A fourth-order finite difference scheme for the

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coefficients, models mixture between diffusion and wave propagation by introducing a term ... analysis for transmission and propagation of electrical signals [24] and also has applications in other fields [6, 5, 26]. ..... ∂x2 −6x(x−1)(5x2 −5x+1)e.
Communications in Numerical Analysis 2013 (2013) 1-11

Available online at www.ispacs.com/cna Volume 2013, Year 2013 Article ID cna-00148, 11 Pages doi:10.5899/2013/cna-00148 Research Article

A fourth-order finite difference scheme for the numerical solution of 1D linear hyperbolic equation Akbar Mohebbi∗ Department of Applied Mathematics, Faculty of Mathematical Science, University of Kashan, Kashan, Iran c Akbar Mohebbi. This is an open access article distributed under the Creative Commons Attribution License, which permits Copyright 2013 ⃝ unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract In this paper, a high-order and unconditionally stable difference method is proposed for the numerical solution of onespace dimensional linear hyperbolic equation. We apply a compact finite difference approximation of fourth-order for discretizing spatial derivative of this equation and a Pade´ approximation of fifth-order for the resulting system of ordinary differential equations. It is shown through analysis that the proposed scheme is unconditionally stable. This new method is easy to implement, produces very accurate results and needs short CPU time. Some numerical examples are included to demonstrate the validity and applicability of the technique. We compare the numerical results of this paper with the numerical results of some methods in the literature. Keywords: Pade´ approximation, compact finite difference scheme, unconditionally stable, linear hyperbolic equation, telegraph

equation, high accuracy.

1 Introduction The hyperbolic partial differential equations model the vibrations of structures (e.g. buildings, beams and machines) and are the basis for fundamental equations of atomic physics. In this paper we consider the second order one-dimensional linear hyperbolic equation

∂ 2u ∂u ∂ 2u (x,t) + 2α (x,t) + β 2 u(x,t) = 2 (x,t) + f (x,t), 2 ∂t ∂t ∂x (x,t) ∈ (0, L) × (0, T ],

(1.1)

with initial conditions u(x, 0) = ϕ1 (x),

∂u (x, 0) = ϕ2 (x), ∂t and boundary conditions u(0,t) = g0 (t), ∗ Corresponding

author. Email address: a [email protected], [email protected], Tel:+983615912373

(1.2)

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u(L,t) = g1 (t) , t ≥ 0.

(1.3)

