Computer Physics Communications 181 (2010) 957–966

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Computer Physics Communications

www.elsevier.com/locate/cpc

Numerical solution of Riccati equation using the cubic B-spline scaling functionsand Chebyshev cardinal functions

Mehrdad Lakestani a, Mehdi Dehghan b,∗a Faculty of Mathematical Sciences, University of Tabriz, Tabriz, Iranb Department of Applied Mathematics, Faculty of Mathematics and Computer Sciences, Amirkabir University of Technology, No. 424 Hafez Avenue, Tehran, Iran

a r t i c l e i n f o a b s t r a c t

Article history:Received 12 November 2009Accepted 29 January 2010Available online 6 February 2010

Keywords:Cubic B-spline functionRiccati equationChebyshev cardinal functionCollocation methodOperational matrix of derivative

Two numerical techniques are presented for solving the solution of Riccati differential equation. Thesemethods use the cubic B-spline scaling functions and Chebyshev cardinal functions. The methods consistof expanding the required approximate solution as the elements of cubic B-spline scaling function orChebyshev cardinal functions. Using the operational matrix of derivative, we reduce the problem toa set of algebraic equations. Some numerical examples are included to demonstrate the validity andapplicability of the new techniques. The methods are easy to implement and produce very accurateresults.

© 2010 Elsevier B.V. All rights reserved.

1. Introduction

In this paper, a numerical method using B-spline scaling functions is presented for solving the following Riccati differential equation:

u′(t) = p(t) + q(t)u(t) + r(t)u2(t), 0 � t � 1, (1.1)

with initial value

u(0) = α, (1.2)

which plays a significant role in many fields of applied science [19]. For example, as is well known, a one-dimensional static Schrödingerequation [8,9] is closely related to the Riccati differential equation. Solitary wave solution of a nonlinear partial differential equation canbe expressed as a polynomial in two elementary functions satisfying the projective Riccati equation [5]. Such type of problems also arisein optimal control. Therefore, the problem has attracted much attention and has been studied by many authors. However, deriving itsanalytical solution in an explicit form seems to be unlikely except for certain special situations. For example, some Riccati differentialequations with constant coefficients can be solved analytically by various methods [20]. Therefore, one has to employ the numericaltechniques or approximate approaches for getting its solution. Recently, Adomian’s decomposition method [13,12] has been proposed forsolving Riccati differential equations in [23]. Abbasbandy [1–3] solved a special Riccati differential equation (quadratic Riccati differentialequation) using He’s variational iteration method [10,11], homotopy perturbation method [15,14] and iterated He’s homotopy perturbationmethod and compared the accuracy of the obtained solution with that derived by Adomian decomposition method [24,25]. Geng et al.[16] introduced a piecewise variational iteration method for the Riccati differential equation, which is a modified variational iterationmethod [26,27]. We refer the interested reader to [21,22] for more works on Riccati differential equation.

In this work we reduce the problem to a set of algebraic equations by expanding the unknown function as cubic B-spline scalingfunctions specially constructed on bounded interval, with unknown coefficients. The operational matrix of derivative is given. This matrixtogether with the cubic B-spline scaling functions are then utilized to evaluate the unknown coefficients. Also we employed the Chebyshevcardinal functions to reduce the problem to a set of algebraic equations.

This paper is organized as follows: In Section 2, we describe the formulation of the cubic B-spline scaling functions on [0,1], andconstruct dual functions and then derive the operational matrix of derivative required for our subsequent development. In Section 3 the

* Corresponding author.E-mail addresses: lakestani@tabrizu.ac.ir (M. Lakestani), mdehghan@aut.ac.ir, mdehghan.aut@gmail.com (M. Dehghan).

0010-4655/$ – see front matter © 2010 Elsevier B.V. All rights reserved.doi:10.1016/j.cpc.2010.01.008

958 M. Lakestani, M. Dehghan / Computer Physics Communications 181 (2010) 957–966

proposed method is used to approximate the solution of the problem in interval [0,1]. As a result a set of algebraic equations is formedand a solution of the considered problem is introduced. In Section 4, we describe the formulation of the Chebyshev cardinal functions, andderive the operational matrix of derivative required for our subsequent development. Then the proposed method is used to approximatethe solution of the problem. As a result a set of algebraic equations is formed and a solution of the considered problem is introduced. InSection 5, we report our computational results and demonstrate the accuracy of the proposed numerical schemes by presenting numericalexamples. Section 6, ends this paper with a brief conclusion. Note that we have computed the numerical results by the well-knownsymbolic software Maple programming.

