Traveling wave solutions and conservation laws for nonlinear evolution equation Dumitru Baleanu, Mustafa Inc, Abdullahi Yusuf, and Aliyu Isa Aliyu
Citation: Journal of Mathematical Physics 59, 023506 (2018); View online: https://doi.org/10.1063/1.5022964 View Table of Contents: http://aip.scitation.org/toc/jmp/59/2 Published by the American Institute of Physics
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JOURNAL OF MATHEMATICAL PHYSICS 59, 023506 (2018)
Traveling wave solutions and conservation laws for nonlinear evolution equation Dumitru Baleanu,1,2,a) Mustafa Inc,3,b) Abdullahi Yusuf,3,4,c) and Aliyu Isa Aliyu3,4,d)
1 Department of Mathematics, Cankaya University, O˘ ¨ gretmenler Cad. 14, 06530 Ankara, Turkey 2 Institute
of Space Sciences, Magurele, Bucharest, Romania Faculty, Department of Mathematics, Firat University, 23119 Elazi˘g, Turkey 4 Science Faculty, Department of Mathematics, Federal University Dutse, 7156 Jigawa, Nigeria 3 Science
(Received 2 May 2017; accepted 18 January 2018; published online 15 February 2018)
In this work, the Riccati-Bernoulli sub-ordinary differential equation and modified tanh-coth methods are used to reach soliton solutions of the nonlinear evolution equation. We acquire new types of traveling wave solutions for the governing equation. We show that the equation is nonlinear self-adjoint by obtaining suitable substitution. Therefore, we construct conservation laws for the equation using new conservation theorem. The obtained solutions in this work may be used to explain and understand the physical nature of the wave spreads in the most dispersive medium. The constraint condition for the existence of solitons is stated. Some three dimensional figures for some of the acquired results are illustrated. Published by AIP Publishing. https://doi.org/10.1063/1.5022964
I. INTRODUCTION
It is well known that most of the phenomena that arise in mathematical physics and engineering fields can be described by nonlinear partial differential equations (NPDEs). In physics, the heat flow and the wave propagation phenomena are well described by partial differential equations (PDEs).1,2 In ecology, most population models are governed by NPDEs.3,4 The dispersion of a chemically reactive material is characterized by NPDEs. In addition, most physical phenomena of fluid dynamics, quantum mechanics, electricity, plasma physics, propagation of shallow water waves, and many other models are controlled within its domain of validity by NPDEs. NPDEs have become a useful tool for the analysis of these physical phenomena in science and engineering models. Several methods have been applied5–19 to reach traveling wave solutions for NPDEs. Moreover, as it is known to all, evolution equations which describe physical phenomena could be found in many fields of sciences. We restrict the pool of equations to classical NPDEs, such as the Korteweg-de Vries (KdV) equation, cylindrical KdV, and generalized KdV20–22 which are well known in modeling the propagation of weakly nonlinear waves in dispersive media. In other words, a well-known type of the KdV equation is known for a long time, the so-called modified KdV (mKdV) equation,23,24 describing nonlinear acoustic waves in anharmonic lattices.25,26 In the present work, we concentrate our intensions to a less studied unnamed type of the mKdV equation,27,28 which differs from the mKdV equation by a first local-derivative term whereby this term changes basically the equation’s property and is given by ut + uxxx − 6u2 ux + 6Sux = 0,
(1)
with u = u(x,t) and S as a nonvanishing parameter.
a)
[email protected] b)
[email protected] c)
[email protected] d)
[email protected]
0022-2488/2018/59(2)/023506/16/$30.00
59, 023506-1
Published by AIP Publishing.
