# Fractal-fractional advection–diffusion–reaction equations by Ritz approximation approach

Fractal-fractional advection–diffusion–reaction equations by Ritz approximation approach 1IntroductionAdvection–diffusion–reaction equation is an important class of partial differential equations that have been used to model various physical processes, such as transport dissipative particle dynamics model for simulating mesoscopic problems [1], in anisotropic media [2], transport of chemical constituents in the Earth’s atmosphere [3], bimolecular chemical reactions [4], Brusselator system [5], rubella epidemic [6], tuberculosis transmission modeling [7], COVID-19 mathematical modeling [8] and many more.This advection–diffusion–reaction equation has been extended to include the fractional derivative, which is called the fractional advection–diffusion–reaction equation. The researchers found that the fractional derivative model was able to describe the transport problems in Earth surface sciences, which include collective behavior of particles in transport [9], heat and mass transfer [10], and the study of the dynamics of cytosolic calcium ion in astrocytes [11]. In this research direction, Caputo fractional derivative is the most common fractional derivative that has been used in this fractional advection–diffusion–reaction equation.Different from most established work, in this work, we intend to study the fractional advection–diffusion–reaction equation in the fractal-fractional sense as follows: (1)Dtα,βFFMu(x,t)=κ1Dxm1u(x,t)+κ2Dxm2u(x,t)+Φ(u(x,t))+f(x,t),0<x,t≤1,u(x,0)=f0(x),ut(x,0)=f1(x),u(0,t)=g0(t),u(1,t)=g1(t),0≤x≤1,\begin{array}{rcl}{}^{{\rm{FFM}}}D_{t}^{\alpha ,\beta }u\left(x,t)& =& {\kappa }_{1}{D}_{x}^{{m}_{1}}u\left(x,t)+{\kappa }_{2}{D}_{x}^{{m}_{2}}u\left(x,t)\\ & & +\Phi \left(u\left(x,t))+f\left(x,t),\hspace{0.33em}0\lt x,\hspace{0.33em}t\le 1,\\ u\left(x,0)& =& {f}_{0}\left(x),\hspace{1em}{u}_{t}\left(x,0)={f}_{1}\left(x),\\ u\left(0,t)& =& {g}_{0}\left(t),\hspace{1em}u\left(1,t)={g}_{1}\left(t),\hspace{0.33em}0\le x\le 1,\end{array}where 0<m1≤10\lt {m}_{1}\le 1, 1<m2≤21\lt {m}_{2}\le 2, κ1{\kappa }_{1}, and κ2>0{\kappa }_{2}\gt 0are the characteristics speed in advection process and diffusion coefficient, respectively. Dtα,βFFM{}^{{\rm{FFM}}}D_{t}^{\alpha ,\beta }is the fractal-fractional differentiation operator of order (α,β)\left(\alpha ,\beta )(where 0<α,β≤10\lt \alpha ,\beta \le 1) with respect to ttin the sense of Atangana–Riemann–Liouville [12,13], and Φ(u(x,t))\Phi \left(u\left(x,t))is some reasonable nonlinear function of u(x,t)u\left(x,t), which may represent reaction process. The fractional derivatives for Dxm1{D}_{x}^{{m}_{1}}and Dxm2{D}_{x}^{{m}_{2}}are defined in Caputo sense. However, to solve the problem as in ref. [14], one can define the Dxm1{D}_{x}^{{m}_{1}}as Dxm1ABC{}^{{\rm{ABC}}}D_{x}^{{m}_{1}}, where ABC denotes Atangana–Baleanu Caputo derivative. This fractional advection–diffusion–reaction equation in the fractal-fractional sense was proposed in ref. [14], while the fractional advection–diffusion–reaction equation’s specific applications are discussed in refs [15,16,17] and the real-world interpretations of this fractal-fractional application are discussed in refs [18,19, 20,21].The fractal-fractional derivatives have been found very useful in many science and engineering applications, such as in modeling anomalous diffusion processes [22]. Researchers found that phenomena that are inherent in abnormal exponential or the phenomena with heavy tail decay processes are best described in the fractal-fractional derivatives [23]. Recently, this combination of fractal-fractional derivative was again shown by Atangana [12,13] that this kind of derivative takes into account not only the memory effect but also other characteristics, such as the heterogeneity, elasco-viscosity of the medium, and the fractal geometry of the dynamic system. In this research direction, fractal-fractional derivatives have been used in many phenomena, such as reaction-diffusion model [18], modeling bank data [19], Shinriki’s oscillator model [20], and malaria transmission model [21].On top of that, numerical methods are always needed to solve fractional calculus problems that arise in engineering applications [24,25,26, 27,28]. Furthermore, the differential equations that arise from the modeling process, especially in science and engineering applications involving fractal-fractional operators, are often very complex, especially when we intend to obtain their analytical solution. Hence, numerical methods are more applicable and suitable for solving fractal-fractional differential equations. Some numerical methods have been derived to tackle this problem, such as Chebyshev polynomials for solving the model of the nonlinear Ginzburg–Landau equation in a fractal-fractional sense [29], the Crank-Nicolson finite difference scheme is extended to solve the fractal-fractional Boussinesq equation [30], and a numerical method based on the Lagrangian piece-wise interpolation is used to obtain the solution of variable-order fractal-fractional time delay equations [31]. Besides that, the wavelet-based approximation method was used for solving the coupled nonlinear 2D Schrödinger equations in a fractal-fractional sense [32]. Different from the existing methods for solving fractal-fractional differential equations, here we extend the Ritz method to obtain the solution of these fractal-fractional differential equations. More specifically, we use the Ritz approximation approach to solve the fractal-fractional advection–diffusion–reaction equation as in Eq. (1). The approach is very easy to use and able to give high accuracy numerical solutions.The Ritz approximation approach had been used in solving considerable problems previously. Among those, Rashedi et al. used the Ritz-Galerkin approach to handle inverse wave problem [33]. Apart from this, the satisfier function was used in Ritz–Galerkin for the identification of a time-dependent diffusivity [34], the Ritz approximation has also been applied to solve some fractional partial differential equations [35,36]. Exact and approximation solutions of the heat equation with nonlocal boundary conditions were found in ref. [37] using the Ritz–Galerkin method with Bernoulli polynomials as the basis. Besides that, Genocchi polynomials had been used in the Ritz–Galerkin method for solving the fractional Klein–Gordon equation and fractional diffusion wave equation [38]. Here, for the first time, we propose to use this Ritz approximation approach to solve the fractal-fractional advection–diffusion–reaction equation as in Eq. (1). Apart from this, two variables-shifted Legendre polynomials, which were derived by Khan and Singh [39], will be used. This is different from two-dimensional-shifted Legendre polynomials. In short, the main objective of this article is to solve the fractal-fractional advection–diffusion–reaction equation using the Ritz approximation via two variables-shifted Legendre polynomials.The rest of the article is organized as follows: Section 2 provides the basic definitions and notations for fractal-fractional derivative and two variables-shifted Legendre polynomials. Section 3 presents the main tool that is used in this article, which is the Ritz approximation and satisfier function for solving fractal-fractional advection–diffusion–reaction equations. Error analysis will also be presented. Section 4 gives some numerical examples to show that the approach is better than some existing methods. Section 5 gives summary and recommendation for future work.2Preliminaries2.1Fractal-fractional derivativeIn this section, we will briefly present some basic definitions related to fractal-fractional derivative.Definition 1The Mittag–Leffler function (2)Ea(x)=∑k=0∞xkΓ(ak+1),whereRe(a)>0,Ea,b(x)=∑k=0∞xkΓ(ak+b),whereRe(a),Re(b)>0.