Nonlinear Dyn DOI 10.1007/s11071-015-2480-8 pro of ORIGINAL PAPER Nonlocal symmetry analysis and conservation laws to an third-order Burgers equation Received: 28 March 2015 / Accepted: 30 October 2015 © Springer Science+Business Media Dordrecht 2015 2 3 4 5 6 7 8 9 10 11 12 13 14 Abstract The nonlocal residual symmetry related to truncated Painlevé expansion of third-order Burgers equation is performed. Then, the potential symmetries of the equation are derived; on the basis of nonlocal symmetries, we linearize the nonlinear third-order Burgers equation to a linear third-order PDE. Furthermore, the exact solutions of the potential equation are presented in terms of the symmetries. In particular, conservation laws are constructed of the equations. Keywords Third-order Burger equation · Nonlocal symmetry analysis · Linearization · The multiplier approach · Conservation laws Mathematics Subject Classification 70S10 1 Introduction 15 It is known that the Burgers hierarchy is of the form [1–7] u t = K m (u) = (D + u + u x D −1 m−1 ) A. H. Kara School of Mathematics, University of the Witwatersrand, Private Bag 3, Wits, Johannesburg 2050, South Africa A. H. Kara · K. Fakhar Department of Mathematics, University of British Columbia, Vancouver V6T 1Z2, Canada K. Fakhar Department of Mathematics, Faculty of Science and Horticulture, Kwantlen Polytechnic University, Surrey, BC V3W 2MB, Canada 21 ut = u x , u t = 2uu x + u x x , u t = 3uu x x + u x x x + 3u 2x + 3u 2 u x , TYPESET DISK LE 22 (3) 23 (4) 24 25 (5) They are of first order, second order, third order and fourth order, respectively. The third-order Burgers equation, similar to the second-order ones, appear in many physical and engineering fields [8–10], such as the plasma physics and fluid mechanics. In addition, these equations play a key role in nonlinear theory and mathematical physics, in particular in the integrable system, soliton theory, nonlinear wave theory, and so on. It is to be noted that Eq. (3) is the dissipative Burgers equation; by using the Hopf–Cole transformation, this equation can be reduced the heat equation u t − u x x = 0. Moreover, the third equation in Eq. (4) is the well-known Sharma–Tasso–Olver (STO) equation. The Burgers equation and the STO equation were investigated in many papers such as [1–19]. In paper CP Disp.:2015/11/6 Pages: 12 Layout: Medium 20 (2) 123 Journal: 11071 MS: 2480 18 Substituting n = 1, 2, 3, 4 into (1), one can get few elements of the hierarchy (1), which are + 12uu 2x + 4u 3 u x + 6u 2 u x x . G. Wang (B) School of Mathematics and Statistics, Beijing Institute of Technology, Beijing 100081, People’s Republic of China e-mail: pukai1121@163.com; wanggangwei@bit.edu.cn 17 19 u t = u x x x x + 10u x u x x + 4uu x x x 22E70 · u x , m = 1, 2, . . . 16 (1) orre 1 unc 1 cted Author Proof Gangwei Wang · A. H. Kara · K. Fakhar 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 G. Wang et al. 46 47 48 49 Author Proof 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 80 u 1t − 3u 1 u 1x x − u 1x x x − 3u 21x − 3u 21 u 1x 83  + φ (−1) − 6u 0 u 1 u 1x − 3u 21 u 0x − 3u 0 u 1x x  − 3u 1 u 0x x − 6u 0x u 1x + u 0t − u 0x x x  + φ (−2) 3φx u 0 u 21 + 6φx u 0 u 1x + 6φx u 1 u 0x TYPESET DISK u 0 = φx , 87 88 89 (7) 90 91 (8) or 92 93 u 0 = 2φx . (9) 94 The compatibility conditions at j = 1, 3 and this equation possesses the Painlevé property [7,11]. Moreover, from the coefficients of φ 0 and φ 1 , one can obtain 96 Lemma 1 u 0 = φx is a symmetry of (4) with a solution of u 1 . In order to proof