Volume 2010, Article ID 451619, 11 pagesdoi:10.1155/2010/451619 Research Article Existence and Nonexistence of Global Solutions of the Quasilinear Parabolic Equations with Inhomogeneous
Trang 1Volume 2010, Article ID 451619, 11 pages
doi:10.1155/2010/451619
Research Article
Existence and Nonexistence of Global
Solutions of the Quasilinear Parabolic Equations with Inhomogeneous Terms
Yasumaro Kobayashi
Faculty of Urban Liberal Arts, Tokyo Metropolitan University, Hachioji, Tokyo 192-0397, Japan
Correspondence should be addressed to Yasumaro Kobayashi,yasumaro@hkg.odn.ne.jp
Received 20 April 2010; Accepted 14 October 2010
Academic Editor: Abdelkader Boucherif
Copyrightq 2010 Yasumaro Kobayashi This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
We consider the quasilinear parabolic equation with inhomogeneous term u t Δu m x σ u p fx,
u x, 0 u0x, where 0 ≤ fx, u0x ∈ CR N , m > 0, p > max{1, m}, and σ > −2,
x : |x|2 11/2 In this paper, we investigate the critical exponents of this equation
1 Introduction
We consider the quasi-linear parabolic equation with inhomogeneous term
u t Δu m x σ u p fx x∈ RN , t > 0
,
u x, 0 u0x x∈ RN
,
1.1
where 0≤ fx, u0x ∈ CR N , m > 0, p > max{1, m}, and σ > −2, x : |x|2 11/2
For the solution ux, t of 1.1, let T∗> 0 be the maximal existence time, that is,
T∗: sup
T > 0; sup
t ∈0,T u·, t∞<∞
If T∗ ∞, we say that ux, t is a global solution; if T∗ < ∞, we say that ux, t blows up in
finite time
Trang 2For quasi-linear parabolic equations, the authors of 1 5 and so on study the homogeneous equations i.e., fx ≡ 0 in 1.1 Baras and Kersner 1 proved that 1.1 with m 1 and fx ≡ 0 has a global solution, two constants c1and c2depending on N and
p exist such that
lim inf
r→ ∞
r −2/p−1
|x|<r
dx
x σ/ p−1
≥ c1
u0dx,
lim inf
|x| → ∞ x σ2u0x p−1≤ c2.
1.3
Mochizuki and Mukai2 and Qi 4 study the case m > 0, σ 0, Pinsky 3 studies the case
m 1, σ > −2, and Suzuki 5 studies the case m ≥ 1, −∞ < σ < ∞ The following two results
are proved by them:
1 if p ≤ p∗
m,σ , then every nontrivial solution ux, t of 1.1 blows up in finite time;
2 if p > p∗
m,σ, then1.1 has a global solution for some initial value u0x, where p∗m,σ m 2 σ/N for N ≥ 2, σ > −2 and for N 1, σ > −1, p∗
m,σ m 1 for N 1,
σ ≤ −1 This p∗
m,σis called the critical exponent
On the other hand,6 9 and so on study the inhomogeneous equations i.e., fx /≡ 0
in1.1 Bandle et al 6 study the case m 1, σ 0, and Zeng 8 and Zhang 9 study the
case σ 0 In this paper, we investigate the critical exponents of 1.1 in the case fx /≡ 0 Our results are as follows
Theorem 1.1 Suppose that N ≥ 3, σ > −2, m > N − 2/N σ, and p > max{1, m} Put
p∗m,σ: m N σ
a If p ≤ p∗
m,σ , then every nontrivial solution u x, t of 1.1 blows up in finite time.
b If p > p∗
m,σ , u0x ≤ C1x −2σ/p−m , and f x ≤ C2x −m2σ/p−m−4 , then1.1 has
a global solution for some constants C1and C2.
Theorem 1.2 Suppose that N 1, 2, σ ≥ −2, m > 0, and p > max{1, m} Then every nontrivial
solution u x, t of 1.1 blows up in finite time.
these theorems, the same results as Theorem 1 in8 are obtained
We will proveTheorem 1.1a and b in Sections3 and4, respectively The proof of Theorem 1.2is included in the proof ofTheorem 1.1a
In the following, R and T are two given positive real numbers greater than 1 C is a positive constant independent of R and T, and its value may change from line to line.
