Bulletin of the Iranian Mathematical Society Vol. 36 No. 2 (2010), pp 239-251.
A QUASILINEAR PARABOLIC EQUATION WITH
INHOMOGENEOUS DENSITY AND ABSORPTION
CH. LIU
Communicated by Henrik Shahgholian
Abstract. We deal with the initial-boundary value problem for
a quasilinear degenerate parabolic equation with inhomogeneous
density and absorption, which appears in a number of applications
to describe the evolution of diffusion processes, in particular nonNewtonian flow in a porous medium. We discuss the extinction of
solution and the finite speed of propagation of perturbations.
1. Introduction
We consider the following equation,
∂u
= div(|∇u|p−2 ∇u) − uq , (x, t) ∈ Q,
(1.1)
ρ(x)
∂t
with the initial-boundary conditions,
(1.2)
(1.3)
u(x, 0) = u0 (x),
u(x, t) = 0,
x ∈ ∂Ω,
where, Q = Ω × (0, ∞), Ω ⊂ Rn is a bounded smooth domain, p > 1,
q ≥ p − 1, u0 (x) ∈ C(Ω) ∩ W01,p (Ω) is a nonzero nonnegative function
and ρ(x) denotes the density. We prefer to consider a typical case of
MSC(2010): Primary: 35B05, 35K55; Secondary: 35K65.
Keywords: Quasilinear parabolic equation, extinction, finite speed of propagation of perturbations.
Received: 3 August 2009, Accepted: 1 October 2009.
c 2010 Iranian Mathematical Society.
239
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Liu
ρ(x), that is, ρ(x) = (1 + |x|)−l , l > 0 [11]. The equation (1.1) is a
prototype of a certain class of degenerate equations and appears to be
relevant to the theory of non-Newtonian fluids [1]. For the case ρ(x) = 1,
there have been many results about the existence, uniqueness and the
regularity of the solutions. We refer the readers to the bibliography
given in [4, 9, 13] and the references therein. Eidus [5], and Eidus and
Kamin [6] considered the following problem:
ρ(x)ut = ∆G(u), (x, t) ∈ QT = Ω × (0, T ),
u(x, 0) = u0 (x),
u|∂Ω = 0.
They proved that the problem had a unique nonnegative solution, satisfying the condition,
Z
Z T
1−n
lim R
G(u(x, t))dtdx = 0,
R→∞
S(R)
0
where, S(R) = {x : x ∈ Ω, |x| = R}.
Recently, Tedeev [10] considered the equation
∂u
= div(um−1 |Du|λ−1 Du),
∂t
where, λ > 0, m + λ − 2 > 0 and with ρ(x) being a positive continuous
function. They examined under which conditions on ρ(x), the corresponding nonnegative solutions of the Cauchy problems possessed the
finite speed of propagations or the interface blow-up phenomena.
The equation (1.1) is degenerate, if p > 2, or singular, if 1 < p <
2. Therefore, problem (1.1)-(1.3) does not admit classical solutions,
in general. So, we study weak solutions in the sense of the following
definition.
ρ(x)
Definition 1.1. A nonnegative function u is said to be a weak solution
of the problem (1.1)-(1.3), if u satisfies following conditions:
(1.4)
u ∈ C(Ω × (0, +∞)) ∩ L∞ (0, ∞; W01,p (Ω)),
Z TZ ∂ϕ
p−2
q
ρ(x)u
− |∇u| ∇u∇ϕ − u ϕ dxdt = 0,
∂t
0
Ω
for all ϕ ∈ C0∞ (QT ), andT ∈ (0, +∞).