this Lemma, we first give the definition of symmetry [12,13] and the Lemma 2. Definition 1 [12,13] Given an evolution equation u t = K [u], (10) here K [u] = K (x, t, u x , u x x . . .), if σ (x, t, u x , u x x . . .) is a symmetry of Eq. (14), then σ (x, t, u x , u x x . . .) satisfy (11) where K ′ means the Fréchet derivative. σt − 6σ uu x − 3u 2 σx − 3σ u x x − 3uσx x − 6σx u x − σx x x = 0, 95 97 98 99 100 101 102 103 104 105 106 107 108 Lemma 2 σ = σ (x, t, u) is a symmetry of Eq. (4) if and only if 109 110 111 (12) 112 where u is a solution of (4). This lemma is easily proved by Definition 1. From the coefficients of φ 0 and φ 1 , we can get 114 u 1t − 3u 1 u 1x x − u 1x x x − 3u 21x − 3u 21 u 1x = 0,  −6u 0 u 1 u 1x − 3u 21 u 0x − 3u 0 u 1x x − 3u 1 u 0x x − 6u 0x u 1x LE 86 Vanishing the coefficients of φ −4 , one can get 123 Journal: 11071 MS: 2480 85 + 6φx u 20 u 1 − 6φx2 u 0x − 6φx φx x u 0  + 12φx u 0 u 0x + 3φx x u 20 − 3u 20 u 0x   + φ (−4) 6φx3 u 0 − 9φx2 u 20 + 3φx u 30 . dσ = K ′ σ, dt The STO equation (4) is Painlevé integrable [7,11]. In order to get residual symmetry via Painlevé analysis, we use the following truncated Painlevé expansion form u0 + u1. (6) u= φ Plugging (7) into (4), one can have 82 − u o φt + 3φx u 0x x + 3φx x u 0x  + φx x x u 0 − 3u 0 u 0x x − 3u 20x  + φ (−3) − 6φx2 u 0 u 1 2 Residual symmetry via Painlevé analysis 79 81 pro of 45 84 cted 44 + 3φx x u 1 u 0 − 3u 20 u 1x − 6u 0 u 1 u 0x orre 43 [1], the authors considered the famous Burgres’ equation and gave infinite series of flows. The authors in [4,5] presented exact solutions, which included selfsimilar solutions for Burgers hierarchy. Also, multiple kink solutions and multiple singular kink solutions of Burgers hierarchy were constructed in [6]. In Ref. [7], the authors considered the Painlev test, generalized symmetries, Bäklund transformations and exact solutions of the STO equation. In paper [18], the authors discussed group classification and exact solutions of generalized Burgers equations with linear damping and variable coefficients in detail. The authors gave the results about self-adjointness and conservation laws of a generalized Burgers equation in [19]. In paper [20], the authors considered the nonclassical potential symmetries of the Burgers equation. The authors analyzed the connection between symmetries and linearization in [21,22]. In this paper, we consider the nonlocal symmetries, exact solutions and derive the conservation laws of third-order Burgers equation (4) and its potential form. The remainder of this paper is organized as follows: In Sect. 2, nonlocal residual symmetry related to truncated Painlevé expansion of this equation is obtained. In Sect. 3, we employ potential symmetry on this equation and present all of the geometric vector fields are constructed. In Sect. 4, linearization by nonlocal symmetries is derived. In Sect. 5, similarity reductions and explicit solutions are presented. In Sect. 6, conservation laws are given. Finally, the main findings of the paper are recapitulated in Sect. 7. unc 42 CP Disp.:2015/11/6 Pages: 12 Layout: Medium 113 115 116 117 118 Nonlocal symmetry analysis and conservation laws 122 123 124 126 127 128 129 Substituting u 0 = φx and u 1 into (16) and (4), one can get the Lemma 1. It is noted that u 0 = φx is a nonlocal symmetry with regard to Bäcklund transformation (6). 3 Point symmetry of the potential system and potential equation In order to get more nonlocal symmetries, we use potential symmetry analysis to deal with (4). PDE (4) is equivalent to the potential system vx = u, 130 vt = 3uu x + u x x + u 3 , 131 and the potential equation 132 vt = 3vx vx x + vx x x + (vx )3 . 