1. Introduction
Fractional analysis, including arbitrary derivatives and integrals, has gained much attention in both theoretical and applied sciences in recent years. In fact, fractional analysis is not a new mathematical theory. The basics are almost as old as the classical analysis.
There have been many studies on the theory of fractional analysis, but in recent years these studies have gained a great deal of momentum thanks to programs such as Mathematica, Matlab and Maple, which are used for symbolic calculations. Thus, the equations that are too complicated to be calculated manually are easily solved. In addition, problems in mathematics, physics and many engineering fields are modeled in more detail using fractional calculus.
Grünwald, Letnikov, Riemann, Liouville and many others have contrived very substantial studies related to the theory of the subject, but it is thought that it cannot be applied to real life problems. However, recent studies in the fields of viscoelasticity and control have shown that the concepts of fractional derivative and integration are more convenient and reliable than the classical derivative and integral concepts that we have been using for modeling and solving many problems. There have been many applications using fractional calculus in real models see [1–3] and the references therein.
Recently, several definitions of fractional derivatives are taken part in the literature. One of them is fractional derivative with exponential kernel defined by Caputo and Fabrizio [4]. Another one is fractional derivative with Mittag-Leffler kernel defined by Atangana and Baleanu [5–8]. Some of the studies which exist in the literature are, Atangana, Aguilar and Jarad et al [9–11] and Al-Refai and Abdeljawad et al [12, 13] studied a fractional operator with exponential kernel for some modeling problems and their solution methods.
In this study, the definition of conformable derivative proposed by Khalil et al [14] has been used instead of the Riemann-Liouville and Caputo fractional derivative definitions, which are frequently used in the solution of fractional differential equations in recent years. This is because the definition of conformable derivative is different from other derivative definitions and allows to obtain the exact solution of the equation. In this direction, the new analytical solutions of the fractional Burgers-K-P equation was first obtained by the sub-equation method and the solutions were compared with the approximate solutions obtained by residual power series method (RPSM) proposed by Arqub [15, 16] by the help of tables and graphs.
After this introductory section, section 2 is devoted to introducing some basic definitions, section 3 is reserved for a brief review of the governing equation, and in section 4 a basic description of the implemented method is illustrated. In section 5 , the analytical and approximate solutions of the Burgers-K-P equation are presented. Finally the paper ends with a conclusion in section 6 .
2. Basic definitions
The conformable fractional derivative of an $f$ function for an arbitrary order of $\mu $ is given by
$\begin{eqnarray}{T}_{\mu }(f)(t)=\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\displaystyle \frac{f(t+\varepsilon {t}^{1-\mu })-(f)(t)}{\varepsilon },\end{eqnarray}$
for $f\,:[0,\infty )\to R$ and for all t > 0, μ ∈ (0,1).The next theorem includes some properties of this new definition [14].
Let $f$ and $g$ are $\mu $-differentiable functions at point $t\gt 0$ for $0\lt \mu \leqslant 1$. Then
| ${T}_{\mu }({mf}+{ng})={{mT}}_{\mu }(f)+{{nT}}_{\mu }(g)$ for all $m,n\in {\mathbb{R}}$ | |
| ${T}_{\mu }({t}^{p})={{pt}}^{p-\mu }$ for all p | |
| ${T}_{\mu }(f.g)={{fT}}_{\mu }(g)+{{gT}}_{\mu }(f)$ | |
| ${T}_{\mu }(\tfrac{f}{g})=\tfrac{{{gT}}_{\mu }(f)-{{fT}}_{\mu }(g)}{{g}^{2}}$ | |
| ${T}_{\mu }(c)=0$ for all $f(t)=c$ constant functions | |
| Moreover, ${T}_{\mu }(f)(t)={t}^{1-\mu }\tfrac{{df}(t)}{{dt}},$ if f is differentiable. |
Let $f$ function is defined with $n$ variables ${x}_{1},\,\ldots ,\,{x}_{n}.$ The conformable partial derivatives of $f$ of order $\mu \in (0,1]$ in ${x}_{i}$ is given by [17]
$\begin{eqnarray}\begin{array}{l}\displaystyle \frac{{{\rm{d}}}^{\mu }}{{\rm{d}}{x}_{i}^{\mu }}f({x}_{1},\,\ldots ,\,{x}_{n})\\ =\,\mathop{\mathrm{lim}}\limits_{\varepsilon \to 0}\displaystyle \frac{f({x}_{1},\,\ldots ,\,{x}_{i-1},{x}_{i}+\varepsilon {x}_{i}^{1-\mu },\,\ldots ,\,{x}_{n})-f({x}_{1},\,\ldots ,\,{x}_{n})}{\varepsilon }.\end{array}\end{eqnarray}$
The conformable integral of an $f$ function for $\mu \geqslant 0$ is defined as [18]:
$\begin{eqnarray}{I}_{\mu }^{a}(f)(s)=\mathop{\mathop{\int }\limits^{s}}\limits_{a}\displaystyle \frac{f(t)}{{t}^{1-\mu }}{\rm{d}}t.\end{eqnarray}$
Next, some necessary theorems and definitions about fractional power series will be presented.
Assume that $f$ is an infinitely $\mu -$ differentiable function at a neighborhood of a point ${t}_{0}$ for some $0\lt \mu \leqslant 1,$ then $f$ has the fractional power series expansion of the form:
$\begin{eqnarray}\begin{array}{rcl}f(t) & = & \mathop{\sum _{k=0}}\limits^{\infty }\displaystyle \frac{{\left({T}_{\mu }^{{t}_{0}}f\right)}^{(k)}({t}_{0}){\left(t-{t}_{0}\right)}^{k\mu }}{{\mu }^{k}k!},\\ & & \times \ {t}_{0}\lt t\lt {t}_{0}+{R}^{\tfrac{1}{\mu }},\ R\gt 0.\end{array}\end{eqnarray}$
Here ${\left({T}_{\mu }^{{t}_{0}}f\right)}^{(k)}({t}_{0})$ represents the application of the fractional derivative $k-$ times [12].$\mathop{\sum _{n=0}}\limits^{\infty }{f}_{n}(x){\left(t-{t}_{0}\right)}^{n\mu }$ is called a multiple fractional power series at ${t}_{0}=0,$ where $0\leqslant m-1\lt \mu \,\lt m,t$ is a variable and ${f}_{n}(x)$ are functions termed the coefficients of the series [19, 20].
