Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 263, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
CRITICAL EXPONENT AND BLOW-UP RATE FOR THE
ω-DIFFUSION EQUATIONS ON GRAPHS WITH DIRICHLET
BOUNDARY CONDITIONS
WEICAN ZHOU, MIAOMIAO CHEN, WENJUN LIU
Abstract. In this article, we study the ω-diffusion equation on a graph with
Dirichlet boundary conditions
ut (x, t) = ∆ω u(x, t) + eβt up (x, t),
u(x, t) = 0,
(x, t) ∈ S × (0, ∞),
(x, t) ∈ ∂S × [0, ∞),
u(x, 0) = u0 (x) ≥ 0,
x ∈ V,
where ∆ω is the discrete weighted Laplacian operator. First, we prove the
existence and uniqueness of the local solution via Banach fixed point theorem.
Then, by the method of supersolutions and subsolutions we prove that the
ω-diffusion problem has a critical exponent pβ : when p > pβ , the solution
becomes global; while when 1 < p < pβ , the solution blows up in finite time.
Under appropriate hypotheses, we estimate the blow-up rate in the L∞ -norm.
Some numerical experiments illustrate our results.
1. Introduction
In this article we are concerned with the blow-up properties of the problem
ut (x, t) = ∆ω u(x, t) + eβt up (x, t),
u(x, t) = 0,
(x, t) ∈ S × (0, ∞),
(x, t) ∈ ∂S × [0, ∞),
u(x, 0) = u0 (x) ≥ 0,
(1.1)
x ∈ V,
where p > 1, β > 0, u0 (x) is nonnegative and nontrivial, and V is the set of
vertices of a graph G(V, E, ω). In general, we can split the set of vertexes V into
two disjoint subjects S and ∂S such that V = S ∪ ∂S, which are called the interior
and the boundary of V , respectively. We write (x, y) ∈ E when two vertices x, y
are adjacent and connected by an edge. Throughout this paper, all the graphs in
our concern are assumed to be simple, finite, connected, undirected and weighted.
Besides, ∆ω u(x, t) is defined as
X
ω(x, y) u(y, t) − u(x, t) , ∀x ∈ V,
y∈V
2000 Mathematics Subject Classification. 35R02, 35B44, 35B51.
Key words and phrases. ω-diffusion equation; critical exponent; blow up; blow-up rate; graph.
c
2014
Texas State University - San Marcos.
Submitted July 10, 2014. Published December 22, 2014.
1
2
W. ZHOU, M. CHEN, W. LIU
EJDE-2014/263
where ω : V × V → R denotes the weighted function, which has the following
properties:
(1) ω(x, x) = 0, for any x ∈ V ,
(2) ω(x, y) = ω(y, x), for any x, y ∈ V ,
(3) ω(x, y) = 0, if and only if (x, y) ∈
/ E.
A function on a graph is understood as a function defined just on the set of
vertices of the graph. The integration of a function f : V → R on a graph G is
defined by
Z
X
f=
f (x).
V
x∈V
As usual, the set C V × (0, ∞) consists of all function u defined on V × (0, ∞)
which satisfies u(x, t) ∈ C(0, ∞) for each x ∈ V .
The ω-diffusion equation of the form ut (x, t) = ∆ω u(x, t) and its variations can
be used to model diffusion process. Recently, more diffusion equations are taken
into account to graphs [2, 11, 14]. In [3, 8, 18], the extinction and positivity of
the solution of the ω-diffusion equation with absorption and its variations was considered. In [4], the decay rates for evolution equations of the averaging operator
∂t − ∆F were studied. In [5], the quenching phenomena for a non-local diffusion
equation with a singular absorption was discussed. Meanwhile, Xin et al [19] considered the blow-up for the ω-heat equation with Dirichlet boundary conditions and
a reaction term up (x, t) on graphs
ut (x, t) = ∆ω u(x, t) + up (x, t),
u(x, t) = 0,
(x, t) ∈ S × (0, ∞),
(x, t) ∈ ∂S × (0, ∞),
u(x, 0) = u0 (x),
(1.2)
x ∈ V.
They proved that when p > 1, the solution of problem (1.2) blows up in finite time
under some suitable conditions and when p ≤ 1, every solution is global.
