Analytical and numerical solutions of a fractional
conformable derivative of the modified Burgers
equation using the Cole-Hopf transformation1
Ilhem MOUSa,b , Abdelhamid LAOUARb
a
    20 Aout 1955 University 21000 Skikda, Algeria
b
    Badji Mokhtar University of Annaba, P.O. Box 12 23000 Annaba, Algeria


                                         Abstract
                                         In this paper, we deal with a study of modified time-fractional Burgers equations. The idea is based on
                                         the use of a Cole-Hopf transformation which transforms the time-fractional modified Burgers equations
                                         into linear parabolic time fractional equations. To solve the latter, we use the Fourier transformation.
                                         Therefore, the solution of the modified time-fractional Burgers equations can be found by using the
                                         solution of parabolic equation and the inverse Cole-Hopf transformation.

                                         Keywords
                                         Cole-Hopf transformation, Conformable derivative, Mod                                   ified Burgers’ equation.




1. Introduction
This work considers the following modified Burgers equations [? ]

                                                                                       𝜕𝛼 𝑢
                                                                                            + (𝜈 + 𝑢)𝑢𝑥 = 𝑟𝑢𝑥𝑥 ,                                            (1.1)
                                                                                       𝜕𝑡 𝛼
where 𝜈 and 𝑟 are nonnegative parameters, 𝛼 is the fractional derivative, 0 < 𝛼 ≤ 1, 𝑥 ∈
[0, 𝑏], 𝑡 > 0, 𝜕 𝛼 𝑢/𝜕𝑡 𝛼 mean conformable derivative of the function 𝑢(𝑥, 𝑡).
When 𝜈 = 0, we get the conformable derivative Burgers equation. So the term 𝜈𝑢 translates
the modifying equation and can provide an interesting improvement concerning the numerical
solution. If the viscosity parameter 𝑟 approaches zero the equation models a inviscid fluid.

Subject to the initial and the boundary conditions
                                                                         ⎧
                                                                         ⎪ 𝑢(𝑥, 0) = 𝑢0 (𝑥), 𝑥 ∈ [0, 𝑏],
                                                                         ⎪
                                                                         ⎨                                                                                  (1.2)
                                                                         ⎪
                                                                         ⎪
                                                                         ⎩ 𝑢(𝑥, 𝑡) = 𝑓 (𝑥, 𝑡), 𝑥 ∈ 𝜕([0, 𝑏]), 𝑡 > 0.



IAM’20: Third conference on informatics and applied mathematics, 21–22 October 2020,Guelma, ALGERIA
" mousilhem@yahoo.fr (I. MOUS)

                                       © 2020 Copyright for this paper by its authors.
                                       Use permitted under Creative Commons License Attribution 4.0 International (CC BY 4.0).
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                  ISSN 1613-0073       CEUR Workshop Proceedings (CEUR-WS.org)
2. Preliminaries
We briefly recall a definition and some properties of fractional derivatives which can be used
in the sequel.
   ([? ? ]) Given a function 𝑓 ∶ [0 ∶ ∞) ⟶ ℝ, then the conformable fractional derivative of 𝑓
of order 𝛼 is defined by:
                                                 𝑓 (𝑡 + 𝜀𝑡 1−𝛼 ) − 𝑓 (𝑡)
                                𝑇𝛼 (𝑓 )(𝑡) = lim                         ,                (2.1)
                                             𝜀→0            𝜀
for all 𝑡 > 0, 𝛼 ∈ (0.1). If f is 𝛼-differentiable in some (0, 𝑎), 𝑎 > 0, and lim+ 𝑓 (𝛼) (𝑡) exists, then
                                                                                   𝑡→0
define
                                          𝑓 (𝛼) (0) = lim+ 𝑓 (𝛼) (𝑡).                               (2.2)
                                                       𝑡→0


Let’s give some properties which are summarized in the above theorem.
  ([? ? ]) Let 0 < 𝛼 ≤ 1 and f, g be 𝛼-differentiable at a point 𝑡 > 0. Then,

