Exploring Fractional Damped Burgers’ Equation: A Comparative Analysis of Analytical Methods

Abstract

This investigation focuses on the study of the fractional damped Burgers’ equation by using the natural residual power series method coupled with the new iteration transform method in the context of the Caputo operator. The equation of Burgers under the damped context is useful when studying one-dimensional nonlinear waves involving damping effect, and is used in fluid dynamics, among other applications. Two new mathematical methods that can be used to obtain an approximate solution to this complex non-linear problem are the natural residual power series method and the new iteration transform method. Therefore, it can be deduced that the Caputo operator aids in modeling of the fractional derivatives, as it provides a better description of the physical realities. Thus, the objective of the present work is to advance the knowledge accumulated on the behavior of solutions to the damped Burgers’ equation, as well as to check the applicability of the proposed approaches to other nonlinear fractional partial differential equations.

1. Introduction

Nonlinear fractional partial differential equations (PDEs) are a significant step in the mathematical modeling of complex systems [1,2]. These equations are continuum generalizations of time-fractional derivatives, and go beyond the standard PDEs in any dimension by accounting for non-integer (fractional) order strains that effectively reproduce preparation memory characteristics specific to numerous physical and biological mechanisms [3,4]. It can be concluded that fractional derivatives are formulated in Riemann–Liouville, Caputo, or another form, which is very flexible and powerful for modeling anomalous diffusion, viscoelastic materials, and other cases where classical models based on integer order fail. Because these equations are nonlinear, they provide a better approximation of the system as a whole; it is even possible to model complex interactions between different parts of the dynamical system, which provides more information on how things work [5,6].

Nonlinear fractional PDEs are used in physics, engineering, finance, and biology. In physics, they describe processes such as anomalous diffusion in porous media and the dynamics of complex fluids [7,8,9,10,11]. In engineering, these equations are used to design and analyze materials with memory effects, such as polymers or biological tissues [12,13,14,15,16]. Fractional PDEs are employed in finance to model market behaviors characterized by long-range dependence and heavy-tailed distributions [17,18,19]. In biology, these fractional derivatives are used to model population dynamics in time and space when it is necessary to measure long-range interactions while accounting for memory effects, for instance when modeling the spread of diseases. The result is not only of theoretical importance but also has important implications in practice, where there is enormous effort to develop numerical schemes and computational methods for practical solutions [20,21,22,23,24,25]. Recent advances in nonlinear dynamics and fractional differential equations have expanded their applicability to diverse scientific and engineering problems. Researchers such as Kai and Yin [26,27] have explored soliton molecules and wave structures in complex equations, revealing intricate behaviors. Xie and Yang have further contributed by analyzing systems with fractional damping and nonlinear flutter in structural dynamics [28,29]. These studies underscore the growing importance of robust mathematical methods for modeling and solving complex physical phenomena [30,31].

The damped Burgers’ equation is similar to the classic Burgers’ equation, except for an additional damping term in the model representing energy dissipation. The dampened Burgers’ equation is obtained by introducing another term, typically au, representing a linear damping. When the damping term is included, the equation becomes more appropriate for modeling physical phenomena with energy dissipation of interest, e.g., fluid mechanics, acoustics, and traffic flow [32,33,34,35,36]. Therefore, the damped Burgers’ equation has all the features of the original Burgers’ equation; however, the damping term makes it more complex. This term affects the development of shock waves and smooth solutions in a manner that is different from what occurs in the undamped case. When viscosity and friction are present, as in the case of fluid dynamic applications, the KdV-Burgers’ equation with a specific damping term can model the transformations of energy dissipation within the flows. In acoustics, it describes the propagation of waves into such media that absorb energy; the amplitude of the waves is reduced due to energy loss. Traffic flow applications employ the damped Burgers’ equation as well; these systems provide definite results by considering the braking effect and other dissipating forces affecting the movement of the vehicles. The work included in the analytical and numerical investigation of the damped Burgers’ equation entails expanding the investigation of the solution of the equation along with its stability and long-term behavior, investigating dissipative systems, and constructing more accurate prediction models [37,38,39,40,41].

The natural residual power series method (NRPSM) is an advanced analytical method that can be applied to solving nonlinear differential equations. This technique combines power series and residual correction approaches, resulting in a higher accuracy method with a reduced computational time. This method starts by describing the solution of a nonlinear differential equation in the form of a power series with unknown coefficients. Then, these terms are determined by substituting the series into the original DE and equating coefficients of similar powers of the independent variable. The main difference with the RPST method is that it considers residual errors due to using a power series solution approximation. Therefore, a correction process systematically minimizes these residuals by gradually adjusting the coefficients to bring them closer and closer to their actual values and ultimately find an accurate solution [42,43,44,45]. The correction is generally obtained by constructing a residual equation, solved by different optimization methods. The method of RPST has many advantages, especially its strong nonlinearity and flexibility in application to various classes/types of differential equations. It is an iterative method that guarantees that the solution will converge to a particular answer of a differential equation; thus, its general use makes it a powerful tool in mathematical analysis and applied mathematics. In addition, the RPST method is computationally efficient in performing iterations, requiring less computational effort than other numerical methods. Therefore, it has been extensively used in many scientific and engineering disciplines to tackle complex differential equations in particular practical cases [46,47,48].

The new iterative transform method (NITM) is a modern computational method for tackling linear and nonlinear equations of differential forms at a leading level. This approach combines a number of of the capabilities of traditional iterative methods and novel transform techniques to improve solution convergence and accuracy. The fundamental idea of NITM is to take a differential equation and reduce it into some more straightforward form by applying transformations. An approximate solution is suggested, typically based on initial or boundary conditions. This leads to a solution that can be further transformed with some integral transform (such as the Laplace or Fourier transform), meaning that this form resembles an algebraic equation. This transformed equation is more straightforward to manipulate and solve, leading to a better solution estimate. This process is repeated iteratively, and every iteration generates a new and improved solution by rectifying the residual errors from the previous step. This rapid convergence of the NITM is due to a very effective reduction in complexity through that transform step. This highly general method can be used for all kinds of differential equations, even those that are extremely nonlinear or have complicated boundary conditions. It is principally beneficial in physics and engineering, where one of the relevant studies involves differential equations. While most NITMs are computationally demanding, a valuable feature in practical applications and theoretical investigations would be achieved by theoretically ideal accuracy with low computational costs. Moreover, it finds expanded usability and efficiency in solving various complex differential equations when combined with another numerical or analytical technique to improve flexibility and robustness [46,47,48].

2. Basic Definitions

First, we provide some background on fractional calculus and other relevant facts from applied mathematics. For further information in this area of research, the reader can see the following references: Hilfer [49], Kilbas et al. [50], Anastassiou [51], Khan and Khan [52], Silambarasn and Belgacem [53], Belgacem and Silambarasan [54].

Definition 1.

The fractional Rieman–Liouville integral of order $p\in {\mathbb{R}}_{+}$ of a function $h\left(\gamma \right)\in L\left(\right[0,1],\mathbb{R})$ is expressed by [55,56]

$$I_0^{\varrho }h\left(\gamma \right)={\displaystyle \frac{1}{\Gamma \left(\varrho \right)}}{\int }_0^t{(t-s)}^{\varrho -1}h\left(s\right)ds$$

on the assumption that the integral on the right side of the equation is convergent.

Definition 2.

For $\mu \in \mathbb{R}$, a function $h:\mathbb{R}\to {\mathbb{R}}^{+}$ is said to be in the space $C_{\mu }$ if it can be written as $h\left(\zeta \right)={\zeta }^q{h}_1\left(\zeta \right)$ with $q>\mu ,h_1\left(\zeta \right)\in C[0,\infty )$, and it is in space $h\left(\zeta \right)\in C_{\mu }^n$ if $h^{\left(n\right)}\in C_{\mu }$ for $n\in \mathbb{N}\cup \left\{0\right\}$ [55,56].

Definition 3.

The fractional Caputo derivative of a function $h\in C_{-1}^n$ with $n\in \mathbb{N}\cup \left\{0\right\}$ is provided as [55,56]

$$D_t^{\varrho }h\left(t\right)=\left\{\begin{array}{cc}{I}^{n-\varrho }{h}^{\left(n\right)},\hfill & n-1<\varrho \le n,n\in \mathbb{N},\hfill \\ {\displaystyle \frac{d^n}{\mathrm{d}{t}^n}}h\left(t\right),\hfill & \varrho =n,n\in \mathbb{N}.\hfill \end{array}\right.$$

Definition 4.

