Boletı́n de la Asociación Matemática Venezolana, Vol. XII, No. 1 (2005)
43
Some properties of solutions for the generalized
thin film equation in one space dimension
Changchun Liu
Abstract
In this paper, the author studies a generalized thin film equation in
one space dimension. Some results on the finite speed of propagation of
perturbations and regularity of solutions are established.
1
Introduction
In this paper, we consider the variant version of the thin film equation, namely
∂u
+ div(|∇∆u|p−2 ∇∆u) = 0,
∂t
x ∈ Ω, t > 0, p > 2,
(1.1)
where Ω ⊂ RN is a bounded domain with smooth boundary.
The equation (1.1) is a typical higher order equations, which have a sharp
physical background and a rich theoretical connotation. It was J.R.King [6]
who first derived the equation. Equation (1.1) describes the surface tension
driven evolution of the height u(x, t) of a thin liquid film on a solid surface in
lubrication approximation [6, 7, 9]. The exponent p is related to the rheological
properties of the liquid: p = 2 corresponds to a Newtonian liquid, whereas p 6= 2
emerges when considering “power-law” liquids. When p > 2 the liquid is said
to be “shear-thinning”.
J.R.King [6] studied Cauchy problem of the equation for one-dimensional,
exploiting local analyses about the edge of the support and special closed form
solutions such as travelling waves, separable solutions, instantaneous source
solutions.
On the basis of physical consideration, as usual the equation (1.1) is supplemented with the natural boundary value conditions
u = ∆u = 0,
x ∈ ∂Ω, t > 0.
(1.2)
The boundary value conditions (1.2) is a reasonable for the thin film equation
or the Cahn-Hilliard equation, (see [1, 2, 4]) and initial value condition
u(x, 0) = u0 (x),
x ∈ Ω.
(1.3)
44
C. Liu
This equation is something quite like the p-Laplacian equation, but many
methods used in the p-Laplacian equation such as those methods based on maximun principle are no longer valid for this equation. Because of the degeneracy,
the problem (1.1)-(1.3) does not admit classical solutions in general. So, we
introduce weak solutions in the following sense
Definition A function u is said to be a weak solution of the problem (1.1)–
(1.3), if the following conditions are satisfied:
1) u ∈ L∞ (0, T ; W 3,p (Ω)) ∩ C(0, T ; L2 (Ω)), u, ∆u ∈ W01,p (Ω),
∂u
∞
−1,p0
(Ω)), where p0 is the conjugate exponent of p;
∂t ∈ L (0, T ; W
∞
2) For any ϕ ∈ C0 (QT ), the following integral equality holds:
ZZ
ZZ
∂ϕ
dxdt +
u
|∇∆u|p−2 ∇∆u∇ϕ dxdt = 0;
∂t
QT
QT
3) u(x, 0) = u0 (x), in L2 (Ω)
In [8] they prove the existence and uniqueness of weak solution for dimension
N ≤ 2. This paper is a further step in the study of the properties of solutions,
we proved the finite speed of propagation of perturbations and regularity of
solutions of the problem (1.1)-(1.3) for one dimensional case.
In addition, through out this paper, we set I = (0, 1).
2
Finite speed of propagation of perturbations
In this section, we are going to prove the following theorem.
Theorem 2.1 Assume p > 2, supp u0 ⊂ [x1 , x2 ], 0 < x1 < x2 < 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, ·) ⊂ [x1 (t), x2 (t)] ∩ [0, 1],
1
1
where x1 (t) = x1 − C1 t 3p−2 , x2 (t) = x2 + C2 t 3p−2 ,
p−2
! 2(3p−2)
Z T Z x1
Z
3 p
C1 = C
|D u| dxdτ
, C2 = C
0
0
0
T
Z
p−2
! 2(3p−2)
1
3
p
|D u| dxdτ
x2
C = 23p−2 (2p + 1)p (p − 1)p−1 (1 + p−p ).