\begin{array}{rcl}{E}_{a}\left(x)& =& \mathop{\displaystyle \sum }\limits_{k=0}^{\infty }\frac{{x}^{k}}{\Gamma \left(ak+1)},\hspace{1em}\hspace{0.1em}\text{where}\hspace{0.1em}\hspace{0.33em}{\rm{Re}}\left(a)\gt 0,\\ {E}_{a,b}\left(x)& =& \mathop{\displaystyle \sum }\limits_{k=0}^{\infty }\frac{{x}^{k}}{\Gamma \left(ak+b)},\hspace{0.1em}\text{where}\hspace{0.1em}\hspace{0.33em}{\rm{Re}}\left(a),{\rm{Re}}\left(b)\gt 0.\end{array}Definition 2The fractal-fractional derivative of order (α,β)\left(\alpha ,\beta )of the continuous function f(x)f\left(x)in the Atangana–Riemann–Liouville sense with Mittag–Leffler kernel [12] is given as follows: (3)Dxα,βFFMf(x)=M(α)1−αddxβ∫x0xf(t)Eα−α(x−t)α1−αdt=M(α)x1−ββ(1−α)ddx∫x0xf(t)Eα−α(x−t)α1−αdt,\begin{array}{l}{}^{{\rm{FFM}}}D_{x}^{\alpha ,\beta }f\left(x)=\frac{M\left(\alpha )}{1-\alpha }\frac{{\rm{d}}}{{\rm{d}}{x}^{\beta }}\underset{{x}_{0}}{\overset{x}{\displaystyle \int }}f\left(t){E}_{\alpha }\left(\frac{-\alpha {\left(x-t)}^{\alpha }}{1-\alpha }\right){\rm{d}}t\\ \hspace{5.2em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\underset{{x}_{0}}{\overset{x}{\displaystyle \int }}f\left(t){E}_{\alpha }\left(\frac{-\alpha {\left(x-t)}^{\alpha }}{1-\alpha }\right){\rm{d}}t,\end{array}where α,β∈(0,1)\alpha ,\beta \in \left(0,1), M(α)M\left(\alpha )is a normalization function with M(0)=M(1)=1M\left(0)=M\left(1)=1, and M(α)=1−α+αΓ(α)M\left(\alpha )=1-\alpha +\frac{\alpha }{\Gamma \left(\alpha )}.Lemma 1The fractal-fractional derivative of order 0<α,β<10\lt \alpha ,\beta \lt 1for f(x)=xpf\left(x)={x}^{p}can be defined as follows: (4)Dxα,βFFMxp=M(α)β(1−α)∑i=0∞(−1)iαip!xαi+p+1−β(1−α)iΓ(iα+p+1).{}^{{\rm{FFM}}}D_{x}^{\alpha ,\beta }{x}^{p}=\frac{M\left(\alpha )}{\beta \left(1-\alpha )}\mathop{\sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}p\&#x0021;{x}^{\alpha i+p+1-\beta }}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +p+1)}.ProofBy using Definition in Eqs. (2) and (3), (5)Dxα,βFFMxp=M(α)x1−ββ(1−α)ddx∫0xtp∑i=0∞−α1−αi(x−t)αiΓ(iα+1)dt=M(α)x1−ββ(1−α)ddx∫0x∑i=0∞(−1)itpαi(x−t)αi(1−α)iΓ(iα+1)dt=M(α)x1−ββ(1−α)ddx∑i=0∞(−1)iαi∫0xtp(x−t)αidt(1−α)iΓ(iα+1),\begin{array}{l}{}^{{\rm{FFM}}}D_{x}^{\alpha ,\beta }{x}^{p}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\underset{0}{\overset{x}{\displaystyle \int }}{t}^{p}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(\frac{-\alpha }{1-\alpha }\right)}^{i}{\left(x-t)}^{\alpha i}}{\Gamma \left(i\alpha +1)}{\rm{d}}t\\ \hspace{4.2em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\underset{0}{\overset{x}{\displaystyle \int }}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{t}^{p}{\alpha }^{i}{\left(x-t)}^{\alpha i}}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +1)}{\rm{d}}t\\ \hspace{4.2em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}{\displaystyle \int }_{0}^{x}{t}^{p}{\left(x-t)}^{\alpha i}{\rm{d}}t}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +1)},\end{array}and knowing that ∫0xtp(x−t)αidt=Γ(αi+1)p!xαi+p+1Γ(αi+p+2){\int }_{0}^{x}{t}^{p}{\left(x-t)}^{\alpha i}{\rm{d}}t=\frac{\Gamma \left(\alpha i+1)p\&#x0021;{x}^{\alpha i+p+1}}{\Gamma \left(\alpha i+p+2)}, we obtain (6)Dxα,βFFMxp=M(α)x1−ββ(1−α)ddx∑i=0∞(−1)iαiΓ(αi+1)p!xαi+p+1Γ(αi+p+2)(1−α)iΓ(iα+1)=M(α)x1−ββ(1−α)ddx∑i=0∞(−1)iαip!xαi+p+1(1−α)iΓ(iα+p+2)=M(α)x1−ββ(1−α)∑i=0∞(−1)iαip!(αi+p+1)xαi+p(1−α)iΓ(iα+p+2)=M(α)β(1−α)∑i=0∞(−1)iαip!xαi+p+1−β(1−α)iΓ(iα+p+1).□\begin{array}{l}{}^{{\rm{FFM}}}D_{x}^{\alpha ,\beta }{x}^{p}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}\frac{\Gamma \left(\alpha i+1)p\&#x0021;{x}^{\alpha i+p+1}}{\Gamma \left(\alpha i+p+2)}}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +1)}\\ \hspace{4.3em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}p\&#x0021;{x}^{\alpha i+p+1}}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +p+2)}\\ \hspace{4.25em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}p\&#x0021;\left(\alpha i+p+1){x}^{\alpha i+p}}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +p+2)}\\ \hspace{4.25em}=\frac{M\left(\alpha )}{\beta \left(1-\alpha )}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}p\&#x0021;{x}^{\alpha i+p+1-\beta }}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +p+1)}.\hspace{1.5em}\square \end{array}Eq. (6) can also be written in Mittag–Leffler form, as shown in ref. [29].2.2Two variables-shifted Legendre polynomialsThe Legendre polynomials, Ln(x){L}_{n}\left(x), can be defined as the coefficients in a formal expansion in powers of ttof the generating function as follows: (7)11−2xt+t2=∑n=0∞Ln(x)tn.\frac{1}{\sqrt{1-2xt+{t}^{2}}}=\mathop{\sum }\limits_{n=0}^{\infty }{L}_{n}\left(x){t}^{n}.There are few different ways for the extension of this Legendre polynomials Ln(x,y){L}_{n}\left(x,y)(or shifted Legendre polynomials, Pn(x,y){P}_{n}\left(x,y)) in several variables, such as (1)Two-variable Legendre polynomials [40](8)11−2xt+yt2=∑n=0∞Ln(x,y)tn.\frac{1}{\sqrt{1-2xt+y{t}^{2}}}=\mathop{\sum }\limits_{n=0}^{\infty }{L}_{n}\left(x,y){t}^{n}.(2)Two-dimensional-shifted Legendre polynomials are defined as [41,42] follows:(9)Pn(x,y)=Pn(x)Pk(y).{P}_{n}\left(x,y)={P}_{n}\left(x){P}_{k}(y).(3)Two variables Legendre polynomials [39](10)11−2xs+s2−2yt+t2=∑n=0∞∑k=0∞Ln,k(x,y)sntk.\frac{1}{\sqrt{1-2xs+{s}^{2}-2yt+{t}^{2}}}=\mathop{\sum }\limits_{n=0}^{\infty }\mathop{\sum }\limits_{k=0}^{\infty }{L}_{n,k}\left(x,y){s}^{n}{t}^{k}.The significant advantage for the two variables Legendre polynomials derived by Khan and Singh [39] in 2010 is that the polynomials can be obtained via generating function in two variables form as shown in Eq. (10). Here, we apply the definition of two variables Legendre polynomials derived by Khan and Singh [39] to define the two variables-shifted Legendre polynomials, Pn,k(x,y){P}_{n,k}\left(x,y), as follows: (11)11−2(2x−1)s+s2−2(2y−1)t+t2=∑n=0∞∑k=0∞Pn,k(x,y)sntk.\begin{array}{l}\frac{1}{\sqrt{1-2\left(2x-1)s+{s}^{2}-2\left(2y-1)t+{t}^{2}}}\\ \hspace{1.0em}=\mathop{\displaystyle \sum }\limits_{n=0}^{\infty }\mathop{\displaystyle \sum }\limits_{k=0}^{\infty }{P}_{n,k}\left(x,y){s}^{n}{t}^{k}.\end{array}As discussed in Section 3 of ref. [39], we have the following analytical expression for two variables-shifted Legendre polynomials.(12)Pn,k(x,y)=∑r=0⌈n2⌉∑j=0k2(1∕2)n+k−r−j(4x−2)n−2r(4y−2)k−2j(−1)r+jr!j!(n−2r)!