Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 112, pp. 1–18.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
PROPAGATION OF PERTURBATIONS FOR A SIXTH-ORDER
THIN FILM EQUATION
ZHENBANG LI, CHANGCHUN LIU
Abstract. We consider an initial-boundary problem for a sixth-order thin
film equation, which arises in the industrial application of the isolation oxidation of silicon. Relying on some necessary uniform estimates of the approximate solutions, we prove the existence of radial symmetric solutions to this
problem in the two-dimensional space. The nonnegativity and the finite speed
of propagation of perturbations of solutions are also discussed.
1. Introduction
This article is devoted to the radial symmetric solutions for a sixth-order thin
film equation
∂u
= div[|u|n ∇∆2 u], x ∈ B,
∂t
with the boundary value conditions
∂u ∂∆u ∂∆2 u =
=
= 0,
∂ν ∂B
∂ν ∂B
∂ν ∂B
and the initial value condition
ut=0 = u0 (x),
where B is the unit ball in R2 , n > 0 is a constant, and ν is the outward unit
normal to ∂B.
The equation is a typical higher order equation, which has a sharp physical
background and a rich theoretical connotation. It was first introduced in [13, 14]
in the case n = 3
∂ 3 ∂5u ∂u
=
u
.
∂t
∂x
∂x5
It describes the spreading of a thin viscous fluid under the driving force of an
elastica (or light plate).
2000 Mathematics Subject Classification. 35D05, 35G25, 35Q99, 76A20.
Key words and phrases. Sixth-order thin film equation; radial solution; existence;
finite speed of propagation.
c
2012
Texas State University - San Marcos.
Submitted January 31, 2012. Published July 3, 2012.
Sponsored by the Scientific Research Foundation for the Returned Overseas Chinese Scholars,
State Education Ministry.
Changchun Liu is the corresponding author.
1
2
Z. LI, C. LIU
EJDE-2012/112
During the past years, only a few works have been devoted to the sixth-order
thin film equation [4, 9, 10, 13, 14]. Bernis and Friedman [4] have studied the initial
boundary value problems to the thin film equation
∂u
∂ ∂ 2m+1 u f (u) 2m+1 = 0,
+ (−1)m−1
∂t
∂x
∂x
n
where f (u) = |u| f0 (u), f0 (u) > 0, n ≥ 1 and proved existence of weak solutions
preserving nonnegativity. Barrett, Langdon and Nuernberg [2] considered the above
equation with m = 2. A finite element method is presented which proves to be well
posed and convergent. Numerical experiments illustrate the theory.
Recently, Jüngel and Milišić [12] studied the sixth-order nonlinear parabolic
equation
i
∂u h 1
1
= u (u(ln u)xx )xx + ((ln u)xx )2
.
∂t
u
2
x x
They proved the global-in-time existence of weak nonnegative solutions in one space
dimension with periodic boundary conditions.
Evans, Galaktionov and King [6, 7] considered the sixth-order thin film equation
containing an unstable (backward parabolic) second-order term
∂u
= div |u|n ∇∆2 u − ∆(|u|p−1 u), n > 0, p > 1.
∂t
By a formal matched expansion technique, they show that, for the first critical
exponent p = p0 = n + 1 + N4 for n ∈ (0, 54 ), where N is the space dimension,
the free-boundary problem with zero-height, zero-contact-angle, zero-moment, and
zero-flux conditions at the interface admits a countable set of continuous Nbranches
of radially symmetric self-similar blow-up solutions uk (x, t) = (T − t)− nN +6 fk (y),
x
k = 1, 2, · · · , y =
, where T > 0 is the blow-up time.
1
(T −t) nN +6
We also refer the following relevant equation
∂u
∂ n ∂3u ,
=−
u
∂t
∂x
∂x3
which has been extensively studied. Bernis and Friedman [4] studied the initial
boundary value problems to the thin film equation n > 0 and proved existence of
weak solutions preserving nonnegativity (see also [3, 15, 16, 18]). They proved that
if n ≥ 2 the support of the solutions u(·, t) is nondecreasing with respect to t.
Our purpose in this paper is to study the radial symmetric solutions for the
equation. We will study the problem in two-dimensional case, which has particular
physical derivation of modeling the oil film spreading over a solid surface, see [17].
After introducing the radial variable r = |x|, we see that the radial symmetric
solution satisfies
∂(ru)
∂ ∂W ∂ ∂V ∂ ∂u =
r|u|n
, rW =
r
, rV =
r
,
(1.1)
∂t
∂r
∂r
∂r
∂r
∂r ∂r
∂u ∂u ∂V ∂V ∂W ∂W =
=
=
=
=
= 0,
(1.2)
r=0
r=1
r=0
r=1
r=0
∂r
∂r
∂r
∂r
∂r r=1
∂r
ut=0 = u0 (r).
(1.3)
It should be noticed that the equation (1.1) is degenerate at the points where r = 0
or u = 0, and hence the arguments for one-dimensional problem can not be applied
directly. Because of the degeneracy, the problem does not admit classical solutions
in general. So, we introduce the weak solutions in the following sense
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PROPAGATION OF PERTURBATIONS
3
Definition 1.1. A function u is said to be a weak solution of the problem (1.1)–
(1.3), if the following conditions are fulfilled:
(1) ru(r, t) is continuous in QT , where QT = (0, 1) × (0, T );
√
2
(2) r|u|n/2 ∂W
∂r ∈ L (QT );
1
(3) For any ϕ ∈ C (QT ), the following integral equality holds
Z 1
Z 1
ru(r, T )ϕ(r, T ) dr −
ru0 (r)ϕ(r, 0) dr
0
ZZ0
ZZ
∂W ∂ϕ
∂ϕ
dr dt +
dr dt = 0.
r|u|n
−
ru
∂t
∂r ∂r
QT
QT
Our interest lies in the existence of weak solutions. Because of the degeneracy,
we will first consider the regularized problem. Based on the uniform estimates for
the approximate solutions, we obtain the existence. Owing to the background, we
are much interested in the nonnegativity of the weak solutions and the solutions
with the property of finite speed of propagation of perturbations. Using weighted
Nirenberg’s inequality and Hardy’s inequality, we proved these properties. This
paper is arranged as follows. We shall prove several preliminary lemmas and obtain
some a priori estimates on the solutions of regularized problem in Section 2, and
then establish the existence in Section 3. Subsequently, we discuss the nonnegativity
of weak solutions in Section 4 and the finite speed of propagation in Section 5.
