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

  • Allaberen Ashyralyev1 and

    Affiliated with

    • Ozgur Yildirim2Email author

      Affiliated with

      Boundary Value Problems20132013:29

      DOI: 10.1186/1687-2770-2013-29

      Received: 30 September 2012

      Accepted: 29 January 2013

      Published: 15 February 2013

      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., [15] and the references given therein). The stability has been an important research topic in the development of numerical techniques for solving these equations (see [627]). 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., [57] 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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ1_HTML.gif
      (1)

      where A is a self-adjoint positive definite linear operator with the domain D ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq1_HTML.gif in a Hilbert space H.

      A function u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq2_HTML.gif is called a solution of problem (1) if the following conditions are satisfied:
      1. (i)

        u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq2_HTML.gif is twice continuously differentiable on the segment [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq3_HTML.gif. The derivatives at the endpoints of the segment are understood as the appropriate unilateral derivatives.

         
      2. (ii)

        The element u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq2_HTML.gif belongs to D ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq4_HTML.gif for all t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq5_HTML.gif and the function A u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq6_HTML.gif is continuous on the segment [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq3_HTML.gif.

         
      3. (iii)

        u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq2_HTML.gif 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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ2_HTML.gif
      (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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ3_HTML.gif
      (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 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ4_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ5_HTML.gif
      (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., [1727] 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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq7_HTML.gif is not only continuous, but also continuously differentiable on [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq8_HTML.gif, φ D ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq9_HTML.gif and ψ D ( A 1 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq10_HTML.gif, 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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ6_HTML.gif
      (6)
      gives a solution of problem (1). Throughout this paper, { c ( t ) , t 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq11_HTML.gif 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ7_HTML.gif
      (7)
      Then, from the definition of the sine operator-function
      s ( t ) u = 0 t c ( s ) u d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ8_HTML.gif
      (8)
      it follows that
      s ( t ) = A 1 / 2 e i t A 1 / 2 e i t A 1 / 2 2 i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ9_HTML.gif
      (9)

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

      A uniform grid is considered on the segment [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq8_HTML.gif
      [ 0 , T ] τ = { t k = k τ , k = 0 , 1 , , N , N τ = T } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ10_HTML.gif
      (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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq12_HTML.gif.

      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:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ11_HTML.gif
      (11)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ12_HTML.gif
      (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 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ13_HTML.gif
      (13)
      Using (11) and (13), the following formula:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ14_HTML.gif
      (14)
      for the third order of approximation of (1), and using (12), (13), the following formula
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ15_HTML.gif
      (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 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ16_HTML.gif
      (16)
      and
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ17_HTML.gif
      (17)