133 The point symmetry 134 t ∗ = t + ετ (x, t, u, v) + O(ε2 ), 135 136 (14) (15) u ∗ = u + εη(x, t, u, v) + O(ε2 ), v ∗ = v + εψ(x, t, u, v) + O(ε2 ), 138 where ε is the group parameter, and 139 X (3) (vx − u) = 0, 141 142 143 144 145 146 147 148 149 150 151 152 153 X (3) (vt − 3uu x − u x x − u 3 ) = 0, Theorem 1 The potential system (18) admits the point symmetry (20) (16) (17) for any (u, v) that solves system (18); ∂ ∂ X = τ (x, t, u, v) + ξ(x, t, u, v) ∂t ∂x ∂ ∂ + ψ(x, t, u, v) , (18) + η(x, t, u, v) ∂u ∂v is the infinitesimal generator of the point symmetry (20); ∂ ∂ X (3) = X + η x + ηt ∂u x ∂u t ∂ ∂ + ψx + ψt + ··· (19) ∂vx ∂vt with t ∂ ∂ , X2 = , ∂x ∂t x ∂ u ∂ ∂ ∂ , X4 = + − , X3 = ∂v ∂t 3 ∂x 3 ∂u   ∂ ∂ X F = e−v u F − e−v Fx − e−v F , ∂u ∂v X1 = η = Dt (η) − u x Dt (ξ ) − u t Dt (τ ), x (23) ψ = Dt (ψ) − vx Dt (ξ ) − vt Dt (τ ), ··· (20) (24) here F = F(x, t), and F satisfies Ft − Fx x x = 0, and hence, the scalar equation (4) admits the corresponding potential symmetry. Proof: One can get the determining equations τu = 0, DISK LE 161 162 165 169 170 171 172 173 174 τx = 0, 175 τv = 0, 176 ψu = 0, 177 ξu = 0, 178 ξv − ηuu = 0, 179 2 η − uψv + u ξv − ψx + uξx = 0, 180 ψv + uξv − ηu − τt + 2ξx = 0, 181 3η + ηv − 3uψv + uψvv + 3uηu + uηuv 182 183 184 (25) Solving these equations, one can obtain (27). 123 TYPESET 160 168 154 Journal: 11071 MS: 2480 159 167 + 3uηx + u 2 ψx − u 3 ξx + 2uηxv + ηx x = 0. ψ x = Dx (ψ) − vx Dx (ξ ) − vt Dx (τ ), 158 166 c2 τ = c1 + c2 t, ξ = x + c3 , 3  1 −v  v e uc2 − 3u F + 3Fx , η=− e 3 ψ = c4 − e−v F, 2u 2 η + 3u 2 ηv + u 2 ηvv − ψt + uξt + u 3 τt t 157 164 + 3uτt − uτtv − 3uξx + 2ηxu − ξx x = 0, η = Dx (η) − u x Dx (ξ ) − u t Dx (τ ), 155 156 163 with x ∗ = x + εξ(x, t, u, v) + O(ε2 ), 137 140 ∂ ∂ ∂ ∂ ∂ + u xt + ux + vx + uxx ∂x ∂u ∂v ∂u x ∂u t ∂ ∂ + vx x + vxt + ··· (21) ∂vx ∂vt ∂ ∂ ∂ ∂ ∂ Dt = + u tt + ut + vt + u xt ∂t ∂u ∂v ∂u x ∂u t ∂ ∂ + vtt + ··· (22) + vxt ∂vx ∂vt Dx = unc Author Proof 125 Here, Dx and Dt denote the total derivative operator and are defined by pro of 121 (13) cted 120 + u 0t − u 0x x x ) = 0. orre 119 CP Disp.:2015/11/6 Pages: 12 Layout: Medium 185 186 G. Wang et al. 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 vx = u, c2 x + c3 , ψ = c4 − e−v F, 3 3 vt = 3uu x + u x x + u , (27) admits the infinite-parameter Lie group of point transformations with infinitesimal generator    ∂ 1 ∂ −v 1 2 , (30) −F uF − F XF = e ∂u ∂v here F = F(x, t), and F also satisfies Ft − Fx x x = 0. Proof: One can get the determining equations ∂ F1 = F 2, ∂x ∂ F2 ∂ F1 . = ∂x2 ∂t Then, one has τv = 0, τx = 0, ξxvv = 0, ψv + ψvv = 0, τt − 3ξx = 0, ψx + ψxv − ξx x = 0. (28) Notes 1. In the above, the vector fields X F are not recoverable point symmetries of the original equation and are potential symmetries (nonlocal). 2. The conservation laws used to construct the potential system are not unique if the there exists more than such conserved forms. Moreover, not all the potential forms may produce potential symmetries. If the respective conserved form is physical, it is purely coincidental. 