Blow-up phenomenon has been widely studied in various evolution equations
[1, 6, 7, 9, 12, 13, 15, 16, 17, 21, 22]. Meier [10] considered the problem
ut = ∆u(x, t) + eβt up (x, t),
u(x, t) = 0,
(x, t) ∈ Ω × (0, ∞),
(x, t) ∈ ∂Ω × [0, ∞),
u(x, 0) = u0 (x),
(1.3)
x ∈ Ω,
and proved that the critical exponent is pβ = 1 + λβ1 , where λ1 is the first Dirichlet eigenvalue of the Laplacian in Ω. Zhang and Wang [20] studied the nonlocal
diffusion equations with Dirichlet boundary condition
Z
ut =
J(x − y) (u(y, t) − u(x, t)) dy + eβt up (x, t), (x, t) ∈ Ω × (0, ∞),
RN
u(x, t) = 0,
(x, t) ∈
/ Ω × [0, ∞),
u(x, 0) = u0 (x) ≥ 0,
(1.4)
x ∈ Ω,
where p > 1, β > 0 and Rthe kernel J ∈ C 1 (RN ) satisfies J ≥ 0 in B1 (the unit ball);
J = 0 in RN \ B1 with B1 J(z) dz = 1. They showed that the critical exponent is
coincident with that of [10].
Motivated by above research, we consider the critical exponent for problem (1.1).
We first study the local existence of the solution via Banach fixed point theorem.
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CRITICAL EXPONENT AND BLOW-UP RATE
3
Then, we deal with the existence of the critical exponent of problem (1.1) by the
method of supersolutions and subsolutions. Meanwhile, we prove that the nonnegative and nontrivial solution blows up in finite time and give the blow-up rate on
L∞ -norm. Finally, we take two examples to support our results.
This paper is organized as follows. In Section 2, we consider the local existence
of the solution. In Section 3, we share many important properties of problem (1.1),
such as critical exponent or blow-up rate. In the end, we take two examples to
check our results theoretically in Section 4.
2. Local existence of solution
In this section we prove the existence of local a solution for problem (1.1) via
Banach fixed point theorem. We first define the following Banach space:
Xt0 = {u(x, t) : u(x, t) ∈ C (V × [0, t0 ]) ; u(x, t) ≡ 0, ∀x ∈ ∂S},
with the norm
kukXt0 = max max |u(x, t)|,
t∈[0,t0 ] x∈S
where t0 > 0 is a fixed constant. And then, we consider the operator D : Xt0 → Xt0
for the fixed v0 (x) which is defined on V
(
Rt
Rt
v0 (x) + 0 ∆ω v(x, τ ) dτ + 0 eβt v p (x, τ ) dτ, x ∈ S,
Dv0 [v](x, t) =
0,
x ∈ ∂S.
In the following lemmas, we prove that the operator Dv0 : Xt0 → Xt0 is well
defined and strictly contractive under some suitable conditions.
Lemma 2.1. The operator Dv0 is well defined, mapping Xt0 to Xt0 . Moreover,
let u0 , v0 be defined on V and u, v ∈ Xt0 , then, there exists a positive constant
C = C(p, ω(x, y), kukXt0 , kvkXt0 , β, |V |) such that
kDu0 [u] − Dv0 [v]kXt0 ≤ max |u0 (x) − v0 (x)| + Ctku − vkXt0 ,
(2.1)
x∈V
where |V | denotes the number of the nodes of the graph G.
Proof. We first show that the operator Dv0 maps Xt0 to Xt0 . On the one hand, for
any (x, t) ∈ S × [0, t0 ], we have
Z
Z t
t
|Dv0 [v](x, t) − Dv0 [v](x, 0)| = ∆ω v(x, τ ) dτ +
eβτ v p (x, τ ) dτ 0
0
(2.2)
≤ 2 max |ω(x, t)||V |kvkXt0 + eβt0 kvkpXt t.
