Bioenergy potential of agricultural crop residues and municipal solid waste in Cameroon

Robinson J. Tanyi; Muyiwa S Adaramola; Robinson J. Tanyi; Muyiwa S Adaramola

doi:10.3934/energy.2023002

AIMS Energy

2023, Volume 11, Issue 1: 31-46. doi: 10.3934/energy.2023002

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Research article Special Issues

Bioenergy potential of agricultural crop residues and municipal solid waste in Cameroon

Robinson J. Tanyi ¹,
Muyiwa S Adaramola ^{2
,
,}

1.
Department of Mechanical Engineering, Kwame Nkrumah University of Science and Technology, Kumasi Ghana
2.
Faculty of Environmental Sciences and Natural Resource Management, Norwegian University of Life Sciences (NMBU), P.O. Box 5003, NO-1432 Ås, Norway

Received: 22 July 2022 Revised: 18 December 2022 Accepted: 22 December 2022 Published: 10 January 2023

Biomass has emerged as an important and promising energy source, particularly in developing countries, owing to continuous research for sustainable energy sources that do not interfere with food, water or land needs. This study introduces the surplus availability factor (SAF), minimum, average and maximum biogas production technique in the assessment of crop production data in 2020 to provide a more precise and current estimate of Cameroon's crop residue and municipal solid waste (MSW) bioenergy potential. Crop residues contributed roughly 96% while MSW contributed the remaining 4% of the total bioenergy potential of 606 PJ per year. The bioenergy potential was calculated using crop production statistics derived from the FAOSTAT database of the Food and Agriculture Organization, while the residue-to-product ratio (RPR) and surplus availability factors (SAF) were found from related studies. The study concludes that crop residues and MSW have significant energy potential capable of meeting the country's electricity, transport fuel and biogas demand while simultaneously mitigating climate change through the capture of about 1.6 billion kg of CO₂ through biogas recovery. It also highlights the lack of accurate and up-to-date data on the country's biomass potential and recommends ground data collection and geospatial mapping of areas with enormous potential for these resources to guide policymakers and investment plans.

Keywords:

Citation: Robinson J. Tanyi, Muyiwa S Adaramola. Bioenergy potential of agricultural crop residues and municipal solid waste in Cameroon[J]. AIMS Energy, 2023, 11(1): 31-46. doi: 10.3934/energy.2023002

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Abstract

1. Introduction

Fractional calculus has been concerned with integration and differentiation of fractional (non-integer) order of the function. Riemann and Liouville defined the concept of fractional order intgro-differential equations ^[1]. Fractional calculus has developed an extensive attraction in current years in applied mathematics such as physics, medical, biology and engineering ^{[2,3,4,5,6,7,8]}. Whenever dealing with the fractional integro-differential equation many authors consider the terms Caputo fractional derivative, Riemann-Liouville and Grunwald-Letnikvo ^{[9,10,11,12,13]}. The subject fractional calculus has many applications in widespread and diverse field of science and engineering such as fractional dynamics in the trajectory control of redundant manipulators, viscoelasticity, electrochemistry, fluid mechanics, optics and signals processing etc.

Fractional integro-differential equations having some uncertainties in the form of boundary conditions, initial conditions and so on ^[14,15,16]. To resolve these type of uncertainties mathematicians introduced some concepts fuzzy set theory is one of them.

Zadeh introduced the concept of fuzzy set theory ^{[17,18,19,20]}. Later on Prade and Dubois ^[21,22], Nahmias ^[23], Tanaka and Mizumoto ^[24]. All of them experienced that the fuzzy number as a location of r-cut $0 \leqslant r \leqslant 1$ .

Many authors investigated some numerical techniques related to these problem which include the existence of the solution for discontinuous ^[25], reproducing kernel algorithm ^[26], integro-differential under generalized Caputo differentiability ^[27], A domain decomposition method ^[28], fractional differential transform method ^[29], Jacobi polynomial operational matrix ^[30], global solutions for nonlinear fuzzy equations ^[31], radioactivity decay model ^[32], Caputo-Katugampola fractional derivative approach ^[33], two-dimensional legendre wavelet method ^[34], fuzzy Laplace transform ^[35], fuzzy sumudu transform ^[36]. Further we can see ^{[37,38,39,40]}

Optimal Homotopy Asymptotic Method (OHAM) is one of the powerful techniques introduced by Marinca at al. ^[41,42,43] for approximate solution of differential equations. OHAM attracted an enormous importance in solving various problems in different field of science. Iqbal et al. applied this technique to Klein-Gordon equations and singular Lane-Emden type equation ^[44]. Sheikholeslami et al. used the proposed method for investigation of the laminar viscous flow and magneto hydrodynamic flow in a permeable channel ^[45]. Hashmi et al. obtained the solution of nonlinear Fredholm integral equations using OHAM ^[46]. Nawaz at al. applied the proposed method for solution of fractional order integro-differential equations ^[47], fractional order partial differential equations ^[48] and three-dimensional integral equations ^[49].

Aim of our study is to extend OHAM for solution of system of fuzzy Volterra integro differential equation of fractional order of the following form

$D_x^\alpha u(x) = h(x) + \mathop \smallint \limits_a^x k(x, t)u(t)dt, \, \, \, \, \, 0 \leqslant \alpha \leqslant 1, \, \, x \in [0, 1],$

(1.1)

with the given initial condition

${u^k}(0), \, \, \, \, {u_0}^k(x), \, \, \, k = 1, 2, 3, ...., \eta - 1, \, \, \, \eta - 1 < \alpha < \eta , \, \, \, \, \eta \in N,$

Where $D_x^\alpha$ represents the fuzzy fractional derivative in Caputo sense for fractional order of $\alpha$ with respect to $x$ , $h:[a, b] \to {\mathbb{R}_{\cal{F}}}$ is fuzzy valued function, $k(x, t)$ is arbitrary kernel ${u_0}(x) \in {\mathbb{R}_{\cal{F}}}$ is an unknown solution. ${\mathbb{R}_{\cal{F}}}$ represent set of all fuzzy valued function on real line.

The remaining paper is structured as follows: A brief overview on some elementary concept, notations and definitions of fuzzy calculus and fuzzy fractional calculus are discussed in section 2. Analysis of the technique is presented in section 3. Proposed method is applied to solve fuzzy fractional order Volterra integro-differential equations in section 4. Result and discussion of the paper is given in section 5 and section 6 is the conclusion of the paper.

2. Preliminaries

In literature there exist various definitions of fuzzy calculus and fuzzy fractional calculus ^[50]. Some elementary concept, notations and definitions of fuzzy calculus and fuzzy fractional calculus related to this study are provided in this section.

Definition 2.1. The Riemann-Liouville fractional integral operator $I_x^\alpha$ of order $\alpha$ is ^[50]:

$I_x^\alpha u(x) = \, \left\{ \begin{array}{l} \frac{1}{{\Gamma (\alpha )}}\mathop \smallint \limits_0^x {(x - t)^{\alpha - 1}}u(t)dt = 0, \, \, \, \, \alpha > 0, \, \, \, \hfill \\ u(x), \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \alpha = 0.\, \, \hfill \\ \end{array} \right. \, \, \, \, \, \, \, \, \,$

(2.1)

Definition 2.2. Caputo partial fractional Derivative operator $D_x^\alpha$ of order $\alpha$ with respect to $x$ is defined as follow ^[50]:

$D_x^\alpha u(x) = \, \left\{ \begin{array}{l} \frac{1}{{\Gamma (\eta - \alpha )}}\mathop \smallint \limits_0^x {(x - t)^{\eta - \alpha - 1}}{u^{(n)}}(t)dt = 0, \, \, \, \, \, \, \, \, \, \, \, \eta - 1 < \alpha \leqslant \eta , \, \, \, \hfill \\ \frac{{{d^\eta }\, u(x)}}{{d{x^\eta }}}, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \alpha = \eta \in N.\, \, \hfill \\ \end{array} \right.$

(2.2)

which clearly shows that

$D_x^\alpha I_x^\alpha u(x) = u(x)$

(2.3)

Definition 2.3. A fuzzy number $\sigma$ is a mapping $\sigma :\mathbb{R} \to [0, 1]$ , satisfy the following property:

a. $\sigma$ is normal that is, $\exists \, \, {x_0} \in \mathbb{R}$ with $u({x_0}) = 1$ ^[51,52].

b. $\sigma$ is a convex fuzzy set that is, $u(\lambda x + (1 - \lambda )y) \geqslant \min \left\{ {u(x), u(y)} \right\}$ for all $x, y \in \mathbb{R}, {\text{ }}\lambda \in [0, 1].$

c. $\sigma$ is upper semi-continuous in $\mathbb{R}.$

d. $\overline {\left\{ {x \in \mathbb{R}:u(x) > 0} \right\}}$ is compact.

