Analytical Approaches for Approximate Solution of the Time-Fractional Coupled Schrödinger–KdV Equation

Abstract

In this article, we use the homotopy perturbation method and the Adomian decomposition method with the Yang transformation to discover analytical solution to the time-fractional coupled Schrödinger–KdV equation. In the Caputo sense, fractional derivatives are described. A convergent series is used to calculate the solutions of fractional PDEs. Analytical results achieved applying the homotopy perturbation and decomposition techniques are numerically calculated and represented in the form of tables and figures. The simplicity, efficacy, and high degree of accuracy of the used method are then demonstrated by comparing these solutions to the actual solutions and the results. Finally, the applied approaches are the most popular and convergent methods for solving nonlinear fractional-order partial deferential problems.

1. Introduction

Fractional calculus is a branch of calculus that extend a function’s derivative to any order. Because fractional calculus is used by many different disciplines of engineering and science to define the characteristics of numerous genuine physical processes, it has recently attracted the attention of several scholars in specialized fields of applied technology and engineering. Due to the convolutions associated with heterogeneity, the concept of the fractional derivative has been developed [1,2,3]. Many scholars began working on FC to illustrate their applications while examining complex models, due to the rapid expansion of mathematical methods with computer packages. Other scholars have recently put forward numerous novel ideas for FC, which established the framework. FC concept and theory are linked to practical projects, and it is widely used in a variety of fields [4,5,6].

Partial differential equations express a specific relationship between an unknown function and its partial derivatives (PDEs). PDEs can be found in almost every field of engineering and research. In recent years, the application of PDEs in fields such as image processing, biology, economics, graphics, and social sciences has increased [7,8]. As a result, when some independent variables interact with one another in each of the fields mentioned earlier, appropriate functions in these variables can be established, allowing the modeling of various processes through equations for the associated functions [9,10,11]. The study of PDEs has various elements. The traditional method, which dominated the nineteenth century, was to develop methods for identifying explicit solutions. It is important to mention that there are some extremely complicated equations that computations cannot solve. Furthermore, it is preferable in most circumstances for the solution to be unique and stable against minor data disruptions. A theoretical understanding of the equation helps determine whether these conditions are met [12,13,14]. Numerous approaches for solving traditional PDEs have been presented, with numerous solutions. Because of this, symmetry evaluation is a great tool for comprehending partial differential equations, especially when looking at equations generated from mathematical concepts connected to accounting. Despite the notion that symmetry is the foundation of nature, the bulk of observations in the natural world lacks it. A clever technique for disguising symmetry is to provide unanticipated symmetry-breaking events. The two categories are finite and infinitesimal symmetry. There are two types of discrete and continuous finite symmetry. Fundamental symmetry such as equality and temporal inversion are discrete while space is a continuous transform [15,16,17]. Mathematicians have always been fascinated by patterns. The identification of spatial and planar patterns truly took off in the seventeenth century. Regrettably, it has become increasingly difficult to solve fractional nonlinear differential equations precisely [18,19,20,21,22,23].

The nonlinear Schrödinger equation (NLSE) is a type of PDE that has been studied a lot for decades. This is because it can be used in a wide range of situations. Different kinds of NLSEs are used in different fields to describe things such as Bose–Einstein condensates [24,25], nonlinear optics [26,27], and fluid dynamics [28], among others. One thing that sets the different types of NLSEs [29,30] apart is their ability to generate more light on the disruption that happens when electromagnetic pulses travel to the optical extreme. Moreover, studies of the soliton and numerical solutions for NLSEs have been published in the literature [31] using different methods. In [32], Kang-Jia Wang proposed a new method named the direct mapping method to study the generalized third-order nonlinear Schrödinger equation. The classical nonlinear Schrödinger–KdV equation is given by [33]

$$\begin{array}{cc}\hfill & \iota {\xi }_{\tau }={\xi }_{\vartheta \vartheta }+\xi \nu ,\hfill \\ \hfill & {\nu }_{\tau }=-6\nu {\nu }_{\vartheta }-{\nu }_{\vartheta \vartheta \vartheta }+{\left(\right|\xi |}^2{)}_{\vartheta },\hfill \end{array}$$

(1)

where $\iota =\sqrt{-1}$. By turning $\xi =\omega +\iota \psi$, it is possible to separate Equation (1) into real and imaginary parts. Here is the model equation for the coupled fractional Schrödinger–KdV equation:

$$\begin{array}{cc}\hfill & \frac{{\partial }^{ß}\omega }{\partial {\tau }^{ß}}=\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau ),\hfill \\ \hfill & \frac{{\partial }^{ß}\psi }{\partial {\tau }^{ß}}=\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau ),\hfill \\ \hfill & \frac{{\partial }^{ß}\nu }{\partial {\tau }^{ß}}=-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta },0<ß\le 1,\hfill \end{array}$$

(2)

In this work, we extend the Adomian decomposition transform method and the homotopy perturbation transform method to find an approximate solution of the given equations. Xiao-Jun Yang presented the Yang transform, which can be utilized to solve various differential equations having constant coefficients. The Adomian decomposition technique (ADM) [34,35] is a well-known systematic method for solving deterministic or stochastic operator equations, such as ordinary, partial, integral and integrodifferential equations. The ADM technique is a sophisticated technique that provides fast algorithms for analytical solutions and numeric simulations in engineering and applied sciences. On the other hand, He [36,37] presented the homotopy perturbation approach in 1998. The implementations of the homotopy perturbation approach, which are fully treated in [38,39,40,41,42], have recently attracted a lot of attention. This method sees the solution as the sum of an infinite series, which usually converges quickly to accurate solutions. This approach has been used for a wide range of mathematical problems.

2. Preliminaries

In this section, we give some definitions of fractional calculus used in our present work.

Definition 1. 

The fractional derivative of order ß in the Caputo sense is defined as [43,44]

$$\begin{array}{cc}\hfill & D_{\tau }^{ß}\omega (\vartheta ,\tau )=\frac{1}{\Gamma (k-ß)}{\int }_0^{\tau }{(\tau -\rho )}^{k-ß-1}{\omega }^{\left(k\right)}(\vartheta ,\rho )d\rho ,k-1<ß\le k,k\in N.\hfill \end{array}$$

(3)

Definition 2. 

For a function $\omega \left(\tau \right)$, the Yang transform is denoted by $Y\left\{\omega \right(\tau \left)\right\}$ or $\mathcal{M}\left(u\right)$ and is defined as in [45,46]

$$Y\left\{\omega \left(\tau \right)\right\}=\mathcal{M}\left(u\right)={\int }_0^{\infty }{e}^{\frac{-\tau }{u}}\omega \left(\tau \right)d\tau ,\tau >0,u\in (-{\tau }_1,{\tau }_2).$$

(4)

and the inverse transform is as

$$Y^{-1}\left\{\mathcal{M}\left(u\right)\right\}=\omega \left(\tau \right).$$

(5)

Definition 3. 

The Yang transformation in terms of fractional-order derivative is as in [45,46]:

$$Y\left\{{\omega }^{ß}\left(\tau \right)\right\}=\frac{\mathcal{M}\left(u\right)}{u^{\gamma }}-\sum _{k=0}^{n-1}\frac{{\omega }^k\left(0\right)}{u^{ß-(k+1)}},0<ß\le n.$$

(6)

Definition 4. 

