Phuong et al.

Boundary Value Problems (2022) 2022:100

https://doi.org/10.1186/s13661-022-01683-1

RESEARCH Open Access

Fractional evolution equation with Cauchy

data in Lp spaces
Nguyen Duc Phuong1 , Dumitru Baleanu2,3 , Ravi P. Agarwal4 and Le Dinh Long5,6*
*
Correspondence:
ledinhlong@vlu.edu.vn Abstract
5
Division of Applied Mathematics,
Science and Technology Advanced In this paper, we consider the Cauchy problem for fractional evolution equations with
Institute, Van Lang University, Ho the Caputo derivative. This problem is not well posed in the sense of Hadamard.
Chi Minh City, Vietnam
6
There have been many results on this problem when data is noisy in L2 and Hs .
Faculty of Applied Technology,
School of Engineering and However, there have not been any papers dealing with this problem with observed
Technology, Van Lang University, Ho data in Lp with p = 2. We study three cases of source functions: homogeneous case,
Chi Minh City, Vietnam inhomogeneous case, and nonlinear case. For all of them, we use a truncation
Full list of author information is
available at the end of the article method to give an approximate solution to the problem. Under different assumptions
on the smoothness of the exact solution, we get error estimates between the
regularized solution and the exact solution in Lp . To our knowledge, Lp evaluations for
the inverse problem are very limited. This work generalizes some recent results on this
problem.
MSC: 60G15; 60G22; 60G52; 60G57
Keywords: Fractional evolution equation; Caputo derivative; Cauchy problem;
Fourier truncation regularization; Sobolev embeddings

1 Introduction
Today, fractional calculus involves the investigation of so-called integral operators and
fractional derivatives over real or complex domains and their applications. There are many
types of mathematical models that require the use of noninteger order derivatives. Re-
cently, it has been widely used in modeling practical models because of its ability to provide
approximation and ignore the influence of external forces such as in physics, engineer-
ing, mechanics science, biology, and some other areas [1–5]. There are several versions of
noninteger derivatives, but perhaps the two types of derivatives, Caputo and Riemann–
Liouville, are of most interest to mathematicians [6–15].
Let  be a bounded domain in Rd (d ≥ 1) with sufficiently smooth boundary ∂. In
this paper, we are interested in the following evolution equation with a time-fractional
derivative:
⎧
⎨ Dα u + u = F(t, x; u), in  × (0, T],
C t
(1.1)
⎩u|∂ = 0, in ,

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Phuong et al. Boundary Value Problems (2022) 2022:100 Page 2 of 22

with the initial value data

u(x, 0) = g(x), ut (x, 0) = ϕ(x), (1.2)

where F is the source function and u describes the distribution of the temperature at po-
sition x and time t. In (1.1), α ∈ (1, 2) is the fractional order and C Dαt denotes the Caputo
fractional derivative with respect to t (time-fractional) and is defined by (see [16, 17])
⎧ t
⎨ Dα u(t, x) = 1
(t
2
– z)1–α ∂∂z2u (z, x) dz, for 1 < α < 2,
C t Γ (2–α) 0
(1.3)
⎩C Dα u(t, x) = α
∂ u
(t, x), for α = 1, 2,
t ∂t α

and Γ is the gamma function. Note that if α = 2, then Eq. (1.1) represents a Cauchy prob-
lem for elliptic equation. There are some main reasons why we are interested in studying
Problem (1.1)–(1.2)

1.1 Our motivation

• The first reason is that Problem (1.1)–(1.2) tends to be the elliptic equation when α → 2.
The Cauchy problem for elliptic equations has many applications in many physical sys-
tems such as plasma physics. There are many interesting papers that have investigated the
Cauchy elliptic equation with classical derivative, for example, [15, 18, 19] and the refer-
ences therein. During the simulation, there are several applied models that are described
by partial differential equations with memory terms attached. For example, the problem
of electrical conduction in biological tissues in the radio frequency range is governed by
an elliptic equation with memory [20]. In many physical phenomena, we need to use non-
integer order derivatives for elliptic equations, to simulate problems related to viscous
models. In [9], the authors studied an elliptic equation with a condition on the bound-
ary associated with a generalized Riemann–Liouville derivative of fractional order. Some
other articles for an elliptic equation with fractional order have been studied in [21, 22].
• We have another detailed explanation of the application of Problem (1.1)–(1.2) when
α → 1. Indeed, if α = 1, we have the following problem:

⎧
⎪
⎨ut + u = F(t, x; u), in  × (0, T],
⎪
u|∂ = 0, in , (1.4)
⎪
⎪
⎩
u(x, 0) = g(x), in .

By changing the variable v(x, t) = u(x, T –t), we turn this problem into a final value problem

⎧
⎪
⎪vt – u = F(t, x; v), in  × (0, T],
⎨
v|∂ = 0, in , (1.5)
⎪
⎪
⎩
v(x, T) = g(x), in .

The backward in time problem for parabolic equation introduced as above is one of the
classic inverse problems involving many applications and has been studied extensively
over the past 50 years. Interesting work on Problem (1.5) can be consulted by (1.5).
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 3 of 22

Let us continue to add justification to our interest in elliptic equations with the Caputo
derivative. We observe that the elliptic problem with the Caputo derivative is a subbranch
of the elliptic equation with a nonlocal condition. According to the work of Bitsadze and
Samarskii [23], the nonlocal elliptic equations have many applications in the theory of
plasma. Furthermore, it is difficult for elliptic equations with classical derivatives to de-
scribe the past process. When the phenomenon needs to involve factors and past informa-
tion, the Caputo derivative plays an important role. In addition, many problems in dynami-
cal processes, electrochemistry, and signal processing lead to elliptic differential equations
of fractional order.
• In the interesting paper, Jin and Rundell [24] considered the ill-posedness of Problem
(1.1)–(1.2) in the simple case F = 0; however, the authors have not provided the approx-
imation and error estimate. The first results investigating the above equation seem to be
of [25] and [26]. Our present paper has generalized the previous results by [25] and [26]
in the sense that the observed data belongs to Lp . It is a strong claim that our paper is part
of a series of investigations on the ill-posedness of fractional diffusion equations, which
have been published in a recent series of works by Baleanu and colleagues [27–29].

