**Analytical Solutions of (2+Time Fractional Order) Dimensional Physical Models, Using Modified Decomposition Method**

#### **Hassan Khan 1, Umar Farooq 1, Rasool Shah 1, Dumitru Baleanu 2,3, Poom Kumam 4,5,\* and Muhammad Arif <sup>1</sup>**


Received: 24 October 2019; Accepted: 21 November 2019; Published: 23 December 2019

**Abstract:** In this article, a new analytical technique based on an innovative transformation is used to solve (2+time fractional-order) dimensional physical models. The proposed method is the hybrid methodology of Shehu transformation along with Adomian decomposition method. The series form solution is obtained by using the suggested method which provides the desired rate of convergence. Some numerical examples are solved by using the proposed method. The solutions of the targeted problems are represented by graphs which have confirmed closed contact between the exact and obtained solutions of the problems. Based on the novelty and straightforward implementation of the method, it is considered to be one of the best analytical techniques to solve linear and non-linear fractional partial differential equations.

**Keywords:** Shehu transformation; Adomian decomposition; analytical solution; Caputo derivatives; (2+time fractional-order) dimensional physical models

#### **1. Introduction**

Fractional calculus is considered to be a powerful tool for modeling complex phenomenon. Recently, the researchers have shown the greatest interest towards fractional calculus because of its numerous applications in different fields of sciences. Despite complicated background of fractional calculus, it came into being from simple question of L'Hospital. The first order represent slope of a function, what will it represent for fractional order ( <sup>1</sup> <sup>2</sup> )? To find the answer of this question, the mathematicians have managed to open a new window of opportunities to improve the mathematical modeling of real world problems, which has given birth to many new questions and intriguing results. These newly established results have numerous implementation in many areas of engineering [1,2], such as fractional-order Buck master and diffusion problems [3], fractional-order telegraph model [4,5], fractional KdV-Burger-Kuramoto equation [6], fractal vehicular traffic flow [7], fractional Drinfeld-Sokolov-Wilson equation [8], fractional-order anomalous sub-diffusion model [9], fractional design of hepatitis B virus [10], fractional modeling chickenpox disease [11], fractional

blood ethanol concentration model [12], fractional model for tuberculosis [13], fractional vibration equation [14], fractional Black-Scholes option pricing equations [15], fractionally damped beams [16], fractionally damped coupled system [17], fractional-order heat, wave and diffusion equations [18,19], fractional order pine wilt disease model [20], fractional diabetes model [21] etc.

Nowadays, the focus of the researchers is to develop different numerical and analytical techniques for the solution of fractional-order models. Therefore, different types of analytical and numerical methods have been developed and used for the solution of different fractional models. The analytical algorithm, the history of integral transform traced back to the time when Laplace started work an integral transform in 1780*s* and Joseph Fourier in 1822. Integral transformations are without question one of the most useful and effective methods in theoretical and applied mathematics, with numerous uses in quantum physics, mechanical engineering and several other areas of science. Moreover, the integral transform is used in chemistry, architecture, and other social sciences to evaluate various models [22]. In recent years, different integral transform such as Laplace transform [23–25] , Fourier transform [26,27], Hankel transform [28], Mellin transform [29], Z-transform [30], Wavelet transform [31], Elzaki transform [32,33], Mahgoub transform [34], Aboodh transform [35], Mohand transform [36], Sumudu transform [37,38], Hermite transform [39] etc have been used for the solution of different physical models.

**Originality of the paper**: In this article, we have applied a new analytical technique, which is based on generalization of sumudu and laplace transform with Adomian decomposition method (ADM) to solve (2+time fractional-order) dimensional physical models. In the present research we have analyzed the fractional view of some important physical problems by using Shehu decomposition method (SDM). Some important fractional-order problems are solved, which provide the best information about the targeted physical problems as compare to integer-order problems solution. The results of the integer-order problem are compared with the fractional-order problems. In conclusion, in the present research work, we provided and improved the existing physical models of integer-order by using the idea of fractional calculus. The modified mathematical models of fractional-order derivative are solved by using a new and sophisticated analytical method. Moreover, the proposed analytical method has provided the solutions of the problems that have a very close contact with the exact solutions of the problems. The methodology can be extended towards other fractional-order partial differential equations, that are frequently occurred in science and engineering.

The rest of the paper is organized as: In Section 2, we presented the basic definitions and theorem of the proposed method. In Section 3, we have discussed the implementation of proposed transformation. In Section 4 we evaluated the numerical examples by using the proposed technique and discussed the plots. In Section 5 we lastly summarized our results.

#### **2. Preliminaries Concepts**

In this section, we present some fundamental and appropriate definitions and preliminary concepts related to the fractional calculus and the Shehu transformation.

#### **Definition 1.** *Shehu transform*

*Shehu transformation is new and similar to other integral transformation which is defined for functions of exponential order [40]. We take a function in the set A define by*

$$A = \{ \mu(\tau) : \exists , \rho\_1, \rho\_2 > 0, \left| \mu(\tau) \right| < M e^{\frac{|\tau|}{\rho\_i}}, \text{ if } \tau \in [0, \infty), \tag{1}$$

*The Shehu transform which is represented by S*(.) *for a function u*(*τ*) *is defined as*

$$S\{u(\tau)\} = V(s,\mu) = \int\_0^\infty u(\tau)e^{\frac{-s\tau}{\mu}}u(\tau)d\tau,\ \ \tau > 0,\ s > 0. \tag{2}$$

*Appl. Sci.* **2020**, *10*, 122

*The Shehu transform of a function u*(*τ*) *is V*(*s*, *μ*)*: then u*(*τ*) *is called the inverse of V*(*s*, *μ*) *which is expressed as*

$$S^{-1}\left\{V(s,\mu)\right\} = u(\tau), \text{ for } \tau \ge 0, \text{ S}^{-1} \text{is inverse Shehu transform.} \tag{3}$$

**Definition 2.** *Shehu transform for nth derivatives*

*The Shehu transformation for nth derivatives is defined as [40]*

$$S\left\{\boldsymbol{u}^{(n)}(\boldsymbol{\tau})\right\} = \frac{\boldsymbol{s}^{n}}{\mu^{n}}V(\boldsymbol{s},\mu) - \sum\_{k=0}^{n-1} \left(\frac{\boldsymbol{s}}{\mu}\right)^{n-k-1} \boldsymbol{u}^{(k)}(0). \tag{4}$$

**Definition 3.** *Caputo operator of fractional partial derivative*

*The fractional Caputo operator is represented as [41]*

$$D\_{\tau}^{\beta}f(\tau) = \begin{cases} \frac{\vartheta^{n}f(\tau)}{\vartheta\tau^{n}}, & \beta = n \in N, \\\frac{1}{\Gamma(n-\beta)}\int\_{0}^{\tau}(\tau-\phi)^{n-\beta-1}f^{(n)}(\phi)\partial\phi, & n-1 < \beta \le n, \quad n \in N. \end{cases} \tag{5}$$

**Definition 4.** *Shehu transform for fractional order derivatives*

*The Shehu transformation for the fractional order derivatives is expressed as*

$$S\left\{\boldsymbol{u}^{(\beta)}(\tau)\right\} = \frac{s^{\beta}}{\mu^{\beta}}V(s,\mu) - \sum\_{k=0}^{n-1} \left(\frac{s}{\mu}\right)^{\beta-k-1} \boldsymbol{u}^{(k)}(0), \quad 0 < \beta \le n,\tag{6}$$

*In Table 1 show different special functions of Shehu transformation.*


**Table 1.** The Shehu transform of some special functions.

**Theorem 1.** *If the function u*(*τ*) *is piecewise continues at every finite interval of* 0 ≤ *τ* ≤ *β and of exponential order α for τ* > *β, then there's the Shehu transform u*(*s*, *μ*) *[40].*

**Proof.** For any natural number *β*, we deduct algebraically:

$$\int\_0^\infty \exp(-\frac{s\tau}{\mu}) u(\tau) d\tau = \int\_0^\beta \exp(-\frac{s\tau}{\mu}) u(\tau) d\tau + \int\_\beta^\infty \exp(-\frac{s\tau}{\mu}) u(\tau) d\tau,\tag{7}$$

since the function *u*(*τ*) continues in a piecewise manner at every finite interval 0 ≤ *τ* ≤ *β*, there's the first integral on the right hand side. We suggest the following situation to validate this statement,

$$\begin{split} &|\int\_{a}^{\infty} \exp(-\frac{s\tau}{\mu}) u(\tau) d\tau| \leq \int\_{a}^{\infty} |\exp(-\frac{s\tau}{\mu}) u(\tau)| d\tau \\ &\leq \int\_{a}^{\infty} \exp(-\frac{s\tau}{\mu}) |u(\tau)| d\tau \\ &\leq \int\_{a}^{\infty} \exp(-\frac{s\tau}{\mu}) N \exp(\beta \tau) d\tau \\ &= N \int\_{a}^{\infty} \exp(-\frac{(s-\beta u)\tau}{u}) d\tau \\ &= -\frac{Nu}{(s-\beta u)} \cdot \lim\_{\gamma \to \infty} \left[ \exp(-\frac{(s-\beta u)\tau}{u}) d\tau \right]\_{0}^{\gamma} \\ &= \frac{Nu}{(s-\beta u)} .\end{split} \tag{8}$$

The proof is complete.

#### **3. Implementation of Shehu Transform**

In this section, we have considered a time fractional (2+time fractional-order) dimensional physical model in the form

$$
\mu\_{\sf T}^{\sf B}(\heartsuit, \pounds, \pi) = \kappa \mu\_{\sf G \heartsuit \heartsuit}(\heartsuit, \pounds, \pi) + \pounds \mu(\heartsuit, \pounds, \pi) + \aleph\_{\sf t} \mu(\heartsuit, \pounds, \pi), \quad \pounds \in [1, 2] \tag{9}
$$

with initial condition

$$
u(\odot,\Re,0) = \mu(\odot,\Re),\tag{10}$$

while *κ* is a non-linear operator and *£* linear operator.

Applying the Shehu transform to both sides of the Equation (9) we obtain

$$\mathcal{S}\left\{\boldsymbol{u}\_{\tau}^{\boldsymbol{\beta}}(\mathcal{G},\mathbb{R},\boldsymbol{\tau})\right\} = \mathcal{S}\left\{\kappa\boldsymbol{u}\_{\mathcal{G}\mathcal{G}}(\mathcal{G},\mathbb{R},\boldsymbol{\tau}) + \xi\boldsymbol{u}(\mathcal{G},\mathbb{R},\boldsymbol{\tau}) + \mathbb{N}\boldsymbol{u}(\mathcal{G},\mathbb{R},\boldsymbol{\tau})\right\}, \quad \boldsymbol{\beta} \in [1,2].\tag{11}$$

Using the differential property of Shehu transformation we have,

$$\frac{\mathfrak{s}^{\beta}}{\mathfrak{s}^{\beta}} \left\{ V(s,\mu) - \frac{\mu}{\mathfrak{s}} \mathfrak{u}(0) - \frac{\mu^{2}}{\mathfrak{s}^{2}} \mathfrak{u}'(0) \right\} = \mathbb{S} \left\{ \mathfrak{x} \mathfrak{u}\_{\mathfrak{J}\odot}(\odot,\mathfrak{R},\tau) + \mathfrak{z} \mathfrak{u}(\odot,\mathfrak{R},\tau) + \mathfrak{N} \mathfrak{u}(\odot,\mathfrak{R},\tau) \right\}. \tag{12}$$

Simplifying Equation (12), we obtain

$$V(\mathbf{s},\boldsymbol{\mu}) = +\frac{\mu\boldsymbol{\delta}}{\mathcal{S}^{\mathsf{f}}}\mathbb{S}\left\{\kappa\boldsymbol{u}\_{\mathbb{S}^{\mathsf{f}}\mathbb{S}}(\mathbb{S},\mathbb{R},\boldsymbol{\tau}) + \mathbb{E}\boldsymbol{u}(\mathbb{S},\mathbb{R},\boldsymbol{\tau}) + \mathbb{N}\boldsymbol{u}(\mathbb{S},\mathbb{R},\boldsymbol{\tau})\right\} + \frac{\mu}{\mathsf{s}}\boldsymbol{u}(0) + \frac{\mu^{2}}{\mathsf{s}^{2}}\boldsymbol{u}'(0). \tag{13}$$

Applying the inverse Shehu transformation, we get

$$u(\mathbb{S}, \mathbb{R}, \tau) = \mathbb{S}^{-1} \left\{ \frac{\mu^{\emptyset}}{s^{\emptyset}} \mathbb{S} \left\{ \pi u\_{\mathbb{S}^{\emptyset} \mathbb{S}}(\mathbb{S}, \mathbb{R}, \tau) + \mathbb{E}u(\mathbb{S}, \mathbb{R}, \tau) + \mathbb{N}u(\mathbb{S}, \mathbb{R}, \tau) \right\} \right\} + u(0) + \tau u^{'}(0). \tag{14}$$

The nonlinear term ℵ*u*(, , *τ*) is evaluated by using the procedure of Adomian polynomial decomposition given by

$$\mathfrak{Nu}(\mathbb{S}, \mathfrak{R}, \mathfrak{r}) = \sum\_{m=0}^{\infty} A\_m(\mathfrak{u}\_{0\prime} \mathfrak{u}\_{1\prime} \cdots)\_{\prime}, \ m = 0, 1\_{\prime} \cdots \tag{15}$$

where,

$$A\_m(u\_0, u\_1, \dots) = \frac{1}{m!} \left[ \frac{d^m}{d\lambda^m} \aleph \left( \sum\_{i=0}^{\infty} \lambda^i u\_i \right) \right]\_{\lambda=0}, \quad m > 0. \tag{16}$$