To prove the Theorem 2.1 we need the following result
Lemma 2.2 The weak solution u of the problem (1.1)-(1.3), satisfying for any
0 ≤ ρ ∈ C 2 (I),
Z
Z
1 1
1 1
ρ(x)|Du(x, t)|2 dx −
ρ(x)|Du0 (x)|2 dx
2 0
2 0
ZZ
= −
|D3 u|p−2 D3 uD2 (ρ(x)Du) dx,
Qt
,
Properties of solutions for thin film equation
where Qt = (0, 1) × (0, t), D =
45
∂
∂x .
Proof. Similar to the discussion in [8], we can also easily prove that for any
0 ≤ ρ ∈ C 2 (Ω),
Z
1
ρ(x)|Du(x, t)|2 dx ∈ C([0, T ]).
fρ (t) =
2 Ω
Consider the functional
1
Φρ [v] =
2
Z
ρ(x)|Dv(x)|2 dx.
Ω
It is easy to see that Φρ [v] is a convex functional on H01 (Ω).
For any τ ∈ (0, T ) and h > 0, we have
Φρ [u(τ + h)] − Φρ [u(τ )] ≥ hu(τ + h) − u(τ ), −D(ρ(x)Du(τ ))i.
δΦ [v]
ρ
= −D(ρ(x)Dv), for any fixed t1 , t2 ∈ [0, T ], t1 < t2 , integrating the
By δv
above inequality with respect to τ over (t1 , t2 ) , we have
Z t2 +h
Z t1 +h
Z t2
hu(τ +h)−u(τ ), −D(ρ(x)Du(τ ))i dτ.
Φρ [u(τ )] dτ −
Φρ [u(τ )] dτ ≥
t2
t1
t1
Multiplying both sides of the above equality by h1 , and letting h → 0, we obtain
Z t2
∂u
Φρ [u(t2 )] − Φρ [u(t1 )] ≥
h , −D(ρ(x)Du(τ ))i dτ.
∂t
t1
Similarly, we have
Φρ [u(τ )] − Φρ [u(τ − h)] ≤ h(u(τ ) − u(τ − h)), −D(ρ(x)Du(τ ))i.
Thus
Z
t2
Φρ [u(t2 )] − Φρ [u(t1 )] ≤
h
∂u
, −D(ρ(x)Du(τ ))i dτ,
∂t
h
∂u
, −D(ρ(x)Du(τ ))i dτ.
∂t
t1
and hence
Z
t2
Φρ [u(t2 )] − Φρ [u(t1 )] =
t1
Taking t1 = 0, t2 = t, we get from the definition of solutions that
Z t
Φρ [u(t)] − Φρ [u(0)] =
h−D(|D3 u|p−2 D3 u), −D(ρ(x)Du(τ ))i dτ
0
Z t
h|D3 u|p−2 D3 u, D2 (ρ(x)Du(τ ))i dτ.
= −
0
46
C. Liu
This completes the proof.
Proof of Theorem 2.1 By Lemma 2.2, take ρ(x) = (x − y)s+ , y ∈ [x2 , 1), we
have
Z
Z tZ 1
1 1
|D3 u|p−2 D3 uD2 [(x − y)s+ Du)] dxdτ.
(x − y)s+ |Du(x, t)|2 dx = −
2 0
0
0
Denote the left side of above equality by J, then we have
Z tZ 1
|D3 u|p−2 D3 uD2 [(x − y)s+ Du] dxdτ
J = −
0
0
Z tZ
1
(x − y)s+ |D3 u|p dxdτ
= −
0
tZ 1
0
Z
−2s
2
3 p−2 3
(x − y)s−1
D u dxdτ
+ D u|D u|
0
0
1
Z tZ
s−2
(x − y)+
|D3 u|p−2 D3 uDu dxdτ
−s(s − 1)
0
Z tZ
≤
(x − y)s+ |D3 u|p dxdτ +
−
0
0
Z tZ
(x −
0
1
+
4
Z tZ
1
2
Z tZ
−
Z tZ
s−p
y)+
|D2 u|p
0
y)s+ |D3 u|p
0
0
(x − y)s+ |D3 u|p dxdτ + C1
0
s−2p
(x − y)+
|Du|p dxdτ
0
Z tZ
0
Z tZ
1
dxdτ + C2
1
1
s−p
(x − y)+
|D2 u|p dxdτ
0
1
s−2p
(x − y)+
|Du|p dxdτ,
+C2
0
(x − y)s+ |D3 u|p dxdτ
Z tZ
0
0
1
dxdτ
0
1
(x −
0
1
4
1
+C1
≤
0
1
0
using Hardy inequality [5], we have
Z 1
Z 1
s−p
p
|D2 u|p dx.