2. Regularized problem
Bernis and Friedman [4] obtained several uniform estimations for the regularized solutions of fourth order thin film equation with the initial boundary value
problems. To discuss the existence of weak solutions of problem (1.1)-(1.3), we
adopt the method of parabolic regularization, namely, the desired solution will be
obtained as the limit of some subsequence of solutions of the following regularized
problem
∂(rε u)
∂W ∂ ∂V ∂ ∂ ∂u rε mε (u)
, rε W =
=
rε
, rε V =
rε
, (2.1)
∂t
∂r
∂r
∂r
∂r
∂r
∂r
∂u ∂u ∂V ∂V ∂W ∂W =
=
=
=
=
= 0,
(2.2)
r=0
r=1
r=0
r=1
r=0
∂r
∂r
∂r
∂r
∂r r=1
∂r
ut=0 = u0ε (r),
(2.3)
where rε = r + ε, mε (u) = (|u|2 + ε)n/2 and u0ε (r) is a smooth approximation of
the initial data u0 (r).
From the classical approach [4], it is not difficult to conclude that the problem
(2.1)–(2.3) admits a global classical solution. We need some uniform estimates on
the classical solutions.
We first introduce some notation. Let I = (0, 1) and for any fixed ε ≥ 0 denote
1,2
by W∗,ε
(I) the class of all functions satisfying
Z 1
1/2 Z 1
1/2
0
2
(r + ε)|u (r)| dr
+
(r + ε)|u(r)|2 dr
< +∞.
kuk∗,ε =
0
0
1,2
It is obvious that W (I) ⊂
but the class W∗,0
(I) is quite different from
1,2
1,2
W (I). In particular, we notice that the functions in W∗,0 (I) may not be bounded.
1,2
1,2
W∗,0
(I),
4
Z. LI, C. LIU
EJDE-2012/112
1,2
For ε > 0 the spaces W 1,2 (I) and W∗,ε
(I) coincide. However, it is not difficult to
1,2
prove that for u ∈ W∗,ε (I), the following properties hold:
1,2
Lemma 2.1. If 0 < α ≤ 1, and u ∈ W∗,ε
(I). Then
sup ((r + ε)α |u(r)|) ≤ Ckuk∗,ε ,
0<r≤1
where C is a constant depending only on α;
Proof. First we consider the case that 0 < r < 1/2. From the mean value theorem,
we obtain immediately
Z 1
u(x)dx = u(ξ)(1 − r)
r
for some ξ ∈ [r, 1]. Thus
Z
Z ξ
1 1
0
u(x)dx
u (x)dx +
|u(r)| ≤ |u(r) − u(ξ)| + |u(ξ)| ≤ 1−r r
r
Z 1
Z 1
≤
|u0 (x)|dx + 2
|u(x)|dx.
r
r
It follows that
1
Z
α
α
(r + ε) |u(r)| ≤
Z
0
1
(r + ε)α |u(x)|dx
(r + ε) |u (x)|dx + 2
r
1
Z
(r + ε)α |u0 (r)|dr + 2
≤
r
Z 1
0
1
Z
≤
(r + ε)α |u(r)|dr
0
(r + ε)|u0 (r)|2 dr
1/2 Z
(r + ε)2α−1 dr
1/2
0
0
+2
1
Z
1
(r + ε)|u(r)|2 dr
1
1/2 Z
(r + ε)2α−1 dr
1/2
0
0
≤ C(α)kuk∗,ε .
Finally, we discuss the case that 1/2 ≤ r ≤ 1, we have
(r + ε)α |u(r)| ≤ (1 + ε)α |u(r)|
Z
hZ 1
|u0 (x)|dx + 2
≤ (1 + ε)α
1/2
1
α
≤ (1 + ε)
h Z
1
|u(y)|dy
i
1/2
0
2
(r + ε)|u (r)| dr
1/2
Z
+2
1
(r + ε)|u(r)|2 dr
1/2 i
0
0
≤ C(α)kuk∗,ε .
The proof is complete.
1,2
Lemma 2.2. If 0 < α ≤ 1/2, and u ∈ W∗,ε
(I). Then for any β < α,
|(r1 + ε)α u(r1 ) − (r2 + ε)α u(r2 )| ≤ C|r1 − r2 |β kuk∗,ε ,
where C is a constant depending only on α and β.
Proof. For fixed 0 < r2 < r1 < 1, we have
Z
r1
|u(r2 ) − u(r1 )| ≤
r2
|u0 (t)|dt.
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PROPAGATION OF PERTURBATIONS
5
It follows that
Z r1
Z r1
(r2 + ε)α |u(r1 ) − u(r2 )| ≤ (r2 + ε)α
|u0 (t)|dt ≤
(t + ε)α |u0 (t)|dt
r2
r2
Z r1
1/2 Z r1
1/2
≤
(t + ε)|u0 (t)|2 dt
(t + ε)2α−1 dt
r2
r2
2α
≤ C(α)kuk∗,ε ((r1 + ε)
− (r2 + ε)2α )1/2
≤ C(α)kuk∗,ε |r1 − r2 |α .
On the other hand, from Lemma 2.1, we have
|((r1 + ε)α − (r2 + ε)α )u(r1 )| ≤ 2|r1 − r2 |α |u(r1 )|
≤ 2|r1 − r2 |β (r1 + ε)α−β |u(r1 )|
≤ C|r1 − r2 |β kuk∗,ε .
Therefore,
|(r1 + ε)α u(r1 ) − (r2 + ε)α u(r2 )|
≤ |((r1 + ε)α − (r2 + ε)α )u(r1 )| + (r2 + ε)α |u(r2 ) − u(r1 )|
≤ C|r1 − r2 |β kuk∗,ε
with C depending only on α and β. The proof is complete.
