Extended Trial Equation Method for Nonlinear Partial Differential Equations

1 Introduction

The effort in finding exact solutions to nonlinear differential equations is important for the understanding of most nonlinear physical phenomena. For instance, the nonlinear wave phenomena observed in fluid dynamics, plasma, and optical fibres are often modelled by bell-shaped sech solutions and kink-shaped tanh solutions. In recent years, the exact solutions of nonlinear partial differential equations (PDEs) have been investigated by many authors (see for example [1–37]) who are interested in nonlinear physical phenomena. Many powerful methods have been presented by these authors, such as inverse scattering transform [1], Backlund transform [2], Darboux transform [3], the generalised Riccati equation [4, 5], the Jacobi elliptic function expansion method [6, 7], the Painlevé expansion method [8], the extended tanh-function method [9, 10], the F-expansion method [11, 12], the exp-function expansion method [13, 14], the sub-ODE method [15, 16], the extended sinh–cosh and sine–cosine methods [17, 18], the (G′/G)-expansion method [19, 20], and so on. Also there are many methods for finding the analytic approximate solutions for nonlinear PDEs such as the homotopy perturbation method [21, 22], the Adomain decomposition method [23, 24], and the variation iteration and homotopy analysis method [25]. There are many other methods for solving the nonlinear PDEs (see, e.g., [26–37]). Recently, Gurefe et al. [38] have presented a direct method namely the extended trial equation method for solving the nonlinear PDEs. The main objective of this paper is to modify the extended trial equation method to construct a series of some new solutions for some nonlinear PDEs in mathematical physics via the Zhiber–Shabat nonlinear differential equations. In this present paper, we will construct the solutions in many different types of the roots of the trial equation. We will obtain many different kinds of solutions in hyperbolic function solutions, trigonometric function solutions, Jacobi elliptic functions solutions, and rational solutions. In this paper, we find that the balance number is not constant and changes by changing the trial equation derivative of the nonlinear PDEs.

2 Description of the Extended Trial Equation Method

Suppose that we have a nonlinear PDE in the following form:

(1)F(u, ut, ux, utt, uxt, uxx, …..) = 0, (1)

where u=u(x,t) is an unknown function and F is a polynomial in u=u(x,t) and its partial derivatives, in which the highest order derivatives and nonlinear terms are involved. Let us now give the main steps for solving (1) using the extended trial equation method as follows [38, 39]:

Step 1. The travelling wave variable

(2)u(x,t) = u(ξ), ξ = x − ωt, (2)

where ω is a nonzero constant. The transformation (2) permits us to reduce (1) to an ODE for u=u(ξ) in the following form:

(3)P(u, −ωu′, u′, ω2u″, −ωu″, u″, ……) = 0, (3)

where P is a polynomial of u=u(ξ) and its total derivatives.

Step 2. Suppose the trial equation takes the form

(4)u(ξ) = ∑i = 0δτiYi, (4)

where Y satisfies the following nonlinear trial differential equation:

(5)(Y′)2 = Λ(Y) = Φ(Y)Ψ(Y) = ξθYθ + ξθ − 1Yθ − 1 + … + ξ1Y + ξ0ζεYε + ζε − 1Yε − 1 + … + ζ1Y + ζ0 (5)

where ξ_i, ζ_j are constants to be determined later. Using (4) and (5), we have

(6)u″(ξ) = Φ′(Y)Ψ(Y) − Φ(Y)Ψ′(Y)2Ψ2(Y)(∑i = 0δi τiYi − 1)  + Φ(Y)Ψ(Y)(∑i = 0δi (i − 1)τiYi − 2). (6)

where Φ(Y), Ψ(Y) are polynomials in Y.

Step 3. Balancing the highest derivative term with the nonlinear term we can find the relations between δ, θ, and ε. We can calculate some values of δ, θ, and ε.

