We consider the simple pendulum equation −u00(t) +f(u0(t)) =λsinu(t), t∈I:= (−1,1), u(t)>0, t∈I, u(±1

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

PRECISE ASYMPTOTIC BEHAVIOR OF SOLUTIONS TO DAMPED SIMPLE PENDULUM EQUATIONS

TETSUTARO SHIBATA

Abstract. We consider the simple pendulum equation

−u⁰⁰(t) +f(u⁰(t)) =λsinu(t), t∈I:= (−1,1), u(t)>0, t∈I, u(±1) = 0,

where 0 < ≤ 1, λ > 0, and the friction term is either f(y) = ±|y| or f(y) = −y. Note that when f(y) = −y and = 1, we have well known original damped simple pendulum equation. To understand the dependance of solutions, to the damped simple pendulum equation with λ 1, upon the termf(u⁰(t)), we present asymptotic formulas for the maximum norm of the solutions. Also we present an asymptotic formula for the time at which maximum occurs, for the casef(u) =−u.

1. Introduction We consider the damped simple pendulum equation

−u⁰⁰(t) +f(u⁰(t)) =λsinu(t), t∈I:= (−1,1), (1.1)

u(t)>0, t∈I, (1.2)

u(±1) = 0, (1.3)

where 0< ≤1,λ >0, and the damping term is eitherf(y) =±|y|orf(y) =−y.

It is known that there exists a solutionu,λto (1.1)–(1.3) for 0< ≤1 andλ1, withku,λk_∞< π; see for example [1].

The purpose of this paper is to study the asymptotic behavior of u,λ(t) as λ → ∞; this is useful for understanding the effect of the damping term on the asymptotic behavior ofu,λ. First, we recall some properties of the solution u0,λ

for the simple pendulum equation without friction (i.e. the case where= 0):

−u⁰⁰(t) =λsinu(t), t∈I, (1.4)

u(t)>0, t∈I, (1.5)

u(±1) = 0. (1.6)

2000Mathematics Subject Classification. 34B15.

Key words and phrases. Damped simple pendulum; asymptotic formula.

c

2009 Texas State University - San Marcos.

Submitted April 21, 2009. Published November 7, 2009.

1

It is well known thatu0,λ→πlocally uniformly in I asλ→ ∞. Furthermore (cf.

Lemma 2.1 in Section 2), asλ→ ∞, ku_0,λk_∞=π−8e⁻

√λ−32√ λe⁻³

√λ+o(√ λe⁻³

√λ). (1.7)

It should be mentioned that the asymptotic behavior of solutions to the original and perturbed simple pendulum problems have been studied in [5, 6, 7]. We also refer the reader to [3] for the basic properties of the solution to simple pendulum problems. As far as the author knows, there are only a few works concerning the precise properties of solutions to (1.1)–(1.3). In particular, an asymptotic formulas such as (1.7) for ku,λk_∞ has not been obtained yet. Therefore, it seems worth considering the precise asymptotic behavior of ku,λk_∞ as λ → ∞, for having a better understanding of the effect of the friction term.

Now we state our main results. We denote byu1,,λ,u2,,λandu3,,λthe solutions of (1.1)–(1.3) withf(y) =−|y|,f(y) =|y|andf(y) =−y, respectively.

Theorem 1.1. Let f(y) =−|y|and let0< ≤1 be fixed. Then, asλ→ ∞, ku1,,λk∞=π−8e⁻e⁻

√λ

+O(λ^−1/2e⁻

√λ

). (1.8)

Since u1,,λ is a super-solution of (1.4)–(1.6), (1.8) is well understood and rea- sonable from a viewpoint of (1.7). Moreover, the formula (1.8) gives us the clear relationship betweenku0,λk∞ andku1,,λk∞.

The following result can be proved by the same arguments as those used in the proof of Theorem 1.1.

Theorem 1.2. Let f(y) =|y| and0< ≤1 be fixed. Then, as λ→ ∞, ku2,,λk_∞=π−8ee⁻

√

λ+O(λ^−1/4e⁻

√

λ). (1.9)

We also note thatu2,,λ is a sub-solution of (1.4)–(1.6), (1.9) is also reasonable result.

Now we consider the case f(y) =−y. Let 0< ≤1 be fixed. Let t,λ ∈I be the unique point satisfying u3,,λ(t,λ) =ku3,,λk_∞. Then we know from [2] that t,λ<0 forλ1.

Theorem 1.3. Let f(y) =−y. Then, asλ→ ∞, t,λ=−

√λ+O λ^−3/4

, (1.10)

ku3,,λk_∞=π−8e⁻

√

λ+O(λ^−1/4e⁻

√

λ). (1.11)

By (1.10), we obtain a precise asymptotic formula fort,λas λ→ ∞. Moreover, since the second term of (1.11) is the same as that of (1.7), the friction term does not have any effect on the second term ofku3,,λk_∞.

