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A High-Order Scheme for Fractional Ordinary Differential Equations with the Caputo–Fabrizio Derivative

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In this paper, we consider numerical solutions of fractional ordinary differential equations with the Caputo–Fabrizio derivative, and construct and analyze a high-order time-stepping scheme for this equation. The proposed method makes use of quadratic interpolation function in sub-intervals, which allows to produce fourth-order convergence. A rigorous stability and convergence analysis of the proposed scheme is given. A series of numerical examples are presented to validate the theoretical claims. Traditionally a scheme having fourth-order convergence could only be obtained by using block-by-block technique. The advantage of our scheme is that the solution can be obtained step by step, which is cheaper than a block-by-block-based approach.


In the last decades the fractional calculus had a remarkable development as shown by many mathematical volumes dedicated to it. We can see for instance the monograph [26] and the references therein. For a general right-hand side function f, it is usually difficult to obtain the analytical solution to a fractional differential equation. Thus there is a need to develop numerical methods for that equations. Cao and Xu [9] considered the fractional ordinary differential equations, and gave a modified block-by-block method for approximation to the fractional-order time derivative. Yang et al. [33] devoted to applications of fractional multistep methods for the fractional diffusion-wave equation. Sun and Wu [30] and Lin and Xu [21] analyzed a finite difference schema for the time discretization of the time-fractional diffusion equation, and proved that the convergence in time is of \(2-\alpha\) order. Gao et al. [14] gave a \(3-\alpha\) order L1-2 formula to approximate the Caputo fractional derivative of order \(\alpha\). Huang et al. [18] proved that the convergence of this method is of order at least three. Some other related work includes fast solvers for time-fractional diffusion equations [2, 7, 8, 11, 20, 22, 25, 31, 34, 35] and special care for treating the starting time singularity of the solutions [17, 19, 29]. We would like also to mention some relevant work for similar problems such as numerical methods for the time-fractional coupled mKdV equation, fractional Fisher’s type equations [15, 27].

However, some issues have been raised for the somewhat cumbersome mathematical expression of fractional operators and the consequent complications in the solutions of the associated equations. Caputo and Fabrizio [10] introduced in 2015 a new definition of fractional derivative with a smooth kernel, which is called the Caputo–Fabrizio fractional derivative. The Caputo–Fabrizio fractional derivative can be useful to understand some phenomena, such as the thermal analysis of a second grade fluid [28], the electrochemical phenomena [16], and the anomalous diffusion [32]. Numerically, Firoozjaee et al. [13] used the Ritz approximation for the Caputo–Fabrizio fractional derivative. In [4,5,6, 12, 23, 24], the authors proposed some second-order finite difference schemes for the Caputo–Fabrizio fractional derivative. Very recently, Akman et al. [1] constructed a third-order finite difference scheme for this derivative. However, to the best of our knowledge, the convergence order of the existing schemes is no more than three.

Inspired by the idea in [9], the current paper aims at constructing and analyzing a higher order numerical method for fractional ordinary differential equations with the Caputo–Fabrizio derivative. The outline of this paper is as follows: in Sect. 2, we present some basic properties of the fractional ordinary differential equation under consideration. In Sect. 3, we describe the detailed construction of the high-order scheme for the Caputo–Fabrizio derivative. The error estimation and stability analysis are given in Sect. 4. We provide some numerical examples in Sect. 5 to support the theoretical results. Finally, some concluding remarks are given in the final section.

Problem and Basic Properties

We consider the following initial value problem: \(\alpha \in (0,1)\),

$$\begin{aligned} D^{\alpha }_t u(t)=f(t),\quad 0< t\le T, \end{aligned}$$

subject to the initial condition \(u(0)=u_0\). In (2.1), the operator \(D^{\alpha }_t\) is the Caputo–Fabrizio derivative, defined by

$$\begin{aligned} D^{\alpha }_tu(t)=\frac{M(\alpha )}{1-\alpha } \int ^t_0u'(\tau )\exp \left( -\alpha \frac{t-\tau }{1-\alpha }\right) {\text{d}}\tau , \end{aligned}$$

where \(M(\alpha )\) is a positive normalization function satisfying \(M(0)=M(1)=1\).

This fractional derivative was first introduced by Caputo–Fabrizio in [10] with the aim of replacing the singular kernel in the traditional Caputo derivative by a regular kernel. It was claimed that the new definition can better describe some class of material heterogeneities, which cannot be well described by classical local theories or by fractional models with a singular kernel. Precisely, the Caputo–Fabrizio derivative is obtained by changing the kernel \((t-\tau )^{-\alpha }\) in the Caputo derivative by the exponential function \(\exp (-\alpha (t-\tau )/(1-\alpha ))\) and \(1/\Gamma (1-\alpha )\) by \({M(\alpha )}/{(1-\alpha )}\). According to the new definition, it is readily seen that if u is a constant function, then \(D^{\alpha }_t u \equiv 0\) as in the Caputo derivative. Contrary to the traditional definition, the main difference of the new definition is that the new kernel has no singularity for \(t = \tau\).

It can be directly verified; see also [10], that the Laplace transform of the Caputo–Fabrizio derivative has the following expression:

$$\begin{aligned} \mathcal {L}[D^{\alpha }_t u(t)](s)=\frac{M(\alpha )(s\mathcal {L}[u(t)](s)-u(0))}{s+\alpha (1-s)}. \end{aligned}$$

This is one of the interesting properties of the Caputo–Fabrizio derivative since the Laplace transform of \(D^{\alpha }_t u(t)\) is linked to the Laplace transform of u(t) in a very simple way. Thus if u is a solution of (2.1), then it follows from the relationship (2.3) that

$$\begin{aligned} \mathcal {L}[u(t)](s) = {1\over s} u(0) + {\alpha \over sM(\alpha )} \mathcal {L}[f(t)](s) + {1-\alpha \over M(\alpha )} \mathcal {L}[f(t)](s). \end{aligned}$$

Using known properties of the inverse Laplace transform, we deduce that

$$\begin{aligned} u(t) = u(0) + {\alpha \over M(\alpha )} \int ^t_0 f(\tau ) {\text{d}}\tau + {1-\alpha \over M(\alpha )} f(t). \end{aligned}$$

It is readily seen from (2.4) that u(t) satisfies the initial condition \(u(0)=u_0\) if and only if \(f(0)=0\). In fact \(f(0)=0\) is a necessary condition for (2.1) to admit a \(C^1\) solution. This can be directly observed by taking limit of the both sides of (2.1) as \(t\rightarrow 0\).

Let us consider the case \(f\equiv 0\), i.e., \(D^{\alpha }_t u(t)=0\). It follows from (2.4) that \(u(t) = u(0)\) for all \(t\ge 0\). This, together with the definition (2.2), proves that \(D^{\alpha }_t u(t)=0\) if and only if u is a constant. This is a good property of this new derivative, which is shared by the fractional Caputo and classical first-order derivatives.

Let us now consider the case \(f= \lambda u\), where \(\lambda\) is a constant. Then (2.1) becomes an eigenvalue problem associated to the Caputo–Fabrizio differential operator \(D^{\alpha }_t\). For integer-order differential equations such a problem has often been served as a standard test case for investigating the stability of numerical methods. We will do the same for the Caputo–Fabrizio differential operator under consideration. Therefore it is interesting to see how the eigenfunctions behave. First we notice that the solution of (2.1) with \(f(t) = \lambda u(t)\) has the following expression:

$$\begin{aligned} u(t)=\frac{M(\alpha )u_0}{M(\alpha )-\lambda (1-\alpha )} \exp \left( \frac{\lambda \alpha t}{M(\alpha )-\lambda (1-\alpha )}\right) . \end{aligned}$$

In fact, by applying the Laplace transform to both sides of (2.1) and using (2.3), we obtain

$$\begin{aligned} \frac{M(\alpha )(s\mathcal {L}[u(t)](s)-u(0))}{s+\alpha (1-s)} = \lambda \mathcal {L}[u(t)](s). \end{aligned}$$

Rearranging it, we arrive at the following equality:

$$\begin{aligned} \mathcal {L}[u(t)](s) = \frac{M(\alpha )u_0}{[M(\alpha )-\lambda (1-\alpha )]s - \lambda \alpha }. \end{aligned}$$

Finally, taking the inverse Laplace transform on both sides of (2.6) gives (2.5). However, we immediately realize that the solution (2.5) does not satisfy the initial condition unless \(\lambda =0\) or \(u_0=0\). In the latter case, we have \(u\equiv 0\). This means that the eigenvalue problem

$$\begin{aligned} \left\{ \begin{array}{l} D^{\alpha }_t u (t) = \lambda u(t), \ t>0, \\ u(0) = u_0 \end{array} \right. \end{aligned}$$

is not well defined since either eigenvalue or eigenfunction has to be zero.

