Piecewise shooting reproducing kernel method for linear singularly perturbed boundary value problems

Accepted Manuscript Piecewise shooting reproducing kernel method for linear singularly perturbed boundary value problems F.Z. Geng, Z.Q. Tang PII: DOI: Reference:

S0893-9659(16)30176-8 http://dx.doi.org/10.1016/j.aml.2016.06.009 AML 5029

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Applied Mathematics Letters

Received date: 9 May 2016 Revised date: 10 June 2016 Accepted date: 10 June 2016 Please cite this article as: F.Z. Geng, Z.Q. Tang, Piecewise shooting reproducing kernel method for linear singularly perturbed boundary value problems, Appl. Math. Lett. (2016), http://dx.doi.org/10.1016/j.aml.2016.06.009 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

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Piecewise shooting reproducing kernel method for linear singularly perturbed boundary value problems F.Z. Geng ∗ Z.Q. Tang Department of Mathematics, Changshu Institute of Technology, Changshu, Jiangsu 215500, China.

Abstract In this letter, a new numerical method is proposed for solving second order linear singularly perturbed boundary value problems with left layers. Firstly a piecewise reproducing kernel method is proposed for second order linear singularly perturbed initial value problems. By combining the method and the shooting method, an effective numerical method is then proposed for solving second order linear singularly perturbed boundary value problems. Two numerical examples are used to show the effectiveness of the present method. Keywords: Reproducing kernel method; singularly perturbed problems; boundary value problems; shooting method

1

Introduction Consider a class of linear singular perturbation problems of the form   εu′′ (x) + a(x)u′ (x) + b(x)u(x) = g(x), 0 < x < 1,  u(0) = α0 , u(1) = α1 ,

(1.1)

where 0 < ε ≪ 1, a(x), b(x) and g(x) are assumed to be sufficiently smooth, and such that (1.1) has a unique solution.

Furthermore, we assume that a(x) ≥ α > 0, α is constant. The above assumption implies that the boundary layer of

the solution to (1.1) will be in the neighborhood of x = 0.

Based on the reproducing kernel theory, a method called the reproducing kernel method (RKM) was developed by Cui, Geng et al. [1,2]. The method has been widely applied to many fields[3-25]. However, the direct application of the RKM to the singularly perturbed problems can not produce accurate numerical solution due to the character of boundary layers. Recently, based on the proposed RKM, some effective numerical methods has been proposed for solving singularly perturbed turning point problems having twin boundary layers, singularly perturbed turning point problems with an interior layer and singularly perturbed delay initial and boundary value problems[20-25]. The rest of the paper is organized as follows. In the next Section, the numerical method for solving (1.1) is introduced. The piecewise reproducing kernel method is proposed for second order linear singularly perturbed initial ∗ Corresponding

author

E-mail: [email protected](F.Geng); [email protected](Z.Q. Tang).

1

2

F.Z. Geng

value problems in Section 3. The numerical examples are provided in Section 4. Section 5 ends this paper with a brief conclusion.

2

Method for singularly perturbed boundary value problem (1.1) Due to the existence of boundary layers, we hope to solve (1.1) by using the RKM in a piecewise fashion. However,

it is difficult to solve (1.1)directly in this way. Since it is possible to solve singularly perturbed initial value problems by piecewise reproducing kernel method, by the idea of shooting method, one natural way to attack (1.1) is to solve the related singularly perturbed initial value problem with a guess as to the appropriate initial value. Consider the following two singularly perturbed initial value problem related to (1.1)   εv ′′ (x) + a(x)v ′ (x) + b(x)v(x) = g(x), 0 < x < 1,  v(0) = α , v ′ (0) = z . 0

and

(2.1)

1

  εw′′ (x) + a(x)w′ (x) + b(x)w(x) = g(x), 0 < x < 1,  w(0) = α , w′ (0) = z . 0

(2.2)

2

Suppose we have solved (2.1) and (2.2) and obtained their solutions v(x), w(x) respectively. Now we form a linear combination of v(x) and w(x) : u(x) = λv(x) + (1 − λ)w(x),

(2.3)

where λ is a parameter to be determined. Clearly, u(x) solves equation (1.1) and meets the first of two boundary conditions, that is, u(0) = α0 . Selecting λ so that u(1) = α1 , we have u(1) = λv(1) + (1 − λ)w(1) = α1 and λ=

α1 − w(1) . v(1) − w(1)

(2.4)

Theorem 2.1. If problem (1.1) has a solution, then either v(x) itself is a solution or v(1) − w(1) 6= 0 (u(x) is a solution).

