Bifurcation Analysis of a Coupled Kuramoto-Sivashinsky- and Ginzburg-Landau-Type Model

Abstract

We study the bifurcation and stability of trivial stationary solution (0,0) of coupled Kuramoto-Sivashinsky- and Ginzburg-Landau-type equations (KS-GL) on a bounded domain (0,𝐿) with Neumann's boundary conditions. The asymptotic behavior of the trivial solution of the equations is considered. With the length 𝐿 of the domain regarded as bifurcation parameter, branches of nontrivial solutions are shown by using the perturbation method. Moreover, local behavior of these branches is studied, and the stability of the bifurcated solutions is analyzed as well.

1. Introduction

The mathematical theory of pattern formation [1–3] has a wide range of applications. In the field of fluid mechanics, Rayleigh-Bénard convection is the most widely studied example. Like the thermally driven Bénard convection, the surface tension-driven Marangoni convection is also an interesting pattern formation of nonlinear system. With mass transfer from liquid phase to gas phase, a typical setup for the Marangoni convection is a liquid layer resting on a rigid surface, with a free deformable upper interface contacting an infinite layer of gas. Nevertheless, at present, we still find it hard to analyze the governing equations for the Marangoni convection, that is, Navier-Stokes equations and mass (surfactant) transfer equation. What we can only use is numerical simulation so far. There is little work that has been done on the nonlinear Marangoni convection.

In this paper, we consider a simplified model, which is proposed to capture important nonlinear features yet more amendable to analyse: 𝐴_𝑡=𝐴+𝐴_𝑥𝑥−𝐴²ℎ𝐴+𝐴ℎ,_𝑡=−ℎ_𝑥𝑥−ℎ_{𝑥𝑥𝑥𝑥}+𝛼𝐴²_𝑥𝑥, (1) where𝐴(𝑥,𝑡)is the amplitude for the Marangoni convection mode andℎ(𝑥,𝑡)is the interface deformation (real function). Equation (1) has been derived by Golovin et al. [4]. The constant 𝛼=𝑘/(𝑀𝑎−𝑀𝑎_𝑐)^1/5is positive, with𝑀𝑎denoting the Marangoni number, and represents the gradient (derivative) of surface tension with respect to surfactant concentration.𝑀𝑎_𝑐is the critical Marangoni number at which the trivial stationary state becomes linearly unstable, and𝑘>0is a constant related to other system parameters.

Equation (1) without the interaction term𝐴ℎis the well-known Ginzburg-Landau equation [5, 6], while the second equation without the term|𝐴²|_𝑥𝑥is the linearized version of the so-called Kuramoto-Sivashinsky equation [7]. Both the G-L equation and K-S equation have been extensively investigated as model examples of infinite dimensional dynamical systems. In [8], Kazhdan et al. have done numerical simulation of this coupled system of Kuramoto-Sivashinsky- and Ginzburg-Landau-type equations (hereafter, KS-GL system). Duan et al. [9] have discussed the existence and uniqueness of global solutions of this coupled system, using the contraction mapping principle and energy estimates. Despite these publications on KS-GL equation, the static bifurcations of the equation have not been thoroughly studied. In this paper, we focus on studying bifurcations of the KS-GL system. In [10], Xiao and Gao analyzed the bifurcations of the 1D Swift-Hohenberg equation with quintic nonlinearity. Two types of structures in the bifurcation diagrams are presented when the bifurcation points are closer, and their stabilities are analyzed. Li and Chen have applied singularity theory and the perturbation method to study the bifurcations of the 1D and 2D K-S equations and get the asymptotic expressions of the steady-state solution branches that have bifurcated from the equilibrium in [11, 12]. In this paper, we will use the methods in [10–12] to discuss the bifurcated solutions. In (1),𝐴is complex. Namely, we can write 𝐴=𝑅exp(𝑖𝜃). The additional phase𝜃makes analysis very complicated. Here, we restrict our attention to invariant subspace in which𝐴is real. We hope to return to the general case in future study. In this paper, we discuss steady solutions of the parabolic partial differential equations 𝐴_𝑡=𝐴+𝐴_𝑥𝑥−𝐴³ℎ+𝐴ℎ,_𝑡=−ℎ_𝑥𝑥−ℎ_{𝑥𝑥𝑥𝑥}+𝛼𝐴²_𝑥𝑥, (2) on the cylindrical domain𝑄=(0,𝐿)×𝑅⁺subject to the boundary conditions 𝐴_𝑥=ℎ_𝑥=𝐴_𝑥𝑥𝑥=ℎ_𝑥𝑥𝑥=0at𝑥=0,𝐿. (3) The steady-state equation of (2) reads as 𝐴+𝐴_𝑥𝑥−𝐴³+𝐴ℎ=0,0<𝑥<𝐿,−ℎ_𝑥𝑥−ℎ_{𝑥𝑥𝑥𝑥}+𝛼𝐴²_𝑥𝑥=0,0<𝑥<𝐿,𝜕𝐴𝜕𝜕𝑥=0,³𝐴𝜕𝑥³=0,𝑥=0,𝐿,𝑡>0,𝜕ℎ𝜕𝜕𝑥=0,³ℎ𝜕𝑥³=0,𝑥=0,𝐿,𝑡>0. (4) We will discuss the bifurcation of the trivial steady state of the equations.