The Yang transform in terms of nth derivative is as in [45,46]

$$Y\left\{{\omega }^n\left(\tau \right)\right\}=\frac{\mathcal{M}\left(u\right)}{u^n}-\sum _{k=0}^{n-1}\frac{{\omega }^k\left(0\right)}{u^{n-k-1}},\forall n=1,2,3,\cdots$$

(7)

3. Homotopy Perturbation Transform Method

Consider the fractional partial differential equation.

$$D_{\tau }^{ß}\omega (\vartheta ,\tau )+{\mathcal{P}}_1\omega (\vartheta ,\tau )+{\mathcal{Q}}_1\omega (\vartheta ,\tau )=00<ß\le 1,$$

(8)

with initial conditions

$$\omega (\vartheta ,0)=h\left(\vartheta \right).$$

Here, $D_{\tau }^{ß}$ represents the fractional differential operator, ${\mathcal{P}}_1$ and ${\mathcal{Q}}_1$ the linear and nonlinear operators, respectively.

Applying Yang’s transformation to (7),

$$Y\left[D_{\tau }^{ß}\omega (\vartheta ,\tau )\right]+Y[{\mathcal{P}}_1\omega (\vartheta ,\tau )+{\mathcal{Q}}_1\omega (\vartheta ,\tau )]=0,$$

(9)

$$\frac{1}{u^{ß}}\{\mathcal{M}\left(u\right)-u\omega \left(0\right)\}+Y[{\mathcal{P}}_1\omega (\vartheta ,\tau )+{\mathcal{Q}}_1\omega (\vartheta ,\tau )]=0.$$

(10)

From (9), we have

$$\mathcal{M}\left(\omega \right)=uh\left(\vartheta \right)-u^{ß}Y[{\mathcal{P}}_1\omega (\vartheta ,\tau )+{\mathcal{Q}}_1\omega (\vartheta ,\tau )].$$

(11)

Using the inverse Yang transform,

$$\omega (\vartheta ,\tau )=H\left(\vartheta \right)-Y^{-1}\left[u^{ß}Y[{\mathcal{P}}_1\omega (\vartheta ,\tau )+{\mathcal{Q}}_1\omega (\vartheta ,\tau )]\right],$$

(12)

Now, by the homotopy perturbation technique

$$\omega (\vartheta ,\tau )=\sum _{k=0}^{\infty }{p}^k{\omega }_k(\vartheta ,\tau ),$$

(13)

The nonlinear functions are defined as

$${\mathcal{Q}}_1\omega (\vartheta ,\tau )=\sum _{k=0}^{\infty }{p}^k{H}_k\left(\omega \right),$$

(14)

for He’s polynomials $H_k$ as

$$\begin{array}{cc}\hfill & H_k({\omega }_0,{\omega }_1,...,{\omega }_k)=\frac{1}{k!}\frac{{\partial }^k}{\partial p^k}{\left[{\mathcal{Q}}_1\left(\sum _{i=0}^{\infty }{p}^i{\omega }_i\right)\right]}_{p=0}.\hfill \\ \hfill & k=0,1,2,3\cdots \hfill \end{array}$$

(15)

Putting (13) and (14) in Equation (11)

$$\sum _{k=0}^{\infty }{p}^k{\omega }_k(\vartheta ,\tau )=H\left(\vartheta \right)-p\times \left(Y^{-1}\left[u^{ß}Y\{{\mathcal{P}}_1\sum _{k=0}^{\infty }{p}^k{\omega }_k(\vartheta ,\tau )+\sum _{k=0}^{\infty }{p}^k{H}_k\left(\omega \right)\}\right]\right).$$

(16)

By comparing the p-terms coefficients, we get

$$\begin{array}{cc}\hfill & p^0:{\omega }_0(\vartheta ,\tau )=H\left(\vartheta \right),\hfill \\ \hfill & p^1:{\omega }_1(\vartheta ,\tau )=-Y^{-1}\left[u^{ß}Y({\mathcal{P}}_1{\omega }_0(\vartheta ,\tau )+H_0\left(\omega \right))\right],\hfill \\ \hfill & p^2:{\omega }_2(\vartheta ,\tau )=-Y^{-1}\left[u^{ß}Y({\mathcal{P}}_1{\omega }_1(\vartheta ,\tau )+H_1\left(\omega \right))\right],\hfill \\ \hfill & .\hfill \\ \hfill & .\hfill \\ \hfill & .\hfill \\ \hfill & p^k:{\omega }_k(\vartheta ,\tau )=-Y^{-1}\left[u^{ß}Y({\mathcal{P}}_1{\omega }_{k-1}(\vartheta ,\tau )+H_{k-1}\left(\omega \right))\right],\hfill \\ \hfill k>0,k\in N.\end{array}$$

(17)

Thus, we simply determinate ${\omega }_k(\vartheta ,\tau )$, with the aid of which we obtain a convergence series. At $p\to 1,$ we get

$$\omega (\vartheta ,\tau )=\underset{M\to \infty }{lim}\sum _{k=1}^M{\omega }_k(\vartheta ,\tau ).$$

(18)

4. Yang’s Transform Decomposition Method

Consider the fractional partial differential equation

$$D_{\tau }^{ß}\omega (\vartheta ,\tau )={\mathcal{P}}_1(\vartheta ,\tau )+{\mathcal{Q}}_1(\vartheta ,\tau )+{\mathcal{R}}_1(\vartheta ,\tau ),0<ß\le 1,$$

(19)

having initial sources

$$\omega (\vartheta ,0)=\omega \left(\vartheta \right).$$

where $D_{\tau }^{ß}=\frac{{\partial }^{ß}}{\partial {\tau }^{ß}}$ represents the fractional differential operator, ${\mathcal{P}}_1$ and ${\mathcal{Q}}_1$ represent the linear and nonlinear functions, respectively, and ${\mathcal{R}}_1$ is the source function.

Applying Yang’s transformation to (18),

$$Y\left[D_{\tau }^{ß}\omega (\vartheta ,\tau )\right]=Y[{\mathcal{P}}_1(\vartheta ,\tau )+{\mathcal{Q}}_1(\vartheta ,\tau )+{\mathcal{R}}_1(\vartheta ,\tau )].$$

(20)

Applying the differentiation property of Yang transform, we get

$$\frac{1}{u^{ß}}\{\mathcal{M}\left(u\right)-u\omega \left(0\right)\}=Y[{\mathcal{P}}_1(\vartheta ,\tau )+{\mathcal{Q}}_1(\vartheta ,\tau )+{\mathcal{R}}_1(\vartheta ,\tau )].$$

(21)

From (20), we have

$$\mathcal{M}\left(\omega \right)=u\omega \left(0\right)+u^{ß}Y[{\mathcal{P}}_1(\vartheta ,\tau )+{\mathcal{Q}}_1(\vartheta ,\tau )+{\mathcal{R}}_1(\vartheta ,\tau )],$$

(22)

Using the inverse Yang transform,

$$\omega (\vartheta ,\tau )=\omega \left(0\right)+Y^{-1}[u^{ß}Y[{\mathcal{P}}_1(\vartheta ,\tau )+{\mathcal{Q}}_1(\vartheta ,\tau )+{\mathcal{R}}_1(\vartheta ,\tau )].$$

(23)

The YTDM series form solution of $\omega (\vartheta ,\tau )$ is defined as

$$\omega (\vartheta ,\tau )=\sum _{m=0}^{\infty }{\omega }_m(\vartheta ,\tau ).$$

(24)

The nonlinear term ${\mathcal{Q}}_1$ is defined as

$${\mathcal{Q}}_1(\vartheta ,\tau )=\sum _{m=0}^{\infty }{\mathcal{A}}_m.$$

(25)