For α > 0 and β = 0, Eq. (1.1) represents a damped wave equation and for α > β > 0, it is called Telegraph equation. We assume that α and β are positive. Eq. (1.1) referred to as second-order telegraph equation with constant coefficients, models mixture between diffusion and wave propagation by introducing a term that accounts for effects of finite velocity to standard heat or mass transport equation [5]. However, Eq. (1.1) is commonly used in signal analysis for transmission and propagation of electrical signals [24] and also has applications in other fields [6, 5, 26]. In recent years, much attention has been given in the literature to the development, analysis and implementation of stable methods for the numerical solution of second-order hyperbolic equations, see for example [14, 16, 22, 27]. These methods are conditionally stable. Mohanty [16, 17, 18] did nice investigations on the one space-dimensional hyperbolic equations. In [17], Mohanty carried over a new technique to solve the linear one-space-dimensional hyperbolic equation (1.1) which is unconditionally stable and is of second-order accurate in both time and space components. Also this author proposed in [18] a three level implicit unconditionally stable difference scheme of second-order accurate in both time and spatial components for the solution of Eq. (1.1) with variable coefficients that fictitious points are not needed at each time step along the boundary. Abbasbandy and Roohani in [1] applied the well-known homotopy analysis method (HAM) and an interesting iterative algorithm for solving the telegraph equation with an integral condition. A numerical scheme is developed in [5] to solve the one-dimensional hyperbolic telegraph equation using collocation points and approximating the solution using a thin plate splines radial basis function. Another numerical method is presented in [6] to solve the one-dimensional hyperbolic telegraph equation using Chebyshev cardinal functions. Authors of [11] applied the Rothe-Wavelet method to the solution of Eq. (1.1) and described in slightly different manner for the discretization of this equation. A numerical scheme based on the shifted Chebyshev tau method is proposed in [26] to solve Eq. (1.1). The combination of fourth order compact finite difference and collocation method for solving Eq. (1.1) is given in [20]. Authors of [7] used a numerical method based on the boundary integral equation and dual reciprocity method. These authors used time stepping scheme to deal with the time derivative and different types of radial basis functions to approximate functions in the dual reciprocity method. The variational iteration method has been applied successfully for the solution linear, variable coefficient, fractional derivative and multi space telegraph equations in [8]. In [15] a new spline difference scheme based on quadratic spline interpolations in space direction and finite difference discretization in time direction is presented for solving the Eq. (1.1). This method is second and fourth-order accurate in time and space components respectively and is unconditionally stable. Rashidinia et al. [25] developed two conditionally stable three level difference schemes of orders O(τ 2 + h2 ) and O(τ 2 + h4 ) based on non-polynomial cubic spline functions for the solution of one-dimensional second-order non-homogeneous hyperbolic equation. Gao and Chi in [12] proposed an explicit difference scheme for the numerical solution of Eq. (1.1). This method is unconditionally stable and has fifth and second order of accuracy in time and space directions respectively. A three level compact difference scheme which is unconditionally stable but is second and fourth order accurate in time and space directions is given in [9]. The aim of this paper is to introduce an unconditionally stable numerical method for the Eq. (1.1) which has high-order of accuracy in both time and space directions. We first discretize the spatial derivative of equation with a fourth-order compact finite difference and then apply Pade´ approximation method of fifth-order for the resulting system of ordinary differential equations. The outline of this paper is as follows. In Section 2, we apply a fourth-order compact finite difference scheme for the second order spatial derivative of partial differential equation (1.1) and obtain a system of ordinary differential equations (ODEs). Then we replace the matrix exponential in recurrence relation with the family of Pade´ approximations of fifth order to solve the resulting system of ODEs. In Section 3, we prove the unconditional stability property of proposed method. In section 4, the numerical results of applying the method of this article on three test problems for equation (1.1) are presented. Also in this section a comparison with the methods of [9, 12, 15, 19] is given. Finally a conclusion is drawn in Section 5. 2 Proposition of new method For a positive integer n let h = t. So we define

L n

denote the step size of spatial variable, x, and △t for step size of time variable,

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xi = ih

, i = 0, 1, 2, ..., n,

tk = k△t

, k = 0, 1, 2, ... .

For derivation of the method, we first discretize Eq. (1.1) in space with a fourth order compact difference scheme to obtain a system of ODEs with unknown function at each spatial grid point. The fourth-order discretization of equation d2y = k(x) , 0 < x < L, dx2

(2.4)

can be written as follows [20] ki = yxx,i = δxx yi −

1 δxx ki h2 + O(h4 ), 12

or 10ki + ki+1 + ki−1 yi+1 − 2yi + yi−1 = + O(h4 ), 12 h2 where ki = k(xi ), yi = y(xi ) and δxx yi =

yi+1 −2yi +yi−1 . h2

(2.5)

Now we rewrite (1.1) as follows

∂ 2u ∂u ∂ 2u (x,t) + 2α (x,t) + β 2 u(x,t) − f (x,t) = 2 (x,t). 2 ∂t ∂t ∂x If we put ut (x,t) = v(x,t) then (2.6) can be written as { ut (x,t) = v(x,t), vt (x,t) + 2α v(x,t) + β 2 u(x,t) − f (x,t) = uxx (x,t).

(2.6)

(2.7)

If we discretize spatial derivative of (2.7) in each grid point by the fourth order scheme (2.5), we will obtain the following relation u′i (t) = vi (t), 1 [10(v′i (t) + 2α vi (t) + β 2 ui (t) − fi (t)) + (v′i+1 (t) + 2α vi+1 (t) + β 2 ui+1 (t) − fi+1 (t)) + 12 1 (v′i−1 (t) + 2α vi−1 (t) + β 2 ui−1 (t) − fi−1 (t))] = 2 (ui−1 (t) − 2ui (t) + ui+1 (t)), h i = 1, 2, ..., n − 1, u′i (t)

(2.8)

v′i (t)

where ui (t) = u(xi ,t), vi (t) = v(xi ,t), = ut (xi ,t), = vt (xi ,t), and fi (t) = f (xi ,t). The initial and boundary conditions (1.2) and (1.3) for equations (2.8) can be used as follows ui (0) = ϕ1 (xi ) , vi (0) = ϕ2 (xi ), u0 (t) = g0 (t)