Proof. Let y0 (x), y1 (x) and y2 (x) be solutions of the following initial value problems, respectively   εy ′′ (x) + a(x)y ′ (x) + b(x)y (x) = g(x), 0 < x < 1, 0 0 0  y (0) = α , y ′ (0) = 0, 0

0

  εy ′′ (x) + a(x)y ′ (x) + b(x)y (x) = 0, 0 < x < 1, 1 1 1  y (0) = 1, y ′ (0) = 0, 1

(2.5)

0

1

(2.6)

3

F.Z. Geng

and   εy ′′ (x) + a(x)y ′ (x) + b(x)y (x) = 0, 0 < x < 1, 2 2 2  y (0) = 0, y ′ (0) = 1, 2

(2.7)

2

It is easy to see that the general solution of equation (1.1) is

y0 (x) + c1 y1 (x) + c2 y2 (x),

(2.8)

where c1 and c2 are arbitrary constants. Naturally, the functions v(x) and w(x) in (2.1) and (2.2) respectively are given by v(x) = y0 (x) + z1 y2 (x), w(x) = y0 (x) + z2 y2 (x),

(2.9)

in which z1 and z2 are arbitrary constants. Under the assumption that equation (1.1) has a solution, c1 and c2 exist such that y0 (0) + c1 y1 (0) + c2 y2 (0) = α0

(2.10)

y0 (1) + c1 y1 (1) + c2 y2 (1) = α1 .

(2.11)

and

From (2.10), we get c1 = 0, and (2.11) reduces to y0 (1) + c2 y2 (1) = α1 .

(2.12)

If v(1) − w(1) 6= 0, then the function u(x) in (2.3) is the solution of (1.1). If v(1) − w(1) = 0, then equation (2.9) yields y2 (1) = 0. From (2.12), it follows that y0 (1) = α1 , that is, v(1) = α1 , therefore, v(x) is a solution of (1.1).

Denote by V (x) and W (x) the approximate solutions of (2.1) and (2.2) respectively. We will obtain them by using the piecewise reproducing kernel method which will be introduced in the next section. Furthermore, we can obtain the approximate solution of singularly perturbed boundary value problem (1.1) by U (x) = λV (x) + (1 − λ)W (x),

(2.13)

where λ=

α1 − W (1) . V (1) − W (1)

(2.14)

4

F.Z. Geng

3

Piecewise reproducing kernel method for second order singularly perturbed initial value problems In this section, we illustrate the piecewise reproducing kernel method for singularly perturbed initial value problems

by solving (2.1). Firstly, we introduce the usual reproducing kernel space W n [a, b], (n ≥ 3) with the reproducing kernel

in the form of piecewise polynomial.

Definition 3.1. W n [a, b] = {u(x) | u(n−1) (x) is an absolutely continuous real value function, u(n) (x) ∈ L2 [a, b], u(a) = 0, u′ (a) = 0}. The inner product and norm in this space are defined as (u, v)n =

n−1 X

u (a)v (a) + (i)

(i)

k u kn =

b

u(n) (x)v (n) (x)dx

a

i=0

and

Z

p (u, u)n , u, v ∈ W n [a, b].

Theorem 3.1. W n [a, b] is a reproducing kernel space and its reproducing kernel is   k (x, y), y ≤ x, 1 k(x, y) =  k (y, x), y > x. 1

(3.1)

1 Especially for n = 3, k1 (x, y) = − 120 (a−y)2 (6a3 −3a2 (5x+y +10)+2a(5x2 +5x(y +6)−y 2 )−10x2 (y +3)+5xy 2 −y 3 ).

For the proof, please refer to [1,2]. Now we solve second order singularly perturbed initial value problem (2.1) by using the RKM in a piecewise fashion. We first divide [0, 1] into M subintervals [xj , xj+1 ], j = 0, 1, · · ·, M − 1, with x0 = 0 and xM = 1. Denote by hi the

length of the ith subinterval, i.e. hi = |xi − xi−1 |. Then we get the approximate solutions on every interval [xi−1 , xi ] by using the RKM.

