In the natural, many physical problems can be solved using the fractional order derivative. For example, in many complicated seepage flow, heat conduction phenomena and so on, they can be solved using the idea of fractional order derivative [1]. Fractional order derivative possesses its broad practice sense. However, solving the fractional order differential equation is very difficult. In this paper, we obtain asymptotic solution for the fractional order differential equation using the singularly perturbed theory, and get its valid estimation using the theory of differential inequalities.

The nonlinear singularly perturbed problem was a very attractive object in the academic circles [2]. During the past decade, many asymptotic methods were developed, including the boundary layer method, the methods of matched asymptotic expansion, the method of averaging and multiple scales. Recently, many scholars such as Hovhannisyan and Vulanovic [3], Abid, Jieli and Trabelsi [4], Graef and Kong [5], Guarguaglini and Natalini [6] and Barbu and Cosma [7] did a great deal of work. Using the method of singular perturbation and others Mo et al. studied also a class of nonlinear boundary value problems for the reaction diffusion equations, a class of activator inhibitor system, the shock wave, the soliton, the laser pulse and the problems of atmospheric physics and so on [8-19]. In this paper, we constructed asymptotic solution for the fractional order differential equation, and proved it's uniformly valid.

The $\alpha$-th order fractional order derivative $D^{\alpha}_{x}u$ of $u(x)$ is defined by

$ D^{\alpha}_{x}u=\frac{1}{\Gamma(1-\alpha)}\frac{d}{dx}\int^{x}_{0}(x-t)^{-\alpha}u(t)dt, $

where $\Gamma$ is the Gamma function, $\alpha$ is a positive fraction less than $1$.

Consider the following singularly perturbed initial value problem,

$ \varepsilon D^{\alpha}_{x}D^{\alpha}_{x}u+a(x)\frac{du}{dx}=f(x, u, \varepsilon), \ \ x\geq 0, $

(1)

$ u(0, \varepsilon)=A(\varepsilon), $

(2)

$ \varepsilon \frac{du}{dx}(0, \varepsilon)=B(\varepsilon), $

(3)

where $\varepsilon$ is a positive small parameters.

We need the following hypotheses:

${\rm [H_{1}]}$$\alpha>0$, $f(x, u, \varepsilon), A(\varepsilon)$ and $B(\varepsilon)$ are sufficiently smooth with respect to their arguments in corresponding domains;

${\rm [H_{2}]}$$f_{\varepsilon}>0, f_{u}\leq -c<0$, where $c$ is a constant.

From the hypotheses, there is a solution $U_{0}(x)$ of the reduced equation for original problem (1)-(3):

$ a(x)\frac{dU_{0}}{dx}=f(x, U_{0}, 0), \ \ x\geq 0, $

(4)

$ U_{0}(0)=A(0). $

(5)

We construct the outer solution $U$ of the original problem (1)-(3). Set

$ U\left( {x,\varepsilon } \right) \sim \sum\limits_{i = 0}^\infty {{U_i}\left( x \right){\varepsilon ^i}.} $

(6)

Substituting (6) into eq. (1), developing the nonlinear term $f(x, U, \varepsilon)$ in $\varepsilon$, and equating coefficients of the same powers of $\varepsilon$ both sides, from the solution $U_{0}(x)$ of the reduced equations (4), (5), we obtain that

$ a(x)\frac{dU_{i}}{dx}=F_{i}-D^{\alpha}_{x}D^{\alpha}_{x}U_{i-1}, \ \ x\geq 0, \ \ i=1, 2, \cdots, $

(7)

$ U_{i}(0)=A_{i}, \ \ i=1, 2, \cdots, $

(8)

where $A_{i}=\frac{1}{i!}[\frac{\partial^{i}A}{\varepsilon^{i}}]_{\varepsilon=0}$ and $F_{i}\ (i=1, 2, \cdots)$ are determined functions, their constructions are omitted.

From the defined of the $\alpha$-th order fractional order derivative $D^{\alpha}_{x}u$, it is easy to see that eq. (7) can translated by the following Volterra integral equation:

$ \int^{x}_{0}(x-t)^{-\alpha}U_{i}(t)dt =\int^{x}_{0}\Gamma(1-\alpha)\frac{F_{i}(t)-D^{\alpha}_{t}D^{\alpha}_{t}U_{i-1}(t)}{a(t)}dt+D_{0}, $

(9)

where $D_{0}$ is an arbitrary constant, which is decided by condition (8). From Volterra integral equation (9) and condition (8), we can obtain the solution $U_{i}(x)\ (i=1, 2, \cdots)$, successively. Substituting $U_{i}(x)$ into eq. (6), then we can obtain the outer solution $U(x, \varepsilon)$ for the original problem (1)-(3). But it may not satisfy initial condition (3), so that we need to construct the initial layer corrective term $V$ near $x=0$.