With the help of Equation (16), Equation (15) can be written as

$$u(\mathbb{S}, \mathbb{R}, \tau) = \mathbb{S}^{-1} \left\{ \frac{\mu^{\mathfrak{G}}}{\mathfrak{s}^{\mathfrak{G}}} \mathcal{S} \left\{ \mathrm{x} u\_{\mathbb{S}\cap\mathfrak{J}}(\mathbb{S}, \mathbb{R}, \tau) + \mathsf{L}u(\mathbb{S}, \mathbb{R}, \tau) + \sum\_{m=0}^{\infty} A\_{m} \right\} \right\} + \mathfrak{u}(0) + \tau \mathsf{u}'(0). \tag{17}$$

Finally, we obtain the recursive relation as

$$\begin{aligned} \mu\_0(\mathcal{G}, \mathcal{R}, \mathbf{r}) &= \mu(0) + \tau \mu'(0), \; m = 0\\ \mu\_m(\mathcal{G}, \mathcal{R}, \mathbf{r}) &= \mathcal{S}^{-1} \left\{ \frac{\mu^6}{\mathfrak{s}^6} \mathcal{S} \left\{ \kappa \mu\_{(m-1)\ominus\mathfrak{l}}(\mathcal{G}, \mathcal{R}, \mathbf{r}) + \mathfrak{L} \mu\_{(m-1)}(\mathcal{G}, \mathcal{R}, \mathbf{r}) + \mathfrak{N} \mu\_{(m-1)}(\mathcal{G}, \mathcal{R}, \mathbf{r}) + \sum\_{m=0}^{\infty} A\_m \right\} \right\}, \; m \ge 1. \end{aligned} \tag{18}$$

#### **4. Applications and Discussion**

**Example 1.** *Consider the (2+time fractional-order) dimensional hyperbolic wave model:*

$$
\mu\_{\rm{\tau}}^{\mathcal{G}}(\mathbb{S}, \mathbb{R}, \tau) = \frac{1}{12} \mathbb{S}^2 \mu\_{\rm{\mathcal{GS}}}(\mathbb{S}, \mathbb{R}, \tau) + \frac{1}{12} \mathbb{R}^2 \mu\_{\rm{\mathcal{RR}}}(\mathbb{S}, \mathbb{R}, \tau), \quad \mathcal{J} \in (1, 2) \tag{19}
$$

*with initial conditions*

$$
\mu(\odot,\mathfrak{R},0) = \mathfrak{Z}^4, \qquad \mu\_\tau(\odot,\mathfrak{R},0) = \mathfrak{R}^4. \tag{20}
$$

*If β* = 2, *then the exact solution of Equation (19) is*

$$
\mu(\odot, \aleph, \tau) = \odot^4 \sinh(t) + \aleph^4 \cosh(t), \tag{21}
$$

*Taking the Shehu transform of Equation (19) we obtain*

$$\frac{s^{\mathfrak{g}}}{\mu^{\mathfrak{g}}} \left\{ V(s,\mu) - \frac{\mu}{s} \mu(0) - \frac{\mu^2}{s^2} \mu'(0) \right\} = S \left\{ \frac{1}{12} \mathfrak{J}^2 u\_{\mathbb{G}\mathfrak{J}}(\mathbb{G}, \mathbb{R}, \tau) + \frac{1}{12} \mathfrak{R}^2 u\_{\mathbb{R}\mathfrak{R}}(\mathbb{G}, \mathbb{R}, \tau) \right\}. \tag{22}$$

*Simplifying Equation (22), we get*

$$\mathcal{V}(\mathbf{s},\boldsymbol{\mu}) = \frac{\mu^{\mathcal{S}}}{\mathfrak{s}^{\mathcal{S}}} \mathcal{S} \left\{ \frac{1}{12} \mathfrak{D}^{2} u\_{\rm GR}(\mathfrak{D}, \mathfrak{R}, \boldsymbol{\tau}) + \frac{1}{12} \mathfrak{R}^{2} u\_{\rm BR}(\mathfrak{D}, \mathfrak{R}, \boldsymbol{\tau}) \right\} + \frac{\mu}{\mathfrak{s}} u(0) + \frac{\mu^{2}}{\mathfrak{s}^{2}} \boldsymbol{\mu}'(0). \tag{23}$$

*Applying inverse Shehu transform, we get*

$$u(\mathbb{S}, \mathbb{R}, \tau) = u(0) + u'(0)\tau + \mathcal{S}^{-1} \left\{ \frac{\mu^6}{\mathfrak{s}^6} \mathcal{S} \left\{ \frac{1}{12} \mathbb{S}^2 u\_{\mathbb{S}\cap\mathbb{S}}(\mathbb{S}, \mathbb{R}, \tau) + \frac{1}{12} \mathcal{R}^2 u\_{\mathbb{R}\mathbb{R}}(\mathbb{S}, \mathbb{R}, \tau) \right\} \right\}. \tag{24}$$

*Thus we get the following recursive scheme*

$$
\mu\_0(\mathbb{S}, \mathbb{R}, \mathfrak{r}) = \mathfrak{u}(0) + \mathfrak{u}'(0)\mathfrak{r} = \mathbb{S}^4 + \mathbb{R}^4 \mathfrak{r},\tag{25}
$$

$$u\_{m+1}(\mathbb{S}, \mathbb{R}, \tau) = \mathbb{S}^{-1} \left\{ \frac{\mu^{\emptyset}}{\text{s}^{\emptyset}} \mathbb{S} \left\{ \frac{1}{12} \mathbb{S}^{2} u\_{m \mathbb{S} \cap \mathbb{S}}(\mathbb{S}, \mathbb{R}, \tau) + \frac{1}{12} \mathbb{R}^{2} u\_{m \mathbb{R} \mathbb{R}}(\mathbb{S}, \mathbb{R}, \tau) \right\} \right\}. \tag{26}$$

*Using Equation (26), for m* = 0, 1, 2, 3, ··· *we get the following values*

$$\begin{aligned} u\_1(\mathbb{S}, \mathbb{R}, \tau) &= \mathbb{S}^4 \frac{\tau^6}{\beta!} + \mathfrak{R}^4 \frac{\tau^{6+1}}{(\beta+1)!}, \\ u\_2(\mathbb{S}, \mathbb{R}, \tau) &= \mathbb{S}^4 \frac{\tau^{2\delta}}{(2\beta)!} + \mathfrak{R}^4 \frac{\tau^{2\delta+1}}{(2\beta+1)!}, \\ u\_3(\mathbb{S}, \mathbb{R}, \tau) &= \mathbb{S}^4 \frac{\tau^{3\delta}}{(3\beta)!} + \mathfrak{R}^4 \frac{\tau^{3\delta+1}}{(3\beta+1)!}, \\ u\_4(\mathbb{S}, \mathbb{R}, \tau) &= \mathbb{S}^4 \frac{\tau^{4\delta}}{(4\beta)!} + \mathfrak{R}^4 \frac{\tau^{4\delta+1}}{(4\beta+1)!}, \\ &\vdots \end{aligned} \tag{27}$$

*Now using the values of u*0, *u*1, *u*2, *u*3, ··· *, we get Shehu transformation solution for example 1*

$$\begin{split} u(\odot,\mathfrak{R},\tau) &= \mathfrak{J}^{4} + \mathfrak{R}^{4}\tau + \mathfrak{J}^{4} \frac{\tau^{\mathfrak{G}}}{\mathfrak{J}!} + \mathfrak{R}^{4} \frac{\tau^{\mathfrak{G}+1}}{(\mathfrak{J}+1)!} + \mathfrak{J}^{4} \frac{\tau^{2\mathfrak{G}}}{(2\mathfrak{J})!} + \mathfrak{R}^{4} \frac{\tau^{2\mathfrak{G}+1}}{(2\mathfrak{J}+1)!} + \mathfrak{J}^{4} \frac{\tau^{3\mathfrak{G}}}{(3\mathfrak{J})!} + \\ \mathfrak{R}^{4} \frac{\tau^{3\mathfrak{G}+1}}{(3\mathfrak{J}+1)!} + \mathfrak{J}^{4} \frac{\tau^{4\mathfrak{G}}}{(4\mathfrak{J})!} + \mathfrak{R}^{4} \frac{\tau^{4\mathfrak{G}+1}}{(4\mathfrak{J}+1)!} + \cdots \end{split} \tag{28}$$

*After simplification, we get*

$$\begin{split} u(\boldsymbol{\Omega}, \boldsymbol{\aleph}, \tau) &= \mathbb{O}^{4} \left\{ 1 + \frac{\tau^{6}}{\beta!} + \frac{\tau^{2\beta}}{(2\beta)!} + \frac{\tau^{3\beta}}{(3\beta)!} + \frac{\tau^{4\beta}}{(4\beta)!} + \cdots \right\} + \mathbb{H}^{4} \left\{ \tau + \frac{\tau^{6+1}}{(\beta+1)!} + \frac{\tau^{2\beta+1}}{(2\beta+1)!} + \cdots \right\} \\ &\frac{\tau^{3\beta+1}}{(3\beta+1)!} + \frac{\tau^{4\beta+1}}{(4\beta+1)!} + \cdots \end{split} \tag{29}$$

*In particular, when β* → 2*, the analytical solution of Shehu transform become as*

$$u(\mathfrak{H}, \mathfrak{R}, \mathfrak{r}) = \mathfrak{H}^{4} \left\{ 1 + \frac{\mathfrak{r}^{2}}{2!} + \frac{\mathfrak{r}^{4}}{(4)!} + \frac{\mathfrak{r}^{6}}{(6)!} + \frac{\mathfrak{r}^{8}}{(8)!} + \cdots \right\} + \mathfrak{R}^{4} \left\{ \mathfrak{r} + \frac{\mathfrak{r}^{3}}{(3)!} + \frac{\mathfrak{r}^{5}}{(5)!} + \frac{\mathfrak{r}^{7}}{(7)!} + \frac{\mathfrak{r}^{9}}{(9)!} + \cdots \right\},\tag{30}$$

*which provide the close form solution as*

$$
\mu(\odot, \aleph, \tau) = \heartsuit^4 \cosh(\tau) + \aleph^4 \sinh(\tau). \tag{31}
$$

*Figures 1 and 2 represent the exact and analytical solutions of Example 1. The solutions-graphs have confirmed the closed contact between the exact solution and the analytical solution obtained by the proposed method. In Figure 3, the solution of Example 1 are calculated at different fractional-order β of the derivative. It is investigated that the solutions at different fractional-orders β are convergent to an integer-order solution of Example 1. Figure 4 represent the solution verses time graph for Example 1. It is observed that as the time fractional-order varies toward time integer-order, the time fractional-order solutions also approaches to the solution of an integer-order problem of Example 1. All the above solution analysis of Example 1 indicate that SDM is an efficient and effective method to solve fractional-order partial differential equations that are frequently arising in science and engineering.*

**Figure 1.** Represents the exact solution of Example 1 at *β* = 2.

**Figure 2.** Represents the analytical solution of Example 1 at *β* = 2.

**Figure 3.** Represents the solution at different fractional order of Example 1.

**Figure 4.** Represents the solution at different fractional order of Example 1.

*In Table 2, the solutions of Shehu transform decomposition method (SDM) and Adomian decomposition method (ADM) are compared with each other. The comparison has shown that the solutions of proposed method are in strong agreement with the solution of ADM.*

**Table 2.** Comparison of SDM and ADM [42] of Example 1 at *τ* = 0.1.


**Example 2.** *Consider the (2+time fractional-order) dimensional Heat model:*

$$
\mu\_{\sf T}^{\sf B}(\heartsuit, \Re, \tr) = \mu\_{\sf G^{\sf G}}(\heartsuit, \Re, \tr) + \mu\_{\sf R}(\heartsuit, \Re, \tr), \quad \beta \in (0, 1] \tag{32}
$$

*with initial condition*

$$
\mu(\odot,\aleph,0) = \sin(\aleph)\cos(\aleph). \tag{33}
$$

*If β* = 1, *then the exact solution of Equation (32) is*

$$
\mu(\odot,\Re,\tau) = e^{-2\tau}\sin(\Im)\cos(\Re).\tag{34}
$$

*Taking Shehu transform of Equation (32)*

$$\frac{s^{\beta}}{\mu^{\beta}} \left\{ V(s,\mu) - \frac{\mu}{s} \mu(0) \right\} = S \left\{ \mu\_{\odot 3}(\odot, \mathfrak{R}, \tau) + \mu\_{\mathfrak{R} \mathfrak{R}}(\odot, \mathfrak{R}, \tau) \right\},\tag{35}$$

*Simplifying Equation (35), we get as*

$$V(s,\mu) = \frac{\mu}{s}\mu(0) + \frac{\mu^{\beta}}{s^{\beta}}\mathcal{S}\left\{\mu\_{\odot \Im \Im}(\Im,\Re,\pi) + \mu\_{\Re \Re}(\Im,\Re,\pi)\right\}.\tag{36}$$

*Applying inverse Shehu transform, we get*

*Appl. Sci.* **2020**, *10*, 122

$$
\mu(\odot,\mathfrak{R},\tau) = \mu(0) + \mathcal{S}^{-1} \left\{ \frac{\mu^{\mathfrak{G}}}{\mathfrak{s}^{\mathfrak{G}}} \mathcal{S} \left\{ \mu\_{\odot \mathfrak{Z}}(\odot,\mathfrak{R},\tau) + \mu\_{\mathfrak{R}\mathfrak{R}}(\odot,\mathfrak{R},\tau) \right\} \right\}.\tag{37}
$$

*Thus we get the following recursive scheme*

$$u\_0(\odot, \aleph, \pi) = u(0) = \sin(\aleph)\cos(\aleph),\tag{38}$$

$$u\_{m+1}(\mathbb{S}, \mathbb{R}, \tau) = \mathcal{S}^{-1} \left\{ \frac{\mu^{\emptyset}}{\mathfrak{s}^{\emptyset}} \mathcal{S} \left\{ \mu\_{m \boxplus \mathbb{S}} \left( \mathbb{S}, \mathbb{R}, \tau \right) + \mu\_{m \boxtimes \mathbb{R}} \left( \mathbb{S}, \mathbb{R}, \tau \right) \right\} \right\},\tag{39}$$

