EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 4, 2021, 1283-1294 ISSN 1307-5543 – ejpam.com Published by New York Business Global A New Spline Method for Solving Linear Two-Point Boundary Value Problems Ahmed Salem Heilat1,∗, Hamzeh Zureigat1, Ra’ed Hatamleh1, Belal Batiha1 1 Department of Mathematics, Jadara University, P.O. Box(733), 21111 Irbid, Jordan Abstract. In this paper, second order linear two-point boundary value problems are treated using new method based on hybrid cubic B-spline. The values of the free parameter, γ, are chosen via optimization. The value of the free parameter plays an important role in giving accurate results. Optimization of this parameter is carried out. This method is tested on four examples and a comparison with cubic B-spline, trigonometric cubic B-spline and extended cubic B-spline methods has been carried out. The examples suggest that this method produces more accurate results than the other three methods. The numerical results are presented to illustrate the efficiency of our method. 2020 Mathematics Subject Classifications: 34B05, 65L10 Key Words and Phrases: Two-point boundary value problems, Cubic B-spline, Trigonometric cubic B-spline, Hybrid cubic B-spline 1. Introduction Boundary value problems occur in various fields of physics, applied mathematics, chem- istry, biology, and engineering. Consider the general form of linear second-order two-point boundary value problem : u′′(x) +m(x)u′(x) + n(x)u(x) = r(x), x ∈ [0, 1], u(0) = β1, u(1) = β2, (1) where m(x), n(x), r(x) ∈ C0[0, 1] and n(x) < 0 on [0, 1]. The existence of solution to such problem was studied in [1, 2]. Linear two-point boundary value problems have been discussed widely and solved numer- ically by many authors [3–9, 14, 21, 24, 26]. Homotopy perturbation method is better than most of the other methods in the literature in giving accurate results with rapid con- vergence [4]. Variational iteration method is successful in dealing with singular problems. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v14i4.4124 Email addresses: ahmed heilat@yahoo.com (A. S. Heilat), hamzeh.zu@jadara.edu.jo (H. Zureigat), raedhat@yahoo.com (R. Hatamleh), belalbatiha@gmail.com (B. Batiha) http://www.ejpam.com 1283 © 2021 EJPAM All rights reserved. A. S. Heilat et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1283-1294 1284 Moreover, this method does not need discretization, interpolation and it can derive the exact solution by using one iteration only. This method is also valid for large coefficients [5]. The extended adomian decomposition method is a very effective algorithm which pro- vides promising results with simple calculations [6]. Spectral method gives highly accurate solutions to boundary value problems [7]. Reproducing kernel gives quite accurate and efficient solution for linear fourth-order multi-point boundary value problems [8, 9]. Finite difference method gives highly accurate results only at the chosen knots whereas in some other methods, the results can be obtained at any point in the domain [3, 15]. The aim of this research is to use hybrid cubic B-spline (HCBSM) to solve equation (1). This equation had already been treated using cubic B-spline method (CBSM), trigono- metric cubic B-spline method (TCBSM), extended cubic B-spline method (ECBSM), and BS2 methods. ECBSM is by