Introduction

The subject of fractional calculus can be considered as a generalization of the classical integer order calculus (derivative and integration), and due its important role in science and engineering, it has been gaining considerable attention of many authors and researchers [1–5]. The reader who is interested with the field of fractional calculus knows that there are many definitions of a fractional order derivative and they have been used to describe several real-life problems in many fields of sciences; the most important and famous ones are Riemann–Liouville, Grunwald–Letnikov, and Caputo derivatives [1, 2, 6–10]. Riemann–Liouville and Caputo use the integral in its construction, particularly the Cauchy integral formula with modifications. Hence, occasionally, we need complex computations to get the fractional derivative in the sense of Riemann–Liouville and Caputo. Furthermore, they do not satisfy the non-linear derivative rules as product, quotient, and chain rules. More recently, a new definition of fractional derivatives which is the so-called conformable fractional derivative has been introduced and attracted the attention of many researchers because it contains many characteristics that correspond to the usual derivative, particularly the Leibniz rules [11–16]. This definition is very simple and more welcome than other fractional definitions since it has been receiving a lot of attention, many applications and phenomena can be modeled based on the CFDs, and it contains many interesting advantages such as the following: it is a local derivative that simulates the normal derivative because it depends on the limit in its formulation and it generalizes all concepts of ordinary calculus and can solve different fractional differential equations with all cases. In addition to this definition, there is another type of local derivatives called non-conformable fractional derivative, and for this purpose, the authors point out the publications [17–22]. In recent years, many authors have handled and studied the Lane–Emden equations because they were used to formulate lots of phenomena in physics and astrophysics [23–28].

The aim of this paper is to find the approximate solution of fractional Lane–Emden equations with conformable fractional derivative (CL-M) using the conformable Homotopy–Adomian decomposition method (CH-A) and conformable residual power series method (CRP); both of these methods are effective and easy to use for solving non-linear CL-M, without linearization or discretization. The benefit of these techniques over the other methods is that they can be performed directly to the given problem by choosing an appropriate value for the initial guess approximations, and they also reduce the difficulty appearing in the computation of the complicated terms [29–36].

This paper is organized as follows: In the “Preliminaries” section, a preliminary introduction of the conformable fractional order derivative is presented. In “The CH-A for solving CL-M” section, we introduced the CH-A for solving CL-M. The basic idea of the CRP technique for solving CL-M was given in the “A CRP for solving CL-M” section. Finally, a conclusion has been drawn.

Preliminaries

In this section, a brief introduction to the definition and properties of the conformable fractional derivative will be given [11, 12, 37–40].

Definition 1

Given a function \(y:[0,+\infty)\to \mathbb {R}\), then the CFD of order α of y is given by:

$$ C{{D}^{\alpha }}(y)(x)=\underset{\in \to 0}{\mathop{\lim }}\,\frac{y(x+\in {{x}^{1-\alpha }})-y(x)}{\in } $$
(1)

for all x>0,α∈(0,1). If y is α-differentiable in some (0,a),a>0 and \(\underset {x\to 0}{\mathop {\lim }}\,{{y}^{(\alpha)}}(x)\) exist, then define \({{y}^{(\alpha)}}(0)=\underset {x\to 0}{\mathop {\lim }}\,{{y}^{(\alpha)}}(x)\).

Definition 2

The conformable integral of order α is defined by:

$$ CI_{t}^{\alpha }(y)(t)=\int\limits_{a}^{t}{y(x){{x}^{\alpha -1}}dx},\,a\ge 0. $$
(2)

where the integral is the usual Riemann improper integral and α∈(0,1).

