Analysis of Homotopy Perturbation Method for Solving Fractional Order Differential Equations

Abstract

The analysis of Homotopy Perturbation Method (HPM) for the solution of fractional partial differential equations (FPDEs) is presented. A unified convergence theorem is given. In order to validate the theory, the solution of fractional-order Burger-Poisson (FBP) equation is obtained. Furthermore, this work presents the method to find the solution of FPDEs, while the same partial differential equation (PDE) with ordinary derivative i.e., for $\alpha =1$, is not defined in the given domain. Moreover, HPM is applied to a complicated obstacle boundary value problem (BVP) of fractional order.

1. Introduction

Fractional models have an important role in many fields of engineering and science, for instance, fluid flows, solute transport, electromagnetic theory, signal processing, biology, economics, physics, and geology, etc. [1,2,3,4,5,6]. Fractional theory has many applications in wireless networks [7,8]. Moreover, fractional modeling has been applied in micro-grids [9], and decentralized wireless networks [10].

Fractional differential equations (FDEs) involve real or complex order derivatives [11]. Various researchers contributed on fractional derivatives during the 18th and 19th centuries, for example, Abel [12], Caputo [13], Euler [14], Fourier [15], Laplace [16], Liouville [17], or Ross [18]. In 1974, Oldham and Spanier presented fractional operators with mass and heat transfer applications [19]. In this paper, the analytical technique to solve fractional order partial differential equations (PDEs) is described. In order to obtain analytical solution, HPM is used to solve fractional order PDEs. Fractional order PDEs do not have closed form exact solutions in most problems, therefore, it is required to develop efficient and accurate analytical and numerical methods. HPM is well known for its accuracy and simplicity [20,21]. HPM has been widely used to obtain approximate series solutions of fractional order linear and nonlinear PDEs [22,23,24,25,26,27].

The Burger Poisson equation is widely used to express different physical phenomena, for instance, mathematical models for shallow water and shock waves in a viscous fluid [28]. Tian and Gao proved the existence of the uni-dimensional viscous equation in 2009 [29]. Moreover, Abidi and Omrani obtained the solution of Burger Poisson equation using homotopy analysis method [30].

The obstacle problem plays a role of bridge in the field of variational inequalities and differential equations. It is originated from the study of elasticity theory. In elasticity theory, it is required to obtain the equilibrium position of elastic membrane with fixed boundaries. Obstacle problems occur in diffusion equation and signals processing while determining heat flux at the boundary of semi-infinite rod [31].

The focus of the paper is to generalize the convergence theorem, in the sense that the mapping is nonself and only Y is complete. Theorem is applied to the solution of FBP equation acquired by HPM. Furthermore, this work presents the method to get the solution of FPDEs, while the same PDE with ordinary derivative i.e., for $\alpha =1$ is not defined in the given domain. Moreover, the proposed HPM method is applied to complex obstacle BVP.

Beside the introduction, the distribution of the article is as under: Section 2 comprises of important definitions and properties, Section 3 contains the implementation of HPM to solve FPDEs and convergence theorem, Section 4 comprises of results and discussion of FBP, Section 5 contains solutions of obstacle BVP, and Section 6 includes the conclusion of the paper.

2. Preliminaries

Fractional calculus is a developing area in mathematical analysis. Several definitions of fractional operators have been propounded like Riemann-Liouvlle, Caputo, and Grunwald-Letnikov [13,17]. These definitions have some limitations, for instance $D_a^{\alpha }\left(1\right)\ne 0$, do not satisfy the product, quotient and chain rules of derivatives. Recently, Khalil et al. published a definition uses for fractional derivatives called conformable [32,33]. Conformable is simpler and natural extension of the usual derivatives as it satisfies the aforementioned properties of derivatives.

Definition 1.

A function $g:[0,\infty )\to \mathbb{R}$. The fractional derivative of g for order α is given below:

$$\begin{array}{c}\hfill D_{\alpha }\left(g\right)\left(t\right)=\underset{ϵ\to 0}{\lim }\frac{g(t+ϵt^{1-\alpha })-g\left(t\right)}{ϵ},\mathit{for}\mathit{all}t>0,\alpha \in (0,1).\end{array}$$

(1)

If g is $\alpha$-differentiable in $(0,b)$, $b>0$. If $li{m}_{t\to 0^{+}}$ exists, then $g^{\alpha }\left(0\right)=li{m}_{t\to 0^{+}}{g}^{\alpha }\left(t\right)$.

The definition given in Equation (1) is known as conformable [33].