On subinterval [x0 , x1 ], (2.1) is reduced to   εu′′ (x) + a(x)u′ (x) + b(x)u(x) = g(x), x ∈ [x0 , x1 ],  u(0) = α , u′ (0) = z . 0

(3.2)

1

Solving (2.2) by the RKM and taking N + 1 equidistant nodes on interval [t0 , t1 ], we obtain the approximate solution v1,N (x) of (2.1) on [x0 , x1 ]. Becausev1,N (x) and its derivative converge uniformly to the exact solution of (3.1) and its derivative, v1,N (t) can provide an effective approximate value of u(x1 ), u′ (x1 ). On subinterval [x1 , x2 ], the approximate form of (2.1) is   εu′′ (x) + a(x)u′ (x) + b(x)u(x) = g(x), x ∈ [x , x ], 1 2  u(x ) = v (x ), u′ (x ) = v ′ (x ). 1

1,N

1

1

1,N

1

(3.3)

5

F.Z. Geng

We can also obtain the approximate solution v2,N (t) of (2.1) on the second subinterval [x1 , x2 ] by using the RKM. In the same manner, on subinterval [xi , xi+1 ](i ≥ 2), (2.1) is approximated by   εu′′ (x) + a(x)u′ (x) + b(x)u(x) = g(x), x ∈ [x , x ], 1 2  u(x ) = v (x ), u′ (x ) = v ′ (x ). i

i,N

i

i

i,N

(3.4)

i

It is easy to obtain the approximate solution vi+1,N (t) of (2.1) on the subinterval [xi , xi+1 ] by using the RKM. After obtaining the approximate solutions of all subintervals [xi−1 , xi ], i = 1, 2, . . . , M , these solutions are combined to obtain the approximate solution V (x) of (2.1) on the entire interval [0, 1]. Clearly, V (x) is continuously differentiable on [0, 1]. Similarly, we can obtain the approximate solution W (x) of (2.2).

4

Error analysis

Theorem 4.1. If V (x), W (x) are obtained by using the RKM in space W 4 , then there exists a positive constant d such that k V (x) − v(x) k∞ = max |v(x) − V (x)| ≤ d h2 , k W (x) − w(x) k∞ = max |w(x) − W (x)| ≤ d h2 , x∈[0,1]

where d is a positive constant and h =

x∈[0,1]

max{ hN1 , hN2 , . . . , hNM

}.

Proof. For the proof, please refer to [4]. Theorem 4.2. If U (x) are obtained by using the RKM in space W 4 , then the error between approximate solution U (x) and exact solution u(x) satisfies k U (x) − u(x) k∞ ≤ d h2 . Proof. From (2.4) and (2.14), one obtains α1 −W (1) V (1)−W (1) | (1)]+W (1)v(1)−w(1)V (1) | | α1 [V (1)−v(1)+w(1)−W [v(1)−w(1)][V (1)−W (1)] (1)v(1)−W (1)V (1)+W (1)V (1)−w(1)V (1) | | α1 [V (1)−v(1)+w(1)−W (1)]+W [v(1)−w(1)][V (1)−W (1)] |α1 |(|V (1)−v(1)|+|w(1)−W (1)|)+|W (1)||v(1)−V (1)|+|W (1)−w(1)||V (1)| . |[v(1)−w(1)][V (1)−W (1)]|

α1 −w(1) |λ − λ| = | v(1)−w(1) −

=

= ≤

(4.1)

Applying Theorem 4.1 to (4.1), there exists a positive constant d2 such that |λ − λ| ≤ d2 h2 .

(4.2)

From (2.3) and (2.13), it follows that k U (x) − u(x) k∞

=k λV (x) − λv(x) + (1 − λ)W (x) − (1 − λ)w(x) k∞

=k λV (x) − λV (x) + λV (x) − λv(x) + (1 − λ)W (x) − (1 − λ)w(x)+ (1 − λ)w(x) − (1 − λ)w(x) k∞

≤k (λ − λ)V (x) k∞ + k λ(V (x) − v(x)) k∞ + k (1 − λ)(W (x) − w(x)) k∞ + k (λ − λ)w(x) k∞ .

