Asymptotic Behavior of Bifurcation Curve for Sine-Gordon-Type Differential Equation

Abstract

We consider the nonlinear eigenvalue problems for the equation −𝑢^′′(𝑡)+sin𝑢(𝑡)=𝜆𝑢(𝑡), 𝑢(𝑡)>0, 𝑡∈𝐼=∶(0,1), 𝑢(0)=𝑢(1)=0, where 𝜆>0 is a parameter. It is known that for a given 𝜉>0, there exists a unique solution pair (𝑢_𝜉,𝜆(𝜉))∈𝐶²(𝐼)×ℝ₊ with ‖𝑢_𝜉‖_∞=𝜉. We establish the precise asymptotic formulas for bifurcation curve 𝜆(𝜉) as 𝜉→∞ and 𝜉→0 to see how the oscillation property of sin𝑢 has effect on the behavior of 𝜆(𝜉). We also establish the precise asymptotic formula for bifurcation curve 𝜆(𝛼)  (𝛼=‖𝑢_𝜆‖₂) to show the difference between 𝜆(𝜉) and 𝜆(𝛼).

1. Introduction

We consider the following nonlinear eigenvalue problem: −𝑢^′′(𝑡)(𝑡)(𝑡)(),𝑢(𝑡)𝑢(0)(1)+sin𝑢=𝜆𝑢,𝑡∈𝐼=∶0,1>0,𝑡∈𝐼,=𝑢=0, (1.1)(1.2)(1.3) where 𝜆>0 is a parameter. This problem comes from sine-Gordon equation and has been investigated from a view point of bifurcation theory in 𝐿^∞-framework. Indeed, by using implicit function theorem, it has been shown in [1] that for 𝜉>0, there exists a continuous function 𝜆=𝜆(𝜉) such that (𝑢_𝜉,𝜆(𝜉))∈𝐶²(𝐼)×ℝ₊ satisfies (1.1)–(1.3) with ‖𝑢_𝜉‖_∞=𝜉. Moreover, the solution set of of (1.1)–(1.3) is given by Γ∶={(𝑢_𝜉,𝜆(𝜉))∈𝐶²(𝐼)×ℝ₊;𝜉>0}. Furthermore, it is well known that 𝑢_𝜉(𝑡)∼𝜉sin𝜋𝑡 for 𝜉≫1 and 0<𝜉≪1. Therefore, we have 𝜆(𝜉)⟶𝜋²(),𝜆(𝜉)𝜉⟶∞⟶𝜋²+1(𝜉⟶0). (1.4)(1.5) Equations (1.1)–(1.3) are the special case of the following semilinear equation: −𝑢^′′(𝑡)(𝑢(𝑡(𝑡)𝑢(𝑡)𝑢(0)(1)+𝑓))=𝜆𝑢,𝑡∈𝐼,>0,𝑡∈𝐼,=𝑢=0. (1.6)(1.7)(1.8) The structures of the global behavior of the bifurcation curves of (1.6)–(1.8) have been studied by many authors in 𝐿^∞-framework. We refer to [2–6] and the references therein. In particular, if 𝑓(𝑢)/𝑢 is strictly increasing as 𝑢→∞, then we know from [3] that 𝜆(𝜉) is also strictly increasing for 𝜉>0 and the asymptotic behavior of 𝜆(𝜉) as 𝜉→∞ is mainly determined by 𝑓(𝜉)/𝜉. For example, if 𝑓(𝑢)=𝑢^𝑝 (𝑝>1) in (1.6), then as 𝜉→∞ (cf. [7]), 𝜆(𝜉)=𝜉^𝑝−1+𝑂𝑒^−𝛿𝜉, (1.9) where 𝛿>0 is a constant. However, since (sin𝑢)/𝑢 is not strictly increasing but oscillating as a function of 𝑢≥0, it is interesting to study whether the oscillation property of sin𝑢 has effect on the asymptotic shape of 𝜆(𝜉) for 𝜉>0 or not.

Motivated by this, we first establish the precise asymptotic formula for 𝜆(𝜉) as 𝜉→∞.