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The function u(x,t) acts as the amplitude and is suitable for describing wave propagation depending upon time t in the sense of an evolution equation in which the steepening effect of the nonlinear term is counterbalanced by the (linear) dispersion term(s). Equation (1) was analyzed and investigated in Refs. 27–29. Furthermore, the Lie symmetry method plays a vital role in the study and finding solution for nonlinear partial differential equations (NPDEs).29 Different techniques are used in the literature for the construction of conservation laws (Cls) for different system of equations, and these Cls are important for the investigation of a physical system.30,31 One of the techniques used for obtaining Cls is the explicit approach used by Laplace.31 In 1915, Noether discovered the relationship between Cls and symmetries of differential equations that are established out of variational principle.32 However, Noether’s theorem cannot be used to the equations like evolution equations and to differential equations whose order is odd.33 Furthermore, these symmetries of Euler-Lagrange equations need to satisfy some additional axioms.33 Fortunately, a new technique based on the formal Lagrangian concept has been introduced for constructing Cls and the new conservation theorem was proved.31,35 The definition of nonlinear selfadjointness of differential equations was introduced.31,35 Summarising, the Ibragimov theorem states that for any system of differential equations that satisfy the conditions of nonlinear self-adjointness, it is possible to use the formal Lagrangian for the construction of local Cls of a system corresponding to the system symmetries. Here, we analyze and investigate symmetries, traveling wave solutions, nonlinear self-adjoint and Cls for Eq. (1) by using the Lie symmetry method, two analytic schemes which are Riccati-Bernoulli (RB) sub-ordinary differential equation (ODE)35,36 and modified tanh-coth (MTC) methods.37 The Cls are constructed using the new conservation theorem.31,35 The rest of the work is organized in the following direction: In Sec. II, we give the description of the two methods. In Sec. III, we present the applications of the two methods for Eq. (1). In Sec. IV, we give symmetries, nonlinear self-adjoint, and Cls for Eq. (1). In Sec. V, we present the physical interpretation of some of the obtained solutions. Some comparison is provided in Sec. VI. Finally, concluding remarks are given in Sec. VII.
II. DESCRIPTION OF THE METHODS A. RB sub-ODE method
Consider a nonlinear PDE as follows: P(q, qt , qx , qxx , qtx , . . .) = 0,
(2)
where P represents a polynomial function. We give the RB sub-ODE method in three steps below: Step 1. Convert x and t to one variable as follows: q(t, x) = q(ξ)
(3)
ξ = k(x + Vt),
(4)
and
where q(ξ) travels with speed V ; by using Eqs. (3) and (4), one can transform Eq. (2) to an ODE, P(q, q ′, q ′′, q ′′′, . . .) = 0.
(5)
Step 2. Suppose Eq. (5) is the solution of the RB equation, then q ′ = aq2−m + bq + cqm .
(6)
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In Eq. (6), a, b, c, and m are constants and will be found later. Taking the second and third derivatives of Eq. (6) yields q ′′ = ab(3 − m)q2−m + a2 (2 − m)q3−2m + mc2 q2m−1
(7)
2
m
+ bc(m + 1)q + (2ac + b )q, q ′′′ = ab(2 − m)(3 − m)q1−m + a2 (2 − m)(3 − 2m)q2−2m 2 2m−2
+ m(2m − 1)c q 2
+ bcm(m + 1)q
(8)
m−1
′
+ (2ac + b )q . Remark. In a situation where ac , 0 and m = 0, Eq. (6) becomes the Riccati equation. In a situation where a , 0, c = 0, and m , 1, Eq. (6) becomes the Bernoulli equation. Obviously, the Riccati equation and Bernoulli equation are special cases of Eq. (6). One can obtain the solutions for Eq. (6) in the following forms: Case 1. If m = 1, Eq. (6) has the solution q(ξ) = Ce(a+b+c)ξ .
(9)
Case 2. If m , 1, b = 0, and c = 0, Eq. (6) has the solution 1
q(ξ) = (a(m − 1)(ξ + C)) m−1 .
(10)
Case 3. If m , 1, b , 0, and c = 0, Eq. (6) has the solution 1 a q(ξ) = (− + Ceb(m−1)ξ ) m−1 . b
(11)
Case 4. If m , 1, a , 0, and b2 ☞ 4ac < 0, Eq. (6) has the solution b q(ξ) = *− + 2a ,
√
1 √ m−1 4ac − b2 (1 − m) 4ac − b2 tan( (ξ + C))+ 2a 2 -
(12)
and b q(ξ) = *− − , 2a
1 √ √ m−1 4ac − b2 (1 − m) 4ac − b2 + cot( (ξ + C)) . 2a 2 -
(13)
Case 5. If m , 1, a , 0, and b2 ☞ 4ac > 0, Eq. (6) has the solution b q(ξ) = *− − , 2a and q(ξ) = (−
b − 2a
√
√
1 √ m−1 b2 − 4ac (1 − m) b2 − 4ac + coth( (ξ + C)) 2a 2 -
√ 1 (1 − m) b2 − 4ac b2 − 4ac tanh( (ξ + C))) m−1 . 2a 2
(14)
(15)
Case 6. If m , 1, a , 0, and b2 ☞ 4ac = 0, Eq. (6) has the solution q(ξ) = (
1 b 1 − ) m−1 , a(m − 1)(ξ + C) 2a
(16)
where C is an arbitrary constant. Step 3. Inserting the derivatives of u in Eq. (5) gives an algebraic equation of u. Noticing the symmetry of the right-hand item of Eq. (6) and setting the highest power exponents of u to equivalence in Eq. (5), m can be found. Comparing the coefficients of ui gives a set of algebraic equations for a, b, c, and V. Solving the set of algebraic equations and putting m, a, b, c, V, and ξ = k(x + Vt) into Eqs. (9)–(16), we can get traveling wave solutions of Eq. (1).