The nonlinear function found with the help of Adomian polynomials is defined as

$${\mathcal{A}}_m=\frac{1}{m!}{\left[\frac{{\partial }^m}{\partial {\delta }^m}\left\{{\mathcal{Q}}_1\left(\sum _{k=0}^{\infty }{\delta }^k{\vartheta }_k,\sum _{k=0}^{\infty }{\delta }^k{\tau }_k\right)\right\}\right]}_{\delta =0},$$

(26)

By putting Equations (23) and (25) into (22), we obtain

$$\sum _{m=0}^{\infty }{\omega }_m(\vartheta ,\tau )=\omega \left(0\right)+Y^{-1}\left[u^{ß}Y\left\{{\mathcal{R}}_1(\vartheta ,\tau )\right\}\right]+Y^{-1}{u}^{ß}\left[Y\left\{{\mathcal{P}}_1(\sum _{m=0}^{\infty }{\vartheta }_m,\sum _{m=0}^{\infty }{\tau }_m)+\sum _{m=0}^{\infty }{\mathcal{A}}_m\right\}\right].$$

(27)

We define the following terms,

$${\omega }_0(\vartheta ,\tau )=\omega \left(0\right)+Y^{-1}\left[u^{ß}Y\left\{{\mathcal{R}}_1(\vartheta ,\tau )\right\}\right],$$

(28)

$${\omega }_1(\vartheta ,\tau )=Y^{-1}\left[u^{ß}Y\{{\mathcal{P}}_1({\vartheta }_0,{\tau }_0)+{\mathcal{A}}_0\}\right],$$

and the general term for $m\ge 1$ is defined as

$${\omega }_{m+1}(\vartheta ,\tau )=Y^{-1}\left[u^{ß}Y\{{\mathcal{P}}_1({\vartheta }_m,{\tau }_m)+{\mathcal{A}}_m\}\right].$$

5. Applications

Example 1. 

Consider the fractional Schrödinger–KdV equation

$$\begin{array}{cc}\hfill & \frac{{\partial }^{ß}\omega }{\partial {\tau }^{ß}}=\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau ),0<ß\le 1,\hfill \\ \hfill & \frac{{\partial }^{ß}\psi }{\partial {\tau }^{ß}}=\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau ),\hfill \\ \hfill & \frac{{\partial }^{ß}\nu }{\partial {\tau }^{ß}}=-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta },\hfill \end{array}$$

(29)

with the initial conditions

$$\left\{\begin{array}{cc}\hfill & \omega (\vartheta ,0)=cos\left(\vartheta \right),\hfill \\ \hfill & \psi (\vartheta ,0)=sin\left(\vartheta \right),\hfill \\ \hfill & \nu (\vartheta ,0)=\frac{3}{4}.\hfill \end{array}\right.$$

At $ß=1$, the exact result of Equation (28) is

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )=cos(\vartheta +\frac{\tau }{4}),\hfill \\ \hfill & \psi (\vartheta ,\tau )=sin(\vartheta +\frac{\tau }{4}),\hfill \\ \hfill & \nu (\vartheta ,\tau )=\frac{3}{4}.\hfill \end{array}$$

(30)

Applying Yang’s transform to (28),

$$\begin{array}{cc}\hfill & Y\left\{\frac{{\partial }^{ß}\omega }{\partial {\tau }^{ß}}\right\}=Y\left[\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & Y\left\{\frac{{\partial }^{ß}\psi }{\partial {\tau }^{ß}}\right\}=Y\left[\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & Y\left\{\frac{{\partial }^{ß}\nu }{\partial {\tau }^{ß}}\right\}=Y\left[-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta }\right],\hfill \end{array}$$

Using the Yang transformation, we get

$$\begin{array}{cc}\hfill & \frac{1}{u^{ß}}\{\mathcal{M}\left(u\right)-u\omega \left(0\right)\}=Y\left[\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & \frac{1}{u^{ß}}\{\mathcal{M}\left(u\right)-u\psi \left(0\right)\}=Y\left[\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & \frac{1}{u^{ß}}\{\mathcal{M}\left(u\right)-u\nu \left(0\right)\}=Y\left[-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta }\right],\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill & \mathcal{M}\left(u\right)=u\omega \left(0\right)+u^{ß}Y\left[\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & \mathcal{M}\left(u\right)=u\psi \left(0\right)+u^{ß}Y\left[\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & \mathcal{M}\left(u\right)=u\nu \left(0\right)+u^{ß}Y\left[-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta }\right].\hfill \end{array}$$

(32)

Using the Yang inverse transform,

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )=\omega \left(0\right)+Y^{-1}\left[u^{ß}\left\{Y\left(\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau )\right)\right\}\right],\hfill \\ \hfill & \psi (\vartheta ,\tau )=\psi \left(0\right)+Y^{-1}\left[u^{ß}\left\{Y\left(\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau )\right)\right\}\right],\hfill \\ \hfill & \nu (\vartheta ,\tau )=\psi \left(0\right)+Y^{-1}\left[u^{ß}\left\{Y\left(-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta }\right)\right\}\right],\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )=cos\left(\vartheta \right)+Y^{-1}\left[u^{ß}\left\{Y\left(\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau )\right)\right\}\right],\hfill \\ \hfill & \psi (\vartheta ,\tau )=sin\left(\vartheta \right)+Y^{-1}\left[u^{ß}\left\{Y\left(\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau )\right)\right\}\right],\hfill \\ \hfill & \nu (\vartheta ,\tau )=\frac{3}{4}+Y^{-1}\left[u^{ß}\left\{Y\left(-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta }\right)\right\}\right],\hfill \end{array}$$

(34)

Now, by the homotopy perturbation method,

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )={\omega }_0+{\omega }_{1}p+{\omega }_2{p}^2+\cdots ,\hfill \\ & \psi (\vartheta ,\tau )={\psi }_0+{\psi }_{1}p+{\psi }_2{p}^2+\cdots ,\hfill \\ & \nu (\vartheta ,\tau )={\nu }_0+{\nu }_{1}p+{\nu }_2{p}^2+\cdots \hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }{p}^k{\omega }_k(\vartheta ,\tau )=cos\left(\vartheta \right)+p\left(Y^{-1}\left[u^{ß}Y\left[{\left(\sum _{k=0}^{\infty }{p}^k{\omega }_k(\vartheta ,\tau )\right)}_{\vartheta \vartheta }+\sum _{k=0}^{\infty }{p}^k{H}_k^1(\vartheta ,\tau )\right]\right]\right),\hfill \\ \hfill & \sum _{k=0}^{\infty }{p}^k{\psi }_k(\vartheta ,\tau )=sin\left(\vartheta \right)+p\left(Y^{-1}\left[u^{ß}Y\left[{\left(\sum _{k=0}^{\infty }{p}^k{\psi }_k(\vartheta ,\tau )\right)}_{\vartheta \vartheta }-\sum _{k=0}^{\infty }{p}^k{H}_k^2(\vartheta ,\tau )\right]\right]\right),\hfill \\ \hfill & \sum _{k=0}^{\infty }{p}^k{\nu }_k(\vartheta ,\tau )=\frac{3}{4}+p\left(Y^{-1}\left[u^{ß}Y\left[-{\left(\sum _{k=0}^{\infty }{p}^k{\nu }_k(\vartheta ,\tau )\right)}_{\vartheta \vartheta \vartheta }-\sum _{k=0}^{\infty }{p}^k{H}_k^3(\vartheta ,\tau )\right]\right]\right),\hfill \end{array}$$

(36)