Theorem 1.1. As 𝜉→∞, 𝜆(𝜉)=𝜋²+22𝜋𝜉^−3/23cos𝜉−4𝜋+22𝜋𝜉^−5/2−383sin𝜉−4𝜋−1√2𝜋²1cos2𝜉−4𝜋+1𝜋²1cos𝜉cos𝜉−4𝜋+𝑜𝜉^−5/2. (1.10)

The local behavior of 𝜆(𝜉) as 𝜉→0 can be obtained formally by the method in [8]. However, it seems rather hard task to obtain the higher terms of the asymptotic expansion of 𝜆(𝜉), since it is necessary to solve the equations derived from the asymptotic expansion of 𝜆(𝜉) step by step.

Here, we introduce a simpler way on how to obtain the asymptotic expansion formula for 𝜆(𝜉) as 𝜉→0.

Theorem 1.2. Let an arbitrary integer 𝑁>0 be fixed. Then as 𝜉→0, 𝜆=𝜋²1+1−8𝜉²+111921+8𝜋²𝜉⁴+^𝑁_𝑛=3𝑎_𝑛𝜉^2𝑛+𝑜𝜉^2𝑁, (1.11) where {𝑎_𝑛}(𝑛=3,4,…) are the constants determined inductively.

Next, since (1.1)–(1.3) is regarded as an eigenvalue problem, we focus our attention on studying the structure of the solution set in 𝐿²-framework. Suppose that 𝑓(𝑢)=𝑢^𝑝(𝑝>1) in (1.6). Then we know from [9] that, for a given 𝛼>0, there exists a unique solution pair (𝑢_𝛼,𝜆(𝛼))∈𝐶²(𝐼)×ℝ₊ of (1.6)–(1.8) satisfying ‖𝑢_𝛼‖₂=𝛼. Furthermore, 𝜆(𝛼) is an increasing function of 𝛼>0 and as 𝛼→∞, 𝜆(𝛼)=𝛼^𝑝−1+𝐶₀𝛼^(𝑝−1)/2+𝑂(1). (1.12)

We see from (1.9) and (1.12) the difference between the asymptotic formulas for 𝜆(𝜉) and 𝜆(𝛼) when 𝑓(𝑢)=𝑢^𝑝 in (1.6). We refer to [4, 7, 9] for the works in this direction.

Motivated by this, it seems interesting to compare the asymptotic behavior of 𝜆(𝛼) and 𝜆(𝜉) of (1.1)–(1.3) when 𝜉≫1 and 𝛼≫1.

Now we consider (1.1)–(1.3) in 𝐿²-framework. Let 𝛼>0 be a given constant. Assume that there exists a solution pair (𝑢_𝛼,𝜆(𝛼))∈𝐶²(𝐼)×ℝ₊ satisfying ‖𝑢_𝛼‖₂=𝛼. Then, it is natural to expect that for 𝑡∈𝐼, as 𝛼→∞, 𝑢_𝛼(𝑡)𝛼⟶√2sin𝜋𝑡. (1.13) Therefore, we expect that ‖𝑢_𝛼‖_∞∼√2‖𝑢_𝛼‖₂ for 𝛼≫1. To obtain the existence, we apply the variational method to our situation, namely, we consider the constrained minimization problem associated with (1.1)–(1.3). Let 𝑀_𝛼∶=𝑣∈𝐻¹₀(𝐼)∶‖𝑣‖₂=𝛼, (1.14) where ‖𝑣‖₂ is the usual 𝐿²-norm of 𝑣, 𝛼>0 is a parameter, and 𝐻¹₀(𝐼) is the usual real Sobolev space. Then consider the following minimizing problem, which depends on 𝛼>0: (𝑣)1Minimize𝐾∶=2𝑣^′2₂+_𝐼(1−cos𝑣(𝑡))𝑑𝑡undertheconstraint𝑣∈𝑀_𝛼. (1.15) Let 𝛽(𝛼)∶=min_{𝑣∈𝑀𝛼}𝐾(𝑣). (1.16) Then by Lagrange multiplier theorem, for a given 𝛼>0, there exists a pair (𝑢_𝛼,𝜆(𝛼))∈𝑀_𝛼×ℝ₊ which satisfies (1.1)–(1.3) with 𝐾(𝑢_𝛼)=𝛽(𝛼). Here, 𝜆(𝛼), which is called the variational eigenvalue, is the Lagrange multiplier. By this variational framework, we parameterize the solution (𝑢,𝜆) of (1.1)–(1.3) by 𝛼, that is, (𝑢,𝜆)=(𝑢_𝛼,𝜆(𝛼))∈𝑀_𝛼×ℝ₊. Then we know from the arguments in [10, 11] that 𝜆(𝛼) is continuous function for 0<𝛼≪1 and 𝛼≫1. Our next aim is to study precisely the asymptotic behavior of 𝜆(𝛼) as 𝛼→∞.