The existense proof for problem (1.1)-(1.3) is similar to the one for
the case ρ(x) = 1. Here, our interest is to investigate the extinction
A quasilinear parabolic equation
241
of solutions and the finite speed of propagation of perturbations. As is
well known, one important property of solutions of the porous medium
equation is the finite speed of propagation of perturbations. So, from
the point of view of physical background, it seems to be natural to investigate this property for the equation (1.1). On the other hand, the
mathematical description of this property is that if supp u0 is bounded,
then for any t > 0, supp u(·, t) is also bounded. So, from a mathematical
viewpoint, this problem seems to be quite interesting. The monotonicity
of support of weak solutions for the p-Laplacian equation was obtained
by Yuan[12]. To prove the extinction of solution, here we use some ideas
in [12]. We first construct a supersolution, and then use the comparison principle. Our method is different from the one given in [5] for
the proof of the finite speed of propagation. We adopt the Bernis energy approach (see [2], [3]) and the main technical tools are weighted
Nirenberg’s inequality and Hardy’s inequality.
2. Comparison principle
Here, we prove some lemmas.
Lemma 2.1. For ϕ ∈ L∞ (t1 , t2 ; W01,p (Ω)) with ϕt ∈ L2 ((t1 , t2 ) × Ω),
the weak solutions u of the problem (1.1)-(1.3) on QT satisfies
Z
Z t2 Z ∂ϕ
ρ(x)u(x, t1 )ϕ(x, t1 )dx +
ρ(x)u
− |∇u|p−2 ∇u∇ϕ dxdt
∂t
Ω
t1
Ω
Z t2 Z
Z
=
uq ϕdxdt +
ρ(x)u(x, t2 )ϕ(x, t2 )dx.
t1
Ω
Ω
In particular, for ϕ ∈ W01,p (Ω), we have
Z
ρ(x)(u(x, t1 ) − u(x, t2 ))ϕdx
Ω
Z t2 Z
Z
p−2
(2.1)
−
|∇u| ∇u∇ϕdxdt −
t1
Ω
t2
t1
Z
uq ϕdxdt = 0 .
Ω
Proof. From ϕ ∈ L∞ (t1 , t2 ; W01,p (Ω)) and ϕt ∈ L2 ((t1 , t2 ) × Ω), it
follows that there exists a sequence of functions {ϕk }, for fixed t ∈
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Liu
(t1 , t2 ), ϕk (·, t) ∈ C0∞ (Ω), and as k → ∞,
kϕkt − ϕt kL2 ((t1 ,t2 )×Ω) → 0,
kϕk − ϕkL∞ (t1 ,t2 ;W 1,p (Ω)) → 0.
0
C0∞ (R)
Choose a function
j(s) ∈
such that j(s) ≥ 0, for s ∈ R; j(s) = 0,
R
∀|s| > 1, and R j(s)ds = 1. For h > 0, define jh (s) = h1 j( hs ) and
Z t−t1 −2h
ηh (t) =
jh (s)ds.
t−t2 +2h
C0∞ (t1 , t2 ),
Clearly, ηh (t) ∈
and limh→0+ ηh (t) = 1, for all t ∈ (t1 , t2 ). In
the definition of weak solutions, choose ϕ = ϕk (x, t)ηh (t). We have
Z t2 Z
Z t2 Z
ρ(x)uϕk jh (t − t1 − 2h)dxdt −
|∇u|p−2 ∇u∇ϕk ηh dxdt
t1
Ω
t2 Z
Z
t1
Z
t1
t2
Ω
Z
ρ(x)uϕk jh (t − t2 + 2h)dxdt
t1
Ω
uq ϕk ηh dxdt.
=
t1
Ω
Z
ρ(x)uϕkt ηh dxdt −
+
Z
t2
Ω
Observe that
Z t2 Z
Z
ρ(x)uϕ
j
(t
−
t
−
2h)dxdt
−
(ρ(x)uϕ
)|
dx
1
t=t
k
h
k
1
t1
Ω
Ω
Z t1 +3h Z
ρ(x)uϕk jh (t − t1 − 2h)dxdt
= t1 +h
Ω
t1 +3h Z
−
(ρ(x)uϕk )|t=t1 jh (t − t1 − 2h)dxdt
t1 +h
Ω
Z
(ρ(x)uϕk )|t − (ρ(x)uϕk )|t dx,
≤
sup
1
Z
t1 +h<t<t1 +3h Ω
and u ∈ C(Q). We see that the right hand side tends to zero, as h → 0.
