An improved homotopy perturbation method for solving local fractional nonlinear oscillators

Abstract

Local fractional nonlinear oscillators are studied by a modification of the homotopy perturbation method coupled with the variational iteration method. In the solution process, the fractional variational iteration method is adopted to transform the nonlinear oscillator equation into an integral equation, which is then decomposed into a series of equations, which are solved by the homotopy perturbation method. Two examples are given to show that the proposed method is simpler and more flexible than the classical homotopy perturbation method. It is also shown that the key to use this method is to select the appropriate decomposition of the initial value series.

Introduction

Most fractional equations are generated by solving specific problems of science and engineering, for example the gas permeability of a nanofiber membrane,^1–6 moisture permeability through a cocoon,^7,8 nanoscale multiphase flow, ⁹ solvent evaporation in the bubble electrospinning,^10–12 oscillators arising in microphysics and tsunami motion,^13,14 and sound transmission in hierarchic porous medium that can be exactly modeled by fractional calculus.^15–17 There are many fractional derivatives in literature, which are incompatible with each other, among which He’s fractional derivative, ¹⁵ the local fractional derivative,^18–20 the Riemann–Liouville fractional derivative, the Caputo fractional derivative, the Hadamard fractional derivative, and the Atangana–Baleanu fractional derivative ²¹ are widely used.

There are many computational methods for handling these fractional differential equations, such as the reduced differential transform method, ²² the transform methods, ²³ the Yang Laplace transform-DJ iteration method, ²⁴ the local fractional Fourier method,^25,26 and others.^15–17

Fractional oscillation equations arise in various areas of engineering and applied sciences, which are used as a powerful tool to vibration isolation and reduction of unnecessary vibration by porous media.

The variational iteration method ²⁷ and the homotopy perturbation method (HPM) ²⁸ were first proposed by Chinese mathematician, Prof. Ji-Huan He, and have been widely used to deal with nonlinear problems.^29–39 In this paper, we attempt to solve nonlinear fractional oscillation equations by coupling the variational iteration method with the HPM, which is simpler than using only the HPM or the variational iteration method.

The organization of the manuscript is as follows. In the next section, the local fractional operators are introduced. In “A modification of the HPM” section, the improved HPM for solving the local fractional oscillation equation is investigated. In “Illustrative examples” section, several examples are considered. In the final section, conclusions are presented.

Local fractional operators

In this section, we introduce the mathematical preliminaries of fractional calculus theory which shall be used in this paper.^18–20

Definition 1. If the function $\varphi \left(x\right)$ is local fractional continuous on the interval $\left(a,b\right)$ , the local fractional derivative of $\varphi \left(x\right)$ of order $\alpha$ at $x=x_0$ is given by

$$D_x{\hspace{0.17em}}^{\left(\alpha \right)}\varphi \left(x_0\right)={\varphi }^{\left(\alpha \right)}\left(x_0\right)=\frac{d^{\alpha }\varphi \left(x\right)}{d{x}^{\alpha }}|{\hspace{0.17em}}_{x=x_0}=\underset{x\to x_0}{\lim }\frac{{\Delta }^{\alpha }\left(\varphi \left(x\right)-\varphi \left(x_0\right)\right)}{{\left(x-x_0\right)}^{\alpha }}$$

(1)where ${\Delta }^{\alpha }\left(\varphi \left(x\right)-\varphi \left(x_0\right)\right)\cong \Gamma \left(1+\alpha \right)\Delta \left(\varphi \left(x\right)-\varphi \left(x_0\right)\right)$ .