Definition 2.4. Parametric form of fuzzy number $\sigma$ represented by an order pair $(\underline \sigma , \bar \sigma )$ of the function $\left( {\underline \sigma (r), \bar \sigma (r)} \right)$ , satisfies the following conditions ^[52,53]:

a. $\underline \sigma (r)$ is bounded monotonic increasing left continuous $\forall \, \, r \in [0, 1].$

b. $\bar \sigma (r)$ is bounded monotonic decreasing left continuous $\forall \, \, r \in [0, 1].$

c. $\underline \sigma (r) \leqslant \bar \sigma (r)\, \, \forall \, \, r \in [0, 1].$

Definition 2.5. Addition and scalar multiplication of fuzzy number is given as:

a. $\left( {{\sigma _1} \oplus {\sigma _2}} \right) = \left( {{{\underline \sigma }_1}(r) + {{\underline \sigma }_2}(r), {{\bar \sigma }_1}(r) + {{\bar \sigma }_2}(r)} \right)$

b. $\left( {k \otimes \sigma } \right) = \left\{ \begin{array}{l} \left( {\underline \sigma (r), \bar \sigma (r)} \right), \, \, \, \, \, k \geqslant 0, \hfill \\ \left( {\underline \sigma (r), \bar \sigma (r)} \right), \, \, \, \, \, k < 0. \hfill \\ \end{array} \right.$

Definition 2.6. A fuzzy real valued function ${\sigma _1}, {\sigma _2}:[a, b] \to \mathbb{R},$ then in ^[54]:

${D_U}({\sigma _1}, {\sigma _2}) = \sup \left\{ {D({\sigma _1}(x), {\sigma _2}(x))|x \in [a, b]} \right\}.$

Definition 2.7. Assume $u:[a, b] \to {\mathbb{R}_{\cal{F}}}.$ For every partition ${\rm P} = \left\{ {{\sigma _0}, {\sigma _1}, {\sigma _2}, {\sigma _3}, ...., {\sigma _n}} \right\}$ and arbitrary ${\ell _i}:{\sigma _{i - 1}} \leqslant {\ell _i} \leqslant {\sigma _i}, {\text{ }}2 \leqslant i \leqslant n$ consider

${\mathbb{R}_p} = \mathop \Sigma \limits_{i = 2}^n u({\ell _j})({\sigma _i} - {\sigma _{i - 1}}).$ The definite integral of $u(x)$ over $[\alpha , \beta ]$ is

$\mathop \smallint \limits_\alpha ^\beta u(x)dx = \lim {\mathbb{R}_\rho },$

which show existence of limit in metric ^[55].

Definite integral exist if $u(x)$ is continuous in metric $D$ ^[51]:

$\begin{gathered} \left( {\underline {\mathop \smallint \limits_\alpha ^\beta u(x)dx} } \right) = \mathop \smallint \limits_\alpha ^\beta \underline u (x)dx, \hfill \\ \left( {\overline {\mathop \smallint \limits_\alpha ^\beta u(x)dx} } \right) = \mathop \smallint \limits_\alpha ^\beta \overline u (x)dxt. \hfill \\ \end{gathered}$

3. Application of OHAM

By considering definition 2.4. as discussed in section 2, Eq (1.1) becomes:

$\left\{ \begin{array}{l} D_x^\alpha \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} (x, r) - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{h} (x, r) - \mathop \smallint \limits_a^x k(x, t)\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} (t, r)dt = 0, \hfill \\ D_x^\alpha \bar u(x, r) - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{h} (x, r) - \mathop \smallint \limits_a^x k(x, t)\bar u(t, r)dt = 0, \hfill \\ \end{array} \right. \, \, \, \, \, 0 \leqslant \alpha \leqslant 1, \, 0 \leqslant r \leqslant 1, \, \, \, x \in [0, 1],$

(3.1)

with the given initial condition

${\left[ {{u^k}(0)} \right]^r}, \, \, \, \, \left( {{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}^k(x, r), {{\bar u}_0}^k(x, r)} \right), \, \, \, k = 1, 2, 3, ...., \eta - 1, \, \, \, \eta - 1 < \alpha < \eta , \, \, \, \, \eta \in N,$

(3.2)

The homotopy of OHAM ^[41,42,43], constructed as follow:

$\left\{ \begin{array}{l} (1 - \rho )\left( {\frac{{{\partial ^\alpha }\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\upsilon } (x, r;\rho )}}{{\partial {t^\alpha }}} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{h} (x, r)} \right) = H(\rho )\left( {\frac{{{\partial ^\alpha }\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\upsilon } (x, r;\rho )}}{{\partial {t^\alpha }}} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{h} (x, r) - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\delta } (\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\upsilon } , r)} \right), \hfill \\ (1 - \rho )\left( {\frac{{{\partial ^\alpha }\bar \upsilon (x, r;\rho )}}{{\partial {t^\alpha }}} - \bar h(x, r)} \right) = H(\rho )\left( {\frac{{{\partial ^\alpha }\bar \upsilon (x, r;\rho )}}{{\partial {t^\alpha }}} - \bar h(x, r) - \bar \delta (\bar \upsilon , r)} \right). \hfill \\ \end{array} \right.$

(3.3)

where $\rho \in [0, 1]$ , $H(\rho ) = \sum\limits_{m\, \, \geqslant 1} {{c_m}{\rho ^m}}$ for all $\rho \ne 0$ is an auxiliary function, if $\rho = 0$ then $H(0) = 0$ where

${\left\{ \begin{array}{l} \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\upsilon } (x, r, 0) = {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}(x, r)\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\upsilon } (x, r;1) = \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} (x, r), \hfill \\ \bar \upsilon (x, r, 0) = {{\bar u}_0}(x, r)\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \bar \upsilon (x, r;1) = \bar u(x, r). \hfill \\ \end{array} \right. _{\, \, }}\,$

and ${c_m}$ represent auxiliary constants. Using Taylor's series to expand $\upsilon (x, r;\rho )$ about $\rho$ we get

$\left\{ \begin{array}{l} \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\upsilon } (x, r;\rho ) = {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}(x, r) + \sum\limits_{m\, \, \geqslant 1} {{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_m}(x, r){\rho ^m}} , \hfill \\ \bar \upsilon (x, r;\rho ) = {{\bar u}_0}(x, r) + \sum\limits_{m\, \, \geqslant 1} {{{\bar u}_m}(x, r){\rho ^m}} . \hfill \\ \end{array} \right.$

(3.4)

Inserting Eq (3.4) into Eq (3.3) we get series of the problems by comparing the like power of $\rho$ given as follow:

${\rho ^0}:\, \, \, \left\{ \begin{array}{l} \, {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}(x, r) - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{h} (x, r) = 0, \hfill \\ \, {{\bar u}_0}(x, r) - \bar h(x, r) = 0. \hfill \\ \end{array} \right.$

(3.5)

