Application Of A Generalized Bernoulli Sub-ODE Method For Finding Traveling Solutions Of Some Nonlinear Equations

Application Of A Generalized Bernoulli Sub-ODE Method For Finding Traveling Solutions Of Some Nonlinear Equations Shandong University of Technology School of Science Zhangzhou Road 2, Zibo, 255049 China zhengbin260@26.com Abstract: In this paper, a generalized Bernoulli sub-ode method is proposed to construct exact traveling solutions of nonlinear evolution equations. We apply the method to establish traveling solutions of the variant Boussinseq equations, (2+)-dimensional NNV equations and (2+)-dimensional Boussinesq and Kadomtsev-Petviashvili equations. As a result, some new exact traveling wave solutions are found. Key Words: Bernoulli sub-ode method, traveling wave solutions, variant Boussinseq equations, NNV equations, Boussinesq and Kadomtsev-Petviashvili equations. Introduction It is well known that nonlinear evolution equations (NLEEs) are widely used to describe many complex physical phenomena such as fluid mechanics, plasma physics, optical fibers, biology, solid state physics, chemical kinematics, chemical physics, and so on. So, the powerful and efficient methods to find analytic solutions of nonlinear equations have drawn a lot of interest by a diverse group of scientists. In the literature, there is a wide variety of approaches to nonlinear problems for constructing traveling wave solutions. Some of these approaches are the homogeneous balance method [, 2], the hyperbolic tangent expansion method [, 4], the trial function method [5], the tanh-method [6-8], the non-linear transform method [9], the inverse scattering transform [0], the Backlund transform [, 2], the Hirota s bilinear method [, 4], the generalized Riccati equation method [5, 6], the Weierstrass elliptic function method [7], the theta function method [8-20], the sine-cosine method [2], the Jacobi elliptic function expansion [22, 2], the complex hyperbolic function method [24-26], the truncated Painleve expansion [27], the F-expansion method [28], the rank analysis method [29], the exp-function expansion method [0], the (G /G)- expansion method [-40] and so on. In [4], we proposed a new Bernoulli sub-ode method to construct exact traveling wave solutions for NLEEs. In this paper, we will apply the Bernoulli sub-ode method to construct exact traveling wave solutions for some special nonlinear equations. First, we reduce the nonlinear equations to ODEs by a traveling wave variable transformation. Second, we suppose the solution can be expressed as an polynomial of single variable G, where G = G(ξ) satisfied the Bernoulli equation. Then the degree of the polynomial can be determined by the homogeneous balance method, and the coefficients can be obtained by solving a set of algebraic equations. The rest of the paper is organized as follows. In Section 2, we describe the Bernoulli sub-ode method for finding traveling wave solutions of nonlinear evolution equations, and give the main steps of the method. In the subsequent sections, we will apply the method to find exact traveling wave solutions of the variant Boussinseq e- quation, (2+)-dimensional NNV equations and (2+)-dimensional Boussinesq and Kadomtsev- Petviashvili equations. In the last Section, some conclusions are presented. 