EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 2, 2021, 493-505
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Finite Rank Solution for Conformable Degenerate
First-Order Abstract Cauchy Problem In Hilbert Spaces

Fahkreddin Seddiki1, Mohammed Al Horani1, Roshdi Khalil1,∗

1 Department of Mathematics, School of Science, The University of Jordan, Amman, Jordan

Abstract. In this paper, we find a solution of finite rank form of fractional Abstract Cauchy
Problem. The fractional derivative used is the Conformable derivative. The main idea of the
proofs are based on theory of tensor product of Banach spaces.

2020 Mathematics Subject Classifications: 26A33, 34A55

Key Words and Phrases: Tensor product of Banach spaces, finite rank function, conformable
derivative, abstract Cauchy problem.

1. Introduction

Let X be a Banach space and I = [0, 1]. Let C(I) be the Banach space of all real
valued continuous functions defined on I under the sup-norm. Let C(I,X) be the Banach
space of all continuous function defined on I with values in X.
A classical and important differential equation is the so called abstract Cauchy problem.
One form such equation is

Bu
′

= Au(t) + f(t)z

u(0) = x0

Here u ∈ C1(I,X) and A,B are densely defined linear operators on the codomain of u.
If f = 0 or z = 0, then the equation is homogeneous otherwise it is called non-homogeneous.
Now in the non-homogeneous problem we have two cases. The first type if u is unknown
and f is given and this is called the direct problem, the second type u and f are unknown
and it is called the inverse problem.
If B is not invertible, then the equation is called degenerate otherwise it is called non-
degenerate.
In this paper we will look for certain solutions called finite rank solutions for the fractional

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i2.3950

Email addresses: fakhrseddik@gmail.com (F. Seddiki),
horani@ju.edu.jo (M. Al Horani), roshdi@ju.edu.jo (R. Khalil)

http://www.ejpam.com 493 c© 2021 EJPAM All rights reserved.



F. Seddiki, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (2) (2021), 493-505 494

abstract Cauchy problem, using the tensor product technique.
First let us present some basic facts on conformable fractional derivative.
For f : [0;∞)→ R and 0 < α ≤ 1, the conformable fractional derivative of f of order α is
defined by

Tα(f)(t) = lim
ε→0

f(t+ εt1−α)− f(t)

ε

for all t > 0, if f is α differentiable on (0, b) where b > 0 and limt→0+ f
(α)(t) exists, then

define f (α)(0) = limt→0+ f (α)(t).
We denote f (α)(t) for Tα(f)(t) and we say f is α differentiable if the conformable fractional
derivative of f of order α exists.
For 0 < α ≤ 1 and f, g be α differentiable at a point t > 0, we have the following properties:
(1) Tα(af + bg) = aTα(f) + bTα(g), for all a, b ∈ R.
(2) Tα(tp) = ptp−α, for all p ∈ R.
(3) Tα(fg) = fTα(g) + gTα(f).

(4) Tα(fg ) = gTα(f)−fTα(g)
g2

.

(5) Tα(λ) = 0, for all λ is constant function.
(6) if f is differentiable, then Tα(f)(t) = t1−α dfdt (t).
The α fractional integral of a function f starting from a ≥ 0 is:

Iaα(f(t)) = Ia1 (tα−1f(t)) =

∫ t

a

f(s)

s1−α
ds

For more on conformable fractional derivative we refer to [1], [6]-[18], [20] and [21].

2. Basic Facts of the Tensor Product of Banach Space

Let X and Y be Banach spaces, X∗ denotes the dual of X. For x ∈ X and y ∈ Y
define the map x⊗ y : X∗ → Y as: x⊗ y(x∗) = 〈x, x∗〉y, for all x∗ ∈ X∗.
Cleaarly, x⊗ y is a bounded linear operator and ‖ x⊗ y ‖=‖ x ‖‖ y ‖[2]. Such an operator
x⊗ y is called an atom. The set X ⊗ Y = span{x⊗ y : x ∈ X and y ∈ Y } is a subspace
of L(X∗, Y ). The following lemma, [5], is needed in our paper.