The above discussion motivates us to consider an alternative eigenvalue problem as follows:

$$\begin{aligned} \left\{ \begin{array}{l} D^{\alpha }_t u (t) = f(t), \ t>0, \\ u(0) = u_0, \end{array} \right. \end{aligned}$$


$$\begin{aligned} f(t) = \lambda \left[ u(t)-u_0\exp \left( -\frac{\alpha }{1-\alpha }t\right) \right] . \end{aligned}$$

We first notice that the function f takes value 0 at \(t=0\), which is a necessary condition for (2.7) to admit a non-trivial solution. In fact, it is not difficult to prove that the problem (2.7) admits the unique solution

$$\begin{aligned} u(t)=u_0\exp \left( \frac{\lambda \alpha t}{M(\alpha )-\lambda (1-\alpha )}\right) . \end{aligned}$$

A direct calculation with (2.9) shows that the solution is decreasing when \(u_0\lambda <0\). This property will be used to analyze the stability of the scheme to be constructed in the next section.

A Finite Difference Approximation to the Caputo–Fabrizio Derivative

In this section, we will construct and analyze an efficient numerical scheme for the problem (2.1). Particularly the scheme to be proposed will be tested for the problem (2.7) to see if the numerical solutions share same monotonic property as the exact solution (2.9).

First we propose a high-order approximation to the Caputo–Fabrizio derivative, and analyze its approximation property. Let us consider the following gird in [0, T]: \(t_j=jh,j=0,1,2,\cdots ,2N\), where N is a positive integer, and \(h=\frac{T}{2N}\) is the grid size. We use \(u_i\) to denote \(u(t_i),i=0,1,2,\cdots ,2N\).

The question is how to efficiently approximate \(D^{\alpha }_tu(t)\) at a given point t. Note that u(t) can be approximated in \([t_0,t_1]\) by using the quadratic interpolation as

$$\begin{aligned} u(t)\approx \varphi _{0,0}(t)u_0+\varphi _{1,0}(t)u_{1/2}+\varphi _{2,0}(t)u_1 \doteq I_{[t_0,t_1]}u(t) \end{aligned}$$

with \(u_{1/2}=u(t_{1/2}),t_{1/2}=t_0+\frac{1}{2}h\), and

$$\begin{aligned} \varphi _{0,0}(t)=\frac{2(t-t_{1/2})(t-t_1)}{h^2}, \varphi _{1,0}(t)=\frac{-4(t-t_{0})(t-t_1)}{h^2}, \varphi _{2,0}(t)=\frac{2(t-t_0)(t-t_{1/2})}{h^2}. \end{aligned}$$

Plugging (3.1) into (2.2), we obtain an approximation

$$\begin{aligned} D^{\alpha }_tu(t_1)= & {} \frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0u'(\tau ) \exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\\approx & {} \frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0[\varphi _{0,0}(\tau )u_0+\varphi _{1,0}(\tau )u_{1/2}+\varphi _{2,0}(\tau )u_1]' \exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\= & {} w_1^{0,0}u_0+w_1^{1,0}u_{1/2}+w_1^{2,0}u_1, \end{aligned}$$


$$\begin{aligned} w_1^{i,0}=\frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0\varphi _{i,0}'(\tau ) \exp \left( -\frac{\alpha }{1-\alpha }(t_1-\tau )\right) {\text{d}}\tau ,\ \ i=0,1,2, \end{aligned}$$

which can be exactly computed. The value of \(u_{1/2}\) will be obtained by the following interpolation:

$$\begin{aligned} u_{1/2}\approx \frac{3}{8}u_0+\frac{3}{4}u_1-\frac{1}{8}u_2. \end{aligned}$$

Inserting the above approximation into (3.2), we get

$$\begin{aligned} D^{\alpha }_tu(t_1)\approx a_1^{0,0}u_0+a_1^{1,0}u_{1}+a_1^{2,0}u_2\doteq D_h^{\alpha }u(t_1), \end{aligned}$$


$$\begin{aligned} a_1^{0,0}=w_1^{0,0}+\frac{3}{8}w_1^{1,0},\quad a_1^{1,0}=\frac{3}{4}w_1^{1,0}+w_1^{2,0},\quad a_1^{2,0}=-\frac{1}{8}w_1^{1,0}. \end{aligned}$$

Similarly, u(t) in \([t_0,t_2]\) can be approximated by

$$\begin{aligned} u(t)\approx \psi _{0,0}(t)u_0+\psi _{1,0}(t)u_{1}+\psi _{2,0}(t)u_2 \doteq I_{[t_0,t_2]}u(t), \end{aligned}$$

where \(\psi _{i,0}(t),i=0,1,2,\) are defined as follows:

$$\begin{aligned} \psi _{0,0}(t)=\frac{(t-t_{1})(t-t_2)}{2h^2}, \psi _{1,0}(t)=\frac{(t-t_{0})(t-t_2)}{-h^2}, \psi _{2,0}(t)=\frac{(t-t_0)(t-t_{1})}{2h^2}. \end{aligned}$$

Using this approximation yields

$$\begin{aligned} D^{\alpha }_tu(t_2)= & {} \frac{M(\alpha )}{1-\alpha }\int ^{t_2}_0u'(\tau )\exp \left( -\alpha \frac{t_2-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\\approx & {} \frac{M(\alpha )}{1-\alpha }\int ^{t_2}_0[\psi _{0,0}(\tau )u_0+\psi _{1,0}(\tau )u_{1}+\psi _{2,0}(\tau )u_2]' \exp \left( -\alpha \frac{t_2-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\= & {} a_2^{0,0}u_0+a_2^{1,0}u_{1}+a_2^{2,0}u_2 \doteq D_h^{\alpha }u(t_2), \end{aligned}$$


$$\begin{aligned} a_2^{i,0}=\frac{M(\alpha )}{1-\alpha }\int ^{t_2}_0\psi _{i,0}'(\tau )\exp \left( -\frac{\alpha }{1-\alpha }(t_2-\tau )\right) {\text{d}}\tau ,\ \ i=0,1,2. \end{aligned}$$