2. Analysis of the Trivial Steady State (0,0)

In this section, we study some properties of the linear problem associated with problem (4). Let 𝑋={(𝐴,ℎ)∈𝐻²[0,𝐿]×𝐻²[0,𝐿]∣𝐴^′(0)=𝐴^′(𝐿)=𝐴^′′′(0)=𝐴^′′′(𝐿)=ℎ^′(0)=ℎ^′(𝐿)=ℎ^′′′(0)=ℎ^′′′(𝐿)=  0}. We linearize the problem at the trivial solution (0,0), and then we have the corresponding differential operator matrix Γ₁=𝜕²𝜕𝑥²𝜕+𝐼00−²𝜕𝑥²−𝜕⁴𝜕𝑥⁴, (5) and the eigenvalue problem is 𝜑_𝑥𝑥+𝜑=𝜆𝜑,0<𝑥<𝐿,−𝜓_𝑥𝑥−𝜓_{𝑥𝑥𝑥𝑥}𝜑=𝜆𝜓,0<𝑥<𝐿,_𝑥(0)=𝜑_𝑥(𝐿)=𝜑_𝑥𝑥𝑥(0)=𝜑_𝑥𝑥𝑥(𝐿)𝜓=0,_𝑥(0)=𝜓_𝑥(𝐿)=𝜓_𝑥𝑥𝑥(0)=𝜓_𝑥𝑥𝑥(𝐿)=0. (6)

The eigenvalues𝜆are given by 𝜆=𝜆_𝑚=−𝑚𝜋𝐿²+1,𝜆=𝛿_𝑛=𝑛𝜋𝐿²−𝑛𝜋𝐿⁴,𝑛,𝑚∈𝑁. (7) If 𝜆_𝑚=𝛿_𝑛, then the corresponding eigenvectors are cos𝑚𝜋𝑥𝐿0,0cos𝑛𝜋𝑥𝐿. (8) If 𝜆_𝑚≠𝛿_𝑛, then for 𝜆_𝑚=−(𝑚𝜋/𝐿)²+1, the corresponding eigenvector is cos𝑚𝜋𝑥𝐿0, (9) and for 𝛿_𝑛=(𝑛𝜋/𝐿)²−(𝑛𝜋/𝐿)⁴, the corresponding eigenvector is 0cos𝑛𝜋𝑥𝐿. (10) It will be convenient to rescale the spatial variable so that the domain(0,𝐿)maps onto the fixed domain (0,1). Thus, we introduce the variables𝑥and 𝑢, 𝐿𝑥=𝑥, 𝑢(𝑥)=𝑢(𝑥), 𝑣(̃𝑥)=𝑣(𝑥), then omit the tildes, and we find that problem (4) becomes 𝐴+𝐿²𝐴_𝑥𝑥−𝐿²𝐴³+𝐿²𝐴ℎ=0,0<𝑥<1,−𝐿²ℎ_𝑥𝑥−ℎ_{𝑥𝑥𝑥𝑥}+𝛼𝐿²𝐴²_𝑥𝑥=0,0<𝑥<1,𝜕𝐴𝜕𝜕𝑥=0,³𝐴𝜕𝑥³=0,𝑥=0,1,𝑡>0,𝜕ℎ𝜕𝜕𝑥=0,³ℎ𝜕𝑥³=0,𝑥=0,1,𝑡>0. (11) The corresponding eigenvalue problem at (0,0) is 𝜑_𝑥𝑥+𝐿²𝜑=𝜆^′𝜑,0<𝑥<1,−𝐿²𝜓_𝑥𝑥−𝜓_{𝑥𝑥𝑥𝑥}=𝜆^′𝜑𝜓,0<𝑥<1,_𝑥(0)=𝜑_𝑥(1)=𝜑_𝑥𝑥𝑥(0)=𝜑_𝑥𝑥𝑥(1)𝜓=0,_𝑥(0)=𝜓_𝑥(1)=𝜓_𝑥𝑥𝑥(0)=𝜓_𝑥𝑥𝑥(1)=0. (12) It is easy to find that𝜆^′=𝐿²𝜆.

3. Bifurcation and Stability Analysis in Different Cases

Let𝐿be a bifurcation parameter;𝐿=𝜋is a bifurcation point. In this section, we discuss how many nontrivial solution branches will be bifurcated from the trivial solution near𝐿=𝜋and their asymptotic expression. Moreover, we will discuss the stability of the solution branches.

When𝐿=𝜋, the eigenvector at (0,0) of (12) is 𝑒₁=𝑐cos𝜋𝑥,𝑒₂=0cos𝜋𝑥,𝑐∈𝑅. (13) We set (𝜀)𝐴ℎ=𝐿=𝜋+𝛾⋅𝜀,_𝑛𝑖=1𝜀𝛼_𝑖𝑒_𝑖+𝜀²𝑎()𝑏()𝑥,𝜀𝑥,𝜀, (14)(15) where𝜀is a small parameter,^{𝑎(𝑥,𝜀)}_{𝑏(𝑥,𝜀)}∈{𝑒₁,𝑒₂}^⟂, 𝛼²₁+𝛼²₂=1.

