New Multiple Solution to The Boussinesq Equation and The Burgers-Like Equation Hasan BULUT , M¨ unevver TUZ and Tolga AKTURK Department of Mathematic, Faculty of Science and Arts,University of Firat ˘ 23119/ELAZIG hbulut@firat.edu.tr,
[email protected],
[email protected]
Abstract In this paper by considering an improved tanh function method,we found some exact solutions of Boussinesq and Burgers -Like equation.The main idea of this method is to take full advantage of the Riccati equation which has more new solutions.We found some exact solutions of the Boussinesq equation and the Burgers- Like equation. 1.Introduction In recent years,nonlinear phenomena play a crucial role in applied mathematics and physics. Directly searching for exact solutions of nonlinear partial differential equations (PDEs) has become more and more attractive partly due to the availability of computer symbolic systems Like mathematica or maple that allow us to perform some complicated and tedious algebraic calculation on a computer as well as helping us to find exact solutions of PDEs [1 − 5] now. Many explicit exact methods have been introduced in literature [6 − 14] .Some of them are : Painleve method, homogeneous balance method, similarity reduction method, sinecosine method, Darboux transformation, Cole-Hopf transformation, generalized Miura transformation,tanh method, Backlund transformation and others methods [15 − 16] . One of the most effectively straightforward method constructing exact solution of PDEs is the extended tanh function method [17].Let us simply describe the tanh function. For doing this, one can consider in two variables general form of nonlinear PDE H (u, utt , ux , uxx , ...) = 0.
(1.1)
and transform Equation (1.1) with u (x, t) = u (ξ)
,
ξ = k (x − λt)
where, k and λ are the wave number and wave speed respectively.After the transformation,we get a nonlinear ODE for u (ξ) ( ) H ′ u′ , u′′ , u′′′ , ... = 0 (1.2) The fact that the solutions of many nonlinear equations can be expressed as a finite series of tanh functions motivates us to seek for the solutions of Eq. (1.2) in the form m
m
i=0
i=0
u (x, t) = u (ξ) = m ai tanhi (ξ) = m ai F i
1
(1.3)
where F i = tanhi (ξ) ,an equation for F (ξ) is obtained. m is a positive integer that can be determined by balancing the linear term of highest order with the nonlinear term in Eq (1.1) and k, λ, a1 , a2 , ..., am are parameters to be determined. Substituting solution (1.3) into Eq (1.2) yields a set of algebraic equations for F i , then all coefficients of F i have to vanish.From these relations k, λ, a1 , a2 , ..., am can be determined. In this work, we will consider to solve general Boussinesq equation and the Burgers -Like equation by using the improved tanh function method which is introduced by Chen and Zhang [19]. 2.Method and Its Applications The main idea of this method is to take full advantage of the Riccati equation that tanh function satisfies and uses its solutions F to replace tanh ξ .The required Riccati equation is given as F ′ = A + BF + CF 2 (2.1) dF and A, B, C are constant. In the following,Chen and Zhang [19] have given dξ several cases to get the solution of Eq (2.1) in the form of finite series of tanh functions (1.3) Case1. If C = 0, B ̸= 0 ,then (2.1) has the solutions
where F =
(exp (Bξ) − A) / B Case 2. If (
) i2 = −1
A = 1/2 B =0 , then (2.1) has the solutions cot hξ ± csc hξ ,tan hξ ± i sec hξ C = −1/2
Case 3. If
A = C = ±1/2 , then (2.1) has the solutions sec ξ ± tan ξ ,csc ξ ± cot ξ . B=0
Case 4. If
A=1 B = 0 , then (2.1) has the solutions tan hξ, cot hξ . C = −1
Case 5. If
A=C=1 , then (2.1) has the solutions tan ξ . B=0
Case 6. If
A = C = −1 , then (2.1) has the solutions cot ξ . B=0
Case 7. If
A=B=0 , then (2.1) has the solutions −1/ (cξ + c0 ). C ̸= 0
2
The solutions of Eq (1.1) can be expressed in the form m
u (x, t) = u (ξ) = m ai F i i=0
where ξ = kx−kωt, k and ω are the wave number and the wave speed respectively, and n is a positive integer that can be determined by balancing the linear term of highest order with the nonlinear term in Eq (1.1) ,and a0 , a1 , ...an are parametres to be determined. Introducing the similarity variable ξ = kx − kωt ,the travelling wave solutions u (ξ) satisfy ( ) H ′ u′ , u′′ , u′′′ , ... = 0. The balancing the highest order of linear of linear term with nonlinear term in Eq (1.2), we can determine n in Eq (1.3) We illustrate the method by considering the Boussinesq equation and Burgers-Like equtions 3. Example 1. The Boussinesq Equation utt = (c1 u + c2 un )xx + uxxtt = 0
(3.0)
If we accept that c1 ≥ 0, c2 ≤ 0 and n = 2 , c = −1 we conclude (3.1) equation by (3.0) equation utt − uxx + 2u.uxx − uxxtt = 0 (3.1) for doing this example. We could use transformation with Eq.(1.1) for the Boussinesq equation. Let us consider the equation Boussinesq u (x, t) = u (ξ)
,
ξ = kx − kwt
(3.2)
Substituting (3.2) into (3.1), we get ( )2 w2 u′′ − u′′ + 2 u′ + 2 u.u′′ − k 2 w2 u4 = 0
(3.3)
an integrating (3.3) we deduce following equation w2 u′ − u′ + 2 u.u′ − k 2 w2 u′′′ = 0.