Similarly,
Z
Z t2 Z
ρ(x)uϕk jh (t − t2 + 2h)dxdt − (ρ(x)uϕk )|t=t2 dx → 0.
t1
Ω
Ω
Letting h → 0 and k → ∞, we obtain:
Z
Z t2 Z ∂ϕ
ρ(x)u(x, t1 )ϕ(x, t1 )dx +
ρ(x)u
− |∇u|p−2 ∇u∇ϕ dxdt
∂t
Ω
t1
Ω
Z t2 Z
Z
=
uq ϕdxdt +
ρ(x)u(x, t2 )ϕ(x, t2 )dx.
t1
Ω
Ω
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243
In particular, for ϕ ∈ W01,p (Ω), we have
Z
ρ(x)(u(x, t1 ) − u(x, t2 ))ϕdx
Ω
Z t2 Z
Z
p−2
−
|∇u| ∇u∇ϕdxdt −
t1
Ω
t2
t1
Z
uq ϕ dxdt = 0,
Ω
which completes the proof.
For a fixed τ ∈ (0, T ), set h satisfying 0 < τ < τ + h < T . Let t1 = τ ,
t2 = τ + h, and then multiply (2.1) by h1 , for ϕ ∈ W01,p (Ω), to obtain:
Z
Z
Z
p−2
(2.2)
ρ(x)(uh (x, τ ))τ ϕdx + (|∇u| ∇u)h ∇ϕdx +
uqh ϕdx = 0,
Ω
Ω
Ω
where,
( R
1 t+h
uh (x, t) =
h
t
u(·, τ )dτ,
0,
t ∈ (0, T − h),
t > T − h.
Lemma 2.2. (Comparison principle) Let u be a weak solution of
(1.1)-(1.3). If v satisfies
∂v
ρ(x)
≥ div |∇v|p−2 ∇v − v q ,
∂t
in the sense of distributions, and
v(x, 0) ≥ u(x, 0),
v(x, t) ≥ u(x, t),
x ∈ ∂Ω,
then we have
v(x, t) ≥ u(x, t),
for all (x, t) ∈ Q.
Proof. By (2.2), we have for ϕ ∈ W01,p (Ω),
Z
Z
ρ(x)(u(x, τ ) − v(x, τ ))hτ ϕ(x)dx + (uq − v q )h (x, τ )ϕdx
Ω
Ω
Z
+ (|∇u|p−2 ∇u − |∇v|p−2 ∇v)h (x, τ )∇ϕdx ≤ 0.
Ω
For a fixed τ , we take ϕ(x) = [(u − v)h ]+ . By the property of the
Steklov mean value and noting that v(x, t) ≥ u(x, t), x ∈ ∂Ω, we see
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Liu
that ϕ(x) = [(u − v)h ]+ ∈ W01,p (Ω). Substituting this function into the
above integral equality, we obtain:
Z
ρ(x)(u(x, τ ) − v(x, τ ))hτ [(u − v)h ]+ dx
Ω
Z
≤ −
[(|∇u|p−2 ∇u − |∇v|p−2 ∇v)h ](x, τ )∇[(u − v)h ]+ dx
ΩZ
− [(uq − v q )h ](x, τ )[(u − v)h ]+ dx.
Ω
Integrating the above equality with respect to τ over (0, t), we have
Z
Z
2
ρ(x)[(u − v)h ]+ (x, t)dx −
ρ(x)[(u − v)h ]2+ (x, 0)dx
Ω
Ω
Z
≤ − [(|∇u|p−2 ∇u − |∇v|p−2 ∇v)h ](x, τ )∇[(u − v)h ]+ dx
ZΩ
− [(uq − v q )h ](x, τ )[(u − v)h ]+ dx,
Ω
It is easily seen that
Z
lim
h→0 Ω
Letting h → 0, we have
Z
ρ(x)[(u − v)h ]+ (x, 0)dx = 0.
