# Analysis of Multiterm Initial Value Problems with Caputo–Fabrizio Derivative

Authors: Mohammed Al-Refai; Muhammed Syam
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we discuss the solvability of a class of multiterm initial value problems involving the Caputo–Fabrizio fractional derivative via the Laplace transform. We derive necessary and sufficient conditions to guarantee the existence of solutions to the problem. We also obtain the solutions in closed forms. We present two examples to illustrate the validity of the obtained results.


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## Body

## 1. Introduction

Recently, there is great interest to develop new types of fractional derivatives of nonsingular kernel. Motivated by applications, Caputo and Fabrizio were the first to introduce such types of fractional derivatives with nonlocal and nonsingular kernel [1]. The Caputo–Fabrizio derivative is connected with a variety of applications (see [2–6]). Stability analysis of fractional differential equations without inputs was studied in [7], where exponential stability is obtained for the Caputo–Fabrizio derivative. Since their kernels are nonlocal, fractional derivatives preserve memories, and therefore, they have been used to model several (SIR) epidemic models (see [8–14]). Several analytical techniques have been implemented to study various fractional equations with fractional derivatives without singular kernels, such as the Laplace transform, reduction to initial value problems with integer derivatives, maximum principles, and fixed point theorems (see [15–21]), just to mention a few out of many in the literature. The following definitions are required to state our problem.Definition 1.
A functionf is said to be absolutely continuous on a,b, if there exists a function f′∈L1a,b such that(1)ft=fa+∫atf′xdx,∀t∈a,b.
In the following, we will use the notationACa,b to denote the space of absolutely continuous functions on a,b.Definition 2.
A functionf:0,+∞⟶ℝ is said to be of exponential order, if there exists three constants T,M,C≥0 such that ft≤CeMt,∀t≥T.
Then, we define the following space.Definition 3.
The spaceℳ1 is defined by(2)ℳ1=f∈Lloc1:fis absolutely continuous on0,T,for anyT>0,andf,f′are of exponential order.
In this paper, we consider the multiterm fractional initial value problem:(3)∑k=1nakDCFC0αkyt=an+1yt+ft,t>0,n≥2,(4)y0=y0,where 1>α1>α2>⋯>αn>0,ak>0,k=1,…,n, and DCFC0α is the Caputo–Fabrizio fractional derivative of Caputo sense. Here, we assume that y,f∈ℳ1, so their Laplace transforms are well defined.Definition 4.
(see [1]). For 0<α<1, t>0 and f∈ℳ1, the Caputo–Fabrizio fractional derivative of Caputo sense is defined by(5)DCFC0αft=Bα1−α∫0tf′τe−α/1−αt−τdτ,where Bα>0 is a normalization function satisfying B0=B1=1.
For the corresponding fractional integral and more properties of the derivative, we refer the readers to [1, 5, 22–24]. In [15, 17], the fractional initial value problems were transformed to equivalent initial value problems with integer derivatives. However, the technique is not valid for the multiterm initial value problems. The single term of the problem with n=1 was discussed in [23], and the solution of the problem was obtained in a closed form. In this paper, we apply the Laplace transform to analyze the solutions of the fractional initial value problems (3) and (4). This paper is organized as follows: in Section 2, we present some preliminary results about the Caputo–Fabrizio fractional derivative and derive necessary and sufficient conditions to guarantee the existence of solutions to problems (3) and (4). We also obtain the exact solutions in closed forms using the Laplace transform. In Section 3, we present two examples to illustrate the validity of our results. Finally, we close up with some concluding remarks in Section 4.


