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(3.1) Integrating (3.1) from 0 to l, we obtain l = rγ2− 1 2 Z µ ρ dη pρ1−γ − η1−γ

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(1)

3 The case 1 < γ < 3

First, consider γ > 1. Multiplying (1.1) by g0(y) and integrating it from 0 to l, we obtain µ2

2 = 1

γ2− 11−γ− µ1−γ).

Hence

ρ = ρ(µ) = γ2− 1

2 µ2+ µ1−γ

1−γ1

and lim

µ→∞ρ(µ) = 0.

Multiplying (1.1) again by g0(y) and integrating it from y < l to l, we obtain (g0(y))2

2 = 1

γ2− 11−γ − (g(y))1−γ) and so

g0(y)

1−γ− (g(y))1−γ = −

r 2

γ2− 1. (3.1)

Integrating (3.1) from 0 to l, we obtain l =

rγ2− 1 2

Z µ ρ

1−γ − η1−γ := G(µ).

We can write

G(µ) =

rγ2− 1

2 · 1

1−γ Z µ

ρ

r

1 − (η ρ)1−γ

.

Let σ =q

1 − (ηρ)1−γ. Then

G(µ) = s

2(γ + 1) γ − 1 ργ+12

Z

1−(µρ)1−γ 0

(1 − σ2)1−γγ dσ.

Let θ =q

1 − (µρ)1−γ. Then

θ2 = 1 − µ1−γ γ2− 1

2 µ2+ µ1−γ

=

γ2− 1 2 µγ+1 γ2− 1

2 µγ+1 + 1

→ 1 as µ → ∞.

From

1 − θ2 = µ1−γ γ2− 1

2 µ2 + µ1−γ

= 1

γ2− 1

2 µγ+1+ 1

(2)

it follows that µ =

 2

γ2− 1

1+γ1  θ2 1 − θ2

1+γ1

and ρ =

 2

γ2− 1

1+γ1  θ2 1 − θ2

1−γ22

θγ−12 , θ ∈ (0, 1).

Thus, we get G(µ) = γ−12 H(θ), where

H(θ) := θ(1 − θ2)γ−11 Z θ

0

(1 − σ2)1−γγ dσ.

Note that lim

θ→1H(θ) = γ − 1

2 and so lim

µ→∞G(µ) = 1. Furthermore, H0(θ) = θ(1 − θ2)−1+ (1 − θ2)γ−11

Z θ 0

(1 − σ2)γ−1−γ

2

γ − 1(1 − θ2)−γ+2γ−1 Z θ

0

(1 − σ2)γ−1−γ dσ. (3.2) Form (3.2) we obtain

H0(θ) = θ(1 − θ2)−1+ (1 − θ2)−γ+2γ−1



1 − γ + 1 γ − 1θ2

 Z θ 0

(1 − σ2)γ−1−γ dσ = (1 − θ2)−1h(θ) (3.3) where

h(θ) := θ + (1 − θ2)γ−11



1 − γ + 1 γ − 1θ2

 Z θ 0

(1 − σ2)γ−1−γ dσ. (3.4) Using the following formula twice

Z θ 0

(1 − σ2)λdσ = 2λ + 3 2λ + 2

Z θ 0

(1 − σ2)λ+1dσ − 1

2λ + 2θ(1 − θ2)λ+1, λ 6= −1, we get that for λ 6= −1, −2

Z θ 0

(1 − σ2)λdσ = (2λ + 3)(2λ + 5) (2λ + 2)(2λ + 4)

Z θ 0

(1 − σ2)λ+2

2λ + 3

(2λ + 4)(2λ + 2)θ(1 − θ2)λ+2 1

2λ + 2θ(1 − θ2)λ+1. We obtain for γ 6= 2

Z θ 0

(1 − σ2)γ−1−γ dσ = (γ − 3)(3γ − 5)

−4(γ − 2) Z θ

0

(1 − σ2)γ−2γ−1 + (γ − 1)(γ − 3)

4(γ − 2) θ(1 − θ2)γ−2γ−1 + γ − 1

2 θ(1 − θ2)γ−1−1 . (3.5)

(3)

Hence for γ 6= 2

h(θ) = θ −(γ − 3)(3γ − 5)

4(γ − 2) (1 − θ2)γ−11



1 − γ + 1 γ − 1θ2

 Z θ 0

(1 − σ2)γ−2γ−1 +(γ − 1)(γ − 3)

4(γ − 2) θ(1 − θ2)



1 − γ + 1 γ − 1θ2



+ γ − 1 2 θ



1 − γ + 1 γ − 1θ2



. (3.6) Form (3.3) we get

H0(r γ − 1

γ + 1) =r γ − 1 γ + 1



1 −γ − 1 γ + 1

−1

=r γ − 1

γ + 1 · γ + 1

2 > 0 for γ > 1. (3.7) Next, we restrict to 1 < γ < 3. Since

limθ→1

Z θ 0

(1 − σ2)

−γ γ−1 (1 − θ2)γ−1−1

= lim

θ→1

1 1 γ − 1· 2θ

= γ − 1 2 ,

we get lim

θ→1h(θ) = 0, if 1 < γ < 3.