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B. Backlund transformation of the RB equation ¨
If qn ☞1 (ξ) and (qn (ξ) = qn (qn ☞1 (ξ))) are the solutions of Eq. (6), we get dqn (ξ) dqn (ξ) dqn−1 (ξ) = dξ dqn−1 ξ dξ dqn (ξ) 2−m m = (aqn−1 + bqn−1 + cqn−1 ), dqn−1 ξ
(17)
namely, dqn−1 (ξ) dqn (ξ) = 2−m . m m + bqn + cqn aqn−1 + bqn−1 + cqn−1
aqn2−m
Taking the integral of Eq. (18) one time with respect to ξ and solving it, we get 1 ! 1−m −cA1 + aA2 (qn−1 (ξ))1−m qn (ξ) = ) , bA1 + aA2 + aA1 (qn−1 (ξ))1−m
(18)
(19)
where A1 and A2 are the arbitrary constants. Equation (19) is the B¨acklund transformation of Eq. (6). This means that we can generate for new infinite sequence of solutions for Eq. (6) by using Eq. (19). C. MTC method
Consider the following nonlinear PDE: qt = G(q, qx , qxx , . . .) = 0.
(20)
If we use ξ = µ(x ☞ ct) and q(ξ) = q(x,t) in Eq. (20), we get the following ODE: −µcq ′(ξ) = G(q(ξ), kq ′(ξ), q2 q ′′(ξ)),
(21)
where q(x,t) travels with speed c. The resulting ODE in Eq. (21) is solved by the MTC method, which admits the use of a finite series of functions of the form N X q(x, t) = q(ξ) = a0 + [aj Y j (ξ) + bj Y −j (ξ)] (22) j=1
and the Riccati equation Y ′ = A + BY + FY 2 ,
(23)
where A, B, and F are constants and will be found later. The parameter N in Eq. (22) is a non-negative constant that is found by balancing the linear term of the highest order with the nonlinear term in Eq. (21). Putting Eq. (22) in the ODE in Eq. (21) and making use of Eq. (23), we have an algebraic equation in powers of Y. This will give a system of algebraic equations with respect to parameters ai , bi , and c. With the use of Mathematica, we can determine ai , bi , µ, and c. The Riccati equation (23) has the following special solutions:37 • A = B = 1, F = 0, and Y (ξ) = eξ ☞ 1, • A = 21 , F = − 21 , B = 0, Y (ξ) = coth(ξ) ± csch(ξ), or Y (ξ) = tanh(ξ) ± isech(ξ), • A = F = ± 21 , B = 0, Y (ξ) = sec(ξ) ± tan(ξ), or Y (ξ) = csc(ξ) ± cot(ξ),
• A = 1, F = ☞1, B = 0, Y (ξ) = tanh(ξ), or Y (ξ) = tanh(ξ) ± coth(ξ),
• A = F = ±1, B = 0, and Y (ξ) = tan(ξ), • A = 1, F = ☞4, B = 0, and Y (ξ) = • A = 1, F = 4, B = 0, and Y (ξ) =
tanh(ξ)
1 + tanh (ξ)2 tan(ξ)
1 − tan (ξ)2
• A = ☞1, F = ☞4, B = 0, and Y (ξ) =
,
,
cot(ξ)
1 − cot (ξ)2
,
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• A = 1, F = 2, B = ±2, and Y (ξ) =
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tan(ξ) 1 ± tan (ξ)2
• A = ☞1, F = ☞2, B = ±2, and Y (ξ) =
,
cot(ξ)
. 1 ± cot (ξ)2 Different values of Y can be derived for different arbitrary values of A, B, and F. III. APPLICATION
In this section, we apply the two methods to obtain the traveling wave solutions for Eq. (1). A. RB sub-ODE method for Eq. (1)
The RB sub-ODE method is applied to obtain the traveling wave solutions for Eq. (1). Using Eqs. (3) and (4), we transform Eq. (1) to the following ordinary differential equation: kVu ′(ξ) − 6kV u2 (ξ)u ′(ξ) + 6SkVu ′(ξ) + u ′′′(ξ) = 0.
