Thin Film Equations with van der Waals Force

63234_7Miloua_ETD_2014.pdf

THIN FILM EQUATIONS WITH VAN DER WAALS

FORCE by

Attou A. Miloua

B.S., Djillali Liabes University, 1984

M.S., Carnegie Mellon University, 1987

Submitted to the Graduate Faculty of

the Kenneth P. Dietrich School of Arts and Sciences in partial fulllment of the requirements for the degree of

Doctor of Philosophy

University of Pittsburgh

2014

UNIVERSITY OF PITTSBURGH

KENNETH P. DIETRICH SCHOOL OF ARTS AND SCIENCES

This dissertation was presented

by

Attou A. Miloua

It was defended on

April 15, 2014

and approved by

Huiqiang Jiang, Ph. D., University of Pittsburgh

Dehua Wang, Ph. D.,University of Pittsburgh

Dejan Slepcev, Ph. D., Carnegie Mellon University

Reza Pakzad, Ph. D., University of Pittsburgh

Ming Chen, Ph. D., University of Pittsburgh

Dissertation Director: Huiqiang Jiang, Ph. D., University of Pittsburgh ii

Copyright

c by Attou A. Miloua 2014
iii

ABSTRACT

THIN FILM EQUATIONS WITH VAN DER WAALS FORCE

Attou A. Miloua, PhD

University of Pittsburgh, 2014

We are interested in the steady states of thin lms in a cylindrical container with van der Waals forces which lead to a singular elliptic equation in a bounded domain with Neumann boundary conditions. Using the prescribed volume of the thin lm as a variable parameter we investigated the structure of radial solutions and their associated energies using rigorous asymptotic analysis and numerical computation. Motivated by the existence of rupture solutions for thin lm equations, we considered elliptic equations with more general non linearity and obtained sucient condition for the existence of weak rupture solutions for a class of generalized elliptic equations. Finally such results can be generalized to a class of quasi-linear elliptic partial dierential equations. Keywords:Thin lm equations, point rupture solutions, elliptic equations, quasi-linear elliptic equations, asymptotic analysis. iv

TABLE OF CONTENTS

PREFACE. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .viii

1.0 INTRODUCTION. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1

1.1 Physical background

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Derivation and formulation of the problem

. . . . . . . . . . . . . . . . . . 4

1.3 Radial steady states of van der Waals force driven thin lms with prescribed

uid volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Rupture solutions of general elliptic equation

. . . . . . . . . . . . . . . . . 8

2.0 RADIAL STEADY STATES OF VAN DER WAALS FORCE DRIVEN

THIN FILMS WITH PRESCRIBED FLUID VOLUME. . . . . . . . .10

2.1 Scaling property of global radial solutions

. . . . . . . . . . . . . . . . . . . 12

2.2 Rupture solution as a limit of smooth solutions!0. . . . . . . . . . . . 26

2.3 Linearization when!1. . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.4 Limiting prole when! 1. . . . . . . . . . . . . . . . . . . . . . . . . .32

3.0 RUPTURE SOLUTIONS OF GENERAL ELLIPTIC EQUATION. .45

3.1 Introduction

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.2 The main result

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.3 Proof of the main result

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.4 Several examples

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.5 Dynamical system formulation

. . . . . . . . . . . . . . . . . . . . . . . . . 55

4.0 POINT RUPTURE SOLUTIONS OF A CLASS OF QUASI-LINEAR

ELLIPTIC EQUATIONS. . . . . . . . . . . . . . . . . . . . . . . . . . . . .56

4.1 Introduction

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
v

4.2 Proof of the main result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.3 Point rupture solution with non-monotonich. . . . . . . . . . . . . . . . .62

5.0 CONCLUSION. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .69

BIBLIOGRAPHY. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .72 vi

LIST OF FIGURES

1 Global rupture solution with 9 oscillations.

. . . . . . . . . . . . . . . . . . . 13

2 Global radial smooth solution with eta=7.

. . . . . . . . . . . . . . . . . . . . 14

3 Radial solution with eta = 11.

. . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 Radial smooth solution with eta = 0.000001

