A statistical approach to the use of control entropy identifies differences in constraints of gait in highly trained versus untrained runners

Rana D. Parshad; Stephen J. McGregor; Michael A. Busa; Joseph D. Skufca; Erik Bollt; Rana D. Parshad; Stephen J. McGregor; Michael A. Busa; Joseph D. Skufca; Erik Bollt

doi:10.3934/mbe.2012.9.123

Mathematical Biosciences and Engineering

2012, Volume 9, Issue 1: 123-145. doi: 10.3934/mbe.2012.9.123

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A statistical approach to the use of control entropy identifies differences in constraints of gait in highly trained versus untrained runners

1.
Department of Mathematics & Computer Science, Clarkson University, Potsdam, NY 13676
2.
Applied Physiology Laboratory, Eastern Michigan University, Ypsilanti, MI 48197

Received: 01 February 2011 Accepted: 29 June 2018 Published: 01 December 2011
MSC : Primary: 37M25, 62H15; Secondary: 92C30.

Control entropy (CE) is a complexity analysis suitable for dynamic, non-stationary conditions which allows the inference of the control effort of a dynamical system generating the signal [4]. These characteristics make CE a highly relevant time varying quantity relevant to the dynamic physiological responses associated with running. Using High Resolution Accelerometry (HRA) signals we evaluate here constraints of running gait, from two different groups of runners, highly trained collegiate and untrained runners. To this end, we further develop the control entropy (CE) statistic to allow for group analysis to examine the non-linear characteristics of movement patterns in highly trained runners with those of untrained runners, to gain insight regarding gaits that are optimal for running. Specifically, CE develops response time series of individuals descriptive of the control effort; a group analysis of these shapes developed here uses Karhunen Loeve Analysis (KL) modes of these time series which are compared between groups by application of a Hotelling $T^{2}$ test to these group response shapes. We find that differences in the shape of the CE response exist within groups, between axes for untrained runners (vertical vs anterior-posterior and mediolateral vs anterior-posterior) and trained runners (mediolateral vs anterior-posterior). Also shape differences exist between groups by axes (vertical vs mediolateral). Further, the CE, as a whole, was higher in each axis in trained vs untrained runners. These results indicate that the approach can provide unique insight regarding the differing constraints on running gait in highly trained and untrained runners when running under dynamic conditions. Further, the final point indicates trained runners are less constrained than untrained runners across all running speeds.

Keywords:

Citation: Rana D. Parshad, Stephen J. McGregor, Michael A. Busa, Joseph D. Skufca, Erik Bollt. A statistical approach to the use of control entropy identifies differences in constraints of gait in highly trained versus untrained runners[J]. Mathematical Biosciences and Engineering, 2012, 9(1): 123-145. doi: 10.3934/mbe.2012.9.123

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Abstract

1. Introduction

In recent decades, importance of fractional order models is well disclosed fact in many fields of engineering and science. Numerous fractional order partial differential equations(FPDEs) have been used by many authors to describe various important biological and physical processes like in the fields of chemistry, biology, mechanics, polymer, economics, biophysics control theory and aerodynamics. In this concern, many researchers have studied various schemes and aspects of PDEs and FPDEs as well, see ^{[1,2,3,4,5,6,7,8,9,10]}. However, the great attention has been given very recently to obtaining the solution of fractional models of the physical interest. Keeping in views, the computation complexities involved in fractional order models is very crucial and is the difficulty in solving these fractional models. Some times, the exact analytic solution of each and every FPDE can not be obtained using the traditional schemes and methods. However, there exists some schemes and methods, which have been proved to be efficient in obtaining the approximation to solution of the fractional order problems. Among them, we bring the attention of readers to these methods and schemes ^{[11,12,13,14,15,16,17,18,19,20,21]} which are used successfully. These methods and schemes have their own demerits and merits. Some of them provide a very good approximation with convenient way. For example, see the methods and schemes in the articles ^{[22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39]}.

The main aim of this work is to develop a new procedure which is easy with respect to application and more efficient as compare with existing procedures. In this concern, we introduced asymptotic homotopy perturbation method (AHPM) to obtain the solution of nonlinear fractional order models. It is a new version of perturbation techniques. In simulation section, we have testified our proposed procedure by considering the test problems of non linear fractional order Zakharov-Kuznetsov $ ZK(m, n, r) $ equations of the form ^[11,12]

$\begin{equation} \begin{split}& \frac{{{\partial ^\alpha }u(x, y, t)}}{{\partial {t^\alpha }}} + {a_0}{\left( {{u^m(x, y, t)}} \right)_x} + {a_1}{\left( {{u^n(x, y, t)}} \right)_{xxx}} + {a_2}{\left( {{u^r(x, y, t)}} \right)_{yyx}} = 0, 0 \lt \alpha \le 1. \end{split} \end{equation}$ $

(1.1)

Where $ a_0 $, $ a_1 $, $ a_2 $ are arbitrary constants and $ m, n, r $ are non zero integers. If $ \alpha = 1, $ then equation (1.1) becomes classical Zakharov-Kuznetsov $ ZK(m, n, r) $ equation given as:

$\begin{equation} \begin{split}& \frac{{{\partial}u(x, y, t)}}{{\partial {t }}}+ {a_0}{\left( {{u^m(x, y, t)}} \right)_x} + {a_1}{\left( {{u^m(x, y, t)}} \right)_{xxx}} + {a_2}{\left( {{u^r(x, y, t)}} \right)_{yyx}} = 0. \end{split} \end{equation}$ $

(1.2)

The ZK equation has been firstly modelized for depicting weakly nonlinear ion-acoustic waves in strongly magnetized lossless plasma ^[40]. The ZK equation governs the behavior of weakly nonlinear ion-acoustic waves in plasma comprising cold ions and hot isothermal electrons in the presence of a uniform magnetic field ^[41,42].

The plan of the rest paper is as follows: Section 2 provides theory of the AHPM; Section 3 provides implementation of AHPM. Finally, a brief conclusion and the further work has been listed.

2. Basic idea of AHPM

Here, we provide that the Caputo type fractional order derivative will be used throughout this paper for the computation of derivative.

Let us consider the nonlinear problem in the form as

$\begin{eqnarray} T\left( {u(x, y, t)} \right) + g(x, y, t) = 0, \end{eqnarray}$ $

(2.1)

$\begin{eqnarray} B\left( {u\left( {x, y, t} \right), \frac{{\partial u\left( {x, y, t} \right)}}{{\partial t}}} \right) = 0. \end{eqnarray}$ $

(2.2)

Where $ T\left({u\left({x, y, t} \right)} \right) $ denotes a differential operator which may consists ordinary, partial or space- fractional or time-fractional differential derivative. $ T\left({u\left({x, y, t} \right)} \right) $ can be expressed for fractional model as follows:

$\begin{eqnarray} \frac{{{\partial ^\alpha }u(x, y, t)}}{{\partial {t^\alpha }}} + N\left( {u(x, y, t)} \right) + g(x, y, t) = 0 \end{eqnarray}$ $

(2.3)

subject to the condition

$\begin{eqnarray} B\left( {u\left( {x, y, t} \right), \frac{{\partial u\left( {x, y, t} \right)}}{{\partial t}}} \right) = 0, \end{eqnarray}$ $

(2.4)

where the operator $ \frac{{{\partial ^\alpha }}}{{\partial {t^\alpha }}} $ denotes the Caputo derivative operator, $ N $ is non linear operator and $ B $ denotes a boundary operator, $ u\left({x, y, t} \right) $ is unknown exact solution of Eq. (2.1), $ g\left({x, y, t} \right) $ denotes known function and $ x, y $ and $ t $ denote special and temporal variables respectively. Let us construct a homotopy $ \Phi (x, y, t;p):\Omega \times \left[{0, 1} \right] \to R $ which satisfies

$\begin{eqnarray} \frac{{{\partial ^\alpha }\Phi (x, y, t;p)}}{{\partial {t^\alpha }}} + g(x, y, t) - p\left[ {N\left( {\Phi (x, y, t;p} \right)} \right] = 0, \end{eqnarray}$ $

(2.5)

where $ p \in \left[{0, 1} \right] $ is said to be an embedding parameter. At this phase of our work it is pertinent that our proposed deformation Eq. (2.5) is an alternate form of the deformation equations as

$\begin{eqnarray} \left( {1 - p} \right)\left[ {L\left( {\Phi (x, y, t;p)} \right) - L\left( {{u_0}(x, y, t)} \right) + g(x, y, t)} \right] + p\left[ {T\left( {\Phi (x, y, t;p} \right) + g(x, y, t)} \right] = 0, \end{eqnarray}$ $

(2.6)

$\begin{eqnarray} \left( {1 - p} \right)\left[ {L\left( {\Phi (x, y, t;p)} \right) - L\left( {{u_0}(x, y, t)} \right)} \right] = hp\left[ {T\left( {\Phi (x, y, t;p} \right) + g(x, y, t)} \right] \end{eqnarray}$ $

(2.7)

and

$\begin{eqnarray} \left( {1 - p} \right)\left[ {L\left( {\Phi (x, y, t;p)} \right) + g(x, y, t)} \right] - H\left( p \right)\left[ {T\left( {\Phi (x, y, t;p} \right) + g(x, y, t)} \right] = 0. \end{eqnarray}$ $

(2.8)

in HPM, HAM, OHAM proposed by Liao in ^[43], He in ^[44] and Marinca in ^[45] respectively.

