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Abstract

Matrix exponentials are widely used to efficiently tackle systems of linear differential equations. To be able to solve systems of fractional differential equations, the Caputo matrix exponential of the index α > 0 was introduced. It generalizes and adapts the conventional matrix exponential to systems of fractional differential equations with constant coefficients. This paper analyzes the most significant properties of the Caputo matrix exponential, in particular those related to its inverse. Several numerical test examples are discussed throughout this exposition in order to outline our approach. Moreover, we demonstrate that the inverse of a Caputo matrix exponential in general is not another Caputo matrix exponential.
mathematics
Article
On the Inverse of the Caputo Matrix Exponential
Emilio Defez 1, Michael M. Tung 1,*, Benito M. Chen-Charpentier 2and José M. Alonso 3
1Instituto de Matemática Multidisciplinar, Universitat Politècnica de València, Camino de Vera, s/n,
46022 Valencia, Spain; edefez@imm.upv.es
2Department of Mathematics, University of Texas at Arlington, Arlington, TX 76019-0408, USA;
bmchen@uta.edu
3Instituto de Instrumentación para Imagen Molecular, Universitat Politècnica de València, Camino de Vera,
s/n, 46022 Valencia, Spain; jmalonso@dsic.upv.es
*Correspondence: mtung@imm.upv.es or mtung@mat.upv.es
Received: 1 October 2019; Accepted: 18 November 2019; Published: 21 November 2019


Abstract:
Matrix exponentials are widely used to efficiently tackle systems of linear differential
equations. To be able to solve systems of fractional differential equations, the Caputo matrix
exponential of the index
α>
0 was introduced. It generalizes and adapts the conventional matrix
exponential to systems of fractional differential equations with constant coefficients. This paper
analyzes the most significant properties of the Caputo matrix exponential, in particular those related
to its inverse. Several numerical test examples are discussed throughout this exposition in order to
outline our approach. Moreover, we demonstrate that the inverse of a Caputo matrix exponential in
general is not another Caputo matrix exponential.
Keywords: Caputo matrix exponential; matrix inverse; fractional derivative
MSC: 15A09; 15A16
1. Introduction and Motivation
Formally a square matrix
ACr×r
can be associated with its exponential matrix function
eAt
.
The traditional matrix exponential takes a prominent position among all matrix functions—ultimately
due to its relevance in the resolution of systems of first-order ordinary differential equations. However,
in practice its efficient numerical computation poses considerable difficulties, see [1] for details.
At the same time, systems of fractional differential equations, which contain derivatives extending
the standard integer-order derivative to arbitrary order
α
0, play an important role in many other
important applications of science and engineering [
2
4
]. Although fractional calculus is factually
known since the end of the 17th century [
5
], only during the recent decades its relevance for practical
modeling and engineering simulations has become evident. Fractional derivatives naturally implement
Volterra’s “principle of the dissipation of hereditary action”, meaning that causality aspects and
memory characteristics of dynamical systems may easily be incorporated. Important applications are
in hydrology, e.g., flow simulations of fluids in porous media, and in civil engineering, e.g., traffic flow
problems on road networks, among many others [6,7].
A great variety of fractional derivatives are proposed and used in the literature. The most
common fractional derivative is the derivative introduced by Caputo [
8
]. It is defined in terms of the
Riemann–Liouville fractional integral of order α0 operating on function f(t):
Jαf(t) = 1
Γ(α)Zt
0(tτ)α1f(τ)dτ,t>0.
Mathematics 2019,7, 1137; doi:10.3390/math7121137 www.mdpi.com/journal/mathematics
Mathematics 2019,7, 1137 2 of 11
Then, provided that
f
is a locally integrable function, the following operation on
f
defines its
fractional derivative of Caputo with order α0:
Dαf(t) = JmαDmf(t),t>0, m1<αm,mN.
Note that, as expected,
Dn=dn/dtn
agrees with the usual derivative of integer order
nN
.