Substituting (14), (15) into (11) leads to 𝜀𝛼₁cos𝜋𝑥+𝜀²𝑎(𝑥,𝜀)^′′+𝜋+𝛾(𝜀)⋅𝜀²×𝜀𝛼₁cos𝜋𝑥+𝜀²𝑎(𝑥,𝜀)−𝜋+𝛾(𝜀)⋅𝜀²×𝜀𝛼₁cos𝜋𝑥+𝜀²𝑎(𝑥,𝜀)³+𝜋+𝛾(𝜀)⋅𝜀²×𝜀𝛼₁cos𝜋𝑥+𝜀²𝑎()×𝑥,𝜀𝜀𝛼₁𝑐+𝜀𝛼₂cos𝜋𝑥+𝜀²𝑏()−(𝜀)𝑥,𝜀=0,𝜋+𝛾⋅𝜀²𝜀𝛼₁𝑐+𝜀𝛼₂cos𝜋𝑥+𝜀²𝑏(𝑥,𝜀)^′′−𝜀𝛼₁𝑐+𝜀𝛼₂cos𝜋𝑥+𝜀²𝑏(𝑥,𝜀)^′′′′+𝛼𝜋+𝛾(𝜀)⋅𝜀²𝜀𝛼₁cos𝜋𝑥+𝜀²𝑎(𝑥,𝜀)^2′′=0, (16)(17) where ′ means𝜕/𝜕𝑥and ′′ means𝜕²/𝜕𝑥².

Letting the coefficient of𝜀²vanish in (16), (17) gives 𝑎₀^′′(𝑥)+𝜋²𝑎₀(𝑥)+2𝜋𝛾(0)𝛼₁cos𝜋𝑥+𝜋²𝛼₁cos𝜋𝑥𝛼₁𝑐+𝛼₂𝑏cos𝜋𝑥=0,_{0′′′′}(𝑥)+𝜋²𝑏₀^′′(𝑥)+2𝜋𝛾(0)𝛼₁cos𝜋𝑥+𝜋²𝛼₁cos𝜋𝑥𝛼₁𝑐+𝛼₂cos𝜋𝑥=0, (18)(19) where^𝑎0_𝑏^(𝑥)_0(𝑥)=^𝑎(𝑥,0)_𝑏(𝑥,0)∈{𝑒₁,𝑒₂}^⟂.

Taking the inner product of (16) withcos𝜋𝑥, we obtain ¹₀𝑎₀^′′(𝑥)cos𝜋𝑥𝑑𝑥+𝜋²¹₀𝑎₀(𝑥)(0)𝛼cos𝜋𝑥𝑑𝑥+2𝜋𝛾₁¹₀cos²𝜋𝑥𝑑𝑥+𝜋²𝛼₁¹₀cos²𝜋𝑥𝛼₁𝑐+𝛼₂cos𝜋𝑥𝑑𝑥=0. (20) For the first term of (20), integrate it by part and from the boundary condition𝑎^′(0)=𝑎^′(1)=0, and we have ¹₀𝑎₀^′′(𝑥)cos𝜋𝑥𝑑𝑥=−𝜋²¹₀𝑎₀(𝑥)cos𝜋𝑥𝑑𝑥. (21) Substituting (21) to (20), we get 𝛾(0)1=−2𝑐𝛼₁𝜋,𝛼₁,𝛾(0)𝛼≠0isanarbitraryconstant,₁=0. (22)

From the aforementioned orthogonality condition, we know that ¹₀𝑎₀(𝑥)cos𝜋𝑥+𝑐𝑏₀(𝑥)𝑑𝑥=0,₁₀𝑏₀(𝑥)cos𝜋𝑥𝑑𝑥=0. (23)(24) Taking the inner product of (19) withcos𝜋𝑥, we obtain ¹₀𝑏₀^′′′′(𝑥)cos𝜋𝑥𝑑𝑥+𝜋²¹₀𝑏₀^′′(𝑥)cos𝜋𝑥𝑑𝑥−2𝛼₂𝜋³𝛾(0)¹₀cos²𝜋𝑥+2𝛼𝛼²₁𝜋⁴¹₀cos2𝜋𝑥cos𝜋𝑥𝑑𝑥=0. (25) Because of the boundary condition𝑏^′(0)=𝑏^′(1)=𝑏^′′′(0)=𝑏^′′′(1)=0, orthogonality condition (24), we get 𝛾(0)=0,𝛼₂,𝛾(0)𝛼≠0isanarbitraryconstant,₂=0, (26) from (25).

Comparing (22) and (26), for𝛼²₁+𝛼²₂=1, we deduce the following important relation: 𝛾(0)1=−2𝑐𝛼₁𝜋. (27)

Substituting (27) to (18), we get 𝑎₀^′′(𝑥)+𝜋²𝑎₀(𝑥)+𝜋²𝛼₁𝛼₂cos²𝜋𝑥=0. (28) For𝑎₀(𝑥)satisfies the boundary condition𝑎^′₀(0)=𝑎^′₁(0)=𝑎₀^′′′(0)=𝑎₀^′′′(1), we calculate from (28) that 𝑎₀(𝑥)=𝑘₁𝛼cos𝜋𝑥+¹𝛼₂6𝛼cos2𝜋𝑥−¹𝛼₂2,𝑘₁∈𝑅. (29)