Now assuming that the values of u at the grid points \(t_j, j=0,1,\cdots ,2m,\) are already known, we want to derive an approximation to \(D^{\alpha }_tu(t_{2m+1})\) and \(D^{\alpha }_tu(t_{2m+2})\). Similar to the previous sub-intervals, we use the following quadratic interpolation functions to approximate u in \([t_{2k-1},t_{2k+1}],k=1,2,\cdots ,m\):

$$\begin{aligned} u(t)\approx \varphi _{0,k}(t)u_{2k-1}+\varphi _{1,k}(t)u_{2k}+\varphi _{2,k}(t)u_{2k+1} \doteq I_{[t_{2k-1},t_{2k+1}]}u(t), \end{aligned}$$

where \(\varphi _{0,k}(t)=\frac{(t-t_{2k})(t-t_{2k+1})}{2h^2}, \varphi _{1,k}(t)=\frac{(t-t_{2k-1})(t-t_{2k+1})}{-h^2}, \varphi _{2,k}(t)=\frac{(t-t_{2k-1})(t-t_{2k})}{2h^2}.\) This suggests the following approach:

$$\begin{aligned} \nonumber&D^{\alpha }_tu(t_{2m+1})\nonumber \\& =\frac{M(\alpha )}{1-\alpha }\int ^{t_{2m+1}}_0u'(\tau )\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\& =\frac{M(\alpha )}{1-\alpha }\left[ \int ^{t_{1}}_0u'(\tau )\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau +\sum _{k=1}^m\int ^{t_{2k+1}}_{t_{2k-1}}u'(\tau )\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right] \nonumber \\& \approx \frac{M(\alpha )}{1-\alpha }\int ^{t_{1}}_0[I_{[t_0,t_1]}u(\tau )]'\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\&\quad +\sum _{k=1}^m\frac{M(\alpha )}{1-\alpha }\int ^{t_{2k+1}}_{t_{2k-1}}[I_{[t_{2k-1},t_{2k+1}]}u(\tau )]'\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\& =a_{2m+1}^{0,0}u_0+a_{2m+1}^{1,0}u_{1}+a_{2m+1}^{2,0}u_2 +\sum _{k=1}^m[a_{2m+1}^{0,k}u_{2k-1}+a_{2m+1}^{1,k}u_{2k}+a_{2m+1}^{2,k}u_{2k+1}]\nonumber \\& \doteq D_h^{\alpha }u(t_{2m+1}) , \end{aligned}$$


$$\begin{aligned} \begin{array}{l} \displaystyle a_{2m+1}^{0,0}=w_{2m+1}^{0,0}+\frac{3}{8}w_{2m+1}^{1,0},\ \ a_{2m+1}^{1,0}=\frac{3}{4}w_{2m+1}^{1,0}+w_{2m+1}^{2,0},\ \ a_{2m+1}^{2,0}=-\frac{1}{8}w_{2m+1}^{1,0},\\ \displaystyle a_{2m+1}^{i,k}=\frac{M(\alpha )}{1-\alpha }\int ^{t_{2k+1}}_{t_{2k-1}}\varphi _{i,k}'(\tau )\exp \left( -\frac{\alpha }{1-\alpha }(t_{2m+1}-\tau )\right) {\text{d}}\tau , i=0,1,2;\ k=1,2,\cdots ,m \end{array} \end{aligned}$$


$$\begin{aligned} w_{2m+1}^{i,0}=\frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0\varphi _{i,0}'(\tau )\exp \left( -\frac{\alpha }{1-\alpha }(t_{2m+1}-\tau )\right) {\text{d}}\tau ,i=0,1,2. \end{aligned}$$

In the sub-intervals \([t_{2k},t_{2k+2}],k=0,1,\cdots ,m\), we approximate u by

$$\begin{aligned} u(t)\approx \psi _{0,k}(t)u_{2k}+\psi _{1,k}(t)u_{2k+1}+\psi _{2,k}(t)u_{2k+2} \doteq I_{[t_{2k},t_{2k+2}]}u(t), \end{aligned}$$

where \(\psi _{0,k}(t)=\frac{(t-t_{2k+1})(t-t_{2k+2})}{2h^2}, \psi _{1,k}(t)=\frac{(t-t_{2k})(t-t_{2k+2})}{-h^2}, \psi _{2,k}(t)=\frac{(t-t_{2k})(t-t_{2k+1})}{2h^2}.\)

As a consequence

$$\begin{aligned}&D^{\alpha }_tu(t_{2m+2})\nonumber \\& =\frac{M(\alpha )}{1-\alpha }\sum _{k=0}^m\int ^{t_{2k+2}}_{t_{2k}}u'(\tau )\exp \left( -\alpha \frac{t_{2m+2}-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\& \approx \frac{M(\alpha )}{1-\alpha }\sum _{k=0}^m\int ^{t_{2k+2}}_{t_{2k}}[I_{[t_{2k},t_{2k+2}]}u(\tau )]'\exp \left( -\alpha \frac{t_{2m+2}-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\& =\sum _{k=0}^m\left(a_{2m+2}^{0,k}u_{2k}+a_{2m+2}^{1,k}u_{2k+1}+a_{2m+2}^{2,k}u_{2k+2}\right) \doteq D_h^{\alpha }u(t_{2m+2}), \end{aligned}$$


$$\begin{aligned}a_{2m+2}^{i,k}=\left\{\begin{array}{l}\frac{M(\alpha )}{1-\alpha }\int ^{t_{2k+2}}_{t_{2k}}\psi _{i,k}'(\tau )\exp \left( -\frac{\alpha }{1-\alpha }(t_{2m+2}-\tau )\right) {\text{d}}\tau ,\\ i=0,1,2;\ k=0,1,\cdots ,m. \end{array}\right.\end{aligned}$$

We are in a position to construct our numerical scheme for the fractional differential equation (2.1) subject to the initial condition \(u_0\). Based on the finite difference operator \(D_h^{\alpha }\) defined in (3.3)–(3.6), we propose the following scheme:

$$\begin{aligned} D_h^{\alpha }u(t_k) = f(t_k), k=1,2,\cdots ,2N \end{aligned}$$

to numerically solve the problem (2.1).

Error Estimate and Stability Analysis

This section is devoted to carry out the stability and convergence analysis, and derive error estimates for the numerical solutions. We start with deriving an error estimate for the finite difference operator \(D^{\alpha }_h\).

Theorem 4.1

Assume\(u(t)\in C^4[0,T]\). Let

$$\begin{aligned} r_k := D^{\alpha }_tu(t_{k})-D^{\alpha }_hu(t_{k}),\ \ \ k=1,2,\cdots ,2N. \end{aligned}$$

Then it holds

$$\begin{aligned} |r_k|\le ch^4,\ \ \ k=1,2,\cdots ,2N, \end{aligned}$$

wherecis a constant independent ofh, but depends onu.


Applying the Taylor theorem, we have

$$\begin{aligned}&\left| u_{1/2}-\left( \frac{3}{8}u_0+\frac{3}{4}u_1-\frac{1}{8}u_2\right) \right| \le ch^3, \end{aligned}$$
$$\left\{\begin{array}{l}u(t)-I_{[t_0,t_1]}u(t)=\frac{u^{(3)}(\xi _0(t))}{6}(t-t_0)(t-t_{1/2})(t-t_1),\xi _0(t)\in (t_0,t_1),\forall t\in [t_0,t_1],\\u(t)-I_{[t_{2k-1},t_{2k+1}]}u(t)=\frac{u^{(3)}(\xi _k(t))}{6}(t-t_{2k-1})(t-t_{2k})(t-t_{2k+1}),\\ \xi _k(t)\in (t_{2k-1},t_{2k+1}),\forall t\in [t_{2k-1},t_{2k+1}],k=1,2,\cdots ,m,\\ u(t)-I_{[t_{2k},t_{2k+2}]}u(t)=\frac{u^{(3)}(\eta _k(t))}{6}(t-t_{2k})(t-t_{2k+1})(t-t_{2k+2}),\\ \eta _k(t)\in (t_{2k},t_{2k+2}),\forall t\in [t_{2k},t_{2k+2}],k=0,1,2,\cdots ,m. \end{array}\right.$$