(3.4)
When balancing (u.u′ ) with (u′′′ ) then gives m = 2. Therefore, we may choose u = a0 + a1 F + a2 F 2
(3.5)
Substituting (3.5) into (3.4) along with Eq(2.1) and using mathematica yields a system of equations w, t, F M . Setting the coefficients of F M in the obtained system of equations to zero, we can deduce the following set of algebraic polynomials with the respect unknowns a0 , a1 , a2 namely. u′ = a1 A + a1 BF + a1 CF 2 + 2a2 AF + 2a2 BF 2 + 2a2 CF 3 = 0 3
u.u′ = a0 a1 A + a0 a1 BF + a0 a1 CF 2 + 2a0 a2 AF + 2a0 a2 BF 2 + 2a0 a2 CF 3 + a21 AF + a21 BF 2 + a21 CF 3 + 2a1 a2 AF 2 + 2a1 a2 BF 3 + 2a1 a2 CF 4 + a1 a2 AF 2 + a1 a2 BF 3 + a1 a2 CF 4 + 2a22 AF 3 + 2a22 BF 4 + 2a22 CF 5 and u′′′ = a1 B 2 A + a1 B 3 F + 7a1 B 2 CF 2 + 8a1 ABCF + 12a1 BC 2 F 3 + 2a1 A2 C + 8a1 AC 2 F 2 + 6a1 C 3 F 4 + 6a2 A2 B + 14a2 AB 2 F + 52a2 ABCF 2 + 16a2 A2 CF + 40a2 AC 2 F 3 + 8a2 B 3 F 2 + 38a2 B 2 CF 3 + 54a2 BC 2 F 4 + 24a2 C 3 F 5 . and
F 0 : w2 a1 A − a1 A + 2a0 a1 A − k 2 w2 a1 B 2 A − 2k 2 w2 a1 A2 C − 6k 2 w2 a2 A2 B = 0
F 1 : w2 a1 B + 2w2 a2 A − a1 B − 2a2 B + 2a0 a1 B + +4a0 a2 A + 2a21 A − k 2 w2 a1 B 3 −8k 2 w2 a1 ABC − 14k 2 w2 a2 AB 2 − 16k 2 w2 a2 A2 C = 0
F2
w2 a1 C + 2w2 a2 B − a1 C − 2a2 B + 2a0 a1 C + 4a0 a2 B + 2a21 B +6a1 a2 A − 7k 2 w2 a1 B 2 C − 8k 2 w2 a1 AC 2 − 52k 2 w2 ABC −8k 2 w2 a2 B 3 = 0
F 3 : 2w2 a2 C − 2a2 C + 4a0 a2 C + 2a21 C + 4a1 a2 B + 2a1 a2 B +4a22 A − 12k 2 w2 a1 BC 2 − 40k 2 w2 a2 AC 2 − 38k 2 w2 B 2 C = 0 F 4 : −6a1 a2 C + 4a22 B − 6k 2 w2 a1 C 3 − 54k 2 w2 a2 BC 2 = 0 F 5 : 4a22 C − 24k 2 w2 a2 C 3 = 0 from the solutions of the system,we can find 1 1 a0 = 4k 2 w2 AC − w2 + 2 2 2 2 2 = 0 , a2 = 6k w C
B = 0 , a1
4
w=
1 4k 2 AC + 1
with the aid of Mathematica, we find (i) when we choose A = 1, B = 0, C = 1 in Eq(3.6) then we can deduce 1 1 a0 = 4k 2 w2 − w2 + , a2 = 6k 2 w2 . 2 2 Therefore, the solution can be found as ) ( [ ] 1 2 1 2 2 + 6k 2 w2 tan2 (kx − kwt) u (x, t) = 4k w − w + 2 2 (ii) In this case, ıf we take A = −1, B = 0, C = −1 in Eq(3.6) then we have 1 1 a0 = 4k 2 w2 − w2 + , a2 = 6k 2 w2 2 2 ( u (x, t) =
1 1 4k 2 w2 − w2 + 2 2
) + 6k 2 w2 cot [kx − kwt]
(iii) Again,when we choose A = 1, B = 0, C = −1 then from Eq(3.6) is 1 1 a0 = −4k 2 w2 − w2 + , a2 = 6k 2 w2 2 2
(
1 u (x, t) = −4k w − w2 + 2 ( 1 u (x, t) = −4k 2 w2 − w2 + 2 2
2
) 1 + 6k 2 w2 tanh2 (kx − kwt) 2 ) 1 + 6k 2 w2 coth2 (kx − kwt) 2
