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.
1. Introduction This work considers the following modified Burgers equations [ ? ]
+ ( + ) = , where and [0, ], > 0, When / mean conformable derivative of the function ( , ) = 0 ⎪ ⎧⎪ ( , 0) = 0( ), ∈ [0, ], ⎪⎩ ( , ) = ( , ), ∈ ([0, ]), > 0. ≤ 1 ∈ (1.2) 2. Preliminaries in the sequel. of order is defined by: We briefly recall a definition and some properties of fractional derivatives which can be used ([? ? ]) Given a function ∶ [0 ∶ ∞) ⟶ ℝ, then the conformable fractional derivative of The Cole-Hopf transformation is performed in two steps: First step: Suppose that = thus Eq.(??) becomes: which can be rewritten as: (1− )
+ ( + ) = . (2( + )2) = ( ) , Let’s give some properties which are summarized in the above theorem.
([? ? ]) Let 0 < ≤ 1 and f, g be -diferentiable at a point > 0. Then, 4. ( ) = ( ) + ( ).
5. ( ) = ( ) − ( ).
2 1. (
+ ) = ( ) + ( ), for all , ∈ ℝ. 2. ( ) = − for all ∈ ℝ 3. ( ) = 0, for all constant functions ( ) = . 6. If, in addition, is diferentiable, then ( )( ) = 1− ( ). this study.
Note that the Property 6 of Theorem 1 is very important and it will often use in the sequel of 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 ( )( ) = l→im0 ( + 1− ) − ( )
, ( )(0) = l→im0+ ( )( ). for all > 0, ∈ (0.1). If f is -diferentiable in some (0, ), > 0, and lim ( )( ) exists, then define we integrate Eq.(??) with respect to , we obtain: where ( ) is arbitrary function depending of .
Second step: Introducing now, the transformation = −2 ln , we obtain = −2 The derivatives of the function are Substituting the derivatives , and in Eq.(??) , we obtain
= −2 , = −2 , = −2 + 2 2 2 . (1− ) (−2
1 ) + ( 2 ( − 2 2
) ) = (−2 2 2 ) + 2 + ( ).
(3.4) (3.5) (3.6) (3.7) (3.8) (3.9) (3.10) then the solutions is independent of ( ).
Let’s give the following theorem which shows that the cancel of function ( ) in Eq.(??) has no efect on the solution of Eq. ( ??). Let (, ) be the solution of Eq.(??), (, ) is defined in (??), − ( ) 1−
Multiply by − ( ) to both sides of Eq.(??), yields By using the Property 6 of Theorem 1, Eq.(??) becomes = − ( ) − − ( ) + − ( ) + ( )
− ( ) − ( ) − ( ) = − ( ) − − ( ) +
− ( ). 2 4 Eq.(??) can be reduced to : where ( ) = − ( ) .
2
Let then,
4
+ ( ), 1 1 ( ) = ∫ 1− ( ), ′( ) =
1− ( ). 1− ( − ( ) )= − ( ) − − ( ) + 2 Let (, ) = − ( )
(, ), then (, ) satisfies the following linear parabolic equation According to Property 6, Eq. (??) becomes Therefore, we have We can see that the diference between the solution of Eq. (??) and Eq.(??) is the factor − ( ). the study, we can take ( ) = 0 in Eq.(??). Then it is written as It is clear that the solutions (, ) and is independent of the function ( ). In order to simplify 1− = − + 4
.
= − + 4
. = − ( ) − ( ) = 4 .
(3.11) (3.12) (3.13) (3.14) (3.15) (3.16) (3.17) (3.18) (3.19) (3.20) (, ) = ( ) exp (, 0) = (0) exp = (, ) −2 −1 ( 2 ∫
0 −1 ( 2 ∫ 0 . (, ) (, 0) ) ) , Integrating both sides of Eq.(??) with respect to , we obtain where ( ) is constant of integration, and at = 0 in Eq.(??), we obtain then the initial condition consider (0) = 1, it yields From Eq.(??), it is clear that (0) has not efect on the final solution of System ( ??). So, we can 0( ) = exp
−1 ( 2 ∫ 0 = −
(, ) (, ), (, ) ∈ ( Ω × (0, )). ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ . ∶ (1− ) Therefore, the time-fractional parabolic equation with the initial and Neumann boundary conditions is given by.
1 2 1 2 −1 =
, 4. Analytical solution of the ?? and ?? ( ) ( ) We introduce the Fourier transform (F.T) Integrating by part with respect to , then we obtain ̂( , ) . = √ (, ) −
, (, ) . ̂( , ) . .
