Symmetries and Large Time Asymptotics of Compressible Euler Flows with Damping By P. L. Sachdev, B. Mayil Vaganan, and G. Sivagami
Large time asymptotics of compressible Euler equations for a polytropic gas with and without the porous media equation are constructed in which the Barenblatt solution is embedded. Invariance analysis for these governing equations are carried out using the classical and the direct methods. A new second order nonlinear partial differential equation is derived and is shown to reduce to an Euler–Painlev´e equation. A regular perturbation solution of a reduced ordinary differential equation is determined. And an exact closed form solution of a system of ordinary differential equations is derived using the invariance analysis.
1. Introduction The compressible Euler equations with damping are ρt + ∇ ◦ (ρ u ) = 0, u ⊗ u ) + ∇ p(ρ) = −αρ u , ( p u )t + ∇ ◦ ρ(
(1) (2)
where u , p, and ρ are the velocity, pressure, and density, respectively, of a gas and α > 0 is the friction constant. Address for correspondence: B. Mayil Vaganan, Department of Applied Mathematics and Statistics, Madurai Kamaraj University, Madurai 625021, India; e-mail:
[email protected]
105 STUDIES IN APPLIED MATHEMATICS 120:105–128 C 2008 by the Massachusetts Institute of Technology Published by Blackwell Publishing, 350 Main Street, Malden, MA 02148, USA, and 9600 Garsington Road, Oxford, OX4 2DQ, UK.
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We assume the gas to be polytropic so that p(ρ) = kρ γ ,
k > 0,
γ > 1.
(3)
Liu [1] justified the Darcy’s law for the compressible flows in the time-asymptotic sense. Indeed, he constructed a class of particular solutions of (1) and (2) and related it, time-asymptotically, to the self-similar solutions (see Barenblatt [2]) of the porous media equation ρt = α −1 δp(ρ),
(4)
when (2) is simplified to the Darcy’s law ∇ p(ρ) = −αρ u .
(5)
Hsiao and Liu [3] showed that (1), (2), and (4) are time-asymptotically equivalent. The singular behavior of the flow near the vacuum from ρ = 0 was studied extensively by Liu [1] and Makino [4, 5]. Nishida [6] showed that the damping term prevents the formation of shocks away from the vacuum. The sound speed c is given by c2 =
dp = kγρ (γ −1) . dρ
(6)
In terms of the characteristic speeds u and c, (1) and (2) become (c2 )t + ∇(c2 )u˙ + (γ − 1)c2 ∇ u˙ = 0, ˙ u+ u t + ( u ∇)
1 ∇(c2 )x = −αu, γ −1
(7) (8)
In spherically symmetric form, (7) and (8) and, (7) and (5), respectively, take the form (n − 1)(γ − 1) 2 (c2 )t + u(c2 )x + c u + (γ − 1)c2 u x = 0, (9) x 1 (10) u t + uu x + (c2 )x = −αu, γ −1 and (c2 )t + u(c2 )x +
where
(n − 1)(γ − 1) 2 c u + (γ − 1)c2 u x = 0, x 1 (c2 )x = −αu, γ −1
x=
n i=0
(11) (12)
1/n xi2
and u = (x/x ) u.
(13)
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Liu [1] looked for solution of (9) and (10) in the form u = a(t)x
c2 = e(t) − b(t)x 2 .
and
(14)
He showed that a(t), b(t), and e(t) are governed by the system of ordinary differential equations b + (nγ − n + 2)ab = 0,
(15)
e + n(γ − 1)ea = 0,
(16)
2 b = 0. γ −1
(17)
a + a 2 + αa −
Liu [1,7] then proved by using the phase-plane analysis that the solutions to (15)–(17) exist for all time t satisfying any given set of initial data a(0), b(0), and e(0). For (11) and (12) Liu [1] inserted the same ansatz ¯ u = a(t)x
and
2 ¯ c2 = e¯ (t) − b(t)x .
(18)
As expected the equations are simpler than (15)–(17): b¯ + (nγ − n + 2)a¯ b¯ = 0,
(19)
e¯ + n(γ − 1)¯ea¯ = 0,
(20)
2 ¯ b = 0, γ −1
(21)
α a¯ −
Lie found the Barenblatt solutions for this system, namely, ¯ = a(t)
1 t −1 , nγ − n + 2
(22)
¯ = b(t)
(γ − 1)α t −1 , 2(nγ − n + 2)
(23)
e¯ (t) = e0 t −n(γ −1)/(nγ −n+2) .
(24)
The purpose of the present work is manifold: We first recover the ansatz (18) using the direct method of Clarkson and Kruskal [8]. Next we solve both the nonlinear, coupled systems of ordinary differential equations (15)–(17) and (19)–(21) in which we embed the Barenblatt solution. Then we study the symmetry properties of both the systems (9)–(10) and (11)–(12) using the classical and the direct methods. Later a new second order nonlinear partial differential equation is derived from (11) to (12) by eliminating u, and this equation is shown to reduce to an Euler–Painlev´e equation. The remaining of paper discusses certain interesting results in which we state that more than one
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partial differential equations of different family reduces to a single ordinary differential equation through various similarity transformations.
2. To recover the ansatz of Liu We insert the following special form of the direct method u = φ(x, t) + θ (x, t)F(z),
c2 = λ(x, t) + µ(x, t)G(z),
z = x.