*Using Equation (39), for m* = 0, 1, 2, 3, ··· *we get the following values*

$$\begin{aligned} u\_1(\odot, \mathfrak{R}, \tau) &= -2\sin(\mathfrak{R})\cos(\mathfrak{R}) \frac{\tau^{\mathfrak{F}}}{(\mathfrak{R})!}, \\ u\_2(\odot, \mathfrak{R}, \tau) &= 4\sin(\mathfrak{R})\cos(\mathfrak{R}) \frac{\tau^{2\mathfrak{F}}}{(2\mathfrak{F})!}, \\ u\_3(\odot, \mathfrak{R}, \tau) &= -8\sin(\mathfrak{R})\cos(\mathfrak{R}) \frac{\tau^{3\mathfrak{F}}}{(3\mathfrak{F})!}, \\ u\_4(\odot, \mathfrak{R}, \tau) &= 16\sin(\mathfrak{R})\cos(\mathfrak{R}) \frac{\tau^{4\mathfrak{F}}}{(4\mathfrak{F})!} \end{aligned} \tag{40}$$

*. . .*

*Now using the values of u*0, *u*1, *u*2, *u*3, ··· *, we get Shehu transformation solution for example 2*

$$\begin{split} u(\clubsuit,\space arc) &= \sin(\heartsuit)\cos(\aleph) - 2\sin(\heartsuit)\cos(\aleph)\frac{\pi^{\p}}{(\beta)!} + 4\sin(\heartsuit)\cos(\aleph)\frac{\pi^{2\p}}{(2\beta)!} + \\ &- 8\sin(\heartsuit)\cos(\aleph)\frac{\pi^{3\p}}{(3\beta)!} + 16\sin(\heartsuit)\cos(\aleph)\frac{\pi^{4\p}}{(4\beta)!} + \cdots \end{split} \tag{41}$$

*After simplification, we get*

$$u(\mathfrak{A}, \mathfrak{R}, \tau) = \sin(\mathfrak{A}) \cos(\mathfrak{R}) \left\{ 1 - 2 \frac{\tau^{\mathfrak{A}}}{(\beta)!} + 4 \frac{\tau^{2\mathfrak{A}}}{(2\beta)!} + \dots - 8 \frac{\tau^{3\mathfrak{A}}}{(3\beta)!} + 16 \frac{\tau^{4\mathfrak{A}}}{(4\beta)!} + \dotsb \right\},\tag{42}$$

*which converge to the solution*

$$
\mu(\odot,\Re,\tau) = \sin(\odot)\cos(\theta)E\_{\beta}(-2\tau^{\beta}),\tag{43}
$$

*For particular case β* → 1*, the Shehu transform solution become as*

$$
\mu(\odot,\aleph,\tau) = \sin(\heartsuit)\cos(\aleph)e^{-2\tau}.\tag{44}
$$

*Figures 5 and 6 show the exact and analytical solution of Example 2 respectively. The graphical representation have confirmed the closed contact of the obtained solution with the exact solution of Example 2. Similarly, Figures 7 and 8 represents the fractional-order solution of Example 2 for two and three space. Both graphs support the convergence phenomena of fractional-order problems to an integer-order problem of Example 2.* *Appl. Sci.* **2020**, *10*, 122

**Figure 5.** Represents the exact solution of Example 2 at *β* = 1.

**Figure 6.** Represents the analytical solution of Example 2 at *β* = 1.

**Figure 7.** *u*(, , *τ*) Represents the solution at different fractional order of Example 2.

**Figure 8.** *u*(, , *τ*) Represents the solution at different fractional order of Example 2.

**Example 3.** *Consider the* (2 + *timef ractional*) *dimensional diffusion model:*

$$
\mu\_{\mathcal{C}}^{\mathcal{B}}(\mathcal{G}, \mathcal{W}, \mathfrak{r}) = \mu\_{\mathcal{C}\mathcal{G}}(\mathcal{G}, \mathcal{W}, \mathfrak{r}) + \mu\_{\mathbb{R}\mathcal{R}}(\mathcal{G}, \mathcal{W}, \mathfrak{r}), \quad \mathcal{S} \in (0, 1] \tag{45}
$$

*with the initial condition*

$$
\mu(\odot,\Re,0) = \mathfrak{e}^{\odot + \Re}.\tag{46}
$$

*If β* = 1, *then the exact solution of Equation (45) is*

$$
\mu(\odot,\Re,\tau) = e^{\odot + \Re + 2\tau} \tag{47}
$$

*Taking Shehu transform of Equation (45)*

$$\frac{\mathbb{S}^{\beta}}{\mu^{\beta}} \left\{ V(s,\mu) - \frac{\mu}{s} \mu(0) \right\} = S \left\{ \mu\_{\mathbb{S}\odot}(\mathbb{S}, \mathbb{R}, \tau) + \mu\_{\mathbb{R}\mathbb{R}}(\mathbb{S}, \mathbb{R}, \tau) \right\}.\tag{48}$$

*Simplifying Equation (46), we get as*

$$V(\mathbf{s},\mu) = \frac{\mu}{\mathbf{s}}u(0) + \frac{\mu^{\beta}}{\mathbf{s}^{\beta}}\mathbb{S}\left\{u\_{\odot 3}(\mathbb{S}, \mathbb{R}, \mathbf{r}) + u\_{\Re \mathsf{R}}(\mathbb{S}, \mathbb{R}, \mathbf{r})\right\}.\tag{49}$$

*Applying inverse operator of Shehu transform, we get*

$$u(\mathbb{S}, \mathbb{R}, \tau) = u(0) + \mathbb{S}^{-1} \left\{ \frac{\mu^{\mathbb{S}}}{\mathsf{s}^{\mathbb{S}}} \mathbb{S} \left\{ \mu\_{\mathbb{S}\cap\mathbb{S}}(\mathbb{S}, \mathbb{R}, \tau) + \mu\_{\mathbb{R}\mathbb{R}}(\mathbb{S}, \mathbb{R}, \tau) \right\} \right\}. \tag{50}$$

*Thus we get the following recursive scheme*

$$
\mu\_0(\odot, \Re \tau) = \mu(0) = e^{\odot + \Re \tau}
$$

$$
\mu\_{m+1}(\mathbb{S}, \mathbb{R}, \tau) = \mathcal{S}^{-1} \left\{ \frac{\mu^{\emptyset}}{\mathfrak{s}^{\emptyset}} \mathcal{S} \left\{ \mu\_{m \heartsuit \heartsuit \heartsuit} (\mathbb{S}, \mathbb{R}, \tau) + \mu\_{m \heartsuit \heartsuit} (\mathbb{S}, \mathbb{R}, \tau) \right\} \right\}, \tag{51}
$$

*Using Equation (51), for m* = 0, 1, 2, 3, ··· *we get the following values*

$$\begin{aligned} u\_1(\mathbb{X}, \mathbb{R}, \tau) &= 2e^{\mathbb{X} + \mathbb{R}} \frac{\tau^{\mathbb{R}}}{(\beta)!}, \\ u\_2(\mathbb{X}, \mathbb{R}, \tau) &= 4e^{\mathbb{X} + \mathbb{R}} \frac{\tau^{2\beta}}{(2\beta)!}, \\ u\_3(\mathbb{X}, \mathbb{R}, \tau) &= 8e^{\mathbb{X} + \mathbb{R}} \frac{\tau^{3\beta}}{(3\beta)!}, \\ u\_4(\mathbb{X}, \mathbb{R}, \tau) &= 16e^{\mathbb{X} + \mathbb{R}} \frac{\tau^{4\beta}}{(4\beta)!}, \\ &\vdots \end{aligned} \tag{52}$$

*Now using the values of u*0, *u*1, *u*2, *u*3, ··· *, we get Shehu transformation solution for Example 3*

$$u(\odot,\mathfrak{R},\tau) = e^{\mathfrak{Z}+\mathfrak{R}} + 2e^{\mathfrak{Z}+\mathfrak{R}} \frac{\tau^{\mathfrak{G}}}{(\beta)!} + 4e^{\mathfrak{Z}+\mathfrak{R}} \frac{\tau^{2\mathfrak{G}}}{(2\beta)!} + 8e^{\mathfrak{Z}+\mathfrak{R}} \frac{\tau^{3\mathfrak{G}}}{(3\beta)!} + 16e^{\mathfrak{Z}+\mathfrak{R}} \frac{\tau^{4\mathfrak{G}}}{(4\beta)!} + \dotsb \dots \tag{53}$$

*After simplification, we get*

$$\ln(\text{3, }\mathfrak{R},\tau) = \sin(\mathfrak{R})\cos(\mathfrak{R}) \left\{ 1 + 2\frac{\mathfrak{r}^{\mathfrak{f}}}{(\mathfrak{f}\mathfrak{)!}} + 4\frac{\mathfrak{r}^{2\mathfrak{f}}}{(2\mathfrak{f})!} + 8\frac{\mathfrak{r}^{3\mathfrak{f}}}{(3\mathfrak{f})!} + 16\frac{\mathfrak{r}^{4\mathfrak{f}}}{(4\mathfrak{f})!} + \cdots \right\}.\tag{54}$$

*The close form solution become as*

$$
\mu(\odot,\aleph,\tau) = \sin(\heartsuit)\cos(\aleph)E\_{\beta}(2\pi^{\beta}).\tag{55}
$$

*When β* → 1 *the calculated result provide the exact solution in the close form*

$$
\mu(\odot,\aleph,\tau) = \sin(\heartsuit)\cos(\aleph)e^{2\tau}.\tag{56}
$$

*Figures 9 and 10 show the exact and analytical solutions of Example 3. Both figures are almost coincident confirming the close contact of both exact and obtained solution. Figures 11 the SDM solutions at different fractional-order β are calculated for Example 3. The convergence phenomena of fractional-order solution towards exact solution is observed. The method is found to be very simple and straightforward to solve fractional-order different equations.*

**Figure 9.** Exact solution of Example 3 at *β* = 1.

**Figure 10.** Represents the analytical solution of Example 3 at *β* = 1.

**Figure 11.** The solution graph at different fractional order *β*.

**Example 4.** *Consider the* (2 + *timef ractional*) *dimensional telegraph model:*

$$u\_{\tau}^{\S}(\mathbb{S}, \mathbb{R}, \tau) = \frac{1}{2} u\_{\mathbb{S}\mathbb{S}\mathbb{S}}(\mathbb{S}, \mathbb{R}, \tau) + \frac{1}{2} u\_{\mathbb{R}\mathbb{R}}(\mathbb{S}, \mathbb{R}, \tau) - 2u\_{l}(\mathbb{S}, \mathbb{R}, \tau) - u(\mathbb{S}, \mathbb{R}, \tau), \quad \mathbb{S} \in (1, 2], \tag{57}$$

*with initial conditions*

$$
\mu(\odot,\aleph,0) = \sinh(\heartsuit)\sinh(\aleph), \qquad \mu\_{\mp}(\odot,\aleph,0) = -2\sinh(\heartsuit)\sinh(\aleph). \tag{58}
$$

*If β* = 2, *then the exact solution of Equation (57) is*

$$
\mu(\odot,\mathfrak{R},\mathfrak{r}) = \sinh(\mathfrak{R})\sinh(\mathfrak{R})e^{-2\mathfrak{r}}.\tag{59}
$$

*Taking Shehu transform of Equation (57)*

$$\frac{s^6}{\mu^6} \left\{ V(s,\mu) - \frac{\mu}{s} u(0) - \frac{\mu^2}{s^2} u'(0) \right\} = S \left\{ \frac{1}{2} u\_{\odot \Omega}(\odot, \mathfrak{R}, \mathfrak{r}) + \frac{1}{2} u\_{\partial \mathbb{R}}(\odot, \mathfrak{R}, \mathfrak{r}) - 2u\_l(\odot, \mathfrak{R}, \mathfrak{r}) - u(\odot, \mathfrak{R}, \mathfrak{r}) \right\},\tag{60}$$

*Simplifying Equation (60), we get as*

$$V(s,\mu) = \frac{\mu^6}{s^6} S\left\{\frac{1}{2}\mu\_{\text{O3}}(\text{\textdegree\text{\textdegree\textdegree\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\}\}{}^4 = \frac{1}{2}\mu\_{\text{\text\textquotedbl}}(\text{\text{\textquotesblle\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\}\}\} \} \} \,\, ^4 = \frac{1}{2}\mu\_{\text{\textquotedbl}}(\text{\textquotedbl\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text$$

*Applying inverse of Shehu transform, we get*

$$u(\mathbb{G}, \mathbb{R}, \boldsymbol{\tau}) = u(0) + \boldsymbol{\tau}u'(0) + \mathbb{S}^{-1} \left\{ \frac{\mu^6}{s^6} \mathbb{S} \left\{ \frac{1}{2} u\_{\mathbb{G}\otimes \mathbb{G}}(\mathbb{G}, \mathbb{R}, \boldsymbol{\tau}) + \frac{1}{2} u\_{\mathbb{R}\mathbb{R}}(\mathbb{G}, \mathbb{R}, \boldsymbol{\tau}) - 2u\_l(\mathbb{G}, \mathbb{R}, \boldsymbol{\tau}) - u(\mathbb{G}, \mathbb{R}, \boldsymbol{\tau}) \right\} \right\}. \tag{62}$$

*Thus we get the following recursive scheme*

$$\begin{aligned} \mu\_0(\mathbb{3}, \mathbb{R}, \tau) &= \mu(0) + \tau \mu'(0) = \\ \sinh(\mathbb{3})\sinh(\mathbb{R}) - 2t\sinh(\mathbb{3})\sinh(\mathbb{R}) \end{aligned}$$

$$u\_{m+1}(\clubsuit,\aleph,\tau) = \operatorname{S}^{-1}\left\{\frac{\mu^{\theta}}{s^{\theta}}\operatorname{S}\left\{\frac{1}{2}u\_{m\circledcirc\Omega}(\clubsuit,\aleph,\tau) + \frac{1}{2}u\_{m\mathsf{R}\aleph \|}(\clubsuit,\aleph,\tau) - 2u\_{m\mathsf{T}}(\clubsuit,\aleph,\tau) - u\_{m}(\clubsuit,\aleph,\tau)\right\}\right\},\tag{63}$$