2 Preliminaries
In this section, we first give the definition of a solution for Problem1.1 and then cite the comparison theorem and a known result
Trang 3Definition 2.1 A continuous function u ux, t is called a solution of Problem 1.1 in Q T ≡
RN × 0, T if the following holds:
i ∇x u m ∈ L2
locRN;
ii for any bounded domain D ⊂ R N and for all ψ ∈ C2D × 0, T and vanishing on
∂D × 0, T,
τ
0
D
u∂tψ − ∇u m ∇ψ x σ u p ψ fψdx dt
D
u x, ·ψx, · τ
0dx, 2.1
for all τ ∈ 0, T.
Lemma 2.2 the comparison theorem Let u, v ∈ C0, T; L2
locΩ, ∇u m , ∇v m ∈ L20, T;
L2
locΩ, and satisfy
ut − Δu m ≤ v t − Δv m , x, t ∈ Ω T ,
u ≤ v, x, t ∈ ∂Ω T 2.2
Ω RN andΩT Ω × 0, T
Lemma 2.3 the monotonicity property Let ux be a nonnegative sub-solution to the stationary
increasing to t.
We first consider the following problem:
ut Δu m x σ u p fx x∈ RN , t > 0
,
u x, 0 0 x∈ RN
.
3.1
It is clear that the positive solution of Problem3.1 is a sub-solution of Problem 1.1 If every positive solution of Problem3.1 blows up in finite time, then, byLemma 2.2, every positive solution of Problem1.1 also blows up in finite time Therefore, we only need to consider Problem3.1
The stationary problem of Problem3.1 is as follows:
Δu m x σ u p fx 0 x∈ RN
It is obvious that 0 is a sub-solution of Problem3.2 and does not satisfy Problem 3.2 Thus,
by making use of Lemmas 2.2and2.3, the positive solution of Problem3.1 is monotone
increasing to t.
Trang 4We argue by contradiction Assume that Problem3.1 has a global positive solution
for p ≤ p∗
m,σ
Let ϕr and ηt be two functions in C∞0, ∞, and satisfy
i 0 ≤ ϕr ≤ 1 in 0, ∞; ϕr ≡ 1 in 0, 1 , ϕr ≡ 0 in 2, ∞; −C ≤ ϕr ≤ 0,
|ϕr| ≤ C;
ii 0 ≤ ηt ≤ 1 in 0, ∞; ηt ≡ 1 in 0, 1 , ηt ≡ 0 in 2, ∞; −C ≤ ηt ≤ 0.
For R > 1 and T > 1, define Q R,T ≡ B 2R ×0, 4T , and let Ψr, t ϕ R rη T t be a cut-off function, where ϕ R r ϕr/R, η T t ηt/2T It is easy to check that
−C
d2ϕ R r
dr2
≤
C
2T ≤ dη T t
dt ≤ 0. 3.3 Let
I R
Q R,T
where s > 1 is a positive number to be determined Then
IR
Q R,T
−u∂ tΨs ∇u m∇Ψs − fΨ s
B 2R
u x, ·Ψr, · s 4T
0 dx
−
Q R,T
uϕ s R dη
s T
Q R,T
∇u m η T s ∇ϕ s
R dx dt−
Q R,T
fΨs dx dt
B 2R
u x, ·ϕ R r s η T·s 4T
0 dx
−
Q R,T
uϕ s R dη
s T
Q R,T
u m η s T Δϕ s
Rdx dt−
Q R,T
fΨs dx dt
4T
0
|x|2R u m η s T ∂ϕ
s R
3.5
Since RN f xdx > 0, there exist δ > 0 and R0> 1 such that B
R f xdx ≥ δ for R > R0:
Q R,T
fΨs dx dt
4T
0
η s T
B 2R
fϕ s
2T
T
B R
Hence, by the definition of ϕ R and η T, we have
I R≤ −
4T
2T
B
uϕ s R
dη s T
4T
0
B \B u m η s
T Δϕ s
Trang 5SinceΔϕ s
R sϕ s−1
R Δϕ R ss − 1ϕ s−2
R |∇ϕ R|2and
Δϕ R r d2ϕ R r
r
dϕ R r
dϕ
R r
dr
2
we obtain from3.3 that
Δϕ s
R ≤ sϕ s−1
R
C
R
ss − 1ϕ s−2
R
C R
2
≤ C
R2ϕ s R−2 3.9
in B 2R \ B Rand
dη s T
dt sη s−1
T
dηT
dt ≥ −sη s−1
T
C
2T ≥ −C
s−1
in2T, 4T Thus, 3.7 becomes
T
4T
2T
B 2R
uΨs−1dx dt C
R2
4T
0
B 2R \B R
Let s be large enough such that s − 1p ≥ s and s − 2p/m ≥ s, and let A σ R be as follows:
A σ R
⎧
⎨
⎩
R N −σ/p−1
σ < N
p− 1,