The nonlinear terms found with the help of He’s polynomials are given as

$$\begin{array}{cc}\hfill & H_0^1\left(\vartheta \right)={\psi }_0{\nu }_0\hfill \\ \hfill & H_1^1\left(\vartheta \right)={\psi }_1{\nu }_0+{\nu }_1{\psi }_0,\hfill \\ \hfill & H_2^1\left(\vartheta \right)={\psi }_1{\nu }_1+{\nu }_0{\psi }_2+{\psi }_0{\nu }_2,\hfill \\ \hfill & ⋮\hfill \\ \hfill & H_0^2\left(\vartheta \right)={\omega }_0{\nu }_0,\hfill \\ \hfill & H_1^2\left(\vartheta \right)={\omega }_1{\nu }_0+{\nu }_1{\omega }_0,\hfill \\ \hfill & H_2^2\left(\vartheta \right)={\omega }_1{\nu }_1+{\nu }_0{\omega }_2+{\omega }_0{\nu }_2,\hfill \\ \hfill & ⋮\hfill \\ \hfill & H_0^3\left(\vartheta \right)=-6{\nu }_0{\nu }_{0\vartheta }+2{\omega }_0{\nu }_{0\vartheta }+2{\psi }_0{\psi }_{0\vartheta },\hfill \\ \hfill & H_1^3\left(\vartheta \right)=-6{\nu }_1{\nu }_{0\vartheta }-6{\nu }_{1\vartheta }{\nu }_0+2{\omega }_1{\nu }_{0\vartheta }+2{\nu }_{1\vartheta }{\omega }_0+2{\psi }_1{\psi }_{0\vartheta }+2{\psi }_{1\vartheta }{\psi }_0,\hfill \\ \hfill & H_2^3\left(\vartheta \right)=-6{\nu }_2{\nu }_{0\vartheta }-6{\nu }_{1\vartheta }{\nu }_1-6{\nu }_2{\nu }_{0\vartheta }+2{\omega }_2{\nu }_{0\vartheta }+2{\nu }_{1\vartheta }{\omega }_1+2{\omega }_0{\nu }_{2\vartheta }+2{\psi }_2{\psi }_{0\vartheta }+2{\psi }_{1\vartheta }{\psi }_1+2{\psi }_0{\psi }_{2\vartheta }.\hfill \end{array}$$

By comparing the coefficients of the p-terms, we achieve

$$\begin{array}{cc}\hfill & p^0:{\omega }_0(\vartheta ,\tau )=cos\left(\vartheta \right),\hfill \\ \hfill & {\psi }_0(\vartheta ,\tau )=sin\left(\vartheta \right),\hfill \\ \hfill & {\nu }_0(\vartheta ,\tau )=\frac{3}{4},\hfill \\ \hfill & p^1:{\omega }_1(\vartheta ,\tau )=-\frac{sin\left(\vartheta \right)}{4\Gamma (ß+1)},\hfill \\ \hfill & {\psi }_1(\vartheta ,\tau )=-\frac{7cos\left(\vartheta \right)}{4\Gamma (ß+1)},\hfill \\ \hfill & {\nu }_1(\vartheta ,\tau )=\frac{2sin\left(x\right)cos\left(\vartheta \right)}{\Gamma (ß+1)},\hfill \\ \hfill & p^2:{\omega }_2(\vartheta ,\tau )=-\frac{cos\left(\vartheta \right)(-39+32co{s}^2\left(\vartheta \right))}{16\Gamma (2ß+1)},\hfill \\ \hfill & {\psi }_2(\vartheta ,\tau )=-\frac{sin\left(\vartheta \right)(-7+32co{s}^2\left(\vartheta \right))}{16\Gamma (2ß+1)},\hfill \\ \hfill & {\nu }_2(\vartheta ,\tau )=\frac{(2co{s}^2\left(\vartheta \right)-1)(8cos\left(\vartheta \right)-9)}{2\Gamma (2ß+1)},\hfill \end{array}$$

$$\begin{array}{cc}\hfill & p^3:{\omega }_3(\vartheta ,\tau )=-\frac{sin\left(\vartheta \right)(-25{(\Gamma (ß+1))}^2-480{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & -\frac{512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)+256{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right)+224\Gamma (2ß+1)co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}-\cdots \hfill \\ \hfill & {\psi }_3(\vartheta ,\tau )=-\frac{cos\left(\vartheta \right)(1329{(\Gamma (ß+1))}^2-1824{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & +\frac{(512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)-256{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right)-32\Gamma (2ß+1)+32\Gamma (2ß+1)co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2},\hfill \\ \hfill & {\nu }_3(\vartheta ,\tau )=-\frac{sin\left(\vartheta \right)(-272{(\Gamma (ß+1))}^2+576{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right)-70{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right))}{8\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & -\frac{512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)+384\Gamma (2ß+1)co{s}^3\left(\vartheta \right)-143\Gamma (2ß+1)cos\left(\vartheta \right)+16\Gamma (2ß+1)co{s}^2\left(\vartheta \right)-8\Gamma (2ß+1)}{8\Gamma (3ß+1){(\Gamma (ß+1))}^2},\hfill \\ \hfill & ⋮\hfill \end{array}$$

At $p\to 1$, we have

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )={\omega }_0+{\omega }_1+{\omega }_2+{\omega }_3+\cdots \hfill \\ \hfill & =cos\left(\vartheta \right)-\frac{sin\left(\vartheta \right)}{4\Gamma (ß+1)}-\frac{cos\left(\vartheta \right)(-39+32co{s}^2\left(\vartheta \right))}{16\Gamma (2ß+1)}-\frac{sin\left(\vartheta \right)(-25{(\Gamma (ß+1))}^2-480{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & -\frac{512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)+256{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right)+224\Gamma (2ß+1)co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}-\cdots \hfill \end{array}$$

$$\begin{array}{cc}\hfill & \psi (\vartheta ,\tau )={\psi }_0+{\psi }_1+{\psi }_2+{\psi }_3+\cdots \hfill \\ \hfill & =sin\left(\vartheta \right)-\frac{7cos\left(\vartheta \right)}{4\Gamma (ß+1)}-\frac{sin\left(\vartheta \right)(-7+32co{s}^2\left(\vartheta \right))}{16\Gamma (2ß+1)}-\frac{cos\left(\vartheta \right)(1329{(\Gamma (ß+1))}^2-1824{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & +\frac{(512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)-256{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right)-32\Gamma (2ß+1)+32\Gamma (2ß+1)co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}-\cdots \hfill \end{array}$$

$$\begin{array}{cc}\hfill & \nu (\vartheta ,\tau )={\nu }_0+{\nu }_1+{\nu }_2+{\nu }_3+\cdots \hfill \\ \hfill & =\frac{3}{4}+\frac{2sin\left(x\right)cos\left(\vartheta \right)}{\Gamma (ß+1)}+\frac{(2co{s}^2\left(\vartheta \right)-1)(8cos\left(\vartheta \right)-9)}{2\Gamma (2ß+1)}\hfill \\ \hfill & -\frac{sin\left(\vartheta \right)(-272{(\Gamma (ß+1))}^2+576{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right)-70{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right))}{8\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & -\frac{512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)+384\Gamma (2ß+1)co{s}^3\left(\vartheta \right)-143\Gamma (2ß+1)cos\left(\vartheta \right)+16\Gamma (2ß+1)co{s}^2\left(\vartheta \right)-8\Gamma (2ß+1)}{8\Gamma (3ß+1){(\Gamma (ß+1))}^2}+\cdots \hfill \end{array}$$

