Constructing two-dimensional optimal system of the group invariant solutions

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1 Constructing two-dimensional optimal system of the group invariant solutions Xiaorui Hu Yuqi Li and Yong Chen Citation: Journal of Mathematical Physics (2016); doi: / View online: View Table of Contents: Published by the AIP Publishing Articles you may be interested in Numerical solution of two-dimensional integral-algebraic systems using Legendre functions AIP Conf. Proc (2012); / Exploding solutions of the complex two-dimensional Burgers equations: Computer simulations J. Math. Phys (2012); / Singularities of complex-valued solutions of the two-dimensional Burgers system J. Math. Phys (2010); / Vortex models based on similarity solutions of the two-dimensional diffusion equation Phys. Fluids (2004); / A class of steady solutions to two-dimensional free convection Phys. Fluids (2003); /

2 JOURNAL OF MATHEMATICAL PHYSICS (2016) Constructing two-dimensional optimal system of the group invariant solutions Xiaorui Hu 1a) Yuqi Li 2 and Yong Chen 2b) 1 Department of Applied Mathematics Zhejiang University of Technology Hangzhou People s Republic of China 2 Shanghai Key Laboratory of Trustworthy Computing East China Normal University Shanghai People s Republic of China (Received 9 February 2015; accepted 3 February 2016; published online 18 February 2016) To search for inequivalent group invariant solutions of two-dimensional optimal system a direct and systematic approach is established which is based on commutator relations adjoint matrix and the invariants. The details of computing all the invariants for two-dimensional algebra are presented which is shown more complex than that of one-dimensional algebra. The optimality of two-dimensional optimal systems is shown clearly for each step of the algorithm with no further proof. To leave the algorithm clear each stage is illustrated with a couple of examples: the heat equation and the Noviov equation. Finally two-dimensional optimal system of the (2+1)-dimensional Navier-Stoes (NS) equation is found and used to generate intrinsically different reduced ordinary differential equations. Some interesting explicit solutions of the NS equation are provided. C 2016 AIP Publishing LLC. [ I. INTRODUCTION The study of group-invariant solutions of differential equations plays an important role in mathematics and physics. The machinery of Lie group theory provides a systematic method to search for these special group invariant solutions. For a system of differential equations with n variables any of its m-dimensional (m < n) symmetry subgroup can transform it into a system of differential equations with n m variables which is generally easier to solve than the original system. By solving these reduced equations rich group-invariant solutions are found. For two group-invariant solutions one may connect them with some group transformation and in this case one calls them equivalent. Naturally it is a significant job to find these inequivalent branches of group-invariant solutions which leads to the concept of the optimal systems. For the classification of group-invariant solutions it is more convenient to wor in the space of Lie algebra and this problem reduces to the problem of finding an optimal system of subalgebras under the adjoint representation. The adjoint representation of a Lie group on its Lie algebra was nown to Lie. The construction of one-dimensional optimal systems of Lie algebra was demonstrated by Ovsianniov 1 using a global matrix for the adjoint transformation. This is also the technique used by Galas 2 and Ibragimov. 3 Then Olver 4 used a slightly different and elegant technique for one-dimensional optimal system which is based on commutator table and adjoint table and presented detailed instructions on the KdV equation and the heat equation. For two-dimensional optimal systems Ovsianniov setched the construction by showing a simple example. Galas refined Ovsianniov s method by removing equivalent subalgebras for the solvable algebra and he also discussed the problem of a nonsolvable algebra which is generally harder. In Ref. 9 the details for constructing two-dimensional optimal systems were shown for the three-dimensional one-temperature hydrodynamic equations. In a fundamental series of papers Patera et al. 5 8 developed a different and a) Electronic mail: lansexiaoer@163.com b) Electronic mail: ychen@sei.ecnu.edu.cn /2016/57(2)/023518/22/$ AIP Publishing LLC

3 Hu Li and Chen J. Math. Phys (2016) powerful method to classify subalgebras and many optimal systems of important Lie algebras arising in mathematical physics are obtained. In this paper we devote ourselves to investigating two-dimensional optimal system of invariant solutions which simultaneously imply two inds of invariance of the differential systems. Using two-dimensional optimal system of a Lie algebra systems of differential equations with n variables would be reduced to some inequivalent systems with n 2 variables. Especially for a given system of differential equations with three variables one-dimensional optimal system can only reduce it to several systems of differential equations with two variables while two-dimensional optimal system would directly lead it to ordinary differential equations which are more easily solved in principle than those partial differential equations. Hence it is a meaningful job to consider two-dimensional optimal systems independently. In almost all of the existing literatures one-dimensional optimal system is required for the calculation of two-dimensional optimal system which taes too much wor and one needs master rich algebraic nowledge before setting about the operation. The purpose of this paper is to introduce a direct and systematic method for constructing two-dimensional optimal system which starts from the Lie algebra itself and only depends on fragments of the theory of Lie algebras without the prior one-dimensional optimal system. For the case of one-dimensional optimal system Olver pointed out that the Killing form of the Lie algebra as an invariant for the adjoint representation is very important since it places restrictions on how far one can expect to simplify the Lie algebra. Chou et al introduced more numerical invariants (which are different from common invariants such as the Casimir operator harmonics and rational invariants) to demonstrate the inequivalence among different elements of the optimal system. However to the best of our nowledge in spite of the importance of the common invariants for the Lie algebra there are few literatures to use more invariants except the Killing form in the process of constructing optimal systems. For this in our early paper 13 we introduced a direct and valid method to compute all the general invariants for a given one-dimensional Lie algebra and then made the best of them to construct one-dimensional optimal system. On the basis of all the invariants the new method can both guarantee the comprehensiveness and the inequivalence of the one-dimensional optimal system. Here we develop the ideas to two-dimensional optimal systems of invariant solutions. The layout of this paper is as follows. Section II provides a theoretical bacground on Lie algebras and all machinery needed to develop the algorithm. In Section III a direct and systematic algorithm is proposed for constructing two-dimensional optimal system of a general symmetry algebra. Since the realization of our new algorithm builds on different invariants a valid method for computing the invariants of two-dimensional subalgebras is also given in this section. To leave our algorithm clear we would illustrate each stage with two examples i.e. the heat equation and the Noviov equation. In Section IV the two-dimensional optimal system of (2+1)-dimensional Navier-Stoes equation is presented and all the corresponding reduced ordinary differential equations with some interesting exact group invariant solutions are obtained. Finally a brief conclusion is given in Section V. II. THEORETICAL BACKGROUND ON LIE ALGEBRA Consider an n-dimensional Lie algebra G of a differential system with p independent variables {x 1 x 2... x p } and q dependent variables {u 1 u 2... u q } which is generated by n vector fields {v 1 v 2... v n }. The corresponding n-parameter symmetry group of G is denoted as G which is the collections of transformations ( n ( x 1... x p ũ 1... ũ q ) = exp a i v i )(x 1... x p u 1... u q ) (1) for all allowed values of the group parameters. The Lie bracet [v i v j ] = v i v j v j v i is the commutator of two of the differential operators. The complete information of the group structure is contained by

4 Hu Li and Chen J. Math. Phys (2016) [v i v j ] = n Ci j v (2) where the Ci j s are called structure constants. The group invariant solutions are large classes of special explicit solutions which are characterized by their invariance under some symmetry group of the system of partial differentials equations. Let H G be an s-parameter subgroup. An H-invariant solution can be transformed into another one by the elements g G not belonging the subgroup H. That is to say two group invariant solutions are essentially different if it is impossible to connect them with any group transformation in (1). In fact if ψ is an H-invariant solution ψ = g ψ is a H-invariant solution with H = ghg 1 = {ghg 1 g G h H}. This group H is called the conjugate subgroup to H under G. For each g G group conjugation K g (h) ghg 1 h G determines a global group action of G on itself. Then the corresponding differential dk g determines a linear map on the Lie algebra G of G called the adjoint representation =1 Ad g (w) dk g (w) (3) for w being any vector field form G. Furthermore if w generates the one-parameter subgroup H = {exp(ϵw) : ϵ R} then Ad g (w) generates the conjugate one-parameter subgroup Ad g H = ghg 1. Let the group element g be generated by the vector field v seen g = exp(ϵv). More simply adjoint representation (3) can be expressed through commutators as Ad exp(ϵv) (w) = w ϵ[v w] + 1 2! ϵ 2 [v[v w]] 1 3! ϵ 3 [v[v[v w]]] +. (4) The infinitesimal adjoint action of (4) is ad v (w) d ( ) Ad exp(ϵv) (w) = [wv]. (5) dϵ ϵ=0 The Killing form is a bilinear form defined on Lie algebra G by K(v w) = trace(ad v ad w ). (6) A real function φ defined on a Lie algebra G is called an invariant if φ(ad g (w)) = φ(w) for all w in G and g in the Lie group G. By the definition of the Killing form we have for all w G and g G. Therefore the function K(Ad g (w) Ad g (w)) = K(w w) (7) f (w) = K(w w) (8) is invariant under the adjoint action. It was shown in our previous paper 13 that the invariants play a very important role in the construction of one-dimensional optimal system. For a given Lie algebra G a family of r-dimensional subalgebras {g α } α A forms an r- parameter optimal system named as O r if (1) any r-dimensional subalgebra is equivalent to some g α and (2) g α and g β are inequivalent for distinct α and β. Each member g α O r is a collection of r linear combinations of generators. In this paper we focus on constructing two-dimensional optimal system O 2. Let G(v w) G( n a i v i n b i v i ) be a general two-dimensional algebra which remains closed under commutation. In G(v w) two subalgebras {w 1 w 2 } and {w 1 w 2 } are called equivalent if one can find some transformation g G and some constants { } so that w 1 = 1Ad g (w 1 ) + 2 Ad g (w 2 ) w 2 = 3Ad g (w 1 ) + 4 Ad g (w 2 ). (9) Since w 1 and w 2 are linearly independent it requires in (9) or else w 1 = cw 2. On the one hand for the above equivalent two-dimensional subalgebras {w 1 w 2 } and {w 1 w 2 } there is