(Our convention is to use Nfor the set of all positive integer numbers, whereas N0=N{0}.)
In 2016, Rodrigo [
9
] introduced the fractional exponential matrix of Caputo of order
α
0.
Similarly, here we are using the following definition for 0 α1:
exp?(tαA;α)=
n0
Antαn
Γ(αn+1),t>0, (1)
in relation with the Mittag–Leffler matrix function [10].
It is well-known that inverse problems [
11
] are among the most basic applications for the inverse
of the conventional exponential matrix. Moreover, mathematical optimisation is another area in which
the inverse of the matrix exponential is frequently encountered and of significant relevance, see, e.g.,
(Reference [
12
], Equations (4.4) and (4.7)). Observe that if in a linear differential system all ordinary
derivatives are replaced by fractional derivatives of Caputo type, then the associated inverse problems
will necessarily involve the inverse of the Caputo matrix exponential.
In the case of the Caputo matrix exponential,
(1)
, there still remains to clarify the existence of its
inverse, in full analogy to the case of the conventional matrix exponential
eAt
with inverse matrix
eAt
.
The final objective of this work will be to study the existence and computation of the inverse of the
Caputo matrix exponential.
The present paper is organized as follows. Section 2first focuses on checking the main properties
of the matrix exponential of Caputo, and also on presenting counterexamples of other questionable
properties which eventually are not satisfied. In Section 3, we will demonstrate that, in general,
the inverse of an exponential matrix of Caputo is not another exponential matrix of Caputo. Finally,
Section 4concludes with the actual computation of the inverse of the Caputo matrix exponential and
gives examples.
In the remainder of this work, we will denote by
Cp×q
the set of rectangular complex matrices.
For a square matrix
ACr×r
, as usual
σ(A)
denotes the spectrum of matrix
A
, i.e., the set of its
eigenvalues. Moreover, we will denote by
kAk
any multiplicative norm of matrix
A
. In particular,
kAk2is the 2-norm, defined by
kAk2=sup
z6=0
kAzk2
kzk2
,
where for any vector
zCq
, the usual Euclidean norm of
z
is
kzk2= (ztz)1/2
. Additionally, it will
be helpful to remember that for a family of matrices
A(k
,
n)Cr×r
with
n
and
k
being positive,
the following identity holds
n0
k0A(k,n) =
n0
n
k=0A(k,nk). (2)
This identity is analogous to the one of the proof for Lemma 11 in (Reference [13], p. 57).
2. Caputo Matrix Exponential
This section first presents some of the fundamental properties of the Caputo matrix exponential
which will be built upon in subsequent parts of this work. Then, the next subsection concentrates on
some striking counterexamples of properties which one could naively intuit but which at the end do
not hold. The final subsection centers on the existence of the inverse of the Caputo matrix exponential
and its conditions.
Mathematics 2019,7, 1137 3 of 11
2.1. Properties
In the following, we list the most important fundamental properties of the Caputo matrix
exponential which unreservedly have to be fulfilled:
(a) For α=1, the Caputo matrix exponential coincides with the conventional matrix exponential:
exp?(tA; 1)=eAt . (3)
(b) If 0r×r,Ir×rare the null and identity matrices of Cr×r, respectively, it is clear that
exp?(0r×r;α)=Ir×r. (4)
(c)
If
ACr×r
, and
σ(A)
denotes the set of its eigenvalues, it is well known that
A
has the Jordan
canonical factorization A=P J P1, where Jis a diagonal block-matrix given by
J=diag {J1,J2, . . . , Jk},Ji=
λi10
λi
...
...1
0λi
,λiσ(A).
Then from Definition (1), it immediately follows that
exp?(tαA;α)=Pdiag exp?(tαJ1;α), exp?(tαJ2;α), . . . , exp?(tαJk;α)P1. (5)
(d)
Avoiding entirely the Jordan canonical form of
A
and only knowing
σ(A)
, Putzer’s method
(see e.g., [9]) allows to explicitly obtain exp?(tαA;α)in fully analytical form.
