Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 216, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
BLOW-UP OF SOLUTIONS FOR A SYSTEM OF NONLINEAR
PARABOLIC EQUATIONS
SHUN-TANG WU
Abstract. The initial boundary value problem for a system of parabolic equations in a bounded domain is considered. We prove that, under suitable conditions on the nonlinearity and certain initial data, the lower bound for the
blow-up time is determined if blow-up does occur. In addition, a criterion for
blow-up to occur and conditions which ensure that blow-up does not occur are
established.
1. Introduction
We consider the initial boundary value problem for the following nonlinear parabolic problems:
ut − div(ρ1 (|∇u|2 )∇u) = f1 (u, v)
in Ω × [0, ∞),
(1.1)
vt − div((ρ2 (|∇v|2 )∇v) = f2 (u, v)
in Ω × [0, ∞),
(1.2)
x ∈ Ω,
(1.3)
u(x, 0) = u0 (x),
v(x, 0) = v0 (x),
u(x, t) = v(x, t) = 0,
x ∈ ∂Ω, t > 0,
(1.4)
where Ω is a bounded domain in RN (N ≥ 1) with a smooth boundary ∂Ω, ρi ,
i = 1, 2, are positive C 1 functions and fi (·, ·) : R2 → R, i = 1, 2, are given functions
which will be specified later. u0 (x), v0 (x) are nonzero and nonnegative functions.
Questions related to the blow-up phenomena of the solutions for the nonlinear
parabolic equations and systems have attracted considerable attention in recent
years. A natural question concerning the blow-up properties is about whether the
solution blows up and, if so, at what time t∗ blow-up occurs. In this direction,
there is a vast literature to deal with the blow-up time when the solution does blow
up at finite time t∗ [1, 2, 3, 4, 5, 6, 7, 8, 10, 12], [15, page 3]. Yet, this blow-up
time can seldom be determined explicitly. Indeed, the methods used in the study
of blow-up very often have yielded only upper bound for t∗ . However, a lower
bound on blow-up time is more important in some applied problems because of
the explosive nature of the solution. To the authors knowledge, some of the first
work on lower bounds for t∗ was by Weissler [16, 17]. Recently, a number of papers
2000 Mathematics Subject Classification. 35K55, 35K60.
Key words and phrases. Blow-up; lower bound of blow-up time; parabolic problem.
c
2013
Texas State University - San Marcos.
Submitted August 21, 2013. Published September 30, 2013.
1
2
S.-T. WU
EJDE-2013/216
deriving lower bounds for t∗ in various problems have appeared, beginning with the
paper of Payne and Schaefer [13]. Payne et al. [14] considered the single equation
ut − div ρ(|∇u|2 )∇u = f (u).
Under certain conditions on the nonlineartities, they obtained a lower bound for
blow-up time if blow-up does occur. Additionally, a criterion for blow-up and
conditions which ensure that blow-up does not occur are obtained.
Motivated by previous works, in this study, we establish the lower bound and
the upper bound for problem (1.1)-(1.4) when blow-up does occur. Besides, the
nonblow-up properties for a class of problem (1.1)-(1.4) are also investigated. Our
proof technique closely follows the arguments of [14], with some modifications being
needed for our problems. The paper is organized as follows. In section 2, under
suitable conditions on ρi , fi , i = 1, 2, the lower bound for the blow-up time is
established if blow-up occurs when Ω is a bounded domain in R3 . In Section 3,
the nonblow-up phenomena are investigated. Finally, the sufficient condition which
guarantees the blow-up occurs is obtained and an upper bound for the blow-up
time is also given.
2. Lower bound for the blow-up time
In this section, we focus our attention to the lower bound time t∗ for the blowup time of the solutions to problem (1.1)-(1.4). For this purpose, we give the
assumptions on ρi and fi , i = 1, 2 as follows.