(24)
Substituting Eqs. (6) and (8) in Eq. (24), we get 12a2 bk 3 u(ξ)4 − 12a2 bk 3 mu(ξ)4 + 3a2 bk 3 m2 u(ξ)4 + 6a3 k 3 u(ξ)5−m − 7a3 k 3 mu(ξ)5−m + 2a3 k 3 m2 u(ξ)5−m + 3bc2 k 3 m2 u(ξ)4m + 7ab2 k 3 u(ξ)3+m + 8a2 ck 3 u(ξ)3+m − 5ab2 k 3 mu(ξ)3+m − 7a2 ck 3 mu(ξ)3+m + ab2 k 3 m2 u(ξ)3+m + 2a2 ck 3 m2 u(ξ)3+m + akVu(ξ)3+m + 6akSVu(ξ)3+m − 6akVu(ξ)5+m + b3 k 3 u(ξ)2+2m + 8abck 3 u(ξ)2+2m − 4abck 3 mu(ξ)2+2m + 2abck 3 m2 u(ξ)2+2m + bkVu(ξ)2+2m + 6bkSVu(ξ)2+2m − 6bkVu(ξ)4+2m + b2 ck 3 u(ξ)1+3m + 2ac2 k 3 u(ξ)1+3m + b2 ck 3 mu(ξ)1+3m − ac2 k 3 mu(ξ)1+3m + b2 ck 3 m2 u(ξ)1+3m + 2ac2 k 3 m2 u(ξ)1+3m + ckVu(ξ)1+3m + 6ckSVu(ξ)1+3m − 6ckVu(ξ)3+3m − c3 k 3 mu(ξ)−1+5m + 2c3 k 3 m2 u(ξ)−1+5m = 0,
(25)
and setting m = 0, we have b2 ck 3 u(ξ) + 2ac2 k 3 u(ξ) + ckVu(ξ) + 6ckSVu(ξ) + b3 k 3 u(ξ)2 + 8abck 3 u(ξ)2 + bkVu(ξ)2 + 6bkSVu(ξ)2 + 7ab2 k 3 u(ξ)3 + 8a2 ck 3 u(ξ)3 + akVu(ξ)3 − 6ckVu(ξ)3 + 6akSVu(ξ)3 + 12a2 bk 3 u(ξ)4 − 6bkVu(ξ)4 + 6a3 k 3 u(ξ)5 − 6akVu(ξ)5 = 0.
Setting each ui (i = 1,2,3,4,5) to zero, we acquire u(ξ) : ck b2 + 2ac k 2 + V + 6SV = 0, u(ξ)2 : bk b2 + 8ac k 2 + V + 6SV = 0, u(ξ)3 : a 7b2 + 8ac k 3 + k(a − 6c + 6aS)V = 0, u(ξ)4 : 6bk 2a2 k 2 − V = 0, u(ξ)5 : 6a3 k 3 − 6akV = 0.
Solving Eqs. (27)–(31) using Mathematica, we obtain ( − a − 6 a S) , V = a2 k 2 . • b = 0, ak , 0, c = 2 Case 1. If m , 1, a , 0, and b2 ☞ 4ac < 0, the solution of Eq. (1) is13 p p u1 (x, t) = 2a −2a(a + 6aS) tan(2 −2a(a + 6aS)(ξ + C))
(26)
(27) (28) (29) (30) (31)
(32)
and
p p u2 (x, t) = 2a −2a(a + 6aS) cot(2 −2a(a + 6aS)(ξ + C)) ,
which exist provided 2a(a + 6aS) < 0.
