Approximate symmetry reduction approach: infinite series reductions to the KdV-Burgers equation

Abstract

For weak dispersion and weak dissipation cases, the (1+1)-dimensional KdV-Burgers equation is investigated in terms of approximate symmetry reduction approach. The formal coherence of similarity reduction solutions and similarity reduction equations of different orders enables series reduction solutions. For weak dissipation case, zero-order similarity solutions satisfy the Painlev\'e II, Painlev\'e I and Jacobi elliptic function equations. For weak dispersion case, zero-order similarity solutions are in the form of Kummer, Airy and hyperbolic tangent functions. Higher order similarity solutions can be obtained by solving linear ordinary differential equations.

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arXiv:0801.0856v2 [nlin.SI] 26 Sep 2008
Approximate symmetry reduction approach: infinite series
reductions to the KdV-Burgers equation
Xiaoyu Jiao1, Ruoxia Yao1,2,3, Shunli Zhang4and S. Y. Lou1,2
1Department of Physics, Shanghai Jiao Tong University, Shanghai, 200240, China
2Department of Physics, Ningbo University, Ningbo, 315211, China
3School of Computer Science, Shaanxi Normal University, Xi’an, 710062, China
4Department of Mathematics, Northwest University, Xi’an, 710069, China
Abstract
For weak dispersion and weak dissipation cases, the (1+1)-dimensional KdV-Burgers equation
is investigated in terms of approximate symmetry reduction approach. The formal coherence of
similarity reduction solutions and similarity reduction equations of different orders enables series
reduction solutions. For weak dissipation case, zero-order similarity solutions satisfy the Painlev´ e
II, Painlev´ e I and Jacobi elliptic function equations. For weak dispersion case, zero-order similarity
solutions are in the form of Kummer, Airy and hyperbolic tangent functions. Higher order similarity
solutions can be obtained by solving linear ordinary differential equations.
PACS numbers: 02.30.Jr
Keywords: KdV-Burgers equation, approximate symmetry reduction, series reduction solutions
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I.INTRODUCTION
Nonlinear problems arise in many fields of science and engineering. Lie group theory
[1, 2, 3] greatly simplifies many nonlinear partial differential equations. Exact analytical
solutions are nonetheless difficult to study in general. Perturbation theory [4, 5, 6] was thus
developed and it plays an essential role in nonlinear science, especially in finding approximate
analytical solutions to perturbed partial differential equations.
The integration of Lie group theory and perturbation theory yields two distinct approxi-
mate symmetry reduction methods. The first method due to Baikov, et al [7, 8] generalizes
symmetry group generators to perturbation forms. For the second method proposed by
Fushchich, et al [9], dependent variables are expanded in perturbation series and approxi-
mate symmetry of the original equation is decomposed into exact symmetry of the system
resulted from perturbation. The second method is superior to the first one from the com-
parison in Refs. [10, 11].
The well known Korteweg-de Vries-Burgers (KdV-Burgers) equation
ut+ 6uux+ µuxxx+ νuxx= 0,
(1)
with µ and ν constant coefficients, is widely used in the many physical fields especially in
fluid dynamics. The effects of nonlinearity (6uux), dispersion (µuxxx) and dissipation (νuxx)
are incorporated in this equation which simulates the propagation of waves on an elastic
tube filled with a viscous fluid [12], and the flow of liquids containing gas bubbles [13] and
turbulence [14, 15], etc.
Johnson [12] inspected the travelling wave solutions to the weak dissipation (ν ≪ 1) KdV-
Burgers equation (1) in the phase plane by a perturbation method and developed formal
asymptotic expansion for the solution. Tanh function method was applied to Eq. (1) in the
limit of weak dispersion (µ ≪ 1) in a perturbative way [16]. In Refs. [17, 18], perturbation
analysis was also applied to the perturbed KdV equations
ηt+ 6ηηx+ ηxxx+ αc1ηxx+ αc2ηxxxx+ αc3(ηηx)x= 0, α ≪ 1,
(2)
and
ut+ uux+ uxxx= ǫαu + ǫβuxx, ǫ ≪ 1,
(3)
respectively.
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Eq. (1) can also be manipulated by means of approximate symmetry reduction approach.
Section II and section III are devoted to applying approximate symmetry reduction approach
to Eq. (1) under the case of weak dissipation (ν ≪ 1) and weak dispersion (µ ≪ 1),