2 Description Of The Bernoulli Sub-ODE Method In this section we describe the Bernoulli Sub-ODE Method. First we present the solutions of the following E-ISSN: 2224-2880 68 Issue 7, Volume, July 202

ODE: G + λg = µg 2, () where λ 0, G = G(ξ). When µ 0, Eq. () is the type of Bernoulli equation, and we can obtain the solution as G = µ, (2) + λ where d is an arbitrary constant. When µ = 0, the solution of Eq. () is given by G = de λξ, () where d is an arbitrary constant. Suppose that a nonlinear equation, say in two or three independent variables x, y, t, is given by P (u, u t, u x, u y,, u tt, u xt, u xx, u xy...) = 0, (4) where u = u(x, y, t) is an unknown function, P is a polynomial in u = u(x, y, t) and its various partial derivatives, in which the highest order derivatives and nonlinear terms are involved. By using the solutions of Eq. (), we can construct a serials of exact solutions of nonlinear equations: Step. We suppose that u(x, y, t) = u(ξ), ξ = ξ(x, y, t). (5) The traveling wave variable (5) permits us reducing (4) to an ODE for u = u(ξ) P (u, u, u,...) = 0. (6) Step 2. Suppose that the solution of (6) can be expressed by a polynomial in G as follows: u(ξ) = a m G m + a m G m +... + a 0, (7) where G = G(ξ) satisfies Eq. (), and a m, a m..., a 0, µ are constants to be determined later with a m 0. The positive integer m can be determined by considering the homogeneous balance between the highest order derivatives and nonlinear terms appearing in (6). Step. Substituting (7) into (6) and using (), collecting all terms with the same order of G together, the left-hand side of (6) is converted to another polynomial in G. Equating each coefficient of this polynomial to zero, yields a set of algebraic equations for a m, a m..., k, c, λ and µ. Step 4. Solving the algebraic equations system in Step, and by using the solutions of Eq. (), we can construct the traveling wave solutions of the nonlinear evolution equation (6). In the following sections, we will apply the method described above to some examples. Application Of The Bernoulli Sub-ODE Method For The Variant Boussinseq Equations In this section, we will consider the variant Boussinseq equations [42, 4]: u t + uu x + v x + αu xxt = 0, (8) v t + (uv) x + βu xxx = 0, (9) where α and β are arbitrary constants, β > 0. Supposing that ξ = k(x ct), (0) by (0), (8) and (9) are converted into ODEs cu + uu + v αk 2 cu = 0 () cv + (uv) + βk 2 u = 0. (2) Integrating () and (2) once, we have cu + 2 u2 + v αk 2 cu = g, () cv + uv + βk 2 u = g 2, (4) where g and g 2 are the integration constants. Suppose that the solution of () and (4) can be expressed by a polynomial in G as follows: u(ξ) = v(ξ) = m a i G i, (5) n b i G i, (6) where a i, b i are