(33)
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Case 2. If m , 1, a , 0, and b2 ☞ 4ac > 0, the solution of Eq. (1) is13 p p u3 (x, t) = 2a 2a(a + 6aS) coth(2 2a(a + 6aS)(ξ + C))
(34)
p p u4 (x, t) = 2a 2a(a + 6aS) tanh(2 2a(a + 6aS)(ξ + C)) ,
(35)
and
which exist provided 2a(a + 6aS) > 0. Case 3. If m , 1, a , 0, and b2 ☞ 4ac = 0, the solution of Eq. (1) is u5 (x, t) = (−
1 ). a(ξ + C)
(36)
B. The MTC for Eq. (1)
Here, we obtain traveling wave solutions for Eq. (1) using the modified tanh-coth method. Using Eqs. (3) and (4) in Eq. (1), we get Eq. (24). Balancing the highest nonlinear term u2 u ′, and the corresponding highest derivative appearing u ′′′, gives N = 1 and thus U(ξ) = a0 + a1 Y (ξ) +
b1 . Y (ξ)
(37)
Substituting Eq. (37) into Eq. (24) and making use of Eq. (23), and since all coefficients of Y i have to perish, we found a system of algebraic equations in the unknown a0 , a1 , and b1 as follows: k ((A + B)a1 − Fb1 ) (1 + 2(A + B)F)k 2 + V + 6SV − 6V a02 + a1 b1 = 0, (38) − 6(A + B)3 k 3 b1 + 6(A + B)kVb13 = 0, − 6kb1 2(A + B)2 k 2 − V b1 (2(A + B)a0 + b1 ) = 0, kb1 (−(A + B) (7 + 8(A + B)F)k 2 + V + 6SV − 6Va02
(39)
(40) (41)
6FVb12 ) = 0,
+ 6V (2a0 + (A + B)a1 ) b1 + kb1 −(1 + 8(A + B)F)k 2 − (1 + 6S)V + 6V a02 + (2Fa0 + a1 ) b1 = 0,
(42)
2
(43)
ka1 (F(7 + 8(A + B)F)k 2 + F(1 + 6S)V − 6V + a1 ((A + B)a1 + Fb1 ))) = 0, 6ka1 2F 2 k 2 − V a1 (2Fa0 + a1 ) = 0, 6F 3 k 3 a1 − 6FkVa13 = 0.
(44)
ka1 ((1 + 8(A + B)F)k + V + 6SV − 6V a0 (a0 + 2(A + B)a1 ) − 6V a1 b1 ) = 0,
(45) (46)
Case 1. Setting A = B = 1 and F = 0 from the above algebraic system of equations and solving the results, we obtain r 1 + 6S k2 • k , 0, a0 = ± , a1 = 0, b1 = 4a0 , 1 + 6 S , 0, V = . 2 2 (1 + 6 S) Substituting these values and Y (ξ) = eξ ☞ 1 in Eqs. (38)–(46), we get13 r 1 + 6S 4 ). (47) u6 (x, t) = (1 + ξ 2 e −1 Case 2. Setting A = 21 , B = 0, and F = − 21 from the above algebraic system of equations and solving the results, we get q k2 • k , 0, a0 = ± −(1+6S) , a1 = ☞a0 , b1 = a0 , 1 + 6 S , 0, V = ☞ , 2 2 (1 + 6 S) √ k2 , and • k , 0, a0 = ± 1 + 6S, a1 = 0, b1 = a0 , 1 + 6 S , 0, V = (1 + 6 S) 2 √ k . • k , 0, a0 = ± 21 1 + 6S, a1 = ☞a0 , b1 = 0, 1 + 6 S , 0, V = (1 + 6 S)
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Substituting these values and Y (ξ) = coth(ξ) ± csch(ξ) or Y (ξ) = tanh(ξ) ± isech(ξ) in Eqs. (38)–(46), we get 13 r −(1 + 6S) 1 u7 (x, t) = (1 − (coth(ξ) ± csch(ξ)) + ) (48) 2 coth(ξ) ± csch(ξ) or r −(1 + 6S) 1 (1 − (tanh(ξ) ± isech(ξ)) + ( )) (49) u8 (x, t) = 2 tanh(ξ) ± isech(ξ) and u9 (x, t) =
1 1√ 1 + 6S(1 + ) 2 coth(ξ) ± csch(ξ)
(50)
u10 (x, t) =
1√ 1 1 + 6S(1 + ) 2 tanh(ξ) ± isech(ξ)
(51)
u11 (x, t) =
1√ 1 + 6S(1 − coth(ξ) ± csch(ξ)) 2
(52)
or
and
