# Qualitative Behaviour of Solutions in Two Models of Thin Liquid Films

Authors: Matthew Michal; Marina Chugunova; Roman Taranets
Journal: International Journal of Differential Equations
 (2016)

---

## Abstract

For the thin-film model of a viscous flow which originates from lubrication approximation and has a full nonlinear curvature term, we prove existence of nonnegative weak solutions. Depending on initial data, we show algebraic or exponential dissipation of an energy functional which implies dissipation of the solution arc length that is a well known property for a Hele-Shaw flow. For the classical thin-film model with linearized curvature term, under some restrictions on parameter and gradient values, we also prove analytically the arc length dissipation property for positive solutions. We compare the numerical solutions for both models, with nonlinear and with linearized curvature terms. In regimes when solutions develop finite time singularities, we explain the difference in qualitative behaviour of solutions.


---

## Body

## 1. Introduction

Fluid flow where advective inertial forces are small compared with viscous forces can be described as a Stokes flow. A Stokes flow model is given by(1)
∇
p
=
μ
∇
2
u
+
f
,
∇
·
u
=
0
,where p is the pressure, u is the velocity of the fluid, μ is the dynamical viscosity, and f is an applied force.In the special case when viscous fluid is moving slowly through a porous medium, one can simplify the Stokes model by introducing a Hele-Shaw regime. Consider a Hele-Shaw flow, wherex and y directions are parallel to flat plates, and the z direction is perpendicular to the plates. Also, there is only a small gap of 2
H between the plates that are at z
=
±
H. Under these assumptions, the Hele-Show flow is guided by the system of equations: (2)
u
=
∇
p
z
2
-
H
2
2
μ
,
∇
2
p
=
0
,
∇
p
·
n
→
=
0
,where the pressure p
=
p
(
x
,
y
,
t
), the velocity of the fluid u
=
u
(
x
,
y
,
t
), the dynamical viscosity μ has a constant value, and n
→ is the normal vector to the fluid surface. The Hele-Shaw model describes viscous flow dynamics when a drop of liquid is pressed between two plates. This model is useful both for oil companies to better understand the process of secondary recovery of oil fields [1] and for environmental agencies attempting to analyze the flow of water moving through the soil [2].It is well known [3] that solutions for the Hele-Shaw model, for some class of initial data, can develop singularities in finite time known as “fingering phenomenon.” An interesting result was obtained by Constantin et al. [4]. It was shown that a thin neck of fluid between two larger droplets will not break in a finite time. Using the lubrication approximation approach, the authors proved that the minimum width of this neck decreases like 1
/
t
4 for large enough time t. Likewise, Almgren [5] showed that singularities can develop in flows driven only by surface tension in Hele-Shaw models.In 2001, Hernández-Machado et al. developed a theory to predict the forced evolution of a liquid-air interface in a Hele-Shall cell [6]. Ruyer-Quil developed a Hele-Shaw model that corrects Darcy’s law for inertial forces [7]. Later in 2002, Lee et al. described the pinch-off in mixtures of fluids and the linear stability of steady states [8]. He also studied the interface dynamics for certain types of fluid mixtures. At about the same time, Crowdy published a method for determining exact solutions in a rotating Hele-Shaw cell when the initial interface is a fluid annulus [9]. Also, at that time, two groups of mathematicians presented two different points of view on applicability of Hele-Shaw models to study instabilities in viscous flow dynamics. Böckmann and Müller showed the example where the slow flow could evolve properly in a two-dimensional Navier-Stokes model exhibiting previously experimentally observed fingering phenomenon but at the same time failed to obey Hele-Shaw model where no finite or infinite time singularity was formed [10]. Martin et al. found the solution for this disagreement in 2002 and showed that the instability actually can be described by a Hele-Shaw model if the equations include a chemical reaction term [11].In lubrication approximation regime, namely, when the film thicknessH is much less than the length of the flow L (i.e., the small parameter is introduced by ϵ
=
H
/
L
≪
1), a coating flow can be modeled by a thin-film equation which, in the general-slip form, is given by(3)
h
t
+
h
n
h
x
x
x
x
=
0
l
i
n
e
a
r
i
z
e
d
curvature model
,where h
=
h
(
x
,
t
) is the time-dependent thickness of the thin fluid film and the value of the order of the nonlinearity n depends on the slip conditions between the viscous liquid and the solid interface. With the order of the nonlinearity n
=
1 (which corresponds to the no-slip regime), this equation was derived for the lubrication limit of Hele-Shaw flow in [12]. The existence of nonnegative weak solutions, their qualitative behaviour, and regularity for (3) were rigorously studied in [13–15]. The asymptotic analysis results for classical and weak solutions of the thin-film equation were obtained in [16, 17]. One of the most well known and still open questions is the uniqueness of strong nonnegative solutions for (3). Some results in this direction, for particular classes of initial data, can be found in [18, 19].In this paper, firstly, we will analyze a general-slip model (3) with nonlinearity n
>
0. Secondly, instead of the linearized curvature term κ
≈
h
x
x (which was used to obtain (3), the linearized curvature model), we will introduce the full nonlinear curvature expression h
t
+
(
h
n
(
(
1
+
h
x
2
)
-
1
/
2
h
x
)
x
x
)
x
=
0 (nonlinear curvature model). This intermediate asymptotic lubrication model allows us to use initial data with large values of the gradient. Recall that the classical thin-film model with linearized curvature term restricts initial data to small gradient values; that is, |
h
x
|
≪
1.Our paper has the following structure. In Section2, using a priori energy estimates, we will prove existence of nonnegative weak solutions for the nonlinear curvature model. In Section 3, we will obtain rate of energy dissipation along weak solutions constructed in Section 2. In Section 4, we will prove arc length dissipation property for some class of positive solutions for the linearized curvature model and, finally, in Section 5, we will compare numerical solutions for two models for the special choice of initial data and parameters which lead to formations of finite time singularities.