Lemma 1

Let α∈(0,1] and f, g be α-differentiable at a point t>0, then:

$$\begin{array}{*{20}l} & 1.\ C{{D}^{\alpha }}{{x}^{p}}=p{{x}^{p-\alpha }}\\ & 2.\ If\,in\,addition\,f\,is\,differentiable,\,then\,C{{D}^{\alpha }}f(x)={{x}^{1-\alpha }}\frac{df}{dx} \end{array} $$
$$\begin{array}{*{20}l} & 3.\,C{{D}^{\alpha }}(\lambda)=0,\,for\,all\,scalar\,function\,y(x)=\lambda \\ & 4.\ C{{I}^{\alpha }}({{x}^{\mu }})=\frac{\Gamma (\alpha +\mu -n)}{\Gamma (\alpha +\mu +1)},\,\mu \in \mathbb{R},\,\alpha \in (n,n+1]\,for\,all\,n\in N \\ & 5.\ C{{I}^{\alpha }}(C{{D}^{\alpha }}y(x))=y(x)-\sum\limits_{k=0}^{n}{\frac{{{y}^{k}}(0)}{k!}{{x}^{k}}} \end{array} $$

The CH-A for solving CL-M

In this section, we present the CH-A for solving CL-M; the CH-A is a hybrid method in which it combines between the HAM and the Adomian decomposition method (ADM). The calculations involved in this technique are more easy than the standard HAM especially when the non-linear term in the CL-M is decomposed by using the Adomian polynomials. To start the procedure, let us consider the general form of CL-M of order α>0:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}+C{{D}^{\alpha }}y+f(y)=0 $$
(3)

where x>0 and 0<α≤1.

To solve this problem using CH-A, first, rewrite Eq. (3), as follows:

$$ C{{D}^{\alpha }}[{{x}^{2\alpha }}C{{D}^{\alpha }}y]=-{{x}^{2\alpha }}f(y) $$
(4)

Then, integrate Eq. (4) twice with respect to x, so the general fractional solution of Eq. (3) is given by:

$$ y(x)={{c}_{2}}+\int{{{c}_{1}}{{x}^{-2\alpha }}{{d}_{\alpha }}x-[\int{{{x}^{-2\alpha }}[\int{{{x}^{2\alpha }}f(y){{d}_{\alpha }}x]{{d}_{\alpha }}x]}}} $$
(5)

where c1 and c2 are constants and dαx=x1−αdx.

In order to solve Eq. (5) by means of HAM, we need to seek the auxiliary linear operator:

$$ \mathcal{L}[\Phi (x,q)]=\Phi (x,q) $$
(6)

We now define the non-linear operator as:

$$ N[\Phi (x,q)]=\Phi (x,q)-{{c}_{2}}-\int{{{c}_{1}}{{x}^{-2\alpha }}{{d}_{\alpha }}x+[\int{{{x}^{-2\alpha }}[\int{{{x}^{2\alpha }}f(y)}}}{{d}_{\alpha }}x]{{d}_{\alpha }}x] $$
(7)

Consequently, the mth-order (m≥1) deformation equations can be expressed using the Adomian polynomials as:

$$ [{{y}_{m}}-{{\text{ }\!\!\chi\!\!\text{ }}_{m}}{{y}_{m-1}}]=\hbar H{{R}_{m}}({{\overset{\scriptscriptstyle\rightharpoonup}{y}}_{m-1}}) $$
(8)

where

$$ {{R}_{m}}({{\overset{\scriptscriptstyle\rightharpoonup}{y}}_{m-1}})={{y}_{m-1}}-(1-{{\text{ }\!\!\chi\!\!\text{ }}_{m}})({{c}_{2}}+\int{{{c}_{1}}{{x}^{-2\alpha }}{{d}_{\alpha }}x-[\int{{{x}^{-2\alpha }}}[\int{{{x}^{2\alpha }}{{A}_{m}}}}{{d}_{\alpha }}x]{{d}_{\alpha }}x]) $$
(9)

Hence

$$ {{y}_{m}}={{\text{ }\!\!\chi\!\!\text{ }}_{m}}{{y}_{m-1}}+\hbar H{{R}_{m}}({{\overset{\scriptscriptstyle\rightharpoonup}{y}}_{m-1}}) $$
(10)

