Electronic Journal of Differential Equations, Vol. 2006(2006), No. 94, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
A BLOW UP CONDITION FOR A NONAUTONOMOUS
SEMILINEAR SYSTEM
AROLDO PÉREZ-PÉREZ
Abstract. We give a sufficient condition for finite time blow up of the nonnegative mild solution to a nonautonomous weakly coupled system with fractal
diffusion having a time dependent factor which is continuous and nonnegative.
1. Introduction
This paper deals with the blow up of nonnegative solutions of the nonautonomous
initial value problem for a weakly coupled system with a fractal diffusion
∂u(t, x)
= k(t)∆α u(t, x) + v β1 (t, x),
∂t
∂v(t, x)
= k(t)∆α v(t, x) + uβ2 (t, x),
∂t
u(0, x) = ϕ1 (x), x ∈ Rd
v(0, x) = ϕ2 (x),
t > 0,
x ∈ Rd
t > 0,
x ∈ Rd
(1.1)
x ∈ Rd ,
where ∆α := −(−∆)α/2 , 0 < α ≤ 2 denotes the α-Laplacian, β1 , β2 > 1 are
constants, 0 ≤ ϕ1 , ϕ2 ∈ B(Rd ) (where B(Rd ) is the space of bounded measurable
functions on Rd ) do not vanish identically, k : [0, ∞) → [0, ∞) is continuous and
satisfies
Z t
ρ
ε1 t ≤
k(r)dr ≤ ε2 tρ , ε1 , ε2 , ρ > 0,
(1.2)
0
for all t large enough.
The associated integral system to (1.1) is given by
Z t
u(t, x) = U (t, 0)ϕ1 (x) +
U (t, r)v β1 (r, x)dr,
0
Z t
v(t, x) = U (t, 0)ϕ2 (x) +
U (t, r)uβ2 (r, x)dr,
t > 0, x ∈ Rd ,
(1.3)
t > 0, x ∈ Rd ,
(1.4)
0
2000 Mathematics Subject Classification. 35B40, 35K45, 35K55, 35K57.
Key words and phrases. Finite time blow up; mild solution; weakly coupled system;
nonautonomous initial value problem; fractal diffusion.
c
2006
Texas State University - San Marcos.
Submitted July 14, 2006. Published August 18, 2006.
1
2
A. PÉREZ
EJDE-2006/94
where {U (t, s)}t≥s≥0 is the evolution family on B(Rd ) that solves the homogeneous
Cauchy problem for the family of generators {k(t)∆α }t≥0 . Clearly
U (t, s) = S(K(t, s)),
t ≥ s ≥ 0,
where {S(t)}t≥0 is the semigroup with infinitesimal generator ∆α , and K(t, s) =
Rt
k(r)dr, t ≥ s ≥ 0.
s
A solution of (1.3)-(1.4) is called a mild solution of (1.1). If there exist a solution
(u, v) of (1.1) in [0, ∞) × Rd such that ku(t, ·)k∞ + kv(t, ·)k∞ < ∞ for any t ≥ 0,
we say that (u, v) is a global solution, and when there exist a number Tϕ1 ,ϕ2 < ∞
such that (1.1) has a bounded solution (u, v) in [0, T ] × Rd for all T < Tϕ1 ,ϕ2 with
limt↑Tϕ1 ,ϕ2 ku(t, ·)k∞ = ∞ or limt↑Tϕ1 ,ϕ2 kv(t, ·)k∞ = ∞ we say that (u, v) blows
up in finite time.
