Linear stability of weak-compacton solutions to the nonlinear dispersive Ostrovsky equation

Nonlinear Analysis: Real World Applications 11 (2010) 1782–1789

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Nonlinear Analysis: Real World Applications journal homepage: www.elsevier.com/locate/nonrwa

Linear stability of weak-compacton solutions to the nonlinear dispersive Ostrovsky equation Jiuli Yin ∗ , Lixin Tian Nonlinear Scientific Research Center, Faculty of Science, Jiangsu University, Zhenjiang, Jiangsu, 212013, PR China

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Article history: Received 15 March 2009 Accepted 3 April 2009 Keywords: Nonlinear dispersive Ostrovsky equation Integrability Compacton solutions Weak solutions Linear stability

abstract Considered herein is the linear stability of compacton solutions to the nonlinear dispersive Ostrovsky equation, which is also a modification of the K (m, n) equation. We show that the equation does not pass the Painlevé test for integrability and has compacton solutions which are different from conventional forms. Furthermore, compactons are proved to be weak solutions. A new equation similar to the nonlinear dispersive Ostrovsky equation is derived from Lagrangian, and some important conservation laws as well as the Hamiltonian structure are given. Finally, the stability of the compacton solutions to the similar equations is studied via using linear stability analysis. The result shows that the linear stability analysis it follows that, unlike compactons to the K (m, n) equation, the weak-compacton solutions are linear stable under certain conditions. © 2009 Elsevier Ltd. All rights reserved.

1. Introduction The interaction of the nonlinear dispersion with nonlinear convection can cause qualitative changes in the nature of genuinely nonlinear phenomena. More non-analytic solitary wave solutions appear in the nonlinear dispersive equations [1–4], which are not possible in the linear dispersive equations. Camassa and Holm discovered the integrable equation with nonlinear dispersion [1] ut − uxxt + 3uux = 2ux uxx + uuxxx ,

(1)

which admits non-analytic solitary wave solutions having a corner at their crest (called peakons). It is interesting that this peakon is characteristic for the waves of great height-waves of largest amplitude [5–7]. The orbital stability of peakons has been proved in [8]. At the same time, Rosenau and Hyman discovered another nonlinear wave equation with nonlinear dispersion, named the K (m, n) equation ut + α(um )x + β(un )xxx = 0,

(2)

which arises as a model for understanding the role of nonlinear dispersion in the formation of patterns in liquid drops [2]. They discovered that Eq. (2) admits compactly supported solitary wave solutions (called compactons) when 2 ≤ m = n ≤ 3. The compacton’s height depends on its speed, but unlike analytic solitary waves, its width is independent of its speed. In fact, the K (m, n) equation with compactons is not integrable [9]. The linear stability of those compactons has been proved in [10]. Considered herein is the nonlinear dispersive Ostrovsky equation (called Os(m, n) equation)

(ut + α(um )x + β(un )xxx )x = γ u, ∗

Corresponding author. E-mail addresses: [email protected] (J. Yin), [email protected] (L. Tian).

1468-1218/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2009.04.002

(3)