*Using Equation (63), for m* = 0, 1, 2, 3, ··· *we get the following values*

$$\begin{split} u\_{1}(\mathbb{G},\mathbb{R},\mathbf{r}) &= 4\sinh(\mathbb{G})\sinh(\mathfrak{R})\frac{\mathbf{r}^{\beta}}{(\beta)!}, \\ u\_{2}(\mathbb{G},\mathbb{R},\mathbf{r}) &= -8\frac{\beta(\beta-1)!\sinh(\mathfrak{J})\sinh(\mathfrak{R})\mathbf{r}^{2\beta}}{(2\beta-1)!(\beta)!}, \\ u\_{3}(\mathbb{G},\mathbb{R},\mathbf{r}) &= 16\frac{\beta(2\beta-1)(\beta-1)!(2\beta-2)!\sinh(\mathfrak{J})\sinh(\mathfrak{J})\mathbf{r}^{3\beta-2}}{(\beta)!(2\beta-1)!(3\beta-2)!}, \\ &\vdots \end{split} \tag{64}$$

*Now using the values of u*0, *u*1, *u*2, *u*3, ··· *, we get Shehu transformation solution for Example 4*

*.*

$$\begin{split} \ln(\mathcal{G},\mathbb{R},\mathbf{r}) &= \sinh(\mathfrak{J})\sinh(\mathfrak{R}) - 2\mathfrak{r}\sinh(\mathfrak{J})\sinh(\mathfrak{R}) + 4\frac{\sinh(\mathfrak{J})\sinh(\mathfrak{R})\mathbf{r}\ell^{\mathfrak{G}}}{(\mathfrak{J})!} - 8\frac{\beta(\mathfrak{J}-1)!\sinh(\mathfrak{J})\sinh(\mathfrak{R})\mathbf{r}^{2\mathfrak{G}}}{(2\mathfrak{J}-1)!(\beta)!} + \\ &\frac{16\beta(2\mathfrak{J}-1)(\mathfrak{J}-1)!(2\mathfrak{J}-2)!\sinh(\mathfrak{J})\sinh(\mathfrak{R})\mathbf{r}^{3\mathfrak{G}-2}}{(\mathfrak{J})!(2\mathfrak{J}-1)!(3\mathfrak{J}-2)!} + \cdots \end{split} \tag{65}$$

*After simplification, we get*

$$u(\left(\clubsuit,\Re,\tau\right)=\sinh(\left\Vert\mathbf{3}\right\Vert)\sinh(\left\Vert\mathbf{8}\right\Vert)\left\{1-2\tau+4\frac{\mathbf{r}^{\beta}}{\beta!}-8\frac{\left\Vert\left(\beta-1\right)!\mathbf{r}^{2\beta}}{\left(2\beta-1\right)!\left(\beta\right)!}+\frac{16\beta(2\beta-1)\left(\beta-1\right)!(2\beta-2)!\mathbf{r}^{3\beta-2}}{\left(\beta\right)!(2\beta-1)!(3\beta-2)!}+\cdots\right\}.\tag{66}$$

*For particular case β* → 2*, the Shehu transform solution become as*

$$\mu(\odot, \mathfrak{R}, \tau) = \sinh(\mathfrak{R}) \sinh(\mathfrak{R}) \left\{ 1 - 2\tau + 4\frac{\tau^2}{2!} - 8\frac{\tau^3}{3!} + 16\frac{\tau^4}{4!} + \dotsb \right\}. \tag{67}$$

*The calculated result provide the exact solution in the close form*

$$
\mu(\odot,\Re,\tau) = \sinh(\odot)\sinh(\Re)e^{-2\tau}.\tag{68}
$$

*Figures 12 and 13, display the exact and analytical solutions of Example 4. The solution graph of SDM is very similarly to the exact solution of Example 4. In Figure 14, we plotted the solutions of Example 4 at different fractional-order β. The fractional-order solutions are found to be convergent towards the exact solution of Example 4. It is investigated from the solution analysis that the present method is a sophisticated technique to solve fractional-order problems.*

**Figure 12.** Exact solution of Example 4 at *β* = 2.

**Figure 13.** analytical solution of Example 4 at *β* = 2.

**Figure 14.** The solution graph at different fractional order *β*. of Example 4.

**Example 5.** *Consider the non-linear* (2 + *timef ractional*) *dimensional Burger's model:*

$$
\mu\_{\mathcal{G}}^{\tau}(\mathcal{G}, \mathcal{W}, \mathsf{r}) = \mu\_{\mathcal{G}\mathcal{G}}(\mathcal{G}, \mathcal{W}, \mathsf{r}) + \mu\_{\mathcal{R}\mathcal{R}}(\mathcal{G}, \mathcal{W}, \mathsf{r}) + \mu\_{\mathcal{G}}(\mathcal{G}, \mathcal{W}, \mathsf{r})\mu(\mathcal{G}, \mathcal{W}, \mathsf{r}), \quad \mathcal{G} \in (0, 1], \tag{69}
$$

*with initial condition*

$$
\mu(\mathcal{G}, \mathcal{Y}, 0) = \mathcal{G} + \mathcal{Y}.\tag{70}
$$

*If β* = 1, *then the exact solution of Equation (69) is*

$$
\mu(\mathbb{S}, \mathbb{R}, \tau) = \frac{\mathbb{S} + \mathbb{R}}{1 - \tau}. \tag{71}
$$

*Taking Shehu transform of Equation (69)*

$$\frac{\varepsilon^{\beta}}{\mu^{\beta}} \left\{ V(s,\mu) - \frac{\mu}{s} \mu(0) \right\} = S \left\{ \mu\_{\text{G}\Im\Im}(\Im/\Re,\tau) + \mu\_{\text{BR}}(\Im/\Re,\tau) + \mu\_{\text{G}}(\Im/\Re,\tau) \mu(\Im/\Re,\tau) \right\},\tag{72}$$

*The simplifying Equation (72), we get as*

$$V(s,\mu) = \frac{\mu}{s}\mu(0) + \frac{\mu^{\beta}}{s^{\beta}}\mathcal{S}\left\{\mu\_{\odot 3}(\mathcal{G},\mathcal{R},\tau) + \mu\_{\text{BR}}(\mathcal{G},\mathcal{R},\tau) + \mu\_{\odot}(\mathcal{G},\mathcal{R},\tau)\mu(\mathcal{G},\mathcal{R},\tau)\right\},\tag{73}$$

*By applying inverse of Shehu transform, we get*

$$u(\odot,\mathfrak{R},\mathfrak{r}) = u(0) + \tau u'(0) + \mathcal{S}^{-1} \left\{ \frac{\mu^6}{s^6} \mathcal{S} \left\{ u\_{\odot \mathcal{Q}}(\odot,\mathfrak{R},\mathfrak{r}) + u\_{\mathsf{R}\mathcal{R}}(\odot,\mathfrak{R},\mathfrak{r}) + u\_{\mathcal{Q}}(\odot,\mathfrak{R},\mathfrak{r}) u(\odot,\mathfrak{R},\mathfrak{r}) \right\} \right\},\tag{74}$$

*Thus we get the following recursive scheme*

$$
\mu\_0(\mathbb{Q}, \mathbb{R}, \pi) = \mathfrak{n}(0) = \mathfrak{J} + \mathfrak{K} \tag{75}
$$

$$\mu\_{m+1}(\odot,\mathfrak{R},\mathfrak{r}) = \mathcal{S}^{-1}\left\{\frac{\mu^{\mathfrak{R}}}{s^{\mathfrak{F}}}\mathcal{S}\left\{\mu\_{m\odot\mathfrak{Z}}(\odot,\mathfrak{R},\mathfrak{r}) + \mu\_{m\boxtimes\mathfrak{R}}(\odot,\mathfrak{R},\mathfrak{r}) + \mu\_{m\odot}(\odot,\mathfrak{R},\mathfrak{r})\mu\_{m}(\odot,\mathfrak{R},\mathfrak{r})\right\}\right\}.\tag{76}$$

*For nonlinear term, use the Equation (12) in recursive scheme (76), we obtain*

$$u\_{\mathfrak{m}+1}(\odot,\mathfrak{R},\tau) = \mathcal{S}^{-1}\left\{\frac{\mu^6}{\mathfrak{s}^6}\mathcal{S}\left\{u\_{\mathfrak{m}\odot\mathfrak{Z}\otimes}(\odot,\mathfrak{R},\tau) + u\_{\mathfrak{m}\mathfrak{R}\otimes}(\odot,\mathfrak{R},\tau) + \sum\_{m=0}^{\infty} A\_{\mathfrak{m}}(u\_{0\varLambda},u\_{1\varLambda},\cdots)\right\}\right\}.\tag{77}$$

*Using Equation (77), for m* = 0, 1, 2, 3, ··· *we get the following values*

$$\begin{aligned} u\_1(\odot, \mathfrak{R}, \tau) &= (\odot + \mathfrak{R}) \frac{t^{\mathfrak{R}}}{(\mathfrak{R})!}, \\ u\_2(\odot, \mathfrak{R}, \tau) &= 2(\odot + \mathfrak{R}) \frac{\tau^{2\mathfrak{R}}}{(2\mathfrak{R})!}, \\ u\_3(\odot, \mathfrak{R}, \tau) &= 4(\odot + \mathfrak{R}) \frac{\tau^{3\mathfrak{R}}}{(3\mathfrak{R})!} + (\odot + \mathfrak{R})(2\mathfrak{R})! \frac{t^{3\mathfrak{R}}}{\beta! \beta! (3\mathfrak{R})!}, \\ &\vdots \end{aligned} \tag{78}$$

*Now using the values of u*0, *u*1, *u*2, *u*3, ··· *, we get Shehu transformation solution for example 5*

*.*

$$u(\mathcal{G}, \mathbb{R}, \mathbb{T}) = \mathcal{G} + \mathbb{R} + (\mathcal{G} + \mathbb{R}) \frac{\mathsf{r}^{\mathfrak{G}}}{(\mathfrak{f})!} + 2(\mathfrak{I} + \mathbb{R}) \frac{\mathsf{r}^{2\mathfrak{G}}}{(2\mathfrak{f})!} + 4(\mathfrak{I} + \mathbb{R}) \frac{\mathsf{r}^{\mathfrak{G}\mathfrak{g}}}{(3\mathfrak{f})!} + (\mathfrak{I} + \mathbb{R})(2\mathfrak{f})! \frac{\mathsf{r}^{\mathfrak{G}\mathfrak{g}}}{\mathfrak{f}! \mathfrak{f}! (3\mathfrak{f})!} + \dotsb \dots \tag{79}$$

*After simplification, we get*

$$u(\mathfrak{J}, \mathfrak{R}, \tau) = (\mathfrak{J} + \mathfrak{R}) \left\{ 1 + \frac{\tau^{\mathfrak{J}}}{(\mathfrak{J})!} + 2 \frac{\tau^{2\mathfrak{J}}}{(2\mathfrak{J})!} + 4 \frac{\tau^{3\mathfrak{J}}}{(3\mathfrak{J})!} + (2\mathfrak{J})! \frac{\tau^{3\mathfrak{J}}}{\mathfrak{J}! \mathfrak{R}! (3\mathfrak{J})!} + \cdots \right\}.\tag{80}$$

*For particular case β* → 1*, the Shehu transform solution become as*

$$\mu(\mathcal{G}, \mathcal{R}, \tau) = (\mathcal{G} + \mathcal{R}) \left\{ 1 + \tau + \tau^2 + \tau^3 + \dotsb \right\}. \tag{81}$$

*The calculated result provide the exact solution in the close form*

$$
\mu(\mathcal{G}, \mathcal{R}, \tau) = \frac{\mathcal{G} + \mathcal{R}}{1 - \tau}. \tag{82}
$$

*Figures 15 and 16 are plotted to discuss the exact and analytical solutions of Example 5. The SDM solutions are in good contact with the exact solution of the Example 5. Figures 17 and 18 are plotted to analyze the fractional-order solutions of Example 5 at fractional-order β* = 0.75 *and* 0.50 *respectively. The graphical analysis has verified the applicability of the proposed method.*

**Figure 15.** Exact solution of Example 5 at *β* = 1.

**Figure 16.** Represents the analytical solution of Example 5 at *β* = 1.

**Figure 17.** The solution of fractional-order *β* = 0.75 of Example 5.

**Figure 18.** The solution of fractional-order *β* = 0.5 of Example 5.

#### **5. Results and Discussion**

In the present research work, we implemented a new analytical technique SDM for the solution of some important problems which are frequently arising in science and engineering, such as hyperbolic wave equation, heat equation, diffusion equation, telegraph and Burgers equations. The Caputo definition of fractional-derivative is used to define fractional-derivative. The proposed method is the combination of Shehu transformation and Adomian decomposition method which is known as Shehu decomposition method. For applicability and novelty of present method, we applied it different physical problems for applied sciences. These problems have been solved by using SDM for both fractional and integer-order of the targeted problems. In this connection some figure analysis have been done to demonstrate the obtained results in a sophisticated manner. It is investigated that SDM solution have a very close contact with the exact solution of the problems. It is also observed that the fractional-order problems are convergent towards the solution of an integer-order problem. Moreover, the high rate of convergence of the current method is noted during the simulation. It is calculated that the SDM can be considered as one of the best analytical technique to solve fractional partial differential equations.

#### **6. Conclusions**

In the present article, we presented some fractional-view analysis of physical problems, arising in science and engineering. A new and sophisticated analytical technique, which is known as Shehu transform decomposition method is implemented for both fractional and integer-orders of the problems. The Caputo definition of fractional derivative is used to express fractional-order derivative. For applicability and reliability of the proposed methods, some illustrative examples are presented from different areas of applied science. It has been investigated through graphical representation that the present technique provides an accurate and deserving analysis about the physical happening of the problems. It is observed through simulations of the present algorithm that as fractional-order of the derivative approaches to integer order of the problem then fractional-order solutions are convergent to integer-order solutions. Moreover, the present method is preferred as compared to other method because of its better rate of convergence. This direction motivates the researchers towards the implementation of the current method for other non-linear fractional partial differential equations.