5 Hu Li and Chen J. Math. Phys (2016) [w 1 w 2 ] = [ 1Ad g (w 1 ) + 2 Ad g (w 2 ) 3 Ad g (w 1 ) + 4 Ad g (w 2 )] = ( )[Ad g (w 1 ) Ad g (w 2 )] = ( )[g(w 1 )g 1 g(w 2 )g 1 ] = ( )g([w 1 w 2 ])g 1. It is clear that [w 1 w 2 ] = 0 if and only if [w 1 w 2 ] = 0; [w 1 w 2 ] 0 if and only if [w 1 w 2 ] 0. On the other hand for any given two-dimensional subalgebra {w 1 w 2 } with [w 1 w 2 ] = λw 1 + µw 2 one can easily find an equivalent one {ŵ 1 ŵ 2 } so that [ŵ 1 ŵ 2 ] = 0 or [ŵ 1 ŵ 2 ] = ŵ 1. Hence to find all the inequivalent elements in the optimal system O 2 without loss of generality we require each member {v w} O 2 satisfy [v w] = 0 or [v w] = v. For the latter case we give out the following remar. Remar 1. If two subalgebras g α = {w 1 w 2 } and g α = {w 1 w 2 } with [w 1 w 2 ] = w 1 and [w 1 w 2 ] = w 1 are equivalent in the form of (9) there must be 2 = 0 and 4 = 1. Proof. If we mae [w 1 w 2 ] = w 1 and [w 1 w 2 ] = w 1 Eq. (10) become [w 1 w 2 ] = ( )g(w 1 )g 1 = ( )Ad g (w 1 ) = w 1 = 1Ad g (w 1 ) + 2 Ad g (w 2 ). (11) For the independence of Ad g (w 1 ) and Ad g (w 2 ) there must be 2 = 0 and 4 = 1. (10) III. A GENERAL ALGORITHM FOR CONSTRUCTING TWO-DIMENSIONAL OPTIMAL SYSTEM In this section we will demonstrate how to construct the adjoint transformation matrix and invariants on the refined two-dimensional algebra and apply them to present an algorithm for two-dimensional optimal system stage by stage. Each step is illustrated by two examples the heat and Noviov equations. A. Construction of the refined two-dimensional algebra To find out all the inequivalent elements in the two-dimensional optimal system O 2 which represent the respective equivalent classes we first require each {v w} O 2 satisfy Let [v w] = δv where δ 01. (12) v = n a i v i w = Requirement (12) will provide a set of restrictive equations for a i and b i. n b i v i. (13) 1. Refined two-dimensional algebra of the heat equation The equation for the conduction of heat in a one-dimensional road is written as u t = u x x. (14) The Lie algebra of infinitesimal symmetries for this equation is spanned by six vector fields v 1 = x v 2 = t v 3 = u u v 4 = x x + 2t t v 5 = 2t x xu u v 6 = 4tx x + 4t 2 t (x 2 + 2t)u u and by the infinitesimal generator of an infinity dimensional subalgebra (15) v h = h(xt) u

6 Hu Li and Chen J. Math. Phys (2016) TABLE I. Commutator table of the heat equation. v 1 v 2 v 3 v 4 v 5 v 6 v v 1 v 3 2v 5 v v 2 2v 1 4v 4 2v 3 v v 4 v 1 2v v 5 2v 6 v 5 v 3 2v 1 0 v v 6 2v 5 2v 3 4v 4 0 2v where h(x t) is an arbitrary solution of the heat equation. Since the infinite-dimensional subalgebra v h does not lead to group invariant solutions it will not be considered in the classification problem. The commutator table and actions of the adjoint representation which are taen from Ref. 13 are given in Tables I and II respectively. For six-dimensional Lie algebra (15) tae w 1 = 6 a i v i w 2 = 6 b j v j. (16) With the help of Table I substituting (16) into (12) leads six restrictive equations: a 1 b 4 + 2a 2 b 5 a 4 b 1 2a 5 b 2 = δa 1 2a 2 b 4 2a 4 b 2 = δa 2 a 1 b 5 2a 2 b 6 + a 5 b 1 + 2a 6 b 2 = δa 3 2a 4 b 6 2a 6 b 4 = δa 6 4a 2 b 6 4a 6 b 2 = δa 4 2a 1 b 6 + a 4 b 5 a 5 b 4 2a 6 b 1 = δa 5. Later on two inequivalent cases of Eqs. (17) with δ = 0 and δ = 1 should be considered respectively. j=1 (17) 2. Refined two-dimensional algebra of the Noviov equation The Noviov equation reads u t u t x x + 4u 2 u x 3uu x u x x u 2 u x x x = 0 (18) which was discovered by Noviov in a recent communication 14 and can be considered as a type of generalization of the nown Camassa-Holm equation. In Ref. 15 the authors gave out a fivedimensional Lie algebra of Eq. (18) which was spanned by the following basis: v 1 = t v 2 = x v 3 = e 2x x + e 2x u u v 4 = e 2x x e 2x u u v 5 = 2t t + u u. (19) In our early paper 13 one-dimensional optimal system of this five-dimensional Lie algebra (19) was constructed and used to find rich group invariant solutions of the Noviov equation. The corresponding commutator and adjoint representation relations are shown by Tables III and IV. TABLE II. Adjoint representation table of the heat equation. Ad v 1 v 2 v 3 v 4 v 5 v 6 v 1 v 1 v 2 v 3 v 4 ϵv 1 v 5 +ϵv 3 v 6 2ϵv 5 ϵ 2 v 3 v 2 v 1 v 2 v 3 v 4 2ϵv 2 v 5 2ϵv 1 v 6 4ϵv 4 +2ϵv 3 +4ϵ 2 v 2 v 3 v 1 v 2 v 3 v 4 v 5 v 6 v 4 e ϵ v 1 e 2ϵ v 2 v 3 v 4 e ϵ v 5 e 2ϵ v 6 v 5 v 1 ϵv 3 v 2 +2ϵv 1 ϵ 2 v 3 v 3 v 4 +ϵv 5 v 5 v 6 v 6 v 1 +2ϵv 5 v 2 2ϵv 3 +4ϵv 4 +4ϵ 2 v 6 v 3 v 4 +2ϵv 6 v 5 v 6

7 Hu Li and Chen J. Math. Phys (2016) TABLE III. Commutator table of the Noviov equation. v 1 v 2 v 3 v 4 v 5 v v 1 v v 3 2v 4 0 v 3 0 2v 3 0 4v 2 0 v 4 0 2v 4 4v v 5 2v In terms of the refined algebra for [w 1 w 2 ] = δw 1 we have the restrictions δa 5 = 0 2(a 5 b 1 a 1 b 5 ) = δa 1 2(a 2 b 3 a 3 b 2 ) = δa 3 2(a 4 b 2 a 2 b 4 ) = δa 4 4(a 4 b 3 a 3 b 4 ) = δa 2. (20) B. Calculation of the adjoint transformation matrix For w 1 = n a i v i its general adjoint transformation matrix A is the product of the matrices of the separate adjoint actions A 1 A 2... A n each corresponding to Ad exp(ϵvi )(w 1 )i = 1 n. For example applying the adjoint action of v 1 to w 1 = n a i v i and with the help of adjoint representation table one has Ad exp(ϵ1 v 1 )(a 1 v 1 + a 2 v a n v n ) = a 1 Ad exp(ϵ1 v 1 )v 1 + a 2 Ad exp(ϵ1 v 1 )v a n Ad exp(ϵ1 v 1 )v n = R 1 v 1 + R 2 v R n v n with R i R i (a 1 a 2... a n ϵ 1 )i = 1... n. To be intuitive formula (21) can be rewritten into the following matrix form: v (a 1 a 2... a n ) Ad exp(ϵ 1 v 1 ) (R 1 R 2... R n ) = (a 1 a 2... a n )A 1. Similarly the matrices A 2 A 3... A n of the separate adjoint actions of v 2 v 3... v n can be constructed respectively. Then the general adjoint transformation matrix A is the product of A 1... A n taen in any order (21) A A(ϵ 1 ϵ 2... ϵ n ) = A 1 A 2 A n. (22) Since only the existence of the element of the group is needed in our algorithm the orders of the product shown in (22) can be arbitrary. Applying the most general adjoint action Ad g = Ad exp(ϵn v n ) Ad exp(ϵ2 v 2 )Ad exp(ϵ1 v 1 ) to w 1 and w 2 we have w 1 (a 1 a 2... a n ) Ad Ad g (w 1 ) (a 1 a 2... a n )A w 2 (b 1 b 2... b n ) Ad Ad g (w 2 ) (b 1 b 2... b n )A. (23) TABLE IV. Adjoint representation table of the Noviov equation. Ad v 1 v 2 v 3 v 4 v 5 v 1 v 1 v 2 v 3 v 4 v 5 +2ϵv 1 v 2 v 1 v 2 e 2ϵ v 3 e 2ϵ v 4 v 5 v 3 v 1 v 2 +2ϵv 3 v 3 v 4 +4ϵv 2 +4ϵ 2 v 3 v 5 v 4 v 1 v 2 2ϵv 4 v 3 4ϵv 2 +4ϵ 2 v 4 v 4 v 5 v 5 e 2ϵ v 1 v 2 v 3 v 4 v 5

