Nonlinear Dyn DOI 10.1007/s11071-013-1149-4

O R I G I N A L PA P E R

The Cauchy problem for the equation of the Burgers hierarchy Nikolai A. Kudryashov · Dmitry I. Sinelshchikov

Received: 17 May 2013 / Accepted: 11 November 2013 © Springer Science+Business Media Dordrecht 2013

Abstract The Cauchy problem for the equation of the Burgers hierarchy is considered. The Green function for the associated linear problem is constructed. Using the Cole–Hopf transformation the solution of the Cauchy problem for the equation of the Burgers hierarchy is given. Several particular cases are considered and discussed. Keywords Burgers hierarchy · Burgers equation · Sharma–Tasso–Olver equation · Cauchy problem · Cole–Hopf transformation

In the recent years the Burgers equation and its generalizations are intensively studied [1–4]. There are a lot of mathematical and physical applications of these equations [5–10]. In the present work we study the equation of the Burgers hierarchy that has the following form: n ∈ N.

N.A. Kudryashov (B) · D.I. Sinelshchikov Department of Applied Mathematics, National Research Nuclear University MEPhI, Kashirskoe sh., 31, 115409 Moscow, Russia e-mail: [email protected]

ut + α1 (uxx + 2uux ) = 0.

(2)

The Burgers equation is one of the simplest models for the flow of viscous incompressible liquid. In the case of n = 2, Eq. (1) transforms to the Sharmo–Tasso–Olver equation [11, 15, 16]:   ut + α2 uxxx + 3u2 ux + 3uuxx + 3u2x = 0. (3)

1 Introduction

 n ∂ ∂ ut + αn + u u = 0, ∂x ∂x

This hierarchy was introduced in [11]. The first members of hierarchy (1) are used for the description of wave processes in hydrodynamics, nonlinear acoustics and other applications. At n = 1 from (1) we have the famous Burgers equation [12–14]:

(1)

Equation (3) is used in nonlinear acoustics [17–19] and for the description of the waves in a liquid with gas bubbles [20, 21]. In [22] various exact solutions of Burgers hierarchy (1) were found. Self-similar solutions of Eq. (1) were constructed in [23]. Nonlinear special polynomials associated with the rational solutions of Eq. (1) were introduced in [24]. However the Cauchy problem for Eq. (1) was not considered. The aim of this work is to construct solution of the Cauchy problem for hierarchy (1) using the Green function method. Our approach is the straightforward. First, we use the Cole–Hopf transformation for the linearization of Eq. (1). Second, we use the Green function method for solving the Cauchy problem for the associated linear problem.

N.A. Kudryashov, D.I. Sinelshchikov

It is known that the Green function for the linear evolution equations is the self–similar solution of the Cauchy problem with special initial condition. In work [23] the general self–similar solution of the Burgers hierarchy was obtained. The special class of rational self–similar solutions of the Burgers hierarchy was considered in [24]. In the present work we construct the other special self–similar solutions of the Burgers hierarchy. These solutions form the Green function for the associated linear Cauchy problem. These solutions were not studied previously. This work is organized as follows. In Sect. 2 we present the statement of the problem and briefly discuss its properties. Section 3 is devoted to the construction of the Green function in the case of odd values of n. In Sect. 4 we construct solution of the Cauchy problem for the even values of n. In Sect. 5 we briefly discuss our results.

In the case of n = 1 and α1 < 0, Eqs. (6), (7) are the well-known Cauchy problem for the linear heat equation. Using solution of this Cauchy problem and formulae (5) we obtain solution of the Cauchy problem for the Burgers equation [13, 27, 28]. Let us study general properties of solutions of Eq. (6). Substituting Ψ = exp{i(kx + ωt)} into Eq. (6) we obtain the dispersion relation ω = −αn i n k n+1 .