The rest of this paper is organized as follows. In Section 2, we prove Theorem 1.1 based on the crucial tool Lemma 2.2, which will be proved in Section 3. We prove Theorem 1.2 in Section 4 by almost the same argument as that to prove Theorem 1.1. We apply the modified argument for the proof of Theorem 1.1 to the proof of Theorem 1.3 in Section 5.

2. Proof of Theorem 1.1

In the following two sections, we letf(y) = −|y|. We fix 0< ≤ 1. Further, we assume that λ1 and we write u_,λ =u_1,,λ for simplicity. We consider the solutionu_,λ(t) withku,λk∞< π. We know

u,λ(t) =u,λ(−t), t∈I, (2.1)

u⁰_,λ(t)>0, t∈[−1,0), (2.2)

u,λ(0) =ku,λk_∞, (2.3)

u_,λ(t)→π asλ→ ∞, (t∈I). (2.4)

Note that (2.1)-(2.3) follow from [2]. (2.4) is a direct consequence of (1.7), (2.3) and (2.6) below.

By (1.1) and (2.2), for−1≤t≤0, we have

{u⁰⁰_,λ(t) +u⁰_,λ(t) +λsinu_,λ(t)}u⁰_,λ(t) = 0.

By this equality and (2.3), for−1≤t≤0, we have 1

2u⁰_,λ(t)²+ Z t

−1

|u⁰_,λ(s)|²ds−λcosu_,λ(t)

= Z 0

−1

|u⁰_,λ(s)|²ds−λcosku,λk∞= constant.

(2.5)

For−1≤t≤0, we obtain 1

2u⁰_,λ(t)²=λ(cosu_,λ(t)−cosku,λk∞) + Z 0

t

|u⁰_,λ(s)|²ds. (2.6) For−1≤t≤0, we put

A(θ) :=Aλ(θ) =λ(cosθ−cosku,λk_∞), (2.7) B(t) :=Bλ(t) =

Z 0

t

|u⁰_,λ(s)|²ds. (2.8) Then by (2.2) and (2.6)–(2.8), for−1≤t≤0,

u⁰_,λ(t) = q

2(A(u_,λ(t)) +B(t)). (2.9)

Then 1 =

Z 0

−1

dt= 1

√2 Z 0

−1

u⁰_,λ(t)

pA(u,λ(t)) +B(t)dt= 1

√2(I+II), (2.10) where

I= Z 0

−1

u⁰_,λ(t)

pA(u,λ(t))dt, (2.11)

II = Z 0

−1

u⁰_,λ(t)

pA(u_,λ(t)) +B(t)dt− Z 0

−1

u⁰_,λ(t) pA(u_,λ(t))dt

= Z 0

−1

−B(t)u⁰_,λ(t) pA(u,λ(t))p

A(u,λ(t)) +B(t)

× 1

pA(u_,λ(t)) +p

A(u_,λ(t)) +B(t)dt.

(2.12)

Lemma 2.1. Let dλ:=π− ku,λk∞. Then, as λ→ ∞, I=

r2 λ

log 4

sin(dλ/2) +1

4(1 +o(1)) log 4 sin(dλ/2)

sin²d_λ 2

. (2.13)

Proof. Putθ=u,λ(t). Then I= 1

√ λ

Z ku,λk∞

0

1

pcosθ−cosku,λk∞

dθ

=

√

√ 2

λsin(ku,λk_∞/2)

Z ku,λk∞/2

0

1 q

1−sin²ϕ/sin²(ku,λk_∞/2) dϕ

= r2

λ Z π/2

0

1 q

1−sin²(ku,λk∞/2) sin²φ dφ

= r2

λK(k),

(2.14)

whereK is the complete elliptic integral of the first kind and k= sin(ku,λk∞/2).

Then by [4], we have

K(k) = log 4 k⁰ +1

4 log 4 k⁰

k⁰²(1 +o(1)), (2.15) wherek⁰=√

1−k²= cos(ku_,λk_∞/2) = cos((π−d_λ)/2) = sin(d_λ/2). By this and (2.14), we obtain (2.13). Thus the proof is complete.

SinceII <0, by (2.10), (2.15) and Lemma 2.1, we have 1< 1

√2I≤ 1

√

λ 1 +Csin²d_λ 2

log 4

sin(dλ/2). (2.16) Then

sindλ

2 ≤4(1 +o(1))e⁻

√ λ, dλ

2 ≤4(1 +o(1))e⁻

√

λ, sinku,λk_∞≤8(1 +o(1))e⁻

√ λ. (2.17) Lemma 2.2. As λ→ ∞,

II =

√ 2

λ log sindλ

2

+O(λ⁻¹). (2.18)

The proof of the above lemma will be given in Section 3. We accept Lemma 2.2 tentatively to prove Theorem 1.1.