(25)
into (11)–(12) to obtain (n − 1)(γ − 1) µt + φµx + φµ + (γ − 1)µφx G x (n − 1)(γ − 1) + θ µx + θ µ + (γ − 1)µθx F G x + φµG + (γ − 1)θ µGF + θµFG = 0,
(26)
(n − 1)(γ − 1) φλ + (γ − 1)λφx x (n − 1)(γ − 1) θλ + (γ − 1)λθx F + θ λx + x
λt + φλx +
+ (γ − 1)λθ F = 0, 1 1 1 αφ + λx + µx G + µG + αθ F = 0, γ −1 γ −1 γ −1
(27) (28)
where we used the method of equation splitting. Here F = d F/d x and G = dG/dx. If the coefficient of FG is used as the normalizing coefficient in (26), then the coefficient of G yields the constraint φ = θ(z) which gives rise to φ = 0, if remark (see Clarkson and Kruskal [8]) concerning φ is used. Equations (28) and (27) now become 1 1 1 λx + µx G + αθ F + µG = 0, γ −1 γ −1 γ −1 (n − 1)(γ − 1)
θ λ + (γ − 1)λθ λt + θλx + x F + (γ − 1)λθ F = 0. x
(29)
(30)
If we treat the coefficient of G as the normalizing coefficient in (29) then it may easily be verified that the constraints µx = µ and λx = µx lead successively to µx = λx = 0, and (29) reduces to
Large Time Asymptotics of Compressible Euler Equations
αθ F +
1 µG = 0. γ −1
109
(31)
It may be noted that if we proceed with θ x = 0, then the resulting solution may be verified to be the same as the one that will be obtained when θ x = 0. So it suffices to consider the case θ x = 0 in Equation (30). Equation (30) then requires that F = kx and reduces to dλ + kn(γ − 1)θλ = 0. dt
(32)
Equation (31), in view of F = kx, demands that G = κ x 2 and then becomes αθ =
−κ 2µ . k γ −1
(33)
Finally (26) takes the form dµ + k(nγ − n + 2)θµ = 0. dt
(34)
Now we insert for F and G into (25) to obtain u = kxθ (t),
c2 = λ(t) + κ x 2 µ(t).
(35)
¯ ¯ λ = e¯ and µ = b, We note that the form (35), with k = 1, κ = −1, θ = a, agrees with (18) used by Liu [1].
3. An exact series solutions of (32)–(34) We may note that the Equations (32)–(34) are equivalent to the system (15)–(17). Now elimination of θ between (33) and (34) results in dµ = µ2 , dt
=
2κ(nγ − n + 2) . α(γ − 1)
(36)
with a solution µ(t) =
−1 α(γ − 1) t −1 . κ 2(nγ − n + 2)
(37)
Equation (37) is the Barenblatt solution (23) obtained by Liu [1]. Now we obtain an exact series solution of the nonlinear equation (36) in which we embed the Barenblatt solution (37) as the first term: The transformation µ(t) = M(τ ), τ = 1/t changes (36) to τ 2 M + M 2 = 0.
(38)
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It is evident that τ = 0 is a regular singular point of (38). Inserting M(τ ) =
∞
aq τ q+1 ,
(39)
q=0
into (38) and equating different powers of τ to 0, we have a0 = −
1 ,
a1 = an arbitrary constant,
=− aq− j−1 a j−1 , q − 1 j=1
(40)
q−1
aq−2
q ≥ 2.
(41)
The relation (41) can be rewritten as aq− j−1 a j−1 , q − 3 j=1 q−2
aq−2 = −
q ≥ 4,
(42)
from which we obtain q−2
aq−2 = (−)q−3 a1
,
q ≥ 4.
(43)
Thus we have M(τ ) =
∞ −1 α(γ − 1) q−2 (−)q−3 a1 τ q−1 , τ+ κ 2(nγ − n + 2) q=3
(44)
where we have used (36) to replace . Correspondingly the function µ(t) is µ(t) =
∞ −1 α(γ − 1) q−2 t −1 + (−)q−3 a1 t 1−q . κ 2(nγ − n + 2) q=3
(45)
It is clear that the Barenblatt solution (23) is contained in (45) as the first term and itself is the general solution of (36) as this solution involves an arbitrary constant a1 . Inserting (45) into (33), we have ∞ 1 1 −2κ q−2 −1 θ= t + (−)q−3 a1 t 1−q , k nγ − n + 2 αk(γ − 1) q=3
(46)
where we used (22) to show that the first term in (46) is the Barenblatt solution. Substituting (46) into (32) and integrating we find the general solution of (32) as
∞ −n(γ −1) q−2 q−3 1−q , (47) (−) a1 t λ = λ0 t (nγ −n+2) + exp q=2
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where λ0 is the constant of integration. We remark that the series solutions (45), (46), and (47) of the system of nonlinear, coupled equations (32), (33), and (34) is an exact asymptotic solutions for large time. Inserting (45), (46), (47) in (35) we get the following exact, large-time asymptotic solutions of (11)–(12): ∞ −2κ q−3 q−2 1−q ¯ + u(x, t) = −κ a(t) x, (−) a1 t (48) αk(γ − 1) q=3
∞ q−3 q−2 1−q 2 (−) a1 t c (x, t) = e¯ (t) + exp q=2
−κ where = (22)–(24).