Theorem 1.3. As 𝛼→∞𝜆(𝛼)=𝜋²+2^3/4𝜋^−1/2𝛼^−3/2cos√32𝛼−4𝜋−𝜋⁻³𝛼⁻²sin√32𝛼−4𝜋cos√32𝛼−4𝜋+2^1/4𝜋^−1/2𝛼^−5/2−38sin√2𝛼−34𝜋−1√2𝜋²cos2√12𝛼−4𝜋+1𝜋²cos√2𝛼cos√2𝛼−14𝜋−14𝜋⁻⁵cos³√32𝛼−4𝜋+𝑜𝛼^−5/2. (1.17)

By Theorems 1.1 and 1.3, we clearly understand the difference between 𝜆(𝜉) and 𝜆(𝛼).

The remainder of this paper is organized as follows. In Section 2, we prove Theorem 1.1. We prove Theorem 1.2 in Section 3. Section 4 is devoted to the proof of Theorem 1.3.

2. Proof of Theorem 1.1

In what follows, 𝐶 denotes various positive constants independent of 𝜉≫1. We write 𝜆=𝜆(𝜉) for simplicity. We know from [1] that if (𝑢_𝜉,𝜆(𝜉))∈𝐶²(𝐼)×ℝ₊ satisfies (1.1)–(1.3), then 𝑢_𝜉(𝑡)=𝑢_𝜉()𝑢1−𝑡,0≤𝑡≤1,_𝜉12=max_0≤𝑡≤1𝑢_𝜉(𝑡)𝑢=𝜉,_′𝜉(𝑡)1>0,0≤𝑡<2. (2.1)(2.2)(2.3) By (1.1), for 𝑡∈𝐼, 𝑢_𝜉^′′(𝑡)+𝜆𝑢_𝜉(𝑡)−sin𝑢_𝜉(𝑡)𝑢^′_𝜉(𝑡)=0. (2.4) This implies that for 𝑡∈𝐼, 𝑑1𝑑𝑡2𝑢^′_𝜉(𝑡)²+12𝜆𝑢_𝜉(𝑡)²+cos𝑢_𝜉(𝑡)=0. (2.5) By this, (2.2) and putting 𝑡=1/2, we obtain 12𝑢^′_𝜉(𝑡)²+12𝜆𝑢_𝜉(𝑡)²+cos𝑢_𝜉(𝑡)1≡constant=2𝜆𝜉²+cos𝜉. (2.6) By this and (2.3), for 0≤𝑡≤1/2, 𝑢^′_𝜉(𝑡)=𝜆𝜉²−𝑢_𝜉(𝑡)²+2cos𝜉−cos𝑢_𝜉(𝑡). (2.7) Then by putting 𝑠=𝑢_𝜉(𝑡)/𝜉, we obtain 12=₀^1/2𝑑𝑡=₀^1/2𝑢^′_𝜉(𝑡)𝜆𝜉²−𝑢_𝜉(𝑡)²+2cos𝜉−cos𝑢_𝜉(𝑡)=1𝑑𝑡√𝜆¹⁰11−𝑠²+2(cos𝜉−cos𝜉𝑠)/𝜆𝜉²=1𝑑𝑠√𝜆¹⁰1√1−𝑠²𝑑𝑠+¹₀1√1−𝑠²+𝐵𝑑𝑠−¹₀1√1−𝑠²=1𝑑𝑠√𝜆𝜋2+𝑉, (2.8) where 𝑉∶=−¹₀𝐵√1−𝑠²+𝐵√1−𝑠²√1−𝑠²+𝐵+√1−𝑠²2𝑑𝑠,𝐵∶=𝜆𝜉²(cos𝜉−cos𝜉𝑠). (2.9)(2.10) We put 𝑉₁1=−𝜆𝜉²¹⁰cos𝜉−cos𝜉𝑠(1−𝑠²)^3/2𝑉𝑑𝑠,₂=𝑉−𝑉₁. (2.11)(2.12)

