Abstract

We consider a nonlinear equation 𝐹(𝜀,𝜆,𝑢)=0, where the parameter 𝜀 is a perturbation parameter, 𝐹 is a differentiable mapping from R×R×𝑋 to 𝑌, and 𝑋, 𝑌 are Banach spaces. We obtain an abstract bifurcation theorem by using the generalized saddle-node bifurcation theorem.

1. Introduction

In [1, 2], Crandall and Rabinowitz proved two celebrated theorems which are now regarded as foundation of the analytical bifurcation theory in infinite-dimensional spaces and both results are based on the implicit function theorem. In [3], we obtained the generalized saddle-node bifurcation theorem by the generalized inverse. In [4], we proved a perturbed problem using Morse Lemma. For a more general introduction to bifurcation theory and other related methods in nonlinear analysis, see, for example, [57]. On the other hand, [811] provide a more detailed introduction to mathematical models in some recent new results in the application of bifurcation theory including chemical reactions, population ecology, and nonautonomous differential equations.

In this paper, we continue the work of [3] and obtain an abstract bifurcation theorem under the opposite condition in [4]. We consider the solution set of 𝐹(𝜀,𝜆,𝑢)=0,(1.1) where 𝜀 indicates the perturbation. Fix 𝜀=𝜀0; let (𝜆0,𝑢0) be a solution of 𝐹(𝜀0,,)=0. From the implicit function theorem, a necessary condition for bifurcation is that 𝐹𝑢(𝜀0,𝜆0,𝑢0) is not invertible; we call (𝜀0,𝜆0,𝑢0) a degenerate solution. In [12], Shi shows the persistence and the bifurcation of degenerate solutions when 𝜀 varies near 𝜀0 by the implicit function theorem and the saddle-node bifurcation theorem. In this paper, we prove a new perturbed bifurcation theorem by the generalized saddle-node bifurcation theorem.

In the paper, we use as the norm of Banach space 𝑋 and , as the duality pair of a Banach space 𝑋 and its dual space 𝑋. For a nonlinear operator 𝐹, we use 𝐹𝑢 as the partial derivative of 𝐹 with respect to argument 𝑢. For a linear operator 𝐿, we use 𝑁(𝐿) as the null space of 𝐿 and 𝑅(𝐿) as the range of 𝐿.

2. Preliminaries

Definition 2.1 (see [13]). Let 𝑋, 𝑌 be Banach spaces, and let 𝐴(𝑋,𝑌) be a linear operator. Then, 𝐴+(𝑌,𝑋) is called the generalized inverse of 𝐴 if it satisfies(i)𝐴𝐴+𝐴=𝐴,(ii)𝐴+𝐴𝐴+=𝐴+.

Definition 2.2 (see [13]). Let 𝑋,𝑌, and 𝐴 be the same as in Definition 2.1. If 𝐴(𝑋,𝑌) has the bounded linear generalized inverse 𝐴+, then 𝐴 is called a generalized regular operator.

Lemma 2.3 (see [13]). Let 𝐴(𝑋,𝑌), then 𝐴 is a generalized regular operator if and only if 𝑁(𝐴),𝑅(𝐴) are topologically complemented in 𝑋,𝑌, respectively. In this case, 𝐼𝐴+𝐴, 𝐴𝐴+ are bounded linear projectors from 𝑋, 𝑌 into 𝑁(𝐴), 𝑅(𝐴), respectively.

We recall the generalized saddle-node bifurcation in [3] and give an alternate proof here using the generalized Lyapunov-Schmidt reduction.

Theorem 2.4 (generalized saddle-node bifurcation). Let 𝑉𝐑×𝑋 be a neighborhood of (𝜆0,𝑢0),𝐹𝐶1(𝑉,𝑌). Suppose that (i)𝐹(𝜆0,𝑢0)=0;(ii)𝐹𝑢(𝜆0,𝑢0)𝑋𝑌 is a generalized regular operator, and 𝐹dim𝑁𝑢𝜆0,𝑢0𝐹codim𝑅𝑢𝜆0,𝑢0=1,(2.1)(iii)𝐹𝜆(𝜆0,𝑢0)𝑅(𝐹𝑢(𝜆0,𝑢0)).
Let 𝑍=𝑅((𝐹𝑢(𝜆0,𝑢0))+), then the subset {(𝜆,𝑢)|𝐹(𝜆,𝑢)=0} contains the curve (𝜆(𝑠),𝑢(𝑠))=(𝜆(𝑠),𝑢0+𝑠𝑤0+𝑧(𝑠)) near (𝜆0,𝑢0), where 𝑤0𝑁(𝐹𝑢(𝜆0,𝑢0)){𝜃}, the mapping 𝑧(𝑠) is continuously differentiable near 𝑠=0, and 𝜆(0)=𝜆0,𝜆(0)=0,𝑧(0)=𝑧(0)=𝜃.