**Author Contributions:** Conceptualization, R.S. and H.K.; Methodology, M.A.; Software, R.S.; Validation, R.S. and U.F.; Formal Analysis, R.S.; Investigation, R.S. and D.B.; Resources, H.K. and P.K.; Writing—Original Draft Preparation, R.S.; Writing—Review and Editing, H.K., U.F. and P.K.; Visualization, M.A.; Supervision, D.B., P.K.; Project Administration, P.K.; Funding Acquisition, P.K.

**Funding:** Center of Excellence in Theoretical and Computational Science (TaCS-CoE) Faculty of Science, King Mongkuts University of Technology Thonburi (KMUTT).

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**


c 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

## *Article* **Fractional View Analysis of Acoustic Wave Equations, Using Fractional-Order Differential Equations**

**Izaz Ali 1, Hassan Khan 1, Rasool Shah 1, Dumitru Baleanu 2,3, Poom Kumam 4,5,\* and Muhammad Arif <sup>1</sup>**


Received: 1 November 2019; Accepted: 15 December 2019; Published: 15 January 2020

**Abstract:** In the present research work, a newly developed technique which is known as variational homotopy perturbation transform method is implemented to solve fractional-order acoustic wave equations. The basic idea behind the present research work is to extend the variational homotopy perturbation method to variational homotopy perturbation transform method. The proposed scheme has confirmed, that it is an accurate and straightforward technique to solve fractional-order partial differential equations. The validity of the method is verified with the help of some illustrative examples. The obtained solutions have shown close contact with the exact solutions. Furthermore, the highest degree of accuracy has been achieved by the suggested method. In fact, the present method can be considered as one of the best analytical techniques compared to other analytical techniques to solve non-linear fractional partial differential equations.

**Keywords:** homotopy perturbation method; variational iteration method; Laplace transform method; acoustic wave equations

#### **1. Introduction**

Recently, fractional calculus and fractional differential equations (FDEs) have attracted the attention of scientists, mathematicians and engineers. A number of important implementations have been evaluated in various fields of sciences and engineering, such as material engineering, viscoelastic, electrochemistry, electromagnetic and dynamics physics which are described by fractional partial differential equations (FPDEs) [1]. Analytical approaches to solve FDEs are of great interest. There is no technique which provides an exact solution to the FDEs. Approximate approaches must be obtained by using techniques of series solution or linearization [2], followed by the application of proper numerical discretization [3–5] and system solvers [6–8]. Non-linear phenomena appear in a number of fields of engineering and sciences, such as solid state physics, chemical kinetics, non-linear spectroscopy, fluid physics, computational biology, quantum mechanics and thermodynamics etc. The concept of non-linearity is designed by various higher-order nonlinear partial differential equations (PDEs). For all of the physical systems, fundamental phenomena are covered by their nonlinear concepts [9,10]. In this paper, Laplace Variational Homotopy Perturbation Method (LVHPM) is implemented to solve the following linear and non-linear fractional-order regularized long wave equations.

$$\frac{\partial^6 \upsilon}{\partial \eta^6} + \frac{1}{2} \frac{\partial \upsilon^2}{\partial \xi^3} - \frac{\partial}{\partial \eta} \left( \frac{\partial^2 \upsilon}{\partial \xi^2} \right) = 0, \qquad 0 < \xi \le 1, \qquad 0 < \beta \le 1, \quad \eta > 0,\tag{1}$$

with initial condition

$$\upsilon(\xi,0) = \overline{\xi},$$

$$\frac{\partial^{\beta}\upsilon}{\partial\eta^{\beta}} + \frac{\partial\upsilon}{\partial\overline{\xi}} + \upsilon\frac{\partial\upsilon}{\partial\eta} - \frac{\partial}{\partial\eta}\left(\frac{\partial^{2}\upsilon}{\partial\overline{\xi}^{2}}\right) = 0, \qquad 0 < \overline{\xi} \le 1, \qquad 0 < \beta \le 1, \quad \eta > 0,\tag{2}$$

with initial condition

$$v(\xi,0) = 3a \sec h^2(\beta \xi), \quad a > 0, \quad \beta = \frac{1}{2} \sqrt{\frac{a}{1+a}}.$$

$$\frac{\partial^\beta v}{\partial \eta^\beta} + \frac{\partial v}{\partial \xi} - 2\frac{\partial}{\partial \eta} \left(\frac{\partial^2 v}{\partial \xi^2}\right) = 0, \qquad 0 < \xi \le 1, \qquad 0 < \beta \le 1, \quad \eta > 0,\tag{3}$$

with initial condition

$$v(\xi,0) = e^{-\xi}\_{\prime}$$

and

$$\frac{\partial^{\beta}\upsilon(\xi,\eta)}{\partial\eta^{\beta}} + \frac{\partial^{4}\upsilon(\xi,\eta)}{\partial\xi^{4}} = 0, \qquad 0 < \xi \le 1, \qquad 0 < \beta \le 1, \quad \eta > 0,\tag{4}$$

with initial condition

$$\upsilon(\not\subset, 0) = \sin \not\subset. \tag{5}$$

Equation (1) is known as the fractional-order non-linear regularized long wave equation (RLWE); Equation (2) is known as the fractional-order non-linear general regularized long wave equation (GRLWE) and Equations (3) and (4) are known as fractional-order linear regularized long wave equations (RLWEs) [11].

The Benjamin Bona Mahony equation (BBME) also identified the regularized long wave (RLW) equation. This equation is the updated version of Korteweg–de Vries equation (KdV) for the modeling of tiny amplitude lengthy surface gravitational waves spreading unidirectionally in two dimensions. RLW equations have several implementations in certain areas of science, such as ion-acoustic waves in plasma, longitudinal dispersive waves in elastic rods, magneto-hydrodynamic waves in plasma, rotating tube flow and stress waves in compressed gas bubble mixes, etc. The RLW equations are described as useful models in applied physics and engineering for many significant physical structures. They also design many liquid flow nature issues where diffusion is significant, either in viscous or shock situations. It can be used to model any dissipation-related non-linear wave diffusion problem. Chemical reaction, heat conduction, mass diffusion, viscosity, thermal radiation or other sources may result from this dissipation, depending on problem modeling [12].

The RLW problem is a family of non-linear growth models that provides excellent designs for predicting natural phenomena. The algorithm was initially introduced to define the undular bore behavior [13]. It was also obtained from the research of acoustic plasma waves of water and ion. An analytical solution for the RLW equation was identified under restricted initial and boundary conditions in [14]. The fractional RLW equations also define numerous significant ocean science and engineering phenomena such as long-wave and small frequency shallow water waves. The non-linear waves modeled on the fractional equations of RLW are of significant interest for several scientists in ocean shallow waves of liquid. The mathematically modeled non-linear waves in the ocean were the fractional RLW equations. Indeed, huge surface waves identified as the tsunami are described as fractional RLW equations. The huge internal waves in the interior of the ocean, resulting from

the difference in temperature, that may destruct marine ships could be defined as fractional RLW equations in the current, highly efficient method.

In recent decades, many researchers and scientists have used analytical methods to solve these types of problems such as homotopy perturbation Sumudu transform method (HPSTM) [11], Adomian decomposition method (ADM) [15,16], least-squares method [17], optimal homotopy perturbation method [18], variational iteration method (VIM) homotopy perturbation method (HPM) [19] and He's homotopy perturbation method [20]. It is observed that these methods have certain deficiencies like calculation of Adomian polynomials, determination of Lagrange multiplier, divergent results and a huge volume of calculations. As a result, a modified analytical technique which is known as VHPTM was introduced to solve differential equations of fractional-order. VHPTM is the combination of three well-known techniques namely, homotopy perturbation method, Laplace transform and variational iteration method. The present method uses the Lagrange multiplier that can limit the consecutive implementation of integral operator and unmanageable computational cost. It is still maintaining higher degree of accuracy. VHPTM [21–24] has an excellent scheme and absorbs all the beneficial characteristics of VIM and HPM.

Finally, He's polynomials have been used in the correction fractional formula to develop the homotopy perturbation method. It is observed that the proposed method is implemented without any use of transformation, discretization and it was found to be free from the generating round off error. Usually, the method of variable separable needs both initial and boundary points to operate, but the present method provides an analytical solution by using initial conditions only. There is a clear advantage of the suggested method that it works without any use of Adomian polynomials, as required by the Adomian decomposition method. Results of the analysis show that the suggested method produces the solution in a series of fast convergence that can result in a closed solution [25–30].

#### **2. Preliminaries Concepts**

**Definition 1.** *Laplace transformation of ρ*(*η*), *η* > 0 *represented as [13]*

$$Q(s) = \mathcal{L}[\rho(\eta)] = \int\_0^\infty e^{-s\eta} \rho(\eta) d\eta.$$

**Theorem 1.** *The convolution of Laplace transform is*

$$\mathcal{L}[\rho\_1 \times \rho\_2] = \mathcal{L}[\rho\_1(\eta)] \times \mathcal{L}[\rho\_2(\eta)]\_{\mathcal{L}}$$

*here ρ*<sup>1</sup> × *ρ*2*, define the convolution between ρ*<sup>1</sup> *and ρ*<sup>2</sup> *,*

$$(\rho\_1 \times \rho\_2)\eta = \int\_0^\tau \rho\_1(\tau)\rho\_2(\eta-\tau)d\eta.$$

*Laplace transform of fractional derivative*

$$\mathcal{L}\left(D\_{\eta}^{\oint}\rho(\eta)\right) = s^{\oint}\mathcal{Q}(s) - \sum\_{k=0}^{n-1} s^{\oint -1-k} \rho^{(k)}(0), \quad n-1 < \beta < n.$$

*where Q*(*s*) *is the Laplace transformation of ρ*(*η*)*.*

**Definition 2.** *The Riemann-Liouville fractional integral operator of order β* ≥ 0 *of a function f* ∈ *Cμ, μ* ≥ −1 *is described in [21]*

$$I\_{\xi}^{\mathfrak{F}}\mathfrak{g}(\xi) = \frac{1}{\Gamma(\mathfrak{F})} \int\_{0}^{\overline{\xi}} (\xi - s)^{\beta - 1} \mathfrak{g}(s) ds\_{\prime}$$

*where* Γ *represent the gamma function as,*

$$\Gamma(\beta) = \int\_0^\infty e^{-\frac{x}{\xi}} \xi^{\beta - 1} d\xi \qquad \beta \in \mathbb{C}.$$

**Definition 3.** *The fractional derivative of g*(*η*) *in the Caputo sense is defined as*

$$D^{\mathcal{S}}g(\eta) = \frac{\partial^{\mathcal{S}}g(\eta)}{\partial \eta^{\mathcal{S}}} = \begin{cases} I^{m-\mathcal{S}} \left[ \frac{\partial^{m}\underline{\mathcal{S}}(\eta)}{\partial \eta^{m}} \right], & \text{if } m-1 < \mathcal{S} < m, \quad m \in \mathbb{N}, \\\frac{\partial^{m}\underline{\mathcal{S}}(\eta)}{\partial \eta^{m}}, & \mathcal{S} = m. \end{cases}$$

**Lemma 1.** *If <sup>m</sup>*˜ <sup>−</sup> <sup>1</sup> <sup>&</sup>lt; *<sup>β</sup>* <sup>≤</sup> *m with* ˜ *<sup>m</sup>*˜ <sup>∈</sup> <sup>N</sup> *and g* <sup>∈</sup> <sup>C</sup>*<sup>η</sup> with <sup>η</sup>* ≥ −1, *then*

$$\begin{aligned} \, \, ^{\mathbb{P}}I^{\mathbb{P}} \mathfrak{g}(\xi) &= I^{\mathbb{P}+\mathbb{a}} \mathfrak{g}(\xi), \qquad a, \beta \ge 0. \\\, \, \, ^{\mathbb{P}}\mathfrak{g}^{\mathbb{A}} &= \frac{\Gamma(\lambda + 1)}{\Gamma(\gamma + \lambda + 1)} \xi^{\mathbb{P} + \lambda}, \qquad \beta > 0, \lambda > -1, \quad \xi > 0. \\\, \, \, \, \, ^{\mathbb{P}}D^{\mathbb{B}} \mathfrak{g}(\xi) &= \mathfrak{g}(\xi) - \sum\_{k=0}^{\mathfrak{M}-1} \mathfrak{g}^{(k)}(0^{+}) \frac{\xi^{\mathbb{B}}}{k!}, \quad \, \, ^{\mathbb{P}}\mathfrak{g} &\quad \, \, \xi > 0, \, \mathfrak{m} - 1 < \beta \le \mathfrak{m}. \end{aligned}$$

**Definition 4.** *Function of Mittag-Leffler, Eα*,*β*(*η*) *for α*, *β* > 0 *is defined as*

$$E\_{\mathfrak{a},\mathfrak{k}}(\eta) = \sum\_{k=0}^{\infty} \frac{\eta^k}{\Gamma(k\mathfrak{a} + \beta)}, \qquad \mathfrak{a}, \beta > 0, \quad \eta \in \mathbb{C}.$$

#### **3. The Procedure of VHPTM**

To demonstrate, the fundamental concept of the present method [21,22], we are considering

$$D\_{\eta}^{\beta}\upsilon(\xi,\eta) + \mathbb{R}\upsilon(\xi,\eta) + \mathbb{N}\upsilon(\xi,\eta) = f(\xi,\eta),\tag{6}$$

with initial condition

$$v(\xi,0) = \mathfrak{g}(\xi)\_{\prime\prime}$$

where *f*(*ξ*, *η*) is an inhomogeneous term, *R*¯ and *N*¯ are particular linear and non-linear differential operators and *D<sup>β</sup> <sup>η</sup> υ*(*ξ*, *η*) is the Caputo fractional derivative of *υ*(*ξ*, *η*).

