1. Introduction
We consider the following nonlinear eigenvalue problem:
(1.1)-u′′(t)+sinu(t)=λu(t), t∈I=:(0,1),(1.2)u(t)>0, t∈I,(1.3)u(0)=u(1)=0,
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 (uξ,λ(ξ))∈C2(I-)×ℝ+ satisfies (1.1)–(1.3) with ∥uξ∥∞=ξ. Moreover, the solution set of of (1.1)–(1.3) is given by Γ:={(uξ,λ(ξ))∈C2(I-)×ℝ+; ξ>0}. Furthermore, it is well known that uξ(t)~ξsinπt for ξ≫1 and 0<ξ≪1. Therefore, we have
(1.4)λ(ξ)⟶π2 (ξ⟶∞),(1.5)λ(ξ)⟶π2+1 (ξ⟶0).
Equations (1.1)–(1.3) are the special case of the following semilinear equation:
(1.6)-u′′(t)+f(u(t))=λu(t), t∈I,(1.7)u(t)>0, t∈I,(1.8)u(0)=u(1)=0.
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)=up (p>1) in (1.6), then as ξ→∞ (cf. [7]),
(1.9)λ(ξ)=ξp-1+O(e-δξ),
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→∞,
(1.10)λ(ξ)=π2+22πξ-3/2cos (ξ-34π)+22πξ-5/2{-38sin(ξ-34π)-12π2cos (2ξ-14π)+22πξ-5/2 +1π2cos ξcos (ξ-14π)}+o(ξ-5/2).

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,
(1.11)λ=π2+1-18ξ2+1192(1+18π2)ξ4+∑n=3Nanξ2n+o(ξ2N),
where {an} (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)=up(p>1) in (1.6). Then we know from [9] that, for a given α>0, there exists a unique solution pair (uα,λ(α))∈C2(I-)×ℝ+ of (1.6)–(1.8) satisfying ∥uα∥2=α. Furthermore, λ(α) is an increasing function of α>0 and as α→∞,
(1.12)λ(α)=αp-1+C0α(p-1)/2+O(1).

We see from (1.9) and (1.12) the difference between the asymptotic formulas for λ(ξ) and λ(α) when f(u)=up 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 (uα,λ(α))∈C2(I-)×ℝ+ satisfying ∥uα∥2=α. Then, it is natural to expect that for t∈I-, as α→∞,
(1.13)uα(t)α⟶2sinπt.
Therefore, we expect that ∥uα∥∞~2 ∥uα∥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
(1.14)Mα:={v∈H01(I):∥v∥2=α},
where ∥v∥2 is the usual L2-norm of v, α>0 is a parameter, and H01(I) is the usual real Sobolev space. Then consider the following minimizing problem, which depends on α>0:
(1.15)Minimize K(v):=12∥v′∥22+∫I(1-cos v(t))dt under the constraint v∈Mα.
Let
(1.16)β(α)≔min v∈MαK(v).
Then by Lagrange multiplier theorem, for a given α>0, there exists a pair (uα,λ(α))∈Mα×ℝ+ which satisfies (1.1)–(1.3) with K(uα)=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)=(uα,λ(α))∈Mα×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→∞(1.17)λ(α)=π2+23/4π-1/2α-3/2cos (2α-34π)-π-3α-2sin(2α-34π)cos (2α-34π)+21/4π-1/2α-5/2{-38sin(2α-34π)-12π2cos (22α-14π) +21/4π-1/2α-5/2+1π2cos (2α)cos (2α-14π)-14π-5cos 3(2α-34π)}+o(α-5/2).

