where 0 < σ ≤ 1 and p > 1. We are concerned with the existence of backward self-similar positive solutions of (1.1) in the form

3 The backward self-similar solutions for a nonlinear parabolic equation

3.1 Introduction

In this chapter, we study the following nonlinear parabolic equation

u t = u ^σ (∆u + u ^p ), x ∈ R ⁿ , t > 0, (1.1)

where 0 < σ ≤ 1 and p > 1. We are concerned with the existence of backward self-similar positive solutions of (1.1) in the form

u(x, t) = (T − t) ^−α ϕ( |x|

(T − t) ^β ), (1.2)

where T > 0 is given and the similarity exponents are necessarily given by

α = 1

p + σ − 1 , β = p − 1

2(p + σ − 1) . (1.3)

Set ξ = |x|/(T − t) ^β . It follows that u satisfies (1.1) if and only if ϕ satisfies the equation

ϕ ⁰⁰ + n − 1

ξ ϕ ⁰ − βξϕ ^−σ ϕ ⁰ + ϕ ^p − αϕ ^1−σ = 0, ϕ ⁰ (0) = 0, ϕ(0) ∈ R ⁺ , ξ > 0. (1.4) Eq. (1.1) has been studied fairly extensively over past years. Indeed, σ = 0 is the case of standard heat equation. For σ < 1, (1.1) can be rewritten as

( 1

1 − σ u ^1−σ ) t = ∆u + u ^p

which is the case of porous medium equation for σ ∈ (0, 1) and is the case of fast diffusion equation for σ < 0.

In all of these cases, much attentions have been paid to the blow-up behaviours of solutions

in order to understand the mechanism of thermal runaway in combustion problems. We refer

the readers to the references listed in the books of Bebernes et al. [3] and Samarskii et al.

[25].

Also, it is well-known that the backward self-similar solutions play an important role in studying the blow-up behaviours of solutions of the parabolic problems. For the existence of backward self-similar solutions, we refer the readers to [13, 26, 6] for the case σ = 0 and [23]

for the case σ < 0. For the case σ ∈ (0, 1), Ad”yutov [1] has shown that when n = 1, there are at least N − 1 distinct nonconstant globally bounded solutions of (1.4), where −N is the largest integer which is less or equal to −1/(2β); when n ≥ 2, Qi [24] has proved that there is a monotone decreasing positive global solution of (1.4) for 1 < p < p c (n), where p c (n) is defined to be

p _c (n) =

( ∞, n ≤ 2

(n + 2)/(n − 2), n ≥ 3 (1.5)

He also discussed the case of p > p _c (n). For the case σ = 1, Guo [16] has recently proved that when n = 1, there are at least N − 1 distinct nonconstant globally bounded solutions of (1.4). See also the paper of Lepin [20] and the references listed in the above mentioned books and papers.

In this chapter we consider the parameter range n ≥ 2, σ ∈ (0, 1] and 1 < p < p c (n). The main result is the following theorem

Theorem 3.1 Suppose that n ≥ 2, σ ∈ (0, 1] and 1 < p < p c (n). Then Eq. (1.4) has at least [ −[−1/(2β)]/2] distinct positive global solutions such that ϕ(0) > κ and ϕ(ξ) → 0 as ξ → ∞, where κ = α ^α and [x] is the largest integer which is less than or equal to x.

Combining Theorem 3.1 with the result of the mentioned references listed in the above,

there always exists a symmetric positive monotone self-similar solution U of (1.1) for σ ∈

(0, 1] and p > 1 with p 6= p c (n) in the form (1.2) such that U blows up only at the single point x = 0 at T for a given finite time T . Also, there are some other self-similar single- point blow-up patterns with different oscillations. Notice that any self-similar solution we constructed has the same number of critical points at any time before the blow-up time; yet the critical points of the self-similar solution merge together to a single point at the blow-up time.

This chapter is organized as follows. In Section 2, we give some preliminary results for the solutions of the initial value problem for Eq. (1.4). Then we construct in Section 4 a finite number of distinct bounded positive global solutions of (1.4). Finally, we give a conjecture in Section 4.

3.2 Preliminary

In this section, we shall study the positive solution of the following initial value problem (P):

ϕ ⁰⁰ + n − 1

ξ ϕ ⁰ − βξϕ ^−σ ϕ ⁰ + ϕ ^p − αϕ ^1−σ = 0, ξ > 0, (2.1)

ϕ ⁰ (0) = 0, ϕ(0) = η, (2.2)

where η > 0 is given. Let κ ≡ α ^α . Then ϕ = κ is the only trivial constant solution of (P).

It is trivial that there is a unique local solution ϕ of (P) for each given η > 0. Let [0, R) be the maximal existence interval of ϕ, where 0 < R ≤ ∞.

For convenience, we define

H(ξ) = 1

2 [ϕ ⁰ (ξ)] ² + G(ϕ(ξ)), ξ ∈ [0, R), where

G(t) =

Z _t

κ (s ^p − αs ^1−σ )ds, t ≥ 0.

