Fractal Fractional Derivative Models for Simulating Chemical Degradation in a Bioreactor

Abstract

A three-differential-equation mathematical model is presented for the degradation of phenol and p-cresol combination in a bioreactor that is continually agitated. The stability analysis of the model’s equilibrium points, as established by the study, is covered. Additionally, we used three alternative kernels to analyze the model with the fractal–fractional derivatives, and we looked into the effects of the fractal size and fractional order. We have developed highly efficient numerical techniques for the concentration of biomass, phenol, and p-cresol. Lastly, numerical simulations are used to illustrate the accuracy of the suggested method.

1. Introduction

In a bioreactor, chemical degradation is the process by which certain chemicals or compounds are broken down or changed by living things in the bioreactor’s controlled environment. Bioreactors are widely used to support biological processes, including fermentation, enzyme manufacturing, and wastewater treatment in various industries, including pharmaceuticals, biotechnology, wastewater treatment, and food production.

Many scientific papers have presented the isolation and work of microbial species with higher-degradation action and abilities to degrade chemical compounds [1]. Many isolated bacteria have been investigated in [2]. The biodegradation of one or all chemical parts hinges on the composition of the specific mixture and the utilized microorganisms [3,4,5,6]. Fractional calculus is an influential extension of the classical derivatives. Fractional differential equations (FDEs) have recently been implemented in different fields. Many authors have worked on these equations, such as the KdV equation [7], advection-dispersion equation [8], telegraph equation [9], Schrodinger equation [10], heat equation [11], convection-diffusion equation [12], Fokker Planck equation [13], and Lambert–Beer equation [14,15]. Some of the FDEs do not have exact solutions. Therefore, it is required to work on numerical methods to solve the mentioned equations, such as solving nonlinear fractional diffusion wave equations with the homotopy analysis technique [16], solving PDEs of fractal order by Adomian decomposition method [17]. In [1], the authors have given a bioreactor model but do not consider the bacteria’s death rate and general configuration of the reactor. We have provided the bioreactor model with the fractal–fractional operators. The model with fractal–fractional derivatives has never been analyzed so far. Our model includes the death rate of bacteria, which is important in the process’s environment. We also consider the general configuration of the reactor, where our model includes a membrane and continuous reactor. Additionally, we fractionalize the model and apply a novel numerical technique to achieve the numerical simulations. In these simulations, we use different fractal dimensions and fractional orders. For more details, see [18,19,20,21,22,23,24,25,26,27,28,29,30].

We organize our manuscript as follows. Problem formulation is performed in Section 2. In Section 3, we discuss the model’s analysis in the classical case and present the equilibrium and stability analysis. Next, we explore the analysis of the model with three different kernels viz. the power-law kernel (Section 4), the exponential-decay kernel (Section 5), and the Mittag–Leffler function (Section 6). Finally, in Section 7, we illustrate the numerical simulations of the proposed models.

2. Preliminaries

The following definitions of fractional differentiation operator and fractal–fractional integral operator with three different kernels are taken from [21].

Definition 1.

The fractional differentiation operator with the power-law-type kernel is described as:

$${ }_c^{FFP}{\mathit{D}}_t^{\alpha ,\eta }f(t)=\frac{1}{1-\alpha }\frac{d}{d{u}^{\eta }}{\int }_c^{t}f(s){(t-s)}^{-\alpha }ds,0<\alpha ,\eta \le 1,$$

(1)

where,

$$\frac{df(s)}{d{s}^{\eta }}=\underset{t\to s}{lim}\frac{f(t)-f(s)}{t^{\eta }-s^{\eta }}$$

(2)

Definition 2.

The fractional differentiation operator with the exponential-decay-type kernel is described as:

$${ }_c^{FFE}{\mathit{D}}_t^{\alpha ,\eta }f(t)=\frac{{\mathit{M}}_{\mathit{1}}(\alpha )}{1-\alpha }\frac{d}{d{t}^{\eta }}{\int }_c^{t}f(s)exp\left(\frac{-\alpha }{1-\alpha }(t-s)\right)ds,0<\alpha ,\eta \le 1.$$

(3)

Definition 3.

The fractional differentiation operator with the Mittag–Leffler-type kernel is described as:

$${ }_c^{FFM}{\mathit{D}}_t^{\alpha ,\eta }f(t)=\frac{\mathit{AB}(\alpha )}{1-\alpha }\frac{d}{d{t}^{\eta }}{\int }_c^{t}f(s){\mathit{E}}_{\alpha }\left(\frac{-\alpha }{1-\alpha }{(t-s)}^{\alpha }\right)ds,0<\alpha ,\eta \le 1,$$

(4)

where $\mathit{AB}(\alpha )=1-\alpha +\frac{\alpha }{\mathsf{\Gamma }(\alpha )}$.

Definition 4.

The fractional integration operator with the power-law-type kernel is described as:

$${ }_0^{FFP}{\mathit{I}}_t^{\alpha ,\eta }f(t)=\frac{\eta }{\mathsf{\Gamma }(\alpha )}{\int }_0^t{(t-s)}^{\alpha -1}{s}^{\tau -1}\varphi (s)ds.$$

(5)

Definition 5.

The fractional integration operator with the exponential-decay-type kernel is described as:

$${ }_0^{FFE}{\mathit{I}}_t^{\alpha ,\eta }f(t)=\frac{\alpha \eta }{{\mathit{M}}_1(\alpha )}{\int }_0^t{s}^{\alpha -1}f(s)ds+\frac{\tau (1-\alpha )t^{\tau -1}}{M_1(\alpha )}\varphi (t).$$

(6)

Definition 6.