Basically, according to homotopy definition, when $ p = 0 $ and $ p = 1 $ we have

$ \Phi \left( {x, y, t;p} \right) = {u_0}\left( {x, y, t} \right), \phi \left( {x, y, t;p} \right) = u\left( {x, y, t} \right). $

Obviously, when the embedding parameter $ p $ varies from $ 0 $ to $ 1 $, the defined homotopy ensures the convergence of $ \phi \left({x, y, t;p} \right) $ to the exact solution $ u\left({x, y, t} \right) $. Consider $ \phi \left({x, y, t;p} \right) $ in the form

$\begin{eqnarray} \Phi \left( {x, y, t;p} \right) = {u_0}\left( {x, y, t} \right) + \sum\limits_{i = 1}^\infty {{u_i}\left( {x, y, t} \right)} {p^i} \end{eqnarray}$ $

(2.9)

and assuming $ N\left({\Phi \left({x, y, t;p} \right)} \right) $ as follows

$\begin{eqnarray} N\left( {\Phi \left( {x, y, t;p} \right)} \right) = {B_1}{N_0} + \sum\limits_{i = 1}^\infty {\left( {\sum\limits_{m = 0}^i {{B_{i + 1 - m}}{N_m}} } \right)} {p^i}, {{B_1} + {B_2} + {B_3} + ...} = -1. \end{eqnarray}$ $

(2.10)

Where

$\begin{eqnarray} {B_i} = {B_i}(x, y, t, {c_i}), for , i = 1, 2, 3, \dots. \end{eqnarray}$ $

(2.11)

are arbitrary auxiliary functions, will be discussed later. Thus, if $ p = 0 $ and $ p = 1 $ in Eq. (2.5), we have

$ \frac{{{\partial ^\alpha }u(x, y, t)}}{{\partial {t^\alpha }}} + f\left( x \right) = 0, \ \text{and}\ \frac{{{\partial ^\alpha }u(x, y, t)}}{{\partial {t^\alpha }}} + N\left( {u(x, y, t)} \right) + g(x, y, t) = 0, $

respectively.

It is obvious that the construction of introduced auxiliary function in Eq. (2.10) is different from the auxiliary functions that are proposed in articles ^[43,44,45]. Hence the procedure proposed in our paper is different from the procedures proposed by Liao, He, Marinca in aforesaid papers ^[43,44,45] as well as Optimal Homotopy perturbation method (OHPM) in ^[46].

Furthermore, when we substitute Eq. (2.9) and Eq. (2.10) in Eq. (2.5) and equate like power of $ p $, the obtained series of simpler linear problems are

$\begin{equation*} \begin{split}& {p^0}: \frac{{{\partial ^\alpha }{u_0}(x, y, t)}}{{\partial {t^\alpha }}} + g = 0, \\& {p^1}: \frac{{{\partial ^\alpha }{u_1}(x, y, t)}}{{\partial {t^\alpha }}} = {B_1}{N_0}, \\& {p^2}: \frac{{{\partial ^\alpha }{u_2}(x, y, t)}}{{\partial {t^\alpha }}} = {B_2}{N_0} + {B_1}{N_1}, \\& {p^3}: \frac{{{\partial ^\alpha }{u_3}(x, y, t)}}{{\partial {t^\alpha }}} = {B_3}{N_0} + {B_2}{N_1} + {B_1}{N_2}, \\& \vdots\\& {p^k}: \frac{{{\partial ^\alpha }{u_k}(x, y, t)}}{{\partial {t^\alpha }}} = \sum\limits_{i = 0}^{k - 1} {{B_{k - i}}{N_i}}. \end{split} \end{equation*}$ $

We obtain the series solutions by using the integral operator $ {J^\alpha } $ on both sides of the above each simple fractional differential equation. The convergence of the series solution Eq. (2.9) to the exact solution depends upon the auxiliary parameters (functions) $ {B_i}(x, y, t, {c_i}) $. The choice of $ {B_i}(x, y, t, {c_i}) $ is purely on the basis of terms appear in nonlinear part of the Eq. (2.1). The Eq.(2.9) converges to the exact solution of Eq. (2.1) at $ p = 1 $:

$\begin{eqnarray} \widetilde u(x, y, t) = {u_0}(x, y, t) + \sum\limits_{k = 1}^\infty {{u_k}(x, y, t;{c_i}), \ i = 1, 2, 3, \dots}. \end{eqnarray}$ $

(2.12)

Particularly, we can truncate the Eq. (2.12) into finite m-terms to obtain the solution of nonlinear problem. The auxiliary convergence control constants $ {c_1}, {c_2}, {c_3}, \dots $ can be found by solving the system

$\begin{eqnarray} R\left( {\partial {x_1}, \partial {y_1}} \right) = R\left( {\partial {x_2}, \partial {y_2}} \right) = R\left( {\partial {x_3}, , \partial {y_3}} \right) = \dots = R\left( {\partial {x_m}, , \partial {y_m}} \right) = 0, \partial {x_i}, \partial {y_i} \in \left[ {a, b} \right]. \end{eqnarray}$ $

(2.13)

It can be verified to observe that HPM is only a case of Eq. (2.5) when $ p = -p $ and

$\begin{eqnarray*} N\left( {\Phi \left( {x, y, t;p} \right)} \right) = {N_0} + \sum\limits_{i = 1}^\infty {{N_i}{p^i}}. \end{eqnarray*}$ $

The HAM is also a case of Eq. (2.5) when $ p = ph $ and

$\begin{eqnarray*} N\left( {\Phi \left( {x, y, t;p} \right)} \right) = {N_0} + \sum\limits_{i = 1}^\infty {{N_i}{p^i}}. \end{eqnarray*}$ $

The OHAM is also another case when

$\begin{eqnarray*} {B_{k - 1}} = {B_{k - 2}} + {h_k}\left( {t, {c_j}} \right) + \sum\limits_{i = 0}^{k - 2} {{h_{\left( {_{k - (i + 1)}} \right)}}\left( {t, {c_j}} \right)} {B_i}, \ \text{and}\ {h_k}\left( {t, {c_j}} \right) = {c_k} \end{eqnarray*}$ $

in Eq. (2.10), we obtain exactly the series problems which are obtained by OHAM after expanding and equating the like power of $ p $ in deformation equation. Furthermore, concerning the Optimal Homotopy Asymptotic Method (OHAM) mentioned in this manuscript and presented in ^[45], that the version of OHAM proposed in 2008 was improved in time and the most recent improvement, which also contains an auxiliary functions, are presented in the papers ^[47,48]. We also have improved the version of OHAM by introducing a very new auxiliary function in Eq. (2.10). Our method proposed in this paper uses a very new and more general form of auxiliary function

$ N\left( {\phi \left( {x, y, t;p} \right)} \right) = {B_1}{N_0} + \sum\limits_{i = 1}^\infty {\left( {\sum\limits_{m = 0}^i {{B_{i + 1 - m}}{N_m}} } \right)} {p^i} $

which depends on arbitrary parameters $ {B_1}, {B_2}, {B_3}, \dots $ and is useful for adjusting and controlling the convergence of nonlinear part as well as linear part of the problem with simple way.

3. Applications

In this portion, we apply AHPM to obtain solution of the following problems to show the accuracy and appropriateness of the new procedure for to solve nonlinear problems.