Lemma 1. Let x1 ⊗ y1 and x2 ⊗ y2 be two nonzero atoms in X ⊗ Y such that

x1 ⊗ y1 + x2 ⊗ y2 = x3 ⊗ y3.

Then either x1, x2 or y1, y2 are linearly dependent.

We can define many norms on X ⊗ Y . The most important one is: the injective norm
For T =

∑n
i=1 xi ⊗ yi ∈ X ⊗ Y define

‖ T ‖∨= sup{|
n∑
i=1

〈xi, x∗〉〈yi, y∗〉 |: x∗ ∈ B1(X
∗) and y∗ ∈ B1(Y

∗)}.



F. Seddiki, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (2) (2021), 493-505 495

So, ‖ . ‖∨ is just the operator norm on L(X∗, Y ). This called the injective norm of T .

The space (X ⊗ Y, ‖ . ‖∨) need not be complete. Let X
∨
⊗Y denote the completion of

(X⊗Y, ‖ . ‖∨) and it is called the completed injective tensor product of X with Y . A nice
result that is used in theory of differential equations is:

Theorem 1. For any compact Housdorff space K, and any Banach space X , C(K,X)

is isometrically isomorphic to C(K)
∨
⊗X.

In particular, for any two compact metric spaces I and J , one has C(I×J) = C(I)
∨
⊗C(J).

3. Main results

3.1. Direct Problem

Let u be an α-differentiable on I = [0, 1] with values in the Hilbert space X = `2,
where `2 = {(xn) :

∑∞
n=1 | xn |2< ∞}. The natural basis of `2 is denotes by {δ1, δ2, ...}.

In `2 we write [x1, x2, ..., xn] to denote the span of {x1, x2, ..., xn}.
Let A : Dom(A) ⊆ `2 → `2, B : Dom(B) ⊆ `2 → `2 be two densely defined linear opera-
tors on `2, where domains of A and B contain the elements of the natural basis of `2.

The homogeneous degenerate fractional abstract Cauchy problem is{
Bu(α)(t) = Au(t)

u(0) = x0
(1)

The nonhomogeneous degenerate fractional abstract Cauchy problem is{
Bu(α)(t) = Au(t) + f(t)z

u(0) = x0
(2)

Where u(t) ∈ Dom(A) ∩Dom(B), u(α)(t) ∈ Dom(B), f ∈ C(I) and z ∈ `2 .
In this section we look for a solution to problems (1) and (2) among finite rank function

of the form u(t) =
∑n

i=1 ui(t)δi, where u
(α)
i ∈ C(I), i = 1, 2, ...n.

Theorem 2. In problem (P1), let u(t) =
∑n

i=1 ui(t)δi, where u
(α)
i ∈ C(I), i = 1, 2, ...n

and assume B = I, then the problem (1) has a unique solution.

Proof. We have, u(t) =
∑n

i=1 ui(t)δi, then u(α)(t) =
∑n

i=1 u
(α)
i (t)δi, thus

n∑
i=1

u
(α)
i (t)δi =

n∑
i=1

ui(t)Aδi. (3)

So, Au(t) ∈ [δ1, δ2, ..., δn], since u(α)(t) is linear combination of δ1, δ2, ..., δn. Hence
[δ1, δ2, ..., δn] is invariant subspace of A.



F. Seddiki, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (2) (2021), 493-505 496

Let  = A |[δ1,δ2,...,δn] be the restriction of A on [δ1, δ2, ..., δn] and so  has a matrix rep-

resentation which is  = [aij ], such that aij = 〈Aδj , δi〉.
Taking the inner product of δj with both sides of equation (3), we get

n∑
i=1

u
(α)
i (t)〈δi, δj〉 =

n∑
i=1

ui(t)〈Aδi, δj〉.

