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https://doi.org/10.53006/rna.962068

Available online at www.resultsinnonlinearanalysis.com Research Article

Note on a Allen-Cahn equation with Caputo-Fabrizio derivative

Phuong Duc Nguyen

Faculty of Fundamental Science, Industrial University of Ho Chi Minh City, Ho Chi Minh City 700000, Vietnam.

Abstract

In this short note, we investigate the Allen-Cahn equation with the appearance of the Caputo-Fabizzio derivative. We obtain a local solution when the initial value is small enough. This is an equation that has many practical applications. The power term in the nonlinear component of the source function and the Caputo-Fabizzio operator combine to make nding the solution space more dicult than the classical prob- lem. We discovered a new technique, connecting Hilbert scale and Lp spaces, to overcome these diculties.

Evaluation of the smoothness of the solution was also performed. The research ideas in this paper can be used for many other models.

Keywords: Allen-Cahn equation, Fractional diusion equation; Caputo-Fabizzio, regularity.

2010 MSC: 35R11, 35B65, 26A33

1. Introduction

Let D be a C2 bounded open set of RN with sucient smooth boundary and T > 0. In this paper, we consider the fractional Sobolev equation





CFDαtu = ∆u + u − u3, (x, t) ∈ D × (0, T ), u = 0, (x, t) ∈ ∂D × (0, T ),

u(x, 0) = u0(x), x ∈ D

(1.1)

where CFDαt is the Caputo-Fabrizio operator for fractional derivatives of order α which is dened as (see [24])

CFDαtv(t) = H(α) 1 − α

Z t 0

Dα(t − s)∂v(s)

∂s ds, for t ≥ 0,

Email address: nguyenducphuong@iuh.edu.vn (Phuong Duc Nguyen)

Received : January 20, 2021; Accepted: August 13, 2021; Online: August 15, 2021.

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where we denote by the kernel Dα(z) = exp



1−αα z and H(α) satises H(0) = H(1) = 1, (see e.g. [22, 23]).

The main goal of this note is to prove the existence of a local solution to the problem given the input data u0 in the space Lp space. If α = 1, Problem (1.1) is called Allen-Cahn equation with classical derivative. If we replace Caputo-Fabizzio derivative by Caputo derivative, Problem (1.1) is studied by [28]. Allen-Cahn model was originally introduced to transform the description of boundaries in coherent solids [29].

Fractional calculus has a long history and has many applications in simulations of physical phenomena or real life for example, mechanics, electricity, chemistry, biology,.. Many mathematical models cannot be expressed in terms of classical derivatives because of the eects of external forces. Therefore, the introduction of fractional calculus has important implications for modeling physical and engineering processes in cases where classical derivatives are not available. Some works are attracting the attention of the community for fractional dierential equations, like A. Debbouche and his group [4, 5, 6], E. Karapnar et al [11, 12, 13, 14, 15, 16, 17, 18]. The Caputo-Fabrizio fractional derivative was rst introduce by [22] which makes sense to avoid singular kernels. It is detemined by the convolution of the exponential function and the rst order derivative. This operator has been widely applied to a number of derivative modes in many elds, such as biology, physics, control systems [19, 20, 26].

There are two main challenges and diculties when we consider this problem. The rst diculty is that it is dicult for us to apply the Lp estimate to the semigroup heat operators because of the appearance of the Caputo-Fabizzio operator. Indeed, in F. Weisler's work [27], they have the advantage of using the Lp evaluation for the half heat group where we do not apply. The second challenge is that we cannot evaluate the function up on Hilbert scale spaces but can only estimates on Lp while the Caputo-Fabizzio operator can only handle in Hilbert scale space. Those are the hard points that we need to overcome. Our novel idea is to connect the evaluations together by embeddings between Lp and Hilbert scales. This new technique can be applied to prove the existence of solutions to a wide range of problems.

2. Main results

Before giving the main result, we recall some knowledge about function spaces and embeddings. Note that A = −∆ is a symmetric uniformly elliptic operator, hence it possesses a non-negative, non-decreasing and discrete spectrum 0 < λ1 ≤ λ2 ≤ ... ≤ λn % ∞. The corresponding eigenvectors of A are denoted by en ∈ D(A), which satisfy Aen(x) = λnen(x) for x ∈ D. Let us introduce the Hilbert scale space, which is given as follows

Hr(D) = (

f ∈ L2(D),

X

n=1

λ2rnhf, eni2L2(D)< ∞ )

,

for any r ≥ 0. Here the symbol h·, ·iL2(M) denotes the inner product in L2(M). It is well-known that Hr(D) is a Hilbert space corresponding to the following norm

kf kHr(D)= v u u t

X

n=1

λ2rnhf, eni2L2(D), f ∈ Hr(D).