Problem 3.1. Let us consider FZK$ (2, 2, 2) $ in the form:

$\begin{eqnarray} \frac{{{\partial ^\alpha }u(x, y, t)}}{{\partial {t^\alpha }}} + {\left( {{u^2}(x, y, t)} \right)_x} + \frac{1}{8}{\left( {{u^2}(x, y, t)} \right)_{xxx}} + \frac{1}{8}{\left( {{u^2}(x, y, t)} \right)_{yyx}} = 0, {\mkern 1mu} {\mkern 1mu} 0 \lt \alpha \le 1, \end{eqnarray}$ $

(3.1)

subject to the condition

$\begin{equation*} u\left( {x, y, 0} \right) = \frac{4}{3}k{\sinh ^2}\left( {x + y} \right). \end{equation*}$ $

When $ \alpha = 1 $. Then the exact solution of Eq. (3.1):

$\begin{equation*} u\left( {x, y, t} \right) = \frac{4}{3}k{\sinh ^2}\left( {x + y-kt} \right). \end{equation*}$ $

As in Eq. (3.1), the non linear part

$\begin{equation*} N\left( {u(x, y, t)} \right) = {\left( {{u^2}(x, y, t)} \right)_x} + \frac{1}{8}{\left( {{u^2}(x, y, t)} \right)_{xxx}} + \frac{1}{8}{\left( {{u^2}(x, y, t)} \right)_{yyx}}. \end{equation*}$ $

Now, follow the procedure of AHPM, we obtain series of the simpler linear problems as:

Zero order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_0}}}{{\partial {t^\alpha }}} = 0, {u_0} = \frac{4}{3}k{\sinh ^2}\left( {x + y} \right). \end{eqnarray}$ $

(3.2)

First order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_1}}}{{\partial {t^\alpha }}} = {B_1}{N_0}, {u_1} = 0. \end{eqnarray}$ $

(3.3)

Second order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_2}}}{{\partial {t^\alpha }}} = {B_2}{N_0} + {B_1}{N_1}, {u_2} = 0. \end{eqnarray}$ $

(3.4)

Third order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_3}}}{{\partial {t^\alpha }}} = {B_3}{N_0} + {B_2}{N_1} + {B_1}{N_2}, {u_3} = 0. \end{eqnarray}$ $

(3.5)

Fourth order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_4}}}{{\partial {t^\alpha }}} = {B_4}{N_0} + {B_3}{N_1} + {B_2}{N_2} + {B_1}{N_3}, {u_4} = 0 \end{eqnarray}$ $

(3.6)

and so on.

Solving the above equations, the respective solutions from Eqs. (3.2)–(3.6) are given as follow:

$\begin{eqnarray} {u_0}\left( {x, y, t} \right) = \frac{4}{3}k{\sinh ^2}(z) \end{eqnarray}$ $

(3.7)

$\begin{eqnarray} {u_1}\left( {x, y, t} \right) = {B_1}{N_0}\frac{{{t^\alpha }}}{{\Gamma \left( {\alpha + 1} \right)}} \end{eqnarray}$ $

(3.8)

$\begin{eqnarray} {u_2}\left( {x, y, t} \right) = {B_2}{N_0}\frac{{{t^\alpha }}}{{\Gamma \left( {\alpha + 1} \right)}} + \frac{{64}}{{27}}{B_1}^2{k^3}\left[ {13\cosh\left( {2z} \right) - 70\cosh\left( {4z} \right) + 75\cosh\left( {6z} \right)} \right] \frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}}, \end{eqnarray}$ $

(3.9)

$\begin{array}{l} {u_3}\left( {x, y, t} \right) = \\ {B_3}{N_0}\frac{{{t^\alpha }}}{{\Gamma \left( {\alpha + 1} \right)}} + \frac{{64}}{{27}}{B_1}{B_2}{k^3}\left[ {13\cosh \left( {2z} \right) - 70\cosh \left( {4z} \right) + 75\cosh \left( {6z} \right)} \right]\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}}\\ + {B_1}\frac{{64}}{{81}}{k^3}\left[ \begin{array}{l} {B_2}\left( {39\cosh \left( {2z} \right) - 210\cosh \left( {4z} \right) + 225\cosh \left( {6z} \right)} \right)\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}}\\ + {B_1}^2k\left( {\begin{array}{*{20}{l}} {160\sinh \left( {2z} \right) + 320\sinh \left( {4z} \right) - 2400\sinh \left( {6z} \right)}\\ { + 3400\sinh \left( {8z} \right)} \end{array}} \right)\frac{{\Gamma \left( {2\alpha + 1} \right){t^{2\alpha }}}}{{{{\left( {\Gamma \left( {\alpha + 1} \right)} \right)}^2}\Gamma \left( {3\alpha + 1} \right)}}\\ + {B_1}^2k\left( {\begin{array}{*{20}{l}} { - 768\sinh \left( {2z} \right) + 9120\sinh \left( {4z} \right) - 26400\sinh \left( {6z} \right)}\\ { + 20400\sinh \left( {8z} \right)} \end{array}} \right)\frac{{{t^{3\alpha }}}}{{\Gamma \left( {3\alpha + 1} \right)}}{\rm{ }} \end{array}$ \right] \end{array}$

(3.10)

$\begin{array}{l} {u_4}\left( {x, y, t} \right) = \\ {B_4}{N_0}\frac{{{t^\alpha }}}{{\Gamma \left( {\alpha + 1} \right)}} + \frac{{64}}{{27}}{B_1}{B_3}{k^3}\left[ {13\cosh \left( {2z} \right) - 70\cosh \left( {4z} \right) + 75\cosh \left( {6z} \right)} \right]\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}} + {B_2}\frac{{64}}{{81}}{k^3}\\ \left[ {\begin{array}{*{20}{l}} {{B_2}\left( {39\cosh \left( {2z} \right) - 210\cosh \left( {4z} \right) + 225\cosh \left( {6z} \right)} \right)\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}}}\\ { + {B_1}^2k\left( {\begin{array}{*{20}{l}} {160\sinh \left( {2z} \right) + 320\sinh \left( {4z} \right) - }\\ {2400\sinh \left( {6z} \right) + 3400\sinh \left( {8z} \right)} \end{array}} \right)\frac{{\Gamma \left( {2\alpha + 1} \right){t^{2\alpha }}}}{{{{\left( {\Gamma \left( {\alpha + 1} \right)} \right)}^2}\Gamma \left( {3\alpha + 1} \right)}}}\\ { + {B_1}^2k\left( {\begin{array}{*{20}{l}} { - 768\sinh \left( {2z} \right) + 9120\sinh \left( {4z} \right) - }\\ {26400\sinh \left( {6z} \right) + 20400\sinh \left( {8z} \right)} \end{array}} \right)\frac{{{t^{3\alpha }}}}{{\Gamma \left( {3\alpha + 1} \right)}}} \end{array}} \right] + {B_1}\frac{{64}}{{243}}{k^3}\\ \left[ \begin{array}{l} {B_1}^3{k^2}\left( {1022400\cosh \left( {6z} \right) - 15200\cosh \left( {4z} \right) - 2502400\cosh \left( {8z} \right) + 1768000\cosh \left( {10z} \right)} \right)\\ \frac{{2\Gamma \left( {2\alpha } \right){t^{4\alpha }}}}{{\alpha {{\left( {\Gamma \left( \alpha \right)} \right)}^2}\Gamma \left( {4\alpha + 1} \right)}} + {B_1}^3{k^2}\left( {\begin{array}{*{20}{l}} {85248\cosh \left( {2z} \right) - 1816320\cosh \left( {4z} \right) + 9878400\cosh \left( {6z} \right)}\\ { - 18278400\cosh \left( {8z} \right) + 10608000\cosh \left( {10z} \right)} \end{array}} \right)\\ \frac{{{t^{4\alpha }}}}{{\Gamma \left( {4\alpha + 1} \right)}} + {B_3}\left( {117\cosh \left( {2z} \right) - 630\cosh \left( {4z} \right) + 675\cosh \left( {6z} \right)} \right)\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}} + {B_1}^3{k^2}\\ \left( {\begin{array}{*{20}{l}} { - 56640\cosh \left( {2z} \right) + 119040\cosh \left( {4z} \right) + 496800\cosh \left( {6z} \right)}\\ { - 2121600\cosh \left( {8z} \right) + 2340000\cosh \left( {10z} \right)} \end{array}} \right)\frac{{\Gamma \left( {3\alpha + 1} \right){t^{4\alpha }}}}{{\Gamma \left( {\alpha + 1} \right)\Gamma \left( {2\alpha + 1} \right)\Gamma \left( {4\alpha + 1} \right)}}\\ + {B_1}{B_2}k\left( {\begin{array}{*{20}{l}} {1440\sinh \left( {2z} \right) + 2880\sinh \left( {4z} \right) - 21600\sinh \left( {6z} \right)}\\ { + 30600\sinh \left( {8z} \right)} \end{array}} \right)\frac{{\Gamma \left( {2\alpha + 1} \right){t^{3\alpha }}}}{{{{\left( {\Gamma \left( {\alpha + 1} \right)} \right)}^2}\Gamma \left( {3\alpha + 1} \right)}}\\ + {B_1}{B_2}k\left( {\begin{array}{*{20}{l}} { - 4608\sinh \left( {2z} \right) + 54720\sinh \left( {4z} \right) - 158400\sinh \left( {6z} \right)}\\ { + 122400\sinh \left( {8z} \right)} \end{array}} \right)\frac{{{t^{3\alpha }}}}{{\Gamma \left( {3\alpha + 1} \right)}} \end{array}$ \right] \end{array}$

(3.11)

so on $\dots$.