Remark 2.3. Let 0 < β < 1/2 and u ∈
1,2
W∗,ε
(I).
Then
|(r1 + ε)u(r1 ) − (r2 + ε)u(r2 )| ≤ C(β)|r1 − r2 |β kuk∗,ε ,
where C(β) is a constant depending only on β.
Lemma 2.4. Let u be a smooth solution of problem (2.1)–(2.3) and for any α ∈
(0, 1/2] and β < α, there is a constant M independent of ε such that
β
|rεα u(r, t) − sα
ε u(s, t)| ≤ M |r − s| ,
|rε u(r, t) − sε u(s, t)| ≤ M |r − s|β
for all r, s ∈ (0, 1), where sε = s + ε.
Proof. Multiplying (2.1) by W and integrating with respect to r over (0, 1), we
obtain
Z 1
∂rε u
∂ ∂W 0=
W−
rε mε (u)
W dr
∂t
∂r
∂r
0
Z 1
1 ∂rε V 2
∂W 2 =
+ rε mε (u)
dr.
2 ∂t
∂r
0
Hence, integrating also with respect to t, we hve
Z 1
rε V 2 dr ≤ C,
(2.4)
0
ZZ
∂W 2
dr ≤ C.
rε mε (u)
(2.5)
∂r
QT
It is easy to see that
Z 1
Z
2
rε V dr =
0
0
1
∂
∂u 1 ∂
∂u rε
·
rε
dr ≤ C.
∂r
∂r rε ∂r
∂r
6
Z. LI, C. LIU
EJDE-2012/112
A simple calculation shows that
Z 1 2 2
Z 1 2
Z 1 2
1 ∂u
∂ u ∂u
∂ u
dr
+
dr
+
2
rε
dr ≤ C.
2
2
∂r
0
0 rε ∂r
0 ∂r ∂r
Using boundary condition (2.2), we have
Z 1 2
∂ u ∂u
dr = 0.
2
0 ∂r ∂r
Hence
Z
1
rε
∂u 2
∂r
Z
0
Z
dr ≤
0
1
rε
0
1
1 ∂u 2
dr ≤ C.
rε ∂r
∂ 2 u 2
∂r2
(2.6)
dr ≤ C,
(2.7)
On the other hand, integrating the equation (2.1) on Qt = (0, 1) × (0, t), we have
Z 1
Z 1
rε u(r, t) dr =
rε u0ε (r) dr.
(2.8)
0
0
Note that, for any ρ ∈ (0, 1),
Z 1
1 + 2ε
u(ρ, t) −
sε u(s, t)ds
2
0
Z 1
=
sε [u(ρ, t) − u(s, t)]ds
0
Z ρZ ρ
Z 1Z ρ
∂u
∂u
∂u
=
sε (r, t) dr ds =
sε (r, t) dr ds +
sε (r, t) dr ds
∂r
∂r
∂r
0
s
0
s
ρ
s
Z ρZ r
Z 1Z 1
∂u
∂u
=
sε (r, t)dsdr +
sε (r, t)dsdr
∂r
∂r
0
0
ρ
r
Z 1h
Z ρ 2
i ∂u
∂u
1
r
+ εr
(r, t) dr +
(1 − r2 ) + ε(1 − r)
(r, t) dr
=
2
∂r
2
∂r
ρ
0
Z ρ
Z 1
∂u
∂u
(r, t)dr.
≤
rε (r, t)dr + 2
∂r
∂r
0
ρ
Z
1
Z
ρ
1/2
Setting ρε = ρ + ε and multiplying the above inequality by 2ρε , we obtain
Z 1
1/2
(1 + 2ε)ρ1/2
u(ρ,
t)
−
2ρ
sε u(s, t)ds
ε
ε
0
ρ
Z 1
∂u
∂u
(r, t)dr
≤
rε (r, t)dr + 4ρ1/2
ε
∂r
∂r
0
ρ
Z ρ
Z 1
∂u
∂u
≤ 2ρ1/2
rε (r, t)dr + 4
rε1/2 (r, t)dr
ε
∂r
∂r
0
ρ
Z 1
1/2
∂u
2
rε (r, t) dr
≤C
.
∂r
0
2ρ1/2
ε
Z
1/2
(2.9)
From (2.6), (2.8) and (2.9), we see that rε u(r, t) is uniformly bounded on QT .
1,2
(I) for any fixed t ∈ (0, T ), with ku(·, t)k∗,ε bounded by a
Furthermore u(·, t) ∈ W∗,ε
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PROPAGATION OF PERTURBATIONS
7
constant C independent of ε. The desired estimates then follow from the properties
1,2
of W∗,ε
(I) mentioned above.
From the above results and using the Remark, we conclude the second inequality.
The proof is complete.
Lemma 2.5. For any α > 0, there is a constant M independent of ε such that
rεα |u(r, t)| ≤ M,
kuk∗,ε ≤ M,
(2.10)
1/16
|rε u(r, t2 ) − rε u(r, t1 )| ≤ M |t2 − t1 |
(2.11)
for all r ∈ (0, 1), t1 , t2 ∈ (0, T ).
Proof. The first two estimates have already been seen from the arguments in Lemma
2.1. Now, we begin to show (2.11). Without loss of generality, we assume that
t1 < t2 and set ∆t = t2 − t1 . Integrating both sides of the equation (2.1) over
(t1 , t2 ) × (y, y + (∆t)α ) and then integrating the resulting relation with respect to
y over (x, x + (∆t)α ), we obtain
Z x+(∆t)α Z 1
h
i
(∆t)α
(y + θ(∆t)α + ε) u(y + θ(∆t)α , t2 ) − u(y + θ(∆t)α , t1 ) dθ dy
Z
x
x+(∆t)α
Z
0
y+(∆t)α
Z
t2
=
x
Z
y
x+(∆t)α
Z
t1
t2
=
x
h
∂
∂r
∂W
rε mε (u)
∂r
dτ dr dy
(y + (∆t)α + ε)mε (u(y + (∆t)α ))
t1
i
∂
∂
W (y + (∆t)α , τ ) − (y + ε)mε (u(y)) W (y, τ ) dτ dy.