constants, G = G(ξ) satisfies Eq. (). Balancing the order of u 2 and v in Eq. (), the order of u and uv in Eq. (4), then we can obtain 2m = n, n + 2 = m + n m =, n = 2, so Eqs. (5) and (6) can be rewritten as u(ξ) = a 2 G 2 + a G + a 0, a 2 0, (7) v(ξ) = b 2 G 2 + b G + b 0, b 2 0, (8) where a, a 0, b 2, b, b 0 are constants to be determined later. Substituting (7) and (8) into () and (4) and collecting all the terms with the same power of G together, equating each coefficient to zero, yields a set of simultaneous algebraic equations as follows: For Eq. (): E-ISSN: 2224-2880 69 Issue 7, Volume, July 202

G 0 : ca 0 g + 2 a2 0 + b 0 = 0. G : b + a 0 a ca αk 2 ca λ 2 = 0. G 2 : ca 2 + b 2 + αk 2 ca µλ + 2 a2 4αk 2 ca 2 λ 2 + a 0 a 2 = 0. G : a a 2 2αk 2 ca µ 2 + 0αk 2 ca 2 µλ = 0. G 4 : 2 a2 2 6αk2 ca 2 µ 2 = 0. For Eq. (4): G 0 : cb 0 g 2 + a 0 b 0 = 0. G : b a 0 + a b 0 cb + βk 2 a λ 2 = 0. G 2 : cb 2 +a b betak 2 a µλ +4βk 2 a 2 λ 2 + a 2 b 0 + a 0 b 2 = 0. G : 2βk 2 a µ 2 +a b 2 0βk 2 a 2 µλ+a 2 b = 0. G 4 : a 2 b 2 + 6βk 2 a 2 µ 2 = 0. Solving the algebraic equations above yields: a 2 = 2αk 2 cµ 2, a = 2αk 2 cµλ, a 0 = 2 ( β+2αc2 +2λ 2 α 2 k 2 c 2 αc ), b 2 = 6βk 2 µ 2, b = 6βk 2 µλ, b 0 = β 4 ( β+2λ2 α 2 k 2 c 2 α 2 c 2 ), g 2 = β 8 ( β2 +4λ 4 α 4 k 4 c 4 α c ), g = 8 ( 4c4 α 2 +β 2 +4λ 4 α 4 k 4 c 4 α 2 c 2 ). (9) Combining with (2) and (), under the conditions µ 0, we can obtain the traveling wave solutions of the variant Boussinseq equations (8) and (9) as follows: u (ξ) = 2 (β + 2αc2 + 2λ 2 α 2 k 2 c 2 ) αc 2αk 2 cµλ λ + ) +2αk 2 cµ 2 λ + )2 (20) v (ξ) = β 4 ( β + 2λ2 α 2 k 2 c 2 α 2 c 2 ) +6βk 2 µλ λ + ) 6βk 2 µ 2 λ + )2 (2) Remark When µ = 0, we obtain the trivial solutions. Remark 2 The exact traveling wave solutions (20)-(2) of the variant Boussinseq equations are different from the results in [42, 4], and have not been reported by other authors to our best knowledge. 4 Application Of The Bernoulli Sub-ODE Method For (2+)- dimensional NNV Equations In this section, we consider the (2+)-dimensional NNV equations [44 46]: u t + au xxx + bu yyy + cu x + du y = a(uv) x + b(uw) y, Suppose that (22) u x = v y, (2) u y = w x. (24) ξ = kx + ly + ωt. (25) By (25), (22), (2) and (24) are converted into ODEs ωu + ak u + bl u + cku + dlu = ak(uv) + bl(uw), (26) ku = lv, (27) lu = kw. (28) Integrating (26), (27) and (28) once times, we have ωu + ak u + bl u + cku + dlu = akuv + bluw + g, (29) ku = lv + g 2, (0) lu = kw + g, () where g, g 2, g are the integration constants. Suppose that the solutions of (29), (0) and () can be expressed by polynomials in G as follows: m u(ξ) = a i G i, (2) v(ξ) = n b i G i, () E-ISSN: 2224-2880 620 Issue 7, Volume, July 202