In similar way, we can compute the solution of the next simpler linear problems which are difficult to compute by using OHAM procedure. we choose $ {{B_1} = {c_1}, B_2} = {c_2}, {B_3} = {c_3}, {B_4} = {c_4} $ and consider

$\begin{eqnarray} \widetilde u(x) = {u_0}(x) + {u_1}(x, {c_1}) + {u_2}(x, {c_1}, {c_2}) + {u_3}(x, {c_1}, {c_2}, {c_3})+{u_4}(x, {c_1}, {c_2}, {c_3}, {c_4}) . \end{eqnarray}$ $

(3.12)

The residual:

$\begin{eqnarray} R\left( {\widetilde u\left( {x, y, t} \right)} \right) = \frac{{{\partial ^\alpha }\widetilde u\left( {x, y, t} \right)}}{{\partial {t^\alpha }}} + {\left( {{{\widetilde u}^2}(x, y, t)} \right)_x} + \frac{1}{8}{\left( {{{\widetilde u}^2}(x, y, t)} \right)_{xxx}} + \frac{1}{8}{\left( {{{\widetilde u}^2}(x, y, t)} \right)_{yyx}}. \end{eqnarray}$ $

(3.13)

We obtain number of optimal values of auxiliary constants by using the Eq. (2.13) and choose those optimal values whose sum is in $ \left[{- 1, } \right.\left. 0 \right) $. Now, substituting the optimal values of auxiliary constants (from the Table 1) into the Eq. (3.12), we obtain the AHPM solutions for different values of $ \alpha $ at $ k = 0.001 $

Table 1. The auxiliary control constants for the problem 3.1.

$ Aux. Const. $	$ \alpha=1 $	$ \alpha=0.75 $	$ \alpha=0.67 $
$ c_1 $	$ -0.03206298594 $	$ 0.02857059949 $	$ 0.02811316381 $
$ c_2 $	$ -0.05626044816 $	$ -0.09201640919 $	$ -0.09255449827 $
$ c_3 $	$ -0.03255192821 $	$ 0.23864503710 $	$ 0.24163106460 $
$ c_4 $	$ 0.09230916402 $	$ -0.58155364530 $	$ -0.59008434920 $
$ c_1+c_2+c_3+ c_4 $	$ -0.0286 $	$ -0.4064 $	$ -0.4129 $

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If $ \alpha = 1, $ then we have

$\begin{equation*} \begin{split}& \widetilde u\left( {x, y, t} \right)\\& = 6.67{e^{ - 4}}\cosh(2z) - 1.5{e^{ - 36}}\cosh(2x - 2y) + 1.52{e^{ - 10}}{t^2}\cosh(2z) - 8.2{e^{ - 10}}{t^2}\cosh(4z) - 9.82{e^{ - 19}}{t^4}\\& \cosh(2z)- 1.96 {e^{ - 17}} {t^4}\cosh(4z) - 3.44 {e^{ - 16}}{t^4}\cosh(8z) + 9.74{e^{ - 15}}{t^3}\sinh(2z) - 2.7 {e^{ - 13}}{t^3}\sinh(4z) -\\& 7.91 {e^{ - 13}}{t^3}\sinh(8z) + 2.45{e^{ - 16}}{t^4}\cosh(10z)+ 8.79{e^{ - 10}}{t^2}\cosh(6z) + 1.56{e^{ - 16}}{t^4}\cosh(6z) + 8.85{e^{ - 13}}{t^3}\\& \sinh(6z) + 1.02{e^{ - 7}}t\sinh(2z) - 1.27{e^{ - 7}}t\sinh(4z) - 6.67{e^{ - 4}}. \end{split} \end{equation*}$ $

If $ \alpha = 0.75, $ then we have

$\begin{equation*} \begin{split}& \widetilde u\left( {x, y, t} \right) \\& = 6.67{e^{ - 4}}\cosh(2z) - 1.5{e^{ - 36}}\cosh(2z) - 9.63{e^{ - 19}}{t^3}\cosh(2z) - 5.12{e^{ - 17}}{t^3}\cosh(4z) + 4.09{e^{ - 10}}{t^{3/2}}\\& \cosh(2z) - 7.79{e^{ - 16}}{t^3}\cosh(8z) - 2.2{e^{ - 9}}{t^{3/2}}\cosh(4z) + 1.57{e^{ - 6}}{t^{3/4}}\sinh(2z) - 1.97{e^{ - 6}}{t^{3/4}}\sinh(4z) \\& + 2.65{e^{ - 14}}{t^{9/4}}\sinh(2z)- 6.14 {e^{ - 13}}{t^{9/4}} \sinh(4z) - 1.74 {e^{ - 12}}{t^{9/4}}\sinh(8z) + 5.34 {e^{ - 16}}{t^3}\cosh(10z) +\\& 3.66 {e^{ - 16}}{t^3}\cosh(6z) + 2.36 {e^{ - 9}}{t^{3/2}}\cosh(6z) + 1.98{e^{ - 12}}{t^{9/4}}\sinh(6z) - 6.67{e^{ - 4}}. \end{split} \end{equation*}$ $

If $ \alpha = 0.67, $ then we have

$\begin{equation*} \begin{split}& \widetilde u\left( {x, y, t} \right) \\& = 6.67{e^{ - 4}}\cosh(2z) - 1.5{e^{ - 36}}\cosh(2x - 2y) - 8.33{e^{ - 19}}{t^{67/25}}\cosh(2z) - 7.12{e^{ - 17}}{t^{67/25}}\cosh(4z) - \\& 1.05{e^{ - 15}}{t^{67/25}}\cosh(8z) + 4.57 {e^{ - 10}}{t^{67/50}}\cosh(2z) - 2.46 {e^{ - 9}}{t^{67/50}}\cosh(4z) + 1.63 {e^{ - 6}}{t^{67/100}}\sinh(2z)\\& - 2.03 {e^{ - 6}}{t^{67/100}}\sinh(4z)+ 3.45 {e^{ - 14}}{t^{201/100}}\sinh(2z) - 7.54 {e^{ - 13}}{t^{201/100}}\sinh(4z) - 2.1 {e^{ - 12}}{t^{201/100}}\\& \sinh(8z) + 7.09 {e^{ - 16}} {t^{67/25}}\cosh(10z) + 4.97 {e^{ - 16}}{t^{67/25}}\cosh(6z) + 2.64 {e^{ - 9}}{t^{67/50}}\cosh(6z) +2.41 {e^{ - 12}}\\& {t^{201/100}}\sinh(6z) - 6.67{e^{ - 4}}. \end{split} \end{equation*}$ $

Tables 2 and 3 show the AHPM solution, VIM solution, exact solution and absolute error of AHPM solution. It is obvious from Tables 2 and 3 that AHPM solution results are more accurate to the exact solution results as compare with VIM ^[11] solution results. The AHPM solution, exact solution and absolute error of AHPM solution are plotted for different values of $ \alpha $, $ x $, $ y $ and $ t $ in Figures 1 and 2. The curves of AHPM and exact solution are exactly matching as compare with homotopy perturbation transform method (HPTM)^[12]. It is obvious from the Tables 2 and 3, Figures 1 and 2, that the AHPM solution of the problem 3.1 is in very good agreement with exact solution.

Table 2. Solution of the problem 3.1 for various values of $ \alpha $, $ x $, $ y $ and $ t $ at $ k = 0.001 $.

$ x $	$ y $	$ t $	VIM ^[1] ($ \alpha=0.67 $)	AHPM ($ \alpha=0.67 $)	VIM ^[1] ($ \alpha=0.75 $)	AHPM ($ \alpha=0.75 $)
$ 0.1 $	$ 0.1 $	$ 0.2 $	$ 5.312992862e-5 $	$ 0.000053661825095 $	$ 5.325267164e-5 $	$ 0.000053719590053 $
		$ 0.3 $	$ 5.285029317e-5 $	$ 0.000053541299605 $	$ 5.297615384e-5 $	$ 0.000053602860321 $
		$ 0.4 $	$ 5.260303851e-5 $	$ 0.000053433636044 $	$ 5.272490734e-5 $	$ 0.000053495694982 $
$ 0.6 $	$ 0.6 $	$ 0.2 $	$ 2.953543396e-3 $	$ 0.0029991909006 $	$ 2.964363202e-3 $	$ 0.0030049612043 $
		$ 0.3 $	$ 2.928652795e-3 $	$ 0.0029871641418 $	$ 2.939926307e-3 $	$ 0.0029932937681 $
		$ 0.4 $	$ 2.905913439e-3 $	$ 0.0029764476067 $	$ 2.917239345e-3 $	$ 0.002982607386 $
$ 0.9 $	$ 0.9 $	$ 0.2 $	$ 1.045289537e-2 $	$ 0.011096367296 $	$ 1.064555345e-2 $	$ 0.011161869914 $
		$ 0.3 $	$ 0.990546789e-2 $	$ 0.010960230827 $	$ 1.017186398e-2 $	$ 0.011029208413 $
		$ 0.4 $	$ 0.927982231e-2 $	$ 0.010839735959 $	$ 0.960539982e-2 $	$ 0.010908463773 $

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Table 3. Solution and absolute error of the problem 3.1 for various values of $ x $, $ y $ and $ t $ at $ k = 0.001, \alpha = 1 $.