   1. 𝑇𝛼 (𝑎𝑓 + 𝑏𝑔) = 𝑎𝑇𝛼 (𝑓 ) + 𝑏𝑇𝛼 (𝑔), for all 𝑎, 𝑏 ∈ ℝ.
   2. 𝑇𝛼 (𝑡 𝑝 ) = 𝑝𝑡 𝑝−𝛼 for all 𝑝 ∈ ℝ.
   3. 𝑇𝛼 (𝜆) = 0, for all constant functions 𝑓 (𝑡) = 𝜆.
   4. 𝑇𝛼 (𝑓 𝑔) = 𝑓 𝑇𝛼 (𝑔) + 𝑔𝑇𝛼 (𝑓 ).
          𝑓       𝑔𝑇𝛼 (𝑓 ) − 𝑓 𝑇𝛼 (𝑔)
   5. 𝑇𝛼 ( ) =                        .
          𝑔               𝑔2
                                                                        𝑑𝑓
   6. If, in addition, 𝑓 is differentiable, then 𝑇𝛼 (𝑓 )(𝑡) = 𝑡 1−𝛼        (𝑡).
                                                                        𝑑𝑡
Note that the Property 6 of Theorem 1 is very important and it will often use in the sequel of
this study.


3. A linearized Cole-Hopf transformation
In this section, we introduce the Cole-Hopf transformation in order to linearize the modified
Burgers equations (??).
Using the Property 6 of Theorem 1, we can rewrite Eq.(??) as follows
                                     𝜕𝑢
                                     𝑡 (1−𝛼)
                                        + (𝜈 + 𝑢)𝑢𝑥 = 𝑟𝑢𝑥𝑥 .                                        (3.1)
                                     𝜕𝑡
The Cole-Hopf transformation is performed in two steps:

First step: Suppose that 𝑢 = 𝜓𝑥 thus Eq.(??) becomes:

                                  𝑡 (1−𝛼) 𝜓𝑥𝑡 + (𝜈 + 𝜓𝑥 )𝜓𝑥𝑥 = 𝑟 (𝜓𝑥𝑥𝑥 ) ,                          (3.2)
which can be rewritten as:
                                                𝜕 1
                              𝑡 (1−𝛼) 𝜓𝑥𝑡 +           (𝜈 + 𝜓𝑥 )2 ) = 𝑟 (𝜓𝑥𝑥𝑥 ) ,                    (3.3)
                                               𝜕𝑥 ( 2
we integrate Eq.(??) with respect to 𝑥, we obtain:
                                                1
                                  𝑡 (1−𝛼) 𝜓𝑡 + ( (𝜈 + 𝜓𝑥 )2 ) = 𝑟 (𝜓𝑥𝑥 ) + 𝜂(𝑡),                  (3.4)
                                                2
where 𝜂(𝑡) is arbitrary function depending of 𝑡.

Second step: Introducing now, the transformation 𝜓 = −2𝑟 ln 𝜙, we obtain

                                                                𝜙𝑥
                                                   𝑢 = −2𝑟         .                              (3.5)
                                                                𝜙

The derivatives of the function 𝜓 are

                                      𝜙𝑡                𝜙𝑥                    𝜙𝑥𝑥     𝜙2
                           𝜓𝑡 = −2𝑟      ,   𝜓𝑥 = −2𝑟      ,      𝜓𝑥𝑥 = −2𝑟       + 2𝑟 𝑥2 .       (3.6)
                                      𝜙                 𝜙                      𝜙      𝜙

Substituting the derivatives 𝜓𝑡 ,𝜓𝑥 and 𝜓𝑥𝑥 in Eq.(??) , we obtain

                             𝜙𝑡   1        𝜙𝑥          𝜙𝑥𝑥     𝜙2
                 𝑡 (1−𝛼) −2𝑟    +   (𝜈 − 2𝑟 )2 = 𝑟 −2𝑟     + 2𝑟 𝑥2 + 𝜂(𝑡).                        (3.7)
                        (    𝜙 ) (2        𝜙 )    (     𝜙      𝜙 )

Eq.(??) can be reduced to :