1 1 (iv) When we choose A = , B = 0, C = then we can find the coefficients of Eq(3.6) 2 2 as 1 1 3 a0 = k 2 w2 − w2 + , a2 = k 2 w2 2 2 2 and using the coefficients, the solutions can be found as ( ) 1 2 1 3 2 2 u (x, t) = k w − w + + k 2 w2 [sec (kx − kwt) + tan (kx − kwt)]2 2 2 2 ( ) 1 1 3 u (x, t) = k2 w2 − w2 + + k 2 w2 [csc (kx − kwt) + cot (kx − kwt)]2 2 2 2
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Figure 1: The solution (3.7) is shown at k = w = 1, C = 0.5, C0 = 0.1 and the second graph represents the exact analytical solution of Eq. (3.7) for t = 0.5. 1 1 (v) When we choose A = − , B = 0, C = − then we can find the coefficients of 2 2 Eq(3.6) 1 1 3 a0 = k 2 w2 − w2 + , a2 = k 2 w2 2 2 2 (
1 u (x, t) = k w − w2 + 2 ( 1 u (x, t) = k2 w2 − w2 + 2 2
2
(vi) When we choose A = Eq(3.6)
) 1 + 2 ) 1 + 2
3 2 2 k w [sec (kx − kwt) + tan (kx − kwt)]2 2 3 2 2 k w [csc (kx − kwt) + cot (kx − kwt)]2 2
1 1 , B = 0, C = − then we can find the coefficients of 2 2
1 1 3 a0 = −k 2 w2 − w2 + , a2 = k 2 w2 2 2 2 (
1 u (x, t) = −k w − w2 + 2 ( 1 u (x, t) = −k 2 w2 − w2 + 2 2
2
) 1 + 2 ) 1 + 2
3 2 2 k w [coth (kx − kwt) ± csc h (kx − kwt)]2 2 3 2 2 k w [tanh (kx − kwt) ± i sec h (kx − kwt)]2 2
(vii) When we choose A = 0, B = 0, C ̸= 0 then we can find the coefficients of Eq(3.6) 1 1 a0 = − w2 + , a2 = 6k 2 w2 C 2 2 2 ( u (x, t) =
1 1 − w2 + 2 2
)
( )2 3 2 2 2 1 + k w C − 2 (kx − kwt) + C0 6
(3.7)
Example 2. Burger- Like Equations ut + ux + u ux + λuxx = 0
(3.8)
where λ = 1, and in order to obtain Burger- Like solution of Equation we get u (x, t) = u (ξ)
,
ξ = k (x − wt)
(3.9)
Substituting (3.8) into (3.7) , we get −wu′ + u′ + u u′ + ku′′ = 0
(3.10)
Balancing (u u′ ) with u′′ gives m = 1. Therefore, we may choose the following ansatz: u = a0 + a1 F
(3.11)
Subsituting (3.11) into (3.10) along with Equation (2.1) and using mathematica yields the following set of algebraic polynomials with the respect unknowns a0 , a1 , a2 , a3 , namely: u′ = a1 A + a1 BF + a1 CF 2 + 2a2 AF + 2a2 BF 2 + 2a2 CF 3 = 0 ( ) u u′ = (a0 + a1 F ) a1 A + a1 BF + a1 CF 2 u′′′ = a1 BA + a1 B 2 F + a1 CAF (1) +3a1 BCF 2 + 2a1 C 2 F 3 and F 0 : w2 a1 A − a1 A + 2a0 a1 A − k 2 w2 a1 B 2 A − 2k 2 w2 a1 A2 C − 6k 2 w2 a2 A2 B = 0 F 1 : w2 a1 B + 2w2 a2 A − a1 B − 2a2 B + 2a0 a1 B + 4a0 a2 A + 2a21 A − k 2 w2 a1 B 3 − 8k 2 w2 a1 ABC − 14k 2 w2 a2 AB 2 − 16k 2 w2 a2 A2 C = 0 F 2 : w2 a1 C + 2w2 a2 B − a1 C − 2a2 B + 2a0 a1 C + 4a0 a2 B + 2a21 B + 6a1 a2 A − 7k 2 w2 a1 B 2 C − 8k 2 w2 a1 AC 2 − 52k 2 w2 a2 B 3 = 0