Assume that $u(x,t)$ has a multiple fractional power series representation at ${t}_{0}=0$ of the form [19]
$\begin{eqnarray}\begin{array}{rcl}u(x,t) & = & \mathop{\sum _{n=0}}\limits^{\infty }{f}_{n}(x){t}^{n\mu },\\ & & 0\leqslant m-1\lt \mu \lt m,\ x\in I,\ 0\leqslant t\leqslant {R}^{\tfrac{1}{\mu }}.\end{array}\end{eqnarray}$
If ${u}_{t}^{(n\mu )}(x,t),$ $n=0,1,2,\ldots $ are continuous on $I\times (0,{R}^{\tfrac{1}{\mu }})$, then ${f}_{n}(x)=\tfrac{{u}_{t}^{(n\mu )}(x,0)}{{\mu }^{n}n!}.$Many studies have been carried out using the above definitions by combining them with different methods for obtaining analytical and approximate solutions of various fractional partial differential equations. In contrast to the definition of Caputo and Riemann-Liouville, the definition of conformable derivative fulfills the chain rule [12]. Thus, using this definition, reliable solutions can be calculated for fractional differential equations.
3. Governing equation
Long waves, also known as shallow water waves, are waves moving at a depth of less than 1/20 of their wavelength in shallow waters of the sea or ocean. Many flow types can be modeled with the help of shallow water equations that are extensively used in flood analysis. These equations offer an attractive and productive field of research. They are very suitable for wave propagation phenomena and describe the interaction of two long waves with different dispersion relations. Eventually, such equations provide strong solutions to the problems arising from hydrodynamic systems.
Kadomtsev and Petviashvili relaxed the restriction that the waves be strictly one-dimensional, so they extended the well-known KdV equation to the (2 + 1)-dimensional K-P equation given by
$\begin{eqnarray}{\left({u}_{t}+6{{uu}}_{x}+{u}_{{xxx}}\right)}_{x}+\lambda {u}_{{yy}}=0,\end{eqnarray}$
where u = u(x, y, t). The K-P equation is used to model shallow water waves with weak nonlinear restoring forces.On the other hand, the Burgers equation is a fundamental second-order partial differential equation arising in fluid mechanics given by
$\begin{eqnarray}{u}_{t}+{{uu}}_{x}+\nu {u}_{{xx}}=0,\end{eqnarray}$
where ν is the kinematic viscosity constant.In this study, we examine the extension sense of the K-P equation, that is a completely integrable dissipative equation, namely the Burgers-K-P equation, with a conformable time-fractional derivative as
$\begin{eqnarray}{T}_{x}[{T}_{t}^{\mu }u+{{uT}}_{x}u+\nu {T}_{x}^{2}u]+\lambda {T}_{y}^{2}u=0.\end{eqnarray}$
In addition to these developments in fractional analysis, in the literature, there exists a number of techniques that have been developed and implemented to acquire analytical and approximate solutions of fractional differential equations, especially in the case of the Burgers-K-P equation. These techniques include the exp-function method [21], Lie symmetry approach [22], $({G}^{{\prime} }/G)$-expansion methods [23], modified direct Clarkson-Kruskal's method [24], the first integral method [25] and Hirota's bilinear method and tanh-coth method [26], the Schwarz function method [27] and q-homotopy analysis method [28]. To the author's knowledge, the conformable fractional version of the Burgers-K-P equation has not been studied. See [29–33] for more methods and applications.4. Description of the methods
4.1. Sub-equation method
Step 1: Suppose the general form of the nonlinear time-fractional partial differential equation is given as
$\begin{eqnarray}P\left(u,{T}_{t}^{\mu }u,{T}_{x}u,{T}_{y}u,{T}_{x}^{2}u,{T}_{y}^{2}u,\ldots \right)=0,\end{eqnarray}$
where ${T}_{t}^{\mu }$ implies a conformable derivative operator with arbitrary order.Step 2: Allow the fractional wave transformation [37]
$\begin{eqnarray}u(x,y,t)=U(\xi ),\ \xi ={kx}+{wy}+c\displaystyle \frac{{t}^{\mu }}{\mu },\end{eqnarray}$
where $c,k,w$ are arbitrary constants to be calculated afterwards.Step 3: Applying the chain rule given in [12], equation ((10 ) reduces to an integer order nonlinear ordinary differential equation of the form
$\begin{eqnarray}G(U(\xi ),{U}^{{\prime} }(\xi ),{U}^{{\prime\prime} }(\xi ),\ \ldots )=0.\end{eqnarray}$
Step 4: Let the solution of equation (12 ) be in the following form12 ) and φ(ξ) is the solution of the Riccati equation (9 ).
$\begin{eqnarray}U(\xi )=\sum _{i=0}^{N}{a}_{i}{\varphi }^{i}(\xi ),\ {a}_{N}\ne 0,\end{eqnarray}$
where ai are stable coefficients for $(0\leqslant i\leqslant N)$ and to be calculated. N is a positive integer to be calculated by the balancing principle [38] in equation (Step 5: A set of the solutions which confirm the equation (9 )) is given below.