J. Yin, L. Tian / Nonlinear Analysis: Real World Applications 11 (2010) 1782–1789

1783

where α, β, γ , m and n are constants. With m = 2 and n = 1, Eq. (3) turns to the Ostrovsky equation [11], which is a model for the unidirectional propagation of weak nonlinear long surface and internal waves of small amplitude in rotating fluid. The Ostrovsky equation is not integrable in the sense that it does not possess an infinite hierarchy of local symmetries [12]. To the Ostrovsky equation, there are no general ways to get exact solitary solutions although Liu and Varlamov proved the existence and stability of the solitary waves [13]. With m = 3 and n = 1, Eq. (3) turns to the modified Ostrovsky equation which describes the propagation of internal waves in the ocean [14]. Note that there are two important aspects. One is that exact solitary waves have not been found in Eq. (3) with linear dispersion. Another is the close relationship between Eqs. (2) and (3). Actually, setting γ = 0 and integrating the result with respect to x, one can find that the Os(m, n) equation turns to the K (m, n) equation. Motivated by the above two points, we are interested in that whether the nonlinear dispersive Ostrovsky equation exists exact compacton solutions as 2 ≤ m = n ≤ 3. If does exist, what is the difference between them and those of the K (m, n) equation. Furthermore, we want to research the integrability of the compacton equation and the linear stability of the compactons. This paper is organized as follows. In Section 2, we study the Painlevé property and obtain compacton solutions to Eq. (3) with 2 ≤ m = n ≤ 3, which are proved to be weak solutions (see Appendix). In Section 3, we derive some important conservation laws and the Hamiltonian structure of a new equation similar to Eq. (3). The last section is devoted to the linear stability of compacton solutions to the similar equation. 2. Painlevé analysis and compacton solutions A model is called Painlevé integrable if it possesses the Painlevé property, which means that the solutions to the model are single, valued about an arbitrary singularity manifold. Let us show that the Os(m, n) equation does not pass the Painlevé test for integrability. We assume the solution u to Eq. (3) about a singularity manifold φ = φ(z ) = 0 as u=

∞ X

uj φ j+ρ .

(4)

j =0

According to Weiss et al., a nonlinear evolution equation will have the Painlevé property if its solutions are single-valued about a movable singularity manifold. That is to say in the expansion (4), the constant ρ should be a negative integer and there should be a primary branch with four arbitrary functions among uj and φ because the Os(m, n) model is a fourth order partial differential equation. In the primary branch of the Os(m, n) equation, in order to including three more arbitrary functions in addition to the arbitrary singularity manifold, there are three resonance conditions which should be satisfied identically. Substituting u = u0 φ ρ

(5)

into (3) and using the leading order analysis, we have only one possible branch with

ρ=−

2

(6)

m−n

for negative ρ . For positive integer m and n, the requirement of being a negative integer leads to only two possible models, that is Os(m, m − 1) equation or Os(m, m − 2) equation. By WTC method [9], we can find that the two models should satisfy resonance conditions if and only if γ = 0, which implies the integrability of two sets of K (m, n) equation [9]. Consequently, the Os(m, n) equation (3) does not pass the Painlevé test for integrability. Next we will search for compacton solutions to the Os(2, 2) and Os(3, 3) equations by the variational iteration method. Integrating (3) with respect to x, and assuming the solution u and all of its derivatives are vanishing at infinity, we obtain 1 ut + α(um )x + β(un )3x = γ D− x u,

(7)

1 where D− x is an integral operator. To solve Eq. (7) by means of variational iteration method [15], we construct a correction functional which reads

ui+1 (x, t ) = ui (x, t ) +

t

Z

1 ˜ i } dτ , λ{(ui )t + α(˜um uni )3x − γ D− i )x + β(˜ x u

(8)

0

where λ is a general Lagrange multiplier, and u˜ i denotes restricted variation, i.e., δ u˜ i = 0. Its stationary conditions can be obtained as follows

λ0 (τ ) = 0, 1 + λ(τ )|t =τ = 0. Eq. (9) is called Lagrange–Euler equation, and Eq. (10) is a natural boundary condition.

(9) (10)

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J. Yin, L. Tian / Nonlinear Analysis: Real World Applications 11 (2010) 1782–1789

The Lagrange multiplier, therefore, can be identified as

λ = −1.

(11)

Substituting Eq. (11) into the correction functional equation (8) results in the following formula ui+1 (x, t ) = ui (x, t ) −

t

Z

n −1 {(ui )t + α(um i )x + β(ui )3x − γ Dx ui }dτ .

(12)

0

First we consider the Os(2, 2) equation 1 ut + α(u2 )x + β(u2 )3x = γ D− x u.

(13)

Its iteration formulation can be constructed as follows ui+1 (x, t ) = ui (x, t ) −

t

Z

1 {(ui )t + α(u2i )x + β(u2i )3x − γ D− x ui }dτ .