The fractional integration operator with the Mittag–Leffler-type kernel is described as:

$${ }_0^{FFM}{\mathit{I}}_t^{\alpha ,\eta }f(t)=\frac{\alpha \eta }{\mathit{AB}(\alpha )}{\int }_0^t{s}^{\alpha -1}f(s){(t-s)}^{\alpha -1}ds+\frac{\tau (1-\alpha )t^{\tau -1}}{\mathit{AB}(\alpha )}f(t).$$

(7)

Here, we present the model to be investigated in this research. We present the model as:

$$\begin{array}{cc}\hfill \frac{\mathrm{d}{S}_{ph}}{\mathrm{d}t}& =D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X,\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill \frac{\mathrm{d}{S}_{cr}}{\mathrm{d}t}& =D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill \frac{\mathrm{d}X}{\mathrm{d}t}& =-D\beta X+\mu \left(S_{ph},S_{cr}\right)X,\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill \mu \left(S_{ph},S_{cr}\right)& =\frac{{\mu }_{max(ph)}{S}_{ph}}{K_{s(ph)}+S_{ph}+\frac{S_{ph}^2}{k_{i(ph)}}+I_{cr/ph}{S}_{cr}}+\frac{{\mu }_{max(cr)}{S}_{cr}}{K_{s(cr)}+S_{cr}+\frac{S_{cr}^2}{k_{i(cr)}}+I_{ph/cr}{S}_{ph}},\hfill \end{array}$$

(11)

The model parameters and variables are detailed in [1]. The parameter $\beta$ is presented in the general configuration. When $\beta =1$, we have continued the reactor. When $\beta =0$, we have a membrane reactor.

3. Analysis of the Model in Classical Sense

Now, we begin with analyzing the properties of the model in classical sense.

We consider the number of equilibrium solutions of the model ((8)–(10)). It is obvious that the model has a branch of the washout given by

$$\begin{array}{cc}\hfill E_0=& (S_{ph},S_{cr},X)=\left(S_{ph0},S_{cr0},0\right).\hfill \end{array}$$

(12)

We obtain the steady state solution of ((8)–(10)) by setting to zero the right side. From the model ((8)–(10)), we have:

$$\begin{array}{cc}\hfill & S_{cr}=\frac{S_{cr0}{k}_{ph}+k_{cr}(S_{ph}-S_{ph0})}{k_{ph}},\hfill \\ \hfill & X=\frac{D\left({\mathit{S}}_{\mathit{ph}\mathit{0}}-{\mathit{S}}_{\mathit{ph}}\right)}{{\mathit{k}}_{\mathit{ph}}\left(\beta D\right)}.\hfill \end{array}$$

(13)

$$\begin{array}{ccc}\hfill f=\left(\genfrac{}{}{0pt}{}{-k_{cr}\mu (s_{ph},s_{cr})X}{\mu (s_{ph},s_{cr})X}\right)& \to \hfill & F=\left[\begin{array}{cc}\frac{\partial \mu (s_{ph},s_{cr})(-k_{cr})X}{\partial (s_{ph},s_{cr})X}& -k_{cr}\mu (s_{ph},s_{cr})\\ \\ \frac{\partial \mu (s_{ph},s_{cr})X}{\partial (s_{ph},s_{cr})X}& \mu (s_{ph},s_{cr})k_{cr}(s_{ph},s_{cr})\end{array}\right]\hfill \end{array}$$

$$V=\left(\genfrac{}{}{0pt}{}{-D(s_{cr0}-s_{cr})}{D\beta X}\right)\to V=\left[\begin{array}{cc}D& 0\\ \\ 0& D\beta \end{array}\right],V^{-1}=\left[\begin{array}{cc}D\beta & 0\\ \\ 0& D\end{array}\right]$$

$$F{V}^{-1}=\left[\begin{array}{cc}0& -k_{cr}\mu (s_{ph},s_{cr})\\ \\ 0& \mu (s_{ph},s_{cr})\end{array}\right]\left[\begin{array}{cc}D& 0\\ \\ 0& D\beta \end{array}\right]=\left[\begin{array}{cc}0& -\beta D{k}_{cr}\mu (s_{ph},s_{cr})\\ \\ 0& \beta D\mu (s_{ph},s_{cr})\end{array}\right]$$

$$det\left[F{V}^{-1}-\lambda I_2\right]=0,\left|\begin{array}{cc}-\lambda & -\beta D{k}_{cr}\mu (s_{ph},s_{cr})\\ \\ 0& \beta D\mu (s_{ph},s_{cr})-\lambda \end{array}\right|=0$$

Thus, we obtain ${\lambda }_1=0$, ${\lambda }_2=\beta D\mu (s_{ph},s_{cr})=R_0$.

Lemma 1.

The steady state solution $E_0$ is locally asymptotically stable when $D>D_{cr}$ and is unstable when $D<D_{cr}$, where

$$D_{cr}=\frac{k_{icr}{k}_{ph}{(s_{cr0}{k}_{ph}-s_{ph0}{k}_{cr})}^{-}{max}_{cr}}{\left[k_{icr}{k}_{ph}(K_{scr}{k}_{ph}+s_{cr0}{k}_{ph}-s_{ph0}{k}_{cr})+{(s_{cr}{k}_{ph}-s_{ph0}{k}_{cr})}^2\right]\beta }.$$

Proof. 