2.2. Counterexamples
However, there are obvious differences between
exp?(tαA;α)
and the matrix exponential
eAt
,
which are straightforward to detect considering the scalar case r=1.
(1)
The matrix exponential
eAt
is a periodic function of period
T=
2
πi Ir×r
, where
i
as usual is the
imaginary unit:
eAt =eAt+2πi Ir×r.
However, this is not the case for the Caputo matrix exponential, even in the scalar case (
r=
1).
In fact, it easily can be checked that for A=1, t=1 and α=1/2, we have
exp?(1; 1/2)5.00898
exp?(1+2πi; 1/2)≈ −0.0144688 +0.0885799i,
so that
exp?(1; 1/2)6=exp?(1+2πi; 1/2).
Thus, we generally conclude that
exp?(A;α)6=exp?(A+2πiIr×r;α). (6)
(2) It is well known that if Aand Bare two commuting matrices, i.e., AB =BA, then
e(A+B)t=eAteBt. (7)
Mathematics 2019,7, 1137 4 of 11
This relation is generally not true for the fractional exponential matrix of Caputo—even for
the simplest scalar case (
r=
1). In fact, we can easily observe that when we take
A=B=1,
t=1, α=1/2, we have
exp?(1; 1/2)5.00898
exp?(2; 1/2)108.941,
so that
exp?(2; 1/2)6=exp?(1; 1/2)exp?(1; 1/2).
Consequently, we generally have
exp?(tα(A+B);α)6=exp?(tαA;α)exp?(tαB;α), (8)
and the Caputo matrix exponential therefore does not satisfy the semigroup property.
(3)
If we denote by
Det(A)
the determinant of the square matrix
A
and by
Tr(A)
its trace, i.e.,
the sum of the elements on the main diagonal, it is well known that the matrix exponential
satisfies
Det eA=eTr(A). (9)
In this way, it becomes obvious that the usual exponential matrix
eA
is always invertible, since
its determinant is always non-zero. Observe that the analogous identity for the Caputo matrix
exponential is not true, i.e.,
Det(exp?(A;α))6=exp?(Tr(A);α)
. To prove that this property is
not true, it is easy to check that
exp? tα 21
43!;α!=
4
3Eα(tα)1
3Eα(2tα)1
3Eα(2tα)1
3Eα(tα)
4
3Eα(tα)4
3Eα(2tα)4
3Eα(2tα)1
3Eα(tα)
,
where Eα(z)is the Mittag-Leffler function defined by
Eα(z) =
j0
zj
Γ(αj+1). (10)
Now, taking t=1, α=1/2 and Tr 21
43!=1, one gets that
Det exp? 21
43!; 1/2!!1.27927 , exp?(1; 1/2)0.427584,
so that in general
Det(exp?(A;α))6=exp?(Tr(A);α).
(4) As a consequence of (7), it follows that for ACr×rit is
eAteAt =Ir×r. (11)
For this reason the exponential matrix
eAt
is always invertible, and its inverse is precisely
eAt
.
On the other hand, for the inverse of the Caputo matrix exponential, it is easy to verify that
property (11) is not fulfilled.
As an example, we consider the two matrix exponentials
exp? tα 21
43!;α!=
4
3Eα(tα)1
3Eα(2tα)1
3Eα(2tα)1
3Eα(tα)
4
3Eα(tα)4
3Eα(2tα)4
3Eα(2tα)1
3Eα(tα)
Mathematics 2019,7, 1137 5 of 11
and
exp? tα 21
43!;α!=
4
3Eα(tα)1
3Eα(2tα)1
3Eα(2tα)1
3Eα(tα)
4
3Eα(tα)4
3Eα(2tα)4
3Eα(2tα)1
3Eα(tα)
.
