In this chapter, we are concerned with the following third order differential equation:

4 A third order differential equation in boundary layer theory

4.1 Introduction

In this chapter, we are concerned with the following third order differential equation:

f

+ [(α + 1)/2]f f

− αf

= 0, (1.1)

where f = f (t) for t ∈ [0, ∞) and α is a real parameter. We notice that (1.1) for α = 0 reduces to the well-known Blasius equation (see, e.g., [4, 10, 18, 19, 31] and references therein).

The equation (1.1) arises in the study of self-similar solutions of the steady free convection problem for a vertical heated impermeable flat plate embedded in a porous medium with the prescribed wall temperature a power function of height. Here the parameter α is just the exponent of the power function. Using the similarity variables in the boundary layer approximation, by assuming the convection takes place in a thin layer around the heating surface, we obtain the following boundary value problem (P

):

f

+ [(α + 1)/2]f f

− αf

= 0, t ∈ (0, ∞), f (0) = 0, f

(0) = 1, f

( ∞) := lim

t→∞

f

(t) = 0.

Here the variable t is the independent similarity variable and the function f is related to the stream function. For the detailed derivation, we refer the readers to [9], [5], and references cited therein. Furthermore, if we assume that the temperature is decreasing with respect to the distance from the wall, then we have the extra constraint

0 ≤ f

(t) ≤ 1 for all t ∈ [0, ∞). (1.2)

The problem (P

) is a very interesting subject for years, see, e.g., [4], [5], and the references cited therein. In particular, Belhachmi, Brighi, and Taous in [5] have shown that the problem (P

) has no solution for α < −1/3, at least one solution for α > −1/3, and exactly one solution for α = −1/3 and α ≥ 0 such that the constraint (1.2) holds. Moreover, the problem (P

) has infinitely many solutions for α = −1/3; and any solution of (P

) satisfies (1.2) if α ∈ [0, 1/3]. On the other hand, Magyari and Keller [22] have considered the equation (1.1) with the boundary conditions f (0) = a, f

(0) = 1, and f

( ∞) = 0 for the cases α = −1/2, −1/3, 1. See also [2, 7, 8].

To study the problem (P

), we consider the following initial value problem (P

):

f

+ [(α + 1)/2]f f

− αf

= 0 on [0, T

), f (0) = 0, f

(0) = 1, f

(0) = c,

where c ∈ R and [0, T

) is the (right) maximal existence interval of the solution. Let f

be the solution of (P

) defined in [0, T

).

For α = 0, Hartman [18] (see also [4]) has shown that there exists c

∈ (−∞, 0) such that f

> 0 on [0, ∞), if c ∈ [c

, ∞), and the limit lim

f

(t) is an increasing function of c ∈ [c

, ∞) onto [0, ∞). Moreover, for each c ∈ (−∞, c

) there exists a t

= t

(c) > 0 such that f

> 0 on [0, t

), f

< 0 on (t

, T

), and f

(t) → −∞ as t → T

. Note that T

< ∞ if and only if c ∈ (−∞, c

).

Thus we shall only consider the case α 6= 0. In [5], the authors proposed many interesting

open questions for the problem (P

). The main purpose of this study is to answer some of

these open questions by studying the structure of solutions of (P

).

Introduce the quantity

E(t) := f

(t) + α + 1

2 f

(t)f

(t).

Then

E

(t) = 3α + 1

2 [f

(t)]

.

Note that f

(t) 6= 0 when f

(t) = 0. Otherwise, we have f

(t) = 0 for all n ≥ 1, by differentiating (1.1). Then by the standard uniqueness theorem for ordinary differential equations, this implies that f

is constant, a contradiction. Therefore, E(t) > 0 if t is a minimum point of f

; and E(t) < 0 if t is a maximum point of f

. Note that the first critical point of f

must be a maximum point.

For α > −1/3, E is non-decreasing. Then f

has at most two critical points. Otherwise, there are three smallest critical points t

, t

, t

of f

with 0 < t

< t

< t

. Then E(t

) < 0, E(t

) > 0, and E(t

) < 0. This contradicts with the fact that E is non-decreasing.

For α < −1/3, E is non-increasing. Then f

has at most one critical point, since E(t

) < 0 and E(t

) > 0 for two smallest critical points t

, t

of f

with 0 < t

< t

(if they exist).

For α = −1/3, E is a constant. Hence f

has at most one critical point.

In any case, f

has at most two critical points. Indeed, we can classify the solution of (P

) into the following six types.

(A) f

> 0 on [0, ∞) and lim

f

(t) = 0.

(B) f

> 0, f

> 0 on (0, T

) and lim

f

(t) = + ∞.

(C) f

> 0 on [0, T

), there exists a t

> 0 such that f

< 0 on [0, t

) and f

> 0 on (t

, T

), and lim

c

f

(t) = + ∞.

(D) There exists a t

> 0 such that f

> 0 on [0, t

), f

< 0 on (t

, ∞), and f

→ 0 as t → ∞.

(E) There exists a t

> 0 such that f

> 0 on [0, t

), f

< 0 on (t

, T

), and f

(t) → −∞ as t → T

.

(F) There exist t

, t

with t

> t

> 0 such that f

> 0 on [0, t

), f

< 0 on (t

, t

), and f

> 0 on (t

, T

).

We remark that if f

is the solution of (P

) and of type (D), then f

is a solution of (P

) such that the temperature is not decreasing with respect to the distance from the wall.

Our first strategy for deriving the structure of solutions of (P

) is to transform the third- order equation into a second-order equation. Such reduction techniques are utilized in [18]

for the analysis of the Blasius equation. With the help of some comparison principles and known results obtained by Belhachmi, Brighi, & Taous in [5], we are able to derive the following structure of solutions of (P

).

(P1) For α ∈ (−∞, −1/2], f

is of type (E) for all c ∈ R. Moreover, T

< ∞ for α ∈ [ −1, −1/2].

(P2) For α ∈ (−1/2, −1/3), f

is either of type (A) or of type (E) for c > 0; and f

is of type (E) with T

< ∞ for c ≤ 0.

