The Adomian Decomposition Method for a Class of First Order Fuzzy Dynamic Equations on Time Scales

1. Introduction

The theory of dynamic equations has many interesting applications in control theory, mathematical economics, mathematical biology, engineering and technology. In some cases, there exists uncertainty, ambiguity or vague factors in such problems, and fuzzy theory and interval analysis are powerful tools for modeling these equations on time scales.

In this paper, we introduce the Adomian decomposition method for the following class of first order fuzzy dynamic equations

$${\delta }_{H}y=f\left(y\right),t\in (t_0,T],$$

(1)

$$y\left(t_0\right)=y_0,$$

(2)

where

(A1) 

$f\in \mathcal{C}([t_0,T]\times F\left(\mathbb{R}\right))$, $f:F\left(\mathbb{R}\right)\to F\left(\mathbb{R}\right)$, $t_0,T\in \mathbb{T}$, $\mathbb{T}$ is an arbitrary time scale with forward jump operator and delta differentiation operator $\sigma$ and $\Delta$, respectively.

Here $F\left(\mathbb{R}\right)$ denotes the set of all real fuzzy numbers, $\tilde{0}$ denotes the zero fuzzy number and ${\delta }_H$ denotes the first type fuzzy delta derivative on $\mathbb{T}$.

To the best of our knowledge, there is a gap in the references for investigations of the Adomian decomposition method for fuzzy dynamic equations on time scales. Here, in this paper we try to fill out this gap.

This paper is organized as follows. In the next section, we give some basic definitions and facts of fuzzy dynamic calculus on time scales. In Section 3, we give an exposition of the Adomian decomposition method on time scales. In Section 4 we introduce the Adomian decomposition method for the problem (1), (2). In Section 5, we give a numerical example.

2. Fuzzy Dynamic Calculus Essentials

In this section, we will give some basic definitions and fact of fuzzy dynamic calculus on time scales. For detailed study of fuzzy dynamic calculus on time scales we refer the reader to the book [2].

Suppose that $\mathbb{T}$ is a time scale with forward jump operator and delta differentiation operator $\sigma$ and $\Delta$, respectively. With $F\left(\mathbb{R}\right)$ we will denote the space of the real fuzzy numbers and with $D(·,·)$ we will denote the Hausdorff distance between the real fuzzy numbers. For more details for fuzzy numbers and Hausdorff distance between the real fuzzy numbers we refer the reader to the appendix of the book [2].

Definition 1.([2]) Assume that $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ is a fuzzy function and $t\in {\mathbb{T}}^{\kappa }$. Then f is said to be first type right fuzzy delta differentiable at t, shortly right ${\delta }_H$-differentiable at t, if there exists an element ${\delta }_H^{+}f\left(t\right)\in F\left(\mathbb{R}\right)$ with the property that, for any given $ϵ>0$, there exists a neighbourhood $U_{\mathbb{T}}$ of t, i.e., $U_{\mathbb{T}}=(t-\delta ,t+\delta )\bigcap \mathbb{T}$ for some $\delta >0$, such that for all $t+h\in U_{\mathbb{T}}$ the H-difference $f(t+h){\ominus }_{H}f\left(\sigma \left(t\right)\right)$ exists and

$$D\left(f(t+h){\ominus }_{H}f\left(\sigma \left(t\right)\right),{\delta }_H^{+}f\left(t\right)(h-\mu \left(t\right))\right)\le ϵ|h-\mu \left(t\right)|$$

with $0\le h<\delta$. In this case, ${\delta }_H^{+}f\left(t\right)$ is said to be first type right fuzzy delta derivative of f at t, shortly right ${\delta }_H$-derivative of f at t.

Definition 2.([2]) Assume that $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ is a fuzzy function and $t\in {\mathbb{T}}^{\kappa }$. Then f is said to be first type left fuzzy delta differentiable at t, shortly left ${\delta }_H$-differentiable at t, if there exists an element ${\delta }_H^{-}f\left(t\right)\in F\left(\mathbb{R}\right)$ with the property that, for any given $ϵ>0$, there exists a neighbourhood $U_{\mathbb{T}}$ of t, i.e., $U_{\mathbb{T}}=(t-\delta ,t+\delta )\bigcap \mathbb{T}$ for some $\delta >0$, such that for all $t-h\in U_{\mathbb{T}}$ the H-difference $f\left(\sigma \left(t\right)\right){\ominus }_{H}f(t-h)$ exists and

$$D\left(f\left(\sigma \left(t\right)\right){\ominus }_{H}f(t-h),{\delta }_H^{-}f\left(t\right)(h+\mu \left(t\right))\right)\le ϵ(h+\mu \left(t\right))$$

with $0\le h<\delta$.

Definition 3.([2]) Let $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ be a fuzzy function and $t\in {\mathbb{T}}^{\kappa }$. Then f is said to be first type fuzzy delta differentiable at t, shortly ${\delta }_H$-differentiable at t, if f is both first type left and right fuzzy delta differentiable at $t\in {\mathbb{T}}^{\kappa }$ and ${\delta }_H^{-}f\left(t\right)={\delta }_H^{+}f\left(t\right)$, and we will denote it by ${\delta }_{H}f\left(t\right)$. We call ${\delta }_{H}f\left(t\right)$ the first type fuzzy delta derivative of f at t, shortly ${\delta }_H$-derivative of f at t. We say that f is first type fuzzy delta differentiable at t, shortly ${\delta }_H$-differentiable at t, if its ${\delta }_H$-derivative exists at t. We say that f is first type fuzzy delta differentiable on ${\mathbb{T}}^{\kappa }$, shortly ${\delta }_H$-differentiable on ${\mathbb{T}}^{\kappa }$, if its ${\delta }_H$-derivative exists at each $t\in {\mathbb{T}}^{\kappa }$. The fuzzy function ${\delta }_{H}f:{\mathbb{T}}^{\kappa }\to F\left(\mathbb{R}\right)$ is then called first type fuzzy delta derivative, shortly ${\delta }_H$-derivative of f on ${\mathbb{T}}^{\kappa }$.

The defined ${\delta }_H$-derivative has the following properties.

Theorem  1.([2]) If the ${\delta }_H$-derivative of f at $t\in {\mathbb{T}}^{\kappa }$ exists, then it is unique. Hence, ${\delta }_H$-derivative is well-defined.

Theorem  2.([2]) Assume that $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ is a continuous function at $t_1\in {\mathbb{T}}^{\kappa }$ and $t_1$ is right-scattered. Then f is ${\delta }_H$-differentiable at $t_1$ and

$${\delta }_{H}f\left(t_1\right)={\displaystyle \frac{f\left(\sigma \left(t_1\right)\right){\ominus }_{H}f\left(t_1\right)}{\mu \left(t_1\right)}}.$$

Theorem  3.([2]) Assume that $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ is ${\delta }_H$-differentiable at $t\in {\mathbb{T}}^{\kappa }$. Then f is continuous at t.