Thus we have the solution of Eq. (6) in the form of the plane wave Ψ = eikx−αn (ik)

αn =

The Cauchy problem for the equation of the Burgers hierarchy takes the form 

n ∂ ∂ + u u = 0, ut + αn ∂x ∂x

(4)

u(x, 0) = ψ(x). We well know that the Burgers hierarchy can be linearized with the help of the Cole–Hopf transformation [25, 26]: u=

Ψx , Ψ

Ψ = Ψ (x, t).

(5)

n+1 t

(9)

.

We require that solutions of Eq. (6) do not grow in time. Consequently we use the following values of the parameter αn : 

2 The Cauchy problem for Eq. (1)

(8)

(−1)m+1 a 2 ,

a ∈ R, n = 2m, m ∈ N,

(−1)m a 2 ,

a ∈ R, n = 2m − 1, m ∈ N. (10)

Let us construct solution of problem (6), (7) using the Fourier transform  ∞ 1 f eiωx dω, Ψ=√ 2π −∞ (11)  ∞ 1 f=√ Ψ e−iωx dx. 2π −∞ Substituting (11) into Eqs. (6), (7) we obtain the Cauchy problem for the ordinary differential equation:

Substituting (5) into problem (4) we get the associated linear Cauchy problem:

ft + αn (iω)n+1 f = 0,  ∞ 1 f (0) = φˆ = √ φe−iωx dx. 2π −∞

Ψt + αn Ψn+1,x = 0,

(6)

From problem (12) we can find f as follows:

(7)

ˆ −αn (iω) f = φe

Ψ (x, 0) = e



ψ(x) dx

= φ(x).

We assume that functions ψ(x) and φ(x) belong to space of rapidly decreasing functions on R. We can see that solution of the Cauchy problem for the Burgers hierarchy (4) can be found with the help of formulae (5) taking into account solution of linear problem (6), (7).

n+1 t

(12)

(13)

.

Using (13) we have solution of problem (6), (7) in the form   ∞  ∞ 1 −iωξ Ψ = φ(ξ )e dξ 2π −∞ −∞ × eixω−αn (iω)

n+1 t

dω

The Cauchy problem for the equation of the Burgers hierarchy

  ∞  ∞ 1 n+1 eiω(x−ξ )−αn (iω) t dω φ(ξ ) dξ 2π −∞ −∞  ∞ = Gn (x, ξ, t)φ(ξ ) dξ, (14)

=

we have  ∞ ζ 2m 1 eizζ − 2m dζ √ 2m 2π 2ma 2 t −∞  ∞ ζ 2m 1 = 2m cos(zζ )e− 2m dζ. √ π 2ma 2 t 0

G2m−1 (z, 0, t) =

−∞

where Gn (x, ξ, t) =

1 2π



∞

−∞

eiω(x−ξ )−αn (iω)

n+1 t

dω

is the Green function. Solution of Cauchy problem (4) for the Burgers hierarchy with initial condition φx u(x, 0) = ψ(x) = φ can be found by the formula ∞ ∂ ∂x ( −∞ Gn (x, ξ, t)φ(ξ ) dξ ) u(x, t) =  ∞ . −∞ Gn (x, ξ, t)φ(ξ ) dξ

(21)

(15)

(16)

(17)

Further we study properties of Green function (15). We find expression for Gn in the explicit form for some values of n. We see that Eq. (6) is invariant under shifts in x. Thus without loss of the generality we can set ξ = 0 in (15):  ∞ 1 n+1 eiωx−αn (iω) t dω. (18) Gn (x, 0, t) = 2π −∞ Below we consider separately the cases of even and odd values of n. We will show that function (18) is connected with the hyper-Airy functions [29–32].

3 Odd values of n (dissipative equations)

Let us consider the integral from the right-hand side of Eq. (21),  I2m =

∞

ζ 2m 2m

cos(zζ )e−

(22)

dζ.

0

Integral (22) is known as the hyper-Airy function of even index [29–32]. Functions I2m are bounded, integrable and infinitely differentiable. The following preposition holds: Proposition 1 Integral (22) is uniformly convergent in domain z ∈ R, is infinitely differentiable function of z and satisfies the following differential equation: d 2m−1 I2m + (−1)m+1 zI2m = 0. dz2m−1

(23)

Proof We see that (22) is uniformly convergent as soon as the following estimate is valid:

ζ 2m

ζ 2m

cos(zζ )e− 2m ≤ e− 2m .