2. Cubic B-spline scaling functions on [0,1]

Scaling functions can be used to expand any function in L2(R). These functions are defined on the entire real lines, so that they canbe outside of the domain of the problem. In order to avoid this occurrence, compactly supported spline scaling functions, constructed forthe bounded interval [0,1], have been taken into account in this paper.

The mth-order cardinal B-spline Nm(t) has the knot sequence {. . . ,−1,0,1, . . .} and consists of polynomials of order m (degree m − 1)between the knots. Let N1(t) = χ[0,1](t) be the characteristic function of [0,1]. Then for each integer m � 2, the mth-order cardinalB-spline is defined inductively by [6,17]

Nm(t) = (Nm−1 ∗ N1)(t) =∞∫

−∞Nm−1(t − x)N1(x)dx =

1∫0

Nm−1(t − x)dx. (2.1)

It can be shown [7] that Nm(t) for m � 2 can be achieved by using the following formula

Nm(t) = t

m − 1Nm−1(t) + m − t

m − 1Nm−1(t − 1),

recursively, and supp[Nm(t)] = [0,m]. Here the cardinal B-splines Nm are chosen to be the scaling functions and we put m = 4 (cubicB-spline scaling functions). It can be shown [17] that a finite-energy sequence g[m,k] exists such that the scaling relation

Nm(t) =∑

k

g[m,k]Nm(2t − k), (2.2)

where

g[m,k] ={

2−m+1(m

k

), 0 � k � m,

0, otherwise

is satisfied and therefore Nm(t) is a scaling function. The explicit expression of N4(t) is [6,7,17]

N4(t) = 1

6

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

t3, t ∈ [0,1],4 − 12t + 12t2 − 3t3, t ∈ [1,2],−44 + 60t − 24t2 + 3t3, t ∈ [2,3],64 − 48t + 12t2 − t3, t ∈ [3,4],0, otherwise.

(2.3)

Suppose φ(t) := N4(t), and put φ j,k(t) = φ(2 jt − k). It is easy to see that

B j,k = supp[φ j,k(t)

] = [2− jk,2− j(m + k)

].

We need that the scaling functions intrinsically defined on [0,1] so we put

φ∗j,k(t) = φ j,k(t)χ[0,1](t). (2.4)

It can be shown [18] that φ∗j,k , k = −3,−2, . . . ,2 j − 1 is a scaling function on [0,1].

2.1. Function approximation

A function f (t) defined over [0,1] may be represented by cubic B-spline scaling functions as

f (t) =2M−1∑k=−3

skφ∗M,k(t) = S T ΦM(t), (2.5)

where

S = [s−3, s−2, . . . , s2M−1]T , (2.6)

ΦM(t) = [φ∗

M,−3(t),φ∗M,−2(t), . . . , φ

∗M,2M−1(t)

]T, (2.7)

with

M. Lakestani, M. Dehghan / Computer Physics Communications 181 (2010) 957–966 959

sk =1∫

0

f (t)φ̃M,k(t)dx, k = −3, . . . ,2M − 1,

where φ̃M,k(t) are dual functions of φ∗M,k(t). These can be obtained by linear combinations of φ∗

M,k(t),k = −3, . . . ,2M − 1, as follows. Let

Φ̃M be the dual functions of ΦM given by

Φ̃M = [φ̃M,−3, . . . , φ̃M,2M−1]T . (2.8)

Eqs. (2.7) and (2.8) give

1∫0

Φ̃M(t)ΦTM(t)dt = I, (2.9)

where I is the (2M + 3) × (2M + 3) identity matrix. Using (2.3)–(2.4) and (2.15) we have

1∫0

ΦM(x)ΦTM(x)dx = P M = 2−M ×

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1252

431680

184

15040

431680

151630

59280

142

15040

184

59280

5991260

3971680

142

15040

15040

142

3971680

151315

3971680

142

15040

. . .. . .

. . .. . .

. . .. . .

. . .

15040

142

3971680

151315

3971680

142

15040

15040

142

3971680

5991260

59280

184

15040

142

59280

151630

431680

15040

184

431680

1252

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

, (2.10)

where P M is a symmetric (2M + 3) × (2M + 3) matrix. From (2.9) and (2.10) we get

Φ̃M = (P M)−1ΦM .