logR 1 σ ≥ Np− 1. 3.12 Then, by making use of Young’s inequality, we have
C
T
4T
2T
B 2R
uΨs−1dx dt
≤
4T
2T
B 2R
1
4p p x σ u pΨs−1p 4q
q x −σq/p C q T −q
dx dt
≤ 1 4
4T
0
B 2R
x σ u pΨs dx dt CT −p/p−14T
2T
B 2R
x −σ/p−1 dx dt
≤ 1
4I R CT1−p/p−1A σ R,
3.13
Trang 6where 1/p 1/q 1 and
C
R2
4T
0
B 2R \B R
u mΨs−2dx dt
≤
4T
0
B 2R \B R
1
4p
px σ u mpΨs−2p
4q
qx −σq/pC qR −2q
dx dt
≤ 1
4
4T
0
B 2R
x σ u pΨs dx dt CR −2p/p−m4T
0
B 2R \B R
x −mσ/p−m dx dt
≤ 1
4I R CTR −2p/p−m R N −mσ/p−m ,
3.14
where p p/m, 1/p 1/q 1 Thus, 3.11 becomes
I R ≤ 1
2I R TCT −p/p−1 A σ R CR N −2pmσ/p−m − δ. 3.15
For N ≥ 3, since σ > −2, 1/p 1/q 1, and max{1, m} < p ≤ mN σ/N − 2, we have
N − 2p − N σm
For N 2, since σ ≥ −2, m > 0, and p > max{1, m}, we have
2−2p mσ
−2 σm
For N 1, since σ ≥ −2, m > 0, and p > max{1, m}, we have
1−2p mσ
−p − 1 σm
−2 σm
p − m ≤ 0. 3.18 Let T ≥ A σ R p−1/p such that T −p/p−1 A σ R ≤ 1, then
that is,
4T
0
B 2R
Thus
2T
T
B
Trang 7By the integral mean value theorem, there exists t1∈ T, 2T such that
2T
T
B R
x σ u p dx dt T
B R
x σ u x, t1p dx ≤ CT, 3.22
that is,
B R
x σ u x, t1p dx ≤ C. 3.23
Since T is a large positive number and a random selection, and ux, t is monotone increasing
to t, there exists a positive number TR > 1 for any fixed R > R0such that, for all t > T R,
B R
x σ u x, t p dx ≤ C. 3.24
By the monotone increasing property of ux, t, B
R x σ u x, t p dx also is increasing to t This,
combined with3.24, yields that the limit I∞
R exists such that
I R∞≡ lim
t→ ∞
B R
x σ u x, t p dx ≤ C. 3.25
Since ux, t is nonnegative, I∞
R is monotone increasing to R This, combined with 3.25, yields that limR→ ∞I R∞exists Thus, for any small ε > 0, there exists a large positive constant which still is denoted by R0, such that, for R > R0,
lim
t→ ∞
B 2R \B R
x σ u x, t p dx ≡ I∞
2R − I∞
Hence, by similar argument as that in3.24, there exists a large positive number TR > 1 such that
B 2R \B R
x σ u x, t p dx < ε, ∀t > TR. 3.27
On the other hand, we argue as in6,10 Let ξx ∈ C2RN be a positive function satisfying
i 0 ≤ ξx ≤ 1 in R N ; ξx ≡ 1 in B1, ξx ≡ 0 in B c
2;
ii ∂ξ/∂ν 0 on ∂B2\ B1;
iii for any α ∈ 0, 1, there exists a positive constant C αsuch that|Δξ| ≤ C α ξ α
Let R and TR be as defined in 3.26 and 3.27 Multiplying 3.1 by ξ R x ξx/R
and then integrating by parts inRN, we have
d
dt
RN
uξ R dx
B \B u m Δξ R dx
RN
x σ u p ξ R dx
RN
fξ R dx. 3.28
Trang 8By the definition of ξ R x, H¨older’s inequality, and 3.27, we have
B 2R \B R
u m Δξ R dx
≤
C α
R2
B 2R \B R
u m ξ R α dx
≤ C α
R2
B 2R \B R
x σ u mpdx
1/p
B 2R \B R
x −σq/pξ R αqdx
1/q
≤ Cα
R2
B 2R \B R
x σ u p dx
m/p
B 2R \B R
x −mσ/p−m dx
p−m/p
≤ Cε m/p R N−mσ/p−mp−m/p−2 ≤ Cε m/p ,
3.29
where p p/m, 1/p 1/q 1, since
p − m
p − 2 N − 2p − N σm
Let F R t RN uξ R dx and G R t RN x σ u p ξ R dx Then, by making use of3.29 and
RN f xdx ≥ δ for R > R0,3.28 becomes
FR t ≥ G R t − Cε m/p δ. 3.31
Thus, let ε be small enough such that Cε m/p ≤ δ/2, then F
R t ≥ G R t δ/2.