$\begin{eqnarray}\varphi (\xi )=\left\{\begin{array}{c}-\sqrt{-\sigma }\tanh \left(\sqrt{-\sigma }\xi \right),\ \sigma \lt 0\\ -\sqrt{-\sigma }\coth \left(\sqrt{-\sigma }\xi \right),\ \ \sigma \lt 0\\ \sqrt{\sigma }\tan \left(\sqrt{\sigma }\xi \right),\ \ \ \ \ \ \ \ \ \sigma \gt 0\\ -\sqrt{\sigma }\cot \left(\sqrt{\sigma }\xi \right),\ \ \ \ \ \ \ \ \ \sigma \gt 0\\ -\tfrac{1}{\xi +\varpi },\ \varpi \ {is}\ a\ \mathrm{cons}.,\ \ \ \sigma =0.\end{array}\right.\end{eqnarray}$
Step 6: Combining all of the obtained results, we achieve a polynomial in terms of $\varphi (\xi )$.
Step 7: Vanishing the coefficients of ${\varphi }^{i}(\xi )$ for $(i=0,1,\,\ldots ,\,N)$ produces a nonlinear algebraic equation system in ${a}_{i},c,k,w,$ $(i=0,1,\,\ldots ,\,N)$.
Step 8: Computing the solution of these nonlinear algebraic equations, we attain the values for ${a}_{i},c,k,w,$ $(i\,=0,1,\,\ldots ,\,N)$.
Step 9: Subrogating all of the acquired values in the formulas of (14 ), we get the exact solutions for the equation (10 ).
4.2. Residual power series method (RPSM)
To illustrate the basic idea of RPSM [15, 16, 39–41], let us consider the following nonlinear fractional differential equation:
$\begin{eqnarray}\begin{array}{l}{T}_{\mu }u(x,y,t)+N[x,y]u(x,y,t)\\ \qquad +\ L[x,y]u(x,y,t)=g(x,y,t),\end{array}\end{eqnarray}$
where $n-1\lt n\mu \leqslant n,$ $t\gt 0,$ $x\in {\mathbb{R}}.$ The initial condition of the equation is $\begin{eqnarray}{f}_{0}(x,y)=u(x,y,0)=f(x,y).\end{eqnarray}$
Here, $N[x,y]$ is a nonlinear and $L[x,y]$ is a linear operator given with the g(x, y, t) continuous functions.Step 1: The RPSM method consists of specifying the solution of (15 ), which is subject to (16 ) as a fractional power series expansion around t = 0.
$\begin{eqnarray}{f}_{(n-1)}(x,y)={T}_{t}^{(n-1)\mu }u(x,y,0)=h(x,y).\end{eqnarray}$
Step 2: The solution can be expressed by a series expansion as
$\begin{eqnarray}u(x,y,t)=f(x,y)+\mathop{\sum _{n=1}}\limits^{\infty }{f}_{n}(x,y)\displaystyle \frac{{t}^{n\mu }}{{\mu }^{n}n!}.\end{eqnarray}$
Thus, the k − th truncated series of u(x, y, t), that is ${u}_{k}(x,y,t)$ can be represented as $\begin{eqnarray}{u}_{k}(x,y,t)=f(x,y)+\mathop{\sum _{n=1}}\limits^{k}{f}_{n}(x,y)\displaystyle \frac{{t}^{n\mu }}{{\mu }^{n}n!}.\end{eqnarray}$
Step 3: Since the 1st approximate solution u1(x, y, t) is
$\begin{eqnarray}{u}_{1}(x,y,t)=f(x,y)+{f}_{1}(x,y)\displaystyle \frac{{t}^{\mu }}{{\mu }^{n}},\end{eqnarray}$
uk(x, y, t) might be reformulated as $\begin{eqnarray}\begin{array}{rcl}{u}_{k}(x,y,t) & = & f(x,y)+{f}_{1}(x,y)\displaystyle \frac{{t}^{\mu }}{{\mu }^{n}}\\ & & +\ \mathop{\sum _{n=2}}\limits^{k}{f}_{n}(x,y)\displaystyle \frac{{t}^{n\mu }}{{\mu }^{n}n!},\ k=2,3,4,\ldots \end{array}\end{eqnarray}$
for $0\leqslant t\lt {{\mathbb{R}}}^{\tfrac{1}{v}},$ $x\in I$ and $0\lt \mu \leqslant 1$.Step 4: Initially we express the residual function and the k − th residual function
$\begin{eqnarray}\begin{array}{rcl}\mathrm{Res}(x,y,t) & = & {T}_{\mu }u(x,y,t)+N[x,y]u(x,y,t)\\ & & +\ L[x,y]u(x,y,t)-g(x,y,t),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\mathrm{Res}}_{k}(x,t) & = & {T}_{\mu }{u}_{k}(x,y,t)+N[x,y]{u}_{k}(x,y,t)\\ & & +L[x,y]{u}_{k}(x,y,t)-g(x,y,t),\\ k & = & 1,2,3,\ldots \end{array}\end{eqnarray}$
respectively.Step 5: Obviously, $\mathrm{Res}(x,y,t)=0$ and $\mathop{\mathrm{lim}}\limits_{k\to \infty }{\mathrm{Res}}_{k}(x,y,t)\,=\mathrm{Res}(x,y,t)$ for each $x\in I$ and $t\geqslant 0.$ Indeed this brings about $\tfrac{{\partial }^{(n-1)\mu }}{\partial {t}^{(n-1)\mu }}{\mathrm{Res}}_{k}(x,y,t)=0$ for $n=1,2,3,\,\ldots ,\,k$ because, the fractional derivative of a constant is zero in the conformable sense [17, 42].
Step 6: Integrating the differential equation $\tfrac{{\partial }^{(n-1)\mu }}{\partial {t}^{(n-1)\mu }}{\mathrm{Res}}_{k}(x,y,0)=0$ produces the needed ${f}_{n}(x,y)$ parameters. Therefore, using these parameters one can achieve the ${u}_{n}(x,y,t)$ approximate solutions in this fashion, respectively.