Next, we want to calculate𝑏₀(𝑥) and substitute (27) to (19); we get the following ODE: 𝑏₀^′′′′(𝑥)+𝜋²𝑏₀^′′(𝑥)+𝑐𝛼₁𝛼₂𝜋⁴cos𝜋𝑥+2𝛼𝛼²₁𝜋⁴cos2𝜋𝑥=0. (30) By lengthy computations and the boundary conditions𝑏^′₀(0)=𝑏^′₁(0)=𝑏₀^′′′(0)=𝑏₀^′′′(1), we deduce that 𝑏₀(𝑥)=𝑐₁+𝑐₂cos𝜋𝑥−𝛼𝛼²₁6cos2𝜋𝑥,𝑐𝛼₁𝛼₂=0. (31)(32) Taking the inner product withcos𝜋𝑥of (31), from (24), we know that ¹₀𝑐₁+𝑐₂cos𝜋𝑥−𝛼𝛼²₁6cos2𝜋𝑥cos𝜋𝑥𝑑𝑥=0, (33) which gives us 𝑐₂𝜋=0,₂𝛼96¹−𝜋²96𝛼𝛼³₁−𝜋²𝛼24³¹cos3𝜋𝑥𝑏₀(𝑥)=𝑐₁−𝛼𝛼²₁6cos2𝜋𝑥. (34)(35) Substituting (29), (35) to (23), we have cos𝜋𝑥𝑘¹₀₁𝛼cos𝜋𝑥+¹𝛼₂6𝛼cos2𝜋𝑥−¹𝛼₂2+𝑐𝑐₁−𝛼𝛼²₁6cos2𝜋𝑥𝑑𝑥=0, (36) from which we get 𝑘₁=−2𝑐𝑐₁,𝑎₀(𝑥)=−2𝑐𝑐₁𝛼cos𝜋𝑥+¹𝛼₂6𝛼cos2𝜋𝑥−¹𝛼₂2. (37)

From the previous discussion, we obtain 𝑐𝛼₁𝛼₂=0, (38) with the normal condition 𝛼²₁+𝛼²₂−1=0. (39)

When𝑐=0, then we have 𝛾₀=0; (40) at this time, 𝑎₀(𝑥)=𝛼₁𝛼₂6𝛼cos2𝜋𝑥−¹𝛼₂2,𝑏₀(𝑥)=𝑐₁−𝛼𝛼²₁6cos2𝜋𝑥. (41)

Letting the coefficient of𝜀³of (16), (17) vanish, we get 𝑎₁^′′(𝑥)+𝜋²𝑎₁(𝑥)+2𝜋𝛼₁𝛾₁(0)cos𝜋𝑥−𝜋²𝛼³₁cos³𝜋𝑥+𝜋²𝛼₁cos𝜋𝑥𝑏₀(𝑥)+𝜋²𝛼₂𝑎₀(𝑥)𝑏cos𝜋𝑥=0,_{1′′′′}(𝑥)+𝜋²𝑏₁^′′(𝑥)−2𝛼₁𝛼𝜋²cos𝜋𝑥𝑎₀^′′(𝑥)+4𝛼₁𝛼𝜋³sin𝜋𝑥𝑎^′₀(𝑥)+2𝛼₁𝛼𝜋⁴cos𝜋𝑥𝑎₀(𝑥)−2𝜋³𝛾₁(0)𝛼₂cos𝜋𝑥=0, (42)(43) where 𝑎₁(𝑥)=(𝜕𝑎(𝑥,𝜀)/𝜕𝑥)|_𝜀=0, 𝑏₁(𝑥)=(𝜕𝑏(𝑥,𝜀)/𝜕𝑥)|_𝜀=0 and 𝛾₁(0)=𝛾^′(0).

Taking the inner product of (42) withcos𝜋𝑥, we obtain ¹₀𝑎₁^′′(𝑥)cos𝜋𝑥𝑑𝑥+𝜋²¹₀𝑎₁(𝑥)cos𝜋𝑥𝑑𝑥+2𝜋𝛼₁𝛾₁(0)¹₀cos²𝜋𝑥𝑑𝑥−𝜋²𝛼³₁¹₀cos⁴𝜋𝑥𝑑𝑥+𝜋²𝛼₁¹₀cos²𝜋𝑥𝑏₀(𝑥)𝑑𝑥+𝜋²𝛼₂¹₀𝑎₀(𝑥)cos²𝜋𝑥𝑑𝑥=0, (44) from which we know that 𝛾₁(0)=38𝜋𝛼²₁+𝛼𝛼²₁𝜋−𝑐24¹𝜋2+524𝜋𝛼²₂,𝛼₁𝛾≠0,₁(0)isanarbitraryconstant,𝛼₁=0. (45)

Next, we discuss the expression and the stability of the bifurcated solutions in different cases.

