and

c ( τ ) = R ( i τ B ) + R ( i τ B ) 2 + o ( τ 4 ) , s ( τ ) = B 1 R ( i τ B ) R ( i τ B ) 2 i + o ( τ 4 ) (22)

for the fourth order of approximation of (1), where

R ( i τ B ) = R = C C ˜ 1 , R ( i τ B ) = R ˜ = C ˜ C 1 . (23)

Here B = A 1 / 2 and

Using (22), (20), (21), (23), we obtain the third order of approximations of s ( τ ) and c ( τ )

(24)

(25)

and the fourth order of approximations of s ( τ ) and c ( τ )

(26)

(27)

Let us remark that in constructing the approximation of u ( 0 ) , it is important to know how to construct f 1 , 1 3 and f 1 , 1 4 such that

(28)

(29)

and formulas of f 1 , 1 3 and f 1 , 1 4 are sufficiently simple. The choice of f 1 , 1 3 and f 1 , 1 4 is not unique. Using Taylor’s formula and integration, we obtain the following formulas for the third order of approximation:

f 1 , 1 3 = { S τ ( τ ) f ( 0 ) + ( C τ ( τ ) f ( 0 ) + S τ ( τ ) f ( 0 ) ) τ 2 2 C τ ( τ ) f ( 0 ) τ 2 6 } = ( I + τ 2 12 A + τ 4 144 A 2 ) 1 { τ f ( 0 ) + ( f ( 0 ) + τ f ( 0 ) ) τ 2 2 f ( 0 ) τ 2 6 } (30)

and for the fourth order of approximation

f 1 , 1 4 = { S τ ( τ ) f ( 0 ) + ( C τ ( τ ) f ( 0 ) + S τ ( τ ) f ( 0 ) ) τ 2 + ( A S τ ( τ ) f ( 0 ) 2 C τ ( τ ) f ( 0 ) + S τ ( τ ) f ( 0 ) ) τ 2 6 + ( A C τ ( τ ) f ( 0 ) 3 C τ ( τ ) f ( 0 ) ) τ 3 24 } = ( I + τ 2 12 A + τ 4 144 A 2 ) 1 { τ ( I τ 2 A 12 ) f ( 0 ) + ( ( I 5 τ 2 A 12 ) f ( 0 ) + τ f ( 0 ) ) τ 2 + ( A τ f ( 0 ) 2 f ( 0 ) + τ f ( 0 ) ) τ 2 6 + ( A f ( 0 ) 3 f ( 0 ) ) τ 3 24 } . (31)

For simplicity, denote

f 1 , 1 = f ( 0 ) + ( f ( 0 ) + τ f ( 0 ) ) 1 2 2 f ( 0 ) τ 6 , (32)

and

f 2 , 2 = { ( I τ 2 A 12 ) f ( 0 ) + ( ( I 5 τ 2 A 12 ) f ( 0 ) + τ f ( 0 ) ) 1 2 + ( A τ f ( 0 ) 2 f ( 0 ) + τ f ( 0 ) ) τ 6 + ( A f ( 0 ) 3 f ( 0 ) ) τ 2 24 } , (33)

where

f 1 , 1 3 = τ ( I + τ 2 12 A + τ 4 144 A 2 ) 1 f 1 , 1 (34)

and

f 1 , 1 4 = τ ( I + τ 2 12 A + τ 4 144 A 2 ) 1 f 2 , 2 . (35)

Thus, we have the following formula for the approximation of u ( 0 ) :

u 1 u 0 τ = c τ ( τ ) I τ φ + s τ ( τ ) τ ψ + f 1 , 1 m , m = 3 , 4 . (36)

Using the approximation formulas above, difference schemes (2) and (3) are constructed.

