Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 41, pp. 1–14. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu LIE SYMMETRY ANALYSIS AND CONSERVATION LAWS FOR THE (2+1)-DIMENSIONAL MIKHALËV EQUATION XINYUE LI, YONGLI ZHANG, HUIQUN ZHANG, QIULAN ZHAO Abstract. Lie symmetry analysis is applied to the (2+1)-dimensional Mikhalëv equation, which can be reduced to several (1+1)-dimensional partial differen- tial equations with constant coefficients or variable coefficients. Then we con- struct exact explicit solutions for part of the above (1+1)-dimensional partial differential equations. Finally, the conservation laws for the (2+1)-dimensional Mikhalëv equation are constructed by means of Ibragimov’s method. 1. Introduction Searching for solutions to partial differential equations (PDEs), which arise from physics, chemistry, economics and other fields, is one of the most fundamental and significant areas. A wealth of solving methods have been developed, such as the Lie symmetry analysis [5, 8, 11, 15], the homogeneous balance method [13, 18], Hirota’s bilinear method [10], the Painlev’s analysis method [6]. The Lie symmetry analysis is one of the most effective tools for solving partial differential equations and it was firstly traced back to the famous Norwegian mathematician Sophus Lie [12], who was influenced and inspired by the Galois theory founded in the early 18th century. Bluman and Cole proposed similarity theory for differential equations in 1970s [?]. Subsequently, the scope of application and theoretical depth of Lie symmetry analysis have been expanded. The (2+1)-dimensional Mikhalëv equation reads [14] uyy + uxt + uxuxy − uyuxx = 0, (1.1) which was first derived by Mikhalëv in 1992. He described a relationship between Poisson-Lie-Berezin-Kirillov brackets and the Mikhalëv system uy = vx, vy + ut + uvx − vux = 0. (1.2) Pavlov adopts the method of extended Hodograph method to study integrability of exceptional hydrodynamic type systems. The corresponding particular solution of Mikhalëv system [16] is constructed under the condition of three-component case. By constructing new integrable hydrodynamic chains, he describes and integrates all their fluid dynamics, and then extracts new (2+1) integrable hydrodynamic sys- tems from them [17]. Derchyi Wu discussed Cauchy problem of Pavlov’s equation and solve the equation by using the backscattering method [19]. Grinevich and 2010 Mathematics Subject Classification. 35Q53, 37K30;,37K40. Key words and phrases. (2+1)-dimensional Mikhalëv equation; Lie symmetry analysis; similarity reduction; conservation law; exact solution. c©2021 Texas State University. Submitted October 28, 2020. Published May 7, 2021. 