The Mittag-Leffler function (MLF) of two parameters is defined as [55,56]

$$E_{\varrho ,\beta }\left(t\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{t^k}{\Gamma (k\varrho +\beta )}}$$

for $\varrho =\beta =1,E_{1,1}\left(t\right)=e^t$ and $E_{1,1}(-t)=e^{-t}$.

Definition 5.

The NT of a function $v(\zeta ,t)$ for $t\ge 0$ is defined by [55,56]

$$\mathcal{N}\left[v(\zeta ,t)\right]=R(\zeta ,s,u)={\int }_0^{\infty }{e}^{-st}v(\zeta ,ut)\mathrm{d}t,$$

where s and u for the transform parameters are taken to be real and positive.

Definition 6.

The MLF of the natural transform (NT) $E_{\varrho ,\beta }$ is provided as [55,56]

$$\mathcal{N}\left[v(\zeta ,t)\right]={\int }_0^{\infty }{e}^{-st}v(\zeta ,ut)\mathrm{d}t=\sum _{k=0}^{\infty }{\displaystyle \frac{u^{k+1}\Gamma (k+1)}{s^{k+1}\Gamma (k\varrho +\beta )}}.$$

Definition 7.

The Miller and Ross sense of the NT of $D^{\varrho }h\left(t\right)$ is expressed by the following [55,56]:

$$\mathcal{N}\left(D^{\varrho }h\left(t\right)\right)={\displaystyle \frac{s^{\varrho }}{u^{\varrho }}}R(s,u)-\sum _{k=0}^{n-1}{\displaystyle \frac{s^{n-k-1}}{u^{n-k}}}{h}^{\left(k\right)}\left(0\right),n-1<\varrho \le n.$$

Lemma 1.

The NT of ${\displaystyle \frac{{\partial }^{\varrho }h(\zeta ,t)}{\partial t^{\varrho }}}$ with respect to t can be defined as follows [55,56]:

$$\mathcal{N}\left[{\displaystyle \frac{{\partial }^{\varrho }h(\zeta ,t)}{\partial t^{\varrho }}}\right]={\displaystyle \frac{s^{\varrho }}{u^{\varrho }}}R(\zeta ,s,u)-\sum _{k=0}^{n-1}{\displaystyle \frac{s^{n-k-1}}{u^{n-k}}}\left[\underset{t\to 0}{\lim }{\displaystyle \frac{{\partial }^{\varrho }h(\zeta ,t)}{\partial t^{\varrho }}}\right].$$

Lemma 2.

The natural transform of the ϱ-order partial derivative of $h(\zeta ,t)$ with respect to ζ is denoted by [55,56]

$$\mathcal{N}\left[{\displaystyle \frac{{\partial }^{\varrho }h(\zeta ,t)}{\partial {\zeta }^{\varrho }}}\right]={\displaystyle \frac{d^{\varrho }}{d{\zeta }^{\varrho }}}R(\zeta ,s,u).$$

Lemma 3.

The dual relationship between the Laplace and natural transforms is expressed by [55,56]

$$\mathcal{N}\left[h(\zeta ,t)\right]=R(\zeta ,s,u)={\displaystyle \frac{1}{u}}{\int }_0^{\infty }{e}^{{\displaystyle \frac{-u}{\zeta }}}h(\zeta ,t)\mathrm{d}t={\displaystyle \frac{1}{u}}\mathcal{L}\left\{h(\zeta ,t)\right\},$$

where $\mathcal{L}$ is the Laplace transform. As a conclusion from the above Lemma, it can be noted that the natural transform can be seen as an extension of both the Sumudu and Laplace transforms. In particular, when $u=1$, then the natural transform reduces to the Laplace transform; in the same way, for $s=1$, the generalization leads to the Sumudu transform.

3. Outline of the Suggested Methodologies

3.1. Introduction of General Implementation of NRPSM

Consider the fractional order’s partial differential equation:

$$\begin{array}{c}\hfill D_t^{\varrho }\mathit{\phi }(\zeta ,t)=N_{\zeta }\left[\mathit{\phi }(\zeta ,t)\right],\\ \hfill \mathit{\phi }(\zeta ,0)=f\left(\zeta \right),\end{array}$$

(1)

where $N_{\zeta }$ is a nonlinear function related to $\zeta$ of degree $r,\zeta \in I,t\ge 0,D_t^{\varrho }$ refers to the $\varrho$-th fractional Caputo operator for $\varrho \in (0,1]$, and $\mathit{\phi }(\zeta ,t)$ is an unknown term.

The following steps may be taken to use the natural RPSM to create the approximate solution of Equation (1):

Step 1: Utilizing the starting data of Equation (1), use the natural transform on each side of Equation (1),

$$\begin{array}{c}\hfill \mathit{\phi }(\zeta ,s)={\displaystyle \frac{f\left(\zeta \right)}{s}}-{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\mathcal{N}\left\{N_{\zeta }\left[\mathit{\phi }(\zeta ,t)\right]\right\},\\ \hfill \mathrm{where}\mathit{\phi }(\zeta ,s)=\mathcal{N}\left[\mathit{\phi }\right(\zeta ,t\left)\right]\left(s\right),s>t.\end{array}$$

(2)

Step 2: Consider the following fractional expansion as the estimated solution of Equation (2):

$$\mathit{\phi }(\zeta ,s)={\displaystyle \frac{f\left(\zeta \right)}{s}}+\sum _{n=1}^{\infty }{\displaystyle \frac{u^{\varrho }{f}_n\left(\zeta \right)}{s^{n\varrho +1}}},x\in I,s>t\ge 0,$$

(3)

The k-th natural series solution is presented as follows:

$${\mathit{\phi }}_k(\zeta ,s)={\displaystyle \frac{f\left(\zeta \right)}{s}}+\sum _{n=1}^k{\displaystyle \frac{u^{\varrho }{f}_n\left(\zeta \right)}{s^{n\varrho +1}}},\zeta \in I,s>t\ge 0.$$

(4)

Step 3: The k-th natural residual fractional function of Equation (2) is defined as

$$\mathcal{N}\left({\mathrm{Res}}_{{\mathit{\phi }}_k}(\zeta ,s)\right)={\mathit{\phi }}_k(\zeta ,s)-{\displaystyle \frac{f\left(\zeta \right)}{s}}+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\mathcal{N}\left\{N_{\zeta }\left[\mathit{\phi }(\zeta ,t)\right]\right\},$$

(5)

and Equation (2)’s natural residual functions are defined as

$$\underset{k\to \infty }{\lim }\mathcal{N}\left({\mathrm{Res}}_{{\mathit{\phi }}_k}(\zeta ,s)\right)=\mathcal{N}\left({\mathrm{Res}}_{\mathit{\phi }}(\zeta ,s)\right)=\mathit{\phi }(\zeta ,s)-{\displaystyle \frac{f\left(\zeta \right)}{s}}+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\mathcal{N}\left\{N_{\zeta }\left[\mathit{\phi }(\zeta ,t)\right]\right\}.$$

(6)

A a few helpful natural residual function facts that are necessary to determine the estimated solution are provided as follows: $-{\lim }_{k\to \infty }\mathcal{N}\left({\mathrm{Res}}_{{\mathit{\phi }}_k}(\zeta ,s)\right)=\mathcal{N}\left({\mathrm{Res}}_{\mathit{\phi }}(\zeta ,s)\right)$, for $\zeta \in I,s>t\ge 0$. $-\mathcal{N}\left({\mathrm{Res}}_{\mathit{\phi }}(\zeta ,s)\right)=0$, for $\zeta \in I,s>t\ge 0$. $-{\lim }_{s\to \infty }{s}^{k\varrho +1}\mathcal{N}\left({\mathrm{Res}}_{{\mathit{\phi }}_k}(\zeta ,s)\right)=0$, for $\zeta \in I,s>t\ge 0$, and $k=1,2,3,\dots$

Step 4: Now, we put the k-th natural series solution of Equation (4) into the k-th natural fractional residual function of Equation (5).