8 Hu Li and Chen J. Math. Phys (2016) Hence the equivalence between {w 1 w 2 } and {w 1 w 2 } shown in (9) can be rewritten as (a 1 a 2... a n) = 1 (a 1 a 2... a n )A + 2 (b 1 b 2... b n )A (b 1 b 2... b n) = 3 (a 1 a 2... a n )A + 4 (b 1 b 2... b n )A. ( ). (24) Remar 2. Eqs. (24) can be regarded as 2n algebraic equations with respect to ϵ 1... ϵ n and which will be taen to judge the equivalence of two given two-dimensional algebras {w 1 w 2 } and {w 1 w 2 }. If Eqs. (24) have the solution it means that { n a i v i n b j v j } is equivalent to { n a i v i n b j v j}. j=1 j=1 1. Adjoint matrix of the heat equation For the heat equation its general adjoint transformation matrix A of (15) is the product of the matrices of the separate adjoint actions A 1 A 2... A 6 each corresponding to Ad exp(ϵvi )(w 1 )i = First under the adjoint action of v 1 and with the help of Table II w 1 can be transformed into Ad exp(ϵ1 v 1 )(a 1 v 1 + a 2 v 2 + a 3 v 3 + a 4 v 4 + a 5 v 5 + a 6 v 6 ) = (a 1 a 4 ϵ 1 )v 1 + a 2 v 2 + (a 3 + a 5 ϵ 1 a 6 ϵ 2 1 )v 3 + a 4 v 4 + (a 5 2ϵ 1 a 6 )v 5 + a 6 v 6. One can rewrite above formula (25) into the following matrix form: where w 1 (a 1 a 2... a 6 ) Ad exp(ϵ 1 v 1 ) (a 1 a 2... a 6 )A 1 (25) A 1 = ϵ (26) 0 0 ϵ ϵ ϵ 1 1 Similarly the rest matrices of the separate adjoint actions of v 2... v 6 are found to be e ϵ e 2ϵ A 2 = ϵ A 4 = (27) 2ϵ e ϵ ϵ 2 2 2ϵ 2 4ϵ e 2ϵ ϵ ϵ 6 0 2ϵ 5 1 ϵ ϵ 6 4ϵ 6 0 4ϵ 2 6 A 5 = ϵ 5 0 A 6 = ϵ 6 (28) with A 3 = E being the identity matrix. Then the general adjoint transformation matrix A is the product of A 1... A 6 which can be taen in any order A (a i j ) 6 6 = A 1 A 2 A 3 A 4 A 5 A 6 (29)

9 Hu Li and Chen J. Math. Phys (2016) with a 11 = e ϵ 4 a 12 = a 16 = 0 a 13 = ϵ 5 e ϵ 4 a 14 = 0 a 15 = 2ϵ 6 e ϵ 4 a 21 = 2ϵ 5 e 2ϵ 4 a 22 = e 2ϵ 4 a 23 = (ϵ ϵ 6)e 2ϵ 4 a 24 = 4ϵ 6 e 2ϵ 4 a 25 = 4ϵ 5 ϵ 6 e 2ϵ 4 a 26 = 4ϵ 2 6 e2ϵ 4 a 31 = a 32 = a 34 = a 35 = a 36 = 0 a 33 = 1 a 41 = e ϵ 4 (ϵ 1 + 4ϵ 2 ϵ 5 e ϵ 4 ) a 42 = 2ϵ 2 e 2ϵ 4 a 43 = 4ϵ 2 ϵ 6 e 2ϵ 4 + ϵ 5 e ϵ 4 (ϵ 1 + 2ϵ 2 ϵ 5 e ϵ 4 ) a 44 = 1 8ϵ 2 ϵ 6 e 2ϵ 4 a 45 = ϵ 5 2ϵ 6 e ϵ 4 (ϵ 1 + 4ϵ 2 ϵ 5 e ϵ 4 ) a 46 = 2ϵ 6 (1 4ϵ 2 ϵ 6 e 2ϵ 4 ) a 51 = 2ϵ 2 e ϵ 4 a 52 = 0 a 53 = ϵ 1 + 2ϵ 2 ϵ 5 e ϵ 4 a 54 = 0 a 55 = e ϵ 4 (1 4ϵ 2 ϵ 6 e 2ϵ 4 ) a 56 = 0 a 61 = 4ϵ 2 e ϵ 4 (ϵ 1 + 2ϵ 2 ϵ 5 e ϵ 4 ) a 62 = 4ϵ 2 2 e2ϵ 4 a 63 = (ϵ 1 + 2ϵ 2 ϵ 5 e ϵ 4 ) 2 + 2ϵ 2 (1 4ϵ 2 ϵ 6 e 2ϵ 4 ) a 64 = 4ϵ 2 (1 4ϵ 2 ϵ 6 e 2ϵ 4 ) a 66 = e 2ϵ 4 (1 4ϵ 2 ϵ 6 e 2ϵ 4 ) 2 a 65 = 8ϵ 2 ϵ 6 e ϵ 4 (ϵ 1 + 2ϵ 2 ϵ 5 e ϵ 4 ) 4ϵ 2 ϵ 5 2ϵ 1 e ϵ Adjoint matrix of the Noviov equation For the Noviov equation its matrices of the separate adjoint actions A 1... A 5 were found in Ref. 13 which are rewritten in the follows: A 1 = A 2 = 0 0 e 2ϵ 0 1 2ϵ A 3 = e 2ϵ ϵ 3 4ϵ ϵ e 2ϵ ϵ 4 0 A 4 = 0 4ϵ 4 1 4ϵ A 5 = The general matrix A is selected as e 2ϵ ϵ 3 ϵ 4 2ϵ 3 2ϵ 4 (4ϵ 3 ϵ 4 1) 0 A = A 1 A 2 A 3 A 4 A 5 = 0 4ϵ 4 e 2ϵ 2 e 2ϵ 2 4ϵ 2 4 e 2ϵ 2 0 (30) 0 4ϵ 3 e 2ϵ 2 (1 4ϵ 3 ϵ 4 ) 4ϵ 2 3 e2ϵ 2 e 2ϵ 2 (1 4ϵ 3 ϵ 4 ) 2 0 2ϵ 1 e 2ϵ C. Calculation of the invariants for the refined two-dimensional algebra For a two-dimensional Lie algebra G(v w) a real function φ is called an invariant if φ(a 11 Ad g (w 1 ) + a 12 Ad g (w 2 ) a 21 Ad g (w 1 ) + a 22 Ad g (w 2 )) = φ(w 1 w 2 ) for any {w 1 w 2 } G(v w) and all g G with a 11 a 12 a 21 a 22 being arbitrary constants. For a general two-dimensional subalgebra { n a i v i n b j v j } G(v w) the corresponding invariant is a function of a 1... a n b 1... b n. j=1