4 Linearization (4) using nonlocal symmetries One of the approaches that involve invariance deal with derivative or integral dependent coefficients in the vector fields/symmetries. The latter is somewhat complicated but may be constructed via the conserved form of the differential equation often referred to as potential form. The symmetries of these systems are nonlocal/potential symmetries of the original equation (see [21–24]). DISK (32) LE 231 232 233 235 238 239 240 (33) In this way, one can get 241 242 ∂Ψ 1 ∂Ψ 1 ∂Ψ 2 − e−v = 1, ue−v ∂u ∂v ∂u 2 1 −v ∂Ψ −v ∂Ψ = 0, −e = 1. −e ∂u ∂u ue−v − e−v ∂Ψ 2 = 0, ∂v 243 (34) 244 It is easy to get a particular solution, which is Ψ 1 = −ev , Ψ 2 = −uev . 245 (35) Then, the invertible mapping µ is given by v 2 (36) which transforms (29) to the following linear system ∂w 1 = w2 , ∂z 1 ∂ 2 w2 ∂w 1 . = 2 ∂z 2 ∂z 1 246 247 v z 1 = x, z 2 = t, w = −e , w = −ue , 248 249 250 (37) Consequently, the noninvertible mapping 251 252 ∂w1 u= w2 ∂z = 11 , w1 w 123 TYPESET 229 230 237 We now identify 1 228 236 α11 = α12 = α21 = β22 = 0, β11 = ue−v , β12 = −e−v , β21 = −e−v . Solving these equations, one can obtain (30). Journal: 11071 MS: 2480 (31) ψ = e−v F 1 . ξt + 2ξx x x = 0, 225 226 234   τ = 0, ξ = 0, η = e−v u F 1 − F 2 , ψt − ψx x x = 0, (29) where F = (F 1 , F 2 ) is an arbitrary solution of the linear system of equations ξv = 0, 224 227 (26) with τ = c1 + c2 t, ξ = pro of Author Proof 191 X1 = cted 190 ∂ ∂ ∂ , X2 = , X3 = , ∂x ∂t ∂v x ∂ ∂ ∂ , X F = −e−v F , X4 = t + ∂t 3 ∂x ∂v In order to linearize (4), we use the Theorems as shown in [21,22]. In the previous section, we showed that the system (17) orre 189 Theorem 2 The potential equation (19) admits the point symmetry (20) unc 187 188 CP Disp.:2015/11/6 Pages: 12 Layout: Medium (38) 253 Nonlocal symmetry analysis and conservation laws Author Proof 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 = ∂z 2 c2 = (39) . ut = u x x x . (40) In this equation, we deal with the invariant solutions of the potential equation (19). Now we consider the following cases Case 1 X 1 . For this case, the invariants is ξ = t, and we get the trivial solution v = c1 . Case 2 X 2 . For this case, the invariants is ξ = x, and one can obtain C1 C2 (C1 C2 )2 + 8c1 C1 , c3 = − , c4 = . 2 2 2C1 (48) λ f ′ + 3 f ′ f ′′ + f ′′′ + ( f ′ )3 = 0. It should be noted that the mapping (38) is the famous Hopf–Cole transformation. Also, we note that this result is the same as in [4,7]. It should be stressed that, however, we analyzed the results from another perspective. 5 Reduction equations and similarity solutions related to potential equation 3 f ′ f ′′ + f ′′′ + ( f ′ )3 = 0. (41) For this case, we get the integrating factor  1  Λ = C1 e2 f + e f C2 ξ 2 + 2C3 ξ + 2C4 , 2 (42) where C1 , C2 , C3 and C4 are arbitrary constants. In this way, (41) can be reduced into following cases:  1 2 f  ′2 e f + 2 f ′′ = c1 . (43) 2   e f f ′2 + f ′′ = c2 . (44)   e f ξ f ′2 + ξ f ′′ − f ′ = c3 . (45)   1 f 2 ′2 (46) e ξ f + ξ 2 f ′′ − 2ξ f ′ + 2 = c4 . 2 Solving the first one, one can obtain  2 2 C1 C2 + 2 C1 2 C2 ξ + C1 2 ξ 2 + 8 c1 f = ln 4C1 . (47) 286 287 Case 3 λX 1 + X 2 . For the linear combination, the invariants is ξ = x − λt and f (ξ ), and one can have For brevity, we rewrite the (39) as follows unc 257 ∂z 13 ∂w 1 pro of ∂ 3 w1 256 Putting (47) into (44), (45), (46), one can get embeds third-order Burgers equation (4) in the following linear equation For this case, we get the integrating factor  √  Λ = C1 e2 f + C2 + C3 sin λξ √  + C4 cos λξ e f , (49) (50) Solving (51), one can get TYPESET DISK LE 290 291 292 293 296 297 298 299 300 301 302 303 304 305 √ √  √ ⎞ λx λC1 − C2 cos λx λ − c2 ⎠. λ 306 Substituting (55) into (52), (53) and (54), one can get 308 ⎛ f = ln ⎝− sin (55) √ C1 2 λ + C2 2 λ − c2 2 , c3 = C1 λ, 2λ √ c4 = −C2 λ. c1 = − 307 309 (56) 310 Case 4 X 4 . 