, v0 (t) = g′0 (t) , v′0 (t) = g′′0 (t),

un (t) = g1 (t) , vn (t) = g′1 (t) , v′n (t) = g′′1 (t). (2.9) We assume that g0 and g1 have second derivatives respect to time component. If we put U(t) = [u1 (t), u2 (t), ..., un−1 (t), v1 (t), v2 (t), ..., vn−1 (t)]T and write equation (2.9) for each interior grid point, we obtain the system of 2n − 2 ordinary differential equations  dU(t)  A dt = B U(t) +C(t), (2.10)  U(0) = U0 ,

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[

in which

In−1 O

A=

O A1

Page 4 of 11

]

[ ,



− 5β6 − h22 2

  − β2 + 1  12 h2   0   .. B1 =  .   ···    0  0

         C1 (t) =        

B=



5 6 1 12

     A1 =     

O B1

1 12 5 6 1 12

0 .. .

..

··· 0 0

2

− 5β6 − h22 2

− β12 + h12 .. . 2

0

··· 0

1 12 5 6

..

.

C(t) = [0, 0, ..., 0,C1 (t)]T ,

,

0

0 ··· ···

1 12

..

.

.

···

5 6 1 12

1 12

0 ··· 0

− β12 + h12

]

In−1 B2

0 ···

0

1 12 5 6 1 12



0 0 0 .. .

     ,  ···   1  12 5 6



0

···

0

0

− β12 + h12

0

···

0

− β12 + h12 .. .

···

0 .. .

2 − 5β6 − h22 2 − β12 + h12

2 − β12 + h12 2 − 5β6 − h22 2 − β12 + h12

2

− 5β6 − h22 .. . 2

2 − β12

+

···

0

0

···

5 1 1 1 12 f 0 (t) + 6 f 1 (t) + 12 f 2 (t) − 12

1 h2

2

0

···

··· 2 − β12 + h12 2 − 5β6 − h22

) ( ′ v0 (t) + 2α v0 (t) + β 2 u0 (t) + h12 u0 (t)

       ,      



       ..  . ,   5 1 1   12 f n−3 (t) + 6 f n−2 (t) + 12 f n−1 (t)   ( ) 1 5 1 1 1 ′ 2  12 f n−2 (t) + 6 f n−1 (t) + 12 f n (t) − 12 vn (t) + 2α vn (t) + β un (t) + h2 un (t) 1 5 1 12 f 1 (t) + 6 f 2 (t) + 12 f 3 (t)

and B2 = −2α A1 and O is zero matrix of size n − 1. Since matrix A1 is a diagonally dominant matrix we can conclude that A is a diagonally dominant matrix so it is invertible and we can write (2.10) as follows  dU(t)  dt = A−1 B U(t) + A−1C(t), (2.11)  U(0) = U0 . For solving system of ODEs (2.11) we replace the matrix exponential in recurrence relation U(t + ∆t) = e∆tDU(t), t = 0, ∆t, 2∆t, ..., where DU(t) =

dU(t) dt

(2.12)

is given by (2.11), with the family of Pade´ approximations. Using (2,2) Pade´ approximation, i.e. ez =

12 + 6z + z2 + O(z5 ), 12 − 6z + z2

we can write (2.12) as follows [12 − 6∆tD + (∆tD)2 ]U(t + ∆t) = [12 + 6∆tD + (∆tD)2 ]U(t).

(2.13)

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Not that D2U(t) =

d 2U(t) dt 2

Page 5 of 11

is given by d 2U(t) = (A−1 B)2U(t) + A−1 B G(t) + G′ (t), dt 2

(2.14)

where G(t) = A−1C(t). Using (2.11), (2.13) and (2.14) we can write Uk+1 = M −1 N Uk + M −1 [(6∆tI − ∆t 2 A−1 B)Gk+1 + (6∆tI + ∆t 2 A−1 B)Gk + ∆t 2 (G′k − G′k+1 )],

(2.15)

where Uk = U(tk ), Gk = G(tk ), G′k = G′ (tk ), M = [12I − 6∆tA−1 B + (∆tA−1 B)2 ], N = [12I + 6∆tA−1 B + (∆tA−1 B)2 ] and I is identity matrix of order 2n − 2. Now we prove that the method is unconditionally stable. 3 Stability analysis For homogeneous boundary conditions, the proposed method (2.15) can be written as Uk+1 = φ Uk , k = 0, 1, 2, ... ,

(3.16)

φ = [12I − 6∆tA−1 B + (∆tA−1 B)2 ]−1 [12I + 6∆tA−1 B + (∆tA−1 B)2 ].