10 Hu Li and Chen J. Math. Phys (2016) Let v = n c v be a general element of G. In conjunction with the commutator table we have =1 Ad g (w 1 ) = Ad exp(ϵv) (w 1 ) = w 1 ϵ[v w 1 ] + 1 2! ϵ 2 [v[v w 1 ]] = (a 1 v a n v n ) ϵ[c 1 v c n v n a 1 v a n v n ] + O(ϵ 2 ) = (a 1 v a n v n ) ϵ(θ a 1 v Θ a nv n ) + O(ϵ 2 ) = (a 1 ϵθ a 1 )v 1 + (a 2 ϵθ a 2 )v (a n ϵθ a n)v n + O(ϵ 2 ) where Θ a i Θ a i (a 1... a n c 1... c n ) can be easily obtained from the commutator table. Similarly applying the same adjoint action v = n c v to w 2 we get =1 Ad g (w 2 ) = Ad exp(ϵv) (w 2 ) = (b 1 ϵθ b 1 )v 1 + (b 2 ϵθ b 2 )v (b 6 ϵθ b n)v n + O(ϵ 2 ) (32) where Θ b i Θ b i (b 1... b n c 1... c n ) is obtained directly by replacing a i with b i in Θ a i (i = 1... n). More intuitively we denote w 1 (a 1 a 2... a n ) w 2 (b 1 b 2... b n ) Ad g (w 1 ) (a 1 ϵθ a 1 a 2 ϵθ a 2... a n ϵθ a n) + O(ϵ 2 ) Ad g (w 2 ) (b 1 ϵθ b 1 b 2 ϵθ b 2... b n ϵθ b n) + O(ϵ 2 ). For the two-dimensional subalgebra {w 1 w 2 } according to the definition of the invariant we have φ(a 11 Ad g (w 1 ) + a 12 Ad g (w 2 ) a 21 Ad g (w 1 ) + a 22 Ad g (w 2 )) = φ(w 1 w 2 ). (34) Further to guarantee a 11 Ad g (w 1 ) + a 12 Ad g (w 2 ) = w 1 and a 21 Ad g (w 1 ) + a 22 Ad g (w 2 ) = w 2 after the substitution of ϵ = 0 we require Then Eq. (34) is modified as (31) (33) a ϵa 11 a 12 ϵa 12 a 21 ϵa 21 a ϵa 22. (35) φ(w 1 w 2 ) = φ((1 + ϵa 11 )Ad g (w 1 ) + ϵa 12 Ad g (w 2 )ϵa 21 Ad g (w 1 ) + (1 + ϵa 22 )Ad g (w 2 )). (36) Remar 3. Since we just need consider the refined two-dimensional algebra two cases in (36) are discussed. (a) (b) When [w 1 w 2 ] = 0 taing the derivative of Eq. (36) with respect to ϵ and setting ϵ = 0 after the substitution of (33) extracting all the coefficients of c i a 11 a 12 a 21 a 22 some linear differential equations of φ are obtained. By solving these equations all the invariants φ on [w 1 w 2 ] = 0 can be found. When [w 1 w 2 ] = w 1 according to Remar 2 first we should mae a 12 = 0 and a 22 = 0 in Eq. (36). Then one does the same procedure just as case (a) to obtain linear differential equations of φ which eep invariable for [w 1 w 2 ] = w Invariant equations of the heat equation For a general two-dimensional subalgebra {w 1 w 2 } of the heat equation the corresponding invariant φ is a real function with twelve independent variables. Let v = 6 c v be a general element of G then in conjunction with the commutator Table I we have Ad g (w 1 ) = Ad exp(ϵv) (w 1 ) = w 1 ϵ[v w 1 ] + 1 2! ϵ 2 [v[v w 1 ]] = (a 1 v a 6 v 6 ) ϵ[c 1 v c 6 v 6 a 1 v a 6 v 6 ] + O(ϵ 2 ) = (a 1 v a 6 v 6 ) ϵ(θ a 1 v Θ a 6 v 6) + O(ϵ 2 ) = (a 1 ϵθ a 1 )v 1 + (a 2 ϵθ a 2 )v (a 6 ϵθ a 6 )v 6 + O(ϵ 2 ) =1 (37)

11 Hu Li and Chen J. Math. Phys (2016) with Θ a 1 = c 4a 1 2c 5 a 2 + c 1 a 4 + 2c 2 a 5 Θ a 2 = 2c 4a 2 + 2c 2 a 4 Θ a 3 = c 5a 1 + 2c 6 a 2 c 1 a 5 2c 2 a 6 Θ a 4 = 4c 6a 2 + 4c 2 a 6 Θ a 5 = 2c 6a 1 c 5 a 4 + c 4 a 5 + 2c 1 a 6 Θ a 6 = 2c 6a 4 + 2c 4 a 6. Similarly applying the same adjoint action v = 6 c v to w 2 we get =1 (38) with Ad g (w 2 ) = Ad exp(ϵv) (w 2 ) = (b 1 ϵθ b 1 )v 1 + (b 2 ϵθ b 2 )v (b 6 ϵθ b 6 )v 6 + O(ϵ 2 ) (39) Θ b 1 = c 4b 1 2c 5 b 2 + c 1 b 4 + 2c 2 b 5 Θ b 2 = 2c 4b 2 + 2c 2 b 4 Θ b 3 = c 5b 1 + 2c 6 b 2 c 1 b 5 2c 2 b 6 Θ b 4 = 4c 6b 2 + 4c 2 b 6 Θ b 5 = 2c 6b 1 c 5 b 4 + c 4 b 5 + 2c 1 b 6 Θ b 6 = 2c 6b 4 + 2c 4 b 6. Following Remar 3 Eq. (36) is separated into two cases. (a) For [w 1 w 2 ] = 0 all the c i (i = ) a 11 a 12 a 21 a 22 in Eq. (36) are arbitrary. Now taing the derivative of Eq. (36) with respect to ϵ and then setting ϵ = 0 extracting the coefficients of all c i a 11 a 12 a 21 a 22 one can directly obtain nine differential equations about φ = φ(a 1... a 6 b 1... b 6 ) a 1 φ a1 + a 2 φ a2 + a 3 φ a3 + a 4 φ a4 + a 5 φ a5 + a 6 φ a6 = 0 a 1 φ b1 + a 2 φ b2 + a 3 φ b3 + a 4 φ b4 + a 5 φ b5 + a 6 φ b6 = 0 b 1 φ b1 + b 2 φ b2 + b 3 φ b3 + b 4 φ b4 + b 5 φ b5 + b 6 φ b6 = 0 2a 2 φ a1 + 2b 2 φ b1 a 1 φ a3 b 1 φ b3 + a 4 φ a5 + b 4 φ b5 = 0 a 4 φ a1 b 4 φ b1 + a 5 φ a3 + b 5 φ b3 2a 6 φ a5 2b 6 φ b5 = 0 a 5 φ a1 b 5 φ b1 a 4 φ a2 b 4 φ b2 + a 6 φ a3 + b 6 φ b3 2a 6 φ a4 2b 6 φ b4 = 0 a 2 φ a3 b 2 φ b3 + 2a 2 φ a4 + 2b 2 φ b4 + a 1 φ a5 + b 1 φ b5 + a 4 φ a6 + b 4 φ b6 = 0 (40) (41) and a 1 φ a1 + b 1 φ b1 + 2a 2 φ a2 + 2b 2 φ b2 a 5 φ a5 b 5 φ b5 2a 6 φ a6 2b 6 φ b6 = 0 b 1 φ a1 + b 2 φ a2 + b 3 φ a3 + b 4 φ a4 + b 5 φ a5 + b 6 φ a6 = 0. Here the subscripts indicate partial derivatives. (b) For [w 1 w 2 ] = w 1 it requires a 12 = a 22 = 0 in Eq. (36) and seven equations about φ which are just Eqs. (41) are obtained. (42) 2. Invariant equations of the Noviov equation For the Noviov equation using the commutator Table III we have ( 5 ) Ad g (w 1 ) = Ad exp(ϵv) a i v i = (a 1 ϵθ a 1 )v a 5 v 5 + O(ϵ 2 ) (43) with Θ a 1 = 2(a 1c 5 a 5 c 1 )Θ a 2 = 4(a 3c 4 a 4 c 3 )Θ a 3 = 2(a 3c 2 a 2 c 3 )Θ a 4 = 2(a 2c 4 a 4 c 2 ). (44) Similarly there is ( 5 ) Ad g (w 2 ) = Ad exp(ϵv) b i v i = (b 1 ϵθ b 1 )v b 5 v 5 + O(ϵ 2 ). (45) Substituting (43) and (45) into Eq. (36) two cases about the invariant φ are obtained.