## 2. Existence of Weak Solutions in Nonlinear Curvature Model

We consider the thin-film equation with full nonlinear expression for the curvature term; namely,(4)
h
t
+
h
n
f
h
x
h
x
x
x
x
=
0
in
Q
Twith initial data(5)
h
x
,
0
=
h
0
x
,coupled with the boundary conditions(6)
h
x
±
a
,
t
=
f
h
x
h
x
x
x
±
a
,
t
=
0
,where f
(
z
)
≔
(
1
+
z
2
)
-
1
/
2 and Ω
=
(
-
a
,
a
), Q
T
=
Ω
×
(
0
,
T
), T
>
0. Obviously, we have the mass conservation(7)
∫
Ω
h
x
,
t
d
x
=
∫
Ω
h
0
x
d
x
=
M
. Formally, multiplying (4) by -
(
f
(
h
x
)
h
x
)
x, after integrating over Ω, we obtain(8)
d
d
t
∫
Ω
1
f
h
x
d
x
+
∫
Ω
h
n
f
h
x
h
x
x
x
2
d
x
=
0
,whence we find the arc length dissipation immediately(9)
∫
Ω
1
+
h
x
2
d
x
⩽
∫
Ω
1
+
h
0
x
2
d
x
.For numerical simulations of this arc length dissipation property (see Figure1), obtained above, we use as initial data h
0
=
s
i
n
(
x
-
π
/
2
)
/
4
+
1
/
2 and apply stable semi-implicit in time pseudospectral method developed in [20].Figure 1
Arc length dissipation for a numerical solution whenn
=
1
/
3.Definition 1.
Letn
>
0. A generalized weak solution of problem (4)–(6) is a function h satisfying (10)
h
∈
C
Q
¯
T
∩
L
∞
0
,
T
;
W
1
1
Ω
,
h
t
∈
L
2
0
,
T
;
H
1
Ω
′
,
h
n
/
2
f
h
x
h
x
x
x
∈
L
2
P
T
,where P
T
≔
Q
¯
T
∖
(
{
h
=
0
}
∪
{
t
=
0
}
) and h satisfies (4) in the following sense: (11)
∫
0
T
h
t
,
ϕ
H
1
′
,
H
1
d
t
-
∬
P
T
h
n
f
h
x
h
x
x
x
ϕ
x
d
x
d
t
=
0 for all ϕ
∈
L
2
(
0
,
T
;
H
1
(
Ω
)
), and (12)
h
·
,
t
⟶
h
0
·
strongly in
L
2
Ω
a
s
t
⟶
0
.Theorem 2 (existence of nonnegative weak solutions).
Assume thatn
>
0 and h
0
∈
W
1
1
(
Ω
), h
0
⩾
0, h
0
α
-
n
+
2
∈
L
1
(
Ω
) for any α
∈
[
-
1
/
2
,
1
). Then, for any time T
>
0, there exists a nonnegative generalized weak solution h in the sense of Definition 1 such that (13)
h
α
-
n
+
2
∈
L
∞
0
,
T
;
L
1
Ω
f
o
r
a
n
y
α
∈
-
1
2
,
1
,
f
3
/
2
h
x
h
α
+
2
/
2
x
x
∈
L
2
Q
T
,
f
3
/
4
h
x
h
α
+
2
/
4
x
∈
L
4
Q
T
f
o
r
a
n
y
α
∈
-
1
2
,
1
,
1
+
h
x
2
·
,
t
⟶
1
+
h
0
x
2
·
s
t
r
o
n
g
l
y
i
n
L
1
Ω
a
s
t
⟶
0
.Proof of Theorem2.
Givenε
>
0, δ
>
0, and σ
>
0, an approximated parabolic problem that we consider is(14)
h
t
+
g
ε
δ
h
f
σ
h
x
h
x
x
x
x
=
0
,with the initial data(15)
h
x
,
0
=
h
0
ε
δ
σ
x
,coupled with the boundary conditions(16)
h
x
±
a
,
t
=
f
h
x
h
x
x
x
±
a
,
t
=
0
,where (17)
g
ε
δ
z
=
z
n
+
4
z
4
+
δ
z
n
+
ε
,
f
σ
z
=
f
z
+
σ
∀
z
∈
R
1
;
0
<
h
0
,
ε
δ
σ
∈
C
4
+
α
Ω
¯
:
h
0
,
ε
δ
σ
⟶
h
0
,
δ
σ
>
0
strongly in
H
1
Ω
as
ε
⟶
0
,
h
0
,
δ
σ
⟶
h
0
,
σ
strongly in
H
1
Ω
as
δ
⟶
0
,
h
0
,
δ
σ
⩾
h
0
,
σ
+
δ
θ
,
θ
∈
0
,
2
5
,
h
0
,
σ
⟶
h
0
strongly in
W
1
1
Ω
,
σ
h
0
σ
,
x
2
2
⟶
0