(A1) ρi (s), i = 1, 2 are nonnegative C 1 function for s > 0 satisfying
ρ1 (s) ≥ b1 + b2 sq1 ,
ρ2 (s) ≥ b3 + b3 sq2 ,
q1 , q2 , bi > 0,
i = 1 − 4.
(A2) Concerning the functions f1 (u, v) and f2 (u, v), we take (see [9])
m+1
m−3
f1 (u, v) = a|u + v|m−1 (u + v) + b|u| 2 |v| 2 u ,
m−3
m+1
f2 (u, v) = a|u + v|m−1 (u + v) + b|v| 2 |u| 2 v ,
(2.1)
(2.2)
where a, b > 0 are constants and m satisfies
N +2
, if N ≥ 3.
m > 1, if N = 1, 2 or 1 < m ≤
N −2
One can easily verify that
uf1 (u, v) + vf2 (u, v) = (m + 1)F (u, v),
∀(u, v) ∈ R2 ,
where
m+1
1 a|u + v|m+1 + 2b|uv| 2 .
m+1
As in [9], we still have the following result.
F (u, v) =
Lemma 2.1. There exists a positive constant β such that, for p > 0 ,
up f1 (u, v) + v p f2 (u, v) ≤ β(|u|p+m + |v|p+m ),
∀(u, v) ∈ R2 .
We define
Z
u2(n−1)(q1 +1)+2 dx +
Ω
Z
Z
σ1
=
u dx +
v σ2 dx,
Z
φ(t) =
Ω
Ω
Ω
v 2(n−1)(q2 +1)+2 dx
(2.3)
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BLOW-UP OF SOLUTIONS
3
where σ1 = 2(n − 1)(q1 + 1) + 2, σ2 = 2(n − 1)(q2 + 1) + 2 and n is a positive
constant satisfying
n > max
n 3(m − 1) − 2q
1 3(m − 1) − 2q2 3(m − 1) − 2(3q1 − 2q2 )
,
,
,
2(q1 + 1)
2(q2 + 1)
2(3q1 − 2q2 + 1)
3(m − 1) − 2(3q2 − 2q1 ) o
.
2(3q2 − 2q1 + 1)
(2.4)
Theorem 2.2. Suppose that (A1), (A2), (2.4) hold and Ω ⊂ R3 is a bounded domain. Assume further that m − 1 > 2 max(q1 , q2 ) > 0 and q1 > 23 q2 > 49 q1 > 0. Let
(u, v) be the nonnegative solution of problem (1.1)-(1.4), which become unbounded
in the measure φ at time t∗ , then t∗ is bounded below as
∗
Z
∞
t ≥
φ(0)
1
P4
i=1
ki φ(s)µi
ds,
where ki > 0 and µi > 0, i = 1 − 4 are constnats given in the proof.
Proof. Differentiating (2.3) and using (1.1)-(1.2), (A1) and Lemma 2.1, we obtain
0
Z
Z
ut dx + σ2
v σ2 −1 vt dx
Ω
Z
Z
= −σ1 (σ1 − 1)
uσ1 −2 ρ1 (|∇u|2 )|∇u|2 dx + σ1
uσ1 −1 f1 (u, v)dx
ZΩ
ZΩ
σ2 −2
2
2
− σ2 (σ2 − 1)
v
ρ2 (|∇v| )|∇v| dx + σ2
v σ2 −1 f2 (u, v)dx
Ω
Ω
Z
≤ −σ1 (σ1 − 1)
uσ1 −2 |∇u|2 (b1 + b2 |∇u|2q1 )dx
Ω
Z
+ βσ1 (um+σ1 −1 + v m+σ1 −1 )dx
Ω
Z
− σ2 (σ2 − 1)
v σ2 −2 |∇v|2 (b3 + b4 |∇v|2q2 )dx
Ω
Z
+ βσ2 (um+σ2 −1 + v m+σ2 −1 )dx.
