New Exact Solutions of Some Nonlinear Systems of Partial Differential Equations Using the First Integral Method

and Applied Analysis 3 P(X, Y) = ∑ m i=0 a i (X)Y i is an irreducible polynomial in the complex domain C[X, Y] such that P [X (ξ) , Y (ξ)] = m ∑ i=0 a i (X (ξ)) Y i (ξ) = 0, (13) where a i (X), (i = 0, 1, 2, . . . , m) are polynomials of X and a m (X) ̸ = 0. Equation (13) is called the first integral to (12a) and (12b). Due to the Division Theorem, there exists a polynomial h(X) + g(X)Y in the complex domain C[X, Y] such that dP dξ = ∂P ∂X dX dξ + ∂P ∂Y dY dξ = [h (X) + g (X)Y] m ∑ i=0 a i (X) Y i . (14) Here, we have considered one case only, assuming that m = 1 in (13). Suppose that m = 1, by equating the coefficients of Y (i = 2, 1, 0) on both sides of (14), we have a 󸀠 1 (X) = g (X) a 1 (X) , (15a) a 󸀠 0 (X) = h (X) a 1 (X) + g (X) a 0 (X) , (15b) a 1 (X) ((− p (r + 2s) 6cqk )X 3 − ( rc 1 − c qk )X − c 2 2cqk ) = h (X) a 0 (X) . (15c) Since a i (X) (i = 0, 1) are polynomials, then from (15a) we have deduced that a 1 (X) is constant and g(X) = 0. For simplicity, take a 1 (X) = 1. Balancing the degrees of h(X) and a 0 (X), we have concluded that deg(h(X)) = 1 only. Suppose that h(X) = AX + B, and A ̸ = 0, then we find a 0 (X) a 0 (X) = A 2 X 2 + BX + D, (16) whereD is an arbitrary integration constant. Substituting a 0 (X), a 1 (X) and h(X) for (15c) and setting all the coefficients of powers X to be zero, then we obtain a system of nonlinear algebraic equations and by solving it, we have obtained c 2 = 0, c 1 = 3c − (√3kDq√−p (r + 2s)) /√cq 3r , A = √−p (r + 2s) √3k√cq , B = 0, (17a) c 2 = 0, c 1 = 3c + (kDq√−p (r + 2s)) /√3√cq r , A = − √−p (r + 2s) √3k√cq , B = 0. (17b) Setting (17a) and (17b) in (13) leads to Y (η) + ( √−p (r + 2s) 2√3k√cq X 2 (η) + D) = 0, Y (η) + (− √−p (r + 2s) 2√3k√cq X 2 (η) + D) = 0. (18) Combining (18) with (12a), a first-order ordinary differential equation is derived, then by solving this derived equation and consideringX = V(ξ) and V(x, t) = V(ξ), we have obtained V 1 (x, t) = i(−3) 1/4 c(q) 1/4 √k√D × tan[((−1 3 ) 1/4 √D(p) 1/4 (r + 2s) 1/4 × [(k (x − ct) + γ) − 3√cqkξ0] ) × (c(q) 1/4 √k) −1 ] × ((p) 3/4 (r + 2s) 1/4 ) −1 , (19) V 2 (x, t) = (−3) 1/4 c(q) 1/4 √k√D × tan [((−1)√D(p)1/4(r + 2s)1/4 × [(k (x − ct) + γ) − 3√cqkξ0] ) × ((3) 1/4 c(q) 1/4 √k) −1 ] × ((p) 1/4 (r + 2s) 1/4 ) −1 , (20) respectively, where ξ 0 is an arbitrary integration constant. Also, by considering the solution u given by the relations (9), we have obtained u 1 (x, t) = ( p 2c ) × [i(−3) 1/4 c(q) 1/4 √k√D × tan[((−1 3 ) 1/4 √D(p) 1/4 (r + 2s) 1/4 × [(k (x − ct) + γ) − 3√cqkξ0] ) × (c(q) 1/4 √k) −1 ] 4 Abstract and Applied Analysis ×((p) 1/4 (r + 2s) 1/4 ) −1 ] 2 + 3c − (√3kDq√−p (r + 2s)) /√cq 3r , (21) u 2 (x, t) = ( p 2c ) × [(−3) 1/4 c(q) 1/4 √k√D × tan [((−1)√D(p)1/4(r + 2s)1/4 × [(k (x − ct) + γ) − 3√cqkξ0] ) × ((3) 1/4 c(q) 1/4 √k) −1 ] ×((p) 1/4 (r + 2s) 1/4 ) −1 ] 2 + 3c + (kDq√−p (r + 2s)) /√3√cq r , (22) respectively, where ξ 0 is an arbitrary integration constant. Thus, two solutions (u 1 , V 1 ) and (u 2 , V 2 ) have been obtained for the system (6). Comparing these results with the results obtained in [36], it can be seen that the solutions here are new. 3.2. (2 + 1)-Dimensional Davey-Stewartson System. The (2 + 1)-dimensional Davey-Stewartson system [37] reads iu t + u xx − u yy − 2|u| 2 u − 2uV = 0, V xx + V yy + 2(|u| 2 ) xx = 0. (23) This equation is completely integrable and used to describe the long-time evolution of a two-dimensional wave packet. Using the wave variables u = e iθ u (ξ) , V = V (ξ) , θ = px + qy + rt + ε, ξ = kx + cy + dt + γ, (24) wherep, q, r, k, c, andd are real constants, converts (23) into the ODE (q 2 − p 2 − r) u + (k 2 − c 2 ) u 󸀠󸀠 − 2u 3 − 2uV = 0, (25) (k 2 + c 2 ) V 󸀠󸀠 + (u 2 ) 󸀠󸀠 = 0. (26) Integrating (26) in the system and neglecting constants of integration, we have found