Theorem  4.([2]) Let $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ be a fuzzy function and let $t\in {\mathbb{T}}^{\kappa }$ be right-dense. Then f is ${\delta }_H$-differentiable at t if and only if the limits

$$\underset{h\to 0+}{lim}{\displaystyle \frac{f(t+h){\ominus }_{H}f\left(t\right)}{h}}\mathrm{and}\underset{h\to 0+}{lim}{\displaystyle \frac{f\left(t\right){\ominus }_{H}f(t-h)}{h}}$$

(3)

exist and satisfy the relations

$$\underset{h\to 0+}{lim}{\displaystyle \frac{f(t+h){\ominus }_{H}f\left(t\right)}{h}}=\underset{h\to 0+}{lim}{\displaystyle \frac{f\left(t\right){\ominus }_{H}f(t-h)}{h}}={\delta }_{H}f\left(t\right).$$

Theorem  5.([2]) Let $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ is ${\delta }_H$-differentiable at $t\in {\mathbb{T}}^{\kappa }$. Then

$$f\left(\sigma \left(t\right)\right)=f\left(t\right)+\mu \left(t\right)·{\delta }_{H}f\left(t\right)$$

or

$$f\left(t\right)=f\left(\sigma \left(t\right)\right)+(-1)·\left(\mu \left(t\right)·{\delta }_{H}f\left(t\right)\right).$$

Theorem  6.([2]) Let $f,g:\mathbb{T}\to F\left(\mathbb{R}\right)$ be ${\delta }_H$-differentiable at $t\in {\mathbb{T}}^{\kappa }$. Then $f+g:\mathbb{T}\to F\left(\mathbb{R}\right)$ is ${\delta }_H$-differentiable at $t\in {\mathbb{T}}^{\kappa }$ and

$${\delta }_H(f+g)\left(t\right)={\delta }_{H}f\left(t\right)+{\delta }_{H}g\left(t\right).$$

Theorem  7.([2]) Let $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ be ${\delta }_H$-differentiable at $t\in {\mathbb{T}}^{\kappa }$. Then for any $\lambda \in \mathbb{R}$ the function $\lambda ·f:\mathbb{T}\to F\left(\mathbb{R}\right)$ is ${\delta }_H$-differentiable at $t\in {\mathbb{T}}^{\kappa }$ and

$${\delta }_H(\lambda ·f)\left(t\right)=\lambda ·{\delta }_{H}f\left(t\right).$$

Theorem  8.([2]) Let $t\in {\mathbb{T}}^{\kappa }$, $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ and $f_{\alpha }\left(t\right)={\left[f\left(t\right)\right]}^{\alpha }$, $\alpha \in [0,1]$. If f is ${\delta }_H$-differentiable at t, then $f_{\alpha }$ is ${\delta }_H$-differentiable at t and

$${\delta }_H{\left[f\left(t\right)\right]}^{\alpha }={\delta }_H{f}_{\alpha }\left(t\right),\alpha \in [0,1].$$

Theorem  9.([2]) Let $t\in {\mathbb{T}}^{\kappa }$, $f:\mathbb{T}\to F\left(\mathbb{R}\right)$ is ${\delta }_H$-differentiable at t. Let also,

$${\left[f\left(t\right)\right]}^{\alpha }=\left[{\underline{f}}^{\alpha }\left(t\right),{\overline{f}}^{\alpha }\left(t\right)\right],\alpha \in [0,1].$$

Then ${\underline{f}}^{\alpha }$ and ${\overline{f}}^{\alpha }$ are Δ-differentiable at t and

$${\left[{\delta }_{H}f\left(t\right)\right]}^{\alpha }=\left[{\underline{f}}^{\alpha \Delta }\left(t\right),{\overline{f}}^{\alpha \Delta }\left(t\right)\right],\alpha \in [0,1].$$

Now, we introduce the conception for the first type fuzzy delta integration on time scales. Let $I\subset \mathbb{T}$.

Definition 4.([2]) A function $f:\mathbb{T}\to \mathbb{R}$ is called a sector of the fuzzy function $F:I\to F\left(\mathbb{R}\right)$ if $f\left(t\right)\in F\left(t\right)$ for all $t\in I$. The set of all rd-continuous sectors of F on I is denoted by $S_{HF}\left(I\right)$.

Theorem  10.([2]) Let $t_0,T\in \mathbb{T}$, $t_0<T$, $F,G:[t_0,T]\to F\left(\mathbb{R}\right)$ be ${\delta }_H$-integrable. Then $F+G:[t_0,T]\to F\left(\mathbb{R}\right)$ is ${\delta }_H$-integrable and

$$\underset{t_0}{\overset{T}{\int }}(F\left(s\right)+G\left(s\right)){\delta }_{H}s=\underset{t_0}{\overset{T}{\int }}F\left(s\right){\delta }_{H}s+\underset{t_0}{\overset{T}{\int }}G\left(s\right){\delta }_{H}s.$$

(4)

Theorem  11.([2]) Let $t_0,T\in \mathbb{T}$, $t_0<T$, $F:[t_0,T]\to F\left(\mathbb{R}\right)$ be ${\delta }_H$-integrable. Then $\lambda ·F:[t_0,T]\to F\left(\mathbb{R}\right)$ is ${\delta }_H$-integrable and

$$\underset{t_0}{\overset{T}{\int }}\lambda ·F\left(s\right){\delta }_{H}s=\lambda ·\underset{t_0}{\overset{T}{\int }}F\left(s\right){\delta }_{H}s$$

for any $\lambda \in \mathbb{R}$.

Theorem  12.([2]) Let $t_0,T\in \mathbb{T}$, $t_0<T$, and $F:[t_0,T]\to F\left(\mathbb{R}\right)$ be ${\delta }_H$-integrable. Then

$$\underset{t_0}{\overset{T}{\int }}F\left(s\right){\delta }_{H}s=\underset{t_0}{\overset{t}{\int }}F\left(s\right){\delta }_{H}s+\underset{t}{\overset{T}{\int }}F\left(s\right){\delta }_{H}s$$

for any $t\in [t_0,T]$.