We begin with estimating \(r_1\). It follows from (3.3), (4.3), and (4.4):

$$\begin{aligned}|r_1|&=| D^{\alpha }_tu(t_{1})-D^{\alpha }_hu(t_{1})|\\ & =\left| \frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0u'(\tau )\exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}\tau -(a_1^{0,0}u_0+a_1^{1,0}u_{1}+a_1^{2,0}u_2)\right| \\ & =\left| \frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0u'(\tau )\exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}\tau -\frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0[I_{[t_0,t_1]}u(\tau )]'\exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}\tau \right. \\&\quad \left. +\frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0\left[ u_{1/2}-\left( \frac{3}{8}u_0+\frac{3}{4}u_1-\frac{1}{8}u_2\right) \right] \varphi _{1,0}'(\tau )\exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \\& \le \frac{M(\alpha )}{1-\alpha }\left| \int ^{t_1}_0[u(\tau )-I_{[t_0,t_1]}u(\tau )]' \exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \\&\quad +\frac{M(\alpha )}{1-\alpha }\left| \int ^{t_1}_0\left[ u_{1/2}-\left( \frac{3}{8}u_0+\frac{3}{4}u_1-\frac{1}{8}u_2\right) \right] \varphi _{1,0}'(\tau )\exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \\& \le \frac{M(\alpha )}{1-\alpha }\left| \int ^{t_1}_0\exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}[u(\tau )-I_{[t_0,t_1]}u(\tau )]\right| \\&\quad +\frac{M(\alpha )}{1-\alpha }ch^3\left| \int ^{t_1}_0\exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) {\text{d}}\varphi _{1,0}(\tau )\right| \\& \le \frac{M(\alpha )}{1-\alpha }\left| \frac{u^{(3)}(\xi _0(\tau ))}{6} (\tau -t_0)(\tau -t_{1/2})(\tau -t_1)\exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) \right| ^{t_1}_{t_0}\\&\quad \left. -\int ^{t_1}_0\frac{u^{(3)}(\xi _0(\tau ))}{6} (\tau -t_0)(\tau -t_{1/2})(\tau -t_1){\text{d}}\left( \exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) \right) \right| \\&\quad +ch^3\frac{M(\alpha )}{1-\alpha }\left| \exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) |\varphi _{1,0}(\tau )|^{t_1}_{t_0} -\int ^{t_1}_0\varphi _{1,0}(\tau ){\text{d}}\left( \exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) \right) \right| \\& \le \frac{M(\alpha )}{1-\alpha }\left| \int ^{t_1}_0\frac{u^{(3)}(\xi _0(t))}{6} (\tau -t_0)(\tau -t_{1/2})(\tau -t_1){\text{d}}\left( \exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) \right) \right| \\&\quad +ch^3\frac{M(\alpha )}{1-\alpha } \left| \int ^{t_1}_0\varphi _{1,0}(\tau ){\text{d}}\left( \exp \left( -\alpha \frac{t_1-\tau }{1-\alpha }\right) \right) \right| \\& \le \frac{cM(\alpha )}{(1-\alpha )^2} \alpha h^4 \exp \left( -\alpha \frac{t_1-\eta }{1-\alpha }\right) +ch^4\frac{|\alpha M(\alpha )|}{(1-\alpha )^2} \exp \left( -\alpha \frac{t_1-\eta }{1-\alpha }\right)\\& \le ch^4, \eta \in (t_0,t_1). \end{aligned}$$

In the above derivation we have used the fact that \(\exp (-\alpha \frac{t_1-\eta }{1-\alpha })<1\). In a similar way we can prove

$$\begin{aligned} |r_2|\le ch^4. \end{aligned}$$

For \(r_{2m+1}\), we have

$$\begin{aligned}|r_{2m+1}|& =\left| \frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0[u(\tau )-I_{[t_0,t_1]}u(\tau )]{'} \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right. \nonumber \\&\quad +\frac{M(\alpha )}{1-\alpha }\int ^{t_1}_0\left[ u_{1/2}-\left( \frac{3}{8}u_0+\frac{3}{4}u_1-\frac{1}{8}u_2\right) \right] \varphi _{1,0}'(\tau )\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\&\quad \left. +\sum _{k=1}^m\frac{M(\alpha )}{1-\alpha }\int ^{t_{2k+1}}_{t_{2k-1}}[u(\tau )-I_{[t_{2k-1},t_{2k+1}]}u(\tau )]' \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \nonumber \\& \le \frac{M(\alpha )}{1-\alpha }\left| \int ^{t_1}_0[u(\tau )-I_{[t_0,t_1]}u(\tau )]{'} \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \nonumber \\&\quad +\frac{M(\alpha )}{1-\alpha }\left| \int ^{t_1}_0\left[ u_{1/2}-\left( \frac{3}{8}u_0+\frac{3}{4}u_1-\frac{1}{8}u_2\right) \right] \varphi _{1,0}'(\tau )\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \nonumber \\&\quad +\sum _{k=1}^m\frac{M(\alpha )}{1-\alpha }\left| \int ^{t_{2k+1}}_{t_{2k-1}}[u(\tau )-I_{[t_{2k-1},t_{2k+1}]}u(\tau )]' \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \nonumber \\& \le ch^4+\frac{M(\alpha )}{1-\alpha }\sum _{k=1}^m\left| \int ^{t_{2k+1}}_{t_{2k-1}} \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}[u(\tau )-I_{[t_{2k-1},t_{2k+1}]}u(\tau )]\right| \nonumber \\& =ch^4+\frac{M(\alpha )}{1-\alpha }\sum _{k=1}^m\left| \frac{u^{(3)}(\xi _k(\tau ))}{6}\prod _{i=0}^{2}(\tau -t_{2k-1+i}) \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) \right| _{t_{2k-1}}^{t_{2k+1}} \nonumber \\&\quad \left. -\int ^{t_{2k+1}}_{t_{2k-1}}\frac{u^{(3)}(\xi _k(\tau ))}{6}\prod _{i=0}^{2}(\tau -t_{2k-1+i}) \frac{\alpha }{1-\alpha }\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \nonumber \\& \le ch^4+\frac{\alpha M(\alpha )}{(1-\alpha )^2}\sum _{k=1}^m \left| \int ^{t_{2k+1}}_{t_{2k-1}} \frac{u^{(3)}(\xi _k(\tau ))}{6}\prod _{i=0}^{2}(\tau -t_{2k-1+i}) \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| . \end{aligned}$$

The second term in the right-hand side, which will be denoted by R hereafter, can be bounded by

$$\begin{aligned} R= & {} c_\alpha \sum _{k=1}^m\left|\int ^{t_{2k+1}}_{t_{2k-1}}\frac{u^{(3)}(\xi _k(\tau ))}{6} \prod _{i=0}^{2}(\tau -t_{2k-1+i}) \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right|\nonumber \\\le & {} c_\alpha \sum _{k=1}^m\left( \left| \int ^{t_{2k+1}}_{t_{2k-1}}\frac{u^{(3)}(\widetilde{\xi }_k)}{6}\prod _{i=0}^{2} (\tau -t_{2k-1+i}) \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \right. \nonumber \\&\left. +\left| \int ^{t_{2k+1}}_{t_{2k-1}}\frac{u^{(3)}(\xi _k(\tau ))-u^{(3)}(\widetilde{\xi }_k)}{6} \prod _{i=0}^{2}(\tau -t_{2k-1+i}) \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \right) \nonumber \\\doteq & {} R_1+R_2, \end{aligned}$$

where \(c_\alpha =\frac{\alpha M(\alpha )}{(1-\alpha )^2}\), \(\widetilde{\xi }_k=t_{2k}\), \(R_1\) can be controlled as follows:

$$\begin{aligned} |R_1|\le & {} c_\alpha B_1\sum _{k=1}^m \left| \int _{t_{2k-1}}^{t_{2k}}\prod _{i=0}^{2}(\tau -t_{2k-1+i}) \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right. \nonumber \\&\left. +\int _{t_{2k}}^{t_{2k+1}}\prod _{i=0}^{2}(\tau -t_{2k-1+i}) \exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \nonumber \\= & {} c_\alpha B_1\sum _{k=1}^m \left| \exp \left( -\alpha \frac{t_{2m+1}-s^k_1}{1-\alpha }\right) \int _{t_{2k-1}}^{t_{2k}}\prod _{i=0}^{2}(\tau -t_{2k-1+i}){\text{d}}\tau \right. \nonumber \\&\left. +\exp \left( -\alpha \frac{t_{2m+1}-s^k_2}{1-\alpha }\right) \int _{t_{2k}}^{t_{2k+1}}\prod _{i=0}^{2}(\tau -t_{2k-1+i}){\text{d}}\tau \right| \nonumber \\= & {} {1\over 4}c_\alpha B_1 h^4 \sum _{k=1}^m\left| \exp \left( -\alpha \frac{t_{2m+1}-s^k_1}{1-\alpha }\right) -\exp \left( -\alpha \frac{t_{2m+1}-s^k_2}{1-\alpha }\right) \right| \nonumber \\= & {} {1\over 4}c_\alpha B_1 h^4 \frac{\alpha }{1-\alpha }\sum _{k=1}^m\left| \int _{s^k_1}^{s^k_2}\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau )\right| \nonumber \\\le & {} c_\alpha B_1\frac{\alpha }{1-\alpha }h^4 \sum _{k=1}^m\left| \int _{t_{2k-1}}^{t_{2k+1}}\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \right| \le c h^4, \end{aligned}$$

where \(B_1\) is an upper bound of \(u^{(3)}, t_{2k-1}\le s_1^{k}\le t_{2k}\le s_2^{k}\le t_{2k+1}\).