or 1√ 1 + 6S(1 − tanh(ξ) ± isech(ξ)). (53) 2 Case 3. Setting A = 1, B = 0, F = ☞1 and from the above algebraic system of equations and solving the results, we get r −(1 + 6S) 7k2 • k , 0, a0 = ± , a1 = ☞2a0 , b1 = ☞a1 , 1 + 6 S , 0, V = ☞ , 2 (1 + 6 S) r 14 5k2 (1 + 6S) , a1 = 0, b1 = 2a0 , 1 + 6 S , 0, V = , and • k , 0, a0 = ± 2 (1 + 6 S) r 10 −(1 + 6S) 5k2 • k , 0, a0 = ± , a1 = ☞2a0 , b1 = 0, 1 + 6 S , 0, V = . 10 2 (1 + 6 S) u12 (x, t) =
Substituting these values and Y (ξ) = tanh(ξ) or Y (ξ) = coth(ξ) in Eqs. (38)–(46), we get13 r −(1 + 6S) 2 (1 − 2 tanh(ξ) + ) u13 (x, t) = 14 tanh(ξ)
(54)
or u14 (x, t) =
r
−(1 + 6S) 2 (1 − 2 coth(ξ) + ) 14 coth(ξ)
(55)
and u15 (x, t) =
r
(1 + 6S) 2 (1 + ) 10 tanh(ξ)
(56)
u16 (x, t) =
r
(1 + 6S) 2 (1 + ) 10 coth(ξ)
(57)
(1 + 6S) (1 − 2 tanh(ξ)) 10
(58)
(1 + 6S) (1 − 2 coth(ξ)). 10
(59)
or
and u17 (x, t) =
r
u18 (x, t) =
r
or
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Case 4. Setting A = 1, B = 0, F = ☞4 and from the above algebraic system of equations and solving the results, we have r 2 −(1 + 6S) −a1 , a1 = ☞8a0 , b1 = ☞ , 1 + 6 S , 0, V = ☞ 2 (131k • k , 0, a0 = ± + 6 S) , 4 r 62 (1 + 6S) 17k2 , a1 = 0, b1 = 2a0 , 1 + 6 S , 0, V = − , and • k , 0, a0 = ± − 10 2 (1 + 6 S) r 17k2 (1 + 6S) • k , 0, a0 = ± , a1 = ☞8a0 , b1 = 0, 1 + 6 S , 0, V = . 34 2 (1 + 6 S) tanh(ξ) Substituting these values and Y (ξ) = 1+tanh in Eqs. (38)–(46), we acquire13 (ξ)2
u19 (x, t) =
r
−(1 + 6S) 8 tanh(ξ) 1 + tanh (ξ)2 (1 − ), − 32 62 tanh(ξ) 1 + tanh (ξ)2 r
(60)
−(1 + 6S) 2(1 + tanh (ξ)2 ) (1 + ), 62 tanh(ξ)
(61)
r −(1 + 6S) (1 + 6S) tanh(ξ) ). −8 ( 62 34 1 + tanh (ξ)2
(62)
u20 (x, t) = and u21 (x, t) =
r
Case 5. Setting A = ☞1, B = ☞2, F = ±2 and from the above algebraic system of equations and solving the results, we get r 23k2 −(1 + 6S) , a1 = 4a0 , b1 = 0, 1 + 6 S , 0, V = ☞ , • k , 0, a0 = ± 2 (1 + 6 S) r 46 (1 + 6S) 49k2 • k , 0, a0 = ± 17 , a1 = 4a0 , b1 = 6a0 , 1 + 6 S , 0, V = , and 2 2 (1 + 6 S) q 3a1 1 (1 + 6S) 49k2 , a = ☞ 4a , b = • k , 0, a0 = ± , 1 + 6 S , 0, V = . 1 0 1 2 7 4 2 (1 + 6 S) cot(ξ) Substituting these values and Y (ξ) = 1±cot in Eqs. (38)–(46), we obtain13 (ξ)2
u22 (x, t) =
1 u23 (x, t) = 7
r
r
−(1 + 6S) cot(ξ) (1 + 4 ), 46 1 ± cot (ξ)2
4 cot(ξ) 6(1 ± cot (ξ)2 ) (1 + 6S) (1 + ), + 2 2 cot(ξ) 1 ± cot (ξ)
(63)
(64)
and 1 u24 (x, t) = 7
r
(1 + 6S) 4 cot(ξ) 6 cot(ξ) (1 + − ). 2 2 1 ± cot (ξ) 1 ± cot (ξ)2
(65)
Case 6. Setting A = 1, B = 2, F = ±2 and from the above algebraic system of equations and solving the results, we obtain r 23k2 −(1 + 6S) , a1 = ☞4a0 , b1 = 0, 1 + 6 S , 0, V = ☞ , • k , 0, a0 = ± 2 (1 + 6 S) r 46 1 (1 + 6S) 49k2 • k , 0, a0 = ± , a1 = ☞4a0 , b1 = ☞6a0 , 1 + 6 S , 0, V = , and 7r 2 2 (1 + 6 S) 1 (1 + 6S) 3a1 49k2 • k , 0, a0 = ± , a1 = ☞4a0 , b1 = , 1 + 6 S , 0, V = . 7 2 4 2 (1 + 6 S)
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tan(ξ) in Eqs. (38)–(46), we have13 Substituting these values and Y (ξ) = 1±tan (ξ)2
u25 (x, t) =
r
−(1 + 6S) 4 tan(ξ) ), (1 − 46 1 ± tan (ξ)2
(66)