Using the above mentioned definition given in Equation (1), we obtain the following useful results:

Let $g,f$ are $\alpha$-differentiable and $\alpha \in (0,1]$, then

• If a function $g:[0,\infty ]\to \mathbb{R}$ is $\alpha$-differentiable at $t_0>0,\alpha \in (0,1]$, then g is continuous at $t_0$.
• $D_{\alpha }(cf+dg)=c{D}_{\alpha }\left(f\right)+d{D}_{\alpha }\left(g\right)$, for all $c,d\in \mathbb{R}$.
• $D_{\alpha }\left(t^p\right)=p{t}^{p-\alpha }$, for all $p\in \mathbb{R}$.
• $D_{\alpha }\left(\xi \right)=0$, for all constant function $f\left(t\right)=\xi$.
• $D_{\alpha }\left(gf\right)=g{D}_{\alpha }\left(f\right)+f{D}_{\alpha }\left(g\right)$.
• $D_{\alpha }(\frac{f}{g})=\frac{g{D}_{\alpha }\left(f\right)-f{D}_{\alpha }\left(g\right)}{g^2}$.
• If g is differentiable, then $D_{\alpha }\left(g\right)\left(t\right)=t^{1-\alpha }\frac{\mathrm{d}g\left(t\right)}{\mathrm{d}t}$.

Fractional integral: The fractional integral of order $\alpha$ can be defined below:

$$\begin{array}{c}\hfill J_{\alpha }^a\left(g\right)\left(t\right)={\int }_a^t\frac{g\left(x\right)}{x^{1-\alpha }}dx,\mathrm{for}\mathrm{all}t>0,\alpha \in (0,1).\end{array}$$

(2)

Here, $J_{\alpha }$ is the Riemann improper integral.

3. Application of HPM to FBP Equation

Consider the time-dependent operator equation

$$B\left(w\left(s,t\right)\right)-h\left(s,t\right)=0,$$

where B denotes a differential operator and $w\left(s,t\right)$ is an unknown function. Moreover, we assume $h(s,t)$ is an analytic function, we can decompose the operator B as;

$$B=L+N$$

where L is linear and N is nonlinear or sometimes the complicated part to handle. Let $\mathsf{\Omega }\subseteq \mathbb{R}\times \mathbb{R}$ the homotopy $H:\mathsf{\Omega }\times \left[0,1\right]\to \mathbb{R},$ is defined as:

$$H\left(w,p\right)=\left(1-p\right)\left[L\left(w\right)-L\left(w_0\right)\right]+p\left[B\left(w\right)-g\right]=0,p\in \left[0,1\right],w\in \mathsf{\Omega }.$$

Note that the function $w_0\left(s,t\right)$ is the initial guess which satisfies the given operator equation. The choice of p from zero to one provide us the deformation from $w_0\left(s,t\right)$ to the solution $w\left(s,t\right)$. Clearly

$$H\left(w,0\right)=0\mathrm{implies}L\left(w\right)-L\left(w_0\right)=0$$

and

$$H\left(w,1\right)=0\mathrm{implies}B\left(w\right)-g=0$$

which is the given operator equation. Using perturbation method, we suppose the solution in power series as;

$$\begin{array}{ccc}\hfill w\left(s,t\right)& =& w_0\left(s,t\right)+p·w_1\left(s,t\right)+p^2·w_2\left(s,t\right)+\dots \hfill \\ & =& \underset{n=0}{\sum ^{\infty }}{p}^n·w_n\left(s,t\right).\hfill \end{array}$$

For $p=1,$

$$w\left(s,t\right)=w_0\left(s,t\right)+w_1\left(s,t\right)+w_2\left(s,t\right)+\dots$$

is an approximate solution to the given operator equation.

Remark 1.

The Banach contraction type theorem about the convergence of the solution stated in [34]. Here the generalized, corrected, and unified form is presented in which the completeness of X is not required and Y must be a subset of X. If $Y⊈X$, we cannot say about the sequence $x_{n+1}=N^n\left(x_0\right)=N\left(x_n\right)$ is contained in Y or not. On the basis of above discussion, a unified theorem is presented as follows:

Theorem 1.

Let X be a normed space and Y $\subseteq X$ be a Banach space, $N:X\to Y$ be a mapping such that for all $x,y\in X$

$${∥N\left(x\right)-N\left(y\right)∥}_Y\le k{∥x-y∥}_X$$

(C)

for some $k\in [0,1),$ then the sequence

$$x_{n+1}=N^n\left(x_0\right)=N\left(x_n\right)$$

for any $x_0\in X$ converges to a unique fixed point of $N.$

Proof. 