                                       𝜕𝛼 𝜙                 𝜈2
                                            = 𝑟𝜙𝑥𝑥 − 𝜈𝜙 𝑥 +    𝜙 + 𝜁 (𝑡)𝜙,                        (3.8)
                                       𝜕𝑡 𝛼                 4𝑟
                −𝜂(𝑡)
where 𝜁 (𝑡) =         .
                 2𝑟
Let’s give the following theorem which shows that the cancel of function 𝜁 (𝑡) in Eq.(??) has no
effect on the solution of Eq. (??). Let 𝜙(𝑥, 𝑡) be the solution of Eq.(??), 𝑢(𝑥, 𝑡) is defined in (??),
then the solutions 𝑢 is independent of 𝜁 (𝑡).
   Let
                                                    1
                                       𝛽(𝑡) = ∫ 1−𝛼 𝜁 (𝑡)𝑑𝑡,
                                                  𝑡
then,
                                                               1
                                                𝛽 ′ (𝑡) =         𝜁 (𝑡).
                                                            𝑡 1−𝛼
Multiply by 𝑒 −𝛽(𝑡) to both sides of Eq.(??), yields

                     𝜕 𝛼 𝜙 −𝛽(𝑡)                                𝜈 2 −𝛽(𝑡)
                          𝑒      = 𝑟𝜙𝑥𝑥 𝑒 −𝛽(𝑡)
                                                − 𝜈𝜙𝑥 𝑒 −𝛽(𝑡)
                                                              +    𝜙𝑒     + 𝜁 (𝑡)𝜙𝑒 −𝛽(𝑡) .       (3.9)
                     𝜕𝑡 𝛼                                       4𝑟
By using the Property 6 of Theorem 1, Eq.(??) becomes

                           𝜕𝜙 −𝛽(𝑡)                                               𝜈2
                   𝑡 1−𝛼      𝑒     − 𝜁 (𝑡)𝜙𝑒 −𝛽(𝑡) = 𝑟𝜙𝑥𝑥 𝑒 −𝛽(𝑡) − 𝜈𝜙𝑥 𝑒 −𝛽(𝑡) + 𝜙𝑒 −𝛽(𝑡) .    (3.10)
                           𝜕𝑡                                                     4𝑟
Then,
                               𝜕 −𝛽(𝑡)                                   𝜈2
                       𝑡 1−𝛼      (𝑒   𝜙 ) = 𝑟𝜙𝑥𝑥 𝑒 −𝛽(𝑡) − 𝜈𝜙𝑥 𝑒 −𝛽(𝑡) + 𝜙𝑒 −𝛽(𝑡) .              (3.11)
                               𝜕𝑡                                        4𝑟

Let 𝜓 (𝑥, 𝑡) = 𝑒 −𝛽(𝑡) 𝜙(𝑥, 𝑡), then 𝜓 (𝑥, 𝑡) satisfies the following linear parabolic equation
                                                  𝜕𝜓            𝜈2
                                          𝑡 1−𝛼      = 𝑟𝜓 − 𝜈𝜓 + 𝜓 .                              (3.12)
                                                  𝜕𝑡            4𝑟
According to Property 6, Eq. (??) becomes
                                            𝜕𝛼 𝜓            𝜈2
                                               𝛼
                                                 = 𝑟𝜓 − 𝜈𝜓 + 𝜓 .                                  (3.13)
                                            𝜕𝑡              4𝑟

We can see that the difference between the solution of Eq.(??) and Eq.(??) is the factor 𝑒 −𝛽(𝑡) .
Therefore, we have
                                                    𝜙𝑥 𝑒 −𝛽(𝑡) 𝜙𝑥 𝜓𝑥
                                        𝑢(𝑥, 𝑡) =     = −𝛽(𝑡) =      .                            (3.14)
                                                    𝜙   𝑒      𝜙  𝜓
It is clear that the solutions 𝑢(𝑥, 𝑡) and is independent of the function 𝜁 (𝑡). In order to simplify
the study, we can take 𝜁 (𝑡) = 0 in Eq.(??). Then it is written as
                                           𝜕𝛼 𝜙                 𝜈2
                                                = 𝑟𝜙𝑥𝑥 − 𝜈𝜙 𝑥 +    𝜙.                             (3.15)
                                           𝜕𝑡 𝛼                 4𝑟

Initial and boundary conditions for Eq. (??)
In the order to determinat the initial condition (IC) and boundary condition (BC), of the Eq.(??),
we use
                                           𝜙𝑥 𝑢(𝑥, 𝑡)
                                               =        .                                  (3.16)
                                           𝜙       −2𝑟
Integrating both sides of Eq.(??) with respect to 𝑥, we obtain