Proof. Since 𝐴=𝐹𝑢(𝜆0,𝑢0) is a generalized regular operator, there exist closed subspaces 𝑍 in 𝑋, 𝑌1 in 𝑌 satisfing 𝑋=𝑍𝑁(𝐴), 𝑌=𝑅(𝐴)𝑌1.
Taking an arbitrary 𝑤0𝑁(𝐴){𝜃}, from Lemma 2.3, 𝐹(𝜆,𝑢)=0 is equivalent to 𝐼𝐴𝐴+𝐹𝜆,𝑢0+𝑠𝑤0+𝑧=0,𝐴𝐴+𝐹𝜆,𝑢0+𝑠𝑤0+𝑧=0,(2.2) where 𝑠𝐑, 𝑧𝑍.
Define 𝐺𝐑×𝐑×𝑍𝑅(𝐴) as 𝐺(𝑠,𝜆,𝑧)=𝐴𝐴+𝐹𝜆,𝑢0+𝑠𝑤0,𝐺+𝑧(𝜆,𝑧)0,𝜆0[(],0𝜏,𝜓)=𝐴𝐴+𝜏𝐹𝜆𝜆0,𝑢0+𝐹𝑢𝜆0,𝑢0[𝜓],=𝐴𝐴+𝐴[𝜓][𝜓],=𝐴(2.3) because of (iii), then 𝐺(𝜆,𝑧)(0,𝜆0,0)𝑅×𝑍𝑅(𝐴) is an isomorphism.
For the equation 𝐺(𝑠,𝜆,𝑧)=0, by the implicit function theorem, there exist 𝜀>0 and (𝜆(𝑠),𝑧(𝑠))𝐶1(𝜀,𝜀), with 𝜆(0)=𝜆0, 𝑧(0)=0 satisfying 𝐺(𝑠,𝜆(𝑠),𝑧(𝑠))=0.(2.4) From (2.2), we have 𝐹𝜆(𝑠),𝑢0+𝑠𝑤0+𝑧(𝑠)=0,𝑠(𝜀,𝜀).(2.5) Differentiating (2.5) with respect to 𝑠, we have 𝐹𝜆𝜆(𝑠),𝑢0+𝑠𝑤0𝜆+𝑧(𝑠)(𝑠)+𝐹𝑢𝜆(𝑠),𝑢0+𝑠𝑤0𝑤+𝑧(𝑠)0+𝑧(𝑠)=0.(2.6) Setting 𝑠=0, 𝐹𝜆𝜆0,𝑢0𝜆(0)+𝐹𝑢𝜆0,𝑢0𝑤0+𝑧(0)=0.(2.7) Thus, 𝜆(0)=0 since (iii) and we have 𝐹𝑢𝜆0,𝑢0𝑧(0)=0,(2.8) that is, 𝑧(0)𝑁(𝐴)𝑍, we have 𝑧(0)=0.

Corollary 2.5. Assume the conditions in Theorem 2.4 are satisfied and 𝐹dim𝑁𝑢𝜆0,𝑢0𝐹=𝑛,𝑁𝑢𝜆0,𝑢0𝑤=span1,𝑤2,,𝑤𝑛,(2.9) then the direction of the solution curves is determined by 𝜆𝑖(0)=𝑙,𝐹𝑢𝑢𝜆0,𝑢0𝑤𝑖,𝑤𝑖𝑙,𝐹𝜆𝜆0,𝑢0,(2.10) where 𝑙𝑅(𝐹𝑢(𝜆0,𝑢0)), 𝑖=1,2,,𝑛. Furthermore, when 𝐹𝑢𝑢𝜆0,𝑢0𝑤𝑖,𝑤𝑖𝐹𝑅𝑢𝜆0,𝑢0(2.11) is satisfied, 𝜆𝑖(0)0, and the solution curve {(𝜆𝑖(𝑠),𝑢𝑖(𝑠))|𝑠|<𝛿} is a parabola-like curve which reaches an extreme point at (𝜆0,𝑢0).