far the efficient spline-based method in producing accurate results [15–18]. Cubic spline is used to obtain the solution at any point in the range [10– 13]. B-spline interpolation is the efficient to interpolate any smooth functions comparing with finite difference, finite element, finite volume methods [15]. Cubic trigonometric B- spline produces more accurate results compared to cubic B-spline if the problems were trigonometric [16]. The ECBSM have a free parameter λ, and this parameter is important to give more accurate results. The value of λ can be obtained by optimization [17]. BS2 methods give more accurate solution [18]. The generalized nonlinear Klien-Gordon equa- tion and nonlinear two-point BVPs have been treated using HCBSM and the results are promising [19, 25]. Therefore, HCBSM can be applied to solve equation (1) . The hybrid cubic B-spline also contains a free parameter, γ. 2. Hybrid Cubic B-spline HCBS is a combination between CBS and TCBS [19]. One free parameter, γ, is introduced within the basis function. For a finite interval [0, 1], suppose that {xi}ni=0 is a uniform partition of a finite interval [0, 1] with n ∈ Z+ such that 0 = x0 < x1 < ... < xn = 1. We can extend the partition using h = 1− 0 n , x0 = 0, xi = x0 + ih, i ∈ Z. Hybrid cubic B-spline basis function, H4 i (x), can be defined as: H4 i (x) = γB4 i (x) + (1− γ)T 4 i (x), γ ∈ R. (2) where B4 i (x) and T 4 i (x) is a basis functions of cubic B-spline and trigonometric cubic B-spline, respectively [15, 19, 20, 22], as shown in (3)-(4). A. S. Heilat et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1283-1294 1285 B4 i (x) = 1 24h4  (x− xi) 3, x ∈ [xi, xi+1], h3 + 3h2(x− xi+1) + 3h(x− xi+1) 2 − 3(x− xi+1) 3, x ∈ [xi+1, xi+2], h3 + 3h2(xi+3 − x) + 3h(xi+3 − x)2 − 3(xi+3 − x)3, x ∈ [xi+2, xi+3], (xi+4 − x)3, x ∈ [xi+3, xi+4]. (3) T 4 i (x) = 1 κ  p3(xi), x ∈ [xi, xi+1], p(xi)[p(xi)q(xi+2) + p(xi+1)q(xi+3)] + p2(xi+1)q(xi+4), x ∈ [xi+1, xi+2], q(xi+4)[p(xi+1)q(xi+3) + p(xi+2)q(xi+4)] + p(xi)q 2(xi+3), x ∈ [xi+2, xi+3], q3(xi+4), x ∈ [xi+3, xi+4], (4) p(xi) = sin( x− xi 2 ), q(xi) = sin( xi − x 2 ), κ = sin( h 2 ) sin(h) sin( 3h 2 ). If γ = 0, this basis function is equal to cubic trigonometric B-spline basis function and if γ = 1, the basis function becomes B-spline basis function. H4 i (x) is a piecewise function of degree 3. The values of Hi and its derivatives H ′ i , H ′′ i at the nodal points are tabulated in Table 1, where a1 = γ 6 + (1− γ) sin2(h2 ) sin(h) sin(3h2 ) , a2 = 4γ 6 + 2(1− γ) sin(h2 ) sin(3h2 ) , a3 = γ 2h + 3(1− γ) 4 sin(3h2 ) , a4 = −γ 2h − 3(1− γ) 4 sin(3h2 ) , a5 = γ h2 + 3(1− γ)[sin(h2 )− 2 sin3(h2 ) + sin(3h2 )] 8 sin(h2 ) sin(h) sin( 3h 2 ) , a6 = −2γ h2 − 3(1− γ)[sin(2h) + 2 sin2(h2 ) sin(h)] 4 sin(h2 ) sin(h) sin( 3h 2 ) . Table 1: Coefficient of Hi, H ′ i and H ′′ i x xi−1 xi xi+1 Hi a1 a2 a1 H ′ i a3 0 a4 H ′′ i a5 a6 a5 From the basis function, an arbitrary spline curve can be generated by the following formula: S(x) = n−1∑ i=−3 CiH 4 i (x), x ∈ [x0, xn], Ci ∈ R. (5) There are only four nonzero basis functions, H4 i−3(x), H 4 i−2(x), H 4 i−1(x), and H4 i (x) on the sub-interval [xi, xi+1]. This is due to the local support property of the B-spline basis. At xi, there are only three nonzero basis functions; namely, H4 i−3(xi), H 4 i−2(xi), and H4 i−1(xi) [19]. So, by finding the first and