1.2 Our novelty and contribution

We note that research results on the inverse and ill-posed problem with observed data on
Lp are very rare. This is a difficult topic attracting the interest of many mathematicians.
The problem will be difficult when we get the observed data in Lp space with p = 2. Note
that Parseval’s equality cannot be applied to our problem with observed data in Lp space
with p = 2. To overcome these challenges, we learn techniques from the article [30] and in
the references [31–43]. This idea can be summed up as the importance of the technique
of using embedding between Lp and Hs (). In this paper, we used the Fourier truncation
method to regularize our problem.
The main contribution of our paper is as follows:
• The first contribution is to investigate the regularized problem for our problem in the
cases: homogeneous case, inhomogeneous case, and nonlinear case.
• We showed the existence of a regularized problem with a nonlinear source by the
Banach fixed point theorem. Complexity occurs when we encounter components that
involve the Mittag-Leffler functions. We need to provide some background and
knowledge about the bound of these functions.
• The last main contribution is to provide the error estimate in Lp space when we
observe the noisy data in Lp space. As we know, problems involving Lp are always
complicated. So we need some new techniques to handle it.
This paper is organized as follows. In Sect. 2, we give some knowledge about some func-
tional spaces and some properties of the bounds of Mittag-Leffler terms. Section 3 con-
siders the homogeneous problem. In Sect. 4, we study the Cauchy problem for the inho-
mogeneous case. Finally, in Sect. 5, we treat the nonlinear case. In each case, we introduce
a regularized solution of the Fourier truncation form. Then, we provide the error between
the regularized solution and the exact solution.
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 4 of 22

2 Preliminary results
This section provides some notation and the functional spaces which will be used through-
out this article. Recall that the spectral problem
⎧
⎨e (x) = –λ e (x), x ∈ ,
j j j
⎩ej (x) = 0, x ∈ ∂,

admits the eigenvalues 0 < λ1 ≤ λ2 ≤ · · · ≤ λj ≤ · · · with λj → ∞ as j → ∞ and corre-

sponding eigenfunctions ej ∈ H01 ().

Definition 2.1 (Hilbert scale space) We recall the Hilbert scale space Hs () given as fol-
lows:

 2
 ∞

H () = f ∈ L () 
s 2 2s
λj f (x)ej (x) dx < ∞
j=1 

for any s ≥ 0. It is well known that Hs () is a Hilbert space corresponding to the norm
 ∞
 2 1/2

f Hs () = λ2s
j f (x)ej (x) dx , f ∈ Hs ().
j=1 

Lemma 2.1 The following statements are true:

d 2d
Lp () → Hμ () if < μ ≤ 0 and p ≥ ,
4 d – 4μ
(2.1)
d 2d
H () → L () if 0 ≤ s < and p ≤
s p
.
4 d – 4s

Also, for M, n > 0, we introduce the n-order Gevrey class GnM () of L2 -functions, see
e.g. [44], defined by the spectrum of the Laplacian as follows:
 ∞  2
 
1 
GnM () := ψ ∈ L () : 2
λnj exp  
2Mλj  ψ(x)ej (x) dx < ∞ ,
α

j=1 

equipped with the following norm:

 ∞  2 1/2
 1  
ψGnM () = λnj exp 2Mλjα  ψ(x)ej (x) dx .
j=1 

Definition 2.2 The Mittag-Leffler function is defined by

∞
zm
Eα,θ (z) = , z ∈ C, (2.2)
m=0
Γ (αm + θ )

where α > 0 and θ ∈ R are arbitrary constants.

The following lemmas provide upper and lower bounds of the Mittag-Leffler functions
Eα,1 (z), Eα,2 (z), Eα,α (z) by the exponential functions.
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 5 of 22

Lemma 2.2 (see [45]) Let 0 < α0 < α1 < 2 and α ∈ [α0 , α1 ]. Then there exist constants
M1 , M2 > 0 and z > 0 such that

M1  1 M2  1
exp z α ≤ Eα,1 (z) ≤ exp z α ,
α α
1 1
M1 exp(z α ) M2 exp(z α )
1 ≤ Eα,2 (z) ≤ 1 ,
α zα α zα
1 1
M1 exp(z α ) M2 exp(z α )
≤ Eα,α (z) ≤ .
α z1– α1 α z1– α1

Lemma 2.3 (see [26]) Let α ∈ [α0 , α1 ] with 1 < α0 < α1 < 2 and z ∈ [0, T]. Then there exists
a positive constant M > 0 independent of z such that

  M  1 
Eα,1 λj zα ≤ exp λjα z , (2.3)
α
  M –1  1 
zEα,2 λj zα ≤ λj α exp λjα z , (2.4)
α
  M 1–α  1 
zα–1 Eα,α λj zα ≤ λj α exp λjα z . (2.5)
α

Next, let us give the explicit fomula of the mild solution to Problem (1.1)–(1.2). Sup-
 
pose that Problem (1.1)–(1.2) has a solution u(t, x) = ∞ j=1 [  u(t, x)ej (x) dx]ej (x). Then the

function uj (t) =  u(t, x)ej (x) dx solves the following ordinary differential equation:

⎧ 
⎪
⎪ D α
u (t) – λ u (t) =  G(t, x, u(t, x))ej (x) dx,
⎨ C t j

j j

uj (0) =  ψ(x)ej (x) dx, (2.6)

⎪
⎪ 
⎩d
u (0) =  ϕ(x)ej (x) dx.
dt j

By applying the method in [17, Sect. 2], we obtain the solution of (2.6) as follows:

   
   
uj (t) = Eα,1 λj t α ψ(x)ej (x) dx + tEα,2 λj t α ϕ(x)ej (x) dx
 
  
t
α–1
 α
  
+ (t – ν) Eα,α λj (t – ν) G ν, x, u(ν, x) ej (x) dx dν.
0 

Consequently, the Fourier series u ∈ L2 () is given as follows:

∞
 
 
u(t, x) = Eα,1 λj t α ψ(x)ej (x) dx ej (x)
j=1 

∞
 
 α

+ tEα,2 λj t ϕ(x)ej (x) dx ej (x) (2.7)
j=1 

∞ 

  
t    
+ (t – ν)α–1 Eα,α λj (t – ν)α G ν, x, u(ν, x) ej (x) dx dν ej (x).
j=1 0 
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 6 of 22

3 Cauchy problem with the homogeneous case

In this section, we consider the following problem:

⎧
⎪
⎪
α
in  × (0, T],
⎨C Dt u + u = 0,
u|∂ = 0, in , (3.1)
⎪
⎪
⎩
u(x, 0) = ψ(x), ut (x, 0) = ϕ(x), in .