Case 1. 𝑐=0,𝛼₁≠0.
Substituting (41), (45) to (42), we have 𝑎₁^′′(𝑥)+𝜋²𝑎₁(𝑥)+𝜋²𝛼12¹−𝜋²12𝛼𝛼³₁−𝜋²3𝛼³₁cos3𝜋𝑥=0. (46) Because of the orthogonality condition ¹₀𝑎₁(𝑥)cos𝜋𝑥𝑑𝑥=0, (47) and boundary condition 𝑎^′₁(0)=𝑎^′₁(1)=0, (48) we get 𝑎₁(𝑥)=𝜋²𝛼96¹−𝜋²96𝛼𝛼³₁−𝜋²𝛼24³¹cos3𝜋𝑥. (49) Substituting (41) to (43), we have 𝑏₁^′′′′(𝑥)+𝜋²𝑏₁^′′(𝑥)−2𝜋³𝛾₁(0)𝛼₂+56𝛼𝛼²₁𝛼₂𝜋⁴+3cos𝜋𝑥2𝛼𝛼²₁𝛼₂𝜋⁴cos3𝜋𝑥=0. (50) Taking the inner product of (50) withcos𝜋𝑥, we obtain ¹₀𝑏₁^′′′′(𝑥)cos𝜋𝑥𝑑𝑥+𝜋²¹₀𝑏₁^′′(𝑥)−cos𝜋𝑥𝑑𝑥2𝜋³𝛾₁(0)𝛼₂+56𝛼𝛼²₁𝛼₂𝜋⁴¹₀cos²+3𝜋𝑥𝑑𝑥2𝛼𝛼²₁𝛼₂𝜋⁴¹₀cos3𝜋𝑥cos𝜋𝑥𝑑𝑥=0, (51) for𝑏₁(𝑥)satisfies the orthogonality condition ¹₀𝑏₁(𝑥)cos𝜋𝑥𝑑𝑥=0, (52) and the boundary condition 𝑏^′₁(0)=𝑏^′₁(1)=𝑏₁^′′′(0)=𝑏₁^′′′(1); (53) we know ¹₀𝑏₁^′′′′(𝑥)cos𝜋𝑥𝑑𝑥+𝜋²¹₀𝑏₁^′′(𝑥)cos𝜋𝑥𝑑𝑥=0, (54) which forces 2𝜋³𝛾₁(0)𝛼₂+56𝛼𝛼²₁𝛼₂𝜋⁴=0. (55) From (45) and (55), we get 𝛾₁(0)5=−12𝛼𝛼²₁3𝜋=8𝜋𝛼²₁+𝛼𝛼²₁𝜋−𝑐24¹𝜋2+524𝜋𝛼²₂,𝛼₂≠0, (56) or 𝛾₁(0)=38𝜋𝛼²₁+𝛼𝛼²₁𝜋−𝑐24¹𝜋2+524𝜋𝛼²₂,𝛼₂=0. (57) Substituting (55) to (50), we get 𝑏₁^′′′′(𝑥)+𝜋²𝑏₁^′′(𝑥)+32𝛼𝛼²₁𝛼₂𝜋⁴cos3𝜋𝑥=0, (58) which gives us 𝑏₁(𝑥)=𝑐₁−148𝛼𝛼²₁𝛼₂cos3𝜋𝑥. (59) As has been discussed previously, we have the following.
For𝐿near𝜋, there are nontrivial steady-state solution branches of (11) bifurcated from the trivial solution: 𝐴(𝜀)=𝜀𝛼₁cos𝜋𝑥+𝜀²𝛼₁𝛼₂6𝛼cos𝜋𝑥−¹𝛼₂2+𝜀³𝜋²𝛼96¹−𝜋²96𝛼𝛼³₁−𝜋²𝛼24³¹𝜀cos3𝜋𝑥+𝑜³,ℎ(𝜀)=𝜀𝛼₂cos𝜋𝑥+𝜀²𝑐₁−𝛼𝛼²₁6cos2𝜋𝑥+𝜀³𝑐₁−148𝛼𝛼²₁𝛼₂cos3𝜋𝑥+𝑜𝜀³,𝐿=𝜋+𝜀²𝛾₁(0)+𝑜𝜀². (60)𝛼₁≠0,𝛼₂,𝛾₁(0)satisfy (56), (57). 𝑐₁,𝑐₁∈𝑅.𝑐₁, 𝑐₁are arbitrary constants.
Next, we wish to consider the stability of the nontrivial solutions given in (60).
Considering the eigenvalue problem 𝐹_𝐴,ℎ(𝐴(𝜀)(𝜀)(𝜀𝜙(𝜀)𝜓(𝜀)(𝜀)𝜙(𝜀)𝜓(𝜀),ℎ,𝐿))=𝜆, (61) since 𝜆(0)=0, 𝜙(0)=cos𝜋𝑥, 𝜓(0)=0, we assume 𝜙(𝜀)𝜓(𝜀)0𝜙&=cos𝜋𝑥+𝜀¹𝜓₁+𝜀²𝜙₂𝜓₂𝜆(𝜀)+⋯,=𝜆₁𝜀+𝜆₂𝜀²𝐹+⋯,_𝐴,ℎ(𝐴(𝜀)(𝜀)(𝜀=𝐿,ℎ,𝐿))₂+𝜕²𝜕𝑥²−3𝐿²𝐴²(𝜀)+𝐿²ℎ(𝜀)𝐿²𝐴(𝜀)2𝛼𝐿²2𝜕𝐴(𝜀)𝜕𝜕𝑥+𝜕𝜕𝑥²𝐴(𝜀)𝜕𝑥²(𝜀)𝜕+𝐴²𝜕𝑥²−𝐿²𝜕²𝜕𝑥²−𝜕⁴𝜕𝑥⁴. (62) Substituting (62) to (61), we