Definition 2. In the interval $\left[c,d\right],$ the local fractional integral of $\varphi \left(x\right)$ of order $\alpha \left(0<\alpha \le 1\right)$ is defined as

$$Ia{{\hspace{0.17em}}_b}^{\left(\alpha \right)}\varphi \left(x\right)=\frac{1}{\Gamma \left(1+\alpha \right)}{\displaystyle {\int }_a^b\varphi \left(x\right){\left(\text{d}x\right)}^{\alpha }}=\frac{1}{\Gamma \left(1+\alpha \right)}\underset{\Delta t\to 0}{\lim }{\displaystyle \sum _{j=0}^{j=N-1}\varphi \left(x_j\right)}{\left(\Delta x_j\right)}^{\alpha }$$

(2)where $\Delta x_j=x_{j+1}-x_j$ , $\Delta x=\max \left\{\Delta x_1,\Delta x_2,\Delta x_j,\mathrm{\dots }\right\}$ and $\left[x_j,x_{j+1}\right]$ , j=0,…,N−1, $x_0=a,x_N=b$ , is a partition of the interval [a,b].

Definition 3. The Yang–Sumudu transform of $\varphi \left(x\right)$ of order α(0 < α ≤ 1) is defined as

$$D_{\alpha }\left\{\varphi \left(x\right)\right\}={\Phi }_{\alpha }\left(z\right)=:\frac{1}{\Gamma \left(1+\alpha \right)}{\displaystyle {\int }_0^{+\infty }{E}_{\alpha }}\left(-z^{-\alpha }{x}^{\alpha }\right)\frac{\varphi \left(x\right)}{z^{\alpha }}{\left(\text{d}x\right)}^{\alpha }$$

(3)

Theorem 1. Assuming $D_{\alpha }\left\{{\varphi }_1\left(x\right)\right\}={\Phi }_{1\alpha }\left(z\right)$ and $D_{\alpha }\left\{{\varphi }_2\left(x\right)\right\}={\Phi }_{2\alpha }\left(z\right)$ , then we have

$$D_{\alpha }\left\{{\varphi }_1\left(x\right)\ast {\varphi }_2\left(x\right)\right\}=z^{\alpha }{\Phi }_{1\alpha }\left(z\right){\Phi }_{2\alpha }\left(z\right)$$

(4)where the convolution ${\varphi }_1\left(x\right)\ast {\varphi }_2\left(x\right)=\frac{1}{\Gamma \left(1+\alpha \right)}I0{{\hspace{0.17em}}_x}^{\left(\alpha \right)}{\varphi }_1\left(t\right){\varphi }_2\left(x-t\right){\left(dt\right)}^{\alpha }$ .

A modification of the HPM

In this section, we shall illustrate the solution procedure of the improved HPM to derive particular solution of some local fractional nonlinear oscillation equations. In order to present this method, we give the following local fractional oscillation equation

$$\frac{{\partial }^{2\alpha }u\left(x\right)}{\partial x^{2\alpha }}=L\left[u\left(x\right)\right]+R\left[u\left(x\right)\right]\text{+}g\left(x\right)$$

(5)associated with the initial conditions $u\left(0\right)={\phi }_0,u^{\left(\alpha \right)}\left(0\right)={\phi }_1,$ $L\left[u\left(x\right)\right]$ represents the linear part and $R\left[u\left(x\right)\right]$ represents the nonlinear terms, $u\left(x\right)$ and $x$ are generalized displacements and time variables, respectively, ${\phi }_0$ and ${\phi }_1$ are the well-known initial conditions, and $g\left(x\right)$ is the inhomogeneous part, often this inhomogeneous part contains excited parameters.

Moreover, α is the value of fractal dimensions of the porous medium. For $\alpha =1$ , the proposed models turn out to be those of the classical oscillation equation without holes.

By means of the initial conditions and according to the rule of fractional variational iteration method, we construct the following initial value $u_0\left(x\right)$ for equation (5)

$$u_0\left(x\right)={\phi }_0+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}{\phi }_1$$

(6)

According to the fractional variational iteration method, ³² we construct a correction functional for equation (5) as follows

$$u_{n+1}\left(x\right)=u_n\left(x\right)+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\lambda \left(x-\tau \right)\left[\frac{{\partial }^{2\alpha }{u}_n}{\partial x^{2\alpha }}-L\left[{\tilde{u}}_n\left(\tau \right)\right]-R\left[{\tilde{u}}_n\left(\tau \right)\right]-g\left(\tau \right)\right]d\tau \right\}$$