We illustrate our result by a simple example.

Example 2.6. Define 𝐹𝑥𝑦𝜆,=𝜆𝑥2𝑦2=0,(2.12) where 𝑈=𝑥𝑦𝐑2, 𝜆𝐑. From simple calculations, we obtain 𝐹𝑈=2𝑥,2𝑦,𝐹𝜆=1,𝐹𝑈𝑈=2002.(2.13) We analyze the bifurcation at (0,00). It is easy to see that 𝑁(𝐹𝑈)=span{𝑤1,𝑤2}, where 𝑤1=10, 𝑤2=01, 𝑅(𝐹𝑈)={0}. So, obviously, dim𝑁(𝐹𝑈)=2, codim𝑅(𝐹𝑈)=1, and 𝐹𝜆𝑅(𝐹𝑈). From the above calculation, 𝐹𝑈𝑈𝑤1,𝑤1=2,𝐹𝑈𝑈𝑤2,𝑤2=2.(2.14) Obviously, 𝐹𝑈𝑈(0,00)[𝑤𝑖,𝑤𝑖]𝑅(𝐹𝑈(0,00)) and 𝜆𝑖(0)=2, 𝑖=1,2. Thus, we can apply Corollary 2.5 to (2.12). In fact, all solution curves for all 𝑤𝑖𝑁(𝐹𝑈) form a surface (see Figure 1).

3. Main Theorems

Applying Theorem 2.4, we discuss the bifurcation of solutions of the perturbed problem. We consider the solution set of 𝐹(𝜀,𝜆,𝑢)=0,(3.1) where the parameter 𝜀 indicates the perturbation, 𝐹𝐶1(𝑀,𝑌), 𝑀𝐑×𝐑×𝑋, and 𝑋, 𝑌 are Banach spaces. Let 𝐹𝐻(𝜀,𝜆,𝑢,𝑤)=𝐹(𝜀,𝜆,𝑢)𝑢[𝑤](𝜀,𝜆,𝑢).(3.2)

Suppose that (𝜀0,𝜆0,𝑢0,𝑤0) is a solution of 𝐻(𝜀,𝜆,𝑢,𝑤)=0. For (𝜀0,𝜆0,𝑢0)𝑀 and 𝑤0𝑋1{𝑥𝑋𝑥=1},(3.3) by Hahn-Banach theorem, there exists a closed subspace 𝑋3 of 𝑋 with codimension 1 such that 𝑋=𝐿(𝑤0)𝑋3, where 𝐿(𝑤0)=span{𝑤0} and 𝑑(𝑤0,𝑋3)=inf{||𝑤0𝑥||𝑥𝑋3}>0. Let 𝑋2=𝑤0+𝑋3={𝑤0+𝑥𝑥𝑋3}. Then, 𝑋2 is a closed hyperplane of 𝑋 with codimension 1. Since 𝑋3 is a closed subspace of 𝑋 and 𝑋3 is also a Banach space in the subspace topology, Hence we can regard 𝑀1=𝑀×𝑋2 as a Banach space with product topology. Moreover, the tangent space of 𝑀1 is homeomorphic to 𝑀×𝑋3 (see [12] for more on the setting).

In the following, we will still use the conditions (𝐹𝑖) on 𝐹 defined in [12].(F1)dim𝑁(𝐹𝑢(𝜀0,𝜆0,𝑢0))=codim𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0))=1, and 𝑁(𝐹𝑢(𝜀0,𝜆0,𝑢0))=span{𝑤0};(F2)𝐹𝜆(𝜀0,𝜆0,𝑢0)𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0));(F3)𝐹𝜆𝑢(𝜀0,𝜆0,𝑢0)[𝑤0]𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0));(F4)𝐹𝑢𝑢(𝜀0,𝜆0,𝑢0)[𝑤0,𝑤0]𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0));(F5)𝐹𝜀(𝜀0,𝜆0,𝑢0)𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0)).

We use the convention that (𝐹𝑖) means that the condition defined in (𝐹𝑖) does not hold.