Lemma 2.1. For 𝜉≫1𝑉₁=𝜋21𝜆𝜉^3/21+15128𝜉²((131+𝑜))cos𝜉−4𝜋−3((138𝜉1+𝑜))sin𝜉−4𝜋. (2.13)

Proof. By putting 𝑠=sin𝜃 in (2.11), integration by parts and l'Hopital's rule, −𝑉₁=1𝜆𝜉^20𝜋/21cos²𝜃((=1cos𝜉−cos𝜉sin𝜃))𝑑𝜃𝜆𝜉^20𝜋/2(tan𝜃)^′((=1cos𝜉−cos𝜉sin𝜃))𝑑𝜃𝜆𝜉²lim_{𝑡→𝜋/2}tan𝑡(cos𝜉−cos(𝜉sin𝑡))^𝑡₀−1𝜆𝜉₀^𝜋/2()1tan𝜃cos𝜃sin𝜉sin𝜃𝑑𝜃=−𝜆𝜉₀^𝜋/2sin𝜃sin(𝜉sin𝜃)𝑑𝜃. (2.14) By [12, page 962], ₀^𝜋/2()𝜋sin𝜃sin𝜉sin𝜃𝑑𝜃=2𝐽₁(𝜉), (2.15) where 𝐽₁(𝜉) is Bessel function of the first kind. For 𝜉≫1, by [12, page 972], we have 𝐽₁(𝜉)=2𝜋𝜉1+15128𝜉²((131+𝑜))cos𝜉−4𝜋−3((138𝜉1+𝑜))sin𝜉−4𝜋. (2.16) By this, (2.14) and (2.15), we obtain (2.13). Thus, the proof is complete.

Remark 2.2. Taking (1.4) into account, (2.13) is written as 𝑉₁=2^−1/2𝜋^−3/2𝜉^−3/2(1+𝑜(1))1+15128𝜉²((131+𝑜))cos𝜉−4𝜋−3((138𝜉1+𝑜))sin𝜉−4𝜋. (2.17) After we obtain (2.31) later, then (2.13) will be improved in the form (2.32).

Lemma 2.3. For 𝜉≫1, 𝑉₂=−2^−1/2𝜋^−7/2(1+𝑜(1))𝜉^−5/21√21cos2𝜉−4𝜋1−cos𝜉cos𝜉−4𝜋+𝑜𝜉^−5/2. (2.18)