The solution of the homogeneous problem has the following series representation:

∞
 
 
u(t, x) = Eα,1 λj t α ψ(x)ej (x) dx ej (x)
j=1 

  (3.2)
∞
 α

+ tEα,2 λj t ϕ(x)ej (x) dx ej (x).
j=1 

For β > 0, we construct a regularized solution as follows:

D
(β)  
 
Wβ (t, x) = Eα,1 λj t α ψβ (x)ej (x) dx ej (x)
j=1 
(3.3)
D
(β)  
 
+ tEα,2 λj t α ϕβ (x)ej (x) dx ej (x),
j=1 

here D(β) depends only on β.

Theorem 3.1 Let β, δ > 0, 0 < p < d4 , and 1 < m < 2. We presume that the Cauchy data
(ψ, ϕ) is disturbed by the noisy data (ψβ , ϕβ ) ∈ Lm () × Lm () such that

ψ – ψβ Lm () + ϕ – ϕβ Lm () ≤ β. (3.4)

Then, if u ∈ L∞ (0, T; Hp+δ ()), we obtain the following estimate:

   1   1   2p – m–2
u(t, ·) – Wβ (t, ·) 2d ≤ (T + 1)C1 C(d, m, p) exp |C| α T D(β) αd D(β) d 2m β
L d–4p ()
 – 2δ
C )–δ D(β) d uL∞ (0,T;Hp+δ ()) ,
+ C(d, p)(

where the constants are defined in the proof.

dm–2d
Proof Since 1 < m < 2, we find that Lm () → H 4m (). This embedding allows us to give
that

ψ – ψβ  dm–2d + ϕ – ϕβ  dm–2d
H 4m () H 4m ()

≤ C(d, m)ψ – ψβ Lm () + C(d, m)ϕ – ϕβ Lm () (3.5)

≤ βC(d, m).
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 7 of 22

It is obvious that
     
u(t, ·) – Wβ (t, ·) p ≤ u(t, ·) – vβ (t, ·) p + vβ (t, ·) – Wβ (t, ·) p , (3.6)
H () H () H ()

where the function vβ is defined by

D
(β)  
 α

vβ (t, x) = Eα,1 λj t ψ(x)ej (x) dx ej (x)
j=1 
(3.7)
D
(β)  
 α

+ tEα,2 λj t ϕ(x)ej (x) dx ej (x).
j=1 

In view of Parseval’s equality, the second quantity on the right-hand side of (3.6) is
bounded by
 
vβ (t, ·) – Wβ (t, ·) p ≤ J1 (t) + J2 (t), (3.8)
H ()

where
D(β)   
  α   
 
J1 (t) =  Eα,1 λj t ψβ (x) – ψ(x) ej (x) dx ej (x) (3.9)
  
j=1 Hp ()

and
D(β)   
  α   
 
J2 (t) =  tEα,2 λj t ϕβ (x) – ϕ(x) ej (x) dx ej (x) . (3.10)
  
j=1 Hp ()

Our upcoming task is to provide the upper bounds for the two components J1 (t) and
J2 (t). By looking at Parseval’s equality for the term J1 (t), we have the following equality:

 2
  D
(β)
2p   2  
J1 (t)2 = λj Eα,1 λj t α  ψβ (x) – ψ(x) ej (x) dx
j=1 

D
 2
(β)
2p– dm–2d    dm–2d  
= λj 2m Eα,1 λj t α 2 λ 2m ψβ (x) – ψ(x) ej (x) dx . (3.11)
j
j=1 

In view of inequality (2.3) as in Lemma (2.3), we find that

 2
2p– dm–2d
2m 
  2 M  1  2p– dm–2d
λj Eα,1 λj t α  ≤   exp 2λjα t λj 2m
. (3.12)
α

It is obvious to see that

 2
λj ≤ Cj2/d ≤ C D(β) d if j ≤ D(β). (3.13)

Thus, we get immediately that

 2
        2   4p – m–2
λj
2p– dm–2d
2m Eα,1 λj t α 2 ≤  M  |C|2p– dm–2d
2m exp 2|C| α T D (β) αd D (β) d
1
m
. (3.14)
α
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 8 of 22

It follows from (3.11) that

     2   4p m–2
J1 (t)2 ≤ |C1 |2 exp 2|C| α1 T D(β) αd D(β) d – m

D
(β)  2
dm–2d  
× λj 2m
ψβ (x) – ψ(x) ej (x) dx
j=1 

   2   4p – m–2
≤ |C1 |2 exp 2|C| α T D(β) αd D(β) d m ψ – ψβ 2
1
dm–2d , (3.15)
H 4m ()

dm–2d
where |C1 |2 = | M
α
|2 |C|2p– 2m . Using (3.5), we find that

 1   1   2p – m–2
J1 (t) ≤ C1 exp |C| α T D(β) αd D(β) d 2m ψ – ψβ  dm–2d
H 4m ()
 1   1   2p – m–2
≤ C1 C(d, m) exp |C| α T D(β) αd D(β) d 2m β. (3.16)

We continue to estimate the second term J2 (t). Indeed, we get that

 2
  D
(β)
2p   2  
J2 (t)2 = λj t 2 Eα,2 λj t α  ϕβ (x) – ϕ(x) ej (x) dx
j=1 

D
 2
(β)
2p– dm–2d
2m 
  2 dm–2d  
=t 2
λj Eα,2 λj t α  λj 2m ϕβ (x) – ϕ(x) ej (x) dx
j=1 

   2   4p – m–2
≤ T 2 |C1 |2 exp 2|C| α T D(β) αd D(β) d m ϕ – ϕβ 2
1
dm–2d . (3.17)
H 4m ()

By the same explanations as above, we claim that

 1   1   2p – m–2
J2 (t) ≤ TC1 C(d, m) exp |C| α T D(β) αd D(β) d 2m β. (3.18)

Combining (3.8), (3.15), and (3.18), we derive that

 
vβ (t, ·) – Wβ (t, ·) p
H ()
(3.19)
 1   1   2p – m–2
≤ (T + 1)C1 C(d, m) exp |C| α T D(β) αd D(β) d 2m β.