Starting with an initial approximation:

$$ {{y}_{0}}={{c}_{2}}+\int{{{c}_{1}}{{x}^{-2\alpha }}{{d}_{\alpha }}x} $$
(11)

So, we have:

$$ {{y}_{m+1}}(x)=-[\int{{{x}^{-2\alpha }}}[\int{{{x}^{2\alpha }}{{A}_{m}}}{{d}_{\alpha }}x]{{d}_{\alpha }}x],\,\,\,\,m=0,1,2,...\,\,. $$
(12)

where

$$ {{\chi}_{m}}=\left\{ \begin{array}{ll} 0, & m\le 1 \\ 1, m>1 \\ \end{array} \right.\\ $$

and \(\left \{ {{A}_{m}} \right \}_{m=0}^{+\infty }\) is the set of Adomian polynomials of f(y) which is defined as follows:

$$ {{A}_{m}}=\frac{1}{m!}\frac{{{d}^{m}}}{d{{\theta }^{m}}}[f(\sum\limits_{i=0}^{+\infty }{{{\theta }^{i}}{{y}_{i}}){{]}_{\theta =0}},\,\,\,\,m=0,1,2,...\,\,.} $$
(13)

Finally, the exact solution of Eq. (3) can be calculated by:

$$ y(x)=\sum\limits_{m=0}^{+\infty }{{{y}_{m}}(x)} $$
(14)

and the nth-order approximate solution of Eq. (3) is given by:

$$ {{y}_{n}}(x)=\sum\limits_{m=0}^{n}{{{y}_{m}}(x)} $$
(15)

Now, we examine some known and charming cases that have been resulting from Eq. (3) as follows:

Case 1

Set f(y)=yk, k=0,1,2,.... in Eq.(3), then the following equation for x>0 is called:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+{{y}^{k}}=0,\,\,\,0<\alpha \le 1,\,\,\,k=0,1,2,...\,\,\,. $$
(16)

The CL-M of the first kind.

To solve this problem, we apply Eq. (13) to compute the set of Adomian polynomials \(\left \{ {{A}_{k}} \right \}_{k=0}^{+\infty }\) of the non-linear function f(y)=yk.

Hence, according to Eq. (14), the general fractional solution of Eq. (16) is given as follows:

$$\ y(x)=\sum\limits_{m=0}^{+\infty }{{{y}_{m}}(x)} $$

where y0 is given by Eq. (11) and ym, m=1,2,3,..., is given by Eq. (12).

Problem 1

Set the initial conditions y(0)=1 and y′(0)=0 into Eq. (16), then the general fractional solution of Eq. (16) in terms of Adomian polynomials is given by Eqs. (11), (12), and (14), respectively.

Problem 2

Set k=0 into Eq. (16), then we have:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+1=0,\,\,\,0<\alpha \le 1 $$
(17)

According to the initial conditions given in problem 1 and by integrating twice, then, the exact solution of this equation (see Fig. 1) can be obtained by :

$$ y(x)=1-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}} $$
(18)
Fig. 1
figure 1

The CH-A solution of Eq. (17)

Full size image

Problem 3

Set k=1 into Eq. (16) and by using the initial conditions y(0)=1 and y′(0)=0, then we have:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+y=0,\,\,\,0<\alpha \le 1 $$
(19)

According to Eqs.(11) and (12), with A0=y0, A1=y1, A2=y2, etc., then the solution of Eq. (19) is (see Fig. 2):

$$ y(x)=1-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}}+\frac{{{x}^{4\alpha }}}{120{{\alpha }^{4}}}-\frac{{{x}^{6\alpha }}}{5040{{\alpha }^{6}}} $$
(20)
Fig. 2
figure 2

The CH-A solution of Eq. (19)

Full size image

Problem 4

Set k=2 into Eq. (16) and by using the initial conditions y(0)=1 and y′(0)=0, then we have:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+{{y}^{2}}=0,\,\,\,0<\alpha \le 1 $$
(21)