Theorem 3.1. Let 𝐹𝐶2(𝑀,𝑌), 𝑇0=(𝜀0,𝜆0,𝑢0,𝑤0)𝑀1 such that 𝐻(𝑇0)=(0,0). Suppose that the operator 𝐹 satisfies (𝐹1), (𝐹2), (𝐹3), (𝐹4), and (𝐹5) at 𝑇0. One also assumes that 𝐹𝑢𝑢𝜀0,𝜆0,𝑢0𝑣1,𝑤0+𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0𝐹𝑅𝑢𝜀0,𝜆0,𝑢0,(3.4) where 𝑣1𝑋3{0} is the unique solution of 𝐹𝜆𝜀0,𝜆0,𝑢0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝑣]=0.(3.5) Then, there exists 𝛿>0 such that all the solutions of 𝐻(𝜀,𝜆,𝑢,𝑤)=(0,0) near 𝑇0 form two 𝐶2 curves: 𝑇1𝜀(𝑠)=1(𝑠),𝜆1(𝑠),𝑢1(𝑠),𝑤1,𝑇(𝑠),𝑠𝐼=(𝛿,𝛿)2(𝜀𝑠)=2(𝑠),𝜆2(𝑠),𝑢2(𝑠),𝑤2(,𝑠),𝑠𝐼=(𝛿,𝛿)(3.6) where 𝜀𝑖(𝑠)=𝜀0+𝜏𝑖(𝑠), 𝑠𝐼; 𝜏𝑖()𝐶2(𝐼,𝐑); 𝜏𝑖(0)=𝜏𝑖(0)=0, and 𝜆1(𝑠)=𝜆0+𝑧11(𝑠),𝜆2(𝑠)=𝜆0+𝑠+𝑧21𝑢(𝑠),𝑠𝐼,1(𝑠)=𝑢0+𝑠𝑤0+𝑧12(𝑠),𝑢2(𝑠)=𝑢0+𝑠𝑣1+𝑧22𝑤(𝑠),𝑠𝐼,1(𝑠)=𝑤0+𝑠𝜓0+𝑧13(𝑠),𝑤2(𝑠)=𝑤0+𝑠𝜓1+𝑧23(𝑠),𝑠𝐼,(3.7) where 𝑧𝑖𝑗()𝐶2(𝐼,𝑍), 𝑧𝑖𝑗(0)=𝑧𝑖𝑗(0)=0  (𝑖=1,2, 𝑗=1,2,3), 𝜓0𝑋3, 𝜓1𝑋3 are the unique solution of 𝐹𝑢𝑢𝜀0,𝜆0,𝑢0𝑤0,𝑤0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝜓]𝐹=0,(3.8)𝑢𝑢𝜀0,𝜆0,𝑢0𝑣1,𝑤0+𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝜓]=0,(3.9) respectively.

Remark 3.2. Theorem 2.4 complements Theorem  3.2 in [4], where the opposite condition (3.4) is imposed.