(7)where $\lambda$ is a fractal Lagrange multipliers and $L\left[{\tilde{v}}_n\left(\tau \right)\right]$ and $R\left[{\tilde{v}}_n\left(\tau \right)\right]$ are considered as restricted local fractional variation, i.e. ${\delta }^{\alpha }L\left[{\tilde{v}}_n\left(\tau \right)\right]=0$ and ${\delta }^{\alpha }R\left[{\tilde{v}}_n\left(\tau \right)\right]\text{=}0.$

Applying the Yang–Sumudu transform^18–20 on both sides of equation (7) results in

$$D_{\alpha }\left\{v_{n+1}\right\}=D_{\alpha }\left\{v_n\right\}+D_{\alpha }\left\{I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\lambda \left(x-\tau \right)\left[\left[\frac{{\partial }^{2\alpha }{u}_n}{\partial x^{2\alpha }}-L\left[{\tilde{u}}_n\left(\tau \right)\right]-R\left[{\tilde{u}}_n\left(\tau \right)\right]-g\left(\tau \right)\right]\right]\right\}d\tau \right\}$$

(8)

Using Theorem 1, we can get

$$D_{\alpha }\left\{v_{n+1}\right\}=D_{\alpha }\left\{v_n\right\}\text{+}{z}^{\alpha }{D}_{\alpha }\left\{\lambda \left(x\right)\right\}{D}_{\alpha }\left\{\left[\frac{{\partial }^{2\alpha }{u}_n}{\partial x^{2\alpha }}-L\left[{\tilde{u}}_n\left(\tau \right)\right]-R\left[{\tilde{u}}_n\left(\tau \right)\right]-g\left(\tau \right)\right]\right\}$$

(9)

It can be easily inferred from equation (9) that

$$\begin{array}{c}{\delta }^{\alpha }\left\{D_{\alpha }\left\{v_{n+1}\right\}\right\}={\delta }^{\alpha }\left\{D_{\alpha }\left\{v_n\right\}\right\}\text{+}{z}^{\alpha }{\delta }^{\alpha }\left\{D_{\alpha }\left\{\lambda \left(x\right)\right\}\right\}{\delta }^{\alpha }\left\{D_{\alpha }\left[\frac{{\partial }^{2\alpha }{u}_n}{\partial x^{2\alpha }}-L\left[{\tilde{u}}_n\left(\tau \right)\right]-R\left[{\tilde{u}}_n\left(\tau \right)\right]-g\left(\tau \right)\right]\right\}\end{array}$$

(10)

By virtue of ${\delta }^{\alpha }L\left[{\tilde{v}}_n\left(\tau \right)\right]=0$ and ${\delta }^{\alpha }R\left[{\tilde{v}}_n\left(\tau \right)\right]\text{=}0,$ we can get

$${\delta }^{\alpha }\left\{D_{\alpha }\left\{u_{n+1}\right\}\right\}={\delta }^{\alpha }\left\{D_{\alpha }\left\{u_n\right\}\right\}\text{+}{z}^{\alpha }{\delta }^{\alpha }\left\{D_{\alpha }\left\{\lambda \left(x\right)\right\}\right\}{\delta }^{\alpha }\left\{D_{\alpha }\left[\frac{{\partial }^{2\alpha }{u}_n}{\partial x^{2\alpha }}\right]\right\}$$

(11)

Using ${\delta }^{\alpha }{u}_{n+1}=0$ , it is easily inferred from equation (11) that

$$1+D_{\alpha }\left\{\lambda \left(x\right)\right\}\frac{1}{z^{\alpha }}=0$$

(12)

According to equation (12) and Theorem 1, we obtain the following fractal Lagrange multiplier

$$\lambda \left(x-\tau \right)=\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}$$

(13)

Substituting equation (13) into equation (7), the variational iteration relationship of equation (5) can be constructed as following

$$\begin{array}{c}{u}_{n+1}\left(x\right)=u_n\left(x\right)+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[\frac{{\partial }^{2\alpha }{u}_n}{\partial x^{2\alpha }}-L\left[u_n\left(\tau \right)\right]-R\left[u_n\left(\tau \right)\right]-g\left(\tau \right)\right]d\tau \right\},\\ u_0\left(x\right)=u\left(0\right)+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}{u}_x{\hspace{0.17em}}^{\left(\alpha \right)}\left(0\right)\end{array}$$

(14)

Wu and Lee ²⁹ gave the proof of the convergences of equation (14).