                                                         −1 𝑥
                                 𝜙(𝑥, 𝑡) = 𝜙(𝑡) exp            𝑢(𝑠, 𝑡)𝑑𝑠 ,                        (3.17)
                                                       ( 2𝑟 ∫0          )
where 𝜙(𝑡) is constant of integration, and at 𝑡 = 0 in Eq.(??), we obtain then the initial condition
                                                          −1 𝑥
                                 𝜙(𝑥, 0) = 𝜙(0) exp             𝑢(𝑠, 0)𝑑𝑠 .                       (3.18)
                                                        ( 2𝑟 ∫0          )

From Eq.(??), it is clear that 𝜙(0) has not effect on the final solution of System (??). So, we can
consider 𝜙(0) = 1, it yields
                                                 −1 𝑥
                                  𝜙0 (𝑥) = exp          𝑢 (𝑠)𝑑𝑠 .                             (3.19)
                                               ( 2𝑟 ∫0 0         )
Now, the transformed boundary condition (BC), can reduced to
                                        1
                               𝜙𝑥 = −      𝑢(𝑥, 𝑡)𝜙(𝑥, 𝑡), (𝑥, 𝑡) ∈ (𝜕Ω × (0, 𝑇 )).               (3.20)
                                        2𝑟
Therefore, the time-fractional parabolic equation with the initial and Neumann boundary con-
ditions is given by.
                    ⎧
                    ⎪              𝜕𝛼 𝜙                 𝜈2
                    ⎪
                    ⎪ 𝐸𝑞. ∶             = 𝑟𝜙𝑥𝑥 − 𝜈𝜙 𝑥 +    𝜙.
                    ⎪              𝜕𝑡 𝛼                 4𝑟
                    ⎪
                    ⎪
                    ⎪
                    ⎪                             −1 𝑥
                    ⎨ 𝐼𝐶 ∶      𝜙0 (𝑥) = exp            𝑢 (𝑠)𝑑𝑠 .                             (3.21)
                    ⎪
                    ⎪                           ( 2𝑟 ∫0 0      )
                    ⎪
                    ⎪
                    ⎪
                    ⎪
                    ⎪                     1
                    ⎩ 𝐵𝐶 ∶
                    ⎪            𝜙𝑥 = −
                                          2𝑟
                                             𝑢(𝑥, 𝑡)𝜙(𝑥, 𝑡), (𝑥, 𝑡) ∈ (𝜕Ω × (0, 𝑇 )).

Reformulating the Problem (??) by using the Property 6 of Theorem ??, it yields
                    ⎧
                    ⎪                        𝜕𝜙               𝜈2
                    ⎪
                    ⎪ 𝐸𝑞. ∶        𝑡 (1−𝛼)      = 𝑟𝜙𝑥𝑥 − 𝜈𝜙𝑥 + 𝜙.
                    ⎪                        𝜕𝑡               4𝑟
                    ⎪
                    ⎪
                    ⎪
                    ⎪                             −1 𝑥
                    ⎨ 𝐼𝐶 ∶      𝜙0 (𝑥) = exp            𝑢 (𝑠)𝑑𝑠 .                             (3.22)
                    ⎪
                    ⎪                           ( 2𝑟 ∫0 0      )
                    ⎪
                    ⎪
                    ⎪
                    ⎪
                    ⎪                     1
                    ⎩ 𝐵𝐶 ∶
                    ⎪            𝜙𝑥 = −
                                          2𝑟
                                             𝑢(𝑥, 𝑡)𝜙(𝑥, 𝑡), (𝑥, 𝑡) ∈ (𝜕Ω × (0, 𝑇 )).


4. Analytical solution of the (??) and (??)
We introduce the Fourier transform (F.T)
                                                    +∞
                               ̂ 𝑥 , 𝑡) = √1
                               𝜙(𝑘
                                         𝐹 .𝑇
                                                         𝜙(𝑥, 𝑡)𝑒 −𝑖𝑘𝑥 𝑥 𝑑𝑥,                   (4.1)
                                           2𝜋 ∫
                                                   −∞

and the inverse Fourier transformation (F.T−1 )
                                                    +∞
                                         1
                                      𝐹 .𝑇 −1
                                            ̂ 𝑥 , 𝑡)𝑒 𝑖𝑘𝑥 .𝑥 𝑑𝑘𝑥 .
                              𝜙(𝑥, 𝑡) = √ ∫ 𝜙(𝑘                                                (4.2)
                                         2𝜋
                                                  −∞