In view of Hr(D) ≡ D((−A)r) is a Hilbert space. Then D((−A)−r) is a Hilbert space with the norm

kvkD((−A)−r)=

X

n=1

|hv, eni|2λ−2rj

!12 ,

where h·, ·i in the latter equality denotes the duality between D((−A)−r) and D((−A)r).

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Lemma 1. The following inclusions hold true:

Lp(Ω) ,→ D(Aσ), if − N

4 < σ ≤ 0, p ≥ 2N N − 4σ, D(Aσ) ,→ Lp(Ω), if 0 ≤ σ < N

4, p ≤ 2N N − 4σ.





(2.1)

Denition 1. The function v is called a mild solution of Problem (1.1) if it satises that v(t) = Sα(t)u0+

Z t 0

Sα(t − s)G(v(s))ds (2.2)

where G(v) = v − v3 and Sα(t) is dened by Sα(t)f = (1 + αλn)−1exp

 −αλn 1 + αλn

t



hf, eniL2(D)en(x), α = 1 − α.

for any w ∈ L2(D).

Lemma 2. Let f ∈ Hm−2(D). Then Sα(t)f

Hm(D) ≤ 1

1 − αkf kHm−2(D). (2.3)

Proof. Using Parseval' s equality, we get that

Sα(t)f

Hm(D)=

X

n=1

λmn (1 + αλn)−2exp −2αλn

1 + αλnt



hf, eni2L2(D)

!1/2

≤ 1 α

X

n=1

λm−2n hf, eni2L2(D)

!1/2

= 1

1 − αkf kHm−2(D) (2.4)

Remark 1. The hardest part about proving the theorem is that we don't immediately get the Lp estimate for the operator Sα(t). For classical problem, we are available for apply Lp estimate since the ideas of [27].

However, we face the operator Sα(t) as above, we have diculty things for considering Lp estimate. The second diculty is that we cannot evaluate the source function on Hilbert scales space.

Theorem 1. Let u0 ∈ Lq/3(D) where 2 ≤ q ≤ 6 and q ≥ 2N. Then problem (1.1) has a local mild solution u ∈ Xβ,q∩ Lp(0, T ; Lq(D)) where 0 < β < 1/3 and 1 < p < β1.

Remark 2. Since the assumption 2 ≤ q ≤ 6 and q ≥ 2N, we can see that 1 ≤ N ≤ 3. Hence, we only study the local existence for the dimensional of the domain D is about 1 to 3.

Proof. It is obvious to see that G(u) − G(v)

=

(u − v) − (u3− v3)

≤ 2(|u − v|)(1 + |u|2+ |v|2). (2.5) Using Hölder inequality, we continue to get the following estimate

G(v1) − G(v2)

Lq3(D)=

Z

D

G(v1) − G(v2)

q 3dx

3q

≤ 2 Z

D

(|v1− v2|)(|v1− v2|2+ |v1− v2|2)

q 3dx

!3q

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≤ 2Z

D



|v1− v2|q33133qZ

D



(|v1|2+ |v2|2)q3

32233q

≤ 2kv1− v2kLρ(D)

"

Z

D

|v1|q2q +

Z

D

|v2|q2q

#

= 2kv1− v2kLq(D)

hkv1k2Lq(D)+ kv2k2Lq(D)

i

. (2.6)

Dene the Banach space Xβ,q of all Bochner integrable functions u : [0, T ] → Lq(D) such that tεu are bounded continuous functions, endowed with the norm

sup

0≤t≤T

tβku(t, .)kLq(D)< ∞.

Let the function J be as follows

J v(t) = Sα(t)u0+ Z t

0

Sα(t − s)G(v(s))ds. (2.7)

Let v, w ∈ Xβ,q. Since Sobolev embedding HN q−2N4q (D) ,→ Lq(D)for any q > 2, we get that

J v − J w Xβ,q

= sup

0≤t≤T

tβkJ v(., t) − J w(., t)kLq(D)

≤ sup

0≤t≤T

tβkJ v(., t) − J w(., t)k

H

N q−2N

4q (D) (2.8)

It is obvious to see that

Z t 0

Sα(t − s)G(v(s))ds − Z t

0

Sα(t − s)G(w(s))ds H

N q−2N 4q (D)

≤ 1

1 − α Z t

0

G(v(s)) − G(w(s)) H

N q−2N 4q −2

(D)

ds

≤ C(N, q) Z t

0

G(v(s)) − G(w(s)) H

N q−6N

4q (D)ds (2.9)

where we note that

HN q−6N4q (D) ,→ HN q−2N4q −2(D) since the condition q ≥ 2N.

