# Some Remarks on Estimate of Mittag-Leffler Function

Authors: Jia Jia; Zhen Wang; Xia Huang; Yunliang Wei
Journal: Journal of Function Spaces
 (2019)

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## Abstract

The estimate of Mittag-Leffler function has been widely applied in the dynamic analysis of fractional-order systems in some recently published papers. In this paper, we show that the estimate for Mittag-Leffler function is not correct. First, we point out the mistakes made in the estimation process of Mittag-Leffler function and provide a counterexample. Then, we propose some sufficient conditions to guarantee that part of the estimate for Mittag-Leffler function is correct. Meanwhile, numerical examples are given to illustrate the validity of the two newly established estimates.


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## Body

## 1. Introduction

Fractional calculus can date back to the seventeenth century, and now it has attracted considerable research interests due to its widespread applications in many fields. There are mainly two types of methods in the dynamic analysis of fractional-order nonlinear systems, that is, Lyapunov function based method and estimation based method. When estimation based method is employed, the solution of the fractional-order system being studied is usually expressed in terms of the Mittag-Leffler function. Obviously, the correctness of the estimate of Mittag-Leffler function is crucial to the whole estimation process and plays an important role if the estimation based method is adopted. Recently, estimation based method has been widely applied to the study of finite-time stability and synchronization of fractional-order memristor-based neural networks [1–7], stability and stabilization of nonlinear fractional-order systems [8–13], finite-time stability of fractional-order neural networks [14, 15], synchronization of fractional-order chaotic systems [16], consensus analysis of fractional-order multiagent systems [17–19], etc. The estimate on Mittag-Leffler function was first proposed in [20]. The definition of Mittag-Leffler function and the estimate of Mittag-Leffler function can be described by Definition 1 and Lemma 2, respectively, as follows.Definition 1 (see [21]).
The Mittag-Leffler function with one parameter is defined as(1)Eαz=∑k=0∞zkΓkα+1, where α>0 and z∈C.
The Mittag-Leffler function with two parameters is defined as(2)Eα,βz=∑k=0∞zkΓkα+β,where α>0 and β>0. When β=1, one has Eα,1(z)=Eα(z), and when α=1 and β=1, one further has E1,1(z)=ez.Lemma 2 (see [20]).
For Mittag-Leffler function, the following properties hold.
(i) There exist constantsM1,M2≥1 such that, for any 0<α<1,(3)Eα,1Atα≤M1eAt,(4)Eα,αAtα≤M2eAt,where A denotes a matrix and · denotes any vector or induced matrix norm.
(ii) Ifα≥1, then for β=1,2,α(5)Eα,βAtα≤eAtα.If A is a diagonal stability matrix, then there exists a constant N>0 such that for t≥0(6)Eα,βAtα≤Ne-ωt,0<α<1,(7)Eα,βAtα≤e-ωt,1≤α<2,where -ω(ω>0) is the largest eigenvalue of the diagonal matrix A.However, we have to point out that Lemma2 is incorrect.In [20], inequalities (3) and (4) are proved as follows:(8)Eα,βAtα=∑k=0∞AtαkΓkα+β=∑k=0∞k!tk1-αΓkα+βAtkk!≤supk=0,1,2,…,∞⁡k!tk1-αΓkα+β∑k=0∞Atkk!=supk=0,1,2,…,∞⁡k!tk1-αΓkα+β∑k=0∞Atkk!≤supτ∈1,∞⁡supk=0,1,2,…,∞⁡k!τk1-αΓkα+βeAt.Actually, there are two problems in the above-mentioned proof. First,(9)∑k=0∞k!tk1-αΓkα+βAtkk!≤supk=0,1,2,…,∞⁡k!tk1-αΓkα+β∑k=0∞Atkk!does not necessarily hold. In fact, (9) holds if all the elements aij(1≤i,j≤n) of matrix A are nonnegative, because matrix norms, such as 1-norm, 2-norm, and ∞-norm, have the property of weak monotony. In other words, (9) may not hold when there exist negative elements in A or Ak. Second, supk=0,1,2,…,∞k!/tk(1-α)Γ(kα+β) does not exist when 0<α<1 and β=1,α. To confirm this point, let N(k)=k!/tk(1-α)Γ(kα+β); when t=2s and α=0.9, the behavior of N(k) with β=1 and β=α is shown in Figures 1(a) and 1(b), respectively. It can be obviously observed from Figure 1 that N(k) goes to infinity as k goes to infinity, so N(k) has no supremum when β=1 or β=α as k goes to infinity for a fixed value of t.Figure 1