## 2. Main Results

We start with the definition and main results concerning the Caputo–Fabrizio derivative. Then, we present necessary and sufficient conditions for the solution of problems (3) and (4). Since DCFC0αft=Bα/1−αf′t∗e−α/1−αt, equation (3) can be written as(6)∑k=1ncke−μkt∗y′t=cn+1yt+ft,t>0,n≥2,where(7)ck=Bαk1−αkak,μk=αk1−αk,k=1,…,n,cn+1=an+1,and ∗ denotes to the convolution of two functions.Remark 1.
LetFs=Pns/Qms where Pns and Qms are polynomials with degrees n and m, respectively, and are with real coefficients. If n<m, then the inverse Laplace transform of Fs exists and can be evaluated using partial fractions.Lemma 1.
Forf∈ℳ1, the following hold:(1)
DCFC0α:AC0,T⟶C0,T(2)
DCFC0αf0=0Proof.
(1)
For any functionf∈AC0,T, DCFC0αf is the convolution product of a continuous function and L1 function, that is continuous(2)
Since for anyf∈AC0,T, DCFC0αf is continuous; then, DCFC0αf0=limt⟶0+DCFC0αft=0Definition 5.
Let the spaceℳ2 be defined by(8)ℳ2=f∈C10,+∞:f′0=f′0+∈ℝ,f′∈AC0,T,for anyT>0,andf,f′,f″are of exponential order.Lemma 2.
The following holds for multiterm fractional initial value problems (3) and (4).(1)
If a solutiony:0,+∞⟶ℝ exists such that y∈AC0,T for any T>0, then cn+1y0+f0=0(2)
Iff∈ℳ1,cn+1y0+f0=0, and cn+1≠∑k=1nck, then a unique solution y∈ℳ1 exists(3)
If a solutiony∈ℳ1 exists for y0≠0 and f∈ℳ2, then cn+1≠∑k=1nckProof.
(1)
Lety∈C0,+∞ be a solution to multiterm fractional initial value problems (3) and (4) with y∈AC0,T for any T>0. By Lemma 1, DCFC0αkyt=0 for any k=1,…,n Then, from equation (3), one gets(9)cn+1y0+f0=0.(2)
Lety0≠0, f∈ℳ1,cn+1y0+f0=0, and cn+1≠∑k=1nck. Then, applying the Laplace transform to equation (6) and using the convolution result, we have(10)∑k=1ncks+μksℒyt−y0=cn+1ℒyt+Fs,whereFs=ℒft. The above equation yields(11)ℒfts∑k=1ncks+μk−cn+1=y0∑k=1ncks+μk+Fs.We have(12)∑k=1ncks+μk=c1s+μ1+c2s+μ2+⋯+cns+μn=∑k=1nck∏j=1,j≠kns+μj∏k=1ns+μk=Qn−1sPns,whereQn−1s=∑k=1nck∏j=1,j≠kns+μj and Pns=∏k=1ns+μk are polynomials of degrees n−1 and n, respectively. Substituting in equation (11), we have(13)ℒysQn−1sPns−cn+1=y0Qn−1sPns+Fs,or(14)ℒys=y0Qn−1ssQn−1s−cn+1Pns+PnssQn−1s−cn+1PnsFs.Now, the leading coefficient of the polynomialQn−1s is ∑k=1nck, and the leading coefficient of the polynomial Pns is 1. If cn+1≠∑k=1nck, then sQn−1s−cn+1Pns is a polynomial of degree n. Let(15)H1s=Qn−1ssQn−1s−cn+1Pns,H2s=1sPnssQn−1s−cn+1Pns,so thatℒ−1H1s=h1tandℒ−1H2s=h2t are well defined. Then,(16)ℒyt=y0H1s+H2ssFs=y0H1s+Hsℒf′t+f0.Using the uniqueness result of the inverse Laplace transform, we have(17)yt=y0h1t+h2t∗f′t+f0h2t,which completes the proof.(3)
Let a solutiony∈ℳ1 exists for y0≠0 and f∈ℳ2. Starting from equation (14), let us define(18)H˜1s=y0Qn−1ssQn−1s−cn+1Pns,g˜s=PnssQn−1s−cn+1PnsFs=Pnss2sQn−1s−cn+1Pnss2Fs.
Define(19)H˜2s=Pnss2sQn−1s−cn+1Pns.
H˜2 admits an inverse Laplace transform since sQn−1s−cn+1Pns is of degree n−1. Moreover, since f∈ℳ2, also s2Fs admits an inverse Laplace transform, thus also g˜s (by the convolution rule). One can rewrite equation (14) as(20)ℒys−g˜s=H˜1s.
However, sincesQn−1s−cn+1Pns is of degree n−1 and y≠0, the right-hand side does not admit an inverse Laplace transform while the left-hand side does, which is a contradiction.Lemma 3.
Lety∈AC0,T,T>0, be a solution to the multiterm fractional initial value problem (3) with y0=0, then f0=0.Proof.
From Lemma1, since y0=y0=0, then f0=0.Remark 2.
Ify0=f0=0,f∈ℳ1, and cn+1≠∑k=1nck, then a solution y∈ℳ1 exists by Lemma 2. If f∈ℳ2, a solution exists even if cn+1=∑k=1nck, and it can be evaluated using the Laplace transform in the following manner.Lemma 4.
Letyt be a possible solution to multiterm fractional initial value problems (3) and (4), with y0=f0=0, cn+1=∑k=1nck, and f∈ℳ2. If Gs=s2ℒft has an inverse Laplace transform, then yt=h3t∗ℒ−1s2Fs, where h3t=ℒ−1H3s and H3 is defined in equation (22).Proof.
Substitutingy0=0, in equation (14), we have(21)ℒyt=PnssQn−1s−cn+1PnsFs.
Sincecn+1=∑k=1nck, we have sQn−1s−cn+1Pns is a polynomial of degree n−1. Let(22)H3s=1s2PnssQn−1s−cn+1Pns,so that ℒ−1H3s=h3t is well defined. Then, equation (14) yields(23)ℒyt=H3ss2Fs.
Iff∈ℳ2, then ℒyt in equation (23) can be written as(24)ℒyt=H3sℒf″t+sf0+f′0=H3sℒf″t+f′0.
Applying the inverse Laplace transform, we have(25)yt=h3∗f″t+f′0h3t.
We summarize the obtained results in the following main theorem.Theorem 1.
Consider multiterm fractional initial value problems (3) and (4).(1)
Fory0=0, if a solution y∈AC0,T,T>0, exists, then f0=0(2)
Iff∈ℳ1,cn+1y0+f0=0, and cn+1≠∑k=1nck, then a solution y∈ℳ1 exists(3)
Iff∈ℳ2 and y0≠0, then a solution y∈ℳ1 exists if and only if cn+1=∑k=1nck(4)
Ify0=0,f∈ℳ2, and f0=0, then a solution y∈ℳ1 exists