φ (t) = σ1
σ1 −1
u
Ω
(2.5)
Ω
R
R
Dropping the terms σ1 (σ1 −1)b1 Ω uσ1 −2 |∇u|2 dx and σ2 (σ2 −1)b3 Ω v σ2 −2 |∇v|2 dx
on the right-hand side of (2.5) and using |∇wn |2 = n2 w2(n−1) |∇w|2 , we deduce that
Z
Z
σ1 (σ1 − 1)b2
n 2(q1 +1)
|∇u |
dx + βσ1 (um+σ1 −1 + v m+σ1 −1 )dx
φ (t) ≤ −
n2(q1 +1)
Ω
Z
ZΩ
σ2 (σ2 − 1)b4
−
|∇v n |2(q2 +1) dx + βσ2 (um+σ2 −1 + v m+σ2 −1 )dx.
n2(q2 +1)
Ω
Ω
0
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EJDE-2013/216
For simplicity, setting w1 = un , w2 = v n and γi = m − 1 − 2qi > 0, i = 1, 2, then
we obtain
Z
σ1 (σ1 − 1)b2
φ0 (t) ≤ −
|∇w1 |2(q1 +1) dx
n2(q1 +1)
Ω
Z
γ
γ
2(q +1)+ n1
2(q1 +1)+ n1
+ βσ1 (w1
+ w2 1
)dx
Ω
(2.6)
Z
Z
γ
σ2 (σ2 − 1)b4
2(q2 +1)+ n2
2(q2 +1)
|∇w2 |
dx + βσ2
w1
−
dx
n2(q2 +1)
Ω
Ω
Z
γ
2(q +1)+ n2
dx.
+ βσ2
w2 2
Ω
Next, we will estimate the right-hand side of (2.6). It follows from [14, (2.12)] that
Z
Z
2/3 Z
2/3
γ
3γ
2(q +1)+ n1
q +1+ 2n1
w1 1
dx ≤ K1
dx
,
(2.7)
|∇w1 |2(q1 +1) dx
w11
Ω
Ω
−
αλ1
4q1 +1
6
Ω
4(q1 +1)
3
, α = 41/3 · 3−1/2 · π −2/3 and λ1 is the first
where K1 =
(q1 + 1)
eigenvalue in the fixed membrane problem
∆w + λw = 0,
w > 0 in Ω,
and w = 0 on ∂Ω.
By using Hölder inequality and (2.3), we obtain
Z
Z
3γ
3γ1
q +1+ 2n1
w11
dx =
un(q1 +1)+ 2 dx
Ω
Ω
µ1
Z
uσ1 dx
· |Ω|1−µ1
≤
Ω
µ1
≤ φ(t)
(2.8)
· |Ω|1−µ1 ,
which together with (2.7) implies
Z
Z
γ
2(1−µ1 )
2µ1
2(q +1)+ n1
w1 1
dx ≤ K1 |Ω| 3 φ(t) 3 ( |∇w1 |2(q1 +1) dx)2/3 .
Ω
Ω
2n(q1 +1)+3γ1
,
2σ1
with µ1 =
the inequality
we note that µ1 < 1 in view of (2.4). Further, thanks to
xr y s ≤ rx + sy,
r + s = 1,
x, y ≥ 0,
(2.9)
we obtain, for α1 > 0,
Z
Z
h 1
i
γ
2(1−µ1 )
2α1
2(q1 +1)+ n1
2µ1
2(q1 +1)
3
w1
dx ≤ K1 |Ω|
φ(t)
+
|∇w
|
dx
. (2.10)
1
3α12
3 Ω
Ω
and similarly
Z
Z
h 1
i
γ
2(1−µ2 )
2α2
2(q +1)+ n2
2µ2
w2 2
dx ≤ K2 |Ω| 3
φ(t)
+
|∇w2 |2(q2 +1) dx , (2.11)
2
3α2
3 Ω
Ω
−
4q2 +1
4(q2 +1)
+1)+3γ2
where α2 > 0, K2 = αλ1 6 (q2 + 1) 3
and µ2 = 2n(q22σ
< 1.