In first, we apply the F.T to the term 𝜙𝑥 ,
                                                        +∞
                                               1             𝜕𝜙 −𝑖𝑘𝑥 𝑥
                                 𝐹 .𝑇 (𝜙𝑥 ) = √ ∫               𝑒      𝑑𝑥,                     (4.3)
                                               2𝜋            𝜕𝑥
                                                    −∞

Integrating by part with respect to 𝑥, then we obtain
                                                                   +∞
                                  1        |+∞  1
                    𝐹 .𝑇 (𝜙𝑥 ) = √ 𝜙(𝑥, 𝑡)|| − √ (𝑖𝑘𝑥 ) ∫               𝜙(𝑥, 𝑡)𝑒 −𝑖𝑘𝑥 𝑥 𝑑𝑥,
                                  2𝜋       |−∞  2𝜋
                                                                  −∞
According to [? ], the boundary conditions for the heat equation on the infinite interval: 𝜙 = 0
as |𝑥| = ∞, so we get,
                                                        +∞
                                             1                                          ̂
                       𝐹 .𝑇 (𝜙𝑥 ) = −(𝑖𝑘𝑥 ) √ ∫              𝜙(𝑥, 𝑡)𝑒 −𝑖𝑘𝑥 𝑥 𝑑𝑥 = (𝑖𝑘𝑥 )𝜙.    (4.4)
                                             2𝜋
                                                      −∞

In same way and by integration twice by part with respect to 𝑥, we have
                                                                   ̂
                                                𝐹 .𝑇 (𝜙𝑥𝑥 ) = −𝑘𝑥2 𝜙.                         (4.5)

Substituting the above results into Eq.(??), we obtain

                                    𝜕 𝛼 𝜙̂         ̂    𝜈2 ̂         ̂
                                           + 𝜈(𝑖𝑘𝑥 𝜙) −    𝜙 = −𝑟𝑘𝑥2 𝜙.                       (4.6)
                                    𝜕𝑡 𝛼                4𝑟
Using the Property 6 of Theorem 1, the Eq.(??) becomes

                                           𝜕 𝜙̂              2
                                                        ̂ − 𝜈 𝜙̂ = −𝑟𝑘 2 𝜙.
                                                                         ̂
                                 𝑡 (1−𝛼)        + 𝜈(𝑖𝑘𝑥 𝜙)            𝑥                       (4.7)
                                           𝜕𝑡               4𝑟
Thus, the solution of Eq.(??) is given by
                                                           2      2 𝛼
                                    𝜙̂ = 𝐴(𝑘𝑥 ) 𝑒 (−𝑖𝑘𝑥 𝜈+𝜈 /4𝑟−𝑟𝑘𝑥 )𝑡 /𝛼 ,                   (4.8)
                                 +∞
                            1
where 𝐴(𝑘𝑥 ) = 𝜙̂0 (𝑘𝑥 ) = √ ∫        𝜙0 (𝑦)𝑒 −𝑖𝑘𝑥 𝑥 𝑑𝑦 is the integration constant.
                            2𝜋
                               −∞
Applying F.T−1 to Eq.(??), then we obtain
                               +∞
                        1                               2      2 𝛼
             𝜙(𝑥, 𝑡) = √ ∫ 𝑒 𝑖𝑘𝑥 .𝑥 𝜙̂0 (𝑘𝑥 )𝑒 −(𝑖𝑘𝑥 𝜈−𝜈 /4𝑟+𝑟𝑘𝑥 )𝑡 /𝛼 𝑑𝑘𝑥
                        2𝜋
                             −∞
                              +∞
                        1         ⎡ 1 +∞                       2      2 𝛼
                                                                                  ⎤
                     = √ ∫ 𝜙0 (𝑦) ⎢ √ ∫ 𝑒 −𝑖𝑘𝑥 .(𝑦−𝑥) −(𝑖𝑘𝑥 𝜈−𝜈 /4𝑟+𝑟𝑘𝑥 )𝑡 /𝛼 𝑑𝑘𝑥 ⎥ 𝑑𝑦.       (4.9)
                        2𝜋        ⎢ 2𝜋                                            ⎥
                           −∞     ⎣    −∞                                         ⎦
By using the program of Maple, the solution of Eq.(??) is given by
                                           +∞
                          1         −𝛼 2 (𝑥 − 𝑦)2 + 2𝛼𝑡 𝛼 𝜈(𝑥 − 𝑦)
              𝜙(𝑥, 𝑡) = √ 𝛼 ∫ exp                                   𝜙 (𝑦)𝑑𝑦.                 (4.10)
                       2 𝜋𝑟𝑡 /𝛼   [             4𝑟𝛼𝑡 𝛼             ] 0
                                      −∞