. . . . . . . . . . . . . . . . . . . 16

5 Radial smooth solution with eta = 20.

. . . . . . . . . . . . . . . . . . . . . . 17

6 Smooth radial rescaled to the unit ball with eta=.5 and k=1

. . . . . . . . . 19

7 Smooth radial rescaled to the unit ball with eta=.5 and k=2

. . . . . . . . . 20

8 Smooth radial rescaled to the unit ball with eta=.5 and k=5

. . . . . . . . . 21

9 Smooth radial rescaled to the unit ball with eta= 2 and k=1

. . . . . . . . . 22

10 Smooth radial rescaled to the unit ball with eta= 2 and k=2

. . . . . . . . . 23

11 Smooth radial rescaled to the unit ball with eta= 2 and k=8

. . . . . . . . . 24

12 Rupture radial rescaled to the unit ball with k=1

. . . . . . . . . . . . . . . . 25

13 Set of smooth radial solution for eta = 0.2 , 0.1 , 0.01 , 0.001 and 0.0001.

. . 27

14 Set of rupture radial solutions rescaled to the unit ball for k=1,2,3,4,5 and 6.

28

15 The Bessel function of rst kind and of order zero.

. . . . . . . . . . . . . . . 31

16 Blow down solution with eta=0.1

. . . . . . . . . . . . . . . . . . . . . . . . 34

17 Asymptotic solution eta=100.

. . . . . . . . . . . . . . . . . . . . . . . . . . 36

18 Average volume hbar.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

19 Average volume hbar and its asymptotic approximation.

. . . . . . . . . . . . 41

20 Energy versus eta. Eta changes from 0 to 100.

. . . . . . . . . . . . . . . . . 42

21 Energy versus hbar.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

22 Average volume for k=1,2 and 3.

. . . . . . . . . . . . . . . . . . . . . . . . . 44
vii

PREFACE

ACKNOWLEDGMENTS

First and foremost, I would like to express my thankfulness and sincere gratefulness to my adviser professor Huiqiang Jiang for his immense help and technical support on my Ph.D study research and the writing of this present thesis. I also would like to express to him my sincere gratitude for his extensive knowledge, perseverance, inspiration and enthusiasm. Similarly, I would like to thank all the members of my thesis committee: professor Dehua Wang, professor Dejan Slepcev, professor Reza Pakzad, and professor Ming Chen, for their myriad encouragement, insightful comments and their constructive critics. I am particularly thankful to professor Xinfu Chen for his help and support on the asymptotic part of this work. I am also grateful to all the faculty members, the graduate students and all the sta of the mathematics department for their collaborations. Last but not least, I would like to thank my family. My wife rst for her help, encouragement and moral support during my studies at the university of Pittsburgh throughout all the great and dicult times of my life as a student. I also would like to thank all my children and particularly my daughters Leena and Jannah for giving me an endless willingness to knowledge and research. Thank you all so much. viii

1.0 INTRODUCTION

1.1 PHYSICAL BACKGROUND

In uid dynamics, lubrication theory describes the ow of uids in a geometry in which one dimension is signicantly smaller than the others. In our situation we have a uid in a solid substrate of bottom a planar region R2and thickness or height usually denoted asuor his considered smaller than the dimensions of the base region . Interior ows are those where the boundaries of the uid volume are known. Here a key goal of lubrication theory is to determine the thickness distribution in the uid volume. The uid in this case is called a lubricant. Free lm lubrication theory is concerned with the case in which one of the uid's surfaces is a free surface in our case the top one. Surface tension due in particular to the walls of the substrate may become important, additional intermolecular forces such as van der Waalls forces may be signicant and issues of wetting and non wetting then arise. An important and major application area is in the coatings industries including the preparation of thin lms, printing, painting and adhesives. Our main interest is in the rupture solutions that occur when the thickness approaches zero. Such problem is motivated by the ruptures in thin lms. In the lubrication model of thin lms,uwill be the thickness of the thin lm over a planar region and the dynamic of the thin lm can be modeled by 1 the general fourth order partial dierential equation. u t=r
(umru) r
(unr
u):(1.1.1) The exponents m, n represent the powers in the destabilizing second-order and the stabi- lizing fourth-order diusive terms, respectively. This class of equations occurs in connection with many physical models involving uid interfaces. Here the fourth-order term in the equation re ects surface tension eects. The second-order term can re ect gravity, van der Waals interactions, or the geometry of the solid substrate. For instance, the equation describes a thin jet in a Hele-Shaw cell forn= 1 andm= 1 discussed in detail in [ 1 ] [ 5 ] [ 7 ] [ 8 ] [ 17 ]. A Hele-Shaw cell is composed by two at superposed plates that are parallel to each other and separated by a very reduced distance. At least one of the plates is transparent. These cells are used for studying many phenomena, including the behavior of granular materials as they are being poured into the space between the plates.

The equation can also model the

uid droplets hanging from a ceiling, this is the case when bothnandmare integers and are equal to three, for this instance a rigorous analysis was done as in [ 9 ]. Water molecules attract each other, in fact every molecule of any liquid attracts each other. In general, every neutral atom or molecule attracts one another. The forces of this attraction repulsion are certainly electrical in nature. They are generally known as van der Waals' forces, and are mainly due to forces between two dipoles meaning imbal- anced charge distributions whose total charge is zero. However, molecules are not polarized by themselves but they generate a dipole moment when they are brought together. Water in free fall will tend to form itself into a sphere the form of least surface area. The sphere of water into contact with another piece of solid material of a planar surface as the ceiling depending on the strength of the attraction between a water molecule and the molecules of the ceiling if stronger than the attraction between water and water the water will tend to stick to the surface and we say that the water causes the wetting of the surface. If water- water attraction is greater it will not stick to the surface and will cause the non-wetting of the surface. If the water can wet the surface it sticks to it and we say that it has found a 2 way to reduce the tension by reducing the surface area further thus it sticks to the surface of the ceiling. The equation has the appellation of the modied Kuramoto-Sivashinsky equation, which describes solidication of a hyper-cooled melt whenn= 0 andm= 1 and it was fully studied in [ 3 ] and [ 4 ]. Hyper-cooled melt is the process of lowering the temperature of a uid below its freezing point without actually observing solidication. A liquid below its standard freezing point will crystallize in the presence of a seed crystal or nucleus around which a crystal structure can form creating a solid. The equation models van der Waals force driven thin lms and this instance is manifested when bothnandmare integers with values 3 and1 respectively. Numerous rigorous anal- ysis were performed as published in [ 6 ] , [ 12 ] , [ 18 ] , [ 19 ] and [ 20 ]. When the space dimension is one R. Laugesen and M. Pugh [ 16 ] considered positive periodic steady states and touch- down steady states in a more general setting. The dynamics of a special type of thin lm equation has been investigated by F. Bernis and A. Friedman [ 2 ]. They established the ex- istence of weak solutions and showed that the support of the thin lm will expand with time. 3

1.2 DERIVATION AND FORMULATION OF THE PROBLEM

Let us consider a cylindrical container with base inR2inside of which is a viscous uid. We are interested in studying the behavior of the thin lm. Our problem is actually motivated by the ruptures in thin lms. In the lubrication model of thin lms,uwill be the thickness of the thin lm over a planar region and the dynamic of the thin lm can be controlled by the fourth order partial dierential equation u t=r
(umru) r
(unr
u):(1.2.1)

Here the fourth-order term in the equation re

ects surface tension eects, and the second- order term can re ect gravity, van der Waals interactions, thermocapillary eects or the geometry of the solid substrate. This class of model equation occurs in connection with many physical systems involving uid interfaces. In our instance, we are dealing with van der Waals force driven thin lm, which means that,n= 3 andm=1 and we are mainly concerned with steady states solutions of (