We have

$$E_0=(s_{ph},s_{cr},x)=(s_{ph0},s_{cr0},0)$$

$$J(E_0)=\left[\begin{array}{ccc}-D-\frac{\partial \mu (s_{ph},s_{cr})}{\partial s_{ph}}{k}_{ph}x& -\frac{\partial \mu (s_{ph},s_{cr})}{\partial s_{ph}}{k}_{ph}x& -\mu (s_{ph},s_{cr})k_{ph}\\ \\ -k_{cr}\frac{\partial \mu (s_{ph},s_{cr})}{\partial s_{ph}}x& -D-\frac{\partial \mu (s_{ph},s_{cr})}{\partial s_{ph}}{k}_{cr}x& -\mu (s_{ph},s_{cr})k_{cr}\\ \\ \frac{\partial \mu (s_{ph},s_{cr})}{\partial s_{ph}}x& -\frac{\partial \mu (s_{ph},s_{cr})}{\partial s_{ph}}x& -D\beta +\mu (s_{ph},s_{cr})\end{array}\right]$$

$$J(E_0)=\left[\begin{array}{ccc}-D& 0& -\mu (s_{ph0},s_{cr0})k_{ph}\\ 0& -D& -\mu (s_{ph0},s_{cr0})k_{cr}\\ 0& 0& -D\beta +\mu (s_{ph0},s_{cr0})\end{array}\right]$$

where

$$\mu (s_{ph},s_{cr})=\frac{{\mu }_{max(ph)}{s}_{ph}}{K_{s(ph)}+s_{ph}+\frac{s_{ph}^2}{K_{i(ph)}}+I_{cr/ph}{s}_{cr}}+\frac{{\mu }_{max(cr)}{s}_{ph}}{K_{s(cr)}+s_{cr}+\frac{s_{cr}^2}{K_{i(cr)}}+I_{ph/cr}{s}_{ph}}$$

$$det[J(E_0)-\lambda I_3]=\left|\begin{array}{ccc}-D-\lambda & 0& -\mu (s_{ph},s_{cr})k_{ph}\\ \\ 0& -D-\lambda & -\mu (s_{ph},s_{cr})k_{cr}\\ \\ 0& 0& \mu (s_{ph},s_{cr})-D\beta -\lambda \end{array}\right|=0$$

$$\begin{array}{cc}\hfill =& (-D-\lambda )(-D-\lambda )(\mu (s_{ph},s_{cr})-D\beta -\lambda )=0\hfill \end{array}$$

$${\lambda }_1=-D,{\lambda }_2=-D,{\lambda }_3=-\beta D+\mu (s_{ph},s_{cr})$$

and

$$\begin{array}{cc}\hfill \mu (s_{ph},s_{cr})& {=}^{-}{max}_{ph}{s}_{ph0}{(k_{sph}+s_{ph0}+\frac{s_{ph0}^2}{K_{i(ph)}}+I_{cr/ph}{s}_{cr0})}^{-1}\hfill \\ \hfill & {+}^{-}{max}_{cr}{s}_{cr0}{(k_{scr}+s_{cr0}+\frac{s_{cr0}^2}{K_{i(ph)}}+I_{ph/cr}{s}_{ph0})}^{-1}\hfill \\ \hfill & D_{cr}=\frac{k_{icr}{k}_{ph}{(s_{cr0}{k}_{ph}-s_{ph0}{k}_{cr})}^{-}{max}_{cr}}{\left[k_{icr}{k}_{ph}(K_{scr}{k}_{ph}+s_{cr0}{k}_{ph}-s_{ph0}{k}_{cr})+{(s_{cr}{k}_{ph}-s_{ph0}{k}_{cr})}^2\right]\beta }\hfill \end{array}$$

If $D>D_{cr},$ then ${\lambda }_3<0.$ Thus, all eigenvalues are negative. This shows that the steady state solution $E_0$ is locally asymptotically stable. □

4. Analysis of the Model with the Power-Law Kernel

Here, we analyze the model with fractional differentiation operator using the power-law kernel as:

$$\begin{array}{cc}\hfill { }_0^{FFP}{D}_t^{\alpha ,\eta }{S}_{ph}& =D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X.\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill { }_0^{FFP}{D}_t^{\alpha ,\eta }{S}_{cr}& =D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X.\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill { }_0^{FFP}{D}_t^{\alpha ,\eta }X& =-D\beta X+\mu \left(S_{ph},S_{cr}\right)X.\hfill \end{array}$$

(16)

We have the following relation between the classical and fractal derivative [21]:

$$D^{\eta }f(t)=\frac{f^{\prime }(t)}{\eta t^{\eta -1}}.$$

(17)

A relation between the classical derivative and the fractal derivative gives

$$\begin{array}{cc}\hfill { }_0^{RL}{D}_t^{\alpha }{S}_{ph}& =\eta t^{\eta -1}\left(D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill { }_0^{RL}{D}_t^{\alpha }{S}_{cr}& =\eta t^{\eta -1}\left(D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill { }_0^{RL}{D}_t^{\alpha }X& =\eta t^{\eta -1}\left(-D\beta X+\mu \left(S_{ph},S_{cr}\right)X\right).\hfill \end{array}$$

(20)

For simplicity, we define

$$\begin{array}{cc}\hfill A(t,S_{ph},S_{cr},X)& =\eta t^{\eta -1}\left(D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill B(t,S_{ph},S_{cr},X)& =\eta t^{\eta -1}\left(D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill C(t,S_{ph},S_{cr},X)& =\eta t^{\eta -1}\left(-D\beta X+\mu \left(S_{ph},S_{cr}\right)X\right).\hfill \end{array}$$

(23)