w(ξ) = s c i G i, (4) where a i, b i, c i are constants, G = G(ξ) satisfies Eq. (). Balancing the order of u and uv in Eq. (29), the order of u and v in Eq. (0), the order of u and w in Eq. (), then we can obtain m + 2 = m + n, m = n, m = s m = n = s = 2, so Eq.(2), () and (4) can be rewritten as u(ξ) = a 2 G 2 + a G + a 0, a 2 0, (5) v(ξ) = b 2 G 2 + b G + b 0, b 2 0, (6) w(ξ) = c 2 G 2 + c G + c 0, c 2 0, (7) where a 2, a, a 0, b 2, b, b 0, c 2, c, c 0 are constants to be determined later. Substituting (5), (6) and (7) into (29), (0) and () and collecting all the terms with the same power of G together, equating each coefficient to zero, yields a set of simultaneous algebraic equations as follows: For Eq. (29): G 0 : cka 0 + dla 0 aka 0 b 0 bla 0 c 0 g 0 + ωa 0 = 0. G : aka 0 b bla 0 c + dla + ak a λ 2 +cka aka b 0 + bl a λ 2 bla c 0 +ωa = 0. G 2 : ak a µλ + cka 2 aka 0 b 2 bla 0 c 2 +4ak a 2 λ 2 aka b bl a µλ + ωa 2 +4bl a 2 λ 2 bla c + dla 2 aka 2 b 0 bla 2 c 0 = 0. G : aka b 2 + 2bl a µ 2 + 2ak a µ 2 aka 2 b bla c 2 0ak a 2 µλ 0bl a 2 µλ bla 2 c = 0. G 4 : aka 2 b 2 bla 2 c 2 + 6ak a 2 µ 2 For Eq.(0): For Eq. (): +6bl a 2 µ 2 = 0. G 0 : ka 0 lb 0 g 2 = 0. G : ka lb = 0. G 2 : ka 2 lb 2 = 0. G 0 : la 0 kc 0 g = 0. G : la kc = 0. G 2 : la 2 kc 2 = 0. Solving the algebraic equations above yields: a 2 = 2lkµ 2, a = 2lµλk, a 0 = a 0, b 2 = 2k 2 µ 2, b = 2µk 2 λ, b 0 = b 0, c 2 = 2µ 2 l 2, c = 2µl 2 λ, k = k, l = l, ω = ω, c 0 = ak a 0 bl a 0 + dl 2 k + ak 4 lλ 2 bl 2 k + ck 2 l ak 2 lb 0 + bl 4 λ 2 k + ωlk bl 2, k ak a 0 bl a 0 + ak 4 lλ 2 + bl 4 λ 2 k g = a 0, lk g 2 = ka 0 lb 0, g = 6bl a 0 ak a 0 + dl 2 k + ak 4 lλ 2 l 2 b ck 2 l ak 2 lb 0 + bl 4 λ 2 k + ωlk l 2, b (8) where k, l, ω, a 0, b 0 are arbitrary constants. Under the condition µ 0, combining with (2) and (), we can obtain the traveling wave solutions of the (2+)-dimensional NNV equations (22)-(24) as follows: u(ξ) = 2lkµ 2 λ + )2 2lµλk λ + ) + a 0 (9) v(ξ) = 2k 2 µ 2 λ + )2 2µk 2 λ λ + ) + b 0 (40) w(ξ) = 2µ 2 l 2 λ + )2 2µl 2 λ λ + ) + ak a 0 bl a 0 + dl 2 k + ak 4 lλ 2 bl 2 k + ck 2 l ak 2 lb 0 + bl 4 λ 2 k + ωlk bl 2 (4) k where ξ = kx + ly + ωt. Remark Some authors have reported some exact solutions for the (2+)-dimensional NNV e- quations in [44-46]. To our best knowledge, our results (9)-(4) have not been reported so far in the literature. Remark 4 When µ = 0, we obtain the trivial solutions. E-ISSN: 2224-2880 62 Issue 7, Volume, July 202

5 Application Of The Bernoulli Sub-ODE Method For (2+) dimensional Boussinesq and Kadomtsev-Petviashvili equations In this section we will consider the following (2+) dimensional Boussinesq and Kadomtsev- Petviashvili equations [47] : u y = q x, (42) v x = q y, (4) q t = q xxx + q yyy + 6(qu) x + 6(qv) y. (44) In order to obtain the traveling wave solutions of (42)-(44), we suppose that u(x, y, t) = u(ξ), v(x, y, t) = v(ξ), q(x, y, t) = q(ξ), ξ = ax + dy ct, (45) where a, d, c are constants that to be determined later. Using the wave variable (45), (42)-(44) can be converted into ODEs du aq = 0, (46) av dq = 0, (47) (a + d )q cq 6auq 6aqu 6dvq 6dqv = 0. (48) Integrating the ODEs above, we obtain du aq = g, (49) av dq = g 2, (50) (a + d )q cq 6auq 6dvq = g.