Let t0> T R By making use of H¨older’s inequality, we obtain that
FR t ≤
RN x σ u p ξRdx
1/p
RN x −σq/p ξRdx
1/q
≤ G R t 1/p
B 2R
x −σ/p−1 dx
p−1/p
≤ CG R t 1/p A σ R p−1/p ,
3.32
where 1/p 1/q 1 Thus, we obtain that
t
t0
F R s p ds ≤ CA σ R p−1t
t0
G R sds ≤ CA σ R p−1t
t0
FR sds
≤ CA σ R p−1F R t − F R t0.
3.33
Trang 9Since F R t ≥ 0 for all t ≥ 0, we have
F R t ≥ CA σ R −p1
t
t0
F R s p ds F R t0
≥ CA σ R −p1
t
t0
F R s p ds.
3.34
Let gt t
t0FR s p ds, then
g t F R t p ≥ CA σ R −pp−1 g t p 3.35
Let t1 > t0such that gt1 > 0 Since p > 1, by solving the differential inequality 3.35 in t1, t ,
we have
t
t1
g s
g s p ds ≥ CA σ R −pp−1
t
t1
ds,
g t1−p≤ gt11−p− Cp− 1Aσ R −pp−1 t − t1,
g t ≥g t11−p− Cp− 1A σ R −pp−1 t − t1−1/p−1
3.36
Thus, there exists T1with t1 < T1 ≤ t1 Cp − 1−1A σ R p p−1 g t11−p, such that limt ↑T1g t
∞, which implies that gt and then u blow up in finite time It contradicts our assumption.
Therefore, every positive solution of Problem 3.1 blows up in finite time Hence, every positive solution of Problem1.1 blows up in finite time
In this section, we prove that for p > mN σ/N −2, there exist some fx and u0x, such
that Problem1.1 admits a global positive solution
We first consider the stationary problem of Problem1.1 as follows:
Δu m x σ u p fx 0 x∈ RN
Let vx C1x −s , where s 2 σ/p − m and the positive constant C1satisfies
C p1−m msN − ms − 2 m 2 σ
N − 2p − N σm
Trang 10
Then, we have
−Δv m ms
2 C
m
1
|x|2 1−ms/2−1Δ|x|2 1
−ms ms 2
m
1
|x|2 1−ms/2−2 ∇
|x|2 1 2
NmsC m
1x −ms−2 − msms 2C m
1|x|2x −ms−4
msN − ms − 2C m
1x −ms−2 msms 2C m
1x −ms−4
4.3
Since C p1−m msN − ms − 2 and −ms − 2 σ − ps, we have
−Δv m C p
1x σ −ps C2x −ms−4 x σ v p C2x −ms−4 , 4.4
where C2 msms 2C m
1 Thus, if f x ≤ C2x −ms−4 and u0x ≤ vx, then v is
a supersolution of Problem 1.1 It is obvious that 0 is s sub-solution of Problem 1.1 Therefore, by the iterative process and the comparison theorem, Problem1.1 admits a global positive solution
Acknowledgments
This paper was introduced to the author by Professor Kiyoshi Mochizuki in Chuo University The author would like to thank him for his proper guidance The author would also like
to thank Ryuichi Suzuki for useful discussions and friendly encouragement during the preparation of this paper
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