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

      In the second step, the approximation of u ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq13_HTML.gif 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ18_HTML.gif
      (18)
      From (18) it is obvious that for the approximation of u ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq14_HTML.gif, it is necessary to approximate the expressions
      s ( τ ) , c ( τ ) , and 1 τ 0 τ s ( τ λ ) f ( λ ) d λ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ19_HTML.gif
      (19)
      Using the definitions of s ( τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq15_HTML.gif, c ( τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq16_HTML.gif, and Padé fractions for the function e z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq17_HTML.gif (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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ20_HTML.gif
      (20)
      for the third order of approximation of (1), where
      R ( i τ B ) = R = D E , R ( i τ B ) = R ˜ = D ˜ E , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ21_HTML.gif
      (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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ22_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ23_HTML.gif
      (23)
      Here B = A 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq18_HTML.gif and
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equa_HTML.gif
      Using (22), (20), (21), (23), we obtain the third order of approximations of s ( τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq15_HTML.gif and c ( τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq19_HTML.gif
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ24_HTML.gif
      (24)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ25_HTML.gif
      (25)
      and the fourth order of approximations of s ( τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq15_HTML.gif and c ( τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq20_HTML.gif
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ26_HTML.gif
      (26)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ27_HTML.gif
      (27)
      Let us remark that in constructing the approximation of u ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq21_HTML.gif, it is important to know how to construct f 1 , 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq22_HTML.gif and f 1 , 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq23_HTML.gif such that
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ28_HTML.gif
      (28)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ29_HTML.gif
      (29)
      and formulas of f 1 , 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq22_HTML.gif and f 1 , 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq23_HTML.gif are sufficiently simple. The choice of f 1 , 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq22_HTML.gif and f 1 , 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq23_HTML.gif 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 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ30_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ31_HTML.gif
      (31)
      For simplicity, denote
      f 1 , 1 = f ( 0 ) + ( f ( 0 ) + τ f ( 0 ) ) 1 2 2 f ( 0 ) τ 6 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ32_HTML.gif
      (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 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ33_HTML.gif
      (33)
      where
      f 1 , 1 3 = τ ( I + τ 2 12 A + τ 4 144 A 2 ) 1 f 1 , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ34_HTML.gif
      (34)
      and
      f 1 , 1 4 = τ ( I + τ 2 12 A + τ 4 144 A 2 ) 1 f 2 , 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ35_HTML.gif
      (35)
      Thus, we have the following formula for the approximation of u ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq24_HTML.gif:
      u 1 u 0 τ = c τ ( τ ) I τ φ + s τ ( τ ) τ ψ + f 1 , 1 m , m = 3 , 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ36_HTML.gif
      (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 ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq25_HTML.gif 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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ37_HTML.gif
      (37)
      and its conjugate R ˜ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq26_HTML.gif
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ38_HTML.gif
      (38)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ39_HTML.gif
      (39)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ40_HTML.gif
      (40)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ41_HTML.gif
      (41)
      and its conjugate R ˜ 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq27_HTML.gif,
      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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ42_HTML.gif
      (42)

      and its conjugate R ˜ 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq28_HTML.gif.

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

      Lemma 1 The 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ43_HTML.gif
      (43)

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

      Theorem 1 Let φ D ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq9_HTML.gif, ψ D ( A 1 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq10_HTML.gif, f 1 , 1 D ( A 1 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq29_HTML.gif. Then, for the solution of difference scheme (2), the following stability estimates hold:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ44_HTML.gif
      (44)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ45_HTML.gif
      (45)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ46_HTML.gif
      (46)

      where M does not depend on τ, φ, ψ, f 1 , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq30_HTML.gif, and f s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq31_HTML.gif, 1 s N 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq32_HTML.gif.

      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 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ47_HTML.gif
      (47)
      and for the solution of (47), the following formula is satisfied (see [11]):
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ48_HTML.gif
      (48)
      where R = q 1 = c + c 2 4 a b 2 b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq33_HTML.gif, R ˜ = q 2 = c c 2 4 a b 2 b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq34_HTML.gif. Here q 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq35_HTML.gif and q 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq36_HTML.gif are roots of equation (47), and a 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq37_HTML.gif, b 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq38_HTML.gif. We can rewrite (47) into the following difference problem:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equb_HTML.gif
      Replacing a with ( I + τ 2 6 A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq39_HTML.gif, b with ( I + τ 2 6 A + τ 4 12 A 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq40_HTML.gif, c with ( 2 2 τ 2 3 A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq41_HTML.gif, 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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equc_HTML.gif
      φ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq42_HTML.gif with τ 2 f k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq43_HTML.gif and applying formula (48), we obtain the following formulas for u k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq44_HTML.gif, 2 k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq45_HTML.gif:
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ49_HTML.gif
      (49)
      Second, we will prove estimates (44), (45), and (46). Using (43) and the formula for u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq46_HTML.gif, and the following simple estimates:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ50_HTML.gif
      (50)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ51_HTML.gif
      (51)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ52_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equd_HTML.gif
      Applying A 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq47_HTML.gif to the formula for u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq46_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Eque_HTML.gif
      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 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ53_HTML.gif
      (53)
      and estimates (43), (52), and the following simple estimates:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ54_HTML.gif
      (54)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ55_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ56_HTML.gif
      (56)
      Applying A 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq47_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ57_HTML.gif
      (57)
      Applying A to the formula for u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq46_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ58_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ59_HTML.gif
      (59)
      Using formula (59), estimates (43), (51), (52), (55), and the following simple estimates:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ60_HTML.gif
      (60)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ61_HTML.gif
      (61)
      we have
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ62_HTML.gif
      (62)
      So, for k = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq48_HTML.gif, the following estimates are proved:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equf_HTML.gif
      Now, we will establish these estimates for any k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq49_HTML.gif. 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 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ63_HTML.gif
      (63)

      for any k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq49_HTML.gif. Combining the estimates u k H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq50_HTML.gif for any k, we obtain (44).