Note that G(0) ∈ (0, ∞), G(∞) = ∞, G(t) > 0 for all t ≥ 0 except t = κ, and there is a unique κ > κ such that G( e κ) = G(0). e

Lemma 3.2 Let ϕ be a nonconstant global solution of (P). Then H is monotone increasing in ξ for all ξ > ^q (n − 1)ϕ ^σ (ξ)/β.

Proof. From

H ⁰ (ξ) = (βξϕ ^−σ (ξ) − n − 1

ξ )[ϕ ⁰ (ξ)] ² ≥ 0

for all ξ > ^q (n − 1)ϕ ^σ (ξ)/β, it follows that H is nondecreasing in ξ for all ξ > ^q (n − 1)ϕ ^σ (ξ)/β.

Note that H ⁰ (ξ) = 0 if and only if ϕ ⁰ (ξ) = 0. Hence H must be monotone increasing in ξ for all ξ > ^q (n − 1)ϕ ^σ (ξ)/β.

Let ϕ be a nonconstant globally bounded solution of (P). It follows from Lemma 3.2 that the limit

L := lim

ξ→∞ H(ξ) (2.3)

exists and L > 0. We also observe that any critical point of ϕ is a maximum point (minimum point) of ϕ if ϕ > κ (ϕ < κ, respectively).

We say that a global solution ϕ is monotone ultimately if ϕ ⁰ (ξ) has a fixed sign for all ξ sufficiently large.

Lemma 3.3 Suppose that ϕ is global and monotone ultimately. Then either ϕ → 0 or ϕ → ∞ as ξ → ∞.

Proof. By assumption, the limit l = lim _ξ→∞ ϕ(ξ) exists.

Suppose that l ∈ (0, +∞). Note that ϕ ⁰ (ξ) has a fixed sign for all ξ ≥ ξ 0 for some ξ 0 > 0.

It follows from

Z _∞

ξ

ϕ ⁰ (ξ) = l − ϕ(ξ ⁰ )

that there is a sequence ξ k → ∞ such that ϕ ⁰ (ξ k ) → 0 as k → ∞.

Now, dividing the equation (2.1) by ξ and integrating it from ξ ₀ to ξ _k , k ≥ 1, we obtain

Z _ξ

ξ

ϕ ⁰⁰

ξ dξ = ϕ ⁰ (ξ k )

ξ _k − ϕ ⁰ (ξ 0 ) ξ ₀ +

Z _ξ

ξ

1 ξ ² ϕ ⁰ dξ,

Z ξ

ξ

n − 1

ξ ² ϕ ⁰ dξ = (n − 1)

Z ξ

ξ

1 ξ ² ϕ ⁰ dξ,

Z _ξ

ξ

βϕ ^−σ ϕ ⁰ dξ =

( β/(1 − σ)[ϕ ^1−σ (ξ k ) − ϕ ^1−σ (ξ 0 )], if 0 < σ < 1 β ln[ϕ(ξ k )/ϕ(ξ 0 )], if σ = 1 .

It is easy to show that the above integrals are uniformly bounded for all k. On the other hand, the integral

Z _ξ

ξ

|ϕ ^p − αϕ ^1−σ |

ξ dξ

tends to + ∞ as k → ∞, if l 6= κ. Therefore, l = κ.

Now, choose Ξ sufficiently large such that

βξϕ ^−σ − n − 1

ξ > 0, ϕ(ξ) < κ, for all ξ e ≥ Ξ,

since ϕ → κ as ξ → ∞. It follows from Lemma 3.2, Eq. (2.3) and the definition of H that the limit lim _ξ→∞ ϕ ⁰ (ξ) exists and is equal to zero. Hence L = 0. This is a contradiction and the lemma is proved.

Let ϕ be the solution of (P) with ϕ(0) = η > 0 and let [0, R) be the maximal existence interval of ϕ, where 0 < R ≤ ∞.

For a given solution ϕ, we define

ρ(y) = ( y ξ 0

) ⁿ⁻¹ exp[ −β

Z _y

ξ

yϕ ^−σ (y)dy], ξ 0 ≤ y < R,

where ξ 0 < R is a fixed number. Then we have

ϕ ⁰ (ξ) = ρϕ ⁰ (ξ 0 ) ρ(ξ) − 1

ρ(ξ)

Z _ξ

ξ

ρ(y)[ϕ ^p (y) − αϕ ^1−σ (y)]dy, ξ 0 ≤ ξ < R. (2.4)

Lemma 3.4 Suppose that ϕ is a solution of (P) such that the limit ϕ(R ⁻ ) = lim _ξ→R

ϕ(ξ) exists. If R < ∞, then ϕ(R ⁻ ) = 0.

