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

Authors: Lei Shi
Journal: Journal of Applied Mathematics
 (2013)

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## 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.


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## Body

## 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:(1)At=A+Axx-|A2|A+Ah,ht=-hxx-hxxxx+α|A2|xx,
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-Mac)1/5is positive, withMadenoting the Marangoni number, and represents the gradient (derivative) of surface tension with respect to surfactant concentration.Macis 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
(2)At=A+Axx-A3+Ah,ht=-hxx-hxxxx+α(A2)xx,
on the cylindrical domainQ=(0,L)×R+subject to the boundary conditions
(3)Ax=hx=Axxx=hxxx=0atx=0,L.
The steady-state equation of (2) reads as
(4)A+Axx-A3+Ah=0,0<x<L,-hxx-hxxxx+α(A2)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.
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
(5)Γ1=(∂2∂x2+I00-∂2∂x2-∂4∂x4),
and the eigenvalue problem is
(6)φxx+φ=λφ,0<x<L,-ψxx-ψxxxx=λψ,0<x<L,φx(0)=φx(L)=φxxx(0)=φxxx(L)=0,ψx(0)=ψx(L)=ψxxx(0)=ψxxx(L)=0.The eigenvaluesλare given by
(7)λ=λm=-(mπL)2+1,λ=δn=(nπL)2-(nπL)4,00000iii000n,m∈N.
If λm=δn, then the corresponding eigenvectors are
(8)(cosmπxL0),(0cosnπxL).
If λm≠δn, then for λm=-(mπ/L)2+1, the corresponding eigenvector is
(9)(cosmπxL0),
and for δn=(nπ/L)2-(nπ/L)4, the corresponding eigenvector is
(10)(0cosnπxL).
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 variablesx~and u~, Lx~=x, u~(x~)=u(x), v~(x~)=v(x), then omit the tildes, and we find that problem (4) becomes
(11)A+L2Axx-L2A3+L2Ah=0,0<x<1,-L2hxx-hxxxx+αL2(A2)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.
The corresponding eigenvalue problem at (0,0) is
(12)φxx+L2φ=λ′φ,0<x<1,-L2ψxx-ψxxxx=λ′ψ,0<x<1,φx(0)=φx(1)=φxxx(0)=φxxx(1)=0,ψx(0)=ψx(1)=ψxxx(0)=ψxxx(1)=0.
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
(13)e1=(cosπxc),e2=(0cosπx),c∈R.
We set
(14)L=π+γ(ε)·ε,(15)(Ah)=∑i=1n‍εαiei+ε2(a(x,ε)b(x,ε)),
whereεis a small parameter,(a(x,ε)b(x,ε))∈{e1,e2}⊥, α12+α22=1.Substituting (14), (15) into (11) leads to
(16)(εα1cosπx+ε2a(x,ε))′′+(π+γ(ε)·ε)2×(εα1cosπx+ε2a(x,ε))-(π+γ(ε)·ε)2×(εα1cosπx+ε2a(x,ε))3+(π+γ(ε)·ε)2×(εα1cosπx+ε2a(x,ε))×(εα1c+εα2cosπx+ε2b(x,ε))=0,(17)-(π+γ(ε)·ε)2(εα1c+εα2cosπx+ε2b(x,ε))′′-(εα1c+εα2cosπx+ε2b(x,ε))′′′′+α(π+γ(ε)·ε)2((εα1cosπx+ε2a(x,ε))2)′′=0,
where ′ means∂/∂xand ′′ means∂2/∂x2.Letting the coefficient ofε2vanish in (16), (17) gives
(18)a0′′(x)+π2a0(x)+2πγ(0)α1cosπx+π2α1cosπx(α1c+α2cosπx)=0,(19)b0′′′′(x)+π2b0′′(x)+2πγ(0)α1cosπx+π2α1cosπx(α1c+α2cosπx)=0,
where(a0(x)b0(x))=(a(x,0)b(x,0))∈{e1,e2}⊥.Taking the inner product of (16) withcosπx, we obtain
(20)∫01‍a0′′(x)cosπxdx+π2∫01‍a0(x)cosπxdx+2πγ(0)α1∫01‍cos2πxdx+π2α1∫01‍cos2πx(α1c+α2cosπx)dx=0.
