# Convergence Analysis of Implicit Euler Method for a Class of Nonlinear Impulsive Fractional Differential Equations

Authors: Yuexin Yu; Sheng Zhen
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

For a class of nonlinear impulsive fractional differential equations, we first transform them into equivalent integral equations, and then the implicit Euler method is adapted for solving the problem. The convergence analysis of the method shows that the method is convergent of the first order. The numerical results verify the correctness of the theoretical results.


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## Body

## 1. Introduction

In recent years, fractional differential equations have become a research hotspot due to their wide application in many fields. We refer the readers to the research papers [1–5] and the monographs by Podlubny [6], Diethelm [7], Kilbas et al. [8], Zhou [9], and the references cited therein. When the fractional differential equations are affected by instantaneous mutation, the impulsive fractional differential equations are obtained. The research of impulsive fractional differential equations can be found in literatures [10–20] and monograph [21]. However, there are few literatures on numerical methods for impulsive fractional differential equations.In this paper, implicit Euler method is constructed for solving a class of nonlinear impulsive fractional differential equations. It is proved that the method is convergent of the first order. The numerical results also verify the correctness of the theoretical results.


## 2. Construction of Numerical Scheme

Consider the following impulsive fractional differential equations:(1)D0Ctαut=ft,ut,t∈J′≔J∖t1,t2,…,tm,J≔0,T,△utk=Ikutk,△u′tk=I¯kutk,k=1,2,…,m,u0=β,u′0=γ,where α∈1,2, β,γ∈ℝ are constants, and Dtα0Cut is the α-order Caputo derivative of solution ut defined by (see [6–8])(2)D0Ctαut=1Γn−α∫0tunst−sα+1−nds,0≤n−1<α<n,0=t0<t1<⋯<tm<tm+1=T, △utk=utk+−utk−, △u′tk=u′tk+−u′tk−, utk−=limε⟶0−utk+ε, and utk+=limε⟶0+utk+ε represent the left and right limits of ut at t=tk, and f:J×ℝ⟶ℝ and Ik,I¯k,:ℝ⟶ℝ are continuous functions and satisfy the following conditions:(3)ft,u−ft,v≤L1u−v,t∈J,u,v∈ℝ,Iku−Ikv≤L2u−v,u,v∈ℝ,k=1,2,…,m,I¯ku−I¯kv≤L3u−v,u,v∈ℝ,k=1,2,…,m,where L1,L2,L3 are nonnegative constants with moderate size.Throughout this paper, letCJ,ℝ be the Banach space of all continuous functions from J into ℝ with the norm uc≔suput:t∈J for u∈CJ,ℝ. We also define PCJ,ℝ≔u:J⟶ℝ,u∈Ctk,tk+1,ℝ,k=0,1,…,m, and utk+ exists, k=1,2,…m. The space PCJ,ℝ is a Banach space equipped with the norm upc≔suput:t∈J.In addition, due to the need of convergence analysis, for functionut∈PCJ,ℝ, there are constants L4 and L5 such that(4)u′t≤L4,∂f∂tt,u≤L5,t∈tk,tk+1,k=0,1,…,m.In order to obtain the numerical scheme for solving problem (1), according to Lemma 3.1 in reference [12], we can express equation (1) as the following equivalent integral equation:(5)ut=u0+u′0t+∑0<tk<tIkutk+∑0<tk<tI¯kutkt−tk+∫0tt−sα−1Γαfs,usds,t∈J.Lethk=tk+1−tk/N, N be a given positive integer, the grid points tk,i=tk+ihk=t0,0+∑j=0k−1Nhj+ihk, k=0,1,…,m, i=1,2,…,N, h=maxh0,h1,…,hm, and utk,i express the true solution of equation (1) at tk,i. Then, an approximation to the integral equation can be attained by right rectangle formula:(6)utk,i=u0+u′0tk,i+∑l=1kIlutl−1,N+∑l=1kI¯lutl−1,Ntk,i−tl−1,N+∑l=1k∑r=1N1Γα∫tl−1,r−1tl−1,rtk,i−sα−1fs,usds+∑r=1i1Γα∫tk,r−1tk,rtk,i−sα−1fs,usds≈u0+u′0tk,i+∑l=1kIlutl−1,N+∑l=1kI¯lutl−1,Ntk,i−tl−1,N+∑l=1k∑r=1N1Γαftl−1,r,utl−1,r∫tl−1,r−1tl−1,rtk,i−sα−1ds+∑r=1i1Γαftk,r,utk,r∫tk,r−1tk,rtk,i−sα−1ds=u0+u′0tk,i+∑l=1kIlutl−1,N+∑l=1kI¯lutl−1,Ntk,i−tl−1,N+∑l=1k∑r=1N1Γα+1ftl−1,r,utl−1,rtk,i−tl−1,r−1α−tk,i−tl−1,rα+∑r=1i1Γα+1ftk,r,utk,rtk,i−tk,r−1α−tk,i−tk,rα.By using the numerical solutionuk,i instead of the true solution utk,i in equation (6), we obtain the implicit Euler method for solving impulsive fractional differential equation (1):(7)uk,i=u0,0+u0,0′tk,i+∑l=1kIlul−1,N+∑l=1kI¯lul−1,Ntk,i−tl−1,N+∑l=1k∑r=1N1Γα+1ftl−1,r,ul−1,rtk,i−tl−1,r−1α−tk,i−tl−1,rα+∑r=1i1Γα+1ftk,r,uk,rtk,i−tk,r−1α−tk,i−tk,rα,k=0,1,…,m,i=1,2,…,N.Remark 1.
It is well known that there are various forms of definition of fractional calculus in the literature, such as Riemann–Liouville fractional calculus, Gru..nwald–Letnikov fractional calculus, and Caputo fractional calculus (see [6–8]). The α-order Riemann–Liouville derivative of function ut is defined by(8)D0Rtαut=1Γn−αdndtn∫0tust−sα+1−nds,0≤n−1<α<n.
We have the following relationship between the Riemann–Liouville derivative and Caputo derivative:(9)D0Rtαut=Dtα0Cut+∑j=0n−1tj−αyj0Γj−α+1,0≤n−1<α<n.
Therefore, the fractional differential equations studied in the present paper only consider Caputo derivative.