Proof. For 𝜉≫1 and 0≤𝑠≤1, by mean value theorem, |𝐵|≤𝐶𝜉⁻¹(1−𝑠)≤𝐶𝜉⁻¹1−𝑠². (2.19) By this and Lebesgue's convergence theorem, we have 𝑉₂2=−𝜆𝜉²¹⁰cos𝜉−cos𝜉𝑠√1−𝑠²×1√1−𝑠²+𝐵√1−𝑠²+𝐵+√1−𝑠²−1√1−𝑠²2√1−𝑠²((12𝑑𝑠=−1+𝑜))𝜆𝜉²¹⁰cos𝜉−cos𝜉𝑠√1−𝑠²×21−𝑠²−1−𝑠²+𝐵+√1−𝑠²√1−𝑠²+𝐵√1−𝑠²+𝐵√1−𝑠²+𝐵+√1−𝑠²√1−𝑠²2√1−𝑠²((12𝑑𝑠=−1+𝑜))𝜆𝜉²¹⁰cos𝜉−cos𝜉𝑠√1−𝑠²⋅1−𝑠²−𝐵−√1−𝑠²√1−𝑠²+𝐵4(1−𝑠²)²((11𝑑𝑠=−1+𝑜))2𝜆𝜉²¹⁰cos𝜉−cos𝜉𝑠√1−𝑠²⋅(1−𝑠²−𝐵)²−1−𝑠²1−𝑠²+𝐵(1−𝑠²)²1−𝑠²−𝐵+√1−𝑠²√1−𝑠²=3+𝐵𝑑𝑠4((111+𝑜))𝜆𝜉²¹⁰cos𝜉−cos𝜉𝑠√1−𝑠²⋅1−𝑠²𝐵(1−𝑠²)³=3𝑑𝑠2((111+𝑜))𝜆²𝜉⁴¹⁰(cos𝜉−cos𝜉𝑠)²(1−𝑠²)^5/2=3𝑑𝑠2((111+𝑜))𝜆²𝜉^40𝜋/2(cos𝜉−cos(𝜉sin𝜃))²cos⁴𝜃=3𝑑𝜃2((111+𝑜))𝜆²𝜉⁴𝑉₃, (2.20) where 𝑉₃∶=₀^𝜋/2(cos𝜉−cos(𝜉sin𝜃))²cos⁴𝜃𝑑𝜃. (2.21) We know 1cos⁴𝜃1𝑑𝜃=31sin𝜃cos³𝜃+2cos𝜃. (2.22) Taking (2.22) into account and integration by parts in 𝑉₃, we obtain that 𝑉₃=lim_{𝜃→𝜋/2}131sin𝜃cos³𝜃+2((cos𝜃cos𝜉−cos𝜉sin𝜃))^2𝜃₀−23𝜉₀^𝜋/21sin𝜃cos²𝜃((()1+2cos𝜉−cos𝜉sin𝜃))sin𝜉sin𝜃𝑑𝜃∶=3𝑉₄−23𝜉𝑉₅+𝑉₆, (2.23) where 𝑉₄∶=lim_{𝜃→𝜋/2}1sin𝜃cos³𝜃+2((cos𝜃cos𝜉−cos𝜉sin𝜃))²𝑉₅∶=₀^𝜋/2sin𝜃cos²𝜃((()𝑉cos𝜉−cos𝜉sin𝜃))sin𝜉sin𝜃𝑑𝜃,₆∶=2₀^𝜋/2sin𝜃(cos𝜉−cos(𝜉sin𝜃))sin(𝜉sin𝜃)𝑑𝜃. (2.24)(2.25)(2.26) Then by l'Hopital's rule, 𝑉₄=lim_{𝜃→𝜋/2}1sin𝜃cos³𝜃+2((cos𝜃cos𝜉−cos𝜉sin𝜃))²=lim_{𝜃→𝜋/2}1+2cos²𝜃(cos𝜃−cos(𝜉sin𝜃))²cos³𝜃=lim_{𝜃→𝜋/2}(cos𝜉−cos(𝜉sin𝜃))²cos³𝜃=lim_{𝜃→𝜋/2}−2𝜉(cos𝜉−cos(𝜉sin𝜃))sin(𝜉sin𝜃)3cos𝜃sin𝜃=lim_{𝜃→𝜋/2}−2𝜉sin𝜉3(cos𝜉−cos(𝜉sin𝜃))cos𝜃sin𝜃=lim_{𝜃→𝜋/2}−2𝜉²sin𝜉3sin(𝜉sin𝜃)cos𝜃cos(2𝜃)=0. (2.27) We next calculate 𝑉₅. We know from [12, pages 442 and 972] that for 𝑧≫1, ₀^𝜋/2()𝜋cos𝑧cos𝜃𝑑𝜃=2𝐽₀(𝑧)=𝜋2(1+𝑜(1))𝑧^−1/21cos𝑧−4𝜋, (2.28) where 𝐽₀(𝑧) is Bessel function. Integration by parts in (2.25), applying the l'Hopital's rule, putting 𝜃=𝜋/2−𝜂 and taking (2.28) into account, we obtain 𝑉₅=lim_{𝜃→𝜋/2}1(((cos𝜃cos𝜉−cos𝜉sin𝜃))sin𝜉sin𝜃)^𝜃₀−𝜉₀^𝜋/2sin²(𝜉sin𝜃)+cos𝜉cos(𝜉sin𝜃)−cos²(𝜉sin𝜃)𝑑𝜃=𝜉₀^𝜋/2cos(2𝜉sin𝜃)𝑑𝜃−𝜉cos𝜉₀^𝜋/2cos(𝜉sin𝜃)𝑑𝜃=𝜉₀^𝜋/2cos2𝜉cos𝜂𝑑𝜂−𝜉cos𝜉₀^𝜋/2=cos𝜉cos𝜂𝑑𝜂𝜋2𝜉^1/2((111+𝑜))√21cos2𝜉−4𝜋1−cos𝜉cos𝜉−4𝜋. (2.29) Clearly, 𝑉₆=𝑂(1). (2.30) By (1.4), (2.20), (2.23), (2.27), (2.29), and (2.30), we obtain (2.18). Thus the proof is complete.