For the first term of (20), integrate it by part and from the boundary conditiona′(0)=a′(1)=0, and we have
(21)∫01‍a0′′(x)cosπxdx=-π2∫01‍a0(x)cosπxdx.
Substituting (21) to (20), we get
(22)γ(0)=-12cα1π,(α1≠0),γ(0)is an arbitrary constant,(α1=0).From the aforementioned orthogonality condition, we know that(23)∫01‍a0(x)cosπx+cb0(x)dx=0,(24)∫01‍b0(x)cosπxdx=0.
Taking the inner product of (19) withcosπx, we obtain
(25)∫01‍b0′′′′(x)cosπxdx+π2∫01‍b0′′(x)cosπxdx-2α2π3γ(0)∫01‍cos2πx+2αα12π4∫01‍cos2πxcosπxdx=0.
Because of the boundary conditionb′(0)=b′(1)=b′′′(0)=b′′′(1)=0, orthogonality condition (24), we get
(26)γ(0)=0,(α2≠0),γ(0)is  an  arbitrary  constant,(α2=0),
from (25).Comparing (22) and (26), forα12+α22=1, we deduce the following important relation:
(27)γ(0)=-12cα1π.Substituting (27) to (18), we get
(28)a0′′(x)+π2a0(x)+π2α1α2cos2πx=0.
Fora0(x)satisfies the boundary conditiona0′(0)=a1′(0)=a0′′′(0)=a0′′′(1), we calculate from (28) that
(29)a0(x)=k1cosπx+α1α26cos2πx-α1α22,k1∈R.Next, we want to calculateb0(x) and substitute (27) to (19); we get the following ODE:
(30)b0′′′′(x)+π2b0′′(x)+cα1α2π4cosπx+2αα12π4cos2πx=0.
By lengthy computations and the boundary conditionsb0′(0)=b1′(0)=b0′′′(0)=b0′′′(1), we deduce that
(31)b0(x)=c1+c2cosπx-αα126cos2πx,(32)cα1α2=0.
Taking the inner product withcosπxof (31), from (24), we know that
(33)∫01‍(c1+c2cosπx-αα126cos2πx)cosπxdx=0,
which gives us
(34)c2=0,(35)(π296α1-π296αα13-π224α13)cos3πxb0(x)=c1-αα126cos2πx.
Substituting (29), (35) to (23), we have
(36)∫01‍cosπx(k1cosπx+α1α26cos2πx-α1α22)+c(c1-αα126cos2πx)dx=0,
from which we get
(37)k1=-2cc1,a0(x)=-2cc1cosπx+α1α26cos2πx-α1α22.From the previous discussion, we obtain(38)cα1α2=0,
with the normal condition
(39)α12+α22-1=0.Whenc=0, then we have
(40)γ0=0;
at this time,
(41)a0(x)=α1α26cos2πx-α1α22,b0(x)=c1-αα126cos2πx.Letting the coefficient ofε3of (16), (17) vanish, we get
(42)a1′′(x)+π2a1(x)+2πα1γ1(0)cosπx-π2α13cos3πx+π2α1cosπxb0(x)+π2α2a0(x)cosπx=0,(43)b1′′′′(x)+π2b1′′(x)-2α1απ2cosπxa0′′(x)+4α1απ3sinπxa0′(x)+2α1απ4cosπxa0(x)-2π3γ1(0)α2cosπx=0,
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
(44)∫01‍a1′′(x)cosπxdx+π2∫01‍a1(x)cosπxdx+2πα1γ1(0)∫01‍cos2πxdx-π2α13∫01‍cos4πxdx+π2α1∫01‍cos2πxb0(x)dx+π2α2∫01‍a0(x)cos2πxdx=0,
from which we know that
(45)γ1(0)=38πα12+αα12π24-c1π2+524πα22,α1≠0,γ1(0)is  an  arbitrary  constant,α1=0.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
(46)a1′′(x)+π2a1(x)+(π212α1-π212αα13-π23α13)cos3πx=0.