## 3. Convergence Analysis

Letzk,i=uk,i−utk,i, where uk,i and utk,i denote the numerical solution and true solution of problem (1) at grid point tk,i, respectively. Then, for the error zk,i, we have(10)zk,i=∑l=1kIlul−1,N−Ilutl−1,N+∑l=1kI¯lul−1,N−I¯lutl−1,Ntk,i−tl−1,N+∑l=1k∑r=1N1Γα+1ftl−1,r,ul−1,r−ftl−1,r,utl−1,r×tk,i−tl−1,r−1α−tk,i−tl−1,rα+∑r=1i1Γα+1ftk,r,uk,r−ftk,r,utk,rtk,i−tk,r−1α−tk,i−tk,rα−Rk,i,where(11)Rk,i=∑l=1k∑r=1N1Γα∫tl−1,r−1tl−1,rtk,i−sα−1fs,usds+∑r=1i1Γα∫tk,r−1tk,rtk,i−sα−1fs,usds−∑l=1k∑r=1N1Γαftl−1,r,utl−1,r∫tl−1,r−1tl−1,rtk,i−sα−1ds−∑r=1i1Γαftk,r,utk,r∫tk,r−1tk,rtk,i−sα−1ds.In order to obtain the convergence result of numerical method (7), we first prove the following lemma.Lemma 1.
Assume the functionsf:J×ℝ⟶ℝ and Ik,I¯k:ℝ⟶ℝ,k=1,2,…,m are continuous and satisfy conditions (3) and (4); then, the truncation error of the discrete scheme (7) satisfies(12)Rk,i≤Ch,k=0,1,…,m,i=1,2,…,N.Throughout the paper,C will denote a positive constant not necessarily the same at different places, which may depend on Lj,j=1,2,…,5, but is independent of h and N.Proof.
From (11), we have(13)Rk,i=∑l=1k∑r=1N1Γα∫tl−1,r−1tl−1,rfs,us−ftl−1,r,utl−1,rtk,i−sα−1ds+∑r=1i1Γα∫tk,r−1tk,rfs,us−ftk,r,utk,rtk,i−sα−1ds≤∑l=1k∑r=1N1Γα∫tl−1,r−1tl−1,rfs,us−ftl−1,r,utl−1,rtk,i−sα−1ds+∑r=1i1Γα∫tk,r−1tk,rfs,us−ftk,r,utk,rtk,i−sα−1ds.
Applying the integral mean value theorem, we know that there existsξl,r∈tl,r−1,tl,r such that(14)Rk,i≤∑l=1k∑r=1N1Γα+1fξl−1,r,uξl−1,r−ftl−1,r,utl−1,r×tk,i−tl−1,r−1α−tk,i−tl−1,rα+∑r=1i1Γα+1fξk,r,uξk,r−ftk,r,utk,rtk,i−tk,r−1α−tk,i−tk,rα≤∑l=1k∑r=1N1Γα+1fξl−1,r,uξl−1,r−fξl−1,r,utl−1,r+fξl−1,r,utl−1,r−ftl−1,r,utl−1,rtk,i−tl−1,r−1α−tk,i−tl−1,rα+∑r=1i1Γα+1fξk,r,uξk,r−fξk,r,utk,r+fξk,r,utk,r−ftk,r,utk,rtk,i−tk,r−1α−tk,i−tk,rα.
Using conditions (2) and (3) and differential mean value theorem, we know that there exists ηl,r,ζl,r∈tl,r−1,tl,r such that(15)Rk,i≤∑l=1k∑r=1N1Γα+1L1uξl−1,r−utl−1,r+hl−1∂f∂tηl−1,r,utl−1,r×tk,i−tl−1,r−1α−tk,i−tl−1,rα+∑r=1i1Γα+1L1uξk,r−utk,r+hk∂f∂tηk,r,utk,r×tk,i−tk,r−1α−tk,i−tk,rα≤C∑l=1k∑r=1Nhl−1u′ζl−1,r+∂f∂tηl−1,r,utl−1,rtk,i−tl−1,r−1α−tk,i−tl−1,rα+C∑r=1ihku′ζk,r+∂f∂tηk,r,utk,rtk,i−tk,r−1α−tk,i−tk,rα≤C∑l=1k∑r=1Nhl−1tk,i−tl−1,r−1α−tk,i−tl−1,rα+C∑r=1ihktk,i−tk,r−1α−tk,i−tk,rα≤Chtk,iα−tk,i−tk−1,Nα+tk,i−tk,0α≤Ch.