(k−2j)!,\begin{array}{l}{P}_{n,k}\left(x,y)\\ \hspace{1.0em}=\mathop{\displaystyle \sum }\limits_{r=0}^{\lceil \frac{n}{2}\rceil }\mathop{\displaystyle \sum }\limits_{j=0}^{&#x2308;\frac{k}{2}&#x2309;}\frac{{\left(1/2)}_{n+k-r-j}{\left(4x-2)}^{n-2r}{\left(4y-2)}^{k-2j}{\left(-1)}^{r+j}}{r\&#x0021;j\&#x0021;\left(n-2r)\&#x0021;\left(k-2j)\&#x0021;},\end{array}where (1∕2)n+k−r−j{\left(1/2)}_{n+k-r-j}denotes the falling factorial. When k = 0, we have Pn,0(x,y)=Pn(x){P}_{n,0}\left(x,y)={P}_{n}\left(x), where Pn(x) is the well known shifted Legendre polynomials. These two variables-shifted Legendre polynomials also can be expressed in terms of hypergeometric function as follows: (13)Pn,k(x,y)=2n+k(1∕2)n+k(2x−1)n(2y−1)kn!k!×F1,0,00,1,1−;−n2,1−n2;−k2,1−k2;1−2n−2k2;−;−;1(2x−1)21(2y−1)2.\begin{array}{l}{P}_{n,k}\left(x,y)\\ \hspace{1.0em}=\frac{{2}^{n+k}{\left(1/2)}_{n+k}{\left(2x-1)}^{n}{\left(2y-1)}^{k}}{n\&#x0021;k\&#x0021;}\\ \hspace{2.0em}\times {F}_{1,0,0}^{0,1,1}\left[\begin{array}{ccc}-;& \frac{-n}{2},\frac{1-n}{2};& \frac{-k}{2},\frac{1-k}{2};\\ \frac{1-2n-2k}{2};& -;& -;\end{array}\frac{1}{{\left(2x-1)}^{2}}\frac{1}{{\left(2y-1)}^{2}}\right].\end{array}For function f(x,y)=xaybf\left(x,y)={x}^{a}{y}^{b}with aaand bbas positive integers, it is easy to see that it can be obtained via (14)f(x,y)=xayb=1(a+1)(b+1)∑n=0∞∑k=0∞cn,kPn,k(x,y),f\left(x,y)={x}^{a}{y}^{b}=\frac{1}{\left(a+1)\left(b+1)}\mathop{\sum }\limits_{n=0}^{\infty }\mathop{\sum }\limits_{k=0}^{\infty }{c}_{n,k}{P}_{n,k}\left(x,y),where (15)cn,k=∫01∫01f(x,y)Pn,k(x,y)dxdy∫01∫01Pn,k(x,y)Pn,k(x,y)dxdy.{c}_{n,k}=\frac{{\int }_{0}^{1}{\int }_{0}^{1}f\left(x,y){P}_{n,k}\left(x,y){\rm{d}}x{\rm{d}}y}{{\int }_{0}^{1}{\int }_{0}^{1}{P}_{n,k}\left(x,y){P}_{n,k}\left(x,y){\rm{d}}x{\rm{d}}y}.3Ritz method3.1Ritz approximation approach and satisfier function for fractal-fractional advection–diffusion–reaction equationIn this Ritz approximation approach, we use two variables-shifted Legendre polynomials Pp,q(x,t){P}_{p,q}\left(x,t), as explained in Section 2 and the approximate solution u(x,t)u\left(x,t)for fractal-fractional advection–diffusion–reaction Eq. (1) is denoted as u˜(x,t)\tilde{u}\left(x,t). Hence, we have (16)u˜(x,t)=∑p=0M∑q=0MKpqωpq(x,t)+ζ(x,t),(x,t)∈[0,L]×[0,T],\begin{array}{rcl}\tilde{u}\left(x,t)& =& \mathop{\displaystyle \sum }\limits_{p=0}^{M}\mathop{\displaystyle \sum }\limits_{q=0}^{M}{K}_{pq}{\omega }_{pq}\left(x,t)+\zeta \left(x,t),\\ & & \left(x,t)\in \left[0,L]\times \left[0,T],\end{array}where ωpq(x,t)=x(x−L)t2Pp,q(x,t){\omega }_{pq}\left(x,t)=x\left(x-L){t}^{2}{P}_{p,q}\left(x,t)and the satisfier function is represented by ζ(x,t)\zeta \left(x,t). Pp,q(x,t){P}_{p,q}\left(x,t)denotes two variables-shifted Legendre polynomials, and Kpq{K}_{pq}is the coefficient that needs to be calculated. The main purpose of using the satisfier function is that the satisfier function satisfies all the initial and boundary conditions.Based on the initial and boundary condition as in Eq. (1), the satisfier equation, ζ(x,t)\zeta \left(x,t)can be found using the following procedure:Step 1: First, we find (17)φ(x,t)=1−xLg0(t)+xLg1(t).\varphi \left(x,t)=\left(1-\frac{x}{L}\right){g}_{0}\left(t)+\frac{x}{L}{g}_{1}\left(t).Step 2: Next, we determine (18)F0(x)=f0(x)−φ(x,0),F1(x)=f1(x)−φt(x,0).\begin{array}{rcl}{F}_{0}\left(x)& =& {f}_{0}\left(x)-\varphi \left(x,0),\\ {F}_{1}\left(x)& =& {f}_{1}\left(x)-{\varphi }_{t}\left(x,0).\end{array}Step 3: Calculate (19)R(x,t)=F0(x)+tF1(x).R\left(x,t)={F}_{0}\left(x)+t{F}_{1}\left(x).Step 4: Finally, we determine (20)ζ(x,t)=R(x,t)+φ(x,t).\zeta \left(x,t)=R\left(x,t)+\varphi \left(x,t).Furthermore, the coefficients Kpq{K}_{pq}in Eq. (16) can also be calculated by using the following inner product as follows: (21)⟨F(u˜),Pp,q(x,t)⟩=0,\langle F\left(\tilde{u}),{P}_{p,q}\left(x,t)\rangle =0,where (22)F(u˜)=Dtα,βFFMu˜(x,t)−κ1Dxm1u˜(x,t)−κ2Dxm2u˜(x,t)−Φ(u˜(x,t))−f(x,t)\begin{array}{rcl}F\left(\tilde{u})& =& {}^{{\rm{FFM}}}D_{t}^{\alpha ,\beta }\tilde{u}\left(x,t)-{\kappa }_{1}{D}_{x}^{{m}_{1}}\tilde{u}\left(x,t)\\ & & -{\kappa }_{2}{D}_{x}^{{m}_{2}}\tilde{u}\left(x,t)-\Phi \left(\tilde{u}\left(x,t))-f\left(x,t)\end{array}and (23)⟨F(u˜),Pp,q(x,t)⟩=∫0L∫0TF(u˜)Pp,q(x,t)dtdx,\langle F\left(\tilde{u}),{P}_{p,q}\left(x,t)\rangle =\underset{0}{\overset{L}{\int }}\underset{0}{\overset{T}{\int }}F\left(\tilde{u}){P}_{p,q}\left(x,t){\rm{d}}t{\rm{d}}x,where Pp,q(x,t){P}_{p,q}\left(x,t)are the two variables-shifted Legendre polynomials. A linear system of equations can be formed by using Eq. (21). Solving this linear system, we can obtain the entries of Kpq{K}_{pq}where p=0,…,M,andq=0,…,Mp=0,\ldots ,M,andq=0,\ldots ,M. Hence, by putting the value obtained for Kpq{K}_{pq}into Eq. (16), we can obtain the approximate solution for fractal-fractional advection–diffusion–reaction equation in (1).3.2Error analysisLemma 2Let the solution of fractal-fractional advection–diffusion–reaction equation is u(x,t)u\left(x,t), where u(x,t)∈Cm+1[0,1]×[0,1]u\left(x,t)\in {C}^{m+1}\left[0,1]\times \left[0,1], suppose Y=span{P0,m(x),P1,m(x),…,Pm,m(x)}⊂L2[0,1]Y=\hspace{0.1em}\text{span}\hspace{0.1em}\left\{{P}_{0,m}\left(x),{P}_{1,m}\left(x),\ldots ,{P}_{m,m}\left(x)\right\}\subset {L}^{2}\left[0,1]and Y′=span{P0,m(t),P1,m(t),…,Pm,m(t)}⊂L2[0,1]Y^{\prime} =\hspace{0.1em}\text{span}\hspace{0.1em}\left\{{P}_{0,m}\left(t),{P}_{1,m}\left(t),\ldots ,{P}_{m,m}\left(t)\right\}\subset {L}^{2}\left[0,1]. We have um(x,t)=Y×Y′{u}_{m}\left(x,t)=Y\times Y^{\prime} is the best approximation of u(x,t)u\left(x,t)by means of two variables-shifted Legendre polynomials, the error bound is given as follows: (24)‖u(x,t)−um(x,t)‖2≤M(m+1)!22m+1(2m+3)(m+2).\Vert u\left(x,t)-{u}_{m}\left(x,t){\Vert }_{2}\le \frac{M}{\left(m+1)\&#x0021;}\frac{\sqrt{2}{2}^{m+1}}{\sqrt{\left(2m+3)\left(m+2)}}.ProofBy using the Taylor series, we have (25)u(x,t)=∑j=0∞1j!(x−a)∂∂x+(t−b)∂∂tju(x,t)u\left(x,t)=\mathop{\sum }\limits_{j=0}^{\infty }\frac{1}{j\&#x0021;}{\left[\left(x-a)\frac{\partial }{\partial x}+\left(t-b)\frac{\partial }{\partial t}\right]}^{j}u\left(x,t)For simplicity, let a=b=0a=b=0, and in practical, we estimate u(x,t)u\left(x,t)up to mmorder as follows: (26)um(x,t)=∑j=0m1j!x∂∂x+t∂∂tju(x,t).{u}_{m}\left(x,t)=\mathop{\sum }\limits_{j=0}^{m}\frac{1}{j\&#x0021;}{\left[x\frac{\partial }{\partial x}+t\frac{\partial }{\partial t}\right]}^{j}u\left(x,t).Since um(x,t){u}_{m}\left(x,t)is the best approximation u(x,t)u\left(x,t)out of Y×Y′Y\times Y^{\prime} , we have (27)‖u(x,t)−um(x,t)‖2≤1(m+1)!x∂∂x+t∂∂tm+1u(ξ,η)2=∫01∫011(m+1)!x∂∂x+t∂∂tm+1u(ξ,η)2dxdt1∕2=∫01∫01∑r=0m+1mr(m+1)!∂m+1u(ξ,η)trxm+1−r∂tr∂xm+1−r2dxdt1∕2≤∫01∫01M(m+1)!(x+t)m+12dxdt1∕2=M(m+1)!