as
σ
⟶
0
.Multiplying (14) by -
f
σ
h
x
h
x
x, after integrating over Ω, we obtain(18)
d
d
t
∫
Ω
1
f
h
x
+
σ
2
h
x
2
d
x
+
∫
Ω
g
ε
δ
h
f
σ
h
x
h
x
x
x
2
d
x
=
0
.Let G
ε
δ
(
z
) be such that G
ε
δ
′
′
(
z
)
=
1
/
g
ε
δ
(
z
)
>
0. Multiplying (14) by G
ε
δ
′
(
h
) (we have enough regularity because our solutions are classical), after integrating over Ω, we obtain(19)
d
d
t
∫
Ω
G
ε
δ
h
d
x
+
σ
∫
Ω
h
x
x
2
d
x
+
∫
Ω
f
3
h
x
h
x
x
2
d
x
=
0
.Using a priori estimates (18) and (19) and local in time existence of a classical solution for (14) (local existence can be shown by application of Galerkin method [21]), following [13], we take ε
→
0. As a result, we obtain a unique positive classical solution h
δ
σ in Q
T for any T
>
0; that is, 0
<
h
δ
σ
∈
C
x
,
t
4,1
(
Q
T
).
LetG
~
δ
(
z
) be such that G
~
δ
′
′
(
z
)
=
z
α
/
g
δ
(
z
)
>
0 for all z
>
0. Multiplying (14) with ε
=
0 by G
~
δ
′
(
h
), after integrating over Ω, we obtain(20)
d
d
t
∫
Ω
G
~
δ
h
d
x
+
σ
∫
Ω
h
α
h
x
x
2
-
α
α
-
1
h
α
-
2
h
x
4
d
x
+
∫
Ω
h
α
f
3
h
x
h
x
x
2
d
x
-
α
α
-
1
∫
Ω
h
α
-
2
F
h
x
h
x
d
x
=
0
,where (21)
F
z
≔
∫
0
z
s
2
f
3
s
d
s
=
ln
⁡
z
+
1
+
z
2
-
z
1
+
z
2
. As (22)
∫
Ω
h
α
-
2
F
h
x
h
x
d
x
=
1
1
-
α
∫
Ω
h
α
-
1
h
x
2
f
3
h
x
h
x
x
d
x
,
∫
Ω
h
α
-
1
h
x
2
f
3
h
x
h
x
x
d
x
⩽
∫
Ω
h
α
f
3
h
x
h
x
x
2
d
x
1
/
2
∫
Ω
h
α
-
2
f
3
h
x
h
x
4
d
x
1
/
2
, then (23)
∫
Ω
h
α
-
2
F
h
x
h
x
d
x
⩽
1
1
-
α
∫
Ω
h
α
f
3
h
x
h
x
x
2
d
x
1
/
2
∫
Ω
h
α
-
2
f
3
h
x
h
x
4
d
x
1
/
2
. Using the inequality (24)
z
4
f
3
z
⩽
3
z
F
z
∀
z
∈
R
1
, we have (25)
∫
Ω
h
α
-
2
F
h
x
h
x
d
x
⩽
3
1
-
α
2
∫
Ω
h
α
f
3
h
x
h
x
x
2
d
x
. As a result, we obtain(26)
d
d
t
∫
Ω
G
~
δ
h
d
x
+
σ
1
+
2
α
1
-
α
∫
Ω
h
α
h
x
x
2
d
x
+
1
+
2
α
1
-
α
∫
Ω
h
α
f
3
h
x
h
x
x
2
d
x
⩽
0
,where α
∈
-
1
/
2
,
1. Using a priori estimate (26), following [22, 23], we take δ
→
0. As a result, we obtain a nonnegative strong solution h
σ in Q
T for any T
>
0; that is, 0
⩽
h
σ
∈
C
1
(
Ω
¯
) for a. e. t
>
0.
From (18)–(26), we obtain that(27)
1
+
h
x
2
σ
>
0
is uniformly bounded in
L
∞
0
,
T
;
L
1
Ω
;
(28)
h
α
-
n
+
2
σ
>
0
is uniformly bounded in
L
∞
0
,
T
;
L
1
Ω
,
α
∈
-
1
2
,
1
;
(29)
f
3
/
2
h
x
h
α
+
2
/
2
x
x
σ
>
0
is uniformly bounded in
L
2
Q
T
,
α
∈
-
1
2
,
1
;
(30)
f
3
/
4
h
x
h
α
+
2
/
4
x
σ
>
0
is uniformly bounded in
L
4
Q
T
,
α
∈
-
1
2
,
1
.
In particular, from (27), it follows that(31)
h
σ
>
0
is uniformly bounded in
L
∞
0
,
T
;
W
1
1
Ω
;
h
σ
>
0
is uniformly bounded in
C
Q
¯
T
.Following [21], we let σ
→
0. As a result, we obtain the existence of a nonnegative weak solution h in Q
T for any T
>
0; that is, 0
⩽
h
∈
C
(
Q
¯
T
).