5. Results and discussion
5.1. Analytical solutions of the Burgers-K-P equation
Consider the time fractional Burgers-K-P equation11 ) to equation (24 ) and integrating the resulting ODE once, we find out the reduced form can be expressed as follows:25 ) is in terms of $\varphi (\xi ),$ while the analytical solutions of equation (9 ) are given as follows:25 ), an algebraic equation system comes to exist with respect to $k,w,c,{a}_{0},{a}_{1}$. Solving the obtain system yields following solution set:14 ) and (11 ) the traveling wave solutions of equation (24 ) can be deducted:
$\begin{eqnarray}{T}_{x}[{T}_{t}^{\mu }u+{{uT}}_{x}u+\nu {T}_{x}^{2}u]+\lambda {T}_{y}^{2}u=0,\end{eqnarray}$
where ${T}_{t}^{\mu }$ is conformable derivative of function u(x, y, t) with arbitrary order. Performing the chain rule [12] and wave transform ( $\begin{eqnarray}{{kcU}}^{{\prime} 2}{{UU}}^{{\prime} 3}\nu {U}^{{\prime} ^{\prime} 2}{U}^{{\prime} }=0.\end{eqnarray}$
Assume that the solution of equation ( $\begin{eqnarray}U(\xi )=\sum _{i=0}^{N}{a}_{i}{\varphi }^{i}(\xi ),\ {a}_{N}\ne 0.\end{eqnarray}$
With the aid of the balancing principle [38], we have N = 1. Bringing together all the obtained data in equation ( $\begin{eqnarray}{a}_{1}=-2k\nu ,c=\displaystyle \frac{-{a}_{0}{k}^{2}-\lambda {w}^{2}}{k},\end{eqnarray}$
and ${a}_{0},k,w$ are free constants. When $\sigma \lt 0$, using ( $\begin{eqnarray}\begin{array}{rcl}{u}_{1}(x,y,t) & = & {a}_{0}+2k\nu \sqrt{-\sigma }\tanh \\ & & \times \left(\sqrt{-\sigma }\left(\displaystyle \frac{{t}^{\mu }\left(-{a}_{0}{k}^{2}-\lambda {w}^{2}\right)}{k\mu }+{kx}+{wy}\right)\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{u}_{2}(x,y,t) & = & {a}_{0}+2k\nu \sqrt{-\sigma }\coth \\ & & \times \left(\sqrt{-\sigma }\left(\displaystyle \frac{{t}^{\mu }\left(-{a}_{0}{k}^{2}-\lambda {w}^{2}\right)}{k\mu }+{kx}+{wy}\right)\right).\end{array}\end{eqnarray}$
In a similar way, for σ > 0, we obtain $\begin{eqnarray}\begin{array}{rcl}{u}_{3}(x,y,t) & = & {a}_{0}-2k\nu \sqrt{\sigma }\tan \\ & & \times \left(\sqrt{\sigma }\left(\displaystyle \frac{{t}^{\mu }\left(-{a}_{0}{k}^{2}-\lambda {w}^{2}\right)}{k\mu }+{kx}+{wy}\right)\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{u}_{4}(x,y,t) & = & {a}_{0}+2k\nu \sqrt{\sigma }\cot \\ & & \times \left(\sqrt{\sigma }\left(\displaystyle \frac{{t}^{\mu }\left(-{a}_{0}{k}^{2}-\lambda {w}^{2}\right)}{k\mu }+{kx}+{wy}\right)\right).\end{array}\end{eqnarray}$
Lastly, when σ = 0 $\begin{eqnarray}{u}_{5}(x,y,t)={a}_{0}+\displaystyle \frac{2k\nu }{\tfrac{{t}^{\mu }\left(-{a}_{0}{k}^{2}-\lambda {w}^{2}\right)}{k\mu }+{kx}+l+{wy}}.\end{eqnarray}$
5.2. Approximate solutions of the Burgers-K-P equation
Consider the nonlinear time-fractional Burgers-K-P equation as follows:
$\begin{eqnarray}{T}_{x}[{T}_{t}^{\mu }u+{{uT}}_{x}u+\nu {T}_{x}^{2}u]+\lambda {T}_{y}^{2}u=0,\end{eqnarray}$
where u = u(x, y, t), t ≥ 0, 0< μ ≤ 1.In this case, take the initial condition
$\begin{eqnarray}f(x,y)={a}_{0}+\displaystyle \frac{2k\nu }{1+{kx}+{wy}},\end{eqnarray}$
that is deduced from the exact solution equation ( $\begin{eqnarray}u(x,y,t)=f(x,y)+\mathop{\sum _{n=1}}\limits^{\infty }{f}_{n}(x,y)\displaystyle \frac{{t}^{n\mu }}{{\mu }^{n}n!},\end{eqnarray}$
and k − th truncated series of u(x, y, t) $\begin{eqnarray}{u}_{k}(x,y,t)=f(x,y)+\mathop{\sum _{n=1}}\limits^{k}{f}_{n}(x,y)\displaystyle \frac{{t}^{n\mu }}{{\mu }^{n}n!},\ k=1,2,3,\ldots \end{eqnarray}$
the general form of the k − th residual functions of the time-fractional Burgers-K-P equation can be expressed by $\begin{eqnarray}\begin{array}{rcl}\mathrm{Res}{u}_{k}(x,y,t) & = & {\left({\partial }_{t}^{\mu }{u}_{k}\right)}_{x}+\lambda {\left({u}_{1}(x,y,t\right)}_{{yy}}\\ & & +\ \nu {\left({u}_{1}(x,y,t\right)}_{{xxx}}+{\left({u}_{1}(x,y,t\right)}_{x}^{2}\\ & & +\ {u}_{1}(x,y,t){\left({u}_{1}(x,y,t\right)}_{{xx}}.\end{array}\end{eqnarray}$
To determine the parameter f1(x, y), in u1(x, y, t), one should replace the 1st truncated series ${u}_{1}(x,y,t)\,=f(x,y)+{f}_{1}(x,y)\tfrac{t\mu }{\mu }$ into the 1st truncated residual function to obtain