We consider the picard sequence $x_{n+1}=N\left(x_n\right)\subseteq Y,$ it will be shown that this sequence $\left(x_n\right)$ is Cauchy in $Y.$ For integers $m\ge n$ consider,

$$∥x_n-x_m∥\le ∥x_n-x_{n+1}∥+∥x_{n+1}-x_{n+2}∥+∥x_{n+2}-x_{n+3}∥+\dots +∥x_{m-1}-x_m∥.$$

Using the contractive condition $\left(C\right),$ and induction on n, is given as

$$∥x_n-x_{n+1}∥\le k^n∥x_0-x_1∥,$$

this implies,

$$\underset{m\to \infty }{\lim }∥x_n-x_m∥\le \frac{k^n}{1+k}∥x_0-x_1∥\to 0,\mathrm{as}n\to \infty .$$

This shows that $\left(x_n\right)$ is Cauchy sequence in $Y,$ completeness of Y allows us to find $z\in Y$, such that

$$\underset{n\to \infty }{\lim }\left(x_n\right)=z\in Y.$$

Clearly, $\left(C\right)$ ensures the continuity of $N,$ thus

$$\underset{n\to \infty }{z=\lim }{x}_{n+1}=\underset{n\to \infty }{\lim }N\left(x_n\right)=N\left(\underset{n\to \infty }{\lim }{x}_n\right)=N\left(z\right).$$

This completes the proof, the uniqueness of z is obvious. The proof of Theorem 1 is similar to the proof given in [34,35], but our case is generalized, in the sense that the mapping is nonself and only Y is complete. Now, the extended HPM using conformable is presented to solve space-time FBP equation. ☐

4. Test Problem 1

The partial differential FBP equation in unidirectional propagation water waves can be described as follows [36,37]:

$$\begin{array}{c}{D}_t^{\alpha }w-D_t^{\alpha }{w}_{xx}+w_x+w{w}_x-(3w_x{w}_{xx}+w{w}_{xxx})=0,\alpha \in (0,1],\hfill \\ \mathrm{with}w(x,0)=-x.\hfill \end{array}$$

(3)

In order to apply HPM, the constructed homotopy is given below:

$$(1-p)D_t^{\alpha }w+p(D_t^{\alpha }w-D_t^{\alpha }{w}_{xx}+w_x+w{w}_{xx}-3(w_x{w}_{xx}+w{w}_{xxx}))=0,$$

or

$$D_t^{\alpha }w+p(-D_t^{\alpha }{w}_{xx}+w_x+w{w}_x-(3w_x{w}_{xx}+w{w}_{xxx}))=0.$$

(4)

Here p is a parameter that lies between 0 and 1. The solution $w(x,t)$ is given as follows:

$$w(x,t)=w_0(x,t)+p{w}_1(x,t)+p^2{w}_2(x,t)+p^3{w}_3(x,t)+\dots .$$

(5)

Setting $p=1$ gives,

$$w(x,t)=w_0(x,t)+w_1(x,t)+w_2(x,t)+w_3(x,t)+\dots ..$$

Now, substitute Equation (5) into Equation (3), and collect the similar powers of p, gives

$$\begin{array}{c}{p}^0:D_t^{\alpha }{w}_0=0,w_0(x,0)=-x;\hfill \\ p^1:D_t^{\alpha }{w}_1-D_t^{\alpha }{w}_{0xx}+w_{0x}+w_0{w}_{0x}-3w_{0x}{w}_{0xx}-w_0{w}_{0xxx}=0,w_1(x,0)=0;\hfill \\ p^2:D_t^{\alpha }{w}_2-D_t^{\alpha }{w}_{1xx}+w_{1x}+w_0{w}_{1x}+w_1{w}_{0x}-3(w_{0x}{w}_{1xx}+w_{1x}{w}_{0xx})\hfill \\ -(w_0{w}_{1xxx}+w_1{w}_{0xxx})=0,w_2(x,0)=0;\hfill \\ p^3:D_t^{\alpha }{w}_3-D_t^{\alpha }{w}_{2xx}+w_{2x}+(w_0{w}_{2x}+w_1{w}_{1x}+w_2{w}_{0x})-3(w_{0x}{w}_{2xx}+w_{1x}{w}_{1xx}+w_{2x}{w}_{0xx})\hfill \\ -(w_0{w}_{2xxx}+w_1{w}_{1xxx}+w_2{w}_{0xxx})=0,w_3(x,0)=0.\hfill \end{array}$$

(6)

Afterwards, the fractional integral operator $J_{\alpha }$ with conformable derivative definition (c.f. Equation (2)) is applied on both sides of Equation (6), we have

$$\begin{array}{cc}\hfill w_0(x,t)& =-x;\hfill \\ \hfill w_1(x,t)& =\frac{t^{\alpha }}{\alpha }(1-x);\hfill \\ \hfill w_2(x,t)& =\frac{t^{2\alpha }}{\alpha }(1-x);\hfill \\ \hfill w_3(x,t)& =\frac{t^{3\alpha }}{\alpha }(1-x)\hfill \\ \hfill ⋮\end{array}$$

(7)