Via calculating, we know that equation (14) is equivalent to the following iteration relationship

$$\begin{array}{c}{u}_{n+1}\left(x\right)=u_0\left(x\right)+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[-L\left[u_n\left(\tau \right)\right]-R\left[u_n\left(\tau \right)\right]-g\left(\tau \right)\right]d\tau \right\},\\ u_0\left(x\right)=u\left(0\right)+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}{u}_x{\hspace{0.17em}}^{\left(\alpha \right)}\left(0\right)\end{array}$$

(15)

The following integral equation can be induced from equation (15)

$$\{\begin{array}{l}u\left(x\right)=f\left(x\right)+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[-L\left[u\left(\tau \right)\right]-R\left[u\left(\tau \right)\right]-g\left(\tau \right)\right]d\tau \right\},\\ f\left(x\right)=u\left(0\right)+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}{u}_x{\hspace{0.17em}}^{\left(\alpha \right)}\left(0\right)\end{array}$$

(16)

Note that both equations (15) and (16) have the same $n$ times approximate solution $v_n\left(x\right)$ , which tends to their exact solution $u\left(x\right)$ of equation (5). Then, solving equation (5) can be reduced to solving equation (16), which actually provides an equivalent equation for equation (5). On the other hand, since there are many choices of the Lagrange multipliers, the equivalent integral equations can be derived by many other forms.

To illustrate the HPM, we consider equation (16) as

$$\{\begin{array}{l}v\left(x\right)=f\left(x\right)+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[-L\left[v\left(\tau \right)\right]-R\left[v\left(\tau \right)\right]-g\left(\tau \right)\right]d\tau \right\},\\ f\left(x\right)=u\left(0\right)+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}{u}_x{\hspace{0.17em}}^{\left(\alpha \right)}\left(0\right)\end{array}$$

(17)

Let the solution $v\left(x\right)$ of equation (17) has the following series form

$$v\left(x\right)={\displaystyle \sum _{i=0}^{\infty }{p}^{i\alpha }}{v}_i\left(x\right),\hspace{1em}n=1,2,\mathrm{\dots }$$

(18)

Substituting equation (18) into equation (17), we can get

$$\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }{p}^{n\alpha }{v}_n\left(x\right)}=u\left(0\right)+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}{u}_x{\hspace{0.17em}}^{\left(\alpha \right)}\left(0\right)-I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}g\left(\tau \right)d\tau \right\}\\ -p^{\alpha }I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[{\displaystyle \sum _{n=0}^{\infty }}L\left[v_n\left(\tau \right)\right]\text{+}R\left[p^{n\alpha }{v}_n\left(x\right)\right]\right]d\tau \right\}\end{array}$$

(19)

This

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }{p}^{n\alpha }{v}_n\left(x\right)}=u\left(0\right)+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}{u}_x{\hspace{0.17em}}^{\left(\alpha \right)}\left(0\right)-I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[g\left(\tau \right)\right]d\tau \right\}\\ -p^{\alpha }{\displaystyle \sum _{n=0}^{\infty }}I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}L\left[v_n\left(\tau \right)\right]d\tau \right\}-{\displaystyle \sum _{n=0}^{\infty }}I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}{H}_n\left(v\right)d\tau \right\}\end{array}$$

(20)where $H_n\left(v\right)$ is the He polynomials and $H_n\left(v\right)$ is given as follows

$$H_n\left(v\right)=\frac{1}{\Gamma \left(1+n\alpha \right)}\frac{d^{n\alpha }}{d{p}^{n\alpha }}{\left\{R\left[\left({\displaystyle \sum _{i=0}^n{p}^{i\alpha }}{v}_i\left(x\right)\right)\right]\right\}}_{\lambda =0}$$