12 Hu Li and Chen J. Math. Phys (2016) (a) For [w 1 w 2 ] = 0 nine differential equations about φ are found a 5 φ a1 + b 5 φ b1 = 0 a 1 φ a1 + b 1 φ b1 = 0 a 2 φ a3 + b 2 φ b3 + 2a 4 φ a2 + 2b 4 φ b2 = 0 a 2 φ a4 + b 2 φ b4 + 2a 3 φ a2 + 2b 3 φ b2 = 0 a 3 φ a3 + b 3 φ b3 a 4 φ a4 b 4 φ b4 = 0 a 1 φ a1 + a 2 φ a2 + a 3 φ a3 + a 4 φ a4 + a 5 φ a5 = 0 a 1 φ b1 + a 2 φ b2 + a 3 φ b3 + a 4 φ b4 + a 5 φ b5 = 0 (46) and b 1 φ b1 + b 2 φ b2 + b 3 φ b3 + b 4 φ b4 + b 5 φ b5 = 0 b 1 φ a1 + b 2 φ a2 + b 3 φ a3 + b 4 φ a4 + b 5 φ a5 = 0. (b) For [w 1 w 2 ] = w 1 one just needs consider seven equations (46). (47) D. Construction of two-dimensional optimal system (1) First step: Present the commutator table and adjoint representation table of the generators {v i } n for a given algebra. Then in terms of [w 1 w 2 ] = δw 1 with δ 01 give out the restrictions about a 1 a 2... a n b 1 b 2... b n. (2) Second step: Following sections B and C compute the adjoint transformation matrix A and determine the general equations about the invariants φ. (3) Third step: For two different cases [w 1 w 2 ] = 0 and [w 1 w 2 ] = w 1 in terms of every restricted condition given by step 1 compute their respective invariants and select the corresponding eligible representative elements {w 1 w 2 }. For ease of calculations we rewrite Eq. (9) as Ad g (w 1 ) = 1 w w 2 Ad g (w 2 ) = 3 w w 2 ( ) (48) which is usually expressed by (a 1 a 2... a n )A = 1 (a 1 a 2... a n) + 2 (b 1 b 2... b n) (b 1 b 2... b n )A = 3 (a 1 a 2... a n) + 4 (b 1 b 2... b n) ( ). (49) Following Remar 2 if Eqs. (49) have the solution with respect to ϵ 1... ϵ n it signifies that the selected representative element {w 1 w 2 } is right; if Eqs. (49) have no solution another new representative element {w 1 w 2 } should be reselected. Repeat the process until all the cases are finished in the restrictions of step Two-dimensional optimal system of the heat equation (a) The case of δ = 0 in restrictive equations (17) Case 1: Not all a 2 a 4 a 6 b 2 b 4 and b 6 are zeroes. Without loss of generality we adopt a 6 0. In fact when only one of a 2 a 4 a 6 b 2 b 4 and b 6 is not zero one can choose appropriate adjoint transformation to transform it into the case ã 6 0. By solving Eqs. (17) with δ = 0 and a 6 0 we have three inds of solutions. (i) a 3 a 4 a 5 a 6 b 3 and b 6 are independent with a 1 = 1 2 a 4 a 5 a 2 = 1 a4 2 b 1 = 1 a 6 4 a 6 2 (ii) a 3 a 4 a 5 a 6 b 3 b 5 and b 6 are arbitrary but with a 1 = 1 2 a 4 a 5 b 6 b a6 2 2 = 1 4 a 4 a 5 a 2 = 1 a4 2 b 1 = 1 a 4 b 5 b 2 = 1 a 6 4 a 6 2 a 6 4 (iii) a 1 a 2 a 3 a 4 a 5 a 6 b 3 and b 6 are arbitrary but with a a 2 4 b 6 a 2 6 a 2 4 b 6 a 2 6 b 4 = a 4b 6 a 6 b 5 = a 5b 6 a 6. (50) b 4 = a 4b 6 a 6 b 5 a 5b 6 a 6. (51) a 4 a 5 or a 2 1 a4 2 b 1 = a 1b 6 b 2 = a 2b 6 b 4 = a 4b 6 b 5 = a 5b 6. (52) a 6 4 a 6 a 6 a 6 a 6 a 6

13 Hu Li and Chen J. Math. Phys (2016) Substituting the above three conditions into Eqs. (41) and (42) we find that φ constant i.e. there is no invariant. Then for each case select the corresponding representative element {w 1 w 2 } and verify whether Eqs. (49) have the solution. For case (i) select a representative element {w 1 w 2 } = {v 6v 3 }. Substituting (50) and w 1 = v 6 w 2 = v 3 into Eqs. (49) we obtain the solution 1 = a 6e 2ϵ 4 2 = 4a 3a 6 + 2a 4 a 6 + a 2 5 4a 6 3 = b 6e 2ϵ 4 4 = 4b 3a6 2 + b 6a a 4a 6 b 6 4a6 2 ϵ 1 = a 5 2a 6 ϵ 2 = a 4. 4a 6 Hence case (i) is equivalent to {v 6 v 3 }. For case (ii) there exist three circumstances in terms of the following expression: Λ 1 2a 6 [a 4 (a 5 b 6 b 5 a 6 ) 2 2a 6 (a 3 b 6 b 3 a 6 ) 2 2a 6 (a 5 b 6 b 5 a 6 )(a 3 b 5 b 3 a 5 )]. (54) (iia) When Λ 1 > 0 case (ii) is equivalent to {v 3 + v 6 v 5 }. After the substitution of (51) with w 1 = v 3 + v 6 w 2 = v 5 Eqs. (49) hold for 1 = Λ 1 4a 6 (a 5 b 6 b 5 a 6 ) 2 ϵ 4 = ln 2 a 6(a 5 b 6 b 5 a 6 ) Λ1 2 = a 5(a 5 b 6 b 5 a 6 ) + 2a 6 (a 3 b 6 b 3 a 6 ) 2 a 6 (a 5 b 6 b 5 a 6 ) 2 Λ1 ϵ 1 = a 3b 6 b 3 a 6 a 5 b 6 b 5 a 6 3 = b 6 a 6 1 ϵ 2 = a 4 4a 6 4 = b 5(a 5 b 6 b 5 a 6 ) + 2b 6 (a 3 b 6 b 3 a 6 ) 2 a 6 (a 5 b 6 b 5 a 6 ) 2 Λ1. (iib) When Λ 1 < 0 case (ii) is equivalent to { v 3 + v 6 v 5 }. The solution for Eqs. (49) is 1 = Λ 1 4a 6 (a 5 b 6 b 5 a 6 ) 2 2 = a 5(a 5 b 6 b 5 a 6 ) + 2a 6 (a 3 b 6 b 3 a 6 ) 2 a 6 (a 5 b 6 b 5 a 6 ) 2 Λ1 3 = b 6 a = b 5(a 5 b 6 b 5 a 6 ) + 2b 6 (a 3 b 6 b 3 a 6 ) 2 a 6 (a 5 b 6 b 5 a 6 ) 2 Λ1 ϵ 1 = a 3b 6 b 3 a 6 a 5 b 6 b 5 a 6 ϵ 2 = a 4 4a 6 ϵ 4 = ln 2 a 6(a 5 b 6 b 5 a 6 ) Λ1. (iic) When Λ 1 = 0 case (ii) is equivalent to {v 6 v 5 }. By solving Eqs. (49) we obtain 1 = a 6e 2ϵ 4 2 = a 5(a 5 b 6 b 5 a 6 ) + 2a 6 (a 3 b 6 b 3 a 6 ) (a 5 b 6 b 5 a 6 )e ϵ 4 3 = b 6e 2ϵ 4 4 = a 5(a 5 b 6 b 5 a 6 ) + 2a 6 (a 3 b 6 b 3 a 6 ) (a 5 b 6 b 5 a 6 )e ϵ 4 ϵ 1 = a 3b 6 b 3 a 6 a 5 b 6 b 5 a 6 ϵ 2 = (a 5b 6 b 5 a 6 )(a 3 b 5 a 5 b 3 ) + (a 3 b 6 b 3 a 6 ) 2 2(a 5 b 6 b 5 a 6 ) 2. For case (iii) it can be divided into the following several types. (iiia) 4a 2 a 6 a 2 4 > 0. Select a representative element {v 2 + v 6 v 3 }. After substituting (52) into Eqs. (49) we have 1 = (a 2 2a 4 ϵ 2 + 4a 6 ϵ 2 2 )e2ϵ 4 2 = a 3 + a a2 1 a 6 a 1 a 4 a 5 + a 2 a5 2 4a 2 a 6 a4 2 3 = b 6(a 2 2a 4 ϵ 2 + 4a 6 ϵ 2 2 ) e 2ϵ 4 4 a = b 3 + b 6 ( a 4 6 a a2 1 a 6 a 1 a 4 a 5 + a 2 a5 2 ) 4a 2 a 6 a4 2 ϵ 1 = 2ϵ 2(2a 1 a 6 a 4 a 5 ) + 2a 2 a 5 a 1 a 4 ϵ 4a 2 a 6 a4 2 5 = a 4a 5 2a 1 a 6 (4a 2 a 6 a4 2)eϵ 4 ϵ 6 = 4a 6 ϵ 2 a 4 4e 2ϵ 4 (a2 2a 4 ϵ 2 + 4a 6 ϵ 2 2 ) e2ϵ 4 = 4a 2 a 6 a a 2 2a 4 ϵ 2 + 4a 6 ϵ 2 2. (53)