Since {δi}ni=1 is orthonormal, we obtain

u
(α)
j (t) =

n∑
i=1

ui(t)〈Aδi, δj〉. (4)

Which is a homogeneous linear system of differential equations

U (α)(t) = ÂU(t), where U(t) = (u1(t), u2(t), ..., un(t))T .

This is system has a unique solution of the form

U(t) = φ(t)c.

Here φ(t) is the fundamental matrix, which is invertible. By the initial condition, we have

ci = 〈φ−1(0)x0, δi〉, i = 1, ..., n.

Consequently, the problem (1) has a unique solution.

Theorem 3. In problem (2), let u(t) =
∑n

i=1 ui(t)δi, where u
(α)
i ∈ C(I), i = 1, 2, ...n and

assume B = I and z ∈ [δ1, ..., δn], then the problem (2) has a unique solution.

Proof. We have, u(t) =
∑n

i=1 ui(t)δi, then u(α)(t) =
∑n

i=1 u
(α)
i (t)δi, thus

n∑
i=1

u
(α)
i (t)δi =

n∑
i=1

ui(t)Aδi + f(t)z. (5)

Let  = A |[δ1,δ2,...,δn] the restriction of A on [δ1, δ2, ..., δn] and so  has a matrix repre-

sentation which is  = [aij ], such that aij = 〈Aδj , δi〉.
Taking the inner product of δj with both sides of equation (5), we get

n∑
i=1

u
(α)
i (t)〈δi, δj〉 =

n∑
i=1

ui(t)〈Aδi, δj〉+ f(t)〈z, δj〉.

Since {δi}ni=1 is orthonormal, we obtain

u
(α)
j (t) =

n∑
i=1

ui(t)〈Aδi, δj〉+ f(t)〈z, δj〉. (6)



F. Seddiki, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (2) (2021), 493-505 497

We set, U(t) = (u1(t), ..., un(t))T and F (t) = f(t)(〈z, δ1〉, ..., 〈z, δn〉)T , so equation (6) can
be written in the form

U (α)(t) = ÂU(t) + F (t).

This system has a unique solution of the form

U(t) = φ(t)c+ φ(t)

∫ t

0

φ−1(s)F (s)

s1−α
ds.

Where φ(t) is the fundamental matrix. This is an invertible matrix. Now we use the initial
condition to find the constant c. Consequently, the problem (2) has a unique solution.

Now, let B 6= I and u(t) is finite rank function. In addition assume that [δ1, δ2, ..., δn] is
invariant under both A and B and let An, Bn be the restriction of A and B to [δ1, δ2, ..., δn].

Theorem 4. In problem (1), let Bn be orthogonally diagonalizable linear operator such
that An |Ker(Bn) is invertible. Then problem (1) has a unique solution.

Proof. Let {θ1, θ2, ..., θn} be an orthonormal basis such that the matrix representation
of Bn with respect this basis is D̃ = diag(λ1, ..., λn), when λ1, ..., λn the corresponding
eigenvalues of Bn. Now, if λi 6= 0 for all i = 1, 2, ...n, then problem (1) becomes u(α)(t) =
B−1n Anu(t) and hence has a unique solution by theorem 3.1.
Suppose λi 6= 0 for i = 1, 2, ...r, and λi = 0 for i = r+1, r+2, ...n. Let u(t) =

∑n
i=1 vi(t)θi:

Then
n∑
i=1

v
(α)
i (t)Bnθi =

n∑
i=1

vi(t)Anθi. (7)

Taking the inner product of θj with both sides of (7), we obtain

n∑
i=1

v
(α)
i (t)〈Bnθi, θj〉 =

n∑
i=1

vi(t)〈Anθi, θj〉.

So, we get the following system

[
D 0
0 0

]

v
(α)
1 (t)
.
.
.

v
(α)
n (t)

 =

[
A1 A2

A3 Ã

]
v1(t)
.
.
.

vn(t)

 . (8)

where D = diag(λ1, ., ., ., λr) and à = An |Ker(Bn)= [〈Anθj , θi〉]i,j=r+1,...,n.

