An Implementation of Haar Wavelet Based Method for Numerical

IEEE/CAA JOURNAL OF AUTOMATICA SINICA, VOL. 6, NO. 1, JANUARY 2019 177

An Implementation of Haar Wavelet Based Method for Numerical Treatment of Time-fractional Schrodinger¨ and Coupled Schrodinger¨ Systems Najeeb Alam Khan and Tooba Hameed

Abstract—The objective of this paper is to solve the time- equations, such as the variational iteration method [3], differ- fractional Schrodinger¨ and coupled Schrodinger¨ differential ential transform method [4], homotopy analysis method [5], equations (TFSE) with appropriate initial conditions by using the Jacobi spectral tau and collocation method [6]−[8], Laplace Haar wavelet approximation. For the most part, this endeavor is made to enlarge the pertinence of the Haar wavelet method transform method [9], homotopy perturbation method [10], to solve a coupled system of time-fractional partial differential Adomian decomposition method [11], [12], high-order finite equations. As a general rule, piecewise constant approximation of element methods [13] and many others [14]−[19]. The scope a function at different resolutions is presentational characteristic and distinct aspects of fractional calculus have been written of Haar wavelet method through which it converts the differential by many authors in [20]−[22]. equation into the Sylvester equation that can be further simplified easily. Study of the TFSE is theoretical and experimental research In the past few years, there has been an extensive attraction and it also helps in the development of automation science, in employing the spectral method [23]−[25] for numerically physics, and engineering as well. Illustratively, several test prob- solving the copious type of differential and integral equations. lems are discussed to draw an effective conclusion, supported The spectral methods have an exponential quota of conver- by the graphical and tabulated results of included examples, to gence and high level of efficiency. Spectral methods are to reveal the proficiency and adaptability of the method. express the approximate solution of the problem in term of Index Terms—Fractional calculus, Haar wavelets, operational a finite sum of certain basis functions and then selection of matrix, wavelets. coefficients in order to reduce the difference between the exact and approximate solutions as much as possible. The spectral collocation method is a distinct type of spectral methods, I.INTRODUCTION that is more relevant and extensively used to solve most of N recent decades, fractional calculus (calculus of integrals differential equations [26]. I and derivatives of any arbitrary real order) has attained For the reason of the distinctive attributes of wavelet theory appreciable fame and importance due to its manifest uses in in representing continuous functions in the form of discontin- apparently diverse and outspread fields of science. Certainly, uous functions [27], its applications as a mathematical tool is it provides potentially helpful tools for solving integral and widely expanding nowadays. Besides image processing and differential equations and many other problems of mathemati- signal decomposition it is also used to assess many other cal physics. The fractional differential equations have become mathematical problems, such as differential and integral equa- crucial research field essentially due to their immense range of tions. Wavelets comprise the incremental conception between utilization in engineering, fluid mechanics, physics, chemistry, two consecutive levels of resolution, called multi-resolution. biology, viscoelasticity, etc. Numerous mathematicians and The first component of multi-resolution analysis is vector physicists have been studying the properties of fractional cal- spaces. For each vector space, another vector space of higher culus [1], [2] and have established several methods for accurate resolution is found and this continues until the final image or analytical and numerical solutions of fractional differential signal is executed. The basis of each of these vector spaces acts as the scaling function for the wavelets. Each vector space Manuscript received November 30, 2015; revised February 9, 2016; ac- having an orthogonal component and a basis function is said cepted March 3, 2016. Recommended by Associate Editor Antonio Visioli. to be the wavelet [28]. (Corresponding author: Najeeb Alam Khan.) Citation: N. A. Khan and T. Hameed, “An implementation of Haar wavelet Up till now, a number of wavelet families have been based method for numerical treatment of time-fractional Schrodinger¨ and presented by different authors, but among all Haar wavelet coupled Schrodinger¨ systems,” IEEE/CAA J. Autom. Sinica, vol. 6, no. 1, are considered to be the easiest wavelets family. Haar wavelet pp. 177−187, Jan. 2019. N. A. Khan and T. Hameed are with the Department of Mathematics, was introduced in 1910 by Hungarian mathematician Alfred University of Karachi, Karachi-75270, Pakistan (e-mail: [email protected]; Haar. These wavelets are obtained from Daubechies wavelets [email protected]). of order 1, which consist of piecewise constant functions on Color versions of one or more of the figures in this paper are available online at http://ieeexplore.ieee.org. the real axis that can take only three values, −1, 0 and 1. Here Digital Object Identifier 10.1109/JAS.2016.7510193 we are using collocation method, by increasing the level of 178 IEEE/CAA JOURNAL OF AUTOMATICA SINICA, VOL. 6, NO. 1, JANUARY 2019