The finite time blow up of (1.1) for α = 2 and k ≡ 1 was initially considered by
Escobedo and Herrero [4]. They proved that when β1 β2 > 1 and (γ+1)/(β1 β2 −1) ≥
d/2 with γ = max{β1 , β2 }, any nontrivial positive solution to (1.1) blows up in
finite time. Related results and more general cases for the Laplacian can be found
for instance in [1, 3, 5, 6, 10, 12, 13, 15, 16]. The case for fractional powers of
the Laplacian when k ≡ 1 for equations with different diffusion operators was
considered in [8, 9]; see also [2, 11] for a probabilistic approach. Sugitani [14] has
considered a scalar version of (1.1) with k ≡ 1 when the nonlinear term is given by
an increasing nonnegative continuous and convex function F (u), defined on [0, ∞),
and Guedda and Kirane [7] have considered a scalar version of (1.1) with k ≡ 1
when the nonlinear term is h(t)uβ , β > 1 with h being a nonnegative continuous
function on [0, ∞) satisfying c0 tσ ≤ h(t) ≤ c1 tσ for sufficiently large t, where
c0 , c1 > 0 and σ > −1 are constants. They proved that in this scalar case, solutions
blow up in finite time if 0 < d(β − 1)/α ≤ 1 + σ for any nontrivial nonnegative
and continuous initial function on Rd . Here we prove that if k satisfies (1.2) and
0 < dρ(βi − 1)/α < 1, i = 1, 2, then any nontrivial positive solution of (1.1) blows
up in finite time. Here solutions will be understood in the mild sense, that is, that
solve (1.3)-(1.4).
2. Blow up condition
Let (u(·, ·), v(·, ·)) be a nonnegative solution of (1.1) and define
Z
Z
u(t) =
p(K(t, 0), x)u(t, x)dx, v(t) =
p(K(t, 0), x)v(t, x)dx,
Rd
t > 0,
Rd
where p(t, x), t > 0, x ∈ Rd denotes the density of the semigroup S(t), t ≥ 0.
Lemma 2.1. For any s, t > 0, and x, y ∈ Rd , p(t, x) satisfies
i)
ii)
iii)
iv)
d
1
p(ts, x) = t− α p(s, t− α x),
p(t, x) ≤ p(t, y) when |x| ≥ |y|,
d
p(t, x) ≥ ( st ) α p(s, x) for t ≥ s,
p(t, τ1 (x − y)) ≥ p(t, x)p(t, y) if p(t, 0) ≤ 1 and τ ≥ 2.
Proof. See Guedda and Kirane [7] or Sugitani [14].
Lemma 2.2. If there exist T0 > 0 such that u(t) = ∞ or v(t) = ∞ for t ≥ T0 ,
then the nonnegative solution of (1.1) blows up in finite time.
EJDE-2006/94
A BLOW UP CONDITION
3
Proof. Due to (1.2) and Lemma 2.1 i), we can assume that
p(K(t, 0), 0) ≤ 1
1/ρ
for all t ≥ T0 .
1/ρ
If T0 ≤ ε1 t and ε1 t ≤ r ≤ (2ε1 )1/ρ t, we have from the conditions of k(t),
h K((10ε )1/ρ t, r) i1/α h K((10ε )1/ρ t, 0) − K(r, 0) i1/α
2
2
τ≡
=
K(r, 0)
K(r, 0)
h K((10ε )1/ρ t, 0)
i1/α
i1/α h ε (10ε )tρ
2
1
2
≥
−
1
≥ 2.
−
1
≥
ε2 (2ε1 )tρ
K((2ε1 )1/ρ t), 0)
1/ρ
t,r) 1/α
2)
,
Hence, using Lemma 2.1 i), iv) with τ = K((10ε
K(r,0)
p(K((10ε2 )1/ρ t, r), x − y)
h K((10ε )1/ρ t, r) i
2
= p(K(r, 0)
, x − y)
K(r, 0)
h
id/α
i1/α
h
K(r, 0)
K(r, 0)
=
p(K(r,
0),
(x − y))
K((10ε2 )1/ρ t, r)
K((10ε2 )1/ρ t, r)
h
id/α
K(r, 0)
p(K(r, 0), x)p(K(r, 0), y), x, y ∈ Rd .
≥
K((10ε2 )1/ρ t), r)
Hence, assuming that u(t) = ∞ for all t ≥ T0 ,
Z
p(K((10ε2 )1/ρ t, r), x − y)u(r, y)dy
Rd
id/α
K(r, 0)
≥
p(K(r, 0), x)u(r) = ∞,
K((10ε2 )1/ρ t, r)
h
(2.1)
d
x∈R .