Proof of Theorem 1.1. By Lemmas 2.1 and 2.2 and (2.17), we have 1 = 1

√

2(I+II) = 1

√ λ

log 4

sin(dλ/2) +1

4(1 +o(1)) sin²dλ

2 log 4 sin(dλ/2)

+

λlog sindλ

2 +O(λ⁻¹)

= 1

√λ log 4−log sindλ

2 +

λlog sindλ

2 +O(λ⁻¹).

(2.19)

This implies

1−

√λ

log sindλ

2 = log 4−√

λ+O(λ^−1/2). (2.20) By this,

log sind_λ

2 = 1 +

√

λ+O(λ⁻¹)

log 4−√

λ+O(λ^−1/2)

=−√

λ+ log 4−+O(λ^−1/2).

(2.21) By this and Taylor expansion,

sindλ

2 = dλ

2 1−d²_λ

3 +o(d²_λ)

= 4e⁻e⁻

√

λ(1 +O(λ^−1/2)).

By this and (2.17), we obtain Theorem 1.1.

3. Proof of Lemma 2.2

In this section, we focus our attention on the proof of Lemma 2.2. Let 0< δ1 be fixed. We definet_δ:=t_λ,δ<0 byu_,λ(t_δ) =ku_,λk_∞−δ. We set

II=II1+II2

:=

Z 0

t_δ

−B(t)u⁰_,λ(t) pA(u_,λ(t))p

A(u_,λ(t)) +B(t)(p

A(u_,λ(t)) +p

A(u_,λ(t)) +B(t))dt +

Z t_δ

−1

−B(t)u⁰_,λ(t) pA(u,λ(t))p

A(u,λ(t)) +B(t) p

A(u,λ(t)) +p

A(u,λ(t)) +B(t)dt.

(3.1) To obtain Lemma 2.2, we estimateII₁ andII₂ by series of lemmas.

Lemma 3.1. For−1≤t≤0, B(t)≤q

2A(u_,λ(t))(ku_,λk_∞−u_,λ(t)) + 2(ku_,λk_∞−u_,λ(t))². (3.2) Proof. Sinceku,λk∞< π, we see from (1.1) thatu⁰⁰_,λ(t)≤0 fort∈I. This along with (2.8) and (2.9) implies that for−1≤t≤0,

0< B(t)

≤ max

t≤s≤0|u⁰_,λ(s)|

Z 0

t

u⁰_,λ(s)ds

=u⁰_,λ(t)(ku,λk∞−u,λ(t))

= q

2A(u,λ(t)) + 2B(t)(ku,λk∞−u,λ(t)).

(3.3)

By (3.3),

B(t)²−2B(t)(ku_,λk_∞−u_,λ(t))²−2A(u_,λ(t))(ku_,λk_∞−u_,λ(t))²≤0.

Since√

a+b≤√ a+√

bfora, b≥0, by this, we obtain B(t)≤(ku,λk_∞−u,λ(t))²

+ q

²(ku,λk_∞−u,λ(t))⁴+ 2A(u,λ(t))(ku,λk_∞−u,λ(t))²

≤2(ku,λk∞−u_,λ(t))²+q

2A(u_,λ(t))(ku,λk∞−u_,λ(t)).

(3.4)

The proof is complete.

By Taylor expansion, fortδ ≤t≤0 and 0< κ1,

cosu_,λ(t)−cosku_,λk_∞≤sinku_,λk_∞(ku_,λk_∞−u_,λ(t)) +1

2(ku,λk_∞−u,λ(t))², (3.5) cosu,λ(t)−cosku,λk∞≥sinku,λk∞(ku,λk∞−u,λ(t))

+1

2(1−κ)(ku,λk∞−u_,λ(t))². (3.6) Lemma 3.2. Forλ1,

|II1| ≤ −

√ 2

λ log sin dλ

2

+O(λ⁻¹). (3.7)

Proof. By (3.1) and Lemma 3.1,

|II₁| ≤ Z 0

t_δ

B(t)u⁰_,λ(t)

2A(u,λ(t))^3/2dt=X₁+X₂

:=

√2λ Z 0

tδ

ku,λk_∞−u,λ(t)

cosu,λ(t)−cosku,λk_∞u⁰_,λ(t)dt +²

Z 0

t_δ

(ku_,λk_∞−u_,λ(t))²

(λ(cosu,λ(t)−cosku,λk_∞)^3/2u⁰_,λ(t)dt

=

√2λ

Z ku,λk∞

ku,λk∞−δ

ku,λk_∞−θ cosθ−cosku,λk∞

dθ

+ ² λ^3/2

Z ku,λk∞

ku_,λk∞−δ

(ku,λk_∞−θ)² (cosθ−cosku,λk∞)^3/2dθ.