( Γ(α+1) tα−ν , m > α ≥ m − 1 resolution, collocation points are also increasing and level of ν Γ(α−ν+1) Dt f(t) = accuracy too. Haar wavelet collocation method is extensively 0, if α ∈ {0, 1, 2, . . . , m − 1}. used due to its constructive ability of being smooth, fast, con- (3) venient and being computationally attractive [29]. In addition, it has the competency to reduce the computations for solving In the following, some main computational properties and differential equations by converting them into some system relations of fractional integral and differential operators are of algebraic equations. The main advantages of the proposed defined as α β α+β β α algorithm are, its simple application and no residual or product 1) It It f(t) = It f(t) = It It f(t) operational matrix is required. The method is well addressed ∂β 2) Iαf(x, t) = Iα−βf(x, t) in [22], [29]−[32]. ∂tβ t t The time-fractional Schrodinger¨ equation (T-FSE) differs n−1 ∂α X tk ∂kf(x, t) | from the standard Schrodinger¨ equation. The first-order time 3) f(x, t) = f(x, t) − t=0 ∂tα k! ∂tk derivative is replaced by a fractional derivative, it makes k=0 the problem overall in time. It describes, how the quantum nX−1 k state (physical situation) of a quantum system changes with = f(x, t) + ζk(x)t (4) time, soliton dispersion, deep water waves, molecular orbital k=0 k theory and the potential energy of a hydrogen-like atom 1 ∂ f(x,t)|t=0 where ζk(x) = − k! ∂tk . For more details see [1]. (fractional “Bohr atom”). The aim of this work is to explore the numerical solutions of the time-fractional Schrodinger¨ B. Haar Wavelets and Function Approximation equations by using Haar wavelet method. Due to the large Basis of Haar wavelets is obtained with a multi-resolution number of applications of the Schrodinger¨ equation in different of piecewise constant functions. Let the interval x ∈ [0, 1) be aspects of quantum mechanics and engineering, many attempts divided into 2m subintervals of equal length, where m = 2j have been exercised on analytical and numerical methods to and J is maximal level of resolution. Next, two parameters are calculate the approximate solution of T-FSE. Some of them introduced, j = 0, 1, 2,...,J and k = 0, 1, 2, . . . , m−1, such are studied [6], [10], [18], [33]−[36], and enumerated here for that the wavelet number i satisfies the relation i = k + m + 1. better perception of the presented analysis. Also, the existence The ith Haar wavelet can be determined as and uniqueness of solutions of fractional Schrodinger¨ equation  1, x ∈ [ϑ , ϑ ) have been proved by multiple authors [35], [37], [38].  1 2 hi(x) = −1, x ∈ [ϑ2, ϑ3) (5)  0, elsewhere II.PRELIMINARIES where ϑ = k/m, ϑ = (k + 0.5)/m, ϑ = (k + 1)/m. In this section, some notations and properties of fractional 1 2 3 For the case i = 1, corresponding scaling function can be calculus, the basis of Haar function approximation for partial defined as: differential equation and solution of Haar by multi-resolution ( analysis are given that will help us in exploring the main theme 1, x ∈ [ϑ1, ϑ3) h1 = (6) of the paper. 0, elsewhere. Here, we consider the wavelet-collocation method, therefore A. Riemann-Liouville Differential and Integral Operator collocation points are generated by using, m l − 0.5 Assume ν > 0, m = dνe and f(x, t) ∈ C ([0, 1] × [0, 1]) xl = , l = 1, 2, 3,..., 2m. (7) then the partial Caputo fractional derivative of f(x, t) with 2m respect to t is defined as The Haar system forms an orthonormal basis for the Hilbert space f(t) ∈ L2(0, 1). We may consider the inner product ( m ν m−ν ∂ expansion of f(t) ∈ L ([0, 1)) in Haar series [31] as ∂ It ∂tm f(x, t) 2 f(x, t) = m (1) ∂tν ∂ f(x, t) JX−1 2Xj −1 ∂tm T f(t) ≈ hf, ϕiϕ(t) + hf, hj,kihj,k(t) = C H(t) ν where It is the Riemann-Liouville fractional integral, given j=0 i=0 as (8) Z t J ν 1 ν−1 where C is 1 × 2 coefficient vector and H(t) = [h0(t), It f(t) = (t − ϕ) f(ϕ)dϕ T Γ(ν) 0 h1(t), . . . , hm−1(t)] . Also, a function of two variables can 0 be expanded by Haar wavelets [32] as It f(t) = f(t) (2) mX−1 mX−1 ν ∂ν T u(x, t) ≈ ui,jhi(x)hj(t) = H (x) × U × H(t) we use the notation Dt in replacement of ∂tν for the Caputo fractional derivative. The Caputo fractional derivative of order i=0 j=0 ν > 0 for f(t) = tα is given as (9) KHAN AND HAMEED: AN IMPLEMENTATION OF HAAR WAVELET BASED METHOD FOR NUMERICAL TREATMENT 179 where U is 2J × 2J coefficient matrix calculated by the inner C. Multi Resolution Analysis (MRA) product u = hh (x), hu(x, t), h ii. The operational matrix i,j i j Any space V can be constructed using a basis function of fractional integration of Haar function is needed to solve h(2mt) as partial differential equation (PDE) of fractional order. A more rigorous derivation for the generalized block pulse operational m Vm = span{h(2 t − n)}n,m∈Z matrices is proposed in [39]. The block pulse function forms a complete set of orthogonal functions which is defined in h(t) is called scaling function, also known as “Father func- interval [a, b) as tion”. The chain of subspaces ...,V ,V ,V ,V ,V ,... ( −2 −1 0 1 2 i−1 i with the following axioms is called multi-resolution analysis 1, m b ≤ t < m b ψi(t) = (10) (MRA) [28]. 0, elsewhere ½[ ¾ for i = 1, 2, . . . , m. It is known that any absolutely integrable 1) Vm = L2(R); function f(t) on [a, b), can be expanded in block pulse n\ o m∈Z functions as 2) Vm = {0}; m∈Z ∼ T f(t) = F ψ(m)(t) (11) 3) There exists h(t) such that, V0 = span{h(t − n)}n∈Z ; so that the mean square error of approximation is minimized. 4) {h(t − n)}n∈Z is an orthogonal set; T T −m Here, F = [f1, f2, f3, . . . , fm] and ψ(m)(t) = [ψ1(t), ψ2(t), 5) If f(t) ∈ Vm then f(2 t) ∈ V0, ∀m ∈ Z); ψ (t), . . . , ψ (t)], where 3 m 6) If f(t) ∈ V then f(t − n) ∈ V , ∀n ∈ Z). (19) Z Z 0 0 m b m (i/m)b fi = f(t)ψi(t)dt = f(t)ψi(t)dt. (12) b a b (i−1)b/m Under the given axioms, there exists a ψ(·) ∈ L2(R), such that {ψ(2mt − n)} spans L (R). The wavelet function The Riemann-Liouville fractional integral is simplified and m,n∈Z 2 ψ(·) is also called “Mother wavelet”. expanded in block pulse functions to yield the generalized block pulse operational matrix F ν as ν ν (I ψm)(t) = F ψm(t) (13) D. Convergence of the Method 3 3 where Let ∂ u(x,t) and ∂ v(x,t) are continuous and bounded func-   ∂t∂x2 ∂t∂x2 1 ξ2 ξ3 . . . ξm tions on (0, 1) × (0, 1), then ∃M1,M2 > 0, ∀x, t ∈ (0, 1) ×   (0, 1)  0 1 ξ2 . . . ξm−1  ν b ν 1   ¯ ¯ ¯ ¯ F = ( )  0 0 1 . . . ξm−2  3 3   ¯∂ u(x, t)¯ ¯∂ v(x, t)¯ m Γ(ν + 2) ...... ¯ ¯ ≤ M and ¯ ¯ ≤ M . (20)  . . . . .  ¯ ∂t∂x2 ¯ 1 ¯ ∂t∂x2 ¯ 2 0 0 0 0 1 Let u (x, t) and v (x, t) are the following approximations with ξ = 1, ξ = pν+1 − 2(p − 1)ν+1 + (p − 2)ν+1 (p = m m 1 p of u(x, t) and v(x, t), 2, 3, 4, . . . , m − i + 1). For further details see [39]. The Haar functions are piecewise constant, so it may be expanded into mX−1 mX−1 an m-term block pulse functions (BPF) as um(x, t) ≈ uijhi(x)hj(t) and i=0 j=0 H (t) = H ψ (t). (14) m m×m m mX−1 mX−1 In [31] Haar wavelets operational matrix of fractional order vm(x, t) ≈ vijhi(x)hj(t) (21) integration is derived by i=0 j=0 ν ν (I Hm)(t) ≈ P Hm(t) (15) then we have where P ν is m × m order Haar wavelets operational matrix X∞ X∞ of fractional order integration. Substituting (2) in (14) we get u(x, t) − um(x, t) = uijhi(x)hj(t) i=m j=m ν ν ν ∞ ∞ (I Hm)(t) ≈ (I Hm×mψm)(t) = Hm×m(I ψm)(t) X X ν = uijhi(x)hj(t) and ≈ Hm×mF ψm(t). (16) i=2p+1 j=2p+1 From (15) and (16), it can be written as X∞ X∞ ν ν v(x, t) − vm(x, t) = vijhi(x)hj(t) P Hm(t) = Hm×mF . (17) i=m j=m Therefore, P ν can be obtained as X∞ X∞ = vijhi(x)hj(t). (22) ν ν −1 P = H × F × H . (18) i=2p+1 j=2p+1 180 IEEE/CAA JOURNAL OF AUTOMATICA SINICA, VOL. 6, NO. 1, JANUARY 2019

