**1. Introduction**

In this paper, we will observe and study the unique existence of approximate solution to the following initial value problem of variable order

$$\begin{cases} D\_{0+}^{p(t)} \mathbf{x}(t) = f(t, \mathbf{x}, D\_{0+}^{q(t)} \mathbf{x}), 0 < t < +\infty, \\ \mathbf{x}(0) = 0, \end{cases} \tag{1}$$

where 0 < *<sup>q</sup>*(*t*) < *<sup>p</sup>*(*t*) < 1, *<sup>f</sup>*(*t*, *<sup>x</sup>*, *<sup>D</sup>q*(*t*) <sup>0</sup><sup>+</sup> *<sup>x</sup>*) are given real functions, and *<sup>D</sup>p*(*t*) <sup>0</sup><sup>+</sup> , *<sup>D</sup>q*(*t*) <sup>0</sup><sup>+</sup> denote derivatives of variable order *p*(*t*) and *q*(*t*) defined by

$$D\_{0+}^{p(t)}\mathbf{x}(t) = \frac{d}{dt} \int\_{0}^{t} \frac{(t-s)^{-p(t)}}{\Gamma(1-p(t))} \mathbf{x}(s)ds, t > 0. \tag{2}$$

$$d\_{0+}(t) = \dots \qquad d\_{t-1}(t-s)^{-q(t)} \tag{3}$$

$$D\_{0+}^{q(t)}x(t) = \frac{d}{dt} \int\_0^t \frac{(t-s)^{-q(t)}}{\Gamma(1-q(t))} x(s)ds, \ t > 0,$$

and <sup>1</sup> Γ(1−*p*(*t*)) *t* <sup>0</sup> (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)−*p*(*t*)*x*(*s*)*ds* is integral of variable order 1 <sup>−</sup> *<sup>p</sup>*(*t*) for function *<sup>x</sup>*(*t*), for details, please refer to [1].

The operators of variable order, which fall into a more complex category, are the derivatives and integrals whose orders are the functions of certain variables. There are several definitions of variable order fractional integrals and derivatives. The following are several definitions of variable order fractional integrals and derivatives, which can be found in [2]. Let −∞ < *a* < *b* < ∞.

**Definition 1.** *Let p* : [*a*, *b*] → (0, +∞)*, the left Riemann–Liouville fractional integral of order α*(*t*) *for function x*(*t*) *are defined as the following two types*

$$\int\_{a+} I\_{a+}^{a(t)} x(t) = \int\_{a}^{t} \frac{(t-s)^{a(t)-1}}{\Gamma(a(t))} x(s)ds, \ t > a,\tag{3}$$

$$\mathbf{1}\_{a+}^{a(t)}\mathbf{x}(t) = \int\_{a}^{t} \frac{(t-s)^{a(s)-1}}{\Gamma(a(s))} \mathbf{x}(s)ds, \ t > a. \tag{4}$$

**Definition 2.** *Let α* : [*a*, *b*] → (*n* − 1, *n*] *(n is a natural number), the left Riemann–Liouville fractional derivative of order α*(*t*) *for function x*(*t*) *are defined as the following two types*

$$D\_{a+}^{a(t)}\mathbf{x}(t) = \left(\frac{d}{dt}\right)^n \int\_a^t \frac{(t-s)^{n-a(t)-1}}{\Gamma(n-a(t))} \mathbf{x}(s)ds, \ t > a,\tag{5}$$

$$D\_{a+}^{a(t)}x(t) = \left(\frac{d}{dt}\right)^n \int\_a^t \frac{(t-s)^{n-a(s)-1}}{\Gamma(n-a(s))} x(s)ds, \ t > a. \tag{6}$$

**Definition 3.** *Let α* : [*a*, *b*] → (*n* − 1, *n*]*(n is a natural number), the left Caputo fractional derivative of order α*(*t*) *for function x*(*t*) *are defined as the following two types*

$${}^{C}D\_{a+}^{a(t)}\mathbf{x}(t) = \int\_{a}^{t} \frac{(t-s)^{n-a(t)-1}}{\Gamma(n-a(t))} \mathbf{x}^{(n)}(s)ds, \ t > a,\tag{7}$$

$$\, ^C \mathbf{D}\_{a+}^{a(t)} \ge (t) = \int\_{a}^{t} \frac{(t-s)^{n-a(s)-1}}{\Gamma(n-a(s))} x^{(n)}(s) ds, \ t > a. \tag{8}$$

The problems denoted by the operator of variable order are apparently more complicated than the ones denoted by the operator of constant order. Recently, some authors have considered the applications of derivatives of variable order in various sciences such as anomalous diffusion modeling, mechanical applications, multi-fractional Gaussian noises. Among these, there have been many works dealing with numerical methods for some class of variable order fractional differential equations, for instance, [1–20].

We notice that, if the order *p*(*t*) is a constant function *q*, then the Riemann–Liouville variable order fractional derivatives and integrals are the Riemann–Liouville fractional derivative and integral, respectively [21]. We know there are some important properties as following. Let −∞ < *b* < ∞.

**Lemma 1.** *[21] The Riemann–Liouville fractional integral defined for function x*(*t*) ∈ *L*(0, *b*) *exists almost everywhere.*

**Lemma 2.** *[21] The equality I γ* 0+*I<sup>δ</sup>* <sup>0</sup>+*x*(*t*) = *<sup>I</sup><sup>δ</sup>* <sup>0</sup>+*I γ* <sup>0</sup>+*x*(*t*) = *I γ*+*δ* <sup>0</sup><sup>+</sup> *x*(*t*)*,* 0 < *γ* < 1, 0 < *δ* < 1 *holds for x* ∈ *L*(0, *b*)*.*

Lemma 2 is semigroup property for the Riemann–Liouville fractional integral, which is very crucial in obtaining the following Lemmas 3–5. In other words, without Lemma 2, one could not have Lemmas 3–5, for details, please refer to [21].

**Lemma 3.** *[21] The equality D<sup>γ</sup>* <sup>0</sup>+*I γ* <sup>0</sup>+*x*(*t*) = *x*(*t*)*,* 0 < *γ* < 1 *holds for x* ∈ *L*(0, *b*)*.* **Lemma 4.** *[21] Let* 0 < *α* < 1*, then the differential equation*

$$D\_{0+}^{\\\alpha} \ge = 0, t > 0$$

*has solution*

$$\mathfrak{x}(t) = ct^{\alpha - 1}, c \in \mathbb{R}.$$

**Lemma 5.** *[21] Let* <sup>0</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>&</sup>lt; <sup>1</sup>*, x* <sup>∈</sup> *<sup>L</sup>*(0, *<sup>b</sup>*)*, D<sup>α</sup>* <sup>0</sup>+*x* ∈ *L*(0, *b*)*. Then the following equality holds*

$$I\_{0+}^{\\\alpha}D\_{0+}^{\\\alpha}\mathfrak{x}(t) = \mathfrak{x}(t) + ct^{\alpha-1}, \\ \mathfrak{c} \in \mathbb{R}.$$

These properties play a very important role in considering the existence of the solutions of differential equations for the Riemann–Liouville fractional derivative, for details, please refer to [22–26]. However, from [15–18], for general functions *h*(*t*), *g*(*t*), we notice that the semigroup property does not hold, i.e., *I h*(*t*) *<sup>a</sup>*<sup>+</sup> *I g*(*t*) *<sup>a</sup>*<sup>+</sup> = *I h*(*t*)+*g*(*t*) *<sup>a</sup>*<sup>+</sup> . Thus, it brings us extreme difficulties, that we cannot get these properties like Lemmas 3–5 for the variable order fractional operators (integral and derivative). Without these properties for variable order fractional derivative and integral, we can hardly consider the existence of solutions of differential equations for variable order derivative by means of nonlinear functional analysis (for instance, some fixed point theorems).

In [18], by means of Banach contraction principle, we considered the uniqueness result of solutions to initial value problems of differential equations of variable order

$$\begin{cases} D\_{0+}^{q(t)} \mathbf{x}(t) = f(t, \mathbf{x}), 0 < t \le T, \\ \mathbf{x}(0) = 0, \end{cases} \tag{9}$$

where 0 < *<sup>T</sup>* < +∞, *<sup>D</sup>q*(*t*) <sup>0</sup><sup>+</sup> denotes derivative of variable order defined by (2), and *q* : [0, *T*] → (0, 1] is a piecewise constant function with partition *P* = {[0, *T*1],(*T*1, *T*2],(*T*2, *T*3], ··· ,(*TN*∗−1, *T*]} (*N*<sup>∗</sup> is a given natural number) of the finite interval [0, *T*], i.e.,

$$q(t) = \sum\_{k=1}^{N^\*} q\_k I\_k(t), t \in [0, T]\_\prime$$

where 0 < *qk* ≤ 1, *<sup>k</sup>* = 1, 2, ··· , *<sup>N</sup>*<sup>∗</sup> are constants, and *Ik* is the indicator of the interval [*Tk*−1, *Tk*], *<sup>k</sup>* = 1, 2, ··· , *<sup>N</sup>*∗(here *<sup>T</sup>*<sup>0</sup> = 0, *TN*<sup>∗</sup> = *<sup>T</sup>*), that is *Ik* = 1 for *<sup>t</sup>* ∈ [*Tk*−1, *Tk*], *Ik* = 0 for elsewhere.

In this paper, we will consider the existence of solutions to the problem (1) for variable orders *p*(*t*), *q*(*t*) are not piecewise constants. Based on some analysis, we will introduce the concept of approximate solution to the problem (1). Then, according to our discussion and analysis, we explore the unique existence of the approximate solution of the problem (1).

This paper is organized as follows. In Section 2, we provide some facts to the variable order integral and derivative through several examples. Also, we state some results which will play a very important role in obtaining our main results. In Section 3, we set forth our main result. Finally, two examples are given.

#### **2. Some Preliminaries on Approximate Solution**

In this section, we give some preliminaries on approximate solutions to the initial value problem (1). First of all, we use an example to illustrate the claim: for general function *p*(*t*), *q*(*t*), the Riemann–Liouville variable order fractional integral does not have the semigroup property.

**Example 1.** *Let p*(*t*) = *<sup>t</sup>* <sup>6</sup> <sup>+</sup> <sup>1</sup> <sup>3</sup> *, <sup>q</sup>*(*t*) = *<sup>t</sup>* <sup>4</sup> <sup>+</sup> <sup>1</sup> <sup>4</sup> *, f*(*t*) = 1, 0 ≤ *t* ≤ 3*. Now, we calculate I p*(*t*) <sup>0</sup><sup>+</sup> *I q*(*t*) <sup>0</sup><sup>+</sup> *f*(*t*)|*t*=<sup>1</sup> *and I p*(*t*)+*q*(*t*) <sup>0</sup><sup>+</sup> *f*(*t*)|*t*=<sup>1</sup> *which are defined in (3).*

*For* 1 ≤ *t* ≤ 3*, we have*

$$\begin{split} I\_{0+}^{p(t)}I\_{0+}^{q(t)}f(t) &= \qquad \int\_{0}^{t} \frac{(t-s)^{\frac{1}{6}+\frac{1}{3}-1}}{\Gamma(\frac{t}{6}+\frac{1}{3})} \int\_{0}^{s} \frac{(s-\tau)^{\frac{4}{4}+\frac{1}{4}-1}}{\Gamma(\frac{t}{4}+\frac{1}{4})} d\tau ds \\ &= \qquad \int\_{0}^{t} \frac{(t-s)^{\frac{1}{6}-\frac{2}{3}}s^{\frac{1}{4}+\frac{s}{4}}}{\Gamma(\frac{t}{6}+\frac{1}{3})\Gamma(\frac{5}{4}+\frac{s}{4})} ds \\ &= \qquad \int\_{0}^{1} \frac{(t-s)^{\frac{1}{6}-\frac{2}{3}}s^{\frac{1}{4}+\frac{s}{4}}}{\Gamma(\frac{t}{6}+\frac{1}{3})\Gamma(\frac{5}{4}+\frac{s}{4})} ds + \int\_{1}^{t} \frac{(t-s)^{\frac{1}{6}-\frac{2}{3}}s^{\frac{1}{4}+\frac{s}{4}}}{\Gamma(\frac{t}{6}+\frac{1}{3})\Gamma(\frac{5}{4}+\frac{s}{4})} ds. \end{split}$$