We know by (1.4) that
Z
p(K(t, 0), x − y)ϕ2 (y)dy
Z tZ
+
p(K(t, r), x − y)uβ2 (r, y)dy dr
Rd
0
Z tZ
≥
p(K(t, r), x − y)uβ2 (r, y)dy dr.
v(t, x) =
Rd
0
Rd
Thus,
1/ρ
v((10ε2 )
(10ε2 )1/ρ t
Z
t, x) ≥
Z
p(K((10ε2 )1/ρ t, r), x − y)uβ2 (r, y)dy dr
Rd
0
and by Jensen’s inequality and (2.1), we get
Z (2ε1 )1/ρ t Z
β2
v((10ε2 )1/ρ t, x) ≥
p(K((10ε2 )1/ρ t, r), x − y)u(r, y)dy
dr = ∞,
1/ρ
ε1
t
Rd
so that v(t, x) = ∞ for any t ≥ (10 εε12 )1/ρ T0 and x ∈ Rd . Similarly, when v(t) = ∞
for all t ≥ T0 , it can be shown that u(t, x) = ∞ for all t ≥ (10 εε12 )1/ρ T0 and
x ∈ Rd .
Theorem 2.3. If 0 < dρ(βi − 1)/α < 1, i = 1, 2, then the nonnegative solution of
system (1.1) blows up in finite time.
4
A. PÉREZ
EJDE-2006/94
Proof. Let t0 ≥ 1 be such that (1.2) holds for all t ≥ t0 and such that p(K(t0 , 0), 0) ≤
1. Using Lemma 2.1 i), iv), we have
1
p(K(t0 , 0), x − y) =p(K(t0 , 0), (2x − 2y)) ≥ p(K(t0 , 0), 2x)p(K(t0 , 0), 2y)
2
=2−d p(2−α K(t0 , 0), x)p(K(t0 , 0), 2y), x, y ∈ Rd .
Therefore (see (1.3))
Z
u(t0 , x) ≥
p(K(t0 , 0), x − y)ϕ1 (y)dy
Z
−d
−α
≥ 2 p(2 K(t0 , 0), x)
p(K(t0 , 0), 2y)ϕ1 (y)dy
Rd
(2.2)
Rd
= N1 p(2−α K(t0 , 0), x),
where N1 = 2−d
R
Rd
x ∈ Rd ,
p(K(t0 , 0), 2y)ϕ1 (y)dy. Notice that
Z
p(K(t + t0 , 0), x − y)ϕ1 (y)dy
Z t+t0 Z
+
p(K(t + t0 , r), x − y)v β1 (r, y)dy dr
Rd
Z 0
=
p(K(t + t0 , t0 ) + K(t0 , 0), x − y)ϕ1 (y)dy
Rd
Z t0 Z
+
p(K(t + t0 , t0 ) + K(t0 , r), x − y)v β1 (r, y)dy dr
0
Rd
Z t+t0 Z
+
p(K(t + t0 , r), x − y)v β1 (r, y)dy dr
Rd
0
Z t
Z
=
p(K(t + t0 , t0 ), x − z)p(K(t0 , 0), z − y)dz ϕ1 (y)dy
Rd
Rd
Z t0 h Z Z
+
p(K(t + t0 , t0 ), x − z)p(K(t0 , r), z − y)dz
d
0
Rd
i R
β1
× v (r, y)dy dr
Z tZ
+
p(K(t + t0 , r + t0 ), x − y)v β1 (r + t0 , y)dy dr,
u(t + t0 , x) =
Rd
0
Rd
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A BLOW UP CONDITION
5
t ≥ 0, x ∈ Rd . From here, by Fubini’s theorem and (1.3) we have
Z
Z
u(t + t0 , x) =
p(K(t + t0 , t0 ), x − z)
p(K(t0 , 0), z − y)ϕ1 (y)dy dz
Rd
Rd
Z
h Z t0 Z
+
p(K(t + t0 , t0 ), x − z)
p(K(t0 , r), z − y)
Rd
0
Rd
i
× v β1 (r, y)dy dr dz
