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

Authors: Tetsutaro Shibata
Journal: Abstract and Applied Analysis
 (2012)

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## 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 (uξ,λ(ξ))∈C2(I¯)×ℝ+ with ∥uξ∥∞=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=∥uλ∥2) to show the difference between λ(ξ) 	and λ(α).


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## Body

## 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 forλ(ξ) as ξ→∞.Theorem 1.1.
Asξ→∞,
(1.10)λ(ξ)=π2+22πξ-3/2cos(ξ-34π)+22πξ-5/2{-38sin(ξ-34π)-12π2cos(2ξ-14π)+22πξ-5/2+1π2cosξcos(ξ-14π)}+o(ξ-5/2).The local behavior ofλ(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 forλ(ξ) as ξ→0.Theorem 1.2.
Let an arbitrary integerN>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 ofλ(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)MinimizeK(v):=12∥v′∥22+∫I‍(1-cosv(t))dtundertheconstraintv∈Mα.
Let
(1.16)β(α)≔minv∈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.
Asα→∞(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-5cos3(2α-34π)}+o(α-5/2).By Theorems1.1 and 1.3, we clearly understand the difference between λ(ξ) and λ(α).The remainder of this paper is organized as follows. In Section2, 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 Theorem1.1

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 (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)=max0≤t≤1uξ(t)=ξ,(2.3)uξ′(t)>0,0≤t<12.
By (1.1), for t∈I-,
(2.4)[uξ′′(t)+λuξ(t)-sinuξ(t)]uξ′(t)=0.
This implies that for t∈I-,
(2.5)ddt[12uξ′(t)2+12λuξ(t)2+cosuξ(t)]=0.
By this, (2.2) and putting t=1/2, we obtain
(2.6)12uξ′(t)2+12λuξ(t)2+cosuξ(t)≡constant=12λξ2+cosξ.
By this and (2.3), for 0≤t≤1/2,
(2.7)uξ′(t)=λ(ξ2-uξ(t)2)+2(cosξ-cosuξ(t)).
Then by putting s=uξ(t)/ξ, we obtain
(2.8)12=∫01/2‍dt=∫01/2‍uξ′(t)λ(ξ2-uξ(t)2)+2(cosξ-cosuξ(t))dt=1λ∫01‍11-s2+2(cosξ-cosξs)/(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(cosξ-cosξs).
We put
(2.11)V1=-1λξ2∫01‍cosξ-cosξs(1-s2)3/2ds,(2.12)V2=V-V1.Lemma 2.1.
Forξ≫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 puttings=sinθ in (2.11), integration by parts and l'Hopital's rule,
(2.14)-V1=1λξ2∫0π/2‍1cos2θ(cosξ-cos(ξsinθ))dθ=1λξ2∫0π/2‍(tanθ)′(cosξ-cos(ξsinθ))dθ=1λξ2limt→π/2[tant(cosξ-cos(ξsint))]0t-1λξ∫0π/2‍tanθcosθsin(ξsinθ)dθ=-1λξ∫0π/2‍sinθsin(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.
Forξ≫1,
(2.18)V2=-2-1/2π-7/2(1+o(1))ξ-5/2{12cos(2ξ-14π)-cosξcos(ξ-14π)}+o(ξ-5/2).Proof.
Forξ≫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∫01‍cosξ-cosξs1-s2×(11-s2+B(1-s2+B+1-s2)-11-s2(21-s2))ds=-(1+o(1))2λξ2∫01‍cosξ-cosξs1-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∫01‍cosξ-cosξs1-s2·1-s2-B-1-s21-s2+B4(1-s2)2ds=-(1+o(1))12λξ2∫01‍cosξ-cosξs1-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∫01‍cosξ-cosξs1-s2·(1-s2)B(1-s2)3ds=32(1+o(1))1λ2ξ4∫01‍(cosξ-cosξs)2(1-s2)5/2ds=32(1+o(1))1λ2ξ4∫0π/2‍(cosξ-cos(ξsinθ))2cos4θdθ=32(1+o(1))1λ2ξ4V3,
where
(2.21)V3:=∫0π/2‍(cosξ-cos(ξsinθ))2cos4θdθ.
We know
(2.22)∫‍1cos4θdθ=13sinθ(1cos3θ+2cosθ).
Taking (2.22) into account and integration by parts in V3, we obtain that
(2.23)V3=limθ→π/2[13sinθ(1cos3θ+2cosθ)(cosξ-cos(ξsinθ))2]0θ-23ξ∫0π/2‍sinθ(1cos2θ+2)(cosξ-cos(ξsinθ))sin(ξsinθ)dθ:=13V4-23ξ(V5+V6),
where
(2.24)V4:=limθ→π/2sinθ(1cos3θ+2cosθ)(cosξ-cos(ξsinθ))2(2.25)V5:=∫0π/2‍sinθcos2θ(cosξ-cos(ξsinθ))sin(ξsinθ)dθ,(2.26)V6:=2∫0π/2‍sinθ(cosξ-cos(ξsinθ))sin(ξsinθ)dθ.
Then by l'Hopital's rule,
(2.27)V4=limθ→π/2sinθ(1cos3θ+2cosθ)(cosξ-cos(ξsinθ))2=limθ→π/2(1+2cos2θ)(cosθ-cos(ξsinθ))2cos3θ=limθ→π/2(cosξ-cos(ξsinθ))2cos3θ=limθ→π/2-2ξ(cosξ-cos(ξsinθ))sin(ξsinθ)3cosθsinθ=limθ→π/2-2ξsinξ3(cosξ-cos(ξsinθ))cosθsinθ=limθ→π/2-2ξ2sinξ3sin(ξsinθ)cosθcos(2θ)=0.
We next calculate V5. We know from [12, pages 442 and 972] that for z≫1,
(2.28)∫0π/2‍cos(zcosθ)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=limθ→π/2[1cosθ(cosξ-cos(ξsinθ))sin(ξsinθ)]0θ-ξ∫0π/2‍(sin2(ξsinθ)+cosξcos(ξsinθ)-cos2(ξsinθ))dθ=ξ∫0π/2‍cos(2ξsinθ)dθ-ξcosξ∫0π/2‍cos(ξsinθ)dθ=ξ∫0π/2‍cos(2ξcosη)dη-ξcosξ∫0π/2‍cos(ξcosη)dη=π2ξ1/2(1+o(1))(12cos(2ξ-14π)-cosξcos(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 Theorem1.1.
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)-cosξcos(ξ-14π))}+o(ξ-5/2).
By this, we obtain (1.10). Thus, the proof is complete.


## 3. Proof of Theorem1.2

We writeλ=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)cosξ-cosξs=∑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 ofλ(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 Theorem1.3

In this section, we assume thatα≫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-cosuα(t))dt.
This along with (4.1) implies that λ(α) is continuous for α≫1.Lemma 4.1.
Forα≫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∥∞-cosuα(t))dt=2∫0∥uα∥∞‍∥uα∥∞2-θ2λ(∥uα∥∞2-θ2)+2(cos∥uα∥∞-cosθ)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.
Forα≫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 Theorem1.3.
By Lemma4.2, 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-5cos3(2α-34π)}+o(α-5/2).
Thus, we obtain (1.17) and the proof is complete.



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