*We set M*<sup>1</sup> <sup>=</sup> max1≤*t*≤<sup>3</sup> <sup>|</sup> <sup>1</sup> <sup>Γ</sup>(*p*(*t*))<sup>|</sup> *and M*<sup>2</sup> <sup>=</sup> max1≤*s*≤<sup>3</sup> <sup>|</sup> <sup>1</sup> Γ( <sup>5</sup> <sup>4</sup> <sup>+</sup> *<sup>s</sup>* 4 ) |*. For* 1 ≤ *t* ≤ 3*, it holds*

$$\begin{split} | \int\_{1}^{t} \frac{(t-s)^{\frac{t}{2}-\frac{s}{3}} 3^{\frac{t}{4}+\frac{s}{4}}}{\Gamma(\frac{t}{6}+\frac{1}{3})\Gamma(\frac{5}{4}+\frac{s}{4})} ds | &=& \int\_{1}^{t} 3^{\frac{t}{6}-\frac{s}{3}} (\frac{t-s}{3})^{\frac{t}{2}-\frac{s}{3}} \frac{s^{\frac{4}{3}+\frac{s}{4}}}{\Gamma(\frac{1}{6}+\frac{1}{3})\Gamma(\frac{5}{4}+\frac{s}{4})} ds | \\ &\leq& M\_{1}M\_{2} \int\_{1}^{t} 3^{\frac{1}{2}-\frac{s}{3}} (\frac{t-s}{3})^{\frac{1}{2}-\frac{s}{3}} ds \\ &\leq& M\_{1}M\_{2} \int\_{1}^{t} 3^{\frac{1}{2}} (t-s)^{-\frac{1}{2}} 3 ds \\ &=& 2 \times 3^{\frac{4}{3}} M\_{1}M\_{2} (t-1)^{\frac{1}{2}}, \end{split}$$

*hence, we have*

$$\left[\int\_{1}^{t} \frac{(t-s)^{\frac{1}{5}-\frac{2}{3}}s^{\frac{1}{4}+\frac{s}{4}}}{\Gamma(\frac{t}{6}+\frac{1}{3})\Gamma(\frac{5}{4}+\frac{s}{4})}ds\right]\_{t=1} = 0.$$

*So, we get*

$$I\_{0+}^{p(t)}I\_{0+}^{q(t)}f(t)|\_{t=1} = \int\_0^1 \frac{(1-s)^{-\frac{1}{2}}s^{\frac{1}{4}+\frac{s}{4}}}{\Gamma(\frac{1}{2})\Gamma(\frac{5}{4}+\frac{s}{4})}ds \approx 1.063$$

*and*

$$\|I\_{0+}^{p(t)+q(t)}f(t)\|\_{t=1} = \int\_0^1 \frac{(1-s)^{p(1)+q(1)-1}}{\Gamma(p(1)+q(1))}ds = \int\_0^1 ds = 1.$$

*Therefore,*

$$I\_{0+}^{p(t)}I\_{0+}^{q(t)}f(t)|\_{t=1} \neq I\_{0+}^{p(t)+q(t)}f(t)|\_{t=1}.$$

Without the semigroup property of the Riemann–Liouville variable order fractional integral, we can assure that the variable order fractional integration operator of non-constant continuous functions *p*(*t*) for *x*(*t*) does not have the properties like Lemmas 3–5. Consequently, we cannot transform differential equations of variable order into an integral equation.

Let *L*[*x*(*t*);*s*], *L*[*I p*(*t*) <sup>0</sup><sup>+</sup> *<sup>x</sup>*(*t*);*s*], *<sup>L</sup>*[*Dp*(*t*) <sup>0</sup><sup>+</sup> *x*(*t*);*s*] denote the Laplace transforms of functions *x*(*t*), *I p*(*t*) <sup>0</sup><sup>+</sup> *<sup>x</sup>*(*t*) and *<sup>D</sup>p*(*t*) <sup>0</sup><sup>+</sup> *x*(*t*). We have not found out the explicit connection between *L*[*x*(*t*);*s*] and *L*[*I p*(*t*) <sup>0</sup><sup>+</sup> *x*(*t*);*s*], as a result, we have not found out the explicit connection between *L*[*x*(*t*);*s*] and *<sup>L</sup>*[*Dp*(*t*) <sup>0</sup><sup>+</sup> *x*(*t*);*s*].

**Example 2.** *Let p*(*t*) = <sup>√</sup> <sup>1</sup> *<sup>t</sup>*+<sup>1</sup> , *t* ≥ 0*. We consider the Laplace transforms of functions t*(*t* ≥ 0*) and I p*(*t*) <sup>0</sup><sup>+</sup> *t*(*t* ≥ 0) *defined in (3). We can know that*

$$L[t;s] = \int\_0^\infty e^{-st}t dt = \frac{1}{s^2},\tag{10}$$

$$L[I\_{0+}^{(t+1)^{-\frac{1}{2}}}t;s] = \qquad \int\_{0}^{\infty} e^{-st} \int\_{0}^{t} \frac{(t-\tau)^{(t+1)^{-\frac{1}{2}}-1}}{\Gamma((t+1)^{-\frac{1}{2}})} \tau d\tau dt$$

$$= \qquad \int\_{0}^{\infty} e^{-st} \int\_{\tau}^{\infty} \frac{(t-\tau)^{(t+1)^{-\frac{1}{2}}-1}}{\Gamma((t+1)^{-\frac{1}{2}})} \tau dt d\tau$$

$$= \qquad \int\_{0}^{\infty} e^{-s(\tau+\tau)} \int\_{0}^{\infty} \frac{r^{(\tau+\tau+1)^{-\frac{1}{2}}-1}}{\Gamma((\tau+\tau+1)^{-\frac{1}{2}})} \tau d\tau d\tau$$

$$= \qquad \int\_{0}^{\infty} e^{-s\tau} \tau \int\_{0}^{\infty} e^{-sr} \frac{r^{(\tau+\tau+1)^{-\frac{1}{2}}-1}}{\Gamma((\tau+\tau+1)^{-\frac{1}{2}})} dr d\tau. \tag{11}$$

By (10) or (11), we do not get the explicit connection between *L*[*t*;*s*] and *L*[*I* (*t*+1) − 1 2 <sup>0</sup><sup>+</sup> *t*;*s*].

In view of this example, the definition of variable order fractional derivative and the connection between the Laplace transforms of function *x*(*t*) and its derivative *x* (*t*), we cannot obtain the Laplace transform formula for variable order fractional derivatives (2). Based on these facts, we cannot get the explicit expression of the solutions for the problem (1).

Throughout this paper, we assume that

(*A*1) Let *p* : [0, +∞) → (0, 1) and *q* : [0, +∞) → (0, 1) be continuous functions, *q*(*t*) < *p*(*t*) for all *t* ∈ [0, +∞), and that *p*(*t*), *q*(*t*) satisfy

$$\lim\_{t \to +\infty} p(t) = \rho\_{1\prime} \lim\_{t \to +\infty} q(t) = \rho\_2, 0 < \rho\_{1\prime}\rho\_2 < 1. \tag{12}$$

The following result is necessary in our next analysis of main result.

**Lemma 6.** *Let condition* (*A*1) *hold. Then there exist positive constant T, natural number n*<sup>∗</sup> *and intervals* [0, *T*1],(*T*1, *T*2], ··· ,(*Tn*∗−1, *T*] (*T*, +∞)*(n*<sup>∗</sup> ∈ *N) and functions α* : [0, +∞) → (0, 1) *and β* : [0, +∞) → (0, 1) *defined by*

$$u(t) = \sum\_{k=1}^{n^\*} p\_k I\_k(t) + \rho\_1 I\_T(t), \ t \in [0, +\infty), \tag{13}$$

$$\beta(t) = \sum\_{k=1}^{n^\*} q\_k I\_k(t) + \rho\_2 I\_T(t), \ t \in [0, +\infty), \tag{14}$$

*where pk*, *qk* ∈ (0, 1)*,* <sup>I</sup>*k*(*t*) *is the indicator of the interval* [*Tk*−1, *Tk*] *(k* = 1, 2, ··· , *<sup>n</sup>*∗, *here <sup>T</sup>*<sup>0</sup> = 0, *Tn*<sup>∗</sup> = *T), i.e.,* <sup>I</sup>*k*(*t*) = 1 *for <sup>t</sup>* ∈ [*Tk*−1, *Tk*]*,* <sup>I</sup>*k*(*t*) = 0 *for <sup>t</sup> lying in elsewhere;* <sup>I</sup>*T*(*t*) *is the indicator of interval* (*T*, +∞)*, i.e.,* I*T*(*t*) = 1 *for t* ∈ (*T*, +∞)*,* I*T*(*t*) = 0 *for t lying in elsewhere, such that for arbitrary small ε* > 0*,*

$$|p(t) - \mathfrak{a}(t)| < \varepsilon,\\ |q(t) - \beta(t)| < \varepsilon, \ 0 \le t < +\infty. \tag{15}$$

**Proof.** By (12), for ∀ *ε* > 0, there exist *T*1, *T*<sup>2</sup> > 0, such that

$$|p(t) - \rho\_1| < \varepsilon, t > \overline{T}\_1; |p(t) - \rho\_2| < \varepsilon, t > \overline{T}\_2.$$

Let *T* = max{*T*1, *T*2}, then, for ∀ *ε* > 0, we have that

$$|p(t) - \rho\_1| < \varepsilon,\\ |p(t) - \rho\_2| < \varepsilon, t > T. \tag{16}$$

We know that *p* : [0, *T*] → (0, 1), *q* : [0, *T*] → (0, 1) are continuous functions. Since *p*(*t*) is right continuous at point 0, then, for arbitrary small *ε* > 0, there is *δ*<sup>01</sup> > 0 such that

$$|p(t) - p(0)| < \varepsilon, \text{ for } 0 \le t \le \delta\_{01}.$$

Since *q*(*t*) is right continuous at point 0, then, for arbitrary small *ε* > 0, there is *δ*<sup>02</sup> > 0 such that

$$|q(t) - q(0)| < \varepsilon, \text{ for } 0 \le t \le \delta\_{02}.$$

Then for arbitrary small *ε* > 0, takeing *δ*<sup>0</sup> = min{*δ*01, *δ*02}, it holds

$$|p(t) - p(0)| < \varepsilon, \ |q(t) - q(0)| < \varepsilon, \text{ for } 0 \le t \le \delta\_0. \tag{17}$$

We take point *<sup>δ</sup>*<sup>0</sup> . = *T*<sup>1</sup> (if *T*<sup>1</sup> < *T*, we consider continuities of *p*(*t*), *q*(*t*) at point *T*1, otherwise, we end this procedure). Since *p*(*t*) is right continuous at point *T*1, so, for arbitrary small *ε* > 0, there is *δ*<sup>11</sup> > 0 such that

$$|p(t) - p(T\_1)| < \varepsilon, \text{ for } T\_1 \le t \le T\_1 + \delta\_{11}\epsilon$$

Since *q*(*t*) is right continuous at point *T*1, then, for arbitrary small *ε* > 0, there is *δ*<sup>12</sup> > 0 such that

$$|q(t) - q(T\_1)| < \varepsilon, \text{ for } T\_1 \le t \le T\_1 + \delta\_{12}.$$

Hence, for arbitrary small *ε* > 0, taking *δ*<sup>1</sup> = min{*δ*11, *δ*12}, it holds

$$|p(t) - p(T\_1)| < \varepsilon, \ |q(t) - q(T\_1)| < \varepsilon, \text{ for } \ T\_1 \le t \le T\_1 + \delta\_1. \tag{18}$$

We take point *T*<sup>1</sup> + *δ*<sup>1</sup> . = *T*<sup>2</sup> (if *T*<sup>2</sup> < *T*, we consider continuities of *p*(*t*), *q*(*t*) at point *T*2, otherwise, we end this procedure). Since *p*(*t*) is right continuous at point *T*2, so, for arbitrary small *ε* > 0, there is *δ*<sup>21</sup> > 0 such that

$$|p(t) - p(T\_2)| < \varepsilon, \text{ for } T\_2 \le t \le T\_2 + \delta\_{21}.$$

Since *q*(*t*) is right continuous at point *T*2, so, for arbitrary small *ε* > 0, there is *δ*<sup>22</sup> > 0 such that

$$|q(t) - q(T\_2)| < \varepsilon, \text{ for } T\_2 \le t \le T\_2 + \delta\_{22}.$$

Thus, for arbitrary small *ε* > 0, taking *δ*<sup>2</sup> = min{*δ*21, *δ*22}, it holds

$$|p(t) - p(T\_2)| < \varepsilon, \ |q(t) - q(T\_2)| < \varepsilon, \text{ for } T\_2 \le t \le T\_2 + \delta\_2. \tag{19}$$

We take point *<sup>T</sup>*<sup>2</sup> <sup>+</sup> *<sup>δ</sup>*<sup>2</sup> . = *T*<sup>3</sup> (if *T*<sup>3</sup> < *T*, we consider continuities of *p*(*t*), *q*(*t*) at point *T*3, otherwise, we end this procedure). Since *p*(*t*) is right continuous at point *T*3, so, for arbitrary small *ε* > 0, there is *δ*<sup>31</sup> > 0 such that

$$|p(t) - p(T\_3)| < \varepsilon, \text{ for } T\_3 \le t \le T\_3 + \delta\_{31}\epsilon$$

Since *q*(*t*) is right continuous at point *T*3, so, for arbitrary small *ε* > 0, there is *δ*<sup>32</sup> > 0 such that

$$|q(t) - q(T\_3)| < \varepsilon, \text{ for } T\_3 \le t \le T\_3 + \delta\_{32}.$$