Now, we will obtain the stability estimates for the solution of difference scheme (2). We consider operators R, R ˜ by (21) and the following operators:

R 1 = ( 5 τ 4 144 A 2 + 7 τ 6 216 A 3 i τ A 1 / 2 ( I + τ 2 12 A + τ 4 144 A 2 ) I + 1 72 τ 4 A 2 ) × ( i τ A 1 / 2 ( I + 1 72 τ 4 A 2 ) ( I + τ 2 12 A + τ 4 144 A 2 ) ) 1 , (37)

and its conjugate R ˜ 1

(38)

(39)

(40)

(41)

and its conjugate R ˜ 5 ,

R 6 = ( I 1 3 τ 2 A + i τ A 1 / 2 I + 1 72 τ 4 A 2 ) × ( τ 2 2 A + τ 4 12 A 2 i τ A 1 / 2 I + 1 72 τ 4 A 2 ) 1 , (42)

and its conjugate R ˜ 6 .

Let us give one lemma, without proof, that will be needed below.

Lemma 1The following estimates hold:

{ R H H 1 , R ˜ H H 1 , R 1 H H 1 , R ˜ 1 H H 1 , A 1 / 2 R 2 H H 1 , τ A 1 / 2 R 3 H H 1 , A 1 / 2 R 4 H H 1 , A 1 / 2 R 5 H H τ , A 1 / 2 R ˜ 5 H H τ , τ A 1 / 2 R 6 H H 1 , τ A 1 / 2 R ˜ 6 H H 1 . (43)

Now, we will give the first main theorem of the present paper on the stability of difference scheme (2).

Theorem 1Let φ D ( A ) , ψ D ( A 1 2 ) , f 1 , 1 D ( A 1 2 ) . Then, for the solution of difference scheme (2), the following stability estimates hold:

(44)

(45)

(46)

whereMdoes not depend onτ, φ, ψ, f 1 , 1 , and f s , 1 s N 1 .

Proof First, we will obtain the formula for the solution of problem (1). It is clear that there exists a unique solution of the initial value problem

a u k 1 c u k + b u k + 1 = φ k , 1 k N 1 , u 0 = φ , u 1 = ψ (47)

and for the solution of (47), the following formula is satisfied (see [11]):

(48)

where R = q 1 = c + c 2 4 a b 2 b , R ˜ = q 2 = c c 2 4 a b 2 b . Here q 1 and q 2 are roots of equation (47), and a 0 , b 0 . We can rewrite (47) into the following difference problem:

Replacing a with ( I + τ 2 6 A ) , b with ( I + τ 2 6 A + τ 4 12 A 2 ) , c with ( 2 2 τ 2 3 A ) , and ψ with

φ + τ ( I + τ 2 12 A + τ 4 144 A 2 ) 1 ( τ 2 A φ + ( I τ 2 12 A ) ψ + τ f 1 , 1 ) ,

φ k with τ 2 f k and applying formula (48), we obtain the following formulas for u k , 2 k N :

u k = 1 2 [ R ˜ 1 R k R 1 R ˜ k ] φ + 1 2 [ R ˜ k R k ] R 2 ψ + 1 2 [ R ˜ k R k ] R 3 τ 2 f 1 , 1 + 1 2 R 4 s = 1 k 1 [ R ˜ k s R k s ] f s τ 2 . (49)

Second, we will prove estimates (44), (45), and (46). Using (43) and the formula for u 1 , and the following simple estimates:

(50)

(51)

(52)

we get

u 1 H ( I 5 12 τ 2 A + τ 4 144 A 2 ) ( I + τ 2 12 A + τ 4 144 A 2 ) 1 H H φ H + τ A 1 / 2 ( I τ 2 12 A ) ( I + τ 2 12 A + τ 4 144 A 2 ) 1 H H A 1 / 2 ψ H + τ A 1 / 2 ( I + τ 2 12 A + τ 4 144 A 2 ) 1 H H τ A 1 / 2 f 1 , 1 H φ H + 12 A 1 / 2 ψ H + 12 11 12 + 11 τ A 1 / 2 f 1 , 1 H M { s = 1 N 1 A 1 / 2 f s H τ + φ H + A 1 / 2 ψ H + τ A 1 / 2 f 1 , 1 H } .

Applying A 1 / 2 to the formula for u 1 and using estimates (43), (50), (51), and (52), we get

A 1 / 2 u 1 H ( I 5 τ 2 A 12 + τ 4 A 2 144 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H A 1 / 2 φ H + τ A 1 / 2 ( I τ 2 12 A ) ( I + τ 2 12 A + τ 4 144 A 2 ) 1 H H ψ H + τ A 1 / 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H τ f 1 , 1 H A 1 / 2 φ H + 12 ψ H + 12 11 12 + 11 τ f 1 , 1 H M { s = 1 N 1 f s H τ + A 1 / 2 φ H + ψ H + τ f 1 , 1 H } .