Proof. If 0 < ϕ(R ⁻ ) < ∞, then ϕ can be continued beyond R. Hence we can assume that ϕ(R ⁻ ) = + ∞, ϕ ⁰ > 0 and ϕ > κ, for all ξ ∈ [ξ 0 , R) for some ξ ₀ > 0. Noting that ϕ > κ in [ξ 0 , R), we have

ρ(ξ) = ( ξ ξ 0

) ⁿ⁻¹ exp[ −β

Z _ξ

ξ

yϕ ^−σ (y)dy]

≥ exp[−β

Z _ξ

ξ

yκ ^−σ dy]

≥ exp[−βκ ^−σ ξ ² ],

for all ξ ∈ [ξ ⁰ , R). From (2.4) and ϕ > κ in [ξ 0 , R), it follows that

ϕ ⁰ (ξ) ≤ ρϕ ⁰ (ξ 0 ) ρ(ξ)

≤ exp[βκ ^−σ ξ ² ]ρ(ξ ₀ )ϕ ⁰ (ξ ₀ )

≤ exp[βκ ^−σ R ² ]ρ(ξ 0 )ϕ ⁰ (ξ 0 ).

This leads to a contradiction and the lemma follows.

Lemma 3.5 Suppose ϕ is a solution of (P) such that lim _ξ→R

ϕ(ξ) = 0. If R < ∞, then ϕ ⁰ (R ⁻ ) < 0.

Proof. By assumption, there is a ξ 0 ∈ [0, R) such that ϕ ⁰ < 0 and ϕ < κ for all ξ ∈ [ξ 0 , R).

Since σ ∈ (0, 1], p > 1 and ϕ(ξ) → 0 as ξ → R ⁻ , there exists ξ 1 > ξ 0 such that

pϕ ^p − n − 1

ξ ² ϕ − βϕ ^{e 1−σ} < 0, (2.5)

βξϕ ^1−σ − n − 1

ξ ϕ > 0, (2.6)

1

2 βξϕ ^−σ > n − 1

ξ (2.7)

for all ξ ∈ [ξ 1 , R) where β = β + α(1 ^e − σ). We claim that there exists ξ 2 ≥ ξ 1 such that ϕ ⁰⁰ (ξ 2 ) ≤ 0. For contradiction, we assume that ϕ ⁰⁰ (ξ) > 0 for all ξ ∈ (ξ 1 , R). Then it follows from (2.1) and (2.7) that

βξϕ ^−σ ϕ ⁰ > ϕ ^p − αϕ ^1−σ + n − 1 ξ ϕ ⁰

> −αϕ ^1−σ + n − 1 ξ ϕ ⁰

> −αϕ ^1−σ + 1

2 βξϕ ^−σ ϕ ⁰ for all ξ ∈ [ξ ¹ , R), which is equivalent to

ϕ ⁰

ϕ > −( 2α β ) 1

ξ in [ξ 1 , R).

By an integration of the above differential inequality, we end up with the estimate

ϕ(ξ) < ϕ(y)( y ξ ) ^2α/β

for all ξ ₁ ≤ ξ < y < R. Letting y → R ⁻ in the above estimate, we obtain that ϕ ≡ 0 in [ξ 1 , R), a contradiction. Hence there is a ξ 2 ≥ ξ ¹ such that ϕ ⁰⁰ (ξ 2 ) ≤ 0.

Now, differentiating Eq. (2.1), we obtain that

ϕ ⁰⁰⁰ (ξ) = −[pϕ ^p − n − 1

ξ ² ϕ − βϕ ^{e 1−σ} ] ϕ ⁰

ϕ − βσξϕ ^−1−σ [ϕ ⁰ ] ² + [βξϕ ^1−σ − n − 1 ξ ϕ] ϕ ⁰⁰

ϕ .

Hence it follows from (2.5) and (2.6) that ϕ ⁰⁰⁰ (ξ) < 0 for all ξ ∈ [ξ ² , R), which turns out that

ϕ ⁰⁰ (ξ) < ϕ ⁰⁰ ( ξ 2 + R 2 ) < 0

for all ξ ∈ [(ξ 2 + R)/2, R). This implies that ϕ ⁰ (R ⁻ ) < 0 and the lemma is proved.

Next, we shall prove the following property of continuous dependence on initial value for which the corresponding solution is decreasing to zero at finite ξ. Notice that this property does not follows from the standard theory of ordinary differential equations, since it is singular when ϕ vanishes.

Lemma 3.6 Suppose that the solution ϕ 0 of (P) with ϕ 0 (0) = η 0 has only finitely many critical points in [0, R 0 ) and is monotone decreasing to zero in a left neighborhood of R 0 for some R 0 < ∞. Then there is δ > 0 small enough such that any solution ϕ of (P) with ϕ(0) = η has the same number of critical points in [0, R) as ϕ ₀ and is monotone decreasing to zero in a left neighborhood of R for some R < ∞, if η ∈ (η ⁰ − δ, η ⁰ + δ).