(5) Supposing that the solutions of the ODEs above can be expressed by a polynomial in G as follows: u(ξ) = v(ξ) = q(ξ) = l a i G i, (52) m b i G i, (5) n c i G i, (54) where a i, b i, c i are constants, and G = G(ξ) satisfies Eq. (). Balancing the order of u and q in Eq. (52), the order of v and q in Eq. (5) and the order of q and vq in Eq. (54), we have l+ = n+, m+ = n +, n + = m + n + l = m = n = 2. So Eq.(52)-(54) can be rewritten as u(ξ) = a 2 G 2 + a G + a 0, a 2 0, (55) v(ξ) = b 2 G 2 + b G + b 0, b 2 0, (56) u(ξ) = c 2 G 2 + c G + c 0, c 2 0, (57) where a i, b i, c i are constants to be determined later. Substituting (55)-(57) into the ODEs (49)- (5), collecting all terms with the same power of G together, equating each coefficient to zero, yields a set of simultaneous algebraic equations as follows: For Eq. (49): G 0 : a 0 d ac 0 g = 0. G : a d ac = 0. G 2 : a 2 d ac 2 = 0. For Eq. (50): G 0 : ab 0 g 2 dc 0 = 0. G : ab dc = 0. G 2 : dc 2 + ab 2 = 0. For Eq. (5): G 0 : g cc 0 6db 0 c 0 6aa 0 c 0 = 0. G : 6aa c 0 6db c 0 + a c λ 2 6db 0 c 6aa 0 c + d c λ 2 cc = 0. G 2 : 4a c 2 λ 2 6aa 0 c 2 6db c 6aa c 6aa 2 c 0 + 4d c 2 λ 2 cc 2 d c µλ 6db 0 c 2 a c µλ 6db 2 c 0 = 0. G : 6aa 2 c 6db 2 c + 2a c µ 2 0d c 2 µλ 6aa c 2 0a c 2 µλ 6db c 2 +2d c µ 2 = 0. G 4 : 6d c 2 µ 2 6aa 2 c 2 +6a c 2 µ 2 6db 2 c 2 = 0. Solving the algebraic equations above yields: E-ISSN: 2224-2880 622 Issue 7, Volume, July 202

Case : a 0 = a 0, a = µλa 2, a 2 = a 2 µ 2, b 0 = b 0, b = d 2 µλ, b 2 = d 2 µ 2, a = a, c 0 = c 0, c = dµλa, c 2 = dµ 2 a, g 2 = ab 0 dc 0, g = da 0 ac 0, d = d, c = 6a c 0 6d c 0 + a 4 dλ 2 6d 2 b 0 a 6a 2 a 0 d + d 4 λ 2 a, ad where a 0, b 0, c 0, a, d are arbitrary constants. Similarly, under the condition µ 0, we can obtain traveling wave solutions of (2+) dimensional Boussinesq and Kadomtsev-Petviashvili e- quations as follows: u 2 (ξ) = d 2 µ 2 λ + )2 + a λ + ) + a 0, (64) v 2 (ξ) = d 2 µ 2 λ + )2 + a λ + ) + b 0, (65) q 2 (ξ) = d 2 µ 2 λ + )2 a λ + ) + c 0, where (66) ξ = dx + dy 6d( b 0 + a 0 )t. (67) 6a c 0 6d c 0 + a 4 dλ 2 + d 4 λ 2 a g = c 0, ad where a 0, b 0, c 0, a, d are arbitrary constants. (58) Assume µ 0, then substituting the results above into (55)-(57), combining with (2) we can obtain the traveling wave solution of (2+) dimensional BKP equation as follows: u (ξ) = a 2 µ 2 λ + )2 µλa 2 λ + ) + a 0, (59) v (ξ) = d 2 µ 2 λ + )2 d 2 µλ λ + ) + b 0, (60) q (ξ) = dµ 2 a λ + )2 dµλa λ + ) + c 0, where Case 2: ξ = ax + dy 6a c 0 6d c 0 + a 4 dλ 2 t ad (6) 6d2 b 0 a 6a 2 a 0 d + d 4 λ 2 a t. (62) ad a 0 = a 0, a = a, a 2 = d 2 µ 2, b 0 = b 0, b = a, b 2 = d 2 µ 2, d = d, c = 6d( b 0 + a 0 ), c 0 = c 0, c = a, c 2 = d 2 µ 2, g 2 = db 0 dc 0, g = da 0 + dc 0, a = d, g = 0, (6) Case : a 0 = a 0, a = 2 d2 µλ( 2 ± 2 a 2 = d 2 µ 2 ( 2 ± 2 b 0 = b 0, b = 2 µλd2, b 2 = d 2 µ 2, c 0 = c 0, c = 2 d2 µλ( 2 ± 2 c 2 = d 2 µ 2 ( 2 ± 2 d = d, a = ( 2 ± 2 i)d, c = 6da 0 ( 2 ± 2 i) 6db0, g = da 0 dc 0 ( 2 ± 2 g 2 = db 0 ( 2 ± 2 i) dc0, g = 0, where a 0, b 0, c 0, d are arbitrary constants. Thus u (ξ) = d 2 µ 2 ( 2 ± 2 µ ) 2 + 2 d2 µλ( 2 ± 2 µ ) + a 0, (68) (69) v (ξ) = d 2 µ 2 ) 2 + 2 µλd2 ) + b 0, ξ = ( 2 ± 2 q (ξ) = d 2 µ 2 ( 2 ± 2 µ ) 2 + 2 d2 µλ( 2 ± 2 µ ) + c 0, where µ 0. i)dx+dy+[6da0 ( 2 ± 2 (70) (7) i)+6db0 ]t, (72) E-ISSN: 2224-2880 62 Issue 7, Volume, July 202

