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− 1(ρ1−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− 1(ρ1−γ − (g(y))1−γ) and so
g0(y)
pρ1−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
dη
pρ1−γ − η1−γ := G(µ).
We can write
G(µ) =
rγ2− 1
2 · 1
pρ1−γ Z µ
ρ
dη 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
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 dσ
− 2θ2
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+2dσ
− 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−1dσ + (γ − 1)(γ − 3)
4(γ − 2) θ(1 − θ2)γ−2γ−1 + γ − 1
2 θ(1 − θ2)γ−1−1 . (3.5)
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−1dσ +(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−1dσ (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].
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
3γ2− 6γ − 1
γ2− 2γ − 3 < θ < 1.
Since r γ − 1 γ + 1 <
s
3γ2− 6γ − 1
γ2− 2γ − 3 , we have h(θ) < 0, if s
3γ2− 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.
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−1dσ
+ 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−1dσ (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.
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−1dσ
− 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.