We continue by dealing with the first term on the right-hand side of (3.6). It is clear to see
that

 2
 
u(t, ·) – vβ (t, ·)2 p = λ –2δ 2p+2δ
λ u(t, x)ej (x) dx . (3.20)
H () j j
j>D (β) 

Because of the fact that

 2/d
λj ≥  C D(β) ,
C j2/d > 
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 9 of 22

≤ (
4δ
we find that λ–2δ
j C)–2δ |D(β)|– d , which allows us to provide that

 
u(t, ·) – vβ (t, ·)2 p
H ()
 2
 – 4δ 2p+2δ
C )–2δ D(β) d
≤ ( λj u(t, x)ej (x) dx
j>D (β) 

 – 4δ  2  – 4δ
C )–2δ D(β) d u(t, ·)Hp+δ () ≤ (
≤ ( C )–2δ D(β) d u2L∞ (0,T;Hp+δ ()) . (3.21)

Therefore, we derive that

   – 2δ
u(t, ·) – vβ (t, ·) p ≤ (
C )–δ D(β) d uL∞ (0,T;Hp+δ ()) . (3.22)
H ()

Combining (3.6), (3.19), and (3.22), we get that

     
u(t, ·) – Wβ (t, ·) p ≤ vβ (t, ·) – Wβ (t, ·) p + u(t, ·) – vβ (t, ·) p
H () H () H ()
 1   1   2p – m–2
≤ (T + 1)C1 C(d, m) exp |C| α T D(β) αd D(β) d 2m β
 – 2δ
C )–δ D(β) d uL∞ (0,T;Hp+δ ()) .
+ ( (3.23)

For 0 < p < d/4, we recall the Sobolev embedding

2d
Hp () → L d–4p ().

Therefore, we obtain the following estimate:

   
u(t, ·) – Wβ (t, ·) 2d ≤ C(d, p)u(t, ·) – Wβ (t, ·)Hp ()
L d–4p ()
 1   1   2p – m–2
≤ (T + 1)C1 C(d, m, p) exp |C| α T D(β) αd D(β) d 2m β
 – 2δ
C )–δ D(β) d uL∞ (0,T;Hp+δ ()) .
+ C(d, p)( 

4 Regularization in Lp space under inhomogeneous source term

In this section, we consider the following problem:

⎧
⎪
⎪
α
in  × (0, T],
⎨C Dt u + u = G(x, t),
u|∂ = 0, in , (4.1)
⎪
⎪
⎩
u(x, 0) = ψ(x), ut (x, 0) = ϕ(x), in .

The solution to Problem (4.1) possesses the following series representation:

∞
 
 
u(t, x) = Eα,1 λj t α ψ(x)ej (x) dx ej (x)
j=1 

∞
 
 
+ tEα,2 λj t α ϕ(x)ej (x) dx ej (x)
j=1 
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 10 of 22

∞ 

  
t  
+ (t – ν)α–1 Eα,α λj (t – ν)α G(ν, x)ej (x) dx dν ej (x).
j=1 0 

In addition, for β > 0, we construct the Fourier regularized solution as follows:

D
(β)  
 α
Vβ (t, x) = Eα,1 λj t ψβ (x)ej (x) dx ej (x)
j=1 

D
(β)  
 
+ tEα,2 λj t α ϕβ (x)ej (x) dx ej (x)
j=1 

D (β) t
  
 
α–1 α
+ (t – ν) Eα,α λj (t – ν) Gβ (ν, x)ej (x) dx dν ej (x).
j=1 0 

Here, D(β) depends only on β.

Theorem 4.1 Let β, δ > 0, 0 < p < d4 , and 1 < m < 2. We presume that the Cauchy data
(ψ, ϕ, G) is disturbed by the noisy data (ψβ , ϕβ , Gβ ) ∈ Lm () × Lm () × L2 (0, T; Lm ())
such that

ψ – ψβ Lm () + ϕ – ϕβ Lm () + Gβ – GL2 (0,T;Lm ()) ≤ β.

Then the following error estimate holds:

   1   1   2p – m–2
u(t, ·) – Vβ (t, ·) 2d  exp |C| α T D(β) αd D(β) d 2m β
L d–4p ()

CM  1   2   2p + 2–2α – m–2
+ exp |C| α T D(β) αd D(β) d dα 2m β
α
 – 2δ
C)–δ D(β) d uL∞ (0,T;Hp+δ ()) .
+ C(d, p)(

Proof Set the following function:

D
(β)  
 
vβ (t, x) = Eα,1 λj t α ψ(x)ej (x) dx ej (x)
j=1 

D
(β)  
 
+ tEα,2 λj t α ϕ(x)ej (x) dx ej (x)
j=1 

D (β) t
  
 
+ (t – ν)α–1 Eα,α λj (t – ν)α G(ν, x)ej (x) dx dν ej (x).
j=1 0 

It is clear to see that

 
vβ (t, ·) – Vβ (t, ·) p ≤ J1 (t) + J2 (t) + J3 (t), (4.2)
H ()
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 11 of 22

where J1 , J2 are defined respectively in (3.9), (3.10), and the third term J3 is defined by
D(β)
 t  

J3 (t) =  (t – ν)α–1 Eα,α λj (t – ν)α
 0
j=1

    (4.3)
  

× Gβ (ν, x) – G(ν, x) ej (x) dx dν ej (x) .
 