∫01∫01(x+t)2m+2dxdt1∕2≤M(m+1)!22m+1(2m+3)(m+2),\begin{array}{l}\Vert u\left(x,t)-{u}_{m}\left(x,t){\Vert }_{2}\\ \hspace{1.0em}\le {\left\Vert \frac{1}{\left(m+1)\&#x0021;}{\left[x\frac{\partial }{\partial x}+t\frac{\partial }{\partial t}\right]}^{m+1}u\left(\xi ,\eta )\right\Vert }_{2}\\ \hspace{1.0em}={\left(\underset{0}{\overset{1}{\displaystyle \int }}\underset{0}{\overset{1}{\displaystyle \int }}{\left(\frac{1}{\left(m+1)\&#x0021;}{\left[x\frac{\partial }{\partial x}+t\frac{\partial }{\partial t}\right]}^{m+1}u\left(\xi ,\eta )\right)}^{2}{\rm{d}}x{\rm{d}}t\right)}^{1/2}\\ \hspace{1.0em}={\left(\underset{0}{\overset{1}{\displaystyle \int }}\underset{0}{\overset{1}{\displaystyle \int }}{\left[\mathop{\displaystyle \sum }\limits_{r=0}^{m+1}\frac{\left(\genfrac{}{}{0ex}{}{m}{r}\right)}{\left(m+1)\&#x0021;}\frac{{\partial }^{m+1}u\left(\xi ,\eta ){t}^{r}{x}^{m+1-r}}{\partial {t}^{r}\partial {x}^{m+1-r}}\right]}^{2}{\rm{d}}x{\rm{d}}t\right)}^{1/2}\\ \hspace{1.0em}\le {\left(\underset{0}{\overset{1}{\displaystyle \int }}\underset{0}{\overset{1}{\displaystyle \int }}{\left(\frac{M}{\left(m+1)\&#x0021;}{\left(x+t)}^{m+1}\right)}^{2}{\rm{d}}x{\rm{d}}t\right)}^{1/2}\\ \hspace{1.0em}=\frac{M}{\left(m+1)\&#x0021;}{\left(\underset{0}{\overset{1}{\displaystyle \int }}\underset{0}{\overset{1}{\displaystyle \int }}{\left(x+t)}^{2m+2}{\rm{d}}x{\rm{d}}t\right)}^{1/2}\\ \hspace{1.0em}\le \frac{M}{\left(m+1)\&#x0021;}\frac{\sqrt{2}{2}^{m+1}}{\sqrt{\left(2m+3)\left(m+2)}},\end{array}where 0≤ξ≤x0\le \xi \le xand 0≤η≤t0\le \eta \le t, we obtain the above error bound.□4Numerical examplesIn this section, we solve two benchmark examples taken from published work. Our calculation shows that the proposed method is comparable with the published work. Here, we conduct numerical experiments using Ritz approximation via satisfier function as explained in Section 3 for solving the fractal-fractional advection–diffusion–reaction equation. Here, we use Maple to perform all the computations.Example 1Consider a fractal-fractional advection–diffusion–reaction equation as in Example 3 [14]: (28)Dt0.65,0.35FFMu(x,t)=DxθABCu(x,t)+Dx1.85Cu(x,t)+sin(u(x,t))+sin(2u(x,t))+f(x,t),0<x,t≤1,u(x,0)=0,ut(x,0)=0,u(0,t)=0,u(1,t)=0.\begin{array}{l}{}^{{\rm{FFM}}}D_{t}^{0.65,0.35}u\left(x,t)\\ \hspace{1.0em}={}^{{\rm{ABC}}}D_{x}^{\theta }u\left(x,t)+{}^{C}D_{x}^{1.85}u\left(x,t)+\sin \left(u\left(x,t))\\ \hspace{2.0em}+\sin \left(2u\left(x,t))+f\left(x,t),\hspace{1em}0\lt x,\hspace{0.33em}t\le 1,\\ \hspace{2.0em}u\left(x,0)=0,\hspace{0.33em}{u}_{t}\left(x,0)=0,\\ \hspace{2.0em}u\left(0,t)=0,\hspace{0.33em}u\left(1,t)=0.\end{array}Here, we refer the reader to Example 3 [14] for the long expression of f(x,t)f\left(x,t). The exact solution is given by u(x,t)=100x4t4(1−x)(1−t)u\left(x,t)=100{x}^{4}{t}^{4}\left(1-x)\left(1-t).When θ=0.5\theta =0.5, we obtained the numerical result as in Table 1. In this example, we are using M=2M=2, which means the two variables-shifted Legendre polynomials is only up to quadratic power. The approximation solution is increasing in terms of accuracy as MMis increased to M=3M=3. The calculation is done by using Maple and applying Lemma 2.4 and Corollary 2.5 in ref. [32] and Corollary 2.5 in ref. [43]. Figures 1 and 2 show the graph for the approximate solution and absolute error using M=3M=3for the Ritz approximation and θ=0.5\theta =0.5for Example 1.Table 1Absolute errors obtained by proposed method with M=2,3M=2,3for Example 1 for θ=0.5\theta =0.5(x,t)\left(x,t)ExactAbs. error, M=2M=2Abs. error, M=3M=3(0.1, 0.1)0.000000819.85884×10−59.85884\times 1{0}^{-5}2.29895×10−82.29895\times 1{0}^{-8}(0.2 ,0.2)0.000163847.45007×10−47.45007\times 1{0}^{-4}1.22948×10−71.22948\times 1{0}^{-7}(0.3 ,0.3)0.003214894.95369×10−34.95369\times 1{0}^{-3}1.56609×10−71.56609\times 1{0}^{-7}(0.4, 0.4)0.023592963.82750×10−33.82750\times 1{0}^{-3}8.62400×10−88.62400\times 1{0}^{-8}(0.5, 0.5)0.097656251.70405×10−21.70405\times 1{0}^{-2}8.87000×10−88.87000\times 1{0}^{-8}(0.6, 0.6)0.268738564.86057×10−24.86057\times 1{0}^{-2}2.29700×10−72.29700\times 1{0}^{-7}(0.7, 0.7)0.518832095.01841×10−25.01841\times 1{0}^{-2}3.15500×10−73.15500\times 1{0}^{-7}(0.8, 0.8)0.671088645.28156×10−35.28156\times 1{0}^{-3}1.92000×10−71.92000\times 1{0}^{-7}(0.9, 0.9)0.430467217.89781×10−37.89781\times 1{0}^{-3}2.74000×10−82.74000\times 1{0}^{-8}Figure 1Diagram of the approximate solution for Example 1 using M=3M=3and θ=0.5\theta =0.5.Figure 2Diagram of the absolute error for Example 1 using M=3M=3and θ=0.5\theta =0.5.Example 2Consider a fractal-fractional advection–diffusion–reaction equation as in Example 1 [14]: (29)Dtα,βFFMu(x,t)=3Dx0.25ABCu(x,t)+2Dx1.95Cu(x,t)+exp(u(x,t))+f(x,t),0<x,t≤1,u(x,0)=0,ut(x,0)=sin(x),u(0,t)=0,u(1,t)=sin(1)sin(t).\begin{array}{l}{}^{{\rm{FFM}}}D_{t}^{\alpha ,\beta }u\left(x,t)\\ \hspace{1.0em}=3{}^{{\rm{ABC}}}D_{x}^{0.25}u\left(x,t)+2{}^{C}D_{x}^{1.95}u\left(x,t)\\ \hspace{1.0em}\hspace{1.0em}+\exp \left(u\left(x,t))+f\left(x,t),\hspace{1em}0\lt x,\hspace{0.33em}t\le 1,\\ u\left(x,0)=0,\hspace{1em}{u}_{t}\left(x,0)=\sin \left(x),\\ u\left(0,t)=0,\hspace{1em}u\left(1,t)=\sin \left(1)\sin \left(t).\end{array}Here, again, we refer the reader to Example 1 [14] for the long expression of f(x,t)f\left(x,t). The exact solution is given by u(x,t)=sin(x)sin(t)u\left(x,t)=\sin \left(x)\sin \left(t). By using the procedure as explained in Section 3.1, the satisfier function, ζ(x,t)\zeta \left(x,t)is xsin(1)sin(t)−txsin(1)+tsin(x).x\sin \left(1)\sin \left(t)-tx\sin \left(1)+t\sin \left(x).In order to obtain the numerical solution, we need Lemma (2.4) as in ref. [32]. From the numerical result as shown in Table 2, with only few terms of two variables-shifted Legendre polynomials via Ritz approximation, our proposed method is comparable with the method via Berstein polynomials and its operational matrix as in ref. [14], using more terms of polynomials. Similar to Example 1, the calculation is done by using Maple and applying Lemma 2.4 and Corollary 2.5 in ref. [32] and Corollary 2.5 in [43].Table 2Comparison of the maximum absolute errors obtained by proposed method with M=2M=2for Example 2 with ref. [14](α,β\alpha ,\beta )mmnnMAE [14]MMMAE (proposed method)(0.75,0.25)775.9238×10−65.9238\times 1{0}^{-6}28.99969×10−68.99969\times 1{0}^{-6}(0.75,0.5)777.0270×10−67.0270\times 1{0}^{-6}29.16790×10−69.16790\times 1{0}^{-6}(0.45,0.35)772.5772×10−62.5772\times 1{0}^{-6}29.15874×10−69.15874\times 1{0}^{-6}(0.65,0.35)774.3460×10−64.3460\times 1{0}^{-6}29.12105×10−69.12105\times 1{0}^{-6}5ConclusionIn this article, we successfully used the Ritz approximation approach to solve fractal-fractional advection–diffusion–reaction equations via two variables-shifted Legendre polynomials. With Ritz approximation, greatly reduces the number of terms of polynomials needed for numerical simulation to obtain the solution for fractal-fractional advection–diffusion–reaction equations. With only few terms of two variables-shifted Legendre polynomials, we were able to obtain the numerical solution with high accuracy. The proposed procedure can be easily extended to solve fractal-fractional advection–diffusion–reaction equations in variable order. Furthermore, we hope to extend the method to tackle the inverse problems related to fractional partial differential equations such as those in ref. [44,45,46], or more complicated scenarios and problems, such as those in ref. [47]. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Open Physics de Gruyter

# Fractal-fractional advection–diffusion–reaction equations by Ritz approximation approach

, Volume 21 (1): 1 – Jan 1, 2023
8 pages

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de Gruyter
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© 2023 the author(s), published by De Gruyter
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2391-5471