Theorem  13.([2]) Let $t_0,T\in \mathbb{T}$, $t_0<T$, $F:[t_0,T]\to F\left(\mathbb{R}\right)$ is rd-continuous. If $X_0\in F\left(\mathbb{R}\right)$ and

$$\begin{array}{c}\hfill f\left(t\right)=X_0+\underset{t_0}{\overset{t}{\int }}F\left(s\right){\delta }_{H}s,t\in [t_0,T],\end{array}$$

then f is ${\delta }_H$-differentiable and

$${\delta }_{H}f\left(t\right)=F\left(t\right),t\in [t_0,T].$$

Theorem  14.([2]) If $f:[a,b]\to F\left(\mathbb{R}\right)$ is ${\delta }_H$-differentiable on $[a,b]$, then

$$\underset{a}{\overset{b}{\int }}{\delta }_{H}f\left(t\right)=f\left(b\right){\ominus }_{H}f\left(a\right).$$

Theorem  15.([2]) Let $f:[a,b]\to F\left(\mathbb{R}\right)$ be ${\delta }_H$-integrable. Then

$$\underset{a}{\overset{b}{\int }}f\left(t\right){\delta }_{H}t=(-1)·\underset{b}{\overset{a}{\int }}f\left(t\right){\delta }_{H}t.$$

Theorem  16

(Integration by Parts). ([2]) Let $f,g:[a,b]\to F\left(\mathbb{R}\right)$ be ${\delta }_H$-differentiable and $f\circ g$ is also ${\delta }_H$-differentiable on $[a,b]$. If

$$I_{f,g}^{\alpha ,1}\left(t\right)\le 0,I_{f^{\sigma },{\delta }_{H}g}^{\alpha ,1}\left(t\right)\le 0,I_{{\delta }_{H}f,g}^{\alpha ,1}\left(t\right)\le 0,t\in {[a,b]}^{\kappa },$$

(5)

then

$$\begin{array}{ccc}\underset{a}{\overset{b}{\int }}f\left(\sigma \left(t\right)\right)\circ {\delta }_{H}g\left(t\right){\delta }_{H}t\hfill & =\hfill & (f\left(b\right)\circ g\left(b\right)){\ominus }_H(f\left(a\right)\circ g\left(a\right))\hfill \\ \\ \hfill & \hfill & {\ominus }_H\underset{a}{\overset{b}{\int }}{\delta }_{H}f\left(t\right)\circ g\left(t\right){\delta }_{H}t.\hfill \end{array}$$

(6)

3. The Adomian Decomposition Method on Time Scales

In this section, we will describe the Adomian decomposition method on arbitrary time scale $\mathbb{T}$ with forward jump operator and delta differentiation operator $\sigma$ and $\Delta$, respectively. The exposition in this section, follows the exposition in [3].

Denote the set consisting of all possible strings of length n, containing exactly k times $\sigma$ and $n-k$ times $\Delta$ operators by ${\displaystyle S_k^{\left(n\right)}}$. Then we have

$$h_n(t,s)h_m(t,s)=\sum _{l=m}^{m+n}\left(\sum _{{\Lambda }_{l,m}\in S_m^{\left(l\right)}}{h}_n^{{\Lambda }_{l,m}}(s,s)\right)h_l(t,s)$$

for every ${\displaystyle t,s\in \mathbb{T}}$. For $s\in \mathbb{T}$, $l,m,n\in {\mathbb{N}}_0$, set

$$A_{l,m,n,s}=\sum _{{\Lambda }_{l,m}\in S_m^{\left(l\right)}}{h}_n^{{\Lambda }_{l,m}}(s,s)$$

and for any $m,n\in {\mathbb{N}}_0$, we have

$$h_n(t,s)h_m(t,s)=\sum _{l=m}^{m+n}{A}_{l,m,n,s}{h}_l(t,s).$$

(7)

For $n\in {\mathbb{N}}_0$, $t,s\in \mathbb{T}$, define the polynomials

$$H_n^1(t,s)={\left(h_1(t,s)\right)}^n,t,s\in \mathbb{T}.$$

Note that

$$H_n^1(t,s)H_m^1(t,s)=H_{n+m}^1(t,s),t,s\in \mathbb{T}.$$

Note also that

$$H_1^1(t,s)=h_1(t,s),$$

(8)

and

$$\begin{array}{ccc}\hfill H_2^1(t,s)& =& A_{1,1,1,s}{H}_1^1(t,s)+A_{2,1,1,s}{h}_2(t,s),\hfill \end{array}$$

whereupon

$$h_2(t,s)=-{\displaystyle \frac{A_{1,1,1,s}}{A_{2,1,1,s}}}{H}_1^1(t,s)+{\displaystyle \frac{1}{A_{2,1,1,s}}}{H}_2^1(t,s),$$

and so on. Below we denote by $B_i^j$, $i,j\in \mathbb{N}$, the constants for which

$$H_n^1(t,s)=B_1^n{h}_1(t,s)+B_2^n{h}_2(t,s)+\cdots +B_n^n{h}_n(t,s),t,s\in \mathbb{T}.$$

(9)

Suppose that $u:\mathbb{T}\to \mathbb{R}$ is a given function which has a convergent series expansion of the form

$$u=\sum _{j=0}^{\infty }{u}_j.$$

(10)

Suppose also that $f:\mathbb{R}\to \mathbb{R}$ is a given analytic function such that

$$f\left(u\right)=\sum _{n=0}^{\infty }{A}_n(u_0,u_1,\dots ,u_n),$$

(11)

where $A_n$, $n\in {\mathbb{N}}_0$, are given by

$$\begin{array}{ccc}\hfill A_0& =& f\left(u_0\right)\hfill \\ \hfill A_n& =& {\displaystyle \sum _{\nu =1}^{n}c(\nu ,n)f^{\left(\nu \right)}\left(u_0\right),n\in \mathbb{N}.}\hfill \end{array}$$

Here the functions $c(\nu ,n)$ denote the sum of products of $\nu$ components $u_j$ of u given in (10), whose subscripts sum up to n, divided by the factorial of the number of repeated subscripts, i.e.,

$$\begin{array}{ccc}\hfill A_0& =& f\left(u_0\right),\hfill \\ \\ \hfill A_1& =& c(1,1)f^{\prime }\left(u_0\right)\hfill \\ \\ & =& u_1{f}^{\prime }\left(u_0\right),\hfill \\ \\ \hfill A_2& =& c(1,2)f^{\prime }\left(u_0\right)+c(2,2)f^{\prime \prime }\left(u_0\right)\hfill \\ \\ & =& u_2{f}^{\prime }\left(u_0\right)+{\displaystyle \frac{u_1^2}{2!}}{f}^{\prime \prime }\left(u_0\right),\hfill \end{array}$$

$$\begin{array}{ccc}\hfill A_3& =& c(1,3)f^{\prime }\left(u_0\right)+c(2,3)f^{\prime \prime }\left(u_0\right)+c(3,3)f^{\prime \prime \prime }\left(u_0\right)\hfill \\ \\ & =& u_3{f}^{\prime }\left(u_0\right)+u_1{u}_2{f}^{\prime \prime }\left(u_0\right)+{\displaystyle \frac{u_1^3}{3!}}{f}^{\prime \prime \prime }\left(u_0\right),\hfill \\ \\ \hfill A_4& =& c(1,4)f^{\prime }\left(u_0\right)+c(2,4)f^{\prime \prime }\left(u_0\right)+c(3,4)f^{\prime \prime \prime }\left(u_0\right)\hfill \\ \\ & & +c(4,4)f^{\left(4\right)}\left(u_0\right)\hfill \\ \\ & =& u_4{f}^{\prime }\left(u_0\right)+\left(u_1{u}_3+{\displaystyle \frac{u_2^2}{2}}\right)f^{\prime \prime }\left(u_0\right)+{\displaystyle \frac{u_1^2{u}_2}{2}}{f}^{\prime \prime \prime }\left(u_0\right)\hfill \\ \\ & & +{\displaystyle \frac{u_1^4}{4!}}{f}^{\left(4\right)}\left(u_0\right)\hfill \end{array}$$

and so on. Suppose now that u is given by the convergent series

$$u=\sum _{n=0}^{\infty }{c}_n{H}_n^1(x,x_0).$$

(12)

