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

2 Existence of Forward Self-similar Solutions

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

ϕ

+ n − 1

ξ ϕ

+ ϕ

+ αϕ

+ βξϕ

ϕ

= 0, ξ > 0, (2.1)

ϕ

(0) = 0, ϕ(0) = η, (2.2)

where η > 0 is given. Note that there is no constant solution of (P ). From the local existence and uniqueness theorem of ordinary differential equations it follows that there is a unique positive local solution ϕ of (P ) for each given η > 0. For convenience, let [0, R) be the maximum existence interval of ϕ such that ϕ > 0, where 0 < R ≤ ∞.

Define

ρ(y) = exp {β

Z _y

0

ξϕ

(ξ)dξ }. (2.3)

From (2.1) it follows that

(ξ

ρ(ξ)ϕ

(ξ))

= −ξ

ρ(ξ)[ϕ

(ξ) + αϕ

(ξ)], (2.4) and so

ϕ

(ξ) = − 1 ξ

ρ(ξ)

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy , ξ > 0. (2.5) Hence ϕ(ξ) is monotone decreasing in [0, R).

Theorem 2.1 For any η > 0, the local solution ϕ(ξ) of (P ) can be continued globally so that R = ∞.

Proof. On the contrary, suppose that R < ∞. Then ϕ(ξ) → 0 as ξ → R

. We shall divide our discussion into three cases. Each one leads a contradiction.

Case 1:

lim

ξ→R⁻

ϕ

(ξ) = −∞. (2.6)

In this case, there exists a sequence {ξ

} in [0, R) such that ξ

→ R

, ϕ(ξ

) → 0

, and ϕ

(ξ

) → −∞ as k → ∞. We note that the limit of ρ(ξ) as ξ → R

exists and is either infinity or positive finite.

First, we suppose that lim

ρ(ξ) = ∞. From (2.5) it follows that

k→∞

lim

Z _ξk

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = ∞.

For convenience, we set g(ξ) :=

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy .

Since g(ξ) is an increasing continuous function in ξ, we obtain that

ξ→R

lim

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = ∞.

Applying L’Hˆopital’s Rule, we compute that

ξ→R

lim

ϕ

(ξ) = − lim

ξ→R⁻

1 ξ

ρ(ξ)

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy .

= − lim

ξ→R⁻

ξ

ρ(ξ)[ϕ

(ξ) + αϕ

(ξ)]

(n − 1)ξ

ρ(ξ) + ξ

ρ(ξ)βξϕ

(ξ)

= − lim

ξ→R⁻

ξ[ϕ

(ξ) + αϕ(ξ)]

(n − 1)ϕ

(ξ) + βξ

= 0.

This contradicts with (2.6).

Next, we consider the case that lim

ρ(ξ) = B, 0 < B < ∞. From (2.5) it follows that

k→∞

lim

Z _ξk

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = ∞, and so

ξ→R

lim

Z _ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = ∞.

From (2.5), we have

ξ→R

lim

ϕ

(ξ) = − lim

ξ→R⁻

1 ξ

ρ(ξ)

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = −∞.

It follows from (2.1) that

ξ→R

lim

ϕ

(ξ) = − lim

ξ→R⁻

{ϕ

(ξ) + ϕ

(ξ)[αϕ(ξ) + (βξ + n − 1

ξ ϕ

(ξ))ϕ

(ξ)] }

= ∞.

This contradicts with (2.6). Hence this is impossible.

Case 2:

lim

ξ→R⁻

ϕ

(ξ) = −M, 0 < M < ∞. (2.7)

In this case, there exists a sequence {ξ

} in [0, R) such that ξ

→ R

, ϕ(ξ

) → 0

,

and ϕ

(ξ

) → −M as k → ∞. If lim

ρ(ξ) = ∞, then, as in Case 1, we have

lim

ϕ

(ξ) = 0. This contradicts with (2.7).