${\rho ^1}:\, \, \, \, \left\{ \begin{array}{l} {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_1}(x, r) + {c_1}\delta ({{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}) + (1 + {c_1}) + {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}(x, r) = 0, \hfill \\ {{\bar u}_1}(x, r) + {c_1}\delta ({{\bar u}_0}) + (1 + {c_1}) + {{\bar u}_0}(x, r) = 0. \hfill \\ \end{array} \right.$

(3.6)

${\rho ^2}:\, \, \, \, \left\{ \begin{array}{l} {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_2}(x, r) + {c_1}\delta ({{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_1}) + {c_2}\delta ({{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}) + {c_2}(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{h} (x, r) - {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}(x, r)) \hfill \\ - (1 + {c_1}){{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_1}(x, r) = 0, \hfill \\ {{\bar u}_2}(x, r) + {c_1}\delta ({{\bar u}_1}) + {c_2}\delta ({{\bar u}_0}) + {c_2}(\bar h(x, r) - {{\bar u}_0}(x, r)) \hfill \\ - (1 + {c_1}){{\bar u}_1}(x, r) = 0. \hfill \\ \end{array} \right.$

(3.7)

${\rho ^n}:\, \, \, \, \left\{ \begin{array}{l} {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_n}(x, r) + {c_1}\delta ({{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_n}) + {c_2}\delta ({{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_{n - 1}}) + {c_3}(h + \delta ({{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}))... - {c_2}{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_{n - 1}}(x, r) - (1 + {c_1}){{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_n}(x, r) = 0, \hfill \\ {{\bar u}_n}(x, r) + {c_1}\delta ({{\bar u}_n}) + {c_2}\delta ({{\bar u}_{n - 1}}) + {c_3}(h + \delta ({{\bar u}_0}))... - {c_2}{{\bar u}_{n - 1}}(x, r) - (1 + {c_1}){{\bar u}_n}(x, r) = 0. \hfill \\ \end{array} \right.$

(3.8)

For calculating the constants ${c_1}, {c_2}, {c_3}...$ , m^th order optimum solution becomes

$\left\{ \begin{array}{l} {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }^m}(x, r, {c_l}) = {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_0}(x, r) + \sum\limits_{k = 1}^m {{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }_k}(x, r, {c_l}), } \, \, \, \, l = 1, 2, 3, ...m, \hfill \\ {{\bar u}^m}(x, r, {c_l}) = {{\bar u}_0}(x, r) + \sum\limits_{k = 1}^m {{{\bar u}_k}(x, r, {c_l}), } \, \, \, \, l = 1, 2, 3, ...m. \hfill \\ \end{array} \right.$

(3.9)

Putting Eq (3.9) into Eq (3.1), we can found our residual given as follow:

$\left\{ \begin{array}{l} \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{R} \left( {x, r;{c_l}} \right) = {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} }^m}(x, r;{c_l}) - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{h} (x, r) - \delta (\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} ), \, \, l = 1, 2, ... \hfill \\ \bar R\left( {x, r;{c_l}} \right) = {{\bar u}^m}(x, r;{c_l}) - \bar h(x, r) - \delta (\bar u), \, \, l = 1, 2, ... \hfill \\ \end{array} \right.$

(3.10)

If $R\left( {x, r;{c_l}} \right) = 0,$ then ${\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{u} ^m}(x, r;{c_l})\, \, \& \, \, {\bar u^m}(x, r;{c_l})$ will be the exact solutions.

Optimum solution contains some auxiliary constants; the optimal values of these constants are obtained through various techniques. In the present work, we have used the least square method ^[56,57]. The method of least squares is a powerful technique for obtaining the values of auxiliary constants. By putting the optimal values of these constants in Eq (8), we obtain the OHAM solution.

4. Application and accuracy

Problem 4.1. Consider system of fuzzy fractional order Volterra integro-differential equation as ^[58]:

$\left\{ \begin{array}{l} D_x^\alpha \underline u (x, r) = \left( {r - 1} \right) + \int\limits_0^x {\underline u (t, r)dt} \hfill \\ D_x^\alpha \bar u(x, r) = \left( {1 - r} \right) + \int\limits_0^x {\bar u(t, r)dt} \hfill \\ \end{array} \right. , \, \, \, \, 0 < \alpha \leqslant 1, \, \, x \in [0, 1],$

(4.1)

subject to the fuzzy initial condition ${\left[ {u(0)} \right]^r} = \left[ {r - 1, 1 - r} \right],$ and for $\alpha = 1$ fuzzy fractional order Volterra integro-differential equations the exact solution is ${\left[ {u(x)} \right]^r} = \left[ {r - 1, 1 - r} \right]Sinh(x)$ and $0 \leqslant r \leqslant 1.$

By follow the technique as discussed in section 3, we get series of problems and their solutions as:

$\left\{ \begin{array}{l} {D_x}^\alpha {\underline u _0}(x, r) + (1 - r) = 0, \hfill \\ {D_x}^\alpha {\overline u _0}(x, r) + (r - 1) = 0. \hfill \\ \end{array} \right.$

(4.2)

$\left\{ \begin{array}{l} {D_x}^\alpha {\underline u _1}(x, r) - 1 + r - {c_1} + r{c_1} + \left( {\mathop \smallint \limits_0^x {{\underline u }_0}(t, r)dt} \right){c_1} - {D_x}^\alpha {\underline u _0}(x, r) - {c_1}{D_x}^\alpha {\underline u _0}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_1}(x, r) + 1 - r + {c_1} - r{c_1} + \left( {\mathop \smallint \limits_0^x {{\bar u}_0}(t, r)dt} \right){c_1} - {D_x}^\alpha {{\bar u}_0}(x, r) - {c_1}{D_x}^\alpha \bar u(x, r) = 0. \hfill \\ \end{array} \right.$

(4.3)

$\left\{ \begin{array}{l} {D_x}^\alpha {\underline u _2}(x, r) + \left( {\mathop \smallint \limits_0^x {{\underline u }_1}(t, r)dt} \right){c_1} - {c_2} + r{c_2} + \left( {\mathop \smallint \limits_0^x {{\underline u }_0}(t, r)dt} \right){c_2} - {c_2}{D_x}^\alpha {\underline u _0}(x, r) - \hfill \\ {D_x}^\alpha {\underline u _1}(x, r) - {c_1}{D_x}^\alpha {\underline u _0}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_2}(x, r) + \left( {\mathop \smallint \limits_0^x {{\bar u}_1}(t, r)dt} \right){c_1} + {c_2} - r{c_2} + \left( {\mathop \smallint \limits_0^x {{\bar u}_0}(t, r)dt} \right){c_2} - {c_2}{D_x}^\alpha {{\bar u}_0}(x, r) - \hfill \\ {D_x}^\alpha {{\bar u}_1}(x, r) - {c_1}{D_x}^\alpha {{\bar u}_0}(x, r) = 0. \hfill \\ \end{array} \right.$

(4.4)

$\left\{ \begin{array}{l} {D_x}^\alpha {\underline u _3}(x, r) + \left( {\mathop \smallint \limits_0^x {{\underline u }_2}(t, r)dt} \right){c_1} + \left( {\mathop \smallint \limits_0^x {{\underline u }_1}(t, r)dt} \right){c_2} - {c_3} + r{c_3} + \left( {\mathop \smallint \limits_0^x {{\underline u }_0}(t, r)dt} \right){c_3} - {c_3}{D_x}^\alpha {\underline u _0}(x, r) \hfill \\ - {c_2}{D_x}^\alpha {\underline u _1}(x, r) - {D_x}^\alpha {\underline u _2}(x, r) - {c_1}{D_x}^\alpha {\underline u _2}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_3}(x, r) + \left( {\mathop \smallint \limits_0^x {{\bar u}_2}(t, r)dt} \right){c_1} + \left( {\mathop \smallint \limits_0^x {{\bar u}_1}(t, r)dt} \right){c_2} + {c_3} - r{c_3} + \left( {\mathop \smallint \limits_0^x {{\bar u}_0}(t, r)dt} \right){c_3} - {c_3}{D_x}^\alpha {{\bar u}_0}(x, r) \hfill \\ - {c_2}{D_x}^\alpha {{\bar u}_1}(x, r) - {D_x}^\alpha {{\bar u}_2}(x, r) - {c_1}{D_x}^\alpha {{\bar u}_2}(x, r) = 0. \hfill \\ \end{array} \right.$