Step 5: The unknown coefficients $h_k\left(\zeta \right)$ for $k=1,2,3,\dots$ might be obtained by solving the system ${\lim }_{s\to \infty }{s}^{ka+1}\mathcal{N}\left({\mathrm{Res}}_{{\mathit{\phi }}_k}(\zeta ,s)\right)=0$. Subsequently, we gather the obtained coefficients using the fractional expansion series in Equation (4): ${\mathit{\phi }}_k(\zeta ,s)$.

Step 6: Finally, we apply the inverse natural transform operator to both sides of the natural series solution to obtain an estimated solution ${\mathit{\phi }}_k(\zeta ,t)$ of the main Equation (1).

3.2. Problem 1

Take the fractional PDE into consideration as follows [57]:

$$D_t^{\varrho }\mathit{\phi }(\zeta ,t)+\mathit{\phi }(\zeta ,t){\displaystyle \frac{{\partial }^3\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^3}}-{\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}-{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}=0,\mathrm{where}0<\varrho \le 1,$$

(7)

with the initial condition

$$\mathit{\phi }(\zeta ,0)={\displaystyle \frac{e^{\zeta /4}}{4}}$$

(8)

and exact result

$$\mathit{\phi }(\zeta ,t)={\displaystyle \frac{1}{4}}{e}^{{\displaystyle \frac{1}{4}}\left({\displaystyle \frac{t}{4}}+\zeta \right)}.$$

Using Equation (8), the NT is used in Equation (7) to obtain

$$\begin{array}{c}\mathit{\phi }(\zeta ,s)-{\displaystyle \frac{\frac{e^{\zeta /4}}{4}}{s^2}}+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}{\mathcal{N}}_t\left[{\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)\times {\displaystyle \frac{{\partial }^3{\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)}{\partial {\zeta }^3}}\right]-{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}{\mathcal{N}}_t\left[{\displaystyle \frac{\partial {\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)}{\partial \zeta }}{\displaystyle \frac{{\partial }^2{\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)}{\partial {\zeta }^2}}\right]\hfill \\ -{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left[{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,s)}{\partial {\zeta }^2}}\right]=0.\hfill \end{array}$$

(9)

Hence, the k-th truncated term series are

$$\begin{array}{c}\hfill \mathit{\phi }\left(\zeta s\right)={\displaystyle \frac{\frac{e^{\zeta /4}}{4}}{s^2}}+\sum _{r=1}^k{\displaystyle \frac{u^{\varrho }{f}_r(\zeta ,s)}{s^{rp+1}}},r=1,2,3,4,\cdots .\end{array}$$

(10)

the natural residual functions (NRFs) are

$$\begin{array}{c}{\mathcal{N}}_{t}Res(\zeta ,s)=\mathit{\phi }(\zeta ,s)-{\displaystyle \frac{\frac{e^{\zeta /4}}{4}}{s^2}}+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}{\mathcal{N}}_t\left[{\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)\times {\displaystyle \frac{{\partial }^3{\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)}{\partial {\zeta }^3}}\right]-{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}{\mathcal{N}}_t\left[{\displaystyle \frac{\partial {\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)}{\partial \zeta }}{\displaystyle \frac{{\partial }^2{\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)}{\partial {\zeta }^2}}\right]\hfill \\ -{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left[{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,s)}{\partial {\zeta }^2}}\right]=0,\hfill \end{array}$$

(11)

and the kth-NRFs are

$$\begin{array}{c}{\mathcal{N}}_{t}Re{s}_k(\zeta ,s)={\mathit{\phi }}_k(\zeta ,s)-{\displaystyle \frac{\frac{e^{\zeta /4}}{4}}{s^2}}+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}{\mathcal{N}}_t\left[{\mathcal{N}}_t^{-1}{\mathit{\phi }}_k(\zeta ,s)\times {\displaystyle \frac{{\partial }^3{\mathcal{N}}_t^{-1}{\mathit{\phi }}_k(\zeta ,s)}{\partial {\zeta }^3}}\right]\hfill \\ -{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}{\mathcal{N}}_t\left[{\displaystyle \frac{\partial {\mathcal{N}}_t^{-1}{\mathit{\phi }}_k(\zeta ,s)}{\partial \zeta }}{\displaystyle \frac{{\partial }^2{\mathcal{N}}_t^{-1}{\mathit{\phi }}_k(\zeta ,s)}{\partial {\zeta }^2}}\right]-{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left[{\displaystyle \frac{{\partial }^2{\mathit{\phi }}_k(\zeta ,s)}{\partial {\zeta }^2}}\right]=0.\hfill \end{array}$$

(12)

To investigate $f_r(\zeta ,s)$, we quickly analyze the relations ${\lim }_{s\to \infty }\left(s^{r\varrho +1}\right)$, multiply the solution equation by $s^{r\varrho +1}$, and put the r-th truncated series in Equation (10) into the r-th natural residual function in Equation (12): $r=1,2,3,\cdots$, and ${\mathcal{N}}_{t}Re{s}_{\mathit{\phi },r}(\zeta ,s))=0$. The first steps is as follows:

$$f_1(\zeta ,s)={\displaystyle \frac{e^{\zeta /4}}{64}},$$

(13)

$$f_2(\zeta ,s)={\displaystyle \frac{e^{\zeta /4}}{1024}},$$

(14)

$$f_3(\zeta ,s)={\displaystyle \frac{e^{\zeta /4}}{16384}},$$

(15)

$$f_4(\zeta ,s)={\displaystyle \frac{e^{\zeta /4}}{262144}},$$

(16)

and so forth.

After placing $f_r(\zeta ,s)$ for $r=1,2,3,\cdots$ in Equation (10), we obtain

$$\mathit{\phi }(\zeta ,s)={\displaystyle \frac{e^{\zeta /4}}{64s^{\varrho +1}}}+{\displaystyle \frac{e^{\zeta /4}}{1024s^{2\varrho +1}}}+{\displaystyle \frac{e^{\zeta /4}}{16384s^{3\varrho +1}}}+{\displaystyle \frac{e^{\zeta /4}}{262144s^{4\varrho +1}}}+{\displaystyle \frac{e^{\zeta /4}}{4s}}+\cdots .$$

(17)

When the inverse natural transform is applied, the results are as follows:

$$\mathit{\phi }(\zeta ,t)={\displaystyle \frac{e^{\zeta /4}}{4}}+{\displaystyle \frac{e^{\zeta /4}{t}^{2p}}{1024\Gamma (2\varrho +1)}}+{\displaystyle \frac{e^{\zeta /4}{t}^{3p}}{16384\Gamma (3\varrho +1)}}+{\displaystyle \frac{e^{\zeta /4}{t}^{4p}}{262144\Gamma (4\varrho +1)}}+{\displaystyle \frac{e^{\zeta /4}{t}^{\varrho }}{64\Gamma (\varrho +1)}}+\cdots .$$

(18)

3.3. Problem 2

Taking into consideration the fractional damped Burger equation [57] as follows:

$$D_t^{\varrho }\mathit{\phi }(\zeta ,t)+{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}+\mathit{\phi }(\zeta ,t){\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}+{\displaystyle \frac{1}{5}}\mathit{\phi }(\zeta ,t)=0,\mathrm{where}0<\varrho \le 1,$$

(19)

with the ICs

$$\mathit{\phi }(\zeta ,0)={\displaystyle \frac{1}{5}}\zeta$$

(20)

and exact solution

$$\mathit{\phi }(\zeta ,t)={\displaystyle \frac{\zeta }{5\left(2e^{\frac{t}{5}}-1\right)}},$$

by using Equation (20) and applying the NT to Equation (19), we obtain the following solution:

$$\mathit{\phi }(\zeta ,s)-{\displaystyle \frac{\frac{1}{5}\zeta }{s^2}}+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left[{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,s)}{\partial {\zeta }^2}}\right]+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}{\mathcal{N}}_t\left[{\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)\times {\displaystyle \frac{\partial {\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)}{\partial \zeta }}\right]+{\displaystyle \frac{u^{\varrho }}{5s^{\varrho }}}\left[\mathit{\phi }(\zeta ,s)\right]=0.$$

(21)

Hence, the k-th truncated term series can be described as follows:

$$\mathit{\phi }(\zeta ,s)={\displaystyle \frac{\frac{1}{5}\zeta }{s^2}}+\sum _{r=1}^k{\displaystyle \frac{u^{\varrho }{f}_r(\zeta ,s)}{s^{r\varrho +1}}},r=1,2,3,4,\cdots .$$