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10.1515/phys-2022-0221
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### Abstract

1IntroductionAdvection–diffusion–reaction equation is an important class of partial differential equations that have been used to model various physical processes, such as transport dissipative particle dynamics model for simulating mesoscopic problems [1], in anisotropic media [2], transport of chemical constituents in the Earth’s atmosphere [3], bimolecular chemical reactions [4], Brusselator system [5], rubella epidemic [6], tuberculosis transmission modeling [7], COVID-19 mathematical modeling [8] and many more.This advection–diffusion–reaction equation has been extended to include the fractional derivative, which is called the fractional advection–diffusion–reaction equation. The researchers found that the fractional derivative model was able to describe the transport problems in Earth surface sciences, which include collective behavior of particles in transport [9], heat and mass transfer [10], and the study of the dynamics of cytosolic calcium ion in astrocytes [11]. In this research direction, Caputo fractional derivative is the most common fractional derivative that has been used in this fractional advection–diffusion–reaction equation.Different from most established work, in this work, we intend to study the fractional advection–diffusion–reaction equation in the fractal-fractional sense as follows: (1)Dtα,βFFMu(x,t)=κ1Dxm1u(x,t)+κ2Dxm2u(x,t)+Φ(u(x,t))+f(x,t),0<x,t≤1,u(x,0)=f0(x),ut(x,0)=f1(x),u(0,t)=g0(t),u(1,t)=g1(t),0≤x≤1,\begin{array}{rcl}{}^{{\rm{FFM}}}D_{t}^{\alpha ,\beta }u\left(x,t)& =& {\kappa }_{1}{D}_{x}^{{m}_{1}}u\left(x,t)+{\kappa }_{2}{D}_{x}^{{m}_{2}}u\left(x,t)\\ & & +\Phi \left(u\left(x,t))+f\left(x,t),\hspace{0.33em}0\lt x,\hspace{0.33em}t\le 1,\\ u\left(x,0)& =& {f}_{0}\left(x),\hspace{1em}{u}_{t}\left(x,0)={f}_{1}\left(x),\\ u\left(0,t)& =& {g}_{0}\left(t),\hspace{1em}u\left(1,t)={g}_{1}\left(t),\hspace{0.33em}0\le x\le 1,\end{array}where 0<m1≤10\lt {m}_{1}\le 1, 1<m2≤21\lt {m}_{2}\le 2, κ1{\kappa }_{1}, and κ2>0{\kappa }_{2}\gt 0are the characteristics speed in advection process and diffusion coefficient, respectively. Dtα,βFFM{}^{{\rm{FFM}}}D_{t}^{\alpha ,\beta }is the fractal-fractional differentiation operator of order (α,β)\left(\alpha ,\beta )(where 0<α,β≤10\lt \alpha ,\beta \le 1) with respect to ttin the sense of Atangana–Riemann–Liouville [12,13], and Φ(u(x,t))\Phi \left(u\left(x,t))is some reasonable nonlinear function of u(x,t)u\left(x,t), which may represent reaction process. The fractional derivatives for Dxm1{D}_{x}^{{m}_{1}}and Dxm2{D}_{x}^{{m}_{2}}are defined in Caputo sense. However, to solve the problem as in ref. [14], one can define the Dxm1{D}_{x}^{{m}_{1}}as Dxm1ABC{}^{{\rm{ABC}}}D_{x}^{{m}_{1}}, where ABC denotes Atangana–Baleanu Caputo derivative. This fractional advection–diffusion–reaction equation in the fractal-fractional sense was proposed in ref. [14], while the fractional advection–diffusion–reaction equation’s specific applications are discussed in refs [15,16,17] and the real-world interpretations of this fractal-fractional application are discussed in refs [18,19, 20,21].The fractal-fractional derivatives have been found very useful in many science and engineering applications, such as in modeling anomalous diffusion processes [22]. Researchers found that phenomena that are inherent in abnormal exponential or the phenomena with heavy tail decay processes are best described in the fractal-fractional derivatives [23]. Recently, this combination of fractal-fractional derivative was again shown by Atangana [12,13] that this kind of derivative takes into account not only the memory effect but also other characteristics, such as the heterogeneity, elasco-viscosity of the medium, and the fractal geometry of the dynamic system. In this research direction, fractal-fractional derivatives have been used in many phenomena, such as reaction-diffusion model [18], modeling bank data [19], Shinriki’s oscillator model [20], and malaria transmission model [21].On top of that, numerical methods are always needed to solve fractional calculus problems that arise in engineering applications [24,25,26, 27,28]. Furthermore, the differential equations that arise from the modeling process, especially in science and engineering applications involving fractal-fractional operators, are often very complex, especially when we intend to obtain their analytical solution. Hence, numerical methods are more applicable and suitable for solving fractal-fractional differential equations. Some numerical methods have been derived to tackle this problem, such as Chebyshev polynomials for solving the model of the nonlinear Ginzburg–Landau equation in a fractal-fractional sense [29], the Crank-Nicolson finite difference scheme is extended to solve the fractal-fractional Boussinesq equation [30], and a numerical method based on the Lagrangian piece-wise interpolation is used to obtain the solution of variable-order fractal-fractional time delay equations [31]. Besides that, the wavelet-based approximation method was used for solving the coupled nonlinear 2D Schrödinger equations in a fractal-fractional sense [32]. Different from the existing methods for solving fractal-fractional differential equations, here we extend the Ritz method to obtain the solution of these fractal-fractional differential equations. More specifically, we use the Ritz approximation approach to solve the fractal-fractional advection–diffusion–reaction equation as in Eq. (1). The approach is very easy to use and able to give high accuracy numerical solutions.The Ritz approximation approach had been used in solving considerable problems previously. Among those, Rashedi et al. used the Ritz-Galerkin approach to handle inverse wave problem [33]. Apart from this, the satisfier function was used in Ritz–Galerkin for the identification of a time-dependent diffusivity [34], the Ritz approximation has also been applied to solve some fractional partial differential equations [35,36]. Exact and approximation solutions of the heat equation with nonlocal boundary conditions were found in ref. [37] using the Ritz–Galerkin method with Bernoulli polynomials as the basis. Besides that, Genocchi polynomials had been used in the Ritz–Galerkin method for solving the fractional Klein–Gordon equation and fractional diffusion wave equation [38]. Here, for the first time, we propose to use this Ritz approximation approach to solve the fractal-fractional advection–diffusion–reaction equation as in Eq. (1). Apart from this, two variables-shifted Legendre polynomials, which were derived by Khan and Singh [39], will be used. This is different from two-dimensional-shifted Legendre polynomials. In short, the main objective of this article is to solve the fractal-fractional advection–diffusion–reaction equation using the Ritz approximation via two variables-shifted Legendre polynomials.The rest of the article is organized as follows: Section 2 provides the basic definitions and notations for fractal-fractional derivative and two variables-shifted Legendre polynomials. Section 3 presents the main tool that is used in this article, which is the Ritz approximation and satisfier function for solving fractal-fractional advection–diffusion–reaction equations. Error analysis will also be presented. Section 4 gives some numerical examples to show that the approach is better than some existing methods. Section 5 gives summary and recommendation for future work.2Preliminaries2.1Fractal-fractional derivativeIn this section, we will briefly present some basic definitions related to fractal-fractional derivative.Definition 1The Mittag–Leffler function (2)Ea(x)=∑k=0∞xkΓ(ak+1),whereRe(a)>0,Ea,b(x)=∑k=0∞xkΓ(ak+b),whereRe(a),Re(b)>0.