(21)

In this paper, we let

$$G=u\left(0\right)+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}{u}_x{\hspace{0.17em}}^{\left(\alpha \right)}\left(0\right)-I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}g\left(\tau \right)d\tau \right\}$$

(22)

And we suppose $G$ can be decomposed as the following series

$$G\text{=}{\displaystyle \sum _{n=0}^{\infty }{G}_n}$$

(23)

According to equations (20) and (23) and the classic HPM, we can construct the following new recursion scheme

$$\{\begin{array}{l}{v}_0=G_0\hfill \\ v_1\text{=}{G}_1-I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[L\left(v_0\right)\right]d\tau \right\}-I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}{H}_{0}d\tau \right\}\hfill \\ \mathrm{\dots \dots },\hfill \\ v_{m+1}=G_n-I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[L\left(v_m\right)\right]d\tau \right\}-I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}{H}_{m}d\tau \right\}\hfill \end{array}$$

(24)

Obviously, similarly to the classic HPM, we can easily verify that

$$u=\underset{p\to 1}{\lim }{\displaystyle \sum _{n=0}^{\infty }{p}^{n\alpha }{v}_n}$$

(25)converges to the exact solution of equation (5).

Illustrative examples

In order to demonstrate the validity of the improved HPM in “A modification of the HPM” section, we give the following local fractional differential equations.

Example 1. Consider the forcing Duffing equation

$$\frac{d^{2\alpha }u}{d{x}^{2\alpha }}+3u-2u^3={\cos }_{\alpha }{x}^{\alpha }{\sin }_{\alpha }{\left(2x\right)}^{\alpha }$$

(26)subject to the initial conditions $u\left(0\right)=0,u_x\left(0\right)=1.$

By means of equation (19), equation (26) can be transformed into the following equation

$$v\left(x\right)=\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}-I0{\hspace{0.17em}}_x^{\alpha }\left[\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}{\cos }_{\alpha }{x}^{\alpha }{\sin }_{\alpha }{\left(2x\right)}^{\alpha }\right]d\tau +I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[3v\left(\tau \right)-2v^3\left(\tau \right)\right]d\tau \right\}$$

(27)

Letting $v\left(x\right)={\displaystyle \sum _{i=0}^{\infty }{p}^{i\alpha }}{v}_i\left(x\right),n=1,2,\mathrm{\dots }$ and substituting it into equation (27), we can derive

$$\begin{array}{c}{\displaystyle \sum _{i=0}^{\infty }{p}^{i\alpha }}{v}_i\left(x\right)=\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}-\frac{{\left(-1\right)}^{n+1}}{2}{\displaystyle \sum _{n=1}^{+\infty }\frac{\left(3^{2n-1}+1\right)x^{\left(2n+1\right)\alpha }{p}^{\left(2n+1\right)\alpha }}{\Gamma \left[1+\left(2n+1\right)\alpha \right]}}\\ +p^{\alpha }I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[3{\displaystyle \sum _{n=0}^{\infty }{p}^{i\alpha }{v}_i\left(x\right)\left(\tau \right)}-2{\displaystyle \sum _{n=0}^{\infty }{H}_n}\right]d\tau \right\}\end{array}$$

(28)where $H_n\left(v\right)$ is the He polynomials for the nonlinear term $v^3$ and $H_n\left(v\right)$ is given as follows

$$H_n\left(v\right)=\frac{1}{\Gamma \left(1+n\alpha \right)}\frac{d^{n\alpha }}{d{p}^{n\alpha }}{\left\{\left[{\left({\displaystyle \sum _{n=0}^{\infty }{p}^{i\alpha }}{v}_i\left(x\right)\right)}^3\right]\right\}}_{\lambda =0.}$$

(29)