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 <xref ref-type="statement" rid="thm1.1">1.1</xref>
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 (uξ,λ(ξ))∈C2(I-)×ℝ+ satisfies (1.1)–(1.3), then
(2.1)uξ(t)=uξ(1-t), 0≤t≤1,(2.2)uξ(12)=max 0≤t≤1 uξ(t)=ξ,(2.3)uξ′(t)>0, 0≤t<12.
By (1.1), for t∈I-,
(2.4) [uξ′′(t)+λuξ(t)-sin uξ(t)]uξ′(t)=0.
This implies that for t∈I-,
(2.5)ddt[12uξ′(t)2+12λuξ(t)2+cos uξ(t)]=0.
By this, (2.2) and putting t=1/2, we obtain
(2.6)12uξ′(t)2+12λuξ(t)2+cos uξ(t)≡ constant =12λξ2+cos ξ.
By this and (2.3), for 0≤t≤1/2,
(2.7)uξ′(t)=λ(ξ2-uξ(t)2)+2(cos ξ-cos uξ(t)).
Then by putting s=uξ(t)/ξ, we obtain
(2.8)12=∫01/2dt=∫01/2uξ′(t)λ(ξ2-uξ(t)2)+2(cos ξ-cos uξ(t))dt=1λ∫0111-s2+2(cos ξ-cos ξs)/(λξ2)ds=1λ{∫0111-s2ds+(∫0111-s2+Bds-∫0111-s2ds)}=1λ(π2+V),
where
(2.9)V:=-∫01B1-s2+B1-s2(1-s2+B+1-s2)ds,(2.10)B:=2λξ2(cos ξ-cos ξs).
We put
(2.11)V1=-1λξ2∫01cos ξ-cos ξs(1-s2)3/2ds,(2.12)V2=V-V1.

Lemma 2.1.
For ξ≫1(2.13)V1=π21λξ3/2[(1+15128ξ2(1+o(1)))cos (ξ-34π)V1=π21λξ3/2 -38ξ(1+o(1)) sin (ξ-34π)].

Proof.
By putting s=sin θ in (2.11), integration by parts and l'Hopital's rule,
(2.14)-V1=1λξ2∫0π/21cos 2θ(cos ξ-cos (ξ sin θ))dθ=1λξ2∫0π/2(tan θ)′(cos ξ-cos (ξsinθ))dθ=1λξ2lim t→π/2[tant(cos ξ-cos (ξsint))]0t -1λξ∫0π/2tanθcos θsin(ξsinθ)dθ=-1λξ∫0π/2sinθ sin(ξsinθ)dθ.
By [12, page 962],
(2.15)∫0π/2sinθsin(ξsinθ)dθ=π2J1(ξ),
where J1(ξ) is Bessel function of the first kind. For ξ≫1, by [12, page 972], we have
(2.16)J1(ξ)=2πξ[(1+15128ξ2(1+o(1)))cos (ξ-34π) -38ξ(1+o(1))sin(ξ-34π)].
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
(2.17)V1=2-1/2π-3/2ξ-3/2(1+o(1))[(1+15128ξ2(1+o(1)))cos (ξ-34π) V1=2-12π-32ξ-32(1+o(1))-38ξ(1+o(1))sin(ξ-34π)].
After we obtain (2.31) later, then (2.13) will be improved in the form (2.32).

Lemma 2.3.
For ξ≫1,
(2.18)V2=-2-1/2π-7/2(1+o(1))ξ-5/2{12cos (2ξ-14π)-cos ξcos (ξ-14π)} +o(ξ-5/2).