In order to calculate next terms, we have

$$\begin{array}{cc}\hfill w_l& =w_{(l-1)xx}+{\int }_a^t\frac{1}{t^{1-\alpha }}\left(3\sum _{i=0}^{l-1}{w}_{ix}{w}_{(l-i-1)xx}+\sum _{i=0}^{l-1}{w}_i{w}_{(l-i-1)xxx}-w_{(l-1)x}-\sum _{i=0}^{l-1}{w}_i{w}_{(l-i-1)x}\right)dt\hfill \\ & w_l(x,0)=0,\mathrm{for}l\ge 4.\hfill \end{array}$$

(8)

If we define $N:C\left(\mathsf{\Omega }\right)\to C\left(\mathsf{\Omega }\right)$ , with iterative sequence as,

$$N\left(w_l\right)=w_{(l-1)xx}+{\int }_a^t\frac{1}{t^{1-\alpha }}\left(3\sum _{i=0}^{l-1}{w}_{ix}{w}_{(l-i-1)xx}+\sum _{i=0}^{l-1}{w}_i{w}_{(l-i-1)xxx}-w_{(l-1)x}-\sum _{i=0}^{l-1}{w}_i{w}_{(l-i-1)x}\right)dt$$

Then for any $w_0\in C\left(\mathsf{\Omega }\right),$ by Theorem 1, the sequence $w_{l+1}=N\left(w_l\right)$ converges to the unique solution w of the given FBP.

The Equation (8) can be calculated with the help of symbolic softwares, for instance, Mathematica and Maple. The HPM solution is given below:

$$\begin{array}{ccc}\hfill w(x,t)& =& -x+\frac{t^{\alpha }}{\alpha }(1-x)+\frac{t^{2\alpha }}{{\alpha }^2}(1-x)+\frac{t^{3\alpha }}{{\alpha }^3}(1-x)+\frac{t^{4\alpha }}{{\alpha }^4}(1-x)+\dots \hfill \end{array}$$

(9)

$$\begin{array}{ccc}& =& \frac{-x\alpha +t^{\alpha }}{\alpha -t^{\alpha }}\hfill \end{array}$$

(10)

The HPM solution of FBP equation when $\alpha =1$ is as follows:

$$\begin{array}{ccc}\hfill w(x,t)& =& -x+(1-x)t+(1-x)t^2-(1-x)t^3+\dots \hfill \end{array}$$

(11)

$$\begin{array}{ccc}& =& \frac{-x+t}{1-t}.\hfill \end{array}$$

(12)

Remark 2.

It is remarked that the exact and HPM solution of FBP equation when $\alpha =1,$ given in Equation (12) does not exist at $t=1$, while for any $\alpha \in \left(0,1\right),$ the solution given in Equation (10) of FBP equation exists. This shows the importance of fractional derivative and its way of dealing these types of the situations where solution of some ordinary PDEs fail to exist.

Convergence of solution: The FBP equation is as follows:

$$\begin{array}{c}{D}_t^{\alpha }w-D_t^{\alpha }{w}_{xx}+w_x+w{w}_x-(3w_x{w}_{xx}+w{w}_{xxx})=0,\alpha \in (0,1],\hfill \\ \mathrm{with}w(x,0)=-x.\hfill \end{array}$$

(14)

The approximate first four term solution of FBP equation for $p=1$ is given by

$$w(x,t)=w_0(x,t)+w_1(x,t)+w_2(x,t)+w_3(x,t),$$

where,

$$\begin{array}{cc}\hfill w_0(x,t)& =-x;\hfill \\ \hfill w_1(x,t)& =\frac{t^{\alpha }}{\alpha }(1-x);\hfill \\ \hfill w_2(x,t)& =\frac{t^{2\alpha }}{{\alpha }^2}(1-x);\hfill \\ \hfill w_3(x,t)& =\frac{t^{3\alpha }}{{\alpha }^3}(1-x).\hfill \end{array}$$

(14)

The sequence generated by HPM will be regarded as

$$V_k=N\left(V_{k-1}\right),V_{k-1}=\sum _{i=1}^{k-1}{w}_i,k=1,2,3\dots .$$

We assume that $V_0=w_0$.