For the choice α=1/2 with t=1, we obtain
exp? 21
43!; 1/2!exp? 21
43!; 1/2! 6.41867 8.56043
34.2417 36.3835 !6=I2×2,
and it clearly is
exp?(tαA;α)exp?(tαA;α)6=Ir×r.
2.3. Existence
In order to guarantee the existence of the inverse of the Caputo matrix exponential
exp?(tαA;α)
for ACr×r, observe that from definition (1), for α>0, t0, it follows that
Ir×rexp?(tαA;α)
=
n1
Antαn
Γ(αn+1)
n1kAktαn
Γ(αn+1)=EαkAktα1.
Then, according to Lemma 2.3.3 in (Ref. [
14
], p. 58), matrix
exp?(tαA;α)
is invertible in the interval
I= [0, t?], where
Eα(kAktα)<2. (12)
Taking into account that the Mittag–Leffler function
g(t) = Eα(kAktα)
satisfies
g(
0
) =
1, and it
is a strictly increasing function for
t(
0,
+)
, we can conclude that there always exists
t?
so that
inequality (12) holds in I= [0, t?]. Therefore, exp?(tαA;α)1also exists, at least for tI.
Example 1.
For the particular case
A= 21
43!
and also index
α=
1
/
4, it is easy to check that
kAk25.46499. Then, it holds
E0.25 5.46499 t1/4<2t[0, 0.0000594].
Thus, if t [0, 0.0000594], the inverse exp? t1/4 21
43!; 1/4!1
exists.
3. A New Inversion Property of the Caputo Matrix Exponential
In Section 2.3, we proved the existence of the inverse of the Caputo matrix exponential. It is
well-known that the inverse of the conventional matrix exponential
eAt
is again an exponential of
the matrix
At
, or simply
eAt1=eAt
. So can we arrive at a similar property for the Caputo
matrix exponential?
For this purpose, let us consider the matrix
A=
0100
0010
0001
0000
. (13)
Mathematics 2019,7, 1137 6 of 11
The square matrix
A
is a nilpotent matrix of index 3, meaning that
A36=
0
4×4
but
An=
0
4×4
for
n4. Thus, applying Definition (1), it is easy to establish
exp?(tαA;α)=
3
n=0
Antαn
Γ(nα+1)=
1tα
Γ(α+1)t2α
Γ(2α+1)t3α
Γ(3α+1)
0 1 tα
Γ(α+1)t2α
Γ(2α+1)
0 0 1 tα
Γ(α+1)
0 0 0 1
.
Now, we proceed to calculate its inverse and obtain
exp?(tαA;α)1=
1tα
Γ(α+1)t2α
Γ(α+1)2t2α
Γ(2α+1)t3α
Γ(α+1)3+2t3α
Γ(α+1)Γ(2α+1)t3α
Γ(3α+1)
0 1 tα
Γ(α+1)t2α
Γ(α+1)2t2α
Γ(2α+1)
0 0 1 tα
Γ(α+1)
0 0 0 1
. (14)
Suppose that there exists a matrix BC4×4such that
exp?(tαA;α)1=exp?(tαB;α)=
n0
Bntαn
Γ(nα+1). (15)
Then, we may recast the expression for the inverse given by (14) in the form
exp?(tαA;α)1=I4×41
Γ(α+1)
0100
0010
0001
0000
tα+1
Γ(α+1)21
Γ(2α+1)
0010
0001
0000
0000
t2α
+1
Γ(α+1)3+2
Γ(α+1)Γ(2α+1)1
Γ(3α+1)
0001
0000
0000
0000
t3α
=I4×41
Γ(α+1)Atα+1
Γ(α+1)21
Γ(2α+1)A2t2α
+1
Γ(α+1)3+2
Γ(α+1)Γ(2α+1)1
Γ(3α+1)A3t3α.