Suppose that lim

ρ(ξ) = B, 0 < B < ∞. Then from (2.5) it follows that

k→∞

lim

Z _ξk

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = BM R

, and so

ξ→R

lim

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = BM R

. Therefore, lim

ϕ

(ξ) = −M. It follows from (2.1) that

ξ→R

lim

ϕ

(ξ) = − lim

ξ→R⁻

{ n − 1

ξ ϕ

(ξ) + ϕ

(ξ) + ϕ

(ξ)[αϕ(ξ) + βξϕ

(ξ)] }

= ∞. (2.8)

Moreover, by (2.1), lim

ϕ(ξ)ϕ

(ξ) = ∞. Differentiating (2.1) we have ϕ

(ξ) = −{ n − 1

ξ ϕ

(ξ) − n − 1

ξ

ϕ

(ξ) + pϕ

(ξ)ϕ

(ξ) + α(1 − σ)ϕ

(ξ)ϕ

(ξ) + βϕ

(ξ)[ϕ(ξ)ϕ

(ξ) − σξ(ϕ

(ξ))

+ ξϕ(ξ)ϕ

(ξ)] }.

Then we compute that

ξ→R

lim

ϕ

(ξ) = −∞, a contradiction to (2.8).

Case 3:

lim

ξ→R⁻

ϕ

(ξ) = 0.

Since ϕ

(ξ) < 0 in [0, R), we have

ξ→R

lim

ϕ

(ξ) = 0. (2.9)

We claim that

ξ→R

lim

ρ(ξ) = ∞ and lim

ξ→R⁻

ρ(ξ)ϕ(ξ) = ∞. (2.10)

If ρ(ξ) → B as ξ → R

for some B ∈ (0, ∞), then from (2.5) it follows that

ξ→R

lim

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = 0.

This is impossible, since y

ρ(y)[ϕ

(y) + αϕ

(y)] > 0 for all y > 0. Therefore,

ξ→R

lim

ρ(ξ) = ∞.

It follows from the definition of ρ(ξ) that β

Z ξ

0

yϕ

(y)dy → ∞ as ξ → R

. Then

ξ→R

lim

β

yϕ

(y)dy

−ln[ϕ(ξ)] = lim

ξ→R⁻

βξϕ

(ξ)

−ϕ

(ξ)ϕ

(ξ) = ∞.

From this and (2.3), we obtain that

ξ→R

lim

ρ(ξ)ϕ(ξ) = lim

ξ→R⁻

exp {β

Z _ξ

0

yϕ

(y)dy + ln[ϕ(ξ)] } = ∞.

Hence (2.10) is proved.

Now, we suppose first that

ξ→R

lim

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = ∞.

Using (2.5) and L’Hˆopital’s Rule, we obtain that

ξ→R

lim

ϕ

(ξ)

ϕ(ξ) = − lim

ξ→R⁻

{ 1 ξ

ρ(ξ)ϕ(ξ)

Z _ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy }

= − lim

ξ→R⁻

{ ξ

ρ(ξ)[ϕ

(ξ) + αϕ

(ξ)]

(n − 1)ξ

ρ(ξ)ϕ(ξ) + ξ

ρ(ξ)βξϕ

(ξ)ϕ(ξ) + ξ

ρ(ξ)ϕ

(ξ) }

= − lim

ξ→R⁻

{ [ϕ

(ξ) + α]

(n−1)

ξ

ϕ

(ξ) + ξβ + ϕ

(ξ)ϕ

(ξ) }

= − α Rβ .

This implies that there exists ξ

> 0 such that ϕ

(ξ)

ϕ(ξ) ≥ − 3α

2Rβ for all ξ ∈ (ξ

, R). (2.11) Integrating (2.11) from ξ

to ξ > ξ

, we have

lnϕ(ξ) − lnϕ(ξ

) ≥ − 3α

2Rβ ξ + 3α 2Rβ ξ

, i.e.,

ϕ(ξ) ≥ Ce

for all ξ ∈ (ξ

, R), where C = ϕ(ξ

)e

> 0. This contradicts with ϕ(ξ) → 0 as

ξ → R

.

Next, we suppose that

ξ→R

lim

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy = B for some B ∈ (0, ∞). Then

ξ→R

lim

ϕ

(ξ)

ϕ(ξ) = − lim

ξ→R⁻

{ 1 ξ

ρ(ξ)ϕ(ξ)

Z _ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy } = 0.