Using the formula

u 1 u 0 τ = ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 × ( τ 2 A φ + ( I τ 2 12 A ) ψ + τ f 1 , 1 ) (53)

and estimates (43), (52), and the following simple estimates:

(54)

(55)

we get

u 1 u 0 τ H 1 2 τ A 1 / 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H A 1 / 2 φ H + ( I τ 2 12 A ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H ψ H + ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H τ f 1 , 1 H 6 11 12 + 11 A 1 / 2 φ H + ψ H + τ f 1 , 1 H M { s = 1 N 1 f s H τ + A 1 / 2 φ H + ψ H + τ f 1 , 1 H } . (56)

Applying A 1 / 2 to (53) and using estimates (43), (52), (54), (55), we obtain

A 1 / 2 u 1 u 0 τ H τ 2 A 1 / 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H A φ H + ( I τ 2 12 A ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H A 1 / 2 ψ H + ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H τ A 1 / 2 f 1 , 1 H 6 11 12 + 11 A φ H + A 1 / 2 ψ H + τ A 1 / 2 f 1 , 1 H M { s = 2 N 1 f s f s 1 H + f 1 H + A φ H + A 1 / 2 ψ H + τ A 1 / 2 f 1 , 1 H } . (57)

Applying A to the formula for u 1 and using estimates (43), (50), (51), (52), we get

A u 1 H ( I 5 τ 2 A 12 + τ 4 A 2 144 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H A φ H + τ A 1 / 2 ( I τ 2 12 A ) ( I + τ 2 12 A + τ 4 144 A 2 ) 1 H H A 1 / 2 ψ H + τ A 1 / 2 ( I + 1 12 τ 2 A + 1 144 τ 4 A 2 ) 1 H H τ A 1 / 2 f 1 , 1 H A φ H + 12 A 1 / 2 ψ H + 12 11 12 + 11 τ A 1 / 2 f 1 , 1 H M { s = 2 N 1 f s f s 1 H + f 1 H + A φ H + A 1 / 2 ψ H + τ A 1 / 2 f 1 , 1 H } . (58)

Using difference scheme (2) and formula (53), we obtain

u 2 2 u 1 + u 0 τ 2 = ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 × { ( I τ 2 A 3 5 τ 4 A 2 72 + τ 6 A 3 1 , 728 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 A φ τ ( A + 1 6 τ 2 A 2 ) ( I τ 2 12 A ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 ψ ( A + τ 2 A 2 6 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ 2 f 1 , 1 + f 1 } . (59)

Using formula (59), estimates (43), (51), (52), (55), and the following simple estimates:

(60)

(61)

we have

(62)

So, for k = 1 , the following estimates are proved:

Now, we will establish these estimates for any k 2 . Using formula (49), estimate (43), and the triangle inequality, we obtain

u k H 1 2 ( R ˜ 1 H H R k H H + R 1 H H R ˜ k H H ) φ H + 1 2 ( A 1 / 2 R 2 H H R ˜ k H H + A 1 / 2 R 2 H H R k H H ) A 1 / 2 ψ H + 1 2 ( τ A 1 / 2 R 3 H H R ˜ k H H + τ A 1 / 2 R 3 H H R k H H ) τ A 1 / 2 f 1 , 1 H + 1 2 τ A 1 / 2 R 4 H H s = 1 k 1 [ R ˜ k s H H + R k s H H ] A 1 / 2 f s H τ M { s = 1 N 1 A 1 / 2 f s H τ + φ H + A 1 / 2 ψ H + τ A 1 / 2 f 1 , 1 H } (63)

for any k 2 . Combining the estimates u k H for any k, we obtain (44).