The behavior ofN(k) with (a) β=1 and (b) β=α.
(a) (b)Next, a counterexample is presented to show thatEα,β(Atα) does not satisfy inequality (3) or (4). Assume that A=122234781; by simple calculation, it has three different eigenvalues, i.e., λ1=9.6135, λ2=-0.3146, and λ3=-4.2990. Hence, a nonsingular matrix P=-0.3085-0.7235-0.1680-0.55890.6641-0.4379-0.7698-0.18850.8832 can be determined to make (10)P-1AP=Λ=diag⁡λ1,λ2,λ3and(11)Eα,βAtα=∑k=0∞PΛtαP-1kΓkα+β=P∑k=0∞ΛtαkΓkα+βP-1=PEα,βλ1tα000Eα,βλ2tα000Eα,βλ3tαP-1.For each fixed value of t, Eα,β(λitα), i=1,2,3 can be calculated by means of the OPC algorithm [22]. Thus, Eα,β(Atα) can be calculated through (11). When α=0.9, the behaviors of Eα,β(Atα)1/eAt1 with β=1 and β=α are displayed in Figures 2(a) and 2(b), respectively. It is obvious that Eα,β(Atα)1/eAt1 goes to infinity as t goes to infinity, so inequalities (3) and (4) are incorrect.Figure 2
The behavior ofE0.9,β(At0.9)1/eAt1 with (a) β=1 and (b) β=α.
(a) (b)The conclusion on inequality (6) is straightforward if inequalities (3) and (4) are correct. Because inequalities (3) and (4) are not correct, inequality (6) is not correct either. For example, let A=diag⁡(-1,-5,-8); the behavior of E0.9,β(At0.9)1/e-t for β=1 and β=α=0.9 is shown in Figures 3(a) and 3(b), respectively. It is clear that E0.9,β(At0.9)1/e-t goes to infinity as t goes to infinity for β=1 and β=α, so inequality (6) does not hold.Figure 3
The behavior ofE0.9,β(At0.9)1/e-t with (a) β=1 and (b) β=α.
(a) (b)Next, we consider the case thatα≥1. In [20], inequality (5) is proved as follows:(12)Eα,βAtα=∑k=0∞AtαkΓkα+β=∑k=0∞k!Γkα+βAtαkk!≤supk=0,1,2,…,∞⁡k!Γkα+β∑k=0∞Atαkk!=supk=0,1,2,…,∞⁡k!Γkα+βeAtα≤supk=0,1,2,…,∞⁡k!Γk+1eAtα=eAtα.With the same argument as stated for (8),(13)∑k=0∞k!Γkα+βAtαkk!≤supk=0,1,2,…,∞k!Γkα+β∑k=0∞Atαkk! does not necessarily hold and it is true if all the elements aij(1≤i,j≤n) of matrix A are nonnegative. The other problem with (12) is that supk=0,1,2,…,∞k!/Γ(kα+β)≤supk=0,1,2,…,∞k!/Γ(k+1) does not hold for 1<α<2 and β=α as supk=0,1,2,…,∞k!/Γ(kα+β)≥1/Γβ=1/Γα>1. For example, when α=β=1.5, supk=0,1,2,…,∞k!/Γ(kα+β)≥1/Γ(1.5)≈1.1284>1. The behavior of Gamma function and its derivative is depicted in Figures 4(a) and 4(b), respectively. From Figure 4, it is clear that 0.8<Γ(z)<1 for 1<z<2. Hence, we can conclude that supk=0,1,2,…,∞k!/Γ(kα+β)≥1/Γβ>1 for β=α when 1<α<2.Figure 4
The behavior of (a)Γ(z) and (b) dΓ(z)/dz.
(a) (b)From the above discussions, we can infer that inequality (5) holds only under some particular conditions; that is, we have to impose some restrictions on matrix A and α,β.Conclusion 3.
Suppose all the elementsaij(1≤i,j≤n) of matrix A are nonnegative; if 1≤α<2, then for β=1,2, inequality (5) holds.Conclusion 4.
Suppose all the elementsaij(1≤i,j≤n) of matrix A are nonnegative; if α≥2, then for β=1,2,α, inequality (5) holds.Now, a counterexample is presented to show that if the conditions in Conclusion3 or Conclusion 4 are not satisfied, inequality (5) may not be correct. For instance, let A=diag⁡(-1,-5,-8); then the behavior of E1.5,β(At1.5)1/eAt1.51 for β=1, β=2, and β=1.5 is shown in Figures 5(a)–5(c), respectively. It is obvious that E1.5,β(At1.5)1/eAt1.51 goes to infinity as t goes to infinity, so inequality (5) does not hold.Figure 5
The behavior ofE1.5,β(At1.5)1/eAt1.51 for (a) β=1, (b) β=2, and (c) β=1.5.
(a) (b) (c)Similarly, (7) is incorrect because (5) is incorrect. The behavior of E1.5,β(At1.5)1/e-t for β=1, β=2, and β=1.5 is shown in Figures 6(a)–6(c), respectively, which is in contradiction to inequality (7).Figure 6
The behavior ofE1.5,β(At1.5)1/e-t for (a) β=1, (b) β=2, and (c) β=1.5.
(a) (b) (c)