second derivatives of S(x) and substituting xi, we obtain S(xi) = A1Ci−3 +A2Ci−2 +A1Ci−1, (6) A. S. Heilat et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1283-1294 1286 S ′ (xi) = A3Ci−3 −A3Ci−1, (7) S ′′ (xi) = A4Ci−3 +A5Ci−2 +A4Ci−1, (8) where Ai = γσi + (1− γ)ηi, for i = 1, 2, ..., 5, (9) with σ1 = 1 6 , σ2 = 4 6 , σ3 = −1 2h , σ4 = 1 h2 , σ5 = −2 h2 , η1 = κ21 κ2κ3 , η2 = 2κ1 κ3 , η3 = −3 4κ3 , η4 = 3(κ1 − 2κ31 + κ3) 8κ1κ2κ3 , η5 = −3(κ4 + 2κ21κ2) 4κ1κ2κ3 , (10) and κ1 = sin( h 2 ), κ2 = sin(h), κ3 = sin( 3h 2 ), κ4 = sin(2h). (11) 3. Hybrid Cubic B-spline interpolation method . In order to solve problem (1), suppose that the solution is S(x), as follows: S ′′ (x) +m(x)S ′ (x) + n(x)S(x) = r(x), x ∈ [0, 1], S(0) = β1, S(1) = β2. (12) Substituting xi in (12) , we have S ′′ (xi) +m(xi)S ′ (xi) + n(xi)S(xi) = r(xi), i = 0, 1, ..., n. (13) By substituting equations (6) to (8) into (1), we have {Ci−3[γσ4 + (1− γ)η4] + Ci−2[γσ5 + (1− γ)η5] + Ci−1[γσ4 + (1− γ)η4]} +m(xi){Ci−3[−γσ3 − (1− γ)η3] + Ci−1[γσ3 + (1− γ)η3]} +n(xi){Ci−3[γσ1+(1−γ)η1]+Ci−2[γσ2+(1−γ)η2]+Ci−1[γσ1+(1−γ)η1]} = r(xi). (14) By collecting the terms that only contain Ci−3, Ci−2, and Ci−1 from (14), we have Ci−3[γσ4 + (1− γ)η4 −m(xi)(γσ3 + (1− γ)η3) + n(xi)(γσ1 + (1− γ)η1)] +Ci−2[γσ5 + (1− γ)η5 + n(xi)(γσ2 + (1− γ)η2)] +Ci−1[γσ4 + (1− γ)η4 +m(xi)(γσ3 + (1− γ)η3) + n(xi)(γσ1 + (1− γ)η1)] = r(xi). (15) A. S. Heilat et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1283-1294 1287 Similarly, the boundary conditions can be simplified as the following: S(0) = S(x0) = C−3[γσ1+(1−γ)η1]+C−2[γσ2+(1−γ)η2]+C−1[γσ1+(1−γ)η1] = β1 (16) S(1) = S(xn) = Cn−3[γσ1+(1−γ)η1]+Cn−2[γσ2+(1−γ)η2]+Cn−1[γσ1+(1−γ)η1] = β2 (17) We can rearrange (15), (16), and (17) to get a matrix equation in the form of [A](n+3)×(n+3)[C](n+3)×1 = [R]1×(n+3), (18) where C = [C−3, C−2, C−1, ..., Cn−1] T is the unknown vector, R = [β1, r(x0), r(x1), ..., r(xn), β2] T , and A is an (n+ 3)× (n+ 3) -dimensional matrix given by A =  γσ1+(1−γ)η1 γσ2+(1−γ)η2 γσ1+(1−γ)η1 0 · · · 0 0 a0(x0) b0(x0) c0(x0) 0 · · · 0 0 0 a1(x1) b1(x1) c1(x1) 0 · · · 0 ... ... ... ... ... ... ... 0 · · · 0 0 an(xn) bn(xn) cn(xn) · · · · γσ1+(1−γ)η1 γσ2+(1−γ)η2 γσ1+(1−γ)η1  (n+3)×(n+3) The coefficients in the matrix A are as the following for i = 0, 1, ..., n: ai(xi) = [γσ4 + (1− γ)η4]−m(xi)[γσ3 + (1− γ)η3] + n(xi)[γσ1 + (1− γ)η1] bi(xi) = [γσ5 + (1− γ)η5] + n(xi)[γσ2 + (1− γ)η2] ci(xi) = [γσ4 + (1− γ)η4] +m(xi)[γσ3 + (1− γ)η3] + n(xi)[γσ1 + (1− γ)η1] Therefore, C can be solved by taking C = A−1R. (19) In order to get the approximate analytical solution of equation (1) we can substitute the values of Ci in equation (5), for i = −3,−2, ..., n− 1 . The numerical solution can be calculated after obtaining the values of γ by trial and error [19]. 4. Optimizing the γ The approximate analytical solution is of the form S(x, γ) = n−1∑ i=−3 CiH 4 i (x, γ), x ∈ [x0, xn], Ci ∈ R. (20) where Ci’s are obtained by solving a linear system of order (n + 1) × (n + 1). Ci’s are functions of x and γ. The approach used is adopted from [17]. Equation (20) has two free A. S. Heilat et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1283-1294 1288 parameters, x and γ. So, S(x) can be written as S(x, γ). S(x, γ) is piecewise polynomials with n intervals, as in equation (21). Each Si(x, γ), for i = 1, 2, ..., n− 1 is a