Let the solution of the initial value problem (1)-(3) is the form

$ u\sim U(x, \varepsilon)+V(\tau, \varepsilon) $

(10)

with

$ V\left( {\tau ,\varepsilon } \right) \sim \sum\limits_{i = 0}^\infty {{V_\tau }{\varepsilon ^i}} , $

(11)

where $\tau=x/\varepsilon$ is a stretched variable [2].

Substituting (10) and (11) into eqs. (1)-(3), developing the nonlinear term $f, A(\varepsilon)$ and $B(\varepsilon)$ in $\varepsilon$ and equating coefficients of the same powers of $\varepsilon$ in both sides of the equations, we obtain

$ D^{\alpha}_{\tau}D^{\alpha}_{\tau}V_{i}+a(x)\frac{dV_{i}}{d\tau}=G_{i}, \ \ \tau\geq 0, \ i=0, 1, 2, \cdots, $

(12)

$ V_{i}|_{\tau=0}=0, \ \ i=0, 1, 2, \cdots, $

(13)

$ V'_{i}|_{\tau=0}=B_{i}-U'_{i}|_{\tau=0}=0, \ \ i=0, 1, 2, \cdots, $

(14)

where $B_{i}=\frac{1}{i!}[\frac{\partial^{i}B}{\varepsilon^{i}}]_{\varepsilon=0}$, and $G_{i}$ are determined functions which constructions are omitted too.

From the fractional order differential equation (12), we can obtain the following Volterra integral systems for $(V_{i}, Z_{i})\ (i=0, 1, 2, \cdots)$:

$ \int^{\tau}_{0}[(\tau-t)^{-\alpha}-\Gamma(1-\alpha)a(t)]Z_{i}(t)dt =\int^{\tau}_{0}\Gamma(1-\alpha)G_{i}(t)dt+D_{1}, $

(15)

$ \int^{\tau}_{0}\Gamma(\tau-t)V_{i}(t)dt\int^{\tau}_{0}\Gamma(1-\alpha)Z_{i}(t)dt+D_{2}, $

(16)

where $D_{i}\ (i=1, 2)$ are arbitrary constants, which are decided by conditions (13) and (14). From the linear Volterra integral system (15), (16) and conditions (13), (14), we can obtain the solutions $(V_{i}, Z_{i})\ (i=0, 1, 2, \cdots)$, successively.

Substituting $V_{i}\ (i=0, 1, 2, \cdots)$, into eq. (11), we can obtain the initial layer corrective term $V(\tau, \varepsilon)$ of the solution $u(x, \varepsilon)$ for the original problem (1)-(3).

From eqs. (6) and (11) and the above obtained $U_{i}$ and $V_{i}\ (i=0, 1, 2, \cdots)$, we obtain the asymptotic expansion of solution for the initial value problem of $2\alpha$-th order fractional order differential equations (1)-(3):

$ u \sim \sum\limits_{i = 0}^\infty {\left[ {{U_i} + {V_i}} \right]} {\varepsilon ^i},x \ge 0,0 < \varepsilon \ll 1. $

(17)

And we can prove inductively that $V_{i}(\tau)\ (i=0, 1, 2, \cdots)$ possess initial layer behavior near $x=0$.

$ V_{i}(\tau)=O(\exp(-k_{i}\tau))=O(\exp(-k_{i}x/\varepsilon)), \ \ 0<\varepsilon\ll 1, \ i=0, 1, 2, \cdots, $

(18)

where $k_{i}\ (i=0, 1, 2, \cdots)$ are positive constants.

Now we prove that the asymptotic solution obtained above is a uniformly valid asymptotic expansion in $\varepsilon$.

Definition  There are two smooth functions $\overline{u}$ and $\underline{u}$, if $\overline{u}\geq \underline{u}$ and they satisfy inequalities, respectively:

$ \varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\overline{u}+a(x)\frac{d\overline{u}}{dx} -f(x, \overline{u}, \varepsilon)\leq 0, \ \overline{u}(0, \varepsilon)\geq A(\varepsilon), \ \varepsilon\overline{u}'(0, \varepsilon)\geq B(\varepsilon), \ \ 0\leq x\leq X_{0}, \\ \varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\underline{u}+a(x)\frac{d\underline{u}}{dx} -f(x, \underline{u}, \varepsilon)\geq 0, \ \underline{u}(0, \varepsilon)\leq A(\varepsilon), \ \varepsilon\underline{u}'(0, \varepsilon)\leq B(\varepsilon), \ \ 0\leq x\leq X_{0}, $

where $X_{0}$ is a constant large enough, then we say that $\overline{u}$ and $\underline{u}$ are upper and lower solutions of the problem (1)-(3), respectively.