## 2. Main Results

In this section, some sufficient conditions are derived to guarantee thatEα,β(Atα) can be bounded by FeAtα for some F>0, which can be formulated by the following two theorems.Theorem 5.
If matrixA is diagonalizable, and the largest real part of eigenvalues λi(i=1,2,…,n) of A is positive, then for 1<α<2 and anyone of the following two conditions:(i)
β>0,(ii)
β=0 and A has no zero eigenvalue,(14)limt→+∞⁡Eα,βAtαeAtα=0,
and further there exists a positive constantF such that for t≥0(15)Eα,βAtα≤FeAtα,where · denotes 1-norm, 2-norm, or ∞-norm of a matrix.To prove Theorem5, another two lemmas are presented as follows, which will be used later.Lemma 6 (see [21]).
Ifα<2, β is an arbitrary real number, μ satisfies πα/2<μ<min⁡{π,πα}, and C1 and C2 are real constants, then(16)Eα,βz≤C11+z1-β/αexp⁡Re⁡z1/α+C21+z,where argz≤μ,z≥0.Lemma 7 (see [21]).
Ifα<2, β is an arbitrary real number, μ satisfies πα/2<μ<min{π,πα}, and C is a real constant, then(17)Eα,βz≤C1+z,where μ≤argz≤π,z≥0.Now, the proof of Theorem5 can proceed.Proof.
BecauseA is diagonalizable, there exists a nonsingular matrix P such that P-1AP=Λ=diag⁡(λ1,λ2,…,λn). Then Eα,β(Atα)=PEα,β(Λtα)P-1 and eAtα=PeΛtαP-1. Since(18)λI-eAtα=λI-PeΛtαP-1=PλIP-1-PeΛtαP-1=PλI-eΛtαP-1=PλI-eΛtαP-1=λI-eΛtα,eAtα and eΛtα are with the same characteristic polynomial and eigenvalues, so eAtα≥ρ(eAtα)=max1≤i≤n{eλitα}. Let m=max1≤i≤nRe⁡λi>0; then eAtα≥emtα.
Letθi be the principal value of the argument of λi. According to the magnitude of the principal value θi of the argument of λi, the cases where θi≤πα/2 and πα/2<θi≤π are considered, separately.
Case 1. If θi≤πα/2, it follows from Lemma 6 that(19)Eα,βλitα≤C11+λitα1-β/αeRe⁡λitα1/α+C21+λitα=C11+λitα1-β/αeRe⁡λi1/αt+C21+λitα=C11+λitα1-β/αeλi1/αcos⁡θi+2kπ/αt+C21+λitα≤C11+min1≤i≤n⁡λitα1-β/αemax1≤i≤nλi1/αt+C21+min1≤i≤nλitα,where k=0,1,2,…,q-1. q is the numerator of α, where α=q/p, (p,q)=1.
Case 2. If πα/2<θi≤π, it follows from Lemma 7 that(20)Eα,βλitα≤C1+λitα=C1+λitα≤C1+min1≤i≤nλitα.Then, it follows from (19) and (20) that(21)Eα,βλitα≤C11+min1≤i≤nλitα1-β/αemax1≤i≤nλi1/αt+Cmax1+min1≤i≤nλitα,where Cmax=max{C2,C}.
Thus, we have(22)limt→+∞⁡Eα,βAtαeAtα≤limt→+∞⁡PEα,βΛtαP-1emtα≤limt→+∞⁡PEα,βΛtαP-1emtα=limt→+∞⁡Pmax1≤i≤nEα,βλitαP-1emtα≤limt→+∞⁡C11+min1≤i≤n⁡λitα1-β/αemax1≤i≤nλi1/α-mtα-1t×PP-1+limt→+∞⁡Cmaxe-mtα1+min1≤i≤nλitαPP-1.When any one of the following two conditions is satisfied: (1)
β≥1,(2)
0<β<1 and ∃i∈1,2,…,n such that λi=0,