## 3. Dissipation Rate of the Energy Functional

Theorem 3 (asymptotic decay).
Assume thath is a nonnegative generalized weak solution from Theorem 2. Then, there exists a constant C
0
>
0 depending only on initial data h
0 and Ω such that the solution h satisfies (32)
∫
Ω
1
+
h
x
2
-
1
d
x
⩽
C
0
1
+
t
-
1
,
h
x
,
t
⟶
M
Ω
i
n
W
1
1
Ω
a
s
t
⟶
+
∞
.Moreover, if M
>
Ω
2, then there exists a constant C
1
>
0 depending only on initial data h
0 and |
Ω
|, such that(33)
∫
Ω
1
+
h
x
2
-
1
d
x
⩽
exp
⁡
-
C
1
t
∫
Ω
1
+
h
0
x
2
-
1
d
x
.Remark 4.
In fact, Theorem3 implies the existence of T
∗
=
T
∗
(
h
0
)
⩾
0 such that h
(
x
,
t
)
>
0 for all t
>
T
∗.Remark 5.
Using Galerkin method we can observe numerically that the arc length has exponential decay rate.
The exponential dissipation rate is illustrated in Figure2 for initial data for h
0
=
s
i
n
(
x
-
π
/
2
)
/
4
+
1
/
2 and the nonlinearity n
=
1
/
3.Figure 2
Exponential decay of the arc length for the numerical solution withC
1
=
2.017.Proof of Theorem3.
We consider a unique positive classical solutionh
=
h
δ
σ. Let us denote(34)
L
σ
z
≔
1
+
z
2
+
σ
2
z
2
,
L
σ
′
z
=
z
f
σ
z
,
L
σ
′
′
z
=
f
3
z
+
σ
.Next, we have (35)
L
σ
h
x
=
L
σ
h
x
x
0
+
∫
x
0
x
L
σ
h
x
x
d
x
=
L
σ
h
x
x
0
+
∫
x
0
x
L
σ
′
h
x
h
x
x
d
x
=
L
σ
h
x
x
0
+
∫
x
0
x
f
σ
h
x
h
x
h
x
x
d
x
=
L
σ
h
x
x
0
+
f
σ
h
x
h
x
2
x
0
x
-
f
σ
h
x
h
x
x
h
x
0
x
+
∫
x
0
x
f
σ
h
x
h
x
x
h
x
d
x
=
f
σ
h
x
h
x
2
-
f
σ
h
x
h
x
x
h
+
∫
x
0
x
f
σ
h
x
h
x
x
h
x
d
x
-
F
x
0
=
2
f
σ
h
x
h
x
2
-
f
σ
h
x
h
x
h
x
+
∫
x
0
x
f
σ
h
x
h
x
x
h
x
d
x
-
F
x
0
,where(36)
F
x
≔
-
L
σ
h
x
+
f
σ
h
x
h
x
2
-
f
σ
h
x
h
x
x
h
. Integrating this equality on Ω, we deduce that (37)
-
∫
Ω
L
σ
h
x
+
F
x
0
d
x
+
2
∫
Ω
f
σ
h
x
h
x
2
d
x
=
-
∫
Ω
∫
x
0
y
f
σ
h
x
h
x
x
x
h
d
x
d
y
.We select a point x
0
∈
Ω
¯ such that h
x
(
x
0
)
=
0 and h
x
x
(
x
0
)
⩾
0. Really, we know that we have at least two such points such that h
x
(
x
0
)
=
0. Of course, if either h
x
x
(
-
a
) or h
x
x
(
a
) is nonnegative, then we are done by choosing x
0 to be that endpoint. If, however, both h
x
x
(
-
a
) and h
x
x
(
a
) are negative, then there exists at least one x
0
∈
Ω for which h
x
(