(3.8)

We first calculateX1. We put X₁=Q₁+Q₂

:=

√2λ

Z ku,λk∞

ku,λk∞−δ

ku,λk_∞−θ cosθ−cosπdθ

+

√2λ

Z ku,λk∞

ku_,λk∞−δ

ku,λk_∞−θ cosθ−cosku,λk∞

−ku,λk_∞−θ cosθ−cosπ

dθ.

(3.9)

We see that

Q₁=Q₁₁+Q₁₂

:=

√2λ

Z ku,λk∞

ku,λk∞−δ

ku,λk_∞−π

cosθ+ 1 dθ+

√2λ

Z ku,λk∞

ku,λk∞−δ

π−θ

cosθ+ 1dθ. (3.10) Then by (2.17),

Q11=−dλ

√2λ

Z ku,λk∞

ku,λk∞−δ

1 cosθ+ 1dθ

=−dλ

√2λ tanθ

2

ku,λk∞

ku,λk∞−δ

=−d_λ

√2λ

cos(d_λ/2)

sin(dλ/2) −sin(π−d_λ−δ)/2) cos(π−dλ−δ)/2)

=O(λ⁻¹).

(3.11)

Next,

Q12=

√2λ

Z ku,λk∞

ku,λk∞−δ

π−θ cosθ+ 1dθ

=

√2λ Z d_λ+δ

d_λ

y 1−cosydy

=

√ 2λ

Z d_λ+δ

d_λ

y

−coty 2

0

dy

=

√2λ

−(d_λ+δ) cotd_λ+δ

2 +d_λcotd_λ 2 + 2 log sin dλ+δ

2

−2 log sin dλ

2

=−

√2

λ log sind_λ

2 +O(λ⁻¹).

(3.12)

Now, we calculateQ₂. Q2=

√2λ(1 + cosku,λk_∞)

Z ku,λk∞

ku,λk∞−δ

1

cosθ+ 1· ku,λk_∞−θ cosθ−cosku_,λk_∞dθ

≤

√2λ(1−cosdλ) 1 sinku_,λk_∞

Z ku,λk∞

ku,λk∞−δ

1 cosθ+ 1dθ

≤Cd_λλ⁻¹ tanθ

2

ku,λk∞

ku,λk∞−δ

=Cλ⁻¹dλ

cos(dλ/2)

sin(d_λ/2) −tanπ−dλ−δ 2

=O(λ⁻¹).

(3.13)

By (3.9)–(3.13), we obtain X₁≤ −

√2

λ log sind_λ

2 +O(λ⁻¹). (3.14)

Finally, we calculateX2. By (2.17) and (3.6), X2=²λ^−3/2

×

Z ku,λk∞

ku,λk∞−δ

(ku,λk∞−θ)²

(sinku_,λk_∞+ (1/2)(1−κ)(ku_,λk_∞−θ))^3/2(ku_,λk_∞−θ)^3/2dθ

≤C²λ^−3/2

Z ku,λk∞

ku,λk∞−δ

(ku,λk∞−θ)^1/2

(sinku_,λk_∞+ (ku_,λk_∞−θ))^3/2dθ

=C²λ^−3/2 Z δ

0

y^1/2

(sinku,λk_∞+y)^3/2dy

≤C²λ^−3/2 Z δ

0

1

sinku_,λk_∞+ydy

≤C²λ^−3/2|log sinku,λk∞|=O(λ⁻¹).

By this and (3.14), we obtain (3.7). Thus the proof is complete.

We estimateII₁ from below. To do this, we need the following lemma.

Lemma 3.3. Forλ1 andtδ < t <0, B(t)≥

√ λ 2

q

−cosu,λ(t)(ku,λk_∞−u,λ(t))²

−1 2

√

λ sinku_,λk_∞

p−cosu_,λ(t)(ku_,λk_∞−u_,λ(t)).

(3.15)

Proof. We recall that for constantsa, b >0, Z p

ax²+bxdx

= 2ax+b 4a

pax²+bx− b² 8a

√1 alog

2ax+b+ 2p

a(ax²+bx) .