Multiplying (8) by

[
I 0

0 Ã−1

]
, we obtain

[
D 0
0 0

]

v
(α)
1 (t)
.
.
.

v
(α)
n (t)

 =

[
A1 A2

Ã−1A3 In−r

]
v1(t)
.
.
.

vn(t)

 .



F. Seddiki, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (2) (2021), 493-505 498

Thus, we get

D


v
(α)
1 (t)
.
.
.

v
(α)
r (t)

 = A1


v1(t)
.
.
.

vr(t)

+A2


vr+1(t)

.

.

.
vn(t)

 , (9)

and

Ã−1A3


v1(t)
.
.
.

vr(t)

+ In−r


vr+1(t)

.

.

.
vn(t)

 = 0. (10)

From equation (10), we have 
vr+1(t)

.

.

.
vn(t)

 = −Ã−1A3


v1(t)
.
.
.

vr(t)

 . (11)

Substitute (11) in equation (9), to get
v
(α)
1 (t)
.
.
.

v
(α)
r (t)

 = D−1(A1 −A2Ã
−1A3)


v1(t)
.
.
.

vr(t)

 .

We put, U1(t) =


v1(t)
.
.
.

vr(t)

, U2(t) =


vr+1(t)

.

.

.
vn(t)

 and M = D−1(A1 −A2Ã
−1A3).

We get the system, U
(α)
1 (t) = MU1(t), which has a unique solution U1(t) = φ(t)c, where

φ(t) is the fundamental matrix. So we have U2(t) = −Ã−1A3U1(t). Therefore u(t) =

[
U1(t)
U2(t)

]T

θ1
.
.
.
θn

 .

We conclude the problem (1) has a unique solution.



F. Seddiki, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (2) (2021), 493-505 499

Theorem 5. In problem (2), let Bn be orthogonally diagonalizable linear operator such
that An |Ker(Bn) is invertible. Then problem (2) has a unique solution.

Proof. Let {θ1, θ2, ..., θn} be an orthonormal basis such that the matrix representation
of Bn with respect this basis is D̃ = diag(λ1, ..., λn), when λ1, ..., λn the corresponding
eigenvalues of Bn. Now, if λi 6= 0, for all i = 1, 2, ...n, then the problem (2) becomes

u(α)(t) = B−1n Anu(t) + f(t)B−1n z.

Hence has a unique solution by theorem 3.2.
Suppose λi 6= 0, for i = 1, 2, ...r, and λi = 0, for i = r+1, r+2, ...n. Let u(t) =

∑n
i=1 vi(t)θi:

Then
n∑
i=1

v
(α)
i (t)Bnθi =

n∑
i=1

vi(t)Anθi + f(t)z. (12)

Taking the inner product of θj with both sides of equation (12), we obtain

n∑
i=1

v
(α)
i (t)〈Bnθi, θj〉 =

n∑
i=1

vi(t)〈Anθi, θj〉+ f(t)〈z, θj〉.

So, we get the following system

[
D 0
0 0

]

v
(α)
1 (t)
.
.
.

v
(α)
n (t)

 =

[
A1 A2

A3 Ã

]
v1(t)
.
.
.

vn(t)

+ f(t)


〈z, θ1〉
.
.
.

〈z, θn〉

 . (13)

where D = diag(λ1, ., ., ., λr) and à = An |Ker(Bn)= [〈Anθj , θi〉]i,j=r+1,...,n.

Multiplying (13) by

[
I 0

0 Ã−1

]
, we obtain

[
D 0
0 0

]

v
(α)
1 (t)
.
.
.

v
(α)
n (t)

 =

[
A1 A2

Ã−1A3 In−r

]
v1(t)
.
.
.

vn(t)

+ f(t)

[
I 0

0 Ã−1

]
〈z, θ1〉
.
.
.

〈z, θn〉

 .