(3.17)

where the amplification matrix is given by

A−1 B

Let λ be an eigenvalue of and z = ∆t λ . For unconditional stability of the new method it is necessary that the absolute values of the eigenvalues of amplification matrix (φ ) be less than or equal to one, i. e. 12 + 6z + z2 (3.18) 12 − 6z + z2 ≤ 1.

Lemma 3.1. ([21]) The relation (3.18) holds iff z be in the left-half complex plane. According to Lemma 3.1 if z or equivalently λ be in the left-half complex plane, the relation (3.18) holds. So we should prove that each eigenvalue of A−1 B has negative real part. Lemma 3.2. ([2]) If the Hermitian matrix Q = [qi j ]2n−2×2n−2 be strictly diagonally dominant with positive (negative) diagonal elements then Q is positive (negative) definite matrix. Lemma 3.3. Each eigenvalue of matrix B has negative real part. Proof. Let η = a + bi be an eigenvalue of B with corresponding nonzero eigenvector ξ = ] ] [ ][ [ O In−1 ξ1 ξ1 =η . B1 B2 ξ2 ξ2

[

ξ1

ξ2

]T

, i.e. (3.19)

From (3.19) we have

ξ2 = ηξ1 , B1 ξ1 + B2 ξ2 = ηξ2 , η 2ξ

(3.20)

− η 2 I)ξ

therefore we have B1 ξ1 + η B2 ξ1 = 1 or (B1 + η B2 1 = 0. If ξ1 = 0 then from (3.20) we see that ξ = 0 which is impossible. So ξ1 ̸= 0 and we should have det(B1 + η B2 − η 2 I) = 0 or det(η 2 I − η B2 − B1 ) = 0. Now we suppose that real part of η (i.e. a) is nonnegative. If we have b = 0 then (η 2 I − η B2 − B1 ) = (a2 I − aB2 − B1 ) which is positive definite and contradiction with det(η 2 I − η B2 − B1 ) = 0 (note that from Lemma 3.2, B1 and B2 are negative definite matrices). So we should have b ̸= 0. In this case we put R = (η 2 I − η B2 − B1 )i = ((a2 − b2 )I − aB2 − B1 )i − b(2aI − B2 ).

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Since a is nonnegative, the symmetric part of R, i.e. 12 (R + R∗ ) = −b(2aI − B2 ) is positive or negative definite matrix depending to sign of b. This gives that R is nonsymmetric positive or negative definite [13] and so it is nonsingular or det(η 2 I − η B2 − B1 ) ̸= 0 which is contradiction. So the real part of η should be negative. Definition 3.1. The matrix A is called generalized positive definite (SPD) if its symmetric part is positive definite, and is called generalized negative definite (GND) if −A is generalized positive definite. Lemma 3.4. ([10]) All eigenvalues of matrix Q have negative real parts, if and only if there exist a generalized negative-definite matrix G and a symmetric positive-definite matrix L such that Q = GL. Theorem 3.1. Each eigenvalue of matrix A−1 B has negative real part. Proof. From Lemma 3, all eigenvalues of matrix B have negative real parts so from Lemma 4 there exists a SPD matrix P in which G = BP is GND matrix. From Lemma 5 and regarding to this point that A−1 is SPD matrix and putting L = P−1 A−1 we can conclude that all eigenvalues of matrix Q = GL = BPP−1 A−1 = BA−1 have negative real parts. Note that eigenvalues of A−1 B and BA−1 are similar. This theorem shows that each eigenvalue of matrix A−1 B is in left-half complex plane and we can conclude that the proposed method is unconditionally stable. 4 Numerical results In this section we present the numerical results of the new method on three test problems. We performed our computations using Matlab 7 software on a PC with Intel Core 2 Duo, 2.8 GHz CPU and 2 GB RAM. We tested the accuracy and stability of the method presented in this paper by performing the mentioned method for different values of ∆t and h. The L∞ and Root-Mean-Square (RMS) errors obtained by new method are shown. Also we calculated the computational orders of the method presented in this article (denoted by C-Order) with the following formula log( EE12 ) log( hh12 )