(3.16)

By Taylor expansion, forku,λk_∞−δ≤u,λ(t)≤θ≤ ku,λk_∞, we have cosθ−cosku,λk_∞≥sinku,λk_∞(ku,λk_∞−θ)−1

2cosu,λ(t)(ku,λk_∞−θ)². By this, (2.7)–(2.9) and (3.16), fortδ ≤t≤0,

B(t) = Z 0

t

q

2A(u,λ(t)) + 2B(t)u⁰_,λ(t)dt

≥√ 2λ

Z ku,λk∞

u,λ(t)

q

cosθ−cosku,λk_∞dθ

≥√ λ

Z ku,λk∞−u,λ(t)

0

q

−x²cosu_,λ(t) + 2xsinku_,λk_∞dx

=√ λ1

2z− sinku,λk_∞ 2 cosu_,λ(t)

q−z²cosu,λ(t) + 2zsinku,λk_∞

−

√λ 2

sin²ku,λk∞

(−cosu,λ(t))^3/2

log(Rλ(u,λ(t)) + 2 sinku,λk_∞)−log(2 sinku,λk_∞) ,

(3.17)

where

Rλ(u,λ(t)) =−2zcosu,λ(t)+2 q

z²cos²u,λ(t)−2zcosu,λ(t) sinku,λk_∞ (3.18) andz:=ku,λk_∞−u,λ(t). We know that log(1 +x)≤xforx≥0. By this,

log(R_λ(u_,λ(t)) + 2 sinku,λk∞)−log(2 sinku,λk∞)

= log

1 + −2zcosu_,λ(t) + 2p

z²cos²u_,λ(t)−2zcosu_,λ(t) sinku_,λk_∞ 2 sinku,λk_∞

≤−zcosu,λ(t) +p

−cosu,λ(t)p

−z²cosu,λ(t) + 2zsinku,λk_∞

sinku,λk_∞ .

By this and (3.17), we obtain (3.15). Thus the proof is complete.

Lemma 3.4. Forλ1,

|II₁| ≥ −

√2

λ log sindλ

2 +O(λ⁻¹). (3.19)

Proof. By (3.1), we have

|II1| ≥ Z 0

t_δ

B(t)u⁰_,λ(t) 2p

A(u_,λ(t)(A(u_,λ(t)) +B(t))dt:=II_1,1−II_1,2

= Z 0

t_δ

B(t)u⁰_,λ(t) 2A(u_,λ(t))^3/2dt +Z 0

t_δ

B(t)u⁰_,λ(t) 2p

A(u_,λ(t))(A(u_,λ(t)) +B(t))dt− Z 0

t_δ

B(t)u⁰_,λ(t) 2A(u,λ(t))^3/2dt

. (3.20) By Lemma 3.3, we put

II1,1=II1,2,1−II1,2,2

= 2

√ λ

Z 0

t_δ

p−cosu_,λ(t)(ku,λk∞−u_λ(t))²u⁰_,λ(t) 2λ^3/2(cosu,λ(t)−cosku,λk_∞)^3/2 dt

−

√ λ

2 sinku,λk_∞ Z 0

tδ

(ku,λk∞−u,λ(t))u⁰_,λ(t) 2p

−cosu,λ(t)λ^3/2(cosu,λ(t)−cosku,λk_∞)^3/2dt.

(3.21) By Taylor expansion, fort_δ ≤t≤0 and 0< η1,

q

−cosku,λk∞−q

−cosu,λ(t)≤ 1 +η

2 sinu,λ(t)(ku,λk∞−u,λ(t)), (3.22) q

−cosku,λk∞= (cosdλ)^1/2= 1−1

4(1 +o(1))d²_λ. (3.23) By (3.5), (3.21) and (3.22),

II1,2,1≥ 4λ

Z ku_,λk∞

ku,λk∞−δ

√−cosθ(ku,λk∞−θ)^1/2

(sinku,λk_∞+ (1/2)(ku,λk_∞−θ))^3/2dθ

=

4λ

Z ku,λk∞

ku,λk∞−δ

p−cosku,λk_∞(ku,λk_∞−θ)^1/2 (sinku_,λk_∞+ (1/2)(ku_,λk_∞−θ))^3/2dθ

− 4λ

Z ku,λk∞

ku,λk∞−δ

(p

−cosku_,λk_∞−√

−cosθ)(ku_,λk_∞−θ)^1/2 (sinku,λk∞+ (1/2)(ku,λk∞−θ))^3/2 dθ

≥ 4λ

q

−cosku,λk∞

Z δ

0

y^1/2

(sinku,λk_∞+ (1/2)y)^3/2dy

−1 +η

8λ sin(ku,λk_∞−δ) Z δ

0

y^3/2

(sinku,λk∞+ (1/2)y)^3/2dy.

(3.24) Then

4λ

Z δ

0

y^1/2

(sinku,λk∞+ (1/2)y)^3/2dy=

√2λ Z δ

0

y^1/2

(2 sinku,λk∞+y)^3/2dy

=

√2 λ

Z

√

δ/(2 sinku_,λk∞) 0

z² (1 +z²)^3/2dz

=−

√2

λ log sinku,λk∞+O(λ⁻¹).