Hp ()

Using (2.5) of Lemma (2.3), we derive

  M 1–α  1 
(t – ν)α–1 Eα,α λj (t – ν)α ≤ λj α exp λjα (t – ν) . (4.4)
α
This inequality together with Parseval’s equality leads to


  2
  D
(β) t    
J3 (t)2 = 2p
λj (t – ν)α–1 Eα,α λj (t – ν)α Gβ (ν, x) – G(ν, x) ej (x) dx dν
j=1 0 

 2 D(β)
 M  2p+ 2–2α
≤   λ α
α j=1 j

  2
t  α1   
× exp 2λj (t – ν) Gβ (ν, x) – G(ν, x) ej (x) dx dν . (4.5)
0 

Since (3.13), we find that

 1    2
exp 2λjα (t – ν) ≤ exp 2|C| α T D(β) αd .
1
(4.6)

Thus, using Hölder inequality, we find that

 2
      2
J3 (t)2 ≤  M  exp 2|C| α1 T D(β) αd
α
D
(β) 
 2
2p+ 2–2α dm–2d
α – 2m
T dm–2d  
× λj λj2m
Gβ (ν, x) – G(ν, x) ej (x) dx dν. (4.7)
j=1 0 

Since (3.13), we know that

2p+ 2–2α dm–2d    2 2p+ 2–2α – dm–2d   4p + 4–4α – m–2

λj α – 2m
≤ C D(β) d α 2m
= C(p, α, m, d)D(β) d dα m ,

which leads to
 2
       2   4p 4–4α m–2
J3 (t)2 ≤  CM  exp 2|C| α1 T D(β) αd D(β) d + dα – m
 α 
 T
 
× Gβ (ν, ·) – G(ν, ·)2 dm–2d dν
0 H 4m ()
 
 CM 2    2   4p 4–4α m–2

≤  exp 2|C| α1 T D(β) αd D(β) d + dα – m
α 

× Gβ – G2 dm–2d . (4.8)

L2 (0,T;H 4m ())
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 12 of 22

Noting that Sobolev embedding

   dm–2d 
L2 0, T; Lm () → L2 0, T; H 4m () ,

we derive that

Gβ – G2 dm–2d ≤ C(d, m)Gβ – G2L2 (0,T;Lm ()) ≤ β 2 C(d, m). (4.9)
L2 (0,T;H 4m ())

Combining (4.8) and (4.9), we obtain that

     2   2p 2–2α m–2
J3 (t)  CM exp |C| α1 T D(β) αd D(β) d + dα – 2m β. (4.10)
α

Here, the hidden constant depends on d, m. Combining (3.16), (3.18), (4.10) and looking
at (4.2), we arrive at

     1   2p m–2
vβ (t, ·) – Vβ (t, ·) p  exp |C| α1 T D(β) αd D(β) d – 2m β
H ()

CM  1   2   2p + 2–2α – m–2
+ exp |C| α T D(β) αd D(β) d dα 2m β. (4.11)
α

Here, the hidden constant depends on d, m, T, C1 .

On the other hand, similar to (3.22), for any u ∈ L∞ (0, T; Hr+δ ()), we also get the fol-
lowing estimate:

   – 2δ
u(t, ·) – vβ (t, ·) p ≤ (
C )–δ D(β) d uL∞ (0,T;Hp+δ ()) . (4.12)
H ()

In conclusion, we obtain the desired estimate

   1   1   2p – m–2
u(t, ·) – Vβ (t, ·) 2d  exp |C| α T D(β) αd D(β) d 2m β
L d–4p ()

CM  1   2   2p + 2–2α – m–2
+ exp |C| α T D(β) αd D(β) d dα 2m β
α
 – 2δ
C )–δ D(β) d uL∞ (0,T;Hp+δ ()) .
+ C(d, p)(

The proof is completed. 

5 Regularization in Lp space under nonlinear source term

In this section, we consider the Cauchy problem for nonlinear problem. Indeed, we will
focus on Problem (2.7). For the sake of brevity, we denote the operators

∞
 
 
P1 (t)v = Eα,1 λj t α v(x)ej (x) dx ej (x) (5.1)
j=1 

and
∞
 
 
P2 (t)v = tEα,2 λj t α v(x)ej (x) dx ej (x), (5.2)
j=1 
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 13 of 22

and
∞
 
 
P3 (t)v = t α–1 Eα,α λj t α v(x)ej (x) dx ej (x) (5.3)
j=1 

for any v ∈ L2 (). We rewrite the mild solution as the following form:
 t  
u(t, x) = P1 (t)ψ + P2 (t)ϕ + (t – ν)α–1 P3 (t – ν)G ν, x, u(ν, x) dν. (5.4)
0

By using the Fourier method, we approximate Problem (2.7) by a new integral equation.
Before providing this integral equation, we need to introduce the operator SR , which is the
orthogonal projection onto the eigenspace span {ej , λj ≤ R} for any R > 0. Let any function
θ ∈ L2 (), then we provide the truncation operator SR θ as follows:

λj ≤R
 

SR θ = θ (x)ej (x) dx ej (x). (5.5)
j=1 

Let us assume that the Cauchy data (ψ, ϕ) is noisy by the observation data (ψβ , ϕβ ) ∈
Lp () × Lp () such that

ψ – ψβ Lp () + ϕ – ϕβ Lp () ≤ β. (5.6)

Since this observation data (ψβ , ϕβ ), we can construct a regularized solution as follows:
 t  
Zβ (t, x) = P1 (t)SRβ ψβ + P2 (t)SRβ ϕβ + P3 (t – ν)SRβ G ν, x, Zβ (ν, x) dν. (5.7)
0

Integral equation as above is called “regularized problem”. Our main purpose in this sec-
tion is to
• Show the existence and uniqueness of the mild solution to regularized problem (5.7);
• Estimate the upper bound of the regularized solution Zβ and the sought solution u on
Lm space.

Theorem 5.1 Let the observed data (ψβ , ϕβ ) ∈ Lp () × Lp (). Then Problem (5.7) has a
2d
unique solution Zβ ∈ L∞
a (0, T; L
d–4r ()). In addition, we get the following bound:

 
M dp–2d  1 
≤ 2 (Rβ ) 4p exp Rβα T ψβ Lp ()
r–
Zβ  ∞ 2d
La (0,T;L d–4r ()) α
 
 
–1/α  M  r–
dp–2d  1 
+ 2λ1  (Rβ ) 4p exp Rβα T ϕβ Lp () (5.8)
α
1
for any a > Rβα and 0 ≤ r < d4 .