Lemma 3.1 H0(θ) < 0 for some θ ∈ (0, 1), if 1 < γ < 3.

Proof. Since H0(θ) = (1 − θ2)−1h(θ). H0(θ) < 0 if h(θ) < 0 for some θ ∈ (0, 1). We divide our discussion with respect to γ into five cases.

Case 1. γ = 5/3. Then (3.6) becomes h(θ) = θ + 2

3θ(1 − θ2)(1 − 4θ2) + 1

3θ(1 − 4θ2)

= 2

3θ(4θ4− 7θ2+ 3)

= 2 3θ



4(θ2 7

8)2 1 16



< 0, if we take θ = r7

8. Case 2. 1 < γ < 5/3. Let θ >q

γ−1

γ+1. From (3.6) we get h(θ) < θ + (γ − 1)(γ − 3)

4(γ − 2) θ(1 − θ2)



1 − γ + 1 γ − 1θ2



+ γ − 1 2 θ



1 − γ + 1 γ − 1θ2



= θ

4(γ − 2) h1, where

h1 := 4(γ − 2) + (γ − 3)(1 − θ2)[(γ − 1) − (γ + 1)θ2] + 2(γ − 2)[(γ − 1) − (γ + 1)θ2].

(4)

Then

h1 = 4γ − 8 + (γ − 3)[(γ + 1)θ4− 2γθ2+ (γ − 1)] + 2(γ2 − 3γ + 2) − 2(γ2− γ − 2)θ2

= (γ2− 2γ − 3)θ4+ (−4γ2+ 8γ + 4)θ2+ (3γ2− 6γ − 1)

= (θ2− 1)[(γ2− 2γ − 3)θ2− (3γ2− 6γ − 1)] > 0, if s

2− 6γ − 1

γ2− 2γ − 3 < θ < 1.

Since r γ − 1 γ + 1 <

s

2− 6γ − 1

γ2− 2γ − 3 , we have h(θ) < 0, if s

2− 6γ − 1

γ2− 2γ − 3 < θ < 1.

Case 3. γ = 2. We know that q1

3 < ee44−1+1 < 3 q11

12. Let θ >p1/3. The (3.4) becomes h(θ) = θ + (1 − θ2)(1 − 3θ2)

Z θ 0

(1 − σ2)−2dσ.

Let σ = sin φ. Then Z θ

0

(1 − σ2)−2dσ =

Z sin−1θ 0

sec3φdφ = 1 2

 θ

1 − θ2 +1

2ln1 + θ 1 − θ



. (3.8)

Hence

h(θ) = θ + 1

2(1 − θ2)(1 − 3θ2)

 θ

1 − θ2 + 1

2ln1 + θ 1 − θ



= θ +1

2θ(1 − 3θ2) + 1

4(1 − θ2)(1 − 3θ2) ln1 + θ 1 − θ

< θ +1

2θ(1 − 3θ2) + (1 − θ2)(1 − 3θ2), if r1

3 < e4− 1 e4+ 1 < θ.

Set

h2(θ) := 6θ4− 3θ3− 8θ2+ 3θ + 2.

Then

h(θ) < 1

2h2(θ), if e4− 1 e4+ 1 < θ.

Since

h02(θ) = 24θ3− 9θ2− 16θ + 3

> 24θ3− 9 − 16 + 3

= 2(12θ3− 11)

> 0, if e4− 1 e4+ 1 < 3

r11

12 < θ < 1.

(5)

Therefore, h2(θ) < h2(1) = 0 for θ ∈ (3 r11

12, 1). Hence h(θ) < 0, if θ ∈ ( 3 r11

12, 1).

Case 4. 5/3 < γ < 2. Let θ >q

γ−1

γ+1. From (3.6) we get h(θ) = θ + (γ − 3)(3γ − 5)

−4(γ − 2) (1 − θ2)γ−11



1 − γ + 1 γ − 1θ2

 Z θ 0

(1 − σ2)γ−2γ−1

+ θ

4(γ − 2)[(3γ2 − 10γ + 7) − (4γ2− 8γ − 4)θ2+ (γ2− 2γ − 3)θ4].