0
(x,y)∈E
From this inequality, we know that the operator Dv0 is continuous at t = 0. Similarly, for any (x, t1 ), (x, t2 ) ∈ S × [0, t0 ], we have
|Dv0 [v](x, t1 ) − Dv0 [v](x, t2 )|
Z
Z t2
t2
=
∆ω v(x, τ ) dτ +
eβτ v p (x, τ ) dτ t
t1
1
≤ 2 max |ω(x, t)||V |kvkXt0 + eβt0 kvkpXt |t2 − t1 |,
(x,y)∈E
(2.3)
0
which shows thatPDu0 is continuous in time for any t ∈ [0, t0 ]. On the other hand,
the convolution
y∈V ω(x, y) (u(y, t) − u(x, t)) is uniformly continuous. So it is
4
W. ZHOU, M. CHEN, W. LIU
EJDE-2014/263
easy to see that the operator Dv0 is continuous as the function of x. Hence, the
operator Dv0 maps Xt0 to Xt0 .
Next, we prove (2.1). For any (x, t) ∈ S × [0, t0 ], we have
|Du0 [u](x, t) − Dv0 [v](x, t)|
Z t
≤ max |u0 (x) − v0 (x)| +
|∆ω (u(x, τ ) − v(x, τ ))| dτ
x∈S
0
Z t
+
eβτ |up (x, τ ) − v p (x, τ )| dτ
0
(2.4)
≤ max |u0 (x) − v0 (x)| + 2 max |ω(x, t)| |V |ku − vkXt0 t
x∈S
+e
βt0
(x,y)∈E
pξ
p−1
ku − vkXt0 t
= max |u0 (x) − v0 (x)| + Cku − vkXt0 t,
x∈S
where ξ = max{kukXt0 , kvkXt0 } and
C = 2 max |ω(x, t)||V |ku − vkXt0 + eβt0 pξ p−1 .
(x,y)∈E
The arbitrariness of (x, t) ∈ S × [0, t0 ] gives the desired estimate (2.1).
Lemma 2.2. Suppose t0 to be small enough, then Dv0 is strictly contractive in the
ball B(u0 , 2ku0 kL∞ (V ) ).
Proof. Let u0 = v0 , (2.1) ensures that Dv0 is a strict contraction in the ball
B u0 , 2ku0 kL∞ (V ) provided that t0 is small enough. In fact, for any u, v ∈
B u0 , 2ku0 kL∞ (V ) , we have
kukXt0 ≤ 3ku0 kL∞ (V ) ,
kvkXt0 ≤ 3ku0 kL∞ (V ) ,
and thus, we obtain
kDu0 [u](x, t) − Du0 [v](x, t)kXt0 ≤ C1 t0 ku − vkXt0 ,
where
p−1
+ 2 max |ω(x, t)||V |.
C1 = eβt0 p 1 + 3ku0 kL
∞ (V )
(x,y)∈E
Therefore, if t0 is small enough such thatC1 t0 < 21 , we obtain that Dv0 is a strict
contraction in the ball B u0 , 2ku0 kL∞ (V ) . We complete the proof.
Theorem 2.3. If p > 1, then problem (1.1) has a unique solution in [0, T ) for
some T > 0 to be sufficiently small.
Proof. By the Banach fixed point theorem, Lemma 2.1 and Lemma 2.2, we can
easily obtain the existence and the uniqueness of solution to (1.1) in the time
interval [0, t0 ]. Thus, if kukXt0 < ∞ and initial data is taken as u(x, t0 ), then, the
solution can be extended to some interval [0, t1 ), where t1 > t0 . Therefore, there
exists a T > 0, which is sufficiently small, such that problem (1.1) has a unique
solution in [0, T ).
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CRITICAL EXPONENT AND BLOW-UP RATE
5
3. Global existence and blow-up phenomenon
To study the global existence and blow-up properties, we have the following
statements.
Definition 3.1. A nonnegative function u(x, t) ∈ C 1 (V × [0, T )) is a supersolution
of problem (1.1) if it satisfies
u(x, t) ≥
X
ω(x, y) (u(y, t) − u(x, t)) + eβt up (x, t),
(x, t) ∈ S × [0, T ),
y∈V
u(x, t) ≥ 0,
(3.1)
(x, t) ∈ ∂S × [0, T ),
u(x, 0) ≥ u0 (x),
x ∈ V.
Similarly, we can define the subsolution u(x, t) by reversing the inequalities.
Now, we introduce the comparison principle for the nonnegative solutions of (1.1)
which plays an important role in the proof of the existence of the critical exponent.