(P3) For α ∈ [−1/3, 0), there is a constant c

∈ (−∞, 0], with c

= 0 and c

< 0 for

α ∈ (−1/3, 0), such that f

is of type (A) for c ∈ [c

, ∞); and f

is of type (E) with

T

< ∞ for c ∈ (−∞, c

). Moreover, for α ∈ (−1/3, 0) there exists d

∈ (c

, 0) such

that f

is bounded for all c ∈ [c

, d

) and unbounded for all c ∈ [d

, ∞). For α = −1/3, the only bounded solution is the solution f

with c = 0.

(P4) For α ∈ (0, 1/3), there is a constant c

∈ (−∞, 0) such that f

is of type (B) with T

= ∞ for c ∈ [0, ∞); f

is of type (C) for c ∈ (c

, 0); f

is of type (A); and f

is either of type (E) or of type (F) for c < c

. Moreover, there exists δ

> 0 such that f

is of type (F) for c ∈ (c

− δ

, c

); and f

is of type (E) for c < c

with −c sufficiently large.

(P5) For α = 1/3, there is a constant c

∈ (−∞, 0) such that f

is of type (B) with T

= ∞ for c ∈ [0, ∞); f

is of type (C) for c ∈ (c

, 0); f

is of type (A); and f

is of type (F) for c ∈ (−∞, c

).

(P6) For α ∈ (1/3, ∞), there is a constant c

∈ (−∞, 0) such that f

is of type (B) for c ∈ [0, ∞); f

is of type (C) for c ∈ (c

, 0); f

is of type (A); and f

is either of type (D) or of type (F) for c < c

.

We remark that the uniqueness of solutions for the problem (P

) with the constraint (1.2) does not hold for α ∈ (−1/3, 0), since, by (P3) and Lemma 3.1 in [5], f

is a solution of the problem (P

) with the constraint (1.2) for any c ∈ (c

, 0). Moreover, for α ∈ (−1/3, 0), f

is bounded if and only if c ∈ [c

, d

). This answers the first two open questions proposed in

§7 of [5].

For further distinctions of solutions, we introduce a change of variables to transform the

third-order equation into a system of two first order equations. Applying the phase plane

analysis, we obtain the following results.

(Q2) There is a constant α

∈ [−1/2, −1/3) such that for each α ∈ (α

, −1/3) the solution f

of (P

) is of type (A) for c ≥ c

; and is of type (E) for c < c

for some c

∈ (0, ∞).

(Q4) For each α ∈ (0, 1/3), there exists b

∈ (−∞, c

) such that the solution f

of (P

) is of type (F) for c ∈ (b

, c

); and is of type (E) for c ∈ (−∞, b

].

(Q6) For any α ∈ (1/3, ∞), there is no type (D) solution of (P

). In other words, the solution f

of (P

) is of type (F) for c < c

for any α ∈ (1/3, ∞).

We remark that (P2), (Q2), and (Q6) answer another two open questions proposed in §7 of [5]. In particular, for α > 0 any solution f of (P

) must satisfy the constraint (1.2).

This chapter is organized as follows. In §2, some preliminary results are given. Then we prove (P1)-(P3) in §3. The case for α > 0 is analyzed in §4 and (P4)-(P6) are derived. In

§5, we obtain the further distinctions of solutions for α ∈ (−1/2, −1/3), (0, 1/3), (1/3, ∞), respectively. Finally, the conclusion is given in §6.

4.2 Preliminary

In the sequel, we shall always assume that α 6= 0. Also, we shall sometimes drop the subscript of f

and simply write f

as f , if there is no danger of ambiguity. First, let us recall a result from [5].

Lemma 4.1 Let f be the solution of (P

) on [0, T ) and t

∈ [0, T ). If α < 0 and f

(t

) ≤ 0 (α > 0 and f

(t

) ≥ 0), then f

< 0 on (t

, T ) (f

> 0 on (t

, T ), respectively).

For any solution f of (1.1) with f

> 0 in [0, r) for some r > 0, we set

y(x) := [f

(t)]

and x := f for t ∈ [0, r). (2.1)

Then

y

:= dy

dx = 2f

(t), y

= 2f

/f

, (2.2) and (P

) is transformed into the initial value problem (Q

) for the second order equation

y

+ α + 1 2

xy

√ y − 2α √

y = 0 (2.3)

with the initial conditions

y(0) = 1, y

(0) = 2c. (2.4)

Lemma 4.2 Let y be a solution of (2.3). Suppose that y is global and monotone ultimately.

Then either y → 0 or y → ∞ as x → ∞.

Proof. By assumption, l := lim

y(x) exists. Suppose that l ∈ (0, ∞). Note that y

(x) has a fixed sign for all x sufficiently large, say, for all x ≥ x

> 0. It follows from

Z

x0

y

(x)dx = l − y(x

)

that there is a sequence x

→ ∞ such that y

(x

) → 0 as k → ∞.

Now, dividing (2.3) by x and integrating it from x

to x

, k ≥ 1, we obtain that

Z

x0

y

x dx +

Z

x0

α + 1 2

y

√ y dx =

Z

x0

2α

√ y x dx.

We compute that

Z

x0

y

x dx = y

(x

)

x

− y

(x

) x

+

Z

x0

y

x

dx and

Z

x0

α + 1 2

y

√ y dx = (α + 1)[ ^q y(x

) − ^q y(x

)].

It is clear that the above two integrals are uniformly bounded for all k. On the other hand, the integral

Z

x0

2α

√ y x dx

tends to ∞ as k → ∞, a contradiction. Therefore, the lemma follows.

Lemma 4.3 Let f be the solution of (P

) defined on [0, T ). Suppose that f

has a unique zero in [0, T ) and f

has a fixed sign on [t

, T ) for some t

∈ [0, T ). Then either f

→ −∞

or f

→ 0 as t → T

.

Proof. By assumption, the limit l := lim

f

exists and l ≤ 0 (l may be −∞). If f is bounded, then T = ∞ and f

→ 0 as t → ∞.