According to theorem for non-linear mapping $N,$ a sufficient condition for convergence of HPM is strictly contraction N. Therefore, we have

$$∥V_0-w∥=∥-x-\frac{-x\alpha +t^{\alpha }}{\alpha -t^{\alpha }}∥=∥\frac{(x-1)t^{\alpha }}{\alpha -t^{\alpha }}∥$$

Now for $\frac{t^{\alpha }}{\alpha }⩽\gamma ,0<\gamma <1,$ we have

$$∥V_1-u∥=∥w_0+w_1-w∥=∥\frac{t^{2\alpha }(x-1)}{\alpha (\alpha -t^{\alpha })}∥⩽\gamma ∥\frac{(x-1)t^{\alpha }}{\alpha -t^{\alpha }}∥=\gamma ∥V_0-w∥$$

$$∥V_2-w∥=∥w_0+w_1+w_2-w∥=∥\frac{t^{3\alpha }(x-1)}{{\alpha }^2(\alpha -t^{\alpha })}∥⩽{\gamma }^2∥\frac{(x-1)t^{\alpha }}{\alpha -t^{\alpha }}∥={\gamma }^2∥V_0-w∥$$

$$∥V_n-w∥=∥\sum _{j=1}^n{w}_j-w∥=∥\frac{{(-1)}^n{t}^{n\alpha +\alpha }(x-1)}{{\alpha }^n(\alpha -t^{\alpha })}∥⩽{\gamma }^n∥\frac{(x-1)t^{\alpha }}{\alpha -t^{\alpha }}∥={\gamma }^n∥V_0-w∥.$$

Therefore, ${\lim }_{n\to \infty }∥V_n-w∥⩽{\gamma }^n∥V_0-w∥=0,$ that is, $w(x,t)={\lim }_{n\to \infty }{V}_n=\frac{-x\alpha +t^{\alpha }}{\alpha -t^{\alpha }},$ which is an exact solution.

Discussion

In this section, results obtained by HPM are discussed. Analytical series solution of space-time FBP equation is given in Equation (9). In Figure 1, a solution is presented for different values of $\alpha ,suchas\left(0.1,0.3,0.5,0.7\right)$. Figure 1 shows a big difference in smaller and larger values of $\alpha$. For larger values of $\alpha$, $w(x,t)$ attains height and for values closer to 0, height of $w(x,t)$ reduces. In Figure 2, results are presented for $\alpha =0.8$ and $\alpha =0.9$, respectively. Exact solution of FBP model for $\alpha =1$ is given in Equation (12). Equation (12) shows discontinuity at $t=1$ which is clearly depicted in Figure 2. As value of $\alpha$ approaches to 1, the shock is produced in the vicinity of $t=1$.

5. Test Problem 2

Consider the fractional second-order obstacle BVP:

$$D^{\alpha }v=\left\{\begin{array}{c}{\displaystyle \frac{v^3}{3!}}+{\displaystyle \frac{v^2}{2!}}+v+1,for1\le x<{\displaystyle \frac{3}{2}}and{\displaystyle \frac{5}{2}}\le x\le 3\hfill \\ {\displaystyle \frac{v^3}{3!}}+{\displaystyle \frac{v^2}{2!}}+2v,for{\displaystyle \frac{3}{2}}\le x<{\displaystyle \frac{5}{2}}\hfill \end{array}\right.$$

(15)

with boundary conditions

$$v(-1)=v\left(1\right)=0$$

In order to apply HPM, we construct homotopy in three different domains:

Case I: $1\le x<{\displaystyle \frac{3}{2}}$

$$(1-p)D^{\alpha }v+p\left(D^{\alpha }v-\left({\displaystyle \frac{v^3}{3!}}+{\displaystyle \frac{v^2}{2!}}+v+1\right)\right)=0$$

or

$$D^{\alpha }v+p\left({\displaystyle \frac{v^3}{3!}}+{\displaystyle \frac{v^2}{2!}}+v+1\right)=0$$

(16)

Here p is parameter that lies between 0 and 1. The solution $v\left(x\right)$ is given as follows:

$$v\left(x\right)=v_0\left(x\right)+p{v}_1\left(x\right)+p^2{v}_2\left(x\right)+p^3{v}_3\left(x\right)+\dots$$

(17)

Setting p = 1 provides,

$$v\left(x\right)=v_0\left(x\right)+v_1\left(x\right)+v_2\left(x\right)+v_3\left(x\right)+\dots$$

(18)

Now, substitute Equation (3) in Equation (2), and comparing the coefficients of similar powers of p, we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill p^0:D^{\alpha }{v}_0& =0,v_0\left(0\right)=0\hfill \\ \hfill p^1:D^{\alpha }{v}_1+{\displaystyle \frac{v_0^3}{3!}}+{\displaystyle \frac{v_0^2}{2!}}+v_0+1& =0,v_1\left(0\right)=0\hfill \\ \hfill p^2:D^{\alpha }{v}_2+{\displaystyle \frac{3v_0^2{v}_1}{3!}}+{\displaystyle \frac{2v_0{v}_1}{2!}}+v_1& =0,v_2\left(0\right)=0\hfill \\ \hfill p^3:D^{\alpha }{v}_3+{\displaystyle \frac{3v_0{v}_1^2}{3!}}+{\displaystyle \frac{3v_0^2{v}_2}{3!}}+{\displaystyle \frac{v_1^2}{2!}}+{\displaystyle \frac{2v_0{v}_2}{2!}}+v_2& =0,v_3\left(0\right)=0\hfill \end{array}\end{array}$$