respectively. Section IV is conclusion and discussion of the results.
II.APPROXIMATE SYMMETRY REDUCTION APPROACH TO WEAK DISSI-
PATION KDV-BURGERS EQUATION
According to the perturbation theory, solutions to perturbed partial differential equations
can be expressed as a series containing a small parameter. Specifically, we suppose that the
weak dissipation (ν ≪ 1) KdV-Burgers equation (1) has the solution
u =
∞
?
k=0
νkuk,
(4)
where ukare functions of x and t, and solve the following system
uk,t+ 6
k
?
i=0
uk−iui,x+ µuk,xxx+ uk−1,xx= 0, (k = 0, 1, ···)(5)
with u−1 = 0, which is obtained by inserting Eq. (4) into Eq. (1) and vanishing the
coefficients of different powers of ν.
The next crucial step is to study symmetry reduction of the above system via the Lie
symmetry approach [19]. To that end, we construct the Lie point symmetries
σk= Xukx+ Tukt− Uk, (k = 0, 1, ···),
(6)
where X, T and Ukare functions with respect to x, t and ui, (i = 0, 1, ···). The linearized
equations for Eq. (5) are:
σk,t+ 6
k
?
i=0
(σk−iui,x+ uk−iσi,x) + µσk,xxx+ σk−1,xx= 0, (k = 0, 1, ···)(7)
with σ−1= 0. Eq. (7) means that Eq. (5) is invariant under the transformations uk →
uk+ εσk(k = 0, 1, ···) with an infinitesimal parameter ε.
There are infinite number of equations in Eqs. (5) and (7) and infinite number of argu-
ments in X, T and Uk(k = 0, 1, ···). To simplify the problem, we begin the discussion
from finite number of equations.
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Confining the range of k to {k|k = 0, 1, 2} in Eqs. (5), (6) and (7), we see that X, T,
U0, U1and U2are functions with respect to x, t, u0, u1and u2. In this case, the determining
equations can be obtained by substituting Eq. (6) into Eq. (7), eliminating u0,t, u1,tand
u2,tin terms of Eq. (5) and vanishing all coefficients of different partial derivatives of u0, u1
and u2. Some of the determining equations read
Tx= Tu0= Tu1= Tu2= 0,
from which we have T = T(t). Considering this condition, we choose the simplest equations
for X
Xu0= Xu1= Xu2= 0,
from which we have X = X(x,t). Considering this condition, we choose the simplest
equations for U0, U1and U2
U0,xu2= U0,u0u0= U0,u0u1= U0,u0u2= U0,u1u1= U0,u1u2= U0,u2u2= 0,
U1,u0u0= U1,u0u1= U1,u0u2= U1,u1u1= U1,u1u2= U1,u2u2= 0,
U2,u0u0= U2,u0u1= U2,u0u2= U2,u1u1= U2,u1u2= U2,u2u2= 0,
of which the solutions are
U0= F1(x,t)u0+ F2(x,t)u1+ F3(t)u2+ F4(x,t),
U1= F5(x,t)u0+ F6(x,t)u1+ F7(x,t)u2+ F8(x,t),
U2= F9(x,t)u0+ F10(x,t)u1+ F11(x,t)u2+ F12(x,t).
Under these relations, the determining equations are simplified to
Xxx= F2= F3= F5= F7= F8= F9= F10= F12= F1,x= F4,xx= F4,t= F6,x= F11,x= 0,
Tt= 3Xx, Xt= 6F4, F1,t= −6F4,x, F6,t= −6F4,x, F11,t= −6F4,x,
Tt= Xx− F1, Tt= 2Xx− F1+ F6, Tt= Xx+ F11− 2F6, Tt= 2Xx− F6+ F11,
which provide us with
X = 6at + cx + x0, T = 3ct + t0, U0= −2cu0+ a, U1= −cu1, U2= 0,
where a, c, x0and t0are arbitrary constants.
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Similarly, limiting the range of k to {k|k = 0, 1, 2, 3} in Eqs. (5), (6) and (7) where
X, T, U0, U1, U2and U3are functions with respect to x, t, u0, u1, u2and u3, we repeat the
calculation as before and obtain
X = 6at + cx + x0, T = 3ct + t0, U0= −2cu0+ a, U1= −cu1, U2= 0, U3= cu3,
where a, c, x0and t0are arbitrary constants.
With more similar calculation considered, we see that X, T and Uk(k = 0, 1, ···) are
formally coherent, i.e.,
X = 6at + cx + x0, T = 3ct + t0, Uk= (k − 2)cuk+ aδk,0, (k = 0, 1, ···) (8)
where a, c, x0 and t0 are arbitrary constants. The notation δk,0 satisfying δ0,0 = 1 and
δk,0= 0 (k ?= 0) is adopted in the following text. Subsequently, solving the characteristic
equations
dx
X=dt
U0
U1
leads to the similarity solutions to Eq. (5) which can be distinguished in the following two
T,du0
=dt
T,du1
=dt
T, ··· ,duk
Uk
=dt
T, ···(9)
subcases.
A.Symmetry reduction of the Painlev´ e II solutions
When c ?= 0, for brevity of the results, we rewrite the constants a, x0and t0as ca, cx0
and ct0, respectively. Solvingdx
X=dt
Tin Eq. (9) leads to the invariant
I(x,t) = ξ = (x − 3at + x0− 3at0)(3t + t0)−1
3.
(10)
In the same way, we get other invariants
I0(x,t,u0) = P0=1
2(3t + t0)
2
3(2u0− a)(11)
and
Ik(x,t,uk) = Pk= uk(3t + t0)−1
3(k−2), (k = 1, 2, ···) (12)
from
du0
U0=
dt
Tand
duk
Uk=
dt
T(k = 1, 2, ···) respectively. Viewing Pk(k = 0, 1, ···) as
functions of ξ, we get the similarity solutions to Eq. (5)
uk= (3t + t0)
1
3(k−2)Pk(ξ) +1
2aδk,0, (k = 0, 1, ···)(13)
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