(14)

0

To search for its compacton solution, we assume the initial solution as u0 (x, t ) = A sin(B(x − Dt )),

(15)

where A, B, D (speed) are constants to be determined. Substituting (15) into (14), and letting

∂k ∂k u ( x , t ) = ui+1 (x, t ), i ∂tk ∂tk

ui (x, t ) = ui+1 (x, t ),

(16)

with the help of Mathematica, we can obtain

∂ u0 (x, t ) = −ABD cos(B(x − Dt )), ∂t ∂ Aγ u1 (x, t ) = − cos(B(x − Dt )) − 2α A2 B cos(B(x − Dt )) sin(B(x − Dt )) ∂t B + 8A2 B3 β cos(B(x − Dt )) sin(B(x − Dt )). Setting

∂ u (x, t ) ∂t 0

−ABD = −

= Aγ B

∂ u (x, t ), ∂t 1

(17)

(18)

and equating the coefficients of like power yields

,

(19)

−2α A2 B + 8A2 B3 β = 0.

(20)

From (19) and (20), we get

r B=±

α 4β

,

D=

4βγ

α

,

(21)

which implies αβ > 0. Substituting (21) into (15), one can obtain the compacton as

 r  α A sin ± (x − Dt ) , u= 4β  0, otherwise,  

r   α 4βγ 4β x − α t ≤ π ,

(22)

where A is an arbitrary constant. Integrating by parts, one can easily prove that (22) is a weak solution to Eq. (3) as m = n = 2 (see Appendix). It is easy to find that (22) is different with conventional compactons in that it has a crest and a valley simultaneously. The reason lies in that the conservation law (31) in the next section must be satisfied. The graph of (22+) q is shown in Fig. 1 with α = β = γ = 1, B = Then we consider Os(3, 3) equation

α

4β

, and its plane graph in Fig. 2 with t = 0.

1 ut + α(u3 )x + β(u3 )3x = γ D− x u.

(23)

And the correction iteration formula of (23) as ui+1 (x, t ) = ui (x, t ) −

t

Z

1 {(ui )t + α(u3i )x + β(u3i )3x − γ D− x ui }dτ .

(24)

0

To search for its compacton solution, we assume the solution of the form u0 (x, t ) = A sin B(x − Dt ).

(25)

With the same method, one can also obtain the weak solution as

r  r  α  3α D − 27βγ ( x − Dt ) , sin ± u= 2α 2 9β  0, otherwise, where D is an arbitrary constant.

r α ≤ π, ( x − Dt ) 9β

(26)

J. Yin, L. Tian / Nonlinear Analysis: Real World Applications 11 (2010) 1782–1789

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1 0.5

2

0 –0.5 –1 –30

1 –20

0

–10 0 10 20

–1

Fig. 1. Compacton solution. 1

0.5

–30

–20

–10

10

20

30

–0.5

1

Fig. 2. Plane graph.

3. Conservation laws and Hamiltonian structure It is easily known that Eq. (3) cannot be derived from Lagrangian and hence does not possess the usual conservation laws of momentum and energy. So we consider another similar equation which contains exactly the same terms as in Eq. (3). The Lagrangian corresponding to the similar equation is given by

(φx )m+1 γ φx φt + α − β(φx )n−1 (φ2x )2 − φ 2 L= ldx = dx 2 m+1 2 which leads to the nonlinear dispersive equation (let φx = u) Z

Z 



1



,

1 ut + α(um )x + β(n − 1)(n − 2)un−3 u3x + 4β(n − 1)un−2 ux u2x + 2β un−1 u3x − γ D− x u = 0.

(27)

(28)

For a traveling wave ξ = x − Dt, Eq. (28) takes the form 1 − Duξ + α(um )ξ + β(n − 1)(n − 2)un−3 u3ξ + 4β(n − 1)un−2 uξ u2ξ + 2β un−1 u3ξ − γ D− ξ u = 0.