Proof. First, we assume n ≥ 2 and consider the quantity

F (ξ) = [α + n − 2

2 β]ϕ(ξ) + βξϕ ⁰ (ξ). (2.8)

Then we compute that

F ⁰ (ξ) − βξϕ ^−σ F (ξ) = − n − 2

2 β ² ξϕ ^1−σ − βξϕ ^p + [α − n − 2

2 β]ϕ ⁰ . (2.9)

Note that α − [(n − 2)/2]β > 0 since 1 < p < p ^c (n). From Eq. (2.8) and (2.9) it follows that if there exists Ξ ∈ [0, R) such that F (Ξ) ≤ 0 and ϕ(Ξ) > 0, then F (ξ) < 0 for all ξ ∈ (Ξ, R). We also have ϕ ⁰ 0 (R ⁻ ₀ ) < 0 by Lemma 3.5. Hence F 0 (R ₀ ⁻ ) < 0, where

F ₀ (ξ) = [α + n − 2

2 β]ϕ ₀ (ξ) + βξϕ ⁰ ₀ (ξ).

Let ξ ∈ [0, R ⁰ ) be the critical point of ϕ 0 such that ϕ ⁰ ₀ < 0 in (ξ, R 0 ). Notice that F 0 (ξ) > 0

and there is a unique ξ 0 > ξ such that F 0 (ξ 0 ) = 0 and F 0 (ξ) < 0 for all ξ ∈ (ξ ⁰ , R 0 ).

Next, we fix any ξ 1 ∈ (ξ 0 , R 0 ). By the standard continuous dependence property, there is a δ > 0 small enough such that ϕ has the same number of critical points as ϕ ₀ in [0, ξ ₁ ], ϕ(ξ) > 0 for all ξ ∈ [0, ξ ¹ ], and F (ξ 1 ) < 0 for any ϕ with initial value η ∈ (η ⁰ − δ, η ⁰ + δ).

Now, we fix any η ∈ (η ⁰ − δ, η ⁰ − δ). Let ϕ be the corresponding solution of (P). We claim that ϕ ⁰ (ξ) < 0 for any ξ > ξ 1 as long as ϕ(ξ) > 0. For contradiction, we suppose that there exists ξ 2 > ξ 1 such that ϕ > 0 in [0, ξ 2 ], ϕ ⁰ < 0 in [ξ 1 , ξ 2 ), and ϕ ⁰ (ξ 2 ) = 0. Then F (ξ 2 ) > 0 by the definition of F . On the other hand, by (2.9), we have

(ρF ) ⁰ (ξ) < 0 for all ξ ∈ (ξ 1 , ξ 2 ), (2.10)

where ρ(ξ) = exp[ −β ^R ξ ^ξ

sϕ ^−σ (s)ds]. It follows that F (ξ 2 ) < 0, a contradiction. Therefore, ϕ ⁰ (ξ) < 0 for any ξ ≥ ξ 1 as long as ϕ(ξ) > 0.

Finally, we claim that ϕ cannot be global. Otherwise, if ϕ is positive globally, then by (2.10) and noting that ρ < 1, there is a positive constant σ such that

F (ξ) < −σ for all ξ ≥ ξ 1 . (2.11)

Recall (2.8). By an integration of (2.11), we obtain that

ϕ(ξ) ≤ (ξ 1 /ξ) ^{e α/β} ϕ(ξ 1 ) − σ/ α + (σ/ ^e α)(ξ e 1 /ξ) ^{α/β e} for all ξ ≥ ξ 1 ,

where α = α + [(n e − 2)/2]β. By passing to infinity, we reach a contradiction. This completes the proof.

Next, let us recall from Lemma 1 on p. 223 of [25] that the solution f to the problem

f ⁰⁰ + n − 1

ξ f ⁰ + f ^p = 0, ξ > 0, (2.12)

f ⁰ (0) = 0, f (0) = 1, (2.13)

is decreasing to zero at some finite ξ _∗ and f ⁰ (ξ _∗ ) < 0 for p ∈ (1, p c (n)).

Lemma 3.7 Suppose that n ≥ 2, σ ∈ (0, 1] and 1 < p < p c (n). If η is sufficiently large, then the solution ϕ of (P) is monotone decreasing to zero at some finite R.

Proof. Define

ψ(ξ) = ϕ(ξ/η ^a )/η, a = (p − 1)/2.

Then ψ satisfies

(ψ ⁰⁰ + n − 1

ξ ψ ⁰ + ψ ^p ) − η ^1−σ−p (αψ ^1−σ + βξψ ^−σ ψ ⁰ ) = 0 (2.14)

ψ ⁰ (0) = 0, ψ(0) = 1. (2.15)

Note that as η → ∞, (2.12)-(2.13) is the limiting problem of (2.14)-(2.15).

We choose ξ 0 ∈ (0, ξ ∗ ) such that α + [(n − 2)/2]β + [βξ ⁰ f ⁰ (ξ 0 )/f (ξ 0 )] < 0. By the theory of the continuous dependence on parameters, for any η >> 1, we have

ψ ⁰ < 0, ψ > 0 in (0, ξ 0 ), and α + n − 2

2 β + βξ 0 ψ ⁰ (ξ 0 )

ψ(ξ 0 ) < 0. (2.16)

Then

F (ξ 0 /η ^a ) = ηψ(ξ 0 )[α + n − 2

2 β + βξ 0 ψ ⁰ (ξ 0 ) ψ(ξ 0 ) ] < 0

By Lemma 3.6, this implies that ϕ is decreasing to zero at some finite R and the lemma follows.