$ x $	$ y $	$ t $	VIM ^[1]	AHPM	Exact	VIM ^[1]	AHPM Error
$ 0.1 $	$ 0.1 $	$ 0.2 $	$ 5.355612471e-5 $	$ 0.00005403406722 $	$ 5.393877159e-5 $	$ 3.83e-7 $	$ 9.53e-8 $
		$ 0.3 $	$ 5.331384269e-5 $	$ 0.000054026996643 $	$ 5.388407669e-5 $	$ 5.7e-7 $	$ 1.43e-7 $
		$ 0.4 $	$ 5.307396595e-5 $	$ 0.000054019939232 $	$ 5.382941057e-5 $	$ 7.55e-7 $	$ 1.91e-7 $
$ 0.6 $	$ 0.6 $	$ 0.2 $	$ 2.991347666e-3 $	$ 0.0030365547595 $	$ 3.036507411e-3 $	$ 4.52e-5 $	$ 4.73e-8 $
		$ 0.3 $	$ 2.969760240e-3 $	$ 0.0030358658613 $	$ 3.035778955e-3 $	$ 6.6e-5 $	$ 8.69e-8 $
		$ 0.4 $	$ 2.948601126e-3 $	$ 0.0030351876618 $	$ 3.035050641e-3 $	$ 8.64e-5 $	$ 1.37e-7 $
$ 0.9 $	$ 0.9 $	$ 0.2 $	$ 1.102746671e-2 $	$ 0.011526053047 $	$ 1.153697757e-2 $	$ 5.1e-4 $	$ 1.09e-5 $
		$ 0.3 $	$ 1.073227877e-2 $	$ 0.011518772848 $	$ 1.153454074e-2 $	$ 8.02e-4 $	$ 1.58e-5 $
		$ 0.4 $	$ 1.035600465e-2 $	$ 0.011511900292 $	$ 1.153210438e-2 $	$ 0.00118 $	$ 2.02e-5 $

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Figure 1. AHPM solution, exact solution and absolute error of AHPM solution of Problem 3.1 at $ \alpha $ = 1 and $ t $ = 0.5 when $ k $ = 0.001.

DownLoad: Full-Size Img PowerPoint

Figure 2. AHPM solution of the Problem 3.1 for various values of $ x $ and $ y $ at $ t $ = 0.5 when $ k $ = 0.001.

DownLoad: Full-Size Img PowerPoint

Problem 3.2. Now, we consider FZK$ (3, 3, 3) $ in the form:

$\begin{eqnarray} \frac{{{\partial ^\alpha }u(x, y, t)}}{{\partial {t^\alpha }}} + {\left( {{u^3}(x, y, t)} \right)_x} + 2{\left( {{u^3}(x, y, t)} \right)_{xxx}} + 2{\left( {{u^3}(x, y, t)} \right)_{yyx}} = 0, {\mkern 1mu} 0 \lt \alpha \le 1, \end{eqnarray}$ $

(3.14)

subject to condition

$ u\left( {x, y, 0} \right) = \frac{3}{2}k\sinh \left( {\frac{1}{6}\left( {x + y} \right)} \right). $

When $ \alpha = 1 $. Then the exact solution of equation (3.14):

$ u\left( {x, y, t} \right) = \frac{3}{2}k\sinh \left( {\frac{1}{6}\left( {x + y-kt} \right)} \right). $

As in Eq. (3.14), the non linear part:

$ N\left( {u(x, y, t)} \right) = {\left( {{u^3}(x, y, t)} \right)_x} + 2{\left( {{u^3}(x, y, t)} \right)_{xxx}} + 2{\left( {{u^3}(x, y, t)} \right)_{yyx}}. $

Now, follow the procedure of AHPM, we obtain series of the simpler linear problems as follow:

Zero order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_0}}}{{\partial {t^\alpha }}} = 0, {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {u_0} = \frac{3}{2}k\sinh \left( {\frac{1}{6}\left( {x + y} \right)} \right). \end{eqnarray}$ $

(3.15)

First order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_1}}}{{\partial {t^\alpha }}} = {B_1}{N_0}, {u_1} = 0. \end{eqnarray}$ $

(3.16)

Second order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_2}}}{{\partial {t^\alpha }}} = {B_2}{N_0} + {B_1}{N_1}, {u_2} = 0. \end{eqnarray}$ $

(3.17)

Third order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_3}}}{{\partial {t^\alpha }}} = {B_3}{N_0} + {B_2}{N_1} + {B_1}{N_2}, {u_3} = 0. \end{eqnarray}$ $

(3.18)

Fourth order problem:

$\begin{eqnarray} \frac{{{\partial ^\alpha }{u_4}}}{{\partial {t^\alpha }}} = {B_4}{N_0} + {B_3}{N_1} + {B_2}{N_2} + {B_1}{N_3}, {u_4} = 0. \end{eqnarray}$ $

(3.19)

The respective solutions of the Eqs. (3.15)–(3.19) are given as follow:

$\begin{eqnarray*} {u_0}\left( {x, y, t} \right) & = & \frac{3}{2}k\sinh \left( {{ {1 \over 6}}z} \right), \\ {u_1}\left( {x, y, t} \right) & = & {B_1}{N_0}\frac{{{t^\alpha }}}{{\Gamma \left( {\alpha + 1} \right)}}, \\ {u_2}\left( {x, y, t} \right) & = & {B_2}{N_0}\frac{{{t^\alpha }}}{{\Gamma \left( {\alpha + 1} \right)}} + \frac{3}{{32}}{k^5}{B_1}^2\left[ \begin{array}{l} 801{{\sinh }^3}\left( {{ {1 \over 6}}z} \right) + 765{{\sinh }^4}\left( {{ {1 \over 6}}z} \right)\\ + 127\sinh \left( {{ {1 \over 6}}z} \right)\end{array} \right]\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}}, \end{eqnarray*}$ $

$\begin{eqnarray*} \begin{array}{l} {u_3}\left( {x, y, t} \right) = {B_3}{N_0}\frac{{{t^\alpha }}}{{\Gamma \left( {\alpha + 1} \right)}} + \frac{3}{{32}}{k^5}{B_1}{B_2}\left[ {801{{\sinh }^3}\left( {{ {1 \over 6}}z} \right) + 765{{\sinh }^4}\left( {{ {1 \over 6}}z} \right) + 127\sinh \left( {{ {1 \over 6}}z} \right)} \right]\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}}\\ + {B_1}\frac{3}{{8192}}{k^5}\left[ \begin{array}{l} {B_2}\left( {1120\sinh \left( {{ {1 \over 6}}z} \right) - 9936\sinh \left( {{ {1 \over 2}}z} \right) + 12240\sinh \left( {{ {5 \over 6}}z} \right)} \right)\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}}\\ {B_1}^2{k^2}\left( \begin{array}{l} 1350\cosh \left( {{ {1 \over 2}}z} \right) + 2770\cosh \left( {{ {1 \over 6}}z} \right) - 29070\cosh \left( {{ {5 \over 6}}z} \right)\\ + 32886\cosh \left( {{ {7 \over 6}}z} \right)\end{array} \right)\frac{{\Gamma \left( {2\alpha + 1} \right){t^{3\alpha }}}}{{{{\left( {\Gamma \left( {\alpha + 1} \right)} \right)}^2}\Gamma \left( {3\alpha + 1} \right)}}\\ {B_1}^2{k^2}\left( \begin{array}{l} 56079\cosh \left( {{ {1 \over 2}}z} \right) - 4155\cosh \left( {{ {1 \over 6}}z} \right) - 182835\cosh \left( {{ {5 \over 6}}z} \right) \\ + 155295\cosh \left( {{ {7 \over 6}}z} \right)\end{array} \right)\frac{{{t^{3\alpha }}}}{{\Gamma \left( {3\alpha + 1} \right)}} \end{array} \right] \end{array}, \end{eqnarray*}$ $