×
∂y
∂y
By the mean value theorem, there exists x∗ = y ∗ +θ∗ (∆t)α , y ∗ ∈ (x, x+(∆t)α ), θ∗ ∈
(0, 1) such that the left hand side of the above equality can be expressed by
Z x+(∆t)α Z 1
α
(∆t)
(y + θ(∆t)α + ε) [u(y + θ(∆t)α , t2 ) − u(y + θ(∆t)α , t1 )] dθ dy
x
0
h
i
= (∆t)2α (y ∗ + θ∗ (∆t)α + ε) u(y ∗ + θ∗ (∆t)α , t2 ) − u(y ∗ + θ∗ (∆t)α , t1 ) .
For the right hand side, we have
Z x+(∆t)α Z t2 h
(y + (∆t)α + ε)mε (u(y + (∆t)α ))
x
t1
i
∂
∂
W (y + (∆t)α , τ ) − (y + ε)mε (u(y)) W (y, τ ) dτ dy
×
∂y
∂y
Z x+2(∆t)α Z t2
∂
=
(r + ε)mε (u(r)) W (r, τ ) dτ dr
∂r
x+(∆t)α
t1
Z x+(∆t)α
∂
−
(r + ε)mε (u(r)) W (r, τ ) dτ dr
∂r
x
Z x+2(∆t)α Z t2
∂
≤
rε mε (u(r)) W (r, τ ) dτ dr.
∂r
x+(∆t)α
t1
By (2.5), we see that
|x∗ε u(x∗ , t2 ) − x∗ε u(x∗ , t1 )| ≤ C(∆t)
1−3α
2
,
8
Z. LI, C. LIU
EJDE-2012/112
which implies, by setting α = 1/4 and using the properties of the functions in
1,2
W∗,ε
(I), that
|rε u(r, t2 ) − rε u(r, t1 )| ≤ C(∆t)1/16 .
The proof is complete.
3. Existence
After the discussion of the regularized problem, we can now turn to the investigation of the existence of weak solutions of the problem (1.1)–(1.3). The main
existence result is the following
Theorem 3.1. If u0 (r) ∈ H 2 (I), then problem (1.1)–(1.3) admits at least one weak
solution.
Proof. Let uε be the approximate solution of (2.1)–(2.3) constructed in the previous section. Using the estimates in Lemma 2.4 and 2.5, for any β < 21 , and
(r1 , t2 ), (r2 , t1 ) ∈ QT , we have
|r1ε uε (r1 , t2 ) − r2ε uε (r2 , t1 )| ≤ C(|r1 − r2 |β + |t1 − t2 |β/4 )
with constant C independent of ε. So, we may extract a subsequence from {rε uε },
denoted also by {rε uε }, such that
rε uε (r, t) → ru(r, t) uniformly in QT ,
and the limiting function ru ∈ C 1/4,1/16 (QT ).
Now we prove that u(r, t) is a weak solution of problem (1.1)-(1.3), let δ > 0 be
fixed and set Pδ = {(r, t) : r|u|n > δ}. We choose ε0 (δ) > 0, such that
rε (|uε |2 + ε)n/2 ≥
δ
,
2
(r, t) ∈ Pδ , 0 < ε < ε0 (δ).
(3.1)
Then from (2.5)
ZZ
∂W 2
C
ε
(3.2)
dr dt ≤ .
∂r
δ
Pδ
To prove the integral equality in the definition of solutions, it suffices to pass the
limit as ε → 0 in
Z 1
Z 1
ZZ
∂ϕ
rε uε (r, T )ϕ(r, T ) dr −
rε u0ε ϕ(r, 0) dr −
rε u ε
dr dt
∂t
0
0
QT
ZZ
∂Wε ∂ϕ
+
rε (u2ε + ε)n/2
dr dt = 0.
∂r ∂r
QT
The limits
Z
1
rε uε (r, T )ϕ(r, T ) =
lim
ε→0
Z
0
1
ru(r, T )ϕ(r, T ) dr,
0
Z 1
rε u0ε (r)ϕ(r, 0) dr =
u0 (r)ϕ(0, r) dr,
ε→0 0
Z Z0
ZZ
∂ϕ
∂ϕ
rε u ε
dr dt =
ru
dr dt,
lim
ε→0
∂t
∂t
QT
QT
Z
1
lim
are obvious. It remains to show that
ZZ
ZZ
∂Wε ∂ϕ
∂W ∂ϕ
lim
dr dt =
r|u|n
dr dt.
rε (u2ε + ε)n/2
ε→0
∂r
∂r
∂r ∂r
QT
QT
(3.3)
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PROPAGATION OF PERTURBATIONS
9
In fact, for any fixed δ > 0,
ZZ
ZZ
∂W ∂ϕ
2
n/2 ∂Wε ∂ϕ
dr dt −
dr dt
r|u|n
rε (uε + ε)
∂r ∂r
∂r ∂r
Q
QT
ZZ
Z ZT
∂Wε ∂ϕ
∂W ∂ϕ
≤
rε (u2ε + ε)n/2
dr dt −
r|u|n
dr dt
∂r
∂r
∂r
∂r
P
Pδ
ZZ
Z Zδ
∂W
∂ϕ
∂W ∂ϕ
ε
2
n/2
+
rε (uε + ε)
dr dt +
r|u|n
dr dt.
∂r ∂r
∂r ∂r
QT \Pδ
QT \Pδ
From the estimates (2.5), we have
ZZ
∂ϕ ∂Wε ∂ϕ
rε (u2ε + ε)n/2
dr dt ≤ Cδ sup , 0 < ε < ε0 (δ),
∂r ∂r
∂r
QT \Pδ
ZZ
∂ϕ ∂W ∂ϕ
r|u|n
dr dt ≤ Cδ sup ,
∂r
∂r
∂r
QT \Pδ
ZZ
ZZ
∂W ∂ϕ
∂Wε ∂ϕ
dr dt −
r|u|n
dr dt
rε (u2ε + ε)n/2
∂r ∂r
∂r ∂r
P
Pδ
Z Zδ
∂W
∂ϕ
ε
dr dt
rε (u2ε + ε)n/2 − r|u|n ≤
∂r ∂r
Pδ
ZZ
∂W
∂W ∂ϕ
ε
+
r|u|n
−
dr dt
∂r
∂r
∂r
Pδ
ZZ
∂W
∂W ∂ϕ
ε
2
n/2
n ∂ϕ C
√ +
≤ sup |rε (uε + ε)
− r|u| |
r|u|n
−
dr dt
∂r
∂r
∂r
∂r
δ
Pδ
and hence
ZZ
lim sup ε→0
QT
∂Wε ∂ϕ
rε
dr dt −
∂r ∂r
ZZ
r
QT
∂ϕ ∂W ∂ϕ
dr dt ≤ Cδ sup .