Theorem 1: Let the functions um(x, t) and vm(x, t) ob- B. The Time-fractional Coupled Schrodinger¨ System tained by using Haar wavelets are the approximation of u(x, t) The time-fractional coupled Schrodinger¨ system (T-FCSS) and v(x, t) then we have the errors bounded as following has the following form M1 ku(x, t) − um(x, t)kE ≤ √ and ν 2 2 3m3 ∂ φ(x, t) ∂ φ(x, t) ∂ φ(x, t) 2 2 i ν + i 2 + 2 + λ1(|φ| + |ψ| ) M2 ∂t ∂x ∂x kv(x, t) − vm(x, t)kE ≤ √ (23) × φ(x, t) + α (x)φ(x, t) + β ψ(x, t) − f (x, t) = 0 3m3 1 1 1 ∂ν ψ(x, t) ∂2ψ(x, t) ∂2ψ(x, t) where i + i + + λ (|φ|2 + |ψ|2) ∂tν ∂x2 ∂x2 2 µZ 1 Z 1 ¶1/2 2 × ψ(x, t) + α2(x)φ(x, t) + β2ψ(x, t) − f2(x, t) = 0 ku(x, t)kE = u (x, t)dxdt and 0 0 (29) µZ 1 Z 1 ¶1/2 2 with initial conditions kv(x, t)kE = v (x, t)dxdt . (24) 0 0 0 See proof in [32]. φ(x, 0) = f3(x), φ(0, t) = g3(t), φ (0, t) = h3(t) 0 ψ(x, 0) = f4(x), ψ(0, t) = g4(t), ψ (0, t) = h4(t) (30) III.THE SCHRODINGER¨ EQUATIONS where 0 < ν ≤ 1, λ1, λ2, α1, α2, β1 and β2 are real constants A. The Time-fractional Schrodinger¨ Equation and φ(x, t), ψ(x, t), f3(x), f4(t), g3(t), g4(t), h3(t), h4(t), The time-fractional Schrodinger¨ equation (T-FSE) has the q1(x, t) and q2(x, t) are complex functions. We can express following form complex functions into their respective real and imaginary ν 2 parts as ∂ φ(x, t) ∂ φ(x, t) 2 i ν + λ 2 + η|φ| φ ∂t ∂x φ(x, t) = u(x, t) + iv(x, t), ψ(x, t) = r(x, t) + is(x, t) + µ(x)φ = q(x, t), 0 < x, t ≤ 1 (25) f3(x) = f5(x) + if6(x), f4(x) = f7(x) + if8(x) with initial conditions φ(x, 0) = f(x), φ(0, t) = g(t), φ0(0, t) g3(t) = g5(t) + ig6(t), g4(t) = g7(t) + ig8(t) = h(t), where 0 < ν ≤ 1, λ and η are real constants, µ(x) is h (t) = h (t) + ih (t), h (t) = h (t) + ih (t) the trapping potential and φ(x, t), f(x), g(t), h(t) and q(x, t) 3 5 6 4 7 8 are complex functions. We can express complex functions q3(x, t) = q5(x, t) + iq6(x, t), q4(x, t) = q7(x, t) + iq8(x, t) φ(x, t), f(x), g(t), h(t) and q(x, t) into their respective real (31) and imaginary parts as Substituting (31) into (29) and equating real and imaginary φ(x, t) = u(x, t) + iv(x, t) parts we get the system of two coupled time-fractional nonlinear partial differential equations. µ(x) = µ1(x) + iµ2 f(x) = f (x) + if (x) ν 2 1 2 ∂ v(x, t) ∂v(x, t) ∂ u(x, t) 2 2 2 2 − ν − + 2 + λ1(u + v + r + s ) g(t) = g1(t) + ig2(t) ∂t ∂x ∂x × u(x, t) + α1u(x, t) + β1r(x, t) − q3(x, t) = 0 h(t) = h1(t) + ih2(t) ∂ν u(x, t) ∂u(x, t) ∂2v(x, t) q(x, t) = q1(x, t) + iq2(x, t) (26) + + + λ (u2 + v2 + r2 + s2) ∂tν ∂x ∂x2 1 Substituting (26) into (25) and collecting real and imaginary × v(x, t) + α1v(x, t) + β1s(x, t) − q4(x, t) = 0 parts, then (25) can be written as coupled time-fractional ∂ν s(x, t) ∂s(x, t) ∂2r(x, t) nonlinear partial differential equations as − + + + λ (u2 + v2 + r2 + s2) ∂tν ∂x ∂x2 2 ν 2 ∂ v(x, t) ∂ u(x, t) 2 2 × r(x, t) + α2u(x, t) + β2r(x, t) − q5(x, t) = 0 − ν + λ 2 + η(u + v )u + µ1(x) ∂t ∂x ∂ν r(x, t) ∂r(x, t) ∂2s(x, t) × u(x, t) − q (x, t) = 0 − + + λ (u2 + v2 + r2 + s2) 1 ∂tν ∂x ∂x2 2 ∂ν u(x, t) ∂2v(x, t) + λ + η(u2 + v2)v + µ (x) × s(x, t) + α2v(x, t) + β2s(x, t) − q6(x, t) = 0 (32) ∂tν ∂x2 2 × v(x, t) − q2(x, t) = 0 (27) IV. THE PROPOSED METHOD with initial conditions A. For Time-fractional Schrodinger¨ Equation u(x, 0) = f1(x), v(x, 0) = f2(x) Any arbitrary function u(x, t) ∈ L ([0, 1) × [0, 1)) and u(0, t) = g (t), v(0, t) = g (t) 2 1 2 v(x, t) ∈ L ([0, 1) × [0, 1)), can be expanded into Haar series 0 0 2 u (0, t) = h1(t), v (0, t) = h2(t) (28) [32] as KHAN AND HAMEED: AN IMPLEMENTATION OF HAAR WAVELET BASED METHOD FOR NUMERICAL TREATMENT 181