1 One can get the invariants and function are ξ = xt − 3 and f (ξ ), and one can have 312 1 ′ ξ f + 3 f ′ f ′′ + f ′′′ + ( f ′ )3 = 0. 3 We get the integrating factor     ξ3 4 5 ,− Λ = e f C1 ξ 2 hypergeom [1], , 3 3 27 (57) 123 Journal: 11071 MS: 2480 289 295 where C1 , C2 , C3 and C4 are arbitrary constants. In this way, one can have  1 2 f  ′2 e f + 2 f ′′ + λ = c1 . (51) 2   (52) e f f ′2 + f ′′ + λ = c2 .  √ √ √ −e f cos( λξ ) λ f ′ − sin( λξ ) f ′2  √ − sin( λξ ) f ′′ = c3 . (53)  √ √ √ e f cos( λξ ) f ′2 + λ sin( λξ ) f ′  √ + cos( λξ ) f ′′ = c4 . (54) √ 288 294 cted 255 orre 254 CP Disp.:2015/11/6 Pages: 12 Layout: Medium 311 313 314 315 316 G. Wang et al. 321 322 323 324 325 pro of Author Proof 320 (59)   √  √  √ −1 2 3 −ξ 3 3 f 3 61 e (−ξ ) , 3ξ I f ′2 3ξ 3 9   √  3  −1 2 3 −ξ , − −ξ 3 I f′ 3 9   √  √ −1 2 3 −ξ 3 (60) + 3ξ I f ′′ = c2 . , 3 9   √ √    √ −1 2 3 −ξ 3 3 f 3 13 , 3I e (−ξ ) f ′2 −ξ 3 2 3ξ 3 9   √  −1 2 3 −ξ 3 , +I f ′ξ 2 3 9   √   √ −1 2 3 −ξ 3 + 3I f ′′ −ξ 3 = c3 . (61) , 3 9 Therefore, one can obtain     ξ3 10 11 ef 6 3ξ hy [3], ,− , 1680 3 3 27    3 ξ 7 8 + 84ξ 4 hy [2], , ,− f′ 3 3 27    ξ3 4 5 2 ,− f ′2 + 1680ξ hy [1], , 3 3 27    ξ3 4 5 ,− f ′′ + 1680ξ 2 hy [1], , 3 3 27    ξ3 4 5 ,− + 560ξ 3 hy [1], , 3 3 27    ξ3 7 8 ,− − 504ξ 3 hy [2], , 3 3 27 cted 319 (58) 326 Solving (60), one can get 1 ⎝ 6 × 6 −ξ 3 ×    −3 ξ 3 −3 ξ 3  ⎞−1  5 2 , +I −3 ξ 3 ⎠ dξ 3 9  √ √ √ 3 −ξ 3 I 23 , 29 −3 ξ 3 −c2 e 1 3 6 I 23 , 9 √ 1 −ξ 3 −3 ξ 3 +I  5 2 3, 9 √ −3 ξ 3  −1 I ⎝6  2 2 3, 9   √ 2   −3 ξ 3 −3 ξ 3 −3 ξ 3 −3 ξ 3 +I   +I  5 2 3, 9 √ −3 ξ 3  −1 ⎞−1 5 2 , −3 ξ 3 ⎠ 3 9 TYPESET DISK LE 332 333 338 dξ C1 339 ξ −1 dξ dξ 340 ξ −1 dξ 341 ⎞ ⎟ dξ + C2 ⎠ , 123 Journal: 11071 MS: 2480 331 dξ c2 −3 ξ   √ √ √ I 2, 3 −ξ 3 I 23 , 29 −3 ξ 3 6 3 √9 × 6 −3 ξ 3      √     −1 √ √ √ √ I 2 , 2 −3 ξ 3 5 2 3 ξ −1 dξ , 3 −ξ 3 I 23 , 29 −3 ξ 3 6 3 √9 +I −3 ξ 3 9 3  ⎛ √ 2  330 337 −3 ξ   329 336  1 3 e   e 2 2 3, 9     √     −1 √ √ √ √ I 2 , 2 −3 ξ 3 5 2 3 , 3 −ξ 3 I 23 , 29 −3 ξ 3 6 3 √9 +I −3 ξ ξ −1 dξ 3 9 3 e − 13  1 3  − I 328 335     √ ⎛ 2 2     −1 3  − 1 √3  √−ξ 3 I 2 , 2 √−3 ξ 3 6 I 3 ,√9 −3 ξ +I 5 , 2 √−3 ξ 3 ξ −1 dξ 3 3 9 3 9 ⎜ −3 ξ 3 f = ln ⎝ e ⎛ 327 334 orre 318    ξ3 4 5 ,− f′ − 3360ξ hy [1], , 3 3 27     ξ3 4 5 ,− = c1 . + 3360hy [1], , 3 3 27 unc 317  √  1 −1 2 3 −ξ 3 , + C2 I (−ξ 3 ) 6 3 9    √  −1 2 3 −ξ 3 3 − 16 , + C3 ξ I . (−ξ ) 3 9  CP Disp.:2015/11/6 Pages: 12 Layout: Medium (62) 342 343 Nonlocal symmetry analysis and conservation laws where I is the Bessel functions of the first kind and hy is the hypergeom functions. 346 6 Conservation law Author Proof 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 372 373 374 Fι (x, u, u (1) , u (2) , . . . , u (k) ) = 0, ι = 1, 2, . . . , r, ∂ 2 uι ∂u ι {u iι }, u (2) ι ∂ xi , u i j = ∂ xi ∂ x j , . . .. (63) {u iι j }, . . ., and u iι = = 1. In this paper, the total derivative operators Di are defined by (21) and (22). 2. Multipliers for PDE system (63) are undetermined functions {Λα [U ]}, which satisfies the following equation Λι [U ]Fι [U ] = Di T i [U ], (64) for a set of functions {T i [U ]} [22–24]. If U σ = u σ (x) is a solution of PDE system, then one can get the conservation law i Di T [U ] = 0, (65) of PDE system (63), where T i [U ] are the conserved densities. 3. The Euler operators defined by [22–25] ∂ ∂ − Di j + · · · Eu j = j ∂u ∂u + (−1)s Di1 · · · Dis ∂ j ∂u i1 ···is + ··· , (66) for each j = 1, 2, . . . , m. By employing the following equation, EU j (Λι [U ]Fι [U ]) ≡ 0, j = 1, . . . , N . 