2κ(nγ − n + 2) , α(γ − 1)
∞ −1 ¯ q−2 b(t) + (−)q−3 a1 t 1−q x 2 , κ q=3
(49)
¯ ¯ and a(t), b(t), and e¯ (t) are the Barenblatt solutions
4. A generalized Barenblatt solution of (15)–(17) We again generalize the Barenblatt solution (22)–(24) by solving the system of nonlinear, coupled equations (15)–(17) which is more general than the system (32)–(34). Elimination of a between (15) and (17) results in bb
−
nγ − n + 3 2 2(nγ − n + 2) 3 b + αbb + b = 0. nγ − n + 2 γ −1
(50)
A series solution of (50) is b(t) =
2(γ − 1)(nγ − n + 1) −2 α(γ − 1) t −1 − t 2(nγ − n + 2) (nγ − n + 2)2 +
4(γ − 1)(nγ − n + 1) −3 t − ···. 13α(nγ − n + 2)2
(51)
Then a(t) is easily obtained from (15) as a(t) =
1 4(nγ − n + 1) −2 t −1 − t nγ − n + 2 α(nγ − n + 2)2 −
16(nγ − n + 1) (12nγ − 12n + 11)t −3 + · · · . 13α 2 (nγ − n + 2)3
(52)
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A series solution of (16) is n(γ −1)
e(t) = e0 t − nγ −n+2 n(γ − 1)(nγ − n + 1) −1 t + ··· , × 1+ α(nγ − n + 2)[n(γ − 1) − (1 + p)(nγ − n + 2)] (53) where p=
n(γ − 1) , nγ − n + 2
(54)
and e0 is an arbitrary constant. We note that the first term in (51), (52), and (53) are nothing but the Barenblatt ¯ ¯ solutions a(t), b(t), and e¯ (t). Therefore the solutions (51), (52), and (53) of (15)–(17) is an another generalization of the Barenblatt solution (22)–(24). The solution of (9)–(10), in the present case is 4(nγ − n + 1) −2 ¯ − u = a(t) t α(nγ − n + 2)2 16(nγ − n + 1)(12nγ − 12n + 11) −3 (55) − t + · · · x, 13α 2 (nγ − n + 2)3 n(γ − 1)(nγ − n + 1) p−1 2 c = e¯ (t) + t + ··· α(nγ − n + 2) [n(γ − 1) − (1 + p)(nγ − n + 2)] ¯ − 2(γ − 1)(nγ − n + 1) t −2 − b(t) (nγ − n + 2)2 4(γ − 1)(nγ − n + 1) −3 (56) + t − · · · x 2, 13α(nγ − n + 2)2 ¯ ¯ where p = n(γ − 1)/(nγ − n + 2) and a(t), b(t), and e¯ (t) are the Barenblatt solution (22)–(24). 5. Lie’s classical symmetries of (9)–(10) We employ the method of Logan and Perez [9] (also see Olver [10], Bluman and Kumei [11]) to seek a one-parameter infinitesimal group of transformations t ∗ = t + T,
x ∗ = x + X,
u ∗ = u + U,
v ∗ = v + V,
(57)
where the generators T, X , U, and V are functions of t, x, u, and v under which the system of Equations (9)–(10) are invariant. We introduce the following notation for the sake of convenience. Let x 1 = t, x 2 = x, u 1 = u, u 2 = v and
Large Time Asymptotics of Compressible Euler Equations
pij =
∂u i , ∂x j
i = 1, 2,
113
j = 1, 2.
The system of differential equations (9)–(10) is said to be constantly conformally invariant under the infinitesimal group (57) if there exist constants α n j (n, j = 1, 2) such that L Hn =
2
αn j H j ,
n = 1, 2,
(58)
j=1
for all smooth surfaces u 1 = u i (x j ). Here L is the derivative in the direction of the extended vector field ∂ ∂ ∂ L = ξxi i + ξuj j + ξ pij i , ∂x ∂u ∂pj where ξx1 = T, i ξ pj =
ξx2 = X,
ξu1 = U,
ξu2 = V,
∂ξui ∂ξui k ∂ξxk i ∂ξxk i m + p − p − p p , ∂x j ∂u k j ∂ x j k ∂u m k j
(59)
are the generators of the transformations of derivatives. Writing out (58) give ∂ Hn j ∂ Hn i ∂ Hn i ξ + ξ + ξ = αn j H j , ∂x j x ∂u i u ∂ pij pj
n = 1, 2.
(60)
If ξ ipj from (59) are substituted into (60), polynomials in pji are obtained. If these equations are made to hold for arbitrary values of pji , thus requiring their coefficients to vanish, the determining equations which are linear partial differential equations in the generators T, X , U, and V are obtained. If we carry out this analysis for (9)–(10) then we may obtain V = 2α22 v,
U = α22 u,
X = α22 x,
T = t0 .
(61)
Integrations of the invariant surface condition du dv dx dt = = , = t0 α22 x α22 u 2α22 v
(62)
give the similarity transformation u = x f (z),
v = x 2 g(z),
z = xe−α22 t/t0 .