(4.3)

6

F.Z. Geng

Combination of theorem 4.1 and (4.2) yields k U (x) − u(x) k∞ ≤ d3 h2 , where d3 is a positive constant.

5

Numerical examples

Example 4.1 Consider the following singular perturbation problem from fluid dynamics for fluid of small viscosity [20,26]   εu′′ (x) + u′ (x) = 1 + 2x, 0 ≤ x < 1,  u(0) = 0, u(1) = 1. −x ε

) It is easy to see that its exact solution is u(x) = x(x + 1 − 2ε) + (2ε−1)(1−e . Using the present method, taking 1 1−e− ε   0.2ε, i ≤ 100, M = 200, N = 80. The numerical results using the present method (PM) are compared hi =  1−20ε , i > 100, 100

with [20,26] in Tables 1 and 2 for ε = 10−3 , 10−4 . The absolute error between the approximate solution and exact

solution for ε = 10−6 is shown in Figure 1.

Table 1: Numerical results of Example 4.1 for ε = 10−3 . Nodes

Exact solution

u(x) [26]

u(x)[20]

Present method

0.000

0.000000

0.000000

0.000000

0.000000

0.001

-0.629857

-0.631119

-0.629684

-0.629852

0.010

-0.987874

-0.989854

-0.987496

-0.987875

0.030

-0.967160

-0.969100

-0.966866

-0.967160

0.100

-0.888200

-0.890000

-0.887957

-0.888200

0.300

-0.608600

-0.610000

-0.608408

-0.608600

0.500

-0.249000

-0.250000

-0.248848

-0.249000

0.700

0.190600

0.190000

0.190706

0.190600

0.900

0.710200

0.709999

0.710243

0.710200

1.000

1.000000

1.000000

1.000000

1.000000

Example 4.2 Consider the following singular perturbation problem[26]   εu′′ (x) + (1 − x )u′ (x) − 1 u(x) = 0, 0 ≤ x < 1, 2 2  u(0) = 0, u(1) = 1.

It is easy to see that its exact solution is u(x) =

1 1 2−x − 2 e

2 x− x − ε4

  0.2ε, i ≤ 100, . Using the present method, taking hi =  1−20ε , i > 100, 100

7

F.Z. Geng

Table 2: Numerical results of Example 4.1 for ε = 10−4 . Nodes

Exact solution

u(x) [26]

u(x)[20]

Present method

0.0000

0.000000

0.000000

0.000000

0.000000

0.0001

-0.631894

-0.632020

-0.633487

-0.631888

0.0010

-0.998753

-0.998953

-0.997079

-0.998754

0.0030

-0.996791

-0.996991

-0.995239

-0.996792

0.1000

-0.889820

-0.890000

-0.888582

-0.889820

0.3000

-0.609860

-0.610000

-0.608887

-0.889820

0.5000

-0.249900

-0.250000

-0.249165

-0.249900

0.7000

0.190060

0.190000

0.190538

0.190060

0.9000

0.710019

0.709999

0.710201

0.710019

1.0000

1.000000

1.000000

1.000000

1.000000

M = 200, N = 80. The numerical results using the present method (PM) are compared with [26] in Table 3 for ε = 10−3 . The absolute error between the approximate solution and exact solution for ε = 10−6 is shown in Figure 2. Table 3: Numerical results of Example 4.1 for ε = 10−3 .

5

Nodes

Exact solution

u(x) [26]

Present method

0.000

0.000000

0.000000

0.000000

0.001

0.3163104

0.3162644

0.3163075

0.010

0.5024899

0.5024893

0.5024898

0.020

0.5050505

0.5050505

0.5050505

0.100

0.5263158

0.5263158

0.5263158

0.300

0.5882353

0.5882353

0.5882353

0.500

0.6666667

0.6666667

0.6666667

0.700

0.7692308

0.7692308

0.7692308

0.900

0.9090909

0.9090909

0.9090909

1.000

1.000000

1.000000

1.000000

Conclusion In this paper, combining the shooting method and piecewise RKM, a new effective method is proposed for solving

singularly perturbed boundary value problems. The comparison of numerical results show that the present method can produce good approximations. Acknowledgments The work was supported by the National Natural Science Foundation of China (No.11201041, No. 11026200), the Special Funds of the National Natural Science Foundation of China (No. 11141003) and Qing Lan Project of Jiangsu

F.Z. Geng

8

Province.