376 6.2 Conservation laws of (4) 379 x Given a system of r PDEs of k-order with x and u [22–25] one can get a set of multipliers. 377 T = u, 3 T = −u − 3uu x − u x x . 375 378 t (67) From the above, we suppose the conservation law is given by Dx T x + Dt T t = 0 on the solutions of (4). For the fourth-order multiplier, one can get (69) TYPESET DISK LE 383 384 385 (70) 386 6.3 Conservation laws of (40) 387 Applying the previous steps, for the fourth-order multiplier, one can get 389 x 2u 9t 2 u x x x x + 3xtu x x + 3tu x + , 2 2 Λ2 = xu + 3tu x x , Λ1 = 388 390 391 Λ3 = u, 392 Λ4 = xu x x + 3tu x x x x + u x , 393 Λ5 = u x x , 394 Λ6 = u x x x x , (71) 395 for which the respective conserved quantities are 396 u(u 2x + 6u x xt x + 9t 2 u x x x x + 6u xt ) 397 T1t = 4 , xuu x 9tuu x x x x 2 uu x x 3tu x u x x − − + 2 2 2 2 9t 2 uu t x x x 9t 2 u x u t x x 3xtu 2x x − + − 2 4 4 9t 2 u x x u t x 9t 2 u t u x x x u2 x 2 u 2x − + − + 4 4 2 4 2 2 3xtu x u t 3xtuu t x 9t u x x x + − . (72) − 4 2 2 u(xu + 3tu x x ) T2t = , 2 xu 2 3tu 2x x uu x + x − T2x = −xuu x x − 2 2 2 3tu t u x 3tuu t x + . (73) − 2 2 u2 , T3t = 2 ux T3x = −uu x x + . (74) 2 u(xu x x + 3tu x x x x + u x ) , T4t = 2 3tu x x x u t xuu t x 3tuu t x x x xu x u t + − − T4x = 2 2 2 2 T1x = − 123 Journal: 11071 MS: 2480 381 382 Therefore, one can yield the fluxes 6.1 Basic concepts i 371 Λ = C1 . In this section, using the multipliers method [23,24], we will handle the conservation law of the (4), (40) and (15). where u (1) = (68) Solving them, one can get cted 348 380 Λu x x = 0, Λu x x x = 0, Λu x x x x = 0. orre 347 Λt = 0, Λx = 0, Λu = 0, Λu x = 0, unc 344 pro of 345 CP Disp.:2015/11/6 Pages: 12 Layout: Medium 398 399 400 401 402 403 404 405 406 407 408 G. Wang et al. Author Proof 413 414 415 416 417 418 419 421 423 424 425 (76) (77) Once again applying the previous steps, for the fourthorder multiplier, one can get 420 422 (75) 6.4 Conservation laws of (15) Λ1 = e2u  2 4 9t u x + 54t 2 u 2x u x x + 6t xu 2x 18 + 36t 2 u x u x x x + 27t 2 u 2x x  + 6t xu x x + 9t 2 u x x x x + 6tu x + x 2 , e2u  4 3tu x + 18tu 2x u x x + xu 2x + 12tu x u x x x 3  + 9tu 2x x + xu x x + 3tu x x x x + u x ,   Λ3 = e2u u 4x + 6u 2x u x x + 4u x u x x x + 3u 2x x + u x x x x , Λ2 =  e2u  2 Λ4 = 3tu x + 3tu x x + x , 3 T1t = Λ6 = e , pro of 412 2u 7 Conclusions 430 In the present paper, we studied the nonlocal symmetries, similarity reduction and explicit solutions of the third-order Burgers equation. In particular, we linearized the equation to the third-order linear PDE via symmetries. Also, some explicit solutions of potential equation are constructed. Furthermore, we also give the conservation laws of the third-order Burgers equation, the reduced equation and the potential equation. In the future, we will deal with the high-order equation, such as the fourth-order Burgers equation, even more highorder equation and then with variable coefficients. Acknowledgments This work was supported by International Graduate Exchange Program of Beijing Institute of Technology (1320012341502), Research Project of China Scholarship Council (No. 201406030057), National Natural Science Foundation of China (NNSFC) (Grant No. 11171022), Graduate Student Science and Technology Innovation Activities of Beijing Institute of Technology (No. 2014cx10037). The authors thank the reviewers for the useful comments and suggestions which have been incorporated into the text. Appendix 123 TYPESET DISK LE (78) 427 429 27t 2 u 4x e2u + 2x 2 u 4 e2u + 6tu 3 u x − 27t 