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On the Lie Point Symmetry Analysis and Solutions of the Inviscid Burgers Equation

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On the Lie Point Symmetry Analysis and Solutions of the Inviscid Burgers Equation PRAMANA c Indian Academy of Sciences Vol. 77, No. 3 — journal of September 2011 physics pp. 407–414 On the Lie point symmetry analysis and solutions of the inviscid Burgers equation MUHAMMAD A ABDULWAHHAB1,∗, ASHFAQUE H BOKHARI1, A H KARA2 and F D ZAMAN1 1Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia 2School of Mathematics and Centre for Differential Equations, Continuum Mechanics and Applications, University of the Witwatersrand, Johannesburg, Private Bag 3, Wits 2050, South Africa ∗Corresponding author. E-mail: [email protected] Abstract. Lie point symmetries of the first-order inviscid Burgers equation in a general setting are studied. Some new and interesting solutions are presented. Keywords. Symmetry analysis; solutions; inviscid Burgers equation. PACS Nos 02.30.Jr; 02.20.Tw; 02.20.Sv; 47.10.ad 1. Introduction The first-order nonlinear evolution equation that arises from the inviscid Burgers equation presents some interesting features. As is well known, the Burgers equation can be obtained as a limiting case of the Navier–Stokes equations [1] and serves as an adequate model for some gas dynamics and transport problems [2]. For the inviscid case, the Burgers equation reduces to ut + uux = 0 where u(x, t) is the particle velocity. However, a more general inviscid Burgers equation arises if the diffusivity is considered to be a general function of u. Towards this end, Ouhadan and El Kinani [3] considered the general inviscid Burgers equation ut + f (u) ux = 0 and used Lie symmetry method [4] for obtaining exact solutions in some cases. However, the analysis presented therein is not complete. The procedure for calculating the Lie point symmetries leads to an underdetermined system of partial differential equations. Ouhadan and El Kinani presented a few simple cases and obtained exact solutions for only those cases. The study was extended by Nadjafikhah [5] to include other cases of interest. In this paper we consider more general cases. 2. Symmetry generators Consider the inviscid Burgers equation, ut(x, t) + f (u)ux (x, t) = 0, (1) DOI: 10.1007/s12043-011-0160-x; ePublication: 26 August 2011 407 Muhammad A Abdulwahhab et al where f (u) is a smooth function having a non-vanishing first derivative. Requiring that eq. (1) be invariant with respect to the first prolonged symmetry generator, ∂ ∂ ∂ ∂ ∂ X = ξ + τ + η + ηx + ηt , (2) ∂x ∂t ∂u ∂ux ∂ut with ξ,τ and η functions of t, x and u, gives [6–8] t x η f (u) ux + η + f (u)η = 0, (3) where x η = ηx + ux ηu − ux (ξx + ux ξu) − ut (τx + ux τu), (4) t η = ηt + ut ηu − ux (ξt + ut ξu) − ut (τt + ut τu). (5) Comparing coefficients of the derivatives of u that occur in the determining equation, ( + ( ) )| = , X ut f u ux ut + f (u) ux =0 0 we obtain the following system of differential equations: ηt + f (u)ηx = 0, (6) 2 − ξt + f (u)τt + η f (u) − f (u)ξx + f (u) τx = 0. (7) Note that the above system is an underdetermined system consisting of only two equations to be solved for four unknown functions, ξ,τ, η and f (u). Hence, to solve this system one needs various ansätze regarding ξ,τ and η. In the earlier paper, Nadjafikhah [5] presented solutions of systems (6) and (7) for only one case by assuming η = η(x, t, u), ξ = ξ(x, t) and τ = τ(x, t). In our classification of solutions we assume in the first case, ξ = k(u)a(x, t), τ = k(u)b(x, t) and in the second case ξ = f (u)a(x, t) + b(x, t), τ = f (u)c(x, t) + d(x, t) with η an arbitrary function of x, t and u. We present a complete reduction of (1) in the first case and briefly state the results in the second case. 2.1 ξ = k(u)a(x, t), τ = k(u)b(x, t) and η = η(x, t, u) In this case (7) becomes η f (u) −a + f (u)b + − f (u) a + f (u)2b = 0. (8) t t k(u) x x From eq. (8) we obtain k(u) η = a + f (u)(a − b ) − f 2(u)b . (9) f (u) t x t x 408 Pramana – J. Phys., Vol. 77, No. 3, September 2011 Lie point symmetry analysis and solutions of the inviscid Burgers equation Substituting for η from (9) in eq. (6) and using the fact that f (u) is arbitrary, we obtain an overdetermined system of partial differential equations for a(x, t) and b(x, t), namely, bxx = 0, 2bxt + axx = 0, btt − 2axt = 0, att = 0. Solving the system of equations gives the components of the symmetry generator as 2 ξ = k(u) [α1x + α2xt + α3t + (α4 + α5) x + α6], 2 τ = k(u) [α2t + α1xt + α5t − α7x + α8], k(u) η = f 2(u) [−α t + α ] + f (u) [α x − α t + α ] + α x + α . f (u) 1 7 1 2 4 2 3 (10) The symmetry generators corresponding to the above infinitesimals are given below. ∂ ∂ xf(u) − tf2(u) ∂ X = k(u) x2 + xt + , 1 ∂ ∂ ( ) ∂ x t f u u ∂ ∂ x − tf(u) ∂ X = k(u) xt + t2 + , 2 ∂ ∂ ( ) ∂ x t f u u ∂ 1 ∂ X = k(u) t + , 3 ∂ ( ) ∂ x f u u ∂ f (u) ∂ X = k(u) x + , 4 ∂x f (u) ∂u ∂ ∂ X = k(u) x + t , 5 ∂x ∂t ∂ X = k(u) , 6 ∂ x ∂ f 2(u) ∂ X = k(u) − x + , 7 ∂t f (u) ∂u ∂ X = k(u) . (11) 8 ∂t When k(u) = 1, eqs (11) reduce to those given by Nadjafikhah [5] and they form a closed algebra as shown in table 1. 2.2 ξ = f (u)a(x, t) + b(x, t), τ = f (u)c(x, t) + d(x, t) and η = η(x, t, u) For this case eq. (7) gives 2 2 3 2 − fat − bt + f ct + fdt + η f − f ax − fbx + f cx + f dx = 0. (12) To determine the unknown functions a, b, c and d in the above expression we differentiate eq. (12) thrice with respect to u to obtain (((η f )u/f )u)/2 f )u)/3 f =−cx . (13) Pramana – J. Phys., Vol. 77, No. 3, September 2011 409 Muhammad A Abdulwahhab et al Table 1. The commutator table for k(u) = 1. X1 X2 X3 X4 X5 X6 X7 X8 X1 00−X2 −X1 −X1 −X4 − X5 0 X7 X2 00 00−X2 −X3 X1 2X5 − X4 X3 X2 00X3 002X4 − X5 −X6 X4 X1 0 −X3 00 −X6 X7 0 X5 X1 X2 000−X6 0 −X8 X6 X4 + X5 X3 0 X6 X6 0 −X8 0 X7 0 −X1 −2X4 + X5 −X7 0 X8 00 X8 −X7 −2X5 + X4 X6 0 X8 000 It follows immediately from eq. (13) that 1 η = f 3 A(x, t) + f 2 B(x, t) + fD(x, t) + E(x, t) , (14) f (u) where A(x, t) =−cx . Upon substituting for η given by (14) into the systems (6) and (7) it is necessary that ct + dx − ax =−B(x, t), (15) dt − at − bx =−D(x, t), (16) bt = E(x, t) (17) and 4 3 2 f Ax + f (At + Bx ) + f (Bt + Dx ) + f (Dt + Ex ) + Et = 0. (18) Since f (u) is arbitrary, solving the above equations gives the components of the symmetry generator as ξ = f (u) d(x, t) − (2α t + α )x2 + (2α x + α )t2 1 2 4 5 + (α7x + α9 − α11)t − (2α10t + α13)x − α14 + (α x3 − α x2t − α xt − α t + α x2 + α x + α ), 1 4 5 6 10 11 12 τ = f (u) −α xt2 − α xt − α x + α t3 + (α − α )t2 1 2 3 4 7 10 + (α8 − α13)t + α15 + d(x, t), 1 η = f 3(u)(α t2 + α t + α ) − f 2(u)(2α xt + α x + α t2 f (u) 1 2 3 1 2 4 + α t + α ) + f (u)(α x2 + 2α xt + α t + α x + α ) 7 8 1 4 5 7 9 2 − (α4x + α5x + α6) (19) where a(x, t) and d(x, t) are two arbitrary functions of x and t that are related via the equation 2 2 d(x, t) − a(x, t) = (2α1t + α2) x − (2α4x + α5) t − (α7x + α9 − α11) t + (2α10t + α13) x + α14. (20) 410 Pramana – J. Phys., Vol. 77, No. 3, September 2011 Lie point symmetry analysis and solutions of the inviscid Burgers equation The corresponding generators are given below. ∂ ∂ Z = −2x2tf(u) + x3 − xt2 f (u) 1 ∂x ∂t 2 3( ) − 2( ) + 2 ( ) ∂ + t f u 2xtf u x f u , f (u) ∂u ∂ ∂ tf3(u) − xf2(u) ∂ Z =−x2 f (u) − xtf(u) + , 2 ∂x ∂t f (u) ∂u ∂ f 3(u) ∂ Z =−xf(u) + , 3 ∂t f (u) ∂u ∂ ∂ Z = 2xt2 f (u) − x2t + t3 f (u) 4 ∂x ∂t 2 2( ) − ( ) + 2 ∂ − t f u 2xtf u x , f (u) ∂u ∂ tf(u) − x ∂ Z = t2 f (u) − xt + , 5 ∂x f (u) ∂u ∂ 1 ∂ Z =−t − , 6 ∂x f (u) ∂u ∂ ∂ xf(u) − tf2(u) ∂ Z = xtf(u) + t2 f (u) + , 7 ∂x ∂t f (u) ∂u ∂ f 2(u) ∂ Z = tf(u) − , 8 ∂t f (u) ∂u ∂ f (u) ∂ Z = tf(u) − , 9 ∂x f (u) ∂u ∂ ∂ Z = 2xtf(u) + x2 − t2 f (u) , 10 ∂x ∂t ∂ Z = (x − tf(u)) , 11 ∂x ∂ Z = , 12 ∂x ∂ ∂ Z = xf(u) − tf(u) , 13 ∂x ∂t ∂ Z = f (u) , 14 ∂x ∂ Z = f (u) .
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