We wish to find the respected transformed series for $f\left(u\right)$. From (10), we have

$$u=\sum _{n=0}^{\infty }{u}_n=\sum _{n=0}^{\infty }{c}_n{H}_n^1(x,x_0),$$

and hence,

$$u_n=c_n{H}_n^1(x,x_0),n\in {\mathbb{N}}_0.$$

Thus,

$$\begin{array}{ccc}\hfill f\left(u\right)& =& \sum _{n=0}^{\infty }{A}_n(u_0,u_1,\dots ,u_n)\hfill \\ \\ & =& f\left(\sum _{n=0}^{\infty }{c}_n{H}_n^1(x,x_0)\right)\hfill \\ \\ & =& \sum _{n=0}^{\infty }{A}^n(c_0,c_1,\dots ,c_n)H_n^1(x,x_0).\hfill \end{array}$$

Hence,

$$A_n(u_0,u_1,\dots ,u_n)=A^n(c_0,c_1,\dots ,c_n)H_n^1(x,x_0).$$

For $n=0$, we have

$$\begin{array}{ccc}\hfill u_0& =& c_0{H}_0^1(x,x_0).\hfill \\ \\ & =& c_0.\hfill \end{array}$$

Thus,

$$\begin{array}{ccc}\hfill A_0\left(u_0\right)& =& A^0\left(c_0\right)H_0^1(x,x_0)\hfill \\ \\ & =& A^0\left(c_0\right).\hfill \end{array}$$

For $n=1$, we find

$$\begin{array}{ccc}\hfill A_1(u_0,u_1)& =& u_1{f}^{\prime }\left(u_0\right)\hfill \\ \\ & =& A^1(c_0,c_1)H_1^1(x,x_0)\hfill \end{array}$$

or

$$c_1{H}_1^1(x,x_0)f^{\prime }\left(u_0\right)=A^1(c_0,c_1)H_1^1(x,x_0),$$

whereupon

$$\begin{array}{ccc}\hfill A^1(c_0,c_1)& =& c_1{f}^{\prime }\left(u_0\right)\hfill \\ \\ & =& c_1{f}^{\prime }\left(c_0\right)\hfill \\ \\ & =& A_1(c_0,c_1).\hfill \end{array}$$

For $n=2$, we have

$$A_2(u_0,u_1,u_2)=A^2(c_0,c_1,c_2)H_2^1(x,x_0)$$

or

$$u_2{f}^{\prime }\left(u_0\right)+{\displaystyle \frac{u_1^2}{2}}{f}^{\prime \prime }\left(u_0\right)=A^2(c_0,c_1,c_2)H_2^1(x,x_0).$$

Then

$$c_2{H}_2^1(x,x_0)f^{\prime }\left(c_0\right)+{\displaystyle \frac{c_1^2{\left(H_1^1(x,x_0)\right)}^2}{2}}{f}^{\prime \prime }\left(c_0\right)=A^2(c_0,c_1,c_2)H_2^1(x,x_0),$$

or

$$\left(c_2{f}^{\prime }\left(c_0\right)+{\displaystyle \frac{c_1^2}{2}}{f}^{\prime \prime }\left(c_0\right)\right)H_2^1(x,x_0)=A^2(c_0,c_1,c_2)H_2^1(x,x_0),$$

whereupon

$$\begin{array}{ccc}\hfill A^2(c_0,c_1,c_2)& =& c_2{f}^{\prime }\left(c_0\right)+{\displaystyle \frac{c_1^2}{2}}{f}^{\prime \prime }\left(c_0\right)\hfill \\ \\ & =& A_2(c_0,c_1,c_2).\hfill \end{array}$$

For $n=3$, we find

$$\begin{array}{ccc}\hfill u_3{f}^{\prime }\left(u_0\right)+u_1{u}_2{f}^{\prime \prime }\left(u_0\right)+{\displaystyle \frac{u_1^3}{3!}}{f}^{\prime \prime \prime }\left(u_0\right)& =& A_3(u_0,u_1,u_2,u_3)\hfill \\ \\ & =& A^3(c_0,c_1,c_2,c_3)H_3^1(x,x_0)\hfill \end{array}$$

or

$$c_3{H}_3^1(x,x_0)f^{\prime }\left(c_0\right)+c_1{c}_2{H}_3^1(x,x_0)f^{\prime \prime }\left(x_0\right)+{\displaystyle \frac{c_1^3}{3!}}{f}^{\prime \prime \prime }\left(c_0\right)H_3^1(x,x_0)=A^3(c_0,c_1,c_2,c_3)H_3^1(x,x_0),$$

whereupon

$$\begin{array}{ccc}\hfill c_3{f}^{\prime }\left(c_0\right)+c_1{c}_2{f}^{\prime \prime }\left(c_0\right)+{\displaystyle \frac{c_1^3}{3!}}{f}^{\prime \prime \prime }\left(c_0\right)& =& A^3(c_0,c_1,c_2,c_3)\hfill \\ \\ & =& A_3(c_0,c_1,c_2,c_3),\hfill \end{array}$$

and so on. We get the following result.

Theorem  17

([3]). Let $u:\mathbb{T}\to \mathbb{R}$ be a function with a convergent expansion given in (12). Let $f:\mathbb{R}\to \mathbb{R}$ be an analytic function having the form (11). Then

$$f\left(u\right)=f\left(\sum _{n=0}^{\infty }{c}_n{H}_n^1(x,x_0)\right)=\sum _{n=0}^{\infty }{A}_n(c_0,c_1,\dots ,c_n)H_n^1(x,x_0).$$

4. The Adomian Decomposition Method for the Problem (1), (2)

In this section, we will introduce the Adomian decomposition method for the problem (1), (2). Firstly, note that the problem (1), (2) can be rewritten in the form

$$\begin{array}{ccc}\hfill \left[{\underline{y}}^{\alpha \Delta },{\overline{y}}^{\alpha \Delta }\right]& =& \left[{\underline{f}}^{\alpha }\left(y\right),{\overline{f}}^{\alpha }\left(y\right)\right],t\in (t_0,T],\hfill \\ \\ \hfill \left[{\underline{y}}^{\alpha }\left(t_0\right),{\overline{y}}^{\alpha }\left(t_0\right)\right]& =& \left[{\underline{y}}_0^{\alpha },{\overline{y}}_0^{\alpha }\right],\alpha \in [0,1].\hfill \end{array}$$