(22)

the natural residual functions (NRFs) are

$$\begin{array}{c}{\mathcal{N}}_{t}Res(\zeta ,s)=\mathit{\phi }(\zeta ,s)-{\displaystyle \frac{\frac{1}{5}\zeta }{s^2}}+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left[{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,s)}{\partial {\zeta }^2}}\right]+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}{\mathcal{N}}_t\left[{\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)\times {\displaystyle \frac{\partial {\mathcal{N}}_t^{-1}\mathit{\phi }(\zeta ,s)}{\partial \zeta }}\right]\hfill \\ +{\displaystyle \frac{u^{\varrho }}{5s^{\varrho }}}\left[\mathit{\phi }(\zeta ,s)\right]=0,\hfill \end{array}$$

(23)

and the k-th NRFs are

$$\begin{array}{c}{\mathcal{N}}_{t}Re{s}_k(\zeta ,s)={\mathit{\phi }}_k(\zeta ,s)-{\displaystyle \frac{\frac{1}{5}\zeta }{s^2}}+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left[{\displaystyle \frac{{\partial }^2{\mathit{\phi }}_k(\zeta ,s)}{\partial {\zeta }^2}}\right]+{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}{\mathcal{N}}_t\left[{\mathcal{N}}_t^{-1}{\mathit{\phi }}_k(\zeta ,s)\times {\displaystyle \frac{\partial {\mathcal{N}}_t^{-1}{\mathit{\phi }}_k(\zeta ,s)}{\partial \zeta }}\right]\hfill \\ +{\displaystyle \frac{u^{\varrho }}{5s^{\varrho }}}\left[{\mathit{\phi }}_k(\zeta ,s)\right]=0.\hfill \end{array}$$

(24)

To investigate $f_r(\zeta ,s)$, we quickly analyze the relations ${\lim }_{s\to \infty }\left(s^{r\varrho +1}\right)$, multiply the solution equation by $s^{r\varrho +1}$, and put the r-th truncated series in Equation (22) into the r-t natural residual function in Equation (24): $r=1,2,3,\cdots$ and ${\mathcal{N}}_{t}Re{s}_{\mathit{\phi },r}(\zeta ,s))=0$. The first steps are as follows:

$$f_1(\zeta ,s)=-{\displaystyle \frac{1}{25}}\left(2\zeta \right),$$

(25)

$$f_2(\zeta ,s)={\displaystyle \frac{6\zeta }{125}},$$

(26)

$$f_3(\zeta ,s)={\displaystyle \frac{2}{625}}\zeta \left(-{\displaystyle \frac{2\Gamma (2\varrho +1)}{\Gamma {(\varrho +1)}^2}}-9\right),$$

(27)

and so on.

Equation (22) is used to obtain the value of $f_r(\zeta ,s)$ for $r=1,2,3,\cdots$:

$$\mathit{\phi }(\zeta ,s)={\displaystyle \frac{6\zeta }{125s^{2\varrho +1}}}-{\displaystyle \frac{2\zeta }{25s^{\varrho +1}}}+{\displaystyle \frac{2\zeta \left(-\frac{2\Gamma (2\varrho +1)}{\Gamma {(\varrho +1)}^2}-9\right)}{625s^{3\varrho +1}}}+{\displaystyle \frac{\zeta }{5s}}+\cdots .$$

(28)

Applying the natural inverse transform, we obtain

$$\mathit{\phi }(\zeta ,t)={\displaystyle \frac{\zeta }{5}}+{\displaystyle \frac{6\zeta t^{2\varrho }}{125\Gamma (2\varrho +1)}}-{\displaystyle \frac{18\zeta t^{3\varrho }}{625\Gamma (3\varrho +1)}}-{\displaystyle \frac{4\zeta t^{3\varrho }\Gamma (2\varrho +1)}{625\Gamma {(\varrho +1)}^2\Gamma (3\varrho +1)}}-{\displaystyle \frac{2\zeta t^{\varrho }}{25\Gamma (\varrho +1)}}+\cdots .$$

(29)

The results of the NRPSM applied to fractional-order problems are presented and discussed with the help of figures and tables. Each figure and table provide information on the behavior of solutions under different fractional orders.

Figure 1: This figure demonstrates the NRPSM solutions for Problem 1 with varying fractional orders over the time interval $t=5$. The comparison highlights the influence of fractional orders on the dynamics of the system, showcasing distinct solution profiles as the fractional parameter changes. These differences emphasize the flexibility of the NRPSM in capturing diverse behaviors in fractional systems. Figure 2: The absolute error analysis for Problem 1 is illustrated in this figure. Subfigure (a) presents the absolute error for $p=1$, subfigure (b) for $p=0.5$, and subfigure (c) for $p=0.1$. The errors decrease as the fractional order approaches $p=1$, indicating the precision of the NRPSM for fractional orders of integers and near-integers. Figure 3: The fractional order comparison for Problem 2 is shown at $t=0.1$. The results reveal how the fractional parameter influences the solution, with distinct trends and amplitudes observed for different orders. These variations provide insights into the adaptability of NRPSM to effectively model fractional-order systems. Figure 4: Similar to Figure 3, this figure shows the fractional order comparison for Problem 2, except at $t=0.01$; the solutions demonstrate a sharper variation at this smaller time step, highlighting the method’s capability to capture rapid changes in dynamics for fractional systems. Figure 5: The absolute error analysis for Problem 2 is provided in this figure. Subfigure (a) corresponds to $p=1$, subfigure (b) to $p=0.5$, and subfigure (c) to $p=0.1$. As with Problem 1, the errors are smaller for higher fractional orders, confirming the robustness and precision of NRPSM.

Table 1. This table shows different fractional orders $\varrho =0.5,0.7$, and 1 of NRPSM, times $\left(t\right)=0.1,0.5$ and $1.0$, and function variables of $\xi =0.2,0.4,0.6,0.8$, and $1.0$ for Problem 1.

t	$\mathit{\zeta }$	${\mathit{NRPSM}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.5}}$	${\mathit{NRPSM}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.7}}$	${\mathit{NRPSM}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$	$\mathit{Exact}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.5}}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.7}}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$
	0.2	0.268655	0.268011	0.266124	0.266124	0.00253169	0.0018872	$2.007703\times {10}^{-9}$
	0.4	0.29691	0.296198	0.294112	0.294112	0.00279795	0.00208568	$2.218855\times {10}^{-9}$
	0.6	0.328136	0.327349	0.325044	0.325044	0.00309222	0.00230503	$2.452214\times {10}^{-9}$
1	0.8	0.362647	0.361777	0.359229	0.359229	0.00341743	0.00254745	$2.710116\times {10}^{-9}$
	1.0	0.400787	0.399825	0.39701	0.39701	0.00377684	0.00281537	$2.995142\times {10}^{-9}$
	0.2	0.262972	0.26089	0.257936	0.257936	0.00503609	0.00295406	$6.241301\times {10}^{-11}$
	0.4	0.290629	0.288328	0.285063	0.285063	0.00556574	0.00326474	$6.897699\times {10}^{-11}$
	0.6	0.321195	0.318652	0.315044	0.315044	0.00615109	0.0036081	$7.623141\times {10}^{-11}$
0.5	0.8	0.354975	0.352165	0.348177	0.348177	0.006798	0.00398756	$8.424871\times {10}^{-11}$
	1.0	0.392308	0.389202	0.384795	0.384795	0.00751296	0.00440694	$9.310924\times {10}^{-11}$
	0.2	0.255675	0.253463	0.251567	0.251567	0.00410712	0.00189519	$1.987299\times {10}^{-14}$
	0.4	0.282564	0.280119	0.278025	0.278025	0.00453907	0.00209451	$2.198241\times {10}^{-14}$
	0.6	0.312282	0.30958	0.307265	0.307265	0.00501645	0.00231479	$2.431388\times {10}^{-14}$
0.1	0.8	0.345124	0.342139	0.33958	0.33958	0.00554403	0.00255824	$2.681188\times {10}^{-14}$
	1.0	0.381422	0.378122	0.375294	0.375294	0.00612711	0.00282729	$2.964295\times {10}^{-14}$

: This table summarizes the NRPSM solutions for Problem 1 under different fractional orders. The numerical values demonstrate the sensitivity of the solution to the fractional parameter, further validating the accuracy and efficiency of the NRPSM.