1 2 X. Y. LI, H. Q. ZHANG, Y. L. ZHANG, Q. L. ZHAO EJDE-2021/41 Santini investigated nonlocality and the inverse scattering transformation for the Mikhalëv equation [9]. Dunajski [7] presented a twistor description of (1.2) and demonstrated that the solutions of (1.2) could be used to construct Lorentzian Einstein-Weyl structures in three dimensions. In this paper, we apply Lie sym- metry analysis to the (2+1)-dimensional Mikhalëv equation to present its exactly explicit solutions and construct its conservation laws. The concept of conserva- tion laws is important in nonlinear science. The famous Noether’s theorem [1] provides a systematic and effective way of determining conservation laws for Euler- Lagrange differential equations once their Noether symmetries are known. Later, researchers made various generalizations of Noether’s theorem. Among these ex- tended methods, the new conservation theorem, also called nonlocal conservation theorem, introduced by Ibragimov, is one of the most frequently used approaches. In this paper we will apply the Ibragimov’s method to construct conservation laws for the (2+1)-dimensional Mikhalëv equation. The paper is organized as follows. In Section 2, we will apply Lie symmetry analysis to the (2+1)-dimensional Mikhalëv equation. In Section 3, we will study some exact explicit solutions for the (2+1)-dimensional Mikhalëv equation based on the similarity reductions. In Section 4, the conservation laws for the (2+1)- dimensional Mikhalëv equation will be established by using Ibragimov’s method. In Section 5, we will give some conclusions and discussions. 2. Lie symmetry analysis for the (2+1)-dimensional Mikhalëv equation First of all, let us consider an one-parameter group of infinitesimal transforma- tion, x→ x+ εξ(x, y, t, u) +O(ε2), t→ t+ ετ(x, y, t, u) +O(ε2), y → y + εη(x, y, t, u) +O(ε2), u→ u+ εφ(x, y, t, u) +O(ε2), (2.1) where ε � 1 is a group parameter. The vector field associated with the above group of transformation (2.1) is presented V = ξ(x, y, t, u) ∂ ∂x + η(x, y, t, u) ∂ ∂y + τ(x, y, t, u) ∂ ∂t + φ(x, y, t, u) ∂ ∂u . (2.2) Thus, the second prolongation pr(2) V is (2) PrV = V + φx ∂ ∂ux + φy ∂ ∂uy + φyy ∂ ∂uyy + φxt ∂ ∂uxt + φxy ∂ ∂uxy + φxx ∂ ∂uxx , (2.3) where φy = Dy(φ− ξux − ηuy − τut) + ξuxy + ηuyy + τuty, φx = Dx(φ− ξux − ηuy − τut) + ξuxx + ηuyx + τutx, φyy = D2 y(φ− ξux − ηuy − τut) + ξuxyy + ηuyyy + τutyy, φxy = DyDx(φ− ξux − ηuy − τut) + ξuxxy + ηuxyy + τuxty, φxx = D2 x(φ− ξux − ηuy − τut) + ξuxxx + ηuxxy + τuxxt, φxt = DtDx(φ− ξux − ηuy − τut) + ξuxxt + ηuxyt + τuxtt, (2.4) EJDE-2021/41 (2+1)-DIMENSIONAL MIKHALËV EQUATION 3 and the operators Dx, Dy, Dt are the total derivatives with respect to x, y, t respec- tively. The determining equation of (1.1) arises from the invariance condition pr(2) V ∣∣ ∆=0 = 0, (2.5) where ∆ = uyy + uxt + uxuxy − uyuxx = 0. Furthermore, we have φyy + φxt + φxuxy + φxyux − φyuxx − φxxuy = 0, (2.6) where the coefficient functions φy, φx, φyy, φxy, φxx and φxt are determined in (2.4). Then, the forms of the coefficient functions by calculating the standard symmetry group are obtained ξ = (F1t(t) + 2c1)x− 1 2 F1tt(t)y 2 + 1 2 (−2F2t(t) + c2)y − F3(t) + c3, η = (F1t(t) + c1)y + F2(t), τ = F1(t), φ = (F1t(t) + 3c1)u− (F1tt(t)y − c2 + F2t(t))x+ 1 6 F1ttt(t)y 3 + 1 2 F2tt(t)y 2 + F3t(t)y + F4(t), (2.7) where ci (i = 1, 2, 3) are arbitrary constants and Fi(t) (i = 1, 2, 3, 4) are arbitrary functions with regard to t. For convenience, we assume that F1(t) = c4t+ c8, F2(t) = c5t+ c9, F3(t) = c6t+ c10, F4(t) = c7t+ c11. (2.8) Therefore, the Lie algebra of infinitesimal symmetries of equation (1.1) is spanned by the vector field V1 = 2x ∂ ∂x + y ∂ ∂y + 3u ∂ ∂u , V2 = 1 2 y ∂ ∂x + x ∂ ∂u , V3 = ∂ ∂x , V4 = x ∂ ∂x + y ∂ ∂y + t ∂ ∂t + u ∂ ∂u , V5 = −y ∂ ∂x + t ∂ ∂y − x ∂ ∂u , V6 = −t ∂ ∂x + y ∂ ∂u , V7 = t ∂ ∂u , V8 = ∂ ∂t , V9 = ∂ ∂y , V10 = − ∂ ∂x , V11 = ∂ ∂u . (2.9) We apply the Lie bracket [Vi, Vj ] = ViVj−VjVi, with the (i, j)-th entry representing [Vi, Vj ] to get the commutator table listed in Table 1. Table 1. Lie bracket of equation (1.1) Lie V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V1 0 −V2 −2V3 0 −V5 −2V6 −3V7 0 −V9 −2V10 −3V11 V2 V2 0 −V11 0 1 2V6 V7 0 0 1 2V10 V11 0 V3 2V3 V11 0 −V10 −V11 0 0 0 0 0 0 V4 0 0 −V3 0 0 0 −V7 −V8 −V9 −V10 −V11 V5 V5 − 1 2V6 V11 0 0 0 0 −V9 −V10 −V11 0 V6 2V6 −V7 0 0 0 0 0 −V10 −V11 0 0 V7 3V7 0 0 V7 0 0 0 −V11 0 0 0 V8 0 0 0 V8 V9 V10 V11 0 0 0 0 V9 V9 − 1 2V10 0 V9 V10 V11 0 0 0 0 0 V10 −2V3 −V11 0 V10 V11 0 0 0 0 0 0 V11 3V3 0 0 V11 0 0 0 0 0 0 0 4 X. Y. LI, H. Q. ZHANG, Y. L. ZHANG, Q. L. ZHAO EJDE-2021/41 Next, using Table 1 and the Lie series Ad(exp(εVi))Vj = Vj − ε[Vi, Vj ] + 1 2 ε2[Vi, [Vi, Vj ]]− . . . , (2.10) where ε is a real number and [·, ·] is the Lie bracket. The adjoint representation is shown in Table 2. Table 2. Adjoint representation of equation (1.1). Ad V1 V2 V3 V4 V5 V6 V1 V1 V2e ε V1e 2ε V4 V5e ε V6e 2ε V2 V1 − εV2 V2 V3 + εV11 V4 V5 − ε 2V6 + ε2 4 V7 V6 − εV7 V3 V1 − 2εV3 V2 − εV11 V3 V4 + εV10 V5 + εV11 V6 V4 V1 V2 V3e ε V4 V5 V6 V5 V1 − εV5 V2 + ε 2V6 V3 − εV11 V4 V5 V6 V6 V1 − 2εV6 V2 + εV7 V3 V4 V5 V6 V7 V1 − 3εV7 V2 V3 V4 − εV7 V5 V6 V8 V1 V2 V3 V4 − εV8 V5 − εV9 V6 + εV3 V9 V1 − εV9 V2 − 1 2 εV3 V3 V4 − εV9 V5 + εV3 V6 − εV11 V10 V1 − 2εV10 V2 + εV11 V3 V4 − εV10 V5 − εV11 V6 V11 V1e −3ε V2 V3 V4 − εV11 V5 V6 Ad V7 V8 V9 V10 V11 V1 V7e 3ε V8 V9e ε V10e 2ε V11e 3ε V2 V7 V8 V9 − ε 2V10 + ε2 4 V11 V10 − εV11 V11 V3 V7 V8 V9 V10 V11 V4 V7e ε V8e ε V9e ε V10e ε V11e ε V5 V7 V8 + εV9 + ε2 2 V10 + ε3 3! V11 V9 + εV10 + ε2 2 V11 V10 + εV11 V11 V6 V7 V8 + εV10 V9 + εV11 V10 V11 V7 V7 V8 + εV11 V9 V10 V11 V8 V7 − εV11 V8 V9 V10 V11 V9 V7 V8 V9 V10 V11 V10 V7 V8 V9 V10 V11 V11 V7 V8 V9 V10 V11 The one-parameter symmetry groups gi (1 ≤ i ≤ 11) generated by the corre- sponding infinitesimal generators Vi (1 ≤ i ≤ 11) will be obtained g1 : (x, y, t, u)→ (e2εx, eεy, t, e3εu), g2 : (x, y, t, u)→ ( 1 2 yε+ x, y, t, 1 4 yε2 + xε+ u), g3 : (x, y, t, u)→ (x+ ε, y, t, u), g4 : (x, y, t, u)→ (eεx, eεy, eεt, eεu), g5 : (x, y, t, u)→ (−ε 2 2 t− εy + x, εt+ y, t, ε3 6 t+ ε2 2 y − εx+ u), g6 : (x, y, t, u)→ (x− tε, y, t, u+ εy), g7 : (x, y, t, u)→ (x, y, t, u+ εt), g8 : (x, y, t, u)→ (x, y, t+ ε, u), g9 : (x, y, t, u)→ (x, y + ε, t, u), g10 : (x, y, t, u)→ (−ε+ x, y, t, u), g11 : (x, y, t, u)→ (x, y, t, u+ ε), (2.11) where g3, g9 are space translations, g8 is a time translation, g11 is a dependent variable translation, g4 is a scaling transformation, and g5 is a generalized Galilean transformation. According to the above one-parameter symmetry groups gi (i = 1, 2, . . . , 11), it implies that if u = f(x, y, t) is a