## 3. Illustrative Examples

We discuss two main examples. The first one is a two-term problem where it holds thatcn+1=∑k=1nck. We show the existence of a solution for a specific case by imposing extra conditions. The second example is a three-term problem where it holds that cn+1≠∑k=1nck. So, the existence of a unique solution is guaranteed. We present the solution in a closed form and discuss several special cases.Example 1.
Consider the two-term fractional initial value problem:(26)e−μ1t+e−μ2t∗y′t=2yt+ft,y0=y0.
Sincec1+c2=c3, then the problem has no solution for y0≠0. For y0=f0=0 and s2ℒft has Laplace inverse, the problem admits a solution. To verify, let us find the solution given by equation (25). We have(27)Q1s=2s+μ1+μ2,P2s=s+μ1s+μ2,H3s=−1s2s+μ1s+μ2μ1+μ2s+2μ1μ2=R1s+R2s2+R3μ1+μ2s+2μ1μ2,where(28)R1=−14μ1+μ2μ1μ2,R2=−12,R3=14μ1−μ22μ1μ2.
Thus,(29)h3t=ℒ−1H3s=R1+R2t+R3μ1+μ2e−2μ1μ2/μ1+μ2t,and the solution yt=h3t∗f″t+f′0h3t is obtained, provided that f∈ℳ2.
As a special case, let us considerft=t2. Then, f0=0, ℒft=2/s3, and s2ℒft=2/s have Laplace inverse. Thus,(30)yt=h3t∗f″t+f′0h3t=h3t∗2=2∫0tR1+R2τ+R3μ1+μ2e−2μ1μ2/μ1+μ2τdτ=2R1t+R2t2−R3μ1μ2e−2μ1μ2/μ1+μ2−1.
Forμ1=2/3 and μ2=1/3, we have R1=−9/8,R2=−1/2,R3=1/8, and(31)yt=94t−12t2−916e−4/9t+916.Example 2.
As a second example, we consider the three-term initial value problem:(32)e−μ1t+e−μ2t+e−μ3t∗y′t=yt+ft,y0=y0.
Sincec1=c2=c3=c4=1 and c1+c2+c3≠c4, the problem has a unique solution given by equation (17) provided that y0+f0=0. We have(33)Q2s=∑k=13ck∏j=1,j≠k3s+μj=s+μ2s+μ3+s+μ1s+μ3+s+μ1s+μ2,P3s=∏k=13s+μk=s+μ1s+μ2s+μ3.
Let(34)T3s=sQ2s−c4P3s=2s3+μ1+μ2+μ3s2−μ1μ2μ3,then(35)H1s=Q2sT3s,H2s=1sP3sT3s,with(36)ℒ−1H1s=h1t,ℒ−1H2s=h2t,and the solution is given by(37)yt=y0h1t+h2t∗f′t+f0h2t.
Forft=t, we have y0=−f0=0, and thus(38)yt=h2t∗1=∫0th2τdτ.
Forft=1, we have y0=−f0=−1, and thus(39)yt=h2t−h1t.
As a specific case, if we considerμ1=7/5,μ2=1/2,andμ3=1/3, we have(40)Q2s=3s2+6715s+43,P3s=s3+6730s2+43s+730,T3s=2s3+6730s2−730,H1s=Q2sP3s=90s2+134s+4060s3+67s2−7=A0s+1+A1s+Δ1+A2s+Δ2,H2s=P3ssT3s=30s3+67s2+40s+7ss+1s+Δ1s+Δ2=−1s+H1s,where Δ1=7+1729/120,Δ2=7−1729/120,A0=−2/23,A1=731729−3113/921729,A2=731729+3113/921729, h1t=A0e−t+A1e−Δ1t+A2e−Δ2t, and h2t=h1t−1.
Forft=t, we have yt=∫0th2τdτ=A0+A1/Δ1+A2/Δ2−t−A0e−t−A1/Δ1e−Δ1t−A2/Δ2e−Δ2t.


## 4. Concluding Remarks

We obtained the solutions of a class of multiterm fractional initial value problems in closed forms using the Laplace transform. We have also discussed several necessary and sufficient conditions to guarantee the existence of solutions to the problem. Whether the results are extendable to wider classes of multiterm initial value problems or systems of fractional equations is left for future work.



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