Proof of Theorem 1.1. By (2.8), Lemmas 2.1 and 2.3, 𝜆=𝜋²+4𝜋𝑉+4𝑉²=𝜋²+4𝜋𝑉₁+𝑂𝜉^−5/2=𝜋²+𝑂𝜉^−3/2. (2.31) By this and Lemma 2.1, 𝑉₁=𝜋21𝜉^3/2𝜋²+𝑂(𝜉^−3/2)⁻¹×3cos𝜉−4𝜋−38(1+𝑜(1))𝜉⁻¹3sin𝜉−4𝜋+𝑂𝜉⁻²=2^−1/2𝜋^−3/2𝜉^−3/23cos𝜉−4𝜋−38𝜉⁻¹3sin𝜉−4𝜋+𝑜𝜉^−5/2. (2.32)
By this, (2.31) and Lemmas 2.1 and 2.3, 𝜆=𝜋²+4𝜋𝑉₁+𝑉₂+𝑂𝑉²=𝜋²+4𝜋2^−1/2𝜋^−3/2𝜉^−3/23cos𝜉−4𝜋−38𝜉⁻¹3sin𝜉−4𝜋−2^−1/2𝜋^−7/2𝜉^−5/21√21cos2𝜉−4𝜋1−cos𝜉cos𝜉−4𝜋+𝑜𝜉^−5/2. (2.33)
By this, we obtain (1.10). Thus, the proof is complete.

3. Proof of Theorem 1.2

We write 𝜆=𝜆(𝜉) for simplicity. We prove (1.11) by showing the calculation to get 𝑎₂. The argument to obtain 𝑎_𝑛(𝑛≥3) is the same as that to obtain 𝑎₂. The argument in this section is a variant used in [11, Section 2]. By (2.8) and (2.10), we have 12=1√𝜆¹⁰1√1−𝑠²+𝐵𝑑𝑠. (3.1) Since 0<𝜉≪1, by Taylor expansion, for 0≤𝑠≤1, we obtain cos𝜉−cos𝜉𝑠=^∞_𝑘=1(−1)^𝑘()!𝜉2𝑘^2𝑘1−𝑠^2𝑘. (3.2) By this and (3.1), √𝜆=2¹₀1√1−𝑠²21+𝜆𝜉²11−𝑠^2∞_𝑘=1(−1)^𝑘()!𝜉2𝑘^2𝑘1−𝑠^2𝑘−1/2𝑑𝑠. (3.3) By using this, direct calculation gives us Theorem 1.2. For completeness, we calculate (1.11) up to the third term.

Step 1. We have 21+𝜆𝜉²11−𝑠^2∞_𝑘=1(−1)^𝑘()!𝜉2𝑘^2𝑘1−𝑠^2𝑘1=1−𝜆+112𝜆1−𝑠⁴1−𝑠²𝜉²+𝑜𝜉². (3.4) By (3.3), (3.4), and Taylor expansion, √𝜆=2¹₀1√1−𝑠²11−𝜆+112𝜆1+𝑠²𝜉²+𝑜𝜉^2−1/2=2√𝑑𝑠𝜆√𝜆−1¹⁰1√1−𝑠²11−24(𝜆−1)1+𝑠²𝜉²+𝑜𝜉²𝑑𝑠. (3.5) By this, (1.5) and direct calculation, we obtain √𝜆−1=𝜋−1𝜉16𝜋²+𝑜𝜉². (3.6) This implies 𝜆=𝜋²1+1−8𝜉²+𝑜𝜉². (3.7)