YTDM Solution:

Apply Yang’s transform to (28),

$$\begin{array}{cc}\hfill & Y\left\{\frac{{\partial }^{ß}\omega }{\partial {\tau }^{ß}}\right\}=Y\left[\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & Y\left\{\frac{{\partial }^{ß}\psi }{\partial {\tau }^{ß}}\right\}=Y\left[\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & Y\left\{\frac{{\partial }^{ß}\nu }{\partial {\tau }^{ß}}\right\}=Y\left[-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta }\right],\hfill \end{array}$$

Using the Yang transformation, we get

$$\begin{array}{cc}\hfill & \frac{1}{u^{ß}}\{\mathcal{M}\left(u\right)-u\omega \left(0\right)\}=Y\left[\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & \frac{1}{u^{ß}}\{\mathcal{M}\left(u\right)-u\psi \left(0\right)\}=Y\left[\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau )\right],\hfill \\ \hfill & \frac{1}{u^{ß}}\{\mathcal{M}\left(u\right)-u\nu \left(0\right)\}=Y\left[-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta }\right].\hfill \end{array}$$

(37)

Using the Yang inverse transform,

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )=\omega \left(0\right)+Y^{-1}\left[u^{ß}\left\{Y\left(\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau )\right)\right\}\right],\hfill \\ \hfill & \psi (\vartheta ,\tau )=\psi \left(0\right)+Y^{-1}\left[u^{ß}\left\{Y\left(\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau )\right)\right\}\right],\hfill \\ \hfill & \nu (\vartheta ,\tau )=\nu \left(0\right)+Y^{-1}\left[u^{ß}\left\{Y\left(-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta }\right)\right\}\right],\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )=cos\left(\vartheta \right)+Y^{-1}\left[u^{ß}\left\{Y\left(\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\psi (\vartheta ,\tau )\nu (\vartheta ,\tau )\right)\right\}\right],\hfill \\ \hfill & \psi (\vartheta ,\tau )=cos\left(\vartheta \right)+Y^{-1}\left[u^{ß}\left\{Y\left(\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\omega (\vartheta ,\tau )\nu (\vartheta ,\tau )\right)\right\}\right],\hfill \\ \hfill & \nu (\vartheta ,\tau )=\frac{3}{4}+Y^{-1}\left[u^{ß}\left\{Y\left(-6\nu \frac{\partial \nu }{\partial \vartheta }-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+2\omega \frac{\partial \nu }{\partial \vartheta }+2\psi \frac{\partial \psi }{\partial \vartheta }\right)\right\}\right].\hfill \\ \hfill \end{array}$$

(39)

The YTDM series form solution of $\omega (\vartheta ,\tau )$, $\psi (\vartheta ,\tau )$ and $\nu (\vartheta ,\tau )$ is defined as:

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )=\sum _{m=0}^{\infty }{\omega }_m(\vartheta ,\tau ),\hfill \\ \hfill & \psi (\vartheta ,\tau )=\sum _{m=0}^{\infty }{\psi }_m(\vartheta ,\tau ),and\hfill \\ \hfill & \nu (\vartheta ,\tau )=\sum _{m=0}^{\infty }{\nu }_m(\vartheta ,\tau )\hfill \end{array}$$

where the Adomian polynomials $\psi \nu ={\sum }_{m=0}^{\infty }{\mathcal{A}}_m$, $\omega \nu ={\sum }_{m=0}^{\infty }{\mathcal{B}}_m$ and $-6\nu {\nu }_{\vartheta }+2\omega {\nu }_{\vartheta }+2\psi {\psi }_{\vartheta }={\sum }_{m=0}^{\infty }{\mathcal{C}}_m$ and the nonlinear terms have been characterized. Thus, Equation (39) can be rewritten as

$$\begin{array}{cc}\hfill & \sum _{m=0}^{\infty }{\omega }_m(\vartheta ,\tau )=\omega (\vartheta ,0)+Y^{-1}\left[u^{ß}Y\left[\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\sum _{m=0}^{\infty }{\mathcal{A}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\psi }_m(\vartheta ,\tau )=\psi (\vartheta ,0)+Y^{-1}\left[u^{ß}Y\left[\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\sum _{m=0}^{\infty }{\mathcal{B}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\nu }_m(\vartheta ,\tau )=\nu (\vartheta ,0)+Y^{-1}\left[u^{ß}Y\left[-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+\sum _{m=0}^{\infty }{\mathcal{C}}_m\right]\right],\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \sum _{m=0}^{\infty }{\omega }_m(\vartheta ,\tau )=cos\left(\vartheta \right)+Y^{-1}\left[u^{ß}Y\left[\frac{{\partial }^2\omega }{\partial {\vartheta }^2}+\sum _{m=0}^{\infty }{\mathcal{A}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\psi }_m(\vartheta ,\tau )=sin\left(\vartheta \right)+Y^{-1}\left[u^{ß}Y\left[\frac{{\partial }^2\psi }{\partial {\vartheta }^2}-\sum _{m=0}^{\infty }{\mathcal{B}}_m\right]\right],\hfill \\ \hfill & \sum _{m=0}^{\infty }{\nu }_m(\vartheta ,\tau )=\frac{3}{4}+Y^{-1}\left[u^{ß}Y\left[-\frac{{\partial }^3\nu }{\partial {\vartheta }^3}+\sum _{m=0}^{\infty }{\mathcal{C}}_m\right]\right].\hfill \end{array}$$

(40)

With the help of Adomian polynomials, the nonlinear term is defined as

$$\begin{array}{cc}\hfill & {\mathcal{A}}_0={\psi }_0{\nu }_0,{\mathcal{A}}_1={\psi }_1{\nu }_0+{\nu }_1{\psi }_0,{\mathcal{A}}_2={\psi }_1{\nu }_1+{\nu }_0{\psi }_2+{\psi }_0{\nu }_2,\hfill \\ \hfill & {\mathcal{B}}_0={\omega }_0{\nu }_0,{\mathcal{B}}_1={\omega }_1{\nu }_0+{\nu }_1{\omega }_0,{\mathcal{B}}_2={\omega }_1{\nu }_1+{\nu }_0{\omega }_2+{\omega }_0{\nu }_2,\hfill \\ \hfill & {\mathcal{C}}_0=-6{\nu }_0{\nu }_{0\vartheta }+2{\omega }_0{\nu }_{0\vartheta }+2{\psi }_0{\psi }_{0\vartheta },{\mathcal{C}}_1=-6{\nu }_1{\nu }_{0\vartheta }-6{\nu }_{1\vartheta }{\nu }_0+2{\omega }_1{\nu }_{0\vartheta }+2{\nu }_{1\vartheta }{\omega }_0+2{\psi }_1{\psi }_{0\vartheta }+2{\psi }_{1\vartheta }{\psi }_0,\hfill \\ & {\mathcal{C}}_2=-6{\nu }_2{\nu }_{0\vartheta }-6{\nu }_{1\vartheta }{\nu }_1-6{\nu }_2{\nu }_{0\vartheta }+2{\omega }_2{\nu }_{0\vartheta }+2{\nu }_{1\vartheta }{\omega }_1+2{\omega }_0{\nu }_{2\vartheta }+2{\psi }_2{\psi }_{0\vartheta }+2{\psi }_{1\vartheta }{\psi }_1+2{\psi }_0{\psi }_{2\vartheta },\hfill \end{array}$$