Lemma 3.8 If η > κ, then ϕ has at least one intersection with κ. In fact, if ϕ ⁰ (ξ 0 ) = 0 for some ξ 0 ≥ 0, then there is ξ ¹ > ξ 0 such that ϕ(ξ 1 ) = κ.

Proof. First, we assume that ϕ(ξ 0 ) > κ. If ϕ(ξ) 6= κ for all ξ > ξ ⁰ , then ϕ(ξ) > κ for all

ξ > ξ 0 . Since ϕ ⁰⁰ (ξ 0 ) < 0, ϕ ⁰ (ξ) < 0 for ξ > ξ 0 with ξ sufficiently close to ξ 0 . We claim that

ϕ ⁰ (ξ) < 0, for all ξ > ξ 0 . Otherwise, there is the first critical point ξ 1 of ϕ after ξ 0 . Since

ϕ(ξ 1 ) > κ, ϕ ⁰⁰ (ξ 1 ) < 0. This implies that ϕ ⁰ (ξ) > 0 for ξ < ξ 1 with ξ sufficiently close to ξ 1 , a contradiction. Hence we have ϕ ⁰ < 0 and ϕ > κ for all ξ > ξ ₀ . This again contradicts to Lemma 3.3. Therefore, there is a ξ 1 > ξ 0 such that ϕ(ξ 1 ) = κ.

The proof for the case ϕ(ξ 0 ) < κ is similar. Hence the lemma follows.

In th sequel, we set d = ^q (n − 1) κ ^{e σ} /β.

Lemma 3.9 Suppose that there is a critical point ξ ₀ ≥ d such that ϕ(ξ 0 ) ≥ κ. Then ϕ is ^e monotone decreasing in [ξ 0 , R) with R < ∞ and ϕ(ξ) → 0 as ξ → R ⁻ .

Proof. Note that ξ 0 must be a maximum point. By Lemma 3.8, we can choose the first ξ 1

such that ξ 1 ≥ ξ 0 , ϕ(ξ 1 ) = e κ and ϕ ⁰ (ξ 1 ) ≤ 0. By Lemma 3.2, H(ξ) is monotone increasing after ξ 1 as long as ϕ ≤ κ. If ϕ is not monotone decreasing after ξ ^e 1 , then ϕ has at least one minimum point ξ 2 > ξ 1 such that ϕ(ξ 2 ) < κ by Lemma 3.8. Hence we have

G( κ) = G(ϕ(ξ e 1 )) ≤ H(ξ ¹ ) < H(ξ 2 ) = G(ϕ(ξ 2 )) < G(0),

a contradiction. Therefore, ϕ is monotone decreasing in [ξ 1 , R).

If ϕ is global, then ϕ → 0 as ξ → ∞ by Lemma 3.3. Hence L = G(0). But, by the property of H, we have L > H(ϕ(ξ 1 )). This is a contradiction, since G(0) = G( κ) e ≤ H(ϕ(ξ ¹ )).

Therefore, R < ∞ and ϕ(ξ) → 0 as ξ → R ⁻ by Lemma 3.4.

3.3 Existence for η > κ

The following result is one of the key lemmas in this chapter.

Lemma 3.10 Suppose that ϕ ⁰ (ξ 0 ) = 0 and ϕ(ξ 0 ) < κ for some ξ 0 ≥ √

2d. Then there is a small λ > 0, independent of ξ 0 for all ξ 0 ≥ √

2d, such that ϕ(ξ _∗ ) ≥ ^e κ for the first critical

point ξ _∗ > ξ 0 (if it exists), provided that ϕ(ξ 0 ) ∈ (0, λ).

Proof. Choose λ ∈ (0, κ/4) such that

B(λ) := _q G(0) − G(λ)

2[G(λ) − G(κ/2)] < βκ ^2−σ

16 ^q 2G(0) (3.1)

Note that the quantity B(t) in (3.1) is increasing in t for t ∈ (0, κ/4).

Suppose that η := ϕ(ξ 0 ) is small and that ϕ(ξ _∗ ) < κ for the first critical point ξ e _∗ > ξ 0 of ϕ. Recall that

H(ξ) = 1

2 [ϕ ⁰ (ξ)] ² + G(ϕ(ξ)).

From ξ 0 ≥ √

2d and Lemma 3.2, it follows that H(ξ) is monotone increasing as long as ξ ∈ [ξ 0 , R) and ϕ(ξ) ≤ κ. Choose ξ ^e 0 < ξ 1 < ξ 2 < ξ _∗ so that ϕ(ξ 1 ) = κ/2 and ϕ(ξ 2 ) = κ.