Theorem 1  Assume that ${\rm [H_{1}], [H_{2}]}$ hold. If $\overline{u}(x, \varepsilon)$ and $\underline{u}(x, \varepsilon)$ are upper and lower solutions of the initial value problem (1)-(3) for the singularly perturbed fractional order differential equation, respectively, then there is a solution $u(x, \varepsilon)$ of the initial value problem (1)-(3) such that

$ \underline{u}(x, \varepsilon)\leq u(x, \varepsilon)\leq \overline{u}(x, \varepsilon). $

Proof  We construct two function sequences which are decided by the following recurrence relations:

$ \varepsilon D^{\alpha}_{x}D^{\alpha}_{x}u_{n+1}+a(x)\frac{du_{n+1}}{dx}=f(x, u_{n}, \varepsilon), \ \ n=0, 1, 2, \cdots, $

(19)

$ u_{n+1}(0, \varepsilon)=A(\varepsilon)\ \ \varepsilon \frac{du_{n+1}}{dx}(0, \varepsilon)=B(\varepsilon), \ \ n=0, 1, 2, \cdots. $

(20)

Let $\overline{u}_{0}=\overline{u}$ and $\underline{u}_{0}=\underline{u}$ are initial iteration functions of eqs. (19) and (20), respectively, and we have $\overline{u}_{n}$ and $\underline{u}_{n}$, successively. Thus we obtain two function sequences $\{ \overline{u}_{n} \}$ and $\{ \underline{u}_{n} \}$. Now we consider their convergence.

Let $\overline{y}_{0}=\overline{u}_{0}-\overline{u}_{1}$, from hypothesis ${\rm [H_{2}]}$, we have

$ \quad\quad\varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\overline{y}_{0}+a(x)\overline{y}'_{0} =\varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\overline{u}_{0}+a(x)\overline{u}'_{0} -\varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\overline{u}_{1}-a(x)\overline{u}'_{1}\\ \leq\quad f(x, \overline{u}_{0}, 0)-f(x, \overline{u}_{0}, 0)=0, \\ \quad\quad\overline{y}_{0}|_{x=0}=\overline{u}_{0}|_{x=0}-\overline{u}_{1}|_{x=0}=0, \ \ \overline{y}'_{0}|_{x=0}=\overline{u}'_{0}|_{x=0}-\overline{u}'_{1}|_{x=0}=0. $

Thus from the extremum principle [2], we have $\overline{y}_{0}\geq 0$. That is $\overline{u}_{0}\geq\overline{u}_{1}$.

If $\overline{u}_{n}\geq \overline{u}_{n+1}$, set $\overline{y}_{0}=\overline{u}_{n}-\overline{u}_{n+1}$, then we have

$ \quad\quad\varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\overline{y}_{n}+a(x)\overline{y}'_{n} =\varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\overline{u}_{n}+a(x)\overline{u}'_{n} -\varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\overline{u}_{n+1}-a(x)\overline{u}'_{n+1}\\ \leq\quad f(x, \overline{u}_{(n-1)}, \varepsilon)-f(x, \overline{u}_{n}, \varepsilon)\leq 0, \\ \quad\quad\overline{y}_{n}|_{x=0}=\overline{u}_{n}|_{x=0}-\overline{u}_{n-1}|_{x=0}, \ \ \overline{y}'_{n}|_{x=0}=\overline{u}'_{n}|_{x=0}-\overline{u}'_{n-1}|_{x=0}). $

Thus $\overline{y}_{n}\geq 0, $ that is $\overline{u}_{n}\geq \overline{u}_{n+1}\ (n\geq 1)$.

From inductive method, we know $\overline{u}=\overline{u}_{0}\geq\overline{u}_{1}\geq\cdots\geq\overline{u}_{n}\geq \overline{u}_{n+1}\geq\cdots. $

Analogously, we have $\underline{u}=\underline{u}_{0}\leq\underline{u}_{1}\leq\cdots\leq\underline{u}_{n}\leq \underline{u}_{n+1}\leq\cdots $ and $\overline{u}_{n}\geq\underline{u}_{n}, \ \ n=0, 1, 2, \cdots. $

From the above, and the Arzela-Ascoli theorem, there is a solution $u(x, \varepsilon)$ of the initial value problem (1)-(3) such that $\underline{u}(x, \varepsilon)\leq u(x, \varepsilon)\leq\overline {u}(x, \varepsilon)$. The proof of Theorem 1 is completed.

Theorem 2  Under hypotheses ${\rm [H_{1}], [H_{2}]}$, there is a solution $u(x.\varepsilon)$ of the initial value problem (1)-(3) for the singularly perturbed fractional order differential equation, which possesses the following uniformly valid asymptotic expansion in $\varepsilon$ on $x\in [0, X_{0}]$.