2.2. The operational matrix of derivative

The differentiation of vectors ΦM in (2.7) can be expressed as

Φ ′M = DΦM , (2.11)

where D is (2M +3)×(2M +3) operational matrix of derivative for cubic B-spline scaling functions on [0,1]. The matrix D can be obtainedby considering

D =1∫

0

Φ ′M(t)Φ̃T

M(t)dt =( 1∫

0

Φ ′M(t)ΦT

M(t)dt

)(P M)−1 = E P (−1)

M , (2.12)

where

E =1∫

0

Φ ′M(t)ΦT

M(t)dt. (2.13)

In (2.13), E is a (2M + 3) × (2M + 3) matrix given by

E =

⎡⎢⎢⎣

∫ 10 ( d

dt φ∗M,−3(t))φ

∗M,−3(t)dt . . .

∫ 10 ( d

dt φ∗M,−3(t))φ

∗M,2M−1

(t)dt

.... . .

...∫ 10 ( d

dt φ∗M,2M−1

(t))φ∗M,−3(t)dt . . .

∫ 10 ( d

dt φ∗M,2M−1

(t))φ∗M,2M−1

(t)dt

⎤⎥⎥⎦ , (2.14)

by calculating the entries of matrix E , we get

960 M. Lakestani, M. Dehghan / Computer Physics Communications 181 (2010) 957–966

E =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

− 172 − 71

720 − 19360 − 1

720

− 180 − 2

9 − 127360 − 7

90 − 1720

140

29120 − 1

72 − 49144 − 7

90 − 1720

1720

790

49144 0 − 49

144 − 790 − 1

720

. . .. . .

. . .. . .

. . .. . .

. . .

1720

790

49144 0 − 49

144 − 790 − 1

7201

7207

9049

1441

72 − 29120 − 1

401

7207

90127360

29

180

1720

19360

71720

172

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (2.15)

3. Description of numerical method

In this section we solve Riccati differential equation (1.1) with the initial value (1.2) by using cubic B-spline scaling function constructedas (2.4). For this purpose, we use Eq. (2.5) to approximate u(t),

u(t) = C T ΦM(t), (3.1)

where ΦM(t) is defined in (2.7), and C is a vector with 2M + 3 unknowns defined similar to (2.6). Eqs. (2.11) and (3.1) yield

u′(t) = C T Φ ′M(t) = C T DΦM(t). (3.2)

From expressions (1.1), (3.1) and (3.2) we have

C T DΦM(t) = p(t) + q(t)C T ΦM(t) + r(t)C T ΦM(t)ΦTM(t)C . (3.3)

Also using the initial value in expressions (1.2) and (3.1) we have

C T ΦM(0) = α. (3.4)

To find the solution in (1.1) we first collocate (3.3) in t j = 2 j+12M+1 , j = 0,1, . . . ,2M − 1, t2M = 1/2 and t2M+1 = 1. So we get

C T DΦM(t j) = p(t j) + q(t j)C T ΦM(t j) + r(t j)C T ΦM(t j)ΦTM(t j)C, j = 0,1, . . . ,2M + 1. (3.5)

By calculating the entries of vector ΦM(t j), we obtain

ΦM(t j) :=[

0, . . . ,0,1

48,

23

48,

23

48,

1

48,0, . . . ,0

]T

, j = 0,1, . . . 2M − 1,

which the ( j + 1)th, ( j + 2)th, ( j + 3)th and ( j + 4)th entries of vector ΦM(t j) are nonzero,

ΦM

(t2M = 1

2

):=

[0, . . . ,0,

1

6,

2

3,

1

6,0, . . . ,0

]T

,

also the (2M−1 + 1)th, (2M−1 + 2)th, and (2M−1 + 3)th entries of vector ΦM(t2M+1) are nonzero,

ΦM(t2M+1 = 1) :=[

0, . . . ,0,1

6,

2

3,

1

6

]T

,

and

ΦM(0) :=[

1

6,

2

3,

1

6,0, . . . ,0

]T

.

Eq. (3.4) together with Eq. (3.5) give (2M + 3) equations, which can be solved for Ci, i = 1,2, . . . ,2M + 3. So the unknown function u(t)can be found.