F 3 : 2w2 a2 C − 2a2 C + 4a0 a2 C + 2a21 C + 4a1 a2 B + 2a1 a2 B + 4a22 A − 12k 2 w2 a1 BC 2 − 40k 2 w2 a2 AC 2 − 38k 2 w2 B 2 C = 0 from the solutions of the system, we can find a0 = w − 1
,
a1 = −2kc
(3.14)
and we obtain the following multiple solution and triangular periodic solutions of Eq (3.8) i. A = C = 1, B = 0 in Eq (3.13) then a1 = −2k
, 7
a0 = w − 1
Therefore, the solution can be found as u (x, t) = (w − 1) − 2k tan [k (x − wt)] ii. In the case if, we A = C = −1, B = 0 in EQ (3.13) then we have a1 = 2k
a0 = w − 1
,
u (x, t) = (w − 1) + 2k cot [k (x − wt)] iii. Again, when we choose A = 1, B = 0, C = −1, a0 = w − 1
a1 = −2k
,
u (x, t) = (w − 1) + 2k tanh [k (x − wt)] u (x, t) = (w − 1) + 2k coth [k (x − wt)] 1 1 iv When we choose A = , B = 0, C = − , 2 2 a0 = w − 1
,
a1 = −k
u (x, t) = (w − 1) − k [sec [k (x − wt)] + tan [k (x − wt)]] u (x, t) = (w − 1) − k [cos ec [k (x − wt)] + cot [k (x − wt)]] 1 v. When we choose A = C = − , B = 0 then we can find the coefficients of Eq (3.13) 2 a0 = k
a0 = w − 1
,
u (x, t) = (w − 1) + k [sec [k (x − wt)] − tan [k (x − wt)]] u (x, t) = (w − 1) + k [cos ec [k (x − wt)] − cot [k (x − wt)]] vi. When we choose A = Eq (3.13)
1 1 , B = 0, C = − then we can find the coefficients of 2 2
u(x, t) = (w − 1) + k [coth [k (x − wt)] ± cos ech [k (x − wt)]] u(x, t) = (w − 1) + k [tanh [k (x − wt)] ± i sec h [k (x − wt)]]
4.Conclusion
8
(3.15)
Figure 2: The solution (3.15) is shown at k = w = 1 and the second graph represents the exact analytical solution of Eq. (3.15) for t = 0.5. We have presented a generalized tanh function method and used it to solve the Boussinesq Equation and the Burgers-Like equation. In fact this method is readily applicable to a large variety of nonlinear PDEs. Firstly, all the nonlinear PDEs which can be solved by other tanh function method can be solved easily by this method. Secondly we have used only the special solutions of Eq (1.3) .If we use only the special solutions of Eq (1.3),we can obtain more solutions. We are also aware of the fact that not all fundemental equations can be treated with the method. We also obtain some new and more general solutions at same time. Furthermore, this method is also computerizable,which allows us to perform complicated and tedious algebraic calculation on a computer.
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