X2m X2m 00 B. For Time-fractional Coupled Schrodinger¨ System u˙ (x, t) = uijhi(x)hj(t) i=1 j=1 Now, we approximate u(x, t), v(x, t), r(x, t) and s(x, t), X2m X2m by the Haar series [32] as 00 v˙ (x, t) = vijhi(x)hj(t) (33) X2m X2m 00 i=1 j=1 u˙ (x, t) = uijhi(x)hj(t) i=1 j=1 where 2M × 2M Haar coefﬁcient matrix of uij and vij in X2m X2m 00 (33) can be written as v˙ (x, t) = vijhi(x)hj(t) i=1 j=1 U˙ 00 = HT (x) × U × H(t) X2m X2m r˙00(x, t) = r h (x)h (t) V˙ 00 = HT (x) × V × H(t). (34) ij i j i=1 j=1 X2m X2m 00 Let dots and primes in (33) represent differentiation with s˙ (x, t) = sijhi(x)hj(t). (39) respect to t and x, respectively. By integrating (34) with i=1 j=1 respect to t from 0 to t, we get Matrix form of (39) can be written as

00 T U00 = HT (x) × U × P 1 × H(t) + u00(x, 0) U˙ = H (x) × U × H(t) 00 T V00 = HT (x) × V × P 1 × H(t) + v00(x, 0) (35) V˙ = H (x) × V × H(t) R˙ 00 = HT (x) × R × H(t) on integrating (35) twice with respect to x from 0 to x, then S˙ 00 = HT (x) × S × H(t). (40) we get Let dots and primes in (40) represent differentiation with U0 = HT (x) × [P 1]T × U × P 1 × H(t) + u0(x, 0) respect to t and x, respectively. By integrating (40) with respect to t from 0 to t, we get − u0(0, 0) + u0(0, t) 00 T 1 00 V0 = HT (x) × [P 1]T × V × P 1 × H(t) + v0(x, 0) U = H (x) × U × P × H(t) + u (x, 0) 00 T 1 00 − v0(0, 0) + v0(0, t) (36) V = H (x) × V × P × H(t) + v (x, 0) R00 = HT (x) × R × P 1 × H(t) + r00(x, 0) and then S00 = HT (x) × S × P 1 × H(t) + s00(x, 0) (41)