$\begin{eqnarray}\begin{array}{rcl}\mathrm{Res}{u}_{1}(x,y,t) & = & {\left({f}_{1}(x,y\right)}_{x}+\lambda \left({\left(f(x,y\right)}_{{yy}}\right.\\ & & +\ \left.\displaystyle \frac{{t}^{\mu }{\left({f}_{1}(x,y\right)}_{{yy}}}{\mu }\right)\\ & & +\ \nu \left({\left(f(x,y\right)}_{{xxx}}+\displaystyle \frac{{t}^{\mu }f{\left({}_{1}(x,y\right)}_{{xxx}}}{\mu }\right)\\ & & +\ \left({\left(f(x,y\right)}_{x}+\displaystyle \frac{{t}^{\mu }{\left({f}_{1}(x,y\right)}_{x}}{\mu }\right){}^{2}\\ & & +\ \left(f(x,y)+\displaystyle \frac{{t}^{\mu }{f}_{1}(x,y)}{\mu }\right)\\ & & \times \ \left({\left(f(x,y\right)}_{{xx}}+\displaystyle \frac{{t}^{\mu }{\left({f}_{1}(x,y\right)}_{{xx}}}{\mu }\right).\end{array}\end{eqnarray}$
Next, by substitution of t = 0 by $\mathrm{Res}{u}_{1}(x,y,t)$, we obtain $\begin{eqnarray}\begin{array}{rcl}{\left({f}_{1}(x,y\right)}_{x} & = & -\lambda {\left(f(x,y\right)}_{{yy}}-\nu {\left(f(x,y\right)}_{{xxx}}\\ & & -\ {\left(f(x,y\right)}_{x}^{2}-f(x,y){\left(f(x,y\right)}_{{xx}}.\end{array}\end{eqnarray}$
Solving this differential equation and evaluating with the initial condition $f(x,y)$ gives the first unknown parameter as $\begin{eqnarray}{f}_{1}(x,y)=\displaystyle \frac{2\nu \left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}{{\left({kx}+l+{wy}\right)}^{2}}.\end{eqnarray}$
Thus, we calculate the first RPSM solution of the time-fractional Burgers-K-P equation as $\begin{eqnarray}{u}_{1}(x,y,t)={a}_{0}+\displaystyle \frac{2k\nu }{{kx}+l+{wy}}+\displaystyle \frac{2\nu {t}^{\mu }\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}{\mu {\left({kx}+l+{wy}\right)}^{2}}.\end{eqnarray}$
Similarly, to calculate f2(x, y), which is the second unknown coefficient, we replace the second truncated series solution ${u}_{2}(x,y,t)=f(x,y)+{f}_{1}(x,y)\tfrac{{t}^{\mu }}{\mu }+{f}_{2}(x,y)\tfrac{{t}^{2\mu }}{2{\mu }^{2}}$ into the second truncated residual function to attain
$\begin{eqnarray}\begin{array}{l}\mathrm{Res}{u}_{2}(x,y,t)={\left({f}_{1}(x,y\right)}_{x}+\displaystyle \frac{{t}^{\mu }{\left({f}_{2}(x,y\right)}_{x}}{\mu }\\ \quad +\,\lambda \left({\left(f(x,y\right)}_{{yy}}+\displaystyle \frac{{t}^{2\mu }{\left({f}_{2}(x,y\right)}_{{yy}}}{2{\mu }^{2}}+\displaystyle \frac{{t}^{\mu }{\left({f}_{1}(x,y\right)}_{{yy}}}{\mu }\right)\\ \quad +\,\nu \left({\left(f(x,y\right)}_{{xxx}}+\displaystyle \frac{{t}^{2\mu }{\left({f}_{2}(x,y\right)}_{{xxx}}}{2{\mu }^{2}}+\displaystyle \frac{{t}^{\mu }{\left({f}_{1}(x,y\right)}_{{xxx}}}{\mu }\right)\\ \quad +\,\left({\left(f(x,y\right)}_{x}+\displaystyle \frac{{t}^{2\mu }{\left({f}_{2}(x,y\right)}_{x}}{2{\mu }^{2}}+\displaystyle \frac{{t}^{\mu }{\left({f}_{1}(x,y\right)}_{x}}{\mu }\right){}^{2}\\ \quad +\,\left(f(x,y)+\displaystyle \frac{{t}^{2\mu }{f}_{2}(x,y)}{2{\mu }^{2}}+\displaystyle \frac{{t}^{\mu }{f}_{1}(x,y)}{\mu }\right)\\ \quad \times \,\left({\left(f(x,y\right)}_{{xx}}+\displaystyle \frac{{t}^{2\mu }{\left({f}_{2}(x,y\right)}_{{xx}}}{2{\mu }^{2}}+\displaystyle \frac{{t}^{\mu }{\left({f}_{1}(x,y\right)}_{{xx}}}{\mu }\right).\end{array}\end{eqnarray}$
Next, applying the conformable derivative Tμ on both sides of $\mathrm{Res}{u}_{2}(x,y,t)$ and solving for t = 0 yields
$\begin{eqnarray}\begin{array}{l}{\left({f}_{2}(x,y\right)}_{x}=-2{\left(f(x,y\right)}_{x}{\left({f}_{1}(x,y\right)}_{x}-{f}_{1}(x,y){\left(f(x,y\right)}_{{xx}}\\ \ \ -\ \lambda {\left({f}_{1}(x,y\right)}_{{yy}}-\nu {\left({f}_{1}(x,y\right)}_{{xxx}}-f(x,y){\left({f}_{1}(x,y\right)}_{{xx}}.\end{array}\end{eqnarray}$
Solving this differential equation and evaluating with the initial condition $f(x,y)$ and first unknown parameter f1(x, y) gives
$\begin{eqnarray}{f}_{2}(x,y)=\displaystyle \frac{4\nu {\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{2}}{k{\left({kx}+l+{wy}\right)}^{3}}.\end{eqnarray}$
So, the second RPSM approximation of the equation is $\begin{eqnarray}\begin{array}{rcl}{u}_{2}(x,y,t) & = & {a}_{0}+\displaystyle \frac{2k\nu }{{kx}+l+{wy}}+\displaystyle \frac{2\nu {t}^{2\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{2}}{k{\mu }^{2}{\left({kx}+l+{wy}\right)}^{3}}\\ & & +\displaystyle \frac{2\nu {t}^{\mu }\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}{\mu {\left({kx}+l+{wy}\right)}^{2}}.\end{array}\end{eqnarray}$
Similarly, following the above scheme for n = 3 and 4 gives the subsequent outcomes.