(24)

Formally, differentiating (22) on z for 2m − 1 times we obtain d 2m−1 I2m = (−1)m dz2m−1



∞

sin(zζ )ζ 2m−1 e−

ζ 2m 2m

dζ.

0

Let us consider the case of n = 2m − 1. Substituting expression for α2m−1 into Eq. (18) we obtain  ∞ 1 m 2 2m G2m−1 (x, 0, t) = eiωx−(−1) a (iω) t dω 2π −∞  ∞ 1 2 2m = eiωx−a ω t dω. (19) 2π −∞

Integral on the right-hand side of (25) is uniformly convergent as well. Because the following estimate holds,

Introducing in (19) the self-similar variables x 2m ζ= , z = 2m 2ma 2 tω, √ 2ma 2 t

the functions under integral sign in (22) and (25) are continuous functions of z and ζ . Thus we can differentiate (22) under the integral sign.

(20)

(25)

ζ 2m

ζ 2m

sin(zζ )ζ 2m−1 e− 2m ≤ ζ 2m−1 e− 2m ,

(26)

N.A. Kudryashov, D.I. Sinelshchikov

Fig. 1 The Green function G2m−1 at t = a = 1 and m = 1, 2, 3

Calculating integral on the right-hand side of (25) by parts we have the formula  ζ 2m ∞ d 2m−1 I2m m+1 sin(zζ )e− 2m 0 = (−1) 2m−1 dz  ∞ 2m − ζ2m −z cos(zζ )e dζ 0

= −(−1)m+1 zI2m .

(27)

We see that differential equation (23) for (22) can be obtained from formula (27). This completes the proof.  In Fig. 1 we present the dependence of Green function (21) on the self-similar variable z. From Fig. 1 we see that Green function (21) rapidly decays at z → ±∞. The amount of oscillations in the Green function tail increases when m increases. The asymptotic expansion for (22) at m > 1 and z → ±∞ has the form [31]:  I2m ∼

1/2

m−1 2 |z|− 2m−1 π(2m − 1)    2m m−1 2m − 1 cos π |z| 2m−1 × exp − 2m 2m − 1    2m 2m − 1 m−1 × cos sin π |z| 2m−1 2m 2m − 1 m−1 π . (28) − 2m − 1 2

Differential equation (23) for I2m is the particular case of the hyper-Bessel equation [30, 33]. This equation is called as the hyper-Airy equation [29–32, 34, 35].

Explicit expression for integral (22) can be found by solving the Cauchy problem for differential equation (23). Initial conditions can be determined by substituting z = 0 in (22) and its derivatives. In the case of m = 1 the general solution of Eq. (23) is expressed via the exponential function. At m > 1 the general solution of Eq. (23) is not expressed via the elementary functions. For example, we can present this solution in the form of the linear combination of the generalized hypergeometric functions. Let us consider the case m = 1. Then we have

π I2 (0) = . (29) 2 We obtain expression for I2 :

π − z2 I2 = e 2. 2

(30)

Thus we find the classic Green function for the heat equation 2 1 − (x−ξ ) G1 (x, ξ, t) = √ e 4a2 t . 2 πa 2 t

(31)

Let us consider the case m = 2. We have the following differential equation for I4 : d 3 I4 − zI4 = 0. dz3

(32)

Calculating I4 and its first two derivatives at z = 0 we obtain initial conditions for Eq. (32):

dI4

π I4 |z=0 = , = 0, 2Γ (3/4) dz z=0 (33)

d 2 I4

Γ (3/4) =− √ . dz2 z=0 2

The Cauchy problem for the equation of the Burgers hierarchy

The general solution of Eq. (32) is expressed via the generalized hypergeometric functions. Using transfor√ mations z = 4 64η, I4 = w, from (32) we have