Step 2. Now we calculate the third term of 𝜆(𝜉). First, we note that ¹₀1+𝑠²+𝑠⁴√1−𝑠²𝑑𝑠=1516𝜋,¹₀(1+𝑠²)²√1−𝑠²𝑑𝑠=1916𝜋. (3.8) By this, (1.5), (3.3), (3.7), Taylor expansion, and the same calculation as that to obtain (3.5), √𝜆−1=2¹₀1√1−𝑠²11−2112(𝜆−1)1+𝑠²𝜉²−1360(𝜆−1)1+𝑠²+𝑠⁴𝜉⁴+381144(𝜆−1)²(1+𝑠²)²𝜉⁴+𝑜𝜉⁴1𝑑𝑠=𝜋−𝜉16𝜋²11+8𝜋²𝜉²+𝑜𝜉²+1360𝜋15𝜉16⁴+1192𝜋³19𝜉16⁴+𝑜𝜉⁴1=𝜋−𝜉16𝜋²+15384𝜋1−8𝜋²𝜉⁴+𝑜𝜉⁴. (3.9) By this, we obtain (1.11) up to the third term. Thus, the proof is complete.

4. Proof of Theorem 1.3

In this section, we assume that 𝛼≫1. We write 𝜆=𝜆(𝛼) for simplicity. We consider the solution pair (𝜆(𝛼),𝑢_𝛼)∈ℝ₊×𝑀_𝛼. We obtain from the same argument as that in [10, Theorem 1.2] that 𝑢_𝛼(𝑡)𝛼⟶√2sin𝜋𝑡 (4.1) uniformly on [0,1] as 𝛼→∞. By this, we have 𝑢_𝛼∞=√2𝛼(1+𝑜(1)). (4.2) Furthermore, by [13, Lemma 2.4], we see that 𝛽(𝛼) is continuous for 𝛼>0. By multiplying 𝑢_𝛼 by (1.1) and integration by parts, we obtain 𝜆(𝛼)𝛼²=𝑢^′_𝛼²₂+¹₀𝑢_𝛼(𝑡)sin𝑢_𝛼(𝑡)𝑑𝑡. (4.3) By this and (1.16), for 𝛼≫1, 𝜆(𝛼)𝛼²=2𝛽(𝛼)+¹₀𝑢_𝛼(𝑡)sin𝑢_𝛼(𝑡)𝑑𝑡−2¹₀1−cos𝑢_𝛼(𝑡)𝑑𝑡. (4.4) This along with (4.1) implies that 𝜆(𝛼) is continuous for 𝛼≫1.

Lemma 4.1. For 𝛼≫1, 𝑢_𝛼²_∞=21−√𝜆𝜋4+𝑈⁻¹𝛼², (4.5) where 𝑈=−¹₀√1−𝑠²𝐵√1−𝑠²+𝐵√1−𝑠²+𝐵+√1−𝑠²𝑑𝑠. (4.6)

Proof. By (2.7), (2.10), and putting 𝜃=𝑢_𝛼 and 𝑠=𝜃/‖𝑢_𝛼‖_∞, 𝑢_𝛼²_∞−𝛼²=2₀^1/2𝑢_𝛼²_∞−𝑢_𝛼(𝑡)²𝑢^′_𝛼(𝑡)𝜆𝑢_𝛼²_∞−𝑢_𝛼(𝑡)²+2cos𝑢_𝛼∞−cos𝑢_𝛼(𝑡)𝑑𝑡=2^{‖𝑢𝛼‖∞}₀𝑢_𝛼²_∞−𝜃²𝜆𝑢_𝛼²_∞−𝜃²+2cos𝑢_𝛼∞𝑢−cos𝜃𝑑𝜃=2^𝛼2∞√𝜆¹⁰1−𝑠²√1−𝑠²𝑢+𝐵𝑑𝑠=2^𝛼2∞√𝜆¹⁰√1−𝑠²𝑑𝑠+¹₀1−𝑠²√1−𝑠²−√+𝐵1−𝑠²𝑢𝑑𝑠=2^𝛼2∞√𝜆𝜋4+𝑈. (4.7) Now, the result follows easily from (4.7). Thus, the proof is complete.