Step 4. Substituting (4)–(6) into (3) and cleaning the denominator yields a polynomial ω(y) of Y as follows:

(7)Ω(y) = ρsYs + … + ρ1Y + ρ0 = 0. (7)

Step 5. Setting the coefficients of the polynomial ω(y) to be zero yields a set of algebraic equations,

(8)ρi = 0, i = 0, …, s. (8)

Solving this system of algebraic equations to determine the values of ξ_θ, ξ_{θ – 1}, …, ξ₁, ξ₀, ζ_ε, ζ_ε, …, ζ₁, ζ₀, and τ_δ, τ_{δ – 1}, …, τ₁, τ₀.

Step 6. Reduce (5) to the elementary integral form

(9)±(ξ − η0) = ∫dYΛ(y) = ∫Ψ(Y)Φ(Y)dY. (9)

where η₀ is an arbitrary constant. Using a complete discrimination system for the polynomial to classify the roots of Φ(Y), we solve (9) with the help of a software package such as Maple or Mathematica and classify the solutions to (3). In addition, we can write the travelling wave solutions to (1).

3 Extended Trial Equation Method for Nonlinear Zhiber–Shabat Nonlinear Differential Equations

We start with the following nonlinear Zhiber–Shabat differential equation:

(10)utx + peu + qe−u + re−2u = 0 (10)

where p, q, and r are nonzero constants. For q=r=0, (10) reduces to the well-known Liouville equation. For q=0, r ≠ 0, (10) gives the well-known Dodd–Bullough–Mikhailov equation (DBM), while for q ≠ 0, r=0, (10) reduces to the Sinh–Gordon equation. Moreover, for p=0, q=–1, r=–1, we obtain the Tzitzeica–Dodd–Bullough equation. These equations play a significant role in many scientific applications such as solid-state physics, nonlinear optics, plasma physics, fluid dynamics, mathematical biology, dislocations in crystals, kink dynamics and chemical kinetics, and quantum field theory [40–42]. The travelling wave variable (2) permits us to convert (10) into the following ODE:

(11)−ωu″ + peu + qe−u + re−2u = 0. (11)

If we use the transformation

(12)v = eu (12)

The transformation (12) leads us to write (11) in the following form:

(13)−ω(vv″ − v′2) + pv3 + qv + r = 0. (13)

From (4) to (9), we can write the following highest order nonlinear terms in order to determine the balance procedure:

(14)v(ξ) = τδYδ + τδ − 1Yδ − 1 + … , (14)

(15)v3(ξ) = τδ3Y3δ + … , (15)

and

(16)(v′)2 = ξθδ2τδ2ζεY2δ − 2(Y′)2 + ⋯ = ξθδ2τδ2ζεY2δ + θ − ε − 2 + ⋯vv″ = W Y2δ + θ − ε − 2 + ⋯. (16)

From (14) to (16) we get the relation between δ, θ, and ε as follows:

(17)θ = ε + δ + 2. (17)

Equation (17) has infinite solutions; consequently, we suppose some of these solutions as follows:

Case 1. In the special case if ε=0 and δ=1, we get θ=3. Equations (4)–(9) lead to

(18)v = τ0 + τ1Y,(v′)2 = τ12(ξ3Y3 + ξ2Y2 + ξ1Y + ξ0)ζ0,vv″ = τ1(3ξ3Y2 + 2ξ2Y + ξ1)2ζ0. (18)

Substituting (18) into (13), we get a system of algebraic equations which can be solved to obtain the following results:

(19)ξ0 = −1ωτ12(2qτ0ζ0 − ωτ02ξ2 + 4pτ03ζ0 + rζ0), ξ3 = 2pζ0τ1ω,ξ1 = −2ωτ1(qζ0 − ωτ0ξ2 + 3p τ02ζ0), (19)

where ζ₀, ξ₂, ω, τ₁, and τ₀ are arbitrary constants. Substituting these results into (5) and (9), we have

(20)±(η − η0) = dYξ3ζ0Y3 + ξ2ζ0Y2 + ξ1ζ0Y + ξoζ0. (20)