Also by using Eqs.(11) and (12), with \({{A}_{0}}=y_{0}^{2},\,\,{{A}_{1}}=2{{y}_{0}}{{y}_{1}},\,\,{{A}_{2}}=2{{y}_{0}}{{y}_{2}}+y_{1}^{2},\,etc.\), then the solution of Eq. (21) is (see Fig. 3):

$$ y(x)=1-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}}+\frac{{{x}^{4\alpha }}}{60{{\alpha }^{4}}}-\frac{11{{x}^{6\alpha }}}{7560{{\alpha }^{6}}} $$
(22)
Fig. 3
figure 3

The CH-A solution of Eq. (21)

Full size image

Case 2

Consider f(y)=ey as in Eq. (3), then the following equation for x>0 is called:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+{{e}^{y}}=0,\,\,\,0<\alpha \le 1 $$
(23)

The CL-M of the second kind.

To solve problem (23) using CH-A and according to Eqs.(11) and (12), with \({{A}_{0}}=1,\,\,{{A}_{1}}={{y}_{1}},\,\,{{A}_{2}}={{y}_{2}}+\frac {1}{2}y_{1}^{2},\,\,{{A}_{3}}={{y}_{3}}+{{y}_{1}}{{y}_{2}}+\frac {1}{3!}y_{1}^{3},\,...,\) the solutions will be drawn as:

$$ {{y}_{0}}=0,\,\,{{y}_{k+1}}(x)=-[\int{{{x}^{-2\alpha }}}[\int{{{x}^{2\alpha }}{{A}_{k}}}{{d}_{\alpha }}x]{{d}_{\alpha }}x],\,\,\,\,k=0,1,2,...\,\, $$
(24)

Therefore, the general fractional power series solution is (see Fig. 4) given by:

$$ y(x)=-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}}+\frac{{{x}^{4\alpha }}}{120{{\alpha }^{4}}}-\frac{{{x}^{6\alpha }}}{1890{{\alpha }^{6}}} $$
(25)
Fig. 4
figure 4

The CH-A solution of Eq. (23)

Full size image

A CRP for solving CL-M

The implementation of the CRP will be described in this segment in order to get the fractional power series solution of the CL-M represented by Eq. (3) subject to the initial conditions:

$$ y({{x}_{0}})=a,\,\,{y}'({{x}_{0}})=b $$
(26)

The solution of Eq. (3) with respect to Eq. (24) is proposed by CRP as a fractional power series expansion about the initial point x=x0 as follows:

$$ y(x)=\sum\limits_{m=0}^{+\infty }{{{y}_{m}}(x)} $$
(27)

where ym is expressed by \({{y}_{m}}={{c}_{m}}\frac {{{(x-{{x}_{0}})}^{m\alpha }}}{{{\alpha }^{m}}m!},\,\,m=0,1,2,...\,\,.\)

Clearly, for the case m=0,1, we have from the initial conditions given by Eq. (26) that c0=a and c1=b.

By truncating the summation into Eq. (27), we get the kth approximate series:

$$ {{y}^{k}}(x)=\sum\limits_{m=0}^{k}{{{c}_{m}}\frac{{{(x-{{x}_{0}})}^{m\alpha }}}{{{\alpha }^{m}}m!}} $$
(28)

For the convergence of the fractional power series Eq. (28), we advise to see [41].

For simplification and before employing the CRP, Eq. (3) can be written in the form:

$$ {{x}^{\alpha }}C{{D}^{2\alpha }}y+2\alpha C{{D}^{\alpha }}y+{{x}^{\alpha }}f(y)=0 $$
(29)