$\begin{eqnarray}{f}_{3}(x,y)=\displaystyle \frac{12\nu {\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{3}}{{k}^{2}{\left({kx}+l+{wy}\right)}^{4}}.\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{u}_{3}(x,y,t) & = & {a}_{0}+\displaystyle \frac{2k\nu }{{kx}+l+{wy}}+\displaystyle \frac{2\nu {t}^{3\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{3}}{{k}^{2}{\mu }^{3}{\left({kx}+l+{wy}\right)}^{4}}\\ & & +\displaystyle \frac{2\nu {t}^{2\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{2}}{k{\mu }^{2}{\left({kx}+l+{wy}\right)}^{3}}+\displaystyle \frac{2\nu {t}^{\mu }\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}{\mu {\left({kx}+l+{wy}\right)}^{2}}.\end{array}\end{eqnarray}$
$\begin{eqnarray}{f}_{4}(x,y)=\displaystyle \frac{48\nu {\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{4}}{{k}^{3}{\left({kx}+l+{wy}\right)}^{5}}.\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{u}_{4}(x,y,t) & = & {a}_{0}+\displaystyle \frac{2k\nu }{{kx}+l+{wy}}+\displaystyle \frac{2\nu {t}^{3\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{3}}{{k}^{2}{\mu }^{3}{\left({kx}+l+{wy}\right)}^{4}}\\ & & +\displaystyle \frac{2\nu {t}^{2\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{2}}{k{\mu }^{2}{\left({kx}+l+{wy}\right)}^{3}}+\displaystyle \frac{2\nu {t}^{\mu }\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}{\mu {\left({kx}+l+{wy}\right)}^{2}}\\ & & +\ \displaystyle \frac{2\nu {t}^{4\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{4}}{{k}^{3}{\mu }^{4}{\left({kx}+l+{wy}\right)}^{5}}.\end{array}\end{eqnarray}$
Figure 1. (Case 5.1) Comparison for a0 = 0.01, k = 0.1, w = 0.5, $\nu =0.01,$ λ = 0.9, l = 5, y = 1, t = 0.1 and μ = 0.35. |
As seen, the differences between approximate and exact solutions in figures 1(a), (b), 2(a), (b), 3(a) and (b) in different cases is not clearly obvious. Therefore, the errors should be computed for the accurate comparison of the numerical and exact solutions. Error results for μ = 0,35, μ = 0,50 and μ = 0,75 in the numerical solutions obtained by exact and the fourth-order RPSM solutions are presented in table 1. The following absolute error formula is used for the numerical solutions;
$\begin{eqnarray*}\mathrm{Error}=\left|u(x,y,t)-{u}_{n}(x,y,t)\right|.\end{eqnarray*}$
In the above formula, u(x, y, t) and un(x, y, t) represent the exact solution and the numerical solution at the point $(x,y,t)$ respectively.
Figure 2. (Case 5.1) Comparison for a0 = 0.01, k = 0.1, w = 0.5, $\nu =0.01,$ λ = 0.9, l = 5, y = 1, t = 0.1 and μ = 0.50. |
Figure 3. (Case 5.1) Comparison for a0 = 0.01, k = 0.1, w = 0.5, $\nu =0.01,$ λ = 0.9, l = 5, y = 1, t = 0.1 and μ = 0.75. |
Table 1. (Case 5.1) RPSM approximate results and comparison with the exact solutions by absolute errors for a0 = 0.01, k = 0.1, w = 0.5, $\nu =0.01,$ λ = 0.9, l = 5, y = 1 and t = 0.1. |
| μ = 0.35 | μ = 0.50 | μ = 0.75 | |||||||
|---|---|---|---|---|---|---|---|---|---|
| x | RPSM | Exact | Abs. Error | RPSM | Exact | Abs. Error | RPSM | Exact | Abs. Error |
| 0.0 | 0.010 731 7 | 0.010 761 3 | 2.959 80E-5 | 0.010 490 1 | 0.010 490 6 | 5.70129E-7 | 0.010 402 7 | 0.010 402 7 | 3.465 43E-9 |
| 0.1 | 0.010 729 2 | 0.010 758 4 | 2.921 92E-5 | 0.010 488 9 | 0.010 489 4 | 5.63592E-7 | 0.010 401 9 | 0.010 401 9 | 3.427 19E-9 |
| 0.2 | 0.010 726 7 | 0.010 755 5 | 2.884 61E-5 | 0.010 487 7 | 0.010 488 2 | 5.57142E-7 | 0.010 401 1 | 0.010 401 1 | 3.389 45E-9 |
| 0.3 | 0.010 724 2 | 0.010 752 7 | 2.847 87E-5 | 0.010 486 5 | 0.010 487 1 | 5.50778E-7 | 0.010 400 3 | 0.010 400 3 | 3.352 19E-9 |
| 0.4 | 0.010 721 7 | 0.010 749 9 | 2.811 67E-5 | 0.010 485 3 | 0.010 485 9 | 5.44499E-7 | 0.010 399 5 | 0.010 399 5 | 3.315 41E-9 |
| 0.5 | 0.010 719 3 | 0.010 747 1 | 2.776 03E-5 | 0.010 484 2 | 0.010 484 7 | 5.38304E-7 | 0.010 398 7 | 0.010 398 7 | 3.279 10E-9 |
| 0.6 | 0.010 716 9 | 0.010 744 3 | 2.740 91E-5 | 0.010 483 0 | 0.010 483 5 | 5.32190E-7 | 0.010 397 9 | 0.010 397 9 | 3.243 25E-9 |
| 0.7 | 0.010 714 5 | 0.010 741 5 | 2.706 33E-5 | 0.010 481 8 | 0.010 482 4 | 5.26158E-7 | 0.010 397 1 | 0.010 397 1 | 3.207 86E-9 |
| 0.8 | 0.010 712 1 | 0.010 738 8 | 2.672 26E-5 | 0.010 480 7 | 0.010 481 2 | 5.20206E-7 | 0.010 396 3 | 0.010 396 3 | 3.172 92E-9 |
| 0.9 | 0.010 709 7 | 0.010 736 1 | 2.638 69E-5 | 0.010 479 5 | 0.010 480 0 | 5.14332E-7 | 0.010 395 5 | 0.010 395 5 | 3.138 42E-9 |
| 1.0 | 0.010 707 3 | 0.010 733 4 | 2.605 63E-5 | 0.010 478 4 | 0.010 478 9 | 5.08535E-7 | 0.010 394 8 | 0.010 394 8 | 3.104 36E-9 |
For μ = 0.35 at x = 0, error is about $2.9E-5$ and that result is decreasing to $2.6E-5$ when x is increasing to 1.