x
0
)
=
0 and h
x
x
(
x
0
)
⩾
0. As a result, F
(
x
0
)
⩽
-
1. Then, we have (38)
∫
Ω
1
+
1
f
h
x
1
-
2
f
2
h
x
+
3
σ
2
h
x
2
d
x
⩽
-
∫
Ω
∫
x
0
y
f
σ
h
x
h
x
x
x
h
d
x
d
y
. In view of the inequality 1
+
1
/
f
z
(
1
-
2
f
2
(
z
)
)
⩾
1
/
f
(
z
)
-
1
∀
z
∈
R
1, we obtain that (39)
∫
Ω
L
σ
h
x
-
1
d
x
⩽
-
∫
Ω
∫
x
0
y
f
σ
h
x
h
x
x
x
h
d
x
d
y
. Using the estimate (40)
-
∫
x
0
x
f
σ
h
x
h
x
x
x
h
d
x
⩽
∫
Ω
g
δ
h
f
σ
h
x
h
x
x
x
2
d
x
1
/
2
∫
Ω
h
2
g
δ
h
d
x
1
/
2 and taking into account ∫
Ω
(
h
2
/
g
δ
(
h
)
)
d
x
⩽
C
0
=
C
0
(
h
0
) due to (19) with ε
=
0, we find that(41)
∫
Ω
L
σ
h
x
-
1
d
x
2
⩽
C
0
∫
Ω
g
δ
h
f
σ
h
x
h
x
x
x
2
d
x
.Thus, from (18), we arrive at (42)
v
σ
t
+
C
0
-
1
∫
0
t
v
σ
2
s
d
s
⩽
v
σ
0
,
where
v
σ
t
≔
∫
Ω
L
σ
h
x
-
1
d
x
.Letting σ
→
0, from this inequality, we obtain (43)
0
⩽
v
0
t
⩽
v
0
0
1
+
C
0
-
1
v
0
0
t
⟶
0
as
t
⟶
+
∞
,and hence (44)
∫
Ω
1
+
h
x
2
-
1
d
x
⟶
0
as
t
⟶
+
∞
.Thus, (45)
h
x
,
t
⟶
M
Ω
in
W
1
1
Ω
a
s
t
⟶
+
∞
.
Now, we show that(46)
if
M
>
Ω
2
then
0
⩽
v
0
t
⩽
v
0
0
exp
⁡
-
C
1
t
⟶
0
as
t
⟶
+
∞
,where C
1 is some positive constant.
Next, letμ
>
0 to be fixed later. Note that (41) implies(47)
∫
Ω
g
δ
h
f
σ
h
x
h
x
x
x
2
d
x
⩾
μ
C
0
-
1
v
σ
t
if
v
σ
t
⩾
μ
.If v
σ
(
t
)
<
μ, then (48)
h
-
M
Ω
⩽
∫
Ω
h
x
d
x
⩽
∫
Ω
1
+
h
x
2
d
x
<
μ
+
Ω
,whence (49)
inf
⁡
h
⩾
h
1
≔
M
Ω
-
μ
-
Ω
>
0
if
μ
<
M
-
Ω
2
Ω
.Thus, if 0
⩽
v
σ
t
<
μ
<
(
M
-
Ω
2
)
/
Ω, then(50)
∫
Ω
g
δ
h
f
σ
h
x
h
x
x
x
2
d
x
⩾
g
δ
h
1
∫
Ω
f
σ
h
x
h
x
x
x
2
d
x
⩾
g
δ
h
1
π
Ω
4
v
σ
t
.Here, we used the fact that g
δ
(
z
) is increasing on 0
,
∞ for any δ
>
0 and the inequalities (51)
v
σ
t
⩽
∫
Ω
1
f
h
x
f
σ
h
x
h
x
2
d
x
⩽
Ω
∫
Ω
1
f
h
x
d
x
∫
Ω
f
σ
h
x
h
x
x
2
d
x
⩽
Ω
3
π
2
∫
Ω
1
f
h
x
d
x
∫
Ω
f
σ
h
x
h
x
x
x
2
d
x
⩽
Ω
3
μ
+
Ω
π
2
∫
Ω
f
σ
h
x
h
x
x
x
2
d
x
;that is,(52)
∫
Ω
f
σ
h
x
h
x
x
x
2
d
x
⩾
π
Ω
4
∫
Ω
f
σ
h
x
h
x
2
d
x
⩾
π
2
Ω
3
μ
+
Ω
v
σ
t
.As a result, from (47) and (50), we find that(53)
∫
Ω
g
δ
h
f
σ
h
x
h
x
x
x
2
d
x
⩾
C
1
v
σ
t
,where (54)
C
1
=
sup
μ
⁡
min
⁡
μ
C
0
-
1
,
g
δ
h
1
π
2
Ω
3
μ
+
Ω
.Thus, from (18), we arrive at (55)
v
σ
t
+
C
1
∫
0
t
v
σ
s
d
s
⩽
v
σ
0
.Letting δ
→
0 and σ
→
0, from this inequality, we obtain (46).