Applying A 1 / 2 to (49), using estimate (43) and the triangle inequality, we get

A 1 / 2 u k H 1 2 ( R ˜ 1 H H R k H H + R 1 H H R ˜ k H H ) A 1 / 2 φ H + 1 2 ( A 1 / 2 R 2 H H R ˜ k H H + A 1 / 2 R 2 H H R k H H ) ψ H + 1 2 ( τ A 1 / 2 R 3 H H R ˜ k H H + τ A 1 / 2 R 3 H H R k H H ) τ f 1 , 1 H + 1 2 τ A 1 / 2 R 4 H H s = 1 k 1 ( R ˜ k s H H + R k s H H ) f s H τ M { s = 1 N 1 f s H τ + A 1 / 2 φ H + ψ H + τ f 1 , 1 H }

for k 2 . Combining the estimates for A 1 / 2 u k H for any k, we obtain

max 1 k N A 1 / 2 u k H M { s = 1 N 1 f s H τ + A 1 / 2 φ H + ψ H + τ f 1 , 1 H } . (64)

Applying formula (49), we get

u k u k 1 τ = 1 τ { 1 2 [ R ˜ 1 R 5 R k 1 R 1 R ˜ 5 R ˜ k 1 ] φ + 1 2 [ R ˜ 5 R ˜ k 1 R 5 R k 1 ] R 2 ψ + 1 2 [ R ˜ 5 R ˜ k 1 R 5 R k 1 ] R 3 τ 2 f 1 , 1 + 1 2 [ R ˜ R ] R 4 τ 2 f k 1 + 1 2 R 4 s = 1 k 2 [ R ˜ 5 R ˜ k 1 s R 5 R k 1 s ] f s τ 2 } . (65)

Using (65), estimate (43), and the triangle inequality, we obtain

u k u k 1 τ H 1 τ { 1 2 ( A 1 / 2 R 5 H H R ˜ 1 H H R k 1 H H + A 1 / 2 R ˜ 5 H H R 1 H H R ˜ k 1 H H ) A 1 / 2 φ H + 1 2 ( A 1 / 2 R 2 H H A 1 / 2 R ˜ 5 H H R ˜ k 1 H H + A 1 / 2 R 2 H H A 1 / 2 R 5 H H R k 1 H H ) ψ H + 1 2 ( τ A 1 / 2 R 3 H H A 1 / 2 R ˜ 5 H H R ˜ k 1 H H + τ A 1 / 2 R 3 H H A 1 / 2 R 5 H H R k 1 H H ) × τ f 1 , 1 H + 1 2 ( τ A 1 / 2 R 4 H H R ˜ H H + τ A 1 / 2 R 4 H H R H H ) A 1 / 2 f k 1 H τ + 1 2 τ A 1 / 2 R 4 H H × s = 1 k 2 ( A 1 / 2 R ˜ 5 H H R ˜ k 1 s H H + A 1 / 2 R 5 H H R k 1 s H H ) f s H τ } M { s = 1 N 1 f s H τ + A 1 / 2 φ H + ψ H + τ f 1 , 1 H } . (66)

Combining the estimates for τ 1 ( u k u k 1 ) H for any k, we obtain

max 1 k N 1 u k u k 1 τ H M { s = 1 N 1 f s H τ + A 1 / 2 φ H + ψ H + τ f 1 , 1 H } . (67)

From estimates (64), (67), estimate (45) follows.

Now, applying Abel’s formula to (65), we obtain

u k u k 1 τ = 1 τ { 1 2 [ R ˜ 1 R 5 R k 1 R 1 R ˜ 5 R ˜ k 1 ] φ + 1 2 [ R ˜ 5 R ˜ k 1 R 5 R k 1 ] R 2 ψ + 1 2 [ R ˜ 5 R ˜ k 1 R 5 R k 1 ] R 3 τ 2 f 1 , 1 + 1 2 [ R ˜ R ] R 4 τ 2 f k 1 + 1 2 R 4 τ 2 ( s = 2 k 2 [ R 5 R 6 R k 1 s R ˜ 5 R ˜ 6 R ˜ k 1 s ] ( f s f s 1 ) × [ R ˜ 5 R ˜ 6 R 5 R 6 ] f k 2 [ R ˜ 5 R ˜ 6 R ˜ k 2 R 5 R 6 R k 2 ] f 1 ) } , 2 k N . (68)

