In [8], a new deﬁnition called α−conformable

fractional derivative was introduced:

Let α∈(0,1), and f:E⊆(0,∞)−→ ℜ.

For x∈Elet:

Dαf(x) = lim

ε−→0

f(x+εx1−α)−f(x)

If the limit exists, then it is called α−

conformable fractional derivative of fat x.

For x= 0, if fis α−diﬀerentiable on (0, r)

for some r > 0, and limx−→0+Dαf(x) exists

then we deﬁne Dαf(0) = limx−→0+Dαf(x).

Solving Non Linear Fractional Newell-Whitehead Equation By Atomic Solution

GHARIB GHARIB1 , MAHA ALSOUDI2, IBTISSEM BENKEMACHE3

1Department of Mathematics, Zarqa University, Amman, JORDAN

2Basic Science Department, Applied Science Private University, Amman, JORDAN

3Department of Mathematics, The University of Jijel, ALGERIA

Abstract: Sometimes, it is not possible to find a general solution for some differential equations using some classical

methods, like separation of variables. In such a case, one can try to use theory of tensor product of Banach spaces to find

certain solutions, called atomic solutions. The goal of this paper is to find atomic solution for Fractional Newell- Witehead

type equation.

Keywords: Conformable derivative, tensor product, atomic solution.

Received: October 15, 2022. Revised: December 16, 2022. Accepted: January 13, 2023. Published: February 16, 2023.

1. Introduction

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Conformable derivative satisﬁes:

1) Tα(af +bg) = aTα(f) + bTα(g), for all

a, b ∈ ℜ.

2) Tα(λ) = 0, For all constant functions

f(t) = λ. Further, for α∈(0,1] and

f, g are α−diﬀerentiable at a point t,

we have:

3) Tα(fg) = f Tα(g) + gTα(f).

4) Tα(f

g) = gTα(f)−fTα(g)

g2, g(t)

= 0.

We list here the fractional derivatives of

certain functions,

1) Tα(tp) = ptp−α, for all p∈ ℜ.

2) Tα(sin 1

αtα) = cos 1

αtα.

3) Tα(cos 1

αtα) = −sin 1

αtα.

4) Tα(e1

αtα) = e1

αtα.

On letting α= 1 in these derivatives, we

get the corresponding classical rules for ordi-

nary derivatives.

One should notice that a function could

be α−diﬀerentiable at a point but not diﬀer-

entiable, for example, take f(t)=2√t, this

2(f)(t) = 1. Hence, T1

2(f)(0) = 1. But

T1(f)(0) does not exist. This is not the case

for the known classical fractional derivatives.

For more on fractional calculus and its ap-

plications we refer to [1] −[17].

Now, let Xand Ybe two Banach spaces

and X∗be the dual of X. Assume x∈Xand

y∈Y.

The operator T:X∗−→ Ydeﬁned by

T(x∗) = x∗(x)y

is bounded one rank linear operator. We

write x⊗yfor T. Such operators are called

atoms.

Atoms are among the main ingredient in

the theory of tensor products.

Atoms are used in theory of the best ap-

proximation in the Banach spaces, see [16].

One of the know result, see [16], that we

need is: if the sum of two atoms is an atom,

then either the ﬁrst components are depen-

dant or the second components are depen-

dant..

For more On tensor products of Banach

spaces we refer to [14] and [16].

Very often, diﬀerential equations can be

solve via separation of variables for the rea-

son or another. One of the main reasons of

the failure of the method of separation of vari-

ables is that the equation is not linear. Here

theory of tensor products comes to give us

certain solutions of the equation.

Our main object in this paper is to ﬁnd an

atomic solution of the equation.

Dβ

tu−D2α

xu=u−u3(1)

With conditions:

u(0, t)=1, uα(0, t)=1

u(x, 0) = √2, uβ(0, t)=1

x, t ≥0

2. The Solution

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This is not a linear fractional partial dif-

ferential equation. Hence the method of sep-

aration of variables does not work. So we are

forced to look for atomic solutions. Hence,

let

u(x, t) = P(x)Q(t) = P⊗Q(2)

be an atomic solution of (1), where P(x)

and Q(t) are not constants

Now, substitute (2) in (1) to get

P(x)Qβ(t)−P2α(x)Q(t) =

P(x)Q(t)−P3(x)Q3(t)(3)

Collecting terms in form of atoms to get

P(x)Qβ(t) = P2α(x) + P(x)

Q(t)−P3(x)Q3(t)(4)

Equation (5) has the tonsorial form:

P⊗Qβ=P2α+P⊗Q−P3⊗Q3(5)

So we have the sum of two atoms is an

atom. Then by a result in [5], we have two

cases:

Case(i)

P2α+P=P

Hence

P2α= 0 (6)

P(x) = c1

xα

α+c2

Conditions implies that c1= 1 and c2= 1.

Hence, the solution of equation is

P(x) = xα

α+ 1 (7)

Now, we go back to (3) and substitute (7)

in (3) to get

xα

α+ 1Qβ(t) =

xα

α+ 12α

+xα

α+ 1!Q(t)

−xα

α+ 13

Q3(t)

(8)

Equation (8) is valid for all x. In particu-

lar, it is true for x= 0. Hence (8) becomes

Qβ(t) = Q(t)−Q3(t) (9)

Simplifying to get

Qβ(t)−Q(t) + Q3(t) = 0 (10)

Since Qβ(t) = t1−βdQ

dt then we get

t1−βdQ

dt −Q(t) + Q3(t) = 0

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Hence

ZdQ

(Q3(t)−Q(t)) +Ztβ−1dt = 0

From which we get

2ln |Q2(t)−1

Q2(t)|+tβ

β=c

Conditions implies that c= 0.

Consequently

Q(t) = 1

s1−e−2tβ

(11)

The atomic solution in this case

u(x, t) = xα

α+ 1

s1−e−2tβ

(12)

This ends the discussion of case i).

Case(ii):

Q=Qβ

Simplifying to get

Qβ−Q= 0 (13)

The auxiliary equation is

r−1 = 0

So, Q(t) = ce

tβ

β.

Conditions implies that c=√2.

Q(y) = √2e

tβ

β(14)

Now, we go back to (3) and substitute (14)

in (3) to

P(x)



√2e

tβ

β





=P2α(x) + P(x)



√2e

tβ

β





−P3(x)



√2e

tβ

β





(15)

Equation (15) is valid for all t. In partic-

ular, it is true for t= 0. Hence (15) becomes

P2α(x)−2P3(x) = 0 (16)

Let Pα(x) = θand P2α(x) = θdθ

Hence

θdθ

dP −2P3= 0

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θ2

2=P4

2+c

Conditions implies that c= 0.

And since θ=Pα(x) we get

(Pα(x))2=P4

Hence

Pα(x) = P2(x)

We have Pα(x) = x1−αdP

dx . Hence

ZdP

P2=Zxα−1dx

−1

P=xα

α+c

Conditions implies that c=−1.

Consequently

P(x) = 1

1−xα

(17)

The atomic solution in this case

u(x, t) = √2e

tβ

1−xα

(18)

This ends the discussion of case ii).

This research is funded by the deanship of

research in Zarqa University/Jordan.

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3. Acknowledgment

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REDUCTION OF ORDER OF FRAC-

TIONAL DIFFERENTIAL EQUATIONS

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This research is funded by the deanship of

research in Zarqa University/Jordan.