Now, we will discuss the roots of the following equation:

(21)2pτ1ωY3 + ξ2ζ0Y2 − 2ωζ0τ1(qζ0 − ωτ0ξ2 + 3p τ02ζ0)Y−1ωζ0τ12(2qτ0ζ0 − ωτ02ξ2 + 4pτ03ζ0 + rζ0) = 0. (21)

To integrate (20) we must discuss the different cases of the roots of (21) as the following families:

Family 1. If (21) has three equal repeated roots α₁, consequently we can write (21) in the following form:

(22)2pτ1ωY3 + ξ2ζ0Y2 − 2ωζ0τ1(qζ0 − ωτ0ξ2 + 3p τ02ζ0)Y −1ωζ0τ12(2qτ0ζ0 − ωτ02ξ2 + 4pτ03ζ0 + rζ0) − (Y − α1)3 = 0. (22)

Equating the coefficients of Y on both sides of (22), we get a system of algebraic equations in ζ₀, ξ₂, ω, τ₁, and τ₀ which can be solved by using the Maple software package to get the following results:

(23)r = ∓2q9p−3qp, ξ2 = −3α1ζ0, τ0 = 12p(−α1ω ± 23−3qp), τ1 = ω2p. (23)

Equations (23), (19), and (20) lead to

(24)​ξ0 = −α13ζ0, ξ1 = 3α12ζ0, ξ3 = ζ0, (24)

where ζ₀ is an arbitrary constant and

(25)±(ξ − η0) = dY(Y − α1)3/2 = −2Y − α1. (25)

or

(26)Y = α1 + 4(x − ωt − η0)2. (26)

Substituting solutions (23), (24), and (26) into (18) we get the travelling wave solution of (13) as follows:

(27)v(x,t) = ±13p−3qp + 4ω2p(x − ωt − η0)2. (27)

Hence the solution of the nonlinear Zhiber–Shabat differential equation (10) takes the following form:

(28)u(x,t) = ln(|±13p−3qp + 4ω2p(x − ωt − η0)2|). (28)

Family 2. If (21) has two equal repeated roots α₁ and the third root is α₂, α₁ ≠ α₂, consequently we can write (21) in the following form:

(29)2pτ1ωY3 + ξ2ζ0Y2 − 2ωζ0τ1(qζ0 − ωτ0ξ2 + 3p τ02ζ0)Y −1ωζ0τ12(2qτ0ζ0 − ωτ02ξ2 + 4pτ03ζ0 + rζ0)  − (Y − α1)2(Y − α2) = 0. (29)

Equating the coefficients of Y on both sides of (29), we get a system of algebraic equations in ζ₀, ξ₂, ω, τ₁, and τ₀ which can be solved by using the Maple software package to get the following results:

(30)r = −118p{D[12pq + 2ω2(α1 − α2)2] + 2pqωα2    + 4pqωα1 + ω3α1(α1 − α2)2},ξ2 = −2α1ζ0 − α2ζ0, τ0 = Dp, τ1 = ω2p, (30)

where D = −α2ω6 − α1ω3 ± 16ω2(α1 − α2)2 − 12pq. Equations (30), (19), and (20) lead to

(31)​ξ0 = −α12α2ζ0, ξ1 = α1(α1 + 2α2)ζ0, ξ3 = ζ0, (31)

where ζ₀ is an arbitrary constant, and if α₂>α₁, we have

(32)±(ξ − η0) = dY(Y − α1)Y − α2 = 2α2 − α1tan−1[Y − α2α2 − α1], α2 > α1 (32)

or

(33)Y = α2 + (α2 − α1)tan2[α2 − α12(ξ − η0)], α2 > α1. (33)

Substituting (33), (31), and (30) into (18), we get the solution of (13) as follows:

(34)v(x,t) = Dp + ω2p{α2 + (α2 − ​α1)tan2[α2 − α12(x − 2kt − η0)]}, (34)

Hence, the solution of the nonlinear Zhiber–Shabat differential equation (10) takes the following form:

(35)u(x,t) = ln(|Dp + ω2p{α2 + (α2 − α1)tan2[α2 − α12(x − 2kt − η0)]}|). (35)

Also when α₁>α₂, we have

(36)Y = α1 + (α1 − α2)csch2[α1 − α22(ξ − η0)], α1 > α2. (36)

Substituting (36), (31), and (30) into (18), we get the solution of (13) as follow:

(37)v(x,t) = Dp + ω2p{α1 + (α1 − ​α2)csch2[α1 − α22(ξ − η0)]}. (37)

Hence, the solution of the nonlinear Zhiber–Shabat differential equation (10) takes the following form:

(38)u(x,t) = ln(|Dp + ω2p{α1 + (α1 − α2)csch2[α1 − α22(ξ − η0)]}|). (38)

Family 3. If (21) has three different roots α₁, α₂, and α₃, consequently, we can write (21) in the following form:

(39)2pτ1ωY3 + ξ2ζ0Y2 − 2ωζ0τ1(qζ0 − ωτ0ξ2 + 3p τ02ζ0)Y −1ωζ0τ12(2qτ0ζ0 − ωτ02ξ2 + 4pτ03ζ0 + rζ0) − (Y − α1) (Y − α2)(Y − α3) = 0. (39)

From equating the coefficients of Y on both sides of (39), we get a system of algebraic equations in ζ₀, ξ₂, ω, τ₁, and τ₀ which can be solved by using the Maple software package to get the following results:

(40)r = −136p2{D[24pq − 4ω2(α1α3 + α2α3 + α2α1 − α12 − α22  − α32)] + 4pqω(α1 + α2 + α3) + ω3(α12α2 + α22α1  + α22α3 + α32α2 + α12α3 + α32α1 − 6α1α2α3)}, ξ2 = −ζ0(α1 + α2 + α3), τ0 = Dp, τ1 = ω2p, (40)

where D = −ω6(α1 + α2 + α3)  ± 16ω2(α12 + α22 + α32 − α1α3 − α2α3 − α2α1) − 12pq.

Equations (40), (19), and (20) lead to

(41)​ξ0 = −α1α3α2ζ0, ξ1 = (α1α3 + α2α3 + α1α2)ζ0, ξ3 = ζ0, (41)

where ζ₀ is an arbitrary constant, and if α₃>α₂>α₁, we have

(42)±(ξ − η0) = dY(Y − α1)(Y − α2)(Y − α3)  = −2α3 − α1EllipticF[Y − α1α2 − α1, α1 − α2α1 − α3], (42)

or

(43)Y = α1 + (α2 − ​α1)sn2[α3 − α12(ξ − η0), α1 − α2α1 − α3]. (43)

Substituting (43), (40), and (41) into (18), we get the exact solution of (13) as follows:

(44)v(x,t) = Dp + ω2p {α1 + (α2 − ​α1)sn2[α3 − α12(ξ − η0), α1 − α2α1 − α3]}. (44)

Hence, the solution of the nonlinear Zhiber–Shabat differential equation (10) takes the following form:

(45)u(x,t) = ln(|Dp + ω2p{α1 + (α2 − ​α1)sn2[α3 − α12(ξ − η0), α1 − α2α1 − α3]}|). (45)

Family 4. If (21) has one real root α₁ and two imaginary roots α₂=N₁ + iN₂, α₃=N₁ – iN₂ (N₁, N₂ are real numbers), consequently, we can write (21) in the following form:

(46)2pτ1ωY3 + ξ2ζ0Y2 − 2ωζ0τ1(qζ0 − ωτ0ξ2 + 3p τ02ζ0)Y  − 1ωζ0τ12(2qτ0ζ0 − ωτ02ξ2 + 4pτ03ζ0 + rζ0) − (Y − α1) (Y2 − 2N1Y + N12 + N22) = 0. (46)