(63)
Putting (63) in (9)–(10) we get the following ordinary differential equations −
α22
1 2 g+ zg + α f = 0, zf + f 2 + zf f + t0 γ −1 γ −1
(64)
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−
α22
zg + z f g + [n(γ − 1) + 2] f g + (γ − 1)zg f = 0. t0
(65)
A simple solution of this system is g(z) = g0 z −2 ,
f (z) = −α,
(66)
provided that α = 2α 22 /(n(γ − 1)t 0 ); here g 0 = 0 is the constant of integration. The solution (66) yields a solution of (9)–(10) in which u t = (c2 )x = 0. In fact this solution is u = −αx,
c2 = g0 e2α22 t/t0 ,
(67)
which clearly satisfy the initial conditions u(x, t = 0) = −αx,
c2 (x, t = 0) = g0 .
(68)
6. Lie’s classical symmetries of (11)–(12) The similarity transformation of (11)–(12) under Lie’s classical method is surprisingly the same transformation (69) obtained for (9)–(10): u = x f (z),
v = x 2 g(z),
z = xe−α22 t/t0 .
(69)
But the equations satisfied by f (z) and g(z) are obviously different:
−
2 1 g+ zg + α f = 0, γ −1 γ −1
(70)
α22
zg + z f g + [n(γ − 1) + 2] f g + (γ − 1)zg f = 0. t0
(71)
Elimination of f between (70) and (71) results in 4 1 2
GG + 2 + n + GG + G
γ −1 γ −1 2 αα22
G2 + +2 n + G = 0, γ −1 t0
(72)
where we used g(z) = G(η),
η = log z.
(73)
Equation (72) is the Euler–Painlev´e equation introduced by Sachdev and collaborators [12–15] and generalized by Mayil Vaganan and Senthilkumaran [16].
Large Time Asymptotics of Compressible Euler Equations
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Equation (72) can be simplified by reducing its order by one through G (η) = H (G) to 4 1 H2 dH + 2+n+ H+ H dG γ −1 γ −1 G αα22 H 2 G+ = 0. +2 n + (74) γ −1 t0 G We now consider the case α 22 = 0 to obtain a solution of a system of ordinary differential equations uv +
(n − 1)(γ − 1) uv + (γ − 1)vu = 0, x 1 v = −αu, (γ − 1)
(75) (76)
which is deduced from (11) to (12) by using u t = 0 = v t , v = c2 , because by (69), z = x and therefore u = u(x), v = v(x). Equation (72) with α 22 = 0 through the transformation G = V 1/(a+1) changes to 4 4 γ V + 2n + V = 0. (77) V
+ 2 + n + γ −1 γ −1 γ −1 Its solution is V (η) = h 1 eb1 η + h 2 eb2 η , where
(78)
4 8 2 − 2+n+ ± 4(1 + n) + n 2 + (1 − γ ) + n + 2 − nγ γ −1 (γ − 1) γ − 1
b1,2 =
2
Here h1 and h2 are arbitrary constants. Then
(γ −1)/γ G(η) = h 1 eb1 η + h 2 eb2 η .
.
(79)
(80)
Thus the solution of the system (75)–(76) is
(γ −1)/γ 2 x h 1 x b 1 + h 2 x b2 α(γ − 1)
(−1)/γ 1 h 1 b1 x b1 + h 2 b2 x b2 . x h 1 x b1 + h 2 x b2 − αγ
(γ −1)/γ . v(x, t) = x 2 h 1 x b1 + h 2 x b2
u(x, t) = −
(81) (82)
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7. Direct method to (9)–(10) Following Clarkson and Kruskal [8] we insert u = φ(x, t) + θ(x, t)F(z),
(83)
c2 = ξ (x, t) + µ(x, t)G(z),
(84)
z = z(x, t),
(85)
into equations (9)–(10) and obtain FF + 6 (z)G + 5 (z)G + 4 (z)F 2 + 3 (z)F
+ 2 (z)F + 1 (z) = 0,
(86)
FG + 7 (z)GF + 6 (z)F + 5 (z)F G + 4 (z)F + 3 (z)G + 2 (z)G + 1 (z) = 0,
(87)
where we have introduced i (z), i = 1, 2, . . . , 6, j (z), j = 1, 2, . . . , 7, such that 1 φt + φφx + αφ + (88) ξx = θ 2 z x 1 (z), γ −1 θt + φθx + αθ + θφx = θ 2 z x 2 (z),
(89)
z t + φz x = θ z x 3 (z),
(90)
θx = θ z x 4 (z),
(91)
1 µx = θ 2 z x 5 (z), γ −1
(92)
1 µ = θ 2 6 (z), γ −1
(93)
(n − 1)(γ − 1) ξ φ + (γ − 1)ξ φx = θµz x 1 (z), x (n − 1)(γ − 1) µt + φµx + µφ + (γ − 1)µφx = θµz x 2 (z), x z t + φz x = θ z x 3 (z), ξt + φξx +
(n − 1)(γ − 1) θ ξ + (γ − 1)ξ θx = θµz x 4 (z), x (n − 1)(γ − 1) θ µx + θ µ + (γ − 1)µθx = θµz x 5 (z), x (γ − 1)ξ = µ6 (z), θ ξx +
(γ − 1) = 7 (z).