   After obtaining the linear time fractional parabolic equations ,the combining of the obtained
solution of parabolic equation and the inverse Cole-Hopf transformation will allow us a solu-
tion of the modified time-fractional Burgers equations.
To calculate the analytical solution of Eq.(??), we calculate first
                                            +∞
                             1                 −𝛼 2 (𝑥 − 𝑦)2 + 2𝛼𝑡 𝛼 𝜈(𝑥 − 𝑦)
              𝜙𝑥 (𝑥, 𝑡) = √            𝑐 exp                                   𝜙 (𝑦)𝑑𝑦,      (4.11)
                           2𝜋𝑟𝑡 𝛼 /𝛼 ∫       [             4𝑟𝛼𝑡 𝛼             ] 0
                                           −∞

            −2𝛼 2 (𝑥 − 𝑦) + 2𝛼𝜈𝑡 𝛼
where 𝑐 =                              .
                     4𝑟𝛼𝑡 𝛼


Once the functions 𝜙(𝑥, 𝑡) and 𝜙𝑥 (𝑥, 𝑡) are known and by using (??), therefore the solutions is


                           +∞      −𝛼 2 (𝑥 − 𝑦)2 + 2𝛼𝑡 𝛼 𝜈(𝑥 − 𝑦) 1 𝑦
                           ∫ 𝑐 ′ exp                              −   ∫ 𝑢0 (𝑠)𝑑𝑠 𝑑𝑦
                       −∞        [             4𝑟𝛼𝑡 𝛼               2𝑟 0         ]
              𝑢(𝑥, 𝑡) = +∞          2        2       𝛼                              ,        (4.12)
                                 −𝛼 (𝑥 − 𝑦) + 2𝛼𝑡 𝜈(𝑥 − 𝑦) 1 𝑦
                         ∫ exp                                  −    ∫ 𝑢0 (𝑠)𝑑𝑠 𝑑𝑦
                        −∞     [             4𝑟𝛼𝑡 𝛼               2𝑟 0         ]

where 𝑐 ′ = [𝛼(𝑥 − 𝑦)𝑡 −𝛼 − 𝜈].


5. Numerical schemes for (??)
We discretize the domain Ω by the finite difference method (FDM) into 𝑛𝑥, each of length
Δ𝑥 = (𝑏 − 𝑎)/𝑛𝑥 along the x-axis, and define the discrete mesh points (𝑥𝑖 , 𝑡𝑛 ) by (𝑎 + 𝑖Δ𝑥, 𝑛Δ𝑡),
where 𝑖 = 0, ..., 𝑛𝑥 and 𝑛 = 0, ..., 𝑇 .

5.1. An explicit scheme
By using a simple forward in time and centered in space discretization at point (𝑥𝑖 , 𝑡𝑛 ), the
explicit scheme for the Eq.(??) is

                          𝜙𝑖𝑛+1 − 𝜙𝑖𝑛     𝜙 𝑛 − 𝜙𝑖−1
                                                 𝑛    𝜈2          𝑛 − 2𝜙 𝑛 + 𝜙 𝑛
                                                                 𝜙𝑖+1
                𝑡𝑛(1−𝛼)               = −𝜈 𝑖+1       + 𝜙𝑖𝑛 + 𝑟          𝑖     𝑖−1
                                                                                    .
                              Δ𝑡             2Δ𝑥      4𝑟       (      Δ𝑥 2        )