Consider the problem

$${\underline{y}}^{\alpha \Delta }={\underline{f}}^{\alpha }\left(y\right),t>t_0,y\left(t_0\right)=0,$$

(13)

where ${\underline{f}}^{\alpha }:\mathbb{R}\to \mathbb{R}$ is an analytic function. We propose a solution of the IVP (13), in the form

$${\underline{y}}^{\alpha }\left(t\right)=\sum _{j=0}^{\infty }{\underline{c}}_j{H}_j^1(t,t_0),t\ge t_0.$$

Like in the general case, we suppose that

$${\underline{f}}^{\alpha }\left(y\right)=\sum _{j=0}^{\infty }{\underline{A}}_j({\underline{c}}_0,\dots ,{\underline{c}}_j)H_j^1(t,t_0),t\ge t_0.$$

We have

$${\underline{y}}^{\alpha }\left(t\right)={\underline{c}}_0+\sum _{j=1}^{\infty }\sum _{k=1}^j{\underline{c}}_j{\underline{B}}_k^j{h}_k(t,t_0),t\ge t_0.$$

(14)

and

$${\underline{f}}^{\alpha }\left(y\right)={\underline{A}}_0\left({\underline{c}}_0\right)+\sum _{j=1}^{\infty }\sum _{k=1}^j{\underline{A}}_j({\underline{c}}_0,\dots ,{\underline{c}}_j){\underline{B}}_k^j{h}_k(t,t_0),t\ge t_0.$$

(15)

Let

$$\mathcal{L}\left({\underline{y}}^{\alpha }\left(t\right)\right)\left(z\right)=Y\left(z\right).$$

Then we have

$$\mathcal{L}\left({\underline{y}}^{\alpha \Delta }\left(t\right)\right)\left(z\right)=zY\left(z\right)-{\underline{y}}^{\alpha }\left(t_0\right)=zY\left(z\right).$$

Taking the Laplace transform of both sides of the dynamic equation (13) we obtain

$$\begin{array}{ccc}\hfill zY\left(z\right)& =& {\displaystyle \mathcal{L}\left({\displaystyle A_0\left(c_0\right)+\sum _{j=1}^{\infty }\sum _{k=1}^j{\underline{A}}_j({\underline{c}}_0,\dots ,{\underline{c}}_j){\underline{B}}_k^j{h}_k(t,t_0)}\right)\left(z\right)}\hfill \\ \hfill & =& {\displaystyle {\underline{A}}_0\left({\underline{c}}_0\right)\frac{1}{z}+\sum _{j=1}^{\infty }\sum _{k=1}^j{\underline{A}}_j({\underline{c}}_0,\dots ,{\underline{c}}_j){\underline{B}}_k^j\frac{1}{z^{k+1}}.}\hfill \end{array}$$

Then we obtain

$${\displaystyle Y\left(z\right)={\underline{A}}_0\left({\underline{c}}_0\right)\frac{1}{z^2}+\sum _{j=1}^{\infty }\sum _{k=1}^j{\underline{A}}_j({\underline{c}}_0,\dots ,{\underline{c}}_j){\underline{B}}_k^j\frac{1}{z^{k+2}}.}$$

Now, by taking inverse Laplace transform of both sides, we get

$${\displaystyle {\underline{y}}^{\alpha }\left(t\right)={\underline{A}}_0\left({\underline{c}}_0\right)h_1(t,t_0)+\sum _{j=1}^{\infty }\sum _{k=1}^j{\underline{A}}_j({\underline{c}}_0,\dots ,{\underline{c}}_j)B_k^j{h}_{k+1}(t,t_0).}$$

Employing (14), we have

$${\displaystyle {\underline{c}}_0+\sum _{j=1}^{\infty }\sum _{k=1}^j{\underline{c}}_j{\underline{B}}_k^j{h}_k(t,t_0)={\underline{A}}_0\left({\underline{c}}_0\right)h_1(t,t_0)+\sum _{j=1}^{\infty }\sum _{k=1}^j{\underline{A}}_j({\underline{c}}_0,\dots ,{\underline{c}}_j){\underline{B}}_k^j{h}_{k+1}(t,t_0).}$$

In order to equate the coefficients of the time scale monomials $h_k(t,t_0)$ on both sides, we reorder the sums as follows.

$$\begin{array}{c}{\displaystyle {\underline{c}}_0+\left(\sum _{j=1}^{\infty }{\underline{c}}_j{\underline{B}}_1^j\right)h_1(t,t_0)+\sum _{k=2}^{\infty }\left(\sum _{j=k}^{\infty }{\underline{c}}_j{\underline{B}}_k^j\right)h_k(t,t_0)}\hfill \\ {\displaystyle ={\underline{A}}_0\left({\underline{c}}_0\right)h_1(t,t_0)+\sum _{k=2}^{\infty }\sum _{j=k-1}^{\infty }{\underline{A}}_j({\underline{c}}_0,\dots ,{\underline{c}}_j){\underline{B}}_{k-1}^j{h}_k(t,t_0).}\hfill \end{array}$$

This results in the following nonlinear system for determining the constants ${\underline{c}}_j$, $j=0,1,\dots$.

$$\begin{array}{ccc}\hfill {\underline{c}}_0& =& 0,\hfill \\ \hfill {\displaystyle \sum _{j=1}^{\infty }{\underline{c}}_j{\underline{B}}_1^j}& =& {\underline{A}}_0\left({\underline{c}}_0\right)={\underline{f}}^{\alpha }\left(0\right)\hfill \\ \hfill {\displaystyle \sum _{j=k}^{\infty }{\underline{c}}_j{\underline{B}}_k^j}& =& {\displaystyle \sum _{j=k-1}^{\infty }{\underline{A}}_j({\underline{c}}_0,\dots ,{\underline{c}}_j){\underline{B}}_{k-1}^j,k\ge 2.}\hfill \end{array}$$

(16)

Notice that the system is infinite and nonlinear in its unknowns. However, the nonlinearity is of polynomial type. This is a results of the nonlinear structure of the function ${\underline{f}}^{\alpha }$.