(94) (95) (96) (97) (98) (99) (100)
Large Time Asymptotics of Compressible Euler Equations
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The remarks listed below play an important role in what follows: Remarks 1 & 2: If φ(x, t) (or ξ (x, t)) is to be obtained from an equation of the form ˜ φ(x, t) = φ(x, t) + β(x, t)(z) (or ξ (x, t) = ξ˜ (x, t) + β(x, t)(z)), then we may set (z) = 0. Remarks 3 and 4: If θ(x, t) (or µ(x, t)) is given by an ˜ t)(z)(or µ(x, t) = µ(x, equation of the form θ(x, t) = θ(x, ˜ t)(z)), then we may choose (z) = 1. Remark 5: If the equation (z) = z˜ (x, t) is to be solved for z, then we may write (z) = z. In view of remark 2, we choose 6 (z) = 0 in (99) and obtain ξ = 0. On using ξ = 0, Equations (94) and (97) give 1 (z) = 4 (z) = 0. Replacing 4 (z) by 4 (z)/4 (z) in (91) and integrating with respect to x, we find that ˆ > 0 is the function of integration; applying remark θ = θˆ (t)4 (z), where θ(t) 3 to get 4 (z) = 0 and θ = θ(t). Next we employ remark 5 to (93) to choose 6 (z) = z and obtain z(x, t) =
1 µθ −2 , γ −1
(101)
A perusal of Equations (90) and (96) reveals that 3 (z) = 3 (z). Using remark 1 to (90) to take 3 (z) = 0 and φ = −z t /z x . Now using θ = θ (t) and 5 (z) = 5 (z)/5 (z) in (98) and integrating the resulting equation with respect to x, we get µ(x, t) = x k s(t),
k = −(n − 1)(γ − 1),
(102)
where s(t) > 0 is the function of integration, provided that remark 4 is used to obtain 5 (z) = 1 implying that 5 (z) = 0. Equations (101) and (92) lead to 5 (z) = 1. Substituting (101) and (102) into φ = −z t /z x , we find that
θ s
x 2 − . (103) φ= k θ s Solving (101) for x after substituting for µ from (102), we get x = [(γ − 1)θ 2 s −1 z]1/k .
(104)
Putting (101), (102), and (104) in (95), we have
1 s
k 1 θ s
−1 2 −1 1−1/k (γ − 1)θ 2 − + = sθ s z 2 (z). k θ s γ −1 s (γ − 1)2 (105) Equation requires that 2 (z) = l 2 z 1/k−1 where l2 is a constant and becomes
1−1/k 1 s
kl2 1 θ s
2 − + = sθ −1 (γ − 1)θ 2 s −1 . (106) 2 k θ s γ −1 s (γ − 1) In a similar fashion, Equations (88)–(89), respectively, force us to choose 1 (z) = l 1 z 2/k−1 and 2 (z) = l 3 z 1/k−1 , where l1 and l3 are constants, and in turn yield
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2 2 θ
θ
1 s
s
2 −2 2 − + 2 k θ θ s s
1 s 2 α θ s
θ + 2 2 − + 2 − = l1 k(γ − 1)−2/k θ 2−4/k s 2/k , k θ s k θ s
θ 1 θ s
l3 k θ 1−2/k s 1/k . + 2 − +α = θ k θ s (γ − 1)1/k
(107) (108)
A solution of the overdetermined system of Equations (106)–(108) is θ(t) = ae pt ,
s(t) = be− p(k−2)t ,
(109)
where
l1 = k −1 (γ − 1)2/k b−2/k a 4/k−2 [k −2 4 p 2 + q 2 − 4 pq) + αk −1 (2 p − q) , l2 = k −1 (γ − 1)1/k b−1/k a 2/k−1 [ p(γ − 1) − p(k − 2)], l3 = k −1 (γ − 1)1/k b−1/k a 2/k−1 [2 p + α].
(110)
The expressions for µ, φ, and z are now obtained by inserting (110) into (102), (103), and (101): µ = x k e− p(k−2)t , z=
φ = px,
1 ba −2 x k e− pkt . γ −1
(111) (112)
The corresponding similarity form of the solution of (9)–(10) is determined if we put (109) and (111)–(112) into (83)–(85): u(x, t) = px + ae pt F(z),
(113)
c2 (x, t) = bx k e− p(k−2)t G(z),
(114)
z(x, t) =
1 ba −2 x k e− pkt . γ −1
(115)
The functions F(z) and G(z) appearing in (113)–(114) are governed by the equations FF + zG + G + c2 z 1/k−1 F + mz 2/k−1 = 0,
(116)
FG + (γ − 1)GF + l2 z 1/k−1 G = 0,
(117)
where we have substituted for i (z), i = 1, 2, 3, 4, 5, 6 and j (z), j = 1, 2, 3, 4, 5, 6, 7. A solution of this system is
Large Time Asymptotics of Compressible Euler Equations
−l2 kz 1/k , F= (γ + 1 − k)
119
k l22 k l2l3 k 2 G= −l1 + − z k −1 . 2 (γ + 1 − k) (γ + 1 − k)2 (118)
8. Direct method to (11)–(12) We insert (83)–(85) into (11)–(12) to obtain FG + 7 (z)GF + 6 (z)F + 5 (z)F G + 4 (z)F + 3 (z)G
+ 2 (z)G + 1 (z) = 0,
(119)
G + 10 (z)G + 9 (z)F + 8 (z) = 0,
(120)
The functions n (z), n = 1, 2, . . . , 10 are introduced according to (n − 1)(γ − 1) ξ φ + (γ − 1)ξ φx = θµz x 1 (z), x (n − 1)(γ − 1) µt + φµx + µφ + (γ − 1)µφx = θµz x 2 (z), x z t + φz x = θ z x 3 (z), ξt + φξx +
(n − 1)(γ − 1) θ ξ + (γ − 1)ξ θx = θµz x 4 (z), x (n − 1)(γ − 1) θµx + θ µ + (γ − 1)µθx = θµz x 5 (z), x (γ − 1)ξ = µ6 (z), θξx +
(γ − 1) = 7 (z), αφ +
(121) (122) (123) (124) (125) (126) (127)
1 1 ξx = µz x 8 (z), (128) γ −1 (γ − 1) αθ =
1 µz x 9 (z), (129) (γ − 1)
µx = µz x 10 (z).