So that, for every interior point (𝑥𝑖 , 𝑡𝑛 ), with 𝑖 = 1, ..., 𝑛𝑥 − 1, we obtain

                            𝜙𝑖𝑛+1 = (𝛼 + 𝛽)𝜙𝑖−1
                                            𝑛
                                                + (1 + 𝛾 − 2𝛽)𝜙𝑖𝑛 − (𝛼 − 𝛽)𝜙𝑖+1
                                                                            𝑛
                                                                                ,             (5.1)

where
                                   𝜈Δ𝑡                  𝑟Δ𝑡                 𝜈 2 Δ𝑡
                            𝛼=         (1−𝛼)
                                             , 𝛽=          (1−𝛼)
                                                                 and 𝛾 =                .
                                 Δ𝑥 𝑡𝑛               Δ𝑥 2 𝑡𝑛               4𝑟 𝑡𝑛(1−𝛼)


Now, let us consider the so-called BC described as
                                                      𝑛 − 𝜙𝑛
                                                     𝜙𝑖+1          1
                                   𝜙𝑥 (𝑥𝑖 , 𝑡𝑛 ) ≃          𝑖−1
                                                                = − 𝑢𝑖𝑛 𝜙𝑖𝑛 ,                 (5.2)
                                                        2Δ𝑥        2𝑟
which can be rewritten as:
                                                  Δ𝑥 𝑛 𝑛
                                           𝑛
                                          𝜙𝑖+1    𝑛
                                               = 𝜙𝑖−1𝑢 𝜙 ,
                                                      −                                          (5.3)
                                                   𝑟 𝑖 𝑖
For 𝑖 = 0 and 𝑖 = 𝑛𝑥, Eq.(??), respectively becomes
                                      Δ𝑥 𝑛 𝑛                                    Δ𝑥 𝑛 𝑛
                        𝜙1𝑛 = 𝜙−1
                               𝑛
                                  −      𝑢 𝜙      and          𝑛
                                                              𝜙𝑛𝑥+1    𝑛
                                                                    = 𝜙𝑛𝑥−1 −      𝑢 𝜙           (5.4)
                                       𝑟 0 0                                     𝑟 𝑛𝑥 𝑛𝑥
Substituting this constraint into Eq.(??) at the boundary points, we obtain respectively

                                                              Δ𝑥 𝑛
                         𝜙0𝑛+1 = 2𝛽𝜙1𝑛 + 1 + 𝛾 − 2𝛽 + (𝛼 + 𝛽)    𝑢 𝜙𝑛 ,
                                        (                      𝑟 0) 0
                                                                                                 (5.5)
                          𝑛+1      𝑛                           Δ𝑥 𝑛
                         𝜙𝑛𝑥  = 2𝛽𝜙𝑛𝑥−1 + 1 + 𝛾 − 2𝛽 + (𝛼 − 𝛽)    𝑢   𝜙𝑛 .
                                         (                      𝑟 𝑛𝑥 ) 𝑛𝑥

5.2. An implicit scheme
By using a simple forward in time and centered in space discretization at point (𝑥𝑖 , 𝑡𝑛 ), the
implicit scheme for Eq.(??) is

                      𝜙𝑖𝑛+1 − 𝜙𝑖𝑛     𝜙 𝑛+1 − 𝜙𝑖−1
                                               𝑛+1  𝜈2            𝑛+1 − 2𝜙 𝑛+1 + 𝜙 𝑛+1
                                                                 𝜙𝑖+1
            𝑡𝑛(1−𝛼)               = −𝜈 𝑖+1         + 𝜙𝑖𝑛+1 + 𝑟            𝑖       𝑖−1
                                                                                         .
                          Δ𝑡              2Δ𝑥       4𝑟         (         Δ𝑥 2          )

which can rewrite as
                                          𝑛+1
                                − (𝛼 + 𝛽)𝜙𝑖−1 + 𝛾 𝜙𝑖𝑛+1 + (𝛼 − 𝛽)𝜙𝑖+1
                                                                  𝑛+1
                                                                      = 𝜙𝑖𝑛                      (5.6)
where
                              𝜈Δ𝑡               𝑟Δ𝑡                       𝜈 2 Δ𝑡
                       𝛼=            ,   𝛽=               ,     𝛾=    1−         + 2𝛽 .
                            2Δ𝑥𝑡𝑛1−𝛼           Δ𝑥 2 𝑡𝑛1−𝛼            (   4𝑟𝑡𝑛1−𝛼     )