Thus, we get

D


v
(α)
1 (t)
.
.
.

v
(α)
r (t)

 = A1


v1(t)
.
.
.

vr(t)

+A2


vr+1(t)

.

.

.
vn(t)

+ f(t)


〈z, θ1〉
.
.
.

〈z, θr〉

 , (14)



F. Seddiki, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (2) (2021), 493-505 500

and

Ã−1A3


v1(t)
.
.
.

vr(t)

+ In−r


vr+1(t)

.

.

.
vn(t)

+ f(t)Ã−1


〈z, θr+1〉

.

.

.
〈z, θn〉

 = 0. (15)

From equation (15), we have
vr+1(t)

.

.

.
vn(t)

 = −Ã−1A3


v1(t)
.
.
.

vr(t)

− f(t)Ã−1


〈z, θr+1〉

.

.

.
〈z, θn〉

 . (16)

Substitute (16) in equation (14), we get
v
(α)
1 (t)
.
.
.

v
(α)
r (t)

 = D−1(A1 −A2Ã
−1A3)


v1(t)
.
.
.

vr(t)

+ f(t)D−1(


〈z, θ1〉
.
.
.

〈z, θr〉

−A2Ã
−1


〈z, θr+1〉

.

.

.
〈z, θn〉

).

We put, U1(t) =


v1(t)
.
.
.

vr(t)

, U2(t) =


vr+1(t)

.

.

.
vn(t)

, M = D−1(A1 − A2Ã
−1A3) and F (t) =

f(t)D−1(


〈z, θ1〉
.
.
.

〈z, θr〉

−A2Ã
−1


〈z, θr+1〉

.

.

.
〈z, θn〉

).

Then we obtain the system

U
(α)
1 (t) = MU1(t) + F (t).

Which is has a unique solution

U1(t) = φ(t)c+ φ(t)

∫ t

0

φ−1(s)F (s)

s1−α
ds,

where φ(t) is the fundamental matrix and we have

U2(t) = −Ã−1A3U1(t)− f(t)Ã−1


〈z, θr+1〉

.

.

.
〈z, θn〉

 .



F. Seddiki, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (2) (2021), 493-505 501

Hence, u(t) =

[
U1(t)
U2(t)

]T

θ1
.
.
.
θn

 .
Therefore, the problem (2) has a unique solution.

3.2. Inverse Problem Case

Let X = `2 be the Hilbert space. Let A : Dom(A) ⊆ `2 → `2, B : Dom(B) ⊆ `2 → `2

be two densely defined linear operators on `2, where domains of A and B contain the
elements of the natural basis of `2.
Consider the two inverse problems (P3) and (P4) respectively{

u(α)(t) = Au(t) + f(t)

u(0) = x0{
Bu(α)(t) = Au(t) + f(t)

u(0) = x0

Where u(α) ∈ C(I,X), f ∈ C(I,X).
In this section we look for a solution to problems (P3) and (P4) among finite rank functions

of the form u(t) =
∑n

i=1 ui(t)δi, and f(t) =
∑n

i=1 fi(t)δi, where, u
(α)
i ∈ C(I) and fi ∈ C(I),

for i = 1, 2, ...n. Here we use a condition similar to that used in [22].

Theorem 6. In problem (P3), let u(t) =
∑n

i=1 ui(t)δi, and f(t) =
∑n

i=1 fi(t)δi where,

u
(α)
i ∈ C(I) and fi ∈ C(I), for i = 1, 2, ...n.

Assume the following two condition are satisfied:

1)There exist, x ∈ `2 such that 〈ui(t)δi, x〉 = gi(t) where g
(α)
i ∈ C(I) and 〈δi, x〉 6= 0.

2) A is diagonal with respect to the basis {δi}ni=1. That is, Aδi = λiδi for all i = 1, ..., n.
Then the problem (P3) has a unique solution.

Proof. Substitute u(t) =
∑n

i=1 ui(t)δi, and f(t) =
∑n

i=1 fi(t)δi, in (P3), we get

n∑
i=1

u
(α)
i (t)δi =

n∑
i=1

ui(t)Aδi +
n∑
i=1

fi(t)δi.