(16)
Equating the powers of
tα
in (15) and (16), we observe that matrix
B
must satisfy the
following system
B=A
B2=Γ(2α+1)1
Γ(α+1)21
Γ(2α+1)A2
B3=Γ(3α+1)1
Γ(α+1)3+2
Γ(α+1)Γ(2α+1)1
Γ(3α+1)A3
. (17)
Eliminating recursively all matrices of the previous system yields
Γ(2α+1)1
Γ(α+1)21
Γ(2α+1)=1
Γ(3α+1)1
Γ(α+1)3+2
Γ(α+1)Γ(2α+1)1
Γ(3α+1)=1
. (18)
If the first equation of (18) holds, i.e., α>0 satisfies
Γ(2α+1)
Γ(α+1)2=2, (19)
Mathematics 2019,7, 1137 7 of 11
then the second equation of (18) also holds. Equation (19) has the unique solution
α=
1, and therefore
also system (17). In consequence—except for the trivial case
α=
1—we affirm that the inverse of the
Caputo matrix exponential generally is not another Caputo matrix exponential.
4. On the Computation of the Inverse of the Caputo Matrix Exponential
We now propose to determine the inverse of the Caputo matrix exponential. For this, we introduce
the following definition:
Definition 1.
Let
ACr×r
be an arbitrary square matrix and
α>
0. We define the sequence of matrices
{Dn(α)}n0as
D0(α) = Ir×r,Dn(α) =
n1
k=0
AnkDk(α)
Γ[(nk)α+1],n1. (20)
We are now in the position to proceed with the following theorem, which is a refinement of the
arguments already presented in Section 2.3, explaining why the inverse of
exp?(tαA;α)
exists for
t>
0,
though sufficiently small, and satisfying inequality (12).
Theorem 1.
Let
ACr×r
be a square matrix and 0
<α
1. Let
t>
0be such that the fractional matrix
function f (t,α), defined by
f(t,α) =
n0
Dn(α)tnα, (21)
converges. Then, it holds
(a) exp?(tαA;α)f(t,α) = Ir×r,
(b) f (t,α)exp?(tαA;α)=Ir×r.
Proof of Theorem 1.For the proof of convergence of f(t,α), recall that asymptotically
Γ(nα+1)2πeαnαnαn+1
2,n,
so that we conclude
lim
n
Γ(nα+1)1/n
nα=eααα.
Hence, the series
n1
znAn
Γ(nα+1)
converges for all
zC
. This convergence occurs uniformly on compact subsets of
C
. Therefore, the set
(zC: sup
λσ(A)
n1
znλn
Γ(nα+1)
<1)
contains a circular disc centered at the origin,
D
0,
r(A
,
α)
, with radius
r(A
,
α)>
0. This radius can be
determined by considering the Mittag-Leffler function already introduced in
(10)
, specifically
Eαzλ
for z,λC, which is analytic.
By the spectral mapping theorem, if
ACr×r
, the spectrum
σEα(zA)
of
EαzACr×r
satisfies
σEα(zA)={Eα(zλ):λσ(A)}.
Mathematics 2019,7, 1137 8 of 11
If Ais not a nilpotent matrix, then sup |λ|:λσ(A)>0. In this case, we choose
rA,α=inf |z|:Eα(zλ) = 0 for some λσ(A)
=inf |w|
|λ|:Eα(w) = 0, wC,λσ(A)
=inf {|w|:Eα(w) = 0, wC}
sup |λ|:λσ(A)>0
because
Eα(
0
) =
1. Moreover, the series
n0Dn(α)zn
converges for
|z|<rA
,
α
, because on the disc
D0, r(A,α)the matrix function EαzAis invertible and analytic in C.
If
A
is a nilpotent matrix, i.e.,
σ(A)={
0
}
, and has index
kN
, then
Dn(α)
defined in
(20)
vanishes when nk. Thus, series (21) only has a finite number of terms and apparently converges.
Now, we proceed with the remainder of the Theorem 1, proving part (a).
(a) Applying the respective Definitions (1) and (21), we compute
exp?(tαA;α)f(t,α) =
n0
Antnα
Γ(αn+1)!
k0
Dk(α)tkα!