Therefore, for arbitrary small *ε* > 0, taking *δ*<sup>3</sup> = min{*δ*31, *δ*32}, it holds

$$|p(t) - p(T\_3)| < \varepsilon, \ |q(t) - q(T\_3)| < \varepsilon, \text{ for } \ T\_3 \le t \le T\_3 + \delta\_3. \tag{20}$$

Since [0, *T*] is a finite interval, then, continuing this analysis procedure, we could obtain that there exist *<sup>δ</sup>n*∗−<sup>2</sup> <sup>&</sup>gt; 0, *<sup>δ</sup>n*∗−<sup>1</sup> <sup>&</sup>gt; 0 (*n*<sup>∗</sup> <sup>∈</sup> *<sup>N</sup>*) such that *Tn*∗−<sup>2</sup> <sup>+</sup> *<sup>δ</sup>n*∗−<sup>2</sup> . = *Tn*∗−<sup>1</sup> < *T*, *Tn*∗−<sup>1</sup> + *δn*∗−<sup>1</sup> ≥ *T*, such that for arbitrary small *ε* > 0, it holds

$$|p(t) - p(T\_{n^\*-1})| < \varepsilon, |q(t) - q(T\_{n^\*-1})| < \varepsilon \text{ for } T\_{n^\*-1} \le t \le T,\tag{21}$$

From (16)–(21), we could let

$$p(0) \doteq p\_1,\\ p(T\_1) \doteq p\_2,\\ p(T\_2) \doteq p\_3,\\ p(T\_3) \doteq p\_4,\\ \cdots,\\ p(T\_{n^\*-1}) \doteq p\_{n^\*}.$$

$$q(0) \doteq q\_1,\\ q(T\_1) \doteq q\_2,\\ q(T\_2) \doteq q\_3,\\ q(T\_3) \doteq q\_4,\\ \cdots,\\ q(T\_{n^\*-1}) \doteq q\_{n^\*}.$$

<sup>=</sup> *<sup>q</sup>*4, ··· , *<sup>q</sup>*(*Tn*∗−1) .

Thus, we define functions *α*, *β* : [0, +∞) → (0, 1) as following

$$a(t) = \begin{cases} p\_{1\prime} & t \in [0, T\_1], \\ \\ p\_{2\prime} & t \in (T\_1, T\_2], \\ \vdots \\ p\_{n^\*\prime} & t \in (T\_{n^\*-1}, T], \\ \\ \rho\_{1\prime} & t \in (T\_\prime, +\infty), \end{cases} \quad \beta(t) = \begin{cases} q\_{1\prime} & t \in [0, T\_1], \\ q\_{2\prime} & t \in (T\_1, T\_2], \\ \vdots \\ q\_{n^\*\prime} & t \in (T\_{n^\*-1}, T], \\ \\ \rho\_{2\prime} & t \in (T, +\infty). \end{cases}$$

Hence, from the previous arguments, for arbitrary small *ε* > 0, we have

$$\begin{cases} |p(t) - p\_1| < \varepsilon, |q(t) - q\_1| < \varepsilon, \text{ for } t \in [0, T\_1], \\\\ |p(t) - p\_2| < \varepsilon, |q(t) - q\_2| < \varepsilon, \text{ for } t \in (T\_1, T\_2], \\\\ \vdots \\ |p(t) - p\_{n^\*}| < \varepsilon, |q(t) - q\_{n^\*}| < \varepsilon, \text{ for } t \in (T\_{n^\*-1}, T], \\\\ |p(t) - p\_1| < \varepsilon, |q(t) - p\_2| < \varepsilon, \text{ for } t \in (T\_\prime, +\infty). \end{cases} \tag{22}$$

Thus, we complete this proof.

The following example illustrates that the semigroup property of the variable order fractional integral does not holds for the piecewise constant functions *p*(*t*) and *q*(*t*) defined in the same partition of finite interval [*a*, *b*].

**Example 3.** *Let <sup>p</sup>*(*t*) =  4, 0 ≤ *t* ≤ 1, 3, 1 <sup>&</sup>lt; *<sup>t</sup>* <sup>≤</sup> 4, *<sup>q</sup>*(*t*) =  3, 0 ≤ *t* ≤ 1, 2, 1 <sup>&</sup>lt; *<sup>t</sup>* <sup>≤</sup> 4, *and <sup>f</sup>*(*t*) = 1, 0 <sup>≤</sup> *<sup>t</sup>* <sup>≤</sup> <sup>4</sup>*. We'll verify I p*(*t*) <sup>0</sup><sup>+</sup> *I q*(*t*) <sup>0</sup><sup>+</sup> *f*(*t*)|*t*=<sup>3</sup> = *I p*(*t*)+*q*(*t*) <sup>0</sup><sup>+</sup> *f*(*t*)|*t*=3*, here, the variable order fractional integral is defined in (3). For* 1 ≤ *t* ≤ 4*, we have*

$$\begin{aligned} &I\_{0+}^{p(t)}I\_{0+}^{q(t)}f(t) \\ &= & \int\_0^1 \frac{(t-s)^{p(t)-1}}{\Gamma(p(t))} \int\_0^s \frac{(s-\tau)^{3-1}}{\Gamma(3)} d\tau ds + \int\_1^t \frac{(t-s)^{p(t)-1}}{\Gamma(p(t))} \int\_0^s \frac{(s-\tau)^{2-1}}{\Gamma(2)} d\tau ds \\ &= & \int\_0^1 \frac{(t-s)^{p(t)-1}s^3}{6\Gamma(p(t))} ds + \int\_1^t \frac{(t-s)^{p(t)-1}s^2}{2\Gamma(2)\Gamma(p(t))} ds, \end{aligned}$$

*thus, we have*

$$I\_{0+}^{p(t)}I\_{0+}^{q(t)}f(t)|\_{t=3} = \quad \int\_0^1 \frac{(3-s)^2s^3}{6\Gamma(3)}ds + \int\_1^3 \frac{(3-s)^2s^2}{2\Gamma(2)\Gamma(3)}ds = \frac{245}{144}.$$

$$d\_{0+}^{p(t)+q(t)}f(t)|\_{t=3} = \int\_0^3 \frac{(3-s)^{p(3)+q(3)-1}}{\Gamma(p(3)+q(3))}ds = \frac{3^{3+2}}{\Gamma(1+3+2)} = \frac{81}{40}.$$

*Therefore, we obtain*

$$|I\_{0+}^{p(t)}I\_{0+}^{q(t)}f(t)|\_{t=3} \neq |I\_{0+}^{p(t)+q(t)}f(t)|\_{t=3,\epsilon}$$

*which implies that the semigroup property of the variable order fractional integral does not hold for the piecewise constant functions p*(*t*) *and q*(*t*) *defined in the same partition* [0, 1],(1, 4] *of finite interval* [0, 4]*.*

**Lemma 7.** *[10] Suppose β* > 0*, a*(*t*) *is a nonnegative nondecreasing function locally integrable on* 0 ≤ *t* < *L (some L* ≤ +∞*) and g*(*t*) *is a nonnegative nondecreasing continuous function defined on* 0 ≤ *t* < *L, g*(*t*) ≤ *M (constant), and suppose u(t) is nonnegative and locally integrable on* 0 ≤ *t* < *L with*

$$u(t) \le a(t) + \mathfrak{g}(t) \int\_0^t (t - s)^{\beta - 1} u(s) ds$$

*on this interval. Then*

$$u(t) \le a(t)E\_{\beta}(\mathcal{g}(t)\Gamma(\beta)t^{\beta}), 0 \le t < L,$$

*where E<sup>β</sup> is the Mittag–Leffler function defined by Eβ*(*z*) = ∑<sup>∞</sup> *<sup>k</sup>*=<sup>0</sup> *<sup>z</sup><sup>k</sup>* <sup>Γ</sup>(*kβ*+1)*.*

## **3. Existence of Approximate Solution**

According to the previous arguments, we do not transform the problem (1) into an integral equation. Here, we consider the unique existence of approximate solution of the problem (1). In this section, we present our main results.

Now we make the following assumptions:

(*A*2) *<sup>f</sup>* : [0, <sup>+</sup>∞) <sup>×</sup> *<sup>R</sup>*<sup>2</sup> <sup>→</sup> *<sup>R</sup>* be a continuous function, and there exist positive constants *<sup>λ</sup>* <sup>&</sup>gt; {*ρ*1, *<sup>ρ</sup>*2}, *<sup>c</sup>*1, *<sup>c</sup>*<sup>2</sup> <sup>&</sup>gt; 0 satisfying *<sup>c</sup>*<sup>1</sup>

$$\frac{c\_1}{\Gamma(1+\rho\_1)} + \frac{c\_2}{\Gamma(1+\rho\_1-\rho\_2)} < 1,$$

such that

$$|f(t,(1+t^{\lambda})\mathbf{x}\_1,(1+t^{\lambda})y\_1) - f(t,(1+t^{\lambda})\mathbf{x}\_2,(1+t^{\lambda})y\_2)| \le c\_1|\mathbf{x}\_1 - \mathbf{x}\_2| + c\_2|y\_1 - y\_2|,\tag{23}$$

where *ρ*1, *ρ*<sup>2</sup> are the constants in (*A*1).

(*A*3) *f*(*t*, 0, 0)(*t* ∈ (0, +∞)) satisfies

$$\lim\_{t \to +\infty} \frac{1}{1 + t^{\lambda}} \int\_{0}^{t} (t - s)^{\rho\_1 - \rho\_2 - 1} |f(s, 0, 0)| ds = 0.$$

Let *Bi* denote the Banach spaces defined as

$$B\_i = \{ \mathbf{x} | \mathbf{x} \in \mathbb{C}[0, T\_i] \}$$

with the norm

$$\|\|\mathbf{x}\|\|\_{\overline{B}\_i} = \max\_{t \in [0, T\_i]} |\mathbf{x}(t)|\_{\prime} \tag{24}$$

where *Ti* is the constant obtained in Lemma 6, *i* = 1, ··· , *n*∗(*Tn*<sup>∗</sup> = *T*). Let

$$E = \left\{ \mathbf{x} \, \middle| \, \mathbf{x} \in \mathbb{C}[0, +\infty), \sup\_{t \ge 0} \frac{|\mathbf{x}(t)|}{1 + t^{\lambda}} < \infty \right\}$$

with the norm

$$\|\|\mathbf{x}\|\|\_{E} = \sup\_{t \ge 0} \frac{|\mathbf{x}(t)|}{1 + t^{\lambda}} \tag{25}$$

where *λ* > {*ρ*1, *ρ*2}. Then, by the same arguments as in Lemma 2.2 of [22], we know that (*E*, · *E*) is a Banach space, here we omit this proof.

Now, we consider the following initial value problem

$$\begin{cases} D\_{0+}^{a(t)}x(t) = f(t, \mathbf{x}, D\_{0+}^{\beta(t)}\mathbf{x}), 0 < t < +\infty, \\ x(0) = 0, \end{cases} \tag{26}$$

where *α*(*t*), *β*(*t*) are defined in (13) and (14).

In order to obtain our main results, we start off by carrying on essential analysis to the equation of (26).