## 4. Arc length Dissipation of Positive Solutions in Linearized Curvature Model

The time evolution behaviour of the arc length for a positive solution of the thin-film equation is bounded from above; namely,∫
Ω
1
+
h
x
2
-
1
d
x
≤
1
/
2
∫
Ω
h
x
2
d
x, which dissipates to zero as time goes to infinity. Our goal is to show that the arc length also dissipates in time. Consider the equation(56)
h
t
+
h
n
h
x
x
x
x
=
0
for
x
,
t
∈
Q
Tcoupled with the initial condition(57)
h
x
,
0
=
h
0
xand the boundary conditions(58)
h
x
=
h
x
x
x
=
0
on
∂
Ω
.For future calculations,(59)
0
≤
f
n
z
≤
1
for any
n
>
0
,
∀
z
∈
R
1
,
1
-
z
2
f
2
z
=
f
2
z
≥
0
⟹
0
≤
z
2
f
2
z
≤
1
,
f
h
x
h
x
x
=
f
3
h
x
h
x
x
,
f
3
h
x
h
x
x
x
=
-
3
f
5
h
x
h
x
h
x
x
2
+
f
3
h
x
h
x
x
x
.Theorem 6.
Assume thath
0
∈
C
4
+
α
Ω
¯, h
0
>
0, and(60)
0
<
n
<
n
∗
≔
3
4
.Then, there exists a classical solution h
∈
C
x
,
t
4,1
(
Q
¯
T
) and a finite δ
=
δ
n
>
0 such that the arc length dissipates; namely,(61)
∫
Ω
1
+
h
x
2
d
x
+
C
δ
∬
Q
T
f
3
h
x
h
n
h
x
x
x
2
d
x
d
t
≤
∫
Ω
1
+
h
0
x
2
d
xprovided(62)
h
x
≤
δ
.Existence of classical solutions which satisfy condition (62) follows from parabolic Schauder estimates for their derivatives [24].Proof of Theorem6.
Ifh
0
>
0, then there is a classical solution that exists locally in time [25]. If we multiply (56) by f
(
h
x
)
h
x
x and integrate on Ω, then we arrive at an equation describing the change in the arc length of the fluid: (63)
∫
Ω
h
t
f
h
x
h
x
x
d
x
+
∫
Ω
h
n
h
x
x
x
x
f
3
h
x
h
x
x
d
x
=
0
,
-
∫
Ω
f
h
x
h
x
h
x
t
d
x
-
∫
Ω
f
3
h
x
h
x
x
x
h
n
h
x
x
x
d
x
=
0
,
-
∫
Ω
h
x
h
x
t
1
+
h
x
2
d
x
-
∫
Ω
-
3
f
5
h
x
h
x
h
x
x
2
+
f
3
h
x
h
x
x
x
h
n
h
x
x
x
d
x
=
0
,
∫
Ω
3
f
5
h
x
h
n
h
x
h
x
x
2
h
x
x
x
d
x
-
∫
Ω
f
3
h
x
h
n
h
x
x
x
2
d
x
=
d
d
t
∫
Ω
1
+
h
x
2
d
x
.Therefore, for classical solutions to be dissipative, the second integral must be greater than the first one.
We rewrite this equality in a more convenient form:(64)
d
d
t
∫
Ω
1
f
h
x
d
x
+
∫
Ω
h
n
f
-
3
h
x
f
h
x
h
x
x
x
2
d
x
=
3
∫
Ω
h
n
f
-
4
h
x
h
x
f
h
x
h
x
x
2
f
h
x
h
x
x
x
d
x
.Using the integration by parts, we find that (65)
d
d
t
∫
Ω
1
f
h
x
d
x
+
∫
Ω
h
n
f
-
3
h
x
f
h
x
h
x
x
x
2
d
x
+
∫
Ω
h
n
f
-
5
h
x
1
+
5
h
x
2
f
h
x
h
x
x
4
d
x
=
-
n
∫
Ω
h
n
-
1
f
-
4
h
x
h
x
2
f
h
x
h
x
x
3
d
x
.Before we continue, notice that (66)
3
∫
Ω
f
5
h
x
h
n
h
x
h
x
x
2
h
x
x
x
d
x
≤
3
∫
Ω
f
3
h
x
h
n
h
x
x
x
2
d
x
1
/
2
∫
Ω
f
7
h
x
h
n
h
x
2
h
x
x
4
d
x
1
/
2
.Next, we show that(67)
∫
Ω
f
7
h
x
h
n
h
x
2
h
x
x
4
d
x
≤
C
0
δ
∫
Ω
f
3
h
x
h
n
h
x
x
x
2
d
x
.Let us denote (68)
G
z
≔
1
3
z
3
f
3
z
-
1
5
z
5
f
5
z
=
1
5
z
3
f
3
z
2
3
+
f
2
z
⟹
G
′
z
=
z
2
f
7
z
.Then, (69)
∫
Ω
G
′
h
x
h
n
h
x
x
4
d
x
=
-
n
∫
Ω
G
h
x
h
x
h
n
-
1
h
x
x
3
d
x
-
3