1:2:1) which is not well understood in the two

dimensional case. Sincenm6= 1, we can dene the quantity p=1mn+ 1umn+1
u; which can be viewed as the pressure of the uid. We can rewrite (

1:2:1) as

u t=r(unrp): Now, we need few assumptions that are physically feasible for the problem. let R2 be the bottom of a cylindrical container occupied by the thin lm uid, we assume that there is no ux across the boundary, which yields the boundary condition @p@ = 0 on@ :(1.2.2) We also ignore the wetting or non-wetting eect, and assume that the uid surface is or- thogonal to the boundary of the container, i.e., @h@ = 0 on@ :(1.2.3) 4 Whenevermn+ 16= 0 andmn+ 26= 0, we may divide by the product of these two terms and associate (

1:2:1) with an energy function

E(u) =Z

 12 jruj21(mn+ 1)(mn+ 2)umn+2 ; and formally, using (

1:2:2);(1:2:3), we have

ddt

E(u) =Z

unjrpj2:

Hence, for a thin lm

uid at rest,phas to be a constant, andusatises
u1mn+ 1umn+1=pin ; which is an elliptic equation. If we further assumemn+ 1<1, which includes the van der Waals force case. We can write the equation as 8< :
u=1
up in ; @u@ = 0on ;(1.2.4) wherepis an unknown constant and =(mn+ 1)>1.

For van der Waals force driven thin lm,= 3.

In physical experiments, usually the total volume of the uid is a known parameter identied by the average thickness of the thin lm dened as u=1j jZ u(x)dx: Therefore for any given u >0 we need to nd a functionuand an unknown constant psatisfying, 8 >>>< > >>:
u=1
up in ; 1j jR u(x)dx= u; @u@ = 0on @ :(1.2.5) 5 This singular elliptic problem modeling steady states solutions for van der Waals force driven thin lm has been rigorously studied in [ 15 ] and a complete description of all contin- uous, radial and positive solutions was given for problem (

1:2:4) . However, the dependence

of solutions with a given uis still not clear. Using rigorous asymptotic analysis and numer- ical computation we would like to investigate the structure and behavior of radial solutions and their associated energies for a prescribed volume of the thin lm. Many of these results are illustrated in section 2.5, where numerical experiments were performed on the particular equation, u 00+1r u01 u+p= 0;(1.2.6) when= 3 andp=13 . Observe that the smooth and the rupture solutions are global oscillatory, positive, continuous, and approaching 1 at innity. 6

1.3 RADIAL STEADY STATES OF VAN DER WAALS FORCE DRIVEN

THIN FILMS WITH PRESCRIBED FLUID VOLUME

In chapter two we would like to understand the dependence of radial solutionsuand the prescribed volume u:We consider solving problem (1:2:4) in the unit ball, thus =B(0;1), and sinceuis radial then it must satisfy the equation (1:2:6) . Our strategy is to x= 3,p=13 and ignore the Neumann boundary condition assuming a given prescribed volume constraint u:Then for any >0 the unique radial solution, denotedu(r) to the dierential equation (1:2:6) with initial conditionsu(0) =and u

0(0) = 0 was shown in [15] that it is global and oscillates around 1. Also, there exists

an increasing sequence of positive radiir k;k= 1;2;


which are the critical points of the solutionu;k(r) dened as, u ;k(x) = (r k)21+u(jxj) = (r k)12 u(jxj): Now, if we restrict the solutionu(r) to the ballBrk(0) then the functionu;kmust satisfy the Neumann boundary problem, 8 < :
u=1
up;kin B1(0); @u@ = 0on @B1(0);(1.3.1) where the unknown constantp;kis dened as, p ;k=13 ((r k)12 ):

Now dene the constant u(;k) by,

u(;k) =1jB1(0)jZ B

1(0)u;k(x)dx=(r

k)12 jB1(0)jZ B rk(0)u(x)dx: 7 All radial solutions to problem (1:2:5) can be obtained in this fashion. Now, to answer the question of behavior of the solution in terms of u, all we need to do is to understand the dependance of uonandk:From the theory of ordinary dierential equations u must be continuous as a function of2[0;1] and2[1;1]. In the case when= 1 the only radial solution is the constant solutionu1 and the critical pointr1is not dened. However, using rigorous asymptotic analysis, we are able to show that uis still continuous at 1. We can also show that whenapproaches zero then the radial solutionu(r) will approach the rupture solutionu0(r) satisfyingu0(0) = 0 andu00(0) =1. Hence for eachk, u(;k) is continuous on [0;1]. For the behavior of u(;k) astends to innity we performed many numerical experiments conrming the asymptotic analysis and we obtained the limiting behavior of u(;k) at innity. Similarly, we studied the energy of the radial solutionuk(r) denoted byE(;k) and we found that it is also continuous on [0;1] and we were also able to derive its limiting prole as approaches innity. Combining the numerical computation of both u(;k) andE(;k) with their limiting prole we have a better understanding of the dependance of the radial solutions and their energies asvaries.

