Fractional Study of the Non-Linear Burgers’ Equations via a Semi-Analytical Technique

Abstract

Most complex physical phenomena are described by non-linear Burgers’ equations, which help us understand them better. This article uses the transformation and the fractional Taylor’s formula to find approximate solutions for non-linear fractional-order partial differential equations. Solving non-linear Burgers’ equations with the right starting data shows that the method utilized is correct and can be utilized. Based on the limit of the idea, a rapid convergence McLaurin series is used to obtain close series solutions for both models with less work and more accuracy. To see how time-Caputo fractional derivatives affect how the results of the above models behave, in three dimension figures are drawn. The results showed that the proposed method is an easy, flexible, and helpful way to solve and understand a wide range of non-linear physical models.

1. Introduction

Fractional differential equations have become more well known in recent years because they are useful in many scientific and engineering fields. For example, a fractional derivative can be used to describe the non-linear oscillations of an earthquake, and the fractional of the fluid traffic dynamic model can solve the insufficiency that comes from assuming that traffic flows continuously [1,2]. Fractional differential problems are also used to model a wide range of mathematical biology, chemical processes and applied sciences models [3,4,5,6]. Different physical, biological, and chemical phenomena can be defined by non-linear partial differential equations (NPDEs). The main goal of current research is to find precise traveling wave solutions for equations like these. Scientists can better understand the complicated physical phenomena and dynamic processes that NPDEs show [7,8,9] when they become clear and exact answers. In the last 40 years, many important methods for obtaining accurate answers to NPDEs [10,11] have been proposed [12,13,14].

Bateman came up with the Burgers’ equation in 1915 [15], later changing it to the Burgers’ equation [16]. The Burgers’ equation is used extensively in engineering and science, especially when solving problems with non-linear equations. More and more important and interesting things are being conducted with Burgers’ equation by mathematicians and researchers. It has been known for a long time that this equation can be used to model things such as dynamics, heat conduction, acoustic waves, turbulence, and many more [17,18,19,20,21,22]. Most of the time, you need to use special methods to solve this kind of non-linear PDE because it cannot be solved analytically. Some researchers and scientists have applied analytical techniques such as the Variational iteration method, the Adomian decomposition method, the Homotopy analysis method, the Homotopy perturbation method and the differential transform method [23,24,25,26,27] to solve these kinds of problems.

A numerical analytic method for solving many forms of ordinary, partial, integrodifferential equations and fractional fuzzy differential equations is known as the residual power series (RPS) approach. Since it offers closed-form solutions of well-known functions, it is an efficient optimization strategy. The RFPS technique solves fuzzy FDEs, various FDEs, and integral equations having fractional order, such as the Newell–Whitehead–Segel equation of fractional order [28], coupled fractional resonant Schrödinger equations [29], time fractional Kundu–Eckhaus and massive Thirring equations [30], time fractional Fokker–Planck models [31], fractional partial differential equations [32], singular initial value problems [33], dractional Fredholm integro-differential equations having order 2b [34], and several classes of fractional fuzzy differential equations [35,36]. The Elzaki transform (ET) is a powerful tool for resolving numerous complex models that appear in various branches of the natural sciences. When analytical methods are combined with the ET operator, non-linear problems can be solved more quickly and precisely.

The main goal of this work is to examine the approximate and analytical solutions for linear and non-linear systems using the Elzaki residual power series (ERPS) method, which was introduced and developed in [37]. The ET and RPS approaches are combined in the ERPS approach, which provides both approximate and accurate solutions as quickly fractional power series (FPS) solutions. The proposed problem is converted to Elzaki space, and the solutions in the form of algebraic equations are created. Lastly, the inverse Elzaki is applied to the proposed problem results. In contrast to the FRPS approach, which depends on the fractional derivative and consumes time to compute the various fractional derivatives in determining the solutions, the unknown coefficients in a modified Elzaki expansion can be identified by employing the limit notion. The ERPS approach takes less time and provides higher accuracy with minor computational requirements.

This work is arranged as follows: Section 2 revisits some valid basic results concerning the Elzaki transformation and fractional power series representations. The proposed technique for obtaining the solutions for the fractional partial differential equation is in Section 3. In Section 4, the simplicity and applicability of RPSM are studied by solving fractional order Burgers’ equations. Last, concluding remarks on our findings are drawn in Section 5.

2. Preliminaries

In this section, we go over some basic concepts and definitions of the Caputo derivative (CD), ET, and theorems that will be useful in this paper.

Definition 1.

Assume that the existence axioms of the ET are satisfied by $\Im (\vartheta ,\tau )$. Then, the ET of $\Im (\vartheta ,\tau )$ is formulated as [38]:

$$\Xi [\Im (\vartheta ,\tau )]=\aleph (\vartheta ,\omega )=\omega {\int }_0^{\infty }\Im (\vartheta ,\tau )e^{-\frac{\tau }{\omega }}d\tau ,\tau \ge 0,e_1\le \omega \le e_2.$$

Definition 2.

The FD of $\Im (\vartheta ,\tau )$ of order ϱ in the CD sense is defined as follows [38]:

$$D_{\tau }^{\varrho }\Im (\vartheta ,\tau )={\mathcal{J}}_{\varrho }^{r-\varrho }{\Im }^{\left(n\right)}(\vartheta ,\tau ),\tau \ge 0,n-1<\varrho \le n,$$

where ${\mathcal{J}}_{\tau }^{n-\varrho }$ is the $R-L$ integral of $\Im (\vartheta ,\tau )$.

Theorem 1.

Assume that the multiple fractional power series (MFPS) representation for the function $\Xi [\Im (\vartheta ,\tau \left)\right]=\aleph (\vartheta ,s)$ is given by [38]

$$\aleph (\vartheta ,\omega )=\sum _{n=0}^{\infty }{\Theta }_n\left(\vartheta \right){\omega }^{n\varrho +2},$$

then we have

$${\Theta }_n\left(\vartheta \right)=D_{\tau }^{n\varrho }\Im (\vartheta ,0),$$

where, $D_{\tau }^{n\varrho }=D_{\tau }^{\varrho }·D_{\tau }^{\varrho }\dots D_{\tau }^{\varrho }(n-$ times). The conditions for the convergence of the MFPS is determined in the following theorem.