For μ = 0.50, the results are about 200 times better than the results obtained when μ = 0.35. For μ = 0.75 at x = 0, error is about $3.4E-9$ and that results is decreasing to $3.1E-9$ when x is increasing to 1.
Now, consider the initial condition
$\begin{eqnarray}f(x,y)={a}_{0}+2k\nu \sqrt{-\sigma }\tanh \left({kx}+{wy})\sqrt{-\sigma }\right),\end{eqnarray}$
that is obtained from the exact solution equation ( $\begin{eqnarray}{f}_{1}(x,y)=2\nu \sigma \left({a}_{0}{k}^{2}+\lambda {w}^{2}\right){{\rm{sech}} }^{2}\left(\sqrt{-\sigma }({kx}+{wy}\right),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{u}_{1}(x,y,t) & = & \displaystyle \frac{2\nu \sigma {t}^{\mu }\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right){{\rm{sech}} }^{2}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{\mu }\\ & & +{a}_{0}+2k\nu \sqrt{-\sigma }\tanh \left(\sqrt{-\sigma }({kx}+{wy}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}{f}_{2}(x,y)=-\displaystyle \frac{4\nu {\sigma }^{2}{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{2}\tanh \left(\sqrt{-\sigma }({kx}+{wy}\right){{\rm{sech}} }^{2}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{k\sqrt{-\sigma }},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{u}_{2}(x,y,t) & = & \displaystyle \frac{2\nu \sigma {t}^{\mu }\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right){\sec }^{2}\left(\sqrt{\sigma }({kx}+{wy}\right)\left(k\mu -\sqrt{\sigma }{t}^{\mu }\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)\tan \left(\sqrt{\sigma }({kx}+{wy}\right)\right)}{k{\mu }^{2}}\\ & & +{a}_{0}-2k\nu \sqrt{\sigma }\tan \left(\sqrt{\sigma }({kx}+{wy}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}{f}_{3}(x,y)=-\displaystyle \frac{4\nu {\sigma }^{2}{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{3}\left(\cosh \left(2\sqrt{-\sigma }({kx}+{wy}\right)-2\right){{\rm{sech}} }^{4}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{{k}^{2}},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{u}_{3}(x,y,t) & = & -\displaystyle \frac{2\nu {\sigma }^{2}{t}^{3\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{3}\left(\cosh \left(2\sqrt{-\sigma }({kx}+{wy}\right)-2\right){{\rm{sech}} }^{4}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{3{k}^{2}{\mu }^{3}}\\ & & -\displaystyle \frac{2\nu {\sigma }^{2}{t}^{2\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{2}\tanh \left(\sqrt{-\sigma }({kx}+{wy}\right){{\rm{sech}} }^{2}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{k{\mu }^{2}\sqrt{-\sigma }}\\ & & +\displaystyle \frac{2\nu \sigma {t}^{\mu }\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right){{\rm{sech}} }^{2}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{\mu }\,+{a}_{0}+2k\nu \sqrt{-\sigma }\tanh \left(\sqrt{-\sigma }({kx}+{wy}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}{f}_{4}(x,y)=\displaystyle \frac{8\nu {\sigma }^{3}{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{4}\left(\cosh \left(2\sqrt{-\sigma }({kx}+{wy}\right)-5\right)\tanh \left(\sqrt{-\sigma }({kx}+{wy}\right){{\rm{sech}} }^{4}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{{k}^{3}\sqrt{-\sigma }},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{u}_{4}(x,y,t) & = & -\displaystyle \frac{2\nu {\sigma }^{2}{t}^{3\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{3}\left(\cosh \left(2\sqrt{-\sigma }({kx}+{wy}\right)-2\right){{\rm{sech}} }^{4}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{3{k}^{2}{\mu }^{3}}\\ & & -\displaystyle \frac{2\nu {\sigma }^{2}{t}^{2\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{2}\tanh \left(\sqrt{-\sigma }({kx}+{wy}\right){{\rm{sech}} }^{2}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{k{\mu }^{2}\sqrt{-\sigma }}\\ & & +\displaystyle \frac{2\nu \sigma {t}^{\mu }\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right){{\rm{sech}} }^{2}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{\mu }\\ & & +\displaystyle \frac{\nu {\sigma }^{3}{t}^{4\mu }{\left({a}_{0}{k}^{2}+\lambda {w}^{2}\right)}^{4}\left(\cosh \left(2\sqrt{-\sigma }({kx}+{wy}\right)-5\right)\tanh \left(\sqrt{-\sigma }({kx}+{wy}\right){{\rm{sech}} }^{4}\left(\sqrt{-\sigma }({kx}+{wy}\right)}{3{k}^{3}{\mu }^{4}\sqrt{-\sigma }}\\ & & +{a}_{0}+2k\nu \sqrt{-\sigma }\tanh \left(\sqrt{-\sigma }({kx}+{wy}\right).\end{array}\end{eqnarray}$
Besides, in figures 4–6, the graphical illustrations of the fourth order RPSM and exact solution $\begin{eqnarray}\begin{array}{rcl}u(x,y,t) & = & 2k\nu \sqrt{-\sigma }\tanh \left(\sqrt{-\sigma }\left(\displaystyle \frac{{t}^{\mu }\left(-{a}_{0}{k}^{2}-\lambda {w}^{2}\right)}{k\mu }\right.\right.\\ & & +\ \left.\left.{kx}+{wy}\right)\right)+{a}_{0},\end{array}\end{eqnarray}$
where a0, k, w, λ , σ and ν are constants. It is clear that they are almost identical in terms of accuracy and are in perfect agreement with each other.