(28a)

Derivating (28) at x once, we get

(ut + α(um )x + β(n − 1)(n − 2)un−3 u3x + 4β(n − 1)un−2 ux u2x + 2β un−1 u3x )x = γ u.

(29)

We find that Eq. (29) has the same terms as Eq. (3), and the only difference is relative parameters. The advantage of (29) lies in that it has Lagrangian which can lead to Hamiltonian structure and some important conversation laws. For a traveling wave ξ = x − Dt, Eq. (29) takes the form

(−Duξ + α(um )ξ + β(n − 1)(n − 2)un−3 u3ξ + 4β(n − 1)un−2 uξ u2ξ + 2β un−1 u3ξ )ξ = γ u.

(29a)

By the same method in Section 2, one can obtain that Eq. (29) does not pass the Painlevé test for integrability, and admits the weak-compacton solutions as

 u=

A sin(Bξ ), |Bξ | ≤ π , 0, otherwise,

(30)

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where ξ = x − Dt. Here A, B and D satisfy q the following two conditions: α

(i) A is an arbitrary constant, B = 2α D−8βγ

(ii) A2 =

α2

,B=

α

q

3β

3βγ

and D = α when m = n = 2.

and D is an arbitrary constant when m = n = 3.

4β

∂Q ∂t

If (29) can be written as + ∂∂Xx = 0, then Q is called a conservation law. Let u(x, t ) and all of its derivatives are vanishing at infinity. Integrating (29) with respect to x over R and we obtain

Z

+∞

udx = 0.

(31)

−∞

Multiplying Eq. (28) by u and integrating it over R yields +∞

Z

1 d 2 dt

u2 dx =

+∞

Z



2

−∞

−∞

γ

1 2 n −1 ((D− (ux )2 + 2β un u2x )x − x u) )x + (β(n − 2)u

 α (um+1 )x dx = 0. m+1

(32)

Hence follows the law of conservation of momentum +∞

Z

u2 dx = const .

P =

(33)

−∞

The law of conservation of total energy can be deduced via Lagrangian, i.e., +∞

Z

[π φ˙ − l]dx,

H = −∞

where φ˙ means φt , π = ∂∂φ˙l =

φx and u = φx . Then we obtain  Z +∞  um+1 γ 1 2 E=H = −α + β un−1 u2x − (D− u ) dx. x m+1 2 −∞ 1 2

(34)

We could not find any other conservation laws, this means that Eq. (29) may not be integrable just like Eq. (3). Furthermore, Eq. (29a) can be obtained from the variation principle

δ(H + DP ) = 0. Eq. (29) can be also written in Hamiltonian structure form ut = ∂ x

δH = {u, H }, δu

where the Poisson bracket structure is defined as

{u(x), u(y)} = ∂x δ(x − y). 4. Linear stability of compactons It is convenient to introduce the notations +∞

Z

um dx,

Im = −∞

+∞

Z

un−1 (ux )2 dx,

J2 =

Z

+∞ 1 2 (D− x u) dx.

K2 =

−∞

(35)

−∞

Then (33) and (34) become H =−

α γ Im+1 + β J2 − J4 , m+1 2

P =

1 2

I2 .

(36)

Now we consider the transformation x → v x, then (35) are transformed to Im (λ) =

Im

λ

,

J2 (λ) = λJ2 ,

K2 (λ) =

1

λ3

K2 .

(37)

Hence H (λ) = −

α γ P Im+1 + βλJ2 − J4 and P = . λ(m + 1) 2λ3 λ

(38)

Using δ(H + DP ) = 0, it follows that ddλ [H (λ) + DP (λ)]λ=1 = 0, i.e., 2α Im+1 + 2β J2 + 3γ K2 − 2DP = 0. (m + 1)

(39)

J. Yin, L. Tian / Nonlinear Analysis: Real World Applications 11 (2010) 1782–1789

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Integrating Eq. (29a) once at ξ = x − Dt and integrating twice by multiplying u, we obtain

− α Im+1 + β(n + 1)J2 − γ K2 + 2DP = 0.