Lemma 4.2. For 𝛼≫1, 𝑢_𝛼∞=√2𝛼−2^−3/4𝜋^−5/2𝛼^−1/2cos√32𝛼−4𝜋+𝑜𝛼^−1/2. (4.8)

Proof. By (2.10) and (4.6), |𝑈|≤𝐶𝜆𝑢_𝛼²_∞¹⁰cos𝑢_𝛼∞−cos𝑢_𝛼∞𝑠√1−𝑠²𝑑𝑠≤𝐶𝑢_𝛼∞⁻². (4.9) By this, (2.8), Lemma 2.1, and Taylor expansion, 21−√𝜆𝜋4+𝑈=1−2(𝜋+2𝑉)⁻¹𝜋4=1+𝑈2−2𝜋𝑉𝑈−2((1=11+𝑜))2+1𝜋𝑉(1+𝑜(1)). (4.10) By this, (4.5), (2.12), (2.13), (2.18), Taylor expansion, and (4.2), 𝑢_𝛼∞=12+1𝜋𝑉(1+𝑜(1))^−1/2𝛼=√211−𝜋𝑉((1𝛼=√1+𝑜))2𝛼−2^−3/4𝜋^−5/2𝛼^−1/2cos√32𝛼−4𝜋+𝑜𝛼^−1/2. (4.11) Thus, the proof is complete.

Proof of Theorem 1.3. By Lemma 4.2, we put 𝑢_𝛼∞=√2𝛼+𝐴𝛼^−1/2+𝑜𝛼^−1/2,𝐴=−2^−3/4𝜋^−5/2cos√32𝛼−4𝜋. (4.12) Then substitute (4.12) for (1.10) and use Taylor expansion to obtain 𝜆=𝜋²+22𝜋√2𝛼+𝐴𝛼^−1/2+𝑜𝛼^−1/2−3/2cos√2𝛼+𝐴𝛼^−1/2+𝑜𝛼^−1/2−34𝜋+22𝜋√2𝛼^−5/2−38sin√2𝛼−34𝜋−1√2𝜋²cos2√12𝛼−4𝜋+1𝜋²cos√2𝛼cos√2𝛼−14𝜋+𝑜𝛼^−5/2=𝜋²+2^3/4𝜋^−1/2𝛼^−3/211+√2𝐴𝛼^−3/2+𝑜𝛼^−3/2−3/2×cos√32𝛼−4𝜋cos𝐴𝛼^−1/2((1√1+𝑜))−sin32𝛼−4𝜋sin𝐴𝛼^−1/2(1+𝑜(1))+22𝜋√2𝛼^−5/2−38sin√2𝛼−34𝜋−1√2𝜋²cos2√12𝛼−4𝜋+1𝜋²cos√2𝛼cos√2𝛼−14𝜋+𝑜𝛼^−5/2=𝜋²+2^3/4𝜋^−1/2𝛼^−3/231−2√2𝐴𝛼^−3/2+𝑜𝛼^−3/2×11−2𝐴²𝛼⁻¹(1+𝑜(1))cos√32𝛼−4𝜋−𝐴𝛼^−1/2(1+𝑜(1))sin√32𝛼−4𝜋+2^1/4𝜋^−1/2𝛼^−5/2−38sin√2𝛼−34𝜋−1√2𝜋²cos2√12𝛼−4𝜋+1𝜋²cos√2𝛼cos√2𝛼−14𝜋+𝑜𝛼^−5/2=𝜋²+2^3/4𝜋^−1/2𝛼^−3/2cos√32𝛼−4𝜋−𝜋⁻³𝛼⁻²sin√32𝛼−4𝜋cos√32𝛼−4𝜋+2^1/4𝜋^−1/2𝛼^−5/2−38sin√2𝛼−34𝜋−1√2𝜋²cos2√12𝛼−4𝜋+1𝜋²cos√2𝛼cos√2𝛼−14𝜋−14𝜋⁻⁵cos³√32𝛼−4𝜋+𝑜𝛼^−5/2. (4.13) Thus, we obtain (1.17) and the proof is complete.

Acknowledgment

The author thanks the referee for the helpful suggestions that improved the paper.