(3.25)

Further, by (2.17), 1 +η

8λ sin(ku,λk∞−δ) Z δ

0

y^3/2

(sinku,λk_∞+ (1/2)y)^3/2dy≤Cλ⁻¹. (3.26) By (3.24)–(3.26),

II1,2,1≥ −

√2

λ log sinku,λk∞+O(λ⁻¹)

=−

√2

λ log sind_λ+O(λ⁻¹)

=−

√2 λ

log 2 + sind_λ

2 + cosd_λ 2

+O(λ⁻¹)

=−

√2

λ log sind_λ

2 +O(λ⁻¹).

(3.27)

Next, by (3.6), II1,2,2≤Cλ⁻¹

Z ku_,λk∞

ku,λk∞−δ

sinku,λk∞

(sinku,λk_∞+ (ku,λk_∞−θ))^3/2(ku,λk_∞−θ)^1/2dθ

≤Cλ⁻¹ Z

√δ/(sinku,λk∞)

0

1

(1 +z²)^3/2dz≤Cλ⁻¹.

(3.28) Finally, by Lemma 3.1 and (3.20), forzλ(u,λ(t)) :=ku,λk_∞−u,λ(t),

|II1,2|=² Z 0

t_δ

B(t)²u⁰_,λ(t) A(u_,λ(t))^5/2dt

≤C² Z 0

t_δ

A(u,λ(t))zλ(u,λ(t))²+²zλ(u,λ(t))⁴

A(u,λ(t))^5/2 u⁰_,λ(t)dt

=C²

Z ku,λk∞

ku,λk∞−δ

zλ(θ)²

A(θ)^3/2dθ+C⁴

Z ku,λk∞

ku,λk∞−δ

zλ(θ)⁴ A(θ)^5/2dθ :=Y1+Y2.

(3.29)

Then by (2.17), (3.6) and (3.25), Y1=C²λ^−3/2

Z ku,λk∞

ku,λk∞−δ

(ku,λk_∞−u,λ(t))² (cosθ−cos|u,λk∞)^3/2dθ

≤C²λ^−3/2

Z ku,λk∞

ku_,λk∞−δ

(ku,λk_∞−u,λ(t))^1/2

(sinku,λk∞+ (1/2)(1−κ)(ku,λk∞−u,λ(t)))^3/2dθ

≤C²λ^−3/2 Z δ

0

y^1/2

(sinku_,λk_∞+y)^3/2dy

≤²λ^−3/2(C+ log sinku,λk∞) =O(λ⁻¹).

(3.30) By the same argument as that just above, by (2.17) and (3.6), we obtain

Y2=C⁴λ^−5/2

Z ku,λk∞

ku,λk∞−δ

(ku,λk∞−u,λ(t))⁴ (cosθ−cos|u_,λk_∞)^5/2dθ

≤C⁴λ^−5/2Z δ 0

1

sinku,λk_∞+ydy+C

≤C⁴λ^−5/2(|log sinku,λk_∞|+C) =O(λ^−3/2).

Thus the proof is complete.

Lemma 3.5. Forλ1, we have |II2| ≤Cλ⁻¹. Proof. By (3.1) and Lemma 3.1,

|II2|

= Z t_δ

−1

B(t)u⁰_,λ(t) pA(u,λ(t)) +B(t)p

A(u,λ(t))(p

A(u,λ(t)) +B(t) +p

A(u,λ(t)))dt

≤C Z t_δ

−1

B(t)u⁰_,λ(t) 2A(u_,λ(t))^3/2dt

≤C

Z ku,λk∞−δ

0

2(ku_,λk_∞−θ)²+p

2λ(cosθ−cosku_,λk_∞)(ku_,λk_∞−θ) (λ(cosθ−cosku,λk∞))^3/2 dθ

≤C(λ^−3/2+λ⁻¹).

The proof is complete.

Now Lemma 2.2 follows from Lemmas 3.2–3.5. The proof is complete.

4. Proof of Theorem 1.2

Letf(y) =|y|in this section. We fix 0< ≤1. Further, we assume that λ1 and we write u,λ = u2,,λ for simplicity. We consider the solution u,λ(t) with ku,λk∞< π. We know that the properties (2.1)–(2.4) are also valid. By the same argument as that to obtain (2.5), we have

1

2u⁰_,λ(t)²=λ(cosu,λ(t)−cosku,λk_∞)− Z 0

t

|u⁰_,λ(s)|²ds. (4.1) Then by (2.6)–(2.8), for−1≤t≤0,

u⁰_,λ(t) = q

2(A(u_,λ(t))−B(t)). (4.2)

By this, 1 =

Z 0

−1

dt= 1

√2 Z 0

−1

u⁰_,λ(t)

pA(u,λ(t))−B(t)dt= 1

√2(I+III), (4.3) where

I= Z 0

−1

u⁰_,λ(t)

pA(u,λ(t))dt, (4.4)

III= Z 0

−1

u⁰_,λ(t)

pA(u_,λ(t))−B(t)dt− Z 0

−1

u⁰_,λ(t) pA(u_,λ(t))dt

= Z 0

−1

B(t)u⁰_,λ(t) pA(u,λ(t))p

A(u,λ(t))−B(t)

× 1

pA(u_,λ(t)) +p

A(u_,λ(t))−B(t)dt.