Case 4: a 0 = a 0, a = 2 d2 µλ( 2 ± 2 a 2 = d 2 µ 2 ( 2 ± 2 i) b 0 = b 0, b = a ( 2 ± 2 b 2 = d 2 µ 2 c 0 = c 0, c = a ( 2 ± 2 c 2 = d 2 µ 2 ( 2 ± 2 i) d = d, a = ( 2 ± 2 i)d, c = 6da 0 ( 2 ± 2 i) 6db0 g = da 0 dc 0 ( 2 ± 2 g 2 = db 0 ( 2 ± 2 i) dc0, g = 0, where a 0, b 0, c 0, d are arbitrary constants. Then u 4 (ξ) = d 2 µ 2 ( 2 ± 2 µ ) 2 + 2 d2 µλ( 2 ± 2 µ ) + a 0, v 4 (ξ) = d 2 µ 2 ) 2 a ( 2 ± 2 µ ) + b 0, q 4 (ξ) = d 2 µ 2 ( 2 ± 2 2 ± 2 a ( 2 ± 2 µ ) + c 0, ξ = ( 2 ± 2 where µ 0. i)dx+dy+[6da0 ( 2 ± 2 (7) (74) (75) µ ) 2 (76) i)+6db0 ]t, (77) Remark 5 When µ = 0, we obtain the trivial solutions. The traveling wave solutions established for (2+)-dimensional Boussinesq and Kadomtsev-Petviashvili equations (59)-(6), (64)-(66), (69)-(7), (74)-(76) have not been reported by other authors to our best knowledge. 6 Comparison with Zayed results In [46,4], Zayed solved the (2+)-dimensional N- NV equations and the variant Boussinseq equations by using the (G /G) expansion method respectively. In this section, we will present some comparisons between the established results in Section -4 and Zayed results. Let µ, λ represent µ, λ in [46] respectively. Then in (9)-(4), considering d, µ, λ, a 0, b 0, l, k, ω are arbitrary constants, if we take d = A + B, µ = B A, λ = λ 2 λ 2 4µ 4µ, a 0 = α 0 2µ, l =, k =, b 0 = 2µ, ω = (a+b)(α 0 2µ ) A + B (a+b)(λ 2 4µ ) λ 2 4µ c 6aµ b(γ 0 2 λ2 ), where A, B, α 0, γ 0 are defined in [46], our solutions (9)-(4) reduce to the solutions derived in [46, (.2 )-(.4)]. Furthermore, under the condition A = 0, B 0, λ > 0, µ = 0, (9)-(4) reduce the solitary solutions in [46, (.4)-(.4)]. If we take d = ia + B i, µ = ia B, λ = i 4µ 4µ λ 2 λ 2, a 0 = α 0 2µ, l =, k =, b 0 = 2µ, ω = (a+b)(α 0 2µ ) ia + B i λ 2 4µ (a+b)(λ 2 4µ ) c 6aµ b(γ 0 2 λ2 ), then our solutions (9)-(4) reduce to the solutions derived in [46, (.5)-(.7)]. So in this way, our results (9)-(4) extend Zayed results for the (2+)-dimensional NNV equations in [46]. For the variant Boussinseq equations, we note that our solutions (20)-(2) are different solutions from the results in [4, ()-(8)]. 7 Conclusions In this paper we have seen that some new traveling wave solutions of the variant Boussinseq e- quations, (2+)-dimensional NNV equations and (2+)-dimensional Boussinesq and Kadomtsev- Petviashvili equations are successfully found by using the Bernoulli sub-ode method. The main points of the method are that assuming the solution of the ODE reduced by using the traveling wave variable as well as integrating can be expressed by an m-th degree polynomial in G, where G = G(ξ) is the general solutions of a Bernoulli sub-ode equation. The positive integer m can be determined by the general homogeneous balance method, and the coefficients of the polynomial can E-ISSN: 2224-2880 624 Issue 7, Volume, July 202

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