Proof Let w ∈ Hq (). Then, for any q ≥ q, we have the following estimate by a simple
computation:

 2  12
  2q –2q   2 2q
P1 (t)SR w q = λj Eα,1 λj t α  λj w(x)ej (x) dx . (5.9)
β H ()
λj ≤Rβ 
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 14 of 22

In view of inequality (2.3) of Lemma (2.3), if λj ≤ Rβ then we find that

 2  
2q –2q   2  M  2q –2q  α1   M 2  1 
λj Eα,1 λj t α   
≤   λj exp 2λj t ≤   (Rβ )2q –2q exp 2Rβα t . (5.10)
α α

From the two observations above, we deduce that

 
     1 
P1 (t)SR w q ≤  M (Rβ )q –q exp R α t wHq () . (5.11)
β H () α β

By a similar techniques as above, for any w ∈ Hq (), we also find that

 
     1 
P2 (t)SR w q ≤ λ–1/α  M (Rβ )q –q exp R α t wHq () (5.12)
β H () 1 α β

and
 
  1–α    1 
P3 (t)SR w q ≤ λ α  M (Rβ )q –q exp R α t wHq () . (5.13)
β H () 1 α β

2d
Let a > 0 be a real constant. Denote by L∞ a (0, T; L
d–4r ()) the subspace of L∞ (0, T;
2d
L d–4r ()) associated with the following norm:
     
f a := max exp –a(T – t) f (·, t)
2d
2d , ∀f ∈ L∞ 0, T; L d–4r () .
0≤t≤T L d–4r ()

2d 2d
Let us define a nonlinear map M : L∞
a (0, T; L
d–4r ()) → L∞ (0, T; L d–4r ()) by
a

 t  
Mz(t, x) = P1 (t)SRβ ψβ + P2 (t)SRβ ϕβ + P3 (t – ν)SRβ G ν, x, z(ν, x) dν. (5.14)
0

By inserting z = 0 into the above expression, we get immediately that

M(z = 0) = P1 (t)SRβ ψβ + P2 (t)SRβ ϕβ .

2d
We aim to show that M(z = 0) ∈ L∞
a (0, T; L
d–4r ()) for 0 ≤ r < d . In view of the embedding
4
2d
Hr () → L d–4r () and (5.11) with q = r ≥ 0 and q = dp–2d
4p
< 0, we find that

   
P1 (t)SR ψβ  2d  P1 (t)SRβ ψβ Hr ()
β
L d–4r ()
 
M dp–2d  1 
≤  (Rβ ) 4p exp Rβα t ψβ 
r–
dp–2d .
α H 4p ()

It is obvious to see that the Sobolev embedding

dp–2d
Lp () → H 4p () (5.15)

since 1 < p < 2. Thus, we find that

 
  M dp–2d  1 
P1 (t)SR ψβ  2d   (Rβ ) 4p exp Rβα t ψβ Lp () .
r–
(5.16)
β
L d–4r () α
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 15 of 22

By a similar argument as above, we also show that

 
     1 
 M (Rβ )r– 4p
dp–2d
P2 (t)SR ψβ   λ–1/α exp Rβα t ϕβ Lp () . (5.17)
β 2d
L d–4r ()
1   α

2d
From the two observations above, we can deduce that M(z = 0) ∈ L∞ a (0, T; L
d–4r ()).
2d
Let any two functions z1 , z2 ∈ L∞
a (0, T; L d–4r ()) for any a > 0. Then we have the follow-

ing identity:
 t     
Mz1 (t, x) – Mz2 (t, x) = P3 (t – ν)SRβ G ν, x, z1 (ν, x) – G ν, x, z2 (ν, x) dν. (5.18)
0

By looking at estimate (5) with q = r and q = 0, we get the following evaluation:

     
P3 (t – ν)SR G ν, x, z1 (ν, x) – G ν, x, z2 (ν, x)  r
β H ()
 
1–α  M   1     
≤ λ1 α  (Rβ )q –q exp Rβα (t – ν) G ν, ·, z1 (ν, ·) – G ν, ·, z2 (ν, ·) L2 ()
α
 
1–α  M   1  
≤ Lg λ1  (Rβ )q –q exp Rβα (t – ν) z1 (ν, ·) – z2 (ν, ·)L2 ()
α 
α
 
1–α  M   1  
≤ Lg λ1 α  (Rβ )q –q exp Rβα (t – ν) z1 (ν, ·) – z2 (ν, ·) 2d , (5.19)
α L d–4r ()

where we have used the global Lipschitz function of G and the Sobolev embedding
2d
L2 () → L d–4r (). It follows from (5.18) that

 
exp(–at)Mz1 (t, ·) – Mz2 (t, ·)Hr ()
1–α
MLg λ1 α (Rβ )q –q
≤
α
 t
 1    
× exp Rβα – a (t – ν) exp(–aν)z1 (ν, ·) – z2 (ν, ·) 2d dν. (5.20)
0 L d–4r ()

Recall that

 
z1 – z2  2d := ess sup exp(–aν)z1 (ν, ·) – z2 (ν, ·) 2d ,
L∞
a (0,T;L d–4r ()) 0<ν<T L d–4r ()

and we get

 
exp(–at)Mz1 (t, ·) – Mz2 (t, ·) 2d
L d–4r ()
 
 exp(–at)Mz1 (t, ·) – Mz2 (t, ·)Hr ()
1–α
 
MLg λ1 α (Rβ )q –q t   1 
≤ exp Rβ – a (t – ν) dν z1 – z2 
α
2d
α 0 L∞
a (0,T;L d–4r ())

1–α
MLg λ1 α (Rβ )q –q
≤ z1 – z2  ∞ 2d ,
aα La (0,T;L d–4r ())
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 16 of 22

where we can verify the following inequality:

 t    1  1
exp –a(t – ν) exp Rβα (t – ν) dν <
0 a

1 2d
for any a > Rβα . Also we apply the embedding Hr () → L d–4r () to find that

1–α
MLg λ1 α (Rβ )q –q
Mz1 – Mz2  ∞ 2d ≤ z1 – z2  ∞ 2d . (5.21)
La (0,T;L d–4r ()) aα La (0,T;L d–4r ())

By choosing a such that

1–α
2MLg λ1 α (Rβ )q –q
a≥ ,
α
1–α
MLg λ1 α (Rβ )q –q
we know that aα
≤ 1/2. Hence, together with the fact that