Proof.
For ξ≫1 and 0≤s≤1, by mean value theorem,
(2.19)|B|≤Cξ-1(1-s)≤Cξ-1(1-s2).
By this and Lebesgue's convergence theorem, we have
(2.20)V2=-2λξ2∫01cos ξ-cos ξs1-s2 ×(11-s2+B(1-s2+B+1-s2)-11-s2(21-s2))ds=-(1+o(1))2λξ2∫01cos ξ-cos ξs1-s2 ×2(1-s2)-(1-s2+B+1-s21-s2+B)1-s2+B(1-s2+B+1-s2)1-s221-s2ds=-(1+o(1))2λξ2∫01cos ξ-cos ξs1-s2·1-s2-B-1-s21-s2+B4(1-s2)2ds=-(1+o(1))12λξ2∫01cos ξ-cos ξs1-s2·(1-s2-B)2-(1-s2)(1-s2+B)(1-s2)2[(1-s2-B)+1-s21-s2+B]ds=34(1+o(1))1λξ2∫01cos ξ-cos ξs1-s2·(1-s2)B(1-s2)3ds=32(1+o(1))1λ2ξ4∫01(cos ξ-cos ξs)2(1-s2)5/2ds=32(1+o(1))1λ2ξ4∫0π/2(cos ξ-cos (ξ sin θ))2cos 4 θdθ=32(1+o(1))1λ2ξ4V3,
where
(2.21)V3:=∫0π/2(cos ξ-cos (ξ sin θ))2cos 4 θdθ.
We know
(2.22)∫1cos 4 θdθ=13sin θ(1cos 3 θ+2cos θ).
Taking (2.22) into account and integration by parts in V3, we obtain that
(2.23)V3=lim θ→π/2[13sinθ(1cos 3 θ+2cos θ)(cos ξ-cos (ξsinθ))2]0θ -23ξ∫0π/2sinθ(1cos 2 θ+2)(cos ξ-cos (ξsinθ))sin(ξsinθ)dθ:=13V4-23ξ(V5+V6),
where
(2.24)V4:=lim θ→π/2sinθ(1cos 3θ+2cos θ)(cos ξ-cos (ξsinθ))2(2.25)V5:=∫0π/2sinθcos 2θ(cos ξ-cos (ξsinθ))sin(ξsinθ)dθ,(2.26)V6:=2∫0π/2sinθ(cos ξ-cos (ξsinθ))sin(ξsinθ)dθ.
Then by l'Hopital's rule,
(2.27)V4=lim θ→π/2sinθ(1cos 3θ+2cos θ)(cos ξ-cos (ξsinθ))2=lim θ→π/2(1+2 cos 2θ)(cos θ-cos (ξsinθ))2cos 3θ=lim θ→π/2(cos ξ-cos (ξsinθ))2cos 3θ=lim θ→π/2-2 ξ(cos ξ-cos (ξsinθ))sin (ξsinθ)3cos θsinθ=lim θ→π/2-2 ξsinξ3(cos ξ-cos (ξsinθ))cos θsinθ=lim θ→π/2-2 ξ2sinξ3sin (ξsinθ)cos θcos (2θ)=0.
We next calculate V5. We know from [12, pages 442 and 972] that for z≫1,
(2.28)∫0π/2cos (zcos θ)dθ=π2J0(z)=π2(1+o(1))z-1/2cos (z-14π),
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
(2.29)V5=lim θ→π/2[1cos θ(cos ξ-cos (ξsinθ))sin(ξsinθ)]0θ -ξ∫0π/2(sin2(ξsinθ)+cos ξcos (ξsinθ)-cos 2(ξsinθ))dθ=ξ∫0π/2cos (2ξsinθ)dθ-ξcos ξ∫0π/2cos (ξsinθ)dθ=ξ∫0π/2cos (2ξcos η)dη-ξcos ξ∫0π/2cos (ξcos η)dη=π2ξ1/2(1+o(1))(12cos (2ξ-14π)-cos ξcos (ξ-14π)).
Clearly,
(2.30)V6=O(1).
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 <xref ref-type="statement" rid="thm1.1">1.1</xref>.
By (2.8), Lemmas 2.1 and 2.3,
(2.31)λ=π2+4πV+4V2=π2+4πV1+O(ξ-5/2)=π2+O(ξ-3/2).
By this and Lemma 2.1,
(2.32)V1=π21ξ3/2(π2+O(ξ-3/2))-1 ×(cos (ξ-34π)-38(1+o(1))ξ-1sin(ξ-34π)+O(ξ-2))=2-1/2π-3/2ξ-3/2(cos (ξ-34π)-38ξ-1sin(ξ-34π))+o(ξ-5/2).