U = HT (x) × [P 2]T × U × P 1 × H(t) + u(x, 0) on integrating (41) twice with respect to x from 0 to x, once we get − u(0, 0) − xu0(0, 0) + xu0(0, t) + u(0, t) 0 T 1 T 1 0 V = HT (x) × [P 2]T × V × P 1 × H(t) + v(x, 0) U = H (x) × [P ] × U × P × H(t) + u (x, 0) 0 0 − v(0, 0) − xv0(0, 0) + xv0(0, t) + v(0, t). (37) − u (0, 0) + u (0, t) V0 = HT (x) × [P 1]T × V × P 1 × H(t) + v0(x, 0) ν 0 0 Applying differential operator Dt on both sides of (37) and − v (0, 0) + v (0, t) using property 2) of (4) R0 = HT (x) × [P 1]T .R.P 1 × H(t) + r0(x, 0) − r0(0, 0) + r0(0, t) Dν U = HT (x) × [P 2]T × U × P 1−ν × H(t) t S0 = HT (x) × [P 1]T .S.P 1 × H(t) + s0(x, 0) + xDν u0(0, t) + Dν u(0, t) t t − s0(0, 0) + s0(0, t) (42) ν T 2 T 1−ν Dt V = H (x) × [P ] × V × P × H(t) ν 0 ν and + xDt v (0, t) + Dt v(0, t). (38) U = HT (x) × [P 2]T × U × P 1 × H(t) + u(x, 0) Substitution of (35), (37) and (38) into (27) may lead to − u(0, 0) − xu0(0, 0) + xu0(0, t) + u(0, t) coupled system of time-fractional differential equations. This V = HT (x) × [P 2]T × V × P 1 × H(t) + v(x, 0) system will have some unknown functions u00(x, 0), u0(x, 0), 0 0 0 ν 0 ν 00 0 − v(0, 0) − xv (0, 0) + xv (0, t) + v(0, t) u (0, 0), u(0, 0), Dt u (0, t), Dt u(0, t), v (x, 0), v (x, 0), 0 ν 0 ν T 2 T 1 v (0, 0), v(0, 0), Dt v (0, t), and Dt v(0, t). With the help R = H (x) × [P ] × R × P × H(t) + r(x, 0) of initial conditions all these functions are calculated. We − r(0, 0) − xr0(0, 0) + xr0(0, t) + r(0, t) solve (27) for unknown Haar coefﬁcients by using collocation S = HT (x) × [P 2]T × S × P 1 × H(t) + s(x, 0) method. Finally, for Haar solution of T-FSE, we substitute 0 0 values of Haar coefﬁcients in (37). − s(0, 0) − xs (0, 0) + xs (0, t) + s(0, t). (43) 182 IEEE/CAA JOURNAL OF AUTOMATICA SINICA, VOL. 6, NO. 1, JANUARY 2019

2−ν ν T 1 Γ(ν + 1)t Applying differential operator Dt on both sides of (43) and + H (x) × V × P × H(t) + Γ(2 − ν + 1) using property 2) of (4) µ ¶ 2t2−ν cos x Dν U = HT (x) × [P 2]T × U × P 1−ν × H(t) − t2 sin x + = 0. (49) t Γ(3 − ν) + xDν u0(0, t) + Dν u(0, t) t t Towards the approximate solution, we first collocate (49) at ν T 2 T 1−ν Dt V = H (x) × [P ] × V × P × H(t) points ν 0 ν + xDt v (0, t) + Dt v(0, t) i − 0.5 j − 0.5 ν T 2 T 1−ν xi = , tj = . (50) Dt R = H (x) × [P ] × R × P × H(t) 2m 2m ν 0 ν + xDt r (0, t) + Dt r(0, t) Equation (49) ⇒ ν T 2 T 1−ν D S = H (x) × [P ] × S × P × H(t) T 2 T 1−ν t − H (xi) × [P ] × V × P × H(tj) ν 0 ν + xDt s (0, t) + Dt s(0, t). (44) + HT (x ) × U × P 1 × H(t ) i Ã j ! Substitution of (41)−(44) into (32) may lead to the two 2−ν 2−ν Γ(ν + 1)xitj 2 2tj sin xi coupled systems of time-fractional differential equations. This + − t cos xi + = 0 Γ(2 − ν + 1) j Γ(3 − ν) system has some unknown functions. With the help of initial conditions all these functions are calculated. We solve system T 2 T 1−ν H (xi) × [P ] × U × P × H(tj) of (32) for unknown Haar coefficients by using collocation + HT (x ) × V × P 1 × H(t ) method. Finally, for Haar solution of time-fractional coupled i Ã j ! 2−ν 2−ν Schrodinger¨ system (T-FCSS), we substitute values of Haar Γ(ν + 1)t 2t cos xi + j − t2 sin x + j = 0. coefficients in (43). Γ(2 − ν + 1) j i Γ(3 − ν) (51) V. NUMERICAL PROBLEMS In this section, four test problems are taken to test the Equation (51) generates two systems of 2M algebraic efficiency and accuracy of the proposed scheme. The com- equations of Haar coefficients. The values of Haar coefficients putations associated with the problems were executed using are obtained from system of (51) by using Newton’s iterative Mathematica 10. method. With the help of these coefficients, Haar solutions Problem 1: Consider the linear T-FSE which is also found are attained from (37), see Fig. 1. Comparison of the Haar in [6] with solutions by Homotopy analysis method in [5] is shown in Table I with different values of ν. λ = 1, η = 0, µ(x) = 0 and µ ¶ 2it2−ν q(x, t) = − t2 eix. (45) Γ(3 − ν) Subjected to initial conditions φ(x, 0) = 0, φ(0, t) = t2, φ0(0, t) = it2 (46) the exact solution for ν = 1 is φ(x, t) = t2eix. (47) Substitute (45) in (25) the determined coupled system of equations is µ ¶ ∂ν v(x, t) ∂2u(x, t) 2t2−ν sin x − + − t2 cos x + = 0 ∂tν ∂x2 Γ(3 − ν) Fig. 1. Haar solutions of Problem 1 at ν = 0.1, 0.3, 0.5. µ ¶ ∂ν u(x, t) ∂2v(x, t) 2t2−ν cos x 2 Problem 2: Consider a nonlinear cubic form of T-FSE [6] ν + 2 − t sin x − = 0. ∂t ∂x Γ(3 − ν) with (48) λ = 1, η = 1, µ(x) = 0 and By using the method given in Section IV-A, (48) can be µ ¶ written as 2t2−ν q(x, t) = − + (−4π2t2 + t6)i e−2πix (52) − HT (x) × [P 2]T × V × P 1−ν × H(t) Γ(3 − ν) Γ(ν + 1)xt2−ν and initial conditions + HT (x) × U × P 1 × H(t) + Γ(2 − ν + 1) φ(x, 0) = 0, φ(0, t) = it2, φ0(0, t) = 2πt2 (53) µ 2−ν ¶ 2 2t sin x − t cos x + = 0 with exact solution for ν = 1 is Γ(3 − ν) HT (x) × [P 2]T × U × P 1−ν × H(t) φ(x, t) = t2ie−2πix. (54) KHAN AND HAMEED: AN IMPLEMENTATION OF HAAR WAVELET BASED METHOD FOR NUMERICAL TREATMENT 183