By taking Laplace transform of Equation (6) on both sides, we get

$$\mathcal{L}\_{\eta} \{ \upsilon(\xi, \eta) \} - \sum\_{k=0}^{m-1} s^{\beta - 1 - k} \frac{\partial^k \upsilon(\xi, \eta)}{\partial^k \eta}|\_{t=0} = -\mathcal{L} \left\{ \mathcal{R} \upsilon(\xi, \eta) + \mathcal{N} \upsilon(\xi, \eta) - f(\xi, \eta) \right\}.$$

where *<sup>υ</sup>*(*s*) = *£η*(*υ*(*ξ*, *<sup>η</sup>*)) = <sup>∞</sup> <sup>0</sup> *<sup>e</sup>*−*<sup>s</sup>ηυ*(*η*)*dη*.

We can build a functional correction according to the variation iteration method

$$\mathcal{L}\_{\eta} \left\{ v\_{j+1}(\boldsymbol{\xi}, \boldsymbol{\eta}) \right\} = \mathcal{L} \{ v\_{j}(\boldsymbol{\xi}, \boldsymbol{\eta}) \} + \lambda(s) \left[ s^{\mathcal{S}} \mathcal{L} \left\{ v\_{\eta}(\boldsymbol{\xi}, \boldsymbol{\eta}) - \sum\_{k=0}^{m-1} s^{\mathcal{S}-1-k} \frac{\delta^{k} v(\boldsymbol{\xi}, \boldsymbol{\eta})}{\delta^{k} \eta} |\_{t=0} + \mathcal{L} \left\{ \mathcal{R} v(\boldsymbol{\xi}, \boldsymbol{\eta}) + \mathcal{N} v(\boldsymbol{\xi}, \boldsymbol{\eta}) - f(\boldsymbol{\xi}, \boldsymbol{\eta}) \right\} \right\} \right], \tag{7}$$

where *λ*(*s*) is the Lagrange multiplier. Here we put *λ*(*s*) = <sup>−</sup><sup>1</sup> *<sup>s</sup><sup>β</sup>* [22].

Applying inverse Laplace of Equation (7)

$$\upsilon\_{j+1}(\underline{\boldsymbol{\xi}},\boldsymbol{\eta}) = \upsilon\_{j}(\underline{\boldsymbol{\xi}},\boldsymbol{\eta}) - \boldsymbol{\xi}^{-1} \left[ \frac{1}{s^{\delta}} \boldsymbol{\mathcal{E}} \left\{ s^{\delta} \frac{\partial \upsilon}{\partial \boldsymbol{\eta}} + \bar{\boldsymbol{R}} \upsilon\_{j}(\underline{\boldsymbol{\xi}},\boldsymbol{\eta}) + \bar{\boldsymbol{N}} \upsilon\_{j}(\underline{\boldsymbol{\xi}},\boldsymbol{\eta}) - f(\underline{\boldsymbol{\xi}},\boldsymbol{\eta}) \right\} \right]. \tag{8}$$

The basic idea in the procedure of homotopy perturbation method is that the solution can be written as a series in powers of *p*:

$$\upsilon(\xi,\eta) = \sum\_{j=0}^{\infty} p^j \upsilon\_j(\xi,\eta) = \upsilon\_0 + p\upsilon\_1 + p^2\upsilon\_2 + p^3\upsilon\_3 + \cdots \,\_ \prime \tag{9}$$

where the non-linear expression can be expressed as

$$
\vec{N}\upsilon(\xi,\eta) = \sum\_{j=0}^{\infty} p^j \vec{H}\_j(\upsilon). \tag{10}
$$

*H*¯*<sup>j</sup>* is He's polynomials,

$$H\_j(\upsilon\_0 + \upsilon\_1 + \dots + \upsilon\_j) = \frac{1}{j!} \frac{\partial^j}{\partial p^j} \left[ N \left( \sum\_{i=0}^{\infty} p^i \upsilon\_i \right) \right]. \tag{11}$$

The technique of fractional VHPTM of Equation (8) with He's polynomials.

$$\sum\_{j=0}^{\infty} p^j \upsilon\_j(\xi, \eta) = \sum\_{j=0}^{\infty} p^j \upsilon\_j(\xi, \eta) + \mathcal{E}^{-1} \left[ \lambda(s) \mathcal{E} \left\{ \sum\_{j=0}^{\infty} p^j \frac{\partial^6 \upsilon\_j}{\partial \eta^6}(\xi, s) + \sum\_{j=0}^{\infty} p^j \mathcal{R} \upsilon\_j(\xi, \eta) + \sum\_{j=0}^{\infty} p^j \mathcal{H}\_j(\upsilon) - f(\xi, \eta) \right\} \right]. \tag{12}$$

By comparing the coefficients of like power of *p* on both sides of Equation (12), we get the VHPTM solution of the given problem.

**Theorem 2.** *Let ξ and* Y *be two Banach spaces and T* : *ξ* → Y *be a contractive nonlinear operator, such that for all υ*; *υ*<sup>∗</sup> ∈; *ξ*, ||*T*(*υ*) − *T*(*υ*∗)|| ≤ *K*||*υ* − *υ*∗||, 0 < *K* < 1 *y [31]. Then, in view of Banach contraction theorem, T has a unique fixed point υ, such that Tυ* = *υ: Let us write the generated series (12), by the Laplace decomposition method as*

$$\xi\_{\mathfrak{m}}^{\mathfrak{r}} = T(\mathfrak{f}\_{\mathfrak{m}-1}), \quad \xi\_{\mathfrak{m}-1}^{\mathfrak{r}} = \sum\_{\mathfrak{m}=1}^{\mathfrak{m}-1} \upsilon\_{\mathfrak{j}}, \quad \mathfrak{m} = 0, 1, 2, \dots$$

*and supposed that ξ*<sup>0</sup> = *υ*<sup>0</sup> ∈ S*p*(*υ*), *where* S*p*(*υ*) = {*υ*<sup>∗</sup> ∈ *ξ* : ||*υ* − *υ*∗|| < *p*} *then, we have*

$$(B\_1)\_{\\\\\xi^m}^{\\\chi} \in \mathcal{S}\_p(\upsilon)$$

$$(B\_2)\_{\\\\\mathfrak{m} \to \infty} \dot{\xi}\_{\\\\\mathfrak{m}} = \upsilon.$$

**Proof.** (*B*1) In view of mathematical induction for *m* = 1, we have

$$||\xi\_1 - \upsilon\_1|| = ||T(\xi\_0 - T(\upsilon))|| \le \mathbb{K}||\upsilon\_0 - \upsilon||.$$

Let the result be true for *m* − 1, then

$$||\zeta\_{\mathfrak{m}-1} - \upsilon|| \le K^{m-1} ||\upsilon\_0 - \upsilon||.$$

We have

$$||\mathfrak{z}\_{m} - \upsilon|| = ||T(\mathfrak{z}\_{m-1} - T(\upsilon))|| \le K||\mathfrak{z}\_{m-1} - \upsilon|| \le K^{m}||\upsilon\_{0} - \upsilon||.$$

Hence, using (*B*1), we have

$$||\zeta\_m - \upsilon|| \le \mathcal{K}^n ||\upsilon\_0 - \upsilon|| \le \mathcal{K}^m p < p\_\prime$$

which implies that *ξ<sup>m</sup>* ∈ S*p*(*υ*).

(*B*2): Since ||*ξ<sup>m</sup>* <sup>−</sup> *<sup>υ</sup>*|| ≤ *<sup>K</sup>m*||*υ*<sup>0</sup> <sup>−</sup> *<sup>υ</sup>*|| and as a lim*m*→<sup>∞</sup> *<sup>K</sup><sup>m</sup>* <sup>=</sup> 0. Therefore; we have lim*m*→<sup>∞</sup> ||*ξ<sup>m</sup>* − *υ*|| = 0 ⇒ lim*m*→<sup>∞</sup> *ξ<sup>m</sup>* = *υ*.

#### **4. Numerical Examples**

#### *4.1. Example*

We consider time fractional-order non-linear RLW equation

$$\frac{1}{2}\frac{\partial^{\beta}v}{\partial\eta^{\beta}} + \frac{1}{2}\frac{\partial v^{2}}{\partial\xi^{\varepsilon}} - \frac{\partial\partial^{2}v}{\partial\eta\partial\xi^{2}} = 0, \qquad 0 < \xi \le 1, \qquad 0 < \beta \le 1, \quad \eta > 0,\tag{13}$$

initial condition is

· ·

$$
v(\xi,0) = \xi,\tag{14}$$

By using Equation (12), the fractional PDE given in Equation (13) can be written as

$$\sum\_{j=0}^{\infty} p^j \upsilon\_{j+1}(\underline{\boldsymbol{\xi}}, \boldsymbol{\eta}) = \sum\_{j=0}^{\infty} p^j \upsilon\_j(\underline{\boldsymbol{\xi}}, \boldsymbol{\eta}) + \mathcal{L}^{-1} \left[ \lambda(\boldsymbol{\xi}) \mathcal{E} \left\{ \mathbf{s}^{\boldsymbol{\beta}} \frac{\partial \upsilon\_j(\underline{\boldsymbol{\xi}}, \boldsymbol{\eta})}{\partial \boldsymbol{\eta}} + \frac{1}{2} \frac{\partial \upsilon\_j^2(\underline{\boldsymbol{\xi}}, \boldsymbol{\eta})}{\partial \boldsymbol{\xi}} - \frac{\partial \boldsymbol{\partial}^2 \upsilon\_j(\underline{\boldsymbol{\xi}}, \boldsymbol{\eta})}{\partial \boldsymbol{\eta} \partial \underline{\boldsymbol{\xi}}^2} \right\} \right], \tag{15}$$

where *λ*(*s*) is the Lagrange multiplier

$$
\lambda(s) = \frac{-1}{s^{\beta}}.
$$

Applying VHPTM using He's polynomials,

$$\begin{split} \sum\_{j=0}^{\infty} p^j \boldsymbol{\upsilon}\_{j+1}(\boldsymbol{\xi}, \boldsymbol{\eta}) &= \sum\_{j=0}^{\infty} p^j \boldsymbol{\upsilon}\_{j}(\boldsymbol{\xi}, \boldsymbol{\eta}) - \sum\_{j=0}^{\infty} p^j \boldsymbol{\mathcal{L}}^{-1} \left[ \frac{1}{s^{\beta}} \boldsymbol{\mathcal{E}} \left\{ s^{\beta} \left( \frac{\partial \boldsymbol{\upsilon}\_{0}}{\partial \boldsymbol{\eta}} + p \frac{\partial \boldsymbol{\upsilon}\_{1}}{\partial \boldsymbol{\eta}} + p^{2} \frac{\partial \boldsymbol{\upsilon}\_{2}}{\partial \boldsymbol{\eta}} + \cdots \right) \right. \\ &+ \frac{1}{2} \frac{\partial}{\partial \boldsymbol{\xi}} \left\{ \boldsymbol{\upsilon}\_{0}^{2} + p(2\boldsymbol{\upsilon}\_{0} \boldsymbol{\upsilon}\_{1}) + p^{2}(2\boldsymbol{\upsilon}\_{0} \boldsymbol{\upsilon}\_{2} + \boldsymbol{\upsilon}\_{1}^{2}) + \cdots \right\} - \left\{ p^{0} \frac{\partial \boldsymbol{\partial}^{2} \boldsymbol{\upsilon}\_{0}}{\partial \boldsymbol{\eta} \partial \boldsymbol{\xi}^{2}} + p^{1} \frac{\partial \boldsymbol{\partial}^{2} \boldsymbol{\upsilon}\_{1}}{\partial \boldsymbol{\eta} \partial \boldsymbol{\xi}\_{5}^{2}} + p^{2} \frac{\partial \boldsymbol{\partial}^{2} \boldsymbol{\upsilon}\_{2}}{\partial \boldsymbol{\eta} \partial \boldsymbol{\xi}^{2}} + \cdots \right\} \right], \end{split} \tag{16}$$

Comparing the coefficients of *p*

*υ*0(*ξ*, *η*) = *ξ*, *<sup>p</sup>*1*υ*1(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*1*υ*0(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*1*£*−<sup>1</sup> 1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*0(*ξ*, *<sup>η</sup>*) *∂η* <sup>+</sup> 1 2 *∂ ∂ξ <sup>υ</sup>*<sup>2</sup> <sup>0</sup>(*ξ*, *<sup>η</sup>*) <sup>−</sup> *∂∂*2*υ*0(*ξ*, *<sup>η</sup>*) *∂η∂ξ*<sup>2</sup> , *<sup>υ</sup>*1(*ξ*, *<sup>η</sup>*) = *<sup>ξ</sup>* <sup>−</sup> *<sup>ξ</sup> <sup>η</sup><sup>β</sup>* Γ(*β* + 1) , *<sup>p</sup>*2*υ*2(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*2*υ*1(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*2*£*−<sup>1</sup> 1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*1(*ξ*, *<sup>η</sup>*) *∂η* <sup>+</sup> 1 2 *∂ ∂ξ* (2*υ*0*υ*1) <sup>−</sup> *∂∂*2*υ*<sup>1</sup> *∂η∂ξ*<sup>2</sup> , *<sup>υ</sup>*2(*ξ*, *<sup>η</sup>*) = *<sup>ξ</sup>* <sup>−</sup> *<sup>ξ</sup> <sup>η</sup><sup>β</sup>* <sup>Γ</sup>(*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>+</sup> <sup>2</sup>*<sup>ξ</sup> <sup>η</sup>*2*<sup>β</sup>* Γ(2*β* + 1) , *<sup>p</sup>*3*υ*3(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*3*υ*2(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*3*£*−<sup>1</sup> 1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*<sup>1</sup> *∂η* <sup>+</sup> 1 2 (2*υ*0*υ*<sup>2</sup> + *υ*<sup>2</sup> <sup>1</sup>) <sup>−</sup> *∂∂*2*υ*<sup>2</sup> *∂η∂ξ*<sup>2</sup> , *<sup>υ</sup>*3(*ξ*, *<sup>η</sup>*) = *<sup>ξ</sup>* <sup>−</sup> *<sup>ξ</sup> <sup>η</sup><sup>β</sup>* <sup>Γ</sup>(*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>+</sup> <sup>2</sup>*<sup>ξ</sup> <sup>η</sup>*2*<sup>β</sup>* <sup>Γ</sup>(2*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>−</sup> *<sup>ξ</sup>* <sup>Γ</sup>(2*<sup>γ</sup>* <sup>+</sup> <sup>1</sup>)*η*3*<sup>β</sup>* (Γ(2*<sup>γ</sup>* <sup>+</sup> <sup>1</sup>))2Γ(3*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>−</sup> <sup>4</sup>*<sup>ξ</sup> <sup>η</sup>*3*<sup>β</sup>* Γ(3*β* + 1) , ·