\begin{array}{rcl}{E}_{a}\left(x)& =& \mathop{\displaystyle \sum }\limits_{k=0}^{\infty }\frac{{x}^{k}}{\Gamma \left(ak+1)},\hspace{1em}\hspace{0.1em}\text{where}\hspace{0.1em}\hspace{0.33em}{\rm{Re}}\left(a)\gt 0,\\ {E}_{a,b}\left(x)& =& \mathop{\displaystyle \sum }\limits_{k=0}^{\infty }\frac{{x}^{k}}{\Gamma \left(ak+b)},\hspace{0.1em}\text{where}\hspace{0.1em}\hspace{0.33em}{\rm{Re}}\left(a),{\rm{Re}}\left(b)\gt 0.\end{array}Definition 2The fractal-fractional derivative of order (α,β)\left(\alpha ,\beta )of the continuous function f(x)f\left(x)in the Atangana–Riemann–Liouville sense with Mittag–Leffler kernel [12] is given as follows: (3)Dxα,βFFMf(x)=M(α)1−αddxβ∫x0xf(t)Eα−α(x−t)α1−αdt=M(α)x1−ββ(1−α)ddx∫x0xf(t)Eα−α(x−t)α1−αdt,\begin{array}{l}{}^{{\rm{FFM}}}D_{x}^{\alpha ,\beta }f\left(x)=\frac{M\left(\alpha )}{1-\alpha }\frac{{\rm{d}}}{{\rm{d}}{x}^{\beta }}\underset{{x}_{0}}{\overset{x}{\displaystyle \int }}f\left(t){E}_{\alpha }\left(\frac{-\alpha {\left(x-t)}^{\alpha }}{1-\alpha }\right){\rm{d}}t\\ \hspace{5.2em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\underset{{x}_{0}}{\overset{x}{\displaystyle \int }}f\left(t){E}_{\alpha }\left(\frac{-\alpha {\left(x-t)}^{\alpha }}{1-\alpha }\right){\rm{d}}t,\end{array}where α,β∈(0,1)\alpha ,\beta \in \left(0,1), M(α)M\left(\alpha )is a normalization function with M(0)=M(1)=1M\left(0)=M\left(1)=1, and M(α)=1−α+αΓ(α)M\left(\alpha )=1-\alpha +\frac{\alpha }{\Gamma \left(\alpha )}.Lemma 1The fractal-fractional derivative of order 0<α,β<10\lt \alpha ,\beta \lt 1for f(x)=xpf\left(x)={x}^{p}can be defined as follows: (4)Dxα,βFFMxp=M(α)β(1−α)∑i=0∞(−1)iαip!xαi+p+1−β(1−α)iΓ(iα+p+1).{}^{{\rm{FFM}}}D_{x}^{\alpha ,\beta }{x}^{p}=\frac{M\left(\alpha )}{\beta \left(1-\alpha )}\mathop{\sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}p\&#x0021;{x}^{\alpha i+p+1-\beta }}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +p+1)}.ProofBy using Definition in Eqs. (2) and (3), (5)Dxα,βFFMxp=M(α)x1−ββ(1−α)ddx∫0xtp∑i=0∞−α1−αi(x−t)αiΓ(iα+1)dt=M(α)x1−ββ(1−α)ddx∫0x∑i=0∞(−1)itpαi(x−t)αi(1−α)iΓ(iα+1)dt=M(α)x1−ββ(1−α)ddx∑i=0∞(−1)iαi∫0xtp(x−t)αidt(1−α)iΓ(iα+1),\begin{array}{l}{}^{{\rm{FFM}}}D_{x}^{\alpha ,\beta }{x}^{p}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\underset{0}{\overset{x}{\displaystyle \int }}{t}^{p}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(\frac{-\alpha }{1-\alpha }\right)}^{i}{\left(x-t)}^{\alpha i}}{\Gamma \left(i\alpha +1)}{\rm{d}}t\\ \hspace{4.2em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\underset{0}{\overset{x}{\displaystyle \int }}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{t}^{p}{\alpha }^{i}{\left(x-t)}^{\alpha i}}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +1)}{\rm{d}}t\\ \hspace{4.2em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}{\displaystyle \int }_{0}^{x}{t}^{p}{\left(x-t)}^{\alpha i}{\rm{d}}t}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +1)},\end{array}and knowing that ∫0xtp(x−t)αidt=Γ(αi+1)p!xαi+p+1Γ(αi+p+2){\int }_{0}^{x}{t}^{p}{\left(x-t)}^{\alpha i}{\rm{d}}t=\frac{\Gamma \left(\alpha i+1)p\&#x0021;{x}^{\alpha i+p+1}}{\Gamma \left(\alpha i+p+2)}, we obtain (6)Dxα,βFFMxp=M(α)x1−ββ(1−α)ddx∑i=0∞(−1)iαiΓ(αi+1)p!xαi+p+1Γ(αi+p+2)(1−α)iΓ(iα+1)=M(α)x1−ββ(1−α)ddx∑i=0∞(−1)iαip!xαi+p+1(1−α)iΓ(iα+p+2)=M(α)x1−ββ(1−α)∑i=0∞(−1)iαip!(αi+p+1)xαi+p(1−α)iΓ(iα+p+2)=M(α)β(1−α)∑i=0∞(−1)iαip!xαi+p+1−β(1−α)iΓ(iα+p+1).□\begin{array}{l}{}^{{\rm{FFM}}}D_{x}^{\alpha ,\beta }{x}^{p}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}\frac{\Gamma \left(\alpha i+1)p\&#x0021;{x}^{\alpha i+p+1}}{\Gamma \left(\alpha i+p+2)}}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +1)}\\ \hspace{4.3em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\frac{{\rm{d}}}{{\rm{d}}x}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}p\&#x0021;{x}^{\alpha i+p+1}}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +p+2)}\\ \hspace{4.25em}=\frac{M\left(\alpha ){x}^{1-\beta }}{\beta \left(1-\alpha )}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}p\&#x0021;\left(\alpha i+p+1){x}^{\alpha i+p}}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +p+2)}\\ \hspace{4.25em}=\frac{M\left(\alpha )}{\beta \left(1-\alpha )}\mathop{\displaystyle \sum }\limits_{i=0}^{\infty }\frac{{\left(-1)}^{i}{\alpha }^{i}p\&#x0021;{x}^{\alpha i+p+1-\beta }}{{\left(1-\alpha )}^{i}\Gamma \left(i\alpha +p+1)}.\hspace{1.5em}\square \end{array}Eq. (6) can also be written in Mittag–Leffler form, as shown in ref. [29].2.2Two variables-shifted Legendre polynomialsThe Legendre polynomials, Ln(x){L}_{n}\left(x), can be defined as the coefficients in a formal expansion in powers of ttof the generating function as follows: (7)11−2xt+t2=∑n=0∞Ln(x)tn.\frac{1}{\sqrt{1-2xt+{t}^{2}}}=\mathop{\sum }\limits_{n=0}^{\infty }{L}_{n}\left(x){t}^{n}.There are few different ways for the extension of this Legendre polynomials Ln(x,y){L}_{n}\left(x,y)(or shifted Legendre polynomials, Pn(x,y){P}_{n}\left(x,y)) in several variables, such as (1)Two-variable Legendre polynomials [40](8)11−2xt+yt2=∑n=0∞Ln(x,y)tn.\frac{1}{\sqrt{1-2xt+y{t}^{2}}}=\mathop{\sum }\limits_{n=0}^{\infty }{L}_{n}\left(x,y){t}^{n}.(2)Two-dimensional-shifted Legendre polynomials are defined as [41,42] follows:(9)Pn(x,y)=Pn(x)Pk(y).{P}_{n}\left(x,y)={P}_{n}\left(x){P}_{k}(y).(3)Two variables Legendre polynomials [39](10)11−2xs+s2−2yt+t2=∑n=0∞∑k=0∞Ln,k(x,y)sntk.\frac{1}{\sqrt{1-2xs+{s}^{2}-2yt+{t}^{2}}}=\mathop{\sum }\limits_{n=0}^{\infty }\mathop{\sum }\limits_{k=0}^{\infty }{L}_{n,k}\left(x,y){s}^{n}{t}^{k}.The significant advantage for the two variables Legendre polynomials derived by Khan and Singh [39] in 2010 is that the polynomials can be obtained via generating function in two variables form as shown in Eq. (10). Here, we apply the definition of two variables Legendre polynomials derived by Khan and Singh [39] to define the two variables-shifted Legendre polynomials, Pn,k(x,y){P}_{n,k}\left(x,y), as follows: (11)11−2(2x−1)s+s2−2(2y−1)t+t2=∑n=0∞∑k=0∞Pn,k(x,y)sntk.\begin{array}{l}\frac{1}{\sqrt{1-2\left(2x-1)s+{s}^{2}-2\left(2y-1)t+{t}^{2}}}\\ \hspace{1.0em}=\mathop{\displaystyle \sum }\limits_{n=0}^{\infty }\mathop{\displaystyle \sum }\limits_{k=0}^{\infty }{P}_{n,k}\left(x,y){s}^{n}{t}^{k}.\end{array}As discussed in Section 3 of ref. [39], we have the following analytical expression for two variables-shifted Legendre polynomials.(12)Pn,k(x,y)=∑r=0⌈n2⌉∑j=0k2(1∕2)n+k−r−j(4x−2)n−2r(4y−2)k−2j(−1)r+jr!j!(n−2r)!