# αmRy 0 gm1 (s)ρ(s)ds ρ(y

## 全文

(1)

4 Asymptotic behavior

We next study the asymptotic behavior of the globally monotone decreasing solution of (P).

Lemma 4.1 If g(y) is globally monotone decreasing to zero, then g(y) → 0 as y → ∞.

Proof. Define ρ(y) = exp{αkRy

0 sgm1−1(s)ds}. From (7), it follows that (ρg)(y) = −αmρ(y)gm1(y)

and so

g(y) = g(0) − αmRy

0 gm1 (s)ρ(s)ds

ρ(y) .

By [−αmRy

0 gm1 (s)ρ(s)ds] = −αmgm1(y)ρ(y) > 0, the integral I = −αm

Z

0

gm1(s)ρ(s)ds

exists and 0 < I ≤ ∞. We claim that g(0) + I = 0. Otherwise |g(y)| → ∞ as y → ∞, since ρ(y) → 0 as y → ∞. A contradiction. Hence g(0) + I = 0.

By L’Hˆopital’s Rule, we get that

y→∞lim g(y) = lim

y→∞

g(0) − αmRy

0 gm1(s)ρ(s)ds ρ(y)

= lim

y→∞

−m kyg−1(y)

= 0.

By L’Hˆopital’s Rule again, we obtain that

y→∞lim g(y)

g(y) = lim

y→∞

g(0) − αmRy

0 gm1(s)ρ(s)ds ρ(y)g(y)

= lim

y→∞

−αm g−1/mg+ αky

= 0 and

y→∞lim yg(y)

g(y) = lim

y→∞

g(0) − αmRy

0 gm1 (s)ρ(s)ds y−1ρ(y)g(y)

= lim

y→∞

−αm

−y−2gm1+1+ y−1g−1/mg+ αk

= −m k.

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(2)

Let z(y) = g(y)/g(y). Then z(y) satisfies the equation

z(y) + αkygm1−1(y)z(y) + αmgm1−1(y) + z2(y) = 0.

It follows that

z(y) = Ry

0[−αmgm1−1(s)ρ(s) − z2(s)ρ(s)]ds + z(0)

ρ(y) . (16)

where ρ(y) = exp{αkRy

0 sgm1−1(s)ds}.

The following proof is similar to the one given in .

Theorem 3 The limit

y→∞lim[ymkg(y)]

exists and is positive.

Proof. From (16), we can write

[yz(y) +m k]yλ =

Ry

0[−αmgm1−1(s)ρ(s) − z2(s)ρ(s)]ds + z(0) + mky−1ρ(y)

y−λ−1ρ(y) ,

where λ ∈ (0, 2). Note that ρ(y) → 0 and z(y) → 0 as y → ∞. From (16), we get Ry

0[−αmgm1−1(s)ρ(s) − z2(s)ρ(s)]ds + z(0) → 0 as y → ∞.

By L’Hˆopital’s Rule, we get that

y→∞lim[yz(y) +m

k]yλ = lim

y→∞

−z2y2g1−m1 mkg1−m1 g1−m1 (−λ − 1)y−λ+ y−λ+2αk

= 0, ∀λ ∈ (0, 2).

By integration, we obtain

g(y) = Aymk[1 + o(y−λ)]

as y → ∞, for some positive constant A. Hence the Theorem follows.

From Theorem 3 and using (4), we have

t→Tlimu(x, t) = A1/m|x|−1/k, x 6= 0, for any globally monotone decreasing solution of (P).

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