Obviously, according to equation (29), the first few He polynomials are, respectively, given by

$$\{\begin{array}{l}{H}_0=v_0^3=\frac{6x^{3\alpha }}{\Gamma \left(1+3\alpha \right)},\hfill \\ H_1{\text{=3}v}_0^2{v}_1=\frac{-60x^{5\alpha }}{\Gamma \left(1+5\alpha \right)},\hfill \\ H_2=3v_0^2{v}_2+3v_1^2{v}_0=\frac{546x^{7\alpha }}{\Gamma \left(1+7\alpha \right)},\hfill \\ \mathrm{\dots \dots },\hfill \\ H_n=\frac{{\left(-1\right)}^{n+1}}{4}\frac{\left(3^{2n+3}-3\right)x^{\left(2n+3\right)\alpha }}{\Gamma \left[1+\left(2n+3\right)\alpha \right]}\hfill \end{array}$$

(30)

In the light of equation (22), we let

$$G\text{=}{\displaystyle \sum _{n=0}^{\infty }{G}_n}=\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}-\frac{{\left(-1\right)}^{n+1}}{2}{\displaystyle \sum _{n=1}^{+\infty }\frac{\left(3^{2n-1}+1\right)x^{\left(2n+1\right)\alpha }}{\Gamma \left[1+\left(2n+1\right)\alpha \right]}}$$

(31)where

$$\begin{array}{l}{G}_0=\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)},\hfill \\ G_1=\frac{-4}{2}\frac{x^{3\alpha }}{\Gamma \left(1+3\alpha \right)},\hfill \\ G_2=\frac{28}{2}\frac{x^{5\alpha }}{\Gamma \left(1+5\alpha \right)},\hfill \\ \mathrm{\dots .},\hfill \\ G_n=-\frac{{\left(-1\right)}^{n+1}}{2}\frac{\left(3^{2n-1}+1\right)x^{\left(2n+1\right)\alpha }}{\Gamma \left[1+\left(2n+1\right)\alpha \right]}\hfill \end{array}$$

(32)

Substituting equation (32) into equation (28) and then in light of equation (23), the recurrence relation of the improved HPM is constructed as follows

$$\{\begin{array}{l}{v}_0=G_0=\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)},\hfill \\ v_1=G_1+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[3v_0\left(\tau \right)-H_0\left(\tau \right)\right]d\tau \right\},\hfill \\ v_2=G_2+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[3v_1\left(\tau \right)-H_1\left(\tau \right)\right]d\tau \right\},\hfill \\ v_3=G_3+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[3v_2\left(\tau \right)-H_2\left(\tau \right)\right]d\tau \right\},\hfill \\ \mathrm{\dots \dots },\hfill \\ v_{m+1}=G_{m+1}+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[3v_m\left(\tau \right)-H_m\left(\tau \right)\right]d\tau \right\}\hfill \end{array}$$

(33)

Substituting equation (30) into equation (33), we can get

$$\{\begin{array}{l}{v}_0=G_0=\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)},\hfill \\ v_1=-\frac{x^{3\alpha }}{\Gamma \left(1+3\alpha \right)},\hfill \\ v_2=\frac{x^{5\alpha }}{\Gamma \left(1+5\alpha \right)},\hfill \\ \mathrm{\dots \dots },\hfill \\ v_m={\left(-1\right)}^m\frac{x^{\left(2m+1\right)\alpha }}{\Gamma \left[1+\left(2m+1\right)\alpha \right]}\hfill \end{array}$$

(34)

Proceeding in this manner, the rest of the components $v_i\left(x\right)$ can also be completely determined and then, making using of equation (25), we can yield the following exact solution of equation (26)

$$u\left(x\right)={\left(-1\right)}^m{\displaystyle \sum _{m=0}^{\infty }\frac{x^{\left(2m+1\right)\alpha }}{\Gamma \left[1+\left(2m+1\right)\alpha \right]}}={\sin }_{\alpha }{x}^{\alpha }$$

(35)