Then, we obtain

$$\begin{array}{cc}\hfill { }_0^{RL}{D}_t^{\alpha }{S}_{ph}& =A(t,S_{ph},S_{cr},X).\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill { }_0^{RL}{D}_t^{\alpha }{S}_{cr}& =B(t,S_{ph},S_{cr},X).\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill { }_0^{RL}{D}_t^{\alpha }X& =C(t,S_{ph},S_{cr},X).\hfill \end{array}$$

(26)

Applying the Riemann–Liouville integral yields:

$$\begin{array}{cc}\hfill S_{ph}(t)-S_{ph}(0)=& \frac{1}{\mathsf{\Gamma }(\alpha )}{\int }_0^{t}A(\tau ,S_{ph},S_{cr},X){(t-\tau )}^{\alpha -1}d\tau .\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill S_{cr}(t)-S_{cr}(0)=& \frac{1}{\mathsf{\Gamma }(\alpha )}{\int }_0^{t}B(\tau ,S_{ph},S_{cr},X){(t-\tau )}^{\alpha -1}d\tau .\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill X(t)-X(0)=& \frac{1}{\mathsf{\Gamma }(\alpha )}{\int }_0^{t}C(\tau ,S_{ph},S_{cr},X){(t-\tau )}^{\alpha -1}d\tau .\hfill \end{array}$$

(29)

Discretizing the above equations at $t_{n+1}$, we receive:

$$\begin{array}{cc}\hfill S_{ph}(t_{n+1})-S_{ph}(0)=& \frac{1}{\mathsf{\Gamma }(\alpha )}{\int }_0^{t_{n+1}}A(\tau ,S_{ph},S_{cr},X){(t_{n+1}-\tau )}^{\alpha -1}d\tau .\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill S_{cr}(t_{n+1})-S_{cr}(0)=& \frac{1}{\mathsf{\Gamma }(\alpha )}{\int }_0^{t_{n+1}}B(\tau ,S_{ph},S_{cr},X){(t_{n+1}-\tau )}^{\alpha -1}d\tau .\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill X(t_{n+1})-X(0)=& \frac{1}{\mathsf{\Gamma }(\alpha )}{\int }_0^{t_{n+1}}C(\tau ,S_{ph},S_{cr},X){(t_{n+1}-\tau )}^{\alpha -1}d\tau .\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill S_{ph}(t_{n+1})-S_{ph}(0)=& \frac{1}{\mathsf{\Gamma }(\alpha )}\sum _{j=0}^n{\int }_{t_j}^{t_{j+1}}A(\tau ,S_{ph},S_{cr},X){(t_{n+1}-\tau )}^{\alpha -1}d\tau .\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill S_{cr}(t_{n+1})-S_{cr}(0)=& \frac{1}{\mathsf{\Gamma }(\alpha )}\sum _{j=0}^n{\int }_{t_j}^{t_{j+1}}B(\tau ,S_{ph},S_{cr},X){(t_{n+1}-\tau )}^{\alpha -1}d\tau .\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill X(t_{n+1})-X(0)=& \frac{1}{\mathsf{\Gamma }(\alpha )}\sum _{j=0}^n{\int }_{t_j}^{t_{j+1}}C(\tau ,S_{ph},S_{cr},X){(t_{n+1}-\tau )}^{\alpha -1}d\tau .\hfill \end{array}$$

(35)

We use two-step Lagrange polynomial as:

$$p_j(\tau ,S_{ph},S_{cr},X)=\frac{\tau -t_{j-1}}{t_j-t_{j-1}}A(t_j,S_{ph},S_{cr},X)-\frac{\tau -t_j}{t_j-t_{j-1}}A(t_{j-1},S_{ph},S_{cr},X).$$

(36)

$$q_j(\tau ,S_{ph},S_{cr},X)=\frac{\tau -t_{j-1}}{t_j-t_{j-1}}B(t_j,S_{ph},S_{cr},X)-\frac{\tau -t_j}{t_j-t_{j-1}}B(t_{j-1},S_{ph},S_{cr},X).$$

(37)

$$s_j(\tau ,S_{ph},S_{cr},X)=\frac{\tau -t_{j-1}}{t_j-t_{j-1}}C(t_j,S_{ph},S_{cr},X)-\frac{\tau -t_j}{t_j-t_{j-1}}C(t_{j-1},S_{ph},S_{cr},X).$$

(38)

Then, we obtain

$$\begin{array}{ccc}\hfill S_{ph}(t_{n+1})-S_{ph}(0)& =& \frac{1}{\mathsf{\Gamma }(\alpha )}\sum _{j=0}^n{\int }_{t_j}^{t_{j+1}}p(\tau ,S_{ph},S_{cr},X){(t_{n+1}-\tau )}^{\alpha -1}d\tau \hfill \\ & =& \sum _{j=0}^n\left[\frac{h^{\alpha }A(t_j,S_{ph},S_{cr},X)}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-j)}^{\alpha }(n-j+2+\alpha )\right.\right.\hfill \\ & & \left.\left.-{(n-j)}^{\alpha }(n-j+2+2\alpha )\right)\right]\hfill \\ & & -\sum _{j=0}^n\left[\frac{h^{\alpha }A(t_{j-1},S_{ph},S_{cr},X)}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-j)}^{\alpha +1}\right.\right.\hfill \\ & & \left.\left.-{(n-j)}^{\alpha }(n-j+1+\alpha )\right)\right]\hfill \end{array}$$