∫
Ω
G
h
x
h
n
h
x
x
2
h
x
x
x
d
x
.By Hölder inequality, we find that (70)
∫
Ω
G
h
x
h
x
h
n
-
1
h
x
x
3
d
x
≤
∫
Ω
G
′
h
x
h
n
h
x
x
4
d
x
3
/
4
∫
Ω
G
h
x
4
G
′
h
x
3
h
n
-
4
h
x
4
d
x
1
/
4
,
∫
Ω
G
h
x
h
n
h
x
x
2
h
x
x
x
d
x
≤
∫
Ω
f
3
h
x
h
n
h
x
x
x
2
d
x
1
/
2
∫
Ω
G
h
x
2
f
3
h
x
h
n
h
x
x
4
d
x
1
/
2
,where(71)
G
z
4
G
′
z
3
=
z
6
625
f
9
z
2
3
+
f
2
z
4
,
G
z
2
f
3
z
=
z
6
f
3
z
25
2
3
+
f
2
z
2
.In particular, due to h
x
≤
δ, we get (72)
G
z
2
f
3
z
≤
a
1
G
′
z
for
a
1
≥
δ
2
2
δ
2
+
5
15
2
.
Let us denote(73)
F
z
≔
G
z
4
G
′
z
3
=
z
6
625
f
9
z
2
3
+
f
2
z
4
.Then,(74)
∫
Ω
F
h
x
h
n
-
4
h
x
4
d
x
=
1
3
-
n
∫
Ω
F
′
h
x
h
n
-
3
h
x
3
h
x
x
d
x
+
3
3
-
n
∫
Ω
F
h
x
h
n
-
3
h
x
2
h
x
x
d
x
.By Hölder inequality, we deduce that(75)
∫
Ω
F
′
h
x
h
n
-
3
h
x
3
h
x
x
d
x
≤
∫
Ω
G
′
h
x
h
n
h
x
x
4
d
x
1
/
4
∫
Ω
F
′
h
x
4
/
3
G
′
h
x
1
/
3
h
n
-
4
h
x
4
d
x
3
/
4
,
∫
Ω
F
h
x
h
n
-
3
h
x
2
h
x
x
d
x
≤
∫
Ω
G
′
h
x
h
n
h
x
x
4
d
x
1
/
4
∫
Ω
F
h
x
4
/
3
G
′
h
x
1
/
3
h
x
4
/
3
h
n
-
4
h
x
4
d
x
3
/
4
.In particular, due to |
h
x
|
≤
δ, we get (76)
F
z
4
/
3
G
′
z
1
/
3
z
4
/
3
≤
a
2
F
z
for
a
2
≥
2
δ
4
+
7
δ
2
+
5
15
4
/
3
,
F
′
z
4
/
3
G
′
z
1
/
3
≤
a
3
F
z
for
a
3
≥
10
δ
4
+
21
δ
2
+
10
5
4
/
3
.Thus, (77)
∫
Ω
F
h
x
h
n
-
4
h
x
4
d
x
≤
a
3
3
/
4
+
3
a
2
3
/
4
3
-
n
4
∫
Ω
G
′
h
x
h
n
h
x
x
4
d
x
.From this, we deduce that (78)
∫
Ω
G
′
h
x
h
n
h
x
x
4
d
x
≤
a
4
∫
Ω
f
3
h
x
h
n
h
x
x
x
2
d
x
,where (79)
a
4
=
9
3
-
n
2
a
1
3
-
n
-
n
a
3
3
/
4
+
3
a
2
3
/
4
2provided (80)
h
x
<
δ
1
≔
1
6
-
7
+
-
11
+
45
/
n
,
0
<
n
<
n
∗
=
3
4
.As a result, we obtain that (81)
d
d
t
∫
Ω
1
+
h
x
2
d
x
+
a
5
∫
Ω
f
3
h
x
h
n
h
x
x
x
2
d
x
≤
0
,where a
5
=
1
-
3
a
4
1
/
2
≥
0 and a
1
<
1
/
81
(
⇒
δ
<
δ
2
≔
1
/
2
-
5
+
115
/
3
≈
0.5458
) provided (82)
h
x
<
δ
=
min
⁡
δ
1
,
δ
2
,
0
<
n
<
n
∗
.We know from [13] that, for n
≥
4 and h
0
>
0, the solution for thin-film equation has the property h
(
·
,
t
)
→
1
/
Ω
∫
h
0
(
x
)
d
x uniformly as t
→
∞; this is why we believe that one can use the approximation h
n
+
4
/
h
4
+
ϵ
h
n for h
n for ϵ
≪
1 to generalize our arc length dissipation result for strong nonnegative solutions.To illustrate analytic results of arc length dissipation (see Figures3 and 4) for the linearized curvature model numerically, we considered the problem on the domain Q
T
=
0,2
π
×
[
0
,
π
/
2
]. We discretized Ω using a uniform distribution with 2
N grid points for N
=
10 and applied Fourier spectral differentiation matrices with periodic boundary conditions to implement a stable semi-implicit in time pseudospectral method described in [20].Figure 3
Arc length dissipation for the numerical solution withn
=
1
/
2 and N
=
32.Figure 4
Arc length dissipation for the numerical solution withn
=
1
/
3 and N
=
32.