2
To estimate the other two terms in the right hand side of (2.6), we use Hölder
inequality and the following result (see [14, (2.7)-(2.10)])
Z
Z
2
4q+1
4(q+1)
3
4(q+1) − 2
w
dx ≤ α (q + 1)
λ1
|∇w|2(q+1) dx , q > 0,
(2.12)
Ω
Ω
EJDE-2013/216
BLOW-UP OF SOLUTIONS
5
to obtain
Z
2(q +1)+
γ1
n
w2 1
dx
Ω
Z
4(q2 +1)
γ
4(q2 +1)
2(q +1)−
+ n1
3
=
dx
w2 3 · w2 1
Ω
Z
Z
3γ
2/3
1/3
3q −2q +1+ 2n1
4(q +1)
≤
w2 2 dx
dx
w2 1 2
Ω
Ω
2/3
Z
2/3 Z
3γ
3q −2q +1+ 2n1
dx
.
≤ K2
|∇w2 |2(q2 +1) dx
w2 1 2
Ω
(2.13)
Ω
As in deriving (2.8), we see that
Z
Z
3γ
3γ1
3q −2q +1+ 2n1
w2 1 2
dx =
v n(3q1 −2q2 +1)+ 2 dx
Ω
ZΩ
≤ ( v σ2 dx)µ3 · |Ω|1−µ3
(2.14)
Ω
≤ φ(t)µ3 · |Ω|1−µ3
2 +1)+3γ1
< 1. Substituting (2.14) into (2.13) and using (2.9)
where µ3 = 2n(3q1 −2q
2σ2
once more, we obtain, for α3 > 0,
Z
Z
h 1
i
γ
2(1−µ3 )
2α3
2(q1 +1)+ n1
2µ3
2(q2 +1)
3
w2
φ(t)
+
|∇w
|
dx
. (2.15)
dx ≤ K2 |Ω|
2
3α32
3 Ω
Ω
and similarly
Z
Z
h 1
i
γ
2(1−µ4 )
2α4
2(q +1)+ n2
2(q1 +1)
2µ4
w1 2
dx ≤ K1 |Ω| 3
|∇w
|
dx
, (2.16)
φ(t)
+
1
3α42
3 Ω
Ω
1 +1)+3γ2
< 1. Combining (2.10), (2.11), (2.15)
where α4 > 0 and µ4 = 2n(3q2 −2q
2σ1
and (2.16) with (2.6), we conclude that
Z
Z
φ0 (t) ≤ −C1
|∇w1 |2(q1 +1) dx − C2
|∇w2 |2(q2 +1) dx
Ω
Ω
+ k1 φ(t)2µ1 + k2 φ(t)2µ2 + k3 φ(t)2µ3 + k4 φ(t)2µ4 ,
where
2(1−µ1 )
2(1−µ4 )
σ1 (σ1 − 1)b2
2α1 K1 βσ1
2α4 K1 βσ2
−
|Ω| 3
−
|Ω| 3 ,
2(q
+1)
1
3
3
n
2(1−µ2 )
2(1−µ3 )
2α3 K2 βσ2
σ2 (σ2 − 1)b4
2α2 K2 βσ1
−
C2 =
−
|Ω| 3
|Ω| 3 ,
2(q
+1)
2
3
3
n
C1 =
k1 =
k3 =
K1 |Ω|
2(1−µ1 )
3
βσ1
3α12
K2 |Ω|
2(1−µ3 )
3
3α32
βσ1
,
k2 =
,
k4 =
K2 |Ω|
2(1−µ2 )
3
βσ2
3α22
K1 |Ω|
2(1−µ4 )
3
βσ2
3α42
,
.
Now, setting α1 = α2 , α3 = α4 , and choosing α1 , α3 such that C1 = 0 and C2 = 0,
hence, we have
φ0 (t) ≤ g(φ),
(2.17)
where
g(s) = k1 s2µ1 + k2 s2µ2 + k3 s2µ3 + k4 s2µ4 .