$$\begin{array}{ccc}\hfill S_{cr}(t_{n+1})-S_{cr}(0)& =& \frac{1}{\mathsf{\Gamma }(\alpha )}\sum _{j=0}^n{\int }_{t_j}^{t_{j+1}}q(\tau ,S_{ph},S_{cr},X){(t_{n+1}-\tau )}^{\alpha -1}d\tau \hfill \\ & =& \sum _{j=0}^n\left[\frac{h^{\alpha }B(t_j,S_{ph},S_{cr},X)}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-j)}^{\alpha }(n-j+2+\alpha )\right.\right.\hfill \\ & & \left.\left.-{(n-j)}^{\alpha }(n-j+2+2\alpha )\right)\right]\hfill \\ & & -\sum _{j=0}^n\left[\frac{h^{\alpha }B(t_{j-1},S_{ph},S_{cr},X)}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-j)}^{\alpha +1}\right.\right.\hfill \\ & & \left.\left.-{(n-j)}^{\alpha }(n-j+1+\alpha )\right)\right]\hfill \end{array}$$

$$\begin{array}{ccc}\hfill X(t_{n+1})-X(0)& =& \frac{1}{\mathsf{\Gamma }(\alpha )}\sum _{j=0}^n{\int }_{t_j}^{t_{j+1}}s(\tau ,S_{ph},S_{cr},X){(t_{n+1}-\tau )}^{\alpha -1}d\tau \hfill \\ & =& \sum _{j=0}^n\left[\frac{h^{\alpha }C(t_j,S_{ph},S_{cr},X)}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-j)}^{\alpha }(n-j+2+\alpha )\right.\right.\hfill \\ & & \left.\left.-{(n-j)}^{\alpha }(n-j+2+2\alpha )\right)\right]\hfill \\ & & -\sum _{j=0}^n\left[\frac{h^{\alpha }C(t_{j-1},S_{ph},S_{cr},X)}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-j)}^{\alpha +1}\right.\right.\hfill \\ & & \left.\left.-{(n-j)}^{\alpha }(n-j+1+\alpha )\right)\right]\hfill \end{array}$$

Thus, the numerical scheme for the model with power law kernel has been obtained. We used this scheme and obtained Figure 1, Figure 2, Figure 3 and Figure 4.

5. Analysis of the Model with the Exponential-Decay Kernel

Next we analyze the model with using the exponential-decay kernel as:

$$\begin{array}{cc}\hfill { }_0^{FFE}{D}_t^{\alpha ,\eta }{S}_{ph}& =D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X.\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill { }_0^{FFE}{D}_t^{\alpha ,\eta }{S}_{cr}& =D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X.\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill { }_0^{FFE}{D}_t^{\alpha ,\eta }X& =-D\beta X+\mu \left(S_{ph},S_{cr}\right)X.\hfill \end{array}$$

(41)

Using the relation between the classical derivative and the fractal derivative yields

$$\begin{array}{cc}\hfill { }_0^{CF}{D}_t^{\alpha }{S}_{ph}& =\eta t^{\eta -1}\left(D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill { }_0^{CF}{D}_t^{\alpha }{S}_{cr}& =\eta t^{\eta -1}\left(D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(43)

$$\begin{array}{cc}\hfill { }_0^{CF}{D}_t^{\alpha }X& =\eta t^{\eta -1}\left(-D\beta X+\mu \left(S_{ph},S_{cr}\right)X\right).\hfill \end{array}$$

(44)

For simplicity, we define

$$\begin{array}{cc}\hfill K(t,S_{ph},S_{cr},X)& =\eta t^{\eta -1}\left(D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill L(t,S_{ph},S_{cr},X)& =\eta t^{\eta -1}\left(D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill M(t,S_{ph},S_{cr},X)& =\eta t^{\eta -1}\left(-D\beta X+\mu \left(S_{ph},S_{cr}\right)X\right).\hfill \end{array}$$

(47)

Then, we obtain

$$\begin{array}{cc}\hfill { }_0^{CF}{D}_t^{\alpha }{S}_{ph}& =K(t,S_{ph},S_{cr},X).\hfill \end{array}$$

(48)

$$\begin{array}{cc}\hfill { }_0^{CF}{D}_t^{\alpha }{S}_{cr}& =L(t,S_{ph},S_{cr},X).\hfill \end{array}$$

(49)

$$\begin{array}{cc}\hfill { }_0^{CF}{D}_t^{\alpha }X& =M(t,S_{ph},S_{cr},X).\hfill \end{array}$$

(50)

Applying the CF integral yields [22]:

$$\begin{array}{ccc}\hfill S_{ph}(t)-S_{ph}(0)& =& \frac{1-\alpha }{M(\alpha )}K(t,S_{ph},S_{cr},X)+\frac{\alpha }{M(\alpha )}{\int }_0^{t}K(\tau ,S_{ph},S_{cr},X)d\tau .\hfill \\ \hfill S_{cr}(t)-S_{cr}(0)& =& \frac{1-\alpha }{M(\alpha )}L(t,S_{ph},S_{cr},X)+\frac{\alpha }{M(\alpha )}{\int }_0^{t}L(\tau ,S_{ph},S_{cr},X)d\tau .\hfill \\ \hfill X(t)-X(0)& =& \frac{1-\alpha }{M(\alpha )}M(t,S_{ph},S_{cr},X)+\frac{\alpha }{M(\alpha )}{\int }_0^{t}M(\tau ,S_{ph},S_{cr},X)d\tau .\hfill \end{array}$$