(130)
Remark 2 and Equation (126) leads to ξ = 6 (z) = 0. Then (121) and (124) give 1 (z) = 4 (z) = 0. Writing 10 (z)/10 (z) for 10 (z) in (130), integrating with respect to x and using remark 4, we arrive at 10 (z) = 0 and µ = µ(t). If we take φ = θ and 8 = 9 , then µt = 0 leading to µ = q, a constant. The condition φ = θ and (122) and (125) give 2 = 5 . Now taking 5 = (γ − 1) 5 (z)/5 (z) in Equation (125), integrating with respect to x and using remark 5, we finally have 5 = (γ − 1)z −1 and
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P. L. Sachdev et al.
z = x n−1 θa(t),
(131)
where the function a(t) is to be determined. We set θ = xb(t). In view of 2−n (131), Equations (123) and (129) require that 9 = c9 z n and 3 − 1 = c3 and we therefore have 2
α(γ − 1) = nc9 qa n b
2−n n
,
(132)
b a
+ = nc3 b(t). b a Solutions of (132) and (133) are n−2 α(γ − 1) n/2 −1 2 2−n a(t) = t 2 , nc9 q 2c3
(133)
b(t) = −
1 −1 t . 2c3
(134)
The expressions for φ, θ, λ, µ, and z thus are 1 −1 xt , λ(x, t) = 0, µ(x, t) = q, 2c3 n −1 2 α(γ − 1) n/2 n −n z(x, t) = x t 2. 2c3 nc9 q
φ(x, t) = θ(x, t) = −
(135)
(136)
The corresponding similarity transformation of (9) and (12) is obtained by putting (135)–(136) into (83)–(84): u(x, t) = −
1 −1 xt (1 + F(z)), 2c3
c2 (x, t) = qG(z).
(137)
The functions F(z) and G(z) appearing in (137) are governed by the equations FG + (γ − 1)GF + (γ − 1)z −1 FG + (1 + c3 )G + (γ − 1)z −1 G = 0, G + c9 z
2−n n
(1 + F) = 0.
(138) (139)
Elimination of F between (138) and (139) leads to GG
+
2n − 2 −1
1 c3 c9 2−n
2 z GG + G − z n G = 0. n γ −1 γ −1
(140)
Multiplying (140) by z 2 and taking G(z) = H (s), s = log z Equation (140) becomes HH
+
n−2 1 c3 c9 2 s
2 HH + H − e n H = 0. n γ −1 γ −1
A solution of (141) is
(141)
Large Time Asymptotics of Compressible Euler Equations
H (s) =
c3 c9
2
2 1 (γ − 1) 1 + nγ −1
e n s,
121
(142)
and the corresponding solution of (140) is
G(z) =
c 3 c9
2
2 1 (γ − 1) 1 + nγ −1
zn .
(143)
Substituting (143) in (139), we find that F(z) = −
2c3 − 1. 2 1 n(γ − 1) 1 + nγ −1
(144)
Putting (143)–(144) in (137) yields the following solution of (9)–(10): 1
xt −1 , 2 1 n(γ − 1) 1 + nγ −1 α t −1 x 2 . c2 (x, t) = − 2 1 2n 1 + nγ −1 u(x, t) =
(145)
(146)
9. A new nonlinear equation If we eliminate u between (11) and (12) and use v = c2 , we have vt −
1 (n − 1) 1 vx2 − vvx − vvx x = 0. α(γ − 1) αx α
(147)
If we apply Lie’s classical method to (147) and seek it to be invariant under v ∗ = v + V (t, x, v),
t ∗ = t + T (t, x, v),
x ∗ = x + X (t, x, v),
(148)
then we find that V = a1 v,
T = ct + d,
x X = (c + a1 ) . 2
(149)
The Lie group generator L is ∂ ∂ ∂ L=X +T +V , ∂x ∂t ∂v x ∂ ∂ ∂ ∂ x ∂ L=c +t + a1 +v +d , 2 ∂x ∂t 2 ∂x ∂v ∂t
(150)
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P. L. Sachdev et al.
where we have used (149). Thus the Lie algebra of infinitesimal symmetries of Equation (147) is spanned by the three vector fields x ∂ x ∂ ∂ ∂ ∂ + t , X2 = + v , X3 = . (151) 2 ∂x ∂t 2 ∂x ∂v ∂t The commutation relations between these vector fields are given by the following table, the entry in row ‘i’ and column ‘j’ representing [X i , X j ]: X1 =
∗ X1 X2 X3
X1 0 0 X3
X2 0 0 0
X3 −X 3 0 0
The corresponding local one-parameter groups are the following: σs1 1 G 1 : (x, t, v) → x exp s, t exp s, v , 2 σs2 1 G 2 : (x, t, v) → x exp s, t, v exp s , 2
(152)
σs3
G 3 : (x, t, v) → (x exp s, exp s, v), j
where σ s = (exp(sX j )), 1 ≤ j ≤ 3. Because each local Lie group G j is a symmetry group, exponentiation shows that if v = v(x, t) is a solution of (147) then so are x v 1 (x, t) = v exp(−s), t exp(−s) , 2 x (153) v 2 (x, t) = v exp(−s), t , 2 v 3 (x, t) = v(x, exp(−s)). We now construct below the solution of (147) for each of the infinitesimal generators: Case (i): X 1 =
x ∂ 2 ∂x
+ t ∂t∂ .