By (13) and (14), we get

$$\int\_{0}^{t} \frac{(t-s)^{-\alpha(t)}}{\Gamma(1-\alpha(t))} \mathbf{x}(s) ds = \sum\_{k=1}^{n^{\*}} I\_{k}(t) \int\_{0}^{t} \frac{(t-s)^{-p\_{k}}}{\Gamma(1-p\_{k})} \mathbf{x}(s) ds + I\_{\Gamma}(t) \int\_{0}^{t} \frac{(t-s)^{-\rho\_{1}}}{\Gamma(1-\rho\_{1})} \mathbf{x}(s) ds,$$

$$\int\_{0}^{t} \frac{(t-s)^{-\beta(t)}}{\Gamma(1-\beta(t))} \mathbf{x}(s) ds = \sum\_{k=1}^{n^{\*}} I\_{k}(t) \int\_{0}^{t} \frac{(t-s)^{-q\_{k}}}{\Gamma(1-q\_{k})} \mathbf{x}(s) ds + I\_{\Gamma}(t) \int\_{0}^{t} \frac{(t-s)^{-\rho\_{2}}}{\Gamma(1-\rho\_{2})} \mathbf{x}(s) ds \doteq h\_{\beta,x}(t),$$

So, the equation of (26) can be written by

$$\frac{d}{dt}\sum\_{k=1}^{n^\*} I\_k(t) \int\_0^t \frac{(t-s)^{-p\_k}}{\Gamma(1-p\_k)} \mathbf{x}(s)ds + I\_T(t) \int\_0^t \frac{(t-s)^{-p\_1}}{\Gamma(1-p\_1)} \mathbf{x}(s)ds = f(t, \mathbf{x}, \frac{d}{dt}h\_{\boldsymbol{\beta}, \mathbf{x}}(t)), \ 0 < t < +\infty. \tag{27}$$

Then, Equation (27) in the interval (0, *T*1] can be written by

$$\frac{d}{dt} \int\_{0}^{t} \frac{(t-s)^{-p\_1}}{\Gamma(1-p\_1)} \mathbf{x}(s) ds = D\_{0+}^{p\_1} \mathbf{x}(t) = f(t, \mathbf{x}, D\_{0+}^{q\_1} \mathbf{x}), 0 < t \le T\_1. \tag{28}$$

The Equation (27) in the interval (*T*1, *T*2] can be written by

$$\frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-p\_2}}{\Gamma(1-p\_2)}\mathbf{x}(s)ds = f(t, \mathbf{x}, \frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-q\_2}}{\Gamma(1-q\_2)}\mathbf{x}(s)ds), \quad T\_1 < t \le T\_2. \tag{29}$$

The Equation (27) in the interval (*T*2, *T*3] can be written by

$$\frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-p\_3}}{\Gamma(1-p\_3)}\mathbf{x}(s)ds = f(t, \mathbf{x}, \frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-q\_3}}{\Gamma(1-q\_3)}\mathbf{x}(s)ds), \; T\_2 < t \le T\_3. \tag{30}$$

The Equation (27) in the interval (*Ti*−1, *Ti*], *<sup>i</sup>* = 4, 5, ··· , *<sup>n</sup>*<sup>∗</sup> (*Tn*<sup>∗</sup> = *<sup>T</sup>*) can be written by

$$\frac{d}{dt}\int\_{0}^{t} \frac{(t-s)^{-p\_i}}{\Gamma(1-p\_i)} \mathbf{x}(s)ds = f(t, \mathbf{x}, \frac{d}{dt}\int\_{0}^{t} \frac{(t-s)^{-q\_i}}{\Gamma(1-q\_i)} \mathbf{x}(s)ds), \quad T\_{i-1} < t \le T\_i. \tag{31}$$

The Equation (27) in the interval (*T*, +∞) can be written by

$$\frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-\rho\_1}}{\Gamma(1-\rho\_1)}\mathbf{x}(s)ds = f(t,\mathbf{x},\frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-\rho\_2}}{\Gamma(1-\rho\_2)}\mathbf{x}(s)ds),\ \ T$$

Now, we present the definition of a solution to the problem (26), which is crucial in our work.

**Definition 4.** *We say the problem (26) exists one unique solution, if there are unique functions ui*(*t*)*, i* = 1, 2, ··· , *n*∗*, such that u*<sup>1</sup> ∈ *C*[0, *T*1] *satisfying Equation (28) and u*1(0) = 0*; u*<sup>2</sup> ∈ *C*[0, *T*2] *satisfying Equation (29) and u*2(0) = 0*; u*<sup>3</sup> ∈ *C*[0, *T*3] *satisfying Equation (30) and u*3(0) = 0*; ui* ∈ *C*[0, *Ti*] *satisfying Equation (31) and ui*(0) = 0 (*i* = 4, 5, ··· , *n*∗)*(Tn*<sup>∗</sup> = *T); uT* ∈ *C*[0, +∞) *satisfying Equation (32) and uT*(0) = 0*.*

The following is the definition of approximate solution of the problem (1).

**Definition 5.** *If there exist T* > 0*, natural number n*<sup>∗</sup> ∈ *N and intervals* [0, *T*1],(*T*1, *T*2], ··· *,* (*Tn*∗−1, *T*]*,* (*T*, +∞) *and functions defined in Equations (13) and (14), such that the problem (26) exists one unique solution, then, we say this solution of the problem (26) is one unique approximate solution of the problem (1).*

Our main result is as follows.

**Theorem 1.** *Let conditions* (*A*1),(*A*2),(*A*3) *hold, then the problem (1) exists one unique approximate solution.*

**Proof of Theorem 1.** From Definitions 4 and 5 and Lemma 6, we only need to consider the unique existence of solution of the problem (26). According to the above analysis, equation of problem (26) can be written as the Equation (27). So Equation (26) in the interval (0, *T*1] can be written as (28). Applying operator *I p*1 <sup>0</sup><sup>+</sup> to both sides of (28), by Lemma 5, we have

$$\mathbf{x}(t) = \mathbf{c}t^{p\_1 - 1} + \frac{1}{\Gamma(p\_1)} \int\_0^t (t - s)^{p\_1 - 1} f(\mathbf{s}, \mathbf{x}(s), D\_{0+}^{q\_1} \mathbf{x}(s)) ds, \ 0 < t \le T\_1.$$

By *x*(0) = 0 and the assumption of function *f* , we get *c* = 0, that is

$$\mathbf{x}(t) = \frac{1}{\Gamma(p\_1)} \int\_0^t (t-s)^{p\_1-1} f(s, \mathbf{x}(s), D\_{0+}^{q\_1} \mathbf{x}(s)) ds, \ 0 \le t \le T\_1. \tag{33}$$

Let *Dq*<sup>1</sup> <sup>0</sup>+*x*(*t*) = *y*(*t*), then, according to *x*(0) = 0 and Lemma 5, we get that

$$x(t) = I\_{0+}^{q\_1} y(t)\_{"\prime}$$

hence we will consider existence of solution to integral equation as following

$$y(t) = \frac{1}{\Gamma(p\_1 - q\_1)} \int\_0^t (t - s)^{p\_1 - q\_1 - 1} f(s, l\_{0+}^{q\_1} y(s), y(s)) ds, \ 0 \le t \le T\_1. \tag{34}$$

Obviously, if *y*<sup>∗</sup> ∈ *B*<sup>1</sup> = *C*[0, *T*1] is a solution of (34), then, applying operator *I q*1 <sup>0</sup><sup>+</sup> on both sides of (34), from Lemma 2, it holds

$$l\_{0+}^{q\_1}y^\*(t) = l\_{0+}^{q\_1}l\_{0+}^{p\_1-q\_1}f(t, l\_{0+}^{q\_1}y^\*(t), y^\*(t)) = l\_{0+}^{p\_1}f(t, l\_{0+}^{q\_1}y^\*(t), y^\*(t)), \ 0 \le t \le T\_{1+}$$

let

$$l\_{0+}^{q\_1}y^\*(t) = x^\*(t), 0 \le t \le T\_{\prime \prime}$$

as a result, we have that

$$\mathbf{x}^\*(t) = I\_{0+}^{p\_1} f(t, \mathbf{x}^\*(t), D\_{0+}^{q\_1} \mathbf{x}^\*(t)), \ 0 \le t \le T\_{1+} $$

that is, *x*<sup>∗</sup> ∈ *B*<sup>1</sup> = *C*[0, *T*1] is a solution of (33), thus, we know that *x*<sup>∗</sup> ∈ *B*<sup>1</sup> = *C*[0, *T*1] is a solution of Equation (28) with zero initial value condition.

Define operator *F* : *B*<sup>1</sup> → *B*<sup>1</sup> by

$$Fy(t) = \frac{1}{\Gamma(p\_1 - q\_1)} \int\_0^t (t - s)^{p\_1 - q\_1 - 1} f(s, l\_{0+}^{q\_1} y(s), y(s)) ds, \ 0 \le t \le T\_1. \tag{35}$$

From the continuity of function *f* and the standard arguments, we know that the operator *F* : *B*<sup>1</sup> → *B*<sup>1</sup> is well defined. Let *M* = max0≤*t*≤*<sup>T</sup>* | *f*(*t*, 0, 0)|. Let Ω<sup>1</sup> be a bounded, convex and closed subset of *B*<sup>1</sup> defined by

$$\Omega\_1 = \{ y|y \in B\_1; |y(t)| \le K\_1 e^{R\_1^2 t^{p\_1 - q\_1}}, 0 \le t \le T\_1 \},$$

where

$$K\_1 = \frac{2MT\_1^{p\_1 - q\_1}}{\Gamma(1 + p\_1 - q\_1)},$$

*R*<sup>1</sup> ∈ *N* satisfying

$$R\_1 > \{ 1, (\frac{2d\_1(1 + T\_1^{p\_1 - q\_1})}{p\_1 - q\_1})^{\frac{1}{p\_1 - q\_1}} \}\_{\prime \prime}$$

here *d*<sup>1</sup> = <sup>1</sup> Γ(*p*1−*q*1) *<sup>c</sup>*1*Tq*<sup>1</sup> 1 <sup>Γ</sup>(1+*q*1) <sup>+</sup> *<sup>c</sup>*<sup>2</sup> (*c*1, *c*<sup>2</sup> are the constants appearing in condition (*A*2)).

By the analogy way as in [23], we could verify that *F* : Ω<sup>1</sup> → Ω<sup>1</sup> is well defined. In fact, for *y* ∈ Ω1, since

$$\begin{split} |I\_{0+}^{\mathcal{q}\_1}y(s)| &\leq \qquad \frac{1}{\Gamma(q\_1)} \int\_0^s (s-\tau)^{q\_1-1} |y(\tau)| d\tau \\ &\leq \qquad \frac{K\_1}{\Gamma(q\_1)} \int\_0^s (s-\tau)^{q\_1-1} e^{R\_1^2 \tau^{p\_1-q\_1}} d\tau \\ &\leq \qquad \frac{K\_1}{\Gamma(q\_1)} \int\_0^s (s-\tau)^{q\_1-1} e^{R\_1^2 s^{p\_1-q\_1}} d\tau \\ &= \qquad \qquad \frac{K\_1}{\Gamma(1+q\_1)} s^{q\_1} e^{R\_1^2 s^{p\_1-q\_1}} \\ &\leq \qquad \qquad \frac{K\_1 T\_1^{q\_1}}{\Gamma(1+q\_1)} \epsilon^{R\_1^2 s^{p\_1-q\_1}} .\end{split}$$

Now, *y* ∈ Ω1, by estimations above and (*A*2), we get

$$\begin{aligned} &|Fy(t)|\\ \leq & \qquad \frac{1}{\Gamma(p\_1 - q\_1)} \int\_0^t (t - s)^{p\_1 - q\_1 - 1} |f(s, l\_{0+}^{q\_1} y(s), y(s))| ds \\ = & \qquad \frac{1}{\Gamma(p\_1 - q\_1)} \int\_0^t (t - s)^{p\_1 - q\_1 - 1} |f(s, l\_{0+}^{q\_1} y(s), y(s)) - f(s, 0, 0) + f(s, 0, 0)| ds \\ \leq & \qquad \frac{MT\_1^{p\_1 - q\_1}}{\Gamma(1 + p\_1 - q\_1)} + \frac{1}{\Gamma(p\_1 - q\_1)} \int\_0^t (t - s)^{p\_1 - q\_1 - 1} (c\_1 \frac{|l\_{0+}^{q\_1} y(s)|}{1 + s^{\lambda}} + c\_2 \frac{|y(s)|}{1 + s^{\lambda}}) ds \\ \leq & \qquad \frac{K\_1}{2} + \frac{1}{\Gamma(p\_1 - q\_1)} \int\_0^t (t - s)^{p\_1 - q\_1 - 1} (c\_1 |l\_{0+}^{q\_1} y(s)| + c\_2 |y(s)|) ds \end{aligned}$$