TABLE I For approximate solution of (56), putting collocation points COMPARISON BETWEEN HAAR SOLUTIONS (J = 1, m = 4) AND of (50) in above equations generates two systems of 2M non HAM [5] OF PROBLEM 1 linear algebraic equations of Haar coefficients. The values ν = 0.1 ν = 0.3 ν = 0.5 of Haar coefficients are obtained by using Newton’s iterative t x HWCM HAM [5] HWCM HAM [5] HWCM HAM [5] method and then with the help of these coefficients Haar solu- 0.125 0.015584 0.015729 0.015588 0.018166 0.015602 0.032192 tions are attained from (37). The Haar solutions comparisons 0.375 0.140258 0.140320 0.140277 0.126725 0.140322 0.18097 with the method in [5] for the different values of ν is shown 0.125 0.625 0.389603 0.391911 0.389639 0.331403 0.389698 0.417840 in Table II. 0.875 0.763621 0.773331 0.763678 0.642092 0.763759 0.735157 TABLE II 0.125 0.015639 0.014928 0.015909 0.015876 0.016599 0.029901 COMPARISON BETWEEN HAAR SOLUTIONS (J = 1, m = 4) AND 0.375 0.140697 0.137242 0.142257 0.117591 0.144873 0.171424 0.375 HAM [5] OF PROBLEM 2 0.625 0.390743 0.387923 0.394153 0.321101 0.398660 0.407056 0.875 0.765763 0.770778 0.771532 0.639027 0.778130 0.736778 ν = 0.1 ν = 0.5 ν = 0.9 t x HWCM HAM [5] HWCM HAM [5] HWCM HAM [5] 0.125 0.015881 0.014367 0.017251 0.013413 0.020637 0.026391 0.125 0.013539 0.013276 0.013578 0.006316 0.013768 0.002593 0.375 0.142657 0.135328 0.150692 0.110280 0.164111 0.159805 0.625 0.375 0.121912 0.132538 0.121985 0.097840 0.122023 0.062856 0.625 0.395831 0.385461 0.413348 0.315020 0.436643 0.396412 0.125 0.625 0.339272 0.385008 0.339565 0.347687 0.340151 0.276693 0.875 0.775363 0.768819 0.805018 0.639973 0.839202 0.740576 0.875 0.668433 0.770812 0.668861 0.793325 0.669170 0.730876 0.125 0.016431 0.014212 0.020145 0.011454 0.028487 0.022180 0.125 0.010136 0.013277 0.012634 0.006316 0.020531 0.002593 0.375 0.147236 0.135101 0.169643 0.107096 0.205320 0.148820 0.875 0.375 0.087449 0.132538 0.091611 0.097839 0.097166 0.062855 0.625 0.407697 0.385159 0.456638 0.314912 0.519962 0.388621 0.375 0.625 0.235941 0.385008 0.240394 0.347687 0.248868 0.276693 0.875 0.797810 0.767939 0.880970 0.644682 0.974802 0.745598 0.875 0.450485 0.770812 0.449529 0.793327 0.445334 0.730872

Substitute (52) in (25) the determine coupled system of 0.125 0.016209 0.013277 0.018019 0.006316 0.037906 0.002600 equations is 0.375 0.144228 0.132538 0.142259 0.097840 0.133228 0.062856 ν 2 0.625 ∂ v(x, t) ∂ u(x, t) ¡ 0.625 0.391331 0.385008 0.383766 0.347687 0.379218 0.276693 − + + (u2 + v2)u − (t6 − 4π2t2) sin 2π ∂tν ∂x2 0.875 0.723229 0.770812 0.709533 0.793325 0.681230 0.730876 2t3−ν ¢ × x − cos 2πx = 0 0.125 0.016724 0.013276 0.011634 0.006316 0.064282 0.002600 Γ(3 − ν) 0.375 0.161766 0.132538 0.166036 0.097840 0.180905 0.062856 ∂ν u(x, t) ∂2v(x, t) ¡ 0.875 + + (u2 + v2)v − (t6 − 4π2t2) cos 2π 0.625 0.448134 0.385008 0.430220 0.347687 0.413375 0.276693 ∂tν ∂x2 0.875 0.856867 0.770812 0.860240 0.793327 0.847833 0.730872 2t3−ν ¢ × x + sin 2πx = 0 (55) Γ(3 − ν) Problem 3: Consider T-FSE with trapping potential [6] for On following the method described in Section IV-A, we λ = 1, η = 1, µ(x) = cos2 x and have µ 3−ν ¶ 6t 1 3 9 3 2 ix −HT (x) × [P 2]T × V × P 1−ν × H(t) + HT (x) × U × P 1 q(x, t) = i − t + t + t cos x e 2 . (57) Γ(4 − ν) 4 T 2 T 1 × H(t) + ((H (x) × [P ] × U × P × H(t) Subjected to initial conditions 2 2 T 2 T 1 + 2πt x) + (H (x) × [P ] × V × P 3 0 t 2 2 T 2 T φ(x, 0) = 0, φ(0, t) = t, φ (0, t) = i (58) × H(t) + t ) ) × (H (x) × [P ] 2 Γ(3) with exact solution for ν = 1 is × U × P 1 × H(t) + 2πt2x) + t2−ν Γ(3 − ν) 3 ix φ(x, t) = t e 2 . (59) ¡ 2t3−ν ¢ − (t6 − 4π2t2) sin 2πx − cos 2πx = 0 Γ(3 − ν) Determine the coupled system of equations by substituting (57) in (25). Next, following the method illustrated in Section HT (x) × [P 2]T × U × P 1−ν × H(t) + HT (x) × V IV-A, by following same steps of Problem 1 Haar solutions are 1 T 2 T 1 × P × H(t) + ((H (x) × [P ] × U × P attained, see Fig. 2. The obtained Haar solutions are compared × H(t) + 2πt2x)2 + (HT (x) × [P 2]T × V × P 1 with the Homotopy analysis method [5] and at different values × H(t) + t2)2) × (HT (x) × [P 2]T × V × P 1 of ν are shown in Table III. Problem 4: Take into consideration non-linear T-FCSS [35] 2πΓ(3) × H(t) + t2) + xt2−ν for Γ(3 − ν) ¡ 2t3−ν ¢ λ1 = 2, λ2 = 4, α1 = α2 = 1, β1 = 1 and β2 = −1 − (t6 − 4π2t2) cos 2πx + sin 2πx = 0. Γ(3 − ν) 2t2−ν q (x, t) = − sin x + 4t6 cos x (56) 1 Γ(3 − ν) 184 IEEE/CAA JOURNAL OF AUTOMATICA SINICA, VOL. 6, NO. 1, JANUARY 2019