(4.5)

Their solutions are

$\left\{ \begin{array}{l} {\underline u _0}(x, r) = \frac{{( - 1 + r){x^\alpha }}}{{\alpha \, \Gamma (\alpha )}} \hfill \\ {{\bar u}_0}(x, r) = - \frac{{( - 1 + r){x^\alpha }}}{{\alpha \, \Gamma (\alpha )}} \hfill \\ \end{array} \right. ,$

(4.6)

$\left\{ \begin{array}{l} {\underline u _1}(x, r) = - \frac{{( - 1 + r){x^{1 + 2\alpha }}{c_1}}}{{\Gamma (2 + 2\alpha )}}, \hfill \\ {{\bar u}_1}(x, r) = \frac{{( - 1 + r){x^{1 + 2\alpha }}{c_1}}}{{\Gamma (2 + 2\alpha )}}. \hfill \\ \end{array} \right.$

(4.7)

$\left\{ \begin{array}{l} {\underline u _2}(x, r) = ( - 1 + r){x^{1 + 2\alpha }}\left( {\frac{{{x^{1 + \alpha }}c_1^2}}{{\Gamma (3 + 3\alpha )}} - \frac{{{c_1} + c_1^2 + {c_2}}}{{\Gamma (2 + 2\alpha )}}} \right), \hfill \\ {{\bar u}_2}(x, r) = ( - 1 + r){x^{1 + \alpha }}\left( { - \frac{{{x^{1 + \alpha }}c_1^2}}{{\Gamma (3 + 3\alpha )}} + \frac{{{c_1} + c_1^2 + {c_2}}}{{\Gamma (2 + 2\alpha )}}} \right). \hfill \\ \end{array} \right.$

(4.8)

$\left\{ \begin{array}{l} {\underline u _3}(x, r) = ( - 1 + r){x^{1 + 2\alpha }}\left( { - \frac{{{x^{2 + 2\alpha }}c_1^3}}{{\Gamma (4 + 4\alpha )}} + \frac{{2{x^{1 + \alpha }}{c_1}({c_1} + c_1^2 + {c_2})}}{{\Gamma (3 + 3\alpha )}} - \frac{{{c_1} + 2c_1^2 + c_1^3 + {c_2} + 2{c_1}{c_2} + {c_3}}}{{\Gamma (2 + 2\alpha )}}} \right), \hfill \\ {{\bar u}_3}(x, r) = ( - 1 + r){x^{1 + 2\alpha }}\left( {\frac{{{x^{2 + 2\alpha }}c_1^3}}{{\Gamma (4 + 4\alpha )}} - \frac{{2{x^{1 + \alpha }}{c_1}({c_1} + c_1^2 + {c_2})}}{{\Gamma (3 + 3\alpha )}} + \frac{{{c_1} + 2c_1^2 + c_1^3 + {c_2} + 2{c_1}{c_2} + {c_3}}}{{\Gamma (2 + 2\alpha )}}} \right). \hfill \\ \end{array} \right.$

(4.9)

Adding (4.6), (4.7), (4.8) and (4.9), one can construct $\underline u (x, r)$ & $\bar u(x, r)$ :

$\left\{ \begin{array}{l} \underline u (x, r) = ( - 1 + r){x^\alpha }\left( \begin{gathered} \frac{1}{{\Gamma (1 + \alpha )}} - \frac{{{x^{3 + 3\alpha }}c_1^3}}{{\Gamma (4 + 4\alpha )}} + \frac{{{x^{2 + 2\alpha }}{c_1}({c_1}(3 + 2{c_1}) + 2{c_2})}}{{\Gamma (3 + 3\alpha )}} - \hfill \\ \frac{{{x^{1 + \alpha }}(2{c_2} + {c_1}(3 + {c_1}(3 + {c_1}) + 2{c_2}) + {c_3})}}{{\Gamma (2 + 2\alpha )}} \hfill \\ \end{gathered} \right), \hfill \\ \bar u(x, r) = ( - 1 + r){x^\alpha }\left( \begin{gathered} - \frac{1}{{\Gamma (1 + \alpha )}} + \frac{{{x^{3 + 3\alpha }}c_1^3}}{{\Gamma (4 + 4\alpha )}} - \frac{{{x^{2 + 2\alpha }}{c_1}({c_1}(3 + 2{c_1}) + 2{c_2})}}{{\Gamma (3 + 3\alpha )}} + \hfill \\ \frac{{{x^{1 + \alpha }}(2{c_2} + {c_1}(3 + {c_1}(3 + {c_1}) + 2{c_2}) + {c_3})}}{{\Gamma (2 + 2\alpha )}} \hfill \\ \end{gathered} \right). \hfill \\ \end{array} \right.$

(4.10)

Values of ${c_1}, \, \, {c_2}$ and ${c_3}$ contain is in Eq (4.10)

Substituting the values from into Eq (4.10), the approximate solutions for $\underline u (x, r)$ & $\bar u(x, r)$ at different values of $\alpha$ taking $r = 0.75$ respectively is as follow

$\alpha = 0.7$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.2751369{x}^{0.7}-0.08602097{x}^{2.4}-0.25{x}^{2.4}(-0.00904995+0.0376716{x}^{1.7})\\ -0.25{x}^{2.4}(0.000425859-0.00198165{x}^{1.7}+0.0021734872{x}^{3.4}), \\ \overline{u}(x, r)\approx 0.2751369{x}^{0.7}+0.08602097{x}^{2.4}-0.25{x}^{2.4}(0.00904995-0.0376716{x}^{1.7})-\\ 0.25{x}^{2.4}(-0.000425859+0.00198165{x}^{1.7}-0.0021734872{x}^{3.4}).\end{array} \right.$

(4.11)

$\alpha = 0.8$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.2684178{x}^{0.8}-0.0685589{x}^{2.6}-0.25{x}^{2.6}(-0.005391327+0.023297809{x}^{1.8})\\ -0.25{x}^{2.6}(0.000197513-0.0009160448{x}^{1.8}+0.001008410504{x}^{3.6}), \\ \overline{u}(x, r)\approx 0.2684178{x}^{0.8}+0.0685589{x}^{2.6}-0.25{x}^{2.6}(0.0053913-0.023297809{x}^{1.8})\\ -0.25{x}^{2.6}(-0.000197513+0.0009160448{x}^{1.8}-0.001008410504{x}^{3.6}).\end{array} \right.$

(4.12)

$\alpha = 0.9$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.2599385{x}^{0.9}-0.0540269{x}^{2.8}-0.25{x}^{2.8}(-0.0031639+0.01418901{x}^{1.9})\\ -0.25{x}^{2.8}(0.00008989-0.0004154745{x}^{1.9}+0.000458881{x}^{3.8}), \\ \overline{u}(x, r)\approx 0.2599385{x}^{0.9}+0.0540269{x}^{2.8}-0.25{x}^{2.8}(0.0031639-0.014189097{x}^{1.9})\\ -0.25{x}^{2.8}(-0.00008989+0.0004154745{x}^{1.9}-0.000458881{x}^{3.8}).\end{array} \right.$

(4.13)

$\alpha = 1$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.25x-0.042114377{x}^{3}-0.25{x}^{3}(-0.001829736+0.0085133796{x}^{2})\\ -0.25{x}^{3}(0.0000401535-0.0001849397{x}^{2}+0.00020487753{x}^{4}), \\ \overline{u}(x, r)\approx 0.25x+0.042114377{x}^{3}-0.25{x}^{3}(0.001829736-0.0085133796{x}^{2})\\ -0.25{x}^{3}(-0.0000401535+0.0001849397{x}^{2}-0.0002048775{x}^{4}).\end{array} \right.$

(4.14)

Table 1. at r = 0.75.