Theorem 3.2. Let u(x, t), u(x, t), be supersolution and subsolution of (1.1), respectively. Meanwhile, there exists a point y ∈ S such that ω(x, y) 6= 0 for any
x ∈ S. Then, for any (x, t) ∈ V × [0, T ), we have u(x, t) ≥ u(x, t).
Proof. Denote v = u − u, and choose T1 < T . In V × [0, T1 ], we have
∂v(x, t)
≤ ∆ω v(x, t) + eβt (up (x, t) − up (x, t))
∂t
= ∆ω v(x, t) + peβt ξ p−1 (x, t)v(x, t),
(3.2)
where ξ(x, t) is bounded in V × [0, T1 ].
Let v+ = max{v, 0}. Multiplying both sides of (3.2) by v+ and integrating on
V , we have
1
2
Z
V
2
v+
(x, t)
Z
Z
≤
t
∆ω v(x, t)v+ (x, t) +
V
2
peβt ξ p−1 v+
(x, t),
(3.3)
V
for all (x, t) ∈ V × [0, T1 ], due to the fact that v+ = 0 on ∂S. Now, set I(t) = {x ∈
V : u > u}. A direct use of definition of ∆w u(x, t) yields
Z
Z
∆ω v(x, t)v+ (x, t) =
V
∆ω v(x, t)v+ (x, t)
I(t)
=
X X
ω(x, y) (v(y, t) − v(x, t)) v+ (x, t)
x∈I(t) y∈V
=
X
X
ω(x, y) (v(y, t) − v(x, t)) v+ (x, t)
x∈I(t) y∈I(t)
+
X
X
x∈I(t) y∈V \I(t)
ω(x, y) (v(y, t) − v(x, t)) v+ (x, t).
(3.4)
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W. ZHOU, M. CHEN, W. LIU
EJDE-2014/263
We first note that
X
X
ω(x, y) (v(y, t) − v(x, t)) v+ (x, t)
x∈I(t) y∈I(t)
=
1 X
2
X
ω(x, y) (v(y, t) − v(x, t)) v(x, t)
x∈I(t) y∈I(t)
1 X
2
X
1 X
=−
2
X
+
(3.5)
ω(x, y) (v(x, t) − v(y, t)) v(y, t)
x∈I(t) y∈I(t)
2
ω(x, y) (v(y, t) − v(x, t)) ≤ 0.
x∈I(t) y∈I(t)
Then, notice that if x ∈ I(t) = {x ∈ V : u > u} and y ∈ V \ I(t) = {y ∈ V : u ≥ u},
then
u(x, t) > u(x, t), u(y, t) ≥ u(y, t),
so we have
u(y, t) − u(x, t) ≤ u(y, t) − u(x, t).
Then
X
X
ω(x, y) (v(y, t) − v(x, t)) v+ (x, t) ≤ 0 .
(3.6)
x∈I(t) y∈V \I(t)
Therefore,
Z
∆ω v(x, t)v+ (x, t) ≤ 0
(3.7)
V
and
Z
2
peβt ξ p−1 v+
(x, t) =
V
X
2
peβt ξ p−1 v+
(x, t)
x∈V
≤ peβT1 ξ p−1
X
2
v+
(x, t) = C
x∈V
X
(3.8)
2
v+
(x, t),
x∈V
where C = peβT1 ξ p−1 . From (3.3), (3.7) and (3.8), we have
Z
Z
2
2
v+
(x, t) ≤ 2C
v+
(x, t).
V
t
V
Since v+ (·, 0) = 0, we arrive at v+ (x, t) = 0 in V × [0, T1 ]. Due to the arbitrariness
of T1 , we obtain
u(x, t) ≥ u(x, t), ∀(x, t) ∈ V × [0, T ).
The proof is complete.
Then, we study the existence of the critical exponent for problem (1.1).
Definition 3.3. pβ is called the critical exponent of problem (1.1), if it satisfies:
(1) when p > pβ , there is a nonnegative and nontrivial global solution u of
equation (1.1);
(2) when 1 < p < pβ , the nontrivial solution u of(1.1) blows up in finite time.