Noticing that

$$\begin{aligned} |u^{(3)}(\xi _k(\tau ))-u^{(3)}(\widetilde{\xi }_k)| \le B_2 h, \forall \tau \in [t_{2k-1}, t_{2k+1}], \end{aligned}$$

where \(B_2\) depends on the upper bound of \(u^{(4)}\), we have

$$\begin{aligned} |R_2|\le & {} \frac{ c_\alpha B_2 h}{6}\sum _{k=1}^m\int _{t_{2k-1}}^{t_{2k+1}}\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) \left|\prod _{i=0}^{2}(\tau -t_{2k-1+i})\right|{\text{d}}\tau \nonumber \\\le & {} c_\alpha B_2 h^4\sum _{k=1}^m\int _{t_{2k-1}}^{t_{2k+1}}\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \nonumber \\\le & {} c_\alpha B_2 h^4 \int _{t_{0}}^{t_{2m+1}}\exp \left( -\alpha \frac{t_{2m+1}-\tau }{1-\alpha }\right) {\text{d}}\tau \le c h^4. \end{aligned}$$

Bringing (4.5)–(4.8) together, we obtain

$$\begin{aligned} |r_{2m+1}|<c h^4. \end{aligned}$$

Similarly, we can prove

$$\begin{aligned} |r_{2m+2}|\le ch^4. \end{aligned}$$

This completes the proof of Theorem 4.1.

In order to simply the stability analysis, we rewrite the scheme in the following form:

$$\begin{aligned}&\widehat{b}_0u_0+\widehat{b}_1u_1+\widehat{b}_2u_2=f(t_1), \end{aligned}$$
$$\begin{aligned}&\sum _{k=0}^{2m+1}b_{k}^{(m)}u_k=f(t_{2m+1}),m=1,2,\cdots ,N-1,\end{aligned}$$
$$\begin{aligned}&\sum _{k=0}^{2m+2}\overline{b}_{k}^{(m)}u_k=f(t_{2m+2}),m=1,2,\cdots ,N-1, \end{aligned}$$


$$\begin{aligned}\left\{\begin{array}{l}\widehat{b}_i=a_{1}^{i,0},\widetilde{b}_i=a_{2}^{i,0},i=0,1,2;\\b_{0}^{(m)}=a_{2m+1}^{0,0},b_{1}^{(m)}=a_{2m+1}^{1,0}+a_{2m+1}^{0,1},b_{2}^{(m)}=a_{2m+1}^{2,0}+a_{2m+1}^{1,1},\\b_{2k}^{(m)}=a_{2m+1}^{1,k},b_{2k-1}^{(m)}=a_{2m+1}^{2,k-1}+a_{2m+1}^{0,k},k=2,3,\cdots ,m; \ \ b_{2m+1}^{(m)}=a_{2m+1}^{2,m};\\ \overline{b}_{0}^{(m)}=a_{2m+2}^{0,0},\overline{b}_{2k}^{(m)}=a_{2m+2}^{2,k-1}+a_{2m+2}^{0,k},k=1,2,\cdots ,m; \\ \overline{b}_{2k+1}^{(m)}=a_{2m+2}^{1,k},k=0,1,2,\cdots ,m;\ \ \overline{b}_{2m+2}^{(m)}=a_{2m+2}^{2,m}. \end{array}\right.\end{aligned}$$

In the following lemma, we study the sign of the coefficients given in (4.13), which plays an important role in analyzing the stability of the scheme.

Lemma 4.1

Some of the coefficients defined in (4.13) have different signs under some conditions as listed in Table 1.

Table 1 Coefficients defined in (4.13)


First, we investigate the sign of \(b_{k}^{(m)}\). A direct computation gives

$$\begin{aligned} b_{0}^{(m)}=\frac{M(\alpha )}{\alpha h^2} F_1(h) \exp \left( \frac{\alpha }{1-\alpha }(-2mh)\right) , \end{aligned}$$

where \(F_1(h):=-\frac{h}{2}-\frac{1-\alpha }{\alpha }+(\frac{3}{2}h+\frac{1-\alpha }{\alpha })\exp (\frac{\alpha }{1-\alpha }(-h)).\) It can be shown that \(F_1(h)<0\) if \(h<\frac{4}{3}\frac{1-\alpha }{\alpha }\). In fact, we have

$$\begin{aligned} F_1'(h)= & {} -\frac{1}{2}+\left( \frac{1}{2}-\frac{3}{2}\frac{h\alpha }{1-\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) ,\\ F_1''(h)= & {} \frac{\alpha }{1-\alpha }\left( -2+\frac{3}{2}\frac{h\alpha }{1-\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) . \end{aligned}$$

Thus, \(F_1''(h) < 0\) if \(h<\frac{4}{3}\frac{1-\alpha }{\alpha }\). Consequently, \(F_1'(h)<F_1'(0)=0\), and furthermore \(F_1(h)<F_1(0)=0\). This gives \(b_{0}^{(m)}<0\).

For \(b_{1}^{(m)}\), we have

$$\begin{aligned} b_{1}^{(m)}=\frac{M(\alpha )}{\alpha h^2} F_2(h) \exp \left( \frac{\alpha }{1-\alpha }(-2m+2)h\right) , \end{aligned}$$

where \(F_2(h):=\frac{h}{2}-\frac{1-\alpha }{\alpha }+(\frac{3}{2}h+\frac{3(1-\alpha )}{\alpha })\exp (\frac{\alpha }{1-\alpha }(-2h)) -2(h+\frac{1-\alpha }{\alpha })\exp (\frac{\alpha }{1-\alpha }(-3h))\), and we calculate the first and second derivatives of \(F_2(h)\) as follows:

$$\begin{aligned} F_2'(h)= & {} \frac{1}{2}-\left( \frac{9}{2}+3\frac{h\alpha }{1-\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) +2\left( 2+3\frac{h\alpha }{1-\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-3h)\right) , \\ F_2''(h)= & {} \frac{-6\alpha }{1-\alpha }\left[ -1-\frac{h\alpha }{1-\alpha }+\left( 1+3\frac{h\alpha }{1-\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) \right] \exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) \\\doteq & {} \frac{-6\alpha }{1-\alpha }G_1(h)\exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) . \end{aligned}$$

It can be checked that

$$\begin{aligned} G_1(h) := -1-\frac{h\alpha }{1-\alpha }+\left( 1+3\frac{h\alpha }{1-\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) >0 \end{aligned}$$

if \(h<\frac{1}{3}\frac{1-\alpha }{\alpha }\). Therefore, \(F_2''(h)<0\), \(F_2'(h)<F_2'(0)=0\), \(F_2(h)<F_2(0)=0\). This gives \(b_{1}^{(m)}<0\).