Next, applying A 1 / 2 to formula (68) and using estimate (43) and the triangle inequality, we get

Combining the estimates for A 1 / 2 τ 1 ( u k u k 1 ) H for any k, we obtain

max 1 k N 1 A 1 / 2 u k u k 1 τ H M { s = 2 N 1 f s f s 1 H + f 1 H + A 1 / 2 ψ H + A φ H + τ A 1 / 2 f 1 , 1 H } . (69)

Now, applying Abel’s formula to (49), we have

u k = 1 2 [ R ˜ 1 R k R 1 R ˜ k ] φ + 1 2 [ R ˜ k R k ] R 2 ψ + 1 2 [ R ˜ k R k ] R 3 τ 2 f 1 , 1 + 1 2 τ 2 R 4 ( s = 2 k 1 [ R 6 R k s R ˜ 6 R ˜ k s ] ( f s f s 1 ) + ( R ˜ 6 R 6 ) f k 1 [ R ˜ 6 R ˜ k 1 R 6 R k 1 ] f 1 ) , 2 k N . (70)

Next, applying A to formula (70) and using (43) and the triangle inequality, we get

A u k H 1 2 ( R ˜ 1 H H R k H H + R 1 H H R ˜ k H H ) A φ H + 1 2 ( A 1 / 2 R 2 H H R ˜ k H H + A 1 / 2 R 2 H H R k H H ) A 1 / 2 ψ H + 1 2 ( τ A 1 / 2 R 3 H H R ˜ k H H + τ A 1 / 2 R 3 H H R k H H ) × τ A 1 / 2 f 1 , 1 H + 1 2 τ A 1 / 2 R 4 H H × ( s = 2 k 1 [ τ A 1 / 2 R 6 H H R k s H H + τ A 1 / 2 R ˜ 6 H H R ˜ k s H H ] × f s f s 1 H + ( τ A 1 / 2 R ˜ 6 H H + τ A 1 / 2 R 6 H H ) f k 1 H + [ τ A 1 / 2 R ˜ 6 H H R ˜ k 1 H H + τ A 1 / 2 R 6 H H R k 1 H H ] f 1 H ) M { s = 2 N 1 f s f s 1 H + f 1 H + A φ H + A 1 / 2 ψ H + τ A 1 / 2 f 1 , 1 H }

for k 2 . Combining the estimates for A u k H for any k, we obtain

max 1 k N A u k H M { s = 2 N 1 f s f s 1 H + f 1 H + A φ H + A 1 / 2 ψ H + τ A 1 / 2 f 1 , 1 H } . (71)

Now, applying formula (49), we get

u k + 1 2 u k + u k 1 τ 2 = 1 τ 2 { 1 2 [ R ˜ 1 R 5 2 R k 1 R 1 R ˜ 5 2 R ˜ k 1 ] φ + 1 2 [ R ˜ 5 2 R ˜ k 1 R 5 2 R k 1 ] R 2 ψ + 1 2 [ R ˜ 5 2 R ˜ k 1 R 5 2 R k 1 ] R 3 τ 2 f 1 , 1 + 1 2 [ R ˜ R ] R 4 τ 2 f k + 1 2 ( R ˜ 5 2 R 5 2 ) R 4 τ 2 f k 1 + 1 2 R 4 s = 1 k 2 [ R ˜ 5 2 R ˜ k 1 s R 5 2 R k 1 s ] f s τ 2 } . (72)

First, applying Abel’s formula to (72), we have

(73)

Second, using formulas (73), (43), and the triangle inequality, we get

(74)

Combining the estimates for τ 1 ( u k + 1 2 u k + u k 1 ) H for any k, we obtain

(75)

From estimates (69), (71), and (75), we obtain (46). Theorem 1 is proved. □

Now, we will obtain the stability estimates for the solution of difference scheme (3). We consider operators R, R ˜ by (23) and the following operators:

which will be used in the sequel. It is obvious that J 1 , J ˜ 1 , and J 2 , J ˜ 2 , and J 3 , J ˜ 3 , and J 4 , J ˜ 4 are conjugates.

Let us give one lemma, without proof, that will be needed below.