Clearly X 1 = X 1 (x, t), X 1 (v) = 0. The invariant surface condition dt/t = 2d x/x = dv/0, when integrated give x2 . t
(154)
4 2 z F + αz F = 0. γ −1
(155)
v = F(z),
z=
Putting (154) in (147), we get 4zFF
+ 2nFF +
We obtain a regular perturbation solution of (155) by taking α = , 0 < 1. Inserting
Large Time Asymptotics of Compressible Euler Equations
F(z) = F0 (z) + F1 (z) + 2 F2 (z) + · · · ≈ F0 (z) + F1 (z),
123
(156)
into (155) with α = , and equating the powers of 0 and 1 to zero, we have 2n 4 2 F0 F0 + F0 = 0, z γ −1
(157)
2n 8 4 F0 F1
+ F1 F0
+ F0 F1 + F1 F0 + F0 F1 + F0 = 0. z γ −1
(158)
4F0 F0
+
A solution of (157)–(158) is 2k5 (2−n)/2 (γ −1)/γ , z F0 (z) = 2−n (γ − 1) . F1 (z) = b1 z, b1 = − 2nγ − 2n + 4) Inserting (160) and (159) to Equation (156), we have 2k5 (2−n)/2 (γ −1)/γ z + b1 z. F(z) = 2−n
(159) (160)
(161)
Thus the corresponding solution of (147) with α = , namely, αvt − 1 v 2 − (n −x 1) vvx − vvx x = 0 is (γ − 1) x
2k5 (2−n) −(2−n)/2 v(x, t) = x t 2−n
(γ −1)/γ
+ b1 x 2 t −1 .
(162)
∂ . Here the similarity transformation is v = Case (ii): X 2 = x2 ∂∂x + v ∂v x F(t), because t = z, and F satisfies 4 + 2n(γ − 1)
F − F 2 = 0. (163) α(γ − 1) 2
Its solution is F(z) = 1/r t + s. The corresponding solution of (147) is v(x, t) = x 2 (r t + s)−1 . Case (iii): X 3 =
∂ . ∂t
(164)
In the present case, we have v = F(z), z = x and
zFF
+ (n − 1)FF +
1 2 z F = 0. γ −1
(165)
A solution of (165) is
k1 2−n z + k2 F(z) = 2−n
(γ γ−1)
.
(166)
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The corresponding solution of (147) is (γ γ−1) k1 2−n v(x, t) = + k2 . x 2−n
(167)
In the general case the invariant surface condition is dt dv dx = = , x ct + d a1 v (c + a1 ) 2 which when integrated yields a1
v = x c1 f (z),
z=
1 c1 = (c + a1 ), 2
(t + d/c)c1 /c . x
(168)
(169)
Here F(z) is any solution of 2
a1 1 1 a1 a1 a1 2
2 2 z F − + (n − 1) + −1 F2 z FF − γ −1 γ − 1 c1 c1 c1 c1 αc1 ac1 −1
2a1 (170) + + n + 1 zFF + z 1 F = 0. c1 (γ − 1) c To solve (170) we consider the following simple cases: Case 1: ac11 − 2 = 0. Equation (170) is changed by inserting F(z) = G(θ ), θ = log z, to an Euler–Painlev´e equation GG
− where
1 αc1
2 G + (A2 − 1)GG + A1 G 2 + G = 0, γ −1 c
(171)
2 1 a1 a1 a1 a1 + (n − 1) + −1 , A1 = − γ − 1 c1 c1 c1 c1 2 a1 +n+1 . A2 = γ − 1 c1
(172)
Case 2: Equation (170) is changed by inserting F(z) = G(θ), θ = log z, to GG
−
1 αc1 ( ac1 −1)θ
2 G + (A2 − 1)GG + A1 G 2 + e 1 G = 0, γ −1 c
(173)
We proceed to determine a perturbation solution of (173) by taking α = , a very small positive real number (See Daniel Zwillinger[17]). Writing G(θ) = G 0 (θ ) + G 1 (θ ) + 2 G 2 (θ ) + · · · ≈ G 0 (θ) + G 1 (θ),
(174)
into the Equation (173), with α = , and equating the coefficient of 0 and 1 to zero, we have
Large Time Asymptotics of Compressible Euler Equations
125
1 G 2 + (A2 − 1)G 0 G 0 + A1 G 20 = 0, (175) γ −1 0 2 G 0 G
1 + G 1 G
0 − G 0 G 1 + (A2 − 1) G 0 G 1 + G 1 G 0 + 2A1 G 0 G 1 γ −1 a c1 ( 1 −1)θ
+ e c1 G 0 = 0. (176) c Solving Equation (175) and we find that G 0 G
0 −
G 0 = Ae Bθ ,
where A is arbitrary and B =
4A (γ −2) (1−A2 )± (1−A2 )2 − 1γ −1 2(γ −2) γ −1
(177) , is a positive real number.