(3.21) (3.22) (4.1) (4.2) (4.3) ⎪ . ∶
= − + ∶ 0( ) = exp
−1 ( 2 ∫
0 ∶ = −
(, ) (, ), (, ) ∈ ( Ω × (0, )).
Reformulating the Problem (??) by using the Property of Theorem ??, it yields and the inverse Fourier transformation (F.T−1 In first, we apply the F.T to the term . ( ) = √ −
, . ( ) = √ (, ) − √ (, ) −
,
1 1 2 √ 1 2 +∞ | | | |−∞ 1 Substituting the above results into Eq.(??), we obtain Using the Property 6 of Theorem 1, the Eq.(??) becomes Thus, the solution of Eq.(??) is given by
. ( ) = − 2.̂ ̂ + ( ̂) − 4 ̂ = − 2.̂ (1− ) ̂ + ( ̂) − ̂ = − 2.̂ 4 ̂ = ( ) (− + 2/4 − 2) /
, +∞ 1 2 ∫ 0( ) −∞ +∞ 1 2 ∫ −∞ −∞ 1
⎣ +∞ as | | = ∞, so we get, According to [? ], the boundary conditions for the heat equation on the infinite interval: = 0 +∞ −∞ In same way and by integration twice by part with respect to , we have By using the program of Maple, the solution of Eq.(??) is given by = √
+∞ 2 ∫ 0( ) ⎢⎡ √1 1
+∞ −∞ ⎢ 2 ∫ − .( − ) −( − 2/4 + 2) / ⎥ . ⎤ ⎥ ⎦ (, ) = √ 2 / ∫ exp [ −∞
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 solution of the modified time-fractional Burgers equations.
To calculate the analytical solution of Eq.(??), we calculate first (, ) = √ 2 1
+∞ / ∫ exp
[ −∞ Once the functions (, ) and (, ) are known and by using (??), therefore the solutions is where ′ = [ ( − ) − − ]. 5. Numerical schemes for (??) Δ = ( − )/ where = 0, ...,
and = 0, ..., .
We discretize the domain Ω by the finite diference method (FDM) into , each of length along the x-axis, and define the discrete mesh points ( , ) by ( + Δ, Δ ), 5.1. An explicit scheme explicit scheme for the Eq.(??) is By using a simple forward in time and centered in space discretization at point ( , ), the (1− ) +1 − = −
Δ
+1 − −1 +
2Δ 2 4 + ( +1 − 2 + −1
Substituting this constraint into Eq.(??) at the boundary points, we obtain respectively 5.2. An implicit scheme implicit scheme for Eq.(??) is By using a simple forward in time and centered in space discretization at point ( , ) , the (1− ) +1 −
Δ = − + . (
(
, Eq.(??), respectively becomes of Eq.(??), for = 1, .., which can rewrite as where 5.3. Calculating the required solution − ( + ) −1 + +1 + ( − ) +1 = +1 = , = ( 1 − be calculated from the first order centered diference formula, for The calculation of solution to the Eq. (??) can be obtained by the inverse Cole-Hopf transformation. Let denote the derivative of at point ( , ) with respect to . Then,
can = 1, ..,
− 1 , = ≃ +1 − −1 2Δ , Note that the derivatives: 0 and
at the end points are known.
Once the approximated values of and
are known at any discrete point ( , ), then the approximated values of at discrete points can be calculated from the following discrete version
= −2
.
(5.7) (5.8) 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 , and . To show the accuracy of the method, both the relative error 1-norm and ∞-norm respectively are given by || || 1 = || − || 1 || || 1
,
Considering modified Burgers equation Eq.( ??), with the initial conditions:
(, ) = sin( ), ∈ [0, 2 ], > 0, and boundary conditions:
(0, ) = (2 , ) = 0.
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, = 0.1, = 0.1, = 0.25, 0.5, 0.75 and 0.9 respectively, It can be see from Figure ??, at diferent time, there is no diference between the numerical and the exact solution curve. In addition, as the time increases, the solution curve approach the -axis and the viscosity value becomes smaller. More, from each of the graphs in Figure ??, we can be observe that as increases, the numerical solution curve are in good agreement with the exact solution curve. (6.1) (6.3) (6.4) 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 nonlinear term in this equation makes solving the problem more dificult. The idea is to use the Cole-Hopf transform to reduce the modified Burgers equation of temporal fractional conformable derivative to a modified linear equation of temporal fractional conformable derivative. 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 solution of the parabolic equation and the inverse Cole-Ho transformation. For illustration, the experimental simulations are given to show the interest of this approach.