$ u=\sum\limits_{i = 0}^m {}[U_{i}(x)+V_{i}(x/\varepsilon)]\varepsilon^{i}+O(\varepsilon^{m+1}), \ \ x\in [0, X_{0}], \ 0<\varepsilon\ll 1. $

(21)

Proof  First, we construct the auxiliary functions $\alpha(x, \varepsilon), \beta(x, \varepsilon)$:

$ \alpha(x, \varepsilon)=\sum\limits_{i = 0}^m {}[U_{i}(x)+V_{i}(x/\varepsilon)]\varepsilon^{i} -r\varepsilon^{m+1}\ \ x\in [0, X_{0}], $

(22)

$ \beta(x, \varepsilon)=\sum\limits_{i = 0}^m {}[U_{i}(x)+V_{i}(x/\varepsilon)]\varepsilon^{i} +r\varepsilon^{m+1}\ \ x\in [0, X_{0}], $

(23)

where $r$ is a positive constant large enough to be chosen below, $m$ is an arbitrary positive integer.

Obviously, we have

$ \alpha(x, \varepsilon)\leq\beta(x, \varepsilon), $

(24)

and for $\varepsilon$ small enough, there is a positive constant $\delta_{1}$ such that

$ \beta(0, \varepsilon)\quad=\quad\sum\limits_{i = 0}^m {}[U_{i}(0)+V_{i}(0)]\varepsilon^{i}+r\varepsilon^{m+1}, \\ \quad\quad\quad\quad\geq\quad A(\varepsilon)-\delta_{1}\varepsilon^{m+1}+r\varepsilon^{m+1} =A(\varepsilon)+(r-\delta_{1})\varepsilon^{m+1}. $

Thus selecting $r\geq\delta_{1}$, we have

$ \beta(0, \varepsilon)\geq A(\varepsilon). $

(25)

Analogously, we have

$ \alpha(0, \varepsilon)\leq A(\varepsilon), \ \ \varepsilon\alpha'(0, \varepsilon)\leq B(\varepsilon)\leq\varepsilon\beta'(0, \varepsilon). $

(26)

Now we prove that

$ \varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\alpha+a(x)\frac{d\alpha}{dx}-f(x, \alpha, \varepsilon)\geq 0, \ \ 0<x\leq X_{0}, $

(27)

$ \varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\beta+a(x)\frac{d\beta}{dx}-f(x, \beta, \varepsilon)\leq 0, \ \ 0<x\leq X_{0}. $

(28)

In fact, from the hypotheses and eq. (18), for $\varepsilon$ small enough, there is a positive constant $\delta_{2}$, such that

$ \quad\quad\quad\quad\varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\beta+a(x)\frac{d\beta}{dx}-f(x, \beta, \varepsilon)\\ \quad=\quad\varepsilon D^{\alpha}_{x}D^{\alpha}_{x}\sum\limits_{i = 0}^m {}[U_{i}(x)+V_{i}(x/\varepsilon)] \varepsilon^{i} +a(x)\sum\limits_{i = 0}^m {}[U'_{i}(x)+V'_{i}(x/\varepsilon)]\varepsilon^{i}\\ \quad\quad\quad-f(x, \sum\limits_{i = 0}^m {}[U_{i}(x)+V_{i}(x/\varepsilon)]\varepsilon^{i}+r\varepsilon^{m+1}, \varepsilon)\\ \quad\leq\quad-f(x, U_{0}, 0)-\sum\limits_{i = 0}^m {}[f_{u}(x, U_{0})U_{i}+F_{i}-D^{\alpha}_{x}D^{\alpha}_{x}U_{i-1} -a(x)U'_{i}]\varepsilon^{i}\\ \quad\quad\quad+\sum\limits_{i = 0}^m {}[D^{\alpha}_{\tau}D^{\alpha}_{\tau}V_{i}+a(x)V'_{i}-G_{i}]\varepsilon^{i} +[f(x, \sum\limits_{i = 0}^m {}[U_{i}(x)+V_{i}(x/\varepsilon)]\varepsilon^{i}, \varepsilon)\\ \quad\quad\quad-f(x, \sum\limits_{i = 0}^m {}[U_{i}(x)+V_{i}(x/\varepsilon)]\varepsilon^{i}+r\varepsilon^{m+1}, \varepsilon)] +\delta_{2}\varepsilon^{m+1} \leq(-cr+\delta_{2})\varepsilon^{m+1}. $

Selecting $r\geq\delta_{2}/c, $ then we have eq. (28). Analogously, we can prove eq. (27). From eqs. (24)-(27), $\alpha$ and $\beta$ are upper and lower solutions respectively. From Theorem 1, there is a solution $u(x, \varepsilon)$ of the initial value problem (1)-(3) such that $\alpha(x, \varepsilon)\leq u(x, \varepsilon)\leq\beta(x, \varepsilon)$. And from eqs. (22) and (23), we have relation (21). The proof of Theorem 2 is completed.