Now, we assume that f is unbounded. Let t

be the unique zero of f

. Then f (t) ≤ f(t

) for all t ∈ [0, T ). Set k = f(t

) + 1, y(x) := [f

(t)]

, and x := k − f for t ∈ (t

, T ). Then

y

:= dy

dx = −2f

(t), y

= 2f

/f

,

and (P

) is transformed into the initial value problem for the equation

y

+ α + 1 2

(k − x)y

√ y + 2α √ y = 0

with the initial conditions

y(x

) = 0, y

(x

) = −2f

(t

) > 0,

where x

= k − f(t

). Since f is unbounded and f

has a fixed sign on [t

, T ) for some t

∈ [0, T ), y is defined for x ∈ [x

, ∞) and y is monotone ultimately.

By a similar argument as Lemma 4.2, we obtain that either y → 0 or y → ∞ as x → ∞.

Returning to the variables t and f , we reach the conclusion.

We recall the following comparison principle from [29].

Lemma 4.4 (Comparison Principle)

(a) Let f (t, y) = (f

, f

, ..., f

) be continuous on the strip S : a ≤ t ≤ b, y ∈ R

and f

be nondecreasing with respect to each of the components y

, i 6= k, of y for k = 1, 2, ..., d. Assume that the solution of the initial value problem y

= f (t, y), y(a) = y

is unique for a fixed y

, and that this solution y = y(t) exists on [a, b]. Let z(t) = (z

(t), z

(t), ..., z

(t)) be continuous on [a, b] such that z

(t) is differentiable, z

(a) ≤ y

(a) and (z

)

(t) ≤ f

(t, z(t)) (or z

(a) ≥ y

(a) and (z

)

(t) ≥ f

(t, z(t))) for a <

t < b and k = 1, 2, 3, ..., d. Then z

(t) ≤ y

(t)(or z

(t) ≥ y

(t)) for a ≤ t ≤ b.

(b) If, in part (a), all initial value problems associated with y

= f (t, y) have unique solu- tions, f

(t, y) is increasing with respect to y

, i 6= k and k = 1, 2, ..., d, and z

(a) < y

(or z

(a) > y

)for at least one index j, then z

(t) < y

(t)(or z

(t) > y

(t)) for a < t ≤ b and k = 1, 2, ..., d.

(c) If, in addition to the assumption of (a), there is an index h such that f

(t, y) is non- decreasing with respect to y

, then y

(t) − z

(t) is nondecreasing on a ≤ t ≤ b.

4.3 The case for α ∈ (−∞, 0)

We start with the case for α ∈ (−∞, 0). Let f be the solution of (P

) defined on [0, T ).

Recall (2.1), (2.2), and the corresponding initial value problem (Q

):

y

+ α + 1 2

xy

√ y − 2α √ y = 0, (3.1)

y(0) = 1, y

(0) = 2c. (3.2)

Let [0, R) be the maximal existence interval of y. It is easy to see that any critical point of a solution of (3.1) is a maximum point. Hence any solution of (3.1) has at most one critical point in its existence interval. Moreover, y

< 0 on (0, R) for the solution y of (Q

) with c ≤ 0. Note that y

(0) < 0 for all c.

Lemma 4.5 If c ≥ 0, then there exists t

≥ 0 such that f

(t

) = 0 and f

< 0 on (t

, T ).

Proof. Let c ≥ 0 be fixed. We divide the proof into two cases.

First, we deal with the case for α ∈ (−1, 0). We argue by contradiction. If not, then f

> 0 on (0, T ). Hence f, f

> 0 and f

> 1 on (0, T ). Thus

f

(t) = − α + 1

2 f (t)f

(t) + αf

(t)

≤ αf

(t)

≤ α

for all t ∈ (0, T ) and this implies that T = ∞. But,

f

(t) = c +

Z

0

f

(s)ds

≤ c + αt

→ −∞ as t → ∞,

a contradiction. Therefore, there exists t

≥ 0 such that f

(t

) = 0. Then, by Lemma 4.1, we have f

< 0 on (t

, T ).

Next, we consider the case for α ∈ (−∞, −1]. If the conclusion were false, then f

> 0 on (0, T ). Hence f, f

are increasing and positive on (0, T ). We claim that f

< 0 on [0, T ).

Indeed, from (1.1) we have f

(0) = α < 0, and so f

< 0 on [0, δ) for some δ < 0. If there

exists δ

> 0 such that f

< 0 on [0, δ

) and f

(δ

) = 0, then f

(δ

) ≥ 0. On the other hand, by differentiating (1.1) and evaluating at δ

, we obtain

f

(δ

) = − α + 1

2 f (δ

)f

(δ

) + 3α − 1

2 f

(δ

)f

(δ

)

= 3α − 1

2 f

(δ

)f

(δ

)

< 0,

a contradiction. Thus f

< 0 on [0, T ) and this implies that T = ∞. Moreover, since f, f

, f

> 0 and f

< 0 on (0, ∞), we have f

< 0 on (0, ∞) by differentiating (1.1).

Hence f

(t) ≤ f

(0) = α for all t ∈ [0, ∞), and so we obtain

f

(t) = f

(0) +

Z

0

f

(s)ds

≤ f

(0) + αt

→ −∞ as t → ∞,

a contradiction. Therefore, there exists t

≥ 0 such that f

(t

) = 0. Also, by Lemma 4.1, we have f

< 0 on (t

, T ). This completes the proof.

Lemma 4.6 Let α ∈ (−∞, 0) and f be the solution of (P

) defined on [0, T ). Then f is either of type (A) or of type (E).

Proof. Suppose that f

> 0 on [0, T ). Then by Lemmas 4.1 and 4.5, there exists t

≥ 0 such that f

< 0 on (t

, T ). Hence T = ∞ and the limit f(∞) := lim

f (t) exists. If f ( ∞) is finite, then lim

f

(t) = 0 and f is of type (A). If f ( ∞) = ∞, let y(x) be the solution of (Q

) with c = f

(0). It follows from (2.1) and (2.2) that y is defined on [0, ∞). Since f

< 0 on (t

, T ), y

< 0 on [f (t

), ∞). Therefore, by Lemma 4.2, we have lim

y(x) = 0.