Discretizing the above equations at $t_{n+1}$ and $t_n$, we receive:

$$\begin{array}{ccc}\hfill S_{ph}^{n+1}& =& S_{ph}^0+\frac{1-\alpha }{M(\alpha )}K(t_n,S_{ph}^n,S_{cr}^n,X^n)\hfill \\ & & +\frac{\alpha }{M(\alpha )}{\int }_0^{t_{n+1}}K(\tau ,S_{ph},S_{cr},X)d\tau \hfill \\ \hfill S_{cr}^{n+1}& =& S_{cr}^0+\frac{1-\alpha }{M(\alpha )}L(t_n,S_{ph}^n,S_{cr}^n,X^n)\hfill \\ & & +\frac{\alpha }{M(\alpha )}{\int }_0^{t_{n+1}}L(\tau ,S_{ph},S_{cr},X)d\tau \hfill \\ \hfill X^{n+1}& =& X^0+\frac{1-\alpha }{M(\alpha )}M(t_n,S_{ph}^n,S_{cr}^n,X^n)\hfill \\ & & +\frac{\alpha }{M(\alpha )}{\int }_0^{t_{n+1}}M(\tau ,S_{ph},S_{cr},X)d\tau \hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill S_{ph}^n& =& S_{ph}^0+\frac{1-\alpha }{M(\alpha )}K(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\hfill \\ & & +\frac{\alpha }{M(\alpha )}{\int }_0^{t_n}K(\tau ,S_{ph},S_{cr},X)d\tau \hfill \\ \hfill S_{cr}^n& =& S_{cr}^0+\frac{1-\alpha }{M(\alpha )}L(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\hfill \\ & & +\frac{\alpha }{M(\alpha )}{\int }_0^{t_n}L(\tau ,S_{ph},S_{cr},X)d\tau \hfill \\ \hfill X^n& =& X^0+\frac{1-\alpha }{M(\alpha )}M(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\hfill \\ & & +\frac{\alpha }{M(\alpha )}{\int }_0^{t_n}M(\tau ,S_{ph},S_{cr},X)d\tau \hfill \end{array}$$

Thus, we reach

$$\begin{array}{ccc}\hfill S_{ph}^{n+1}& =& S_{ph}^n+\frac{1-\alpha }{M(\alpha )}\left(K(t_n,S_{ph}^n,S_{cr}^n,X^n)-K(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\right)\hfill \\ & & +\frac{\alpha }{M(\alpha )}{\int }_{t_n}^{t_{n+1}}K(\tau ,S_{ph},S_{cr},X)d\tau \hfill \\ \hfill S_{cr}^{n+1}& =& S_{cr}^n+\frac{1-\alpha }{M(\alpha )}\left(L(t_n,S_{ph}^n,S_{cr}^n,X^n)-L(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\right)\hfill \\ & & +\frac{\alpha }{M(\alpha )}{\int }_{t_n}^{t_{n+1}}L(\tau ,S_{ph},S_{cr},X)d\tau \hfill \\ \hfill X^{n+1}& =& X^n+\frac{1-\alpha }{M(\alpha )}\left(M(t_n,S_{ph}^n,S_{cr}^n,X^n)-M(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\right)\hfill \\ & & +\frac{\alpha }{M(\alpha )}{\int }_{t_n}^{t_{n+1}}M(\tau ,S_{ph},S_{cr},X)d\tau \hfill \end{array}$$

Using the two-step Lagrange polynomial yields, we receive:

$$\begin{array}{ccc}\hfill S_{ph}^{n+1}& =& S_{ph}^n+\frac{1-\alpha }{M(\alpha )}\left(K(t_n,S_{ph}^n,S_{cr}^n,X^n)-K(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\right)\hfill \\ & & +\frac{\alpha }{M(\alpha )}\left(\frac{3h}{2}K(t_n,S_{ph}^n,S_{cr}^n,X^n)-\frac{h}{2}K(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\right)\hfill \\ \hfill S_{cr}^{n+1}& =& S_{cr}^n+\frac{1-\alpha }{M(\alpha )}\left(L(t_n,S_{ph}^n,S_{cr}^n,X^n)-L(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\right)\hfill \\ & & +\frac{\alpha }{M(\alpha )}\left(\frac{3h}{2}L(t_n,S_{ph}^n,S_{cr}^n,X^n)-\frac{h}{2}L(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\right)\hfill \\ \hfill X^{n+1}& =& X^n+\frac{1-\alpha }{M(\alpha )}\left(M(t_n,S_{ph}^n,S_{cr}^n,X^n)-M(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\right)\hfill \\ & & +\frac{\alpha }{M(\alpha )}\left(\frac{3h}{2}M(t_n,S_{ph}^n,S_{cr}^n,X^n)-\frac{h}{2}M(t_{n-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})\right)\hfill \end{array}$$

Thus, the numerical scheme for the model with exponential decay kernel has been obtained. We used this scheme and obtained Figure 5, Figure 6, Figure 7 and Figure 8.

6. Analysis of the Model with the Mittag–Leffler Kernel

Now, we analyze the model with fractional differentiation operator using the Mittag–Leffler kernel as:

$$\begin{array}{cc}\hfill { }_0^{FFM}{D}_t^{\alpha ,\eta }{S}_{ph}& =D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X.\hfill \end{array}$$

(51)

$$\begin{array}{cc}\hfill { }_0^{FFM}{D}_t^{\alpha ,\eta }{S}_{cr}& =D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X.\hfill \end{array}$$

(52)

$$\begin{array}{cc}\hfill { }_0^{FFM}{D}_t^{\alpha ,\eta }X& =-D\beta X+\mu \left(S_{ph},S_{cr}\right)X.\hfill \end{array}$$

(53)

Then, we obtain

$$\begin{array}{cc}\hfill { }_0^{AB}{D}_t^{\alpha }{S}_{ph}& =\eta t^{\eta -1}\left(D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(54)

$$\begin{array}{cc}\hfill { }_0^{AB}{D}_t^{\alpha }{S}_{cr}& =\eta t^{\eta -1}\left(D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(55)

$$\begin{array}{cc}\hfill { }_0^{AB}{D}_t^{\alpha }X& =\eta t^{\eta -1}\left(-D\beta X+\mu \left(S_{ph},S_{cr}\right)X\right).\hfill \end{array}$$