(40)

From (39) and (40), we obtain J2 and Im+1 (when m = n)

 1  α Im+1 = [2(m + 3)DP − (3m + 5)γ K2 ], 4

 β J2 =

1

4(m + 1)

(41)

[2(m − 1)DP − (3m + 1)γ K2 ].

Consider the general scaling transformation 1

u → µ 2 u(λx), then H and P are transformed to H (λ, µ) and P (λ, µ), i.e., m+1

H (λ, µ) = −

µ 2 µ n+1 α Im+1 + λµ 2 β J2 + 3 γ K2 , λ(m + 1) 2λ

P =

λ P, µ

(42)

and m+1

µ 2 µ λ n+1 φ(λ, µ) = H (λ, µ) + DP (λ, µ) = − α Im+1 + λµ 2 β J2 + 3 γ K2 + DP , λ(m + 1) 2λ µ ∂φ

(43)

∂φ

where φ(λ, µ) = H (λ, µ) + DP (λ, µ). The equations ∂λ = ∂µ = 0 give the stationary point at λ = µ = 1 (then φ(1, 1) is 1

Eq. (29a)). The most important, when λ = µ, the transformation u → µ 2 u(λx) does not change the momentum P near the point λ = µ = 1. Hence we get

φ(λ) = −

λ

m−1 2

m+1

α Im+1 + λ

n+3 2

β J2 +

1 2λ2

γ K2 + DP .

(44)

Using Taylor series for λ = µ = 1, we obtain

δ 2 φ(λ) = δ 2 H (λ) ≈

(λ − 1)2



8

 (m − 1)(m − 3) γ Im+1 + (n + 1)(n + 3)β J2 + K2 , m+1 3

(45)

which has a definite sign. If it is positive (negative), the expression H (λ) = H (λ, µ) λ=µ = −

λ



m−1 2

m+1

α Im+1 + λ

n+3 2

β J2 +

1 2λ2

γ K2

(46)

has a minimum (maximum) at λ = 1. Assume that u = uD + v, |v| << 1,

(uD , v) = 0,

where uD is a solution to Eq. (29a). Substituting u into (28a) and after linearization, we have

∂T v = ∂ξ Lˆ v,

(47)

where ξ = x − dt, T = t are independent variables and Lˆ is given by Lˆ = d − α mum−1 − 2β(n − 1)un−2 uξ ∂ξ − β(n − 1)(n − 2)un−3 u2ξ 2 − 2β(n − 1)un−2 u2ξ − 2β un−1 ∂ξ2 + γ D− ξ .

(48)

It is easy to know that (48) has a solution of the form as

v(ξ , T ) = e−iwT ϕ(ξ ) + eiw T ϕ ∗ (ξ ), ∗

(49a)

where ϕ(ξ ) satisfies the equation

wϕ(ξ ) = i∂ξ Lˆ ϕ(ξ ).

(49b)

Integrating (28a) once w.r.t. ξ , leads to Lˆ ∂ξ uD = 0.

(50)

From above, we can find that ∂ξ uD is the eigenfunction of the operator Lˆ corresponding to the eigenvalue equal to zero.

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J. Yin, L. Tian / Nonlinear Analysis: Real World Applications 11 (2010) 1782–1789

Differentiating (28a) w.r.t. D, we obtain

  ˆL ∂ uD = −uD . ∂d

(51)

From (50), one of the solutions to (49b) is ϕ(ξ ) ∝ ∂ξ uD corresponding to w = 0. We can also find that, if w 6= 0, (49b) has a solution ϕ(ξ ). Then (49b) also has the solution (−w, ϕ(−ξ )). From (49a), we know that the solution is stable if w is real and the solution is unstable if w is complex. The full operator in (49b) is a product of two Hermitian operators. In this case all eigenvalues w are real if one of the operators is positive [10]. From this, it follows that a sufficient condition of real eigenvalues w is