5.3. Calculating the required solution
The calculation of solution to the Eq. (??) can be obtained by the inverse Cole-Hopf transfor-
mation. Let 𝐷𝑥 𝜙𝑖𝑛 denote the derivative of 𝜙, at point (𝑥𝑖 , 𝑡𝑛 ) with respect to 𝑥. Then, 𝐷𝑥 𝜙𝑖𝑛 can
be calculated from the first order centered difference formula, for 𝑖 = 1, .., 𝑛𝑥 − 1
                                                         𝑛 − 𝜙𝑛
                                                    𝜕𝜙 𝜙𝑖+1
                                         𝐷𝑥 𝜙𝑖𝑛 =      ≃      𝑖−1
                                                                  ,                              (5.7)
                                                    𝜕𝑥    2Δ𝑥
Note that the derivatives: 𝐷𝑥 𝜙0𝑛 and 𝐷𝑥 𝜙𝑛𝑥
                                          𝑛
                                             at the end points are known.

Once the approximated values of 𝜙 and 𝜙𝑥 are known at any discrete point (𝑥𝑖 , 𝑡𝑛 ), then the ap-
proximated values of 𝑢 at discrete points can be calculated from the following discrete version
of Eq.(??), for 𝑖 = 1, .., 𝑛𝑥,
                                                 𝐷𝑥 𝜙 𝑛
                                        𝑢𝑖𝑛 = −2𝑟 𝑛 𝑖 .                                      (5.8)
                                                   𝜙𝑖
Figure 1: Numerical and analytical solution of modified Burger’s equation at 𝑇 = 0.5, 2, 4 for 𝛼 = 0.75.


6. Numerical experiment and discussion
In this section, we discuss a example to test the performance and accuracy of the method.
The numerical results arrived by this method are compared with analytic solution for various
values of 𝛼, and 𝑇 . To show the accuracy of the method, both the relative error 𝐿1 -norm and
𝐿∞ -norm respectively are given by
                                                       ||𝑢𝑎 − 𝑢𝑛 ||𝐿1
                                     ||𝐸𝑟𝑟𝑒𝑢𝑟𝑢||𝐿1 =                  ,                            (6.1)
                                                          ||𝑢𝑎 ||𝐿1
                                                       ||𝑢𝑎 − 𝑢𝑛 ||𝐿∞
                                    ||𝐸𝑟𝑟𝑒𝑢𝑟𝑢||𝐿∞ =                   ,                            (6.2)
                                                          ||𝑢𝑎 ||𝐿∞
where 𝑢𝑎 represents the analytical solution (??) and 𝑢𝑛 represents the computed solution (??)
for Eq.(??). We use the Matleb program to calculate the 𝑢𝑛 .

Considering modified Burgers equation Eq.(??), with the initial conditions:
                                 𝑢(𝑥, 𝑡) = sin(𝑥), 𝑥 ∈ [0, 2𝜋], 𝑡 > 0,                             (6.3)
and boundary conditions:
                                        𝑢(0, 𝑡) = 𝑢(2𝜋, 𝑡) = 0.                                    (6.4)
After computing, let’s give in Figure ?? and Figure ?? respectively the graphs of the numerical
solution and the exact solution. For simulation, we take the following data, 𝑟 = 0.1, 𝜈 =
0.1, 𝛼 = 0.25, 0.5, 0.75 and 0.9 respectively,

It can be see from Figure ??, at different time, there is no difference between the numerical and
the exact solution curve. In addition, as the time increases, the solution curve approach the 𝑥
-axis and the viscosity value becomes smaller. More, from each of the graphs in Figure ??, we
can be observe that as 𝛼 increases, the numerical solution curve are in good agreement with
the exact solution curve.




7. Conclusion
In this paper, we deal with a study of the modified Burgers equation with fractional conformal
derivatives with respect to time. The presence of both the fractional time derivative and the
Figure 2: Numerical and analytical solution of modified Burger’s equation at 𝑇 = 1 for different value
of 𝛼.


nonlinear term in this equation makes solving the problem more difficult. The idea is to use
the Cole-Hopf transform to reduce the modified Burgers equation of temporal fractional con-
formable derivative to a modified linear equation of temporal fractional conformable deriva-
tive. Then we can solve the latter using the Fourier transformation. Therefore, the solution
of the time-fractional conformable modified Burgers equation can be found using both the so-
lution of the parabolic equation and the inverse Cole-Ho transformation. For illustration, the
experimental simulations are given to show the interest of this approach.


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