Table 2. This table shows different fractional orders $\varrho =0.5,0.7$, and 1 of NRPSM for Problem 2.

t	$\mathit{\zeta }$	${\mathit{NRPSM}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.5}}$	${\mathit{NRPSM}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.7}}$	${\mathit{NRPSM}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$	$\mathit{Exact}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.5}}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.7}}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$
	0.2	0.07015	0.073533	0.0768932	0.0768933	0.00674328	0.00336034	$7.775646\times {10}^{-8}$
	0.4	0.1403	0.147066	0.153786	0.153787	0.0134866	0.00672067	$1.555129\times {10}^{-7}$
0.1	0.6	0.21045	0.220599	0.23068	0.23068	0.0202298	0.010081	$2.332694\times {10}^{-7}$
	0.8	0.2806	0.294132	0.307573	0.307573	0.0269731	0.0134413	$3.110258\times {10}^{-7}$
	1.0	0.35075	0.367665	0.384466	0.384467	0.0337164	0.0168017	$3.887823\times {10}^{-7}$
	0.2	0.0765701	0.078622	0.079681	0.079681	0.00311089	0.00105894	$7.976980\times {10}^{-12}$
	0.4	0.15314	0.157244	0.159362	0.159362	0.00622178	0.00211787	$1.595396\times {10}^{-11}$
0.01	0.6	0.22971	0.235866	0.239043	0.239043	0.00933267	0.00317681	$2.393094\times {10}^{-11}$
	0.8	0.30628	0.314488	0.318724	0.318724	0.0124436	0.00423574	$3.190792\times {10}^{-11}$
	1.0	0.38285	0.39311	0.398405	0.398405	0.0155544	0.00529468	$3.988487\times {10}^{-11}$

: Similar to Table 1, this table provides the NRPSM solutions for Problem 2 in various fractional orders. The results exhibit consistency with the graphical findings, reinforcing the reliability of the method in addressing fractional-order problems. In summary, the figures and tables collectively illustrate the efficiency, accuracy, and adaptability of the NRPSM in solving fractional-order problems. The method’s ability to handle diverse fractional dynamics and provide precise solutions with minimal error highlights its potential as a robust tool for analyzing fractional differential equations.

3.4. Idea of the Natural Iterative Transform Method

Let us examine a general space–time fractional order PDE:

$$D_t^{\varrho }\mathit{\phi }(\zeta ,t)=\Phi \left(\mathit{\phi }(\zeta ,t),D_{\zeta }^t\mathit{\phi }(\zeta ,t),D_{\zeta }^{2t}\mathit{\phi }(\zeta ,t),D_{\zeta }^{3t}\mathit{\phi }(\zeta ,t)\right)0<\varrho ,t\le 1,$$

(30)

with initial scenarios

$${\mathit{\phi }}^{\left(k\right)}(\zeta ,0)=h_k,k=0,1,2,\cdots ,m-1,$$

(31)

Assuming $\mathit{\phi }(\zeta ,t)$ as the unknown function, while $\Phi \left(\mathit{\phi }(\zeta ,t),D_{\zeta }^t\mathit{\phi }(\zeta ,t),D_{\zeta }^{2t}\mathit{\phi }(\zeta ,t),D_{\zeta }^{3t}\mathit{\phi }(\zeta ,t)\right)$ may be a nonlinear or linear operator of $\mathit{\phi }(\zeta ,t),D_{\zeta }^t\mathit{\phi }(\zeta ,t),D_{\zeta }^{2t}\mathit{\phi }(\zeta ,t)$ and $D_{\zeta }^{3t}\mathit{\phi }(\zeta ,t)$, applying the Natural transform to both sides of Equation (30) yields the following equation, where $\mathit{\phi }(\zeta ,t)$ is represented by $\mathit{\phi }$ for simplicity:

$$\mathcal{N}\left[\mathit{\phi }(\zeta ,t)\right]={\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\sum _{k=0}^{m-1}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}+\mathcal{N}\left[\Phi \left(\mathit{\phi }(\zeta ,t),D_{\zeta }^t\mathit{\phi }(\zeta ,t),D_{\zeta }^{2t}\mathit{\phi }(\zeta ,t),D_{\zeta }^{3t}\mathit{\phi }(\zeta ,t)\right)\right]\right).$$

(32)

The following result may be obtained by applying the inverse natural transform to the problem:

$$\mathit{\phi }(\zeta ,t)={\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\sum _{k=0}^{m-1}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}+\mathcal{N}\left[\Phi \left(\mathit{\phi }(\zeta ,t),D_{\zeta }^t\mathit{\phi }(\zeta ,t),D_{\zeta }^{2t}\mathit{\phi }(\zeta ,t),D_{\zeta }^{3t}\mathit{\phi }(\zeta ,t)\right)\right]\right)\right].$$

(33)

The iterative processing of the natural transform approach utilizes an infinite series to describe the accomplished result:

$$\mathit{\phi }(\zeta ,t)=\sum _{i=0}^{\infty }{\mathit{\phi }}_i.$$

(34)

Because $\Phi \left(\mathit{\phi },D_{\zeta }^t\mathit{\phi },D_{\zeta }^{2t}\mathit{\phi },D_{\zeta }^{3t}\mathit{\phi }\right)$ is a linear or nonlinear operator that has the decomposition

$$\begin{array}{c}\Phi \left(\mathit{\phi },D_{\zeta }^t\mathit{\phi },D_{\zeta }^{2t}\mathit{\phi },D_{\zeta }^{3t}\mathit{\phi }\right)=\Phi \left({\mathit{\phi }}_0,D_{\zeta }^t{\mathit{\phi }}_0,D_{\zeta }^{2t}{\mathit{\phi }}_0,D_{\zeta }^{3t}{\mathit{\phi }}_0\right)\hfill \\ +\sum _{i=0}^{\infty }\left(\Phi \left(\sum _{k=0}^i\left({\mathit{\phi }}_k,D_{\zeta }^t{\mathit{\phi }}_k,D_{\zeta }^{2t}{\mathit{\phi }}_k,D_{\zeta }^{3t}{\mathit{\phi }}_k\right)\right)-\Phi \left(\sum _{k=1}^{i-1}\left({\mathit{\phi }}_k,D_{\zeta }^t{\mathit{\phi }}_k,D_{\zeta }^{2t}{\mathit{\phi }}_k,D_{\zeta }^{3t}{\mathit{\phi }}_k\right)\right)\right),\hfill \end{array}$$

(35)

Equations (35) and (33) must be substituted into Equation (34) in order to obtain the subsequent equation:

$$\begin{array}{c}\sum _{i=0}^{\infty }{\mathit{\phi }}_i(\zeta ,t)={\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\sum _{k=0}^{m-1}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}+\mathcal{N}[\Phi ({\mathit{\phi }}_0,D_{\zeta }^t{\mathit{\phi }}_0,D_{\zeta }^{2t}{\mathit{\phi }}_0,D_{\zeta }^{3t}{\mathit{\phi }}_0)]\right)\right]\hfill \\ +{\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\mathcal{N}\left[\sum _{i=0}^{\infty }\left(\Phi \sum _{k=0}^i({\mathit{\phi }}_k,D_{\zeta }^t{\mathit{\phi }}_k,D_{\zeta }^{2t}{\mathit{\phi }}_k,D_{\zeta }^{3t}{\mathit{\phi }}_k)\right)\right]\right)\right]\hfill \\ -{\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\mathcal{N}\left[\left(\Phi \sum _{k=1}^{i-1}({\mathit{\phi }}_k,D_{\zeta }^t{\mathit{\phi }}_k,D_{\zeta }^{2t}{\mathit{\phi }}_k,D_{\zeta }^{3t}{\mathit{\phi }}_k)\right)\right]\right)\right]\hfill \end{array}$$