,

in which E1 and E2 are errors correspond to grids with mesh size h1 and h2 respectively. Table 1: Absolute errors obtained for Test problem 1 with h = π /30, ∆t = 0.1 x

Method of [12]

Method of [15]

Present method

CPU time (s)

T =1 π /30

0.0904 × 10−4

0.0321 × 10−3

1.0815 × 10−8

0.0160

8π /30 15π /30

0.6479 × 10−4 0.8928 × 10−4

0.6995 × 10−3 0.2033 × 10−3

7.6892 × 10−8 1.0347 × 10−7

22π /30 29π /30

0.7069 × 10−4 0.0904 × 10−4

0.6995 × 10−3 0.0412 × 10−3

7.6892 × 10−8 2.1512 × 10−8

T =2 π /30

0.0884 × 10−4

0.0532 × 10−3

9.2830 × 10−9

8π /30 15π /30

0.6337 × 10−4 0.8731 × 10−4

0.5065 × 10−3 0.3128 × 10−3

6.5997 × 10−8 8.8808 × 10−8

22π /30 29π /30

0.6913 × 10−4 0.0884 × 10−4

0.5065 × 10−3 0.0331 × 10−3

6.5997 × 10−8 1.8464 × 10−8

0.0310

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4.1 Test Problem 1 We consider the following hyperbolic equation

∂ 2u ∂u ∂ 2u (x,t) + 4 (x,t) + 2u(x,t) = 2 (x,t), 2 ∂t ∂t ∂x where the exact solution is given by u(x,t) = e−t sin(x).

0 ≤ x ≤ π,

t > 0,

Figure 1: Surface plots of approximate solutions for Test problems 1 (left panel) and 2 (right panel). The boundary and initial conditions can be obtained from exact solution. We compare the numerical results of new method presented in this paper with the results of [12, 15]. In Table 1 we compare the absolute error obtained in solving Test problem 1 with h = π /30, ∆t = 0.1, T = 1 and T = 2. As we see the new method has better results. Table 2 shows the L∞ and Root-Mean-Square (RMS) errors, C-Order and CPU time of new method for solving Test problem 1 with T = 1, ∆t = 0.02 and several values of h. As we see the method achieve an accuracy of order 10−9 in under 0.6 seconds. Also the computational orders in Table 2 shows the fourth-order accuracy of method in spatial variable. Table 2: Numerical results obtained for Test problem 1 with ∆t = 0.02 h

RMS

L∞

C-Order

CPU time(s)

π /5 π /10 π /20 π /40 π /80

5.8341 × 10−5 3.3312 × 10−6

6.5473 × 10−5 4.2545 × 10−6

− 3.9438

0.0150 0.016

1.9744 × 10−7 1.2070 × 10−8

2.6521 × 10−7 1.6640 × 10−8

4.0038 3.9944

0.0310 0.0930

7.9948 × 10−10

1.1165 × 10−9

3.8976

0.5780

4.2 Test Problem 2 We consider the following hyperbolic equation

∂ 2u ∂u ∂ 2u + 3 + 2u = − 6x(x − 1)(5x2 − 5x + 1)e−t , ∂ t2 ∂t ∂ x2

0 ≤ x ≤ 1,

t > 0,

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where the exact solution is given by

Page 8 of 11

u(x,t) = x3 (x − 1)3 e−t .

The boundary and initial conditions can be obtained from exact solution. We compare the numerical results of proposed method with the results of [9, 19]. Table 3 shows the L∞ and Root-Mean-Square (RMS) errors, C-Order and CPU time of new method for solving Test problem 2 with T = 1, ∆t = 1/20 and several values of h. As we see the method achieve an accuracy of order 10−10 in under 1 second. Also the computational orders in Table 3 shows the fourth-order accuracy of method in spatial variable. Table 3: Numerical results obtained for Test problem 2 with ∆t = 1/20 h

RMS

L∞

C-Order

CPU time(s)