Since A is diagonal with respect to {δi}ni=1, we have

n∑
i=1

u
(α)
i (t)δi =

n∑
i=1

λiui(t)δi +
n∑
i=1

fi(t)δi. (17)

Taking the inner product of δj with both sides of equation (17), we obtain

u
(α)
j (t) = λjuj(t) + fj(t). (18)



F. Seddiki, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 14 (2) (2021), 493-505 502

Multiplying equation (18) by δj and use condition (1), we obtain

g
(α)
j (t) = λjgj(t) + 〈fj(t)δj , x〉.

Thus, we have

fj(t) =
g
(α)
j (t)− λjgj(t)
〈δj , x〉

.

Hence, fj(t) is determined uniquely for j = 1, ...n and thus f(t) is determined uniquely.
Now to find u(t). Since f(t) is determined, then we have

uj(t) = uj(0)eλj
tα

α + eλj
tα

α

∫ t

0

e−λj
sα

α fj(s)

s1−α
ds.

Consequently, the problem (P3) has a unique solution.

Now, to solve problem (P4) we need to assume the following satisfy:
Assumption 1. Bn = B |[δ1,...,δn] is orthogonally diagonalizable linear operator with
respect to the orthonormal basis {θ1, ..., θn} and corresponding eigenvalues λ1, ..., λn such
that An |Ker(Bn) is invertible, where An = A |[δ1,...,δn].
Assumption 2. An is diagonal with respect to {θ1, ..., θn} ie Anθj = µjθj for j = 1, ..., n.
Now, let u(t) =

∑n
i=1 ui(t)θi

Assumption 3. There exist, x ∈ `2 such that 〈ui(t)θi, x〉 = gi(t) where g
(α)
i ∈ C(I).

Assumption 4. M = [〈δi, θj〉〈θj , x〉]i,j=1,...,nis invertible.

Theorem 7. Under assumptions 1,2, 3 and 4, problem (P4) has a unique solution.

Proof. Since u(t) =
∑n

i=1 ui(t)θi, then we substitute in problem (P4), we have

n∑
i=1

u
(α)
i (t)Bnθi =

n∑
i=1

ui(t)Anθi +

n∑
i=1

fi(t)δi.

This implies
n∑
i=1

u
(α)
i (t)λiθi =

n∑
i=1

ui(t)µiθi +
n∑
i=1

fi(t)δi. (19)

Taking the inner product of θj with both sides of equation (19), we obtain

λju
(α)
j (t) = µjuj(t) +

n∑
i=1

fi(t)〈δi, θj〉. (20)

Multiplying equation (20) by θj and using assumption 3, we obtain

λjg
(α)
j (t) = µjgj(t) +

n∑
i=1

fi(t)〈(δi, θj〉〈θj , x〉.



REFERENCES 503

Hence, we get the following system:
λ1g

(α)
1 (t)− µ1g1(t)

.

.

.

λng
(α)
n (t)− µngn(t)

 = MT


f1(t)
.
.
.

fn(t)

 .

Where, M = [〈δi, θj〉〈θj , x〉]i,j=1,...,n. By assumption 4 M is invertible, then MT is also
invertible and (MT )−1 = (M−1)T , thus

f1(t)
.
.
.

fn(t)

 = (M−1)T


λ1g

(α)
1 (t)− µ1g1(t)

.

.

.

λng
(α)
n (t)− µngn(t)

 .

Therefore f is determined uniquely.
Now to find u(t), we have

• If λj = 0, then uj(t) =
∑n

i=1
fi(t)〈δi,θj〉
−µj .

• If λj 6= 0, then

uj(t) = uj(0)e
µjt

α

λjα +

n∑
i=1

e
µjt

α

λjα 〈δi, θj〉
∫ t

0

fi(s)e
−µjs

α

λjα

s1−α
ds.

Consequently, the problem (P4) has a unique solution.

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