The analytical expression is therefore obtained in the following way

$$\upsilon(\xi,\eta) = \overline{\xi} - \overline{\xi}\frac{\eta^{\mathfrak{G}}}{\Gamma(\mathfrak{G}+1)} + 2\overline{\xi}\frac{\eta^{2\mathfrak{G}}}{\Gamma(2\mathfrak{G}+1)} - \overline{\xi}\frac{\Gamma(2\gamma+1)\eta^{3\mathfrak{G}}}{(\Gamma(\gamma+1))^2\Gamma(3\mathfrak{G}+1)} - 4\overline{\xi}\frac{\eta^{3\mathfrak{G}}}{\Gamma(3\mathfrak{G}+1)} + \dotsb \tag{17}$$

If *β* = 1 the series form is

$$\upsilon(\xi,\eta) = \xi \left(1 - \eta + \eta^2 - \eta^3 + \dotsb\right). \tag{18}$$

The exact solution at *β* = 1

$$
\upsilon(\xi,\eta) = \frac{\xi}{1+\eta}.\tag{19}
$$

Figure 1, shows the plot of exact and VHPTM solutions of example 4.1, at integer-order *β* = 1. It is confirmed from the figure that both exact and VHPTM solutions are in good contact with each other. In Figure 2, the VHPTM solutions at different fractional-orders *β* = 1, 0.8, 0.6 and 0.4 are calculated. Investigations show that the solutions at different fractional-orders are convergent to the solution of an integer problem as the fractional-order approaches to an integer-order.

**Figure 1.** Variational homotopy perturbation transform method solution of example 4.1 for *β* = 1.

In Table 1, we compared the solutions of VHPTM and VIM at an integer-order *β* = 1 for example 4.1. In addition, the solutions at fractional-orders *β* = 0.55 and *β* = 0.75 are listed in the table. It is observed that VHPTM solutions are almost identical with each other. The results given in the table support the applicability of the VHPTM.

**Figure 2.** Variational homotopy perturbation transform method solution of example 4.1 at different fractional order *β* = 1, 0.8, 0.6, 0.4.

**Table 1.** Comparison of VHPTM and variational iteration method (VIM) [19] of example 1 at *η* = 0.1.


#### *4.2. Example*

We consider time fractional-order non-linear GRLW equation

$$\frac{\partial^{\beta}\upsilon}{\partial\eta^{\beta}} + \frac{\partial\upsilon}{\partial\xi^{\tau}} + \upsilon\frac{\partial\upsilon}{\partial\xi^{\tau}} - \frac{\partial\partial^{2}\upsilon}{\partial\eta\partial\xi^{2}} = 0, \qquad 0 < \xi \le 1, \qquad 0 < \beta \le 1, \quad \eta > 0,\tag{20}$$

with initial condition

$$\upsilon(\xi,0) = 3a \sec h^2(\beta \xi), \quad a > 0, \quad \beta = \frac{1}{2} \sqrt{\frac{a}{1+a}}.\tag{21}$$

By using Equation (12), the fractional PDE given in Equation (20) can be written as

$$\sum\_{j=0}^{\infty} p^j v\_{j+1}(\boldsymbol{\xi}, \boldsymbol{\eta}) = \sum\_{j=0}^{\infty} p^j v\_j(\boldsymbol{\xi}, \boldsymbol{\eta}) + p^j \boldsymbol{\xi}^{-1} \left[ \lambda(s) \mathcal{L} \left\{ \mathcal{s}^\beta \frac{\partial v\_j(\boldsymbol{\xi}, \boldsymbol{\eta})}{\partial \boldsymbol{\eta}} + \frac{\partial v\_j(\boldsymbol{\xi}, \boldsymbol{\eta})}{\partial \boldsymbol{\xi}} + v\_j(\boldsymbol{\xi}, \boldsymbol{\eta}) \frac{\partial v\_j(\boldsymbol{\xi}, \boldsymbol{\eta})}{\partial \boldsymbol{\xi}} - \frac{\partial \partial^2 v\_j(\boldsymbol{\xi}, \boldsymbol{\eta})}{\partial \boldsymbol{\eta} \partial \boldsymbol{\xi}^2} \right\} \right],\tag{22}$$

where *λ*(*s*) is the Lagrange multiplier

$$
\lambda(s) = \frac{-1}{s^{\circledast}}.
$$

Applying VHPTM using He's polynomials,

$$\begin{split} \sum\_{j=0}^{\infty} p^{j} v\_{j}(\xi,\eta) &= \sum\_{j=0}^{\infty} p^{j} v\_{j}(\xi,\eta) - p\mathcal{L}^{-1} \left[ \frac{1}{\mathcal{g}^{\emptyset}} \mathcal{E} \left\{ \mathcal{s}^{\emptyset} \left( \frac{\partial v\_{0}}{\partial \eta} + p \frac{\partial v\_{1}}{\partial \eta} + p^{2} \frac{\partial v\_{1}}{\partial \eta} + \cdots \right) + \left\{ \frac{\partial v\_{0}}{\partial \xi} + p^{1} \frac{\partial v\_{1}}{\partial \xi} + p^{2} \frac{\partial v\_{2}}{\partial \xi} + \cdots \right\} \right. \\ &+ \left\{ v\_{0} \frac{\partial v\_{0}}{\partial \xi} + p(v\_{0} \frac{\partial v\_{1}}{\partial \xi} + v\_{1} \frac{\partial v\_{0}}{\partial \xi}) + p^{2} (v\_{0} \frac{\partial v\_{2}}{\partial \xi} + v\_{1} \frac{\partial v\_{1}}{\partial \xi} + v\_{2} \frac{\partial v\_{0}}{\partial \xi}) + \cdots \right\} - \left\{ \frac{\partial \partial^{2} v\_{0}}{\partial \eta \partial \xi^{2}} + p^{1} \frac{\partial \partial^{2} v\_{1}}{\partial \eta \partial \xi^{2}} + p^{2} \frac{\partial \partial^{2} v\_{2}}{\partial \eta \partial \xi^{2}} + \cdots \right\} \end{split} \tag{23}$$

Comparing the coefficients of *p*

*υ*0(*ξ*, *η*) = 3*α* sec *h*2(*βξ*), *<sup>p</sup>*1*υ*1(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*1*υ*0(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*1*£*−<sup>1</sup> 1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*<sup>0</sup> *∂η* <sup>+</sup> *∂υ*<sup>0</sup> *∂ξ* <sup>+</sup> *<sup>υ</sup>*<sup>0</sup> *∂υ*<sup>0</sup> *∂ξ* <sup>−</sup> *∂∂*2*υ*<sup>0</sup> *∂η∂ξ*<sup>2</sup> , *<sup>υ</sup>*1(*ξ*, *<sup>η</sup>*) = <sup>3</sup>*<sup>α</sup>* sec *<sup>h</sup>*2(*βξ*) + <sup>3</sup>*αβ*{<sup>1</sup> <sup>+</sup> <sup>6</sup>*αβ* <sup>+</sup> cosh(2*βξ*)} sec *<sup>h</sup>*4(*βξ*)tanh(*βξ*) *<sup>η</sup><sup>β</sup>* Γ(*β* + 1) , *<sup>p</sup>*2*υ*2(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*2*υ*1(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*2*£*−<sup>1</sup> 1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*<sup>1</sup> *∂η* <sup>+</sup> *∂υ*<sup>1</sup> *∂ξ* <sup>+</sup> *<sup>υ</sup>*<sup>0</sup> *∂υ*<sup>1</sup> *∂ξ* <sup>+</sup> *<sup>υ</sup>*<sup>1</sup> *∂υ*<sup>0</sup> *∂ξ* <sup>−</sup> *∂∂*2*υ*<sup>1</sup> *∂η∂ξ*<sup>2</sup> , *<sup>υ</sup>*2(*ξ*, *<sup>η</sup>*) = <sup>3</sup>*<sup>α</sup>* sec *<sup>h</sup>*2(*βξ*) + <sup>3</sup>*αβ*{<sup>1</sup> <sup>+</sup> <sup>6</sup>*αβ* <sup>+</sup> cosh(2*βξ*)} sec *<sup>h</sup>*4(*βξ*)tanh(*βξ*) *<sup>η</sup><sup>β</sup>* Γ(*β* + 1) − 3 <sup>32</sup> *αβ*2{−<sup>8</sup> <sup>−</sup> <sup>96</sup>*<sup>α</sup>* <sup>−</sup> <sup>576</sup>*α*<sup>2</sup> <sup>+</sup> <sup>3</sup>(−<sup>3</sup> <sup>−</sup> <sup>16</sup>*<sup>α</sup>* <sup>+</sup> <sup>144</sup>*α*2) cosh(2*βξ*) + <sup>48</sup>*<sup>α</sup>* cosh(4*βξ*) + cosh(6*βξ*)} sec *<sup>h</sup>*8(*βξ*) *<sup>η</sup>*2*<sup>β</sup>* Γ(2*β* + 1) , · · ·

The analytical expression is therefore obtained in the following way

$$\begin{split} v(\xi,\eta) &= 3a\sec h^2(\beta\xi) + 3a\{1 + 6a\beta + \cosh(2\beta\xi)\}\sec h^4(\beta\xi)\tanh(\beta\xi)\frac{\eta^6}{\Gamma(\beta+1)} \\ &- \frac{3}{32}a\beta^2\{-8 - 96a - 576a^2 + 3(-3 - 16a + 144a^2)\cosh(2\beta\xi) + 48a\cosh(4\beta\xi) + \cosh(6\beta\xi)\} \\ &+ \sec h^6(\beta\xi)\frac{\eta^{2\delta}}{\Gamma(2\beta+1)} + \frac{1}{32}a\beta^3\{-8\xi - 1416a - 8496\beta^2 - 2937\beta^3 + 4(-31 - 432a - 1584a^2 + 3456a^3)\} \\ &- \cosh(2\beta\xi) - 4(11 + 54a - 540a^2)\cosh(4\beta\xi) - 4\cosh(6\beta\xi) + 96a\cosh(6\beta\xi) + \cosh(8\beta\xi) + \sec h^8(\beta\xi) \\ &+ \tanh(\beta\xi)\frac{\eta^{3\delta}}{\Gamma(3\beta+1)} + \cdots \end{split} \tag{24}$$

The exact solution at *β* = 1

$$\upsilon(\xi,\eta) = 3a \sec h^2(\beta(\xi - (1+a)\eta)).\tag{25}$$

In Figure 3, we compared the analytical solution of VHPTM with the exact solution of example 4.2. The comparison has shown the close contact between VHPTM solution and exact solution of the problems. Figure 4, represents VHPTM solution at different fractional-orders *β* = 1, 0.8, 0.6 and 0.4 The convergence analysis of fractional-order problems are convergent towards the integer-order problem of example 4.2, as observed.

**Figure 3.** Variational homotopy perturbation transform method solution of example 4.2 for *β* = 1.

**Figure 4.** Variational homotopy perturbation transform method solution of example 4.2 at different fractional order *β* = 1, 0.8, 0.6, 0.4.

#### *4.3. Example*

We consider time fractional-order linear RLW equation

$$2\frac{\partial^{\beta}v}{\partial \eta^{\beta}} + \frac{\partial v}{\partial \xi^{\varepsilon}} - 2\frac{\partial \partial^{2}v}{\partial \eta \partial \xi^{2}} = 0, \qquad 0 < \xi \le 1, \qquad 0 < \beta \le 1, \quad \eta > 0,\tag{26}$$

initial condition is

$$
\sigma(\emptyset, 0) = e^{-\frac{\pi}{6}},
\tag{27}
$$

By using Equation (12), the fractional PDE given in Equation (26) can be written as

$$\sum\_{j=0}^{\infty} p^j \upsilon\_{j+1}(\xi, \eta) = \sum\_{j=0}^{\infty} p^j \upsilon\_j(\xi, \eta) + p^j \mathcal{L}^{-1} \left[ \lambda(s) \mathcal{L} \left\{ s^{\oint} \frac{\partial^{\delta} \upsilon\_j}{\partial \eta^{\delta}}(\xi, \eta) + \frac{\partial \upsilon\_j}{\partial \xi} - 2 \frac{\partial \partial^2 \upsilon\_j}{\partial \eta \partial \xi^2} \right\} \right], \tag{28}$$

where *λ*(*s*) is the Lagrange multiplier

$$
\lambda(s) = \frac{-1}{s^{\circledast}}.
$$

Applying VHPTM using He's polynomials,

$$\begin{split} \sum\_{j=0}^{\infty} p^j \upsilon\_j(\mathbb{S}, \boldsymbol{\eta}) &= \sum\_{j=0}^{\infty} p^j \upsilon\_j(\mathbb{S}, \boldsymbol{\eta}) - p\mathcal{L}^{-1} \left[ \frac{1}{s^{\beta}} \mathcal{E} \left\{ s^{\beta} \left( \frac{\partial \upsilon\_0}{\partial \boldsymbol{\eta}} + p \frac{\partial \upsilon\_1}{\partial \boldsymbol{\eta}} + p^2 \frac{\partial \upsilon\_2}{\partial \boldsymbol{\eta}} + \cdots \right) \right. \\ &+ \frac{\partial}{\partial \boldsymbol{\xi}} \left\{ \upsilon\_0 + p\upsilon\_1 + p^2 \upsilon\_2 + \cdots \right\} - 2 \left\{ \frac{\partial \partial^2 \upsilon\_0}{\partial \boldsymbol{\eta} \partial \boldsymbol{\xi}^2} + p^1 \frac{\partial \partial^2 \upsilon\_1}{\partial \boldsymbol{\eta} \partial \boldsymbol{\xi}^2} + p^2 \frac{\partial \partial^2 \upsilon\_2}{\partial \boldsymbol{\eta} \partial \boldsymbol{\xi}^2} + \cdots \right\} \right\}. \end{split} \tag{29}$$