ρ(x)|(u − v)+ |2 (x, t)dx ≤ 0,
Ω
that is,
R
2
Ω |(u−v)+ | dx
= 0. Therefore, v ≥ u, and the proof is complete.
3. Extinction and monotonicity of support
We now to prove the following theorem.
Theorem 3.1. Let u be a nonnegative weak solution of the problem
(1.1)-(1.3), and p > 2. Then,
suppu(·, s) ⊂ suppu(·, t),
for all s, t with 0 < s < t.
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245
Lemma 3.2. Let u be a nonnegative weak solution of the problem (1.1)(1.3) If p > 2, then
∂u
u
≥−
,
∂t
(p − 2)t
in the sense of distributions.
Proof. Denote:
ur (x, t) = ru(x, rp−2 t), for all (x, t) ∈ Q,
r > 1.
By p − 1 ≤ q and r > 1, we have
ρ(x)
(3.1)
(3.2)
∂ur
≥ div(|∇ur |p−2 ∇ur ) − uqr ,
∂t
ur (x, 0) = ru0 (x),
ur (x, t) = 0,
x ∈ ∂Ω.
Noting that r > 1, and using (1.2), (3.1), and (3.2), we get
(3.3)
(3.4)
ur (x, 0) ≥ u0 (x),
ur (x, t) = u(x, t),
x ∈ ∂Ω.
Applying the comparison principle, we have
(3.5)
ur (x, t) ≥ u(x, t).
For p > 2, by (3.5), we obtain:
[u(x, λt)]p−2 − [u(x, t)]p−2
(1/λ − 1)[u(x, t)]p−2
≥
,
λt − t
λt − t
where, λ = rp−2 . Letting λ → 1+ , we get
∂
1
[u(x, t)]p−2 ≥ − [u(x, t)]p−2 ,
∂t
t
in the distribution, which implies that lemma holds. Thus the proof is
now complete.
Proof of Theorem 3.1. For p > 2, from Lemma 3.2 we obtain:
∂(t1/(p−2) u)
≥ 0.
∂t
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Liu
Theorem 3.3. Let u be a weak solution of (1.1)-(1.3). If 1 < p < 2,
then there exists a time T such that
u(x, t) = 0,
for all (x, t) ∈ Ω × (T, ∞).
Proof. Denote s+ = max{s, 0}, for all s ∈ (−∞, +∞).
Define an auxiliary function,
1/(2−p)
v(x, t) = k(T − t)+
(3.6)
ln(m + x1 + · · · + xn ),
where,
"
(3.7)
(p − 1)(2 − p)np/2
k=
(2m)p ln(2m)
#1/(2−p)
,
T =
max ku0 k∞
k ln 2
2−p
and
(3.8)
m = sup{|x1 | + · · · + |xn |} + 2.
x∈Ω
Compute
(3.9)
∂v
k
(p−1)/(2−p)
=−
(T − t)+
ln(m + x1 + · · · + xn ),
∂t
2−p
div |∇v|p−2 ∇v
(p−1)/(2−p) (p−2)/2
=div{k p−1 (T − t)+
n
((m + x1 + · · · + xn )1−p , · · · , (m + x1 + · · · + xn )1−p )}
(3.10)
(p−1)/(2−p)
= − (p − 1)k p−1 (T − t)+
·
· n(p−2)/2 (m + x1 + · · · + xn )−p ,
and
q/(2−p)
v q = k q (T − t)+
lnq (m + x1 + · · · + xn ).
By ρ(x) = (1 + |x|)−l , l > 0, and using (3.6)-(3.10), we get
!
∂v
∂ ∂v p−2 ∂v
ρ(x)
≥
− vq ,
∂t
∂x ∂x ∂x
and
v(x, 0) ≥ u(x, 0),
v(x, t) ≥ u(x, t) = 0, x ∈ ∂Ω.