Then by the monotonicity of H we have

G(η) < 1

2 [ϕ ⁰ (ξ 1 )] ² + G( κ

2 ) = H(ξ 1 ) < H(ξ 2 ) = 1

2 [ϕ ⁰ (ξ 2 )] ² < G(ϕ(ξ _∗ )) < G(0). (3.2)

Since H(ξ) is monotone increasing for ξ ∈ [ξ 1 , ξ 2 ] and G(t) is decreasing in [κ/2, κ], we have

ϕ ⁰ (ξ 2 ) > ϕ ⁰ (ξ) > ϕ ⁰ (ξ 1 ) > 0 for all ξ ∈ [ξ 1 , ξ 2 ]. (3.3)

Hence we obtain that

κ = κ/2 +

Z _ξ

ξ

ϕ ⁰ (ξ)dξ ≤ κ/2 + ϕ ⁰ (ξ 2 )(ξ 2 − ξ ¹ )

and so

ξ ₂ − ξ 1 ≥ κ 2

1

ϕ ⁰ (ξ 2 ) . (3.4)

Since G(ϕ(ξ _∗ )) < G(0), it follows from (3.2) and (3.4) that

ξ 2 − ξ 1 ≥ κ 2

q 1

2G(0) . (3.5)

Similarly, from

κ/2 = η +

Z _ξ

ξ

ϕ ⁰ (ξ)dξ ≤ η + ϕ ⁰ (ξ ₁ )ξ ₁ ,

it follows that

ξ 1 ≥ κ 4

1

ϕ ⁰ (ξ 1 ) (3.6)

if η ≤ κ/4.

Now, recall that

H ⁰ (ξ) = (βξϕ ^−σ − n − 1

ξ )[ϕ ⁰ (ξ)] ² ≥ 1

2 βξϕ ^−σ [ϕ ⁰ (ξ)] ²

as long as ξ ≥ ξ 0 and ϕ(ξ) ≤ κ. Hence by (3.2), (3.3), (3.5), and (3.6) we have ^e

H(ξ 2 ) − H(ξ ¹ ) =

Z _ξ

ξ

(βξϕ ^−σ − n − 1

ξ )[ϕ ⁰ (ξ)] ² dξ

≥

Z _ξ

ξ

1

2 βξϕ ^−σ [ϕ ⁰ (ξ)] ² dξ

≥ 1

2 βκ ^−σ [ϕ ⁰ (ξ 1 )] ²

Z _ξ

ξ

ξdξ

= β

4κ ^σ [ϕ ⁰ (ξ 1 )] ² (ξ 2 − ξ 1 )(ξ 2 + ξ 1 )

≥ βκ ^2−σ 16 ^q 2G(0)

q 2[G(η) − G(κ/2)]

≥ G(0) − G(η),

if η ≤ λ. Hence

H(ξ 2 ) = [H(ξ 2 ) − H(ξ 1 )] + H(ξ 1 )

≥ [G(0) − G(η)] + G(η)

= G(0).

From the monotonicity of H, it follows that ϕ(ξ _∗ ) ≥ κ. This leads to a contradiction and ^e

the lemma is proved.

Now, we shall study the solution for η near κ. Write η = κ + ε. We shall consider the case when η > κ so that ε > 0 and is small. If we write ϕ(ξ) = κ + εv(ξ), then, as ε → 0, v satisfies the limiting equation

v ⁰⁰ + n − 1

ξ v ⁰ − β

κ ^σ ξv ⁰ + 1

κ ^σ v = 0, ξ > 0, v ⁰ (0) = 0, v(0) = 1.

It is well-known (cf. [25]) that v has exactly N zeros in (0, ∞), where −N is the largest integer which is less or equal to −1/(2β) = −(p + σ − 1)/(p − 1). Notice that N ≥ 2, since p > 1 and σ ∈ (0, 1]. By the standard theory on continuous dependence, it is easy to show that for η > κ which is sufficiently close to κ, the solution ϕ of (P) intersects κ at least N times.

Now, we ready to prove the main result of this paper as follows.

Theorem 3.11 Suppose that n ≥ 2, σ ∈ (0, 1] and 1 < p < p ^c (n). The problem (P) has at least N 1 distinct positive global solutions such that ϕ(0) > κ and ϕ(ξ) → 0 as ξ → ∞, where N 1 is the largest integer which is less than or equal to N/2.

Proof. For each kappa ∈ {0, 1, ..., N ¹ − 1}, let

I k = {η > κ | ϕ has at least k + 1 minimum in (0, R)}.

Notice that I _k 6= ∅, since the solution ϕ of (P) intersects κ at least N times if η > κ and η is sufficiently close to κ. Also, it follows from Lemma 3.7 that I k is bounded above. Set µ k = sup I k . Then µ k is well-defined. Let ϕ k be the solution of (P) with ϕ k (0) = µ k .