1.4 RUPTURE SOLUTIONS OF GENERAL ELLIPTIC EQUATION

In chapter three we are interested in rupture solutions of the general elliptic equation for their special properties and would like to understand the mechanism of producing ruptures for elliptic partial dierential equations. We also conjecture that the ruptures are discrete for nite energy solutions, and expect that the radial point rupture solutions will serve as the blow up prole of the solution near any point rupture. The general elliptic equation we are interested in has the form
u=f(u) 8 in a region R2and assuming Neumann boundary conditions@u@n = 0 on@ . The functionfis positive continuous and satisfying lim u!0+f(u) =1: Motivated by the thin lm equations, a solutionuis said to be a point rupture solution if for somep2 ,u(p) = 0 andu(p)>0 in nfpg. Our main result is a sucient condition onffor the existence of radial point rupture solutions. Actually, we are only concerned with the local solutions in a neighborhood of the point rupture. Since the equation has no singularity away from the point rupture, the possible extension of point rupture solution to a global solution could be carried out using similar arguments as in [ 15 ] where the case f(u) =u1. In our context, the functionfis not required to be monotone decreasing as we do not impose on it to cross the x-axis with negative slope. In chapter four we apply our results of rupture solutions for general elliptic equations to show existence of point rupture solutions to a class of quasi-linear elliptic equations of the form div(a(u)ru) =a0(u)2 jruj2+g(u) wherea2C1andg2C0are positive functions of a real variable. 9

2.0 RADIAL STEADY STATES OF VAN DER WAALS FORCE DRIVEN

THIN FILMS WITH PRESCRIBED FLUID VOLUME

Let R2be a bounded smooth domain. We consider the thin lms in a cylindrical container with base which is governed by the fourth order partial dierential equation h t=r(hnrp) (2.0.1) wherehis thickness of the thin lm and the pressure p=
h+1 h(2.0.2) is a sum of linearized surface tension and van der Waals force. Heren >0 and >1 are physical constants and in the van der Waals force driven thin lm, we haven= 3 and= 3.

We assume that there is no

ux across the boundary, which yields the boundary condition @p@ = 0 on@ :(2.0.3) We also ignore the wetting or non-wetting eect, and assume that the uid surface is or- thogonal to the boundary of the container, i.e., @h@ = 0 on@ :(2.0.4)

We associate (

2:0:1) with an energy functional

E(h) =Z

 12 jrhj21(1)h1 ; 10 and formally, using (2:0:1) and the boundary conditions (2:0:3);(2:0:4), we have ddt

E(h) =Z

rhrht+1 hht = Z 
h+1 h h t = Z pr(hnrp) =Z hnjrpj20:

Hence, for a thin lm

uid at rest, the pressurephas to be a constant, andhsatises the elliptic equation
h+1 h=pin with the Neumann boundary condition (

2:0:4).

In physical experiments, usually the total volume of the uid is a known parameter. i.e.,  h=1j jZ h(x)dx is given. Therefore for any given h >0, we need to nd a functionhand an unknown constantpsatisfying 8 >>>< > >>:
h=1 hpin ; 1j jR h(x)dx=h; @u@ = 0 on@ :(2.0.5) Our goal in this chapter is to understand the structure of the radial solutions and their associated energy when the prescribed volume hchanges. 11

2.1 SCALING PROPERTY OF GLOBAL RADIAL SOLUTIONS

Let =B1(0), a unit disk inR2. We are interested in the radial solutions of (2:0:5) when  his given which leads to 8>>>< > >>:h rr+1r hr=1 hpinB1(0); 2 R1

0rh(r)dr=h;

h

0(1) = 0:(2.1.1)

From the elliptic theory,his smooth whenever it is positive, hence we also require that h

0(0) = 0 ifh(0)>0.

We rst ignore the volume constraint and the Neumann boundary condition. Fixing p=1 , we consider the ordinary dierential equation h rr+1r hr=1 h1 (2.1.2) dened on [0;1). It has been shown in [15] that for any >0, 8>>>< > >>:h rr+1r hr=1 h1 ; h(0) =; h

0(0) = 0(2.1.3)

has a unique positive solutionhdened on [0;1). And when= 0, there exists a unique rupture solutionh0which is continuous on [0;1) such thath0(0) = 0 andh0is positive and satises (

2:1:2) on (0;1). We remark here thath0r(0) =1. Some plots for the radial

rupture solution and smooth radial solutions follow in the next page. 12 Figure 1: Global rupture solution with 9 oscillations. 13 Figure 2: Global radial smooth solution with eta=7. 14

Figure 3: Radial solution with eta = 11.

15 Figure 4: Radial smooth solution with eta = 0.000001 16

Figure 5: Radial smooth solution with eta = 20.

17 Obviouslyh1 if= 1. When0,6= 1,hoscillates around 1 and there exists an increasing sequence of critical radiusr k! 1such that (h)0(r k) = 0. Given0,6= 1 and a positive integerk,h(r) satises the Neumann boundary condition atr=r k. We now dene h ;k(r) = (r k)21+h(r kr): One can easily verify thath;k(x) =h;k(jxj) satises the elliptic equation
h=1
hp;kin B1(0) with Neumann boundary condition @h@ = 0on @B1(0):

Herep;kis dened by

p ;k=1 (r k)21+: We can also calculate the average thickness forh;k,  h;k=1jB1(0)jZ B

1(0)h;k(x)dx=(r

k)21+jBr k(0)jZ B r k(0)h(r)dr = 2(r k)21+2Zr k

0rh(r)dr:

Its associated energy is given by

E ;k=Z B 1(0) 12 rh;k21(1)h;k
1 = (r k)41+Z B r k(0) 12 jrhj21(1)(h)1 = 2(r k)41+Z r k 0 12  dhdr  2 1(1)(h)1! rdr:

Hence, we constructed a solution to (

2:1:1) with

 h=h;k:

Actually, all nonconstant solutions to (

2:1:1) can be obtained in this fashion.