Z tZ
+
p(K(t + t0 , r + t0 ), x − y)v β1 (r + t0 , y)dy dr
Rd
Z 0
=
p(K(t + t0 , t0 ), x − y)u(t0 , y)dy
Rd
Z tZ
+
p(K(t + t0 , r + t0 ), x − y)v β1 (r + t0 , y)dy dr,
Rd
0
d
t ≥ 0, x ∈ R . Thus, using (2.2) gives
Z
u(t + t0 , x) ≥ N1
p(K(t + t0 , t0 ), x − y)p(2−α K(t0 , 0), y)dy
Rd
Z tZ
p(K(t + t0 , r + t0 ), x − y)v β1 (r + t0 , y)dy dr
+
Rd
0
(2.3)
= N1 p(K(t + t0 , t0 ) + 2−α K(t0 , 0), x)
Z tZ
+
p(K(t + t0 , r + t0 ), x − y)v β1 (r + t0 , y)dy dr,
Rd
0
d
t ≥ 0, x ∈ R . Multiplying both sides of (2.3) by p(K(t + t0 , 0), x) and integrating,
we have
u(t + t0 ) = N1 p(2K(t + t0 , t0 ) + (2−α + 1)K(t0 , 0), 0)
Z tZ
+
p(2K(t + t0 , 0) − K(r + t0 , 0), y)v β1 (r + t0 , y)dy dr,
Rd
0
t ≥ 0. Applying Lemma 2.1 i), iii), we get
u(t + t0 ) ≥ N1 [2K(t + t0 , t0 ) + (2−α + 1)K(t0 , 0)]−d/α p(1, 0)
Z t
K(r + t0 , 0) d/α β1
v (r + t0 )dr.
+
2K(t + t0 , 0)
0
For a suitable choice of θ > 0 given below, we define f1 (t) = K d/α (t + t0 , 0)u(t +
t0 ), g1 (t) = K d/α (t + t0 , 0)v(t + t0 ), t ≥ θ. Then
Z t
−d/α
f1 (t) ≥ N 1 + 2
K −d(β1 −1)/α (r + t0 , 0)g1β1 (r)dr, t ≥ θ,
θ
where N 1 = p(1, 0)N1
that
h
K(θ,0)
2K(θ+t0 ,0)+(2−α +1)K(t0 ,0)
−d/α
Z
g1 (t) ≥ N 2 + 2
id/α
. Similarly, it can be shown
t
K −d(β2 −1)/α (r + t0 , 0)f1β2 (r)dr,
t ≥ θ,
θ
where N 2 = p(1, 0)N2
2y)ϕ2 (y)dy.
h
K(θ,0)
2K(θ+t0 ,0)+(2−α +1)K(t0 ,0)
id/α
with N2 = 2−d
R
Rd
p(K(t0 , 0),
6
A. PÉREZ
EJDE-2006/94
Letting N = min{N 1 , N 2 }, we get
Z t
f1 (t) ≥ N + 2−d/α
min K −d(βi −1)/α (r + t0 , 0)g1β1 (r)dr,
t ≥ θ,
min K −d(βi −1)/α (r + t0 , 0)f1β2 (r)dr,
t ≥ θ.
Let (f2 (t), g2 (t)) be the solution of the system integral equations
Z t
min K −d(βi −1)/α (r + t0 , 0)g2β1 (r)dr,
f2 (t) = N + 2−d/α
t ≥ θ,
min K −d(βi −1)/α (r + t0 , 0)f2β2 (r)dr,
t ≥ θ,
g1 (t) ≥ N + 2−d/α
g2 (t) = N + 2−d/α
θ i∈{1,2}
Z t
θ i∈{1,2}
θ i∈{1,2}
Z t
θ i∈{1,2}
whose differential expression is
f20 (t) = 2−d/α min K −d(βi −1)/α (t + t0 , 0)g2β1 (t),
t > θ,
g20 (t) = 2−d/α min K −d(βi −1)/α (t + t0 , 0)f2β2 (t),
t > θ,
i∈{1,2}
(2.4)
i∈{1,2}
f2 (θ) = N,
g2 (θ) = N.