(4.5)

Let 0< δ1 be fixed. Further, let−1< tδ <0 satisfyu,λ(tδ) =ku,λk∞. We put

III=III1+III2

:=

Z 0

tδ

B(t)u⁰_,λ(t) pA(u,λ(t))p

A(u,λ(t))−B(t)(p

A(u,λ(t)) +p

A(u,λ(t))−B(t))dt +

Z tδ

−1

B(t)u⁰_,λ(t) pA(u_,λ(t))p

A(u_,λ(t))−B(t) p

A(u_,λ(t)) +p

A(u_,λ(t))−B(t)dt.

(4.6) Lemma 4.1. Forλ1 and−1< t <0,

B(t)≤q

2A(u,λ(t))(ku,λk_∞−u,λ(t)). (4.7) Proof. By (2.2), (2.8) and (4.2),

B(t) = Z 0

t

u⁰_,λ(s)²ds

≤ Z 0

t

q

2A(u,λ(s))u⁰_,λ(s)ds

≤q

2A(u_,λ(t)) Z 0

t

u⁰_,λ(s)ds

= q

2A(u,λ(t))(ku,λk_∞−u,λ(t)).

Thus the proof if complete.

By (3.6) and Lemma 4.1, we obtain the estimate ofIII1 from above as follows.

Indeed, fortδ ≤t≤0,

A(u,λ(t))−B(t)≥A(u,λ(t))− q

2A(u,λ(t)(ku,λk∞−u,λ(t))

=A(u,λ(t)) 1−

√2(ku,λk∞−u,λ(t)) pA(u,λ(t))

≥A(u,λ(t))

1− 2(ku,λk_∞−u,λ(t)) p(λ(1−κ)/2)(ku_,λk_∞−u_,λ(t))

≥A(u,λ(t)) 1− C

√λ

.

(4.8)

By this and (4.6), III1≤

Z 0

tδ

B(t)u⁰_,λ(t) 2p

A(u,λ(t))(A(u,λ(t))−B(t))

≤ Z 0

tδ

B(t)u⁰_,λ(t) 2p

A(u_,λ(t))A(u_,λ(t))(1−C/√ λ)dt

≤ 1 + C

√ λ

Z 0

t_δ

B(t)u⁰_,λ(t) 2p

A(u,λ(t))A(u,λ(t))dt.

By this, Lemma 4.1 and the same argument as that used in Lemma 3.2, forλ1, we obtain

III₁≤ −√ 2

λ log sind_λ

2 +O(λ⁻¹). (4.9)

Furthermore, by (4.3) and (4.8),

1≤ 1

√2 Z 0

−1

u⁰_,λ(t) pA(u,λ(t))(1−C/√

λ)dt

≤ 1

√ 2

1 + C

√ λ

Z 0

−1

u⁰_,λ(t) pA(u,λ(t))dt.

By this and (2.14)–(2.16), we obtain

sindλ

2 ≤Ce⁻

√

λ, sinku,λk∞≤Ce⁻

√

λ. (4.10)

Lemma 4.2. Forλ1 andtδ < t <0,

B(t)≥√ 2

Z ku,λk∞

u,λ(t)

q

λ(cosθ−cosku,λk_∞)dθ

−2√

A(u_,λ(t))^1/4(ku,λk∞−u_,λ(t))^3/2.

(4.11)

Proof. By (2.8), (4.2) and (4.7),

B(t) = Z 0

t

u⁰_,λ(s)²ds

≥ Z 0

t

q

2A(u,λ(s))u⁰_,λ(s)ds− Z 0

t

p2B(s)u⁰_,λ(s)ds

≥ Z 0

t

q

2A(u,λ(s))u⁰_,λ(s)ds−p 2B(t)

Z 0

t

u⁰_,λ(s)ds

≥ Z 0

t

q

2A(u,λ(s))u⁰_,λ(s)ds−p

2B(t)(ku,λk_∞−u,λ(t))

≥ Z 0

t

q

2A(u,λ(s))u⁰_,λ(s)ds−2√

A(u,λ(t))^1/4(ku,λk_∞−u,λ(t))^3/2 :=W1−W2.

(4.12)

The proof is complete.