Theorem 2.

Let $\Xi [\Im (\vartheta ,\tau \left)\right]=\aleph (\vartheta ,\omega )$ can be represented as the new form of MFPS explained in Theorem 1. If $\left|\frac{1}{{\omega }^2}\Xi \left[D_{\tau }^{(n+1)\varrho }\Im (\vartheta ,\tau )\right]\right|\le \mathcal{Z}$, then the remainder ${\mathcal{R}}_n(\vartheta ,\omega )$ of MFPS satisfies the following inequality [38]:

$$\left|{\mathcal{R}}_j(\vartheta ,\omega )\right|\le \mathcal{Z}{\omega }^{(n+1)\varrho +2}.$$

3. General Implementation of Elzaki Residual Power Series Method

Consider the general fractional partial differential equation

$$D_{\tau }^{\varrho }\nu (\vartheta ,\tau )=c{D}_{\vartheta }^2\nu (\vartheta ,\tau )+a\nu (\vartheta ,\tau )-b{\nu }^4(\vartheta ,\tau ),$$

(1)

with the initial condition

$$\nu (\vartheta ,\tau )=f_0\left(\vartheta \right).$$

(2)

First, we use the Elzaki transformation to (1),

$$\Xi \left[D_{\tau }^{\varrho }\nu (\vartheta ,\tau )\right]=c\Xi \left[D_{\vartheta }^2\nu (\vartheta ,\tau )\right]+a\Xi \left[\nu {(\vartheta ,\tau )}^2\right]-b\Xi \left[{\nu }^4(\vartheta ,\tau )\right].$$

(3)

By the fact that $\Xi \left[D_{\tau }^{\varrho }w(\vartheta ,\tau )\right]=s^a\Xi \left[w(\vartheta ,\tau )\right]-s^{a-1}\nu (\vartheta ,0)$ and using the initial condition (2), we rewrite (3) as

$$V(\vartheta ,s)=s^2{f}_0\left(\vartheta \right)+s^{\varrho }c{D}_{\vartheta }^{2}V(\vartheta ,s)+s^{\varrho }aV(\vartheta ,s)-s^{\varrho }b\Xi \left[{\left[{\Xi }^{-1}{\left[V(\vartheta ,s)\right]}^2\right]}^{\varrho }\right],$$

(4)

where $V(\vartheta ,s)=\Xi \left[w\right(\vartheta ,\tau \left)\right]$.

Second, we define the transform term $V(\vartheta ,s)$ as the following expression

$$V(\vartheta ,s)=\sum _{n=0}^{\infty }{s}^{n\varrho +1}{f}_s\left(\vartheta \right).$$

(5)

The series form of kth-truncated of Equation (5)

$$V_k(\vartheta ,s)=\sum _{n=0}^k{s}^{n\varrho +1}{f}_s\left(\vartheta \right)=s^2{f}_o\left(\vartheta \right)+\sum _{n=1}^k{s}^{n\varrho +1}{f}_k\left(\vartheta \right).$$

(6)

The Elzaki residual function to (5) is

$$\begin{array}{cc}\hfill {\Xi Res}_k(\vartheta ,s)=& V_k(\vartheta ,s)-s{f}_0\left(\vartheta \right)-s^{\varrho }c{D}_s^2{V}_k(\vartheta ,s)-s^{\varrho }a{V}_k(\vartheta ,s)+s^{\varrho }b\Xi \left[{\left[{\Xi }^{-1}\left[V_k(\vartheta ,s)\right]\right]}^q\right].\hfill \end{array}$$

(7)

Third, we use a few properties that come up in the standard RPSM [25,26] to point out certain facts:

$\Xi Res(\vartheta ,s)=0$ and ${lim}_{k\to \infty }\Xi Res{s}_k(\vartheta ,s)=\Xi Res(\vartheta ,s)$ for each $s>0$.

${lim}_{s\to \infty }s\Xi Res(\vartheta ,s)=0\Rightarrow {lim}_{s\to \infty }s\Xi Res(\vartheta ,s)=0.$

${lim}_{s\to \infty }{s}^{k\varrho +1}\Xi Res(\vartheta ,s)={lim}_{s\to \infty }{s}^{k\varrho +1}\Xi {Res}_k(\vartheta ,s)=0,0<\varrho \le$$1,k=1,2,3,\dots$

So, to find the co-efficient functions $f_n\left(\vartheta \right),$ we solve the following scheme successively:

$$\underset{s\to \infty }{lim}\left(s^{ka+1}{\Xi Res}_k(\vartheta ,s)\right)=0,0<\varrho \le 1,k=1,2,3,\dots$$

Finally, we use the inverse Elzaki to $V_k(\vartheta ,s)$, to achieved the kth approximated supportive result ${\nu }_k(\vartheta ,\tau )$.

4. Numerical Examples

Example 1.