$\alpha$	${\underline c _1}$ & ${\overline c _1}$	${\underline c _2}$ & ${\overline c _2}$	${\underline c _3}$ & ${\overline c _3}$
0.7	−1.0257850714449026	5.298291106236844×10⁻⁴	−3.040859671410477×10⁻⁵
0.8	−1.0193406988378892	3.249294721058776×10⁻⁴	−1.5364294422415488×10⁻⁵
0.9	−1.014446487354385	1.9694362983845834×10⁻⁴	−7.63055570435551×10⁻⁶
1	−1.0107450504316333	1.1791102776455743×10⁻⁴	−3.779171763451589×10⁻⁶

| Show Table

DownLoad: CSV

Substituting the values from into Eq (4.10), the approximate solutions for $\underline u (x, r)$ & $\bar u(x, r)$ at different values of $\alpha$ taking $r = 0.5$ respectively is as follow

$\alpha = 0.7$

$\left\{ \begin{array}{l} \underline u (x, r) \approx - 0.5502737{x^{0.7}} - 0.172041940{x^{2.4}} - 0.5{x^{2.4}}( - 0.00904995 + 0.0376716{x^{1.7}}) \hfill \\ - 0.5{x^{2.4}}(0.0004258594 - 0.0019816476{x^{1.7}} + 0.0021734872{x^{3.4}}), \hfill \\ \bar u(x, r) \approx 0.5502737{x^{0.7}} + 0.172041940{x^{2.4}} - 0.5{x^{2.4}}(0.00904995 - 0.03767164{x^{1.7}}) \hfill \\ - 0.5{x^{2.4}}( - 0.0004258594 + 0.0019816476{x^{1.7}} - 0.0021734872{x^{3.4}}). \hfill \\ \end{array} \right.$

(4.15)

$\alpha = 0.8$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.53683564{x}^{0.8}-0.137117858{x}^{2.6}-0.5{x}^{2.6}(-0.00539133+0.02329781{x}^{1.8})\\ -0.5{x}^{2.6}(0.00019751-0.0009160448{x}^{1.8}+0.0010084105{x}^{3.6}), \\ \overline{u}(x, r)\approx 0.53683564{x}^{0.8}+0.137117858{x}^{2.6}-0.5{x}^{2.6}(0.00539133-0.02329781{x}^{1.8})\\ -0.5{x}^{2.6}(-0.00019751+0.0009160448{x}^{1.8}-0.0010084105{x}^{3.6}).\end{array} \right.$

(4.16)

$\alpha = 0.9$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5198771{x}^{0.9}-0.10805378{x}^{2.8}-0.5{x}^{2.8}(-0.0031640+0.0141891{x}^{1.9})\\ -0.5{x}^{2.8}(0.0000898945-0.0004154745{x}^{1.9}+0.000458881{x}^{3.8}), \\ \overline{u}(x, r)\approx 0.5198771{x}^{0.9}+0.10805378{x}^{2.8}-0.5{x}^{2.8}(0.0031640-0.0141891{x}^{1.9})\\ -0.5{x}^{2.8}(-0.0000898945+0.0004154745{x}^{1.9}-0.000458881{x}^{3.8}).\end{array} \right.$

(4.17)

$\alpha = 1$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5x-0.0842288{x}^{3}-0.5{x}^{3}(-0.0018298+0.008513{x}^{2})-0.5{x}^{3}\\ (0.0000401612-0.00018494622{x}^{2}+0.0002048777{x}^{4}), \\ \overline{u}(x, r)\approx 0.5x+0.0842288{x}^{3}-0.5{x}^{3}(0.0018298-0.008513{x}^{2})-0.5{x}^{3}\\ (-0.000040162+0.00018494622{x}^{2}-0.0002048777{x}^{4}).\end{array} \right.$

(4.18)

Table 2. at r = 0.5.

$\alpha$	${\underline c _1}$ & ${\overline c _1}$	${\underline c _2}$ & ${\overline c _2}$	${\underline c _3}$ & ${\overline c _3}$
0.7	−1.0257850714449026	5.298291106236844×10⁻⁴	−3.040859671410477×10⁻⁵
0.8	−1.0193406988378892	3.249294721058776×10⁻⁴	−1.5364294422415488×10⁻⁵
0.9 1	−1.014446487354385 −1.0107453381292266	1.9694362983845834×10⁻⁴ 1.1800167363027721×10⁻⁴	−7.630555570435551×10⁻⁶ −3.726389827252244×10⁻⁶

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DownLoad: CSV

Problem 4.2. Consider system of fuzzy fractional order Volterra integro-differential equation as ^[59]:

$D_x^\alpha u(x, r) + \int\limits_0^t {u(t, r)dt = 0} , \, \, \, \, 0 < \alpha \leqslant 1, \, \, x \in [0, 1],$

(4.19)

subject to the fuzzy initial condition ${\left[ {u(0)} \right]^r} = \left[ {r - 1, 1 - r} \right],$ and the exact solution is $\underline u (x, r) = \left( {r - 1} \right){E_{\alpha + 1}}\left( { - {t^{\alpha + 1}}} \right), \bar u(x, r) = \left( {1 - r} \right){E_{\alpha + 1}}\left( { - {t^{\alpha + 1}}} \right),$

where ${E_{\alpha + 1}}$ is a Mittag-Leffler function and $0 \leqslant r \leqslant 1.$

By follow the technique as discussed in section 3, we get series of problems and their solutions as:

$\left\{ \begin{array}{l} {D_x}^\alpha {\underline u _0}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_0}(x, r) = 0. \hfill \\ \end{array} \right.$

(4.20)

$\left\{ \begin{array}{l} {D_x}^\alpha {\underline u _1}(x, r) + \left( { - \mathop \smallint \limits_0^x {{\underline u }_0}(t, r)dt} \right){c_1} - {D_x}^\alpha {\underline u _0}(x, r) - {c_1}{D_x}^\alpha {\underline u _0}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_1}(x, r) + \left( { - \mathop \smallint \limits_0^x {{\bar u}_0}(t, r)dt} \right){c_1} - {D_x}^\alpha {{\bar u}_0}(x, r) - {c_1}{D_x}^\alpha \bar u(x, r) = 0. \hfill \\ \end{array} \right.$

(4.21)

$\left\{ \begin{array}{l} {D_x}^\alpha {\underline u _2}(x, r) + \left( {\mathop \smallint \limits_0^x {{\underline u }_1}(t, r)dt} \right){c_1} - \left( {\mathop \smallint \limits_0^x {{\underline u }_0}(t, r)dt} \right){c_2} - {c_2}{D_x}^\alpha {\underline u _0}(x, r) - {D_x}^\alpha {\underline u _1}(x, r) - {c_1}{D_x}^\alpha {\underline u _0}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_2}(x, r) + \left( { - \mathop \smallint \limits_0^x {{\bar u}_1}(t, r)dt} \right){c_1} - \left( {\mathop \smallint \limits_0^x {{\bar u}_0}(t, r)dt} \right){c_2} - {c_2}{D_x}^\alpha {{\bar u}_0}(x, r) - {D_x}^\alpha {{\bar u}_1}(x, r) - {c_1}{D_x}^\alpha {{\bar u}_0}(x, r) = 0. \hfill \\ \end{array} \right.$

(4.22)