18 Figure 6: Smooth radial rescaled to the unit ball with eta=.5 and k=1 19 Figure 7: Smooth radial rescaled to the unit ball with eta=.5 and k=2 20 Figure 8: Smooth radial rescaled to the unit ball with eta=.5 and k=5 21
Figure 9: Smooth radial rescaled to the unit ball with eta= 2 and k=1 22
Figure 10: Smooth radial rescaled to the unit ball with eta= 2 and k=2 23
Figure 11: Smooth radial rescaled to the unit ball with eta= 2 and k=8 24
Figure 12: Rupture radial rescaled to the unit ball with k=1 25
To obtain the structure of nontrivial radial solutions when his given, we need to nd ;kso thath=h;k. Hence, we need to completely understand the dependence ofh;kon andk. Fixing a positive integerk, from the continuous dependence of ordinary dierential equa- tions on the initial data, h(;k) =h;kis a continuous function onon (0;1)[(1;1). In the next three sections, we want to investigate the behavior of h(;k) as!0,!1 and! 1. Also we would like to show that for a xedthenh(;k) approaches zero as k! 1

2.2 RUPTURE SOLUTION AS A LIMIT OF SMOOTH SOLUTIONS!0

As!0+,hconverges uniformly to the rupture solutionh0on [0;1). Hence,h(;k) is continuous at= 0. See the following plots for illustration. 26
Figure 13: Set of smooth radial solution for eta = 0.2 , 0.1 , 0.01 , 0.001 and 0.0001. This gure proves that as!0 the corresponding smooth radial solution approaches the rupture solution. 27
Figure 14: Set of rupture radial solutions rescaled to the unit ball for k=1,2,3,4,5 and 6. From this plot we see that for xed, here it is zero since ruptures are considered only,  h;ktends to zero ask! 1in a decreasing manner. This result is in fact true for any. 28
Inspired by the numerical suggestions, we would like to intend to prove that for xed the average volume h;ktends to zero ask! 1in a decreasing manner. We will prove that indeed h;k!0 ask! 1however, we are unable to prove monotonicity here.

Proposition 2.1.For any xed0thenh;k!0ask! 1.

Proof.We know that ask! 1the critical pointr

k! 1. We also know that asr! 1 the radial solutionh(r) tends to 1. Therefore for anyR >0 there existsN >0 such that fork > Nthenr k> Rand for someR >0 the functionh(r)<2. Thus since we can write,  h;k= 2(r k)21+2 ZR 0 rh(r)dr+Z r k

Rrh(r)dr!

; then the result follows.2.3 LINEARIZATION WHEN!1  h(;k) is not dened when= 1. To understand the behavior ofh(;k) as!1, we need to understand the behavior ofhas!1. Recall thathis a solution to (2:1:2) with h (0) =and (h)0(0) = 0. We dene "=1; and w (r) =h(r)1" :

Thenwis a solution to the dierential equation

w rr+1r wr=1"  1 (1 +"w)1  (2.3.1) with initial condition w(0) = 1; w0(0) = 0: As!1,"!0, formally, (2:3:1) converges to the Bessel's dierential equation with order 0, w rr+1r wr+w= 0 29
with the initial date w(0) = 1; w0(0) = 0: Such limiting initial value problem has a unique solution which is the Bessel's function of the rst kind with order 0 and is given by J

0(x) =1X

n=0(1)n(n!)2 x2  2n: We remark here thatJ0(x) is oscillating around 0. See plot of Bessel function of the rst kind of order zero in the next page. 30
Figure 15: The Bessel function of rst kind and of order zero. This is the plot of the Bessel function of rst kind and of order zero ploted over the interval [0,70]. It oscillates around zero. One can see its similarity with the radial solutions. 31
We can show that as!1,wconverges uniformly toJ0. Since bothwandJ0are oscillating around 0,r k!rkas!1 whererk,k= 1;2;


is an increasing sequence of the critical radius of the Bessel's functionJ0(x). Sinceh!1 uniformly as!1, we have  h;k=(r k)21+jBr k(0)jZ B r k(0)h(r)dr!(rk)21+ as!1 and E ;k= (r k)41+Z B r k(0) 12 jrhj21(1)(h)1 !

1(1)(rk)41+Brk(0)=(1)(rk)241+:

as!1.

Hence, we can dene

h;kandE;kso that they are both continuous functions on [0;1).

2.4 LIMITING PROFILE WHEN! 1

In this section, we want to understand the behavior of h;kandE;kas! 1. Let >1 andhbe the solution to (2:1:3). we dene the blow down solutionzby z(x) =1 h(r) withr=px. Then we have z x=pp hr(r); z xx=hrr(r) and hence z 00+1x z0= h rr+1r hr =h1 = z 1: 32

Denoting"=1

, we have"!0 as! 1. The blow down functionzis a solution to the initial value problem 8 < :z 00+1x z0="z 1; z(0) = 1;andz0(0) = 0:(2.4.1) The following plot illustrates the blow down functionzfor"= 10. 33

Figure 16: Blow down solution with eta=0.1

This plot show an instance of blow down solution when"= 0:10. 34
Formally, as"!0, (2:4:1) converges to the limiting equation 8< :z 00+1x z0=1; z(0) = 1;andz0(0) = 0:(2.4.2) The following plot shows the prole of the asymptotic solution. 35

Figure 17: Asymptotic solution eta=100.

This plot illustrates how the asymptotic solution behaves and appears as the parameter ! 1 36
which has a unique global solution z(x) = 114 x2:

However, we can't expect

lim "!0z"(x) = 114 x2 since the function 114 x2becomes negative whenx >2. Nonetheless, we can establish the following theorem: Theorem 2.2.For every" >0, letz"(x)be the unique solution of the initial value problem (

2:4:1).