|Du|
dx
≤
C
(x − y)+
(x − y)s−2p
+
0
0
Hence
Z
1
2
1
y)s+ |Du|p
1
dx +
2
Z tZ
1
(x −
(x − y)s+ |D3 u|p dxdτ
0
0
0
Z tZ 1
2 p
≤ C1
(x − y)s−p
+ |D u| dxdτ.
0
(2.1)
0
Thus
Z
(x −
sup
0<τ ≤t
1
0
y)s+ |Du|p
Z tZ
dx ≤ C1
0
0
1
s−p
(x − y)+
|D2 u|p dxdτ,
(2.2)
47
Properties of solutions for thin film equation
and
1
Z tZ
1
Z tZ
s−p
(x − y)+
|D2 u|p dxdτ.
(2.3)
For (2.2) again using Hardy inequality, we have
Z tZ 1
Z 1
s
p
(x − y)s+ |D3 u|p dxdτ.
sup
(x − y)+ |Du| dx ≤ C
(2.4)
y)s+ |D3 u|p
(x −
dxdτ ≤ C1
0
0
0
0
0<τ ≤t
0
0
0
Let
Z tZ
Es (y) =
Z tZ
1
(x − y)s+ |D3 u|p dxdτ,
E0 (y) =
|D3 u|p dxdτ.
y
0
0
0
1
In (2.3), set s = 2p + 1 and using Nirenberg inequality [3], we have
E2p+1 (y)
ZZ
2 p
≤ C1
(x − y)p+1
+ |D u| dxdτ
Qt
Z t Z
1
≤ C
(x −
0
where
1
p
=
3 p
y)p+1
+ |D u|
a Z
(1−a)p/2
p+1
2
dx
(x − y)+ |Du| dx
dτ,
0
1
p+2
Ω
a( p1
+
−
2
p+2 )
+ (1 −
0<a=
a) 21 ,
1
p
1
p
−
−
therefore
1
p+2
2
p+2
−
−
1
2
1
2
< 1.
Using (2.4) we obtain
E2p+1 (y)
ZZ
(1−a)p/2 Z t Z 1
3 p
3 p
a
≤ C
(z − z0 )p+1
|D
u|
dxdτ
((x − y)p+1
+
+ |D u| dx) dτ
Qt
0
(1−a)p/2
ZZ
≤ C[Ep+1 (y)]
(x −
3 p
y)p+1
+ |D u|
0
a
dxdτ
t1−a
Qt
≤ CEp+1 (y)(1−a)p/2+a t1−a .
Denote λ = 1 − a, γ = a + (1 − a)p/2. Applying Hölder’s inequality, we have
E2p+1 (y)
ZZ
γ
p+1
λ
3 p
≤ Ct
(x − y)+ |D u| dxdτ
Qt
≤ Ct
λ
ZZ
(x −
y)2p+1
|D3 u|p
+
(p+1)γ
Z t Z
(2p+1)
≤ Ctλ [E2p+1 (y)](p+1)γ/(2p+1) [E0 (y)]pγ/(2p+1) .
3
p
|D u| dxdτ
dxdτ
Qt
1
0
y
pγ
(2p+1)
48
C. Liu
Therefore
E2p+1 (y) ≤ Ctλ/σ [E0 (y)]pγ/((2p+1)σ) ,
σ =1−
p+1
γ > 0.
2p + 1
Using Hölder’s inequality again, we get
E1 (y) ≤ [E2p+1 (y)]1/(2p+1) [E0 (y)]2p/(2p+1) ≤ Ctγ1 [E0 (y)]1+θ ,
where
γ1 =
λ
,
σ(2p + 1)
θ=
1
pγ
−
> 0.