From equating the coefficients of Y on both sides of (46), we get a system of algebraic equations in ζ₀, ξ₂, ω, τ₁, and τ₀ which can be solved by using the Maple software package to get the following results:

(47)r = −118p2{D[12pq − ω2(4N1α1 + 6N22 − 2N12 − 2α12]  + 2pqω(2N1 + α1) + ω3(N13 + N22N2 + α12N1  − 4N22α1 − 2N12α1)},ξ2 = −ζ0(2N1 + α1), τ0 = Dp, τ1 = ω2p, (47)

where D = −ω6(2N1 + α1) ± 16ω2(N12 + α12 − 3N22 − 2α1N1) − 12pq.

Equations (47), (19), and (20) lead to

(48) ξ0 = −α1ζ0(N12 + N22), ξ1 = (N12 + N22 + 2α1N1)ζ0, ξ3 = ζ0, (48)

where ζ₀ is an arbitrary constant. With the help of the Maple software package, the integration of (20) in this family takes the following form:

(49)±(ξ − η0) = dY(Y − α1)(Y2 − 2N1Y + N12 + N22) = 2N1 + iN2 − α1EllipticF[Y − α1N1 − iN2 − α1, N1 − iN2 − α1N1 + iN2 − α1], (49)

or

(50)Y = α1 + (N1 − iN2 − ​α1)sn2[N1 + iN2 − ​α12(ξ − η0), N1 − iN2 − α1N1 + iN2 − α1]. (50)

Substituting (50), (48), and (47) into (18), we get the solution of (13) as follows:

(51)v(x,t) = Dp + ω2p{α1 + (N1 − iN2 − ​α1)sn2[N1 + iN2 − ​α12(ξ − η0), N1 − iN2 − α1N1 + iN2 − α1]}. (51)

Hence, the solution of the nonlinear Zhiber–Shabat differential equation (10) takes the following form:

(52)u(x,t) = ln(|Dp + ω2p{α1 + (N1 − iN2 − ​α1)sn2[N1 + iN2 − ​α12(ξ − η0), N1 − iN2 − α1N1 + iN2 − α1]}|). (52)

Case 2. In the special case if ε=0 and θ=4, we get δ=2. Equations (4)–(9) lead to

(53)v(ξ) = τ0+τ1Y+τ2Y2,(v′)2 = (τ1+2τ2Y)2(ξ4Y4+ξ3Y3+ξ2Y2+ξ1Y+ξ0)ζ0,v″ = τ1(4ξ4Y3+3ξ3Y2+2ξ2Y+ξ1)2ζ0+τ2ζ0(6ξ4Y4+5ξ3Y3+4ξ2Y2+3ξ1Y+3ξ0). (53)

Substituting (53) into (13) and setting the coefficient Y to be zero, we get a system of algebraic equations which can be solved to obtain the following results:

(54)ξ0 = p2ζ024ξ4ω2(−ωξ32 + 8τ0ξ4pζ0)2[ω2τ02ξ34 − 16ωτ03ξ32ξ4pζ0 − 8τ0ξ4ζ0ωξ32q + 64τ02ξ42pζ02q + 64τ04ξ42p2ζ02 + 64τ0ξ42pζ02r− 4ωξ32rζ0ξ4],ξ1 = ξ3pζ02ξ4ω(−ωξ32+8τ0ξ4pζ0)2[ω2τ0ξ34 − 4ωξ32qξ4ζ0− 16 ωξ32ξ4τ02pζ0 + 32qτ0ζ02pξ42 + 64p2τ03ζ02ξ42 + 16rζ02ξ42p],ξ2 = 14ξ4ω2(−ωξ32 + 8τ0ξ4pζ0)2[−12ω2ξ34τ0pξ4ζ0 + ω3ξ36− 16ξ42pζ02ωξ32q + 128ξ43p2ζ03qτ0​​​​ + 256ξ43p3ζ03τ03 + 64ξ43p2ζ03r],τ1 = ξ3ωpζ0, τ2 = 2ξ4ωpζ0, (54)

where ζ₀, ξ₃, ξ₄, p, q, and r are arbitrary constants. Substituting these results into (5) and (9), we have