(36)

$$\begin{array}{c}{\mathit{\phi }}_0(\zeta ,t)={\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\sum _{k=0}^{m-1}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}\right)\right],\hfill \\ {\mathit{\phi }}_1(\zeta ,t)={\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\mathcal{N}[\Phi ({\mathit{\phi }}_0,D_{\zeta }^t{\mathit{\phi }}_0,D_{\zeta }^{2t}{\mathit{\phi }}_0,D_{\zeta }^{3t}{\mathit{\phi }}_0)]\right)\right],\hfill \\ ⋮\hfill \\ {\mathit{\phi }}_{m+1}(\zeta ,t)={\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\mathcal{N}\left[\sum _{i=0}^{\infty }\left(\Phi \sum _{k=0}^i({\mathit{\phi }}_k,D_{\zeta }^t{\mathit{\phi }}_k,D_{\zeta }^{2t}{\mathit{\phi }}_k,D_{\zeta }^{3t}{\mathit{\phi }}_k)\right)\right]\right)\right]\hfill \\ -{\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\mathcal{N}\left[\left(\Phi \sum _{k=1}^{i-1}({\mathit{\phi }}_k,D_{\zeta }^t{\mathit{\phi }}_k,D_{\zeta }^{2t}{\mathit{\phi }}_k,D_{\zeta }^{3t}{\mathit{\phi }}_k)\right)\right]\right)\right],m=1,2,\cdots .\hfill \end{array}$$

(37)

The following formulation of Equation (30) yields the analytically approximate solution for the m-term expression:

$$\mathit{\phi }(\zeta ,t)=\sum _{i=0}^{m-1}{\mathit{\phi }}_i.$$

(38)

3.4.1. Problem of Fractional PDE with NITM

Consider the following fractional PDE [57]:

$$D_t^{\varrho }\mathit{\phi }(\zeta ,t)=-\mathit{\phi }(\zeta ,t){\displaystyle \frac{{\partial }^3\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^3}}+{\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}+{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}},\mathrm{where}0<\varrho \le 1$$

(39)

with ICs

$$\mathit{\phi }(\zeta ,0)={\displaystyle \frac{e^{\zeta /4}}{4}}$$

(40)

and exact solution

$$\mathit{\phi }(\zeta ,t)={\displaystyle \frac{1}{4}}{e}^{{\displaystyle \frac{1}{4}}\left({\displaystyle \frac{t}{4}}+\zeta \right)}.$$

The following result is obtained by applying the natural transform to both sides of Equation (39):

$$\mathcal{N}\left[D_t^{\varrho }\mathit{\phi }(\zeta ,t)\right]={\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\sum _{k=0}^{m-1}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}+\mathcal{N}\left[-\mathit{\phi }(\zeta ,t){\displaystyle \frac{{\partial }^3\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^3}}+{\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}+{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}\right]\right).$$

(41)

The inverse natural transform is applied to both sides of Equation (41) to generate

$$\mathit{\phi }(\zeta ,t)={\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\sum _{k=0}^{m-1}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}+\mathcal{N}\left[-\mathit{\phi }(\zeta ,t){\displaystyle \frac{{\partial }^3\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^3}}+{\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}+{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}\right]\right)\right].$$

(42)

The equation that results from iteratively applying the natural transform may be presented as follows:

$$\begin{array}{cc}\hfill {\mathit{\phi }}_0(\zeta ,t)=& {\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left({\displaystyle \sum _{k=0}^{m-1}}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}\right)\right]\hfill \\ & ={\mathcal{N}}^{-1}\left[{\displaystyle \frac{\mathit{\phi }(\zeta ,0)}{s^2}}\right]\hfill \\ & ={\displaystyle \frac{e^{\zeta /4}}{4}}.\hfill \end{array}$$

The equivalent form can be obtained by applying the RL integral to Equation (39):

$$\mathit{\phi }(\zeta ,t)={\displaystyle \frac{e^{\zeta /4}}{4}}-\mathcal{N}\left[-\mathit{\phi }(\zeta ,t){\displaystyle \frac{{\partial }^3\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^3}}+{\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}+{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}\right].$$

(43)

Some of the terms that may be acquired by using the NITM approach are as follows:

$$\begin{array}{c}{\mathit{\phi }}_0(\zeta ,t)={\displaystyle \frac{e^{\zeta /4}}{4}},\hfill \\ {\mathit{\phi }}_1(\zeta ,t)={\displaystyle \frac{e^{\zeta /4}{t}^{\varrho }}{64\Gamma (\varrho +1)}},\hfill \\ {\mathit{\phi }}_2(\zeta ,t)={\displaystyle \frac{\sqrt{\pi }{4}^{-p-5}{e}^{\zeta /4}{t}^{2\varrho }}{\Gamma \left(\varrho +\frac{1}{2}\right)\Gamma (\varrho +1)}},\hfill \\ {\mathit{\phi }}_3(\zeta ,t)={\displaystyle \frac{e^{\zeta /4}{t}^{3\varrho }}{16384\Gamma (3\varrho +1)}},\hfill \\ {\mathit{\phi }}_4(\zeta ,t)={\displaystyle \frac{e^{\zeta /4}{t}^{4\varrho }}{262144\varrho \Gamma \left(\varrho \right)\Gamma (3\varrho +1)}}.\hfill \end{array}$$

(44)

The final solution is

$$\mathit{\phi }(\zeta ,t)={\mathit{\phi }}_0(\zeta ,t)+{\mathit{\phi }}_1(\zeta ,t)+{\mathit{\phi }}_2(\zeta ,t)+{\mathit{\phi }}_3(\zeta ,t)+\cdots .$$

(45)

$$\begin{array}{cc}\hfill \mathit{\phi }(\zeta ,t)=& {\displaystyle \frac{e^{\zeta /4}{t}^{\varrho }}{64\Gamma (\varrho +1)}}+{\displaystyle \frac{\sqrt{\pi }{4}^{-\varrho -5}{e}^{\zeta /4}{t}^{2\varrho }}{\Gamma \left(\varrho +\frac{1}{2}\right)\Gamma (\varrho +1)}}+{\displaystyle \frac{e^{\zeta /4}{t}^{3\varrho }}{16384\Gamma (3\varrho +1)}}\hfill \\ & +{\displaystyle \frac{e^{\zeta /4}{t}^{4\varrho }}{262144\varrho \Gamma \left(\varrho \right)\Gamma (3\varrho +1)}}+\cdots .\hfill \end{array}$$

(46)

3.4.2. Problem of Fractional Damped Burger Equation with NITM

Taking into consideration the fractional damped Burger equation [57] as follows:

$$D_t^{\varrho }\mathit{\phi }(\zeta ,t)=-{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}-\mathit{\phi }(\zeta ,t){\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}-{\displaystyle \frac{1}{5}}\mathit{\phi }(\zeta ,t),\mathrm{where}0<\varrho \le 1,$$

(47)

with ICs

$$\mathit{\phi }(\zeta ,0)={\displaystyle \frac{1}{5}}\zeta$$

(48)

and the exact solution

$$\mathit{\phi }(\zeta ,t)={\displaystyle \frac{\zeta }{5\left(2e^{\frac{t}{5}}-1\right)}},$$

by applying the natural transform to both sides of Equation (47), we obtain the following equation:

$$\mathcal{N}\left[D_t^{\varrho }\mathit{\phi }(\zeta ,t)\right]={\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\sum _{k=0}^{m-1}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}+\mathcal{N}\left[-{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}-\mathit{\phi }(\zeta ,t){\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}-{\displaystyle \frac{1}{5}}\mathit{\phi }(\zeta ,t)\right]\right).$$

(49)

When the inverse natural transform is applied to Equation (49), the equation that results is as follows:

$$\mathit{\phi }(\zeta ,t)={\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\sum _{k=0}^{m-1}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}+\mathcal{N}\left[-{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}-\mathit{\phi }(\zeta ,t){\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}-{\displaystyle \frac{1}{5}}\mathit{\phi }(\zeta ,t)\right]\right)\right].$$

(50)

When the iterative procedure of the natural transform is used, the following equation is obtained:

$$\begin{array}{cc}\hfill {\mathit{\phi }}_0(\zeta ,t)=& {\mathcal{N}}^{-1}\left[{\displaystyle \frac{u^{\varrho }}{s^{\varrho }}}\left(\sum _{k=0}^{m-1}{\displaystyle \frac{{\mathit{\phi }}^{\left(k\right)}(\zeta ,0)}{s^{2-\varrho +k}}}\right)\right]\hfill \\ & ={\mathcal{N}}^{-1}\left[{\displaystyle \frac{\mathit{\phi }(\zeta ,0)}{s^2}}\right]\hfill \\ & ={\displaystyle \frac{1}{5}}\zeta .\hfill \end{array}$$