Hence lim

f

(t) = 0 and f is of type (A).

If f

has one first zero, say t

, then f

(t

) ≤ 0. Hence, by Lemma 4.1, we have f

< 0 on (t

, T ). Therefore, f

< 0 on (t

, T ) and f

→ −∞ as t → T

by Lemma 4.3. Thus f is of type (E).

The following proposition gives the non-global existence for the case α ∈ [−1, 0).

Proposition 4.7 Let α ∈ [−1, 0). If f is of type (E) solution of (P

) defined on [0, T ), then T < ∞.

Proof. By assumption, there exists t

> 0 such that f

(t

) = 0 and f

(t

) ≤ 0. Hence f

, f

< 0 on (t

, T ) by Lemma 4.1. If T = ∞, then there exists t

> t

such that f, f

, f

< 0 on (t

, ∞). Now, we rewrite (1.1) as

f

= −[(α + 1)/2]ff

+ αf

Notice that the right hand side of the above equation is a nondecreasing function of f, f

for (f, f

, f

) ∈ (−∞, 0) × (−∞, 0) × (−∞, 0). Choose k ∈ (t

, ∞) such that

−6/(k − t

) ≥ f(t

), −6/(k − t

)

≥ f

(t

), −12/(k − t

)

≥ f

(t

).

Noting that z(t) := −6/(k − t) is a solution of (1.1) and using Lemma 4.4, we have

z(t) ≥ f(t) for t ≥ t

,

a contradiction. Therefore, f is only defined in a finite interval. This proves the proposi- tion.

4.4 The case for α ∈ (−∞, −1/2]

Theorem 4.8 Let α ∈ (−∞, −1/2] and f be the solution of (P

). Then f is of type (E).

Proof. Suppose that f

> 0 on [0, T ). Then by the same argument as the proof of Lemma 4.6 we see that f is a solution of (P

). This contradicts Theorem 4.1 of [5]. Hence there exists t

> 0 such that f

(t

) = 0. Therefore, f is of type (E) by Lemma 4.6.

4.5 The case for α ∈ (−1/2, −1/3)

Theorem 4.9 Let α ∈ (−1/2, −1/3) and f be the solution of (P

) with c ≤ 0. Then f is of type (E).

Proof. Suppose that f

> 0 on [0, T ). Then, by the same argument as Lemma 4.6, we conclude that f is a solution of (P

). By Lemma 4.1, f

< 0 in (0, ∞) if c ≤ 0. Hence f is a solution of (P

) satisfying (1.2). This contradicts Theorem 4.2 of [5]. Hence there exists t

> 0 such that f

(t

) = 0 and it follows from Lemma 4.6 that f is of type (E). This proves the theorem.

Remark 4.10 It follows from Lemma 4.6 that f is either of type (A) or of type (E) for any c > 0.

4.6 The case for α ∈ [−1/3, 0)

We first consider the case for c > 0.

Lemma 4.11 Let α ∈ [−1/3, 0) and f be the solution of (P

) with c > 0. Then f is of type (A) and satisfies that

t→∞

lim f (t) = ∞.

Proof. Suppose that c > 0. By Lemma 4.5, there exists t

≥ 0 such that f

(t

) = 0 and f

<

0 on [t

, T ). Let y be the solution of (Q

) and set x

= f (t

). Then y

(x

) = 2f

(t

) = 0

and so x

is a maximal point of y. Hence y

< 0 on (x

, R).

We first claim that R = ∞. For contradiction, we suppose that R < ∞. Consider the quantity

G(x) = y

2 + α + 1 2 x √ y.

By (3.1), we obtain

G

(x) = 3α + 1 2

√ y ≥ 0 for all x ∈ (0, R),

and so G(t) is non-decreasing in x. However, G(0) = y

(0)/2 = c > 0 and G(R

) = y

(R

)/2 ≤ 0, a contradiction. Therefore, we conclude that R = ∞.

By Lemma 4.2, we have y( ∞) = 0. Returning to the variables t and f, we obtain

t→T

lim

f (t) = ∞ and lim

t→T⁻

f

(t) = 0.

This implies that T = ∞ and the lemma follows.

Now, let us consider the case for c ≤ 0. We have the following comparison principle.

Lemma 4.12 Let α ∈ [−1/3, 0). Let y

be the solution of (P

) defined on [0, R

) with c

≤ 0, i = 1, 2. If c

< c

, then y

< y

for all x ∈ (0, R), where R := min{R

, R

}.

Moreover, if R

< ∞ (so that y

(R

) = 0), then R

< ∞, R

< R

, and y

(R

), y

(R

) exist such that y

(R

) ≤ y

(R

).

Proof. Note that y

< 0 on (0, R

) for i = 1, 2. Hence the inverse function of y

exists. Let X

(1 − y) be the inverse function of y

for i = 1, 2. Following [18], we set

V

(y) = y

(X

(y))

for i = 1, 2. Then it follows from (3.1) that X = X

, V = V

satisfies the following system dX

dy = −1/V (6.1)

dV

dy = [(α + 1)/2](X/ ^q 1 − y) − 2α( ^q 1 − y/V ). (6.2)

Note that the function on the right hand side of (6.1) is an increasing function of V and a nondecreasing function of X; the function on the right hand side of (6.2) is an increasing function of X. Therefore, by Lemma 4.4, X

> X

and V

> V

for all y > 0 for which X

and V

are defined for i = 1, 2. Moreover, X

− X

is positive and nondecreasing. Returning to the variables x and y and noting that y

< 0 on [0, R

) for i = 1, 2, we have y

< y

for all x ∈ (0, R). Using that X

− X

is nondecreasing, we obtain that R

< R

. By considering the function G defined in the proof of Lemma 4.11, we obtain that y

(R

) exists, since R

< ∞ for i = 1, 2. Since V

> V

for all y ∈ (0, R

), we have y

(R

) ≤ y

(R

). The lemma follows.