This periodic solution is nondifferentiable and it is stair shaped, which can be seen in the works of Chinese mathematician, Prof. Xiao-jun Yang.^18–20 Since α is the value of fractal dimensions of the fractal medium, with the help of the geometric meaning of the Cantor fractal space, we can quantitatively analyze the intensity of sound wave propagation in porous medium.^18–20

Example 2. Consider the following fractional differential equation

$$\frac{d^{2\alpha }u}{d{x}^{2\alpha }}-u+u^{\text{2}}=E_{\alpha }{\left(2x\right)}^{\alpha }$$

(36)subject to the initial conditions $u\left(0\right)=1,u_x\left(0\right)=1.$

By means of equation (19), equation (36) can be transformed into the following equation

$$v\left(x\right)=\left(1+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}\right)-I0{\hspace{0.17em}}_x^{\alpha }{E}_{\alpha }{\left(2\tau \right)}^{\alpha }d\tau -I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[v\left(\tau \right)-v^{\text{2}}\left(\tau \right)\right]d\tau \right\}$$

(37)

Letting $v\left(x\right)={\displaystyle \sum _{i=0}^{\infty }{p}^{i\alpha }}{v}_i\left(x\right),n=1,2,\mathrm{\dots }$ and substituting it into equation (37), we can derive

$$\begin{array}{c}{\displaystyle \sum _{i=0}^{\infty }{p}^{i\alpha }}{v}_i\left(x\right)=\left(1+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}\right)-{\displaystyle \sum _{n=0}^{+\infty }\frac{2^{n\alpha }{x}^{\left(n+2\right)\alpha }}{\Gamma \left[1+\left(n+2\right)\alpha \right]}}\\ -p^{\alpha }I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[{\displaystyle \sum _{n=0}^{\infty }{p}^{i\alpha }{v}_i\left(x\right)\left(\tau \right)}-{\displaystyle \sum _{n=0}^{\infty }{H}_n\left(\tau \right)}\right]d\tau \right\}\end{array}$$

(38)where $H_n\left(v\right)$ is the He polynomials for the nonlinear term $v^{\text{2}}$ and $H_n\left(v\right)$ is given as follows

$$H_n\left(v\right)=\frac{1}{\Gamma \left(1+n\alpha \right)}\frac{d^{n\alpha }}{d{p}^{n\alpha }}{\left\{\left[{\left({\displaystyle \sum _{n=0}^{\infty }{p}^{i\alpha }}{v}_i\left(\tau \right)\right)}^2\right]\right\}}_{\lambda =0}$$

(39)

Obviously, according to equation (39), the first few He polynomials are, respectively, given by

$$\{\begin{array}{l}{H}_0=v_0^2=1+\frac{2x^{\alpha }}{\Gamma \left(1+\alpha \right)}+\frac{2x^{2\alpha }}{\Gamma \left(1+2\alpha \right)},\hfill \\ H_1\text{=}2v_0{v}_1=\frac{2x^{2\alpha }}{\Gamma \left(1+2\alpha \right)}+\frac{8x^{3\alpha }}{\Gamma \left(1+3\alpha \right)}+\frac{8x^{4\alpha }}{\Gamma \left(1+4\alpha \right)},\hfill \\ H_2=v_1^2+2v_0{v}_2=\frac{8x^{4\alpha }}{\Gamma \left(1+4\alpha \right)}+\frac{32x^{5\alpha }}{\Gamma \left(1+5\alpha \right)}+\frac{32x^{6\alpha }}{\Gamma \left(1+6\alpha \right)},\hfill \\ \mathrm{\dots \dots },\hfill \\ H_{m+1}=2^{2m-1}\frac{x^{2m\alpha }}{\Gamma \left(1+2m\alpha \right)}+2^{m+3}\frac{x^{\left(2m+1\right)\alpha }}{\Gamma \left[1+\left(2m+1\right)\alpha \right]}+2^{m+3}\frac{x^{\left(2m+2\right)\alpha }}{\Gamma \left[1+\left(2m+2\right)\alpha \right]}\hfill \end{array}$$

(40)