$\begin{eqnarray*} \begin{array}{l} {u_4}\left( {x, y, t} \right) = \\ {B_4}{N_0}\frac{{{t^\alpha }}}{{\Gamma \left( {\alpha + 1} \right)}} + \frac{3}{{32}}{k^5}{B_1}{B_3}\left[ {801{{\sinh }^3}\left( {{ {1 \over 6}}z} \right) + 765{{\sinh }^4}\left( {{ {1 \over 6}}z} \right) + 127\sinh \left( {{ {1 \over 6}}z} \right)} \right]\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}} + {B_2}\frac{3}{{8192}}{k^5}\\ \left[ \begin{array}{l} {B_2}\left( {1120\sinh \left( {{ {1 \over 6}}z} \right) - 9936\sinh \left( {{ {1 \over 2}}z} \right) + 12240\sinh \left( {{ {5 \over 6}}z} \right)} \right)\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}}\\ {B_1}^2{k^2}\left( \begin{array}{l} 1350\cosh \left( {{ {1 \over 2}}z} \right) + 2770\cosh \left( {{ {1 \over 6}}z} \right) - 29070\cosh \left( {{ {5 \over 6}}z} \right)\\ + 32886\cosh \left( {{ {7 \over 6}}z} \right)\end{array} \right)\frac{{\Gamma \left( {2\alpha + 1} \right){t^{3\alpha }}}}{{{{\left( {\Gamma \left( {\alpha + 1} \right)} \right)}^2}\Gamma \left( {3\alpha + 1} \right)}}\\ {B_1}^2{k^2}\left( \begin{array}{l} 56079\cosh \left( {{ {1 \over 2}}z} \right) - 4155\cosh \left( {{ {1 \over 6}}z} \right) - 182835\cosh \left( {{ {5 \over 6}}z} \right) \\ + 155295\cosh \left( {{ {7 \over 6}}z} \right)\end{array} \right)\frac{{{t^{3\alpha }}}}{{\Gamma \left( {3\alpha + 1} \right)}} \end{array} \right] + {B_1}\frac{1}{{131072}}{k^5}\\ \left[ \begin{array}{l} {B_1}^3{k^3}\left( {937040\cosh \left( {{ {1 \over 3}}z} \right) - 36000\cosh \left( {{ {2 \over 3}}z} \right) + 16083360\cosh \left( {{ {4 \over 3}}z} \right)} \right)\frac{{\Gamma \left( {3\alpha + 1} \right){t^{4\alpha }}}}{{\Gamma \left( {\alpha + 1} \right)\Gamma \left( {2\alpha + 1} \right)\Gamma \left( {4\alpha + 1} \right)}}\\ + {B_3}\left( { - 476928\sinh \left( {{ {1 \over 2}}z} \right) + 53760\sinh \left( {{ {1 \over 6}}z} \right) + 587520\sinh \left( {{ {5 \over 6}}z} \right)} \right)\frac{{{t^{2\alpha }}}}{{\Gamma \left( {2\alpha + 1} \right)}}\\ + {B_1}^3{k^4}\left( \begin{array}{l} 552744\sinh \left( {{ {1 \over 2}}z} \right) + 1771470\sinh \left( {{ {3 \over 2}}z} \right) - 63900\sinh \left( {{ {1 \over 6}}z} \right) \\ - 275400\sinh \left( {{ {5 \over 6}}z} \right) - 1479870\sinh \left( {{ {7 \over 6}}z} \right) \end{array} \right)\frac{{\Gamma \left( {3\alpha + 1} \right){t^{4\alpha }}}}{{{{\left( {\Gamma \left( {\alpha + 1} \right)} \right)}^2}\Gamma \left( {4\alpha + 1} \right)}}\\ + {B_1}^3{k^4}\left( \begin{array}{l} - 2349000\sinh \left( {{ {1 \over 2}}z} \right) + 39956490\sinh \left( {{ {3 \over 2}}z} \right) - 21300\sinh \left( {{ {1 \over 6}}z} \right)\\ + 23555880\sinh \left( {{ {5 \over 6}}z} \right) - 57758778\sinh \left( {{ {7 \over 6}}z} \right) \end{array} \right)\frac{{\Gamma \left( {2\alpha + 1} \right){t^{4\alpha }}}}{{{{\left( {\Gamma \left( {\alpha + 1} \right)} \right)}^2}\Gamma \left( {4\alpha + 1} \right)}}\\ + {B_1}^3{k^4}\left( \begin{array}{l} 2948400\sinh \left( {{ {1 \over 2}}z} \right) + 33461100\sinh \left( {{ {3 \over 2}}z} \right) - 540600\sinh \left( {{ {1 \over 6}}z} \right)\\ + 9510480\sinh \left( {{ {5 \over 6}}z} \right) - 39704364\sinh \left( {{ {7 \over 6}}z} \right) - 10167120\cosh \left( z \right) \end{array} \right)\frac{{\Gamma \left( {3\alpha + 1} \right){t^{4\alpha }}}}{{\Gamma \left( {\alpha + 1} \right)\Gamma \left( {2\alpha + 1} \right)\Gamma \left( {4\alpha + 1} \right)}}\\ + {B_1}^3{k^4}\left( \begin{array}{l} - 24230988\sinh \left( {{ {1 \over 2}}z} \right) + 188683425\sinh \left( {{ {3 \over 2}}z} \right) + 903510\sinh \left( {{ {1 \over 6}}z} \right) + \\ 147146220\sinh \left( {{ {5 \over 6}}z} \right) - 300495825\sinh \left( {{ {7 \over 6}}z} \right) \end{array} \right)\frac{{{t^{4\alpha }}}}{{\Gamma \left( {4\alpha + 1} \right)}}\\ + {B_1}{B_2}{k^2}\left( \begin{array}{l} 5383584\cosh \left( {{ {1 \over 2}}z} \right) - 398880\cosh \left( {{ {1 \over 6}}z} \right) - 17552160\cosh \left( {{ {5 \over 6}}z} \right) \\+ 14908320\cosh \left( {{ {7 \over 6}}z} \right) \end{array} \right)\frac{{{t^{3\alpha }}}}{{\Gamma \left( {3\alpha + 1} \right)}}\\ + {B_1}{B_2}k\left( \begin{array}{l} 1399680\sinh \left( z \right) - 108160\sinh \left( {{ {1 \over 3}}z} \right) - 576000\sinh \left( {{ {2 \over 3}}z} \right) \end{array} \right)\frac{{\Gamma \left( {2\alpha + 1} \right){t^{3\alpha }}}}{{{{\left( {\Gamma \left( {\alpha + 1} \right)} \right)}^2}\Gamma \left( {3\alpha + 1} \right)}}\\ + {B_1}{B_2}{k^2}\left( \begin{array}{l} 129600\cosh \left( {{ {1 \over 2}}z} \right) + 265920\cosh \left( {{ {1 \over 6}}z} \right) - 2790720\cosh \left( {{ {5 \over 6}}z} \right) \\+ 3157056\cosh \left( {{ {7 \over 6}}z} \right) \end{array} \right)\frac{{\Gamma \left( {2\alpha + 1} \right){t^{3\alpha }}}}{{{{\left( {\Gamma \left( {\alpha + 1} \right)} \right)}^2}\Gamma \left( {3\alpha + 1} \right)}} \end{array} \right] \end{array} \end{eqnarray*}$ $

and so on.

In similar way, we can compute the solution of the next simpler linear problems. We choose $ {{B_1} = {c_1}, B_2} = {c_2}, {B_3} = {c_3}, {B_4} = {c_4} $ and consider

$\begin{eqnarray} \widetilde u(x) = {u_0}(x) + {u_1}(x, {c_1}) + {u_2}(x, {c_1}, {c_2}) + {u_3}(x, {c_1}, {c_2}, {c_3})+{u_3}(x, {c_1}, {c_2}, {c_3}, {c_4}). \end{eqnarray}$ $

(3.20)

We compute the residual;

$\begin{equation*} R\left( {\widetilde u\left( {x, y, t} \right)} \right) = \frac{{{\partial ^\alpha }\widetilde u(x, y, t)}}{{\partial {t^\alpha }}} + {\left( {{{\widetilde u}^3}(x, y, t)} \right)_x} + 2{\left( {{{\widetilde u}^3}(x, y, t)} \right)_{xxx}} + 2{\left( {{{\widetilde u}^3}(x, y, t)} \right)_{yyx}}. \end{equation*}$ $

We obtain number of optimal values of auxiliary constants by using the Eq. (2.13) and choose those optimal values whose sum is in $ \left[{- 1, } \right.\left. 0 \right) $. Now, substituting the optimal values of auxiliary constants (from Table 4) into the Eq. (3.20), we obtain the AHPM solutions for different values of $ \alpha $ at $ k = 0.001. $

Table 4. The auxiliary control constants for the problem 3.2.