Consider the fractional two-dimensional Burger’s equation

$$\begin{array}{c}\hfill D_{\tau }^{\varrho }\nu (\vartheta ,\zeta ,\tau )=\nu (\vartheta ,\zeta ){\nu }_{\vartheta }(\vartheta ,\zeta )+{\nu }_{\vartheta \vartheta }(\vartheta ,\zeta )+{\nu }_{\zeta \zeta }(\vartheta ,\zeta ),0<\varrho \le 1\end{array}$$

(8)

with the initial condition (IC)

$$\begin{array}{c}\hfill \nu (\vartheta ,\zeta ,0)=\vartheta +\zeta =f_0\left(\vartheta \right)\end{array}$$

(9)

The exact solution of (8) $(\varrho =1)$ is

$$\begin{array}{c}\hfill \nu (\vartheta ,\zeta ,\tau )=\frac{\vartheta +\zeta }{1-\tau }\end{array}$$

(10)

Using ET to (8) and applying the IC (9), we get

$$\begin{array}{cc}\hfill V(\vartheta ,s)& =s^2\left(\vartheta +\zeta \right)+s^{\varrho }{\Xi }_{\tau }\left[\left[{\Xi }_{\tau }^{-1}\left\{V_{\vartheta }(\vartheta ,\zeta ,s)\right\}\right]\left[{\Xi }_{\tau }^{-1}\left\{V_{\vartheta }(\vartheta ,\zeta ,s)\right\}\right]\right]\hfill \\ \hfill & +s^{\varrho }{\Xi }_{\tau }\left[{\Xi }_{\tau }^{-1}\left\{V_{\vartheta \vartheta }(\vartheta ,\zeta ,s)\right\}\right]+s^{\varrho }{\Xi }_{\tau }\left[{\Xi }_{\tau }^{-1}\left\{V_{yy}(\vartheta ,\zeta ,s)\right\}\right].\hfill \end{array}$$

(11)

The k-th truncated term series of (11) is

$$\begin{array}{c}\hfill V_k(\vartheta ,\zeta ,s)=s^2\left(\vartheta +\zeta \right)+\sum _{n=1}^k{s}^{n\varrho +1}{f}_n\left(\vartheta \right),\end{array}$$

(12)

and the k-th Elzaki residual function is

$$\begin{array}{cc}\hfill {\Xi }_{\tau }Re{s}_k(\vartheta ,\zeta ,s)& =V_k(\vartheta ,\zeta ,s)-s^2(\vartheta +\zeta )-s^{\varrho }{\Xi }_{\tau }[[{\Xi }_{\tau }^{-1}\{V(\vartheta ,\zeta ,s)\}][{\Xi }_{\tau }^{-1}\{\frac{\partial }{\partial \vartheta }{V}_{\vartheta }(\vartheta ,\zeta ,s)\}]]\hfill \\ \hfill & -s^{\varrho }{\Xi }_{\tau }[{\Xi }_{\tau }^{-1}\{\frac{{\partial }^2}{\partial {\vartheta }^2}V(\vartheta ,\zeta ,s)\}]-s^{\varrho }{\Xi }_{\tau }[{\Xi }_{\tau }^{-1}\{\frac{{\partial }^2}{\partial {\zeta }^2}{V}_{yy}(\vartheta ,\zeta ,s)\}].\hfill \end{array}$$

(13)

Now, to calculate $f_k(\vartheta ,\zeta )$, $k=1,2,3,\cdots$, we put the kth-truncate series (12) into the kth-Elzaki residual term (13), multiply the solution equation by $s^{k\varrho +1}$, and then the relation is solved recursively ${lim}_{s\to \infty }\left[s^{k\varrho +1}Re{s}_k(\vartheta ,\zeta ,s)\right]=0$, $k=1,2,3,\cdots$ for $f_k$. The following are the some terms $f_k(\vartheta ,\zeta )$

$$\begin{array}{cc}\hfill f_1(\vartheta ,\zeta )& =(\vartheta +\zeta )\hfill \\ \hfill f_2(\vartheta ,\zeta )& =2(\vartheta +\zeta )\hfill \\ \hfill f_3(\vartheta ,\zeta )& =(\vartheta +\zeta )\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}\right]\hfill \\ \hfill f_4(\vartheta ,\zeta )& =2(\vartheta +\zeta )\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}+\frac{2\Gamma (1+3\varrho )}{\Gamma (1+\varrho )\Gamma (1+2\varrho )}\right]\hfill \\ \hfill & ⋮.\hfill \end{array}$$

(14)

Putting the value of $f_n(\vartheta ,\zeta )$, $(n\ge 1)$ in Equation (12), we have

$$\begin{array}{cc}\hfill V(\vartheta ,\zeta ,s)& =s^2\vartheta +\zeta +s^{\varrho +1}(\vartheta +\zeta )+s^{2\varrho +1}2(\vartheta +\zeta )+s^{3\varrho +1}(\vartheta +\zeta )\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}\right]\hfill \\ \hfill & +s^{4\varrho +1}(\vartheta +\zeta )\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}+\frac{2\Gamma (1+3\varrho )}{\Gamma (1+\varrho )\Gamma (1+2\varrho )}\right]+\cdots .\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill V(\vartheta ,\zeta ,s)& =(\vartheta +\zeta )[s^2+s^{\varrho +1}+2s^{2\varrho +1}+s^{3\varrho +1}\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}\right]\hfill \\ \hfill & +s^{4\varrho +1}\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}+\frac{2\Gamma (1+3\varrho )}{\Gamma (1+\varrho )\Gamma (1+2\varrho )}\right]+\cdots ].\hfill \end{array}$$

(16)

Applying inverse Elzaki transform, we get

$$\begin{array}{cc}\hfill \nu (\vartheta ,\zeta ,\tau )& =(\vartheta +\zeta )[1+\frac{{\tau }^{\varrho }}{\Gamma (\varrho +1)}+\frac{2{\tau }^{\varrho }}{\Gamma (2\varrho +1)}+\frac{[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}]}{\Gamma (3\varrho +1)}{\tau }^{3\varrho }\hfill \\ \hfill & +\frac{[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}+\frac{2\Gamma (1+3\varrho )}{\Gamma (1+\varrho )\Gamma (1+2\varrho )}]}{\Gamma (4\varrho +1)}{\tau }^{4\varrho }+\cdots ].\hfill \end{array}$$

(17)

Now, if we substitute $\varrho =1$ in Equation (17), it gives

$$\begin{array}{c}\hfill \nu (\vartheta ,\zeta ,\tau )=(\vartheta +\zeta )\left[1+\tau +\frac{{\tau }^2}{2!}+\frac{{\tau }^3}{3!}+\frac{{\tau }^4}{4!}+\cdots \right]\end{array}$$

(18)

The result of Equation (18) closed contact with the Maclaurin series of

$$\begin{array}{cc}\hfill \nu (\vartheta ,\zeta ,\tau )& =\frac{\vartheta +\zeta }{1-\tau }.\hfill \end{array}$$

(19)

In Figure 1, the exact and analytical solution of $\nu (\vartheta ,\zeta ,\tau )$ at $\varrho =1$ and $0.8$ of Example 1. In Figure 2, the analytical solutions of $\nu (\vartheta ,\zeta ,\tau )$ at $\varrho =0.6$ and $0.4$ and similarly in Figure 3, the different fractional order of $\varrho$ of $\nu (\vartheta ,\zeta ,\tau )$ of Example 1, which show that the close contact with the exact solution.