∂r ∂r
∂r
By the arbitrariness of δ, we see that the limit (3.3) holds. The proof is complete.
4. Nonnegativity
Just as mentioned by several authors, it is much interesting to discuss the physical solutions. For the two-dimensional problem (1.1)–(1.3), a very typical example
is the modeling of oil films spreading over a solid surface, where the unknown
function u denotes the height from the surface of the oil film to the solid surface.
Motivated by this idea, we devote this section to the discussion of the nonnegativity
of solutions.
Theorem 4.1. The weak solution u obtained in Section 3 satisfies u(r, t) ≥ 0, if
u0 (r) ≥ 0.
Proof. Suppose the contrary, that is, the set
E = {(r, t) ∈ QT ; u(r, t) < 0}
∞
(4.1)
is nonempty. For any fixed δ > 0, choose a C function Hδ (s) such that Hδ (s) =
−δ for s ≥ −δ, Hδ (s) = −1, for s ≤ −2δ and that Hδ (s) is nondecreasing for
−2δ < s < −δ. Also, we extend the function u(r, t) to be defined in the whole
plane R2 such that the extension ū(r, t) = 0 for t ≥ T + 1 and t ≤ −1. Let α(s) be
10
Z. LI, C. LIU
EJDE-2012/112
the kernel of mollifier in one-dimension, that is, α(s) ∈ C ∞ (R), suppα = [−1, 1],
R1
α(s) > 0 in (−1, 1), and −1 α(s)ds = 1. For any fixed k > 0, δ > 0, define
Z
uh (r, t) =
ū(s, r)αh (t − s)ds,
Z
R
+∞
βδ (t) =
α
t
s −
T
2
T
2
−δ
T
2
1
ds,
−δ
1
h α(s/h).
where αh (s) =
The function
ϕhδ (r, t) ≡ [βδ (t)Hδ (uh )]h
is clearly an admissible test function, that is, the following integral equality holds
Z 1
Z 1
ru0 (r)ϕhδ (r, 0) dr
ru(r, T )ϕhδ (T, r) dr −
0
0
(4.2)
ZZ
ZZ
h
∂ϕhδ
n ∂W ∂ϕδ
−
ru
dr dt +
r|u|
dr dt = 0.
∂t
∂r ∂r
QT
QT
To proceed further, we give an analysis on the properties of the test function ϕhδ (r, t).
The definition of βδ (t) implies that
δ
δ
t≥T − , h< .
2
2
Since ū(r, t) is continuous, for fixed δ, there exists η1 (δ) > 0, such that
ϕhδ (r, t) = 0,
δ
uh (r, t) ≥ − , t ≤ η1 (δ), 0 ≤ r ≤ 1, h < η1 (δ),
2
which together with the definition of βδ (t) and Hδ (s) imply
Hδ (uh (r, t)) = −δ,
t ≤ η1 (δ),
0 ≤ r ≤ 1,
h < η1 (δ)
(4.3)
(4.4)
(4.5)
and hence
1
1
η1 (δ), 0 ≤ r ≤ 1, h < η1 (δ).
(4.6)
2
2
We note also that for any functions f (t), g(t) ∈ L2 (R),
Z
Z
Z
Z
Z
h
f (t)g (t)dt =
f (t)dt g(s)αn (t − s)ds =
f (t) g(s)αn (s − t)ds
R
R
ZR
ZR
ZR
=
g(s)ds f (t)αn (s − t)dt =
f h (t)g(t)dt.
ϕhδ = −δ,
R
t≤
R
R
Taking this into account and using (4.3), (4.5), (4.6), we have
ZZ
Z +∞ Z 1
∂
h
∂
dt
ru
βδ (t)Hδ (uh ) dr
ru ϕhδ dr dt =
∂t
∂t
−∞
0
QT
ZZ
∂
(ru)h
βδ (t)Hδ (uh ) dr dt
=
∂t
QT
and hence by integrating by parts
ZZ
∂
(ru)h
βδ (t)Hδ (uh ) dr dt
∂t
QT
Z 1
Z 1
h
h
=
(ru) (r, T )βδ (T )Hδ (u (r, T )) dr −
(ru)h (r, 0)βδ (0)Hδ (uh (r, 0)) dr
0
0
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PROPAGATION OF PERTURBATIONS
11
∂(ru)h
dr dt
∂t
QT
Z 1
ZZ
∂
h
=δ
(ru) (r, 0) dr −
rβδ (t) Fδ (uh ) dr dt,
∂t
0
QT
Rs
where Fδ (s) = 0 Hδ (σ)dσ.
Again by (4.5)
Z uh (r,0)
Z 1
h
Fδ (u (r, 0)) =
Hδ (σ)dσ =
Hδ (λuh (r, 0))dλ · uh (r, 0)
ZZ
−
βδ (t)Hδ (uh )
0
0
= −δuh (r, 0)
and hence
ZZ
∂
(ru)h (βδ (t)Hδ (uh )) dr dt
∂t
QT
Z 1
Z 1
ZZ
h
h
=δ
(ru) (r, 0) dr +
rβδ (0)Fδ (u (r, 0)) dr +
0
=−T
2
0
1
−δ
ZZ
rFδ (uh )α
QT
t−
T
2
T
2
−δ
dr dt.
From (4.3), (4.6) it is clear that
Z 1
ru(r, T )ϕhδ (T, r) dr = 0,
0<h<
0
Z
−
rFδ (uh )βδ0 (t) dr dt (4.7)
QT
1
ru0 (r)ϕhδ (r, 0) dr = δ
0
Z
1
η1 (δ),
2
(4.8)
1
ru0 (r) dr.
