Bäcklund Transformation and New Exact Solutions of the Sharma-Tasso-Olver Equation

and Applied Analysis 3 At resonance levels, ur should be arbitrary, and then we are deducing a nonlinear equation g ur−1 · · ·u0, φ · · · 0. If the equation f implies g, then the compatibility condition is unconditionally satisfied. An equation for which these three steps can be carried out consistently and unambiguously passes the Painlevé test. Equation 1.3 is called the truncated Painlevé expansion. 2. The Compatibility Conditions and Bäcklund Transform Let u φ ∞ ∑ j 0 ujφ j , 2.1 where φ φ x, t , uj uj x, t are analytic functions on x, t in the neighborhood of the manifold M { x, t : φ x, t 0}. Substituting 2.1 into 1.1 , we can get the α −1. Thus, 2.1 becomes u ∞ ∑ j 0 ujφ j−1. 2.2 Expressing 1.1 , we have that ut 3αuux 3αux 3αuuxx αuxxx 0. 2.3 Differentiating 2.2 , we obtain that ut ( j − 1) ∞ ∑ j 0 ujφ j−2φt ∞ ∑ j 0 uj,tφ j−1, ux ( j − 1) ∞ ∑ j 0 ujφ j−2φx ∞ ∑ j 0 uj,xφ j−1, uxx ( j − 1) ∞ ∑ j 0 ujφ j−2φxx 2 ( j − 1) ∞ ∑ j 0 uj,xφ j−2φx ( j − 1)(j − 2) ∞ ∑ j 0 ujφ j−3φ2 x ∞ ∑ j 0 uj,xxφ j−1, uxxx 3 ( j − 1) ∞ ∑ j 0 uj,xφ j−2φxx ( j − 1) ∞ ∑ j 0 ujφ j−2φxxx 3 ( j − 1)(j − 2) ∞ ∑ j 0 ujφ j−3φxφxx 3 ( j − 1) ∞ ∑ j 0 uj,xxφ j−2φx ∞ ∑ j 0 uj,xxxφ j−1 3 ( j − 1)(j − 2) ∞ ∑ j 0 uj,xφ j−3φ2 x ( j − 1)(j − 2)(j − 3) ∞ ∑ j 0 ujφ j−4φ3 x. 2.4 4 Abstract and Applied Analysis Substituting 2.4 into 2.3 , we get the cycle formula on uj ( j − 3uj−2φt uj−3,t 3α j−n ∑ m 0 j ∑ n 0 uj−m−num [ n − 1 unφx un−1,x ] 3α j ∑ m 0 [( j −m − 1uj−mφx uj−m−1,x ][ m − 1 umφx um−1,x ] 3α j ∑ m 0 uj−m [ m − 2 um−1φxx 2 m − 2 um−1,xφx m − 1 m − 2 umφ x um−2,xx ] α [ 3 ( j − 3uj−2,xφxx ( j − 3uj−2φxxx 3 ( j − 2)(j − 3uj−1φxφxx 3 ( j − 2)(j − 3uj−1,xφ2 x uj−3,xxx 3 ( j − 3uj−2,xxφx ( j − 1)(j − 2)(j − 3ujφ3 x ] 0. 2.5 Taking j 0 in 2.5 , we deduce that u0 φx or u0 2φx. We will get the Bäcklund transformations, according to the above two cases, respectively. Case 1 u0 φx . Substituting u0 φx into 2.5 , we have the following equation on uj : ( j 1 )( j − 1)(j − 3αujφ3 x Fj ( uj−1 · · ·u0, φt, φx, φxx, φxxx · · · ) ( j 1, 2, . . . ) . 2.6 By 2.6 we see that u−1, u1 and u3 are arbitrary. Hence j 3 are resonances. Moreover the compatibility conditions will be deduced by 2.5 or 2.6 . Setting j 2 and j 3, we infer that φt 3α ( u1φx u2φ 2 x u1,xφx u1φxx ) αφxxx 0, 2.7 ∂ ∂x [ φt 3α ( u1φx u2φ 2 x u1,xφx u1φxx ) αφxxx ] 0. 2.8 It is easy to see that 2.8 does so if 2.7 holds. So u3 is arbitrary. Taking j 4, u3 0 and letting u2 u4 0 in 2.6 , we have that u1,t 3αu1u1,x 3αu 2 1,x 3αu1u1,xx αu1,xxx 0. 2.9 By 2.6 , noting that u2 0, u3 0 and u4 0, we can infer that uj 0 ( j ≥ 2). 2.10 Abstract and Applied Analysis 5 Thus, we obtain from the Bäckland transformation of the STO equation thatand Applied Analysis 5 Thus, we obtain from the Bäckland transformation of the STO equation that