$ Aux. Const. $	$ \alpha=1 $	$ \alpha=0.75 $	$ \alpha=0.67 $
$ c_1 $	$ -0.5877657093 $	$ -0.6018150968 $	$ -0.2216842316 $
$ c_2 $	$ -0.1517357373 $	$ -0.1655454563 $	$ -1.594288158 $
$ c_3 $	$ -0.3855312838 $	$ -0.3415880346 $	$ 5.172517408 $
$ c_4 $	$ 0.1250311669 $	$ 0.10894720410 $	$ -4.35654487 $
$ c_1+c_2+c_3+ c_4 $	$ -1.0000 $	$ -1.0000 $	$ -1.0000 $

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If $ \alpha = 1, $ then we have

$\begin{equation*} \begin{split}& \widetilde u\left( {x, y, t} \right) = \\& 1.5 {e^{ - 3}}\sinh \left( {0.17z} \right) - 1.3 {e^{ - 21}}{t^3}\cosh \left( {0.5z} \right) - 6.2 {e^{ - 16}}{t^2}\sinh \left( {0.5z} \right) - 6.4 {e^{ - 28}}{t^4}\sinh \left( {0.5z} \right) - 3.4 {e^{ - 21}} {t^3}\\& \cosh \left( {1.2z} \right) - 1.4{e^{ - 21}}{t^3}\cosh \left( {1.2z} \right) - 2.1{e^{ - 26}}{t^4}\sinh \left( {1.2z} \right) + 1.8{e^{ - 24}}{t^4}\cosh \left( {1.3z} \right) + 1.1{e^{ - 25}}{t^4}\cosh \left( {0.33z} \right) \\& - 2.2{e^{ - 23}}{t^3} \cosh \left( {0.17z} \right) - 4.1 {e^{ - 27}}{t^4}\cosh \left( {0.67z} \right) - 8.9{e^{ - 24}}{t^3}c\cosh \left( {0.17z} \right) + 1.4{e^{ - 20}}{t^3}\sinh \left( {0.33z} \right) + 2.3\\& {e^{ - 17}}{t^2}\sinh \left( {0.17z} \right) + 7.7{e^{ - 20}}{t^3}\sinh \left( {0.67z} \right) + 4.6 {e^{ - 17}}{t^2}\sinh \left( {0.17z} \right) - 4.3 {e^{ - 29}}{t^4}\sinh \left( {0.17z} \right) + 1.4 {e^{ - 26}}{t^4}\\& \sinh \left( {1.5z} \right) + 5.3 {e^{ - 21}}{t^3} \cosh \left( {0.83z} \right) + 5.1 {e^{ - 16}} {t^2} \sinh \left( {0.83z} \right) + 8.4 {e^{ - 27}} {t^4} \sinh \left( {0.83z} \right) + 2.5 {e^{ - 16}} {t^2}\\& \sinh \left( {0.83z} \right) - 1.2 {e^{ - 24}}{t^4}\cosh \left( z \right) - 1.9 {e^{ - 19}}{t^3}\sinh \left( z \right) - 3.8 {e^{ - 10}}t\cosh \left( {0.17z} \right)(9.0{\cosh ^2}\left( {0.17z} \right) - 8) + 3.1\\& {e^{ - 17}}{t^2}\sinh \left( {0.17z} \right)(800.0{\sinh ^2}\left( {0.17z} \right) + 766 {\sinh ^4}\left( {0.17z} \right) + 133). \end{split} \end{equation*}$ $

If $ \alpha = 0.75, $ then we have

$\begin{equation*} \begin{split}& \widetilde u\left( {x, y, t} \right) = \\& 1.5{e^{ - 3}}\sinh \left( {0.17z} \right) - 3.3{e^{ - 21}}{t^{2.25}}\cosh \left( {0.5z} \right) - 3.3{e^{ - 27}}{t^3}\sinh \left( {0.5z} \right) - 9.1{e^{ - 16}}{t^{1.5}}\sinh \left( {0.5z} \right) - 8.3{e^{ - 21}}{t^{2.25}}\\& \cosh \left( {1.2z} \right) - 3.6{e^{ - 21}}{t^{2.25}}\cosh \left( {1.2z} \right) - 8{e^{ - 26}}{t^3}\sinh \left( {1.2z} \right) + 5.6{e^{ - 24}}{t^3}\cosh \left( {1.3z} \right) + 3.3{e^{ - 25}}{t^3}\cosh \left( {0.33z} \right) \\& - 8.2{e^{ - 24}}{t^{2.25}}\cosh \left( {0.17z} \right) - 1.3{e^{ - 26}}{t^3}\cosh \left( {0.67z} \right) - 3.5{e^{ - 24}}{t^{2.25}}\cosh \left( {0.17z} \right) + 3.1 {e^{ - 20}}{t^{(2.25}}\sinh \left( {0.33z} \right)\\& + 3.9 {e^{ - 17}}{t^{1.5}}\sinh \left( {0.17z} \right) + 1.6 {e^{ - 19}}{t^{2.25}}\sinh \left( {0.67z} \right) - 7.8 {e^{ - 29}}{t^3}\sinh \left( {0.17z} \right) + 6.3{e^{ - 17}} {t^{1.5}} \sinh \left( {0.17z} \right) + \\& 5.5 {e^{ - 26}}{t^3}\sinh \left( {1.5z} \right) + 13 {e^{ - 21}}{t^{2.25}}\cosh \left( {0.83z} \right) + 3.4{e^{ - 26}}{t^3}\sinh \left( {0.83z} \right) + 11.2 {e^{ - 16}}{t^{1.5}}\sinh \left( {0.83z} \right) - 3.5\\& {e^{ - 24}} {t^3}\cosh \left( z \right) - 4 {e^{ - 19}} {t^{2.25}}\sinh \left( z \right) - 4.1 {e^{ - 10}} {t^{0.75}}\cosh \left( {0.17z} \right) (9{\cosh ^2}\left( {0.17z} \right) - 8) + 4.7 {e^{ - 17}}{t^{1.5}}\\& \sinh \left( {0.17z} \right) (800{\sinh ^2}\left( {0.17z} \right) + 766{\sinh ^4}\left( {0.17z} \right) + 133). \end{split} \end{equation*}$ $

If $ \alpha = 0.67, $ then we have

$\begin{equation*} \begin{split}& \widetilde u\left( {x, y, t} \right) = \\& 1.5 {e^{ - 3}} \sinh \left( {0.17z} \right) - 2.6 {e^{ - 21}}{t^{2.01}} \cosh \left( {0.5z} \right) - 9.3 {e^{ - 29}}{t^{2.68}}\sinh \left( {0.5z} \right) - 5.3 {e^{ - 15}}{t^{1.34}}\sinh \left( {0.5z} \right) - \\& 3.3{e^{ - 21}}{t^{2.01}}\cosh \left( {1.2z} \right) - 5.8{e^{ - 21}}{t^{2.01}} \cosh \left( {1.2z} \right) - 2.1{e^{ - 27}}{t^{2.68}} \sinh \left( {1.2z} \right) + 1.4{e^{ - 25}}{t^{2.68}}\cosh \left( {1.3z} \right)\\& + 7.9 {e^{ - 27}}{t^{2.68}}\cosh \left( {0.33z} \right) + 1.6 {e^{ - 24}} {t^{2.01}}\cosh \left( {0.17z} \right) - 3.0 {e^{ - 28}} {t^{2.68}}\cosh \left( {0.67z} \right) + 2.7 {e^{ - 24}} {t^{2.01}}\\& \cosh \left( {0.17z} \right) + 4.7 {e^{ - 20}}{t^{2.01}}\sinh \left( {0.33z} \right) + 9.9 {e^{ - 16}}{t^{1.34}} \sinh \left( {0.17z} \right) + 2.5 {e^{ - 19}}{t^{2.01}} \sinh \left( {0.67z} \right) - 1.4\\& {e^{ - 30}}{t^{2.68}}\sinh \left( {0.17z} \right) - 3.9 {e^{ - 16}} {t^{1.34}}\sinh \left( {0.17z} \right) + 1.4 {e^{ - 27}}{t^{2.68}}\sinh \left( {1.5z} \right) + 10 {e^{ - 21}}{t^{2.01}}\cosh \left( {0.83z} \right)\\& + 9.0{e^{ - 28}}{t^{2.68}}\sinh \left( {0.83z} \right) - 4.3{e^{ - 15}}{t^{1.34}} \sinh \left( {0.83z} \right) + 1.1{e^{ - 14}} {t^{1.34}} \sinh \left( {0.83z} \right) - 8.6{e^{ - 26}}{t^{2.68}} \cosh \left( z \right) \\& - 6.1{e^{ - 19}}{t^{2.01}} \sinh \left( z \right) - 4.2{e^{ - 10}}{t^{0.67}} \cosh \left( {0.17z} \right) (9{\cosh ^2}\left( {0.17z} \right) - 8) - 5.8{e^{ - 17}}{t^{1.34}}\sinh \left( {0.17z} \right)\\& (800{\sinh ^2}\left( {0.17z} \right) + 766{\sinh ^4}\left( {0.17z} \right) + 133). \end{split} \end{equation*}$ $

Tables 5–7 show the AHPM solution, VIM solution, exact solution and absolute error of AHPM solution. The AHPM solution, exact solution and absolute error of AHPM solution are plotted for different values of $ \alpha $, $ x $, $ y $ and $ t $ in Figures 3 and 4. It is obvious from the Tables 5–7, Figures 3 and 4, that the AHPM solution of the problem 3.2 is in very good agreement with exact solution.

Table 5. Solution of the problem 3.2 for varios values of $ \alpha $, $ x $, $ y $ and $ t $ at $ k = 0.001 $.