Example 2.

Consider the fractional two-dimensional Burger’s equation

$$\begin{array}{c}\hfill D_{\tau }^{\varrho }\nu =\nu {\nu }_{\vartheta }+{\nu }_{\vartheta \vartheta }+{\nu }_{yy}+{\nu }_{zz},0<\varrho \le 1\end{array}$$

(20)

with the IC

$$\begin{array}{c}\hfill \nu (\vartheta ,\zeta ,z,0)=\vartheta +\zeta +z=f_0\left(\vartheta \right)\end{array}$$

(21)

The exact solution of (20) $(\varrho =1)$ is

$$\begin{array}{c}\hfill \nu (\vartheta ,\zeta ,z,\tau )=\frac{\vartheta +\zeta +z}{1-\tau }\end{array}$$

(22)

Using Elzaki transformation to (20) and applying the IC (21), we obtain

$$\begin{array}{cc}\hfill V(\vartheta ,\zeta ,z,s)& =s^2(\vartheta +\zeta +z)+s^{\varrho }{\Xi }_{\tau }\left[\left[{\Xi }_{\tau }^{-1}\left\{V_{\vartheta }(\vartheta ,\zeta ,z,s)\right\}\right]\left[{\Xi }_{\tau }^{-1}\left\{V_{\vartheta }(\vartheta ,\zeta ,z,s)\right\}\right]\right]\hfill \\ \hfill & +s^{\varrho }{\Xi }_{\tau }\left[{\Xi }_{\tau }^{-1}\left\{V_{\vartheta \vartheta }(\vartheta ,\zeta ,z,s)\right\}\right]+s^{\varrho }{\Xi }_{\tau }\left[{\Xi }_{\tau }^{-1}\left\{V_{yy}(\vartheta ,\zeta ,z,s)\right\}\right]+s^{\varrho }{\Xi }_{\tau }\left[{\Xi }_{\tau }^{-1}\left\{V_{zz}(\vartheta ,\zeta ,z,s)\right\}\right].\hfill \end{array}$$

(23)

The k-th truncated term series of (23) is

$$\begin{array}{c}\hfill V_k(\vartheta ,\zeta ,z,s)=s^2\left(\vartheta +\zeta +z\right)+\sum _{n=1}^k{s}^{n\varrho +1}{f}_n(\vartheta ,\zeta ,z).\end{array}$$

(24)

and the k-th Elzaki residual function is

$$\begin{array}{cc}\hfill & {\Xi }_{\tau }Re{s}_k(\vartheta ,\zeta ,z,s)=V_k(\vartheta ,\zeta ,z,s)-s^2(\vartheta +\zeta +z)-s^{\varrho }{\Xi }_{\tau }[[{\Xi }_{\tau }^{-1}\{V(\vartheta ,\zeta ,z,s)\}][{\Xi }_{\tau }^{-1}\{\frac{\partial }{\partial \vartheta }{V}_k(\vartheta ,\zeta ,z,s)\}]]\hfill \\ \hfill & -s^{\varrho }{\Xi }_{\tau }[{\Xi }_{\tau }^{-1}\{\frac{{\partial }^2}{\partial {\vartheta }^2}{V}_k(\vartheta ,\zeta ,z,s)\}]-s^{\varrho }{\Xi }_{\tau }[{\Xi }_{\tau }^{-1}\{\frac{{\partial }^2}{\partial {\zeta }^2}{V}_k(\vartheta ,\zeta ,z,s)\}]+s^{\varrho }{\Xi }_{\tau }[{\Xi }_{\tau }^{-1}\{\frac{{\partial }^2}{\partial z^2}{V}_k(\vartheta ,\zeta ,z,s)\}].\hfill \end{array}$$

(25)

Now, to calculate $f_k(\vartheta ,\zeta )$, $k=1,2,3,\cdots$, we put the kth-truncate series (24) into the kth-Elzaki residual term (25), multiply the solution equation by $s^{k\varrho +1}$, and then the relation is solved recursively ${lim}_{s\to \infty }\left[s^{k\varrho +1}Re{s}_k(\vartheta ,\zeta ,s)\right]=0$, $k=1,2,3,\cdots$ for $f_k$. The following are the some terms $f_k(\vartheta ,\zeta )$

$$\begin{array}{cc}\hfill f_1(\vartheta ,\zeta ,z)& =(\vartheta +\zeta +z)\hfill \\ \hfill f_2(\vartheta ,\zeta ,z)& =2(\vartheta +\zeta +z)\hfill \\ \hfill f_3(\vartheta ,\zeta ,z)& =(\vartheta +\zeta +z)[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}]\hfill \\ \hfill f_4(\vartheta ,\zeta ,z)& =2(\vartheta +\zeta +z)\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}+\frac{2\Gamma (1+3\varrho )}{\Gamma (1+\varrho )\Gamma (1+2\varrho )}\right]\hfill \\ \hfill & ⋮.\hfill \end{array}$$