we have(23)limt→+∞⁡1+min1≤i≤nλitα1-β/αemax1≤i≤nλi1/α-mtα-1t=0. When 0≤β<1 and λi≠0(i=1,2,…,n), the following equality can be derived by L'Hospital's rule,(24)limt→+∞⁡1+min1≤i≤nλitα1-β/αemax1≤i≤nλi1/α-mtα-1t=limt→+∞⁡1-βmin1≤i≤nλi1+min1≤i≤nλitα1-β+α/αtα-1emtα-1-max⁡1≤i≤nλi1/αtαmtα-1-max⁡1≤i≤nλi1/α.As(25)limt→+∞1+min1≤i≤n⁡λitα1-β+α/αtα-1=limt→+∞⁡tα-11+min1≤i≤nλitαβ+α-1/α=limt→+∞⁡t-βt-α+min1≤i≤nλiβ+α-1/α=0,β≠0,1min1≤i≤nλiα-1/α,β=0, and(26)limt→+∞⁡emtα-1-max⁡1≤i≤nλi1/αtαmtα-1-max1≤i≤nλi1/α=+∞, it can be obtained from (24) that limt→+∞1+min1≤i≤nλitα(1-β)/αemax1≤i≤nλi1/α-mtα-1t=0 when 0≤β<1 and λi≠0(i=1,2,…,n). To summarize, (27)limt→+∞1+min1≤i≤nλitα1-β/αemax1≤i≤nλi1/α-mtα-1t=0 if conditions i or ii in Theorem 5 are satisfied. It is obvious that limt→+∞Cmaxe-mtα/1+min1≤i≤nλitα=0. According to (22), we have limt→+∞Eα,βAtα/eAtα=0. Due to the continuity of matrix norms, there exists a positive constant F such that Eα,β(Atα)/eAtα≤F, which implies that inequality (15) holds. This completes the proof.Now, an example is presented to verify the correctness of the newly established Theorem5. Assume that A=122234781; it is diagonalizable and the eigenvalues are λ1=9.6135,λ2=-0.3146,λ3=-4.2990, respectively. Hence, the largest real part of the eigenvalues is m=9.6135>0; then the behavior of E1.5,β(At1.5)1/eAt1.51 for β=0, β=0.3, β=0.8, β=1, β=2, and β=1.5 is shown in Figures 7(a)–7(f), respectively. It can be seen from Figure 7 that E1.5,β(At1.5)1/eAt1.51 converges to zero, which indicates the validity of Theorem 5.Figure 7
The behavior ofE1.5,β(At1.5)1/eAt1.51 for (a) β=0, (b) β=0.3, (c) β=0.8, (d) β=1, (e) β=2, and (f) β=1.5.
(a) (b) (c) (d) (e) (f)Remark 8.
Note that the condition that1<α<2 is needed in Theorem 5. If 0<α<1, then 1+min1≤i≤nλitα(1-β)/αemax1≤i≤nλi1/α-mtα-1t may go to infinity as t goes to infinity, so the process of the proof cannot be carried out and the conclusion in Theorem 5 may not be obtained. To get the similar estimate of Mittag-Leffler function for 0<α<1, an extra restriction has to be imposed on the eigenvalues of matrix A, which is given in the following Theorem 9.Theorem 9.
If matrixA is diagonalizable and satisfies the following two conditions:(i)
the largest real part of eigenvaluesλii=1,2,…,n of A is positive,(ii)
the principal valueθi of the argument of λi satisfies πα/2<θi≤π, ∀i=1,2,…,n,
then for0<α<1 and β∈R,(28)limt→+∞⁡Eα,βAtαeAtα=0,and further there exists a positive constant G such that for t≥0(29)Eα,βAtα≤GeAtα,where · denotes 1-norm, 2-norm, or ∞-norm of a matrix.Proof.
According to conditionii in Theorem 9, (20) holds for each λi, i=1,2,…,n. Thus we have(30)limt→+∞⁡Eα,βAtαeAtα≤limt→+∞⁡PEα,βΛtαP-1emtα≤limt→+∞⁡PEα,βΛtαP-1emtα=limt→+∞⁡Pmax1≤i≤nEα,βλitαP-1emtα≤limt→+∞⁡Ce-mtα1+min1≤i≤nλitαPP-1=0.