2 u 2 u 2x x + 9t 2 u 3 u x x x x − 2x 2 u 4 72u 4 2u 27t 2 e u 2 u 2x x − 27t 2 u 4x − 9t 2 e2u u 3 u x x x x + 12te2u u x u 4 + 54t 2 e2u u 4 u 2x x + 72u 4 2 2u 4 2 2u 3 2 18t e u u x x x x − 54t e u u x x + 18t 2 e2u u 4 u 4x − 36t 2 e2u u 3 u 4x + 54t 2 e2u u 2 u 4x + 72u 4 3 2 2u 4 2 2 6tu u x x x − 54t e uu x − 36t u u x u x x x + 81t 2 uu 2x u x x − 6t xu 2 u 2x − 6te2u u 3 u x + 72u 4 2 2u 4 2 2u 4 2 72t e u u x u x x x + 108t e u u x u x x − 162t 2 e2u u 3 u 2x u x x + 12t xe2u u 4 u 2x + 72u 4 2u 4 2 2u 3 12t xe u u x x − 72t e u u x u x x x + 162t 2 e2u u 2 u 2x u x x − 12xte2u u 3 u 2x + 72u 4 2 2u 2 2u 3 36t e u u x u x x x − 6t xe u u x x − 81t 2 e2u uu x x u 2x + 6t xe2u u 2 u 2x , + 72u 4 Journal: 11071 MS: 2480 426 for which the respective conserved quantities are given in Appendix. cted 411 Λ5 = e2u (u 2x + u x x ), orre 410 3tu t x u x x 3uu x x x 3tu x u t x x − − 2 2 2 ux uxx xu 2x x 3tu 2x x x + − . − 2 2 2 uu x x T5t = , 2 uu t x u x ut u2 + . T5x = − x x − 2 2 2 uu x x x x T6t = , 2 u x ut x x u x x ut x uu t x x x T6x = − + − 2 2 2 u 2x x x u x x x ut − . + 2 2 + unc 409 CP Disp.:2015/11/6 Pages: 12 Layout: Medium 428 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 Nonlocal symmetry analysis and conservation laws 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 pro of Author Proof 463 cted 462 orre 461 4e2u u 4 − 4u 4 + 27t 2 e2u u 2 u x u t x x + 27t 2 e2u u 2 u x x u t x + 54te2u u 2 u x u x x −72u 4 2u 4 2 2u 4 12t xe u u x x + 96te u u x u x x − 108te2u u 3 u x u x x + 36t 2 e2u u 4 u x u t x x + −72u 4 2 2u 4 4 2 2u 3 108t e u u x u x x − 18t e u u t u x x x − 54t 2 e2u u 3 u x u t x x − 54t 2 e2u u 3 u t x u x x + −72u 4 12t xe2u u 4 u 4x + 162t 2 e2u u 4 u 2x u 2x x + 36t 2 e2u u 4 u 3x u x x x − 54t 2 e2u u 3 u 2x u t x + −72u 4 2 2u 2 2 2 2u 4 54t e u u t x u x + 72t e u u t x u x x + 12t xe2u u 4 u t x − 36t 2 e2u u 3 u t u 3x + −72u 4 2 2u 2 3 2 2u u 3 54t e u u t u x − 54t e u t u x + 54t 2 uu t u x u x x − 6xtu 2 u t u x + −72u 4 2 2u 2 2 2u 9t e u u t u x x x − 27t e uu t x u 2x − 6t xe2u u 3 u t x + 4x 2 e2u u 4 u x x + −72u 4 3 2 2 6t xu u t x − 9t u u t u x x x − 27t 2 u 2 u t x u x x − 27t 2 u 2 u x u t x x − 54tu 2 u x u x x + −72u 4 2 2 2u 4 27t uu t x u x + 36te u u x x x + 24te2u u 4 u 3x + 18t 2 e2u u 4 u 6x + 27t 2 e2u u t u 3x + −72u 4 2 2u 4 2 2u 3 3 2 2x e u u x − 36te u u x + 18t e2u u 4 u 2x x x + 36te2u u 2 u 3x − 6xe2u u 3 u x + −72u 4 2u 4 2u 3 2u 4xe u u x − 18te uu x − 18te u 3 u x x x − 9t 2 e2u u 3 u t x x x + 18t 2 e2u u 4 u t x x x + −72u 4 18tu 3 u x x x + 9t 2 u 3 u t x x x + 6xu 3 u x − 27t 2 u t u 3x + 108t 2 e2u u 4 u x u x x u x x x + −72u 4 3 2 2u 18tuu x − 54t e uu t u x u x x + 6t xe2u u 2 u t u x − 108t 2 e2u u 3 u t u x u x x + −72u 4 2 2u 2 2u 108t e u u t u x u x x − 12t xe u 3 u t u x + 24t xe2u u 4 u 2x u x x + . −72u 4 −3te2u u 3 u x x x x + xe2u u 2 u 2x u x x + 9te2u u 2 u 2x x − xe2u u 3 u x x + 2xe2u u 4 u 2x T2t = 12u 4 2u 4 2u 4 2 24te u u x u x x x + 36te u u x u x x − 54te2u u 3 u 2x u x x − 24te2u u 3 u x u x x x + 12u 4 −27te2u uu x x u 2x + 12te2u u 2 u x u x x x + u 3 u x + 18te2u u 4 u 2x x − 18te2u u 3 u 2x x + 12u 4 2u 4 4 2u 3 4 2u 6te u u x − 12te u u x + 18te u 2 u 4x − 18te2u uu 4x + 6te2u u 4 u x x x x + 12u 4 2 4 2u 4 −12tu u x u x x x − 9tu x + 2e u u x + 9te2u u 4x − e2u u 3 u x + 3tu 3 u x x x x − xu 2 u 2x + 12u 4 2u 2 2 2u 4 54te u u x u x x + 2xe u u x x + 27tuu 2x u x x − 9tu 2 u 2x x + xu 3 u x x + , 12u 4 T1x = unc 460 480 123 Journal: 11071 MS: 2480 TYPESET DISK LE