(26)

put the value of $f_n(\vartheta ,\zeta ,z)$, $(n\ge 1)$ in Equation (24), we obtain

$$\begin{array}{cc}\hfill V(\vartheta ,\zeta ,z,s)& =s^2\vartheta +\zeta +z+s^{\varrho +1}(\vartheta +\zeta +z)+s^{2\varrho +1}2(\vartheta +\zeta +z)+s^{3\varrho +1}(\vartheta +\zeta +z)\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}\right]\hfill \\ \hfill & +s^{4\varrho +1}2(\vartheta +\zeta +z)\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}+\frac{2\Gamma (1+3\varrho )}{\Gamma (1+\varrho )\Gamma (1+2\varrho )}\right]+\cdots .\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill V(\vartheta ,\zeta ,z,s)& =(\vartheta +\zeta +z)[s^2+s^{\varrho +1}+2s^{2\varrho +1}+s^{3\varrho +1}\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}\right]\hfill \\ \hfill & +s^{4\varrho +1}\left[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}+\frac{2\Gamma (1+3\varrho )}{\Gamma (1+\varrho )\Gamma (1+2\varrho )}\right]+\cdots ].\hfill \end{array}$$

(28)

Applying inverse Elzaki transform, we obtain

$$\begin{array}{cc}\hfill \nu (\vartheta ,\zeta ,z,\tau )& =(\vartheta +\zeta +z)[1+\frac{{\tau }^{\varrho }}{\Gamma (\varrho +1)}+\frac{2{\tau }^{\varrho }}{\Gamma (2\varrho +1)}+\frac{[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}]}{\Gamma (3\varrho +1)}{\tau }^{3\varrho }\hfill \\ \hfill & +\frac{[4+\frac{\Gamma (1+2\varrho )}{\Gamma {(1+\varrho )}^2}+\frac{2\Gamma (1+3\varrho )}{\Gamma (1+\varrho )\Gamma (1+2\varrho )}]}{\Gamma (4\varrho +1)}{\tau }^{4\varrho }+\cdots ].\hfill \end{array}$$

(29)

Now, if we substitute $\varrho =1$ in Equation (29), it gives

$$\begin{array}{cc}\hfill \nu (\vartheta ,\zeta ,z,\tau )& =(\vartheta +\zeta +z)[1+\tau +{\tau }^2+{\tau }^3+{\tau }^4+\cdots ]\hfill \end{array}$$

(30)

The solution of Equation (30) agrees with the Maclaurin series is

$$\begin{array}{cc}\hfill \nu (\vartheta ,\zeta ,z,\tau )& =\frac{\vartheta +\zeta +z}{1-\tau }.\hfill \end{array}$$

(31)

In Figure 4, the exact and analytical solution of $\nu (\vartheta ,\zeta ,\tau )$ at $\varrho =1$ and $0.8$ of Example 2. In Figure 5, the analytical solutions of $\nu (\vartheta ,\zeta ,\tau )$ at $\varrho =0.6$ and $0.4$ and similarly in Figure 6, the different fractional order of ϱ of $\nu (\vartheta ,\zeta ,\tau )$ of Example 2, which show that the close contact with the exact solution.

Example 3.

Consider the fractional one dimensional system Burger’s equation

$$\begin{array}{cc}\hfill D_{\tau }^{\varrho }\nu & =2\nu {\nu }_{\vartheta }-{\nu }_{\vartheta \vartheta }+{\left(\nu \psi \right)}_{\vartheta },\hfill \\ \hfill D_{\tau }^{\varrho }\psi & =-2\psi {\psi }_{\vartheta }-{\psi }_{\vartheta \vartheta }-{\left(\nu \psi \right)}_{\vartheta },0<\varrho \le 1\hfill \end{array}$$

(32)

with the ICs

$$\begin{array}{c}\hfill \nu (\vartheta ,0)=sin\left(\vartheta \right),\psi (\vartheta ,0)=-sin\left(\vartheta \right).\end{array}$$

(33)

Using Elzaki transformation to (32) and the ICs given in (32), we obtain

$$\begin{array}{cc}\hfill V(\vartheta ,s)& =s^{2}sin\left(\vartheta \right)+2s^{\varrho }{\Xi }_{\tau }\left[{\Xi }_{\tau }^{-1}\left\{V(\vartheta ,s)\frac{\partial }{\partial \vartheta }V(\vartheta ,s)\right\}\right]\hfill \\ \hfill & +s^{\varrho }{\Xi }_{\tau }{\left(\left[{\Xi }_{\tau }^{-1}V(\vartheta ,s)\right]\left[{\Xi }_{\tau }^{-1}V(\vartheta ,s)\right]\right)}_{\vartheta }-s^{\varrho }\left\{{\Xi }_{\tau }^{-1}\left[\frac{{\partial }^2}{\partial {\vartheta }^2}V(\vartheta ,s)\right]\right\},\hfill \\ \hfill \Psi (\vartheta ,s)& =s^2-sin\left(\vartheta \right)-2s^{\varrho }{\Xi }_{\tau }\left[{\Xi }_{\tau }^{-1}\{\Psi (\vartheta ,s)\frac{\partial }{\partial \vartheta }\Psi (\vartheta ,s)\}\right]\hfill \\ \hfill & -s^{\varrho }{\Xi }_{\tau }{\left(\left[{\Xi }_{\tau }^{-1}V(\vartheta ,s)\right][{\Xi }_{\tau }^{-1}\Psi (\vartheta ,s)]\right)}_{\vartheta }-s^{\varrho }\left\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\vartheta }^2}\Psi (\vartheta ,s)]\right\}.\hfill \end{array}$$

(34)

The k-th truncated term series of Equation (34) is

$$\begin{array}{cc}\hfill V_k(\vartheta ,s)& =s^{2}sin\left(\vartheta \right)+\sum _{n=0}^{\infty }{s}^{n\varrho +1}{f}_n\left(\vartheta \right),\hfill \\ \hfill {\Psi }_k(\vartheta ,s)& =-s^{2}sin\left(\vartheta \right)s+\sum _{n=0}^{\infty }{s}^{n\varrho +1}{g}_n\left(\vartheta \right).\hfill \end{array}$$