From (2.4) it follows that
Z t
f2β2 (r)f20 (r)dr =
θ
Z
t
g2β1 (r)g20 (r)dr,
θ
that is,
1
1
[f2β2 +1 (t) − N β2 +1 ] =
[g β1 +1 (t) − N β1 +1 ].
β2 + 1
β1 + 1 2
Fix θ > 0 such that 0 < N ≤ 1. This is possible due to N 1 , N 2 → 0 when θ → 0.
We assume without loss of generality that β2 ≥ β1 . Then
g β1 +1 (t)
f2β2 +1 (t)
≤ 2
β2 + 1
β1 + 1
or, equivalently
β + 1 β 1+1 β2 +1
1
1
f2β1 +1 (t), t ≥ θ.
β2 + 1
Substituting this in the first equation of (2.4), we have
1
β1 (β2 +1)
β + 1 β β+1
d(βi −1)
1
1
f20 (t) ≥ 2−d/α min K − α (t + t0 , 0)
f2 β1 +1 (t),
β2 + 1
i∈{1,2}
g2 (t) ≥
t ≥ θ,
that is,
−β1 (β2 +1)
β1 +1
f2
(t)f20 (t) ≥ 2−d/α
1
β + 1 β β+1
d(βi −1)
1
1
min K − α (t + t0 , 0),
β2 + 1
i∈{1,2}
Integrating from θ to t yields
1 β2
1−β1 β2 i
β1 + 1 h 1−β
f2 β1 +1 (t) − N β1 +1
1 − β1 β2
β1 Z t
d(βi −1)
−d/α β1 + 1 β1 +1
≥2
min K − α (r + t0 , 0)dr.
β2 + 1
i∈{1,2}
θ
t ≥ θ.
EJDE-2006/94
A BLOW UP CONDITION
7
Thus (remember that β1 , β2 > 1)
β1 +1
1
1 − β β β + 1 β β+1
h 1−β1 β2
i 1−β
1 2
1
1
1 β2
f2 (t) ≥ N β1 +1 − 2−d/α
H(t)
,
β1 + 1
β2 + 1
where
Z
t
min K −
H(t) ≡
θ i∈{1,2}
d(βi −1)
α
(r + t0 , 0)dr,
t ≥ θ.
From (1.2) we have
Z
H(t) ≥
t
min (ε2 (r + t0 )ρ )−
d(βi −1)
α
dr.
θ i∈{1,2}
Using the fact that 0 < dρ (β2 − 1) /α < 1 we get
Z t
d(βi −1)
dρ(β2 −1)
−
(r + t0 )− α dr
H(t) ≥ min ε2 α
i∈{1,2}
θ
i
d(βi −1) h
α−dρ(β2 −1)
α−dρ(β2 −1)
α
−
α
α
=
min ε2 α
− (θ + t0 )
.
(t + t0 )
α − dρ(β2 − 1) i∈{1,2}
Thus H(t) → ∞ when t → ∞. So, we have that there exists T0 ≥ θ such that
f2 (t) = ∞ for t = T0 . By comparison we have
K d/α (t + t0 , 0)u(t + t0 ) = f1 (t) ≥ f2 (t) = ∞ for t = T0 ,
which implies by Lemma 2.2 that v(t, x) = ∞ for all t ≥ (10 εε12 )1/ρ (T0 + t0 ) and
x ∈ Rd .
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Aroldo Pérez-Pérez
División Académica de Ciencias Básicas, Universidad Juárez Autónoma de Tabasco, Km.
1 Carretera Cunduacán-Jalpa de Méndez, C.P. 86690 A.P. 24, Cunduacán, Tabasco,
México
E-mail address: aroldo.perez@dacb.ujat.mx