Because of the orthogonality condition
(47)∫01‍a1(x)cosπxdx=0,
and boundary condition
(48)a1′(0)=a1′(1)=0,
we get
(49)a1(x)=(π296α1-π296αα13-π224α13)cos3πx.
Substituting (41) to (43), we have
(50)b1′′′′(x)+π2b1′′(x)-(2π3γ1(0)α2+56αα12α2π4)cosπx+32αα12α2π4cos3πx=0.
Taking the inner product of (50) withcosπx, we obtain
(51)∫01‍b1′′′′(x)cosπxdx+π2∫01‍b1′′(x)cosπxdx-(2π3γ1(0)α2+56αα12α2π4)∫01‍cos2πxdx+32αα12α2π4∫01‍cos3πxcosπxdx=0,
forb1(x)satisfies the orthogonality condition
(52)∫01‍b1(x)cosπxdx=0,
and the boundary condition
(53)b1′(0)=b1′(1)=b1′′′(0)=b1′′′(1);
we know
(54)∫01‍b1′′′′(x)cosπxdx+π2∫01‍b1′′(x)cosπxdx=0,
which forces
(55)2π3γ1(0)α2+56αα12α2π4=0.
From (45) and (55), we get
(56)γ1(0)=-512αα12π=38πα12+αα12π24-c1π2+524πα22,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhα2≠0,
or
(57)γ1(0)=38πα12+αα12π24-c1π2+524πα22,hhhhhhhhhhhhhhhhhhhhα2=0.
Substituting (55) to (50), we get
(58)b1′′′′(x)+π2b1′′(x)+32αα12α2π4cos3πx=0,
which gives us
(59)b1(x)=c1~-148αα12α2cos3πx.
As has been discussed previously, we have the following.
ForLnearπ, there are nontrivial steady-state solution branches of (11) bifurcated from the trivial solution:
(60)A(ε)=εα1cosπx+ε2(α1α26cosπx-α1α22)+ε3(π296α1-π296αα13-π224α13)cos3πx+o(ε3),h(ε)=εα2cosπx+ε2(c1-αα126cos2πx)+ε3(c1~-148αα12α2cos3πx)+o(ε3),L=π+ε2γ1(0)+o(ε2).α1≠0,α2,γ1(0)satisfy (56), (57). c1,c1~∈R.c1, c1~are arbitrary constants.
Next, we wish to consider the stability of the nontrivial solutions given in (60).
Considering the eigenvalue problem(61)FA,h(A(ε),h(ε),L(ε))(ϕ(ε)ψ(ε))=λ(ε)(ϕ(ε)ψ(ε)),
since λ(0)=0, ϕ(0)=cosπx, ψ(0)=0, we assume
(62)(ϕ(ε)ψ(ε))&=(cosπx0)+ε(ϕ1ψ1)+ε2(ϕ2ψ2)+⋯,λ(ε)=λ1ε+λ2ε2+⋯,FA,h(A(ε),h(ε),L(ε))=(L2+∂2∂x2-3L2A2(ε)+L2h(ε)L2A(ε)2αL2(2∂A(ε)∂x∂∂x+∂2A(ε)∂x2+A(ε)∂2∂x2)-L2∂2∂x2-∂4∂x4).
Substituting (62) to (61), we get
(63)∂2∂x2ϕ(ε)+L2ϕ(ε)-3L2A2(ε)ϕ(ε)+L2h(ε)ϕ(ε)+L2A(ε)ψ(ε)-λ(ε)ϕ(ε)=0,2αL2(2∂A(ε)∂x∂ϕ(ε)∂x+∂2A(ε)∂x2ϕ(ε)+A(ε)∂2ϕ(ε)∂x2)-L2∂2ψ∂x2-∂4ψ∂x4-λ(ε)ψ(ε)=0.