CP Disp.:2015/11/6 Pages: 12 Layout: Medium G. Wang et al. 485 486 487 488 489 490 491 492 493 494 495 496 unc Author Proof 484 pro of 483 cted 482 36te2u u 4 u x u x x u x x x − 18te2u uu t u x u x x − 36te2u u 3 u t u x u x x + 36te2u u 2 u t u x u x x −12u 4 3u 3 u x x x + 3uu 3x − xu t u x u 2 − 3tu 2 u t u x x x − 9tu 2 u t x u x x − 9tu 2 u t x x u x + −12u 4 2xe2u u 4 u 4x + 6tu 4 e2u u 6x + 9te2u u t u 3x + 2xe2u u 4 u 2x x + 16e2u u 4 u x u x x + −12u 4 2u 3 2u 2 −xe u u t x + 9e u u x u x x − 3te2u u 3 u t x x x − 18e2u u 3 u x u x x + 6te2u u 4 u t x x x + −12u 4 2u 4 2u 4 2xe u u t x + 24te u u t x u x x + 12te2u u 4 u x u t x x − 6te2u u t u 3 u x x x − 2xe2u u 3 u x u t + −12u 4 2 2u 3 2u 4 2 54te u u x u x x − 18te u u t x x u x − 18te2u u 3 u t x u x x + 12te2u u 4 u 3x u x x x + −12u 4 2u 2 2 2u 3 2 18te u u x u t x − 18te u u x u t x + 4xe2u u 4 u x x u 2x − 12te2u u 3 u t u 3x + −12u 4 2u 2 3 2u 3 18te u u t u x − 18te uu t u x + 18tuu t u x u x x − 9te2u uu t x u 2x + xe2u u 2 u t u x + −12u 4 2u 2 2u 2 9te u u x u t x x + 9te u u x x u t x + 3te2u u 2 u t u x x x − 3e2u u 3 u x x x − 3e2u uu 3x + −12u 4 6e2u u 4 u x x x − 6e2u u 3 u 3x + 6e2u u 2 u 3x + 4e2u u 4 u 3x + 12te2u u 4 u 2x u t x + 6te2u u 4 u 2x x x + −12u 4 2u 4 4 3 2 36te u u x u x x 3tu u t x x x − 9u u x u x x + xu 3 u t x − 9tu t u 3x + 9tuu t x u 2x + . −12u 4 2e2u u 4 u 4x + 12e2u u 4 u 2x u x x − 4e2u u 3 u 4x + 8e2u u 4 u x u x x x + 6e2u u 4 u 2x x T3t = 4u 4 2u 3 2 2u 2 4 −18e u u x u x x + 6e u u x + 2e2u u 4 u x x x x − 8e2u u 3 u x u x x x − 6e2u u 3 u 2x x + , 4u 4 18e2u u 2 u 2x u x x − 6e2u uu 4x − e2u u 3 u x x x x + 4e2u u 2 u x u x x x + 3e2u u 2 u 2x x , + 4u 4 9e2u uu 2x u x x + 3e2u uu 4x + u 3 u x x x x − 4u 2 u x u x x x − 3u 2 u 2x x + 9uu 2x u x x − 3u 4x + , 4u 4 T2x = orre 481 123 Journal: 11071 MS: 2480 TYPESET DISK LE CP Disp.:2015/11/6 Pages: 12 Layout: Medium Nonlocal symmetry analysis and conservation laws 501 502 503 504 505 506 507 508 509 510 511 512 513 514 pro of Author Proof 500 cted 499 orre 498 −3u t u 3x + u 3 u t x x x − 4e2u u 3 u t u 3x + 4e2u u 4 u t x u 2x + 12e2u u 4 u 4x u x x + 18e2u u 4 u 2x u 2x x 4u 4 2u 3 2u −6e u x u t uu x x − 12u e u x u t u x x + 12u 2 e2u u x u t u x x + 12e2u u 4 u x u x x u x x x + 4u 4 2 2u 3 2u 3 3uu t x u x − e u u t x x x + 3e u t u x + 2e2u u 4 u 2x x x + 2e2u u 4 u t x x x + 2e2u u 4 u 6x + 4u 4 2 2 2 −3u u t x x u x − 3u u t x u x x − u u t u x x x + 4e2u u 4 u 3x u x x x − 6e2u u 3 u t x u 2x + 4u 4 2u 2 3 2u 4 6e u u t u x + 4e u u x u t x x + 8e2u u 4 u t x u x x − 3e2u uu t x u 2x + 3e2u u 2 u x u t x x + 4u 4 3e2u u 2 u t x u x x + e2u u 2 u t u x x x − 6e2u uu t u 3x − 6e2u u 3 u t x x u x − 6e2u u 3 u t x u x x + 4u 4 2u 3 2u 2 2 −2e u u t u x x x + 6e u u t x u x + 6uu x u t u x x . + 4u 4 6te2u u 2 u 2x + 6te2u u 2 u x x − 6te2u uu 2x + 2xe2u u 2 − 3tue2u u x x T4t = 12u 2 2u 2 2 3te u x − 2xu + 3tuu x x − 3tu 2x + , 12u 2 2xe2u u 2 u 2x + 12te2u u 2 u 2x u x x − 3te2u uu t x + 6te2u u 2 u t x − 3tu t u x T4x = 12u 2 2u 3uu x + 3tuu t x + 3tu t u x e − 6tue2u u x u t + 6te2u u 2 u 2x x + 12u 2 2u 2 2u 4 2 4e u u x x x + 6te u x u + 2e2u u 2 u x − 3e2u uu x + . 12u 2 2e2u u 2 u 2x + 2e2u u 2 u x x − 2e2u uu 2x − e2u uu x x + e2u u 2x + uu x x − u 2x , T5t = 4u 2 2e2u u 2 u 4x + 4e2u u 2 u 2x u x x + 2e2u u 2 u 2x x + 2e2u u 2 u t x T5x = − 4u 2 2u 2u −2e uu t u x − e uu t x + e2u u x u t + uu t x − u x u t + . 4u 2 e2u − 1 , T6t = 2  e2u  2 u x + 2u x x . 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