,
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247
Applying Lemma 2.2, we obtain:
u(x, t) ≤ v(x, t),
for all (x, t) ∈ Q. By the definition of v(x, t), we have
u(x, t) ≤ v(x, t) = 0,
for all (x, t) ∈ Ω × (T, +∞). Thus, the proof is complete.
4. Finite speed of propagation of perturbations
Here, we study the property of finite speed of perturbations.
Theorem 4.1. Assume p > 2, |ξn (0)| ≤ a, |σn (0)| ≤ b, and u is the
weak solution of the problem (1.1)-(1.3). Then, for any fixed t > 0, we
have
Z T Z
β
σn (t) − σn (0) ≤ C1 tµ
|∇u|p dxdt ,
0
ξn (t) − ξn (0) ≥ −C2 t
µ
Z
0
Ω
T
Z
β
|∇u| dxdt ,
p
Ω
where, C1 , C2 are constants depending on n, p, a, b, u0 (x), σn (t) = sup{z; x ∈
suppu(·, t)}, ξn (t) = inf{z; x ∈ suppu(·, t)}, z = xn , β > 0, µ > 0, and
a, b > 0 are constants independent of t.
Proof. First, we discuss the following Dirichlet problem,
p−2
∂u
(4.1)
= div((|∇u|2 + ε) 2 ∇u) − uq ,
ρ(x)
∂t
(4.2)
u(x, t) = 0, x ∈ ∂Ω,
(4.3)
u(x, 0) = u0ε (x).
It is well known that (4.1)-(4.3) has a nonnegative classical solution uε
[8].
Multiplying (4.1) by (z − y)s+ uε (x), s ≥ 2p, y ≥ σn (0), and letting
ε → 0, we have,
Z
Z tZ
1
s
2
(z − y)s+ uq+1 dxdτ
I=
ρ(x)(z − y)+ |u(x, t)| dx +
2 Ω
0
Ω
Z tZ
=−
|∇u|p−2 ∇u∇((z − y)s+ u)dxdτ.
0
Ω
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Liu
Then, we have,
Z tZ
I =−
0
Z tZ
=−
0
(z − y)s+ |∇u|p dxdτ
Ω
Z tZ
−
0
|∇u|p−2 ∇u∇((z − y)s+ u)]dxdτ
Ω
∇[(z − y)s+ ]u|∇u|p−2 ∇udxdτ.
Ω
Using Hölder inequality,
Z Z
Z tZ
1 t
s
p
I ≤−
(x − y)s+ |∇u|p dxdτ
(z − y)+ |∇u| dxdτ +
2
0
Ω
0
Ω
Z tZ
s−p
+ C1
(z − y)+ |u|p dxdτ
0
Ω
Z Z
Z tZ
1 t
p
≤−
(x − y)s+ |∇u|p dxdτ + C1
(z − y)s−p
+ |u| dxdτ.
2 0 Ω
0
Ω
Applying Hardy inequality [7], we obtain:
p Z
Z
p
s−p
p
(z − y)+ |u| dx ≤
(z − y)s+ |Dz u|p dx.
s−p+1
Ω
Ω
Hence,
Z
1
2
ρ(x)(z −
Ω
Z tZ
≤C
0
y)s+ |u|2 dx
1
+
2
Z tZ
0
(z − y)s+ |∇u|p dxdτ
Ω
p
(z − y)s−p
+ |u| dxdτ.
Ω
Thus,
Z
(4.4)
sup
0<τ ≤t Ω
ρ(x)(z − y)s+ |u|2 dx ≤ C
ZZ
p
(z − y)s−p
+ |u| dxdτ,
Qt
and
ZZ
(z −
(4.5)
Qt
y)s+ |∇u|p dxdτ
ZZ
p
(z − y)s−p
+ |u| dxdτ.
≤C
Qt
For (4.4), using Hardy inequality, again we have
Z
ZZ
s
2
(4.6)
sup
ρ(x)(z − y)+ |u| dx ≤ C
(z − y)s+ |∇u|p dxdτ.