      Applying A 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq47_HTML.gif 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 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equg_HTML.gif
      for k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq49_HTML.gif. Combining the estimates for A 1 / 2 u k H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq51_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ64_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ65_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ66_HTML.gif
      (66)
      Combining the estimates for τ 1 ( u k u k 1 ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq52_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ67_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ68_HTML.gif
      (68)
      Next, applying A 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq47_HTML.gif to formula (68) and using estimate (43) and the triangle inequality, we get
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equh_HTML.gif
      Combining the estimates for A 1 / 2 τ 1 ( u k u k 1 ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq53_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ69_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ70_HTML.gif
      (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 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equi_HTML.gif
      for k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq49_HTML.gif. Combining the estimates for A u k H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq54_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ71_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ72_HTML.gif
      (72)
      First, applying Abel’s formula to (72), we have
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ73_HTML.gif
      (73)
      Second, using formulas (73), (43), and the triangle inequality, we get
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ74_HTML.gif
      (74)
      Combining the estimates for τ 1 ( u k + 1 2 u k + u k 1 ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq55_HTML.gif for any k, we obtain
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ75_HTML.gif
      (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 ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq25_HTML.gif by (23) and the following operators:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equj_HTML.gif

      which will be used in the sequel. It is obvious that J 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq56_HTML.gif, J ˜ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq57_HTML.gif, and J 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq58_HTML.gif, J ˜ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq59_HTML.gif, and J 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq60_HTML.gif, J ˜ 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq61_HTML.gif, and J 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq62_HTML.gif, J ˜ 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq63_HTML.gif are conjugates.

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

      Lemma 2 The 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ76_HTML.gif
      (76)

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

      Theorem 2 Let φ D ( A ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq9_HTML.gif, ψ D ( A 1 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq10_HTML.gif, f 2 , 2 D ( A 1 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq64_HTML.gif. Then, for the solution of difference scheme (3), the following stability estimates hold:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ77_HTML.gif
      (77)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ78_HTML.gif
      (78)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ79_HTML.gif
      (79)

      where M does not depend on τ, φ, ψ, f 2 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq65_HTML.gif, and f s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq31_HTML.gif, 1 s N 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq32_HTML.gif.

      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:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ80_HTML.gif
      (80)
      Replacing a with ( I + τ 2 A 12 + τ 4 A 2 144 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq66_HTML.gif, b with ( I + τ 2 A 12 + τ 4 A 2 144 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq67_HTML.gif, c with
      ( 2 5 τ 2 A 6 + τ 4 A 2 72 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equk_HTML.gif
      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 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equl_HTML.gif
      φ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq42_HTML.gif with τ 2 f k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq43_HTML.gif and applying formula (48), we obtain the following formula for u k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq44_HTML.gif, 2 k N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq45_HTML.gif:
      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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ81_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ82_HTML.gif
      (82)
      Using estimates (76), the formula for u 1 + u 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq68_HTML.gif, and the following simple estimates:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ83_HTML.gif
      (83)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ84_HTML.gif
      (84)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ85_HTML.gif
      (85)
      we get
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equm_HTML.gif
      Applying A 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq47_HTML.gif to the formula for u 1 + u 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq68_HTML.gif and using estimates (76), (83), (84), and (85), we obtain
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equn_HTML.gif
      Applying A to the formula for u 1 + u 0 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq68_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ86_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ87_HTML.gif
      (87)
      Using formula (87), estimates (76), (83), (85), and the following simple estimates:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ88_HTML.gif
      (88)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ89_HTML.gif
      (89)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ90_HTML.gif
      (90)
      we get
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ91_HTML.gif
      (91)
      Applying A 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq47_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ92_HTML.gif
      (92)
      So, for k = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq48_HTML.gif, the following estimates are proved:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equo_HTML.gif
      Now, we will establish these estimates for any k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq49_HTML.gif. 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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ93_HTML.gif
      (93)
      Using formula (93), estimate (76), and the triangle inequality, we get
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equp_HTML.gif

      for any k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq49_HTML.gif. Combining the estimates u k + u k 1 2 H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq69_HTML.gif for any k, we obtain (77).