$\left\{ \begin{array}{l} {D_x}^\alpha {\underline u _3}(x, r) - \left( {\mathop \smallint \limits_0^x {{\underline u }_2}(t, r)dt} \right){c_1} - \left( {\mathop \smallint \limits_0^x {{\underline u }_1}(t, r)dt} \right){c_2} - \left( {\mathop \smallint \limits_0^x {{\underline u }_0}(t, r)dt} \right){c_3} - {c_3}{D_x}^\alpha {\underline u _0}(x, r) \hfill \\ - {c_2}{D_x}^\alpha {\underline u _1}(x, r) - {D_x}^\alpha {\underline u _2}(x, r) - {c_1}{D_x}^\alpha {\underline u _2}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_3}(x, r) - \left( {\mathop \smallint \limits_0^x {{\bar u}_2}(t, r)dt} \right){c_1} - \left( {\mathop \smallint \limits_0^x {{\bar u}_1}(t, r)dt} \right){c_2} - \left( {\mathop \smallint \limits_0^x {{\bar u}_0}(t, r)dt} \right){c_3} - {c_3}{D_x}^\alpha {{\bar u}_0}(x, r) \hfill \\ - {c_2}{D_x}^\alpha {{\bar u}_1}(x, r) - {D_x}^\alpha {{\bar u}_2}(x, r) - {c_1}{D_x}^\alpha {{\bar u}_2}(x, r) = 0. \hfill \\ \end{array} \right.$

(4.23)

And their solutions are

$\left\{ \begin{array}{l} {\underline u _0}(x, r) = r - 1, \hfill \\ {{\bar u}_0}(x, r) = 1 - r. \hfill \\ \end{array} \right.$

(4.24)

$\left\{ \begin{array}{l} {\underline u _1}(x, \underline r ) = \frac{{( - 1 + r){x^{1 + \alpha }}{c_1}}}{{(\alpha + {\alpha ^2})\Gamma (\alpha )}}, \hfill \\ {\overline u _1}(x, \overline r ) = - \frac{{( - 1 + r){x^{1 + \alpha }}{c_1}}}{{\alpha (1 + \alpha )\Gamma (\alpha )}}, \hfill \\ \end{array} \right.$

(4.25)

$\left\{ \begin{array}{l} {\underline u _2}(x, r) = ( - 1 + r){x^{1 + \alpha }}\left( {\frac{{{x^{1 + \alpha }}c_1^2}}{{\Gamma (3 + 2\alpha )}} + \frac{{{c_1} + c_1^2 + {c_2}}}{{\Gamma (2 + \alpha )}}} \right), \hfill \\ {{\bar u}_2}(x, r) = ( - 1 + r){x^{1 + \alpha }}\left( { - \frac{{{x^{1 + \alpha }}c_1^2}}{{\Gamma (3 + 2\alpha )}} - \frac{{{c_1} + c_1^2 + {c_2}}}{{\Gamma (2 + \alpha )}}} \right). \hfill \\ \end{array} \right.$

(4.26)

$\left\{ \begin{array}{l} {\underline u _3}(x, r) = \frac{{( - 1 + r){x^{1 + \alpha }}\left( {{c_2} + {c_1}\left( {1 + \frac{{{x^{2 + 2\alpha }}c_1^2}}{{\Gamma (4 + 2\alpha )}} + {c_1}(2 + {c_1}) + 2{c_2} + \frac{{2{x^{1 + \alpha }}({c_1} + c_1^2 + {c_2})}}{{\Gamma (3 + 3\alpha )}}} \right) + {c_3}} \right)}}{{\Gamma (1 + \alpha )}}, \hfill \\ {{\bar u}_3}(x, r) = ( - 1 + r){x^{1 + \alpha }}\left( { - \frac{{{x^{2 + 2\alpha }}c_1^3}}{{\Gamma (4 + 3\alpha )}} - \frac{{2{x^{1 + \alpha }}{c_1}({c_1} + c_1^2 + {c_2})}}{{\Gamma (3 + 2\alpha )}} - \frac{{{c_2} + {c_1}({{(1 + {c_1})}^2} + 2{c_2}) + {c_3}}}{{\Gamma (2 + \alpha )}}} \right), \hfill \\ \end{array} \right.$

(4.27)

Adding (4.24), (4.25), (4.26) and (4.27), one can construct $\underline u (x, r)$ & $\bar u(x, r)$ :

$\left\{ \begin{array}{l} \underline u (x, r) = - 1 + r + ( - 1 + r){x^{1 + \alpha }}\left( \begin{gathered} \frac{{{x^{1 + \alpha }}c_1^2}}{{\Gamma (3 + 2\alpha )}} + \frac{{{x^{2 + 2\alpha }}c_1^3}}{{\Gamma (1 + \alpha )\Gamma (4 + 2\alpha )}} + \frac{{2{x^{1 + \alpha }}{c_1}({c_1} + c_1^2 + {c_2})}}{{\Gamma (1 + \alpha )\Gamma (3 + \alpha )}} \hfill \\ + \frac{{{c_1}(2 + {c_1}) + {c_2}}}{{\Gamma (2 + \alpha )}} + \frac{{{c_2} + {c_1}({{(1 + {c_1})}^2} + 2{c_2}) + {c_3}}}{{\Gamma (1 + \alpha )}} \hfill \\ \end{gathered} \right), \hfill \\ \bar u(x, r) = 1 - r + ( - 1 + r){x^{1 + \alpha }}\left( \begin{gathered} - \frac{{{x^{2 + 2\alpha }}c_1^3}}{{\Gamma (4 + 3\alpha )}} - \frac{{{x^{1 + \alpha }}{c_1}({c_1}(3 + 2{c_1}) + 2{c_2})}}{{\Gamma (3 + 2\alpha )}} - \hfill \\ \frac{{2{c_2} + {c_1}(3 + {c_1}(3 + {c_1}) + 2{c_2}) + {c_3}}}{{\Gamma (2 + \alpha )}} \hfill \\ \end{gathered} \right). \hfill \\ \end{array} \right.$

(4.28)

Values of ${c_1}, \, \, {c_2}$ and ${c_3}$ contain in Eq (4.28)

Substituting the values from and into Eq (4.28), the approximate solutions for $\underline u (x, r)$ & $\bar u(x, r)$ at different values of $\alpha$ taking $r = 0.5$ is as follow

$\alpha = 0.2$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{1.2}(-0.905948+0.325139{x}^{1.2}-0.055807{x}^{2.4}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{1.2}(0.905948-0.325139{x}^{1.2}+0.055807{x}^{2.4}).\end{array} \right.$

(4.29)

$\alpha = 0.4$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{1.4}(-0.804545+0.210093{x}^{1.4}-0.025570{x}^{2.8}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{1.4}(0.80454545-0.210093{x}^{1.4}+0.025570{x}^{2.8}).\end{array} \right.$

(4.30)

$\alpha = 0.6$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{1.6}(-0.699349+0.128177{x}^{1.6}-0.010450{x}^{3.2}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{1.6}(0.699349-0.128177{x}^{1.6}+0.010450{x}^{3.2}).\end{array} \right.$

(4.31)

$\alpha = 0.8$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{1.8}(-0.596450+0.074561{x}^{1.8}-0.0038863{x}^{3.6}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{1.8}(0.5964503-0.074561{x}^{1.8}+0.0038863{x}^{3.6}).\end{array} \right.$

(4.32)

$\alpha = 1$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{2}(-0.499992+0.04163089{x}^{2}-0.001336253{x}^{4}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{2}(0.4999914-0.04163076{x}^{2}+0.001336247{x}^{4}).\end{array} \right.$

(4.33)

Table 3. at r = 0.5.

$\alpha$	${\underline c _1}$	${\underline c _2}$	${\underline c _3}$
0.2	−0.8038238618267683	7.6178003377104005×10⁻³	−1.334733828882206×10⁻³
0.4	−0.739676946061329	0.02530379950927192	−3.78307542584476×10⁻³
0.6	−0.6725325561865596	0.04771922184372877	−8.140310215705777×10⁻³
0.8	−0.6062340661192892	0.06889015391501904	−0.015746150088743013
1.0	−0.5432795308983783	0.08615024033359142	−0.026582381449582644

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Table 4. at r = 0.5.