Proof. We apply Theorem 2.4 to the operator 𝐻, so we need to verify all the conditions. We define a differential operator 𝐾𝐑×𝑋×𝑋3𝑌×𝑌,  𝐾[]𝜏,𝑣,𝜓=𝐻(𝜆,𝑢,𝑤)𝜀0,𝜆0,𝑢0,𝑤0[]=𝜏,𝑣,𝜓𝜏𝐹𝜆𝜀0,𝜆0,𝑢0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝑣]𝜏𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0+𝐹𝑢𝑢𝜀0,𝜆0,𝑢0𝑣,𝑤0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝜓].(3.10)
(1) dim𝑁(𝐾)=2. Suppose that (𝜏,𝑣,𝜓)𝑁(𝐾) and (𝜏,𝑣,𝜓)0. If 𝜏=0, from 𝐹𝑢(𝜀0,𝜆0,𝑢0)[𝑣]=0 and (𝐹1), then we have 𝑣=𝑘𝑤0 and 𝑘𝐹𝑢𝑢𝜀0,𝜆0,𝑢0𝑤0,𝑤0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝜓]=0.(3.11) From (𝐹4), we can define 𝜓0𝑋3 is the unique solution of (3.8). Thus, (0,𝑤0,𝜓0)𝑁(𝐾) and (𝜏,𝑣,𝜓)=𝑘(0,𝑤0,𝜓0).
Next, we consider 𝜏0. Without loss of generality, we assume that 𝜏=1. Notice that 𝐹𝜆(𝜀0,𝜆0,𝑢0)𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0)) from (𝐹2), we can define that 𝑣1𝑋3{0} is unique solution of (3.5). Substituting 𝜏=1, 𝑣=𝑣1 into (3.10), we have 𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0+𝐹𝑢𝑢𝜀0,𝜆0,𝑢0𝑣1,𝑤0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝜓]=0.(3.12) From (3.4), there exists a unique 𝜓1𝑋3 satisfies (3.9). Then, 𝑁(𝐾)=span0,𝑤0,𝜓0,1,𝑣1,𝜓1,(3.13) that is, dim𝑁(𝐾)=2.
(2) codim𝑅(𝐾)=1. We only claim that 𝑅𝐹(𝐾)=𝑅𝑢𝜀0,𝜆0,𝑢0×𝑌.(3.14) Let (,𝑔)𝑅(𝐾) and (𝜏,𝑣,𝜓)𝐑×𝑋×𝑋3 satisfy 𝜏𝐹𝜆𝜀0,𝜆0,𝑢0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝑣]=,(3.15)𝜏𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0+𝐹𝑢𝑢𝜀0,𝜆0,𝑢0𝑣,𝑤0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝜓]=𝑔.(3.16) Using (3.15) and (𝐹2), then (,𝑔)𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0))×𝑌 and 𝑅(𝐾)𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0))×𝑌.
Conversely, for any (,𝑔)𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0))×𝑌, from (𝐹3), set 𝜏1=𝑙,𝑔𝑙,𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0,(3.17) where 𝑙𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0))𝑌. From (𝐹2), we have 𝜏1𝐹𝜆𝜀0,𝜆0,𝑢0𝐹𝑅𝑢𝜀0,𝜆0,𝑢0.(3.18) Set 𝑣2=[𝐹𝑢|𝑋3]1[𝜏1𝐹𝜆(𝜀0,𝜆0,𝑢0)]𝑋3, we obtain that 𝜏1𝐹𝜆𝜀0,𝜆0,𝑢0+𝐹𝑢𝜀0,𝜆0,𝑢0𝑣2=.(3.19) Substituting 𝜏=𝜏1, 𝑣=𝑣2 into (3.16), we have 𝜏1𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0+𝐹𝑢𝑢𝜀0,𝜆0,𝑢0𝑣2,𝑤0+𝐹𝑢𝜀0,𝜆0,𝑢0[𝜓]=𝑔.(3.20) Using (𝐹1), (𝐹3), then there exists 𝑣3𝑋3, 𝜏2𝐑 satisfies 𝐹𝑢𝑢𝜀0,𝜆0,𝑢0𝑣2,𝑤0=𝜏2𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0+𝐹𝑢𝜀0,𝜆0,𝑢0𝑣3.(3.21) Substituting (3.21) into (3.20), we have 𝜏1+𝜏2𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0+𝐹𝑢𝜀0,𝜆0,𝑢0𝜓+𝑣3=𝑔.(3.22) Applying 𝑙 to (3.22), we have 𝜏2=0 because of the definition of 𝜏1 and 𝑔𝜏1𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0𝐹𝑅𝑢𝜀0,𝜆0,𝑢0.(3.23) Thus we can define 𝜓2=𝐹𝑢𝑋31𝑔𝜏1𝐹𝜆𝑢𝜀0,𝜆0,𝑢0𝑤0𝑣3𝑋3.(3.24) Therefore, 𝐾(𝜏1,𝑣2,𝜓2)=(,𝑔), that is, 𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0))×𝑌𝑅(𝐾). Hence, 𝑅(𝐾)=𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0))×𝑌. That is, codim𝑅(𝐾)=1.
(3) 𝐻𝜀(𝜀0,𝜆0,𝑢0,𝑤0)𝑅(𝐾). Since 𝑅(𝐾)=𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0))×𝑌, we need only to show that 𝐹𝜀(𝜀0,𝜆0,𝑢0)𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0)) but that is exactly assumed in (𝐹5). So, the statement of the theorem follows from Theorem 2.4.