=
n0
k0
AnDk(α)tkαtαn
Γ(αn+1), taking A(n,k) = AnDk(α)tkαtαn
Γ(αn+1)of (2),
=
n0
n
k=0
AnkDk(α)tkαt(nk)α
Γ[(nk)α+1]=
n0 n
k=0
AnkDk(α)
Γ[(nk)α+1]!tnα
=D0(α) +
n1 n
k=0
AnkDk(α)
Γ[(nk)α+1]!tnα
=Ir×r+
n1 n1
k=0
AnkDk(α)
Γ[(nk)α+1]+Dn(α)!tnα
=Ir×r+
n1 n1
k=0
AnkDk(α)
Γ[(nk)α+1]
n1
k=0
AnkDk(α)
Γ[(nk)α+1]!tnα
=Ir×r.
(b)
This equality is equivalent to (a), because the matrix operators
exp?(tαA;α)
and
f(t
,
α)
commute.
In fact, let
f(z)
and
g(z)
be holomorphic functions of the complex variable
z
, both defined on an
open set
C
. Further, let matrix
ACr×r
be such that
σ(A)
. Then, from the properties
of the matrix functional calculus ([15], p. 558), it follows f(A)g(A) = g(A)f(A).
Remark 1.
Following the same line of argument of this proof, a result similar to that of Theorem 1is also valid
for bounded operators
A
acting in a complex Banach space. One requires the well-known fact that
σ(A)C
is
non-void and compact together with the equality
sup{|λ|:λσ(A)}=lim sup
nkAnk1/n.
Then, if A is quasi-nilpotent, it follows sup{|λ|:λσ(A)}=0, and thus r(A,α) = .
Mathematics 2019,7, 1137 9 of 11
Remark 2.
Note that the sequence
Dn(α)nN0
in Formula
(20)
may be recast into the following
compact expression
Dn(α) = Ann
`=1
(1)`
n1+...+n`=n,nj1
1
`
j=1
Γ(αnj+1)
,
providing a closed form and thereby avoiding recurrence relations.
Obviously, for α=1, we have that (21) is the inverse of the usual matrix exponential:
Theorem 2. Let A Cr×rbe a square matrix and α=1. Then, the matrix sequence defined by (20) satisfies
Dn(1) = (1)nAn
n!,n0. (22)
Proof of Theorem 2.
We proceed by mathematical induction on. For
α=
1, from definition (20),
one obtains for the base case (n=0):
D0(1) = Ir×r=(1)0A0
0! .
In the same way, taking n=1, the definition of sequence Dn(1)immediately yields
D1(1) = AD0(1)
1! =A
1! =(1)1A1
1! .
Finally, in the induction step, we suppose that for
k=
0, 1, 2,
. . .
,
n
1, property (22) is true. Then,
for n, we conclude
Dn(1) =
n1
k=0
AnkDk(1)
(nk)!=
n1
k=0
Ank(1)kAk
(nk)!k!=An
n!
n1
k=0
(1)kn!
(nk)!k!=An
n!
n1
k=0n
k(1)k
=An
n!"n
k=0n
k(1)kn
n(1)n#=An
n!n
n(1)n=(1)nAn
n!.
Now, we move on to the numerical computation of the inverse of the Caputo matrix exponential
evaluated in the previous example.
Example 2.
In Example 1, we have shown that the matrix inverse of the Caputo matrix exponential
exp?(tαA;α)
for
A= 21
43!
and
α=
1
/
4exists at least for
t
within the interval
I= [
0, 0.0000594
]
.
It is easy to verify that t =4×105I produces the following numerical result
exp? 4×1051/4 21
43!; 1/4!= 1.177537946538906 0.08207192162719279
0.32828768650877116 0.767178338402942 !,
and then
exp? 4×1051/4 21
43!; 1/4!1
= 0.8246349372113879 0.08821856737867267
0.3528742695146907 1.2657277741047512 !.