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[16] O. Abu Arqub, M. AL-Smadi, S. Momani, T. Hayat, Numerical solutions of fuzzy differential equations using reproducing kernel Hilbert space method, Soft Computing (2015). DOI: 10.1007/s00500-0151707-4. [17] S. Abbasbandy, B. Azarnavid, M.S. Alhuthali, A shooting reproducing kernel Hilbert space method for multiple solutions of nonlinear boundary value problems, J. Comput. Appl. Math. 279 (2015) 293-305. [18] R. Ketabchi, R. Mokhtari, E. Babolian, Some error estimates for solving Volterra integral equations by using the reproducing kernel method, J. Comput. Appl. Math. 273 (2015) 245-250. [19] M. Ghasemi, M. Fardi, R.K. Ghaziani, Numerical solution of nonlinear delay differential equations of fractional order in reproducing kernel Hilbert space, Appl. Math. Comput. 268 (2015) 815-831. [20] F.Z. Geng, M.G. Cui, A novel method for solving a class of singularly perturbed boundary value problems based on reproducing kernel method, Appl. Math. Comput. 218 (2011) 4211-4215. [21] F.Z. Geng, S.P. Qian, S. Li, A numerical method for singularly perturbed turning point problems with an interior layer, J. Comput. Appl. Math. 255 (2013) 97C105. [22] F.Z. Geng, S.P. Qian, Reproducing kernel method for singularly perturbed turning point problems having twin boundary layers, Appl. Math. Lett. 26 (2013) 998-1004. [23] F.Z. Geng, S.P. Qian, Piecewise reproducing kernel method for singularly perturbed delay initial value problems, Appl. Math. Lett. 37 (2014) 67-71. [24] F.Z. Geng, S.P. Qian, Modified reproducing kernel method for singularly perturbed boundary value problems with a delay, Appl. Math. Model. 39 (2015) 5592-5597. [25] F.Z. Geng, S.P. Qian, A Numerical Method for Solving Fractional Singularly Perturbed Initial Value Problems Based on the Reproducing Kernel Method, J. Comput. Complex. Appl., 1 (2015)89-94. [26] M. Kumar, H.K. Mishra, P. Singh, A boundary value approach for a class of linear singularly perturbed boundary value problems, Adv. Eng. Softw. 40 (2009) 298-304.

F.Z. Geng

List of figure captions Figure 1: Absolute errors for ε = 10−5 for Example 4.1 (The left: [0, 20ε]; the right: [20ε, 1]). Figure 1: Absolute errors for ε = 10−6 for Example 4.2 (The left: [0, 20ε]; the right: [20ε, 1]).

10

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F.Z. Geng

5. ´ 10-6

1. ´ 10-6

4. ´ 10-6

8. ´ 10-7

3. ´ 10-6

6. ´ 10-7

2. ´ 10-6

4. ´ 10-7

1. ´ 10-6

2. ´ 10-7

0.00005

0.00010

0.00015

0.00020

0.2

0.4

0.6

0.8

1.0

Figure 1: Absolute errors for ε = 10−5 for Example 4.1 (The left: [0, 20ε]; the right: [20ε, 1])

5. ´ 10-8 2.5 ´ 10-6

4. ´ 10-8

2. ´ 10-6 3. ´ 10-8 1.5 ´ 10-6 2. ´ 10-8 1. ´ 10-6 1. ´ 10-8

5. ´ 10-7

5. ´ 10-6

0.00001

0.000015

0.00002

0.2

0.4

0.6

0.8

Figure 2: Absolute errors for ε = 10−6 for Example 4.2 (The left: [0, 20ε]; the right: [20ε, 1])

1.0