(56)

For simplicity, we define

$$\begin{array}{cc}\hfill Y(t,S_{ph},S_{cr},X)& =\eta t^{\eta -1}\left(D\left(S_{ph0}-S_{ph}\right)-k_{ph}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(57)

$$\begin{array}{cc}\hfill Z(t,S_{ph},S_{cr},X)& =\eta t^{\eta -1}\left(D\left(S_{cr0}-S_{cr}\right)-k_{cr}·\mu \left(S_{ph},S_{cr}\right)·X\right).\hfill \end{array}$$

(58)

$$\begin{array}{cc}\hfill T(t,S_{ph},S_{cr},X)& =\eta t^{\eta -1}\left(-D\beta X+\mu \left(S_{ph},S_{cr}\right)X\right).\hfill \end{array}$$

(59)

Then, we receive

$$\begin{array}{cc}\hfill { }_0^{AB}{D}_t^{\alpha }{S}_{ph}& =Y(t,S_{ph},S_{cr},X).\hfill \end{array}$$

(60)

$$\begin{array}{cc}\hfill { }_0^{AB}{D}_t^{\alpha }{S}_{cr}& =Z(t,S_{ph},S_{cr},X).\hfill \end{array}$$

(61)

$$\begin{array}{cc}\hfill { }_0^{AB}{D}_t^{\alpha }X& =T(t,S_{ph},S_{cr},X).\hfill \end{array}$$

(62)

Applying the AB integral gives:

$$\begin{array}{ccc}\hfill S_{ph}(t)-S_{ph}(0)& =& \frac{1-\alpha }{AB(\alpha )}Y(t,S_{ph},S_{cr},X)+\frac{\alpha }{AB(\alpha )\mathsf{\Gamma }(\alpha )}{\int }_0^t{(t-p)}^{\alpha -1}Y(p,S_{ph},S_{cr},X)dp.\hfill \\ \hfill S_{cr}(t)-S_{cr}(0)& =& \frac{1-\alpha }{AB(\alpha )}Z(t,S_{ph},S_{cr},X)+\frac{\alpha }{AB(\alpha )\mathsf{\Gamma }(\alpha )}{\int }_0^t{(t-p)}^{\alpha -1}Z(p,S_{ph},S_{cr},X)dp.\hfill \\ \hfill X(t)-X(0)& =& \frac{1-\alpha }{AB(\alpha )}T(t,S_{ph},S_{cr},X)+\frac{\alpha }{AB(\alpha )\mathsf{\Gamma }(\alpha )}{\int }_0^t{(t-p)}^{\alpha -1}T(p,S_{ph},S_{cr},X)dp.\hfill \end{array}$$

Discretizing the above equations at $t_{n+1}$, we receive:

$$\begin{array}{ccc}\hfill S_{ph}^{n+1}& =& S_{ph}^0+\frac{1-\alpha }{AB(\alpha )}Y(t_{n+1},S_{ph}^n,S_{cr}^n,X^n)\hfill \\ & & +\frac{\alpha }{AB(\alpha )\mathsf{\Gamma }(\alpha )}{\int }_0^{t_{n+1}}{(t_{n+1}-p)}^{\alpha -1}Y(p,S_{ph},S_{cr},X)dp\hfill \\ \hfill S_{cr}^{n+1}& =& S_{cr}^0+\frac{1-\alpha }{AB(\alpha )}Z(t_{n+1},S_{ph}^n,S_{cr}^n,X^n)\hfill \\ & & +\frac{\alpha }{AB(\alpha )\mathsf{\Gamma }(\alpha )}{\int }_0^{t_{n+1}}{(t_{n+1}-p)}^{\alpha -1}Z(p,S_{ph},S_{cr},X)dp\hfill \\ \hfill X^{n+1}& =& X^0+\frac{1-\alpha }{AB(\alpha )}T(t_{n+1},S_{ph}^n,S_{cr}^n,X^n)\hfill \\ & & +\frac{\alpha }{AB(\alpha )\mathsf{\Gamma }(\alpha )}{\int }_0^{t_{n+1}}{(t_{n+1}-p)}^{\alpha -1}T(p,S_{ph},S_{cr},X)dp\hfill \end{array}$$

Then, we obtain

$$\begin{array}{ccc}\hfill S_{ph}^{n+1}& =& S_{ph}^0+\frac{1-\alpha }{AB(\alpha )}Y(t_{n+1},S_{ph}^n,S_{cr}^n,X^n)\hfill \\ & +& \frac{\alpha }{AB(\alpha )}\sum _{i=0}^n\left[\frac{h^{\alpha }Y(t_i,S_{ph}^n,S_{cr}^n,X^n)}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-i)}^{\alpha }(n-i+2+\alpha )\right.\right.\hfill \\ & & \left.\left.-{(n-i)}^{\alpha }(n-i+2+2\alpha )\right)\right]\hfill \\ & -& \frac{\alpha }{AB(\alpha )}\sum _{i=0}^n\left[\frac{h^{\alpha }Y(t_{i-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-i)}^{\alpha +1}\right.\right.\hfill \\ & & \left.\left.-{(n-i)}^{\alpha }(n-i+1+\alpha )\right)\right]\hfill \end{array}$$