Given a fixed α ∈ [−1/3, 0). Set A

= {c < 0 | f

> 0 on (0, T ) } and c

= inf A

. Then by the argument of Theorem 5.1 of [5], we have c

∈ (−∞, 0) for α ∈ (−1/3, 0) and c

= 0 for α = −1/3. Moreover, f

> 0, f

( ∞) = 0, f

( ∞) = 0, and 0 < f

≤ (2/ √

α + 1).

Note that by (2.1) and (2.2) we have R

< ∞ and y

(R

) = 0. For α = −1/3, the only bounded solution is the solution f

with c = 0.

Theorem 4.13 Let α ∈ (−1/3, 0). Then we have

(1) For all c ∈ [c

, ∞), the solution f

of (P

) is of type (A). Moreover, for α ∈ (−1/3, 0) there exists d

∈ (c

, 0) such that f

is bounded for all c ∈ [c

, d

) and unbounded for all c ∈ [d

, ∞).

(2) For all c ∈ (−∞, c

), the solution f

of (P

) is of type (E).

Proof. By the definition of c

, for any c ∈ (−∞, c

) fixed, there exists t

> 0 such that the

solution f

of (P

) satisfying f

> 0 on [0, t

) and f

(t

) = 0. Hence f

is of type (E) by

Lemma 4.6.

Next, we consider the case for c ∈ (c

, 0]. Noting that every critical point of y

is a maximal point and that y

(0) = 2α < 0, we obtain that y

< 0 on (0, R

) for all c ∈ (c

, 0].

Therefore, by Lemma 4.12, the limit y

(R

) exists and is equal to 0, if R

< ∞ . Moreover, if there exists d ∈ (c

, 0] such that R

= ∞, then R

= ∞ for all c ∈ [d, 0]. To show that there exists d

∈ (c

, 0) such that R

= ∞ for all c ∈ [d

, 0] and R

< ∞ for all c ∈ [c

, d

), it suffices to show that R

= ∞ for c < 0 and −c is sufficiently small.

Now, given any fixed α ∈ (−1/3, 0). Consider the quantity

G

(x) = y

(x)

2 + α + 1

2 x ^q y

(x).

By (3.1), we obtain

G

(x) = 3α + 1 2

q y

(x) > 0 for all x ∈ (0, R

)

and so G

(x) is increasing in x. Let y

be the solution of (Q

) defined on [0, R

). By a similar argument as in the proof of Lemma 4.11, we can show that R

= ∞ and A :=

R

0

[(3α + 1)/2] ^q y

(x)dx > 0. Note that A may be + ∞. Let σ := min{1, A/2}. Then by the standard theory of continuous dependence on initial condition, there exists δ ∈ (0, σ) and R > 0 such that R

> R and ^R

[(3α + 1)/2] ^q y

(x)dx > σ for all c ∈ (−δ, 0). Thus for c ∈ (−δ, 0)

G

(R) = G

(0) +

Z

0

G

(x)dx

= y

(0)/2 +

Z

0

[(3α + 1)/2] ^q y

(x)dx

> c + σ

> 0.

If R

< ∞ for some c ∈ (−δ, 0), then

G

(R

) = y

(R

)/2 ≤ 0,

a contradiction with the fact that G

(x) is increasing. Hence R

= ∞ for all c ∈ (−δ, 0).

Therefore, there exists d

∈ (c

, 0) such that R

= ∞ for all c ∈ [d

, 0] and R

< ∞ for all c ∈ [c

, d

).

Given any c ∈ [c

, d

). Note that R

< ∞. Then f

> 0 on [0, T

), lim

c

f

(t) = 0, and lim

f

(t) = 0, where T

is defined by f

(T

) = R

. Recall that f

6= 0 when f

= 0.

Hence T

= ∞ and so f

is of type (A) by Lemma 4.6. Notice that f

is bounded in this case.

Finally, R

= ∞ for all c ∈ [d

, 0]. Therefore, y

( ∞) = 0 by Lemma 4.2. Returning to the variables f and t, we have f

> 0 on [0, T

), lim

c

f

(t) = ∞, and lim

f

(t) = 0.

Hence T

= ∞ and f

is of type (A) by Lemma 4.6. Notice that f

is unbounded in this case.

Combining with Lemma 4.11, we obtain that the solution f

of (P

) is of type (A) for all c ∈ [c

, ∞) and the theorem follows.

4.7 The case for α ∈ (0, ∞)

We now turn to the case for α ∈ (0, ∞). Let f be the solution of (P

) defined on [0, T ).

Recall (2.1), (2.2), and the corresponding initial value problem (Q

):

y

+ α + 1 2

xy

√ y − 2α √ y = 0, (7.1)

y(0) = 1, y

(0) = 2c. (7.2)

Let [0, R) be the maximal existence interval of y. It is easy to see that any critical point of

a solution of (7.1) is a minimum point. Hence any solution of (7.1) has at most one critical

point in its existence interval. Moreover, y

> 0 on (0, R) for the solution y of (Q

) with c ≥ 0. Note that y

(0) > 0 for all c.

For c < 0, there are only two kinds of solutions of (Q

) as follows.

Lemma 4.14 Let y be the solution of (Q

) defined on [0, R) with c < 0. Then y must be one of the following two types:

(I) y

< 0, y

> 0 on [0, R) and R is finite.

(II) There exists x

> 0 such that y

< 0 on [0, x

) and y

> 0 on (x

, R).

Proof. Firstly, we note that y

> 0 on [0, R) if y

< 0 on [0, R).

We argue by contradiction. Since y

(0) = 2c < 0, we have y > 0, y

< 0 and y

> 0 on [0, b] for some b > 0. If y is not of type (I) or (II), then y is defined on [0, ∞) with y

< 0 on [0, ∞). Therefore, l := lim

y(x) exists and is finite. By Lemma 4.2, l = 0. Let f be the solution of (P

). By (2.1) and (2.2), we have f > 0, 0 < f

≤ 1 and f

< 0 on [0, T ).

Moreover, f → ∞ as t → T

, since y is defined on [0, ∞). From l = 0 and f

< 0 on [0, T ), it follows that T = ∞ and f

→ 0 as t → ∞. Therefore, f is an unbounded solution of (P

) with the constraint (1.2), contradicting Proposition 3.2 of [5].