Then for γ 6= 2

h0(θ) = 1 + f1(θ) + f2(θ) + f3(θ) + f4(θ) + f5(θ), (3.9) where

f1(θ) := (γ − 3)(3γ − 5)

−4(γ − 2) · 1

γ − 1(1 − θ2)

−γ+2 γ−1 (−2θ)



1 −γ + 1 γ − 1θ2

 Z θ 0

(1 − σ2)γ−2γ−1dσ, f2(θ) := (γ − 3)(3γ − 5)

−4(γ − 2) (1 − θ2)γ−11



2(γ + 1) γ − 1 θ

 Z θ 0

(1 − σ2)γ−2γ−1dσ, f3(θ) := (γ − 3)(3γ − 5)

−4(γ − 2) (1 − θ2)



1 − γ + 1 γ − 1θ2

 , f4(θ) := 1

4(γ − 2)[(3γ2− 10γ + 7) − (4γ2− 8γ − 4)θ2+ (γ2− 2γ − 3)θ4], f5(θ) := θ

4(γ − 2)[−2(4γ2− 8γ − 4)θ + 4(γ2− 2γ − 3)θ3].

By L’Hˆopital’s Rule, we get that

limθ→1

Z θ 0

(1 − σ2)γ−2γ−1 (1 − θ2)γ−2γ−1

= lim

θ→1

(1 − θ2)γ−2γ−1

γ−2

γ−1(1 − θ2)γ−1−1 (−2θ)

= 0

and

lim

θ→1

Z θ 0

(1 − σ2)γ−2γ−1 (1 − θ2)γ−1−1

= lim

θ→1

(1 − θ2)γ−2γ−1

1

γ−1(1 − θ2)γ−1−γ · 2θ

= 0.

Then we get

θ→1limf1 = 0, lim

θ→1f2(θ) = 0, lim

θ→1f3(θ) = 0, lim

θ→1f4(θ) = −1, lim

θ→1f5(θ) = (γ − 1)2 2 − γ . So that

limθ→1h0(θ) = (γ − 1)2 2 − γ > 0.

(6)

It means that h(θ) is increasing near 1. We have obtained lim

θ→1h(θ) = 0. Hence h(θ) < 0 for θ near 1.

Case 5. 2 < γ < 3. Let θ > q

γ−1

γ+1. Follow the same argument as in Case 4. From (3.9), we get

θ→1limf1(θ) = ∞, lim

θ→1f2(θ) = 0, lim

θ→1f3(θ) = 0, lim

θ→1f4(θ) = −1, lim

θ→1f5(θ) = (γ − 1)2 2 − γ .

So that h0(θ) → ∞ as θ → 1. Then h(θ) is increasing near 1. We have obtained lim

θ→1h(θ) = 0.

Hence h(θ) < 0 for θ near 1.

Thus it follows from (3.7) and Lemma 3.1 that H has a critical point θ0 ∈ (0, 1).

Lemma 3.2 Any critical point of H(θ) must be a maximal point.

Proof. Assume H0(θ) = 0 for some θ ∈ (q

γ−1

γ+1, 1). Since H0(θ) = (1 − θ2)−1h(θ), we have h(θ) = 0 and

Z θ 0

(1 − σ2)γ−1−γ dσ = −θ(1 − θ2)γ−1−1



1 − γ + 1 γ − 1θ2

−1

. We compute the derivative of h(θ):

h0(θ) = 1 − 2

γ − 1θ(1 − θ2)

−γ+2 γ−1



1 −γ + 1 γ − 1θ2

 Z θ 0

(1 − σ2)

−γ γ−1

2(γ + 1)

γ − 1 θ(1 − θ2)γ−11 Z θ

0

(1 − σ2)

−γ

γ−1dσ + (1 − θ2)−1



1 − γ + 1 γ − 1θ2



= 1 + 2

γ − 1θ2(1 − θ2)−1+2(γ + 1) γ − 1 θ2



1 − γ + 1 γ − 1θ2

−1

+ (1 − θ2)−1



1 − γ + 1 γ − 1θ2



= 1 + (1 − θ2)−1

 2

γ − 1θ2+ 1 − γ + 1 γ − 1θ2



+2(γ + 1) γ − 1 θ2



1 −γ + 1 γ − 1θ2

−1

= 2

"

1 + γ + 1 γ − 1θ2



1 − γ + 1 γ − 1θ2

−1#

= 2



1 − γ + 1 γ − 1θ2

−1

< 0.

Since H00(θ) = 2θ(1 − θ2)−2h(θ) + (1 − θ2)−1h0(θ), H00(θ) < 0. The lemma follows.

Since H has a critical point θ0 and any critical point of H must be a maximum point, it follows that H is monotone increasing in (0, θ0) and monotone decreasing in (θ0, 1). Therefore, we conclude that for 1 < γ < 3 there exists 1 < l < ∞ such that

(1) (1.1)-(1.2) has a unique solution if 0 < l ≤ 1 or l = l; (2) (1.1)-(1.2) has two solutions if 1 < l < l;

(3) (1.1)-(1.2) has no solution if l > l.

Therefore, the proof of Theorem 1 is completed.

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