The following theorems imply that if p > 1, problem (1.1) admits a critical
exponent pβ = 1+ λβ1 , where λ1 is the principle eigenvalue of the eigenvalue problem
−∆ω ϕ(x) = λ1 ϕ(x),
ϕ(x) = 0,
x ∈ S,
x ∈ ∂S.
(3.9)
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CRITICAL EXPONENT AND BLOW-UP RATE
7
Theorem 3.4. Suppose p > pβ = 1 + λβ1 and the initial value u0 ≤ z0 φ1 (x), then
the solution of (1.1) is global and positive. Here φ1 (x) which corresponds to λ1
is the positive eigenfunction with kφ1 (x)kL∞ = 1 and z0 is a constant satisfying
1
p−1
β
0 < z0 < λ1 − p−1
.
Proof. Let v(x, t) = φ1 (x)e−λ1 t , then it is the solution of the problem
∂v(x, t)
= ∆ω v(x, t), (x, t) ∈ S × (0, ∞),
∂t
v(x, t) = 0, (x, t) ∈ ∂S × [0, ∞),
v(x, 0) = φ1 (x),
(3.10)
x ∈ V.
In addition, let z(t) be the solution of the initial-value problem
dz
p−1 p
= eβt kv(·, t)kL
∞ z (t),
dt
z(0) = z0 > 0,
where z0 is a constant satisfying z0 < λ1 −
(3.11) with
kv(·, t)kp−1
L∞
=e
−λ1 (p−1)t
z(t) = z01−p −
β
p−1
1
p−1
(3.11)
. Solve the ODE problem of
, then we have
1
1−p
p−1
1 − e(β−(p−1)λ1 )t
,
(p − 1)λ1 − β
which is bounded uniformly for t ∈ [0, ∞).
Now, let u(x, t) = z(t)v(x, t) for (x, t) ∈ S × (0, ∞), then
∂u(x, t)
dz(t)
∂v
=
v(x, t) + z(t)
∂t
dt
∂t
p−1 p
βt
= e kv(·, t)kL∞ z (t)v(x, t) + z(t)∆ω v(x, t)
(3.12)
≥ ∆ω u(x, t) + eβt v p z p (t)
= ∆ω u(x, t) + eβt up (x, t),
and
u(x, t) = 0,
∀(x, t) ∈ ∂S × [0, ∞),
(3.13)
furthermore,
u(x, 0) = z0 φ1 (x) ≥ u0 (x).
(3.14)
From (3.12), (3.13) and (3.14), we obtain that u(x, t) is a supersolution of problem
(1.1). By Theorem 3.2, we conclude that (1.1) admits a global positive solution
provided that the initial value u0 is small.
Theorem 3.5. Suppose that u0 (x) is nonnegative and nontrivial. If 1 < p < pβ =
1 + λβ1 , then the corresponding solution to (1.1) on graph G blows up in the sense
P
of limt→T ∗− x∈S u(x, t)ϕ(x) = +∞ and the blow-up time T ∗ satisfies
β−(p−1)λ1 1−p 1
−
G0
ln
1−p
T∗ =
.
β − (p − 1)λ1
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W. ZHOU, M. CHEN, W. LIU
EJDE-2014/263
Proof. Consider the eigenvalue problem
−∆ω ϕ(x) = λ1 ϕ(x),
x ∈ S,
x ∈ ∂S,
ϕ(x) = 0,
(3.15)
kϕ(x)kL∞ = 1,
where λ1 is the principle eigenvalue of the eigenvalue problem. Multiplying ϕ(x)
on the both sides of (1.1) and summing on S, we have
X
X
X
ut (x, t)ϕ(x) −
∆ω u(x, t)ϕ(x) =
eβt up (x, t)ϕ(x).