∂ν v(x, t) ∂v(x, t) ∂2u(x, t) − − + ∂tν ∂x ∂x2 + 2(u2 + v2 + r2 + s2)u(x, t) + u(x, t) + r(x, t) 2t2−ν + sin x + 4t6 cos x = 0 Γ(3 − ν) ∂ν u(x, t) ∂u(x, t) ∂2v(x, t) + + ∂tν ∂x ∂x2 + 2(u2 + v2 + r2 + s2)v(x, t) + v(x, t) + s(x, t) 2t2−ν − ( cos x + 4t6 sin x) = 0 Γ(3 − ν) Fig. 2. Haar solutions of Problem 3 at ν = 0.1, 0.3, 0.5. ∂ν s(x, t) ∂s(x, t) ∂2r(x, t) − + + TABLE III ∂tν ∂x ∂x2 2 2 2 2 COMPARISON BETWEEN HAAR SOLUTIONS (J = 1, m = 4) AND + 2(u + v + r + s )r(x, t) + u(x, t) − r(x, t) HAM [5] OF PROBLEM 3 2t2−ν + sin x + 8t6 cos x = 0 Γ(3 − ν) ν = 0.1 ν = 0.3 ν = 0.5 t x HWCM HAM [5] HWCM HAM [5] HWCM HAM [5] ∂ν r(x, t) ∂r(x, t) ∂2s(x, t) − + 0.125 0.002666 0.001164 0.005127 0.001923 0.009737 0.003186 ∂tν ∂x ∂x2 0.375 0.064478 0.032624 0.099565 0.043939 0.151781 0.100635 2 2 2 2 0.125 + 2(u + v + r + s )s(x, t) + v(x, t) − s(x, t) 0.625 0.283647 0.155864 0.395466 0.219652 0.544319 0.438394 2t2−ν 0.875 0.752604 0.482083 0.981003 0.706422 1.262370 1.267630 − ( cos x + 8t6 sin x) = 0. (63) Γ(3 − ν) 0.125 0.002335 0.001064 0.004453 0.001859 0.008360 0.002993 Next, following the method illustrated in Section IV-B, we 0.375 0.056487 0.029282 0.086504 0.040986 0.130392 0.082892 0.375 have 0.625 0.248554 0.138363 0.343701 0.159805 0.467832 0.388469 ³ 0.875 0.660306 0.425678 0.853338 0.297603 1.085680 1.080400 − HT (x) × [P 2]T × V × P 1−ν × H(t) 0.125 0.001737 0.000940 0.003234 0.001782 0.005880 0.002751 ´ Γ(ν + 1) 2−ν T 1 T 0.375 0.042093 0.025192 0.062894 0.046640 0.091797 0.073477 + xt − H (x) × [P ] 0.625 Γ(2 − ν + 1) 0.625 0.185563 0.117006 0.250376 0.240900 0.330180 0.329875 × V × P 1 × H(t) + HT (x) × U × P 1 0.875 0.498019 0.355007 0.625716 0.854561 0.769576 0.869481 × H(t) + U + 2(U2 + V2 + R2 + S2)U 0.125 0.000973 0.000833 0.001714 0.001823 0.002981 0.002530 2t2−ν 0.375 0.023635 0.021729 0.033100 0.051878 0.045350 0.065040 + R + sin x + 4t6 cos x = 0 0.875 Γ(3 − ν) 0.625 0.105052 0.098823 0.131553 0.315074 0.158672 0.279519 T 2 T 1−ν 0.875 0.299206 0.291128 0.337855 0.257690 0.368943 0.300179 H (x) × [P ] × U × P × H(t) Γ(ν + 1) + xt2−ν + HT (x) × [P 1]T µ 2−ν ¶ Γ(2 − ν + 1) 2t 6 + i cos x + 4t sin x and 1 T 1 Γ(3 − ν) × U × P × H(t) + H (x) × V × P 2 2 2 2 2−ν × H(t) + V + 2(U + V + R + S )V + S 2t 6 µ ¶ q2(x, t) = − sin x + 8t cos x 2t2−ν Γ(3 − ν) − cos x + 4t6 sin x = 0 µ ¶ Γ(3 − ν) 2t2−ν ³ + i cos x + 8t6 sin x (60) Γ(3 − ν) − HT (x) × [P 2]T × S × P 1−ν × H(t) with initial conditions Γ(ν + 1) ´ + xt2−ν + HT (x) × [P 1]T Γ(2 − ν + 1) φ(x, 0) = ψ(x, 0) = 0 × S × P 1 × H(t) + HT (x) × R × P 1 φ(0, t) = ψ(0, t) = t × H(t) + U + 4(U2 + V2 + R2 + S2)R − R φ0(0, t) = ψ0(0, t) = it2 (61) 2t2−ν + sin x + 8t6 cos x = 0 and with exact solution for these values of λ, α, β and ν = Γ(3 − ν) 1 is (HT (x) × [P 2]T × R × P 1−ν × H(t) φ(x, t) = ψ(x, t) = t2eix. (62) Γ(ν + 1) + xt2−ν ) − HT (x) × [P 1]T Γ(2 − ν + 1) Determine the coupled system of two equations by substi- 1 T 1 tuting (60) in (29) as × R × P × H(t) + H (x) × S × P KHAN AND HAMEED: AN IMPLEMENTATION OF HAAR WAVELET BASED METHOD FOR NUMERICAL TREATMENT 185