For \(b_{2}^{(m)}\), it holds

$$\begin{aligned} b_{2}^{(m)}=-\frac{M(\alpha )}{\alpha h^2} F_3(h) \exp \left( \frac{\alpha }{1-\alpha }(-2m+2)h\right) , \end{aligned}$$

where \(F_3(h):=2(h-\frac{1-\alpha }{\alpha })+(\frac{3}{2}h+\frac{3(1-\alpha )}{\alpha })\exp (\frac{\alpha }{1-\alpha }(-2h)) -(\frac{h}{2}+\frac{1-\alpha }{\alpha })\exp (\frac{\alpha }{1-\alpha }(-3h))\).

It can be proved \(F_3(h)>0\) by following the same lines. Thus \(b_{2}^{(m)}<0\).

For \(k=2,\cdots ,m\), we have

$$\begin{aligned} b_{2k-1}^{(m)}=\frac{M(\alpha )}{\alpha h^2} F_4(h) \exp \left( \frac{\alpha }{1-\alpha }(2k-2-2m)h\right) ,\\ b_{2k}^{(m)}=-\frac{2M(\alpha )}{\alpha h^2} F_5(h) \exp \left( \frac{\alpha }{1-\alpha }(2k-2m)h\right) , \end{aligned}$$


$$\begin{aligned} F_4(h)= & {} \left( \frac{h}{2}-\frac{1-\alpha }{\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(2h)\right) +3h +\left( \frac{h}{2}+\frac{1-\alpha }{\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) , \end{aligned}$$
$$\begin{aligned} F_5(h)= & {} h-\frac{1-\alpha }{\alpha }+\left( h+\frac{1-\alpha }{\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) . \end{aligned}$$

It can be directly verified that \(F_4(h)<0\) if \(h<\frac{1}{2}\frac{1-\alpha }{\alpha }\) and \(F_5(h)>0\) for all \(h>0\). This proves that for all \(k=2,\cdots ,m\), \(b_{2k-1}^{(m)}<0\) if \(h<\frac{1}{2}\frac{1-\alpha }{\alpha }\), and \(b_{2k}^{(m)}<0\) for all \(h>0\).

Some more calculation gives

$$\begin{aligned} b_{2m+1}^{(m)} =\frac{M(\alpha )}{\alpha h^2}\left( \frac{3}{2}h-\frac{1-\alpha }{\alpha }+\left( \frac{h}{2}+\frac{1-\alpha }{\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) \right) >0. \end{aligned}$$

Now we turn to check the sign of the coefficients \(\overline{b}_{k}^{(m)}\).

For \(\overline{b}_{0}^{(m)}\), we have

$$\begin{aligned} \overline{b}_{0}^{(m)}=\frac{M(\alpha )}{\alpha h^2} F_6(h) \exp \left( \frac{\alpha }{1-\alpha }(-2mh)\right) , \end{aligned}$$

where \(F_6(h):=\frac{1}{2}h-\frac{1-\alpha }{\alpha }+(\frac{3}{2}h+\frac{1-\alpha }{\alpha })\exp \left( \frac{\alpha }{1-\alpha }(-2h)\right)\), which is negative if \(h<\frac{1}{3}\frac{1-\alpha }{\alpha }\). Therefore, \(\overline{b}_{0}^{(m)}<0\) under the same condition.

Furthermore, it holds

$$\begin{aligned} \overline{b}_{2k}^{(m)}= & {} \frac{M(\alpha )}{\alpha h^2} F_4(h)\exp \left( \frac{\alpha }{1-\alpha }(2k-2-2m)h\right)<0, \ \text{ if } h<\frac{1}{2}\frac{1-\alpha }{\alpha }, \ \ k=1,\cdots ,m;\\ \overline{b}_{2k+1}^{(m)}= & {} -\frac{2M(\alpha )}{\alpha h^2} F_5(h)\exp \left( \frac{\alpha }{1-\alpha }(2k-2m)h\right) <0,\ \ k=0,1,\cdots ,m; \\ \overline{b}_{2m+2}^{(m)}= & {} \frac{M(\alpha )}{\alpha h^2} \left( \frac{3}{2}h-\frac{1-\alpha }{\alpha }+\left( \frac{h}{2}+\frac{1-\alpha }{\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) \right) >0. \end{aligned}$$

The proof is completed.

Some more properties of the coefficients in (4.13) are given in Lemma 4.2.

Lemma 4.2

It holds

$$\begin{aligned} \beta _1:= & {} \widehat{b}_1+\widetilde{b}_2>0, \end{aligned}$$
$$\begin{aligned} \beta _2:= & {} \exp \left( \frac{\alpha }{1-\alpha }(-t_1)\right) \widetilde{b}_2-\widehat{b}_0-\exp \left( \frac{\alpha }{1-\alpha }(-t_2)\right) \widehat{b}_2>0. \end{aligned}$$

Furthermore, if \(h<\frac{2}{3}\frac{1-\alpha }{\alpha }\) , it holds

$$\begin{aligned} \beta _3:=\exp \left( \frac{\alpha }{1-\alpha }(-t_2)\right) \widehat{b}_1-\widetilde{b}_0-\exp \left( \frac{\alpha }{1-\alpha }(-t_1)\right) \widetilde{b}_1>0. \end{aligned}$$


A direct computation gives

$$\begin{aligned} \widehat{b}_1= & {} \frac{2M(\alpha )}{\alpha h^2} \left[ \frac{1-\alpha }{\alpha }-\left( h+\frac{1-\alpha }{\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) \right] ,\\ \widetilde{b}_2= & {} \frac{M(\alpha )}{\alpha h^2} \left[ \frac{3}{2}h-\frac{1-\alpha }{\alpha }+\left( \frac{h}{2}+\frac{1-\alpha }{\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) \right] . \end{aligned}$$

Summing these two equalities, we obtain

$$\begin{aligned} \beta _1 =\frac{M(\alpha )}{\alpha h^2} \left[ \left( \frac{h}{2}+\frac{1-\alpha }{\alpha }\right) \left( \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) -1\right) ^2 -h\left( \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) -1\right) \right] >0. \end{aligned}$$

This proves (4.16). To prove (4.17), noticing

$$\begin{aligned} \widehat{b}_0= & {} \frac{M(\alpha )}{\alpha h^2} \left[ -\frac{1}{2}h-\frac{1-\alpha }{\alpha }+\left( \frac{3}{2}h+\frac{1-\alpha }{\alpha }\right) \exp \left(\frac{\alpha }{1-\alpha }(-h)\right)\right] , \\ \widehat{b}_2= & {} \frac{M(\alpha )}{\alpha h^2} \left[ \frac{1}{2}h-\frac{1-\alpha }{\alpha }+\left( \frac{1}{2}h+\frac{1-\alpha }{\alpha }\right) \exp \left(\frac{\alpha }{1-\alpha }(-h)\right)\right] . \end{aligned}$$

We have

$$\begin{aligned} \beta _2= & {} \frac{M(\alpha )}{\alpha h^2} \left[ \left( \frac{h}{2}+\frac{1-\alpha }{\alpha }\right) \left( \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) -1\right) ^2\right. \\&\left. -h\exp \left( \frac{\alpha }{1-\alpha }(-h)\right) \left( \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) -1\right) \right] >0. \end{aligned}$$

Finally, using the equalities as follows:

$$\begin{aligned} \widetilde{b}_0= & {} \frac{M(\alpha )}{\alpha h^2} \left[ \frac{1}{2}h-\frac{1-\alpha }{\alpha }+\left( \frac{3}{2}h+\frac{1-\alpha }{\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) \right] ,\\ \widetilde{b}_1= & {} -\frac{2M(\alpha )}{\alpha h^2} \left[ h-\frac{1-\alpha }{\alpha }+\left( h+\frac{1-\alpha }{\alpha }\right) \exp \left( \frac{\alpha }{1-\alpha }(-2h)\right) \right] , \end{aligned}$$

we get

$$\begin{aligned} \beta _3= & {} \frac{M(\alpha )}{\alpha h^2} \left[ \left( -\frac{3}{2}h+\frac{1-\alpha }{\alpha }\right) \left( \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) -1\right) ^2\right. \\&\left. -h\left( \exp \left( \frac{\alpha }{1-\alpha }(-h)\right) -1\right) \right] , \end{aligned}$$

which is positive if \(h<\frac{2}{3}\frac{1-\alpha }{\alpha }\). This proves the lemma.