We have the following comparison principle.

Lemma 4.15 Let y

be the solution of (Q

) defined on [0, R

) with c

< 0, i = 1, 2. If c

< c

and y

is of type (I), then the following hold:

(1) y

is of type (I).

(2) y

< y

and y

< y

for all x ∈ (0, R), R := min{R

, R

}.

(3) R

< R

and y

(R

) < y

(R

).

Proof. Since c

< 0, y

< 0 on [0, R

) for some R

> 0 for i = 1, 2. Hence the inverse function of y

(i = 1, 2) exists on [b, 1] for some b ∈ (0, 1). Let X

(1 − y) be the inverse function of y

, i = 1, 2. We also set

V

(y) = y

(X

(y)), i = 1, 2.

Then it follows from (7.1) that X = X

, V = V

satisfy the following system dX

dy = −1/V (7.3)

dV

dy = [(α + 1)/2](X/ ^q 1 − y) − 2α( ^q 1 − y/V ). (7.4) Note that the function on the right hand side of (7.3) is an increasing function of V and a non-decreasing function of X; the function on the right hand side of (7.4) is an increasing function of X and an increasing function of V . Therefore, by Lemma 4.4, X

> X

and V

> V

for all y > 0 for which X

and V

are defined for i = 1, 2. Moreover, X

− X

and V

− V

are positive and non-decreasing. Returning to the variables x and y and noting that y

< 0, we obtain that y

< y

for all x ∈ (0, R) and that y

(x) < 0 as long as y

(y

(x)) < 0. Hence y

is of type (I). Using that X

− X

is increasing, we obtain that R

< R

. From (7.1) it follows that y

> 0 on [0, R

) for i = 1, 2, and so we obtain that y

(R

) exists for i = 1, 2. Noting that V

− V

is non-decreasing for all y ∈ (0, 1), we have y

(R

) < y

(R

). This completes the proof of the lemma.

Lemma 4.16 Let α > 0 and y

be the solution of (Q

) defined on [0, R

) with c < 0. Then there exists c

∈ (−∞, 0) such that the following hold:

(1) For c < c

, y

is of type (I) and y

(R

) < 0.

(2) y

is of type (I) and y

(R

) = 0.

(3) For c > c

, y

of is of type (II).

Proof. Firstly, we claim that for c < k

:= min {−(3α + 1)/2, −(3α + 3)/4} the solution y

of (Q

) is of type (I). Indeed, we shall claim that R

≤ 1 and y

< 0 on [0, R

) for all c < k

.

To this end, we first note that (setting y = y

)

y

(x) = 2c − [(α + 1)/2]

Z

0

(s/ √

y)y

ds + 2α

Z

0

√ yds

= 2c − (α + 1)x √ y + (3α + 1)

Z

√ yds,

by using (7.1) and an integrating by parts. Hence we obtain that

y

(x) ≤ 2c + (3α + 1)x (7.5)

if 0 < y ≤ 1 on [0, x]. Integrating the inequality (7.5), we obtain

y(x) ≤ 1 + 2cx + [(3α + 1)/2]x

, (7.6)

if 0 < y ≤ 1 on [0, x].

Now, we fix any c < k

. Suppose for contradiction that R

> 1. Then, by (7.5), y

(x) < 0 for all x ∈ [0, 1]. But, from (7.6) it follows that y(1) ≤ 1 + 2c + (3α + 1)/2 < 0. This contradiction leads that R

≤ 1. Moreover, y

< 0 on [0, R

). Otherwise, if y

< 0 on [0, x

) and y

(x

) = 0 for some x

∈ [0, R

), then 0 ≤ y ≤ 1 on [0, x

] and by (7.5)

y

(x

) ≤ 2c + (3α + 1)x

≤ 2c + (3α + 1) < 0,

a contradiction. Hence y

< 0 on [0, R

). Therefore, by Lemma 4.14, y is of type (I) and

y

(R

) < 0 for all c < k

.

On the other hand, since for c = 0 the corresponding solution y

of (Q

) satisfying y

> 0 in (0, R) for some R > 0, it follows from the standard theory of the continuous dependence on initial data that y

is of type (II) for 0 < −c << 1. Set

c

:= sup {c | y

is of type (I) and y

(R

) < 0 }.

Note that c

∈ [k

, 0). Then the lemma follows from Lemma 4.15.

We now state one of the key results of this section as follows.

Theorem 4.17 Let α > 0 and f

be the solution of (P

) on [0, T

). Then there exists c

∈ (−∞, 0) such that the following hold:

(1) For c ∈ (−∞, c

), f

has at least one zero.

(2) f

is of type (A).

(3) For c ∈ (c

, 0), f is of type (C).

(4) For c ∈ [0, ∞), f is of type (B).

Proof. Let y

be the solution of (Q

) defined on [0, R

). First, (1) follows from Lemma 4.16 easily. Recall that f

6= 0 when f

= 0. Hence (2) follows from Lemma 4.16.

By Lemmas 4.16 and 4.2, for c ∈ (c

, 0), there exists x

> 0 such that y

< 0 on [0, x

), y

> 0 on (x

, R

) and lim

c

y(x) = ∞. Back to the origin variables t and f, f

is of type (C).

Since any critical point of a solution of (7.1) is a minimum point, we have y

> 0 on (0, R

)

for c ∈ [0, ∞). Then from Lemma 4.2 it follows that lim

y(x) = ∞. This implies that

f

is of type (B). This completes the proof of the theorem.

Note that the fact f

is of type (A) has been derived in [5]. The following proposition gives the global existence for α ∈ (0, 1/3] and c ≥ 0.

Proposition 4.18 Let α ∈ (0, 1/3] and f be the solution of (P

) defined on [0, T

). If c ≥ 0, then T

= ∞.

Proof. We argue by contradiction. Suppose T

< ∞. Since c ≥ 0, it follows from Lemma 4.1 that f

> 0 on (0, T

). Hence f, f

> 0 on (0, T

) and f → ∞ as t → T

. This implies that there exists t

> 0 such that f

(t

) > 0. On the other hand, by differentiating (1.1), we obtain

f

(t) = [(3α − 1)/2]f

(t)f

(t) − [(α + 1)/2]f(t)f

(t).