TABLE V × H(t) + V + 4(U2 + V2 + R2 + S2)S − S µ ¶ COMPARISON BETWEEN HAAR SOLUTIONS 2t2−ν − cos x + 8t6 sin x = 0 (64) (J = 1, m = 4, ν = 0.5) AND HAM [5] OF PROBLEM 4 Γ(3 − ν) |φ(x, t)| |ψ(x, t)| t x where HWCM HAM [5] HWCM HAM [5] 0.125 0.015386 0.015609 0.015323 0.008203 T 2 T 1 2 U = H (x) × [P ] × U × P × H(t) + t 0.375 0.138522 0.139708 0.137974 0.074131 0.125 V = HT (x) × [P 2]T × V × P 1 × H(t) + xt2 0.625 0.385897 0.376762 0.384668 0.225031 0.875 0.763618 0.919085 0.763415 0.811099 R = HT (x) × [P 2]T × R × P 1 × H(t) + t2 S = HT (x) × [P 2]T × S × P 1 × H(t) + xt2. 0.125 0.013951 0.015608 0.013522 0.008203 0.375 0.125873 0.139708 0.122165 0.074132 0.375 Using (50), the values of Haar coefficients are obtained and 0.625 0.357309 0.376762 0.348789 0.225031 finally with the help of these coefficients Haar solutions are 0.875 0.748433 0.919085 0.746600 0.811099 attained from (43). The Haar solutions are compared with the 0.125 0.011025 0.015609 0.009723 0.008203 method in [5] with ν = 0.1 and results are shown in Table 0.375 0.100259 0.139708 0.088875 0.074132 0.625 IV, and for ν = 0.5 and ν = 0.9 in Table V and Table VI, 0.625 0.296585 0.376762 0.272878 0.225031 respectively. 0.875 0.696443 0.919085 0.705993 0.811099

TABLE IV 0.125 0.008579 0.015609 0.004118 0.008202 COMPARISON BETWEEN HAAR SOLUTIONS 0.375 0.079810 0.139708 0.040065 0.074131 0.875 (J = 1, m = 4, ν = 0.1) AND HAM [5] OF PROBLEM 4 0.625 0.235819 0.376762 0.162549 0.225031 0.875 0.623663 0.919085 0.660364 0.811099 |φ(x, t)| |ψ(x, t)| t x HWCM HAM [5] HWCM HAM [5] 0.125 0.015383 0.014440 0.0153167 0.004286 TABLE VI 0.375 0.138501 0.128015 0.137928 0.038537 COMPARISON BETWEEN HAAR SOLUTIONS 0.125 0.625 0.385851 0.365216 0.384604 0.143183 (J = 1, m = 4, ν = 0.9) AND HAM [5] OF PROBLEM 4 0.875 0.763499 0.949120 0.763355 0.811326 |φ(x, t)| |ψ(x, t)| t x 0.125 0.013949 0.014439 0.013495 0.004286 HWCM HAM [5] HWCM HAM [5] 0.375 0.125927 0.128015 0.121958 0.038536 0.125 0.015395 0.013101 0.015357 0.013101 0.375 0.625 0.357238 0.365216 0.348492 0.143183 0.375 0.138552 0.117724 0.138256 0.118102 0.125 0.875 0.747914 0.949120 0.746334 0.811326 0.625 0.385995 0.322206 0.384873 0.336649 0.875 0.763808 0.631637 0.763550 0.834047 0.125 0.011086 0.014440 0.009704 0.004285 0.375 0.100695 0.128015 0.088716 0.038536 0.125 0.013964 0.013101 0.013688 0.013101 0.625 0.625 0.297240 0.365216 0.272767 0.143183 0.375 0.125649 0.117724 0.123396 0.118102 0.375 0.875 0.697857 0.949120 0.706649 0.811326 0.625 0.357809 0.322206 0.349879 0.336649 0.875 0.749377 0.631637 0.747267 0.834047 0.125 0.008487 0.014440 0.004160 0.004286 0.375 0.077508 0.128015 0.040194 0.038537 0.125 0.010878 0.013101 0.009803 0.013101 0.875 0.625 0.236292 0.365216 0.163674 0.143183 0.375 0.099177 0.117725 0.089536 0.118102 0.625 0.875 0.634398 0.949120 0.664360 0.811326 0.625 0.296418 0.322206 0.273663 0.336649 0.875 0.694211 0.631637 0.701977 0.834047

VI.CONCLUSION 0.125 0.008446 0.013101 0.003678 0.013101 0.375 0.085571 0.117725 0.040710 0.118102 0.875 In this paper, we extended the capability of the Haar 0.625 0.231591 0.322206 0.157683 0.336649 wavelet collocation method (HWCM) for the solution of time- 0.875 0.604013 0.631637 0.640823 0.834047 fractional coupled system of partial differential equations. The main advantage of HWCM is the ability to achieve a good solution and rapid convergence with small number problems of partial differential equations of fractional order. of collocation points. The presence of maximum zeros in The problem discussed here is just for showing the appli- the Haar matrices reduces the number of unknown wavelet cability of the proposed computational technique to handle coefﬁcients that is to be determined, which as a result dimin- the complex system of differential equation in fractional- ishes the computation time as well. The scheme is tested on order problems in a straight forward way. Also, the Haar some examples of time fractional Schrodinger¨ equations. The wavelet method proves to be capable to efﬁciently handle presented procedure may very well be extended to solve two the nonlinearity of partial differential equations of fractional dimensional Schrodinger¨ equation and other similar nonlinear order. The main advantages of the proposed algorithm are, its 186 IEEE/CAA JOURNAL OF AUTOMATICA SINICA, VOL. 6, NO. 1, JANUARY 2019

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