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EJDE-2013/216
An integration of (2.17) from 0 to t leads to
Z φ(t)
ds
≤ t,
g(s)
φ(0)
so that if (u, v) blows up in the measure of φ as t → t∗ , we derive the lower bound
Z ∞
ds
≤ t∗ ,
g(s)
φ(0)
and Theorem 2.2 is proved. Clearly, the integral is bounded since 2µ1 > 1.
3. Non blow-up case
In this section, we consider the non blow-up property of problem (1.1)-(1.4) when
2 max(q1 , q2 ) > m − 1 > 0. To achieve this, we define the auxiliary function
Z
Z
1
1
u2 dx +
v 2 dx.
(3.1)
φ(t) =
2 Ω
2 Ω
Theorem 3.1. Suppose that (A1), (A2) hold and that 2 max(q1 , q2 ) > m − 1 > 0.
Let (u, v) be the nonnegative solution of problem (1.1)-(1.4), then (u, v) can not
blow up in the measure φ in finite time.
Proof. From(3.1), (1.1), (1.2) and (A2), we have
Z
Z
0
φ (t) =
uut dx +
vvt dx
Ω
Ω
Z
Z
≤−
|∇u|2 (b1 + b2 |∇u|2q1 )dx −
|∇v|2 (b3 + b4 |∇v|2q2 )dx
Ω
Ω
Z
+ β (um+1 + v m+1 )dx
Ω
Z
Z
m+1
2(q1 +1)
≤ (βu
− b2 |∇u|
)dx + (βv m+1 − b4 |∇v|2(q2 +1) )dx
Ω
Ω
Z λ1
q1 +1 2(q1 +1)
m+1
)
u
≤
βu
− b2 (
dx
(q1 + 1)2
Ω
Z
λ1
)q2 +1 v 2(q2 +1) dx,
+
βv m+1 − b4 (
2
(q2 + 1)
Ω
(3.2)
where the last inequality is obtained by using [14, (2.10]. For q > 0,
Z
Z
(q + 1)2 q+1
2(q+1)
w
dx ≤ (
)
|∇w|2(q+1) dx,
λ1
Ω
Ω
where λ1 is the first eigenvalue in the fixed membrane problem, as defined in Section
2. Employing Hölder inequality, we have
Z
Z
2(qm+1
2q1 −m+1
1 +1)
m+1
u
dx ≤
u2(q1 +1) dx
· |Ω| 2(q1 +1) ,
(3.3)
Ω
Ω
Z
Z
2(qm+1+1)
2q2 −m+1
21
v m+1 dx ≤
v 2(q2 +1) dx
· |Ω| 2(q2 +1) ,
(3.4)
Ω
Ω
Z
Z
2
m+1
m−1
· |Ω| m+1 .
u2 dx ≤
um+1 dx
(3.5)
Ω
Ω
EJDE-2013/216
BLOW-UP OF SOLUTIONS
7
Inserting (3.3)-(3.5) into (3.2), we see that
Z
Z
2q1 −m+1
0
m+1
φ (t) ≤
)dx
u
dx(β − M1 ( u2 dx) 2
Ω
Ω
Z
Z
2q2 −m+1
)dx
+2
v m+1 dx(β − M2 ( v 2 dx) 2
Ω
(3.6)
Ω
where
M 1 = b2 (
2q1 −m+1
λ1
)q1 +1 |Ω|− 2
,
(q1 + 1)2
M2 = b4 (
2q2 −m+1
λ1
)q2 +1 |Ω|− 2
.
(q2 + 1)2
Apparently, if (u, v) blows up in the φ measure at some time t then φ0 (t) would be
negative which leads to a contradiction. Thus, the solution (u, v) can not blow up
in the measure φ. The proof is complete.
4. Criterion for blow-up
In this section, we investigate the blow up properties of solutions for (1.1)-(1.4)
with
ρ1 (s) = b1 + b2 sq1 ,
ρ2 (s) = b3 + b3 sq2 ,
q1 , q2 , bi > 0,
i = 1 − 4.