solution of (1.1), then u(j) (1 ≤ EJDE-2021/41 (2+1)-DIMENSIONAL MIKHALËV EQUATION 5 j ≤ 11) are also solutions of (1.1) u(1) = e3εf(xe−2ε, ye−ε, t), u(2) = −ε 2 4 y + xε+ f(x− ε 2 y, y, t), u(3) = f(x− ε, y, t), u(4) = eεf(xe−ε, ye−ε, te−ε), u(5) = −εx− ε2 2 y + ε3 6 t+ f(x+ εy − ε2 2 t, y − εt, t), u(6) = εy + f(x+ tε, y, t), u(7) = εt+ f(x, y, t), u(8) = f(x, y, t− ε), u(9) = f(x, y − ε, t), u(10) = f(x+ ε, y, t), u(11) = ε+ f(x, y, t), (2.12) where ε is an arbitrary real number. 3. Similarity reductions and exact solutions The similarity reductions of the given equations can be identified by solving the characteristic equation dt F1(t) = dx (F1t(t) + 2c1)x− 1 2F1tt(t)y2 + 1 2 (−2F2t(t) + c2)y − F3(t) + c3 = dy (F1t(t) + c1) · y + F2(t) = ( (F1t(t) + 3c1)u− (F1tt(t)y − c2 + F2t(t))x+ 1 6 F1ttt(t)y 3 + 1 2 F2tt(t)y 2 + F3t(t)y + F4(t) )−1 du. (3.1) Here, we give the corresponding similarity reduction and provide some exact solu- tions of the original equation (1.1). Case 1. Taking F1(t) = 0, F2(t) = 0, F3(t) = 0, F4(t) = 0, c1 6= 0, c2 = 0, c3 = 0 in (3.2) yields dt 0 = dx 2c1x = dy c1y = du 3c1u , (3.2) where the expression dt 0 means that the first integral of time t is a constant. Solving (3.2) provides v = t, w = yx−1/2, u = f(v, w)x3/2. (3.3) Substituting (3.3) into (1.1), we obtain the following (1+1)-dimensional nonlinear PDE with variable coefficients 4fww + 6fv − 2wfwv + 3ffw − 3wffww + wf2 w = 0. (3.4) Case 2. If we take F1(t) = 0, F2(t) = 0, F3(t) = 0, F4(t) = 0, c1 = 0, c2 6= 0, c3 = 0 in (3.2), then we obtain dt 0 = dx 1 2c2y = dy 0 = du c2x . (3.5) Solving this equation, we obtain the similarity variables and the group-invariant solution v = t, w = y, u = f(w, v) + x2 y . (3.6) 6 X. Y. LI, H. Q. ZHANG, Y. L. ZHANG, Q. L. ZHAO EJDE-2021/41 Substituting (3.6) into (1.1), we derive reduced PDE with variable coefficients fww − 2w−1fw = 0. (3.7) Solving this equation, we obtain f = F2(v)w3 + F1(v), (3.8) where F1(v), F2(v) are arbitrary functions of v. Based on (3.6) and (3.8), we obtain the exact solution of (1.1) u = F2(t)y3 + F1(t) + x2 y , (3.9) where F1(t), F2(t) are arbitrary functions of t. Case 3. Letting F1(t) = d1, F2(t) = d2, F3(t) = 0, F4(t) = d4, c1 = 0, c2 = 0, c3 6= 0, where d1, d2, and d4 are nonzero constants and we have dt d1 = dx c3 = dy d2 = du d4 . (3.10) Solving (3.10), we obtain the similarity variables and group-invariant solution v = d2x− c3y, w = d1x− c3t, u = d4 c3 x+ f(w, v). (3.11) Substituting (3.11) into (1.1) yields (c23 + d2d4)fvv − c3d1fww + (d1d4 + c3d2)fvw + d2 1c3fwfwv − d2 1c3fvfww − d1d2c3fvfwv + d1d2c3fwfvv = 0. (3.12) Letting d1 = d2 = d4 = c3 = 1, we obtain a reduced equation − fww + 2fvv + fwfwv − fvfww − fvfwv + fwfvv = 0. (3.13) Solving (3.13), the result is obtained f = k3 tanh ( − 1 2 k2v + k2w + k1 )3 + k4 tanh ( − 1 2 k2v + k2w + k1 ) + k5, (3.14) where k1, k2, k3, k4, k5 are arbitrary constants. Combining (3.11) and (3.14), one can obtain u = x+ k3 tanh (k2 2 x+ k2 2 y − k2t+ k1 )3 + k4 tanh (k2 2 x+ k2 2 y − k2t+ k1 ) + k5, (3.15) where k1, k2, k3, k4, and k5 are arbitrary constants. Case 4. If we take F1(t) = F3(t) = 