This implies that there exists ξ

> 0 such that ϕ

(ξ)

ϕ(ξ) ≥ −1 for all ξ ∈ (ξ

, R). (2.12) Similarly, integrating (2.12) from ξ

to ξ > ξ

, we obtain that

ϕ(ξ) ≥ Ce

for all ξ ∈ (ξ

, R), for some positive constant C. Again, this is a contradiction.

This complete the proof of the theorem.

Next, we shall study that the limits of ϕ(ξ) and ϕ

(ξ) as ξ → ∞.

Lemma 2.2 There holds ϕ(ξ) → 0 as ξ → ∞.

Proof. Since ϕ(ξ) is monotone decreasing, the limit l = lim

ξ→∞

ϕ(ξ) exists and l ≥ 0. Suppose l > 0, we define

H(ξ) = 1

2 [ϕ

(ξ)]

+ G(ϕ(ξ)), ξ > 0, (2.13) where

G(ϕ) =

Z ϕ

η

(s

+ αs

)ds, l ≤ ϕ < η. (2.14) Since

H

(ξ) = −( n − 1

ξ + βξϕ

)(ϕ

(ξ)) ≥ 0 and |G(ϕ)| ≤ (η

+ αl

)(η − l), the limit

L = lim

ξ→∞

H(ξ) exists and is finite. By the definition of H(ξ), the limit

−K = lim

ξ→∞

ϕ

(ξ)

exists and K ≥ 0. Since

0

ϕ

(ξ)dξ = l − η,

there exists a sequence {ξ

} such that ξ

→ ∞ and ϕ

(ξ

) → 0 as k → ∞. Hence K = 0.

Therefore, ∃ M > 0 such that | ϕ(ξ) |≤ M and | ϕ

(ξ) |≤ M , ∀ξ ≥ 0. Dividing (2.1) by ξ, we have

ϕ

(ξ)

ξ + n − 1

ξ

ϕ

(ξ) = − ϕ

(ξ)

ξ − α ϕ

(ξ)

ξ − βϕ

(ξ)ϕ

(ξ).

Integrating it from 1 to ξ

, we obtain

|

Z ξk

1

ϕ

(ξ)

ξ + n − 1

ξ

ϕ

(ξ) | = | ϕ

(ξ

)

ξ

− ϕ

(1) +

Z ξk

1

n

ξ

ϕ

(ξ) |≤ M+ | ϕ

(1) | +nM,

|

Z ξk

1

βϕ

(ξ)ϕ

(ξ) | = | β

1 − σ (ϕ

(ξ

) − ϕ

(1)) |≤ β

σ − 1 (l

+ ϕ

(1)) But,

Z _ξk

1

ϕ

(ξ) + αϕ

(ξ)

ξ ≥ α l

Z _ξk

1

1

ξ → ∞ as k → ∞, a contradiction. Hence, l = 0.

Lemma 2.3 We have that ϕ

(ξ) → 0 as ξ → ∞.

Proof. Using (2.5) and applying L’Hˆopital’s Rule, we have

ξ→∞

lim ϕ

(ξ) = − lim

1 ξ

ρ(ξ)

Z ξ

0

y

ρ(y)[ϕ

(y) + αϕ

(y)]dy .

= − lim

ξ→∞

ξ

ρ(ξ)[ϕ

(ξ) + αϕ

(ξ)]

(n − 1)ξ

ρ(ξ) + ξ

ρ(ξ)βξϕ

(ξ)

= − lim

ξ→∞

[ϕ

(ξ) + αϕ(ξ)]

(n−1)ϕ^σ(ξ)

ξ

+ ξβ

= 0.

The lemma follows.

3 Blow-up Patterns for One-dimensional Case

In this section, we shall study the the positive solution of (1.5) with n = 1 as the following initial value problem (Q) :

ϕ

(ξ) + ϕ

(ξ) − αϕ

(ξ) − βξϕ

(ξ)ϕ

(ξ) = 0, ξ > 0, (3.1)

ϕ

(0) = 0, ϕ(0) = η.