14 Hu Li and Chen J. Math. Phys (2016) (iiib) 4a 2 a 6 a 2 4 < 0. Choose a representative element { v 2 + v 6 v 3 }. Now Eqs. (49) have the solution 1 = (a 2 2a 4 ϵ 2 + 4a 6 ϵ 2 2 )e2ϵ 4 2 = a 3 + a a2 1 a 6 a 1 a 4 a 5 + a 2 a5 2 4a 2 a 6 a4 2 3 = b 6(a 2 2a 4 ϵ 2 + 4a 6 ϵ 2 2 ) e 2ϵ 4 ϵ 1 = 2ϵ 2(2a 1 a 6 a 4 a 5 ) + 2a 2 a 5 a 1 a 4 a 6 4a 2 a 6 a4 2 4 = b 3 + b 6 ( a 4 a a2 1 a 6 a 1 a 4 a 5 + a 2 a5 2 ) ϵ 4a 2 a 6 a4 2 5 = a 4a 5 2a 1 a 6 (4a 2 a 6 a4 2)eϵ 4 4a 6 ϵ 2 a (4a 2 a 6 a 4 4 2) ϵ 6 = 4e 2ϵ 4 (a2 2a 4 ϵ 2 + 4a 6 ϵ 2 2 ) e2ϵ 4 = 2 a 2 2a 4 ϵ 2 + 4a 6 ϵ 2 2. (iiic) 4a 2 a 6 a4 2 = 0. In this case two conditions should be considered. When 2a 1 a 6 a 4 a 5 > 0 adopt the representative element {v 1 + v 6 v 3 }. Then the solution for Eqs. (49) is 1 = a 6 Z 2 2 = (ϵ ϵ 5)(2a 1 a 6 a 4 a 5 ) 2a 6 Z + a 3 + a a2 5 4a 6 3 = b 6(2a 1 a 6 a 4 a 5 ) Z ϵ 2a6 2 2 = a 4 4a 6 ϵ 4 = ln(z) 4 = b 6 [ 2(ϵ 2 4a ϵ 5)(2a 1 a 6 a 4 a 5 )Z + (a a 4a 6 )] + b 3 ϵ 1 = ϵ 6(2a 1 a 6 a 4 a 5 ) Z 2 + a 5. (Z = ( ) 1 2a6 2 3 ) 2a 6 2a 1 a 6 a 4 a 5 When 2a 1 a 6 a 4 a 5 < 0 adopt the representative element { v 1 + v 6 v 3 }. By solving Eqs. (49) we find 1 = a 6 Z 2 2 = (ϵ 5 ϵ 2 6 )(2a 1a 6 a 4 a 5 ) Z + a 3 + a 4 2a a2 5 4a 6 2a = b 6(2a 1 a 6 a 4 a 5 ) Z ϵ 2a6 2 2 = a 4 ϵ 4 = ln(z ) 4a 6 4 = b 6 [ 2(ϵ 4a6 2 5 ϵ 2 6 )(2a 1a 6 a 4 a 5 )Z + (a a 4a 6 )] + b 3 ϵ 1 = ϵ 6(2a 1 a 6 a 4 a 5 ) Z 2 + a 5. (Z 2a6 2 = ( ) 1 2a6 2 3 ). 2a 6 2a 1 a 6 a 4 a 5 Case 2: a 2 = a 4 = a 6 = b 2 = b 4 = b 6 = 0. Now determined equations (17) become a 1 b 5 + a 5 b 1 = 0. (55) Here we just need consider not all a 1 a 5 b 1 and b 5 are zeroes. Without loss of generality let a 5 0. Similarly if one of a 1 a 5 b 1 b 5 is not zero one can choose appropriate adjoint transformation to transform it into the case a 5 0. By solving Eq. (55) we obtain b 1 = a 1b 5 a 5. (56) Adopt a representative element {v 5 v 3 }. Then one can verify that all the {a 1 v 1 + a 3 v 3 + a 5 v 5 b 1 v 1 + b 3 v 3 + b 5 v 5 } with condition (56) are equivalent to {v 5 v 3 } since the solution for Eqs. (49) is 1 = a 5e ϵ 4 2 = a 3 + a 5 ϵ 1 3 = b 5e ϵ 4 4 = b 3 + b 5 ϵ 1 ϵ 2 = a 1 2a 5. (b) The case of δ = 1 in restrictive equations (17) Case 3: Not all a 2 a 4 and a 6 are zeroes.

15 Hu Li and Chen J. Math. Phys (2016) Without loss of generality we adopt a 6 0 and it can also guarantee b 6 0 through transformation (9). Hence Let b 6 0 first and it leads Eqs. (17) to a 1 = a 6(2b 4 + 1)(4b 1 b 6 b 5 2b 4 b 5 ) 4b 2 6 a 2 = a 6(2b 4 + 1) 2 16b 2 6 a 3 = 4a 6 b a 6(2b 4 + 1)(8b 1 b 5 1) a 6b 2 5 (2b 4 + 1) 2 4b 6 a 4 = a 6(2b 4 + 1) a 5 = a 6(b 5 + 2b 4 b 5 4b 1 b 6 ) b 2 = 4b b 6 b 6 16b 6 Substituting (57) into Eqs. (41) it yields an invariant for {w 1 w 2 } 4b 2 6 (57) φ = 1 16b 6 b 2 1 2b 4 16b 1 b 4 b 5 4b 3 + b2 5 (4b2 4 1) b 6. (58) In condition of (57) and 1 = c choose the corresponding representative element {v 6 ( 1 4 c 4 )v v 4 + v 6 } and Eqs. (49) have the solution 1 = a 6 2 = 0 3 = b = 1 ϵ 1 = b 5 + 2b 4 b 5 4b 1 b 6 2b 6 ϵ 2 = 2b b 6 ϵ 5 = 8b 1 b 6 4b 4 b 5 ϵ 4 = ϵ 6 = 0. For simplicity one can tae {v 6 v 4 + βv 3 }( β R) instead of {v 6 ( 1 4 c 4 )v v 4 + v 6 }. Case 4: a 2 = a 4 = a 6 = 0. Not all a 1 and a 5 are zeroes or else there must be a 3 = 0 shown in Eqs. (17). Let a 5 0 Taing a 2 = a 4 = a 6 = 0 with a 5 0 into (17) we have a 3 = b 1 a 5 a 1 b 5 b 2 = a 1(a 1 b 6 a 5 ) b a5 2 4 = 2a 1a 6 a 5. (59) a 5 Substituting relations (59) into Eqs. (41) we obtain an invariant for {w 1 w 2 } i.e. φ = 2 b 1 b 5 + b 6 b 2 1 b 3 a 1(b b 6 + 2b 1 b 5 b 6 ) + b 6b 2 5 a2 1. (60) a 5 In this case choose a representative element {v 5 cv 3 v 4 )} for 2 = c. Then solving Eqs. (49) one get 1 = a 5 2 = 0 3 = b 5 + 2b 1 b 6 2a 1b 5 b 6 a 5 4 = 1 ϵ 1 = b 1 + a 1b 5 a 5 ϵ 2 = a 1 2a 5 ϵ 6 = b 6 2 ϵ 4 = ϵ 5 = 0. In summary we have completed the construction of the two-dimensional optimal system O 2 g 1 = {v 6 v 3 } g 2 = {v 3 + v 6 v 5 } g 2 = { v 3 + v 6 v 5 } g 4 = {v 6 v 5 } g 5 = {v 2 + v 6 v 3 } g 6 = { v 2 + v 6 v 3 } g 7 = {v 1 + v 6 v 3 } g 8 = { v 1 + v 6 v 3 } g 9 = {v 5 v 3 } g 10 = {v 6 v 4 + βv 3 } g 11 = {v 5 v 4 + βv 3 } ( β R). Remar 4. The process of the construction ensures that all g i (i = ) are mutually inequivalent since each case is closed. One can also easily find this inequivalence from the incompatibility of Eqs. (49). In Ref. 10 Chou et al. gave a two-parameter optimal system {M i } 10 1 for the same Lie algebra (15) of the heat equation and showed their inequivalences by sufficient numerical invariants. One can see that {M i } 10 1 in Ref. 10 are equivalent to our {g 11g 10 g 6 g 4 g 1 g 2 g 3 g 8 g 5 g 9 } respectively. Furthermore we realize that g 7 is inequivalent to any of the elements in {M i } Hence here the two-dimensional optimal system O 2 given by (61) is complete and really optimal. a 5 1 (61)

16 Hu Li and Chen J. Math. Phys (2016) 2. Two-dimensional optimal system of the Noviov equation (a) The case of δ = 0 in restrictive equations (20) Case 1: Not all a 5 and b 5 are zeroes. Let a 5 0. Case 1.1: Not all a 2 a 3 a 4 b 2 b 3 and b 4 are zeroes. One can mae a 4 0 and it leads restrictive equations (20) to Denote b 1 = a 1b 5 a 5 b 2 = a 2b 4 a 4 b 3 = a 3b 4 a 4. (62) For Λ 2 = 0 select {v 4 + v 5 v 4 } and one solution for Eqs. (49) is Λ 2 a 2 2 4a 3a 4. (63) 1 = a 5 2 = a 4 a 5 3 = b 5 4 = b 4 b 5 ϵ 1 = a 1 2a 5 ϵ 2 = 0 ϵ 3 = a 2 4a 4. For Λ 2 > 0 select {v 2 + v 4 + v 5 v 5 } and Eqs. (49) have the solution Λ2 Λ2 1 = Λ 2 2 = a 5 Λ 2 3 = b 4 4 a = a 4b 5 b 4 4 a 4 ϵ 1 = a 1 Λ2 a 2 Λ2 a 4 ϵ 2 = 0 ϵ 3 = ϵ 4 = 2a 5 4a 4 2. Λ 2 For Λ 2 < 0 choose {v 3 + v 4 + v 5 v 5 } and Eqs. (49) have the solution 1 = a 4e 2ϵ 2 2 = a 5 a 4 e 2ϵ 2 3 = b 4e 2ϵ 2 4 = b 5 b 4 e 2ϵ 2 ϵ 1 = a 1 e 2ϵ 2 = 1 Λ 2 ϵ 2a 5 2 a4 2 3 = a 2 e 2ϵ 2 ϵ 4 = 0. 4a 4 Case 1.2: a 2 = a 3 = a 4 = b 2 = b 3 = b 4 = 0. This case is meaningless for w 2 = 0 or w 2 = cw 1 in terms of Eqs. (20). Case 2: a 5 = b 5 = 0. Case 2.1: Not all a 1 and b 1 are zeroes. Let a 1 0. Then not all a 2 a 3 a 4 b 2 b 3 and b 4 = 0 are zeros and we tae a 4 0. Restrictive equations (20) become b 2 = a 2b 4 a 4 b 3 = a 3b 4 a 4. (64) In accordance with Λ 2 = 0 Λ 2 > 0 and Λ 2 < 0 we adopt {v 1 + v 4 v 4 } {v 1 + v 2 + v 4 v 1 } and {v 1 + v 3 + v 4 v 1 } respectively and the corresponding solutions of Eqs. (49) read 1 = a 1 2 = a 4 a 1 3 = b 1 4 = b 4 b 1 ϵ 2 = 0 ϵ 3 = a 2 ϵ 5 = 0 4a 4 Λ2 Λ2 1 = Λ 2 2 = a 1 Λ 2 3 = b 4 4 a = a 4b 1 b 4 4 a 4 and ϵ 2 = ϵ 5 = 0 ϵ 3 = Λ2 a 2 Λ2 a 4 ϵ 4 = 4a 4 2 Λ 2 1 = a 4e 2ϵ 2 2 = a 1 a 4 e 2ϵ 2 3 = b 4e 2ϵ 2 4 = b 1 b 4 e 2ϵ 2 e 2ϵ 2 = 1 Λ 2 ϵ 2 a4 2 3 = a 2 e 2ϵ 2 ϵ 4 = ϵ 5 = 0. 4a 4