Note that f

(t) < 0 if f

(t) = 0. Hence either f

(t) > 0 for all t ∈ [t

, T

) or there is a unique t

> t

such that f

(t

) = 0 and f

(t) < 0 for all t ∈ (t

, T

). In the first case, we have f

(t) < 0 for all t ∈ [t

, T

). However, both cases all imply that f is bounded on [0, T

), a contradiction. Therefore, T

= ∞.

From now on we shall focus on the case for c < c

.

Lemma 4.19 Let α > 0 and f

be the solution of (P

) on [0, T

). For c ∈ (−∞, c

), f

is either of type (D), of type (E) or of type (F).

Proof. Suppose that f

has a unique zero. Then, by Lemmas 4.1 and 4.3, f

is either of type (D) or of type (E).

Suppose that f

has at least two zeros, say t

and t

with t

< t

. Then f

(t

) ≥ 0 and

thus f

> 0 on (t

, T

) by Lemma 4.1. Therefore, f

is of type (F).

Lemma 4.20 Let α > 0 and f

be the solution of (P

) with c < c

. Then f

(t

) → 0 as c → c

, where t

is the first zero of f

.

Proof. Let y

be the solution of (Q

) defined on [0, R

). By Lemma 4.16, we have R

= f

(t

) and y

(R

) = f

(t

). Hence it suffices to show that lim

α

y

(R

) = 0. For this purpose, we divide the proof into three steps.

Step 1. We claim that the limit k := lim

y

(R

) exists and k ≤ 0. In fact, by Lemma 4.16, for any c ∈ (−∞, c

), y

is strictly decreasing to zero with y

(R

) < 0.

Furthermore, by Lemma 4.15, we have y

(R

) < y

(R

) and R

< R

for c

< c

< c

. Therefore, y

(R

) is an increasing function of c on ( −∞, c

). Hence the limit k exists and k ≤ 0.

Step 2. We claim that the limit l := lim

R

exists and l = f

( ∞). Since R

is an increasing function on ( −∞, c

), l exists and l ≤ f

( ∞). If f

( ∞) > l, then by the standard theory of the continuous dependence on initial condition, there exists c

< c

and a > 0 such that (l + f

( ∞))/2 < f

(a) < f

( ∞) and f

(a) > 0. Then R

= f

(t

) ≥ f

(a) > (l + f

( ∞))/2 > l, a contradiction. Hence l = f

( ∞).

Step 3. We claim that k = 0. If not, then k < 0. Using (7.1), y

> 0 on [0, R

) and so y

(x) < k on [0, R

) for any c < c

. By passing to the limit, we have y

(x) ≤ k on [0, R

).

Hence

0 = f

( ∞) = lim

x→R⁻cα

y

(x) ≤ k < 0,

a contradiction.

This completes the proof.

Lemma 4.21 Let α > 0 and f

be the solution of (P

) with c < c

. Then there exists δ

> 0 such that f

has a zero in (0, T

) for all c ∈ (c

− δ

, c

).

Proof. First, by the standard theory of the continuous dependence on initial condition, Lemma 4.15 and Remark 5.1 of [5], there exists δ

> 0 such that for all c ∈ (c

− δ

, c

),

1/ √

4α + 2 < f

(t

) < 2/ √

α + 1 (7.7)

where t

is the first zero of f

. Note that δ

depends on α.

Next, for contradiction we suppose that f

does not vanish on (0, T

). Then f

< 0 on [0, T

). We divide the proof into several steps.

Step 1. We claim that there exists ^b t

∈ (t

, T

) such that f

( t ^b

) = f

(t

)/2 and ( t ^b

−t

) → ∞ as c → c

. By assumption, f

(t) < 0 on [t

, T

) and f

(t

) = 0, then there exists ^b t

∈ (t

, T

) such that f

( ^b t

) = f

(t

)/2. Since f

> 0, f

< 0, and f

< 0 on [t

, ^b t

], by (1.1) we have f

> 0 on [t

, ^b t

]. This implies that f

(t) > f

(t

) for all t ∈ (t

, t ^b

]. We compute that

f

(t

)/2 = −f

( ^b t

) + f

(t

)

= − ^{Z b}

tc

f

(t)dt

= − ^{Z b}

tc

[

Z

tc

f

(s)ds]dt

< − ^{Z b}

tc

[

Z

tc

f

(t

)ds]dt

= −[( t ^b

− t

)

/2]f

(t

),

which is equivalent to

b t

− t

> ^q −f

(t

)/f

(t

). (7.8)

By Lemma 4.20, we have f

(t

) → 0 as c → c

. Hence from (7.7) and (7.8) it follows that

t b

− t

→ ∞ as c → c

.

Step 2. We claim that there exists a

∈ (t

, ^b t

) such that f

(a

) = f

(t

)/2 and a

− t

≤ (8 ln 2)/ √

α + 1 for all c ∈ (c

− δ

, c

) for some δ

∈ (0, δ

). Indeed, by (1.1) and (7.7), we have

f

(t) ≥ −[(α + 1)/2]f

(t)f

(t)

≥ −[(α + 1)/2][1/(2 √

4α + 2)]f

(t)

≥ −[ √

α + 1f

(t)]/8 on [t

, ^b t

].

By an integration of the above inequality, we have

f

(t) ≥ e

f

(t

) on [t

, t ^b

]. (7.9)

On the other hand, by Step 1, we can choose δ

∈ (0, δ

) such that ( t ^b

−t

) ≥ (16 ln 2)/ √ α + 1 for all c ∈ (c

− δ

, c

). Then t

+ (8 ln 2)/ √

α + 1 ∈ (t

, t ^b

) and by (7.9) we have

f

(t

+ (8 ln 2)/ √

α + 1) ≥ f

(t

)/2. (7.10)

Noting that f

< 0 on [t

, t ^b

], it follows from (7.10) that there exists a

∈ (t

, ^b t

) such that f

(a

) = f

(t

)/2 and a

− t

≤ (8 ln 2)/ √

α + 1.