By this, (2.31) and Lemmas 2.1 and 2.3,
(2.33)λ=π2+4π(V1+V2)+O(V2)=π2+4π{2-1/2π-3/2ξ-3/2(cos (ξ-34π)-38ξ-1sin (ξ-34π)) =π2+4π-2-1/2π-7/2ξ-5/2(12cos (2ξ-14π)-cos ξcos (ξ-14π))} +o(ξ-5/2).

By this, we obtain (1.10). Thus, the proof is complete.

3. Proof of Theorem <xref ref-type="statement" rid="thm1.2">1.2</xref>
We write λ=λ(ξ) for simplicity. We prove (1.11) by showing the calculation to get a2. The argument to obtain an (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
(3.1)12=1λ∫0111-s2+Bds.
Since 0<ξ≪1, by Taylor expansion, for 0≤s≤1, we obtain
(3.2)cos ξ-cos ξs=∑k=1∞(-1)k(2k)!ξ2k(1-s2k).
By this and (3.1),
(3.3)λ=2∫0111-s2(1+2λξ211-s2∑k=1∞(-1)k(2k)!ξ2k(1-s2k))-1/2ds.
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
(3.4)1+2λξ211-s2∑k=1∞(-1)k(2k)!ξ2k(1-s2k)=1-1λ+112λ1-s41-s2ξ2+o(ξ2).
By (3.3), (3.4), and Taylor expansion,
(3.5)λ=2∫0111-s2(1-1λ+112λ(1+s2)ξ2+o(ξ2))-1/2ds=2λλ-1∫0111-s2(1-124(λ-1)(1+s2)ξ2+o(ξ2))ds.
By this, (1.5) and direct calculation, we obtain
(3.6)λ-1=π-116πξ2+o(ξ2).
This implies
(3.7)λ=π2+1-18ξ2+o(ξ2).

Step 2.
Now we calculate the third term of λ(ξ). First, we note that
(3.8)∫011+s2+s41-s2ds=1516π, ∫01(1+s2)21-s2ds=1916π.
By this, (1.5), (3.3), (3.7), Taylor expansion, and the same calculation as that to obtain (3.5),
(3.9)λ-1=2∫0111-s2{381-12(112(λ-1)(1+s2)ξ2 =2∫0111-s2-1360(λ-1)(1+s2+s4)ξ4)+381144(λ-1)2(1+s2)2ξ4+o(ξ4)}ds=π-116πξ2(1+18π2ξ2+o(ξ2))+1360π1516ξ4+1192π31916ξ4+o(ξ4)=π-116πξ2+1384π(1-58π2)ξ4+o(ξ4).
By this, we obtain (1.11) up to the third term. Thus, the proof is complete.

4. Proof of Theorem <xref ref-type="statement" rid="thm1.3">1.3</xref>
In this section, we assume that α≫1. We write λ=λ(α) for simplicity. We consider the solution pair (λ(α),uα)∈ℝ+×Mα. We obtain from the same argument as that in [10, Theorem 1.2] that
(4.1)uα(t)α⟶2sinπt
uniformly on [0,1] as α→∞. By this, we have
(4.2)∥uα∥∞=2α(1+o(1)).
Furthermore, by [13, Lemma 2.4], we see that β(α) is continuous for α>0. By multiplying uα by (1.1) and integration by parts, we obtain
(4.3)λ(α)α2=∥uα′∥22+∫01uα(t)sinuα(t)dt.
By this and (1.16), for α≫1,
(4.4)λ(α)α2=2β(α)+∫01uα(t)sinuα(t)dt-2∫01(1-cos uα(t))dt.
This along with (4.1) implies that λ(α) is continuous for α≫1.

Lemma 4.1.
For α≫1,
(4.5)∥uα∥∞2=(1-2λ(π4+U))-1α2,
where
(4.6)U=-∫011-s2B1-s2+B(1-s2+B+1-s2)ds.