(19)

Now, the fractional integral operator $J_{\alpha }$ with comfortable derivative definition is applied on both sides of Equation (5), we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill v_0\left(x\right)& =c_{1}x+c_2,\hfill \\ \hfill v_1\left(x\right)& =c_{1}x+c_2+{\displaystyle \frac{1}{24}}{x}^{4+\alpha }{c}_1^3+{\displaystyle \frac{1}{6}}{x}^{3+\alpha }{c}_1^2+{\displaystyle \frac{1}{6}}{x}^{3+\alpha }{c}_1^2{c}_2+{\displaystyle \frac{1}{2}}{x}^{2+\alpha }{c}_1{c}_2+{\displaystyle \frac{1}{2}}{x}^{2+\alpha }{c}_1\hfill \\ & +{\displaystyle \frac{1}{4}}{x}^{2+\alpha }{c}_1{c}_2^2+{\displaystyle \frac{1}{2}}{x}^{1+\alpha }{c}_2^2+x^{1+\alpha }+{\displaystyle \frac{1}{6}}{x}^{1+\alpha }{c}_2^3+{\displaystyle \frac{1}{6}}{x}^{4+\alpha }{c}_1^3+x^{1+\alpha }{c}_2,\hfill \\ & .\hfill \\ & .\hfill \\ & .\hfill \end{array}\end{array}$$

(20)

Case II: ${\displaystyle \frac{3}{2}}\le x<{\displaystyle \frac{5}{2}}$

$$(1-p)D^{\alpha }v+p\left(D^{\alpha }v-\left({\displaystyle \frac{v^3}{3!}}+{\displaystyle \frac{v^2}{2!}}+2v\right)\right)=0$$

or

$$D^{\alpha }v+p\left({\displaystyle \frac{v^3}{3!}}+{\displaystyle \frac{v^2}{2!}}+2v\right)=0$$

(21)

Now, substitute Equation (3) in Equation (7), and comparing the coefficients of similar powers of p, we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill p^0:D^{\alpha }{v}_0& =0,v_0\left(0\right)=0\hfill \\ \hfill p^1:D^{\alpha }{v}_1+{\displaystyle \frac{v_0^3}{3!}}+{\displaystyle \frac{v_0^2}{2!}}+2v_0& =0,v_1\left(0\right)=0\hfill \\ \hfill p^2:D^{\alpha }{v}_2+{\displaystyle \frac{3v_0^2{v}_1}{3!}}+{\displaystyle \frac{2v_0{v}_1}{2!}}+2v_1& =0,v_2\left(0\right)=0\hfill \\ \hfill p^3:D^{\alpha }{v}_3+{\displaystyle \frac{3v_0{v}_1^2}{3!}}+{\displaystyle \frac{3v_0^2{v}_2}{3!}}+{\displaystyle \frac{v_1^2}{2!}}+{\displaystyle \frac{2v_0{v}_2}{2!}}+2v_2& =0,v_3\left(0\right)=0\hfill \end{array}\end{array}$$

(22)

Now, the fractional integral operator $J_{\alpha }$ with comfortable derivative definition is applied on both sides of Equation (8), we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill v_0\left(x\right)& =c_{3}x+c_4\hfill \\ \hfill v_1\left(x\right)& =c_{3}x+c_4+{\displaystyle \frac{1}{24}}{x}^{4+\alpha }{c}_3^3+{\displaystyle \frac{1}{6}}{x}^{3+\alpha }{c}_3^2+{\displaystyle \frac{1}{6}}{x}^{3+\alpha }{c}_3^2{c}_4+{\displaystyle \frac{1}{2}}{x}^{2+\alpha }{c}_3{c}_4+{\displaystyle \frac{1}{2}}{x}^{2+\alpha }{c}_3\hfill \\ & +{\displaystyle \frac{1}{4}}{x}^{2+\alpha }{c}_3{c}_4^2+{\displaystyle \frac{1}{2}}{x}^{1+\alpha }{c}_4^2+x^{1+\alpha }+{\displaystyle \frac{1}{6}}{x}^{1+\alpha }{c}_4^3+{\displaystyle \frac{1}{6}}{x}^{4+\alpha }{c}_3^3+x^{1+\alpha }{c}_4\hfill \\ & .\hfill \\ & .\hfill \\ & .\hfill \end{array}\end{array}$$

(23)