Similarly, we can prove that there exists a positive constantG such that Eα,β(Atα)/eAtα≤G, which implies that inequality (15) holds. This completes the proof.Now, an example is presented to verify the correctness of Theorem9. Assume that A=-3000120-11; it is diagonalizable and the eigenvalues are λ1=-3, λ2=1-j2, λ3=1+j2, respectively, where j=-1. It is obvious that the largest real part of the eigenvalues is m=1>0. Choose α=0.5; then θ1=π>πα/2=0.25π, θ2=θ3=0.9553∈0.25π,π. Hence the two conditions in Theorem 9 are all satisfied, and Figures 8(a)–8(c) depict the behavior of E0.5,β(At0.5)1/eAt0.51 for β=0.5, β=1, and β=3, respectively, which shows that E0.5,β(At0.5)1/eAt0.51 converges to zero and indicates the validity of Theorem 9.Figure 8
The behavior ofE0.5,β(At0.5)1/eAt0.51 for (a) β=0.5, (b) β=1, and (c) β=3.
(a) (b) (c)Remark 10.
Note that the condition that matrixA is diagonalizable is needed in Theorems 5 and 9; otherwise, there exists a nonsingular matrix P such that P-1AP=J=diag⁡J1,J2,…,Js, where Ji=diag⁡Ji1,Ji2,…,Jiri and(31)Jik=λi1λi⋱⋱1λiuik×uik.One can obtain that Eα,βAtα=PEα,βJtαP-1 and eAtα=PeJtαP-1. In the proof of Theorems 5 and 9, the lower bound of eAtα=PeJtαP-1 and the upper bound of Eα,βAtα=PEα,βJtαP-1 are needed. When A is not diagnosable, eAtα≥emtα still holds but the upper bound of Eα,βAtα=PEα,βJtαP-1 is difficult to obtain. The difficulties are described as follows.
For the Jordan block (31), we have(32)Eα,βJiktα=QikEα,βλitαdEα,βλitαdλitα12!d2Eα,βλitαdλitα2…1uik-1!duik-1Eα,βλitαdλitαuik-1Eα,βλitαdEα,βλitαdλitα…1uik-2!duik-2Eα,βλitαdλitαuik-2Eα,βλitα…1uik-3!duik-3Eα,βλitαdλitαuik-3⋱⋮Eα,βλitαuik×uikQik-1,where Qik=111…11/tα1/tα…1/tα1/t2α…1/t2α⋱⋮1/tuik-1α is a nonsingular matrix such that Qik-1JiktαQik=λitα1λitα⋱⋱1λitαuik×uik. As Jtα is composed of Jiktα, it is difficult calculate Eα,βJtα via (32), so the upper bound of Eα,βAtα is difficult to obtain and Eα,βAtα/eAtα is difficult to estimate.
To the best of our knowledge, the estimate of Mittag-Leffler function by the exponential function is still an open problem due to the complexity of Mittag-Leffler function and deserves further research.


## 3. Conclusion

In this paper, several counterexamples are presented to numerically show that the estimate for Mittag-Leffler function used in some recently published papers is not completely correct and the mistakes made in the estimation process are mainly due to the misuse of the properties of matrix norms. Besides, some sufficient conditions are developed to guarantee that the estimateEα,β(Atα)≤FeAtα holds for some F>0 and numerical examples are given to verify the correctness of the newly developed results.



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