4. Calculations of Bifurcation Directions

In Theorem 3.1, we have 𝜀1(0)=𝜀2(0)=𝜀0, 𝜀1(0)=𝜀2(0)=0, 𝜆1(0)=𝜆2(0)=𝜆0, 𝑢1(0)=𝑢2(0)=𝑢0, 𝑤1(0)=𝑤2(0)=𝑤0, 𝜆1(0)=0, 𝑢1(0)=𝑤0, 𝑤1(0)=𝜓0, 𝜆2(0)=1, 𝑢2(0)=𝑣1, 𝑤2(0)=𝜓1.

To completely determine the turning direction of curve of degenerate solutions, we need some calculations.

Let {𝑇𝑖(𝑠)=(𝜀𝑖(𝑠),𝜆𝑖(𝑠),𝑢𝑖(𝑠),𝑤𝑖(𝑠))𝑠(𝛿,𝛿)} be a curve of degenerate solutions which we obtain in Theorem 3.1. Differentiating 𝐻(𝜀𝑖(𝑠),𝜆𝑖(𝑠),𝑢𝑖(𝑠),𝑤𝑖(𝑠))=0 with respect to 𝑠, we obtain 𝐹𝜀𝜀𝑖(𝑠)+𝐹𝜆𝜆𝑖(𝑠)+𝐹𝑢𝑢𝑖𝐹(𝑠)=0,𝜀𝑢𝑤𝑖(𝜀𝑠)𝑖(𝑠)+𝐹𝜆𝑢𝑤𝑖(𝜆𝑠)𝑖(𝑠)+𝐹𝑢𝑢𝑤𝑖(𝑠),𝑢𝑖(𝑠)+𝐹𝑢𝑤𝑖(𝑠)=0.(4.1) Setting 𝑠=0 in (4.1), we get exactly 𝐹𝑢[𝑤0]=0, (3.5), (3.8), and (3.9). We differentiate (4.1) again, and we have (omit the subscript 𝑖 in the equation) 𝐹𝜀𝜀𝜀(𝑠)2+𝐹𝜀𝜀(𝑠)+𝐹𝜆𝜆𝜆(𝑠)2+𝐹𝜆𝜆(𝑠)+𝐹𝑢𝑢𝑢(𝑠),𝑢(𝑠)+𝐹𝑢𝑢(𝑠)+2𝐹𝜀𝜆𝜀(𝑠)𝜆(𝑠)+2𝐹𝜀𝑢𝑢𝜀(𝑠)(𝑠)+2𝐹𝜆𝑢𝑢𝜆(𝑠)𝐹(𝑠)=0,(4.2)𝜀𝜀𝑢[]𝜀𝑤(𝑠)(𝑠)2+𝐹𝜀𝜆𝑢[]𝜀𝑤(𝑠)(𝑠)𝜆(𝑠)+𝐹𝜀𝑢𝑢𝑢𝜀(𝑠),𝑤(𝑠)(𝑠)+𝐹𝜀𝑢𝑤𝜀(𝑠)(𝑠)+𝐹𝜀𝑢[]𝜀𝑤(𝑠)(𝑠)+𝐹𝜀𝜆𝑢[]𝜀𝑤(𝑠)(𝑠)𝜆(𝑠)+𝐹𝜆𝜆𝑢[]𝜆𝑤(𝑠)(𝑠)2+𝐹𝜆𝑢𝑢𝑢𝜆(𝑠),𝑤(𝑠)(𝑠)+𝐹𝜆𝑢𝑤𝜆(𝑠)(𝑠)+𝐹𝜆𝑢[]𝜆𝑤(𝑠)(𝑠)+𝐹𝜀𝑢𝑢𝑢(𝑠),𝑤(𝑠)𝜀(𝑠)+𝐹𝜆𝑢𝑢𝑢𝜆(𝑠),𝑤(𝑠)(𝑠)+𝐹𝑢𝑢𝑢𝑢(𝑠),𝑢(𝑠),𝑤(𝑠)+𝐹𝑢𝑢𝑤(𝑠),𝑢(𝑠)+𝐹𝑢𝑢𝑤(𝑠),𝑢(𝑠)+𝐹𝜀𝑢𝑤𝜀(𝑠)(𝑠)+𝐹𝜆𝑢𝑤𝜆(𝑠)(𝑠)+𝐹𝑢𝑢𝑤(𝑠),𝑢(𝑠)+𝐹𝑢𝑤𝐹(𝑠)=0,(4.3)𝜀𝜀𝑢[]𝜀𝑤(𝑠)(𝑠)2+𝐹𝜀𝑢[]𝜀𝑤(𝑠)(𝑠)+𝐹𝜆𝑢[]𝜆𝑤(𝑠)(𝑠)+𝐹𝜆𝜆𝑢[]𝜆𝑤(𝑠)(𝑠)2+𝐹𝑢𝑢𝑢𝑢(𝑠),𝑢(𝑠),𝑤(𝑠)+𝐹𝑢𝑢𝑤(𝑠),𝑢(𝑠)+𝐹𝑢𝑤(𝑠)+2𝐹𝜀𝜆𝑢[]𝜀𝑤(𝑠)(𝑠)𝜆(𝑠)+2𝐹𝜀𝑢𝑢𝑢𝜀(𝑠),𝑤(𝑠)(𝑠)+2𝐹𝜆𝑢𝑢𝑢𝜆(𝑠),𝑤(𝑠)(𝑠)+2𝐹𝜀𝑢𝑤𝜀(𝑠)(𝑠)+2𝐹𝜆𝑢𝑤𝜆(𝑠)(𝑠)+2𝐹𝑢𝑢𝑤(𝑠),𝑢(𝑠)=0.(4.4)