17 Hu Li and Chen J. Math. Phys (2016) Case 2.2: a 1 = b 1 = 0. Now not all a 2 a 3 a 4 b 2 b 3 and b 4 = 0 are zeros and let a 4 0. By solving Eqs. (20) we get b 2 = a 2b 4 a 4 b 3 = a 3b 4 a 4. (65) We just need consider Λ 2 > 0 and select {v 2 + v 4 v 4 }. (b) The case of δ = 1 in restrictive equations (20) Solving Eqs. (20) it first requires a 5 = 0 and b 5 = 1 2. Case 1: a 1 0. Case 1.1: Not all a 2 a 3 a 4 are zeroes and then one can mae b 4 0 which leads Eqs. (20) to a 2 = a 4(2b 2 1) a 3 = a 4(2b 2 1) 2 b 2b 4 16b 2 3 = 4b (66) 16b 4 4 For a 1 a 4 > 0 select {v 1 + v v 2 + v v 5} and Eqs. (49) have the solution 1 = a 4 2 = 0 3 = a 4b 1 4 a = 1 ϵ 1 = ϵ 2 = 0 1 ϵ 3 = 1 2b 2 ϵ 4 = b 4 a 4b 1 1 ϵ 5 = 1 ( 8b 4 a 1 2 ln a4 ). a 1 For a 1 a 4 < 0 select {v 1 v v 2 + v v 5} and Eqs. (49) have the solution 1 = a 4 2 = 0 3 = a 4b 1 4 a = 1 ϵ 1 = ϵ 2 = 0 1 ϵ 3 = 1 2b 2 ϵ 4 = b 4 a 4b 1 1 ϵ 5 = 1 ( 8b 4 a 1 2 ln a ) 4. a 1 Case 1.2: a 2 = a 3 = a 4 = 0. Case 1.2.1: Not all b 2 b 3 b 4 are zeros and let b 4 0. Now we have an invariant by solving Eqs. (46) saying K 1 b 2 2 4b 3b 4. (67) For K 1 = c we choose {v 1 c 4 v 3 v v 5} and {v 1 c 4 v 3 + v v 5}. Case 1.2.2: b 2 = b 3 = b 4 = 0. In this case it only remains {v v 5}. Case 2: a 1 = 0. Now not all a 2 a 3 a 4 are zeros and this can yield b 4 0. Solving Eqs. (20) we obtain a 2 = a 4(2b 2 1) a 3 = a 4(2b 2 1) 2 b 2b 4 16b 2 3 = 4b (68) 16b 4 4 Choose {v v 2 + v v 5} and one solution of Eqs. (49) is 1 = a 4 2 = 0 3 = b = 1 ϵ 2 = ϵ 4 = 0 ϵ 1 = b 1 ϵ 3 = 1 2b 2 8b 4. Recapitulating a two-dimensional optimal system O 2 of the Noviov equation contains thirteen elements g 1 = {v 5 v 4 } g 2 = {v 2 + v 4 v 5 } g 2 = {v 3 + v 4 v 5 } g 4 = {v 1 v 4 } g 5 = {v 2 + v 4 v 1 } g 6 = {v 3 + v 4 v 1 } g 7 = {v 2 v 4 } g 8 = {v 1 + v 4 v 2 + 2v 4 v 5 } g 9 = {v 1 v 4 v 2 + 2v 4 v 5 } g 10 = {v 1 βv 3 2v 4 v 5 } (69) g 11 = {v 1 βv 3 + 2v 4 v 5 }. g 12 = {v 1 v 5 } g 13 = {v 4 v 2 + 2v 4 v 5 } ( β R). IV. TWO-DIMENSIONAL OPTIMAL SYSTEM AND INVARIANT SOLUTIONS OF (2+1)-DIMENSIONAL NAVIER-STOKES EQUATION One of the most important open problems in fluid is the existence and smoothness problem of the Navier-Stoes (NS) equation which has been recognized as the basic equation and the

18 Hu Li and Chen J. Math. Phys (2016) TABLE V. Commutator table of the NS equation. v 1 v 2 v 3 v 4 v 1 0 v 2 v 3 0 v 2 v 2 0 v 4 0 v 3 v 3 v v very starting point of all problems in fluid physics In Ref. 18 by means of the classical Lie symmetry method we investigated the (2+1)-dimensional Navier-Stoes equation One can rewrite Eq. (70) into ω = ψ x x + ψ y y ω t + ψ x ω y ψ y ω x γ(ω x x + ω y y ) = 0. ψ x xt + ψ y yt + ψ x ψ x x y + ψ x ψ y y y ψ y ψ x x x ψ y ψ x y y γ(ψ x x x x + 2ψ x x y y + ψ y y y y ) = 0. The associated vector fields for the one-parameter Lie group of NS equation (71) are given by (70) (71) v 1 = x 2 x + y 2 y + t t v 2 = t v 3 = yt x + xt y + x2 + y 2 ψ 2 v 4 = y x + x y v 5 = f (t) x f (t)y ψ v 6 = g(t) y + g (t)x ψ v 7 = h(t) ψ. Here ignoring the discussion of the infinite dimensional subalgebra we apply the new approach to construct the two-dimensional optimal system and the corresponding invariant solutions for the four-dimensional Lie algebra spanned by v 1 v 2 v 3 v 4 in (72). The commutator table and the adjoint representation table for {v 1 v 2 v 3 v 4 } are given in Tables V and VI respectively. (72) A. Adjoint transformation matrix and the invariant equations Applying the adjoint action of v 1 to w 1 = 4 a i v i we have Ad exp(ϵ1 v 1 )(a 1 v 1 + a 2 v 2 + a 3 v 3 + a 4 v 4 ) = a 1 v 1 + a 2 e ϵ 1 v 2 + a 3 e ϵ 1 v 3 + a 4 v 4. Hence the corresponding adjoint transformation matrix A 1 is saying A 1 = 0 e ϵ e ϵ 1 0. (73) TABLE VI. Adjoint representation table of the NS equation. Ad v 1 v 2 v 3 v 4 v 1 v 1 e ϵ v 2 e ϵ v 3 v 4 v 2 v 1 ϵv 2 v 2 v 3 ϵv 4 v 4 v 3 v 1 +ϵv 3 v 2 +ϵv 4 v 3 v 4 v 4 v 1 v 2 v 3 v 4