0<τ ≤t Ω
Qt
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249
Set
ZZ
(z −
fs (y) =
y)s+ |∇u|p dxdτ,
Z tZ
|∇u|p dxdτ.
f0 (y) =
Qt
{x∈Ω,xn >y}
0
From (4.5) and weighted Nirenberg inequality, we have
f2p+1 (y)
ZZ
p
≤C1
(z − y)p+1
+ |u| dxdτ
Qt
Z t Z
≤C
(z −
0
where,
1
p
p
y)p+1
+ |∇u| dx
a Z
(z −
Ω
2
y)p+1
+ |u| dx
(1−a)p
2
dτ,
Ω
= a( p1 −
1
n+p+1 )
+ (1 − a) 12 . Therefore,
a=
1
p
1
p
−
−
1
2
1
n+p+1
−
1
2
< 1.
Using (4.6), we obtain:
f2p+1 (y)
Z Z
(1−a)p Z t Z
a
2
p+1
p+1
p
p
≤C
(z − y)+ |∇u| dxdτ
(z − y)+ |∇u| dx dτ
0
Qt
≤C[fp+1 (y)](1−a)p/2
Z Z
Ω
p
(z − y)p+1
+ |∇u| dxdτ
a
t1−a
Qt
≤Cfp+1 (y)(1−a)p/2+a t1−a .
Denote λ = 1 − a and µ = a + (1 − a)p/2. Then, λ > 0 and 1 < µ.
Applying Hölder’s inequality, we have
f2p+1 (y)
Z Z
µ
p
≤Ctλ
(z − y)p+1
|∇u|
dxds
+
Qt
≤Ct
λ
(p+1)µ Z t Z
2p+1
Z Z
(z −
y)2p+1
|∇u|p dxds
+
Qt
≤Ctλ [f2p+1 (y)]
0
(p+1)µ
2p+1
[f0 (y)]
pµ
2p+1
.
Ω
pµ
2p+1
|∇u| dxds
p
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Liu
Therefore,
pµ
λ
σ =1−
f2p+1 (y) ≤ Ct σ [f0 (y)] (2p+1)σ ,
p+1
µ > 0.
2p + 1
Using Hölder’s inequality again, we get
2p
1
f1 (y) ≤ [f0 (y)] 2p+1 [f2p+1 (y)] 2p+1 ≤ Ctγ [f0 (y)]1+θ ,
where,
γ=
λ
,
(2p + 1)σ
θ=
pµ
1
−
> 0.
2
(2p + 1) σ 2p + 1
Noticing that f10 (y) = −f0 (y), we obtain:
f10 (y) ≤ −Ct−γ/(θ+1) [f1 (y)]1/(θ+1) .
If f1 (σn (0)) = 0, then σn (t) ≤ b. If f1 (σn (0)) > 0, then there exists a
maximal interval (σn (0), x∗ ), in which f1 (y) > 0, and
h
i0
θ
f10 (y)
f1 (y)θ/(θ+1) =
≤ −Ct−γ/(θ+1) .
θ + 1 [f1 (y)]1/(θ+1)
Integrating the above inequality over (σn (0), x∗ ), we have
f1 (x∗ )θ/(θ+1) − f1 (σn (0))θ/(θ+1) ≤ −Ct−γ/(θ+1) (x∗ − σn (0)),
which implies that
x∗ ≤ σn (0) + Ctγ (f0 (y))θ ,
noticing that f0 (y) can be controlled by a constant C independent of y.
Similarly, we have
Z T Z
β
µ
p
ξn (t) ≥ ξn (0) − C2 t
|∇u| dxdt .
0
Ω
The proof is now complete.
Acknowledgments
The author thanks Professor Eiji Yanagida and Professor Kazuhiro Ishige
for careful reading of and valuable suggestions for the my manuscript.
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251
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Changchun Liu
Department of Mathematics, Jilin University, Changchun 130012, P. R. China and
Mathematical Institute, Tohoku University, Sendai 980-8578, Japan.
Email: [email protected]