Setting 𝑠=0 in (4.2), we obtain 𝐹𝜀𝜀1(0)+𝐹𝜆𝜆1(0)+𝐹𝑢𝑢𝑤0,𝑤0+𝐹𝑢𝑢1(0)=0,(4.5)𝐹𝜀𝜀2(0)+𝐹𝜆𝜆+𝐹𝜆𝜆2(0)+𝐹𝑢𝑢𝑣1,𝑣1+𝐹𝑢𝑢2(0)+2𝐹𝜆𝑢𝑣1=0.(4.6) And applying 𝑙 to it, we have 𝜀1(0)=0,(4.7)𝜀2(0)=𝑙,𝐹𝜆𝜆+𝐹𝑢𝑢𝑣1,𝑣1+2𝐹𝜆𝑢𝑣1𝑙,𝐹𝜀,(4.8) Using (𝐹2), (𝐹4), (𝐹5). From (4.7), (4.5) implies 𝑢1(0)=𝜆1(0)𝑣1+𝜓0+𝑘𝑤0. Setting 𝑠=0 in (4.4), 𝐹𝜆𝑢𝑤0𝜆1(0)+𝐹𝑢𝑢𝑢𝑤0,𝑤0,𝑤0+𝐹𝑢𝑢𝑤0,𝑢1(0)+𝐹𝑢𝑤1(0)+2𝐹𝑢𝑢𝜓0,𝑤0𝐹=0,(4.9)𝜀𝑢𝑤0𝜀2(0)+𝐹𝜆𝑢𝑤0𝜆2(0)+𝐹𝜆𝜆𝑢𝑤0+𝐹𝑢𝑢𝑢𝑣1,𝑣1,𝑤0+𝐹𝑢𝑢𝑤0,𝑢2(0)+𝐹𝑢𝑤2(0)+2𝐹𝜆𝑢𝑢𝑣1,𝑤0+2𝐹𝜆𝑢𝜓1+2𝐹𝑢𝑢𝜓1,𝑣1=0.(4.10) Substituting the expression of 𝑢1(0) into (4.9), we have 𝜆1𝐹(0)𝜆𝑢𝑤0+𝐹𝑢𝑢𝑣1,𝑤0+3𝐹𝑢𝑢𝜓0,𝑤0+𝐹𝑢𝑢𝑢𝑤0,𝑤0,𝑤0+𝑘𝐹𝑢𝑢𝑤0,𝑤0+𝐹𝑢𝑤1(0)=0.(4.11) And applying 𝑙 to it, we obtain 𝑙,𝐹𝑢𝑢𝑢[𝑤0,𝑤0,𝑤0]+3𝐹𝑢𝑢[𝜓0,𝑤0]=0, that is, 𝐹𝑢𝑢𝑢𝑤0,𝑤0,𝑤0+3𝐹𝑢𝑢𝜓0,𝑤0𝐹𝑅𝑢𝜀0,𝜆0,𝑢0.(4.12) Assume 𝐹𝜀𝑢[𝑤0]𝑅(𝐹𝑢(𝜀0,𝜆0,𝑢0)) and applying 𝑙 to (4.10), 𝜆2(0)=𝑙,𝐹𝜆𝜆𝑢𝑤0+𝐹𝑢𝑢𝑢𝑣1,𝑣1,𝑤0+𝐹𝑢𝑢𝑤0,𝑢2(0)+2𝐹𝜆𝑢𝑢𝑣1,𝑤0+2𝐹𝜆𝑢𝜓1+2𝐹𝑢𝑢𝜓1,𝑣1𝑙,𝐹𝜆𝑢𝑤0.