≤ *K*1 <sup>2</sup> <sup>+</sup> 1 Γ(*p*<sup>1</sup> − *q*1) *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−<sup>1</sup> *<sup>K</sup>*1*c*1*Tq*<sup>1</sup> 1 Γ(1 + *q*1) *eR*2 <sup>1</sup>*sp*1−*q*<sup>1</sup> <sup>+</sup> *<sup>c</sup>*2*K*1*eR*<sup>2</sup> <sup>1</sup>*sp*1−*q*<sup>1</sup> *ds* ≤ *K*1 <sup>2</sup> <sup>+</sup> *<sup>K</sup>*1*d*<sup>1</sup> *R*1−1 ∑ *i*=1 *it R*1 (*i*−1)*t R*1 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−1*eR*<sup>2</sup> <sup>1</sup>*sp*1−*q*<sup>1</sup> *ds* + *t* (*R*1−1)*t R*1 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−1*eR*<sup>2</sup> <sup>1</sup>*sp*1−*q*<sup>1</sup> *ds* ≤ *K*1 <sup>2</sup> <sup>+</sup> *<sup>K</sup>*1*d*<sup>1</sup> *R*1−1 ∑ *i*=1 *it R*1 (*i*−1)*t R*1 *<sup>R</sup>*1−*p*1+*q*<sup>1</sup> <sup>1</sup> (*R*<sup>1</sup> <sup>−</sup> *<sup>i</sup>*)*p*1−*q*1−1*<sup>t</sup> <sup>p</sup>*1−*q*1−1*eR*<sup>2</sup> <sup>1</sup>*sp*1−*q*<sup>1</sup> *ds* + *t* (*R*1−1)*t R*1 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−1*eR*<sup>2</sup> <sup>1</sup>*tp*1−*q*<sup>1</sup> *ds* ≤ *K*1 <sup>2</sup> <sup>+</sup> *<sup>K</sup>*1*d*<sup>1</sup> *R*1−1 ∑ *i*=1 *it R*1 (*i*−1)*t R*1 *<sup>R</sup>*1−*p*1+*q*<sup>1</sup> <sup>1</sup> *t <sup>p</sup>*1−*q*1−1*eR*<sup>2</sup> <sup>1</sup>*sp*1−*q*<sup>1</sup> *ds* + *t* (*R*1−1)*t R*1 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−1*eR*<sup>2</sup> <sup>1</sup>*tp*1−*q*<sup>1</sup> *ds* <sup>=</sup> *<sup>K</sup>*<sup>1</sup> <sup>2</sup> <sup>+</sup> *<sup>K</sup>*1*d*1*R*1−*p*1+*q*<sup>1</sup> 1 (*R*1−1)*t R*1 0 *t <sup>p</sup>*1−*q*1−1*eR*<sup>2</sup> <sup>1</sup>*sp*1−*q*<sup>1</sup> *ds* + *<sup>K</sup>*1*d*1*Rq*1−*p*<sup>1</sup> <sup>1</sup> *<sup>T</sup>p*1−*q*<sup>1</sup> 1 *p*<sup>1</sup> − *q*<sup>1</sup> *eR*2 <sup>1</sup>*tp*1−*q*<sup>1</sup> ≤ *K*1 <sup>2</sup> <sup>+</sup> *<sup>K</sup>*1*d*1*R*1−*p*1+*q*<sup>1</sup> 1 (*R*1−1)*t R*1 0 *sp*1−*q*1−1*eR*<sup>2</sup> <sup>1</sup>*sp*1−*q*<sup>1</sup> *ds* + *<sup>K</sup>*1*d*1*Rq*1−*p*<sup>1</sup> <sup>1</sup> *<sup>T</sup>p*1−*q*<sup>1</sup> 1 *p*<sup>1</sup> − *q*<sup>1</sup> *eR*2 <sup>1</sup>*tp*1−*q*<sup>1</sup> ≤ *K*1 <sup>2</sup> <sup>+</sup> *<sup>K</sup>*1*d*1*R*1−*p*1+*q*<sup>1</sup> 1 *R*2 <sup>1</sup>(*p*<sup>1</sup> <sup>−</sup> *<sup>q</sup>*1) *<sup>e</sup> R*2 1( (*R*1−1)*t <sup>R</sup>*<sup>1</sup> )*p*1−*q*<sup>1</sup> + *<sup>K</sup>*1*d*1*Rq*1−*p*<sup>1</sup> <sup>1</sup> *<sup>T</sup>p*1−*q*<sup>1</sup> 1 *p*<sup>1</sup> − *q*<sup>1</sup> *eR*2 <sup>1</sup>*tp*1−*q*<sup>1</sup> ≤ *K*1 <sup>2</sup> <sup>+</sup> *<sup>K</sup>*1*d*1*R*−1−*p*1+*q*<sup>1</sup> 1 *p*<sup>1</sup> − *q*<sup>1</sup> *eR*2 <sup>1</sup>*tp*1−*q*<sup>1</sup> + *<sup>K</sup>*1*d*1*Rq*1−*p*<sup>1</sup> <sup>1</sup> *<sup>T</sup>p*1−*q*<sup>1</sup> 1 *p*<sup>1</sup> − *q*<sup>1</sup> *eR*2 <sup>1</sup>*tp*1−*q*<sup>1</sup> ≤ *K*1 <sup>2</sup> *<sup>e</sup>R*<sup>2</sup> <sup>1</sup>*tp*1−*q*<sup>1</sup> + *<sup>K</sup>*1*d*1(<sup>1</sup> <sup>+</sup> *<sup>T</sup>p*1−*q*<sup>1</sup> <sup>1</sup> ) *p*<sup>1</sup> − *q*<sup>1</sup> *Rq*1−*p*<sup>1</sup> <sup>1</sup> *<sup>e</sup>R*<sup>2</sup> <sup>1</sup>*tp*1−*q*<sup>1</sup> ≤ *K*1 <sup>2</sup> *<sup>e</sup>R*<sup>2</sup> <sup>1</sup>*tp*1−*q*<sup>1</sup> + *K*1 <sup>2</sup> *<sup>e</sup>R*<sup>2</sup> <sup>1</sup>*tp*1−*q*<sup>1</sup> <sup>=</sup> *<sup>K</sup>*1*eR*<sup>2</sup> <sup>1</sup>*tp*1−*q*<sup>1</sup> ,

which implies that *F* : Ω<sup>1</sup> → Ω<sup>1</sup> is well defined. By the standard arguments, we could know that *F* : Ω<sup>1</sup> → Ω<sup>1</sup> is a completely operator. Hence, the Schauder fixed point theorem assures that operator *F* has at least one fixed point *y*1(*t*) ∈ Ω1. Obviously, *y*1(0) = 0. Now, we will verify the uniqueness of solution to the integral Equation (34). We notice that: for 0 ≤ *s* ≤ *t* ≤ *T*1, if 0 ≤ *t* − *s* ≤ 1, then (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−<sup>1</sup> <sup>≤</sup> (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1<sup>−</sup>1; if *<sup>t</sup>* <sup>−</sup> *<sup>s</sup>* <sup>≥</sup> 1, then (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−<sup>1</sup> <sup>≤</sup> (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1<sup>−</sup>1. As a result, we take

$$\max\{(t-s)^{p\_1-1}, (t-s)^{p\_1-q\_1-1}\} \doteq (t-s)^{a-1},$$

where *α* denotes *p*<sup>1</sup> or *p*<sup>1</sup> − *q*1. Now, let *u*1(*t*), *u*2(*t*) zre two solutions of the integral Equation (34), by expression above and (*A*2), we get

$$\begin{aligned} &|u\_1(t) - u\_2(t)| \\ \leq & \frac{1}{\Gamma(p\_1 - q\_1)} \int\_0^t (t - s)^{p\_1 - q\_1 - 1} (c\_1 \frac{|I\_{0+}^{q\_1}(u\_1(s) - u\_2(s))|}{1 + s^\lambda} + c\_2 \frac{|u\_1(s) - u\_2(s)|}{1 + s^\lambda}) ds \end{aligned}$$

≤ 1 Γ(*p*<sup>1</sup> − *q*1) *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−1(*c*1|*<sup>I</sup> q*1 <sup>0</sup>+(*u*1(*s*) − *u*2(*s*))| + *c*2|*u*1(*s*) − *u*2(*s*)|)*ds* <sup>≤</sup> *<sup>c</sup>*<sup>1</sup> Γ(*p*<sup>1</sup> − *q*1)Γ(*q*1) *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−<sup>1</sup> *s* 0 (*<sup>s</sup>* <sup>−</sup> *<sup>τ</sup>*)*q*1−1|*u*1(*τ*) <sup>−</sup> *<sup>u</sup>*2(*τ*)|*dτds* <sup>+</sup> *<sup>c</sup>*<sup>2</sup> Γ(*p*<sup>1</sup> − *q*1) *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−1|*u*1(*s*) <sup>−</sup> *<sup>u</sup>*2(*s*)|*ds* <sup>=</sup> *<sup>c</sup>*<sup>1</sup> Γ(*p*<sup>1</sup> − *q*1)Γ(*q*1) *t* 0 *t τ* (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−1(*<sup>s</sup>* <sup>−</sup> *<sup>τ</sup>*)*q*1−1|*u*1(*τ*) <sup>−</sup> *<sup>u</sup>*2(*τ*)|*dsd<sup>τ</sup>* <sup>+</sup> *<sup>c</sup>*<sup>2</sup> Γ(*p*<sup>1</sup> − *q*1) *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−1|*u*1(*s*) <sup>−</sup> *<sup>u</sup>*2(*s*)|*ds* <sup>=</sup> *<sup>c</sup>*<sup>1</sup> Γ(*p*1) *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>τ</sup>*)*p*1−1|*u*1(*τ*) <sup>−</sup> *<sup>u</sup>*2(*τ*)|*d<sup>τ</sup>* <sup>+</sup> *<sup>c</sup>*<sup>2</sup> Γ(*p*<sup>1</sup> − *q*1) *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*p*1−*q*1−1|*u*1(*s*) <sup>−</sup> *<sup>u</sup>*2(*s*)|*ds* <sup>≤</sup> *<sup>c</sup>*<sup>1</sup> Γ(*p*1) *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>τ</sup>*)*α*−1|*u*1(*τ*) <sup>−</sup> *<sup>u</sup>*2(*τ*)|*d<sup>τ</sup>* <sup>+</sup> *<sup>c</sup>*<sup>2</sup> Γ(*p*<sup>1</sup> − *q*1) *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*)*α*−1|*u*1(*s*) <sup>−</sup> *<sup>u</sup>*2(*s*)|*ds* = [ *<sup>c</sup>*<sup>1</sup> <sup>Γ</sup>(*p*1) <sup>+</sup> *<sup>c</sup>*<sup>2</sup> Γ(*p*<sup>1</sup> − *q*1) ] *t* 0 (*<sup>t</sup>* <sup>−</sup> *<sup>τ</sup>*)*α*−1|*u*1(*τ*) <sup>−</sup> *<sup>u</sup>*2(*τ*)|*dτ*,

by Lemma 7, we obtain that *u*1(*t*) = *u*2(*t*), 0 ≤ *t* ≤ *T*1, this assures the uniqueness of solution of (34). As a result, by some arguments above, *x*1(*t*) = *I q*1 <sup>0</sup>+*y*1(*t*) is one unique solution of the Equation (28) with zero initial value condition.

Also, we have obtained that the Equation (27) in the interval (*T*1, *T*2] can be written by (29). In order to consider the existence result of solutions to (29), we may discuss the following equation defined on interval (0, *T*2]

$$\frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-p\_2}\mathbf{x}(s)}{\Gamma(1-p\_2)}ds = D\_{0+}^{p\_2}\mathbf{x}(t) = f(t, \mathbf{x}, \frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-q\_2}\mathbf{x}(s)}{\Gamma(1-q\_2)}ds) = f(t, \mathbf{x}, D\_{0+}^{q\_2}\mathbf{x}).\tag{36}$$

It is clear that if function *x* ∈ *C*[0, *T*2] satisfies the Equation (36), then *x*(*t*) must satisfy the Equation (29). In fact, if *x*<sup>∗</sup> ∈ *C*[0, *T*2] with *x*∗(0) = 0 is a solution of the Equation (36) with initial value condition *x*(0) = 0, that is

$$\begin{aligned} &D\_{0+}^{\mathbb{P}\_2^2} \mathbf{x}^\*(t) \\ &= & \frac{d}{dt} \int\_0^t \frac{(t-s)^{-p\_2} \mathbf{x}^\*(s)}{\Gamma(1-p\_2)} ds \\ &= & f(t, \mathbf{x}^\*(t), D\_{0+}^{\mathbb{P}\_2^\*} \mathbf{x}^\*(t)) = f(t, \mathbf{x}^\*(t), \frac{d}{dt} \int\_0^t \frac{(t-s)^{-q\_2} \mathbf{x}^\*(s)}{\Gamma(1-q\_2)} ds), \ 0 < t \le T\_2; \ \mathbf{x}^\*(0) = 0. \end{aligned}$$

Hence, from the equality above, we have that *x*<sup>∗</sup> ∈ *C*[0, *T*2] with *x*∗(0) = 0 satisfies the equation

$$\frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-p\_2}x^\*(s)}{\Gamma(1-p\_2)}ds = f(t,x^\*(t),\frac{d}{dt}\int\_{0}^{t}\frac{(t-s)^{-q\_2}x^\*(s)}{\Gamma(1-q\_2)}ds),\ \ T\_1 \le t \le T\_2.$$

which means the function *x*<sup>∗</sup> ∈ *C*[0, *T*2] with *x*∗(0) = 0 is a solution of the Equation (29).

Based on this fact, we consider the existence of solutions to the Equation (36) with initial value condition *x*(0) = 0.