Now, consider the problem

$${\overline{y}}^{\alpha \Delta }={\overline{f}}^{\alpha }\left(y\right),t>t_0,y\left(t_0\right)=0,$$

(17)

where ${\overline{f}}^{\alpha }:\mathbb{R}\to \mathbb{R}$ is an analytic function. We will search a solution of the IVP (17), in the form

$${\overline{y}}^{\alpha }\left(t\right)=\sum _{j=0}^{\infty }{\overline{c}}_j{H}_j^1(t,t_0),t\ge t_0.$$

Assume that

$${\overline{f}}^{\alpha }\left(y\right)=\sum _{j=0}^{\infty }{\overline{A}}_j({\overline{c}}_0,\dots ,{\overline{c}}_j)H_j^1(t,t_0),t\ge t_0.$$

We have

$${\overline{y}}^{\alpha }\left(t\right)={\overline{c}}_0+\sum _{j=1}^{\infty }\sum _{k=1}^j{\overline{c}}_j{\overline{B}}_k^j{h}_k(t,t_0),t\ge t_0.$$

(18)

and

$${\overline{f}}^{\alpha }\left(y\right)={\overline{A}}_0\left({\overline{c}}_0\right)+\sum _{j=1}^{\infty }\sum _{k=1}^j{\overline{A}}_j({\overline{c}}_0,\dots ,{\overline{c}}_j){\overline{B}}_k^j{h}_k(t,t_0),t\ge t_0.$$

(19)

As above, we get the following system for the constants ${\overline{c}}_j$, $j=0,1,\dots$.

$$\begin{array}{ccc}\hfill {\overline{c}}_0& =& 0,\hfill \\ \hfill {\displaystyle \sum _{j=1}^{\infty }{\overline{c}}_j{\overline{B}}_1^j}& =& {\overline{A}}_0\left({\overline{c}}_0\right)={\overline{f}}^{\alpha }\left(0\right)\hfill \\ \hfill {\displaystyle \sum _{j=k}^{\infty }{\overline{c}}_j{\overline{B}}_k^j}& =& {\displaystyle \sum _{j=k-1}^{\infty }{\overline{A}}_j({\overline{c}}_0,\dots ,{\overline{c}}_j){\overline{B}}_{k-1}^j,k\ge 2.}\hfill \end{array}$$

(20)

5. A Numerical Example

Consider the initial value problem associated with the first order nonlinear fuzzy dynamic equation of the form

$$\left[{\underline{y}}^{\alpha \Delta }\left(t\right),{\overline{y}}^{\alpha \Delta }\left(t\right)\right]=\left[e^{\alpha y\left(t\right)},e^{2\alpha y\left(t\right)}\right],t\ge 0,y\left(t_0\right)=[0,0],$$

(21)

$\alpha \in [0,1]$. Consider the IVP

$${\underline{y}}^{\alpha \Delta }\left(t\right)=e^{\alpha y\left(t\right)},{\overline{y}}^{\alpha }\left(0\right)=0.$$

Assume that the solution has the series representation

$$y\left(t\right)=\sum _{j=0}^{\infty }{c}_j{H}_j^1(t,0),t\ge 0,$$

where $c_j$, $j\in {\mathbb{N}}_0$ are the coefficients to be determined.

$$f\left(y\right)=e^{\alpha y\left(t\right)}=\sum _{j=0}^{\infty }{A}_j(c_0,\dots ,c_j)H_j^1(t,0),t\ge 0,$$

where

$$\begin{array}{ccc}\hfill A_0& =& f\left(c_0\right)\hfill \\ \hfill & =& e^{\alpha c_0}\hfill \\ \hfill A_1& =& c_1{f}^{\prime }\left(c_0\right)\hfill \\ \hfill & =& \alpha c_1{e}^{\alpha c_0}\hfill \\ \hfill A_2& =& c_2{f}^{\prime }\left(c_0\right)+{\displaystyle \frac{c_1^2}{2!}}{f}^{\prime \prime }\left(c_0\right)\hfill \\ \hfill & =& \left(\alpha c_2+{\displaystyle \frac{{\left(\alpha c_1\right)}^2}{2!}}\right)e^{\alpha c_0}\hfill \\ \hfill A_3& =& c_3{f}^{\prime }\left(c_0\right)+c_1{c}_2{f}^{\prime \prime }\left(c_0\right)+{\displaystyle \frac{c_1^3}{3!}}{f}^{\prime \prime \prime }\left(c_0\right)\hfill \\ \hfill & =& \left(\alpha c_3+{\alpha }^2{c}_1{c}_2+{\displaystyle \frac{{\left(\alpha c_1\right)}^3}{3!}}\right)e^{\alpha c_0}\hfill \\ \hfill A_4& =& c_4{f}^{\prime }\left(c_0\right)+\left(c_1{c}_3+{\displaystyle \frac{c_2^2}{2}}\right)f^{\prime \prime }\left(c_0\right)+{\displaystyle \frac{c_1^2{c}_2}{2}}{f}^{\prime \prime \prime }\left(c_0\right)+{\displaystyle \frac{c_1^4}{4!}}{f}^{\left(4\right)}\left(c_0\right)\hfill \\ \hfill & =& \left(\alpha c_4+{\alpha }^2{c}_1{c}_3+{\displaystyle \frac{{\left(\alpha c_2\right)}^2}{2}}+{\displaystyle \frac{{\alpha }^3{c}_1^2{c}_2}{2}}+{\displaystyle \frac{{\left(\alpha c_1\right)}^4}{4!}}\right)e^{\alpha c_0}\hfill \\ \hfill & \cdots \end{array}$$

(22)

The infinite nonlinear system for this example has the form

$$\begin{array}{ccc}\hfill c_0& =& 0,\hfill \\ \hfill c_1{B}_1^1+c_2{B}_1^2+c_3{B}_1^3+\cdots & =& 1\hfill \\ \hfill c_2{B}_2^2+c_3{B}_2^3+c_4{B}_2^4+\cdots & =& \alpha c_1{B}_1^1+\left(\alpha c_2+{\displaystyle \frac{{\left(\alpha c_1\right)}^2}{2!}}\right)B_1^2+\cdots \hfill \\ \hfill c_3{B}_3^3+c_4{B}_3^4+c_5{B}_3^5+\cdots & =& \left(\alpha c_2+{\displaystyle \frac{{\left(\alpha c_1\right)}^2}{2!}}\right)B_2^2+\cdots \hfill \\ \hfill \cdots & \end{array}$$

(23)

Solving this nonlinear system one can approximately obtain $c_i$, $i\in \mathbb{N}$, and hence, the approximate solution of the initial value problem which is

$${\underline{y}}^{\alpha }\left(t\right)=c_1{H}_1^1(t,0)+c_2{H}_2^1(t,0)+c_3{H}_3^1(t,0)+\cdots$$

(24)

As above,

$${\overline{y}}^{\alpha }\left(t\right)={\overline{c}}_1{H}_1^1(t,0)+{\overline{c}}_2{H}_2^1(t,0)+{\overline{c}}_3{H}_3^1(t,0)+\cdots$$