## 5. Discussion

Modeling time evolution for different initial data,h
0, we noticed that the solutions of two different models, with nonlinear and linearized curvature terms, developed different shapes during the time evolution. When we applied our numerical simulations to the initial data for the large enough gradient value, namely, for(83)
h
0
x
=
1
0.09
2
π
e
-
x
-
π
2
/
2
0.09
2
, we observed that at the same time t
=
1
0
-
4 the numerical solution for linearized model had lower height and more triangular shape to compare to the numerical solution for the nonlinear models which developed almost parallel sides of the node (see Figure 5), where m
t
=
h
(
·
,
t
)
1, m
0
=
m
(
0
), and h
=
h
x
,
t for the linearized curvature model and h
c
=
h
x
,
t for nonlinear curvature model.Figure 5
Comparison of solutions to (4) as the solid line and (56) as the dashed line for n
=
2 and N
=
2
8 when the solutions are at the maximum difference in their L
2 norms. Both numerical solutions are mass preserving.It was also observed numerically that solutions of linearized and nonlinear curvature models approach zero (touchdown point) with a different speed; namely, the numerical solution for the linearized curvature model loses uniform positivity faster to compare to the numerical solution that corresponds to the nonlinear curvature model. To study the numerical time evolution of solutions near touchdown points, we implemented the numerical method suggested in [26] and used initial data h
0
(
x
)
=
0.8
-
cos
⁡
π
x
+
0.25
cos
⁡
(
2
π
x
) (see Figure 6).Figure 6
This time, the comparison of solutions to the equations is shown at the time when dashed line solution of (56) reaches touchdown point and at the same time the solid line solution of (4) retains its positivity for n
=
1
/
2 and N
=
2
6. Likewise, the L
2 norm of the difference of the solutions grows over time. Both numerical solutions are mass preserving.



---

*Source: 4063740-2016-05-05.xml*