      Applying A 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq47_HTML.gif to (93), using estimate (76) and the triangle inequality, we get
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equq_HTML.gif
      for k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq49_HTML.gif. Combining the estimates for A 1 / 2 u k + u k 1 2 H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq70_HTML.gif for any k, we get
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ94_HTML.gif
      (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 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ95_HTML.gif
      (95)
      Applying A to formula (95), using estimate (76) and the triangle inequality, we get
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equr_HTML.gif
      Combining the estimates for A u k + u k 1 2 H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq71_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ96_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ97_HTML.gif
      (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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ98_HTML.gif
      (98)
      Combining the estimates for u k + 1 u k 1 2 τ H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq72_HTML.gif 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 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ99_HTML.gif
      (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 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ100_HTML.gif
      (100)
      Next, applying A 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq47_HTML.gif to formula (100), using estimate (76) and the triangle inequality, we obtain
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equs_HTML.gif
      Combining the estimates for A 1 / 2 u k + 1 u k 1 2 τ H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq73_HTML.gif for any k, we get
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ101_HTML.gif
      (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 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ102_HTML.gif
      (102)

      for a one-dimensional hyperbolic equation. The exact solution of this problem is u ( t , x ) = e t sin x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq74_HTML.gif.

      We consider the set [ 0 , 1 ] τ × [ 0 , π ] h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq75_HTML.gif 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 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq76_HTML.gif. 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 ( u n 1 u n 0 ) h 4 4 ( u n 1 1 u n 1 0 ) + u n 2 1 u n 2 0 h 4 ) = φ n N , φ n N = τ 2 2 A sin ( x n ) + τ ( I τ 2 12 A ) ( sin ( x n ) ) + f 1 , 1 τ , 2 n M 2 , f 1 , 1 = ( f ( 0 ) + ( f ( 0 ) + τ f ( 0 ) ) 1 2 2 f ( 0 ) τ 6 ) , u 0 k = u M k = 0 , 0 k N , u 3 k = 4 u 2 k 5 u 1 k , u M 3 k = 4 u M 2 k 5 u M 1 k , 0 k N , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ103_HTML.gif
      (103)
      for the approximate solution of Cauchy problem (102). We have ( N + 1 ) × ( N + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq77_HTML.gif system of linear equations in (103) which can be written in the matrix form as follows:
      A u n + 2 + B u n + 1 + C u n + D u n 1 + E u n 2 = R φ n , 2 n M 2 , u 0 = u M = 0 ˜ , u 3 k = 4 u 2 k 5 u 1 k , u M 3 k = 4 u M 2 k 5 u M 1 k , 0 k N , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ104_HTML.gif
      (104)
      where
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ105_HTML.gif
      (105)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ106_HTML.gif
      (106)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ107_HTML.gif
      (107)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ108_HTML.gif
      (108)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ109_HTML.gif
      (109)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ110_HTML.gif
      (110)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ111_HTML.gif
      (111)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ112_HTML.gif
      (112)
      and
      u s k = [ u s 0 u s 1 u s N ] ( N + 1 ) × 1 , 0 k N , s = n 2 , n 1 , n , n + 1 , n + 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ113_HTML.gif
      (113)
      Note that
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ114_HTML.gif
      (114)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ115_HTML.gif
      (115)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ116_HTML.gif
      (116)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ117_HTML.gif
      (117)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ118_HTML.gif
      (118)
      For a solution of difference equation (103), we have applied the procedure of the modified Gauss elimination method with respect to n with matrix coefficients. Therefore, we seek a solution of the matrix equation by using the following iteration formula:
      u n = α n + 1 u n + 1 + β n + 1 u n + 2 + γ n + 1 , n = M 2 , , 2 , 1 , 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ119_HTML.gif
      (119)
      where α j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq78_HTML.gif, β j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq79_HTML.gif ( j = 1 , , M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq80_HTML.gif) are ( N + 1 ) × ( N + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq81_HTML.gif square matrices and γ j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq82_HTML.gif are ( N + 1 ) × 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq83_HTML.gif column matrices. Now, we obtain the formulas of α n + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq84_HTML.gif, β n + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq85_HTML.gif, γ n + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq86_HTML.gif as
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ120_HTML.gif
      (120)
      for n = 2 : M 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq87_HTML.gif and α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq88_HTML.gif, β 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq89_HTML.gif, γ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq90_HTML.gif, α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq91_HTML.gif, β 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq92_HTML.gif, γ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq93_HTML.gif
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ121_HTML.gif
      (121)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ122_HTML.gif
      (122)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ123_HTML.gif
      (123)
      and u M = 0 ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq94_HTML.gif. Applying the formulas
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ124_HTML.gif
      (124)
      we get
      ( 4 I α M 2 ) ( α M 1 u M 1 + γ M 1 ) ( 5 I + β M 1 ) u M 1 = γ M 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ125_HTML.gif
      (125)
      From that, it follows
      u M 1 = [ ( β M 2 + 5 I ) ( 4 I α M 2 ) α M 1 ] 1 [ ( 4 I α M 2 ) γ M 1 γ M 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ126_HTML.gif
      (126)