Proof of Theorem 1.2. We calculate the estimate of III1 from below. By (3.6), (4.6), (4.8), (4.10) and (4.12)

Z 0

tδ

W2u⁰_,λ(t) 2p

A(u,λ(t))(A(u,λ(t))−B(t))dt

≤C Z 0

tδ

A(u,λ(t))^1/4(ku,λk_∞−u,λ(t))^3/2 pA(u_,λ(t))(A(u_,λ(t))(1−C/√

λ)u⁰_,λ(t)dt

≤C 1 + C

√ λ

λ^−5/4

Z ku,λk∞

ku,λk∞−δ

(ku,λk∞−θ)^3/2 (cosθ−cosku_,λk_∞)^5/4dθ

≤C 1 + C

√λ

λ^−5/4 Z δ

0

y^1/4

(sinku,λk∞+y)^5/4dy

≤C 1 + C

√ λ

λ^−5/4

Z δ/sinku_,λk∞

0

z^1/4 (1 +z)^5/4dz

≤C 1 + C

√λ

λ^−5/4|log sinku,λk_∞|

≤Cλ^−3/4.

(4.13)

By (4.8), (4.12) and Lemmas 3.3 and 3.4, forλ1,

Z 0

tδ

W1u⁰_,λ(t) 2p

A(u,λ(t))(A(u,λ(t))−B(t))dt≤ −√ 2

λ log sindλ

2 +O(λ⁻¹). (4.14) Further, by (4.6), (4.8) and Lemma 3.5, forλ1, we obtain

III2≤Cλ⁻¹. (4.15)

By (4.6), (4.9) and (4.13)–(4.15), we obtain III =−√

2

λ log sindλ

2 +O(λ^−3/4). (4.16)

By this, (2.14) and the same argument as (2.18)–(2.20), we obtain (1.9). The proof

is complete.

5. Proof of Theorem 1.3

In this section, letf(y) =−y. We writetλ=t,λ<0 for simplicity. The proof of the Theorem 1.3 is almost the same as those of Theorems 1.1 and 1.2. We begin with the fundamental properties ofu_,λ.

Lemma 5.1. (i) u⁰_,λ(t)>0 for−1≤t < t_λ andu⁰_,λ(t)>0 fort_λ< t <1.

(ii) u_,λ(t)→π locally uniformly inI asλ→ ∞.

(iii) t_λ<0 andt_λ→0 asλ→ ∞.

Since the proof of Lemma 5.1 is quite easy, we omit it. To prove Theorem 1.3, we repeat the same arguments as those in Sections 3 and 4. We see that

1 +tλ= Z t_λ

−1

dt= 1

√ 2

Z t_λ

−1

u⁰_,λ(t)

pA(u,λ(t) +B(t)dt= 1

√

2(P+Q), (5.1) where

P = Z t_λ

−1

u⁰_,λ(t)

pA(u,λ(t)dt, (5.2)

Q= Z t_λ

−1

u⁰_,λ(t)

pA(u,λ(t) +B(t)dt− Z t_λ

−1

u⁰_,λ(t) pA(u,λ(t)dt

= Z tλ

−1

−B(t)u⁰_,λ(t) pA(u_,λ(t)p

A(u_,λ(t) +B(t)(p

A(u_,λ(t) +p

A(u_,λ(t) +B(t))dt.

(5.3) Then it is clear thatP=I in (2.11). Furthermore, by using the same argument as that in Section 3, it is easy to show thatQ=II+O(λ⁻¹). We also find that

1−tλ= Z 1

t_λ

dt= 1

√ 2

Z 1

t_λ

−u⁰_,λ(t)

pA(u,λ(t)−B(t)dt= 1

√

2(R+S), (5.4) whereR=IandS =III+O(λ^−3/4). By (5.1) and (5.4), we obtain

2tλ=√

2Q+O(λ^−3/4) =√

2II+O(λ^−3/4) = 2

λ log sindλ

2

+O(λ^−3/4), (5.5) which implies

tλ=

λlog sindλ

2

+O(λ^−3/4). (5.6)

By this and (5.1), we obtain

1 = 1

√2I+O(λ^−3/4). (5.7)

By this and Lemma 2.1, we obtain (1.11). By (1.11) and (5.6), we obtain (1.10).

The proof is complete.

References

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[2] H. Berestycki and L. Nirenberg;Monotonicity, symmetry and antisymmetry of solutions of semilinear elliptic equations, J. Geom. Phys.5(1988), 237–275.

[3] D. G. Figueiredo;On the uniqueness of positive solutions of the Dirichlet problem−4u= λsinu, Pitman Res. Notes in Math.122(1985), 80–83.

[4] I. S. Gradshteyn and I. M. Ryzhik; Table of integrals, series and products (fifth edition), Academic Press, San Diego, 1994.

[5] T. Shibata, Precise spectral asymptotics for the Dirichlet problem −u⁰⁰(t) +g(u(t)) = λsinu(t), J. Math. Anal. Appl.267(2002), 576–598.

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Tetsutaro Shibata

Department of Applied Mathematics, Graduate School of Engineering, Hiroshima Uni- versity, Higashi-Hiroshima, 739-8527, Japan

E-mail address:[email protected]