This means the proof of Lemma1 is completed.Theorem 1.
Letuk,i and utk,i denote the numerical solution and true solution of problem (1) at grid point tk,i, respectively. Then, the convergence inequality(16)uk,i−utk,i≤Ch,k=0,1,…,m,i=1,2,…,N,holds when h is small enough and the conditions of Lemma 1 are satisfied. This means the numerical method (7) is convergent of the first order.Proof.
From (10), we can obtain that(17)1−1Γα+1L1hkαzk,i≤∑l=1kIlul−1,N−Ilutl−1,N+∑l=1kI¯lul−1,N−I¯lutl−1,Ntk,i−tl−1,N+∑l=1k∑r=1N1Γα+1ftl−1,r,ul−1,r−ftl−1,r,utl−1,r×tk,i−tl−1,r−1α−tk,i−tl−1,rα+∑r=1i−11Γα+1ftk,r,uk,r−ftk,r,utk,r×tk,i−tk,r−1α−tk,i−tk,rα+Rk,i≤∑l=1kzl−1,NL2+L3tk,i−tl−1,N+∑l=1k∑r=1N1Γα+1L1zl−1,rtk,i−tl−1,r−1α−tk,i−tl−1,rα+∑r=1i−11Γα+1L1zk,rtk,i−tk,r−1α−tk,i−tk,rα+Rk,i.
Whenh is small enough, i.e.,(18)Γα+1−L1hkα>0,we have(19)zk,i≤pk,i+wk∑l=1k∑r=1Nvl−1,rzl−1,r+wk∑r=1i−1ak,rzk,r,where(20)pk,i=Γα+1Rk,iΓα+1−L1hkα,wk=1Γα+1−L1hkα,k=0,1,…,m,ak,r=L1tk,i−tk,m−1α−tk,i−tk,rα,r=1,2,…,N−1,vl−1,r=L1tk,i−tl−1,r−1α−tk,i−tl−1,rα,l=1,2,…,k,vl−1,N=Γα+1L2+L3tk,i−tl−1,N+L1tk,i−tl−1,N−1α−tk,i−tl−1,Nα.
By using the discrete analogue of Gronwall’s inequality (see Theorem 2 in [22]), it follows that(21)zk,i≤pk,i+wk∏l=1k∏r=1N1+wl−1vl−1,r∏r=1i−11+wkak,r×∑q=1k∑j=1Npq−1,jvq−1,j∏l=1q∏r=1j1+wl−1vl−1,r−1+∑g=1i−1ak,gpk,g∏l=1k∏r=1N1+wl−1vl−1,r−1∏r=1g1+wkak,r−1.
According to Lemma1, when h is small enough, we have 0<wk≤C,k=0,1,…,m, and(22)0<pk,i≤Ch.
Hence, we obtain(23)∑q=1k∑j=1Npq−1,jvq−1,j∏l=1q∏r=1j1+wl−1vl−1,r−1+∑g=1i−1ak,gpk,g∏l=1k∏r=1N1+wvl−1,r−1∏r=1g1+wkak,r−1≤∑q=1k∑j=1Npq−1,jvq−1,j+∑g=1i−1ak,gpk,g≤ChL1tk,iα−tk,i−tk−1,Nα+∑q=1kΓα+1L2+L3tk,i−tq−1,N+ChL1tk,i−tk,0α−tk,i−tk,i−1α≤ChL1tk,iα−tk,i−tk,i−1α+∑q=1kΓα+1L2+L3tk,i−tq−1,N≤Ch,(24)∏l=1k∏r=1N1+wl−1vl−1,r∏r=1i−11+wkak,r≤exp∑l=1k∑r=1Nwl−1vl−1,r+∑r=1i−1wkak,r≤expCL1tk,iα−tk,i−tk−1,Nα+∑l=1kΓα+1L2+L3tk,i−tl−1,N+CL1tk,i−tk,0α−tk,i−tk,i−1α≤expCL1tk,iα−tk,i−tk,i−1α+∑l=1kΓα+1L2+L3tk,i−tl−1,N≤C,where we have used inequality (22) and 1+x≤ex for x≥−1. Therefore, substituting (22)–(24) into (21) leads to(25)zk,i=uk,i−utk,i≤Ch,k=0,1,…,m,i=1,2,…,N,which means the method is convergent of the first order.