(35)

and the k-th Elzaki residual function is

$$\begin{array}{cc}\hfill {\Xi }_{\tau }Re{s}_k(\vartheta ,s)& =V_k(\vartheta ,s)-\frac{sin(\vartheta )}{s}-2s^{\varrho }{\Xi }_{\tau }[{\Xi }_{\tau }^{-1}\{V_k(\vartheta ,s)\frac{\partial }{\partial \vartheta }{V}_k(\vartheta ,s)\}]\hfill \\ \hfill & -s^{\varrho }{\Xi }_{\tau }{([{\Xi }_{\tau }^{-1}{V}_k(\vartheta ,s)][{\Xi }_{\tau }^{-1}{\Psi }_k(\vartheta ,s)])}_{\vartheta }+s^{\varrho }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\vartheta }^2}{V}_k(\vartheta ,s)]\},\hfill \\ \hfill {\Xi }_{\tau }Re{s}_k(\vartheta ,s)& =\Psi (\vartheta ,s)-\frac{-sin(\vartheta )}{s}+2s^{\varrho }{\Xi }_{\tau }[{\Xi }_{\tau }^{-1}\{{\Psi }_k(\vartheta ,s)\frac{\partial }{\partial \vartheta }{\Psi }_k(\vartheta ,s)\}]\hfill \\ & -s^{\varrho }{\Xi }_{\tau }{([{\Xi }_{\tau }^{-1}{V}_k(\vartheta ,s)][{\Xi }_{\tau }^{-1}{\Psi }_k(\vartheta ,s)])}_{\vartheta }-s^{\varrho }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\vartheta }^2}{\Psi }_k(\vartheta ,s)]\}.\hfill \end{array}$$

(36)

Now, to determine $f_k\left(\vartheta \right)$ and $g_k\left(\vartheta \right)$, $k=1,2,3,\cdots$, we substitute the kth-truncated series (35) into the kth-Elzaki residual function (36), multiply the resulting equation by $s^{k\varrho +1}$, and then solve recursively the relation ${lim}_{s\to \infty }\left[s^{k\varrho +1}Re{s}_k(\vartheta ,s)\right]=0$, $k=1,2,3,\cdots$ for $f_k$ and $g_k$. Some terms of the sequences $f_k\left(\vartheta \right)$ and $g_k\left(\vartheta \right)$

$$\begin{array}{cc}\hfill f_1\left(\vartheta \right)& =sin\left(\vartheta \right),\hfill \\ \hfill g_1\left(\vartheta \right)& =-sin\left(\vartheta \right).\hfill \\ \hfill f_2\left(\vartheta \right)& =sin\left(\vartheta \right),\hfill \\ \hfill g_2\left(\vartheta \right)& =-sin\left(\vartheta \right).\hfill \\ \hfill f_3\left(\vartheta \right)& =sin\left(\vartheta \right),\hfill \\ \hfill g_3\left(\vartheta \right)& =-sin\left(\vartheta \right).\hfill \\ \hfill f_4\left(\vartheta \right)& =sin\left(\vartheta \right),\hfill \\ \hfill g_4\left(\vartheta \right)& =-sin\left(\vartheta \right).\hfill \\ \hfill & ⋮.\hfill \end{array}$$

(37)

Putting the values of $f_n\left(\vartheta \right)$ and $g_n\left(\vartheta \right)$ for $(n\ge 1)$ in Equation (35), we obtain

$$\begin{array}{cc}\hfill V(\vartheta ,s)& =s^{2}sin\left(\vartheta \right)+s^{\varrho +1}sin\left(\vartheta \right)+s^{2\varrho +1}sin\left(\vartheta \right)+s^{3\varrho +1}sin\left(\vartheta \right)+\cdots ,\hfill \\ \hfill \Psi (\vartheta ,s)& =-s^{2}sin\left(\vartheta \right)-s^{\varrho +1}sin\left(\vartheta \right)-s^{2\varrho +1}sin\left(\vartheta \right)-s^{3\varrho +1}sin\left(\vartheta \right)+\cdots .\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill V(\vartheta ,s)& =sin\left(\vartheta \right)\left[s+s^{\varrho +1}+s^{2\varrho +1}+s^{3\varrho +1}+\cdots \right],\hfill \\ \hfill \Psi (\vartheta ,s)& =-sin\left(\vartheta \right)\left[s+s^{\varrho +1}+s^{2\varrho +1}+s^{3\varrho +1}+\cdots \right].\hfill \end{array}$$

(39)

Applying the inverse Elzaki transform to Equation (39), we obtain

$$\begin{array}{cc}\hfill \nu (\vartheta ,\tau )& =sin\left(\vartheta \right)\left[1+\frac{{\tau }^{\varrho }}{\Gamma (\varrho +1)}+\frac{{\tau }^{2\varrho }}{\Gamma (2\varrho +1)}+\frac{{\tau }^{3\varrho }}{\Gamma (3\varrho +1)}+\cdots \right],\hfill \\ \hfill \psi (\vartheta ,\tau )& =-sin\left(\vartheta \right)\left[1+\frac{{\tau }^{\varrho }}{\Gamma (\varrho +1)}+\frac{{\tau }^{2\varrho }}{\Gamma (2\varrho +1)}+\frac{{\tau }^{3\varrho }}{\Gamma (3\varrho +1)}+\cdots \right].\hfill \end{array}$$

(40)

Putting $\varrho =1$, we obtain the solution of Equation (32) in closed form

$$\begin{array}{cc}\hfill \nu (\vartheta ,\tau )& =e^{\tau }sin\left(\vartheta \right),\hfill \\ \hfill \psi (\vartheta ,\tau )& =-e^{\tau }sin\left(\vartheta \right).\hfill \end{array}$$

(41)

Figure 7 shows the exact and analytical solution of $\nu (\vartheta ,\zeta ,z,\tau )$ at $\varrho =1$ and $0.8$ of Example 3. Figure 8 shows the analytical solutions of $\nu (\vartheta ,\zeta ,z,\tau )$ at $\varrho =0.6$ and $0.4$. Similarly, Figure 9, shows the different fractional-order of ϱ of $\nu (\vartheta \zeta ,z,\tau )$ of Example 3, which show the close contact with the exact solution. Figure 10, Figure 11 and Figure 12 show the close relation with respect to $\nu (\vartheta ,\zeta ,z,\tau )$ of Example 3.