(55)±(ξ − η0) = dYξ4ζ0Y4 + ξ3ζ0Y3 + ξ2ζ0Y2 + ξ1ζ0Y + ξ0ζ0. (55)

Now we will discuss the roots of the following equation:

(56)ξ4ζ0Y4+ξ3ζ0Y3+14ξ4ω2ζ0(−ωξ32+8τ0ξ4pζ0)2[−12ω2ξ34τ0pξ4ζ0 + ω3ξ36−16ξ42pζ02ωξ32q +128ξ43p2ζ03qτ0​​​​ +256ξ43p3ζ03τ03 + 64ξ43p2ζ03r]Y2 + ξ3p2ξ4ω(−ωξ32+8τ0ξ4pζ0)2[ω2τ0ξ34 − 4ωξ32qξ4ζ0−16ωξ32ξ4τ02pζ0​​​​  +32qτ0ζ02pξ42 + 64p2τ03ζ02ξ42 + 16rζ02ξ42p]Y + p2ζ024ξ4ω2(−ωξ32+8τ0ξ4pζ0)2[ω2τ02ξ34 − 16ωτ03ξ32ξ4pζ0 − 8τ0ξ4ζ0ωξ32q + 64τ02ξ42pζ02q + 64τ04ξ42p2ζ02 + 64τ0ξ42pζ02r − 4ωξ32rζ0ξ4]=0. (56)

To integrate (55), we must discuss the different cases of the roots of (56) as the following families.

Family 5. If (56) has four equal repeated roots α₁, consequently we can write (56) in the following form:

(57)ξ4ζ0Y4+ξ3ζ0Y3+14ξ4ω2ζ0(−ωξ32+8τ0ξ4pζ0)2[−12ω2ξ34τ0pξ4ζ0 + ω3ξ36−16ξ42pζ02ωξ32q + 128ξ43p2ζ03qτ0​​​​ + 256ξ43p3ζ03τ03 + 64ξ43p2ζ03r]Y2 + ξ3p2ξ4ω(−ωξ32+8τ0ξ4pζ0)2[ω2τ0ξ34 − 4ωξ32qξ4ζ0−16ωξ32ξ4τ02pζ0​​​​ + 32qτ0ζ02pξ42 + 64p2τ03ζ02ξ42 + 16rζ02ξ42p]Y + p2ζ024ξ4ω2(−ωξ32+8τ0ξ4pζ0)2[ω2τ02ξ34 − 16ωτ03ξ32ξ4pζ0 − 8τ0ξ4ζ0ωξ32q + 64τ02ξ42pζ02q + 64τ04ξ42p2ζ02 + 64τ0ξ42pζ02r − 4ωξ32rζ0ξ4] − (Y−α1)4= 0. (57)

From equating the coefficients of Y on both sides of (57), we get a system of algebraic equations in ζ₀, ξ₃, ξ₃, τ₀, and r, ω which can be solved by using the Maple software package to get the following results:

(58)r = −2q3p(−2α12ω + D), ξ3 = −4α1ζ0, ξ4 = ζ0, τ0 = Dp, (58)

where D = 2α12ω ± −3pq3.

Equations (58), (54), and (55) lead to

(59)​ξ0 = ζ0α14, ξ1 = −4ζ0α13, ξ2 = 6ζ0α12, τ1 = −4α1ωp, τ2 = 2ωp, (59)

where ζ₀ is an arbitrary constant and

(60)±(η − η0) = dY(Y − α1)2 = −1Y − α1, (60)

or

(61)Y = α1∓4(x − ωt − η0). (61)

Substituting (61), (59), and (58) into (53), we get the solution of (13) as follows:

(62)v(ξ) = Dp − 4α1ωp[α1 ∓ 4(x − ωt − η0)] + 2ωp[α1 ∓ 4(x − ωt − η0)]2. (62)