We claim that ϕ k has exactly k minima and ϕ k is a positive global solution of (2.1) with

ϕ k (ξ) → 0 as ξ → ∞. It is clear that ϕ k cannot have more than k minima by the theory of

continuous dependence. Then from Lemma 3.4 and 3.6 it follows that ϕ k is globally defined.

Suppose that ϕ _k is global and unbounded. Since ϕ ⁰ _k > 0 in (ξ ₀ , ∞) for some ξ 0 ≥ 0 and ϕ(ξ) → ∞ as ξ → ∞, there is K ∈ (ξ ⁰ , ∞) such that K > √

2d and ϕ k (K) > κ + 1. Then e

by the continuous dependence there is a δ > 0 small enough such that any solution ϕ of (P) with η ∈ (µ k − δ, µ k ) has the same number of minima as ϕ k in (0, K) and ϕ ⁰ (K) > 0, ϕ(K) > κ + 1/2. If ϕ e ⁰ > 0 in (K, ∞), then ϕ has at most k minima, a contradiction.

Otherwise, there is the first critical point ξ ₁ > K. Since ϕ(ξ ₁ ) > ϕ(K) > κ + 1/2, ϕ has at e

most k minima by Lemma 3.9. It is again a contradiction. Therefore, ϕ k is bounded.

Suppose that ϕ k has fewer than k minima. Then by Lemma 3.3 we have ϕ k (ξ) → 0 as ξ → ∞. Choose K sufficiently large such that K > √

2d, ϕ ⁰ _k (K) < 0 and ϕ k (K) < λ/2, where λ is the positive constant given in Lemma 3.10. By the continuous dependence on initial value, there is δ > 0 such that any solution ϕ of (P) with η ∈ (µ ^k − δ, µ ^k ) has the same number of minima as ϕ k in (0, K) and ϕ(K) < λ, ϕ ⁰ (K) < 0. Then, by Lemmas 3.9 and 3.10, ϕ has at most one more minima after K, a contradiction. Hence ϕ k has exactly k minima.

We have shown that ϕ k is a global positive solution of (2.1) such that it has exactly k minima. Moreover, ϕ k (ξ) → 0 as ξ → ∞ by Lemma 3.3. The proof is complete.

In particular, since N 1 ≥ 1 for any p > 1, we obtain the following result.

Corollary 3.12 Suppose that n ≥ 2 and 1 < p < p ^c (n). There is an η > κ such that the

solution of (P) exists globally and is monotone decreasing to zero.

3.4 Conclusion

In this section, we shall show that when n = 1, Eq. (2.1) has N − 1 − N ¹ distinct positive global solutions such that ϕ(0) ∈ (0, κ) and ϕ(ξ) → 0 as ξ → ∞. Combining with Theorem 3.11, this implies that when n = 1, Eq. (2.1) has at least N − 1 distinct positive global solutions such that ϕ(ξ) → 0 as ξ → ∞. Although this result is the same as the one in [1], our method is different.

Lemma 3.13 Every solution of (P) is bounded.

Proof. Suppose first that ϕ is nonglobal, i.e., R < ∞. We claim that ϕ is bounded. This is trivial when ϕ has only finitely many critical points. Indeed, from Lemma 3.4 it follows that ϕ(ξ) → 0 as ξ → R ⁻ . For contradiction, we assume that there is an unbounded solution ϕ of (P). Then there exist a monotone increasing sequence {ξ k } in (0, R) such that ξ k → R ⁻ as k → ∞ and

ϕ(ξ _2i−1 ) = κ + 1, ϕ(ξ 2i ) = κ, ϕ ⁰ < 0 in [ξ _2i−1 , ξ 2i ] for all i ≥ 1.

Then it follows from the mean value theorem that

ϕ ⁰ (y i ) = −1 ξ _2i − ξ 2i−1

(4.1)

for some y i ∈ (ξ 2i−1 , ξ 2i ), i = 1, 2, . . .. Hence there is an i 0 sufficiently large such that F (y i

) <

0, where F is defined in (2.8). Hence ϕ is decreasing to zero after y _i

, a contradiction.

Therefore, ϕ is bounded if R < ∞.

Next, it follows from Lemma 3.9 that ϕ is bounded, if ϕ is global and has infinitely many critical points. For contradiction, suppose that there is an unbounded global solution ϕ.

Then we can choose ξ 0 > 1 such that ϕ > 1 and ϕ ⁰ > 0 in (ξ 0 , ∞).

We claim that there exists ξ 1 > ξ 0 such that ϕ ⁰⁰ > 0 in (ξ 1 , ∞). Indeed, since σ ∈ (0, 1], p > 1 and ϕ(ξ) → ∞ as ξ → ∞, there exists ξ 1 > ξ ₀ such that

pϕ ^p − n − 1

ξ ² ϕ − βϕ ^{e 1−σ} > 0 (4.2)

for all ξ ≥ ξ 1 where β = β + α(1 ^e − σ). Differentiating (2.1), we obtain that

ϕ ⁰⁰⁰ (ξ) = −[pϕ ^p − n − 1

ξ ² ϕ − βϕ ^{e 1−σ} ] ϕ ⁰

ϕ − βσξϕ ^−1−σ [ϕ ⁰ ] ² + [βξϕ ^1−σ − n − 1 ξ ϕ] ϕ ⁰⁰

ϕ . (4.3)

Hence it follows that ϕ ⁰⁰⁰ (ξ) < 0 if ξ ≥ ξ ¹ and ϕ ⁰⁰ (ξ) = 0.