(4.9)
0
Substituting (4.7), (4.8) and (4.9) into (4.2), we have
Z 1
ZZ
t− T 2
h
2
rFδ (u )α T
dr dt + δ
ru0 (r) dr
T − 2δ
QT
0
2 −δ
ZZ
∂W ∂ϕhδ
dr dt = 0.
+
r|u|n
∂r ∂r
P
(4.10)
By the uniform continuity of u(r, t) in QT , there exists η2 (δ) > 0, such that
u(r, t) ≥ −
δ
2
∀(r, t) ∈ P δ ,
(4.11)
where P δ = {(r, t); dist((r, t), P ) < η2 (δ)}. Here we have used the fact that u(r, t) >
0 in P . Thus
1
Hδ (uh (r, t)) = −δ, ∀(r, t) ∈ P δ/2 , 0 < h < η2 (δ)
2
where P δ/2 = {(r, t); dist((r, t), P ) < 21 η2 (δ)}.
By this and the definition of uh , Hδ (s) shows that the function ϕhδ (r, t) is only a
function of t in P , whenever h < 12 η2 (δ). Therefore
Dϕhδ (r, t) = 0,
∀(r, t) ∈ P,
0<h<
1
η2 (δ)
2
(4.12)
12
Z. LI, C. LIU
and so (4.10) becomes
Z 1
δ
ru0 (r) dr +
0
2
T − 2δ
ZZ
EJDE-2012/112
rFδ (uh )α
QT
2t − T T − 2δ
dr dt = 0,
where η(δ) = min(η1 (δ), η2 (δ)). Letting h tend to zero, we have
Z 1
ZZ
2t − T 2
δ
dr dt = 0.
ru0 (r) dr +
rFδ (u)α
T − 2δ
T − 2δ
0
QT
(4.13)
(4.14)
From the definition of Fδ (s) and Hδ (s), it is easily seen that
Fδ (u(r, t)) → −χE (r, t)u(r, t)
(δ → 0)
and so by letting δ tend to zero in (4.14), we have
ZZ
2t − T |u(r, t)|α
dr dt = 0,
T
E
which contradicts the fact that α 2t−T
> 0 for 0 < t < T . We have thus proved
T
the theorem.
Lemma 4.2. Let u be the limit function of the approximate solutions obtained
above. Then the following integral inequality holds
Z 1
ZZ
Z 1
∂V 2
2−n
ru
dr +
r
dr ds ≤
ru2−n
dr.
0
∂r
0
Qt
0
Proof. Let uε be the solution of the problem (2.1)-(2.3). Denote
Z u
Z u
dr
,
G
(u)
=
gε (r) dr.
gε (u) =
ε
2
n/2
0
0 (|r| + ε)
Multiplying both sides of the equation (2.1) by gε (uε ), and then integrating over
Qt , we obtain
Z 1
Z 1
ZZ
∂W ∂uε
dr ds =
(r + ε)Gε (u0ε (r)) dr.
(r + ε)Gε (uε (r, t)) dr +
(r + ε)
∂r ∂r
0
0
Qt
(4.15)
Integrating by parts, we obtain
Z 1
Z 1
ZZ
∂V 2
dr ds =
Gε (u0ε (r)) dr.
(r + ε)Gε (uε (r, t)) dr +
(r + ε)
∂r
0
Qt
0
Letting ε → 0 and using the fact that Gε (uε ) → u2−n /(1 − n)(2 − n) and uε → u
pointwise and the lower semi-continuity of the integrals, we immediately get the
conclusion of the lemma. The proof is complete.
Theorem 4.3. Suppose that u0 (r) > 0 and n ≥ 6, then the weak solution u satisfies
u(r, t) > 0.
Proof. Since we have proved that u(r, t) ≥ 0, if the conclusion were false, then
there would exist a point (r0 , t0 ) ∈ QT , such that u(r0 , t0 ) = 0. From the Hölder
continuity of ru, we see that
ru(r, t0 ) ≤ C|r − r0 |1/4 .
Since n ≥ 6, we have
Z 1
Z
2−n
ru(r, t0 )
dr ≥ C
0
0
1
|r − r0 |(2−n)/4 dr = ∞.
EJDE-2012/112
PROPAGATION OF PERTURBATIONS
13
On the other hand, by Lemma 4.2,
Z 1
ru(r, t0 )2−n dr ≤ C,
0
which is a contradiction. The proof is complete.
5. Finite speed of propagation of perturbations
As is well known, one of the important properties 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 thin film equation. Bernis and Friedman [4], Bernis [5] considered this
property for thin film equation. 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 the point of view of mathematics, this problem seems to be
quite interesting. We adopt the weighted energy method and the main technical
tools are weighted Nirenberg’s inequality and Hardy’s inequality.
Theorem 5.1. Assume 0 < n < 1, u0 ∈ H01 (I) ∩ H 2 (I), u0 ≥ 0, supp u0 ⊂ [r1 , r2 ],
0 < r1 < r2 < 1, and u is the weak solution of the problem (1.1)-(1.3), then for any
fixed t > 0, we have
supp u(x, ·) ⊂ [r1 (t), r2 (t)] ∩ [0, 1],
where r1 (t) = r1 − C1 tγ , r2 (t) = r2 + C2 tγ , C1 , C2 , γ > 0.
We need some uniform estimates on such approximate solutions uε .
Lemma 5.2. Let u be the limit function of the approximate solutions obtained
above. Then for any y ∈ R+ , the following integral inequality holds
ZZ
Z 1
∂ 3 u 2
1
2−n
dr ds
r(r − y)α
r(r − y)α
u
dr
+
+
+
2
∂r3
Qt
0
ZZ
ZZ
2
∂ 2 u 2
α−4 ∂u
≤C
r(r − y)+
dr ds
dr ds + C
r(x − y)α−2
+
∂r
∂r2
Qt
Qt
Z 1
ZZ
1/2
2−n
2−n
+C
r(r − y)α
|u
|
dr
+
C
r|u
|
dr
ds
,
0
+ 0
0
Qt
where C depends only on n, u0 and α ≥ 2p − 1, where (r − y)+ denotes the positive
part of r − y.