get 𝜕²𝜕𝑥²𝜙(𝜀)+𝐿²𝜙(𝜀)−3𝐿²𝐴²(𝜀)𝜙(𝜀)+𝐿²ℎ(𝜀)𝜙(𝜀)+𝐿²𝐴(𝜀)𝜓(𝜀)−𝜆(𝜀)𝜙(𝜀)=0,2𝛼𝐿²2𝜕𝐴(𝜀)(𝜀)𝜕𝑥𝜕𝜙+𝜕𝜕𝑥²𝐴(𝜀)𝜕𝑥²𝜙(𝜀)(𝜀)𝜕+𝐴²𝜙(𝜀)𝜕𝑥²−𝐿²𝜕²𝜓𝜕𝑥²−𝜕⁴𝜓𝜕𝑥⁴−𝜆(𝜀)𝜓(𝜀)=0. (63) On the branches given in (60), equating the coefficient of𝜀in (63) to 0, we obtain 𝜙₁^′′+𝜋²𝜙₁+𝛼₂𝜋²cos²𝜋𝑥−𝜆₁cos𝜋𝑥=0,4𝛼𝛼₁𝜋⁴cos2𝜋𝑥+𝜋²𝜓₁^′′+𝜓₁^′′′′=0. (64) From (64) and the boundary conditions 𝜙^′₁(0)=𝜙^′₁(1)=𝜙₁^′′′(0)=𝜙₁^′′′(1),𝜓^′₁(0)=𝜓^′₁(1)=𝜓₁^′′′(0)=𝜓₁^′′′(1), (65) we get 𝜆₁𝜙=0,₁𝜓₁=𝑝₁𝛼cos𝜋𝑥+²6𝛼cos2𝜋𝑥−²2𝑞₁+𝑞₂cos𝜋𝑥−𝛼𝛼₁3cos2𝜋𝑥⋅𝑝₁,𝑞₁,𝑞₂∈𝑅. (66) Using the second-order term of𝜀in (63), we obtain 𝜙₂^′′+𝜋²𝜙₂−−32𝛼²₁+𝛼₁𝑞₁−𝛼6𝛼²₁𝜋²−𝜆₂+𝑝cos𝜋𝑥₁𝛼₂2+𝑞₂𝛼₁2𝜋²+−3cos2𝜋𝑥4𝛼²₁+1𝛼12²¹−𝜋²4𝛼𝛼²₁+𝜋cos3𝜋𝑥₂2𝑝₁𝛼₂+𝜋²2𝑞₂𝛼₁𝜓=0,_{2′′′′}+𝜋²𝜓₂^′′−53𝛼₁𝛼₂𝛼𝜋⁴cos𝜋𝑥+4𝛼𝛼₁𝑝₁𝜋⁴cos2𝜋𝑥+3𝛼₁𝛼₂𝛼𝜋⁴cos3𝜋𝑥=0, (67) and considering the boundary condition 𝜙^′₂(0)=𝜙^′₂(1)=𝜙₁^′′′(0)=𝜙₂^′′′(1),𝜓^′₂(0)=𝜓^′₂(1)=𝜓₂^′′′(0)=𝜓₂^′′′(1), (68) we deduce that (67) has solution if and only if𝛼₂=0; in this case, we have𝛼₁=±1 (𝛼²₁+𝛼²₂=1). Taking the sign of𝜀into account in (15), we consider𝛼₁=1only. Therefore, we have 𝜆₂=−32−𝛼6+𝑞₁𝜋²,𝜙₂𝜓₂=𝑝₂𝑞cos𝜋𝑥+²63cos2𝜋𝑥−+𝛼32𝑞32cos3𝜋𝑥−²2𝑞₁+𝑞₂𝑝cos𝜋𝑥−¹𝛼3cos2𝜋𝑥⋅𝑝₂,𝑞₁,𝑞₂∈𝑅. (69) So in Case 1, on the branches (60), the eigenvalue of𝐹_𝐴,ℎ(𝐴(𝜀),ℎ(𝜀),𝐿(𝜀))is 𝜆(𝜀)=−32−𝛼6+𝑞₁𝜋²𝜀²+𝑜𝜀². (70) The corresponding eigenfunction is 𝜙𝜓=0+cos𝜋𝑥𝑝₁𝑞cos𝜋𝑥₁+𝑞₂𝛼cos𝜋𝑥−3𝜀+𝑝cos2𝜋𝑥₂3cos𝜋𝑥−+𝛼32𝑞32cos3𝜋𝑥−²2+𝑞₂6𝑞cos2𝜋𝑥₁+𝑞₂𝑝cos𝜋𝑥−¹𝛼3𝜀cos2𝜋𝑥²+𝑜𝜀². (71) From (70), we know that the eigenvalue of the linearized operator𝐹_𝐴,ℎ(𝐴(𝜀),ℎ(𝜀),𝐿(𝜀))at the nontrivial solutions given in (60) is negative if𝑞₁<(3/2+𝛼/6), and positive if𝑞₁>(3/2+𝛼/6). Finally, for small 𝜀, if𝑞₁<(3/2+𝛼/6), then the corresponding solution branches are stable; if𝑞₁>(3/2+𝛼/6), then the corresponding solution branches are unstable; if𝑞₁=(3/2+𝛼/6), we need higher-order items of𝜆to determine the result.
Thus, we have proved the following theorems.