Using of the RL integral leads to the acquisition of an equivalent form derived from Equation (19):

$$\mathit{\phi }(\zeta ,t)={\displaystyle \frac{1}{5}}\zeta -\mathcal{N}\left[-{\displaystyle \frac{{\partial }^2\mathit{\phi }(\zeta ,t)}{\partial {\zeta }^2}}-\mathit{\phi }(\zeta ,t){\displaystyle \frac{\partial \mathit{\phi }(\zeta ,t)}{\partial \zeta }}-{\displaystyle \frac{1}{5}}\mathit{\phi }(\zeta ,t)\right].$$

(51)

The NITM approach yields the following few terms:

$$\begin{array}{c}{\mathit{\phi }}_0(\zeta ,t)={\displaystyle \frac{1}{5}}\zeta ,\hfill \\ {\mathit{\phi }}_1(\zeta ,t)=-{\displaystyle \frac{2\zeta t^{\varrho }}{25\Gamma (\varrho +1)}},\hfill \\ {\mathit{\phi }}_2(\zeta ,t)={\displaystyle \frac{2\zeta t^{2\varrho }\left(15-\frac{2t^{\varrho }\Gamma {(2\varrho +1)}^2}{\Gamma {(\varrho +1)}^2\Gamma (3\varrho +1)}\right)}{625\Gamma (2\varrho +1)}},\hfill \\ {\mathit{\phi }}_3\left(\zeta t\right)={\displaystyle \frac{2\zeta t^{3\varrho }}{390625}}\left(2t^{\varrho }\right(-{\displaystyle \frac{4t^{3\varrho }\Gamma {(2\varrho +1)}^2\Gamma (6\varrho +1)}{\Gamma {(\varrho +1)}^4\Gamma {(3\varrho +1)}^2\Gamma (7\varrho +1)}}+{\displaystyle \frac{\frac{60t^{2\varrho }\Gamma (5\varrho +1)}{\Gamma (3\varrho +1)\Gamma (6\varrho +1)}+\frac{125\sqrt{\pi }{2}^{-4\varrho }}{\Gamma \left(2\varrho +\frac{1}{2}\right)}}{\Gamma {(\varrho +1)}^2}}\hfill \\ -{\displaystyle \frac{225t^{\varrho }\Gamma (4\varrho +1)}{\Gamma {(2\varrho +1)}^2\Gamma (5\varrho +1)}}-{\displaystyle \frac{100t^{\varrho }\Gamma (2\varrho +1)\Gamma (4\varrho +1)}{\Gamma {(\varrho +1)}^3\Gamma (3\varrho +1)\Gamma (5\varrho +1)}}+{\displaystyle \frac{750\Gamma (3\varrho +1)}{\Gamma (\varrho +1)\Gamma (2\varrho +1)\Gamma (4\varrho +1)}})-{\displaystyle \frac{1875}{\Gamma (3\varrho +1)}}).\hfill \end{array}$$

(52)

The NITM algorithm produces the following final result:

$$\mathit{\phi }(\zeta ,t)={\mathit{\phi }}_0(\zeta ,t)+{\mathit{\phi }}_1(\zeta ,t)+{\mathit{\phi }}_2(\zeta ,t)+{\mathit{\phi }}_3(\zeta ,t)+\cdots .$$

(53)

$$\begin{array}{c}\mathit{\phi }(\zeta ,t)={\displaystyle \frac{1}{5}}\zeta -{\displaystyle \frac{2\zeta t^{\varrho }}{25\Gamma (\varrho +1)}}+{\displaystyle \frac{2\zeta t^{2\varrho }\left(15-\frac{2t^{\varrho }\Gamma {(2\varrho +1)}^2}{\Gamma {(\varrho +1)}^2\Gamma (3\varrho +1)}\right)}{625\Gamma (2\varrho +1)}}\hfill \\ +{\displaystyle \frac{2\zeta t^{3\varrho }}{390625}}\left(2t^{\varrho }\right(-{\displaystyle \frac{4t^{3\varrho }\Gamma {(2\varrho +1)}^2\Gamma (6\varrho +1)}{\Gamma {(\varrho +1)}^4\Gamma {(3\varrho +1)}^2\Gamma (7\varrho +1)}}+{\displaystyle \frac{\frac{60t^{2\varrho }\Gamma (5\varrho +1)}{\Gamma (3\varrho +1)\Gamma (6\varrho +1)}}{+}}{\displaystyle \frac{125\sqrt{\pi 2^{-4\varrho }}\Gamma \left(2\varrho +\frac{1}{2}\right)}{\Gamma {(\varrho +1)}^2}}\hfill \\ -{\displaystyle \frac{225t^{\varrho }\Gamma (4\varrho +1)}{\Gamma {(2\varrho +1)}^2\Gamma (5\varrho +1)}}-{\displaystyle \frac{100t^{\varrho }\Gamma (2\varrho +1)\Gamma (4\varrho +1)}{\Gamma {(\varrho +1)}^3\Gamma (3\varrho +1)\Gamma (5\varrho +1)}}+{\displaystyle \frac{750\Gamma (3\varrho +1)}{\Gamma (\varrho +1)\Gamma (2\varrho +1)\Gamma (4\varrho +1)}})-{\displaystyle \frac{1875}{\Gamma (3\varrho +1)}}).\hfill \end{array}$$

(54)

Below, we present the numerical results and provide a comprehensive discussion of the solutions obtained using the numerical iterative transform method (NITM). The performance and accuracy of the method are evaluated through graphical and tabular representations, with comparisons across different fractional orders. Figure 6 illustrates the comparison of various fractional-order solutions of Problem 1 for $t=0$ to $t=5$ using NITM. The results demonstrate the sensitivity of the solutions to changes in the fractional order, highlighting the method’s ability to accurately capture the dynamic behavior of the system over time. Figure 7 presents the absolute error analysis for Problem 1, where Figure 7a shows the absolute error for $\varrho =1$, Figure 7b shows the absolute error for $\varrho =0.5$, and Figure 7c shows the absolute error for $\varrho =0.1$. The decreasing error with smaller fractional orders demonstrates the robustness of the NITM approach in approximating solutions with high precision. Figure 8 and Figure 9 depict the solutions of Problem 2 for different fractional orders at $t=0.1$ and $t=0.01$, respectively. These results confirm that the NITM is capable of handling variations in fractional orders and capturing subtle differences in the system’s behavior. Figure 10 shows the absolute error analysis for Problem 2; analogously to Figure 7, Figure 10a corresponds to $\varrho =1$, Figure 10b corresponds to $\varrho =0.5$, and Figure 10c corresponds to $\varrho =0.1$. The results highlight the effectiveness of the NITM in minimizing errors, particularly at lower fractional orders.

Table 3. Various fractional-order solutions of NITM for Problem 1.

t	$\mathit{\zeta }$	${\mathit{NITM}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.5}}$	${\mathit{NITM}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.7}}$	${\mathit{NITM}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$	$\mathit{Exact}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.5}}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.7}}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$
	0.2	0.268012	0.268012	0.266124	0.266124	0.0018883	0.0018883	$4.748294\times {10}^{-7}$
	0.4	0.296199	0.296199	0.294113	0.294112	0.00208689	0.00208689	$5.247677\times {10}^{-7}$
	0.6	0.32735	0.32735	0.325045	0.325044	0.00230637	0.00230637	$5.799580\times {10}^{-7}$
1	0.8	0.361778	0.361778	0.35923	0.359229	0.00254893	0.00254893	$6.409527\times {10}^{-7}$
	1.0	0.399827	0.399827	0.39701	0.39701	0.00281701	0.00281701	$7.083623\times {10}^{-7}$
	0.2	0.26089	0.26089	0.257936	0.257936	0.00295422	0.00295422	$2.973990\times {10}^{-8}$
	0.4	0.288328	0.288328	0.285063	0.285063	0.00326492	0.00326492	$3.286768\times {10}^{-8}$
	0.6	0.318652	0.318652	0.315044	0.315044	0.00360829	0.00360829	$3.632440\times {10}^{-8}$
0.5	0.8	0.352165	0.352165	0.348177	0.348177	0.00398778	0.00398778	$4.014467\times {10}^{-8}$
	1.0	0.389202	0.389202	0.384795	0.384795	0.00440718	0.00440718	$4.436673\times {10}^{-8}$
	0.2	0.253463	0.253463	0.251567	0.251567	0.00189519	0.00189519	$4.766387\times {10}^{-11}$
	0.4	0.280119	0.280119	0.278025	0.278025	0.00209451	0.00209451	$5.267669\times {10}^{-11}$
	0.6	0.30958	0.30958	0.307265	0.307265	0.00231479	0.00231479	$5.821670\times {10}^{-11}$
0.1	0.8	0.342139	0.342139	0.33958	0.33958	0.00255824	0.00255824	$6.433947\times {10}^{-11}$
	1.0	0.378122	0.378122	0.375294	0.375294	0.0028273	0.0028273	$7.110606\times {10}^{-11}$

and

Table 4. Various fractional-order solution of NITM for Problem 2.