(25)

where

$$\begin{array}{ccc}\hfill {\overline{c}}_0& =& 0,\hfill \\ \hfill {\overline{c}}_1{B}_1^1+{\overline{c}}_2{B}_1^2+{\overline{c}}_3{B}_1^3+\cdots & =& 1\hfill \\ \hfill {\overline{c}}_2{B}_2^2+{\overline{c}}_3{B}_2^3+{\overline{c}}_4{B}_2^4+\cdots & =& 2\alpha {\overline{c}}_1{B}_1^1+\left(2\alpha {\overline{c}}_2+{\displaystyle \frac{{\left(2\alpha {\overline{c}}_1\right)}^2}{2!}}\right)B_1^2+\cdots \hfill \\ \hfill {\overline{c}}_3{B}_3^3+{\overline{c}}_4{B}_3^4+{\overline{c}}_5{B}_3^5+\cdots & =& \left(2\alpha {\overline{c}}_2+{\displaystyle \frac{{\left(2\alpha {\overline{c}}_1\right)}^2}{2!}}\right)B_2^2+\cdots \hfill \\ \hfill \cdots & \end{array}$$

(26)

Let $\mathbb{T}=2^{{\mathbb{N}}_0}$ and $t_0=1$. Then $\sigma \left(t\right)=2t$, $t\in \mathbb{T}$, and

$$\begin{array}{ccc}\hfill h_1(t,t_0)& =& h_1(t,1)\hfill \\ \\ & =& t-1,t\in \mathbb{T}.\hfill \end{array}$$

Next,

$$h_2(t,t_0)={\displaystyle \frac{t^2}{3}}-t+{\displaystyle \frac{2}{3}},t\in \mathbb{T}.$$

Really,

$$\begin{array}{ccc}\hfill h_2^{\Delta }(t,t_0)& =& {\displaystyle \frac{\sigma \left(t\right)+t}{3}}-1\hfill \\ \\ & =& {\displaystyle \frac{2t+t}{3}}-1\hfill \\ \\ & =& t-1\hfill \\ \\ & =& h_1(t,t_0),t\in \mathbb{T}.\hfill \end{array}$$

Moreover,

$$h_3(t,t_0)={\displaystyle \frac{t^3}{21}}-{\displaystyle \frac{t^2}{3}}+{\displaystyle \frac{2}{3}}t-{\displaystyle \frac{8}{21}},t\in \mathbb{T}.$$

Indeed,

$$\begin{array}{ccc}\hfill h_3^{\Delta }(t,t_0)& =& {\displaystyle \frac{{\left(\sigma \left(t\right)\right)}^2+t\sigma \left(t\right)+t^2}{21}}-{\displaystyle \frac{\sigma \left(t\right)+t}{3}}+{\displaystyle \frac{2}{3}}\hfill \\ \\ & =& {\displaystyle \frac{{\left(2t\right)}^2+t\left(2t\right)t^2}{21}}-{\displaystyle \frac{2t+t}{3}}+{\displaystyle \frac{2}{3}}\hfill \\ \\ & =& {\displaystyle \frac{4t^2+2t^2+t^2}{21}}-{\displaystyle \frac{3t}{3}}+{\displaystyle \frac{2}{3}}\hfill \\ \\ & =& {\displaystyle \frac{7t^2}{21}}-t+{\displaystyle \frac{2}{3}}\hfill \\ \\ & =& {\displaystyle \frac{1}{3}}{t}^2-t+{\displaystyle \frac{2}{3}}\hfill \\ \\ & =& h_2(t,t_0),t\in \mathbb{T}.\hfill \end{array}$$

Note that

$$\begin{array}{ccc}\hfill H_n^1(t,t_0)& =& {(t-t_0)}^n\hfill \\ \\ & =& {(t-1)}^{n}t\in \mathbb{T}.\hfill \end{array}$$

Then

$$\begin{array}{ccc}\hfill H_2^1(t,t_0)& =& {(t-1)}^2\hfill \\ \\ & =& t^2-2t+1,\hfill \\ \\ \hfill H_3^1(t,t_0)& =& {(t-1)}^3\hfill \\ \\ & =& t^3-3t^2+3t-1,t\in \mathbb{T}.\hfill \end{array}$$

For $n=1$, we get

$$H_1^1(t,t_0)=B_1^1{h}_1(t,t_0),t\in \mathbb{T},$$

whereupon

$$t-1=B_1^1(t-1),t\in \mathbb{T}.$$

Therefore $B_1^1=1$. For $n=2$, we find

$$H_2^1(t,t_0)=B_1^2{h}_1(t,t_0)+B_2^2{h}_2(t,t_0),t\in \mathbb{T},$$

or

$${(t-1)}^2=B_1^2(t-1)+B_2^2\left({\displaystyle \frac{t^2}{3}}-t+{\displaystyle \frac{2}{3}}\right),t\in \mathbb{T},$$

or

$$\begin{array}{ccc}\hfill t^2-2t+1& =& B_1^{2}t-B_1^2+{\displaystyle \frac{B_2^2}{3}}{t}^2-B_2^{2}t+{\displaystyle \frac{2}{3}}{B}_2^2\hfill \\ \\ & =& {\displaystyle \frac{B_2^2}{3}}{t}^2+(B_1^2-B_2^2)t+{\displaystyle \frac{2}{3}}{B}_2^2-B_1^2,t\in \mathbb{T},\hfill \end{array}$$

whereupon we get the system

$$\begin{array}{ccc}\hfill {\displaystyle \frac{B_2^2}{3}}& =& 1\hfill \\ \\ \hfill B_1^2-B_2^2& =& -2\hfill \\ \\ \hfill {\displaystyle \frac{2}{3}}{B}_2^2-B_1^2& =& 1,\hfill \end{array}$$

whose solutions are

$$\begin{array}{ccc}\hfill B_1^2& =& 1\hfill \\ \\ \hfill B_2^2& =& 3.\hfill \end{array}$$

Next,

$$H_3^1(t,t_0)=B_1^3{h}_1(t,t_0)+B_2^3{h}_2(t,t_0)+B_3^3{h}_3(t,t_0),t\in \mathbb{T},$$

or

$${(t-1)}^3=B_1^3(t-1)+B_2^3\left({\displaystyle \frac{t^2}{3}}-t+{\displaystyle \frac{2}{3}}\right)+B_3^3\left({\displaystyle \frac{t^3}{21}}-{\displaystyle \frac{t^2}{3}}+{\displaystyle \frac{2}{3}}t-{\displaystyle \frac{8}{21}}\right),$$