Now, applying operator *I p*2 <sup>0</sup><sup>+</sup> on both sides of (36), by Lemma 5, we have

$$\mathbf{x}(t) = ct^{p\_2 - 1} + \frac{1}{\Gamma(p\_2)} \int\_0^t (t - s)^{p\_2 - 1} f(s, \mathbf{x}(s), D\_{0+}^{q\_2} \mathbf{x}(s)) ds, \ 0 < t \le T\_2.$$

By initial value condition *x*(0) = 0, we have *c* = 0, that is

$$\mathbf{x}(t) = \frac{1}{\Gamma(p\_2)} \int\_0^t (t-s)^{p\_2-1} f(s, \mathbf{x}(s), D\_{0+}^{q\_2} \mathbf{x}(s)) ds, \ 0 \le t \le T\_2. \tag{37}$$

Let *Dq*<sup>2</sup> <sup>0</sup>+*x*(*t*) = *y*(*t*), then, according to *x*(0) = 0 and Lemma 5, we get that

$$x(t) = I\_{0+}^{\\\\\oplus 2}y(t),$$

hence we will consider existence of solution to integral equation as following

$$y(t) = \frac{1}{\Gamma(p\_2 - q\_2)} \int\_0^t (t - s)^{p\_2 - q\_2 - 1} f(s, l\_{0+}^{q\_2} y(s), y(s)) ds, \ 0 \le t \le T\_2. \tag{38}$$

Obviously, if *y*<sup>∗</sup> ∈ *B*<sup>2</sup> = *C*[0, *T*2] is a solution of (38), then, by (38) and Lemma 2, it holds

$$I\_{0+}^{q\_2}y^\*(t) = I\_{0+}^{q\_2}I\_{0+}^{p\_2-q\_2}f(t, I\_{0+}^{q\_2}y^\*(t), y^\*(t)) = I\_{0+}^{p\_2}f(t, I\_{0+}^{q\_2}y^\*(t), y^\*(t)), \ 0 \le t \le T\_{2+}$$

let

$$I\_{0+}^{\\\\\ell^2}y^\*(t) = x^\*(t), \\\\0 \le t \le T\_{2\prime}$$

as a result, we have that

$$\mathbf{x}^\*(t) = I\_{0+}^{p\_2} f(t, \mathbf{x}^\*(t), D\_{0+}^{q\_2} \mathbf{x}^\*(t)), \ 0 \le t \le T\_{2\nu}$$

that is, *x*<sup>∗</sup> ∈ *B*<sup>2</sup> = *C*[0, *T*2] is a solution of (37), hence, *x*<sup>∗</sup> ∈ *B*<sup>2</sup> = *C*[0, *T*2] is a solution of Equation (29) with zero initial value condition.

Define operator *F* : *B*<sup>2</sup> → *B*<sup>2</sup> by

$$Fy(t) = \frac{1}{\Gamma(p\_2 - q\_2)} \int\_0^t (t - s)^{p\_2 - q\_2 - 1} f(s, l\_{0+}^{q\_2} y(s), y(s)) ds, \ 0 \le t \le T\_2.$$

From the continuity of function *f* and the standard arguments, we know that the operator *F* : *B*<sup>1</sup> → *B*<sup>2</sup> is well defined. Let Ω<sup>2</sup> be a bounded, convex and closed subset of *B*<sup>2</sup> defined by

$$\Omega\_2 = \{ y|y \in B\_2 \colon |y(t)| \le K\_2 e^{R\_2^2 t^{\nu\_2 - q\_2}}, 0 \le t \le T\_2 \},$$

where

$$K\_2 = \frac{2MT\_2^{p\_2 - q\_2}}{\Gamma(1 + p\_2 - q\_2)},$$

*R*<sup>2</sup> ∈ *N* satisfying

$$R\_2 > \{1, (\frac{2d\_2(1 + T\_2^{p\_2 - q\_2})}{p\_2 - q\_2})^{\frac{1}{p\_2 - q\_2}}\}\_{\prime \prime}$$

here *d*<sup>2</sup> = <sup>1</sup> Γ(*p*2−*q*2) *<sup>c</sup>*1*Tq*<sup>2</sup> 2 <sup>Γ</sup>(1+*q*2) <sup>+</sup> *<sup>c</sup>*<sup>2</sup> (*c*1, *c*<sup>2</sup> are the constants appearing in condition (*A*2)). By the same arguments above, there exists *y*<sup>2</sup> ∈ Ω<sup>2</sup> such that *x*2(*t*) = *I q*2 <sup>0</sup>+*y*2(*t*) is one unique solution of the Equation (29) with zero initial value condition.

In a similar way, for *<sup>i</sup>* = 3, ··· , *<sup>n</sup>*∗, we get that the Equation (31) defined on (*Ti*−1, *Ti*] (*Tn*<sup>∗</sup> = *<sup>T</sup>*) has one solution *xi*(*t*) ∈ Ω*<sup>i</sup>* ⊂ *Bi* with *xi*(0) = 0 , where

$$\Omega\_i = \{ y \vert y \in B\_i \colon \vert y(t) \vert \le K\_i e^{R\_i^2 t^{p\_i - q\_i}}, 0 \le t \le T\_i \},$$

$$K\_i = \frac{2MT\_i^{p\_i - q\_i}}{\Gamma(1 + p\_i - q\_i)},$$

*Ri* ∈ *N* satisfying

$$R\_i > \left\{ 1, \left( \frac{2d\_i(1 + T\_i^{p\_i - q\_i})}{p\_i - q\_i} \right)^{\frac{1}{p\_i - q\_i}} \right\},$$

here *di* = <sup>1</sup> Γ(*pi*−*qi*) *<sup>c</sup>*1*Tqi i* <sup>Γ</sup>(1+*qi*) <sup>+</sup> *<sup>c</sup>*<sup>2</sup> (*c*1, *c*<sup>2</sup> are the constants appearing in condition (*A*2)), *i* = 3, 4, ··· , *n*∗, *Tn*<sup>∗</sup> = *T*.

Finally, we get that the Equation (27) in the interval (*T*, +∞) can be written by (32). In order to consider the existence result of solutions to (32), we may discuss the following equation defined on interval (0, +∞)

$$\frac{d}{dt} \int\_{0}^{t} \frac{(t-s)^{-\rho\_1}}{\Gamma(1-\rho\_1)} \mathbf{x}(s) ds = D\_{0+}^{\rho\_1} \mathbf{x}(t) = f(t, \mathbf{x}, D\_{0+}^{\rho\_2} \mathbf{x}), \ 0 < t < +\infty. \tag{39}$$

We see that, if function *x* ∈ *C*[0, +∞) satisfies the Equation (39), then *x*(*t*) must satisfy the Equation (32). In fact, if *x*<sup>∗</sup> ∈ *C*[0, +∞) with *x*∗(0) = 0 is a solution of the Equation (39) with initial value condition *x*(0) = 0, that is

$$D\_{0+}^{\rho\_1} \mathbf{x}^\*(t) = \qquad \frac{d}{dt} \int\_0^t \frac{(t-s)^{-\rho\_1} \mathbf{x}^\*(s)}{\Gamma(1-\rho\_1)} ds = f(t, \mathbf{x}^\*(t), D\_{0+}^{\rho\_2} \mathbf{x}^\*)$$

$$= \qquad f(t, \mathbf{x}^\*(t), \frac{d}{dt} \int\_0^t \frac{(t-s)^{-\rho\_2} \mathbf{x}^\*(s)}{\Gamma(1-\rho\_2)} ds), \; 0 < t < +\infty; \; \mathbf{x}^\*(0) = 0.$$

Hence, from the equality above, we have *x*<sup>∗</sup> ∈ *C*[0, +∞) with *x*∗(0) = 0 satisfying the equation

$$\frac{d}{dt} \int\_0^t \frac{(t-s)^{-\rho\_1} \mathbf{x}(s)}{\Gamma(1-\rho\_1)} ds = f(t, \mathbf{x}(t), \frac{d}{dt} \int\_0^t \frac{(t-s)^{-\rho\_2} \mathbf{x}(s)}{\Gamma(1-\rho\_2)} ds), \quad T < t < +\infty,$$

which means the function *x*<sup>∗</sup> ∈ *C*[0, +∞) with *x*∗(0) = 0 is a solution of the Equation (32).

Based on this fact, we will consider the existence of solutions to the Equation (39) with initial value condition *x*(0) = 0.

Now, applying operator *I ρ*1 <sup>0</sup><sup>+</sup> on both sides of (39), by Lemma 5, we have that

$$\mathbf{x}(t) = ct^{\rho\_1 - 1} + \frac{1}{\Gamma(\rho\_1)} \int\_0^t (t - s)^{\rho\_1 - 1} f(s, \mathbf{x}(s), D\_{0+}^{\rho\_2} \mathbf{x}(s)) ds, \ 0 < t < +\infty.$$

By initial value condition *x*(0) = 0, we have *c* = 0, that is

$$\mathbf{x}(t) = \frac{1}{\Gamma(\rho\_1)} \int\_0^t (t-s)^{\rho\_1-1} f(\mathbf{s}, \mathbf{x}(\mathbf{s}), D\_{0+}^{\rho\_2} \mathbf{x}(\mathbf{s})) ds, \ 0 \le t < +\infty. \tag{40}$$

Similar to arguments above, we let *Dρ*<sup>2</sup> <sup>0</sup>+*x*(*t*) = *y*(*t*), then, according to *x*(0) = 0 and Lemma 5, we get that

$$x(t) = I\_{0+}^{\\\\\rho\_2}y(t),$$

hence we will consider existence of solution to integral equation as following

$$y(t) = \frac{1}{\Gamma(\rho\_1 - \rho\_2)} \int\_0^t (t - s)^{\rho\_1 - \rho\_2 - 1} f(s, l\_{0+}^{\rho\_2} y(s), y(s)) ds, \ 0 \le t < +\infty. \tag{41}$$

Obviously, if *y*<sup>∗</sup> ∈ *E* is a solution of (41), then, by (41) and Lemma 2, it holds

$$l\_{0+}^{\rho\_2}y^\*(t) = l\_{0+}^{\rho\_2}l\_{0+}^{\rho\_1-\rho\_2}f(t, l\_{0+}^{\rho\_2}y^\*(t), y^\*(t)) = l\_{0+}^{\rho\_1}f(t, l\_{0+}^{\rho\_2}y^\*(t), y^\*(t)), \ 0 \le t < +\infty.$$

Let

$$d\_{0+}^{\\\\\rho\_2}y^\*(t) = x^\*(t), \\\\
0 \le t < +\infty.$$

As a result, we have that

$$\mathbf{x}^\*(t) = I\_{0+}^{\rho\_1} f(t, \mathbf{x}^\*(t), D\_{0+}^{\rho\_2} \mathbf{x}^\*(t)), \ 0 \le t < +\infty,$$

that is, *x*<sup>∗</sup> ∈ *E* is a solution of (40), hence, *x*<sup>∗</sup> ∈ *E* is a solution of Equation (32) with zero initial value condition.

Defining operator *F* : *E* → *E* as follows

$$Fy(t) = \frac{1}{\Gamma(\rho\_1 - \rho\_2)} \int\_0^t (t - s)^{\rho\_1 - \rho\_2 - 1} f(s, l\_{0+}^{\rho\_2} y(s), y(s)) ds, \\ 0 \le t < +\infty.$$

To get the operator *F* : *E* → *E* is well defined. First, we verify that *Fy* ∈ *C*[0, +∞) for *x* ∈ *E*. In fact, for the case of *t*<sup>0</sup> ∈ (0, +∞), take *t* > *t*0, *t* − *t*<sup>0</sup> < 1, then

$$(t\_0 - s)^{\rho\_1 - 1} > (t - s)^{\rho\_1 - 1}, 0 \le s < t\_0.$$

Now, for *y* ∈ *E*, it holds

$$\begin{array}{ll} \frac{|l\_{0}^{\rho\_{2}}y(s)|}{1+s^{\lambda}} \leq& \frac{\int\_{0}^{s} (s-\tau)^{\rho\_{2}-1} |y(\tau)| d\tau}{\Gamma(\rho\_{2})(1+s^{\lambda})}\\ \leq& \frac{\int\_{0}^{s} (s-\tau)^{\rho\_{2}-1} (1+\tau^{\lambda}) ||y||\_{E} d\tau}{\Gamma(\rho\_{2})(1+s^{\lambda})}\\ \leq& \frac{\int\_{0}^{s} (s-\tau)^{\rho\_{2}-1} (1+s^{\lambda}) ||y||\_{E} d\tau}{\Gamma(\rho\_{2})(1+s^{\lambda})}\\ =& \frac{||y||\_{E} s^{\rho\_{2}}}{\Gamma(1+\rho\_{2})} \end{array}$$

thus, for *y* ∈ *E*, we have

$$|Fy(t)| \leq \qquad \frac{1}{\Gamma(\rho\_1 - \rho\_2)} \int\_0^{t\_0} ((t\_0 - s)^{\rho\_1 - \rho\_2 - 1} - (t - s)^{\rho\_1 - \rho\_2 - 1}) (c\_1 \frac{|I\_{0+}^{\rho\_1} y(s)|}{1 + s^\lambda} + c\_2 \frac{|y(s)|}{1 + s^\lambda}) ds$$

$$+ \frac{1}{\Gamma(\rho\_1 - \rho\_2)} \int\_{t\_0}^t (t - s)^{\rho\_1 - \rho\_2 - 1} (c\_1 \frac{|I\_{0+}^{\rho\_2} y(s)|}{1 + s^\lambda} + c\_2 \frac{|y(s)|}{1 + s^\lambda}) ds$$

$$+ \frac{1}{\Gamma(\rho\_1 - \rho\_2)} \int\_0^{t\_0} ((t\_0 - s)^{\rho\_1 - \rho\_2 - 1} - (t - s)^{\rho\_1 - \rho\_2 - 1}) |f(s, 0, 0)| ds$$

$$+ \frac{1}{\Gamma(\rho\_1 - \rho\_2)} \int\_{t\_0}^t (t - s)^{\rho\_1 - \rho\_2 - 1} |f(s, 0, 0)| ds$$

$$\begin{split} \leq & \qquad \frac{||y||\_{E}}{\Gamma(\rho\_{1}-\rho\_{2})} \int\_{0}^{t\_{0}} ((t\_{0}-s)^{\rho\_{1}-\rho\_{2}-1}-(t-s)^{\rho\_{1}-\rho\_{2}-1}) \Big(c\_{1}\frac{s^{\rho\_{2}}}{\Gamma(1+\rho\_{2})}+c\_{2}\Big)ds \\ & + \frac{||y||\_{E}}{\Gamma(\rho\_{1}-\rho\_{2})} \int\_{t\_{0}}^{t} (t-s)^{\rho\_{1}-\rho\_{2}-1} \Big(c\_{1}\frac{s^{\rho\_{2}}}{\Gamma(1+\rho\_{2})}+c\_{2}\Big)ds \\ & + \frac{\max\_{0\leq t\leq t\_{0}+1}|f(t,0,0)|}{\Gamma(\rho\_{1}-\rho\_{2})} \int\_{0}^{t\_{0}} ((t\_{0}-s)^{\rho\_{1}-\rho\_{2}-1}-(t-s)^{\rho\_{1}-\rho\_{2}-1})ds \\ & + \frac{\max\_{0\leq t\leq t\_{0}+1}|f(t,0,0)|}{\Gamma(\rho\_{1}-\rho\_{2})} \int\_{t\_{0}}^{t} (t-s)^{\rho\_{1}-\rho\_{2}-1}ds. \end{split}$$