Lemma 2The following estimates hold:

{ R H H 1 , R ˜ H H 1 , J 1 H H 1 , J ˜ 1 H H 1 , τ A 1 / 2 J 2 H H 1 , τ A 1 / 2 J ˜ 2 H H 1 , A 1 / 2 J 3 H H τ , A 1 / 2 J ˜ 3 H H τ , τ A 1 / 2 J 4 H H 1 , τ A 1 / 2 J ˜ 4 H H 1 . (76)

Now, we will give the first main theorem of the present paper on the stability of difference scheme (3).

Theorem 2Let φ D ( A ) , ψ D ( A 1 2 ) , f 2 , 2 D ( A 1 2 ) . Then, for the solution of difference scheme (3), the following stability estimates hold:

(77)

(78)

(79)

whereMdoes not depend onτ, φ, ψ, f 2 , 2 , and f s , 1 s N 1 .

Proof First, we will obtain the formula for the solution of problem (1). In exactly the same manner as in Theorem 1, one establishes formula (48) for the solution of initial value problem (47).

We can rewrite (47) into the following difference problem:

(80)

Replacing a with ( I + τ 2 A 12 + τ 4 A 2 144 ) , b with ( I + τ 2 A 12 + τ 4 A 2 144 ) , c with

( 2 5 τ 2 A 6 + τ 4 A 2 72 ) ,

and ψ with

φ + τ ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 ( τ 2 A φ + ( I τ 2 A 12 ) ψ + τ f 2 , 2 ) ,

φ k with τ 2 f k and applying formula (48), we obtain the following formula for u k , 2 k N :

u k = 1 2 [ R k + R ˜ k ] φ + 1 2 [ R ˜ k R k ] i A 1 / 2 ψ + ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 ( R ˜ R ) 1 [ R ˜ k R k ] τ 2 f 2 , 2 + ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 ( R ˜ R ) 1 s = 1 k 1 [ R ˜ k s R k s ] τ 2 f s . (81)

Now, we will prove estimates (77), (78), and (78). First, using the formula of difference scheme (3), we obtain the formula

u 1 + u 0 2 = ( I τ 2 A 6 + τ 4 A 2 144 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 φ + 1 2 ( I τ 2 A 12 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ ψ + 1 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ 2 f 2 , 2 . (82)

Using estimates (76), the formula for u 1 + u 0 2 , and the following simple estimates:

(83)

(84)

(85)

we get

Applying A 1 / 2 to the formula for u 1 + u 0 2 and using estimates (76), (83), (84), and (85), we obtain

Applying A to the formula for u 1 + u 0 2 and using estimates (76), (83), (84), and (85), we have

A u 1 + u 0 2 H ( I τ 2 A 6 + τ 4 A 2 144 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H × A φ H + 1 2 τ A 1 / 2 ( I τ 2 A 12 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H × A 1 / 2 ψ H + 1 2 τ A 1 / 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H τ A 1 / 2 f 2 , 2 H A φ H + 6 A 1 / 2 ψ H + 6 11 12 + 11 τ A 1 / 2 f 2 , 2 H M { s = 2 N 1 f s f s 1 H + f 1 H + A φ H + A 1 / 2 ψ H + τ A 1 / 2 f 2 , 2 H } . (86)

Using the formula for the solution of difference scheme (3), we obtain

u 2 u 0 2 τ = τ A 2 ( I τ 2 A 6 + τ 4 A 2 144 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 2 φ + ( I 5 τ 2 A 12 + τ 4 A 2 144 ) ( I τ 2 12 A ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 2 ψ + ( I 5 τ 2 A 12 + τ 4 A 2 144 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 2 τ f 2 , 2 + 1 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ f 1 . (87)

Using formula (87), estimates (76), (83), (85), and the following simple estimates:

(88)

(89)

(90)

we get

(91)