$ x $	$ y $	$ t $	VIM ^[1] ($ \alpha=0.67 $)	AHPM ($ \alpha=0.67 $)	VIM ^[1] ($ \alpha=0.75 $)	AHPM ($ \alpha=0.75 $)
$ 0.1 $	$ 0.1 $	$ 0.2 $	$ 5.000911707e-5 $	$ 0.000050009117063 $	$ 5.000913646e-5 $	$ 0.000050009136457 $
		$ 0.3 $	$ 5.000907252e-5 $	$ 0.000050009072517 $	$ 5.000909264e-5 $	$ 0.000050009092629 $
		$ 0.4 $	$ 5.000903274e-5 $	$ 0.000050009032711 $	$ 5.000905240e-5 $	$ 0.00005000905238 $
$ 0.6 $	$ 0.6 $	$ 0.2 $	$ 3.020038072e-4 $	$ 0.00030200380721 $	$ 3.020038341e-4 $	$ 0.00030200383392 $
		$ 0.3 $	$ 3.020037458e-4 $	$ 0.00030200374584 $	$ 3.020037735e-4 $	$ 0.00030200377354 $
		$ 0.4 $	$ 3.020036910e-4 $	$ 0.000302003691 $	$ 3.020037181e-4 $	$ 0.00030200371809 $
$ 0.9 $	$ 0.9 $	$ 0.2 $	$ 4.567801693e-4 $	$ 0.00045678016935 $	$ 4.567802061e-4 $	$ 0.00045678020615 $
		$ 0.3 $	$ 4.567800847e-4 $	$ 0.00045678008481 $	$ 4.567801231e-4 $	$ 0.00045678012298 $
		$ 0.4 $	$ 4.567800092e-4 $	$ 0.00045678000927 $	$ 4.567800467e-4 $	$ 0.0004567800466 $

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Table 6. Solution and absolute error of the problem 3.2 for various values of $ x $, $ y $ and $ t $ at $ k = 0.001 $ and $ \alpha = 1 $.

$ x $	$ y $	$ t $	VIM ^[1]	AHPM	Exact
$ 0.1 $	$ 0.1 $	$ 0.2 $	$ 5.000918398e-5 $	$ 0.000050009183981 $	$ 4.995923204e-5 $
		$ 0.3 $	$ 5.000914609e-5 $	$ 0.000050009146085 $	$ 4.993421817e-5 $
		$ 0.4 $	$ 5.000910820e-5 $	$ 0.000050009108189 $	$ 4.990920434e-5 $
$ 0.6 $	$ 0.6 $	$ 0.2 $	$ 3.020038992e-4 $	$ 0.0003020038994 $	$ 3.019530008e-4 $
		$ 0.3 $	$ 3.020038472e-4 $	$ 0.00030200384719 $	$ 3.019274992e-4 $
		$ 0.4 $	$ 3.020037950e-4 $	$ 0.00030200379498 $	$ 3.019019978e-4 $
$ 0.9 $	$ 0.9 $	$ 0.2 $	$ 4.567802964e-4 $	$ 0.00045678029634 $	$ 4.567281735e-4 $
		$ 0.3 $	$ 4.567802242e-4 $	$ 0.00045678022442 $	$ 4.567020404e-4 $
		$ 0.4 $	$ 4.567801525e-4 $	$ 0.00045678015251 $	$ 4.566759074e-4 $

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Table 7. AHPM solution and exact solution and absolute error of AHPM solution of the problem 3.2 for various values of $ \alpha $, $ x $, $ y $ and $ t $ at $ k = 0.001 $.

$ x $	$ y $	$ t $	AHPM ($ \alpha=0.67 $)	AHPM ($ \alpha=0.75 $)	AHPM ($ \alpha=1 $)	Exact ($ \alpha=1 $)	Error
$ 1 $	$ 1 $	$ 0.2 $	$ 0.00050931053201 $	$ 0.0005093105733 $	$ 0.0005093106745 $	$ 0.00050925803257 $	$ 5.26e-8 $
		$ 0.4 $	$ 0.00050931035239 $	$ 0.00050931039427 $	$ 0.00050931051311 $	$ 0.00050920522983 $	$ 1.05e-7 $
		$ 0.6 $	$ 0.00050931020147 $	$ 0.00050931023732 $	$ 0.00050931035172 $	$ 0.00050915242765 $	$ 1.58e-7 $
		$ 0.8 $	$ 0.00050931006661 $	$ 0.00050931009319 $	$ 0.00050931019033 $	$ 0.00050909962604 $	$ 2.11e-7 $
		$ 1 $	$ 0.00050930994256 $	$ 0.00050930995788 $	$ 0.00050931002895 $	$ 0.00050904682499 $	$ 2.63e-7 $
$ 5 $	$ 5 $	$ 0.2 $	$ 0.003829187741 $	$ 0.0038291908792 $	$ 0.00382919857 $	$ 0.0038290737537 $	$ 1.25e-7 $
		$ 0.4 $	$ 0.0038291740912 $	$ 0.0038291772737 $	$ 0.0038291863046 $	$ 0.003828936676 $	$ 2.5e-7 $
		$ 0.6 $	$ 0.0038291626226 $	$ 0.0038291653465 $	$ 0.0038291740396 $	$ 0.0038287996024 $	$ 3.74e-7 $
		$ 0.8 $	$ 0.0038291523746 $	$ 0.0038291543934 $	$ 0.003829161775 $	$ 0.0038286625332 $	$ 4.99e-7 $
		$ 1 $	$ 0.0038291429482 $	$ 0.003829144112 $	$ 0.0038291495107 $	$ 0.0038285254682 $	$ 6.24e-7 $
$ 10 $	$ 10 $	$ 0.2 $	$ 0.020993470055 $	$ 0.020993944094 $	$ 0.02099510678 $	$ 0.020996261505 $	$ 1.15e-6 $
		$ 0.4 $	$ 0.020991408888 $	$ 0.020991888681 $	$ 0.02099325193 $	$ 0.020995559858 $	$ 2.31e-6 $
		$ 0.6 $	$ 0.020989679031 $	$ 0.020990088717 $	$ 0.020991398624 $	$ 0.020994858233 $	$ 3.46e-6 $
		$ 0.8 $	$ 0.020988134787 $	$ 0.020988437313 $	$ 0.020989546856 $	$ 0.020994156633 $	$ 4.61e-6 $
		$ 1 $	$ 0.020986715582 $	$ 0.02098688854 $	$ 0.020987696623 $	$ 0.020993455055 $	$ 5.76e-6 $

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Figure 3. AHPM solution, exact solution and absolute error of AHPM solution of Problem 3.2 at $ \alpha $ = 1 and $ t $ = 0.2 when $ k $ = 0.001.

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Figure 4. AHPM solution, exact solution and absolute error of AHPM solution of Problem 3.2 at $ \alpha $ = 1 and $ k $ = 0.0001.

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4. Discussion and conclusions

In this article, asymptotic homotopy perturbation method (AHPM) is developed to solve non-linear fractional models. It is a different procedure from the procedures of HAM, HPM and OHAM. The two special cases, $ ZK(2, 2, 2) $ and $ ZK(3, 3, 3) $ of fractional Zakharov-Kuznetsov model are considered to illustrate a very simple procedure of the homotopy methods. The numerical results in simulation section of AHPM solutions are more accurate to the exact solutions as comparing with fractional complex transform (FCT) using variational iteration method (VIM). In the field of fractional calculus, it is necessary to introduce various procedures and schemes to compute the solution of non-linear fractional models. In this concern, we expect that this new proposed procedure is a best effort. The best improvement and the application of this new procedures to the solution of advanced non-linear fractional models with computer software codes will be our further consideration.

Acknowledgments

The authors would like to thank anonymous referees for their careful corrections to and valuable comments on the original version of this paper.

Conflict of interest

The authors declare no conflict of interest.

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1.	Elvise Berchio, Alessio Falocchi, Maurizio Garrione, On the Stability of a Nonlinear Nonhomogeneous Multiply Hinged Beam, 2021, 20, 1536-0040, 908, 10.1137/20M1374109
2.	Filippo Gazzola, Mohamed Jleli, Bessem Samet, A new detailed explanation of the Tacoma collapse and some optimization problems to improve the stability of suspension bridges, 2022, 5, 2640-3501, 1, 10.3934/mine.2023045

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Mathematical Biosciences and Engineering

A statistical approach to the use of control entropy identifies differences in constraints of gait in highly trained versus untrained runners

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Abstract

1. Introduction

2. Basic idea of AHPM

3. Applications

4. Discussion and conclusions

Acknowledgments

Conflict of interest

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Mathematical Biosciences and Engineering

A statistical approach to the use of control entropy identifies differences in constraints of gait in highly trained versus untrained runners

Related Papers:

Abstract

1. Introduction

2. Basic idea of AHPM

3. Applications

4. Discussion and conclusions

Acknowledgments

Conflict of interest

This article has been cited by:

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