0, F2(t) = d2, F4(t) = t, c1 = c2 = 0, c3 6= 0 where d2 and c3 are nonzero constants. The defining equation is dt 0 = dx c3 = dy d2 = du t . (3.16) Solving (3.16), we can obtain the similarity variables and the group-invariant solu- tion v = t, w = d2x− c3y, u = t c3 x+ f(w, v). (3.17) Substituting (3.17) into (1.1), we obtain the following reduced PDE with variable coefficients c23fww + d2fvv − d2vfwv − c3d2 2 ( fvfw,v − fwfvv ) + 1 c3 = 0 (3.18) EJDE-2021/41 (2+1)-DIMENSIONAL MIKHALËV EQUATION 7 Case 5. Taking F1(t) = d1, F2(t) = d2, F3(t) = d3, F4(t) = 0, c1 = 0, c2 = 0, c3 6= 0, where d1 and d2, d3 are nonzero constants, the characteristic equation becomes dt d1 = dx −d3 + c3 = dy d2 = du 0 . (3.19) Solving this equation, we obtain the corresponding similarity variables and a group- invariant solution v = d2t− d1y, w = (c3 − d3)t− d1x, u = f(w, v). (3.20) Substituting (3.20) into (1.1), we have d1fvv + (d3 − c3)fww − d2fwv − d2 1fwfvw + d2 1fvfww = 0. (3.21) Solving this equation, we obtain f = k7 tanh (1 2 ( d2 + √ −4d1d3 + 4d1c3 + d2 2 ) k2v d1 + k2w + k1 )3 + k5 tanh (1 2 ( d2 + √ −4d1d3 + 4d1c3 + d2 2 ) k2v d1 + k2w + k1 ) + k4, (3.22) where k1, k2, k4, k5, k7 are arbitrary constants. Combining (3.20) and (3.22), we obtain the exact solution of (1.1), u = k3 tanh (1 2 (d2 + √ −4d1d3 + 4d1c3 + d2 2)k2(d2t− d1y) d1 + k2[(c3 − d3)t− d1x] + k1 )3 + k5 tanh (1 2 (d2 + √ −4d1d3 + 4d1c3 + d2 2)k2(d2t− d1y) d1 + k2[(c3 − d3)t− d1x] + k1 ) + k4, (3.23) where k1, k2, k3, k4, k5 are arbitrary constants. Case 6. Setting F1(t) = 0, F2(t) = 0, F3(t) = 0, F4(t) = 0, c1 6= 0, c2 = 0, c3 = 0, the characteristic equation is dt 0 = dx 2c1x = dy c1y = du 3c1u . (3.24) Solving this equation, the similarity variables and a group-invariant solution can be obtained. They are v = xy−2, w = t, u = y3f(w, v), (3.25) Substituting (3.25) into (1.1), it is obvious that the reduced nonlinear PDE with variable coefficients is 6f − 6vfv + 4v2fvv + fvw + f2 v − 3ffvv = 0. (3.26) Case 7. Letting F1(t) = 0, F2(t) = d2, F3(t) = d3, F4(t) = d4, c1 = 0, c2 = 0, c3 6= 0, where d2, d3, d4 are nonzero constants, then the characteristic equation becomes dt 0 = dx −d3 + c3 = dy d2 = du d4 . (3.27) 8 X. Y. LI, H. Q. ZHANG, Y. L. ZHANG, Q. L. ZHAO EJDE-2021/41 Solving this equation, we obtain v = (c3 − d3)y − d2x, w = t, u = d4 d2 y + f(w, v). (3.28) Substituting (3.28) into (1.1) yields a reduced PDE of (1.1) with constant coeffi- cients ( (c3 − d3)2 − d2d4 ) fvv − d2fvw = 0. (3.29) Case 8. Letting F1(t) = c4t + c5, F2(t) = 0, F3(t) = 0, F4(t) = 0, c1 = 0, c2 = 0, c3 = 0, c4 6= 0, c5 6= 0 in (3.1), then we obtain dt c4t+ c5 = dx c4x = dy c4y = du c4u . (3.30) Solving (3.30), we can get the similarity variables and the group-invariant solution v = xy−1, w = (c4t+ c5)x−1, u = f(v, w)x. (3.31) Substituting (3.31) into (1.1), it is easily to obtain the reduced nonlinear PDE with variable coefficients through a straight calculation 2v3fv + v4fvv + c4vfvw − c4wfww − 2v2ffv + wv2ffwv − v3ffvv + 2wv2fwfv − v2w2fwfwv − wv3fwfvv − wv3fvfwv + w2v2fvfww = 0. (3.32) (a) (b) (c) Figure 1. Propagation of the exact