$t\in \mathbb{T}$, or

$$\begin{array}{ccc}\hfill t^3-3t^2+3t-1& =& B_1^{3}t-B_1^3+{\displaystyle \frac{B_2^3}{3}}{t}^2-B_2^{3}t+{\displaystyle \frac{2}{3}}{B}_2^3\hfill \\ \\ & & +{\displaystyle \frac{B_3^2}{21}}{t}^3-{\displaystyle \frac{B_3^3}{3}}{t}^2+{\displaystyle \frac{2}{3}}{B}_3^{3}t-{\displaystyle \frac{8}{21}}{B}_3^3\hfill \\ \\ & =& {\displaystyle \frac{B_3^3}{21}}{t}^3+\left({\displaystyle \frac{B_2^3}{3}}-{\displaystyle \frac{B_3^3}{3}}\right)t^2\hfill \\ \\ & & +\left(B_1^3-B_2^3+{\displaystyle \frac{2}{3}}{B}_3^3\right)t\hfill \\ \\ & & +\left(-B_1^3+{\displaystyle \frac{2}{3}}{B}_2^3-{\displaystyle \frac{8}{21}}{B}_3^3\right),t\in \mathbb{T},\hfill \end{array}$$

whereupon we get the system

$$\begin{array}{ccc}\hfill {\displaystyle \frac{B_3^3}{21}}& =& 1\hfill \\ \\ \hfill {\displaystyle \frac{B_2^3}{3}}-{\displaystyle \frac{B_3^3}{3}}& =& 1\hfill \\ \\ \hfill B_1^3-B_2^3+{\displaystyle \frac{2}{3}}{B}_3^3& =& 3\hfill \\ \\ \hfill -B_1^3+{\displaystyle \frac{2}{3}}{B}_2^3-{\displaystyle \frac{8}{21}}{B}_3^3& =& -1,\hfill \end{array}$$

whose solutions are

$$\begin{array}{ccc}\hfill B_1^3& =& 1\hfill \\ \\ \hfill B_2^3& =& 12\hfill \\ \\ \hfill B_3^3& =& 21.\hfill \end{array}$$

Now, we consider the first four equations of (23) with the following approximations

$$\begin{array}{ccc}\hfill c_0& =& 0\hfill \\ \\ \hfill c_1{B}_1^1+c_2{B}_1^2+c_3{B}_1^3& =& 1\hfill \\ \\ \hfill c_2{B}_2^2+c_3{B}_2^3& =& \alpha c_1{B}_1^1+\left(\alpha c_2+{\displaystyle \frac{{\left(\alpha c_1\right)}^2}{2}}\right)B_1^2\hfill \\ \\ \hfill c_3{B}_3^3& =& \left(\alpha c_2+{\displaystyle \frac{{\left(\alpha c_1\right)}^2}{2}}\right)B_2^2\hfill \end{array}$$

or

$$\begin{array}{ccc}\hfill c_0& =& 0\hfill \\ \\ \hfill c_1+c_2+c_3& =& 1\hfill \\ \\ \hfill 3c_2+12c_3& =& \alpha c_1+\left(\alpha c_2+{\displaystyle \frac{{\alpha }^2{c}_1^2}{2}}\right)\hfill \\ \\ \hfill 21c_3& =& 3\left(\alpha c_2+{\displaystyle \frac{{\alpha }^2{c}_1^2}{2}}\right),\hfill \end{array}$$

whereupon we get

$$\begin{array}{ccc}\hfill {\left(c_1\right)}_{1,2}& =& {\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}},\hfill \\ \\ \hfill {\left(c_2\right)}_{1,2}& =& {\displaystyle \frac{1}{2(\alpha +7)}}(-{\alpha }^2{\left({\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}}\right)}^2\hfill \\ \\ & & -14\left({\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}}\right)+14)\hfill \\ \\ \hfill {\left(c_3\right)}_{1,2}& =& {\displaystyle \frac{1}{2(\alpha +7)}}({\alpha }^2{\left({\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}}\right)}^2\hfill \\ \\ & & -2\alpha \left({\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}}\right)+2\alpha ).\hfill \end{array}$$

Replacing $\alpha$ with $2\alpha$, we find

$$\begin{array}{ccc}\hfill {\left({\overline{c}}_1\right)}_{1,2}& =& {\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}},\hfill \\ \\ \hfill {\left({\overline{c}}_2\right)}_{1,2}& =& {\displaystyle \frac{1}{2(2\alpha +7)}}(-4{\alpha }^2{\left({\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}}\right)}^2\hfill \\ \\ & & -14\left({\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}}\right)+14)\hfill \\ \\ \hfill {\left({\overline{c}}_3\right)}_{1,2}& =& {\displaystyle \frac{1}{2(2\alpha +7)}}(4{\alpha }^2{\left({\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}}\right)}^2\hfill \\ \\ & & -4\alpha \left({\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}}\right)+2\alpha )\hfill \end{array}$$

Therefore approximative solutions are

$$\begin{array}{ccc}{\underline{y}}^{\alpha }\left(t\right)\hfill & =\hfill & c_0+c_1{H}_1^1(t,1)+c_2{H}_2^1(t,1)+c_3{H}_3^1(t,1)\hfill \\ \\ \hfill & =\hfill & {\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}}(t-1),\hfill \\ \\ \hfill & \hfill & +{\displaystyle \frac{1}{2(\alpha +7)}}(-{\alpha }^2{\left({\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}}\right)}^2\hfill \\ \\ \hfill & \hfill & -14\left({\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}}\right)+14){(t-1)}^2\hfill \\ \\ \hfill & \hfill & +{\displaystyle \frac{1}{2(\alpha +7)}}({\alpha }^2{\left({\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}}\right)}^2\hfill \\ \\ \hfill & \hfill & -2\alpha \left({\displaystyle \frac{{\alpha }^2+12\alpha +21\pm \sqrt{{({\alpha }^2+12\alpha +21)}^2-4{\alpha }^2(5\alpha +21)}}{2{\alpha }^2}}\right)+2\alpha ){(t-1)}^3\hfill \end{array}$$

and

$$\begin{array}{ccc}{\overline{y}}^{\alpha }\left(t\right)\hfill & =\hfill & {\overline{c}}_0+{\overline{c}}_1{H}_1^1(t,1)+{\overline{c}}_2{H}_2^1(t,1)+{\overline{c}}_3{H}_3^1(t,1)\hfill \\ \\ \hfill & =\hfill & {\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}}(t-1)\hfill \\ \\ \hfill & \hfill & +{\displaystyle \frac{1}{2(2\alpha +7)}}(-4{\alpha }^2{\left({\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}}\right)}^2\hfill \\ \\ \hfill & \hfill & -14\left({\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}}\right)+14){(t-1)}^2\hfill \\ \\ \hfill & \hfill & +{\displaystyle \frac{1}{2(2\alpha +7)}}(4{\alpha }^2{\left({\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}}\right)}^2\hfill \\ \\ \hfill & \hfill & -4\alpha \left({\displaystyle \frac{4{\alpha }^2+24\alpha +21\pm \sqrt{{(4{\alpha }^2+24\alpha +21)}^2-16{\alpha }^2(10\alpha +21)}}{8{\alpha }^2}}\right)+2\alpha ){(t-1)}^3.\hfill \end{array}$$

In figures below are shown the solutions for $\alpha ={\displaystyle \frac{1}{8}}$, $\alpha ={\displaystyle \frac{1}{4}}$ and $\alpha ={\displaystyle \frac{1}{2}}$, respectively, at $t=1,2,4,8,16,32,64$.