Suppose that there is a ξ 2 ≥ ξ ¹ such that ϕ ⁰⁰ (ξ 2 ) = 0. Then ϕ ⁰⁰⁰ (ξ 2 ) < 0 and so ϕ ⁰⁰ (ξ) < 0 for ξ > ξ 2 with ξ − ξ 2 sufficiently small. Hence

ϕ ⁰⁰ (ξ) < 0 for all ξ > ξ ₂ . (4.4)

Otherwise, ϕ ⁰⁰⁰ (ξ) ≥ 0 for the first ξ > ξ ² such that ϕ ⁰⁰ (ξ) = 0, a contradiction. Thus (4.4) holds.

It follows from (4.4) that ϕ ⁰ (ξ) ≤ ϕ ⁰ (ξ 2 ) for all ξ ≥ ξ 2 and so ϕ grows to infinity at most linearly. Noting that σ ∈ (0, 1], we have

βξϕ ^1−σ − n − 1

ξ ϕ ≥ βξ − n − 1

ξ ϕ > 0 for all ξ ≥ ξ ³ (4.5)

for some ξ 3 > ξ 2 . From (4.2)-(4.5), it follows that ϕ ⁰⁰⁰ (ξ) < 0 for all ξ ≥ ξ 3 , and so

ϕ ⁰⁰ (ξ) < ϕ ⁰⁰ (ξ 3 ) < 0 for all ξ ≥ ξ 3 .

This implies that ϕ ⁰ (ξ) → −∞ as ξ → ∞, a contradiction. Therefore we conclude that ϕ ⁰⁰ (ξ)

has a fixed sign for all ξ ≥ ξ ¹ .

If ϕ ⁰⁰ (ξ) < 0 for all ξ ≥ ξ 1 , then it follows the same argument that ϕ ⁰ (ξ) → −∞ as ξ → ∞.

This is again a contradiction. Therefore we obtain that

ϕ ⁰⁰ (ξ) > 0 in (ξ 1 , ∞). (4.6)

Hence ϕ ⁰ (ξ) ≥ ϕ ⁰ (ξ 1 ) > 0 for all ξ ≥ ξ 1 and so ϕ grows to infinity at least linearly.

On the other hand, by (2.1) and (4.6), we have

βξϕ ^−σ ϕ ⁰ > ϕ ^p − αϕ ^1−σ + n − 1

ξ ϕ ⁰ > 1

2 ϕ ^p (4.7)

for all ξ ≥ ξ 4 for some ξ ₄ > ξ ₁ , since p > 1, σ ∈ (0, 1], ϕ ⁰ > 0 on [ξ ₁ , + ∞) and ϕ(ξ) → ∞ as ξ → ∞. Choose ε ∈ (0, p + σ − 1). From (4.7) it follows that

( ϕ ^−p−σ+1

−p − σ + 1 ) > 1

2βξ ≥ 1

2βξ ^1+ε , (4.8)

for all ξ ≥ ξ ⁴ . By an integration of (4.8), we end up with the estimate

ϕ(ξ) ≤ ( p + σ − 1

2βε ) ^{−1/(p+σ−1)} ξ ^{ε/(p+σ−1)} , for all ξ ≥ ξ 4 . This leads to a contradiction and the lemma follows.

Theorem 3.14 Assume n = 1. Eq. (2.1) has at least N − 1 − N 1 distinct positive global solutions such that ϕ(0) ∈ (0, κ) and ϕ(ξ) → 0 as ξ → ∞, where N ¹ is the integer defined in Theorem 3.11.

Proof. For each k ∈ {0, 1, ..., N − 2 − N 1 }, let

J _k = {η ∈ (0, κ) | ϕ has at least k + 1 minimum in (0, R)}.

Note that J k 6= ∅. Set ν ^k = inf J k . Then ν k is well-defined and ν k ≥ λ > 0, since η 6∈ J ^k for

all η ∈ (0, λ) by Lemmas 3.9, 3.10 and 3.13. Thus the proof follows by the same reasoning

as the one for Theorem 3.11.

From the proof of Theorem 3.14, it follows that if we can show that the constant λ in Lemma 3.10 is independent of ξ ₀ for all ξ ₀ ≥ 0, then we have the following conjecture.

Conjecture 3.15 Assume n ≥ 2. Eq. (2.1) has at least N − 1 − N 1 distinct positive global

solutions such that ϕ(0) ∈ (0, κ) and ϕ(ξ) → 0 as ξ → ∞, where N 1 is the integer defined

in Theorem 3.11.