solutions of (1.1) via (3.15) with parameters: k1 = 4, k2 = 1, k3 = 3, k4 = −3, k5 = 0. Perspective of the solutions with: (a) t = 0, (b) x = 0, (c) y = 0. Case 9. If we set F1(t) = c4, F2(t) = c5, F3(t) = 0, F4(t) = 0, c1 = 0, c2 = 0, c3 = 0, c4 6= 0, c5 6= 0, the defining equation is dt c4 = dx 0 = dy c5 = du 0 . (3.33) Solving this equation, we obtain the similarity variables and the group-invariant solution v = c5t− c4y, w = x, u = f(w, v). (3.34) Then, we obtain the reduced nonlinear PDE with constant coefficients c24fvv + c5fwv − c4fwfwv + c4fvfww = 0. (3.35) EJDE-2021/41 (2+1)-DIMENSIONAL MIKHALËV EQUATION 9 (a) (b) (c) Figure 2. Propagation of the exact solutions of (1.1) via (3.35) with parameters: k1 = 0, k2 = −1, k3 = 4, k4 = 2, k5 = 1, c4 = 1, c5 = 2. Perspective of the solutions with: (a) t = 0, (b) x = 0, (c) y = 0. (a) (b) (c) (d) (e) (f) Figure 3. Propagation of the exact solutions of (1.1) via (3.48) with parameters: k1 = 1, k2 = 4, k3 = −1, k4 = 2, k5 = 1, c4 = −2, c5 = 1, c6 = 2. Perspective of the solutions with: (a) t = 0, (b) x = 0, (c) y = 0. Wave propagation pattern of the wave along with: (d) the t axis, (e) the x axis, (f) the y axis. Solving this equation gives f = k2 tanh ( k3v − k3c 2 4 c5 w + k1 )3 + k5 tanh ( k3v − k3c 2 4 c5 w + k1 ) + k4, (3.36) 10 X. Y. LI, H. Q. ZHANG, Y. L. ZHANG, Q. L. ZHAO EJDE-2021/41 where k1, k2, k3, k4, k5 are arbitrary constants. Combining (3.34) and (3.36), the exact solution of (1.1) is presented, u = k2 tanh ( k3(−c4y+c5t)− k3c 2 4x c5 + k1 )3 + k5 tanh ( k3(−c4y+c5t)− k3c 2 4x c5 + k1 ) + k4, (3.37) where k1, k2, k3, k4, k5 are arbitrary constants. Case 10. If taking F1(t) = 0, F2(t) = t, F3(t) = 0, F4(t) = 0, c1 = 0, c2 = 0, c3 = 0 in (3.1), then the characteristic equation becomes dt 0 = dx −y = dy t = du −x . (3.38) Solving this equation, the similarity variables and the group-invariant solution are presented as follows v = tx+ 1 2 y2, w = t, u = t−1f(w, v) + 1 6 t−2y3 − t−1xy − 1 2 t−2y3. (3.39) Then, we obtain the PDE with variable coefficients wfvw + 2vfvv = 0. (3.40) Solving (3.40), we obtain f = F2(w) + F1 ( v w2 ) w2, (3.41) where F1( v w2 ), F2(w) are arbitrary functions of variables v and w. Combining (3.39) and (3.41), we obtain the exact solution of (1.1) u = F2(t)t−1 + F1 (2tx+ y2 2t2 ) t− xyt−1 − 1 3 y3t−2, (3.42) where F1 and F2 are arbitrary functions of variables x, t and y. Case 11. Taking F1(t) = c4, F2(t) = t, F3(t) = 0, F4(t) = 0, c1 = 0, c2 = 0, c4 6= 0 in (3.1) yields dt c4 = dx −y = dy t = du −x . (3.43) Solving (3.43), we obtain the similarity variables and the group-invariant solution v = t3 3c4 − yt− c4x, w = t2 2 − c4y, u = f(w, v) + v c24 t+ t4 24c34 − wt2 2c34 . (3.44) Substituting (3.44) into (1.1) yields c24fww − wfvv − c34fvfvw + c34fwfvv − 1 c4 = 0. (3.45) Case 12. Letting F1(t) = c4, F2(t) = 0, F3(t) = c5t+ c6, F4(t) = 0, c1 = 0, c2 = 0, c3 = 0, c4 6= 0, c5 6= 0, c6 6= 0 in (3.1), we can obtain dt c4 = dx −c5t− c6 = dy 0 = du c5y . (3.46) Solving this equation we obtain the similarity variables and the group-invariant solution v = −c5 2 t2 − c6t− c4x, w = y, u = f(w, v) + c5 c4 