Then as"tends to zero positively,z"(x)converges uniformly on[0;1)toz(x), the solution of the limiting initial value problem 8 >>>< > >>:z

00+1x

z0=1; z>0inS1 j=0(aj;aj+1): z (0) = 1;andz0(0) = 0; z (aj) = 0; z0(aj+) =z0(aj)(2.4.3) wherea0= 0,2 =a1< a2<


are inductively computed by solving the IVP(2:4:3). Hence, in particular, we havez"(x) converges uniformly to 114 x2on [0;2] as"!0 and r 1p converges to 2 as! 1. More generally, we have fork= 1;2;3;


, lim !1r 

2k1p

=ak and lim !1r  2kp =bk wherebkis the maximum point ofzin (ak;ak+1). 37

Given a positive integerkand given >1, we have

 h;k= 2(r k)21+2Zr k

0rh(r)dr

= 2(r k)21+2Z r k

0rzrp

 dr = 2(r k)21+22Z r kp

0sz(s)ds

= 211+1+r kp  21+2Zr kp

0sz(s)ds

Hence, we have fork= 1;2;3;


,

lim !1 h;2k1 1+= 211+a21+2 kZ ak

0sz(s)ds

and lim !1 h;2k 1+= 211+b21+2 kZ bk

0sz(s)ds

We remark here that for each positive integerk,h;k! 1as! 1. Next we investigate the energy of radial solutions as! 1. Since z(x) =1 h(r); dh dr =z0(x)dxdr =pp z0(x): Hence E ;k= 2(r k)41+Z r k 0 12  dhdr  2 1(1)(h)1! rdr = 2(r kp )41+()121+Z r kp 0 12  (z0(x))21(1)(z(x))1 xdx =21+(r kp )41+221+Z r kp

0xjz0j2dx+O



221+

:

Hence, we have fork= 1;2;3;


,

lim !1E ;2k1

221+=21+a41+

kZ ak

0s(z)02ds:

38
and lim !1E ;2k

221+=21+b41+

kZ bk

0s(z)02ds:

The following plots illustrate how the average volume hof the uid changes aschanges from 0 to 100. 39

Figure 18: Average volume hbar.

This gure shows how for xedkthe parameterhbehaves as! 1. Here we let eta varies from 0 to 100. 40
Figure 19: Average volume hbar and its asymptotic approximation.

This plot shows that the plot of

hcoincides with its asymptotic behavior as! 1. We did the experiment withchanging from 0 to 100. We note the match of the approximation since5: 41
Figure 20: Energy versus eta. Eta changes from 0 to 100. This plot illustrates the behavior of the energy function as a function of. 42

Figure 21: Energy versus hbar.

This plot shows the behavior of the energy in terms of h. 43

Figure 22: Average volume for k=1,2 and 3.

This plot shows the variation of

hfork= 1;2;3 andin [0, 25]. This graph suggests that we do have bifurcations of solutions for some values of h. 44

3.0 RUPTURE SOLUTIONS OF GENERAL ELLIPTIC EQUATION

3.1 INTRODUCTION

In this chapter we will prove that under a light integrability condition the general two dimensional semi linear elliptic equation has a continuous weak radial point rupture solution which is locally monotone increasing near the origin.
u=f(u) in (3.1.1) This equation is obtained for steady states solutions of the modeling equation for thin lms over a planar region. u t=r
(umru) r
(unr
u):(3.1.2) This equation is only assumed to be valid whenu >0 and the set whereu= 0 is called the rupture set of the thin lm. Many rigorous works have been done on the above equation. However, when the space dimension is two, the physically realistic dimension, the dynamics of this equation is still not well understood. For Van der Waals the equation becomes 4u=1 upin ;(3.1.3) wherepis an unknown constant to be determined and= 3: 45

3.2 THE MAIN RESULT

The following is the statement of our main result for sucient conditions for existence of a rupture solution and its a priori bounds. Theorem 3.1.Let>0andfbe a continuous, monotone decreasing positive function on (0;]such that lim v!0+f(v) =1: Let

G(v) =Z

v

01f(s)ds:(3.2.1)

Assume in addition that

vf(v)G(v)2L1[0;]:(3.2.2) Then there existsr>0and a radial point rupture solutionu0to(3:1:1)inBr(0)such that u

0=u0(r)is continuous on[0;r],

u

0(0) = 0,u0(r)>0for anyr2(0;r];

anduis a weak solution to(3:1:1)inBr(0). Moreover,u0is monotone increasing and G 114 r2 u0(r)Z

G1(14

r2)

0vf(v)G(v)dsfor anyr2[0;r]:

Remark3.2.Here the technical assumption (3:2:2) is not very strong, for example, iff(v) = v , for some >0, we would have vf(v)G(v)=vv 11+v1+
= 1 +2L1[0;]: Such assumption also holds for some singularity of exponential growth, for example, if f(v) =vp+1e1v p, 0< p <1; 46
we have vf(v)G(v)=pv p2L1[0;]: Remark3.3.The assumption thatfis monotone decreasing can be replaced by the assumption thatfis a product of a monotone decreasing function and a uniformly bounded positive function, one may check how this idea is developed in chapter four in an ndimensional vector space wheren3.

3.3 PROOF OF THE MAIN RESULT

For any2(0;), we useuto denote the unique solution to the initial value problem 8 < :u rr+1r ur=f(u); u(0) =;u0(0) = 0:(3.3.1) Lemma 3.4.There existsr>0such thatuis dened on[0;r]withu(r) =.

Moreover,u0(r)>0on(0;r]and

G 114 r2 u(r)+Z

G1(14

r2)

0vf(v)G(v)dson[0;r]:(3.3.2)

Proof.For simplicity, we suppress thesubscript in this proof. We write u rr+1r ur=f(u) 47
in the form of (rur)r=rf(u)0; so we have ru r=Z r 0 sf(u(s))ds0: In particular,uis monotone increasing anducan be extended wheneverf(u) is dened and bounded. Hence, there existsr>0 such thatuis dened on [0;r] withu(r) =. Sinceuis monotone increasing andfis monotone decreasing, we have ru r=Z r 0 sf(u(s))dsf(u(r))Z r 0 sds=12 r2f(u(r)); hence, u rf(u)12 r:

Integrating again, we have

G(u(r))G() +14

r214 r2:

SinceGis strictly monotone increasing, we have

u(r)G114 r2 :

On the other hand,

ru r=Z r 0 sf(u(s))dsZ r 0 f G 114 s2 sds:

Letv=G114

s2
, we haveG(v) =14 s2, and

1f(v)dv=12

sds:

Hence,

Z r 0 f G 114 s2 sds=Z

G1(14

r2) 0

2dv= 2G114

r2 : 48

Hence,

u r2r

G114

r2 which yields u(r)+Z r 02s

G114

s2 ds =+Z

G1(14

r2)

0vf(v)G(v)dv:The bounds onucan be explored to nd a uniform lower bound for allr, more accurately

this is stated in the following corollary. Corollary 3.5.There existsr>0such that for any20;2 , r r:

We can take

r = 2sG  H 12  where

H(u) =Z

u

0vf(v)G(v)dv:

49

Proof.For any20;2

,  =u(r)+Z

G1(14

r2)

0vf(v)G(v)dv

 2 +Z

G1(14

r2)

0vf(v)G(v)dv:

Hence,

Z

G1(14

r2)

0vf(v)G(v)dv2

: Since vf(v)G(v)is integrable, the function

H(u) =Z

u

0vf(v)G(v)dv

is strictly monotone increasing, so H  G 114 r2 2 implies r 2sG  H 12  :The point rupture solutions can be constructed as a uniform limit of sequences of smooth solutions ofukask!0.

Proposition 3.6.There exists a sequencefkg1

k=10;2 satisfyinglimk!1k= 0, such thatuk!u0uniformly inB r(0)ask! 1, for some functionu02C0B r(0) \ C 2B r(0)nf0g . Moreover,u0is a classical solution to(3:1:1)inBr(0)nf0gand G 114 r2 u0(r)Z

G1(14

r2)

0vf(v)G(v)dson[0;r]:

50

Proof.For any" >0,u,20;2

is a family of uniformly bounded classical solutions to
u=f(u) inB r(0)nB"(0); hence by a diagonal argument, there exists a sequencefkg1 k=10;2 satisfying limk!1k=

0, such thatuk!u0locally uniformly inB

r(0)nf0gask! 1. Now (3:3:2) implies G 114 r2 u0(r)Z

G1(14

r2)

0vf(v)G(v)dson [0;r]:

Since lim r!0Z

G1(14

r2)

0vf(v)G(v)ds= 0;

it is not diculty to see, from the bounds ofuandu0, thatuk!u0uniformly inB r(0) ask! 1.Remark3.7.The above limit should be independent of the choice offkg1 k=1. Actually, we expect thatu!u0uniformly on [0;r] as!0. Unfortunately, we are unable to provide a proof here and the question remains open. The following lemma is crucial in the remaining part in order to show that the function u

0is actually a Sobolev solution to the original problem.

Lemma 3.8.

lim r!0+ru00(r) = 0:(3.3.3)

Proof.For anyr2(0;r), we have

(ru00(r))0=rf(u0)>0: 51
Hence,ru00(r) is monotone increasing in (0;r). Sinceru00(r)0 in (0;r), = lim r!0+ru00(r)0 is well dened. If >0, we have forrsuciently small, sayr2(0;~r], ru

00(r)2

hence, for anyr2(0;~r], u

0(r) =u0(~r)Z

~r r u00(r)dru0(~r)Z ~r r2rdr: which contradicts to the fact thatu0is continuous at 0 if we letr!0+. Hence= 0 and (

3:3:3) holds.Proposition 3.9.f(u0)2L1(Br(0))andu0is a weak solution to(3:1:1)inBr(0).

Proof.For any test function'2C1c(Br(0)), we have

Z B r(0)u

0
'dx= lim

"!0+Z B r(0)nB "(0)u 0
'dx = lim "!0+ Z B r(0)nB "(0)
u0'dxZ @B "(0) u 0@'@n '@u0@n  ds x = lim "!0+ Z B r(0)nB "(0)f(u0)'dxZ @B "(0)u 0@'@n dsx+Z @B "(0)'@u0@n dsx : Now for any"2(0;r), sinceu0(")u0(r), we have  Z @B "(0)u 0@'@n dsxu0(")kr'kL1(Br(0))j@B"(0)j 2"u0(")kr'kL1(Br(0))!0 as"!0+. On the other hand, (3:3:3) implies that 52
 Z @B "(0)'@u0@n dsx2"u00(")k'kL1(Br(0))!0 as"!0+. Hence, we have for any'2C1c(Br(0)), Z B r(0)u

0
'dx= lim

"!0+Z B r(0)nB "(0)f(u0)'dx: Choosing'such that'1 near the origin, the above limit implies thatf(u0) is integrable near the origin. Sincef(u0) is a positive continuous function inBr(0)nf0g, we concludef(u0)2L1(Br(0)). So we have for any test function'2C1c(Br(0)) Z B r(0)u

0
'dx= lim

"!0+Z B r(0)nB "(0)f(u0)'dx=Z B r(0)f(u0)'dx; i.e.,u0is a weak solution to (3:1:1) inBr(0).3.4 SEVERAL EXAMPLES In this section, we discuss several examples in order to get a better understanding of the technical assumption.

Example 3.10.

f(u) =u where >0 is a constant. Here we have

G(v) =Z

v

0dsf(s)=Z

v 0dss =Z v 0 sds=v+1+ 1 therefore vG(v)f(v)=vv +1+1v=+ 12L1[0;]: 53

Example 3.11.For some 0< p <1,

f(v) =vp+1e1v p

We have

G(v) =Z

v

0dsf(s)=Z

v 0dss p+1e1s p=pe1v p therfore, vG(v)f(v)=1pv p2L1[0;1] sinceZ 1 0dvpv p=1p(1p): Example 3.12.This example shows that our result is not optimal. Let f(u) = 2u3e2u which is monotone decreasing near the origin and lim u!0+f(u) =1: we have vf(v)G(v)62L1(0;] for any>0. However, let u=1lnr; we have u r=1rln2r; urr=1r

2ln2r21r

2ln3r;

and so u rr+1r ur=21r

2ln3r= 2u3e2u

=f(u) Henceu=1lnr;is a rupture solution even if the technical assumption is not satised. 54