Comparing the coefficients of *p*

*υ*0(*ξ*, *η*) = *e* <sup>−</sup>*<sup>ξ</sup>* , *<sup>p</sup>*1*υ*1(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*1*υ*0(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*1*£*−<sup>1</sup> 1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*<sup>0</sup> *∂η* (*ξ*, *<sup>η</sup>*) + *∂υ*<sup>0</sup> *∂ξ* <sup>−</sup> <sup>2</sup> *∂∂*2*υ*<sup>0</sup> *∂η∂ξ*<sup>2</sup> , *υ*1(*ξ*, *η*) = *e* <sup>−</sup>*<sup>ξ</sup>* + *e* <sup>−</sup>*<sup>ξ</sup> <sup>η</sup><sup>β</sup>* Γ(*β* + 1) , *<sup>p</sup>*2*υ*2(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*2*υ*1(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*2*£*−<sup>1</sup> 1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*<sup>1</sup> *∂η* (*ξ*, *<sup>η</sup>*) + *∂υ*<sup>1</sup> *∂ξ* <sup>−</sup> <sup>2</sup> *∂∂*2*υ*<sup>1</sup> *∂η∂ξ*<sup>2</sup> , *υ*2(*ξ*, *η*) = *e* <sup>−</sup>*<sup>ξ</sup>* + *e* <sup>−</sup>*<sup>ξ</sup> <sup>η</sup><sup>β</sup>* <sup>Γ</sup>(*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>+</sup> *<sup>e</sup>* <sup>−</sup>*<sup>ξ</sup> <sup>η</sup>*2*<sup>β</sup>* Γ(2*β* + 1) , *<sup>p</sup>*3*υ*3(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*3*υ*2(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*3*£*−<sup>1</sup> −1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*<sup>2</sup> *∂η* (*ξ*, *<sup>η</sup>*) + *∂υ*<sup>2</sup> *∂ξ* <sup>−</sup> <sup>2</sup> *∂∂*2*υ*<sup>2</sup> *∂η∂ξ*<sup>2</sup> , *υ*3(*ξ*, *η*) = *e* <sup>−</sup>*<sup>ξ</sup>* + *e* <sup>−</sup>*<sup>ξ</sup> <sup>η</sup><sup>β</sup>* <sup>Γ</sup>(*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>+</sup> *<sup>e</sup>* <sup>−</sup>*<sup>ξ</sup> <sup>η</sup>*2*<sup>β</sup>* <sup>Γ</sup>(2*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>+</sup> *<sup>e</sup>* <sup>−</sup>*<sup>ξ</sup> <sup>η</sup>*3*<sup>β</sup>* Γ(3*β* + 1) . · ·

The analytical expression is therefore obtained in the following way

$$\upsilon(\xi,\eta) = e^{-\overline{\xi}} + e^{-\overline{\xi}} \frac{\eta^{\beta}}{\Gamma(\beta+1)} + e^{-\overline{\xi}} \frac{\eta^{2\beta}}{\Gamma(2\beta+1)} + e^{-\overline{\xi}} \frac{\eta^{3\beta}}{\Gamma(3\beta+1)} + \cdots \tag{30}$$

If *β* = 1 the series form is

·

$$\upsilon(\zeta,\eta) = e^{-\frac{\pi}{5}} \left( 1 + \eta + \frac{\eta^2}{2!} + \frac{\eta^3}{3!} + \dotsb \right) \tag{31}$$

The exact solution at *β* = 1

$$
\sigma(\emptyset, \eta) = \mathfrak{e}^{\eta - \mathfrak{F}}.\tag{32}
$$

In Table 2, the analytical solutions of VHPTM and HPSTM are compared in terms of absolute error. The accuracy has been measured for both the methods. By comparison it has shown that the proposed method VHPTM has a higher degree of accuracy than HPSTM.

In Figure 5, the graphs of exact and approximate solutions of example 4.3 are plotted. The graphical representation has confirmed that exact and VHPTM solutions are coincident. The exact and approximate solutions are closed to each other and verify the validity of the proposed method. The solution of example 4.3 at different fractional-orders *β* = 1, 0.8, 0.6 and 0.4 are shown graphically in Figure 6. The obtained solutions support the convergence phenomena of the solution of fractional-order problems to the solution of integer-order problem for the example 4.3.


**Table 2.** Comparison of VHPTM and HPSTM [11] of example 4.3 at *η* = 0.0000001.

**Figure 5.** Variational homotopy perturbation transform method solution of example 4.3 for *β* = 1.

**Figure 6.** Variational homotopy perturbation transform method solution of example 4.3 at different fractional order *β* = 1, 0.8, 0.6, 0.4.

#### *4.4. Example*

We consider time fractional-order linear RLW equation

$$\frac{\partial^{\beta}\upsilon(\xi,\eta)}{\partial\eta^{\beta}} + \frac{\partial^{4}\upsilon(\xi,\eta)}{\partial\xi^{4}} = 0, \qquad 0 < \xi \le 1, \qquad 0 < \beta \le 1, \quad \eta > 0,\tag{33}$$

with initial condition

$$\upsilon(\not\xi,0) = \sin\not\xi,\tag{34}$$

By using Equation (12), the fractional PDE given in Equation (33) can be written as

$$\sum\_{j=0}^{\infty} p^j \upsilon\_{j+1}(\underline{\boldsymbol{\xi}}, \boldsymbol{\eta}) = \sum\_{j=0}^{\infty} p^j \upsilon\_j(\underline{\boldsymbol{\xi}}, \boldsymbol{\eta}) + \sum\_{j=0}^{\infty} p^j \boldsymbol{\xi}^{-1} \left[ \lambda(\boldsymbol{s}) \boldsymbol{\xi} \left\{ \boldsymbol{s}^{\boldsymbol{\beta}} \frac{\partial^{\boldsymbol{\beta}} \upsilon\_j(\underline{\boldsymbol{\xi}}, \boldsymbol{\eta})}{\partial \boldsymbol{\eta}^{\boldsymbol{\beta}}} (\underline{\boldsymbol{\xi}}, \boldsymbol{\eta}) + \frac{\partial^{\boldsymbol{4}} \upsilon\_j(\underline{\boldsymbol{\xi}}, \boldsymbol{\eta})}{\partial \underline{\boldsymbol{\xi}}^{\boldsymbol{4}}} \right\} \right],\tag{35}$$

where *λ*(*s*) is the Lagrange multiplier

$$
\lambda(\mathbf{s}) = \frac{-1}{\mathbf{s}^{\beta}},
$$

Applying VHPTM using He's polynomials,

$$\begin{split} \sum\_{j=0}^{\infty} p^j v\_j(\xi, \eta) &= \sum\_{j=0}^{\infty} p^j v\_j(\xi, \eta) - p \mathcal{L}^{-1} \left[ \frac{1}{s^{\delta}} \mathcal{E} \left\{ s^{\delta} \left( \frac{\partial v\_0}{\partial \eta} + p \frac{\partial v\_1}{\partial \eta} + p^2 \frac{\partial v\_2}{\partial \eta} + \dots \right) \right. \\ &+ \frac{\partial^4}{\partial \xi^4} \left\{ v\_0 + p v\_1 + p^2 v\_2 + \dots \right\} \right] \Bigg) \end{split} \tag{36}$$

Comparing the coefficients of *p*

*υ*0(*ξ*, *η*) = sin *ξ*, *<sup>p</sup>*1*υ*1(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*1*υ*0(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*1*£*−<sup>1</sup> 1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*0(*ξ*.*η*) *∂η* (*ξ*, *<sup>η</sup>*) + *<sup>∂</sup>*4*υ*0(*ξ*.*η*) *∂ξ*<sup>4</sup> , *<sup>υ</sup>*1(*ξ*, *<sup>η</sup>*) = sin *<sup>ξ</sup>* <sup>−</sup> sin *<sup>ξ</sup> <sup>η</sup><sup>β</sup>* Γ(*β* + 1) , *<sup>p</sup>*2*υ*2(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*2*υ*1(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*2*£*−1[ 1 *<sup>s</sup><sup>β</sup> £*{*s<sup>β</sup> ∂υ*1(*ξ*.*η*) *∂η* (*ξ*, *<sup>η</sup>*) + *<sup>∂</sup>*4*υ*1(*ξ*.*η*) *∂ξ*<sup>4</sup> }], *<sup>υ</sup>*2(*ξ*, *<sup>η</sup>*) = sin *<sup>ξ</sup>* <sup>−</sup> sin *<sup>ξ</sup> <sup>η</sup><sup>β</sup>* <sup>Γ</sup>(*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>+</sup> sin *<sup>ξ</sup> <sup>η</sup>*2*<sup>β</sup>* Γ(2*β* + 1) , *<sup>p</sup>*3*υ*3(*ξ*, *<sup>η</sup>*) = *<sup>p</sup>*3*υ*2(*ξ*, *<sup>η</sup>*) <sup>−</sup> *<sup>p</sup>*3*£*−<sup>1</sup> 1 *sβ £ <sup>s</sup><sup>β</sup> ∂υ*2(*ξ*.*η*) *∂η* (*ξ*, *<sup>η</sup>*) + *<sup>∂</sup>*4*υ*2(*ξ*.*η*) *∂ξ*<sup>4</sup> , *<sup>υ</sup>*3(*ξ*, *<sup>η</sup>*) = sin *<sup>ξ</sup>* <sup>−</sup> sin *<sup>ξ</sup> <sup>η</sup><sup>β</sup>* <sup>Γ</sup>(*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>+</sup> sin *<sup>ξ</sup> <sup>η</sup>*2*<sup>β</sup>* <sup>Γ</sup>(2*<sup>β</sup>* <sup>+</sup> <sup>1</sup>) <sup>−</sup> sin *<sup>ξ</sup> <sup>η</sup>*3*<sup>β</sup>* Γ(3*β* + 1) . · ·

The analytical expression is therefore obtained in the following way

$$\upsilon(\zeta,\eta) = \sin\zeta - \sin\zeta \frac{\eta^{\beta}}{\Gamma(\beta+1)} + \sin\zeta \frac{\eta^{2\beta}}{\Gamma(2\beta+1)} - \sin\zeta \frac{\eta^{3\beta}}{\Gamma(3\beta+1)} + \cdots \,\,\_{\times}\tag{37}$$

If *β* = 1 the series form is

·

$$w(\xi,\eta) = \sin\xi \left(1 - \eta + \frac{\eta^2}{2!} - \frac{\eta^3}{3!} + \dotsb\right). \tag{38}$$

The exact solution at *β* = 1

$$\upsilon(\xi,\eta) = \sin\xi e^{-\eta}.\tag{39}$$

In Figure 7, the exact and VHPTM solution for example 4.4 are plotted. It can be seen from the figure that exact and VHPTM solutions are in closed contact with each other. In Figure 8, the VHPTM solutions for the example 4.4 at different fractional-orders are calculated. The convergence of fractional-order solutions towards integer-order solution has proved the applicability of the proposed method.

**Figure 7.** Variational homotopy perturbation transform method solution of example 4.4 for *β* = 1.

**Figure 8.** Variational homotopy perturbation transform method solution of example 4.4 at different fractional order *β* = 1, 0.8, 0.6, 0.4.

#### **5. Results and Discussion**

Several numerical examples are considered checking the applicability and reliability of the proposed method. The graphical representations of the solutions of examples 1 to 4, have provided the information about the accuracy and reliability of the proposed method. All the results of examples 1–4, confirmed strong agreement of VHPTM solutions with the exact solutions of the problems. The graphs represented the solutions for each problems at different fractional-order *β* = 1, 0.8, 0.6 and 0.4. Investigations show that solutions of fractional-order problem are convergent to the solution of integer-order problems. Moreover, the simple and straightforward implementation of the suggested method is also observed throughout the simulation. From the above properties of the present method, we expect that it can be modified for other fractional-order differential equations which arise in science and engineering.

#### **6. Conclusions**

In this article, the fractional view of acoustic wave equation is discussed by using a modified analytical technique. The solution graphs are plotted to provide clear pictures and analysis of the obtained results. The graphical representation has suggested the greatest rate of convergence as compared to other analytical methods. The fractional-order analysis of the acoustic wave equation is important to investigate the behaviour of the dynamics as compared to the classical one. Therefore, in the present application scenario, the proposed method has played a significant role to describe sophisticated solutions of fractional-order partial differential equations arising in different areas of sciences and engineering. Moreover, the present method uses the variational parameters which reduces the calculations' complexity. Also, the He's polynomials have been used to obtain the solutions in an accurate way as compared to Adomian polynomials. The rate of convergence of the suggested method is found to be higher than other existing methods. Hence, it is concluded that the present method can be extended to solve other fractional non-linear partial differential equations.

**Author Contributions:** Conceptualization, I.A. and H.K.; Methodology, R.S.; Software, H.K.; Validation, R.S. and I.A.; Formal Analysis, R.S.; Investigation, R.S. and D.B.; Resources, H.K. and P.K.; Writing—Original Draft Preparation, R.S.; Writing—Review and Editing, H.K., M.A. and P.K.; Visualization, M.A.; Supervision, D.B., P.K.; Project Administration, P.K.; Funding Acquisition, P.K. All authors have read and agreed to the published version of the manuscript.

**Funding:** Center of Excellence in Theoretical and Computational Science (TaCS-CoE) Faculty of Science, King Mongkuts University of Technology Thonburi (KMUTT).

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**


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