yt. (3.47) EJDE-2021/41 (2+1)-DIMENSIONAL MIKHALËV EQUATION 11 Substituting (3.47) into (1.1) yields nonlinear PDE with constant coefficients fww + c4c6fvv + c24fvfvw − c24fwfvv = 0. (3.48) Solving this equation we have f = k3 tanh ( − k2v√ −c4c6 + k2w + k1 )3 + k5 tanh ( − k2v√ −c4c6 + k2w + k1 ) + k4, (3.49) where k1, k2, k3, k4, k5 are arbitrary constants. Combining (3.47) and (3.48), we obtain the exact solutions of (1.1) u = k3 tanh ( − k2(− c52 t 2 − c6t− c4x) √ −c4c6 + k2y + k1 )3 + k5 tanh ( − k2(− c52 t 2 − c6t− c4x) √ −c4c6 + k2y + k1 ) + k4 + c5 c4 yt, (3.50) where k1, k2, k3, k4, k5 are arbitrary constants. The illustrative examples of exact solutions to case 3, case 9 and case 12 are presented graphically. 4. Construction of conservation laws In this section, we will construct conservation laws for the (2+1)-dimensional Mikhalëv equation (1.1). The formal Lagrangian form of (1.1) is present by ψ = v(uyy + uxt + uxuxy − uyuxx). (4.1) Furthermore, the adjoint equation is written in this form F ∗ = −2vxuxy + 2vyuxx + vxyux − vxxuy + vyy + vxt = 0. (4.2) Let us consider a Lie point symmetry generator, X = 7x ∂ ∂x + 6y ∂ ∂y + 5t ∂ ∂t + 8u ∂ ∂u . (4.3) Thus, the extension of (4.3) to v has the form Y = 7x ∂ ∂x + 6y ∂ ∂y + 5t ∂ ∂t + 8u ∂ ∂u − 14v ∂ ∂v . (4.4) Theorem 4.1. Any infinitesimal symmetry X = ξi(x, u, u(1), . . .) ∂ ∂xi + ηα(x, u, u(1), . . .) ∂ ∂uα (4.5) of a nonlinearly self-adjoint system to differential equation (1.1) produces a conser- vation law for this system, [Di(C i)](1.1) = 0 (4.6) The components of the conserved vector are given by Ci = ξiψ +Wα [ ∂ψ ∂uαi −Dj ( ∂ψ ∂uαij ) +DjDk( ∂ψ ∂uαijk )− · · · ] +Dj(W α) [ ∂ψ ∂uαij −Dk ( ∂ψ ∂uαijk ) + · · · ] +DjDk(Wα) [ ∂ψ ∂uαijk − · · · ] , (4.7) where Wα = ηα − ξjuαj , (4.8) 12 X. Y. LI, H. Q. ZHANG, Y. L. ZHANG, Q. L. ZHAO EJDE-2021/41 and ψ is the formal Lagrangian. In this case, we obtain the conservation laws Dx(C1) +Dt(C 2) +Dy(C3) = 0, (4.9) with the components of conserved vector C = (C1, C2, C3), where C1 = 7xv(uyy + uxt + uxuxy − uyuxx) + (3ut − 7xuxt − 5tutt − 6yuty)v − (ux − 7xuxx − 5tuxt − 6yuxy)(vuy) + (2uy − 7xuxy − 5tuty − 6yuyy)(vux) + (8u− 7xux − 5tut − 6yuy)(vuxy + vxuy − vyux − vt), (4.10) C2 = 5tv(uyy + uxt + uxuxy − uyuxx)− 8uvx + 7xuxvx + 5tutvx + 6yuyvx + vux − 7xvuxx − 5tvuxt − 6yvuxy, (4.11) C3 = 6yv(uyy + uxt + uxuxy − uyuxx) + (2uy − 7xuxy − 5tuyt − 6yuyy)(v) + (8u− 7xux − 5tut − 6yuy)(−2vuxx + vy − vxux) + (ux − 7xuxx − 5tutx − 6yuyx)(vux). (4.12) This conserved vector includes an arbitrary solution v of the adjoint equation F ∗ = −2vxuxy +2vyuxx+vxyux−vxxuy +vyy +vxt = 0, and it can derive infinitely many conservation laws. For convenience, let us take v = t, then the components of the conserved vector are simplified to the form C1 = 7xt(uyy + uxt + uxuxy − uyuxx) + (8u− 7xux − 5tut − 6yuy)(tuxy) − (ux − 7xuxx − 5tuxt − 6yuxy)(tuy) + (2uy − 7xuxy − 5tuty − 6yuyy)(tux) + (3ut − 7xuxt − 5tutt − 6yuty)t, (4.13) C2 = 5t2(uyy + uxt + uxuxy − uyuxx) + tux − 7xtuxx − 5t2uxt − 6ytuxy, (4.14) C3 = 6yt(uyy + uxt + uxuxy − uyuxx) + (8u− 7xux − 5tut − 6yuy)(−2tuxx) + (t)(2uy − 7xuxy − 5tuyt − 6yuyy) + (ux − 7xuxx − 5tutx − 6yuyx)(tux). (4.15) Then, we consider the point symmetry for the (2+1)-dimensional Mikhalëv equa- tion (1.1), X = ∂ ∂y + ∂ ∂t , (4.16) and we obtain the conserved vector C1 = (−uy − ut)(vxuy + vuxy − vyux − vt) + (uxy + uxt)(vuy) − (uyy + uyt)(vux)− v(uty + utt), (4.17) C2 = (uy + ut)(vx) + (uyy + uxuxy − uyuxx − uyx)(v), (4.18) C3 = uyvy + utvy + (uxuy + uxut)vx + (uyuxx + 2utuxx − uxuxt − uty + uxt)v. (4.19) Similarly, we take v = −1 and get simplified conserved vector C1 = utuxy − uxtuy + uyt + utt + (uty + uyy)ux, (4.20) C2 = uxt + uyx, (4.21) C3 = −uyuxx − 2utuxx + uxuxt + uty − uxt. (4.22) EJDE-2021/41 (2+1)-DIMENSIONAL MIKHALËV EQUATION 13 We study a point symmetry for the (2+1)-dimensional Mikhalëv equation (1.1) X = ∂ ∂x , (4.23) and the conserved vector C1 = (−2uxuxy − uxt + uyuxx)v + uxvt + u2 xvy − uxuyvx, (4.24) C2 = uxvx − uxxv, (4.25) C3 = (uxuxx − uxy)v + u2 xvx + uxvy. (4.26) Taking the solution v = −1 of (4.2), the following vector can be obtained C1 = (2uxuxy + uxt − uyuxx) = uxuxy − uyy, (4.27) C2 = uxx, (4.28) C3 = (uxuxx + uxy − 2uxuxx) = −uxuxx + uxy. (4.29) Specially, the conservation laws for the vector (4.27)-(4.29) have the form Dx(C1) +Dt(C 2) +Dy(C3) = uxuxxy + 2uxxt − uyuxxx + uxyy = (F )x + uxxt = 0. (4.30) 5. Conclusions and discussions In this paper, we have presented the Lie symmetry analysis for the (2+1)- dimensional Mikhalëv equation and applied the Ibragimov’s method to construct its conservation laws. We have taken F1(t), F2(t), F3(t) and F4(t) as linear func- tions and systematically shown the Lie bracket and the adjoint representation to the Mikhalëv equation. Compared with [2], we have obtained several partial dif- ferential equations with variable coefficients, such as, (3.7), (3.18), (3.40) and get their solutions. Meanwhile, we also have derived the solutions of partial differential equations with constant coefficients such as equations (3.12), (3.21), (3.35), (3.48). Illustrative examples of solutions for the (2+1)-dimensional Mikhalëv equation are exhibited. Acknowledgments. This work was supported by the National Nature Science Foundation of China (No. 11701334) and the “Jingying” Project of Shandong Uni- versity of Science and Technology. References [1] E. D. Avdonina, N. 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Xinyue Li College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao 266590, China Email address: xyli@sdust.edu.cn Yongli Zhang Department of Mathematics and Statistics, Qingdao University, Qingdao, Shandong 266071, China Email address: zhangyonglisumili@163.com Huiqun Zhang Department of Mathematics and Statistics, Qingdao University, Qingdao, Shandong 266071, China Email address: qddxzhanghq@163.com Qiulan Zhao (corresponding author) College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao 266590, China Email address: qlzhao@sdust.edu.cn 1. Introduction 2. Lie symmetry analysis for the (2+1)-dimensional Mikhalëv equation 3. Similarity reductions and exact solutions 4. Construction of conservation laws 5. Conclusions and discussions Acknowledgments References