(4.1)
For this purpose, we first define
1
φ(t) =
2
Z
1
u dx +
2
Ω
2
Z
v 2 dx
(4.2)
Ω
and
Z
b1
b2
b3
2
ψ(t) = − k∇uk2 −
|∇u|2(q1 +1) dx − k∇vk22
2
2(q1 + 1) Ω
2
Z
Z
b4
|∇v|2(q2 +1) dx +
F (u, v)dx,
−
2(q2 + 1) Ω
Ω
(4.3)
where k · k2 is the L2 (Ω)-norm.
Theorem 4.1. Suppose that (4.1) and (A2) hold. Assume further that m − 1 >
2 max(q1 , q2 ) ≥ 0 and ψ(0) > 0. If (u,v) is the non-negative solution of problem
(1.1)-(1.4), then the solution blows up at finite time t∗ with
t∗ ≤
φ(0)−2m−1
.
(2m + 1)(m + 1)
Proof. From (4.1)-(4.3), we have
Z
Z
0
2
2q1
φ (t) = −
|∇u| (b1 + b2 |∇u| )dx −
|∇v|2 (b3 + b4 |∇v|2q2 )dx
Ω
Ω
Z
+ (m + 1)
F (u, v)dx
Ω
Z
h b Z
b2
1
≥ (m + 1) −
|∇u|2 dx −
|∇u|2(q1 +1) dx
2 Ω
2(q1 + 1) Ω
Z
Z
i
b3
b4
2
2(q2 +1)
− k∇vk2 −
|∇v|
dx +
F (u, v)dx
2
2(q2 + 1) Ω
Ω
= (m + 1)ψ(t),
(4.4)
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EJDE-2013/216
and
Z
Z
|∇u|2q1 ∇u · ∇ut dx − b3
∇v · ∇vt dx
Ω
Ω
Z
ZΩ
− b4
a|u + v|m−1 (u + v)(ut + vt )dx
|∇v|2q2 ∇v · ∇vt dx −
Ω
Ω
Z
m+1
m−3
m+1
m−3
−b
|u| 2 |v| 2 uut + |v| 2 |u| 2 vvt dx
Z Ω
= (u2t + vt2 )dx ≥ 0.
ψ 0 (t) = −b1
Z
∇u · ∇ut dx − b2
(4.5)
Ω
This, together with ψ(0) > 0, implies that ψ(t) ≥ ψ(0) > 0, for t ≥ 0. By using
Hölder inequality, Schwarz inequality, (4.2) and (4.5), we obtain
Z
Z
2
(φ0 (t))2 =
uut dx +
vvt dx
≤
Ω
kuk22 kut k22
Ω
+ kvk22 kvt k22 + kuk22 kvt k22 + kuk22 kut k22
(4.6)
1
φ(t)ψ 0 (t).
2
Then, using (4.4) and (4.6), we deduce that
=
φ0 (t)ψ(t) ≤
1
1
(φ0 (t))2 ≤
φψ 0 (t),
m+1
2(m + 1)
which implies that
(ψ(t)φ(t)−2m−2 )0 ≥ 0.
(4.7)
An integration of (4.7) from 0 to t gives to
ψ(t)φ(t)−2m−2 ≥ ψ(0)φ(0)−2m−2 ≡ M.
(4.8)
Combining (4.4) with (4.8) and integrating the resultant differential inequality, we
have
φ(t)−2m−1 ≤ φ(0)−2m−1 − (2m + 1)(m + 1)M t
(4.9)
Since φ(0) > 0, (4.9) shows that φ becomes infinite in a finite time
t∗ ≤ T =
φ(0)−2m−1
.
(2m + 1)(m + 1)
This completes the proof.
Acknowledgments. The authors would like to thank the anonymous referees for
their valuable comments and useful suggestions on this work.
References
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Shun-Tang Wu
General Education Center, National Taipei University of Technology, Taipei, 106 Taiwan
E-mail address: stwu@ntut.edu.tw