2
σ(2p + 1)
2p + 1
Noticing that E10 (y) = −E0 (y), we obtain
E10 (y) ≤ −Ct−γ1 /(θ+1) [E1 (y)]1/(θ+1) .
If E1 (x2 ) = 0, then suppu ⊂ [0, x2 ]. If E1 (x2 ) > 0, then there exists a maximal
interval (x2 , x∗2 ) in which E1 (y) > 0, E1 (x∗2 ) = 0 and
h
E1 (y)θ/(θ+1)
i0
=
E10 (y)
θ
≤ −Ct−γ1 /(θ+1) .
θ + 1 [E1 (y)]1/(θ+1)
Integrating the above inequality over (x2 , x∗2 ), we have
E1 (x∗2 )θ/(θ+1) − E1 (x2 )θ/(θ+1) ≤ −Ct−γ1 /(θ+1) (x∗2 − x2 ),
which implies that
x∗2 ≤ x2 + Ctµ (E0 (x2 ))θ/(θ+1) ≡ x2 (t),
µ=
1
γ1
=
> 0.
θ+1
3p − 2
Results in [8] imply that E0 (y) can be controlled by a constant C independent
of y. Therefore
suppu(·, t) ⊂ [0, x2 (t)].
Similarly, we have
suppu(·, t) ⊂ [x1 (t), 1].
We have thus completed the proof of Theorem 2.1.
3
Regularity of solutions
Theorem 3.1 If u is weak solution of the problem(1.1)-(1.3), then for any
(x1 , t1 ), (x2 , t2 ) ∈ QT , we have
|u(x1 , t1 ) − u(x2 , t2 )| ≤ C(|x1 − x2 | + |t1 − t2 |1/2 ),
where C is a constant depending only on p.
49
Properties of solutions for thin film equation
Proof. Let
Z
T
Z
jε (x − y, t − s)u(y, s) dyds
uε (x, t) = Jε u(x, t) =
0
|x−y|<ε
where jε (x, t) is a mollifier.
For any x1 , x2 ∈ I, we have
uε (x1 , t) − uε (x2 , t)
Z TZ
Z
=
jε (x1 − y, t − s)u(y, s)dyds −
0
R
T
Z
jε (x2 − y, t − s)u(y, s) dyds
R
∂jε (zx1 + (1 − z)x2 − y, t − s)
u(y, s) dzdyds
∂z
R
Z Z 1
Dx jε (zx1 + (1 − z)x2 − y, t − s)(x1 − x2 )u(y, s) dzdyds
0
T
=
0
R 0
T Z Z 1
Z
= −
Dy jε (zx1 + (1 − z)x2 − y, t − s)(x1 − x2 )u(y, s) dzdyds
0
T
Z
Z
Z
=
Z
0
T
R
Z Z
0
1
jε (zx1 + (1 − z)x2 − y, t − s)Dy u(y, s) dzdyds(x1 − x2 ).
=
0
R
0
Therefore
|uε (x1 , t) − uε (x2 , t)|
Z
T
Z Z
≤
1
|jε (zx1 + (1 − z)x2 − y, t − s)||Dy u(y, s)| dzdyds|x1 − x2 |,
0
R
0
by u ∈ L∞ (0, T ; W 3,p (Ω)), hence using Sobolev embedding theorem, we have
∂u
∞
∞
∂x ∈ L (QT ) and u ∈ L (QT ). Thus we obtain
|uε (x1 , t) − uε (x2 , t)| ≤ C|x1 − x2 |.