t	$\mathit{\zeta }$	${\mathit{NITM}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.5}}$	${\mathit{NITM}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.7}}$	${\mathit{NITM}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$	$\mathit{Exact}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.5}}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{0.7}}$	${\mathit{Error}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$
	0.2	0.0703566	0.0735629	0.0768945	0.0768933	0.00653675	0.00333041	$1.244183\times {10}^{-6}$
	0.4	0.140713	0.147126	0.153789	0.153787	0.0130735	0.00666082	$2.488366\times {10}^{-6}$
0.1	0.6	0.21107	0.220689	0.230684	0.23068	0.0196103	0.00999123	$3.732549\times {10}^{-6}$
	0.8	0.281426	0.294252	0.307578	0.307573	0.026147	0.0133216	$4.976732\times {10}^{-6}$
	1.0	0.351783	0.367814	0.384473	0.384467	0.0326838	0.016652	$6.220915\times {10}^{-6}$
	0.2	0.0765761	0.0786222	0.079681	0.079681	0.00310486	0.00105871	$1.276282\times {10}^{-9}$
	0.4	0.153152	0.157244	0.159362	0.159362	0.00620971	0.00211742	$2.552564\times {10}^{-9}$
0.01	0.6	0.229728	0.235867	0.239043	0.239043	0.00931457	0.00317614	$3.828847\times {10}^{-9}$
	0.8	0.306304	0.314489	0.318724	0.318724	0.0124194	0.00423485	$5.105129\times {10}^{-9}$
	1.0	0.382881	0.393111	0.398405	0.398405	0.0155243	0.00529356	$6.381411\times {10}^{-9}$

provide the numerical solutions of NITM for Problems 1 and 2, respectively, for various fractional orders. The results showcase the method’s consistency and accuracy in producing reliable approximations across different scenarios.

Table 5. Analysis of the absolute error between NRPSM and NITM for Problem 1.

t	$\mathit{\zeta }$	$\mathit{EXACT}$	${\mathit{NITM}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$	${\mathit{NRPSM}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$	$\mathit{NITMError}$	$\mathit{NRPSMError}$
	0.2	0.266124	0.266124	0.266124	$4.748294\times {10}^{-7}$	$2.007703\times {10}^{-9}$
	0.4	0.294112	0.294113	0.294112	$5.247677\times {10}^{-7}$	$2.218855\times {10}^{-9}$
	0.6	0.325044	0.325045	0.325044	$5.799580\times {10}^{-7}$	$2.452214\times {10}^{-9}$
1	0.8	0.359229	0.35923	0.359229	$6.409527\times {10}^{-7}$	$2.710116\times {10}^{-9}$
	1.0	0.39701	0.39701	0.39701	$7.083623\times {10}^{-7}$	$2.995142\times {10}^{-9}$
	0.2	0.257936	0.257936	0.257936	$2.973990\times {10}^{-8}$	$6.241301\times {10}^{-11}$
	0.4	0.285063	0.285063	0.285063	$3.286768\times {10}^{-8}$	$6.897699\times {10}^{-11}$
	0.6	0.315044	0.315044	0.315044	$3.632440\times {10}^{-8}$	$7.623141\times {10}^{-11}$
0.5	0.8	0.348177	0.348177	0.348177	$4.014467\times {10}^{-8}$	$8.424871\times {10}^{-11}$
	1.0	0.384795	0.384795	0.384795	$4.436673\times {10}^{-8}$	$9.310924\times {10}^{-11}$
	0.2	0.251567	0.251567	0.251567	$4.766381\times {10}^{-11}$	$1.987299\times {10}^{-14}$
	0.4	0.278025	0.278025	0.278025	$5.267669\times {10}^{-11}$	$2.192690\times {10}^{-14}$
	0.6	0.307265	0.307265	0.307265	$5.821670\times {10}^{-11}$	$2.431388\times {10}^{-14}$
0.1	0.8	0.33958	0.33958	0.33958	$6.433942\times {10}^{-11}$	$2.681188\times {10}^{-14}$
	1.0	0.375294	0.375294	0.375294	$7.110606\times {10}^{-11}$	$2.964295\times {10}^{-14}$

and

Table 6. Analysis of the absolute error between NRPSM and NITM for Problem 2.

t	$\mathit{\zeta }$	$\mathit{EXACT}$	${\mathit{NITM}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$	${\mathit{NRPSM}}_{\mathit{\varrho }\mathbf{=}\mathbf{1.0}}$	$\mathit{NITMError}$	$\mathit{NRPSMError}$
	0.2	0.0768933	0.0768945	0.0768932	$1.244183\times {10}^{-6}$	$7.775646\times {10}^{-8}$
	0.2	0.153787	0.153789	0.153786	$2.488366\times {10}^{-6}$	$1.555129\times {10}^{-7}$
0.1	0.6	0.23068	0.230684	0.23068	$3.732549\times {10}^{-6}$	$2.332694\times {10}^{-7}$
	0.8	0.307573	0.307578	0.307573	$4.976732\times {10}^{-6}$	$3.110258\times {10}^{-7}$
	1.0	0.384467	0.384473	0.384466	$6.220915\times {10}^{-6}$	$3.887823\times {10}^{-7}$
	0.2	0.079681	0.079681	0.079681	$1.276282\times {10}^{-9}$	$7.976985\times {10}^{-12}$
	0.4	0.159362	0.159362	0.159362	$2.552564\times {10}^{-9}$	$1.595397\times {10}^{-11}$
0.01	0.6	0.239043	0.239043	0.239043	$3.828847\times {10}^{-9}$	$2.393095\times {10}^{-11}$
	0.8	0.318724	0.318724	0.318724	$5.105129\times {10}^{-9}$	$3.190794\times {10}^{-11}$
	1.0	0.398405	0.398405	0.398405	$6.381411\times {10}^{-9}$	$3.988492\times {10}^{-11}$

compare the absolute errors between the NRPSM and NITM for Problems 1 and 2, respectively. The data reveal that NITM outperforms NRPSM regarding error minimization, affirming its superiority in solving fractional-order systems.

The NITM results demonstrate its efficiency and reliability for solving fractional-order differential equations. The graphical and tabular data show the method’s ability to adapt to different fractional orders and initial conditions. The error analysis further underscores its accuracy, significantly reducing absolute errors observed across various scenarios.

The comparative analysis with NRPSM highlights NITM’s competitive advantage. It consistently delivers better approximations with lower errors, establishing NITM as a powerful tool for addressing complex fractional-order problems in mathematical modeling and applied sciences.

The graphical and tabular discussions provided here confirm that the proposed NITM provides accurate and efficient solutions, making it a valuable approach for fractional-order systems across diverse applications.

4. Conclusions

In this study, we have analyzed the damped Burger’s equation by employing advanced mathematical techniques such as the natural residual power series method and the new iteration transform method within the context of the Caputo operator. The damped Burger’s equation is a fundamental model for comprehending the behavior of nonlinear waves under damping effects, with applications in fluid dynamics and related fields. The natural residual power series method and the new iteration transform method are demonstrated to be effective tools in providing approximate solutions to this intricate equation, offering innovative approaches for handling nonlinear partial differential equations. Incorporating the Caputo operator enriches the modeling by considering fractional derivatives, contributing to a more accurate representation of the underlying dynamics. This research provides valuable insights into the dynamic behavior of solutions to the damped Burgers’ equation, showcasing the robustness of the proposed methods in addressing the challenges posed by nonlinear fractional partial differential equations. Overall, this investigation contributes to the advancement of mathematical modeling and numerical analysis, emphasizing the applicability of the suggested methods in solving complex equations across various scientific domains.