On the branches given in (60), equating the coefficient ofεin (63) to 0, we obtain
(64)ϕ1′′+π2ϕ1+α2π2cos2πx-λ1cosπx=0,4αα1π4cos2πx+π2ψ1′′+ψ1′′′′=0.
From (64) and the boundary conditions
(65)ϕ1′(0)=ϕ1′(1)=ϕ1′′′(0)=ϕ1′′′(1),ψ1′(0)=ψ1′(1)=ψ1′′′(0)=ψ1′′′(1),
we get
(66)λ1=0,(ϕ1ψ1)=(p1cosπx+α26cos2πx-α22q1+q2cosπx-αα13cos2πx)·p1,hhhhhhhhhhhhhhhhhhhhhlhhq1,q2∈R.
Using the second-order term ofεin (63), we obtain
(67)ϕ2′′+π2ϕ2-[(-32α12+α1q1-α6α12)π2-λ2]cosπx+(p1α22+q2α12)π2cos2πx+(-34α12+112α12-π24αα12)cos3πx+π22p1α2+π22q2α1=0,ψ2′′′′+π2ψ2′′-53α1α2απ4cosπx+4αα1p1π4cos2πx+3α1α2απ4cos3πx=0,
and considering the boundary condition
(68)ϕ2′(0)=ϕ2′(1)=ϕ1′′′(0)=ϕ2′′′(1),ψ2′(0)=ψ2′(1)=ψ2′′′(0)=ψ2′′′(1),
we deduce that (67) has solution if and only ifα2=0; in this case, we haveα1=±1 (α12+α22=1). Taking the sign ofεinto account in (15), we considerα1=1only. Therefore, we have
(69)λ2=(-32-α6+q1)π2,(ϕ2ψ2)=(p2cosπx+q26cos2πx-(332+α32)cos3πx-q22q1~+q2~cosπx-p1α3cos2πx)·p2,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhq1~,q2~∈R.
So in Case 1, on the branches (60), the eigenvalue ofFA,h(A(ε),h(ε),L(ε))is
(70)λ(ε)=(-32-α6+q1)π2ε2+o(ε2).
The corresponding eigenfunction is
(71)(ϕψ)=(cosπx0)+(p1cosπxq1+q2cosπx-α3cos2πx)ε+(p2cosπx-(332+α32)cos3πx-q22+q26cos2πxq1~+q2~cosπx-p1α3cos2πx)ε2+o(ε2).
From (70), we know that the eigenvalue of the linearized operatorFA,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:
(72)A(ε)=εα1cosπxL+ε2(α1α26cos2πxL-α1α22)+ε3(π296α1-π296αα13-π224α13)cos3πxL+o(ε3),h(ε)=εα2cosπxL+ε2(c1-αα126cos2πxL)+ε3(c1~-148αα12α2cos3πx)+o(ε3),L=π+ε2γ1(0)+o(ε2),
whereα1=1,α2=0,γ1(0)=3/8+α/24-c1/2,c1,c1~∈R.Remark 2.
Sinceγ1(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 operatorFA,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
(73)a0(x)=0,b0(x)=c1,γ(0)=0,a1(x)=0,b1(x)=c1~,γ1(0)=0.
Similarly discussed as in Case 1 previously, we get
(74)γ2(0)=γ3(0)=γ4(0)=⋯=0.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
(75)a0(x)=-2cc1cosπx,b0(x)=c1-α6cos2πx,γ(0)=-12cπ.
Letting the coefficient ofε3of (16) vanish, we get
(76)a1′′(x)+π2a1(x)+(c1π2+2πγ1-34c2π2hhhhhhhhhhhhhhhh-34π2-α12π2)cosπx+(α12π2-π24)cos3πx=0.