We will consider the four terms above, respectively. For 0 < *η* < *ρ*<sup>1</sup> − *ρ*2, it is easy to show that

$$\int\_0^t (t-s)^{\rho\_1-\rho\_2-1} s^{\eta} ds = \frac{\Gamma(1+\eta)\Gamma(\rho\_1-\rho\_2)t^{\rho\_1-\rho\_2+\eta}}{\Gamma(1+\rho\_1-\rho\_2+\eta)}.$$

Hence, for any given *ε* > 0, there exists a *δ*<sup>1</sup> > 0, such that, when 0 ≤ *t*<sup>0</sup> ≤ *δ*1, it holds that

$$\frac{c\_1 \|y\|\_E}{\Gamma(\rho\_1 - \rho\_2)\Gamma(1 + \rho\_2)} \int\_0^{t\_0} (t\_0 - s)^{\rho\_1 - \rho\_2 - 1} s^{\rho\_2} ds < \frac{\varepsilon}{4'} \frac{c\_2 \|y\|\_E}{\Gamma(\rho\_2 - \rho\_2)} \int\_0^{t\_0} (t - s)^{\rho\_1 - \rho\_2 - 1} ds < \frac{\varepsilon}{4}.\tag{42}$$

Moreover, we get

$$\begin{aligned} &\int\_{\delta\_1}^{t\_0} ((t\_0 - s)^{\rho\_1 - \rho\_2 - 1} - (t - s)^{\rho\_1 - \rho\_2 - 1}) s^{\rho\_2} ds \\ &\leq \qquad \qquad t\_0^{\rho\_2} \int\_{\delta\_1}^{t\_0} ((t\_0 - s)^{\rho\_1 - \rho\_2 - 1} - (t - s)^{\rho\_1 - \rho\_2 - 1}) ds \\ &= \qquad \qquad \qquad \qquad \frac{t\_0^{\rho\_2}}{\rho\_1 - \rho\_2} ((t\_0 - \delta\_1)^{\rho\_1 - \rho\_2} - (t - \delta\_1)^{\rho\_1 - \rho\_2} + (t - t\_0)^{\rho\_1 - \rho\_2}) \\ &\leq \qquad \qquad \qquad \frac{t\_0^{\rho\_2}}{\rho\_1 - \rho\_2} (t - t\_0)^{\rho\_1 - \rho\_2} ,\end{aligned}$$

$$\int\_{\delta\_1}^{t\_0} ((t\_0 - s)^{\rho\_1 - \rho\_2 - 1} - (t - s)^{\rho\_1 - \rho\_2 - 1}) ds \le \frac{1}{\rho\_1 - \rho\_2} (t - t\_0)^{\rho\_1 - \rho\_2},$$

hence, we know that there exists *δ*<sup>2</sup> > 0 such that for 0 < *t* − *t*<sup>0</sup> < *δ*2, we have

$$\frac{c\_1 \|\|y\|\|\_E}{\Gamma(\rho\_1 - \rho\_2)\Gamma(1 + \rho\_2)} \int\_{\delta\_1}^{t\_0} ((t\_0 - s)^{\rho\_1 - \rho\_2 - 1} - (t - s)^{\rho\_1 - \rho\_2 - 1}) s^{\rho\_2} ds < \frac{\varepsilon}{4},$$

$$\frac{c\_2 \|\|y\|\|\_E}{\Gamma(\rho\_2 - \rho\_2)} \int\_{\delta\_1}^{t\_0} ((t\_0 - s)^{\rho\_1 - \rho\_2 - 1} - (t - s)^{\rho\_1 - \rho\_2 - 1}) ds < \frac{\varepsilon}{4},$$

together with (42), it leads to

$$\int\_{0}^{t\_0} ((t\_0 - s)^{\rho\_1 - \rho\_2 - 1} - (t - s)^{\rho\_1 - \rho\_2 - 1}) (\frac{c\_1 \|y\|\_E \mathfrak{s}^{\rho\_2}}{\Gamma(\rho\_1 - \rho\_2)\Gamma(1 + \rho\_2)} + \frac{c\_2 \|y\|\_E}{\Gamma(\rho\_2 - \rho\_2)}) ds < \varepsilon.$$

By the direct calculation, we have

$$\int\_{t\_0}^t (t-s)^{\rho\_1-\rho\_2-1} s^{\rho\_2} ds \le (t\_0+1)^{\rho\_2} \frac{(t-t\_0)^{\rho\_1-\rho\_2}}{\rho\_1-\rho\_2}.$$


$$\int\_{t\_0}^{t} (t - s)^{\rho\_1 - \rho\_2 - 1} ds \le \frac{(t - t\_0)^{\rho\_1 - \rho\_2}}{\rho\_1 - \rho\_2}.$$

which implies that there exists *δ*<sup>3</sup> > 0 such that for 0 < *t* − *t*<sup>0</sup> < *δ*3, we get

$$\int\_{t\_0}^t (t-s)^{\rho\_1-\rho\_2-1} (\frac{c\_1 \|\underline{y}\|\\_{s} \ell^{\rho\_2}}{\Gamma(\rho\_1-\rho\_2)\Gamma(1+\rho\_2)} + \frac{c\_2 \|\underline{y}\|\\_{E}}{\Gamma(\rho\_2-\rho\_2)}) ds < \varepsilon.$$

By the same arguments, we get that these estimations still hold for the last two terms above. Hence, we obtain *Fx*(*t*) is continuous on point *t*0. In view of the arbitrariness of *t*0, we have *Fx* ∈ *C*(0, +∞).

For the case of *t*<sup>0</sup> = 0, by (*A*2), for *y* ∈ *E*, take *t* < 1, then

$$|Fy(t)| = \qquad |\frac{1}{\Gamma(\rho\_1 - \rho\_2)} \int\_0^t (t - s)^{\rho\_1 - \rho\_2 - 1} f(s, l\_{0+}^{\rho\_2} y(s), y(s)) ds|$$

$$\leq \qquad \frac{||y||\_E}{\Gamma(\rho\_1 - \rho\_2)} \int\_0^t (t - s)^{\rho\_1 - \rho\_2 - 1} ((c\_1 \frac{s^{\rho\_2}}{\Gamma(1 + \rho\_2)} + c\_2) ds|)$$

$$+ \frac{\max\_{0 \leq t \leq 1} |f(t, 0, 0)|}{\Gamma(\rho\_1 - \rho\_2)} \int\_0^t (t - s)^{\rho\_1 - \rho\_2 - 1} ds,$$

From the previous arguments, we could know that *Fy*(*t*) is continuous on point 0. As a result, we have *Fy* ∈ *C*[0, +∞) for *x* ∈ *E*.

By the similar arguments, for *y* ∈ *E*, by (*A*2), we have

$$\begin{split} |\frac{\partial y(t)}{1+t^{\lambda}}| &\leq \qquad \frac{1}{\Gamma(\rho\_{1}-\rho\_{2})(1+t^{\lambda})} \int\_{0}^{t} (t-s)^{\rho\_{1}-\rho\_{2}-1} (c\_{1}\frac{|\boldsymbol{I}\_{0+}^{\rho\_{1}}y(s)|}{1+s^{\lambda}} + c\_{2}\frac{|y(s)|}{1+s^{\lambda}}) ds \\ &+ \frac{1}{\Gamma(\rho\_{1}-\rho\_{2})(1+t^{\lambda})} \int\_{0}^{t} (t-s)^{\rho\_{1}-\rho\_{2}-1} |f(s,0,0)| ds \\ &\leq \qquad \frac{\|y\|\_{\mathbb{E}}}{\Gamma(\rho\_{1}-\rho\_{2})(1+t^{\lambda})} \int\_{0}^{t} (t-s)^{\rho\_{1}-\rho\_{2}-1} (c\_{1}\frac{s^{\rho\_{2}}}{\Gamma(1+\rho\_{2})} + c\_{2}) ds \\ &+ \frac{1}{\Gamma(\rho\_{1}-\rho\_{2})(1+t^{\lambda})} \int\_{0}^{t} (t-s)^{\rho\_{1}-\rho\_{2}-1} |f(s,0,0)| ds \\ &= \qquad \frac{\|y\|\_{\mathbb{E}}}{1+t^{\lambda}} \left[ \frac{c\_{1}t^{\rho\_{1}}}{\Gamma(1+\rho\_{1})} + \frac{c\_{2}t^{\rho\_{1}-\rho\_{2}}}{\Gamma(1+\rho\_{1}-\rho\_{2})} \right] \\ &+ \frac{1}{\Gamma(\rho\_{1}-\rho\_{2})(1+t^{\lambda})} \int\_{0}^{t} (t-s)^{\rho\_{1}-\rho\_{2}-1} |f(s,0,0)| ds, \end{split}$$

according to these estimations and (*A*2), we ge that lim*t*→+<sup>∞</sup> *Fy*(*t*) <sup>1</sup>+*t<sup>λ</sup>* = 0. Hence, *<sup>F</sup>* : *<sup>E</sup>* → *<sup>E</sup>* is well defined. Now, for *x*, *y* ∈ *E*, by a similar way, we get

$$\begin{aligned} &\quad \frac{|Fx(t) - Fy(t)|}{1 + t^{\lambda}} \\ \leq & \qquad \frac{1}{\Gamma(\rho\_1 - \rho\_2)} \int\_0^t (t - s)^{\rho\_1 - \rho\_2 - 1} (c\_1 \frac{t^{\rho\_2} |x(s) - y(s)|}{1 + s^{\lambda}} + c\_2 \frac{|x(s) - y(s)|}{1 + s^{\lambda}}) ds \\ \leq & \qquad \frac{||\!| x - y||\_E}{1 + t^{\lambda}} [\frac{c\_1 t^{\rho\_1}}{\Gamma(1 + \rho\_1)} + \frac{c\_2 t^{\rho\_1 - \rho\_2}}{\Gamma(1 + \rho\_1 - \rho\_2)}] \\ \leq & \qquad [\frac{c\_1}{\Gamma(1 + \rho\_1)} + \frac{c\_2}{\Gamma(1 + \rho\_1 - \rho\_2)}] ||x - y||\_{E\_\rho} \end{aligned}$$

which implies that the operator *F* : *E* → *E* is a contraction operator, so the Banach contraction principle assures that the operator *F* has a unique fixed point *yT*(*t*) ∈ *E*. According to some arguments above, we obtain that *xT*(*t*) = *I ρ*2 <sup>0</sup>+*yT*(*t*) is one unique solution of the Equation (32) with zero initial value condition. Thus, according to Definition 5, we obtain that the problem (1) has one unique approximate solution.