1 u26 (x, t) = 7
r
(1 + 6S) 4 tan(ξ) 6 tan(ξ) (1 − − ), 2 2 1 ± tan (ξ) 1 ± tan (ξ)2
(67)
1 u27 (x, t) = 7
r
6 tan(ξ) (1 + 6S) 4 tan(ξ) + ). (1 − 2 2 1 ± tan (ξ) 1 ± tan (ξ)2
(68)
and
IV. LIE SYMMETRY ANALYSIS FOR EQ. (1)
In this section, we compute the Lie point symmetry group for Eq. (1). Consider one parameter Lie symmetry point as follows: x˜ = x + εξ 1 (x, t, u) + O(ε 2 ), 2
(69)
2
(70)
2
(71)
˜t = t + εξ (x, t, u) + O(ε ), u˜ = u + εη(x, t, u) + O(ε ). Let the vector field of the infinitesimal generators of Eq. (1) be X = ξ1
∂ ∂ ∂ + ξ2 + η , ∂x ∂t ∂u
(72)
where ξ 1 , ξ 2 , η are the function of x, t, u. First, we compute the determining equations of Eq. (1) by the use of the SYM package introduced in Ref. 17; a very huge amount of a system of PDE are obtained. Solving for η(x,t,u), ξ 2 (x,t,u), and ξ 1 (x,t,u) from the obtained determining equations, we have that ξ 1 = c1 + xc2 ,
(73)
2
ξ = 3tc2 + c3 , η = −uc2 ,
(74) (75)
where c1 , c2 , and c3 are arbitrary constants. The Lie symmetry algebra admitted by Eq. (1) is spanned by following three infinitesimals generators: X1 = ∂x ,
(76)
X2 = ∂t ,
(77)
X3 = 3t∂t − u∂u + x∂x .
(78)
The corresponding commutator table of the infinitesimal generators is given as Table I.
TABLE I. Commutator table for Eq. (1). [Xi ,Xj ]
X1
X2
X3
X1 X2 X3
0 0 ☞X1
0 0 ☞3X2
X1 3X 2 0
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FIG. 1. Plot of solution (40) with a = ☞0.1, S = 0.8, k = 0.5.
A. Nonlinear self-adjoint
There are many methods for the construction of Cls for evolution equations in the literature. For instance, a more systematical direct approach for the construction of Cls without the existence of Lagrangian was proposed in Refs. 40 and 41. The techniques therein40,41 used pairs of symmetries and adjoint symmetries, whereby adjoint symmetries made up for the disadvantage of non-Lagrangian structures in depicting a correspondence between symmetries and Cls. Herein, we use the concept of nonlinear self-adjoint proposed in Ref. 31 to construct the Cls. The formal Lagrangian of Eq. (1) is computed according to Ref. 31 which is given by L = v(t, x)(ut − 6u2 ux + 6Su2 ux + uxxx ).
FIG. 2. Plot of solution (41) with a = ☞0.1, S = 0.8, k = 0.5.
(79)
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FIG. 3. Plot of solution (55) with a = 0.8, S = ☞8, k = 1.5.
The adjoint equation for Eq. (1) is given by F ∗ = −υt − 6(−1 + S)u(x, t)2 υx − υxxx = 0.
(80)
Next, we find a substitution for nonlinear self-adjoint. Let v(x, t) = φ(x, t, u).
(81)
Substituting Eq. (81) in Eq. (80) using the SYM package,29 we get ux3 φuuu + φt + 6Su(x, t)2 φx + 3uxx φxu + 3ux2 φxuu + 3ux (uxx φuu + φxxu ) + φxxx − 6u(x, t)2 φx = 0.
FIG. 4. Plot of solution (55) with a = 0.8, S = 8, k = 1.5.
(82)
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FIG. 5. Plot of solution (60) with a = 0.8, S = ☞8, k = 1.5.