Based on Sobolev embedding Lq3(D) ,→ H

N q−6N

4q (D)for any q < 6, we derive that Z t

0

G(v(s)) − G(w(s)) H

N q−6N 4q (D)

ds ≤ C(N, q) Z t

0

G(v(s)) − G(w(s))

Lq3(D)ds (2.10) Set the following ball

B(R) :=

n

w : [0, T ] → Lq(D), w

Xβ,q ≤ Ro

(2.11) If v, w ∈ Xβ,q then we get

kv(., s)kLq(D) ≤ s−β v

Xβ,q ≤ s−βR. (2.12)

In view of (2.6), we obtain Z t

0

G(v(s)) − G(w(s))

Lq3(D)ds

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≤ 2 Z t

0

kv(., s) − w(., s)kLq(D)

h

kv(., s)k2Lq(D)+ kw(., s)k2Lq(D)

i

≤ 4R2 Z t

0

s−2βkv(., s) − w(., s)kLq(D)ds

= 4R2 Z t

0

s−3βsβkv(., s) − w(., s)kLq(D)ds ≤ 4R2

Z t 0

s−3βds

 v − w

Xβ,q

(2.13)

Under the condition β < 1/3, we know that R0ts−3βds is convergent. So, we get that Z t

0

G(v(s)) − G(w(s))

Lq3(D)ds ≤ 4R2 t1−3β 1 − 3β

v − w

Xβ,q

. (2.14)

Combining (2.8), (2.9), (2.14), we arrive at

J v − J w

Xβ,q ≤ sup

0≤t≤T

tβ+1−3β

!

C(N, q)4R2 1 1 − 3β

v − w

Xβ,q

≤ 4C(N, q)R2T1−2β 1 − 3β

v − w

Xβ,q

. (2.15)

Let us choose R, T such that

4C(N, q)R2T1−2β

1 − 3β < 1/2.

We next evaluate Sαu0

Xβ,q

= sup

0≤t≤T

tβkSα(t)u0kLq(D)≤ C(N, q) sup

0≤t≤T

tβkSα(t)u0k

H

N q−2N

4q (D). (2.16) By looking back Lemma (2), we nd that

kSα(t)u0k

H

N q−2N 4q (D)

≤ 1

1 − αku0k

H

N q−2N −8q

4q (D) (2.17)

Since the condition q ≥ 2N, we remind the Sobolev embedding HN q−6N4q (D) ,→ HN q−2N −8q4q (D) . This allows us to provide that the following estimate

Sαu0

Xβ,q

≤ C(N.q) 1 − α ku0k

H

N q−6N

4q (D) (2.18)

The condition q < 6 give us the embedding

Lq3(D) ,→ H

N q−6N 4q (D) which leads to

ku0k

H

N q−6N 4q (D)

≤ C(N, q)ku0k

Lq3(D) (2.19)

Hence, we deduce that

J (0)

Xβ,q

= Sαu0

Xβ,q

≤ TβC(N, q)ku0k

Lq3(D). (2.20)

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Let us choose T such that

TβC(N, q)ku0k

Lq3(D)≤ R 2. It follows from (2.15) that

J v

Xβ,q

J v − J (0) Xβ,q

+ J (0)

Xβ,q

≤ 1 2

v

Xβ,q+ TβC(N, q)ku0k

Lq3(D)

for any v ∈ Xβ,q. This says that

J v

Xβ,q

< R (2.21)

which allows us to deduce that J is a mapping from B(R) to itself B(R). Using the Banach mapping theorem, we conclude that J have a xed point u in B(R).

Our aim is to investigate the regularity of the mild solution u. Indeed, we get

u Xβ,q

J u − J (0) Xβ,q

+ J (0)

Xβ,q

≤ 1 2 u

Xβ,q + TβC(N, q)ku0k

Lq3(D). (2.22) This implies that

u

Xβ,q ≤ 2TβC(N, q)ku0k

Lq3(D). Hence, we nd that

u(., t)

Lq(D)≤ 2Tβt−βC(N, q)ku0k

Lq3(D) (2.23)

The above expression allows us to obtain that

Z T 0

u(., t)

p Lq(D)dt

1/p

≤ 2TβC(N, q)ku0k

Lq3(D)

Z T 0

t−βpdt

1/p

(2.24)

Since 1 < p < 1β, we deduce that the proper integral R0T t−βpdt is convergent. Therefore, we can say that u ∈ Lp(0, T ; Lq(D)).

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