By comparing both sides of Equation (40), we get

$$\begin{array}{cc}\hfill & {\omega }_0(\vartheta ,\tau )=cos\left(\vartheta \right),\hfill \\ \hfill & {\psi }_0(\vartheta ,\tau )=sin\left(\vartheta \right),\hfill \\ \hfill & {\nu }_0(\vartheta ,\tau )=\frac{3}{4}.\hfill \end{array}$$

For $m=0$:

$$\begin{array}{cc}\hfill & {\omega }_1(\vartheta ,\tau )=-\frac{sin\left(\vartheta \right)}{4\Gamma (ß+1)},\hfill \\ \hfill & {\psi }_1(\vartheta ,\tau )=-\frac{7cos\left(\vartheta \right)}{4\Gamma (ß+1)},\hfill \\ \hfill & {\nu }_1(\vartheta ,\tau )=\frac{2sin\left(x\right)cos\left(\vartheta \right)}{\Gamma (ß+1)}.\hfill \end{array}$$

For $m=1$:

$$\begin{array}{cc}\hfill & {\omega }_2(\vartheta ,\tau )=-\frac{cos\left(\vartheta \right)(-39+32co{s}^2\left(\vartheta \right))}{16\Gamma (2ß+1)},\hfill \\ \hfill & {\psi }_2(\vartheta ,\tau )=-\frac{sin\left(\vartheta \right)(-7+32co{s}^2\left(\vartheta \right))}{16\Gamma (2ß+1)},\hfill \\ \hfill & {\nu }_2(\vartheta ,\tau )=\frac{(2co{s}^2\left(\vartheta \right)-1)(8cos\left(\vartheta \right)-9)}{2\Gamma (2ß+1)}.\hfill \end{array}$$

For $m=2$:

$$\begin{array}{cc}\hfill & p^3:{\omega }_3(\vartheta ,\tau )=-\frac{sin\left(\vartheta \right)(-25{(\Gamma (ß+1))}^2-480{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & -\frac{512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)+256{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right)+224\Gamma (2ß+1)co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & {\psi }_3(\vartheta ,\tau )=-\frac{cos\left(\vartheta \right)(1329{(\Gamma (ß+1))}^2-1824{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & +\frac{(512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)-256{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right)-32\Gamma (2ß+1)+32\Gamma (2ß+1)co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2},\hfill \\ \hfill & {\nu }_3(\vartheta ,\tau )=-\frac{sin\left(\vartheta \right)(-272{(\Gamma (ß+1))}^2+576{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right)-70{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right))}{8\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & -\frac{512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)+384\Gamma (2ß+1)co{s}^3\left(\vartheta \right)-143\Gamma (2ß+1)cos\left(\vartheta \right)+16\Gamma (2ß+1)co{s}^2\left(\vartheta \right)-8\Gamma (2ß+1)}{8\Gamma (3ß+1){(\Gamma (ß+1))}^2},\hfill \\ \hfill & ⋮\hfill \end{array}$$

The series form solution is given as

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )=\sum _{m=0}^{\infty }{\omega }_m(\vartheta ,\tau )={\omega }_0(\vartheta ,\tau )+{\omega }_1(\vartheta ,\tau )+{\omega }_2(\vartheta ,\tau )+{\omega }_3(\vartheta ,\tau )+\cdots \hfill \\ \hfill & \psi (\vartheta ,\tau )=\sum _{m=0}^{\infty }{\psi }_m(\vartheta ,\tau )={\psi }_0(\vartheta ,\tau )+{\psi }_1(\vartheta ,\tau )+\psi (\vartheta ,\tau )+{\psi }_3(\vartheta ,\tau )+\cdots \hfill \\ \hfill & \nu (\vartheta ,\tau )=\sum _{m=0}^{\infty }{\nu }_m(\vartheta ,\tau )={\nu }_0(\vartheta ,\tau )+{\nu }_1(\vartheta ,\tau )+\nu (\vartheta ,\tau )+{\nu }_3(\vartheta ,\tau )+\cdots \hfill \end{array}$$

$$\begin{array}{cc}\hfill & \omega (\vartheta ,\tau )=cos\left(\vartheta \right)-\frac{sin\left(\vartheta \right)}{4\Gamma (ß+1)}-\frac{cos\left(\vartheta \right)(-39+32co{s}^2\left(\vartheta \right))}{16\Gamma (2ß+1)}-\frac{sin\left(\vartheta \right)(-25{(\Gamma (ß+1))}^2-480{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & -\frac{512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)+256{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right)+224\Gamma (2ß+1)co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}-\cdots \hfill \\ \hfill & \psi (\vartheta ,\tau )=sin\left(\vartheta \right)-\frac{7cos\left(\vartheta \right)}{4\Gamma (ß+1)}-\frac{sin\left(\vartheta \right)(-7+32co{s}^2\left(\vartheta \right))}{16\Gamma (2ß+1)}-\frac{cos\left(\vartheta \right)(1329{(\Gamma (ß+1))}^2-1824{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & +\frac{(512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)-256{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right)-32\Gamma (2ß+1)+32\Gamma (2ß+1)co{s}^2\left(\vartheta \right))}{64\Gamma (3ß+1){(\Gamma (ß+1))}^2}-\cdots \hfill \\ \hfill & \nu (\vartheta ,\tau )=\frac{3}{4}+\frac{2sin\left(x\right)cos\left(\vartheta \right)}{\Gamma (ß+1)}+\frac{(2co{s}^2\left(\vartheta \right)-1)(8cos\left(\vartheta \right)-9)}{2\Gamma (2ß+1)}\hfill \\ \hfill & -\frac{sin\left(\vartheta \right)(-272{(\Gamma (ß+1))}^2+576{(\Gamma (ß+1))}^{2}co{s}^2\left(\vartheta \right)-70{(\Gamma (ß+1))}^{2}cos\left(\vartheta \right))}{8\Gamma (3ß+1){(\Gamma (ß+1))}^2}\hfill \\ \hfill & -\frac{512{(\Gamma (ß+1))}^{2}co{s}^3\left(\vartheta \right)+384\Gamma (2ß+1)co{s}^3\left(\vartheta \right)-143\Gamma (2ß+1)cos\left(\vartheta \right)+16\Gamma (2ß+1)co{s}^2\left(\vartheta \right)-8\Gamma (2ß+1)}{8\Gamma (3ß+1){(\Gamma (ß+1))}^2}+\cdots \hfill \end{array}$$

In Figure 1, the actual and approximate solutions for $\omega (\vartheta ,\tau )$. Figure 2, the suggested techniques result at $ß=0.9,0.8$ for $\omega (\vartheta ,\tau )$.

Figure 3, the three and two-dimensional behavior of the suggested technique’s result at various fractional orders for $\omega (\vartheta ,\tau )$. Similarly in Figure 4, actual and approximate solutions for $\psi (\vartheta ,\tau )$ and Figure 5, suggested technique’s result at $ß=0.9,0.8$ for $\psi (\vartheta ,\tau )$. In Figure 6, three and two-dimensional behavior of the suggested technique’s result at various fractional orders for $\psi (\vartheta ,\tau )$. In

Table 1. HPTM and YTDM corresponding absolute errors of Example 1 for $\omega (\vartheta ,\tau )$.