Proof. Let gε (u) and Gε (u) be defined as in the proof of Lemma 4.2. Let uε be
the approximate solutions derived from the problem (2.1)–(2.3). Then, using the
equation (2.1) and integrating by parts, we obtain
Z 1
Z 1
α
r(r − y)+ Gε (uε ) dr −
r(r − y)α
+ Gε (u0 ) dr
0
0
ZZ
∂W ∂ rε (|uε |2 + ε)n/2
=−
(r − y)α
+ gε (uε ) dr ds
∂r ∂r
Q
ZZ t
∂W
∂uε
(r − y)α
dr ds
=−
rε
+
∂r
∂r
Qt
ZZ
∂W −
rε (|uε |2 + ε)n/2
α(r − y)α−1
+ gε (uε ) dr ds
∂r
Qt
14
Z. LI, C. LIU
EJDE-2012/112
≡ I1 + I2 .
As for I1 , integrating by parts, we have
I1
ZZ
∂uε
∂W
=−
(r − y)α
dr ds
rε
+
∂r
∂r
Qt
ZZ
ZZ
∂uε
∂ ∂uε =
α(r − y)α−1
dr
ds
+
rε
(r − y)α
W rε
W
+ dr ds
+
∂r
∂r
∂r
Qt
Qt
ZZ
ZZ
∂V ∂uε
∂V ∂ 2 uε
α−1
α(r − y)+ dr ds −
α(α − 1)(r − y)α−2
dr ds
rε
=−
rε
+
2
∂r
∂r
∂r ∂r
Qt
Qt
ZZ
Z
Z
∂V 2
∂V
dr ds −
rε α(r − y)α−1
−
rε (r − y)α
V dr ds.
+
+
∂r
∂r
Qt
Qt
Therefore,
Z 1
Z
α
rε (r − y)+ Gε (uε ) dr −
0
1
rε (r −
y)α
+ Gε (u0 ) dr
ZZ
+
0
rε (r − y)α
+
Qt
∂V 2
∂r
dr ds
ZZ
∂V ∂ 2 uε
∂V ∂uε
α−1
α(r
−
y)
α(α − 1)(r − y)α−2
dr
ds
−
rε
dr ds
+
+
2
∂r ∂r
∂r ∂r
Qt
Qt
ZZ
α−1 ∂V
V dr ds
−
rε α(r − y)+
∂r
Qt
ZZ
∂W −
rε (|uε |2 + ε)n/2
α(r − y)α−1
+ gε (uε ) dr ds
∂r
Qt
≡ Ia + Ib + Ic + Id .
ZZ
=−
rε
Hölder’s inequality yields
ZZ
ZZ
∂V 2
2 2
1
α−2 ∂ uε
dr ds,
|Ia | ≤
rε (r − y)α
dr
ds
+
C
r
(r
−
y)
ε
+
+
8
∂r
∂r2
Qt
Qt
2
ZZ
ZZ
∂u 2
∂V
1
ε
α
dr ds,
rε (r − y)+
dr ds + C1
rε (r − y)α−4
|Ib | ≤
+
8
∂r
∂r
Qt
Qt
Z
Z
ZZ
∂V 2
1
2
rε (r − y)α
dr ds + C2
rε (r − y)α−2
|Ic | ≤
+
+ V dr ds.
4
∂r
Qt
Qt
Noticing that
(|uε |2 + ε)n/2 |gε (uε )| ≤
2
|uε |,
1−n
using (2.5), we have
|Id | ≤ C
ZZ
rε (|uε |2 + ε)n/2
Qt
×
ZZ
∂W 2
∂r
1/2
rε (|uε |2 + ε)n/2 gε (uε )2 dr ds
Qt
≤C
ZZ
Qt
1/2
dr ds
1/2
rε |uε |2−n dr ds
.
EJDE-2012/112
PROPAGATION OF PERTURBATIONS
15
Summing up, we have
Z 1
Z 1
rε (r − y)α
G
(u
)
dr
−
rε (r − y)α
ε
ε
+
+ Gε (u0 ) dr
0
0
ZZ
∂V 2
1
dr ds
+
rε (r − y)α
+
2
∂r
Qt
ZZ
ZZ
2
2 2
α−4 ∂uε
α−2 ∂ uε
dr
ds
+
C
dr ds
≤C
r
(r
−
y)
rε (r − y)+
1
ε
+
∂r2
∂r
Qt
Qt
ZZ
ZZ
1/2
2
2−n
+ C2
rε (r − y)α−2
V
dr
ds
+
C
r
|u
|
dr
ds
.
3
ε
ε
+
Qt
Qt
A simple calculation shows that
ZZ
∂V 2
1
rε (r − y)α
dr ds
+
2
∂r
Qt
3 ZZ
∂ 3 uε ∂ 1 ∂uε ∂ 1 ∂uε 2
∂ uε 2
1
α
+2 3
rε (r − y)+
+
=
dr ds
2
∂r3
∂r ∂r rε ∂r
∂r rε ∂r
Qt
ZZ
ZZ
∂ 3 u 2
∂ 1 ∂u 2
1
1
ε
ε
α
rε (r − y)α
dr
ds
+
r
(r
−
y)
=
dr ds
ε
+
+
2
∂r3
2
∂r rε ∂r
Qt
Qt
ZZ
∂ 3 uε ∂ 1 ∂uε dr ds.