Setting the coefficients of the derivatives of ux and uxx to zero, we get 3φuu = 0, φuuu = 0,
(83) (84)
3φxu = 0, 3φxuu = 0,
(85) (86)
3φxxu = 0,
(87) 2
2
φt − 6u(x, t) φx + 6Su(x, t) φx + φxxx = 0.
(88)
Solving Eqs. (83)–(88), we obtain that φ = uc1 + c2 .
FIG. 6. Plot of solution (61) with a = 0.8, S = ☞8, k = 1.5.
(89)
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FIG. 7. Plot of solution (64) with a = 0.8, S = 8, k = 1.5.
Hence Eq. (1) is a non-linear self-adjoint. This property gives us the liberty to construct Cls for Eq. (1). B. Cls for Eq. (1)
Since Eq. (1) is a non-linear self-adjoint, using its point symmetries, we can use the Noether operator N to obtain its conserved vectors, (C 1 ,C 2 ).30,31 The obtained vectors will satisfy the conservation equation Dx C 1 + Dt C 2 = 0. Therefore for this fact, the non local variables appearing in that formula must be substituted according to Eq. (89). Now, we make use of the obtained symmetries to obtain their conserved vectors. The following are the conserved vectors obtained from the symmetries X 1 , X 2 , and X 3 , respectively:
FIG. 8. Plot of solution (65) with a = 0.8, S = 8, k = 1.5.
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FIG. 9. Plot of solution (66) with a = 0.8, S = 8, k = 1.5.
C1t = (c2 + c1 u(x, t))ut , C1x = −(c2 + c1 u(x, t))ux , C2t = c1 ux uxt − ut 6(−1 + S)u(x, t)2 (c2 + c1 u(x, t)) + c1 uxx − (c2 + c1 u(x, t))uxxt , C2x = (c2 + c1 u(x, t)) 6(−1 + S)u(x, t)2 ux + uxxx ,
C3t = −6c1 (−1 + S)u(x, t)4 − 18c2 (−1 + S)tu(x, t)2 ut − 6(−1 + S)u(x, t)3 (c2 + 3c1 tut ) + c1 ux (2ux + 3tuxt ) + ut (c2 x − 3c1 tuxx ) + c1 u(x, t) × (xut − 4uxx − 3tuxxt ) − 3c2 (uxx + tuxxt ) ,
C3x = (c2 + c1 u(x, t)) (−xux + u(x, t) (−1 + 18(−1 + S)tu(x, t)ux ) + 3tuxxx ) .
FIG. 10. Plot of solution (70) with a = 0.8, S = ☞8, k = 1.5.
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V. GRAPHICAL AND PHYSICAL EXPLANATION OF THE OBTAINED RESULTS
In this section, we present some three dimensional figures for some of the obtained solutions presented in this work. The construction of the figures is carried out by taking suitable values of the parameters in order to see the mechanism of the original Eq. (1). From Figs. 1–10, one can see that the obtained solutions possess kink-type, bell-shape, singular solitons, and periodic wave solutions. VI. COMPARISON OF THE OBTAINED RESULTS
This section will provide some comparison of the obtained results with the earlier studies. We studied a class of mKdV equations given by Eq. (1) by means of two integration schemes: Lie symmetry analysis and Cls. We obtain traveling wave solutions which include trigonometric, hyperbolic, singular periodic, and algebraic solutions. In Ref. 38, the authors presented a solutions to the class of mKdv equations by using lax pairs; in Ref. 13, the general solution to the Riccati equation was given; in Ref. 39, the authors studied and analyzed a class of mKdv equations by means of a transformed rational function method where they obtained solutions, namely, exponential function solutions, hyperbolic function solutions, trigonometric function solutions, and rational solutions. However, in our paper, by using RB-sub ODE and MTC methods, we extracted new solutions to the class of mKdV equations. We also investigated the symmetries of the governing equation and computed its Cls via the nonlinear self-adjointness. VII. CONCLUDING REMARK
In this work, the RB sub-ODE and MTC methods are applied to carry out the traveling wave solutions for Eq. (1). We acquire lots of traveling wave solutions which includes singular, periodic wave, kink-type, bell-shaped, and algebraic solutions. The equation is shown to be a nonlinear selfadjoint with the obtained substitution in Eq. (89); therefore, we constructed Cls for Eq. (1) using the new Cl theorem presented by Ibragimov. The obtained solutions in this paper could be used for the analysis and interpretation of physical phenomena and other solitary waves. Finally, our obtained results in this work have been verified by the use of Mathematica 9 by putting them back into the original equation. 1 D.
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