Example 4.

Consider the fractional two dimensional system of Burger’s equation

$$\begin{array}{cc}\hfill D_{\tau }^{\varrho }\nu & ={\nu }_{\vartheta \vartheta }+{\nu }_{\zeta \zeta }-\nu {\nu }_{\vartheta }-\psi {\nu }_{\zeta },\hfill \\ \hfill D_{\tau }^{\varrho }\psi & ={\psi }_{\vartheta \vartheta }+{\nu }_{\zeta \zeta }-\nu {\psi }_{\vartheta }-\psi {\psi }_{\zeta },0<\varrho \le 1\hfill \end{array}$$

(42)

with the ICs

$$\begin{array}{c}\hfill \nu (\vartheta ,\zeta ,0)=\vartheta +\zeta ,\psi (\vartheta ,\zeta ,0)=\vartheta -\zeta .\end{array}$$

(43)

Applying Elzaki transform to Equation (42) and applying the ICs in Equation (43), we obtain

$$\begin{array}{cc}\hfill V(\vartheta ,\zeta ,s)& =s^2\vartheta +\zeta +s^{\varrho }{\xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\vartheta }^2}V(\vartheta ,\zeta ,s)]\}+s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\zeta }^2}V(\vartheta ,\zeta ,s)]\}\hfill \\ \hfill & -s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[V(\vartheta ,\zeta ,s)]{\Xi }_{\tau }^{-1}[\frac{\partial }{\partial \vartheta }V(\vartheta ,\zeta ,s)]\}-s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\Psi (\vartheta ,\zeta ,s)]{\Xi }_{\tau }^{-1}[\frac{\partial }{\partial \zeta }V(\vartheta ,\zeta ,s)]\},\hfill \\ \hfill \Psi (\vartheta ,\zeta ,s)& =s^2\vartheta +\zeta +s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\vartheta }^2}\Psi (\vartheta ,\zeta ,s)]\}+s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\zeta }^2}\Psi (\vartheta ,\zeta ,s)]\}\hfill \\ \hfill & -s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[V(\vartheta ,\zeta ,s)]{\Xi }_{\tau }^{-1}[\frac{\partial }{\partial \vartheta }\Psi (\vartheta ,\zeta ,s)]\}-s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\Psi (\vartheta ,\zeta ,s)]{\Xi }_{\tau }^{-1}[\frac{\partial }{\partial \zeta }\Psi (\vartheta ,\zeta ,s)]\},\hfill \end{array}$$

(44)

The k-th truncated functions series of Equation (44) is

$$\begin{array}{cc}\hfill V_k(\vartheta ,\zeta ,s)& =\frac{\vartheta +\zeta }{s}+\sum _{n=0}^{\infty }\frac{f_n(\vartheta ,\zeta )}{s^{n\varrho +1}},\hfill \\ \hfill {\Psi }_k(\vartheta ,\zeta ,s)& =\frac{\vartheta -\zeta }{s}+\sum _{n=0}^{\infty }\frac{g_n(\vartheta ,\zeta )}{s^{n\varrho +1}}.\hfill \end{array}$$

(45)

and k-th Elzaki residual function is

$$\begin{array}{cc}\hfill {\Xi }_{\tau }Re{s}_k(\vartheta ,\zeta ,s)& =V_k(\vartheta ,\zeta ,s)-s^2\vartheta +\zeta +s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\vartheta }^2}{V}_k(\vartheta ,\zeta ,s)]\}-s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\zeta }^2}{V}_k(\vartheta ,\zeta ,s)]\}\hfill \\ \hfill & +s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[V_k(\vartheta ,\zeta ,s)]{\Xi }_{\tau }^{-1}[\frac{\partial }{\partial \vartheta }{V}_k(\vartheta ,\zeta ,s)]\}+s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[{\Psi }_k(\vartheta ,\zeta ,s)]{\Xi }_{\tau }^{-1}[\frac{\partial }{\partial \zeta }{V}_k(\vartheta ,\zeta ,s)]\},\hfill \\ \hfill {\Xi }_{\tau }Re{s}_k(\vartheta ,\zeta ,s)& =\Psi (\vartheta ,\zeta ,s)-s^2\vartheta +\zeta +s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\vartheta }^2}{\Psi }_k(\vartheta ,\zeta ,s)]\}-s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[\frac{{\partial }^2}{\partial {\zeta }^2}{\Psi }_k(\vartheta ,\zeta ,s)]\}\hfill \\ \hfill & +s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[V_k(\vartheta ,\zeta ,s)]{\Xi }_{\tau }^{-1}[\frac{\partial }{\partial \vartheta }{\Psi }_k(\vartheta ,\zeta ,s)]\}+s^{\varrho }{\Xi }_{\tau }\{{\Xi }_{\tau }^{-1}[{\Psi }_k(\vartheta ,\zeta ,s)]{\Xi }_{\tau }^{-1}[\frac{\partial }{\partial \zeta }{\Psi }_k(\vartheta ,\zeta ,s)]\},\hfill \end{array}$$

(46)