**Example 4.** *Now, we consider the initial value problem as following*

$$\begin{cases} D\_{0+}^{\frac{1}{2}+\frac{t}{200(1+t^2)}}x(t) = \frac{\Gamma(\frac{3}{2})x^4}{12(1+t^2)^4(1+x^4)} + \frac{\Gamma(\frac{7}{60})(D\_{0+}^{\frac{3}{2}+\frac{t}{600(1+t^2+t^3)}}x)^2}{12(1+t^2)^2(1+(D\_{0+}^{\frac{3}{2}+\frac{t}{600(1+t^2+t^3)}}x)^2)}, 0 < t < +\infty, \\ x(0) = 0. \end{cases} \tag{43}$$

*We let*

$$p(t) = \frac{1}{2} + \frac{t}{200(1+t^2)},\\q(t) = \frac{1}{3} + \frac{t}{600(1+t^2+t^3)},\\0 \le t < +\infty,$$

$$f(t, \mathbf{x}(t), y(t)) = \frac{\Gamma(\frac{3}{2})\mathbf{x}^4(t)}{12(1+t^2)^4(1+\mathbf{x}^4(t))} + \frac{\Gamma(\frac{7}{6})y^2(t)}{12(1+t^2)^2(1+y^2(t))},\\0 < t < +\infty,\\x(t), y(t) \in \mathbb{R}.$$

*Obviously, we get* lim*t*→+<sup>∞</sup> *<sup>p</sup>*(*t*) = <sup>1</sup> <sup>2</sup> *and* lim*t*→+<sup>∞</sup> *<sup>q</sup>*(*t*) = <sup>1</sup> <sup>3</sup> *, thus, <sup>p</sup> satisfies* (*A*1) *with <sup>ρ</sup>*<sup>1</sup> <sup>=</sup> <sup>1</sup> 2 *, ρ*<sup>2</sup> = <sup>1</sup> <sup>3</sup> *. That f*(*t*, 0, 0) = 0*. In addition, for all* 0 ≤ *t* < +∞, *x*(*t*), *y*(*t*) ∈ *R, from the differentiation mean theorem, we get*

$$|f(t,(1+t^2)\mathbf{x}\_{1\prime}(1+t^2)\mathbf{y}\_1) - f(t,(1+t^2)\mathbf{x}\_{2\prime}(1+t^2)\mathbf{y}\_2)|$$

$$\begin{aligned} &\leq \qquad \frac{\Gamma(\frac{3}{2})}{12} | \frac{\mathbf{x}\_1^4}{1 + (1 + t^2)^4 \mathbf{x}\_1^4} - \frac{\mathbf{x}\_2^4}{1 + (1 + t^2)^4 \mathbf{x}\_2^4}| \\ &\qquad + \frac{\Gamma(\frac{7}{6})}{12} | \frac{\mathbf{y}\_1^2(t)}{1 + (1 + t^2)^2 \mathbf{y}\_1^2} - \frac{\mathbf{y}\_2^2}{1 + (1 + t^2)^2 \mathbf{y}\_2^2}| \\ &\leq \qquad \frac{\Gamma(\frac{3}{2})}{3} | \mathbf{x}\_1 - \mathbf{x}\_2| + \frac{\Gamma(\frac{7}{6})}{3} | \mathbf{y}\_1 - \mathbf{y}\_1| \end{aligned}$$

*which implies that f satisfies* (*A*2) *with c*<sup>1</sup> <sup>=</sup> <sup>Γ</sup>( <sup>3</sup> 2 ) <sup>3</sup> *, c*<sup>2</sup> <sup>=</sup> <sup>Γ</sup>( <sup>7</sup> 6 ) <sup>3</sup> *, which satisfies*

$$\frac{c\_1}{\Gamma(1+\rho\_1)} + \frac{c\_2}{\Gamma(1+\rho\_1-\rho\_2)}$$

$$=\qquad\frac{\Gamma(\frac{2}{2})}{3}\frac{1}{\Gamma(1+\frac{1}{2})} + \frac{\Gamma(\frac{7}{6})}{3}\frac{1}{\Gamma(1+\frac{1}{2}-\frac{1}{3})}$$

$$=\qquad\frac{2}{3} < 1.$$

*For given arbitrary small ε* = 1.1 <sup>100</sup> *, there exists T* <sup>=</sup> <sup>22</sup> *<sup>ε</sup>* = 2000*, such that*

$$|p(t) - \frac{1}{2}| = \frac{t}{200(1 + t^2)} < \frac{1}{t} \le \frac{1}{T} = \frac{\varepsilon}{22} < \varepsilon, t \ge T,$$

$$|q(t) - \frac{1}{3}| = \frac{t}{600(1 + t^2 + t^3)} < \frac{1}{t} \le \frac{1}{T} = \frac{\varepsilon}{22} < \varepsilon, t \ge T.$$

*Now, we consider function p*(*t*) *restricted on interval* [0, *T*]=[0, 2000]*. By the right continuity of function p*(*t*) *at point* 0*, for ε* = 1.1 <sup>100</sup> *, taking δ*<sup>0</sup> = 2*, when* 0 ≤ *t* ≤ *δ*<sup>0</sup> = 2*, we have*

$$|p(t) - p(0)| = |\frac{t}{200(1 + t^2)}| \le \frac{t}{200} < \frac{\delta\_0}{200} = \frac{1}{100} < \frac{1.1}{100} = \varepsilon.$$

$$|q(t) - q(0)| = |\frac{t}{600(1 + t^2 + t^3)}| \le \frac{t}{600} < \frac{\delta\_0}{200} = \frac{1}{100} < \frac{1.1}{100} = \varepsilon.$$

*We get t*<sup>1</sup> = *δ*<sup>0</sup> = 2*. By the right continuity of functions p*(*t*)*, q*(*t*) *at the point t*1*, for ε* = 1.1 <sup>100</sup> *, taking δ*<sup>1</sup> = 2*, when* 0 ≤ *t* − *t*<sup>1</sup> ≤ *δ*1*, by differential mean value theorem, we have*

$$|p(t) - p(t\_1)| = \qquad |\frac{t}{200(1 + t^2)} - \frac{t\_1}{200(1 + t\_1^2)}|$$

$$\leq \qquad |\frac{1 - \xi^2}{200(1 + \xi^2)^2}||t - t\_1||$$

$$\leq \qquad \frac{1 + \xi^2}{200(1 + \xi^2)^2}|t - t\_1|$$

$$\leq \qquad \frac{1}{200}|t - t\_1|$$

$$ < \qquad \frac{\delta\_1}{200} = \frac{1}{100} < \frac{1.1}{100} = \varepsilon\_r$$

$$|q(t) - q(t\_1)| = \qquad |\frac{t}{600(1 + t^2 + t^3)} - \frac{t\_1}{600(1 + t\_1^2 + t\_1^3)}|$$

$$\leq \qquad |\frac{1 - \eta^2 - 2\eta^3}{600(1 + \eta^2 + \eta^3)^2}| |t - t\_1|$$

$$\leq \qquad |\frac{1 + \eta^2 + 2\eta^3}{600(1 + \eta^2 + \eta^3)^2}| |t - t\_1|$$

$$\leq \qquad \frac{3}{600}|t - t\_1|$$

$$\leq \qquad \frac{\delta\_1}{200} = \frac{1}{100} < \frac{1.1}{100} = \varepsilon\_r$$

*where t*<sup>1</sup> < *ξ* < *t, t*<sup>1</sup> < *η* < *t. We let t*<sup>2</sup> = *t*<sup>1</sup> + *δ*<sup>1</sup> = 4*. By the right continuity of function p*(*t*) *at point t*2*, for ε* = 1.1 <sup>100</sup> *, taking δ*<sup>2</sup> = 2*, when* 0 ≤ *t* − *t*<sup>1</sup> ≤ *δ*2*, by the same reasons above, we have*

$$|p(t) - p(t\_1)| = \qquad |\frac{t}{200(1 + t^2)} - \frac{t\_2}{20(1 + t\_2^2)}| < \frac{\delta\_2}{200} = \frac{1}{100} < \frac{1.1}{100} = \varepsilon\_r$$

$$|q(t) - q(t\_1)| = \qquad |\frac{t}{600(1 + t^2 + t^3)} - \frac{t\_2}{600(1 + t\_2^2 + t^3)}| < \frac{\delta\_2}{200} = \frac{1}{100} < \frac{1.1}{100} = \varepsilon\_r$$

*Continuing this procession, from tn*−<sup>1</sup> = 2(*n* − 1) < 2000, *tn* = *tn*−<sup>1</sup> + *δn*−<sup>1</sup> = 2(*n* − 1) + 2 = 2000*, we get n* = 1000*. Thus, let*

$$p\_1 \doteq p(0) = \frac{1}{2}, \\ p\_2 \doteq p(t\_1) = p(2) = \frac{1}{2} + \frac{2}{200(1+4)}\sqrt{2}$$

$$p\_3 \doteq p(t\_2) = p(4) = \frac{1}{2} + \frac{4}{200 \times (1+16)}, \dots, p\_{1000} = p(t\_{999}) = p(1998) = \frac{1}{2} + \frac{1998}{200 \times (1+1998^2)},$$

$$q\_1 \doteq q(0) = \frac{1}{3}, q\_2 \doteq q(t\_1) = q(2) = \frac{1}{3} + \frac{2}{600(1+4+8)},$$

$$q\_3 \doteq q(t\_2) = q(4) = \frac{1}{3} + \frac{4}{600 \times (1+16+64)}, \dots,$$

$$q\_{1000} = q(t\_{999}) = q(1998) = \frac{1}{3} + \frac{1998}{600 \times (1+1998^2+1998^3)}.$$

*As a result, we get intervals* [0, 2],(2, 4], ··· ,(1998, 2000],(2000, +∞) *and function α*(*t*) *defined by*

$$x(t) = \begin{cases} p\_1 = \frac{1}{2}, & \text{for } t \in [0, 2], \\\\ p\_2 = \frac{1}{2} + \frac{2}{200 \times (1 + 4)}, & \text{for } t \in (2, 4], \\\\ p\_3 = \frac{1}{2} + \frac{4}{200 \times (1 + 16)}, & \text{for } t \in (4, 6], \\\\ \dots, & \\\\ p\_{1000} = \frac{1}{2} + \frac{198}{200 \times (1 + 1988^2)}, & \text{for } t \in (1998, 2000] \\\\ p\_1 = \frac{1}{2}, & \text{for } t \in (2000, +\infty). \end{cases} \tag{44}$$

$$\beta(t) = \begin{cases} q\_1 = \frac{1}{5}, & \text{for } t \in [0, 2], \\\\ q\_2 = \frac{1}{5} + \frac{4}{600 \times (1 + 4 + 8)}, & \text{for } t \in (2, 4], \\\\ q\_3 = \frac{1}{5} + \frac{4}{600 \times (1 + 16 + 64)}, & \text{for } t \in (4, 6], \\\\ \dots, & \\\\ q\_{1000} = \frac{1}{5} + \frac{1988}{6000 \times (1 + 1988 + 1988^2)}, & \text{for } t \in (1998, 2000] \end{cases} \tag{45}$$

*By Definitions 4 and 5 and the arguments of Theorem 1, the problem (43) has one unique approximate solution.*

**Remark 1.** *From Lemma 6 and Definition 5, we may take arbitrary small ε, such that the problem (43) has one unique approximate solution. This means that the proximity is very high.*

**Example 5.** *Finally, we calculate the approximate solution of the following initial value problem for linear equation*

$$D\_{0+}^{\frac{1}{2} + \frac{t}{200(1+t^2)}}\mathbf{x}(t) = t^{\frac{1}{4}}, \mathbf{x}(0) = 0, 0 < t < +\infty,\tag{45}$$

*According to analysis in Example 4, we get intervals* [0, 2],(2, 4], ··· ,(1998, 2000]*,* (2000, +∞) *and function α*(*t*) *defined in (44). By Definitions 4 and 5, we calculate out the approximate solution of the problem (45) as following*

$$\begin{cases} \begin{aligned} \mathbf{x}\_{1}(t) &= \frac{\Gamma(\frac{5}{4})}{\Gamma(\frac{4}{4})} t^{\frac{3}{4}} \in \mathbb{C}[0,2], \\\\ \mathbf{x}\_{2}(t) &= \frac{\Gamma(\frac{5}{4})}{\Gamma(\frac{4}{4} + \frac{200\times(1+4)}{200\times(1+4)})} t^{\frac{3}{4} + \frac{200\times(1+4)}{200\times(1+4)}} \in \mathbb{C}[0,4], \\\\ \mathbf{x}\_{3}(t) &= \frac{\Gamma(\frac{5}{4})}{\Gamma(\frac{4}{4} + \frac{4}{200\times(1+16)})} t^{\frac{3}{4} + \frac{4}{200\times(1+16)}} \in \mathbb{C}[0,6], \\\\ \vdots \\\\ \mathbf{x}\_{1000} &= \frac{\Gamma(\frac{5}{4})}{\Gamma(\frac{4}{4} + \frac{1008}{200\times(1+1998)})} t^{\frac{3}{4} + \frac{1998}{200\times(1+1998^{2})}} \in \mathbb{C}[0,2000], \\\\ \mathbf{x}\_{2000}(t) &= \frac{\Gamma(\frac{5}{4})}{\Gamma(\frac{4}{4})} t^{\frac{3}{4}} \in \mathbb{C}[0,+\infty). \end{aligned} \end{cases}$$

**Remark 2.** *By the characters of variable order derivative, we cannot get accurate solution of the problem (45). Hence, the approximate solution given by us is significative.*