$\mathit{\tau }$	$\mathit{\omega }$	$|\mathit{Exact}-\mathit{HPTM}|$	$|\mathit{Exact}-\mathit{HPTM}|$	$|\mathit{Exact}-\mathit{YTDM}|$	$|\mathit{Exact}-\mathit{YTDM}|$
		$\mathbf{\sigma }=\mathbf{0}.\mathbf{6}$	$\mathbf{\sigma }=\mathbf{1}$	$\mathbf{\sigma }=\mathbf{0}.\mathbf{8}$	$\mathbf{\sigma }=\mathbf{1}$
	−3	5.2714952$\times {10}^{-4}$	3.2020000$\times {10}^{-10}$	3.8360200$\times {10}^{-7}$	3.2020000$\times {10}^{-10}$
	−2	2.4281644$\times {10}^{-3}$	2.4670000$\times {10}^{-9}$	2.3747700$\times {10}^{-6}$	2.4670000$\times {10}^{-9}$
	−1	2.0967361$\times {10}^{-3}$	2.2620000$\times {10}^{-9}$	2.1825200$\times {10}^{-6}$	2.2620000$\times {10}^{-9}$
0.01	0	1.6242170$\times {10}^{-4}$	1.0200000$\times {10}^{-10}$	1.6300000$\times {10}^{-7}$	1.0200000$\times {10}^{-10}$
	1	2.2722498$\times {10}^{-3}$	2.1620000$\times {10}^{-9}$	2.2000200$\times {10}^{-6}$	2.1620000$\times {10}^{-9}$
	2	2.2929819$\times {10}^{-3}$	2.2670000$\times {10}^{-9}$	2.3610700$\times {10}^{-6}$	2.2670000$\times {10}^{-9}$
	3	2.0555702$\times {10}^{-4}$	4.2020000$\times {10}^{-10}$	3.5130200$\times {10}^{-7}$	4.2020000$\times {10}^{-10}$
	−3	5.4601850$\times {10}^{-3}$	2.9405000$\times {10}^{-9}$	3.1974050$\times {10}^{-6}$	2.9405000$\times {10}^{-9}$
	−2	1.9783029$\times {10}^{-2}$	1.8934000$\times {10}^{-8}$	1.9049740$\times {10}^{-5}$	1.8934000$\times {10}^{-8}$
	−1	1.5917448$\times {10}^{-2}$	1.7524000$\times {10}^{-8}$	1.7387740$\times {10}^{-5}$	1.7524000$\times {10}^{-8}$
0.02	0	2.5825619$\times {10}^{-3}$	9.7811000$\times {10}^{-10}$	2.6040000$\times {10}^{-7}$	9.7811000$\times {10}^{-10}$
	1	1.8708176$\times {10}^{-2}$	1.7524000$\times {10}^{-8}$	1.7669140$\times {10}^{-5}$	1.7524000$\times {10}^{-8}$
	2	1.7633579$\times {10}^{-2}$	1.8934000$\times {10}^{-8}$	1.8832940$\times {10}^{-5}$	1.8934000$\times {10}^{-8}$
	3	3.4675125$\times {10}^{-4}$	2.9405000$\times {10}^{-9}$	2.6819050$\times {10}^{-6}$	2.9405000$\times {10}^{-9}$
	−3	2.2456521$\times {10}^{-2}$	1.0061000$\times {10}^{-8}$	1.1224610$\times {10}^{-5}$	1.0061000$\times {10}^{-8}$
	−2	6.7545169$\times {10}^{-2}$	6.4001000$\times {10}^{-8}$	6.4465610$\times {10}^{-5}$	6.4001000$\times {10}^{-8}$
	−1	5.0533099$\times {10}^{-2}$	5.9086000$\times {10}^{-8}$	5.8437160$\times {10}^{-5}$	5.9086000$\times {10}^{-8}$
0.03	0	1.2938868$\times {10}^{-2}$	1.0000000$\times {10}^{-10}$	1.3181000$\times {10}^{-6}$	1.0000000$\times {10}^{-10}$
	1	6.4514900$\times {10}^{-2}$	5.9186000$\times {10}^{-8}$	5.9861360$\times {10}^{-5}$	5.9186000$\times {10}^{-8}$
	2	5.6776230$\times {10}^{-2}$	6.3901000$\times {10}^{-8}$	6.3368410$\times {10}^{-5}$	6.3901000$\times {10}^{-8}$
	3	3.1622443$\times {10}^{-3}$	9.8610000$\times {10}^{-9}$	8.6148100$\times {10}^{-6}$	9.8610000$\times {10}^{-9}$
	−3	6.2270596$\times {10}^{-2}$	2.3881000$\times {10}^{-8}$	2.7631810$\times {10}^{-5}$	2.3881000$\times {10}^{-8}$
	−2	1.6092170$\times {10}^{-1}$	1.5176800$\times {10}^{-7}$	1.5320728$\times {10}^{-4}$	1.5176800$\times {10}^{-7}$
	−1	1.1162213$\times {10}^{-1}$	1.4004800$\times {10}^{-7}$	1.3792458$\times {10}^{-4}$	1.4004800$\times {10}^{-7}$
0.04	0	4.0302305$\times {10}^{-2}$	4.0000000$\times {10}^{-10}$	4.1653000$\times {10}^{-6}$	4.0000000$\times {10}^{-10}$
	1	1.5517299$\times {10}^{-1}$	1.4044800$\times {10}^{-7}$	1.4242548$\times {10}^{-4}$	1.4044800$\times {10}^{-7}$
	2	1.27378348$\times {10}^{-1}$	1.5136800$\times {10}^{-7}$	1.4974038$\times {10}^{-4}$	1.5136800$\times {10}^{-7}$
	3	1.7527364$\times {10}^{-2}$	2.3181000$\times {10}^{-8}$	1.9384610$\times {10}^{-5}$	2.3181000$\times {10}^{-8}$
	−3	1.3808520$\times {10}^{-1}$	4.7001000$\times {10}^{-8}$	5.5967110$\times {10}^{-5}$	4.7001000$\times {10}^{-8}$
	−2	3.1390079$\times {10}^{-1}$	2.9654000$\times {10}^{-7}$	2.9999530$\times {10}^{-4}$	2.9654000$\times {10}^{-7}$
	−1	2.0111743$\times {10}^{-1}$	2.7341000$\times {10}^{-7}$	2.6820900$\times {10}^{-4}$	2.7341000$\times {10}^{-7}$
0.05	0	9.6572362$\times {10}^{-2}$	1.0000000$\times {10}^{-9}$	1.0167200$\times {10}^{-5}$	1.0000000$\times {10}^{-9}$
	1	3.0547397$\times {10}^{-1}$	2.7441000$\times {10}^{-7}$	2.7919580$\times {10}^{-4}$	2.7441000$\times {10}^{-7}$
	2	2.3352422$\times {10}^{-1}$	2.9554000$\times {10}^{-7}$	2.9153300$\times {10}^{-4}$	2.9554000$\times {10}^{-7}$
	3	5.3126621$\times {10}^{-2}$	4.4901000$\times {10}^{-8}$	3.5836110$\times {10}^{-5}$	4.4901000$\times {10}^{-8}$

and

Table 2. YTDM and YTDM corresponding absolute errors of Example 1 for $\psi (\vartheta ,\tau )$.