(k−2j)!,\begin{array}{l}{P}_{n,k}\left(x,y)\\ \hspace{1.0em}=\mathop{\displaystyle \sum }\limits_{r=0}^{\lceil \frac{n}{2}\rceil }\mathop{\displaystyle \sum }\limits_{j=0}^{&#x2308;\frac{k}{2}&#x2309;}\frac{{\left(1/2)}_{n+k-r-j}{\left(4x-2)}^{n-2r}{\left(4y-2)}^{k-2j}{\left(-1)}^{r+j}}{r\&#x0021;j\&#x0021;\left(n-2r)\&#x0021;\left(k-2j)\&#x0021;},\end{array}where (1∕2)n+k−r−j{\left(1/2)}_{n+k-r-j}denotes the falling factorial. When k = 0, we have Pn,0(x,y)=Pn(x){P}_{n,0}\left(x,y)={P}_{n}\left(x), where Pn(x) is the well known shifted Legendre polynomials. These two variables-shifted Legendre polynomials also can be expressed in terms of hypergeometric function as follows: (13)Pn,k(x,y)=2n+k(1∕2)n+k(2x−1)n(2y−1)kn!k!×F1,0,00,1,1−;−n2,1−n2;−k2,1−k2;1−2n−2k2;−;−;1(2x−1)21(2y−1)2.\begin{array}{l}{P}_{n,k}\left(x,y)\\ \hspace{1.0em}=\frac{{2}^{n+k}{\left(1/2)}_{n+k}{\left(2x-1)}^{n}{\left(2y-1)}^{k}}{n\&#x0021;k\&#x0021;}\\ \hspace{2.0em}\times {F}_{1,0,0}^{0,1,1}\left[\begin{array}{ccc}-;& \frac{-n}{2},\frac{1-n}{2};& \frac{-k}{2},\frac{1-k}{2};\\ \frac{1-2n-2k}{2};& -;& -;\end{array}\frac{1}{{\left(2x-1)}^{2}}\frac{1}{{\left(2y-1)}^{2}}\right].\end{array}For function f(x,y)=xaybf\left(x,y)={x}^{a}{y}^{b}with aaand bbas positive integers, it is easy to see that it can be obtained via (14)f(x,y)=xayb=1(a+1)(b+1)∑n=0∞∑k=0∞cn,kPn,k(x,y),f\left(x,y)={x}^{a}{y}^{b}=\frac{1}{\left(a+1)\left(b+1)}\mathop{\sum }\limits_{n=0}^{\infty }\mathop{\sum }\limits_{k=0}^{\infty }{c}_{n,k}{P}_{n,k}\left(x,y),where (15)cn,k=∫01∫01f(x,y)Pn,k(x,y)dxdy∫01∫01Pn,k(x,y)Pn,k(x,y)dxdy.{c}_{n,k}=\frac{{\int }_{0}^{1}{\int }_{0}^{1}f\left(x,y){P}_{n,k}\left(x,y){\rm{d}}x{\rm{d}}y}{{\int }_{0}^{1}{\int }_{0}^{1}{P}_{n,k}\left(x,y){P}_{n,k}\left(x,y){\rm{d}}x{\rm{d}}y}.3Ritz method3.1Ritz approximation approach and satisfier function for fractal-fractional advection–diffusion–reaction equationIn this Ritz approximation approach, we use two variables-shifted Legendre polynomials Pp,q(x,t){P}_{p,q}\left(x,t), as explained in Section 2 and the approximate solution u(x,t)u\left(x,t)for fractal-fractional advection–diffusion–reaction Eq. (1) is denoted as u˜(x,t)\tilde{u}\left(x,t). Hence, we have (16)u˜(x,t)=∑p=0M∑q=0MKpqωpq(x,t)+ζ(x,t),(x,t)∈[0,L]×[0,T],\begin{array}{rcl}\tilde{u}\left(x,t)& =& \mathop{\displaystyle \sum }\limits_{p=0}^{M}\mathop{\displaystyle \sum }\limits_{q=0}^{M}{K}_{pq}{\omega }_{pq}\left(x,t)+\zeta \left(x,t),\\ & & \left(x,t)\in \left[0,L]\times \left[0,T],\end{array}where ωpq(x,t)=x(x−L)t2Pp,q(x,t){\omega }_{pq}\left(x,t)=x\left(x-L){t}^{2}{P}_{p,q}\left(x,t)and the satisfier function is represented by ζ(x,t)\zeta \left(x,t). Pp,q(x,t){P}_{p,q}\left(x,t)denotes two variables-shifted Legendre polynomials, and Kpq{K}_{pq}is the coefficient that needs to be calculated. The main purpose of using the satisfier function is that the satisfier function satisfies all the initial and boundary conditions.Based on the initial and boundary condition as in Eq. (1), the satisfier equation, ζ(x,t)\zeta \left(x,t)can be found using the following procedure:Step 1: First, we find (17)φ(x,t)=1−xLg0(t)+xLg1(t).\varphi \left(x,t)=\left(1-\frac{x}{L}\right){g}_{0}\left(t)+\frac{x}{L}{g}_{1}\left(t).Step 2: Next, we determine (18)F0(x)=f0(x)−φ(x,0),F1(x)=f1(x)−φt(x,0).\begin{array}{rcl}{F}_{0}\left(x)& =& {f}_{0}\left(x)-\varphi \left(x,0),\\ {F}_{1}\left(x)& =& {f}_{1}\left(x)-{\varphi }_{t}\left(x,0).\end{array}Step 3: Calculate (19)R(x,t)=F0(x)+tF1(x).R\left(x,t)={F}_{0}\left(x)+t{F}_{1}\left(x).Step 4: Finally, we determine (20)ζ(x,t)=R(x,t)+φ(x,t).\zeta \left(x,t)=R\left(x,t)+\varphi \left(x,t).Furthermore, the coefficients Kpq{K}_{pq}in Eq. (16) can also be calculated by using the following inner product as follows: (21)⟨F(u˜),Pp,q(x,t)⟩=0,\langle F\left(\tilde{u}),{P}_{p,q}\left(x,t)\rangle =0,where (22)F(u˜)=Dtα,βFFMu˜(x,t)−κ1Dxm1u˜(x,t)−κ2Dxm2u˜(x,t)−Φ(u˜(x,t))−f(x,t)\begin{array}{rcl}F\left(\tilde{u})& =& {}^{{\rm{FFM}}}D_{t}^{\alpha ,\beta }\tilde{u}\left(x,t)-{\kappa }_{1}{D}_{x}^{{m}_{1}}\tilde{u}\left(x,t)\\ & & -{\kappa }_{2}{D}_{x}^{{m}_{2}}\tilde{u}\left(x,t)-\Phi \left(\tilde{u}\left(x,t))-f\left(x,t)\end{array}and (23)⟨F(u˜),Pp,q(x,t)⟩=∫0L∫0TF(u˜)Pp,q(x,t)dtdx,\langle F\left(\tilde{u}),{P}_{p,q}\left(x,t)\rangle =\underset{0}{\overset{L}{\int }}\underset{0}{\overset{T}{\int }}F\left(\tilde{u}){P}_{p,q}\left(x,t){\rm{d}}t{\rm{d}}x,where Pp,q(x,t){P}_{p,q}\left(x,t)are the two variables-shifted Legendre polynomials. A linear system of equations can be formed by using Eq. (21). Solving this linear system, we can obtain the entries of Kpq{K}_{pq}where p=0,…,M,andq=0,…,Mp=0,\ldots ,M,andq=0,\ldots ,M. Hence, by putting the value obtained for Kpq{K}_{pq}into Eq. (16), we can obtain the approximate solution for fractal-fractional advection–diffusion–reaction equation in (1).3.2Error analysisLemma 2Let the solution of fractal-fractional advection–diffusion–reaction equation is u(x,t)u\left(x,t), where u(x,t)∈Cm+1[0,1]×[0,1]u\left(x,t)\in {C}^{m+1}\left[0,1]\times \left[0,1], suppose Y=span{P0,m(x),P1,m(x),…,Pm,m(x)}⊂L2[0,1]Y=\hspace{0.1em}\text{span}\hspace{0.1em}\left\{{P}_{0,m}\left(x),{P}_{1,m}\left(x),\ldots ,{P}_{m,m}\left(x)\right\}\subset {L}^{2}\left[0,1]and Y′=span{P0,m(t),P1,m(t),…,Pm,m(t)}⊂L2[0,1]Y^{\prime} =\hspace{0.1em}\text{span}\hspace{0.1em}\left\{{P}_{0,m}\left(t),{P}_{1,m}\left(t),\ldots ,{P}_{m,m}\left(t)\right\}\subset {L}^{2}\left[0,1]. We have um(x,t)=Y×Y′{u}_{m}\left(x,t)=Y\times Y^{\prime} is the best approximation of u(x,t)u\left(x,t)by means of two variables-shifted Legendre polynomials, the error bound is given as follows: (24)‖u(x,t)−um(x,t)‖2≤M(m+1)!22m+1(2m+3)(m+2).\Vert u\left(x,t)-{u}_{m}\left(x,t){\Vert }_{2}\le \frac{M}{\left(m+1)\&#x0021;}\frac{\sqrt{2}{2}^{m+1}}{\sqrt{\left(2m+3)\left(m+2)}}.ProofBy using the Taylor series, we have (25)u(x,t)=∑j=0∞1j!(x−a)∂∂x+(t−b)∂∂tju(x,t)u\left(x,t)=\mathop{\sum }\limits_{j=0}^{\infty }\frac{1}{j\&#x0021;}{\left[\left(x-a)\frac{\partial }{\partial x}+\left(t-b)\frac{\partial }{\partial t}\right]}^{j}u\left(x,t)For simplicity, let a=b=0a=b=0, and in practical, we estimate u(x,t)u\left(x,t)up to mmorder as follows: (26)um(x,t)=∑j=0m1j!x∂∂x+t∂∂tju(x,t).{u}_{m}\left(x,t)=\mathop{\sum }\limits_{j=0}^{m}\frac{1}{j\&#x0021;}{\left[x\frac{\partial }{\partial x}+t\frac{\partial }{\partial t}\right]}^{j}u\left(x,t).Since um(x,t){u}_{m}\left(x,t)is the best approximation u(x,t)u\left(x,t)out of Y×Y′Y\times Y^{\prime} , we have (27)‖u(x,t)−um(x,t)‖2≤1(m+1)!x∂∂x+t∂∂tm+1u(ξ,η)2=∫01∫011(m+1)!x∂∂x+t∂∂tm+1u(ξ,η)2dxdt1∕2=∫01∫01∑r=0m+1mr(m+1)!∂m+1u(ξ,η)trxm+1−r∂tr∂xm+1−r2dxdt1∕2≤∫01∫01M(m+1)!(x+t)m+12dxdt1∕2=M(m+1)!