Taking the inner product of (76) withcosπx, we obtain
(77)∫01‍a1′′(x)cosπxdx+π2∫01‍a1(x)cosπxdx+∫01‍(c1π2+2πγ1-34c2π2hhhhhhhhh-34π2-α12π2)cos2πxdx+(α12π2-π24)∫01‍cos3πxcosπxdx,
from which we get
(78)γ1(0)=38c2π+38π-c1π2+α24π,a1(x)=k2cosπx+(α96-132)cos3πx.
Letting the coefficient ofε3of (17) vanish, we get
(79)b1′′′′(x)+π2b1′′(x)-(8c1+83)π4αccos2πx=0,
from which we get
(80)b1(x)=(23c1αc+29αc)cos2πx+c2.
Substituting (78), (80) to the orthogonality condition
(81)∫01‍cosπxa1(x)+cb1(x)dx=0,
we get
(82)k2=-2cc2.
Thus, we have that forLnearπ, there are nontrivial steady-state solution branches of (11) bifurcated from the trivial solution:
(83)A(ε)=εcosπx+ε2(-2cc1)cosπx+ε3[-2cc2cosπx+(α96-132)cos3πx]+o(ε3),h(ε)=ε2(c1-α6cos2πx)+ε3(c2+23c1αc+29αc)cos2πx+o(ε3),L=π-12cπε+(38c2π+38πhhhhhhhhhhhhh-c1π2+α24π)ε2+o(ε2).
Next, we wish to consider the stability of the nontrivial solutions given in (83).
Considering the eigenvalue problem(84)FA,h(A(ε),h(ε),L(ε))(ϕ(ε)ψ(ε))=λ(ε)(ϕ(ε)ψ(ε)),
since λ(0)=0, ϕ(0)=cosπx, ψ(0)=c, we assume that
(85)(ϕ(ε)ψ(ε))=(cosπxc)+ε(ϕ1ψ1)+ε2(ϕ2ψ2)+⋯,λ(ε)=λ1ε+λ2ε2+⋯,FA,h(A(ε),h(ε),L(ε))=(L2+∂2∂x2-3L2A2(ε)+L2h(ε)L2A(ε)2αL2(2∂A(ε)∂x∂∂x+∂2A(ε)∂x2+A(ε)∂2∂x2)-L2∂2∂x2-∂4∂x4).
Substituting (85) to (84), we get
(86)∂2∂x2ϕ(ε)+L2ϕ(ε)-3L2A2(ε)ϕ(ε)+L2h(ε)ϕ(ε)+L2A(ε)ψ(ε)-λ(ε)ϕ(ε)=0,2αL2(2∂A(ε)∂x∂ϕ(ε)∂x+∂2A(ε)∂x2ϕ(ε)+A(ε)∂2ϕ(ε)∂x2)-L2∂2ψ∂x2-∂4ψ∂x4-λ(ε)ψ(ε)=0.
On the branches given in (83), equating the coefficient ofεin (86) to 0, we obtain
(87)ϕ1′′+π2ϕ1-cπ2cosπx-λ1cosπx=0,4απ4cos2πx+π2ψ1′′+ψ1′′′′+λ1c=0.
From (87) and the boundary conditions
(88)ϕ1′(0)=ϕ1′(1)=ϕ1′′′(0)=ϕ1′′′(1),ψ1′(0)=ψ1′(1)=ψ1′′′(0)=ψ1′′′(1),
we get
(89)c=0,λ1=-cπ2=0,(ϕ1ψ1)=(p1~~cosπxq1~~+q2~~cosπx-α3cos2πx)·p1~~,hhhhhhhhhhhhhhhhhhhhhq1~~,q2~~∈R.
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
(90)a0(x)=-2cc1cosπx,b0(x)=c1,γ(0)=0.
Similarly discussed as in Case 1 previously, we get
(91)γ1(0)=γ2(0)=γ3(0)=γ4(0)=⋯=0.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.



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*Source: 926512-2013-08-12.xml*