Step 3. We claim that there exists δ

∈ (0, δ

) such that a

− t

≥ ln (4/3)/ √

α + 1 for all c ∈ (c

− δ

, c

). Set A = (8 ln 2)/ √

α + 1. Noting that f

> 0 on [t

, t ^b

] and using Step 2, we have

−f

(t) = −

Z

tc

f

(s)ds

≤ −f

(t

)(t − t

)

≤ −f

(t

)(a

− t

)

≤ −Af

(t

) on [t

, a

].

It follows from the above inequality, (7.7), and (1.1) that

f

(t) = −[(α + 1)/2]f

(t)f

(t) + αf

(t)

≤ −[(α + 1)/2](2/ √

α + 1)f

(t) + αA

f

(t

)

= − √

α + 1f

(t) + αA

f

(t

) on [t

, a

].

By an integration, we obtain that

f

(t) ≤ f

(t

)e

+ [αA

f

(t

)/ √

α + 1](1 − e

) on [t

, a

].

Setting t = a

in the above inequality, we obtain

f

(t

)/2 ≤ f

(t

)e

+ [αA

f

(t

)/ √

α + 1](1 − e

),

which is equivalent to

[1/2 − (αA

f

(t

)/ √

α + 1)]/[1 − (αA

f

(t

)/ √

α + 1] ≥ e

. (7.11)

By Lemma 4.20, we have f

(t

) → 0 as c → c

. It follows that the left hand side of (7.11) approaches 1/2 as c → c

. Hence there exists δ

∈ (0, δ

) such that

3/4 ≥ e

for all c ∈ (c

− δ

, c

). This implies that a

− t

≥ ln (4/3)/ √

α + 1 for all c ∈ (c

− δ

, c

).

Step 4. We reach the conclusion by a contradiction. Set B = ln (4/3)/(2 √

α + 1). Recall that f

> 0 on [t

, ^b t

]. Then by Step 2 and Step 3 we have

−f

(a

) = −

Z

tc

f

(s)ds

≥ −

Z

tc

f

(a

)ds

= −[f

(t

)(a

− t

)]/2

≥ −Bf

(t

).

Since f

< 0 on [t

, T

], we have −f

(t) ≥ −Bf

(t

) on [a

, t ^b

]. From the fact that f

≥ 1/(2 √

4α + 2) on [t

, ^b t

], it follows that

f

(t) = −[(α + 1)/2]f

(t)f

(t) + αf

(t)

≥ −[(α + 1)/2][1/(2 √

4α + 2)]f

(t) + αB

f

(t

)

≥ −( √

α + 1/8)f

(t) + αB

f

(t

) on [a

, ^b t

].

By an integration, we obtain

f

(t) ≥ (f

(t

)/2)e

+ [8αB

f

(t

)/ √

α + 1](1 − e

) (7.12)

on [a

, ^b t

]. Noting that {a

− t

} is bounded for c ∈ (c

− δ

, c

) and using (7.7) and (7.8), we have

exp {− √

α + 1( t ^b

− a

)/8 } = exp{ √

α + 1(a

− t

)/8 } exp{− √

α + 1( ^b t

− t

)/8 }

< exp { √

α + 1(a

− t

)/8 } exp{− ^q −(α + 1)f

(t

)/f

(t

)/8 }

< M

exp {−M

[ −f

(t

)]

}

where M

, M

are positive constants independent of c for c ∈ (c

− δ

, c

). This implies that

lim

c→c⁻α

e

^b

/[ −f

(t

)] = 0 (7.13)

On the other hand, since {a

− t

} is bounded for c ∈ (c

− δ

, c

) and ( t ^b

− t

) → ∞ as c → c

, we can choose δ

∈ (0, δ

) such that

1 − e

^b

= 1 − e

e

^b

≥ 1/2 (7.14)

for all c ∈ (c

− δ

, c

). Therefore, it follows from (7.12), (7.13) and (7.14) that

f

( t ^b

) ≥ (f

(t

)/2)e

^b

+ [8αB

f

(t

)/ √

α + 1](1 − e

^b

)

≥ (f

(t

)/2)e

^b

+ 4αB

f

(t

)/ √ α + 1

= [f

(t

)]

 

 − 1 2

e

^b

−f

(t

) + 4αB

/ √ α + 1

 



> 0

for c ∈ (c

− δ

, c

) for some δ

∈ (0, δ

). We reaches a contradiction. This finishes the proof of the lemma.

4.8 The case for α ∈ (0, 1/3)

It follows from Theorem 4.17, Proposition 4.18, Lemma 4.19, and Theorem 6.2 of [5] that f

is of type (B) with T

= ∞ for c ∈ [0, ∞); f

is of type (C) for c ∈ (c

, 0); f

is of type (A);

and f

is either of type (E) or of type (F) for c < c

. We shall distinguish the type of f

for any c < c

.

Lemma 4.22 Let α ∈ (0, 1/3) and f

be the solution of (P

) with c < c

. Then there exist δ

> 0 such that f

is of type (F) for all c ∈ (c

− δ

, c

).

Proof. By Lemma 4.21, there exists δ

> 0 such that f

has a zero in (0, T

) for all c ∈ (c

− δ

, c

). Then from Lemma 4.1 it follows that there exists t ^e

> t

such that f

> 0 for all t ∈ ( ^e t

, T

), and thus f

is not of type (E) for all c ∈ (c

− δ

, c

). By Theorem 6.2 of [5], there is no solution of type (D) for α ∈ (0, 1/3). Hence f

is of type (F) for all c ∈ (c

− δ

, c

) by Lemma 4.19.

The following property is an interesting phenomenon for solutions with two critical points.

Proposition 4.23 Let α ∈ (0, 1/3) and f be the solution of (P

) defined on [0, T ) with

c < 0. Suppose that there exist t

and t

such that f

> 0 on [0, t

), f

< 0 on (t

, t

), and

f

> 0 on (t

, T ). Then f (t

) ≤ 0.