∫01∫01(x+t)2m+2dxdt1∕2≤M(m+1)!22m+1(2m+3)(m+2),\begin{array}{l}\Vert u\left(x,t)-{u}_{m}\left(x,t){\Vert }_{2}\\ \hspace{1.0em}\le {\left\Vert \frac{1}{\left(m+1)\&#x0021;}{\left[x\frac{\partial }{\partial x}+t\frac{\partial }{\partial t}\right]}^{m+1}u\left(\xi ,\eta )\right\Vert }_{2}\\ \hspace{1.0em}={\left(\underset{0}{\overset{1}{\displaystyle \int }}\underset{0}{\overset{1}{\displaystyle \int }}{\left(\frac{1}{\left(m+1)\&#x0021;}{\left[x\frac{\partial }{\partial x}+t\frac{\partial }{\partial t}\right]}^{m+1}u\left(\xi ,\eta )\right)}^{2}{\rm{d}}x{\rm{d}}t\right)}^{1/2}\\ \hspace{1.0em}={\left(\underset{0}{\overset{1}{\displaystyle \int }}\underset{0}{\overset{1}{\displaystyle \int }}{\left[\mathop{\displaystyle \sum }\limits_{r=0}^{m+1}\frac{\left(\genfrac{}{}{0ex}{}{m}{r}\right)}{\left(m+1)\&#x0021;}\frac{{\partial }^{m+1}u\left(\xi ,\eta ){t}^{r}{x}^{m+1-r}}{\partial {t}^{r}\partial {x}^{m+1-r}}\right]}^{2}{\rm{d}}x{\rm{d}}t\right)}^{1/2}\\ \hspace{1.0em}\le {\left(\underset{0}{\overset{1}{\displaystyle \int }}\underset{0}{\overset{1}{\displaystyle \int }}{\left(\frac{M}{\left(m+1)\&#x0021;}{\left(x+t)}^{m+1}\right)}^{2}{\rm{d}}x{\rm{d}}t\right)}^{1/2}\\ \hspace{1.0em}=\frac{M}{\left(m+1)\&#x0021;}{\left(\underset{0}{\overset{1}{\displaystyle \int }}\underset{0}{\overset{1}{\displaystyle \int }}{\left(x+t)}^{2m+2}{\rm{d}}x{\rm{d}}t\right)}^{1/2}\\ \hspace{1.0em}\le \frac{M}{\left(m+1)\&#x0021;}\frac{\sqrt{2}{2}^{m+1}}{\sqrt{\left(2m+3)\left(m+2)}},\end{array}where 0≤ξ≤x0\le \xi \le xand 0≤η≤t0\le \eta \le t, we obtain the above error bound.□4Numerical examplesIn this section, we solve two benchmark examples taken from published work. Our calculation shows that the proposed method is comparable with the published work. Here, we conduct numerical experiments using Ritz approximation via satisfier function as explained in Section 3 for solving the fractal-fractional advection–diffusion–reaction equation. Here, we use Maple to perform all the computations.Example 1Consider a fractal-fractional advection–diffusion–reaction equation as in Example 3 [14]: (28)Dt0.65,0.35FFMu(x,t)=DxθABCu(x,t)+Dx1.85Cu(x,t)+sin(u(x,t))+sin(2u(x,t))+f(x,t),0<x,t≤1,u(x,0)=0,ut(x,0)=0,u(0,t)=0,u(1,t)=0.\begin{array}{l}{}^{{\rm{FFM}}}D_{t}^{0.65,0.35}u\left(x,t)\\ \hspace{1.0em}={}^{{\rm{ABC}}}D_{x}^{\theta }u\left(x,t)+{}^{C}D_{x}^{1.85}u\left(x,t)+\sin \left(u\left(x,t))\\ \hspace{2.0em}+\sin \left(2u\left(x,t))+f\left(x,t),\hspace{1em}0\lt x,\hspace{0.33em}t\le 1,\\ \hspace{2.0em}u\left(x,0)=0,\hspace{0.33em}{u}_{t}\left(x,0)=0,\\ \hspace{2.0em}u\left(0,t)=0,\hspace{0.33em}u\left(1,t)=0.\end{array}Here, we refer the reader to Example 3 [14] for the long expression of f(x,t)f\left(x,t). The exact solution is given by u(x,t)=100x4t4(1−x)(1−t)u\left(x,t)=100{x}^{4}{t}^{4}\left(1-x)\left(1-t).When θ=0.5\theta =0.5, we obtained the numerical result as in Table 1. In this example, we are using M=2M=2, which means the two variables-shifted Legendre polynomials is only up to quadratic power. The approximation solution is increasing in terms of accuracy as MMis increased to M=3M=3. The calculation is done by using Maple and applying Lemma 2.4 and Corollary 2.5 in ref. [32] and Corollary 2.5 in ref. [43]. Figures 1 and 2 show the graph for the approximate solution and absolute error using M=3M=3for the Ritz approximation and θ=0.5\theta =0.5for Example 1.Table 1Absolute errors obtained by proposed method with M=2,3M=2,3for Example 1 for θ=0.5\theta =0.5(x,t)\left(x,t)ExactAbs. error, M=2M=2Abs. error, M=3M=3(0.1, 0.1)0.000000819.85884×10−59.85884\times 1{0}^{-5}2.29895×10−82.29895\times 1{0}^{-8}(0.2 ,0.2)0.000163847.45007×10−47.45007\times 1{0}^{-4}1.22948×10−71.22948\times 1{0}^{-7}(0.3 ,0.3)0.003214894.95369×10−34.95369\times 1{0}^{-3}1.56609×10−71.56609\times 1{0}^{-7}(0.4, 0.4)0.023592963.82750×10−33.82750\times 1{0}^{-3}8.62400×10−88.62400\times 1{0}^{-8}(0.5, 0.5)0.097656251.70405×10−21.70405\times 1{0}^{-2}8.87000×10−88.87000\times 1{0}^{-8}(0.6, 0.6)0.268738564.86057×10−24.86057\times 1{0}^{-2}2.29700×10−72.29700\times 1{0}^{-7}(0.7, 0.7)0.518832095.01841×10−25.01841\times 1{0}^{-2}3.15500×10−73.15500\times 1{0}^{-7}(0.8, 0.8)0.671088645.28156×10−35.28156\times 1{0}^{-3}1.92000×10−71.92000\times 1{0}^{-7}(0.9, 0.9)0.430467217.89781×10−37.89781\times 1{0}^{-3}2.74000×10−82.74000\times 1{0}^{-8}Figure 1Diagram of the approximate solution for Example 1 using M=3M=3and θ=0.5\theta =0.5.Figure 2Diagram of the absolute error for Example 1 using M=3M=3and θ=0.5\theta =0.5.Example 2Consider a fractal-fractional advection–diffusion–reaction equation as in Example 1 [14]: (29)Dtα,βFFMu(x,t)=3Dx0.25ABCu(x,t)+2Dx1.95Cu(x,t)+exp(u(x,t))+f(x,t),0<x,t≤1,u(x,0)=0,ut(x,0)=sin(x),u(0,t)=0,u(1,t)=sin(1)sin(t).\begin{array}{l}{}^{{\rm{FFM}}}D_{t}^{\alpha ,\beta }u\left(x,t)\\ \hspace{1.0em}=3{}^{{\rm{ABC}}}D_{x}^{0.25}u\left(x,t)+2{}^{C}D_{x}^{1.95}u\left(x,t)\\ \hspace{1.0em}\hspace{1.0em}+\exp \left(u\left(x,t))+f\left(x,t),\hspace{1em}0\lt x,\hspace{0.33em}t\le 1,\\ u\left(x,0)=0,\hspace{1em}{u}_{t}\left(x,0)=\sin \left(x),\\ u\left(0,t)=0,\hspace{1em}u\left(1,t)=\sin \left(1)\sin \left(t).\end{array}Here, again, we refer the reader to Example 1 [14] for the long expression of f(x,t)f\left(x,t). The exact solution is given by u(x,t)=sin(x)sin(t)u\left(x,t)=\sin \left(x)\sin \left(t). By using the procedure as explained in Section 3.1, the satisfier function, ζ(x,t)\zeta \left(x,t)is xsin(1)sin(t)−txsin(1)+tsin(x).x\sin \left(1)\sin \left(t)-tx\sin \left(1)+t\sin \left(x).In order to obtain the numerical solution, we need Lemma (2.4) as in ref. [32]. From the numerical result as shown in Table 2, with only few terms of two variables-shifted Legendre polynomials via Ritz approximation, our proposed method is comparable with the method via Berstein polynomials and its operational matrix as in ref. [14], using more terms of polynomials. Similar to Example 1, the calculation is done by using Maple and applying Lemma 2.4 and Corollary 2.5 in ref. [32] and Corollary 2.5 in [43].Table 2Comparison of the maximum absolute errors obtained by proposed method with M=2M=2for Example 2 with ref. [14](α,β\alpha ,\beta )mmnnMAE [14]MMMAE (proposed method)(0.75,0.25)775.9238×10−65.9238\times 1{0}^{-6}28.99969×10−68.99969\times 1{0}^{-6}(0.75,0.5)777.0270×10−67.0270\times 1{0}^{-6}29.16790×10−69.16790\times 1{0}^{-6}(0.45,0.35)772.5772×10−62.5772\times 1{0}^{-6}29.15874×10−69.15874\times 1{0}^{-6}(0.65,0.35)774.3460×10−64.3460\times 1{0}^{-6}29.12105×10−69.12105\times 1{0}^{-6}5ConclusionIn this article, we successfully used the Ritz approximation approach to solve fractal-fractional advection–diffusion–reaction equations via two variables-shifted Legendre polynomials. With Ritz approximation, greatly reduces the number of terms of polynomials needed for numerical simulation to obtain the solution for fractal-fractional advection–diffusion–reaction equations. With only few terms of two variables-shifted Legendre polynomials, we were able to obtain the numerical solution with high accuracy. The proposed procedure can be easily extended to solve fractal-fractional advection–diffusion–reaction equations in variable order. Furthermore, we hope to extend the method to tackle the inverse problems related to fractional partial differential equations such as those in ref. [44,45,46], or more complicated scenarios and problems, such as those in ref. [47].

### Journal

Open Physicsde Gruyter

Published: Jan 1, 2023

Keywords: fractal-fractional derivative; Ritz approximation; satisfier function; fractional advection–diffusion–reaction equations; two variables-shifted Legendre polynomials

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