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−2g−m1+1+ y−1g−1/mg′+ αk
= −m k.
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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 [8].
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) = Ay−mk[1 + o(y−λ)]
as y → ∞, for some positive constant A. Hence the Theorem follows.
From Theorem 3 and using (4), we have
t→Tlim−u(x, t) = A1/m|x|−1/k, x 6= 0, for any globally monotone decreasing solution of (P).
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