(3.1)
Set 0 < ε < t1 < t2 < T . Let ∆t = t2 −t1 , Iρ = I(∆t)1/2 (x0 )= (x0 −(∆t)1/2 , x0 +
(∆t)1/2 ), x0 ∈ I, choose ρ sufficiently small, such that Iρ ⊂ I, ϕ ∈ C01 (Iρ ), we
50
C. Liu
can obtain
Z
ϕ(x)(uε (x, t2 ) − uε (x, t1 )) dx
Z 1
∂uε (x, st2 + (1 − s)t1 )
dsdx
ϕ(x)
∂s
Iρ
0
Z
Z 1Z T Z
∆t
u(y, τ )·
ϕ(x)
ZIρ
=
=
Iρ
(3.2)
|x−y|<ε
0
0
·j (x − y, st2 + (1 − s)t1 − τ ) dydτ dsdx
Z εt
1Z T Z
u(y, τ )·
ϕ(x)
Z
= −∆t
Iρ
|x−y|<ε
0
0
·jετ (x − y, st2 + (1 − s)t1 − τ ) dydτ dsdx.
Fixed (x, t) ∈ QT , 0 < ε < t < T − ε, we have jε (x − y, t − τ ) ∈ C01 (QT ), from
definition of weak solution
Z TZ
jετ (x − y, st2 + (1 − s)t1 − τ )u(y, τ ) dydτ
|x−y|<ε
0
Z
T
Z
|Dy3 u|p−2 Dy3 uDy jε (x − y, st2 + (1 − s)t1 − τ )u(y, τ ) dydτ,
= −
0
|x−y|<ε
hence (3.2) is converted into
Z
ϕ(x)(uε (x, t2 ) − uε (x, t1 )) dx
Iρ
Z
Z
=
Z
T
Z
|Dy3 u|p−2 Dy3 u ·
ϕ(x)
∆t
Iρ
Z
=
1
∆t
0
1
0
Z
|x−y|<ε
0
·Dy jε (x − y, st2 + (1 − s)t1 − τ )u(y, τ ) dydτ dsdx
Z TZ
Dx ϕ(x)
|Dy3 u|p−2 Dy3 u ·
Iρ
|x−y|<ε
0
·jε (x − y, st2 + (1 − s)t1 − τ )u(y, τ ) dydτ dxds.
Taking
Z
(∆t)1/2 −|x−x0 |−2h
ϕ(x) = ϕh (x) =
δh (s) ds,
−h
R
where δ(s) ∈ C01 (R); δ(s) ≥ 0; δ(s) = 0, as |s| ≥ 1; R δ(s) ds = 1. For h > 0
define δh (s) = h1 δ( hs ).
Hence
Z
ϕh (x)(uε (x, t2 ) − uε (x, t1 )) dx
Iρ
Properties of solutions for thin film equation
Z
=
1
Z
∆t
0
δh ((∆t)1/2 − |x − x0 | − 2h)
Iρ
51
x0 − x
Jε (|D3 u|p−2 D3 u) dxds,
|x − x0 |
Noting that for x ∈ Iρ , lim ϕh (x) = 1, and if |x − x0 | < (∆t)1/2 − h, then
h→0
δh ((∆t)1/2 − |x − x0 | − 2h) = 0. δh ≤
C
h,
and
m(Iρ \I(∆t)1/2 −|x−x0 |−2h ) ≤ Ch.
By Jε (|D3 u|p−2 D3 u) ≤ C, therefore
Z
ϕh (x)(uε (x, t2 ) − uε (x, t1 )) dx ≤ C∆t.
Iρ
Letting h → 0, we obtain
Z
(uε (x, t2 ) − uε (x, t1 )) dx ≤ C∆t.
Iρ
Applying the mean value theorem, we see that for some x∗ ∈ Iρ such that
|uε (x∗ , t2 ) − uε (x∗ , t1 )| ≤ C(∆t)1/2 .
Taking this into account and using (3.1), it follows that
|uε (x, t2 ) − uε (x, t1 )|
≤ |uε (x, t2 ) − uε (x∗ , t2 )| + |uε (x∗ , t2 ) − uε (x∗ , t1 )| + |uε (x∗ , t1 ) − uε (x, t1 )|
≤ C(∆t)1/2 ,
letting ε → 0, we known that u is Hölder continuous. This completes the proof.
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Changchun Liu
Department of Mathematics, Jilin University,
Changchun 130012, China
lcc@email.jlu.edu.cn