In the light of equation (22), we let

$$G\text{=}{\displaystyle \sum _{n=0}^{\infty }{G}_n}=1+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)}-{\displaystyle \sum _{n=0}^{+\infty }\frac{2^{n\alpha }{x}^{\left(n+2\right)\alpha }}{\Gamma \left[1+\left(n+2\right)\alpha \right]}}$$

(41)where

$$\begin{array}{l}{G}_0=1+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)},\hfill \\ G_1=-\frac{x^{2\alpha }}{\Gamma \left(1+2\alpha \right)}-2\frac{x^{3\alpha }}{\Gamma \left(1+3\alpha \right)}-2\frac{x^{4\alpha }}{\Gamma \left(1+4\alpha \right)},\hfill \\ G_2=-\frac{2x^{4\alpha }}{\Gamma \left(1+4\alpha \right)}-\frac{8x^{5\alpha }}{\Gamma \left(1+5\alpha \right)}-\frac{8x^{6\alpha }}{\Gamma \left(1+6\alpha \right)},\hfill \\ \begin{array}{l}\mathrm{\dots .},\\ G_{n+1}=-2^{2m-1}\frac{x^{\left(2m+2\right)\alpha }}{\Gamma \left[1+\left(2m+2\right)\alpha \right]}-2^{m+3}\frac{x^{\left(2m+3\right)\alpha }}{\Gamma \left[1+\left(2m+3\right)\alpha \right]}-2^{m+3}\frac{x^{\left(2m+4\right)\alpha }}{\Gamma \left[1+\left(2m+4\right)\alpha \right]}\end{array}\hfill \end{array}$$

(42)

Substituting equation (42) into equation (38) and then in light of equation (23), the recurrence relation of the improved HPM is constructed as follows

$$\{\begin{array}{l}{v}_0=G_0=1+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)},\hfill \\ v_1=G_1+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[v_0\left(\tau \right)-H_0\right]d\tau \right\},\hfill \\ v_2=G_2+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[v_1\left(\tau \right)-H_1\right]d\tau \right\},\hfill \\ v_3=G_3+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[v_2\left(\tau \right)-H_2\right]d\tau \right\},\hfill \\ \mathrm{\dots \dots },\hfill \\ v_{m+1}=G_{m+1}+I0{\hspace{0.17em}}_x^{\left(\alpha \right)}\left\{\frac{{\left(x-\tau \right)}^{\alpha }}{\Gamma \left(1+\alpha \right)}\left[v_m\left(\tau \right)-H_m\right]d\tau \right\}\hfill \\ .\hfill \end{array}$$

(43)

Substituting equation (40) into equation (43), we can get

$$\{\begin{array}{l}{v}_0=G=1+\frac{x^{\alpha }}{\Gamma \left(1+\alpha \right)},\hfill \\ v_1=\frac{x^{2\alpha }}{\Gamma \left(1+2\alpha \right)}+\frac{x^{3\alpha }}{\Gamma \left(1+3\alpha \right)},\hfill \\ v_2=\frac{x^{4\alpha }}{\Gamma \left(1+4\alpha \right)}+\frac{x^{5\alpha }}{\Gamma \left(1+5\alpha \right)},\hfill \\ \mathrm{\dots \dots },\hfill \\ v_m=\frac{x^{2m\alpha }}{\Gamma \left(1+2m\alpha \right)}+\frac{x^{\left(2m+1\right)\alpha }}{\Gamma \left[1+\left(2m+1\right)\alpha \right]}\hfill \end{array}$$

(44)

Proceeding in this manner, the rest of the components of $v_i\left(x\right)$ can also be completely determined and then, making using of equation (25), we can yield the following exact solution of equation (36)

$$u\left(x\right)={\displaystyle \sum _{m=0}^{\infty }\frac{x^{m\alpha }}{\Gamma \left(1+m\alpha \right)}}=E_{\alpha }{\left(x\right)}^{\alpha }$$

(45)

Conclusions

In this work we proposed the improved HPM. The test examples showed that the suggested new technique can be regarded as an efficient tool for solving the local fractional differential equations, which avoids cumbersome computational works.