Applying A 1 / 2 to (87) and using estimates (76), (85), (89), (90), we obtain

A 1 / 2 u 2 u 0 2 τ H ( I τ 2 A 6 + τ 4 A 2 144 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 × τ A 1 / 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H A φ H + ( I 5 τ 2 A 12 + τ 4 A 2 144 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H × ( I τ 2 12 A ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H A 1 / 2 ψ H + ( I 5 τ 2 A 12 + τ 4 A 2 144 ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H × ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H τ A 1 / 2 f 2 , 2 H + 1 2 τ A 1 / 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H f 1 H 12 11 12 + 11 A φ H + 6 11 12 + 11 f 1 H + A 1 / 2 ψ H + τ A 1 / 2 f 2 , 2 H M { s = 2 N 1 f s f s 1 H + f 1 H + A φ H + A 1 / 2 ψ H + τ A 1 / 2 f 2 , 2 H } . (92)

So, for k = 1 , the following estimates are proved:

Now, we will establish these estimates for any k 2 . Using formula (81), we obtain

u k + u k 1 2 = 1 2 [ J 1 R k 1 + J ˜ 1 R ˜ k 1 ] φ + 1 2 [ J ˜ 1 R ˜ k 1 J 1 R k 1 ] i A 1 / 2 ψ + 1 2 [ J ˜ 2 R ˜ k 1 J 2 R k 1 ] τ 2 f 2 , 2 + 1 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ 2 f k 1 + 1 2 s = 1 k 2 [ J ˜ 2 R ˜ k 1 s J 2 R k 1 s ] τ 2 f s . (93)

Using formula (93), estimate (76), and the triangle inequality, we get

for any k 2 . Combining the estimates u k + u k 1 2 H for any k, we obtain (77).

Applying A 1 / 2 to (93), using estimate (76) and the triangle inequality, we get

for k 2 . Combining the estimates for A 1 / 2 u k + u k 1 2 H for any k, we get

(94)

Now, applying Abel’s formula to (93), we obtain

u k + u k 1 2 = 1 2 [ J 1 R k 1 + J ˜ 1 R ˜ k 1 ] φ + 1 2 [ J ˜ 1 R ˜ k 1 J 1 R k 1 ] i A 1 / 2 ψ + 1 2 [ J ˜ 2 R ˜ k 1 J 2 R k 1 ] τ 2 f 2 , 2 + 1 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ 2 f k 1 + 1 2 A 1 ( s = 2 k 2 [ R ˜ k s + R k s ] ( f s f s 1 ) + ( R ˜ + R ) f k 2 ( R ˜ k 1 + R k 1 ) f 1 ) . (95)

Applying A to formula (95), using estimate (76) and the triangle inequality, we get

Combining the estimates for A u k + u k 1 2 H for any k, we obtain

max 1 k N 1 A u k + u k 1 2 H M { s = 2 N 1 f s f s 1 H + f 1 H + A 1 / 2 ψ H + A φ H + τ A 1 / 2 f 2 , 2 H } . (96)

Using formula (81), we get

u k + 1 u k 1 2 τ = 1 2 τ [ J 3 R k 1 + J ˜ 3 R ˜ k 1 ] φ + 1 2 τ [ J ˜ 3 R ˜ k 1 J 3 R k 1 ] i A 1 / 2 ψ + 1 2 τ [ R ˜ k + R k ] ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ 2 f 2 , 2 + 1 2 τ ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ 2 f k + 1 2 τ ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 [ R ˜ + R ] τ 2 f k 1 + 1 2 τ ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 s = 1 k 2 [ R ˜ k s + R k s ] f s τ 2 . (97)

Next, using formula (97), estimate (76), and the triangle inequality, we obtain

u k + 1 u k 1 2 τ H 1 2 τ ( A 1 / 2 J 3 H H R k 1 H H + A 1 / 2 J ˜ 3 H H R ˜ k 1 H H ) A 1 / 2 φ H + 1 2 | i | ( A 1 / 2 J ˜ 3 H H R ˜ k 1 H H + A 1 / 2 J 3 H H R k 1 H H ) ψ H + 1 2 ( R ˜ k H H + R k H H ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H × f 2 , 2 H τ + 1 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H f k H τ + 1 2 ( R ˜ H H + R H H ) ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H × f k 1 H τ + 1 2 ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 H H × s = 1 k 2 ( R ˜ k s H H + R k s H H ) f s H τ M { s = 1 N 1 f s H τ + A 1 / 2 φ H + ψ H + τ f 2 , 2 H } . (98)

Combining the estimates for u k + 1 u k 1 2 τ H for any k, we get

max 1 k N 1 u k + 1 u k 1 2 τ H M { s = 1 N 1 f s H τ + A 1 / 2 φ H + ψ H + τ f 2 , 2 H } . (99)

From estimates (94) and (99), we obtain (78).