$\alpha$	${\overline c _1}$	${\overline c _2}$	${\overline c _3}$
0.2	−0.9072542694138958	3.5727393445527333×10⁻³	3.637119200533134×10⁻⁴
0.4	−0.9409187563211361	1.9803045412863643×10⁻³	1.7807647588483674×10⁻⁴
0.6	−0.9636043521097131	9.808463578935658×10⁻⁴	7.369503633679291×10⁻⁵
0.8	−0.9781782948007163	4.4492367269725435×10⁻⁴	2.6729929232143615×10⁻⁵
1.0	−0.9872029432879605	1.896941835601795×10⁻⁴	1.021455544474314×10⁻⁵

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Substituting the values from and into Eq (4.28), the approximate solutions for $\underline u (x, r)$ & $\bar u(x, r)$ at different values of $r$ taking $\alpha = 0.5$ is as follow

$r = 0$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -1-{x}^{1.5}(-0.75199032254+0.165168588631{x}^{1.5}-0.016559344247{x}^{3.}), \\ \overline{u}(x, r)\approx 1-{x}^{1.5}(0.75199032256-0.165168588653{x}^{1.5}+0.016559344262{x}^{3.}).\end{array} \right.$

(4.34)

$r = 0.2$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.8-0.8{x}^{1.5}(-0.751990322550+0.165168588{x}^{1.5}-0.01655934425{x}^{3.}), \\ \overline{u}(x, r)\approx 0.8-0.8{x}^{1.5}(0.751990322554-0.165168588{x}^{1.5}+0.01655934426{x}^{3.}). \end{array} \right.$

(4.35)

$r = 0.4$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.6-0.6{x}^{1.5}(-0.7519903225+0.1651685886{x}^{1.5}-0.016559344250{x}^{3.}), \\ \overline{u}(x, r)\approx 0.6-0.6{x}^{1.5}(0.7519903225-0.1651685886{x}^{1.5}+0.0165593442496{x}^{3.}).\end{array} \right.$

(4.36)

$r = 0.6$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.4-0.4{x}^{1.5}(-0.751990322550+0.16516858863{x}^{1.5}-0.01655934425{x}^{3.}), \\ \overline{u}(x, r)\approx 0.4-0.4{x}^{1.5}(0.751990322554-0.16516858865{x}^{1.5}+0.01655934426{x}^{3.}).\end{array} \right.$

(4.37)

$r = 0.8$

$\left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.1999910-0.1999910{x}^{1.5}(-0.7519903+0.16516859{x}^{1.5}-0.0165593{x}^{3.}), \\ \overline{u}(x, r)\approx 0.19999910-0.19999910{x}^{1.5}(0.7519903-0.16516859{x}^{1.5}+0.0165593{x}^{3.}).\end{array} \right.$

(4.38)

Table 5. at

$\alpha$ = 0.5.

$r$	${\underline c _1}$	${\underline c _2}$	${\underline c _3}$
$0$	$- 0.7062087686601037$	$0.03638991875243272$	$- 5.6065011933001474 \times {10^{ - 3}}$
$0.2$	$- 0.7062087687083373$	$0.03638991875755982$	$- 5.606501189519195 \times {10^{ - 3}}$
$0.4$	$- 0.7062087686911187$	$0.03638991875566347$	$- 5.606501190876118 \times {10^{ - 3}}$
$0.6$	$- 0.7062087687083373$	$0.03638991875755982$	$- 5.606501189519195 \times {10^{ - 3}}$
$0.8$	$- 0.7062087686312865$	$0.036389918749298394$	$- 5.606501195566148 \times {10^{ - 3}}$

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Table 6. at

$\alpha$ = 0.5.

$r$	${\overline c _1}$	${\overline c _2}$	${\overline c _3}$
$0$	$- 0.9534544876637709$	$1.4110239362094313 \times {10^{ - 3}}$	$1.1669890178958486 \times {10^{ - 4}}$
$0.2$	$- 0.9534544876175205$	$1.4110239407705756 \times {10^{ - 3}}$	$1.1669890245071123 \times {10^{ - 4}}$
$0.4$	$- 0.9534544874104965$	$1.4110239614371703 \times {10^{ - 3}}$	$1.1669890547775563 \times {10^{ - 4}}$
$0.6$	$- 0.9534544876175205$	$1.4110239407705756 \times {10^{ - 3}}$	$1.1669890245071123 \times {10^{ - 4}}$
$0.8$	$- 0.9534544876289686$	$1.4110239398903034 \times {10^{ - 3}}$	$1.1669890235330204 \times {10^{ - 4}}$

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5. Result and discussion

– show the values of auxiliary constant at different values of $r$ & $\alpha$ for both lower and upper solution of OHAM for the solved problems. and show the comparison of absolute error of 3^rd order OHAM with Fractional Residual Power Series (FRPS) Method for 5-approximated solution and $k = 5$ for both lower and upper solutions of OHAM at different value of $\alpha$ for problem 1. Comparison of absolute error of 3^rd orders OHAM for both lower and upper solution of OHAM are shown in and . Numerical result show that OHAM provide more accuracy as compared to the other method and as $\alpha \to 1$ the approximate solution become very close to the exact solution. Graphical representation confirmed the convergence of fractional order solution towards the integer order solution. In graphical representation of OHAM at $\alpha = 0.7, \, \, 0.8, \, \, 0.9\, , \, \, 1, \, \, r = 0.75$ and $\alpha = 0.7, \, \, 0.8, \, \, 0.9\, , \, \, 1, \, \, r = 0.50$ are discussed for both $\underline u (x, r)$ & $\bar u(x, r)$ for problem 1. Figures 2 and 3 show the comparison of OHAM with the exact solution at different values of and taking r = 0.75 & r = 0.5 respectively for problem 1. represent the comparison of OHAM at $\alpha = 0.2, \, 0.4, \, \, 0.6, \, \, 0.8, \, \, 1, \, \, r = 0.5$ and $r = 0, \, \, 0.2, \, 0.4, \, \, 0.6, \, \, 0.8, \, \, \alpha \, = 0.5$ for both $\underline u (x, r)$ and $\bar u(x, r)$ for problem 2. Figure 5 shows the comparison of OHAM with the exact solution at different values of and r = 0.5 while Figure 6 shows the comparison of OHAM with the exact solution at different values of r and = 0.5 for problem 2.

Table 7. Comparison of Absolute Error (Abs Err.) of 3^rd order OHAM for

$\underline u (x, r)$ and Fractional Residual Power Series (FRPS) ^[54] Method for 5-approximated solution and

$k = 5$ for problem 1.

$r$	$x$	FRPS ^[58] $\alpha = 0.7$	OHAM	FRPS ^[58] $\alpha = 0.8$	OHAM	FRPS ^[58] $\alpha = 0.9$	OHAM	FRPS ^[58] $\alpha = 1$	OHAM
0.75	0.2	0.042797	0.040621	0.025514	0.024764	0.011512	0.011321	6.35273×10⁻¹⁰	2.14676×10⁻⁹
	0.4	0.059664	0.051698	0.035840	0.032584	0.016392	0.015405	8.14507×10⁻⁸	1.00832×10⁻⁸
	0.6	0.075769	0.058997	0.045171	0.037635	0.020545	0.018031	1.39554×10⁻⁶	1.11336×10⁻⁸
	0.8	0.094364	0.066136	0.055863	0.04232	0.025182	0.020362	1.04955×10⁻⁵	6.21567×10⁻⁹
0.50	0.2	0.085595	0.081241	0.051027	0.049528	0.011321	0.022643	1.27055×10⁻⁹	4.25946×10⁻⁹
	0.4	0.119328	0.103396	0.071680	0.065167	0.015405	0.030811	1.62901×10⁻⁷	1.98877×10⁻⁸
	0.6	0.151537	0.117994	0.090342	0.075269	0.018031	0.036062	2.79107×10⁻⁶	2.12923×10⁻⁸
	0.8	0.188728	0.132271	0.111723	0.084640	0.020362	0.040725	2.09911×10⁻⁵	1.00120×10⁻⁸

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Table 8. Comparison of Absolute Error (Abs Err.) of 3rd order OHAM for

$\bar u(x, r)$ and Fractional Residual Power Series (FRPS) ^[58] Method for 5-approximated solution and

$k = 5$ for problem 1.