4. Riccati equation and Chebyshev cardinal functions

Chebyshev cardinal functions of order N in [−1,1] are defined as [4]:

C j(x) = T N+1(x)

T N+1,x(x j)(x − x j), j = 1,2, . . . , N + 1, (4.1)

where T N+1(x) is the second kind Chebyshev function of order N + 1 in [−1,1] defined by

T N+1(x) = cos((N + 1)arccos(x)

),

M. Lakestani, M. Dehghan / Computer Physics Communications 181 (2010) 957–966 961

subscript x denotes x-differentiation and x j , j = 1,2, . . . , N + 1, are the zeros of T N+1(x) defined by cos((2 j − 1)/(2N + 2)), j = 1,2, . . . ,

N + 1. We change the variable t = (x + 1)L/2 to use these functions on [0, L]. Now any function g(t) on [0, L] can be approximated as

g(t) ≈N+1∑j=1

g(t j)C j(t) = G T ΘN(t), (4.2)

where t j , j = 1,2, . . . , N + 1, are the shifted points of x j , j = 1,2, . . . , N + 1, by transforming t = (x + 1)L/2 (here we choose t j so that,t1 < t2 < · · · < tN+1),

G = [g(t1), g(t2), . . . , g(tN+1)

]T, (4.3)

and

ΘN(t) = [C1(t), C2(t), . . . , CN+1(t)

]T. (4.4)

4.1. The operational matrix of derivative

The differentiation of vector ΘN in (4.4) can be expressed as

Θ ′N = DΘN , (4.5)

where D is (N + 1) × (N + 1) operational matrix of derivative for Chebyshev cardinal functions.The matrix D can be obtained by the following process. Let

Θ ′N(t) = [

C ′1(t), C ′

2(t), . . . , C ′N+1(t)

]T.

Using Eq. (4.2) any function C ′j(t) can be approximated as

C ′j(t) =

N+1∑k=1

C ′j(tk)Ck(t). (4.6)

Comparing Eqs. (4.5) and (4.6) we get

D =

⎡⎢⎢⎣

C ′1(t1) . . . C ′

1(tN+1)

......

C ′N+1(t1) . . . C ′

N+1(tN+1)

⎤⎥⎥⎦ . (4.7)

To calculate the entries C ′j(tk), j,k = 1,2, . . . , N + 1, we have

T N+1(t)

t − t j= β ×

N+1∏k=1k �= j

(t − tk), (4.8)

where β = 22N+1/LN+1 is the coefficient of xN+1 in the shifted Chebyshev polynomial function T N+1(t). Using Eq. (4.8) we get

d

dt

(T N+1(t)

t − t j

)= β ×

N+1∑i=1i �= j

N+1∏k=1

k �=i, j

(t − tk) =N+1∑i=1i �= j

T N+1(t)

(t − t j)(t − ti), (4.9)

so we have

C ′j(t) = 1

T N+1,t(t j)× d

dt

(T N+1(t)

t − t j

)= 1

T N+1,x(t j)

N+1∑i=1i �= j

T N+1(t)

(t − t j)(t − ti)=

N+1∑i=1i �= j

1

t − tiC j(t). (4.10)

For j = k using Eq. (4.10) we get

C ′j(tk) =

N+1∑i=1i �= j

1

t j − ti, (4.11)

and for j �= k using Eq. (4.10) we have

C ′j(tk) = β

T N+1,t(t j)

N+1∏�=1

��=k, j

(tk − t�). (4.12)

So the entries of the matrix D can be found using Eqs. (4.11) and (4.12).

962 M. Lakestani, M. Dehghan / Computer Physics Communications 181 (2010) 957–966

4.2. Description of numerical method using Chebyshev cardinal functions

In this section we solve Riccati differential equation (1.1) with the initial value (1.2) by using Chebyshev cardinal functions on [0, L].For this purpose, we use Eq. (4.2) to approximate u(t),

u(t) = U T ΘN(t), (4.13)

where ΘM(t) is defined in (4.4), and U is a vector with N + 1 unknowns. Using (4.11) and (4.13) gives

u′(t) = U T Θ ′N(t) = U T DΘN(t). (4.14)

Also using Eq. (4.2) the functions p(t),q(t) and r(t) can be expanded as:

p(t) ≈N+1∑j=1

p(t j)C j(t) = P T ΘN(t),

q(t) ≈N+1∑j=1

q(t j)C j(t) = Q T ΘN(t),

r(t) ≈N+1∑j=1

r(t j)C j(t) = RT ΘN(t), (4.15)

where

P = [p(t1), p(t2), . . . , p(tN+1)

]T,

Q = [q(t1),q(t2), . . . ,q(tN+1)

]T,

R = [r(t1), r(t2), . . . , r(tN+1)

]T.