Figure 4. (Case 5.2) Comparison for a0 = 0.01, k = 0.1, w = 0.5, $\nu =0.01,$ λ = 0.9, l = 5, y = 1, σ = −0.5, $t=0.1$ and μ = 0.45. |
Figure 5. (Case 5.2) Comparison for a0 = 0.01, k = 0.1, w = 0.5, $\nu =0.01,$ λ = 0.9, l = 5, y = 1, σ = −0.5, $t=0.1$ and μ = 0.55. |
Figure 6. (Case 5.2) Comparison for a0 = 0.01, k = 0.1, w = 0.5, $\nu =0.01,$ λ = 0.9, l = 5, y = 1, σ = −0.5, $t=0.1$ and μ = 0.75. |
Also, in table 2, the absolute errors are offered for μ = 0,45, μ = 0,55 and μ = 0,75 respectively. The outcomes reveal that absolute errors decrease with increasing x values, as they decrease with increasing μ values.
Table 2. (Case 5.2) RPSM approximate results and comparison with the exact solutions by absolute errors for a0 = 0.01, k = 0.1, w = 0.5, $\nu =0.01,$ λ = 0.9, l = 5, y = 1, σ = −0.5 and t = 0.1. |
| μ = 0.45 | μ = 0.55 | μ = 0.75 | |||||||
|---|---|---|---|---|---|---|---|---|---|
| x | RPSM | Exact | Abs. Error | RPSM | Exact | Abs. Error | RPSM | Exact | Abs. Error |
| 0.0 | 0.009 361 2 | 0.008 986 0 | 3.752 06E-4 | 0.009 428 7 | 0.009 389 3 | 3. 93727E-5 | 0.009 966 8 | 0.009 966 2 | 5.890 58E-7 |
| 0.1 | 0.009 358 7 | 0.008 990 8 | 3.678 53E-4 | 0.009 435 6 | 0.009 397 5 | 3.81430E-5 | 0.009 976 8 | 0.009 976 2 | 5.569 00E-7 |
| 0.2 | 0.009 356 1 | 0.008 995 8 | 3.602 89E-4 | 0.009 442 6 | 0.009 405 7 | 3.68946E-5 | 0.009 986 8 | 0.009 986 2 | 5.246 58E-7 |
| 0.3 | 0.009 353 3 | 0.009 000 8 | 3.525 22E-4 | 0.009 449 6 | 0.009 414 0 | 3.56291E-5 | 0.009 996 7 | 0.009 996 2 | 4.923 68E-7 |
| 0.4 | 0.009 350 3 | 0.009 005 8 | 3.445 62E-4 | 0.009 456 6 | 0.009 422 3 | 3.43478E-5 | 0.010 006 7 | 0.010 006 3 | 4.600 65E-7 |
| 0.5 | 0.009 347 3 | 0.009 010 9 | 3.364 17E-4 | 0.009 463 7 | 0.009 430 6 | 3.30521E-5 | 0.010 016 7 | 0.010 016 3 | 4.277 82E-7 |
| 0.6 | 0.009 344 1 | 0.009 016 0 | 3.280 97E-4 | 0.009 470 8 | 0.009 439 0 | 3.17437E-5 | 0.010 026 7 | 0.010 026 3 | 3.955 54E-7 |
| 0.7 | 0.009 340 8 | 0.009 021 2 | 3.196 12E-4 | 0.009 477 9 | 0.009 447 5 | 3.04238E-5 | 0.010 036 6 | 0.010 036 3 | 3.634 15E-7 |
| 0.8 | 0.009 337 4 | 0.009 026 4 | 3.109 72E-4 | 0.009 485 1 | 0.009 456 0 | 2.90941E-5 | 0.010 046 6 | 0.010 046 3 | 3.313 97E-7 |
| 0.9 | 0.009 333 9 | 0.009 031 7 | 3.021 85E-4 | 0.009 492 3 | 0.009 464 5 | 2.77559E-5 | 0.010 056 5 | 0.010 056 2 | 2.995 31E-7 |
| 1.0 | 0.009 330 3 | 0.009 037 0 | 2.932 61E-4 | 0.009 499 5 | 0.009 473 1 | 2.64106E-5 | 0.010 066 5 | 0.010 066 2 | 2.678 50E-7 |
6. Conclusion
In this paper, two reliable methods, namely the residual power series method and the sub-equation method, have been implemented to the time-fractional Burgers-K-P partial differential equation which arise in the shallow water waves. Using conformable derivative definition, numerous new exact solutions of the equation have been obtained which do not exist in the literature. Then approximate solutions obtained by RPSM have been compared with these exact solutions by surface plots and tables. Thus, it is shown that the methods are very effective and could be used for different types of FPDEs arising in different areas of mathematical physics.