The Green function of problem (6), (7) at n = 3 has the form

9 3 η2 wηηη + ηwηη + wη − w = 0. 4 8

G3 (x, ξ, t) (34)

The fundamental solution of Eq. (34) has the following form [36]:     1 3 3 5 1/2   w = C10 F2 , , η + C2 η 0 F2 , , η 2 4 2 4   3 η1/4 0 F2 5 , 3 , η , (35) +C 4 4 where 0 F2 (a, b, η) is the generalized hypergeometric i , i = 1, 2, 3, are integration constants. function, and C The general expression for I4 is   4 1 3 z4 2 z2 0 F2 3 , 5 , z , , +C 2 4 64 2 4 64   4 3 z0 F2 5 , 3 , z . +C (36) 4 4 64

10 F2 I4 = C



With the help of (33) we find   1 3 z4 π , , I4 = 0 F2 2Γ (3/4) 2 4 64   5 3 z4 Γ (3/4) 2 . − √ z 0 F2 , , 4 2 64 2 2 Fig. 2 Solution of problem (6), (7) at n = 3, a = 1, 2 φ = 1 + e−x and at time t = 0, 1, 5, 20

=

   1 3 (x − ξ )4 1 π , , F √ 0 2 4 2 4 4a 2 t π 4a 2 t Γ (3/4)   5 3 (x − ξ )4 Γ (3/4) (x − ξ )2 , , . − √ √ 0 F2 4 2 4a 2 t 2 4a 2 t (38)

Using Green function (38) and formula (14) we can obtain solution of problem (6), (7) at n = 3. Let us consider problem (6), (7) at n = 3 and 2 φ(x) = 1 + e−x . Substituting (38) and the initial condition into (14) we obtain solution of problem (6), (7). The plot of this solution at a = 1 and at various values of t is presented in Fig. 2. Using the Cole–Hopf transformation we find solution of the fourth-order equation (Eq. (1) at n = 3) with the initial condition u(x, 0) = ψ(x) = −

(37)

2xe−x

2

1 + e−x

2

.

(39)

The plot of solution of problem (4) at n = 3 and a = 1 corresponding to initial condition (39) at various values of t is presented in Fig. 3.

N.A. Kudryashov, D.I. Sinelshchikov Fig. 3 Solution of problem (4) at n = 3, a = 1 with initial condition (39) and at various values of time t = 0, 1, 5, 20

4 Even values of n (dispersive equations) Let us consider the case of even n. Substituting from (10) expression for αn at even values of n, from (18) we have  ∞ 1 m+1 2 2m+1 t G2m (x, 0, t) = eiωx−(−1) a (iω) dω 2π −∞  ∞ 1 2 2m+1 t) = ei(ωx+a ω dω 2π −∞  ∞    1 = cos ωx + a 2 ω2m+1 t 2π −∞   + i sin ωx + a 2 ω2m+1 t dω    1 ∞ = cos ωx + a 2 ω2m+1 t dω. (40) π 0 Using in (40) the self-similar variables 2m+1 ζ= (2m + 1)a 2 tω, z=



2m+1

(41)

x (2m + 1)a 2 t



1

(2m + 1)a 2 t    ∞ ζ 2m+1 dζ. cos zζ + × 2m + 1 0

π

The family of integrals (43) is called as the hyper-Airy function of odd index [31, 32]. Asymptotic expansion for (43) at z → +∞ has the following form [31]:  2m−1 z− 4m 2m I2m+1 ∼ √ exp − 2m + 1 πm   m − 1 π 2m+1 2m z × cos m 2    2m m − 1 π 2m+1 × cos sin z 2m 2m + 1 m 2 m−1π − , z → +∞. 2m 2

(44)

In the case of z → −∞ the integral (43) has the asymptotic expansion [31]:   2m−1 2m+1 z− 4m π 2m 2m z , I2m+1 ∼ √ − cos 2m + 1 4 πm

we obtain G2m (z, 0, t) =

Thus for the construction of the Green function in the explicit form we have to calculate the following integral:    ∞ ζ 2m+1 dζ. (43) cos zζ + I2m+1 = 2m + 1 0

z → −∞.