Case III: ${\displaystyle \frac{5}{2}}\le x\le 3$

In this case, the constructed homotopy will be same as in Case I. After substituting Equation (3) in Equation (2), we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill v_0\left(x\right)& =c_{5}x+c_6\hfill \\ \hfill v_1\left(x\right)& =c_{5}x+c_6+{\displaystyle \frac{1}{24}}{x}^{4+\alpha }{c}_5^3+{\displaystyle \frac{1}{6}}{x}^{3+\alpha }{c}_5^2+{\displaystyle \frac{1}{6}}{x}^{3+\alpha }{c}_5^2{c}_6+{\displaystyle \frac{1}{2}}{x}^{2+\alpha }{c}_5{c}_6+{\displaystyle \frac{1}{2}}{x}^{2+\alpha }{c}_5\hfill \\ & +{\displaystyle \frac{1}{4}}{x}^{2+\alpha }{c}_5{c}_6^2+{\displaystyle \frac{1}{2}}{x}^{1+\alpha }{c}_6^2+x^{1+\alpha }+{\displaystyle \frac{1}{6}}{x}^{1+\alpha }{c}_6^3+{\displaystyle \frac{1}{6}}{x}^{4+\alpha }{c}_5^3+x^{1+\alpha }{c}_6\hfill \\ & .\hfill \\ & .\hfill \\ & .\hfill \end{array}\end{array}$$

(24)

We calculate the results by taking $\alpha =1.25,1.5$ and $1.75$.

1. For $\alpha =1.25$

$$v\left(x\right)=\left\{\begin{array}{c}{c}_{1}x+c_2+0.04166666667x^{\frac{21}{4}}{c}_1^3+0.1666666666x^{\frac{17}{4}}{c}_1^2+0.1666666666x^{\frac{17}{4}}{c}_1^2{c}_2\hfill \\ +0.500000000x^{\frac{13}{4}}{c}_1{c}_2+0.500000000x^{\frac{13}{4}}{c}_1+.250000000x^{\frac{13}{4}}{c}_1{c}_2^2+0.5000000000x^{\frac{9}{4}}{c}_2^2\hfill \\ +x^{\frac{9}{4}}+0.1666666667x^{\frac{9}{4}}{c}_2^3+x^{\frac{9}{4}}{c}_2,for1\le x<{\displaystyle \frac{3}{2}}\hfill \\ c_{3}x+c_4+0.04166666667x^{\frac{21}{4}}{c}_3^3+0.1666666666x^{\frac{17}{4}}{c}_3^2+0.1666666666x^{\frac{17}{4}}{c}_3^2{c}_4\hfill \\ +0.500000000x^{\frac{13}{4}}{c}_3{c}_4+1.000000000x^{\frac{13}{4}}{c}_3+0.250000000x^{\frac{13}{4}}{c}_3{c}_4^2+0.5000000000x^{\frac{9}{4}}{c}_4^2\hfill \\ +0.1666666667x^{\frac{9}{4}}{c}_4^3+2x^{\frac{9}{4}}{c}_4,for{\displaystyle \frac{3}{2}}\le x<{\displaystyle \frac{5}{2}}\hfill \\ c_{5}x+c_6+0.04166666667x^{\frac{21}{4}}{c}_5^3+0.1666666666x^{\frac{17}{4}}{c}_5^2+0.1666666666x^{\frac{17}{4}}{c}_5^2{c}_6\hfill \\ +0.500000000x^{\frac{13}{4}}{c}_5{c}_6+0.500000000x^{\frac{13}{4}}{c}_5+0.250000000x^{\frac{13}{4}}{c}_5{c}_6^2+0.5000000000x^{\frac{9}{4}}{c}_6^2\hfill \\ +x^{\frac{9}{4}}+0.1666666667x^{\frac{9}{4}}{c}_6^3+x^{\frac{9}{4}}{c}_6,for{\displaystyle \frac{5}{2}}\le x\le 3\hfill \end{array}\right.$$

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Now by applying the continuity conditions at $x={\displaystyle \frac{3}{2}}$ and $x={\displaystyle \frac{5}{2}}$ and BCs, we get a system of six nonlinear equations. By using Newton’s method for nonlinear system, we obtain the values of constants:

$$c_1=-1.3295886370,c_2=0.4580256818,c_3=-0.00096246893$$

$$c_4=0.0247772323,c_5=-0.4962384877,c_6=-0.5257196774$$

(26)

By substituting values of constants from Equation (12) into Equation (11), we get the analytical solution of system of second-order fractional BVPs subject to obstacle problem given in Equation (1).