 2d 
M(z = 0) ∈ L∞
a 0, T; L
d–4r () ,

2d
we can say that M is a contraction in L∞a (0, T; L
d–4r ()). By applying the Banach fixed
2d
point theorem, we can deduce that M has a fixed point Zβ ∈ L∞ a (0, T; L
d–4r ()). In the

following, we continue to estimate the upper bound of Zβ . By letting z1 = Zβ and z2 = 0

into (5.21), we find that

  1
MZβ – P1 (t)SR ψβ – P2 (t)SR ϕβ  2d ≤ Zβ  ∞ 2d . (5.22)
L∞
β β
a (0,T;L d–4r ()) 2 La (0,T;L d–4r ())

Due to the fact that MZβ = Zβ and combining (5.16) and (5.17), we find that

Zβ  2d
L∞
a (0,T;L d–4r ())

= MZβ  2d
L∞
a (0,T;L d–4r ())

  1
≤ P1 (t)SRβ ψβ + P2 (t)SRβ ϕβ  2d + Zβ  ∞ 2d
L∞
a (0,T;L d–4r ()) 2 La (0,T;L d–4r ())
 
M dp–2d  1 
≤  (Rβ ) 4p exp Rβα T ψβ Lp ()
r–
α
 
 
–1/α  M  r–
dp–2d  1  1
+ λ1  (Rβ ) 4p exp Rβα T ϕβ Lp () + Zβ  ∞ 2d .
α 2 La (0,T;L d–4r ())

Thus, we arrive at the desired result. 

The following theorem provides the error between the regularized solution and the ex-
act solution in the space of Lp type when the noisy data in Lp .

Theorem 5.2 Let us assume that Problem (1.1)–(1.2) has a unique solution u ∈ L∞ (0, T;
Hr+δ ()) ∩ L∞ (0, T; GsMT ()) for any δ > 0 and 0 ≤ r < d4 and s > r. Let the observed data
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 17 of 22

(ψβ , ϕβ ) ∈ Lp () × Lp () satisfy (5.6) and 1 < p < 2. Let us choose Rβ such that

dp–2d
r– 4p  1 
lim Rβ = ∞, lim (Rβ ) exp Rβα T β = 0. (5.23)
β→0 β→0

Then the following estimate holds:

   dp–2d  1  
Zβ (t, ·) – u(t, ·) 2d
r–
 max (Rβ ) 4p exp Rβα T β, (Rβ )r–s , (Rβ )–δ . (5.24)
L d–4r ()

Remark 5.1 Let us choose Rβ = ( 1–μ

T
)α logα ( β1 ) for any 0 < μ < 1. Then the error between
2d
the regularized solution Zβ and the exact solution u in L d–4r () is of order


–δ 
r–s 
α(r– dp–2d
4p )

1 1 1
max log , log , log βμ .
β β β

Proof It follows from (5.4) that

 t  
SR u(t, x) = P1 (t)SR ψ + P2 (t)SR ϕ + (t – ν)α–1 P3 (t – ν)SR G ν, x, u(ν, x) dν. (5.25)
0

This implies that

     
Zβ (t, ·) – u(t, ·) 2 ≤ Zβ (t, ·) – SR u(t, ·) 2 + SR u(t, ·) – u(t, ·) 2
L () β L () β L ()

= (I) + (II). (5.26)

Let us first treat the term (I). Indeed, by the definition of Zβ as in (5.7) and (5.25), we arrive
at

 
Zβ (t, ·) – SR u(t, ·) 2
β L ()
   
≤ P1 (t)SRβ ψβ – P1 (t)SRβ ψ L2 () + P2 (t)SR ϕβ – P2 (t)SR ϕ L2 ()
 t
    
+ (t – ν)α–1 P3 (t – ν)SRβ G ν, x, Zβ (ν, x) – G ν, x, u(ν, x) dν. (5.27)
0

In view of the estimate (5.11) with q = 0 and q = dp–2d

4p
< 0, we find that

 
P1 (t)SR ψβ – P1 (t)SR ψ  2
β β L ()
 
≤ P1 (t)SRβ ψβ – P1 (t)SRβ ψ  2d
L d–4r ()
 
  M dp–2d  1 
 P1 (t)SRβ (ψβ – ψ)Hr () ≤  (Rβ ) 4p exp Rβα t ψβ – ψ dp–2d
–
α H 4p ()
   
M dp–2d  1  M dp–2d  1 
  (Rβ ) 4p exp Rβα t ψβ – ψLp ()   (Rβ ) 4p exp Rβα t β,
– –
α α
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 18 of 22

where we have used (5.15) to get the last estimate. By a similar explanation, we obtain the
estimate for the second term on the right-hand side of (5.27) as follows:
 
P2 (t)SR ϕβ – P2 (t)SR ϕ  2
β β L ()
 
≤ P2 (t)SRβ ϕβ – P2 (t)SRβ ϕ  2d
L d–4r ()
 
 P2 (t)SRβ ϕβ – P2 (t)SRβ ϕ Hr ()
 
 
–1/α  M  –
dp–2d  1 
≤ λ1  (Rβ ) 4p exp Rβα t ϕβ – ϕ dp–2d
α H 4p ()
   
 
–1/α  M 
dp–2d
– 4p  α1   
–1/α  M  –
dp–2d  1 
 λ1  (Rβ ) exp Rβ t ϕβ – ϕL ()  λ1  (Rβ ) 4p exp Rβα t β.
p
α α

We continue to consider the third term on the right-hand side of (5.27). Indeed, we get
 t 
      
 (t – ν)α–1 P3 (t – ν)SR G ν, x, Zβ (ν, x) – G ν, x, u(ν, x) dν 
 
0 L2 ()
  t
1–α  M   1     
≤ λ1 α   exp R α (t – ν) G ν, ·, Zβ (ν, ·) – G ν, ·, u(ν, ·)  2 dν
 β L ()
α 0
  t
1–α  M   1  
≤ Lg λ1 α   exp Rβα (t – ν) Zβ (ν, ·) – u(ν, ·)L2 () dν. (5.28)
α 0

From two above observations and (5.27), we find that

 
Zβ (t, ·) – SR u(t, ·) 2
L ()
 
  M  dp–2d  1 
+ 1  (Rβ ) 4p exp Rβα t β
–
≤ λ–1/α1
α
  t
1–α  M   1  
+ Lg λ1 α   exp Rβα (t – ν) Zβ (ν, ·) – u(ν, ·)L2 () dν. (5.29)
α 0

Next, we continue to provide the following estimate:

 12
   1  1  2
u(t) – SR u(t, ·) 2 = exp –2(T – t)λjα λ–2s α 2s
β L () j exp 2(T – t)λj λj u(t), ej
λj >Rβ

 1
≤ (Rβ )–s exp –(T – t)Rβα B, (5.30)

where we remind that

 ∞
 12
 1  2
B = ess sup λ2s
j exp 2(T – t)λj
α
u(t), ej ≤ uL∞ (0,T;GsMT ()) . (5.31)
0≤t≤T j=1

Combining (5.29) and (5.30), we arrive at the following estimate:

 
Zβ (t, ·) – u(t, ·) 2
L ()
   
≤ Zβ (t, ·) – SRβ u(t, ·)L2 () + u(t) – SRβ u(t, ·)L2 ()
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 19 of 22

 
  M  dp–2d  1   1
≤ λ–1/α + 1  (Rβ )– 4p exp R α t β + (Rβ )–s exp –(T – t)R α B
1 α β β

  t
1–α  M   1  
+ Lg λ1 α   exp Rβα (t – ν) Zβ (ν, ·) – u(ν, ·)L2 () dν. (5.32)
α 0

1
Multiplying both sides of the above expression by exp(Rβα (T – t)), we obtain that

 1  
exp Rβα (T – t) Zβ (t, ·) – u(t, ·)L2 ()

M(λ–1/α + 1) –
dp–2d  1 
≤ 1
(Rβ ) 4p exp Rβα T β + (Rβ )–s B
α
1–α 
MLg λ1 α t  1  
+ exp Rβα (T – ν) Zβ (ν, ·) – u(ν, ·)L2 () dν. (5.33)
α 0

By using Gronwall’s inequality, we get that

 1  
exp Rβα (T – t) Zβ (t, ·) – u(t, ·)L2 ()
 
1–α 
M(λ–1/α + 1) dp–2d
– 4p  α1  MLg λ1 α
≤ 1
(Rβ ) –s
exp Rβ T β + (Rβ ) B exp t . (5.34)
α α

This implies that

 
Zβ (t, ·) – u(t, ·) 2
L ()
 
 1  M(λ–1/α + 1) –
dp–2d  1 
 exp Rβα (t – T) 1
(Rβ ) 4p exp Rβα T β + (Rβ )–s B . (5.35)
α

Our next aim is to considering the quantity Zβ (t, ·) – u(t, ·)Hr () . Using the triangle
inequality, we arrive at

     
Zβ (t, ·) – u(t, ·) r ≤ Zβ (t, ·) – SR u(t, ·) r + SR u(t, ·) – u(t, ·) r . (5.36)
H () β H () β H ()

We control the first term on the right above as follows:


λ ≤R
 2
   j β
 
Zβ (t, ·) – SR u(t, ·) r ≤  λ 2r
Z β (x, t) – u(x, t) ej (x) dx
β H () j
j=1 

 
≤ (Rβ )r Zβ (t, ·) – SRβ u(t, ·)L2 () . (5.37)

The second quantity on the right-hand side of (5.36) is bounded by


 2
  
SR u(t, ·) – u(t, ·) r =  2r
λj u(x, t)ej (x) dx
β H ()
λj >Rβ 


 2


=  –2δ 2r+2δ
λj λj u(x, t)ej (x) dx
λj >Rβ 
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 20 of 22

≤ (Rβ )–δ uL∞ (0,T;Hr+δ ()) , (5.38)

where


 2
 ∞ 2r+2δ
uL∞ (0,T;Hr+δ ()) ≥  λj u(x, t)ej (x) dx .
j=1 

Combining (5.36), (5.37), and (5.38) together with (5.35), we get that

 
Zβ (t, ·) – u(t, ·) r
H ()
 
≤ (Rβ )r Zβ (t, ·) – SRβ u(t, ·)L2 () + (Rβ )–δ uL∞ (0,T;Hr+δ ())
 
 α1  M(λ–1/α + 1) dp–2d
r– 4p  α1 
≤ exp Rβ (t – T) 1
(Rβ ) r–s
exp Rβ T β + (Rβ ) B
α
+ (Rβ )–δ uL∞ (0,T;Hr+δ ()) . (5.39)

2d
Using the embedding Hr () → L d–4r (), we deduce that

   
Zβ (t, ·) – u(t, ·) 2d  Zβ (t, ·) – u(t, ·)Hr ()
L d–4r ()
 r–
dp–2d  1  
 max (Rβ ) 4p exp Rβα T β, (Rβ )r–s , (Rβ )–δ . (5.40)


6 Conclusion
We consider the Cauchy problem for an evolution equation with the Caputo fractional
derivative. The most important feature of the paper is that the problem is not well posed
in the sense of Hadamard. Throughout this work, the homogeneous case, the inhomoge-
neous case, and the nonlinear case are investigated in turn. For each case, we construct
a regularized solution by using the Fourier truncation method. In addition, using Cauchy
data in Lp spaces, p = 2, by some embeddings linking the Hilbert scale-spaces and the
Lebesgue spaces, we derive error estimates between regularized solutions and exact solu-
tions.

Acknowledgements
The author Le Dinh Long is thankful to Van Lang University. This research is supported by Industrial University of Ho Chi
Minh City (IUH) under Grant Number 130/HD-DHCN.

Funding
Not applicable.

Availability of data and materials

Not applicable.

Declarations
Competing interests
The authors declare no competing interests.

Author contributions
All authors contributed equally. All the authors read and approved the final manuscript.
Phuong et al. Boundary Value Problems (2022) 2022:100 Page 21 of 22

Author details
1
Industrial University of Ho Chi Minh City, Ho Chi Minh City, Vietnam. 2 Department of Mathematics, Cankaya University,
06530 Balgat, Ankara, Turkey. 3 Institute of Space Sciences, R76900 Magurele-Bucharest, Romania. 4 Department of
Mathematics, Texas AM University-Kingsville, 700 University Blvd., MSC 172, Kingsville, TX, USA. 5 Division of Applied
Mathematics, Science and Technology Advanced Institute, Van Lang University, Ho Chi Minh City, Vietnam. 6 Faculty of
Applied Technology, School of Engineering and Technology, Van Lang University, Ho Chi Minh City, Vietnam.

Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 17 March 2022 Accepted: 6 December 2022

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