Now, to determine $f_k(\vartheta ,\zeta )$ and $g_k(\vartheta ,\zeta )$,$k=1,2,3,\cdots$, we substitute the kth-truncated series (45) into the kth-Elzaki residual function (46), multiply the resulting equation by $s^{k\varrho +1}$, and then solve recursively the relation ${lim}_{s\to \infty }\left[s^{k\varrho +1}Re{s}_k(\vartheta ,s)\right]=0$, $k=1,2,3,\cdots$ for $f_k$ and $g_k$. The first some terms of the sequences $f_k(\vartheta ,\zeta )$ and $g_k(\vartheta ,\zeta )$

$$\begin{array}{cc}\hfill f_1(\vartheta ,\zeta )& =-2\vartheta ,\hfill \\ \hfill g_1(\vartheta ,\zeta )& =-2\zeta .\hfill \\ \hfill f_2(\vartheta ,\zeta )& =4(\vartheta +\zeta ),\hfill \\ \hfill g_2(\vartheta ,\zeta )& =4(\vartheta -\zeta ).\hfill \\ \hfill f_3(\vartheta ,\zeta )& =-16\vartheta -4\vartheta \frac{\Gamma (2\varrho +1)}{\Gamma {(\varrho +1)}^2},\hfill \\ \hfill g_3(\vartheta ,\zeta )& =-16\zeta -4\zeta \frac{\Gamma (2\varrho +1)}{\Gamma {(\varrho +1)}^2}.\hfill \\ \hfill & ⋮.\hfill \end{array}$$

(47)

substituting the value of $f_n(\vartheta ,\zeta )$ and $g_n(\vartheta ,\zeta )$, for $n\ge 1$ in Equation (45), we have

$$\begin{array}{cc}\hfill U(\vartheta ,\zeta ,s)& =s^2\vartheta +\zeta -2s^{\varrho +1}\vartheta +4s^{2\varrho +1}(\vartheta +\zeta )+s^{\varrho +1}\left(-16\vartheta -4\vartheta \frac{\Gamma (2\varrho +1)}{\Gamma {(\varrho +1)}^2}\right)+\cdots ,\hfill \\ \hfill \Psi (\vartheta ,\zeta ,s)& =s^2\vartheta -\zeta -2s^{\varrho +1}\zeta +4s^{\varrho +1}(\vartheta +\zeta )+s^{\varrho +1}\left(-16\zeta -4\zeta \frac{\Gamma (2\varrho +1)}{\Gamma {(\varrho +1)}^2}\right)+\cdots .\hfill \end{array}$$

(48)

Applying inverse Elzaki transform to Equation (48), we have

$$\begin{array}{cc}\hfill \nu (\vartheta ,\zeta ,\tau )& =\vartheta +\zeta +\frac{-2\vartheta }{\Gamma (\varrho +1)}{\tau }^{\varrho }+\frac{4(\vartheta +\zeta )}{\Gamma (2\varrho +1)}{\tau }^{2\varrho }+\frac{-16\vartheta -4\vartheta \frac{\Gamma (2\varrho +1)}{\Gamma {(\varrho +1)}^2}}{\Gamma (3\varrho +1)}{\tau }^{3\varrho }+\cdots ,\hfill \\ \hfill \psi (\vartheta ,\zeta ,\tau )& =\vartheta -\zeta +\frac{-2\zeta }{\Gamma (\varrho +1)}{\tau }^{\varrho }+\frac{4(\vartheta +\zeta )}{\Gamma (2\varrho +1)}{\tau }^{2\varrho }+\frac{-16\zeta -4\zeta \frac{\Gamma (2\varrho +1)}{\Gamma {(\varrho +1)}^2}}{\Gamma (3\varrho +1)}{\tau }^{3\varrho }+\cdots .\hfill \end{array}$$

(49)

Putting $\varrho =1$, we obtain the solution of Equation (42) in closed form

$$\begin{array}{cc}\hfill \nu (\vartheta ,\zeta ,\tau )& =\vartheta +\zeta -2\vartheta \tau +2(\vartheta +\zeta ){\tau }^2-4\vartheta {\tau }^3+4(\vartheta +\zeta ){\tau }^4-8\vartheta {\tau }^5+\cdots ,\hfill \\ \hfill \psi (\vartheta ,\zeta ,\tau )& =\vartheta -\zeta -2\zeta \tau +2(\vartheta +\zeta ){\tau }^2-4\zeta {\tau }^3+4(\vartheta -\zeta ){\tau }^4-8\vartheta {\tau }^5+\cdots .\hfill \\ \hfill \nu (\vartheta ,\zeta ,\tau )& =\frac{\vartheta +\zeta -2\vartheta \tau }{1-2{\tau }^2},\hfill \\ \hfill \psi (\vartheta ,\zeta ,\tau )& =\frac{\vartheta -\zeta -2\zeta \tau }{1-2{\tau }^2}.\hfill \end{array}$$

(50)

Figure 13 shows the exact and analytical solution of $\nu (\vartheta ,\zeta ,z,\tau )$ at $\varrho =1$ and $0.8$ of Example 4. Figure 14 shows the analytical solutions of $\nu (\vartheta ,\zeta ,z,\tau )$ at $\varrho =0.6$ and $0.4$. Figure 15 similarly shows the different fractional-order of ϱ of $\nu (\vartheta ,\zeta ,z,\tau )$ of Example 4, which show the close contact with the exact solution. Figure 16, Figure 17 and Figure 18 show the close relation with respect to $\nu (\vartheta ,\zeta ,z,\tau )$ of Example 4.

5. Conclusions

In this paper, the analytical solutions are constructed and analyzed for Burgers’ equations with suitable initial conditions utilizing the RPSTM under time-Caputo differentiability. The fractional RPSTM is modified in the current method by tying it to the Yang transform operator. The advantage of use the RPSTM is that it gives more accurate convergence McLaurin series and needs only a small size of calculation without involving the perturbation, discretization, or any other physical restrictive condition. Two well-known physical applications are tested in order to show the viability and superiority of the proposed technique. Graphics and numerical simulation are used to discuss the resulting approximations. The resulting results are contrasted with those of other widely used, published methodologies. In light of these findings, it is clear that the RPSTM is a simple and practical method for handling the wide spectrum of non-linear fractional partial differential equations that might emerge in engineering and scientific situations. Future research can use the RPSTM to discover both precise and approximative answers to fractional partial differential equations. As a result, physical models and dynamical models can be handled by the RPSTM application.