polynomials of degree four. Si(x, γ), x ∈ [xi, xi+1], (21) Equation (1) can be written as in Equation (22). u′′(x) +m(x)u′(x) + n(x)u(x)− r(x) ≈ 0. (22) Substituting the approximate solutions, S(x, γ) and its derivatives into (22), we have S ′′ (x, γ) +m(x)S ′ (x, γ) + n(x)S(x, γ)− r(x) ≈ 0 (23) Equation (23) can be written as of error formula. From this equation, we have D(x, γ) = S ′′ (x, γ) +m(x)S ′ (x, γ) + n(x)S(x, γ)− r(x), x ∈ [x0, xn] which can be expanded into equation (24). S ′′ i (x, γ) +m(x)S ′ i(x, γ) + n(x)Si(x, γ)− r(x), x ∈ [xi, xi+1], (24) Since equation (24) is piecewise functions with n equations, it is wise to have some represen- tatives from every sub-interval. Suppose we have a sequence , {x∗i } n−1 i=1 , where x ∗ i ∈ [x0, xn] and m ∈ Z+ such that x∗i = xi+xi+1 2 , for i = 1, 2, ..., n− 1. Evaluating D(x, γ) at {x∗i } n−1 i=1 would produce a sequence of 2n expressions containing γ, D(x∗i , γ), x ∈ [xi, xi+1] (25) By handling equations (25) as the error at collocation points, the expressions are combined using the two-norm formula resulting equation (26). We hope to minimize d1(γ) norm as follows: d1(γ) = √√√√n−1∑ i=1 (D(x∗i ), γ) 2 (26) Also, from equation (26) we can obtain d2(γ) which is assumed to be easier to calculate than the former. d2(γ) = n−1∑ i=1 (D(x∗i , γ)) 2 (27) On the other hand, we can combine the expressions using one-norm formula, as in equation (28). d3(γ) = n−1∑ i=1 |D(x∗i , γ)| (28) This is done to make comparisons between results of d1(γ), d2(γ), and d3(γ) in terms of computational time and accuracy. d3(γ) is significantly more simplified that the other A. S. Heilat et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1283-1294 1289 two. By using Mathematica, we can find the value of γ depends on equations (26), (27), and (28) as follows: d1(γ) = √√√√n−1∑ i=1 (D(x∗i ), γ) 2 = 0 d2(γ) = n−1∑ i=1 (D(x∗i , γ)) 2 = 0 d3(γ) = n−1∑ i=1 |D(x∗i , γ)| = 0 Finally, the values of γ and Ci can be obtained for i = 1, 2, ..., n− 1. Thus, the solutions for each knot, xi, can be obtained from Equation (1). 5. Numerical Examples and Discussions HCBSM was applied on Examples 1-4 with n = 10 and the results are compared with the analytical solutions. Norms of the results are found numerically by the following formulas: L∞ = n−1 max i=1 | S(xi)− u(xi) |, L2 = √√√√n−1∑ i=1 [S(xi)− u(xi)]2 Example 1. [15]{ u′′(x)− u′(x) = −ex−1 − 1, 0 ≤ x ≤ 1 u(0) = 0, u(1) = 0 (29) The analytical solution is u(x) = x(1− ex−1). The results for examples 1, 2, 3, and 4 are shown in Tables 2, 3, 4, and 5 and Figures 1, 2, 3, and 4, respectively. We found that our method produced much more accurate results than the CBSM [15], TCBSM [16], and ECBSM [23]. Table 2: Absolute errors and norms of CBSM [15], TCBSM [16], ECBSM [23], and HCBSM for Example 1 with n = 10 x CBSM TCBSM ECBSM HCBSM (λ = 2.9375E − 03) (γ = 1.73125) 0.1 7.531E − 05 1.756E − 04 2.996E − 06 1.568E − 07 0.2 1.439E − 04 3.369E − 04 4.896E − 06 2.264E − 07 0.3 2.031E − 04 4.772E − 04 5.739E − 06 2.062E − 07 0.4 2.499E − 04 5.890E − 04 5.611E − 06 1.022E − 07 0.5 2.803E − 04 6.633E − 04 4.668E − 06 6.743E − 07 0.6 2.900E − 04 6.890E − 04 3.144E − 06 2.696E − 07 0.7 2.736E − 04 6.528E − 04 1.375E − 06 4.527E − 07 0.8 2.249E − 04 5.390E − 04 1.776E − 07 5.425E − 07 0.9 1.366E − 04 3.288E − 04 9.013E − 07 4.369E − 07 L∞ 2.900E − 04 6.890E − 04 5.739E − 06 5.425E − 07 L2 6.609E − 04 1.568E − 03 1.148E − 05 9.466E − 07 A. S. Heilat et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1283-1294 1290 Figure 1: Numerical solution S(x) and exact solution u(x) for Example 1 using HCBSM with n = 10 and γ = 1.73125 Example 2. [15]{ u′′(x) + (x+ 1)u′(x)− 2u(x) = (1− x2)e−x, 0 ≤ x ≤ 1 u(0) = −1, u(1) = 0 (30) The analytical solution is u(x) = (x− 1)e−x. Table 3: Absolute errors and norms of CBSM [15], TCBSM [16], ECBSM [23], and HCBSM for Example 2 with n = 10 x CBSM TCBSM ECBSM HCBSM (λ = 2.9375E − 03) (γ = 1.6775) 0.1 1.161E − 04 2.922E − 04 1.004E − 06 3.618E − 07 0.2 1.872E − 04 4.686E − 04 3.920E − 07 4.115E − 07 0.3 2.229E − 04 5.547E − 04 1.030E − 06 2.802E − 07 0.4 2.311E − 04 5.719E − 04 2.655E − 06 6.830E − 07 0.5 2.185E − 04 5.376E − 04 4.038E − 06 1.495E − 07 0.6 1.906E − 04 4.661E − 04 4.875E − 06 3.199E − 07 0.7 1.152E − 04 3.687E − 04 4.969E − 06 4.071E − 07 0.8 1.053E − 04 2.543E − 04 4.208E − 06 3.890E − 07 0.9 5.406E − 05 1.297E − 04 2.551E − 06 2.544E − 07 L∞ 2.311E − 04 5.719E − 04 4.969E − 06 4.071E − 07 L2 5.222E − 04 1.290E − 03 9.912E − 06 9.433E − 07 Figure 2: Numerical solution S(x) and exact solution u(x) for Example 2 using HCBSM with n = 10 and γ = 1.6775 A. S. Heilat et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1283-1294 1291 Example 3. [2]{ u′′(x)− π2u(x) = −2π2 sin(πx), 0 ≤ x ≤ 1 u(0) = 0, u(1) = 0 (31) The analytical solution is u(x) = sin(πx). Table 4: Absolute errors and norms of CBSM [15], TCBSM [16], ECBSM [23], and HCBSM for Example 3 with n = 10 x CBSM TCBSM ECBSM HCBSM (λ = −1.6500E − 02) (γ = −3.0919) 0.1 1.270E − 03 9.568E − 04 2.769E − 07 5.184E − 09 0.2 2.415E − 03 1.820E − 03 5.267E − 07 9.861E − 09 0.3 3.324E − 03 2.505E − 03 7.249E − 07 1.357E − 08 0.4 3.908E − 03 2.945E − 03 8.522E − 07 1.596E − 08 0.5 4.109E − 03 3.096E − 03 8.960E − 07 1.678E − 08 0.6 3.908E − 03 2.945E − 03 8.522E − 07 1.596E − 08 0.7 3.324E − 03 2.505E − 03 7.249E − 07 1.357E − 08 0.8 2.415E − 03 1.820E − 03 5.267E − 07 9.862E − 09 0.9 1.270E − 03 9.568E − 04 2.769E − 07 5.182E − 09 L∞ 4.109E − 03 3.096E − 03 8.960E − 07 1.678E − 08 L2 9.188E − 03 6.923E − 03 2.004E − 06 3.751E − 08 Figure 3: Numerical solution S(x) and exact solution u(x) for Example 3 using HCBSM with n = 10 and γ = −3.0919 Example 4. [16]{ u′′(x)− u(x) = 0, 0 ≤ x ≤ 1 u(0) = 0, u(1) = sinh(1) (32) The analytical solution is u(x) = sinh(x). REFERENCES 1292 Table 5: Absolute errors and norms of CBSM [15], TCBSM [16], ECBSM [23], and HCBSM for Example 4 with n = 10 x CBSM TCBSM ECBSM HCBSM (λ = 1.6875E − 03) (γ = 1.32475) 0.1 1.294E − 05 5.269E − 05 1.660E − 07 7.773E − 10 0.2 2.518E − 05 1.025E − 04 3.230E − 07 1.512E − 09 0.3 3.598E − 05 1.465E − 04 4.615E − 07 2.161E − 09 0.4 4.460E − 05 1.816E − 04 5.721E − 07 2.679E − 09 0.5 5.023E − 05 2.045E − 04 6.444E − 07 3.017E − 09 0.6 5.201E − 05 2.118E − 04 6.672E − 07 3.124E − 09 0.7 4.899E − 05 1.995E − 04 6.284E − 07 2.942E − 09 0.8 4.012E − 05 1.634E − 04 5.146E − 07 2.409E − 09 0.9 2.423E − 05 9.866E − 05 3.107E − 07 1.455E − 09 L∞ 5.201E − 05 2.118E − 04 6.672E − 07 3.124E − 09 L2 1.179E − 04 4.802E − 04 1.513E − 06 7.083E − 09 Figure 4: Numerical solution S(x) and exact solution u(x) for Example 4 using HCBSM with n = 10 and γ = 1.32475 6. 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