Now, applying Abel’s formula to (97), we have

u k + 1 u k 1 2 τ = 1 2 τ [ J 3 R k 1 + J ˜ 3 R ˜ k 1 ] φ + 1 2 τ [ J ˜ 3 R ˜ k 1 J 3 R k 1 ] i A 1 / 2 ψ + 1 2 τ [ R ˜ k + R k ] ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ 2 f 2 , 2 + 1 2 τ ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 τ 2 f k + 1 2 τ ( I + τ 2 A 12 + τ 4 A 2 144 ) 1 [ R ˜ + R ] τ 2 f k 1 + 1 2 τ s = 2 k 2 [ J ˜ 2 R ˜ k s + J 2 R k s ] ( f s f s 1 ) 1 2 τ [ J ˜ 2 R ˜ + J 2 R ] f k 2 + 1 2 τ [ J ˜ 2 R ˜ k 1 + J 2 R k 1 ] f 1 } , 2 k N . (100)

Next, applying A 1 / 2 to formula (100), using estimate (76) and the triangle inequality, we obtain

Combining the estimates for A 1 / 2 u k + 1 u k 1 2 τ H for any k, we get

(101)

From estimates (96) and (101), estimate (78) follows. Theorem 2 is proved. □

3 Numerical analysis

In the present section, finite difference method is used and two numerical examples and some numerical results are presented in order to support our theoretical statements.

Generally, we have not been able to determine a sharp estimate for the constant figuring in the stability inequalities. However, the numerical results are presented by considering the Cauchy problem

{ 2 u ( t , x ) t 2 2 u ( t , x ) x 2 = 2 e t sin x , 0 < t < 1 , 0 < x < π , u ( 0 , x ) = sin x , 0 x π , u t ( 0 , x ) = sin x , 0 x π , u ( t , 0 ) = u ( t , π ) = 0 , 0 t 1 , (102)

for a one-dimensional hyperbolic equation. The exact solution of this problem is u ( t , x ) = e t sin x .

We consider the set [ 0 , 1 ] τ × [ 0 , π ] h of a family of grid points depending on the small parameters τ and h: [ 0 , 1 ] τ × [ 0 , π ] h = { ( t k , x n ) : t k = k τ , 0 k N , N τ = 1 , x n = n h , 0 n M , M h = π } . First, for the approximate solution of problem (102), we have applied the third and fourth orders of accuracy difference schemes respectively.

First, let us consider the third order of accuracy in time difference scheme for the approximate solution of Cauchy problem (102). Using difference scheme (2), we obtain

{ u n k + 1 2 u n k + u n k 1 τ 2 2 3 ( u n + 1 k 2 u n k + u n 1 k h 2 ) 1 6 ( u n + 1 k + 1 2 u n k + 1 + u n 1 k + 1 h 2 + u n + 1 k 1 2 u n k 1 + u n 1 k 1 h 2 ) + τ 2 12 ( u n + 2 k + 1 4 u n + 1 k + 1 + 6 u n k + 1 4 u n 1 k + 1 + u n 2 k + 1 h 4 ) = φ n k , φ n k = ( 4 3 exp ( t k ) + 1 3 ( exp ( t k + 1 ) + exp ( t k 1 ) ) ) sin ( x n ) , t k = k τ , 1 k N 1 , N τ = 1 , x n = n h , 2 n M 2 , M h = π , u n 0 = φ n 0 , φ n 0 = sin ( x n ) , 0 n M , u n 1 u n 0 τ 2 12 ( ( u n + 1 1 u n + 1 0 ) 2 ( u n 1 u n 0 ) + ( u n 1 1 u n 1 0 ) h 2 ) + τ 4 144 ( u n + 2 1 u n + 2 0 4 ( u n + 1 1 u n + 1 0 ) + 6 (