(3.16)
x∈S
x∈S
x∈S
Let G(t) = x∈S u(x, t)ϕ(x), then G0 (t) =
X
∆ω u(x, t)ϕ(x)
P
P
x∈S
ut (x, t)ϕ(x), and
x∈S
=
XX
ω(x, y) (u(y, t) − u(x, t)) ϕ(x)
x∈S y∈V
=
XX
ω(x, y) (u(y, t) − u(x, t)) ϕ(x) −
x∈S y∈S
X X
ω(x, y)u(x, t)ϕ(x)
x∈S y∈∂S
1 XX
ω(x, y) (u(y, t) − u(x, t)) (ϕ(y) − ϕ(x))
2
x∈S y∈S
X X
−
ω(x, y)u(x, t)ϕ(x)
=−
(3.17)
x∈S y∈∂S
=
XX
ω(x, y) (ϕ(y) − ϕ(x)) u(x, t) −
x∈S y∈S
=
XX
X X
ω(x, y)u(x, t)ϕ(x)
x∈S y∈∂S
ω(x, y) (ϕ(y) − ϕ(x)) u(x, t)
x∈S y∈V
=
X
∆ω ϕ(x)u(x, t) = −λ1 G(t).
x∈S
Moreover, applying Jensen’s inequality, we obtain
p
X
X
eβt up (x, t)ϕ(x) ≥ eβt
u(x, t)ϕ(x) = eβt Gp (t).
x∈S
(3.18)
x∈S
Substituting (3.17) and (3.18) into (3.16), we have
G0 (t) ≥ −λ1 G(t) + eβt Gp (t).
(3.19)
Thus, using the comparison for the linear ODE, we have
1−p
1−p
β−(p−1)λ1 t
+
e
e(p−1)λ1 t , (3.20)
G1−p (t) ≤ G1−p
−
0
β − λ1 (p − 1) β − λ1 (p − 1)
and then
Gp−1 (t) ≥
1
G1−p
0
−
1−p
β−λ1 (p−1)
+
1−p
(β−(p−1)λ1 )t
β−λ1 (p−1) e
e(p−1)λ1 t
1−p
Since 1 < p < pβ = 1 + λβ1 , we have β−(p−1)λ
< 0 and G1−p (0) −
1
P
Thus, G(t) = x∈S u(x, t)ϕ(x) cannot be global.
.
(3.21)
1−p
β−(p−1)λ1
> 0.
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9
From the right side of (3.21), we have the blow-up time is
1
ln 1 − β−(p−1)λ
G1−p
0
1−p
∗
T =
.
β − (p − 1)λ1
Next we have a blow-up rate in the L∞ -norm.
Theorem 3.6. Let 1 < p < pβ and u(x, t) be a solution of problem (1.1) which
blows up at time T , then
1
e−βT p−1
1
.
(3.22)
lim (T − t) p−1 max u(x, t) =
x∈V
p−1
t→T −
Proof. Let U (t) = u x(t), t = maxx∈V u(x, t). On the one hand,
X
U 0 (t) =
ω(x, y) (U (y, t) − U (x, t)) + eβt U p (x, t).
(3.23)
y∈V
Integrating over (t, T ), we obtain
1
1
U 1−p (T ) − U 1−p (t) ≤ (eβT − eβt ),
1−p
β
then
1
1
β p−1
max u(x, t) = U (t) ≥
(eβT − eβt )− p−1 ,
x∈V
p−1
so
1
1
e−βT p−1
(T − t) β p−1
1
p−1
=
.
lim− (T − t) p−1 max u(x, t) ≥ lim−
x∈V
eβT − eβt
p−1
t→T
t→T
On the other hand,
X
ut (x, t) =
ω(x, y) (u(y, t) − u(x, t)) + eβt up (x, t)
(3.24)
y∈V
(3.25)
≥ −ku(x, t) + eβt up (x, t)
= up (x, t) eβt − ku1−p (x, t) ,
where k = maxx∈S
P
ω(x, y). In particular, we have
U 0 (t) ≥ U p (x, t) eβt − kU 1−p (x, t)
p − 1 βT
≥ U p (x, t) eβt − k
(e − eβt ) .
β
Integrating as before over (t, T ), we have
U 1−p (t) ≥
y∈V
(3.26)
(p − 1)(β + k(p − 1)) βT
k(p − 1)2
(e − eβt ) −
(T − t)eβT ,
2
β
β
then
U (t) ≤
(p − 1) (β + k(p − 1))
β2
(eβT − eβt ) −
so
1
lim− (T − t) p−1 max u(x, t) ≤
t→T
x∈V
From (3.24) and (3.27), we complete the proof.
1
1−p
k(p − 1)2
(T − t)eβT
,
β
1
e−βT p−1
p−1
.