$r$	$x$	FRPS ^[58] $\alpha = 0.7$	OHAM	FRPS ^[58] $\alpha = 0.8$	OHAM	FRPS ^[58] $\alpha = 0.9$	OHAM	FRPS ^[58] $\alpha = 1$	OHAM
0.75	0.2	0.085595	0.081242	0.025514	0.024764	0.011512	0.011321	6.35273×10⁻¹⁰	2.14676×10⁻⁹
	0.4	0.119328	0.103396	0.035840	0.032584	0.016392	0.015405	8.14507×10⁻⁸	1.00832×10⁻⁸
	0.6	0.151537	0.117994	0.045171	0.037635	0.020545	0.018031	1.39554×10⁻⁶	1.11336×10⁻⁸
	0.8	0.188728	0.132271	0.055862	0.04232	0.025182	0.020362	1.04955×10⁻⁵	6.21567×10⁻⁹
0.50	0.2	0.042797	0.040621	0.051027	0.049528	0.023025	0.022643	1.27055×10⁻⁹	4.25946×10⁻⁹
	0.4	0.059664	0.051698	0.071680	0.065167	0.032784	0.030811	1.62901×10⁻⁷	1.98877×10⁻⁸
	0.6	0.075769	0.058997	0.090342	0.075269	0.041090	0.036062	2.79107×10⁻⁶	2.12923×10⁻⁸
	0.8	0.094364	0.066136	0.111723	0.084640	0.050364	0.040725	2.09911×10⁻⁵	1.0012×10⁻⁸

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Table 9. Comparison of Absolute Error (Abs Err.) of 3^rd order OHAM for

$\underline u (x, \underline r )$ &

$\bar u(x, r)$ at different values of

$\alpha$ taking

$r = 0.5$ for problem 2.

$x$	$\underline u (x, r)$ $\alpha = 0.4$	$\bar u(x, r)$	$\underline u (x, r)$ $\alpha = 0.6$	$\bar u(x, r)$	$\underline u (x, r)$ $\alpha = 0.8$	$\bar u(x, r)$	$\underline u (x, r)$ $\alpha = 1$	$\bar u(x, r)$
0.2	1.2736×10⁻⁵	1.2736×10⁻⁵	3.2527×10⁻⁶	3.2527×10⁻⁶	6.93075×10⁻⁷	6.93076×10⁻⁷	1.29476×10⁻⁷	1.44585×10⁻⁷
0.4	2.3337×10⁻⁶	2.3337×10⁻⁶	2.4426×10⁻⁶	2.4426×10⁻⁶	9.50573×10⁻⁷	9.50573×10⁻⁷	2.67538×10⁻⁷	3.26771×10⁻⁷
0.6	9.9993×10^-6	9.9993×10⁻⁶	1.8451×10⁻⁶	1.8451×10⁻⁶	3.8544×10⁻⁸	3.85438×10⁻⁸	1.10198×10⁻⁷	2.39031×10⁻⁷
0.8	4.1251×10⁻⁶	4.1251×10⁻⁶	1.3416×10⁻⁷	1.3416×10⁻⁷	6.55017×10⁻⁸	6.55017×10⁻⁸	9.64628×10⁻⁹	2.27883×10⁻⁷
1.0	1.312×10⁻⁵	1.312×10⁻⁵	1.8373×10⁻⁶	1.8373×10⁻⁶	5.49341×10⁻⁸	5.4934×10⁻⁸	7.7356×10⁻⁸	3.9735×10⁻⁷

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Table 10. Comparison of Absolute Error (Abs Err.) of 3^rd order OHAM for

$\underline u (x, r)$ &

$\bar u(x, r)$ at different values of

$r$ taking

$\alpha = 0.5$ for problem 2.

$x$	$\underline u (x, r)$ $r = 0.4$	$\bar u(x, r)$	$\underline u (x, r)$ $r = 0.6$	$\bar u(x, r)$	$\underline u (x, r)$ $r = 0.8$	$\bar u(x, r)$	$\underline u (x, r)$ $r = 1$	$\bar u(x, r)$
0.	0.	0.	0.	0.	0.	0.	0.	0.
0.2	1.0578×10⁻⁵	1.0578×10⁻⁵	7.9338×10⁻⁶	7.9338×10⁻⁶	5.28921×10⁻⁶	5.28921×10⁻⁶	2.64461×10⁻⁶	2.64461×10⁻⁶
0.4	4.8940×10⁻⁶	4.8940×10⁻⁶	3.6705×10⁻⁶	3.6705×10⁻⁶	2.44698×10⁻⁶	2.44698×10⁻⁶	1.22349×10⁻⁶	1.22349×10⁻⁶
0.6	7.4825×10⁻⁶	7.4825×10⁻⁶	5.6119×10⁻⁶	5.6119×10⁻⁶	3.74125×10⁻⁶	3.74125×10⁻⁶	1.87062×10⁻⁶	1.87062×10⁻⁶
1.8	1.8038×10⁻⁶	1.8038×10⁻⁶	1.35291×10⁻⁶	1.35291×10⁻⁶	9.01939×10⁻⁷	9.01939×10⁻⁷	4.5097×10⁻⁷	4.5097×10⁻⁷

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Figure 1. Solution plot of OHAM for

$\underline u (x, r)$ &

$\bar u(x, r)$ at different values of

$r$ &

$\alpha$ for problem 1.

DownLoad: Full-Size Img PowerPoint

Figure 2. Solution plot of OHAM and Exact for

$\underline u (x, r)$ &

$\bar u(x, r)$ at different values of

$\alpha$ taking

$r = 0.75$ for problem 1.

DownLoad: Full-Size Img PowerPoint

Figure 3. Solution plot of OHAM and Exact for

$\underline u (x, r)$ &

$\bar u(x, r)$ at different values of

$\alpha$ taking

$r = 0.50$ for problem 1.

DownLoad: Full-Size Img PowerPoint

Figure 4. Solution plot of OHAM for

$\underline u (x, r)$ &

$\bar u(x, r)$ at different values of

$r$ &

$\alpha$ for problem 2.

DownLoad: Full-Size Img PowerPoint

Figure 5. Solution plot of OHAM and Exact for

$\underline u (x, r)$ &

$\bar u(x, r)$ at different values of

$\alpha$ taking

$r = 0.5$ for problem 2.

DownLoad: Full-Size Img PowerPoint

Figure 6. Solution plot of OHAM and Exact for

$\underline u (x, r)$ &

$\bar u(x, r)$ at different values of

$r$ taking

$\alpha = 0.5$ for problem 2.

DownLoad: Full-Size Img PowerPoint

6. Conclusions

In the research paper, a powerful technique known as Optimal Homotopy Asymptotic Method (OHAM) has been extended to the solution of system of fuzzy integro differential equations of fractional order. The obtained results are quite interesting and are in good agreement with the exact solution. Two numerical equations are taken as test examples which show the behavior and reliability of the proposed method. The extension of OHAM to system of fuzzy integro differential equations of fractional order is more accurate and as a result this technique will more appealing for the researchers for finding out optimum solutions of system of fuzzy integro differential equations of fractional order.

Conflict of interest

The authors declare no conflict of interest.

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Bioenergy potential of agricultural crop residues and municipal solid waste in Cameroon

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5. Result and discussion

6. Conclusions

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