From expressions (1.1), (4.13), (4.14) and (4.15) we have

U T DΘN(t) = P T ΘN (t) + Q T ΘN(t)Θ TN (t)U + RT ΘN(t)U T ΘN(t)Θ T

N (t)U . (4.16)

Also using the initial value given in (1.2) and (4.13) we have

U T ΘN(0) = α. (4.17)

To find the solution in (1.1) we first collocate (4.16) in t j , j = 2,3, . . . , N + 1. So we get

U T DΘN(t j) = P T ΘN(t j) + Q T ΘN(t j)ΘTN (t j)U + RT ΘN(t j)U T ΘN(t j)Θ

TN (t j)U .

Because of the cardinal property of the functions C j(t), we get

U T De j = P T e j + Q T e jeTj U + RT e j U

T e jeTj U , j = 2,3, . . . , N + 1, (4.18)

where e j := [0, . . . ,0,1,0, . . . ,0]T is an (N + 1) vector, which it’s jth entry is equal to 1.Eq. (4.18) together with Eq. (4.17) give a system of nonlinear equations with N + 1 unknowns and N + 1 equations, which can be

solved to find the entries of vector U . So the function u(t) can be found.

5. Numerical examples

In this section we give some computational results of numerical experiments with methods based on two preceding sections, to supportour theoretical discussion. The nonlinear systems obtained by methods are solved by the well-known Newton method.

Note that in Examples 1 and 2 in order to solve the problem using the B-spline functions (2.4), we decompose the interval [0,4] into4 subintervals: [0,1], [1,2], [2,3] and [3,4], then we solve the problem on these subintervals respectively.

Example 1. Consider the following Riccati differential equation [1–3,16,23]:{u′(t) = 1 + 2u(t) − u2(t), 0 � t � 4,

u(0) = 0.

The exact solution of this problem is

u(t) = 1 + √2 tanh

(√2t +

log(−1+√2

1+√2

)

2

).

Figs. 1 and 2 show the plot of error with M = 3,4, and M = 5,6, using B-spline method, respectively. Fig. 3 shows the plot of error forthe modified variational iteration method [16] and Fig. 4 shows the plot of error using Chebyshev cardinal functions with N = 19 and 29.By the comparison of the errors obtained using the methods presented in the current paper with the method of [16], it is easily foundthat the present approximations are more efficient.

M. Lakestani, M. Dehghan / Computer Physics Communications 181 (2010) 957–966 963

Fig. 1. Plot of error for B-spline method; left: M = 3, right: M = 4.

Fig. 2. Plot of error for B-spline method; left: M = 5, right: M = 6.

Fig. 3. Plot of error for the method presented in [16].

Example 2. Consider the following Riccati differential equation [16]:

{u′(t) = 1 + t2 − u2(t), 0 � t � 4,

u(0) = 1.

The exact solution of this problem is

u(t) = t + e−t2

1 + ∫ t e−x2 dt.

0

964 M. Lakestani, M. Dehghan / Computer Physics Communications 181 (2010) 957–966

Fig. 4. Plot of error for cardinal method; left: N = 19, right: N = 29.

Fig. 5. Plot of error for B-spline method; left: M = 3, right: M = 6.

Fig. 6. Plot of error for cardinal method; left: N = 19, right: N = 29.

Fig. 5 shows the plot of error using B-spline method with M = 3, and 6 respectively and Fig. 6 shows the plot of error using Chebyshevcardinal functions with N = 19 and 29.

Example 3. Consider (1.1) and (1.2) with

⎧⎪⎨⎪⎩

p(t) = −1/(1 + t),q(t) = 1,

r(t) = −1,

α = 1.