Theorem 1. For𝐿near𝜋, there are nontrivial steady-state solution branches of (4) bifurcated from the trivial solution: 𝐴(𝜀)=𝜀𝛼₁cos𝜋𝑥𝐿+𝜀²𝛼₁𝛼₂6cos2𝜋𝑥𝐿−𝛼₁𝛼₂2+𝜀³𝜋²𝛼96¹−𝜋²96𝛼𝛼³₁−𝜋²𝛼24³¹cos3𝜋𝑥𝐿+𝑜𝜀³,ℎ(𝜀)=𝜀𝛼₂cos𝜋𝑥𝐿+𝜀²𝑐₁−𝛼𝛼²₁6cos2𝜋𝑥𝐿+𝜀³𝑐₁−148𝛼𝛼²₁𝛼₂cos3𝜋𝑥+𝑜𝜀³,𝐿=𝜋+𝜀²𝛾₁(0)+𝑜𝜀², (72) where𝛼₁=1,𝛼₂=0,𝛾₁(0)=3/8+𝛼/24−𝑐₁/2,𝑐₁,𝑐₁∈𝑅.

Remark 2. Since 𝛾₁(0)=3/8+𝛼/24−𝑐₁/2, when 𝑐₁<3/4+𝛼/12, 𝐿=𝜋 is a supercritical bifurcation point; when 𝑐₁>(3/4+𝛼/12),𝐿=𝜋is a supercritical bifurcation point.

Theorem 3. The eigenvalue of the linearized operator 𝐹_𝐴,ℎ(𝐴(𝜀),ℎ(𝜀),𝐿(𝜀))at the nontrivial solutions given in Theorem 1 is negative if 𝑞₁<(3/2+𝛼/6), positive if 𝑞₁>(3/2+𝛼/6). Finally, for small 𝜀, if 𝑞₁<(3/2+𝛼/6), then the corresponding solution branches are stable; if 𝑞₁>(3/2+𝛼/6), then the corresponding solution branches are unstable.

Case 2. 𝑐=0,𝛼₁=0.
If𝑐=0,𝛼₁=0, then𝛼₂=±1; taking the sign of𝜀into account in (15), we consider𝛼₂=1only. In this case, from (40), (41), (45), (49), (56), and (59), we have 𝑎₀(𝑥)=0,𝑏₀(𝑥)=𝑐₁(0)𝑎,𝛾=0,₁(𝑥)=0,𝑏₁(𝑥)=𝑐₁,𝛾₁(0)=0. (73) Similarly discussed as in Case 1 previously, we get 𝛾₂(0)=𝛾₃(0)=𝛾₄(0)=⋯=0. (74)

Remark 4. Since𝛾(0)=𝛾₁(0)=𝛾₂(0)=𝛾₃(0)=𝛾₄(0)=⋯=0, there are no solution branches of (4) bifurcated from the trivial solution in Case 2.