$$v\left(x\right)=\left\{\begin{array}{c}-1.3295886370x+0.4580256818-0.09793561230x^{\frac{21}{4}}+0.4295844110x^{\frac{17}{4}}\hfill \\ -1.039019967x^{\frac{13}{4}}+1.578934123x^{\frac{9}{4}},for1\le x<{\displaystyle \frac{3}{2}}\hfill \\ -0.0096246893x+0.0247772323-3.714915284\times {10}^{-8}{x}^{\frac{21}{4}}\hfill \\ +1.582164570\times {10}^{-5}{x}^{\frac{17}{4}}-0.009745403058x^{\frac{13}{4}}\hfill \\ +0.0498639553x^{\frac{9}{4}},for{\displaystyle \frac{3}{2}}\le x<{\displaystyle \frac{5}{2}}\hfill \\ -0.4962384877x-0.5257196774-0.005091668167x^{\frac{21}{4}}+0.01946546333x^{\frac{17}{4}}\hfill \\ -0.1519658196x^{\frac{13}{4}}+0.5882544081x^{\frac{9}{4}},for{\displaystyle \frac{5}{2}}\le x\le 3\hfill \end{array}\right.$$

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In similar manners, we can find the solution of problem mentioned in Equation (1) for $\alpha =1.5$ and $\alpha =1.75$.

2. For $\alpha =1.5$

We get the following analytical solution of system given in Equation (1) for $\alpha =1.5$.

$$v\left(x\right)=\left\{\begin{array}{c}-1.5493863800+0.6139530788-0.15497725350x^{\frac{11}{2}}+0.6457421302x^{\frac{9}{2}}\hfill \\ -1.396324258x^{\frac{7}{2}}+1.840992684x^{\frac{5}{2}},for1\le x<{\displaystyle \frac{3}{2}}\hfill \\ -0.0191553809x+0.0400302957-2.928607263\times {10}^{-7}{x}^{\frac{11}{2}}\hfill \\ +6360281305\times {10}^{-5}{x}^{\frac{9}{2}}-0.01954645244x^{\frac{7}{2}}\hfill \\ +0.08087249461x^{\frac{5}{2}},for{\displaystyle \frac{3}{2}}\le x<{\displaystyle \frac{5}{2}}\hfill \\ -0.4940057367x-0.5879890892-0.005023249330x^{\frac{11}{2}}+0.01675797164x^{\frac{9}{2}}\hfill \\ -0.1444661720x^{\frac{7}{2}}+0.5509954694x^{\frac{5}{2}},for{\displaystyle \frac{5}{2}}\le x\le 3\hfill \end{array}\right.$$

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3. For $\alpha =1.75$

We obtain the following analytical solution of the considered problem given in Equation (1) for $\alpha =1.75$ is as follows:

$$v\left(x\right)=\left\{\begin{array}{c}-1.7503932530x+0.7531819483-0.2234578677x^{\frac{23}{4}}+0.8952555066x^{\frac{19}{4}}\hfill \\ -1.782621032x^{\frac{15}{4}}+2.108034697x^{\frac{11}{4}},for1\le x<{\displaystyle \frac{3}{2}}\hfill \\ -0.0345683146x+0.0637056320-1.721168446\times {10}^{-6}{x}^{\frac{23}{4}}\hfill \\ +0.000211849098x^{\frac{19}{4}}-0.03570448584x^{\frac{15}{4}}\hfill \\ +0.1294835584x^{\frac{11}{4}},for{\displaystyle \frac{3}{2}}\le x<{\displaystyle \frac{5}{2}}\hfill \\ -0.4976869784x-0.6267857182-0.005136385271x^{\frac{23}{4}}+0.01540705241x^{\frac{19}{4}}\hfill \\ -0.1417523126x^{\frac{15}{4}}+0.5286045753x^{\frac{11}{4}},for{\displaystyle \frac{5}{2}}\le x\le 3\hfill \end{array}\right.$$

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Discussion

Figure 3, Figure 4 and Figure 5 present the solution of obstacle problem for different values of $\alpha =1.25$, $\alpha =1.50$, $\alpha =1.75$, respectively. Figure 6 shows the comparison between different values of $\alpha$. In Figure 6, obstacle achieves more height for larger values of $\alpha$ and vice versa.

6. Conclusions

The analysis of HPM for the solution of FPDEs was given. A unified convergence theorem was proved and results were validated for the solution of FBP equation. The method to solve FPDEs was presented, while the same partial differential equation with ordinary derivative i.e., for $\alpha =1$ fails to exist. This study demonstrated the importance of fractional derivative and the technique of dealing with these types of PDEs where solution of some ordinary PDEs does not exist. Moreover, HPM was applied to solve complex obstacle BVP. The suggested method can be applied to find solutions of other PDEs (both linear and nonlinear) of fractional order. The present study can be useful to analyze other traditional analytical techniques, such as Adomian Decomposition Method and Homotopy Analysis Method, to solve nonlinear differential equation of non- integer order. Furthermore, the present work may be extended to solve practical fractional models, for example wireless networks and nonlinear obstacle problems.