Mathematics 2019,7, 1137 10 of 11
Evaluating only the first 13 terms of the series (21) with f (0.00004, 0.25), we obtain
f13(4×105, 1/4) =
13
k=0
Dk(1/4)(4×105)k/4 = 0.8246349372113879 0.08821856737867273
0.3528742695146909 1.2657277741047515 !,
with an approximation error
exp? 4×1051/4 21
43!; 1/4!1
f13(4×105, 1/4)
2
=3.1650 ×1016.
Example 3.
Consider again the matrix
A
given in (13). In this case, the elements of the matrix sequence
{Dn(α)}n0can be calculated explicitly:
D0(α) = I4×4,
D1(α) = A
Γ(α+1),
D2(α) = A21
Γ(α+1)21
Γ(2α+1),
D3(α) = A31
Γ(α+1)3+2
Γ(2α+1)Γ(α+1)1
Γ(3α+1),
Dn(α) = 0for n 4.
Taking into account Definition (21), one simplifies
f(t,α) =
n0
Dn(α)tnα=
3
n=0
Dn(α)tnα=D0(α) + D1(α)tα+D2(α)t2α+D3(α)t3α
=I4×4A
Γ(α+1)tα+A21
Γ(α+1)21
Γ(2α+1)t2α
+A31
Γ(α+1)3+2
Γ(2α+1)Γ(α+1)1
Γ(3α+1)t3α
=
1tα
Γ(α+1)t2α
Γ(α+1)2t2α
Γ(2α+1)t3α
Γ(α+1)3+2t3α
Γ(α+1)Γ(2α+1)t3α
Γ(3α+1)
0 1 tα
Γ(α+1)t2α
Γ(α+1)2t2α
Γ(2α+1)
0 0 1 tα
Γ(α+1)
0 0 0 1
,
which gives the same result as the matrix inverse already calculated in (14).
5. Conclusions
The starting point of this discussion was the observation and well-known fact that the conventional
matrix exponential always possesses an inverse due to its semigroup property. On the other hand,
Caputo’s matrix exponential carries a leading role in fractional calculus.
In this work, we have shown that for the Caputo matrix exponential the inverse does not
necessarily exist per se. Nevertheless, its existence is guaranteed in a specific interval, which we
have determined to relate to an uncomplicated inequality, viz. Equation (12). Furthermore, we have
established that this inverse is generally not again a Caputo matrix exponential.
Additionally, several explicit procedures have been outlined to calculate the inverse of the Caputo
matrix exponential, and it is hoped that they will open up novel pathways for the development of
future numerical methods for its efficient computation.
Mathematics 2019,7, 1137 11 of 11
Author Contributions:
E.D. contributed in funding acquisition, project administration, conceptualization,
investigation, formal analysis, methodology, validation, editing and writing; M.M.T. contributed in
conceptualization, investigation, methodology, validation, writing the original draft, writing and editing; B.M.C.-C.
and J.M.A. contributed in conceptualization, supervision, validation, visualization. All authors agree and approve
the final version of this manuscript.
Funding:
This work has been partially supported by Spanish Ministerio de Economía y Competitividad and
European Regional Development Fund (ERDF) grants TIN2017-89314-P and by the Programa de Apoyo a la
Investigación y Desarrollo 2018 of the Universitat Politècnica de València (PAID-06-18) grant SP20180016.
Acknowledgments:
We wish to express our gratitude to the anonymous referees for their exceptionally useful
comments to improve the quality of the manuscript. Further, Emilio Defez acknowledges the kind reception by
the University of Texas at Arlington during a research stay in May 2019.
Conflicts of Interest:
The authors declare that there are no conflicts of interest regarding the publication of
this paper.
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2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access
article distributed under the terms and conditions of the Creative Commons Attribution
(CC BY) license (http://creativecommons.org/licenses/by/4.0/).
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