$\mathit{\tau }$	$\mathit{\omega }$	$|\mathit{Exact}-\mathit{HPTM}|$	$|\mathit{Exact}-\mathit{HPTM}|$	$|\mathit{Exact}-\mathit{YTDM}|$	$|\mathit{Exact}-\mathit{YTDM}|$
		$\mathbf{\sigma }=\mathbf{0}.\mathbf{6}$	$\mathbf{\sigma }=\mathbf{1}$	$\mathbf{\sigma }=\mathbf{0}.\mathbf{8}$	$\mathbf{\sigma }=\mathbf{1}$
	−3	2.5471399$\times {10}^{-3}$	2.6420000$\times {10}^{-9}$	2.5758200$\times {10}^{-6}$	2.6420000$\times {10}^{-9}$
	−2	9.3264440$\times {10}^{-4}$	9.9100000$\times {10}^{-10}$	1.0689100$\times {10}^{-6}$	9.9100000$\times {10}^{-10}$
	−1	1.5393200$\times {10}^{-3}$	1.4650000$\times {10}^{-9}$	1.4207500$\times {10}^{-6}$	1.4650000$\times {10}^{-9}$
0.01	0	2.5960407$\times {10}^{-3}$	2.6040000$\times {10}^{-9}$	2.6040900$\times {10}^{-6}$	2.6040000$\times {10}^{-9}$
	1	1.2659736$\times {10}^{-3}$	1.3650000$\times {10}^{-9}$	1.3932500$\times {10}^{-6}$	1.3650000$\times {10}^{-9}$
	2	1.2280238$\times {10}^{-3}$	1.1910000$\times {10}^{-9}$	1.0984100$\times {10}^{-6}$	1.1910000$\times {10}^{-9}$
	3	2.5929819$\times {10}^{-3}$	2.5420000$\times {10}^{-9}$	2.5803200$\times {10}^{-6}$	2.5420000$\times {10}^{-9}$
	−3	2.0004111$\times {10}^{-2}$	2.0683000$\times {10}^{-8}$	2.0585530$\times {10}^{-5}$	2.0683000$\times {10}^{-8}$
	−2	6.2136800$\times {10}^{-3}$	8.6820000$\times {10}^{-9}$	8.4318200$\times {10}^{-6}$	8.6820000$\times {10}^{-9}$
	−1	1.3289579$\times {10}^{-2}$	1.1330000$\times {10}^{-8}$	1.1474000$\times {10}^{-5}$	1.1330000$\times {10}^{-8}$
0.02	0	2.0574461$\times {10}^{-2}$	2.0833000$\times {10}^{-8}$	2.0830730$\times {10}^{-5}$	2.0833000$\times {10}^{-8}$
	1	8.9432781$\times {10}^{-3}$	1.1230000$\times {10}^{-8}$	1.1035800$\times {10}^{-5}$	1.1230000$\times {10}^{-8}$
	2	1.0910313$\times {10}^{-2}$	8.6820000$\times {10}^{-9}$	8.9054200$\times {10}^{-6}$	8.6820000$\times {10}^{-9}$
	3	2.0733013$\times {10}^{-2}$	2.0583000$\times {10}^{-}$	2.0658960$\times {10}^{-5}$	2.0583000$\times {10}^{-8}$
	−3	6.5851181$\times {10}^{-2}$	6.9624000$\times {10}^{-8}$	6.9403340$\times {10}^{-5}$	6.9624000$\times {10}^{-8}$
	−2	1.6683033$\times {10}^{-2}$	2.9074000$\times {10}^{-8}$	2.8053540$\times {10}^{-5}$	2.9074000$\times {10}^{-8}$
	−1	4.7823418$\times {10}^{-2}$	3.8094000$\times {10}^{-8}$	3.9088440$\times {10}^{-5}$	3.8094000$\times {10}^{-8}$
0.03	0	6.8361240$\times {10}^{-2}$	7.0312000$\times {10}^{-10}$	7.0292730$\times {10}^{-5}$	7.0312000$\times {10}^{-10}$
	1	2.6048052$\times {10}^{-2}$	3.7794000$\times {10}^{-8}$	3.6870240$\times {10}^{-5}$	3.7794000$\times {10}^{-8}$
	2	4.0213594$\times {10}^{-2}$	2.9374000$\times {10}^{-8}$	3.0450640$\times {10}^{-5}$	2.9374000$\times {10}^{-8}$
	3	6.9503047$\times {10}^{-2}$	6.9624000$\times {10}^{-9}$	6.9775220$\times {10}^{-5}$	6.9624000$\times {10}^{-9}$
	−3	1.5125507$\times {10}^{-1}$	1.6496600$\times {10}^{-7}$	1.6432856$\times {10}^{-4}$	1.6496600$\times {10}^{-7}$
	−2	2.9324565$\times {10}^{-2}$	6.8965000$\times {10}^{-8}$	6.5535650$\times {10}^{-5}$	6.8965000$\times {10}^{-8}$
	−1	1.1956681$\times {10}^{-1}$	9.0359000$\times {10}^{-8}$	9.3510290$\times {10}^{-5}$	9.0359000$\times {10}^{-8}$
0.04	0	1.5852901$\times {10}^{-1}$	1.6666600$\times {10}^{-7}$	1.6658335$\times {10}^{-4}$	1.6666600$\times {10}^{-7}$
	1	5.1740371$\times {10}^{-2}$	8.9759000$\times {10}^{-8}$	8.6500390$\times {10}^{-5}$	8.9759000$\times {10}^{-8}$
	2	1.0261813$\times {10}^{-1}$	6.9765000$\times {10}^{-8}$	7.3110550$\times {10}^{-5}$	6.9765000$\times {10}^{-8}$
	3	1.6262999$\times {10}^{-1}$	1.6506600$\times {10}^{-7}$	1.6550399$\times {10}^{-4}$	1.6506600$\times {10}^{-7}$
	−3	2.8437467$\times {10}^{-1}$	3.2221000$\times {10}^{-7}$	3.2057670$\times {10}^{-4}$	3.2221000$\times {10}^{-7}$
	−2	3.7453597$\times {10}^{-2}$	1.3445600$\times {10}^{-7}$	1.2611366$\times {10}^{-4}$	1.3445600$\times {10}^{-7}$
	−1	2.4390214$\times {10}^{-1}$	1.7672400$\times {10}^{-7}$	1.8429774$\times {10}^{-4}$	1.7672400$\times {10}^{-7}$
0.05	0	3.0101538$\times {10}^{-1}$	3.2552000$\times {10}^{-7}$	3.2526660$\times {10}^{-4}$	3.2552000$\times {10}^{-7}$
	1	8.1376463$\times {10}^{-2}$	1.7502400$\times {10}^{-7}$	1.6718684$\times {10}^{-4}$	1.7502400$\times {10}^{-7}$
	2	2.1307959$\times {10}^{-1}$	1.3645600$\times {10}^{-7}$	1.4460366$\times {10}^{-4}$	1.3645600$\times {10}^{-7}$
	3	3.1163126$\times {10}^{-1}$	3.2241000$\times {10}^{-7}$	3.2344630$\times {10}^{-4}$	3.2241000$\times {10}^{-7}$

, HPTM and YTDM corresponding absolute errors of Example 1 for $\omega (\vartheta ,\tau )$ and $\psi (\vartheta ,\tau )$.

6. Conclusions

The current research focused on a fractional view analysis of heat-related equations employing effective analytical techniques. A complex method was used to provide approximate analytical solutions for nonlinear fractional Schrödinger–KdV equations having fractional and integer orders. The modified approach, which computed a distinct initial approximation for each iteration, also improved the accuracy of the methods. The graphical examination of the solutions obtained was completed satisfactorily. The results of the research revealed a high level of agreement among the proposed and exact solutions. The solutions, which were generated for various fractional orders of the problems, revealed diverse dynamical patterns as the fractional order varied. The tables provided a number of fractional-order solutions, demonstrating the applicability of the suggested method. In comparison to other analytical and numerical techniques, the current method proved to be an effective and simple procedure. Furthermore, the proposed method needed less calculations and could thus be used to solve other fractional-order problems.