M. Lakestani, M. Dehghan / Computer Physics Communications 181 (2010) 957–966 965

Table 1Absolute values of errors for u(t) using B-spline method.

t M = 3 M = 4 M = 5 M = 6

0.1 1.5×10−5 1.6×10−8 1.5×10−9 1.6 × 10−9

0.2 5.9×10−6 2.2×10−7 1.4×10−8 1.0 × 10−9

0.3 7.7×10−6 1.1×10−7 1.5×10−8 1.2 × 10−9

0.4 7.7×10−6 4.0×10−8 1.1×10−8 1.5 × 10−9

0.5 8.3×10−6 1.6×10−7 5.8×10−9 9.2 × 10−10

0.6 6.3×10−6 1.3×10−8 1.3×10−8 1.4 × 10−9

0.7 5.7×10−6 3.3×10−8 1.4×10−8 1.3 × 10−9

0.8 5.9×10−6 2.5×10−8 1.4×10−8 1.4 × 10−9

0.9 5.9×10−6 1.4×10−8 1.3×10−8 1.4 × 10−9

1.0 5.7×10−6 1.7×10−8 1.4×10−8 1.4 × 10−9

Table 2Absolute values of errors for u(t) using cardinal method.

t N = 9 N = 19 N = 29

0.1 3.3 × 10−9 1.3 × 10−16 4.1 × 10−24

0.2 1.8 × 10−9 1.7 × 10−16 4.8 × 10−24

0.3 2.8 × 10−8 7.0 × 10−16 4.3 × 10−24

0.4 3.3 × 10−8 5.9 × 10−16 5.3 × 10−24

0.5 5.9 × 10−9 1.3 × 10−17 9.7 × 10−25

0.6 6.4 × 10−8 8.9 × 10−16 3.2 × 10−24

0.7 4.9 × 10−8 1.1 × 10−15 1.6 × 10−24

0.8 2.1 × 10−8 5.2 × 10−16 4.2 × 10−23

0.9 2.1 × 10−9 5.2 × 10−16 1.2 × 10−22

1.0 8.7 × 10−8 3.8 × 10−15 1.2 × 10−22

Table 3Absolute values of errors for u(t) using B-spline method.

t M = 4 M = 6

0.1 5.6 × 10−4 7.0 × 10−5

0.2 5.8 × 10−4 7.2 × 10−5

0.3 1.2 × 10−4 7.4 × 10−5

0.4 6.0 × 10−4 7.5 × 10−5

0.5 6.1 × 10−4 7.6 × 10−5

0.6 6.0 × 10−4 7.5 × 10−5

0.7 5.9 × 10−4 7.4 × 10−5

0.8 5.6 × 10−4 7.0 × 10−5

0.9 5.3 × 10−4 6.6 × 10−5

1.0 2.1 × 10−7 6.0 × 10−5

The exact solution of this problem is

u(t) = 1

1 + t.

Table 1 shows the absolute errors using B-spline method with M = 3,4,5, and 6 and Table 2 shows the absolute errors using Chebyshevcardinal method with N = 9,19 and 29.

Example 4. Consider (1.1) and (1.2) with⎧⎪⎪⎪⎨⎪⎪⎪⎩

p(t) = t3 − t2 + 32

√t,

q(t) = √t,

r(t) = −1,

α = 0.

The exact solution of this problem is

u(t) = t√

t.

Table 3 shows the absolute errors for M = 3 and 6 and Table 4 shows the absolute errors using Chebyshev cardinal method with N = 9and 29.

6. Conclusion

In this paper we presented two numerical schemes for solving the Riccati differential equation. The cubic B-spline scaling function andChebyshev cardinal functions were employed to solve the Ricatti differential equation. The two new methods developed in the currentpaper were tested on several examples. The obtained results show that these approaches can solve the problem effectively.

966 M. Lakestani, M. Dehghan / Computer Physics Communications 181 (2010) 957–966

Table 4Absolute values of errors for u(t) using cardinal method.

t N = 9 N = 29

0.1 7.9 × 10−5 2.4 × 10−6

0.2 6.7 × 10−5 2.5 × 10−6

0.3 1.7 × 10−4 3.5 × 10−6

0.4 1.2 × 10−5 3.6 × 10−6

0.5 9.4 × 10−5 3.3 × 10−6

0.6 1.8 × 10−4 3.1 × 10−6

0.7 4.3 × 10−6 3.0 × 10−6

0.8 1.2 × 10−4 3.8 × 10−6

0.9 8.0 × 10−5 3.7 × 10−6

1.0 1.4 × 10−4 5.0 × 10−6

Acknowledgements

The authors are very grateful to both reviewers for their comments.

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