A byproduct of the above lemmas is the well-posedness of the discrete problem (3.8), which is given in the following theorem.

Theorem 4.2

For any given functionf(t) and initial condition\(u_0\), the linear system (3.8) admits a unique solution\((u_1,u_2,\cdots ,u_{2N})^{\rm T}\).


First, it follows from (3.3) and (3.4):

$$\begin{aligned} S_2\varvec{u}_2 =\varvec{f}_2, \end{aligned}$$

where \(S_2=(a_i^{j,0})_{i,j=1}^2, \varvec{u}_2 =(u_1,u_2)^{\rm T}, \varvec{f}_2=(f(t_1)-a_1^{0,0}u_0,f(t_2)-a_2^{0,0}u_0)^{\rm T}\).

In virtue of the proof of Lemma 4.2, we have

$$\begin{aligned} a_1^{1,0}a_2^{2,0}-a_1^{2,0}a_2^{1,0} =\widehat{b}_1\widetilde{b}_2-\widehat{b}_2\widetilde{b}_1 = \left( \frac{M(\alpha )}{h\alpha }\right) ^2\left( \exp \left( -\frac{\alpha }{1-\alpha }h\right) -1\right) ^2 \ne 0. \end{aligned}$$

Thus \(S_2\) is invertible, and consequently the system (4.19) admits a unique solution \((u_1,u_2)^{\rm T}\).

Furthermore, we deduce from (3.5) and (3.6),

$$\begin{aligned} S_N \varvec{u}_N= \varvec{f}_N, \end{aligned}$$


$$\begin{aligned} S_N={ \left( \begin{array}{ccccccc} a_3^{2,1} &{} 0 &{} \cdots &{} 0 &{} 0 \\ a_4^{1,1} &{} a_4^{2,1}&{}\cdots &{} 0 &{} 0 \\ \vdots &{} \vdots &{} &{}\vdots &{} \vdots \\ a_{2N-1}^{2,1}+a_{2N-1}^{0,2} &{} a_{2N-1}^{1,2}&{}\cdots &{}a_{2N-1}^{2,N-1}&{}0\\ a_{2N}^{1,1} &{} a_{2N}^{2,1}+a_{2N}^{0,2}&{}\cdots &{}a_{2N}^{1,N-1}&{} a_{2N}^{2,N-1} \end{array} \right) }, \end{aligned}$$

\(\varvec{u}_N=(u_3,u_4,\cdots ,u_{2N})^{\rm T}, \varvec{f}_N=(\widehat{f}_3,\widehat{f}_4,\cdots ,\widehat{f}_{2N})^{\rm T}\) with \(\widehat{f}_{2m+1}=f(t_{2m+1})-a_{2m+1}^{0,0}u_0-(a_{2m+1}^{1,0}+a_{2m+1}^{0,1})u_{1} -(a_{2m+1}^{2,0}+a_{2m+1}^{1,1})u_2\), \(\widehat{f}_{2m+2}=f(t_{2m+2})-a_{2m+2}^{0,0}u_{0}-a_{2m+2}^{1,0}u_{1}-(a_{2m+2}^{2,0}+a_{2m+2}^{0,1})u_{2}, m=1,\cdots ,N-1\).

We see that \(S_N\) is a lower triangular matrix. According to Lemma  4.1, all diagonal entries of \(S_N\) are positive. Therefore, the linear system (4.20) admits a unique solution \((u_3,u_4,\cdots ,u_{2N})^{\rm T}\). This proves the theorem.

Now we turn to analyze the stability property of the scheme (4.9)–(4.12).

Theorem 4.3

If\(u_0>0,\lambda <0\), the scheme (4.9)–(4.12) applied to (2.1) withfgiven in (2.8) is stable with respect to the initial value under the condition

$$\begin{aligned} h<\frac{1-\alpha }{3\alpha }. \end{aligned}$$

That is, the numerical solution\(\{u_j\}_{j=0}^{2N}\)satisfies

$$\begin{aligned} 0<u_j< u_0,\ \ j=1,2,\cdots ,2N. \end{aligned}$$


First, inserting (2.8) into (4.9) and (4.10) gives

$$\begin{aligned} \widehat{b}_0u_0+\widehat{b}_1u_1+\widehat{b}_2u_2= & {} \lambda \left[ u_1-u_0\exp \left( -\frac{\alpha }{1-\alpha }t_1\right) \right] , \\ \widetilde{b}_0u_0+\widetilde{b}_1u_1+\widetilde{b}_2u_2= & {} \lambda \left[ u_2-u_0\exp \left( -\frac{\alpha }{1-\alpha }t_2\right) \right] . \end{aligned}$$

The solution of this equation can be expressed by

$$\begin{aligned} u_1= {G_1\over G_2} u_0,\ \ \ u_2= {G_3\over G_2} u_0, \end{aligned}$$


$$\begin{aligned} G_1= & {} G-\lambda \beta _2+\lambda ^2\exp \left( -\frac{\alpha }{1-\alpha }t_1\right) , \\ G_2= & {} G-\lambda \beta _1+\lambda ^2, \\ G_3= & {} G-\lambda \beta _3+\lambda ^2\exp \left( -\frac{\alpha }{1-\alpha }t_2\right) \end{aligned}$$

with \(G = \left( \frac{M(\alpha )}{h\alpha }\right) ^2\left( \exp (-\frac{\alpha }{1-\alpha }h)-1\right) ^2 > 0\) and \(\beta _1, \beta _2\), and \(\beta _3\) defined in Lemma  4.2.

In virtue of Lemma 4.2, we have \(\beta _1>0,\beta _2>0, \beta _3>0\) under the condition \(h<\frac{2}{3}\frac{1-\alpha }{\alpha }\). This means \(G_1>0, G_2>0, G_3>0\). Furthermore, a direct calculation shows

$$\begin{aligned} G_1 - G_2= & {} \displaystyle \lambda \frac{M(\alpha )}{\alpha h}\left( \exp \left( -\frac{\alpha }{1-\alpha }t_1\right) -1\right) ^2 +\lambda ^2\left( \exp \left( -\frac{\alpha }{1-\alpha }t_1\right) -1\right)< 0,\\ G_3 - G_2= & {} \displaystyle \lambda \frac{2M(\alpha )}{\alpha h}\left( \exp \left( -\frac{\alpha }{1-\alpha }t_1\right) -1\right) ^2 +\lambda ^2\left( \exp (-\frac{\alpha }{1-\alpha }t_2)-1\right) < 0. \end{aligned}$$

As a result, we get

$$\begin{aligned} 0<u_1<u_0,\ \ \ 0<u_2<u_0. \end{aligned}$$

It follows from (4.11),

$$\begin{aligned} u_3=\displaystyle \left. \left[ \left( -b_{0}^{(1)}-\lambda \exp \left( -\frac{\alpha }{1-\alpha }t_3\right) \right) u_0-b_{1}^{(1)}u_1-b_{2}^{(1)}u_2\right] \right/(b_{3}^{(1)}-\lambda ). \end{aligned}$$

According to Lemma 4.1, it holds

$$\begin{aligned} -b_{0}^{(1)}>0,-b_{1}^{(1)}>0,-b_{2}^{(1)}>0,b_{3}^{(1)}>0. \end{aligned}$$

This, together with (4.22), leads to

$$\begin{aligned} \left. 0<u_3<\left[ -b_{0}^{(1)}-\lambda \exp \left( -\frac{\alpha }{1-\alpha }t_3\right) -b_{1}^{(1)}-b_{2}^{(1)}\right] u_0 \right/(b_{3}^{(1)}-\lambda ). \end{aligned}$$

Furthermore, it can be directly verified that

$$\begin{aligned} 0< -b_{0}^{(1)}-\lambda \exp \left( -\frac{\alpha }{1-\alpha }t_3\right) -b_{1}^{(1)}-b_{2}^{(1)} < b_{3}^{(1)}-\lambda . \end{aligned}$$