+
rε (r − y)α
+
∂r3 ∂r rε ∂r
Qt
Note that
ZZ
∂ 3 uε ∂ 1 ∂uε dr
ds
rε (r − y)α
+
3 ∂r r ∂r
∂r
ε
Qt
ZZ
ZZ
3
2
3
1
α ∂ uε ∂uε
α ∂ uε ∂ uε
dr
ds
−
(r
−
y)
dr
ds
=
(r − y)+ 3
+
2
3 ∂r
∂r
∂r
r
∂r
Qt ε
Qt
ZZ
ZZ
2 2
∂ 3 u 2
1
rε
ε
α ∂ uε
≤
dr
ds
+
C
(r
−
y)
dr ds
rε (r − y)α
+
+
2
8
∂r3
∂r2
Qt rε
Qt
ZZ
ZZ
∂ 3 u 2
2
1
rε2
ε
α ∂uε
+
rε (r − y)α
dr
ds
+
C
(r
−
y)
dr ds
+
+
4
8
∂r3
∂r
Qt
Qt rε
ZZ
ZZ
3 2
2 2
1
α−2 ∂ uε
α ∂ uε
dr
ds
+
C
r
(r
−
y)
dr ds
rε (r − y)+
≤
ε
+
4
∂r3
∂r2
Q
Qt
Z Zt
∂u 2
ε
rε (r − y)α−4
+C
dr ds.
+
∂r
Qt
On the other hand, we have
ZZ
2
rε (r − y)α−2
+ V dr ds
Qt
ZZ
≤2
Qt
rε (r − y)α−2
+
∂ 2 u 2
ε
∂r2
ZZ
dr ds + 2
Qt
rε (r − y)α−4
+
∂u 2
ε
∂r
dr ds.
Letting ε → 0, we immediately get the desired conclusion and complete the proof
of the lemma.
16
Z. LI, C. LIU
EJDE-2012/112
Proof of Theorem 5.1. For any y ≥ r2 , Lemma 5.2 and Hardy’s inequality [11]
imply that for any t ∈ [0, T ],
Z 1
ZZ
∂ 3 u 2
1
2−n
dr ds
r(r − y)α
u
dr
+
r(r − y)α
+
+
2
∂r3
Qt
0
ZZ
ZZ
∂u 2
∂ 2 u 2
dr ds
(5.1)
≤C
r(r − y)α−4
dr
ds
+
C
r(r − y)α−2
+
+
∂r
∂r2
Qt
Qt
ZZ
∂ 2 u 2
dr ds.
≤C
(r − y)α−2
+
∂r2
Qt
For any positive number m, define
Z tZ 1
∂ 3 u 2
dr ds,
fm (y) =
r(r − y)m
+
∂r3
0
0
Z tZ
f0 (y) =
0
y
1
∂ 3 u 2
r 3 dr ds.
∂r
By y > r2 > 0, then, weighted Nirenberg’s inequality [1] and estimate (5.1) imply
that
f2p+1 (y)
ZZ
∂ 2 u 2
dr ds
≤C
(r − y)2p−1
+
∂r2
Qt
Z tZ 1
Z
2(1−a)/q
∂ 3 u 2 a 1
(r − y)2p−1
≤ C1
(r − y)2p−1
ds
dr
|u|q dr
+
+
3
∂r
0
0
0
a
Z 1
2(1−a)/q
ZZ
∂ 3 u 2
2p−1
1−a
q
dr ds .
r(r − y)+ |u| dr
(r − y)2p−1
≤ C sup
t
+
∂r3
0<t<T
0
Qt
Using (5.1) and Hardy’s inequality, we have
ZZ
Z 1
2p−1
q
r(r − y)+ |u| dr ≤ C
sup
0<t<T
∂ 3 u 2
dr ds
(r − y)2p−1
+
∂r3
Qt
0
and hence
f2p+1 (y) ≤ Ct1−a
a+2(1−a)/q
∂ 3 u 2
(r − y)2p−1
dr
ds
,
+
∂r3
Qt
ZZ
where q = 2 − n and
a=
1
1
1
2 − p − q
1
3
1
2 − 2p − q
.
Denote λ = 1 − a, µ = a + 2(1 − a)/q, then λ > 0, 1 < µ. Applying Hölder’s
inequality, we have
h ZZ
iµ
∂ 3 u 2
f2p+1 (y) ≤ Ctλ
(r − y)2p−1
dr
ds
+
∂r3
Qt
Z
Z
h
iµ
∂ 3 u 2
dr ds
≤ Ctλ
r(r − y)2p−2
+
∂r3
Qt
i(2p−2)µ/(2p+1)
h ZZ
∂ 3 u 2
dr ds
≤ Ctλ
r(r − y)2p+1
+
∂r3
Qt
Z
Z
h t 1 ∂3u i3µ/(2p+1)
2
×
r 3 dr ds
∂r
0
y
EJDE-2012/112
PROPAGATION OF PERTURBATIONS
17
≤ Ctλ [f2p+1 (y)](2p−2)µ/(2p+1) [f0 (y)]3µ/(2p+1) .
Therefore,
f2p+1 (y) ≤ Ctλ/σ [f0 (y)]3µ/(2p+1)σ ,
σ =1−
2p − 2
µ > 0.
2p + 1
Using Hölder’s inequality again, we obtain
f1 (y) ≤ [f0 (y)]2p/2p+1 [f2p+1 (y)]1/2p+1 ≤ Ctγ [f0 (y)]1+θ ,
where
2µ
1
λ
, θ=
−
> 0.
(2p + 1)σ
(2p + 1)2 σ 2p + 1
Noticing that f10 (y) = −f0 (y), we obtain
γ=
f10 (y) ≤ −Ct−γ/(θ+1) [f1 (y)]1/(θ+1) .
If f1 (r2 ) = 0, then suppu ⊂ [0, r2 ]. If f1 (r2 ) > 0, then there exists a maximal
interval (r2 , r2∗ ) 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 (r2 , r2∗ ), we have
f1 (r2∗ )θ/(θ+1) − f1 (r2 )θ/(θ+1) ≤ −Ct−γ/(θ+1) (r2∗ − r2 ),
which implies
r2∗ ≤ r2 + Ctγ (f0 (r2 ))θ .
Therefore.
sup supp u(·, t) ≤ r2 + Ctγ ≡ r2 (t).
We have thus completed the proof of Theorem 5.1.
Acknowledgment. The authors would like to express their gratitude to the anonymous referees for their valuable suggestions which lead to improvements in this
article.
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Zhenbang Li
Department of Mathematics, Jilin University, Changchun 130012, China
E-mail address: jamesbom23@yahoo.com.cn
Changchun Liu
Department of Mathematics, Jilin University, Changchun 130012, China
E-mail address: liucc@jlu.edu.cn