Neural basis of topographical disorientation in the primate posterior cingulate gyrus based on a labeled graph

Yang Yu; Tsuyoshi Setogawa; Jumpei Matsumoto; Hiroshi Nishimaru; Hisao Nishijo; Yang Yu; Tsuyoshi Setogawa; Jumpei Matsumoto; Hiroshi Nishimaru; Hisao Nishijo

doi:10.3934/Neuroscience.2022021

AIMS Neuroscience

2022, Volume 9, Issue 3: 373-394. doi: 10.3934/Neuroscience.2022021

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Theory article Topical Sections

Neural basis of topographical disorientation in the primate posterior cingulate gyrus based on a labeled graph

1.
System Emotional Science, Faculty of Medicine, University of Toyama, Toyama, Japan
2.
Research Center for Idling Brain Science (RCIBS), University of Toyama, Toyama, Japan

Received: 23 June 2022 Revised: 29 August 2022 Accepted: 06 September 2022 Published: 09 September 2022

Patients with lesions in the posterior cingulate gyrus (PCG), including the retrosplenial cortex (RSC) and posterior cingulate cortex (PCC), cannot navigate in familiar environments, nor draw routes on a 2D map of the familiar environments. This suggests that the topographical knowledge of the environments (i.e., cognitive map) to find the right route to a goal is represented in the PCG, and the patients lack such knowledge. However, theoretical backgrounds in neuronal levels for these symptoms in primates are unclear. Recent behavioral studies suggest that human spatial knowledge is constructed based on a labeled graph that consists of topological connections (edges) between places (nodes), where local metric information, such as distances between nodes (edge weights) and angles between edges (node labels), are incorporated. We hypothesize that the population neural activity in the PCG may represent such knowledge based on a labeled graph to encode routes in both 3D environments and 2D maps. Since no previous data are available to test the hypothesis, we recorded PCG neuronal activity from a monkey during performance of virtual navigation and map drawing-like tasks. The results indicated that most PCG neurons responded differentially to spatial parameters of the environments, including the place, head direction, and reward delivery at specific reward areas. The labeled graph-based analyses of the data suggest that the population activity of the PCG neurons represents the distance traveled, locations, movement direction, and navigation routes in the 3D and 2D virtual environments. These results support the hypothesis and provide a neuronal basis for the labeled graph-based representation of a familiar environment, consistent with PCG functions inferred from the human clinicopathological studies.

Keywords:

Citation: Yang Yu, Tsuyoshi Setogawa, Jumpei Matsumoto, Hiroshi Nishimaru, Hisao Nishijo. Neural basis of topographical disorientation in the primate posterior cingulate gyrus based on a labeled graph[J]. AIMS Neuroscience, 2022, 9(3): 373-394. doi: 10.3934/Neuroscience.2022021

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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 ) 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 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 ) 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

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

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

| Show Table

DownLoad: CSV

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.

DownLoad: Full-Size Img PowerPoint

Figure 4. AHPM solution, exact solution and absolute error of AHPM solution of Problem 3.2 at

$\alpha$ = 1 and

$k$ = 0.0001.

DownLoad: Full-Size Img PowerPoint

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.

Acknowledgments

This research was supported by the Takeda Science Foundation and a research grant from the University of Toyama.

Conflict of interest

The authors declare no conflict of interest.

Author contributions

HisN and HirN conceived the study and designed the experiment. YY, TS, and HisN performed the experiment. YY, TS, HisN and JM analyzed data and wrote the paper. HisN, HirN, JM, and AB revised the paper. All the authors discussed the results and commented on the manuscript, and read and approved the final manuscript.

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