## 4. Numerical Experiments

In the section, we utilize the following example to verify the theoretical results obtained in the previous section. Here, we will give error estimates and convergence rates for the numerical scheme.Example 1.
Consider the following impulsive fractional differential equation:(26)D0Ctαut=−3ut+0.2sinut+5tα+2,t=0,3∖0.5,2,1<α<2,Δu1=−12u0.5−,Δu2=−23u2−,Δu′1=14u0.5−,Δu′2=14u2−,u0=1,u′0=1.
Because it is difficult to obtain the true solution of the equation, we take the numerical solutionuk,i as the true solution utk,i at t=tk,i with N=6400. The numerical scheme (7) is solved by using the Newton iteration method with uk,i0=0k=0,1,…,m,i=1,2,…,N as an initial value. We iteratively compute uk,ij until(27)max0≤k≤m,1≤i≤Nuk,ij−uk,ij−1<10−8.
TheL∞ norm of the global error is denoted as(28)em,N=max0≤k≤m,1≤i≤Nuk,i−utk,i.
Then, the convergence order of the numerical method is(29)rm,N=log2em,Nem,2N.
Whenα takes different values, the error and convergence order of numerical method (7) are shown in Table 1.
Table1 shows that the numerical method (7) is convergent of the first order, which supports the convergence estimate of Theorem 1.
In Figure1, we can see that the numerical solution is discontinuous due to the existence of the impulses.Table 1
Error and convergence order of numerical method (7) with different α.
N = 40N = 80N = 160N = 320N = 640α = 1.14.197e − 22.108e − 21.047e − 25.112e − 32.425e − 30.9931.0101.0331.076—α = 1.34.849e − 22.430e − 21.205e − 25.881e − 32.789e − 30.9971.0121.0341.076—α = 1.56.299e − 23.157e − 21.566e − 27.644e − 33.625e − 30.9971.0121.0341.076—α = 1.78.186e − 24.096e − 22.029e − 29.901e − 34.694e − 30.9991.0131.0351.077—α = 1.98.612e − 24.283e − 22.115e − 21.031e − 24.882e − 31.0081.0181.0371.078—Figure 1
Numerical solution of equation (26) with α=1.5 and N=100.


## 5. Conclusion

In this paper, we focus on the numerical solution of impulsive fractional differential equations. For a class of nonlinear impulsive fractional differential equations, the implicit Euler method is adapted for solving the problem. After careful convergence analysis, we prove that the method is convergent of the first order. For future work, we will study the higher-order methods for solving impulsive fractional differential equations and analyze their convergence.



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