Now, to evaluate cm that appears in the series expansion given by Eq. (28), substituting yk(x) into Eq. (29), we obtain the following kth residual function:

$$\begin{array}{*{20}l} {{\operatorname{Res}}^{k}}(x)=&{{x}^{\alpha }}\sum\limits_{m=2}^{k}{{{\alpha }^{2}}m(m-1){{c}_{m}}\frac{{{(x-{{x}_{0}})}^{(m-2)\alpha }}}{{{\alpha }^{m}}m!}+2\alpha }\sum\limits_{m=1}^{k}\alpha m{{c}_{m}}\frac{{{(x-{{x}_{0}})}^{(m-1)\alpha }}}{{{\alpha }^{m}}m!}\\&+{{x}^{\alpha }}f\left(\sum\limits_{m=0}^{k}{{{c}_{m}}\frac{{{(x-{{x}_{0}})}^{m\alpha }}}{{{\alpha }^{m}}m!}}\right) \end{array} $$

Then, \({{\operatorname {Res}}^{\infty }}(x)=\underset {k\to \infty }{\mathop {\lim }}\,{{\operatorname {Res}}^{k}}(x)\) clearly Res∞(x)=0 for each x∈[x0, x0+a].

Furthermore, CD(k−1)αResk(x0)=0, k=1,2,3,...,k which is coming from lemma 1 (property 3) and this is a basic rule to compute the coefficients cm, m=2,3,...,k.

However, finding cm demands to solve the algebraic expression:

$$ C{{D}^{(k-1)\alpha }}{{\operatorname{Res}}^{k}}({{x}_{0}})=0,\,x\in [{{x}_{0}},\,{{x}_{0}}+a] $$
(30)

In this manner, all the desired coefficients cm will be found.

For solving Eq. (3) using CRP, different cases for Eq. (3) will be discussed as follows:

  • Consider Eq. (17), subject to the initial conditions y(0)=1 and y′(0)=0, according to the initial conditions given above, and by using Eq. (30), hence, we have \({{c}_{0}}=1,\,{{c}_{1}}=0,\,{{c}_{2}}=\frac {-1}{3},\,{{c}_{m}}=0\,for\,all\,m\ge 3,\) substituting these values into Eq. (28) then the general fractional series solution in this case can be obtained by:

    $$\ {{y}^{\infty }}(x)=1-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}} $$

    The nature of CRP solution of Eq.(17) is given in Fig. 1.

  • In this case, the problem under consideration combines Eq. (21) together with the initial conditions y(0)=1 and y′(0)=0 also according to the given initial conditions and Eq.(30); then, we have \({{c}_{0}}=1,\,{{c}_{1}}=0,\,{{c}_{2}}=\frac {-1}{3},\,{{c}_{3}}=0,\,{{c}_{4}}= \frac {2}{5},\,{{c}_{5}}=0,\,...\); then, the fractional power series solution is:

    $$\ {{y}^{6}}(x)=1-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}}+\frac{{{x}^{4\alpha }}}{60{{\alpha }^{4}}}-\frac{11{{x}^{6\alpha }}}{7560{{\alpha }^{6}}} $$

    Also, the nature of the CRP solution of Eq. (21) is presented in Fig. 3.

  • Consider the problem given by Eq. (23) with respect to the initial conditions y(0)=0 and y′(0)=0, then after taking the initial conditions in consideration and by employing Eq. (30), we get \({{c}_{0}}={{c}_{1}}=0,\,{{c}_{2}}=\frac {-1}{3},\,{{c}_{3}}=0,\,{{c}_{4}}=\frac {1}{5}, \,{{c}_{5}}=0,\,...\); thus, the fractional power series solution in this case is:

    $$\ {{y}^{6}}(x)=-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}}+\frac{{{x}^{4\alpha }}}{120{{\alpha }^{4}}}-\frac{{{x}^{6\alpha }}}{1890{{\alpha }^{6}}} $$

    The quality of the CRP solution of Eq. (23) is given in Fig. 4.

Conclusions

In this paper, we applied CH-A and CRP to find the approximate analytic solution of some classes of CL-M in terms of infinite fractional power series. The proposed methods introduced an easy way to compute the components of the solution that have been converging rapidly to the exact solution. The results obtained by CH-A and CRP prove that these algorithms are highly effective and convenient in non-linear cases of the CL-M and can be employed to examine a wide class of non-linear fractional mathematical models.