Using (177), then the solution of (176) is G 1 = l 3 e m 3 θ + l 4 e m 4 θ + k 3 e k4 θ ,
(178)
where l 3 , l 4 are arbitrary constants and a1 − 1, c1 2 2B 2B − (A2 − 1) + − (A2 − 1) − 4(B 2 − 2B − A2 B + 2A1 ) γ −1 γ −1 m3 = , 2 2 2B 2B − (A2 − 1) − − (A2 − 1) − 4(B 2 − 2B − A2 B + 2A1 ) γ −1 γ −1 m4 = , 2 −1 c1 B 2 2B 2 k4 + A2 − 1 − k4 + B − 2B + A2 B + 2A1 . k3 = − c γ −1 k4 =
(179)
Then the solution of (173) is
G(θ) = Ae Bθ + l3 em 3 θ + l4 em 4 θ + k3 ek4 θ ,
(180)
and the corresponding solution of (170) is a1 −1 F(z) = Az B + l3 z m 3 + l4 z m 4 + k3 z c1 ,
(181)
finally the solution of Equation (147) is B m a1 (t + d/c)c1 /c (t + d/c)c1 /c 3 c1 v(x, t) = x A + l3 x x m a1 −1 (t + d/c)c1 /c 4 (t + d/c)c1 /c c1 + l4 + k3 . x x
(182)
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10. Nonclassical symmetries of (147) To apply the nonclassical method to (147) we require that (147) and the invariant condition to be invariant under the infinitesimal generator. In the case T = 0, without loss of generality, we may set T (x, t, v) = 1. The nonclassical method applied to (147) gives rise to the following determining equations for the infinitesimals 1 v X vv − X v = 0, (183) γ −1 Vt −
(n − 1) 1 vVx − vVx x + 2V X x − v −1 V 2 = 0. αx α
1 2(n − 1) 1 Vv − v X v − vVvv α(γ − 1) αx α 2 1 v −1 V = 0, + X xv + α α(γ − 1) (n − 1) 2 (n − 1) V v[Vv − X x ] v X − X + 2V X − − t v x αx 2 α(γ − 1) αx
(184)
−2X X v −
2(n − 1) 1 v X x + v −1 X V = 0, − v[2Vxv − X x x ] − α αx Solving these equations, we obtain 1 + p1 T = 1, X = [ p1 t + p2 ]−1 x, V = [ p1 t + p2 ]−1 v. 2
(185)
(186)
(187)
These infinitesimals are same as in the classical method. A special case: Suppose V is a function of v only then it is easy to verify that infinitesimals are c1 (188) T = 1, X = x, V = c1 v. 2 Integrations of the invariant surface conditions dv dx dt = c x = . 1 1 c1 v 2 give the similarity solution in the form v = x 2 F(z),
z = log x
(189)
c1 t. 2
This form of solution to (9) and (12) is new.
1 2x F − x F (z) . u=− α(γ − 1)
(190)
(191)
Large Time Asymptotics of Compressible Euler Equations
127
Putting (190) in (147), we get the equation for F(z) as 4 4 1 αc1
2
2 F + 2+n+ FF + F + F = 0, FF + 2n + γ −1 γ −1 γ −1 2 (192) which is an Euler–Painlev´e equation.
11. Results and discussion We have reduced the PDE (147) through the nonclassical similarity transformation (190) to the ODE of the form H
+ A1 H (1−γ )/γ H + A2 H + A3 H = 0.
(193)
Interestingly the following PDEs ut + um u x +
r j u − (at + bt jn/2 )−1 u x x = 0, 2t 2
δ vt + Av m vx + σ v − vx x = 0, 2 also reduce through the respective similarity reductions u = t −1/m (at + bt jn/2 )−1/n f (z), c t jm/2−1 z=x− log , a( jm/2 − 1) a + bt jm/2−1 v = f (z),
z = x − ct,
(194) (195)
(196) (197)
to ODEs of the same form given in (193). We generalized the Barenblatt solution (22)–(24) by solving the two systems of nonlinear, coupled system of ODEs. We remark that the series solution (45)–(47) of the system of nonlinear, coupled equations (32), (33), and (34) is an exact asymptotic for large time. We now infer the following facts from the two solutions (48)–(49) and (55)–(56). We note that the former is a solution of (9) and (12) while the latter is that of (9)–(10). 1. The dependence of u(x, t) on x and t is the same. 2. The coefficients of x2 in c2 (x, t) agree in form. 3. The coefficients of x0 in c2 (x, t), apart from e¯ (t), are entirely different. Indeed it is evident that (49) contains an exponential of an exact series while (56) includes only a series. We prove that the large-time behaviors of solutions of the two systems (9)–(10) and (11)–(12) are asymptotically equivalent by establishing the fact
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that the solutions of the two systems of ODEs (15)–(17) and (22)–(24) are asymptotically equivalent for large time. We also have obtained an exact closed form solutions (75)–(76) of the nonlinear coupled system of ordinary differential equations (168)–(170) using the invariance analysis. References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.
17.
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