Multiplying (2.1) by g 0 (y) and integrating it from 0 to l, we obtain µ 2 = ln µ − ln ρ = ln µ

2 The case γ = 1

For γ = 1, (1.1) becomes

g ⁰⁰ (y) = 1

2 g ⁻¹ (y). (2.1)

Multiplying (2.1) by g ⁰ (y) and integrating it from 0 to l, we obtain µ ² = ln µ − ln ρ = ln µ

ρ . Hence

ρ = ρ(µ) = µ

exp µ ² and lim

µ→∞ ρ(µ) = 0.

Multiplying (2.1) again by g ⁰ (y) and integrating it from y < l to l, we have g ⁰ (y)

pln g(y) − ln ρ = −1. (2.2)

Integrating (2.2) from 0 to l, we obtain l =

Z µ ρ

√ dη

ln η − ln ρ := G(µ).

Let σ ² = ln η − ln ρ. Then

G(µ) = 2µ exp µ ²

Z µ 0

exp σ ² dσ, (2.3)

G ⁰ (µ) = 2 − 4µ ² exp µ ²

Z µ 0

exp σ ² dσ + 2µ, (2.4)

G ⁰⁰ (µ) = 8µ ³ − 12µ exp µ ²

Z µ 0

exp σ ² dσ + 4 − 4µ ² . (2.5)

From (2.4), G ⁰ (p1/2) = √

2 > 0. Since the power series for e ^σ

is e ^σ

= 1 + σ ² + σ ⁴

2! + σ ⁶ 3! + σ ⁸

4! + σ ¹⁰ 5! + σ ¹²

6! + σ ¹⁴

7! + · · · + σ ²ⁿ

n! + · · · .

G ⁰ (2) < −14 e ⁴

 2 + 2 ³

3 + 2 ⁵

5 · 2! + 2 ⁷

7 · 3! + 2 ⁹

9 · 4! + 2 ¹¹

11 · 5! + 2 ¹³

13 · 6! + 2 ¹⁵ 15 · 7!



+ 4 < 0.

Therefore, there is a µ ₀ > 0 such that G ⁰ (µ ₀ ) = 0.

Lemma 2.1 Any critical point of G(µ) must be a maximal point.

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Proof. If G ⁰ (µ) = 0, then Z µ

0

exp σ ² dσ = µ exp µ ²

2µ ² − 1 and 1 − 2µ ² < 0. Form (2.5) we have G ⁰⁰ (µ) = 8µ ³ − 12µ

2µ ² − 1 · µ + 4 − 4µ ² = 4

1 − 2µ ² < 0.

This proves the lemma.

From (2.3), we obtain lim

µ→0 G(µ) = 0 and lim

µ→∞ G(µ) = 1. Since G has a critical point µ ₀ and any critical point of G must be a maximum point, it follows that G is monotone increasing in (0, µ ₀ ) and monotone decreasing in (µ ₀ , ∞). Therefore, we conclude that for γ = 1 there exists l ^∗ ∈ (1, ∞) 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 ^∗ .

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