(ψ, Lˆ ψ) > 0,

(52)

where ψ is a function in the subspace orthogonal to uD as

(ψ, uD ) = (ψ, ∂ξ uD ) = 0. Moreover, condition (52) is also associated with the extremum of H + DP. Using the equation δ(H + DP ) = 0, we can find that the second variation of H + DP is given by

δ 2 (H + DP )uc =

1

Z

+∞

2 −∞

(v, Lˆ v)dξ ,

(53)

where the operator Lˆ is defined as (48). This means that, if δ 2 (H + DP )uD > 0 (namely H + DP has a minimum at uD ), then

(ψ, Lˆ ψ) > 0. That is, the minimum of H + DP is a sufficient condition of linear stability of compacton solutions with respect to small perturbations. From (46), letting m = n, we can obtain the condition for the minimum of perturbed Hamiltonian H (λ) at λ = 1 by 1 4(m + 1)

[−8DP + 2(m + 3)γ K2 ].

(54)

From (30), (35) and (36), it is easy to prove that the compacton solutions to Eq. (29) are linear stable when either m = n = 2, γ > 0 or m = n = 3, 9γ − 34βα D > 0. Acknowledgements The research was supported by the National Nature Science Foundation of China (No: 10771088), the high-level talented person special subsidizes of Jiangsu University (No. 07JDG082) and the post-doctoral Foundation of Jiangsu Province (No. 0801028C). Appendix For Eq. (3) as m = n = 2, we define its weak solutions as any functions u(t , x) in the following sense. In what follows, we assume m = n = 2. 1 (R1 )), the following definitions are therefore natural. If u ∈ L1local (R+ , Hlocal 1 (R1 )) is a solitary wave solution to Eq. (3) if φ satisfies Eq. (7) in distribution Definition 1. A function u ∈ L1local (R+ , Hlocal + 1 sense, where R = (0, +∞), R = (−∞, +∞). 1 Definition 2. A solitary wave u(t , x) ∈ C [R+ × R1 ] is a weak solution to Eq. (3), if and only if u ∈ L1local (R+ , Hlocal (R1 )) and

hL, ϕi =:

∞

Z 0

Z

+∞ 1 [ut ϕ + α(u2 )x ϕ + β(u2 )x ϕxx − γ (D− x u)ϕ]dxdt = 0

−∞

for any test function ϕ(t , x) ∈ C0∞ [R+ × R1 ]. To verify (22) a weak solution to Eq. (3) as m = n = 2, we consider u = A sin(B(x − Dt )).

(A.1)

J. Yin, L. Tian / Nonlinear Analysis: Real World Applications 11 (2010) 1782–1789

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q For |B(x − Dt )| ≤ π and u = 0 otherwise. We will substitute u into (A.1). If u satisfies (A.1) for B = ± 4αβ and D =

4βγ

α

,

then the verification is done. Substituting u into the left-hand side of (A.1), we have

hL, ϕi =

∞

Z

Z dt

− πB +Dt

0

∞

Z =

− πB +Dt

Z dt

0

− πB +Dt − πB +Dt

1 [ut ϕ + α(u2 )x ϕ + β(u2 )x ϕxx − γ (D− x u)ϕ]dx



[−ABD cos(B(x − Dt ))]ϕ

+ [−A BD sin(B(x − Dt )) cos(B(x − Dt ))](αϕ + βϕxx ) + 2

Aγ B



cos(B(x − Dt ))ϕ dx

= · · · (integration by parts)  Z ∞ Z − π +Dt  B A 1 2 dt = A BD(4β B2 − α) sin(2B(x − Dt )) + (γ − B2 D) cos(B(x − Dt )) ϕ dx. 0

q Thus, if B = ± 4αβ , D =

− πB +Dt 4βγ

α

2

B

and hL, φi = 0. This proves that (22) is a weak solution to Eq. (3) as m = n = 2.

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