(4.13) We differentiate (4.2) again: 𝐹𝜀𝜀𝜀𝜀(𝑠)3+𝐹𝜀𝜀(𝑠)+𝐹𝜆𝜆𝜆𝜆(𝑠)3+𝐹𝜆𝜆(𝑠)+𝐹𝑢𝑢𝑢𝑢(𝑠),𝑢(𝑠),𝑢(𝑠)+𝐹𝑢𝑢(𝑠)+3𝐹𝜀𝜀𝜀(𝑠)𝜀(𝑠)+3𝐹𝜆𝜆𝜆(𝑠)𝜆(𝑠)+3𝐹𝑢𝑢𝑢(𝑠),𝑢(𝑠)+3𝐹𝜀𝜆𝜀(𝑠)𝜆(𝑠)+3𝐹𝜀𝜆𝜀(𝑠)𝜆(𝑠)+3𝐹𝜀𝑢𝑢𝜀(𝑠)(𝑠)+3𝐹𝜀𝑢𝑢𝜀(𝑠)(𝑠)+3𝐹𝜆𝑢𝑢𝜆(𝑠)(𝑠)+3𝐹𝜆𝑢𝑢𝜆(𝑠)(𝑠)+3𝐹𝜀𝜆𝜆𝜀𝜆(𝑠)(𝑠)2+3𝐹𝜀𝜀𝜆𝜀(𝑠)2𝜆(𝑠)+3𝐹𝜀𝜀𝑢𝑢𝜀(𝑠)(𝑠)2+3𝐹𝜀𝑢𝑢𝑢(𝑠),𝑢𝜀(𝑠)(𝑠)+3𝐹𝜆𝜆𝑢𝑢𝜆(𝑠)(𝑠)2+3𝐹𝜆𝑢𝑢𝑢(𝑠),𝑢𝜆(𝑠)(𝑠)+6𝐹𝜀𝜆𝑢𝑢𝜀(𝑠)(𝑠)𝜆(𝑠)=0.(4.14) Setting 𝑠=0 in (4.14), we obtain 𝐹𝜀𝜀1(0)+𝐹𝜆𝜆1(0)+𝐹𝑢𝑢𝑢𝑤0,𝑤0,𝑤0𝑢+𝐹𝑢1(0)+3𝐹𝑢𝑢𝑢1(0),𝑤0+3𝐹𝜆𝑢𝑤0𝜆1(𝐹0)=0,(4.15)𝜆𝜆2(0)+𝐹𝜀𝜀2(0)+3𝜆2𝐹(0)𝜆𝜆+𝐹𝜆𝑢𝑣1+3𝜀2𝐹(0)𝜀𝜆+𝐹𝜀𝑢𝑣1+𝐹𝜆𝜆𝜆+𝐹𝑢𝑢𝑢𝑣1,𝑣1,𝑣1+𝐹𝑢𝑢2(0)+3𝐹𝑢𝑢𝑢2(0),𝑣1+3𝐹𝜆𝑢𝑢2(0)+3𝐹𝜆𝜆𝑢𝑣1+3𝐹𝜆𝑢𝑢𝑣1,𝑣1=0.(4.16) Substituting the expression of 𝑢1(0) into (4.15) and applying 𝑙 to it, we have 𝜀1(0)=0 using (3.4), (4.12), and (𝐹5).

Acknowledgments

The authors the referee for very careful reading and helpful suggestions on the paper. The paper was partially supported by NSFC (Grant no. 11071051), Youth Science Foundation of Heilongjiang Province (Grant no. QC2009C73), Harbin Normal University academic backbone of youth project, and NCET of Heilongjiang Province of China (1251–NCET–002).