19 Hu Li and Chen J. Math. Phys (2016) Similarly one can get 1 ϵ ϵ 3 0 A 2 = ϵ 2 A 3 = ϵ A 4 = E Then the most general adjoint matrix A can be taen as Let 1 ϵ 2 ϵ 3 ϵ 2 ϵ 3 A = A 1 A 2 A 3 A 4 = 0 e ϵ 1 0 e ϵ 1 ϵ e ϵ 1 e ϵ 1 ϵ 2. (74) w 1 = 4 a i v i w 2 = 4 b j v j. (75) For the general two-dimensional subalgebra {w 1 w 2 } the corresponding invariant φ is a real function of a 1... a 4 b 1... b 4. Let v = 4 c v be a general element of G then in conjunction with Table V we have Ad g (w 1 ) = Ad exp(ϵv) (w 1 ) =1 j=1 = w 1 ϵ[v w 1 ] + 1 2! ϵ 2 [v[v w 1 ]] = (a 1 v a 4 v 4 ) ϵ[c 1 v c 4 v 4 a 1 v a 4 v 4 ] + O(ϵ 2 ) (76) = a 1 v 1 + (a 2 ϵ(c 2 a 1 c 1 a 2 ))v 2 + (a 3 ϵ(c 1 a 3 c 3 a 1 ))v 3 + (a 4 ϵ(c 2 a 3 c 3 a 2 ))v 4 + O(ϵ 2 ). Similarly applying the same adjoint action v = 4 c v to w 2 we get =1 Ad g (w 2 ) = b 1 v 1 + (a 2 ϵ(c 2 b 1 c 1 b 2 ))v 2 + (b 3 ϵ(c 1 b 3 c 3 b 1 ))v 3 + (b 4 ϵ(c 2 b 3 c 3 b 2 ))v 4 + O(ϵ 2 ). Two cases are considered in the follows. (a) When [w 1 w 2 ] = 0 the invariant function φ = φ(a 1 a 2 a 3 a 4 b 1 b 2 b 3 b 4 ) is determined by seven equations and a 1 φ a1 + a 2 φ a2 + a 3 φ a3 + a 4 φ a4 = 0 a 1 φ b1 + a 2 φ b2 + a 3 φ b3 + a 4 φ b4 = 0 a 2 φ a2 a 3 φ a3 + b 2 φ b2 b 3 φ b3 = 0 a 1 φ a2 + a 3 φ a4 + b 1 φ b2 + b 3 φ b4 = 0 a 1 φ a3 + a 2 φ a4 + b 1 φ b3 + b 2 φ b4 = 0 b 1 φ b1 + b 2 φ b2 + b 3 φ b3 + b 4 φ b4 = 0 b 1 φ a1 + b 2 φ a2 + b 3 φ a3 + b 4 φ a4 = 0. (79) (b) When [w 1 w 2 ] = w 1 the invariant function φ = φ(a 1 a 2 a 3 a 4 b 1 b 2 b 3 b 4 ) only needs to meet Eqs. (78) in terms of a 12 = 0 and a 22 = 0. (77) (78) B. Construction of two-dimensional optimal system for the NS equation Substituting (75) into [w 1 w 2 ] = δw 1 the determined equations are found as follows: δa 1 = 0 a 2 b 1 a 1 b 2 = δa 2 a 1 b 3 a 3 b 1 = δa 3 a 2 b 3 a 3 b 2 = δa 4. (80)

20 Hu Li and Chen J. Math. Phys (2016) 1. The case of δ = 0 in restrictive equations (80) We consider two cases. Case 1: Not all a 1 and b 1 are zeroes. Without loss of generality we adopt a 1 0. Solving (80) we get b 2 = a 2b 1 a 1 b 3 = a 3b 1 a 1 (81) with a 1 a 2 a 3 a 4 b 1 and b 4 being arbitrary. Substituting condition (81) into Eqs. (78) and (79) we find that φ = constant. Hence according to (81) select the corresponding representative element {v 1 v 4 }. Since Eqs. (49) have the solution 1 = a 1 2 = a 1a 4 a 2 a 3 a 1 3 = b 1 4 = b 4a1 2 a 2a 3 b 1 ϵ a1 2 2 = eϵ 1a 2 ϵ 3 = a 3 a 1 a 1 e ϵ 1 case (1) is equivalent to {v 1 v 4 }. Case 2: a 1 = b 1 = 0. Now determined equations (80) become a 2 b 3 a 3 b 2 = 0. (82) Case 2.1: Not all a 2 and b 2 are zeroes and we let a 2 0. From Eq. (82) we get b 3 = a 3b 2 a. Then by solving Eqs. (78) and (79) we find φ constant. In 2 this case there exist three circumstances in terms of the sign of a 2 a 3. (i). When a 3 = 0 choose the representative element {v 2 v 4 } and Eqs. (49) have the solution 1 = eϵ 1 a 2 2 = eϵ 1 ϵ 3 a 2 + a 4 3 = eϵ 1 b 2 4 = eϵ 1 ϵ 3 b 2 + b 4. (ii). For a 2 a 3 > 0 we select {v 2 + v 3 v 4 } as a representative element. Eqs. (49) hold for 1 = a3 a 2 a 2 2 = 3 = a3 a 2 b 2 4 = a3 (ϵ 3 ϵ 2 )a 2 + a 4 a 2 a3 (ϵ 3 ϵ 2 )b 2 + b 4. a 2 (iii). For a 2 a 3 < 0 we select {v 2 v 3 v 4 } as a representative element and Eqs. (49) have the solution 1 = a 3 a 2 2 a = a 3 (ϵ 3 + ϵ 2 )a 2 + a 4 2 a 2 3 = a 3 a 2 b 2 4 = a3 a 2 (ϵ 3 + ϵ 2 )b 2 + b 4. Case 2.2: For a 2 = b 2 = 0 Eqs. (49) always stand up and the general two-dimensional Lie algebra becomes {a 3 v 3 + a 4 v 4 b 3 v 3 + b 4 v 4 }. Then if not all a 3 and b 3 are zeroes (and let a 3 0) it will equivalent to {v 3 v 4 } since that Eqs. (49) have the solution 1 = e ϵ 1 a 3 2 = e ϵ 1 ϵ 2 a 3 + a 4 3 = e ϵ 1 b 3 4 = e ϵ 1 ϵ 2 b 3 + b 4. For the case of a 3 = b 3 = 0 the general two-dimensional Lie algebra {a 4 v 4 b 4 v 4 } is trivial. 2. The case of δ = 1 in restrictive equations (80) Substituting δ = 1 into Eqs. (80) there must be a 1 = 0. Case 3: a 2 0. Now Eqs. (80) require Substituting (83) into Eqs. (78) it leads to an invariant for {w 1 w 2 } a 3 = 0 a 4 = a 2 b 3 b 1 = 1. (83) φ = 3 b 4 b 2 b 3. (84)

21 Hu Li and Chen J. Math. Phys (2016) In condition of (83) and 3 = c choose the corresponding representative element {v 2 v 1 + cv 4 } and Eqs. (49) have the solution 1 = a 2 2 = 0 3 = b 2 4 = 1 ϵ 3 = b 3 ϵ 1 = ϵ 2 = 0. (85) Case 4: a 2 = 0. By solving Eqs. (80) we get b 1 = 1 a 4 = a 3 b 2. (86) Substituting (86) with a 1 = a 2 = 0 into Eqs. (78) one can obtain an invariant as follows: φ = 4 b 4 + b 2 b 3. (87) In condition of (86) and 4 = c select a representative element {v 3 v 1 + cv 4 } and Eqs. (49) have the solution 1 = a 3 2 = 0 3 = b 3 4 = 1 ϵ 2 = b 2 ϵ 1 = ϵ 3 = 0. In summary a two-dimensional optimal system O 2 for the four-dimensional Lie algebra spanned by v 1 v 2 v 3 v 4 in (72) is shown as follows: g 1 = {v 1v 4 } g 2 = {v 2v 4 } g 3 = {v 2 + v 3 v 4 } g 4 = {v 2 v 3 v 4 } g 5 = {v 3v 4 } g 6 = {v 2v 1 + cv 4 } g 7 = {v 3 v 1 + cv 4 } (c R). (88) C. Two-dimensional reductions for the NS equation Using two-dimensional optimal system (88) one can reduce the (2+1)-dimensional NS equation to some ordinary differential equations and further get rich group invariant solutions. For the case of g 1 = {v 1v 4 } and g 2 = {v 2v 4 } one can refer to Ref. 18. The case of g 5 leads to no group invariant solutions. Then we just consider the rest elements in (88). (a) g 3 = {v 2 + v 3 v 4 } and g 4 = {v 2 v 3 v 4 }. By solving (v 2 ± v 3 )(ψ) = 0 and v 4 (ψ) = 0 we have ψ = F(x 2 + y 2 ) ± 1 2 t(x2 + y 2 ). Substituting it into Eq. (71) one can get 8γ[ξ 2 F (4) (ξ) + 4ξF (ξ) + 2F (ξ)] 1 = 0 (89) with ξ = x 2 + y 2. By solving Eq. (89) we find g 3 and g 4 lead to the same group invariant solution ψ = c 1 + c 2 (x 2 + y 2 ) + c 3 ln(x 2 + y 2 ) + c 4 (x 2 + y 2 )[ln(x 2 + y 2 ) 1] γ (x2 + y 2 ) t(x2 + y 2 ). (b) g 6 = {v 2v 1 + cv 4 }. From v 2 (ψ) = 0 and (v 1 + cv 4 )(ψ) = 0 one can get ψ = F(arctan( y x ) c ln(x 2 + y 2 )). Substituting it into Eq. (71) and integrating the reduced equation once we have γ[(4c 2 + 1)G (ξ) + 8cG (ξ) + 4G(ξ)] G 2 (ξ) = 0 (G(ξ) = F (ξ)) (90) with ξ = arctan( y x ) c ln(x2 + y 2 ). Specially when c = 0 in Eq. (90) there is a solution Then it leads to the solution of the NS equation G(ξ) = 6γsech 2 (ξ + c 0 ) + 4γ. (91) ψ = 6γ tanh(arctan( y x ) + c 0) + 4γ arctan( y x ) + c 1. (92) (c) g 7 = {v 3v 1 + cv 4 }. In this case we have ψ = arctan( x y ) + c ln(t) + F( x2 +y 2 t ). The reduced equation for Eq. (71) is with Z = x2 +y 2 t. 4γZ 2 G (Z) + Z(Z + 8γ 2)G (Z) + (Z 2)G(Z) = 0(F (Z) = G(Z)) (93)