1/4

7.6340 × 10−4

8.6522 × 10−4



0

1/8 1/16

4.4429 × 10−5 2.6172 × 10−6

5.3900 × 10−5 3.3408 × 10−6

4.0047 4.0120

0.015 0.016

1/32 1/64

1.5732 × 10−7 9.4494 × 10−9

2.0370 × 10−7 1.1481 × 10−8

4.0357 4.1491

0.0310 0.1400

1/128

5.9225 × 10−10

8.6262 × 10−10

3.7344

0.8750

In Table 4 we compare the absolute error obtained in solving Test problem 2 with ∆t = 1/64, T = 1 and several values of h. As we see the new method has better results. Also Figure 1 shows the surface plots of approximate solutions of Test problems 1 and 2. Table 4: Absolute errors obtained for Test problem 2 with ∆t = 1/64 h

Method of [19]

Method of [9]

Present method

CPU time(s)

1/16 1/32

1.16767 × 10−5 2.91918 × 10−6

1.16695 × 10−5 2.91738 × 10−6

3.3694 × 10−6 2.1004 × 10−7

0.0320 0.0780

1/64

7.29796 × 10−7

7.29344 × 10−7

1.2984 × 10−8

0.4220

1/128 1/256

1.82449 × 10−7

1.82336 × 10−7

4.56122 × 10−8

4.55840 × 10−8

7.7731 × 10−10 4.1534 × 10−11

2.7030 24.4070

4.3 Test Problem 3 We consider the following hyperbolic equation

∂ 2u ∂ 2u ∂u 2 u = + 2 + + β 2 sin(x) cos(t) − 2α sin(x) sin(t), α β ∂ t2 ∂t ∂ x2

0 ≤ x ≤ 2,

t > 0,

where the exact solution is given by u(x,t) = sin(x) cos(t). The boundary and initial conditions can be obtained from exact solution. This test problem is given in [20]. We show the RMS error obtained in solving this test problem at T = 1 with h = 0.05, ∆t = 0.02 and several values of α and β in Table 5. Table 6 presents the numerical results in solving Test problem 3 with α = 10, β = 5, h = 1/80 and several values of ∆t. As we see with the above parameters the new method achieve an accuracy of order 10−9 in under 7 seconds. Figure 2 shows the surface plot of approximate solution of Test problem 3.

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Figure 2: Surface plot of approximate solution for Test problems 3.

Table 5: RMS error obtained for Test problem 3 with h = 0.05 and ∆t = 0.02 α α α α α

T=0.5

T = 1.0

T = 2.0

T = 5.0

= 10, β = 5 = 20, β = 10

4.8138 × 10−8 1.0226 × 10−7

8.9352 × 10−8 1.8206 × 10−7

9.3459 × 10−8 1.9143 × 10−7

1.0029 × 10−7 2.0296 × 10−7

= 50, β = 2 = 100, β = 5

2.3896 × 10−7 4.9369 × 10−7

4.2999 × 10−7 8.7782 × 10−7

4.6418 × 10−7 9.4931 × 10−7

5.2234 × 10−7 1.0329 × 10−6

= 100, β = 5

5.1789 × 10−7

9.0263 × 10−7

9.6904 × 10−7

1.0229 × 10−6

Table 6: Numerical results obtained for Test problem 2 with α = 10, β = 5 and h = 1/80 ∆t

L∞

RMS

C-Order

CPU time(s)

1/5 1/10

8.5635 × 10−4

6.9347 × 10−4

2.7277 × 10−4

2.7863 × 10−5

− 4.6374

0.4690 0.8900

1/20 1/40

1.5954 × 10−5 1.0557 × 10−6

1.3717 × 10−6 8.4767 × 10−8

4.3443 4.0163

1.7030 3.3130

1/80

6.7967 × 10−8

5.8279 × 10−9

3.8625

6.5320

5 Conclusion In this paper we proposed a class of high order compact schemes for solving the 1D linear hyperbolic equation. We combined a high-order compact finite difference scheme of fourth-order to approximate the spatial derivative and a fifth order Pade´ approximation for the time integration of the resulted linear system of ordinary differential equations. The proposed method for solving the mentioned equation has high order of accuracy and is unconditionally stable. Computational experiments confirmed the unconditional stability and high accuracy of the proposed method and presented that the new method can achieve good accuracy in short CPU time.

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