Therefore we get

$$\begin{aligned} 0<u_3<u_0. \end{aligned}$$

Next we deduce from (4.12),

$$\begin{aligned} \left. u_4=\displaystyle \left[ \left( -\overline{b}_{0}^{(1)}-\lambda \exp \left( -\frac{\alpha }{1-\alpha }t_4\right) \right) u_0-\overline{b}_{1}^{(1)}u_1-\overline{b}_{2}^{(1)}u_2 -\overline{b}_3^{(1)}u_3\right] \right/(\overline{b}_{4}^{(1)}-\lambda ). \end{aligned}$$

Once again, it follows from Lemma 4.1,

$$\begin{aligned} -\overline{b}_{0}^{(1)}>0,\ -\overline{b}_{1}^{(1)}>0,\ -\overline{b}_{2}^{(1)}>0,\ -\overline{b}_{3}^{(1)}>0,\ \overline{b}_{4}^{(1)}>0. \end{aligned}$$

Then using some relationships between these coefficients and (4.22)–(4.23), we obtain

$$\begin{aligned} 0<u_4<u_0. \end{aligned}$$

Now we prove the remaining results by using mathematical induction. Assume (4.21) holds for all \(j=1,2,\cdots ,2m;\, m=1,2,\cdots ,N-1\), we want to prove that it also holds for \(j=2m+1\) and \(j=2m+2\).

It follows from (4.11),

$$\begin{aligned} u_{2m+1}= & {} \left[ \left( -b_{0}^{(m)}-\lambda \exp \left( -\frac{\alpha }{1-\alpha }t_{2m+1}\right) \right) u_0 -b_{1}^{(m)}u_1-b_{2}^{(m)}u_2\right. \nonumber \\&\left. \left. -\sum _{k=2}^{m}b_{2k}^{(m)}u_{2k}-\sum _{k=2}^{m}b_{2k-1}^{(m)}u_{2k-1} \right] \right/(b_{2m+1}^{(m)}-\lambda ). \end{aligned}$$

It is not difficult to see, by using Lemma 4.1, that all coefficients in the right-hand side of (4.25) are positive. Thus we deduce from the induction assumption

$$\begin{aligned} u_{2m+1} < {G_4\over G_5}u_0, \end{aligned}$$


$$\begin{aligned} G_4= & {} -b_{0}^{(m)}-\lambda \exp \left( -\frac{\alpha }{1-\alpha }t_{2m+1}\right) -b_{1}^{(m)}-b_{2}^{(m)} -\sum _{k=2}^{m}b_{2k}^{(m)}-\sum _{k=2}^{m}b_{2k-1}^{(m)}>0, \\ G_5= & {} b_{2m+1}^{(m)}-\lambda >0. \end{aligned}$$

Using the fact that the scheme (4.9)–(4.12) is accurate for the constant solution, we have

$$\begin{aligned} \displaystyle b_{0}^{(m)}+b_{1}^{(m)}+b_{2}^{(m)}+\sum _{k=2}^{m}b_{2k}^{(m)}+\sum _{k=2}^{m}b_{2k-1}^{(m)}+b_{2m+1}^{(m)}=0. \end{aligned}$$


$$\begin{aligned} G_4-G_5=-\lambda \left[ \exp \left( -\frac{\alpha }{1-\alpha }t_{2m+1}\right) -1\right] <0. \end{aligned}$$

This proves

$$\begin{aligned} 0<u_{2m+1}<u_0. \end{aligned}$$

In an exactly same way, we can deduce from (4.12) and Lemma 4.1:

$$\begin{aligned} 0<u_{2m+2}<u_0. \end{aligned}$$

The proof is completed.

Numerical Results

We present several numerical examples to verify the theoretical results obtained in the previous sections. Precisely, our main purpose is to check the convergence order of the numerical solution with respect to the step size h.

We consider the initial value problem (2.1) with several right-hand side f(tu(t)) as follows:

  1. i)


  2. ii)


  3. iii)



$$\begin{aligned} G(t) = M(\alpha ) \left[ \frac{3}{\alpha }t^2-\frac{6(1-\alpha ) }{\alpha ^2}t +\frac{6(1-\alpha )^2 }{\alpha ^3} -\frac{6(1-\alpha )^2 }{\alpha ^3}\exp \left( -\frac{\alpha }{1-\alpha }t\right) \right] . \end{aligned}$$

In the first case, f is independent of u, while in the second and third cases f is function of u. In particular, f is nonlinear with respect to u. However, it can be verified that the exact solution is \(u(t)=t^3\) for all three cases. Note the main difference between these examples is that the first example considers a right-hand side function which is independent of the solution, while the second example addresses a right-hand side function linearly dependent of the solution, and the third one is a nonlinear function of the solution.

All the results presented in this example correspond to the numerical solution captured at \(T=1\). As in [1, 3], we choose a special normalization function \(M(\alpha )=1\) such that \(M(0)=M(1)=1\) in the numerical tests. In Tables 2, 3, 4, we list the maximum errors, i.e., \(\displaystyle \max _i|u(t_i)-u_i|\) as a function of h for several \(\alpha\). Also shown are the corresponding error decay rates. It is observed from these tables that for all tested right-hand side functions and values of \(\alpha\), the convergence rate is close to four. This is in a good agreement with the theoretical prediction. In particular, it is worthy to emphasize that the non-linearity of f seems to have no impact on the accuracy of the scheme.

Table 2 Maximum errors and decay rates with \(\alpha =0.3\) and 0.7 for \(f(t,u(t))=G(t)\)
Table 3 Maximum errors and decay rates with \(\alpha =0.3\) and 0.7 for \(f(t,u(t))=G(t)-t^3+u(t)\)
Table 4 Maximum errors and decay rates with \(\alpha =0.3\) and 0.7 for \(f(t,u(t))=G(t)-t^6+u^2(t)\)

The next test concerns the stability investigation. To this end, the scheme is applied to the problem (2.1) with the fabricated right-hand side function \(f(t,u(t))=\frac{M(\alpha )}{\alpha ^2+(1-\alpha )^2}[(1-\alpha )\sin t+\alpha \cos t-\alpha \exp (-\frac{\alpha }{1-\alpha }t)]\), so that the exact solution is \(u(t)=\sin t\). The calculation is run up to \(T=1\,000\), long enough to study the stability of the scheme. Table  5 shows the error behavior as a function of t, computed with fixed \(h=0.01\). It is observed that the numerical solution remains to be good approximation to the exact solution even after long time computation. This test demonstrates good stability property of the proposed scheme.

Table 5 Errors \(\displaystyle |u(t_i)-u_i|\) for \(\alpha =0.3,0.5\) and 0.7

Concluding Remarks

We have proposed an efficient high-order scheme for fractional ordinary differential equations with the Caputo–Fabrizio derivative. The stability and convergence analysis was carried out to prove that the proposed scheme is stable under a slight restriction on the step size, which only depends on the fractional order. The obtained error estimate shows that the proposed scheme is of order 4. The carried out numerical tests confirmed the theoretical prediction.


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Correspondence to Chuanju Xu.

Additional information

This research was supported by the National Natural Science Foundation of China (Grant numbers 11501140, 51661135011, 11421110001, and 91630204) and the Foundation of Guizhou Science and Technology Department (No. [2017]1086). The first author would like to acknowledge the financial support by the China Scholarship Council (201708525037).

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Cao, J., Wang, Z. & Xu, C. A High-Order Scheme for Fractional Ordinary Differential Equations with the Caputo–Fabrizio Derivative. Commun. Appl. Math. Comput. 2, 179–199 (2020). https://doi.org/10.1007/s42967-019-00043-8

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  • Caputo–Fabrizio derivative
  • Fractional differential equations
  • High-order numerical scheme

Mathematics Subject Classification

  • 26A33
  • 34A08
  • 65M12
  • 65M06