(3.27)
10
W. ZHOU, M. CHEN, W. LIU
EJDE-2014/263
4. Examples and numerical experiments
In this section, we give two examples to illustrate our results from section 3.
First, we consider the special graph G1 for problem (1.1). The graph G1 has two
nodes x1 and x2 , where x1 is boundary and x2 is interior. Then, problem (1.1) can
be rewritten as
ut (x2 , t) = −ωu(x2 , t) + eβt up (x2 , t), t > 0,
(4.1)
u(x2 , 0) = u0 > 0,
where ω, β > 0 and p > 1 are real constants.
Figure 1. Graph G1
Equation (4.1) is the well-known Bernoulli equation. The explicit solution of
(4.1) is
1
1−p
1−p
1−p
+
e(β−ω(p−1))t
. (4.2)
u(x2 , t) = eω(p−1)t u1−p
−
0
β − (p − 1)ω β − (p − 1)ω
Consider 1 < p < pβ and ω = λ1 , then
u1−p
−
0
1−p
β−(p−1)ω
< 0, so we can get that
1−p
1−p
+
e(β−ω(p−1))t
β − (p − 1)ω β − (p − 1)ω
1−p
is decreasing on t. Notice that u0 > 0 and β−(p−1)ω
< 0, then the solution u(t) will
∞
blow up in finite time. The blow-up time on L -norm to problem (4.1) is
(β−ω(p−1))u1−p
0
ln 1 −
1−p
T =
.
β − ω(p − 1)
Then we have the limit,
1
−βT p−1
1
e
lim− (T − t) p−1 u(x2 , t) =
.
p−1
t→T
1−p
1−p
− β−(p−1)ω
We remark that if β−(p−1)ω
> 0 and u1−p
> 0, then the only solution
0
(4.2) to problem (4.1) is global.
Now, we consider a complicated graph G2 which has six nodes, where x1 , x4 and
x6 are boundary and x2 , x3 and x5 are interior. Moreover, we only consider the
weight function ω=1. Thus, the problem can be rewritten as
ut (x2 , t) = u(x3 , t) + u(x5 , t) − 3u(x2 , t) + eβt up (x2 , t),
ut (x3 , t) = u(x2 , t) + u(x5 , t) − 3u(x3 , t) + eβt up (x3 , t),
ut (x5 , t) = u(x2 , t) + u(x3 , t) − 3u(x5 , t) + eβt up (x5 , t),
u(x2 , 0) = α > 0,
u(x3 , 0) = ζ > 0,
u(x5 , 0) = γ > 0.
(4.3)
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CRITICAL EXPONENT AND BLOW-UP RATE
11
Figure 2. Graph G2
T
Let U = (u(x2 , t), u(x3 , t), u(x5 , t)) , and the coefficient matrix is


−3 1
1
A =  1 −3 1  .
1
1 −3
Thus, (4.3) can be rewritten as
Ut = A ∗ U + eβt U p ,
U0 = (α, ζ, γ)T .
(4.4)
Because of nonlinearity, it is hard to handle system (4.4) by exact analysis technique. Instead, we calculate the solution by difference method. Then the explicit
scheme is
U n+1 − U n
= A ∗ U n + (U n )p ,
∆t
(4.5)
U 0 = (α, ζ, γ)T ,
T
where U n = (u(x2 , n∆t), u(x3 , n∆t), u(x5 , n∆t)) and ∆t is the time step. Moreover, we set α = 0.4, ζ = 0.6, γ = 0.7, β = 3 and p = 2, respectively. The numerical
experiment result is shown in Figure 3. We observe that the solution blows up in
finite time.
Acknowledgments. This work was partly supported by the National Natural
Science Foundation of China (Grant No. 11301277, 41475091) and the Qing Lan
Project of Jiangsu Province.
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Weican Zhou
College of Mathematics and Statistics, Nanjing University of Information Science and
Technology, Nanjing 210044, China
E-mail address: 000496@nuist.edu.cn
Miaomiao Chen
College of Mathematics and Statistics, Nanjing University of Information Science and
Technology, Nanjing 210044, China
E-mail address: mmchennuist@163.com
Wenjun Liu (corresponding author)
College of Mathematics and Statistics, Nanjing University of Information Science and
Technology, Nanjing 210044, China
E-mail address: wjliu@nuist.edu.cn