      Thus, using the formulas and matrices above, the third order of approximation difference scheme with matrix coefficients for the approximate solution of Cauchy problem (102) is obtained.

      Second, let us consider the fourth order of accuracy in time difference scheme (3) for the approximate solution of Cauchy problem (102).

      Using difference scheme (3), we obtain
      { u n k + 1 2 u n k + u n k 1 τ 2 5 6 ( u n + 1 k 2 u n k + u n 1 k h 2 ) 1 12 ( 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 72 ( u n + 2 k 4 u n + 1 k + 6 u n k 4 u n 1 k + u n 2 k h 4 ) + τ 2 144 ( 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 + 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 = ( 5 3 ( exp ( t k ) ) + 1 6 ( 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 , 1 n M 1 , 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 ( u n 1 u n 0 ) h 4 4 ( u n 1 1 u n 1 0 ) + u n 2 1 u n 2 0 h 4 ) = φ n N , φ n N = τ 2 2 A sin ( x n ) + τ ( I τ 2 12 A ) ( sin ( x n ) ) + τ f 2 , 2 , f 2 , 2 = ( I τ 2 A 12 ) f ( 0 ) τ ( ( I 5 τ 2 A 12 ) τ f ( 0 ) ) τ 2 f 2 , 2 = + ( A τ f ( 0 ) 2 f ( 0 ) + τ f ( 0 ) ) τ 2 6 + ( A f ( 0 ) 3 f ( 0 ) ) τ 3 24 , 2 n M 2 , u 0 k = u M k = 0 , 0 k N , u 3 k = 4 u 2 k 5 u 1 k , u M 3 k = 4 u M 2 k 5 u M 1 k , 0 k N , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ127_HTML.gif
      (127)
      for an approximate solution of Cauchy problem (102). We have again the same ( N + 1 ) × ( N + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_IEq77_HTML.gif system of linear equations (104), where
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ128_HTML.gif
      (128)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ129_HTML.gif
      (129)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ130_HTML.gif
      (130)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ131_HTML.gif
      (131)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ132_HTML.gif
      (132)
      and
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ133_HTML.gif
      (133)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ134_HTML.gif
      (134)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ135_HTML.gif
      (135)
      Note that
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ136_HTML.gif
      (136)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ137_HTML.gif
      (137)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ138_HTML.gif
      (138)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-29/MediaObjects/13661_2012_Article_287_Equ139_HTML.gif
      (139)

      For the solution of difference equation (127), we use exactly the same method that we have used for the solution of difference equation (103). Thus, using the formulas and matrices above, the fourth order of approximation difference scheme with matrix coefficients for the approximate solution of Cauchy problem (102) is obtained. The implementations of the numerical experiments are carried out by Matlab.

      The errors are computed by the following formula:
      E M N = max 1 k N