Case 3. 𝑐≠0,𝛼₁≠0.
If𝑐≠0,𝛼₁≠0, then𝛼₂=0,𝛼₁=±1; taking the sign of𝜀into account in (15), we consider𝛼₂=1only. In this case, from (27), (31), and (37), we have 𝑎₀(𝑥)=−2𝑐𝑐₁𝑏cos𝜋𝑥,₀(𝑥)=𝑐₁−𝛼6𝛾(0)1cos2𝜋𝑥,=−2𝑐𝜋. (75) Letting the coefficient of𝜀³of (16) vanish, we get 𝑎₁^′′(𝑥)+𝜋²𝑎₁(𝑥)+𝑐₁𝜋²+2𝜋𝛾₁−34𝑐²𝜋²−34𝜋²−𝛼𝜋12²+𝛼cos𝜋𝑥𝜋12²−𝜋²4cos3𝜋𝑥=0. (76) Taking the inner product of (76) withcos𝜋𝑥, we obtain ¹₀𝑎₁^′′(𝑥)cos𝜋𝑥𝑑𝑥+𝜋²¹₀𝑎₁(𝑥)+cos𝜋𝑥𝑑𝑥₁₀𝑐₁𝜋²+2𝜋𝛾₁−34𝑐²𝜋²−34𝜋²−𝛼𝜋12²cos²+𝛼𝜋𝑥𝑑𝑥𝜋12²−𝜋²4¹₀cos3𝜋𝑥cos𝜋𝑥𝑑𝑥, (77) from which we get 𝛾₁(0)=38𝑐²3𝜋+8𝑐𝜋−¹𝜋2+𝛼𝑎24𝜋,₁(𝑥)=𝑘₂𝛼cos𝜋𝑥+−19632cos3𝜋𝑥. (78) Letting the coefficient of𝜀³of (17) vanish, we get 𝑏₁^′′′′(𝑥)+𝜋²𝑏₁^′′(𝑥)−8𝑐₁+83𝜋⁴𝛼𝑐cos2𝜋𝑥=0, (79) from which we get 𝑏₁(𝑥)=23𝑐₁2𝛼𝑐+9𝛼𝑐cos2𝜋𝑥+𝑐₂. (80) Substituting (78), (80) to the orthogonality condition ¹₀cos𝜋𝑥𝑎₁(𝑥)+𝑐𝑏₁(𝑥)𝑑𝑥=0, (81) we get 𝑘₂=−2𝑐𝑐₂. (82) Thus, we have that for𝐿near𝜋, there are nontrivial steady-state solution branches of (11) bifurcated from the trivial solution: 𝐴(𝜀)=𝜀cos𝜋𝑥+𝜀²−2𝑐𝑐₁cos𝜋𝑥+𝜀³−2𝑐𝑐₂𝛼cos𝜋𝑥+−196𝜀32cos3𝜋𝑥+𝑜³,ℎ(𝜀)=𝜀²𝑐₁−𝛼6cos2𝜋𝑥+𝜀³𝑐₂+23𝑐₁2𝛼𝑐+9𝛼𝑐cos2𝜋𝑥+𝑜𝜀³,1𝐿=𝜋−23𝑐𝜋𝜀+8𝑐²3𝜋+8𝜋−𝑐₁𝜋2+𝛼𝜋𝜀24²+𝑜𝜀². (83)
Next, we wish to consider the stability of the nontrivial solutions given in (83).
Considering the eigenvalue problem 𝐹_𝐴,ℎ(𝐴(𝜀)(𝜀)(𝜀𝜙(𝜀)𝜓(𝜀)(𝜀)𝜙(𝜀)𝜓(𝜀),ℎ,𝐿))=𝜆, (84) since 𝜆(0)=0, 𝜙(0)=cos𝜋𝑥, 𝜓(0)=𝑐, we assume that 𝜙(𝜀)𝜓(𝜀)=𝑐cos𝜋𝑥𝜙+𝜀¹𝜓₁+𝜀²𝜙₂𝜓₂𝜆(𝜀)+⋯,=𝜆₁𝜀+𝜆₂𝜀²𝐹+⋯,_𝐴,ℎ(𝐴(𝜀)(𝜀)(𝜀=𝐿,ℎ,𝐿))₂+𝜕²𝜕𝑥²−3𝐿²𝐴²(𝜀)+𝐿²ℎ(𝜀)𝐿²𝐴(𝜀)2𝛼𝐿²2𝜕𝐴(𝜀)𝜕𝜕𝑥+𝜕𝜕𝑥²𝐴(𝜀)𝜕𝑥²(𝜀)𝜕+𝐴²𝜕𝑥²−𝐿²𝜕²𝜕𝑥²−𝜕⁴𝜕𝑥⁴. (85) Substituting (85) to (84), we get 𝜕²𝜕𝑥²𝜙(𝜀)+𝐿²𝜙(𝜀)−3𝐿²𝐴²(𝜀)𝜙(𝜀)+𝐿²ℎ(𝜀)𝜙(𝜀)+𝐿²𝐴(𝜀)𝜓(𝜀)−𝜆(𝜀)𝜙(𝜀)=0,2𝛼𝐿²2𝜕𝐴(𝜀)(𝜀)𝜕𝑥𝜕𝜙+𝜕𝜕𝑥²𝐴(𝜀)𝜕𝑥²𝜙(𝜀)(𝜀)𝜕+𝐴²𝜙(𝜀)𝜕𝑥²−𝐿²𝜕²𝜓𝜕𝑥²−𝜕⁴𝜓𝜕𝑥⁴−𝜆(𝜀)𝜓(𝜀)=0. (86) On the branches given in (83), equating the coefficient of𝜀in (86) to 0, we obtain 𝜙₁^′′+𝜋²𝜙₁−𝑐𝜋²cos𝜋𝑥−𝜆₁cos𝜋𝑥=0,4𝛼𝜋⁴cos2𝜋𝑥+𝜋²𝜓₁^′′+𝜓₁^′′′′+𝜆₁𝑐=0. (87) From (87) and the boundary conditions 𝜙^′₁(0)=𝜙^′₁(1)=𝜙₁^′′′(0)=𝜙₁^′′′(1),𝜓^′₁(0)=𝜓^′₁(1)=𝜓₁^′′′(0)=𝜓₁^′′′(1), (88) we get 𝜆𝑐=0,₁=−𝑐𝜋²𝜙=0,₁𝜓₁=𝑝₁𝑞cos𝜋𝑥₁+𝑞₂𝛼cos𝜋𝑥−3⋅𝑝cos2𝜋𝑥¹,𝑞₁,𝑞₂∈𝑅. (89) We find that𝑐=0is a contradiction to𝑐≠0, so there are no solution branches of (4) bifurcated from the trivial solution in Case 3.

Case 4. 𝑐≠0,𝛼₁=0.
If𝑐≠0,𝛼₁=0, then𝛼₂=±1; taking the sign of𝜀into account in (15), we consider𝛼₂=1only. In this case, from (27), (31), and (37), we have 𝑎₀(𝑥)=−2𝑐𝑐₁cos𝜋𝑥,𝑏₀(𝑥)=𝑐₁,𝛾(0)=0. (90) Similarly discussed as in Case 1 previously, we get 𝛾₁(0)=𝛾₂(0)=𝛾₃(0)=𝛾₄(0)=⋯=0. (91)

Remark 5. Since𝛾(0)=𝛾₁(0)=𝛾₂(0)=𝛾₃(0)=𝛾₄(0)=⋯=0, there were no solution branches of (4) bifurcated from the trivial solution in Case 4.

Remark 6. In this paper, trivial stationary solution and bifurcations at it are considered for the KS-GL equations with Neumann’s boundary conditions on a finite domain(0,𝐿). The results we get in this paper are new and original. It would be very interesting and much more complicated to try and extend our study to stationary patterns and the corresponding bifurcations. Another interesting but challenging direction would be the case in several space dimensions.

Acknowledgments

The work is supported by the Youth Science and Technology Innovation Fund of Nanjing Agriculture University (no. KJ2010026), the College Graduate Student Scientific Research Innovation Plan of Jiangsu Province (CXLX12_0390) and the Fundamental Research Funds for the Central Universities (no. Y0201100265).