(45)

2m+1

(42)

We can calculate the values of (42) as the function of parameter z. The plot of (42) at various values of m is demonstrated in Fig. 4.

The Cauchy problem for the equation of the Burgers hierarchy

Fig. 4 The Green function G2m at t = a = 1 and m = 1, 3, 4

Let us consider convergence of integral (43). Integrating (43) by parts we obtain  I2m+1 = 0

=

ζ 2m+1 ∞ 2m+1 )



z + ζ 2m 0

(49)

Thus we obtain the differential equation for calculating the explicit form of integral (43):

 + 0

∞

d 2m I2m+1 + (−1)m zI2m+1 = 0. dz2m

2mζ 2m−1 (z + ζ 2m )2

 ζ 2m+1 dζ × sin zζ + 2m + 1    ∞ 2mζ 2m−1 ζ 2m+1 dζ. = sin zζ + 2m + 1 (z + ζ 2m )2 0 

lim Iα = 0.

α→0+0

2m+1 ∞ d sin(zζ + ζ 2m+1 ) z + ζ 2m

sin(zζ +

We see that

Let us consider the case of m = 1. Then from (50) we have the equation d 2 I3 − zI3 = 0. dz2 (46)

(51)

Bounded solution of (51) is the Airy function I3 = π Ai(z).

We see that integral on the right-hand side of (46) is absolutely convergent. Thus integral (43) is absolutely convergent as well. Also one can show that (46) is infinitely differentiable function [31]. Differentiating (43) on z for 2m times we have d 2m I2m+1 = (−1)m dz2m



∞

cos η dη − (−1)m zI2m+1 .

0

(47) Let us show that the first integral on the right-hand side of (47) is zero. Using the convergence factor e−αη , α > 0, we find  Iα =

∞

1 +1 0  ∞  −αη −αη sin η − ke cos η 0 × e e−αη cos η dη =

α . =− 2 α +1

(50)

α2

(52)

Thus the Green function at m = 1 has the form   x −ξ 1 G3 (x, ξ, t) = √ Ai √ . 3 3 3a 2 t 3a 2 t

Let us solve problem (6) at n = 2 and φ(x) = 2 1 + e−x . The Green function in this case has the form (53). Substituting (53) and the initial condition into (14) we obtain the solution of this problem in the form   x −ξ Ψ=√ 1+e Ai √ dξ. (54) 3 3 3a 2 t −∞ 3a 2 t We demonstrate solution (54) at various time moments in Fig. 5. Using the Cole–Hopf transformation (5) we find solution of the Sharmo–Tasso–Olver equation with the initial condition: 1



∞

u(x, 0) = ψ(x) = − (48)

(53)

−ξ 2

2xe−x



2

1 + e−x

2

.

The plot of this solution is presented in Fig. 6.

(55)

N.A. Kudryashov, D.I. Sinelshchikov Fig. 5 Solution (54) of problem (6), (7) at n = 2, a = 1 with initial condition 2 φ(x) = 1 + e−x and at time moments t = 0, 0.2, 1, 4

Fig. 6 Solution of problem (4) at n = 2, a = 1 with initial condition (55) and at time moments t = 0, 0.2, 1, 4

5 Conclusion We considered the Cauchy problem for the equation of the Burgers hierarchy (1). We constructed the Green function for the Cauchy problem (6), (7) for family of linear evolution equations. This family of linear evolution equations associated with the Burgers hierarchy by means of the Cole–Hopf transformation. We found the explicit formulas for the Green function in cases of n = 2 and n = 3. We gave some examples of solutions for the equation of the Burgers hierarchy.

Acknowledgements This research was partially supported by Federal Target Programm Research and Scientific–Pedagogical Personnel of Innovation in Russian Federation on 2009-2013, by RFBR Grant 12-01-31329 and Grant for Scientific Schools 16.120.11.6148.

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