$$\begin{array}{ccc}\hfill S_{cr}^{n+1}& =& S_{cr}^0+\frac{1-\alpha }{AB(\alpha )}Z(t_{n+1},S_{ph}^n,S_{cr}^n,X^n)\hfill \\ & +& \frac{\alpha }{AB(\alpha )}\sum _{i=0}^n\left[\frac{h^{\alpha }Z(t_i,S_{ph}^n,S_{cr}^n,X^n)}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-i)}^{\alpha }(n-i+2+\alpha )\right.\right.\hfill \\ & & \left.\left.-{(n-i)}^{\alpha }(n-i+2+2\alpha )\right)\right]\hfill \\ & -& \frac{\alpha }{AB(\alpha )}\sum _{i=0}^n\left[\frac{h^{\alpha }Z(t_{i-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-i)}^{\alpha +1}\right.\right.\hfill \\ & & \left.\left.-{(n-i)}^{\alpha }(n-i+1+\alpha )\right)\right]\hfill \end{array}$$

$$\begin{array}{ccc}\hfill X^{n+1}& =& X^0+\frac{1-\alpha }{AB(\alpha )}T(t_{n+1},S_{ph}^n,S_{cr}^n,X^n)\hfill \\ & +& \frac{\alpha }{AB(\alpha )}\sum _{i=0}^n\left[\frac{h^{\alpha }T(t_i,S_{ph}^n,S_{cr}^n,X^n)}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-i)}^{\alpha }(n-i+2+\alpha )\right.\right.\hfill \\ & & \left.\left.-{(n-i)}^{\alpha }(n-i+2+2\alpha )\right)\right]\hfill \\ & -& \frac{\alpha }{AB(\alpha )}\sum _{i=0}^n\left[\frac{h^{\alpha }T(t_{i-1},S_{ph}^{n-1},S_{cr}^{n-1},X^{n-1})}{\mathsf{\Gamma }(\alpha +2)}\left({(n+1-i)}^{\alpha +1}\right.\right.\hfill \\ & & \left.\left.-{(n-i)}^{\alpha }(n-i+1+\alpha )\right)\right].\hfill \end{array}$$

Thus, the numerical scheme for the model with Mittag–Leffler kernel has been obtained. We used this scheme and obtained Figure 9, Figure 10, Figure 11 and Figure 12.

Remark 1.

A valuable and huge benefit of fractional differentiation operator is that we can formulate models better defining the systems with memory effects. It is known that the use of integro-differential kernels of a certain type in integro-differential equations leads us to the fractional derivative operator [31]. The kernels with degree functions in integro-differential equations of the Voltaire type [32], allow us to describe this memory effect [33,34].

Fractal–fractional operators with different memories are related to the non-local dynamical systems’ different types of relaxation processes. Thus, models with fractional differentiation operators are more effective and valuable.

7. Results and Discussions

In this section, we present numerical simulations for different fractional order and fractal dimension values. We also add the classical derivative with the integer fractal dimension equal to 1.

We chose fractal dimension as the integer and noninteger in the figures. We discuss the results with the three kernels described in Section 5, Section 6 and Section 7. The figures $\alpha ,\beta$, and $\eta$ are between zero and one. In these simulations, $\beta$ is the parameter given on the model, $\eta$ is the fractal dimension, and $\alpha$ is the fractional order. We see the effect of the fractional order $\alpha$ under different kernels and values of the parameter $\beta$ and the fractal dimension $\eta$. Figure 1 and Figure 2 show the numerical simulations for $\beta =1$, the fractal dimensions $\eta =1$ and $\eta =0.8$, and for different fractional order $\alpha$ values with the power-law kernel. We also show how this kernel behaves for $\beta =0.5$ and the fractal dimension $\eta =1$ and $\eta =0.9$ in Figure 3 and Figure 4. We see that the convergence is faster for the case $\beta =1$ than to the case $\beta =0.5$, as long as the fractal dimension is close to 1. The concentrations $S_{ph}(t)$ and $S_{cr}(t)$ decrease as long as $\alpha$ decreases. In all the cases, the concentration $X(t)$ decreases to 0.

The results for the exponential-decay kernel are shown for $\beta =1$ in Figure 5 (with fractal dimension $\eta =1$) and Figure 6 (with fractal dimension $\eta =0.7$). We demonstrate the results for $\beta =0.8$ and $\eta =1$ (Figure 7) and $\eta =0.7$ (Figure 8). Despite varying the parameters $\beta$, $\alpha$, and the fractal dimension, there are fewer differences in the concentrations with respect to the results shown by the power-law kernel.

Finally, in Figure 9, Figure 10, Figure 11 and Figure 12, we show the results for the Mittag–Leffler kernel. The numerical simulations for $\beta =1$ are shown in Figure 9 ($\eta =1$) and Figure 10 ($\eta =0.5$). We also see the behavior of the solution for $\beta =0.5,\eta =1$ in Figure 11 and $\beta =0.5,\eta =0.6$ in Figure 12.

We have seen that the exponential-decay kernel is the one that converges faster to the equilibrium, with the smaller difference among concentrations of the substances.

8. Conclusions

This work provides a mathematical model for breaking down a phenol and p-cresol mixture in a bioreactor with continuous stirring. Three nonlinear ordinary differential equations served as the foundation for the model. The equilibrium points of the model were identified, and their stability was examined and shown. Additionally, we used the fractional differentiation operator to examine the model and three distinct kernels to examine the effects of the fractal dimension and fractional order. We developed very efficient numerical algorithms for biomass, phenol, and p-cresol concentrations. To demonstrate the accuracy of the suggested approach, we offered numerical simulations for different $\alpha$ and $\beta$ values. The right choice of model parameters would require validation with experimental data.