Proof.
By (2.7), (2.10), and putting θ=uα and s=θ/∥uα∥∞,
(4.7)∥uα∥∞2-α2=2∫01/2(∥uα∥∞2-uα(t)2)uα′(t)λ(∥uα∥∞2-uα(t)2)+2(cos ∥uα∥∞-cos uα(t))dt=2∫0∥uα∥∞∥uα∥∞2-θ2λ(∥uα∥∞2-θ2)+2(cos ∥uα∥∞-cos θ)dθ=2∥uα∥∞2λ∫011-s21-s2+Bds=2∥uα∥∞2λ[∫011-s2ds+∫01(1-s21-s2+B-1-s2)ds]=2∥uα∥∞2λ(π4+U).
Now, the result follows easily from (4.7). Thus, the proof is complete.

Lemma 4.2.
For α≫1,
(4.8)∥uα∥∞=2α-2-3/4π-5/2α-1/2cos (2α-34π)+o(α-1/2).

Proof.
By (2.10) and (4.6),
(4.9)|U|≤Cλ∥uα∥∞2|∫01cos ∥uα∥∞-cos (∥uα∥∞s)1-s2ds|≤C(∥uα∥∞-2).
By this, (2.8), Lemma 2.1, and Taylor expansion,
(4.10)1-2λ(π4+U)=1-2(π+2V)-1(π4+U)=12-2π(U-V2(1+o(1)))=12+1πV(1+o(1)).
By this, (4.5), (2.12), (2.13), (2.18), Taylor expansion, and (4.2),
(4.11)∥uα∥∞=(12+1πV(1+o(1)))-1/2α=2(1-1πV(1+o(1)))α=2α-2-3/4π-5/2α-1/2cos (2α-34π)+o(α-1/2).
Thus, the proof is complete.

Proof of Theorem <xref ref-type="statement" rid="thm1.3">1.3</xref>.
By Lemma 4.2, we put
(4.12)∥uα∥∞=2α+Aα-1/2+o(α-1/2),A=-2-3/4π-5/2cos (2α-34π).
Then substitute (4.12) for (1.10) and use Taylor expansion to obtain
(4.13)λ=π2+22π(2α+Aα-1/2+o(α-1/2))-3/2cos (2α+Aα-1/2+o(α-1/2)-34π) +22π(2α)-5/2{-38sin(2α-34π)-12π2cos (22α-14π) +22π(2α)-5/2 +1π2cos (2α)cos (2α-14π)}+o(α-5/2)=π2+23/4π-1/2α-3/2(1+12Aα-3/2+o(α-3/2))-3/2 ×{cos (2α-34π)cos (Aα-1/2(1+o(1))) -sin(2α-34π)sin(Aα-1/2(1+o(1)))} +22π(2α)-5/2{-38sin(2α-34π)-12π2cos (22α-14π) +22π(2α)-5/2 +1π2cos (2α)cos (2α-14π)}+o(α-5/2)=π2+23/4π-1/2α-3/2(1-322Aα-3/2+o(α-3/2)) ×{(1-12A2α-1(1+o(1)))cos (2α-34π) -Aα-1/2(1+o(1))sin(2α-34π)} +21/4π-1/2α-5/2{-38sin(2α-34π)-12π2cos (22α-14π) +21/4π-1/2α-5/2 +1π2cos (2α)cos (2α-14π)}+o(α-5/2)=π2+23/4π-1/2α-3/2cos (2α-34π) -π-3α-2sin(2α-34π)cos (2α-34π) +21/4π-1/2α-5/2{-38sin(2α-34π)-12π2cos (22α-14π)+21/4π-1/2α-5/2 +1π2cos (2α)cos (2α-14π)-14π-5cos 3(2α-34π)} +o(α-5/2).
Thus, we obtain (1.17) and the proof is complete.