PRAMANA c Indian Academy of Sciences Vol. 77, No. 3 — journal of September 2011 physics pp. 407–414

On the Lie point 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 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) . (21) 15 ∂t Note that a number of symmetries exist for the inviscid Burgers equation (1) for which ξ and τ depend on u which were not found or mentioned in earlier works. Inter alia,we have ∂ ∂ V = L(u) , V = M(u) , 1 ∂x 2 ∂t ∂ ∂ V = Q(u)a(x, t) , V = R(u)b(x, t) , 3 ∂x 4 ∂t where L(u), M(u), Q(u) and R(u) are arbitrary smooth functions of u and

at + f (u)ax = bt + f (u)bx = 0.

Pramana – J. Phys., Vol. 77, No. 3, September 2011 411 Muhammad A Abdulwahhab et al

3. Reductions for the case ξ = k(u)a(x, t), τ = k(u)b(x, t), η = η(x, t, u)

As a first example, we perform a reduction of (1) using the generator ∂ ∂ xf − tf2 ∂ X = x2 + xt + 1 ∂x ∂t f  ∂u by choosing k(u) = 1. From this generator we find the invariants to be y = x/t and w = f/(x − tf) so that

 x x  2 f ut = − w − w (1 + tw)2 t2 and

 1 2  x f ux = w + tw + w (1 + tw)2 t which, when substituted into (1), leads to the reduced equation w = 0. The invariant solution is therefore kx u(x, t) = f −1 1 + kt for some constant k.Nowif f (u) = u2 or f (u) = u/(1 + u2), then we have the solutions  kx u(x, t) = 1 + kt and  (1 + kt) ± (1 + kt)2 − 4k2x2 u(x, t) = 2kx respectively. Similarly, the boost-type symmetry given by ∂ 1 ∂ X = t + 3 ∂x f  ∂u with f (u) = u has invariants y = t and w = f (u) − x/t that reduces (1) to the ordinary differential equation w/w =−1/y. The invariant solution is therefore

x + k u(x, t) = f −1 . t

Following the same procedure, the generators X2, X4, X5, X7 and X1 + X2 yield the respective solutions,

x − k u(x, t) = f −1 , t x u(x, t) = f −1 , t − k x u(x, t) = f −1 , t x u(x, t) = f −1 , t + k x + k u(x, t) = f −1 . t − k

412 Pramana – J. Phys., Vol. 77, No. 3, September 2011 Lie point symmetry analysis and solutions of the inviscid Burgers equation

From all the above solutions, we conclude that

αx + β u(x, t) = f −1 γ t + δ is an invariant solution of the generalized inviscid Burgers equation (1) resulting from any of the generators Xi , i = 1, 2, ..., 8, or any of their linear combinations if and only if α = γ where α, β, γ and δ are arbitrary constants. The linear combination, ∂ ∂ f (u) ∂ X + X = 2x + t + , 4 5 ∂x ∂t f (u) ∂u with invariants y = x/t2 and w = g(u)/t reduce (1) to the ordinary differential equation w w + . w − 2y Using the substitution w = 2yz(y)/(z(y) − 1), we obtain the invariant solution

2x u(x, t) = f −1 √ . t + t2 − kx We now perform a class of reductions leading to invariant solutions using transformations of x and t that depend on u. The combination of V1, X6, and the boost lead to the symmetry generator ∂ 1 ∂ (t + 1 + L(u)) + , ∂x f  ∂u which has invariants  y = t,w=−x + (t + 1) f + Lf du.

Then, w − f f  + 1 u = and u = , t (t + 1 + L) f  x (t + 1 + L) f  which, when substituted into (1), leads to the invariant solution  k + x = (t + 1) f + Lfdu.

This corresponds to an infinite number of solutions ranging from the simple to the complex depending on the choices for f and L. For instance, if L = f , then the invariant solution is of the form    u(x, t) = f −1 −(t + 1) ± (t + 1)2 + 2x + 2k .

If f (u) = u/(u + 1), u =−1, then the solution of (1) is  −(t + 1) ± (t + 1)2 + 2x + 2k u(x, t) =  . (t + 2) ∓ (t + 1)2 + 2x + 2k

Pramana – J. Phys., Vol. 77, No. 3, September 2011 413 Muhammad A Abdulwahhab et al

Similarly, the generator, ∂ f 2 (u) ∂ V + X = (M (u) − x) + , 2 7 ∂t f  (u) ∂u with invariants,  x M y = x,w=−t + + f du, f f 2 leads to the solution  x M k + t = + f du. f f 2 √ As an example, suppose M = f = cosh u. Then    k + t ± (k + t)2 − 4x u(x, t) = cosh−1 2 is an invariant solution of eq. (1).

4. Conclusions

We find some interesting exact solutions for the inviscid Burgers equation corresponding to the situation when the space and time transformation are velocity dependent. These more general cases have not been considered in the earlier works cited above.

References

[1] J Constant, http://www.coolissues.com-mathematics/Navier-Stokes (2008) [2] B J Cantwell, Introduction to symmetry analysis (Cambridge University Press, Cambridge, 2002) [3] A Ouhadan and E H El Kinani, Adv. Appl. Clifford Alg. 17, 95 (2007) [4] S Lie, Theories Der Transfermationgruppen Dritter und Letzter Abschnitt (Teubner, Leipzig, 1893) [5] M Nadjafikhah, Adv. Appl. Clifford Alg. 20, 71 (2010) [6] G Bluman and S C Anco, Symmetry and integration methods for differential equations (Springer, New York, 2002) [7] P J Olver, Application of Lie groups to differential equations (Springer, New York, 1993) [8]NHIbragimov,CRC handbook of analysis of differential equations, Volume 1, Symmetries, exact solutions and conservation laws (CRC Press, Boca Raton, 1994)

414 Pramana – J. Phys., Vol. 77, No. 3, September 2011