3.5 DYNAMICAL SYSTEM FORMULATION

In this section, we give a formulation of our problem in the dynamical system setting which might be helpful in proving the uniqueness of rupture solutions. We are interested in the rupture solutions to u rr+1r ur=f(u):(3.5.1)

Letr=et, we have

dr=rdt and u r=1r ut;urr=1r

2ut+1r

2utt; hence u tt=r2f(u): Let

R=r2;v=ut;

we obtain an autonomous system

8>>>>>>>>><

> >>>>>>>>:u t=v; v t=Rf(u); R t=2R:(3.5.2) It is not dicult to see that a radial rupture solution to (

3:5:1) is equivalent to a trajectory

of (

3:5:2) in the regionf(u;v;R) :u;v;R >0gwhich approaches the origin ast! 1. Hence

givenf, we are interested in the existence and uniqueness of such trajectory. 55

4.0 POINT RUPTURE SOLUTIONS OF A CLASS OF QUASI-LINEAR

ELLIPTIC EQUATIONS

4.1 INTRODUCTION

Let be a region inR2, andfbe a smooth function dened on (0;1) satisfying lim s!0+f(s) =1;(4.1.1) we consider the quasi-linear elliptic equations of the form div(a(u)ru) =a0(u)2 jruj2+f(u) (4.1.2) where the terms depending uponaare formally associated with the functional Z a(u)jruj2 which can be viewed as a minimizing problem in presence of a Riemannian metric tensor depending upon the unknownuitself. Motivated by the studies of thin lm equations, a solution to (

4:1:2) is said to be a point

rupture solution if for somep2 ,u(p) = 0 andu(x)>0 for anyx2 nfpg. Our main result is the existence of a radial rupture solution: 56
Theorem 4.1.Assume that for some>0,a2C1[0;],f2C1(0;]are positive functions such that for some positive constantsm < M, ma(u)M holds for anyu2[0;]andfis monotone decreasing function on(0;]satisfying uG(u)f(u)2L1[0;] (4.1.3) where

G(u) =Z

u

01f(s)ds:

Then there existsr>0and a radial point rupture solutionu0to(4:1:2)inBr(0)such that u

0=u0(r)is continuous on[0;r],

u

0(0) = 0,u0(r)>0for anyr2(0;r]:

Moreover,u0is a weak solution to(4:1:2)in the sense that for any'2C10(B1(r)), Z B

1(r)A(u0)
'=Z

B

1(r)f(u0)pa(u0)'

where

A(u) =Z

u

0pa(s)ds:

Whena1, (4:1:2) is reduced to the simpler form

u=f(u) and its rupture solution has been investigated in [ 13 ],[ 15 ]when f(u) =u1, >1 which has application to the van der Waals force driven thin lms, in [ 14 ] withfsatisfying the growth condition ( 4.1.3 ) and in [ 11 ] when the space dimension3. (4:1:2) has also been studied by F. Gladiali and M. Squassina [ 10 ] where they are interested in the so called explosive solutions. 57

4.2 PROOF OF THE MAIN RESULT

We consider the quasi-linear equations of the form div(a(u)ru) =a0(u)2 jruj2+f(u) (4.2.1) in a region RNwhere for some>0,a2C1[0;] andf2C1(0;] are positive functions.

Following [

10 ], letgbe the unique solution to the Cauchy problem g

0=1pa(g); g(0) = 0;

and letvbe a solution to
v=h(v) (4.2.2) where h(v) =f(g(v))pa(g(v)): Dene u=g(v):

We have

ru=g0(v)rv=1pa(g)rv; hence rv=pa(u)ru; which leads to
v=pa(u)
u+12

1pa(u)a0(u)jruj2:

Hence (

4:2:2) implies

pa(u)
u+12

1pa(u)a0(u)jruj2=f(u)pa(u)

which is equivalent to (

4:2:1). Hence, (4:2:1) admits a point rupture solution if and only if

(

4:2:2) has a point rupture solution.

58
Noticing thath(v) =f(g(v))pa(g(v))is not necessary monotone decreasing inv, we can't apply the result in [ 14 ] directly. However, the boundedness ofaand the monotone properties off andgimplies that 1pM f(g(v))h(v)1pm f(g(v)); i.e.,his bounded by two monotone decreasing functions.

We need to following theorem:

Theorem 4.2.Let>0andh1,h22C1(0;]be monotone decreasing functions such that

0< h1h2on(0;]

and lim v!0+h1(v) = lim v!0+h2(v) =1:

Leth2C1(0;]satisfy

h

1hh2on(0;]:

Let G

1(v) =Z

v 01h

1(s)ds:(4.2.3)

Assume in addition that

h 2h

12L1[0;]andR

u 0h 2(v)h

1(v)dvG

1(u)h1(u)2L1[0;]:(4.2.4)

Then there existsr>0and a radial point rupture solutionu0to
u=h(u) (4.2.5) inBr(0)such thatu0=u0(r)is continuous on[0;r], u

0(0) = 0,u0(r)>0for anyr2(0;r];

anduis a weak solution to(4:2:5)inBr(0). Moreover,u0is monotone increasing and G 1114 r2 u0(r)Z G1 1(14 r2) 0R u 0h 2(v)h

1(v)dvG

1(u)h1(u)dufor anyr2[0;r]:

59

We leave its proof to the next section.

Let h

1(v) =1pM

f(g(v)) andh2(v) =1pm f(g(v)):

We have

h

1(v)h(v)h2(v)

on (0;]. Moreover,h1;h22C1(0;] are monotone decreasing functions such that

0< h1h2on (0;]

and lim v!0+h1(v) = lim v!0+h2(v) =1:

Next we check the growth condition. We have

h 2h

1=pMpm

2L1[0;]:

Now since

g

0=1pa(g);

we have G

1(v) =Z

v 01h

1(s)ds:=pM

Z v

01f(g(s))ds

= pM Z v

0pa(g)f(g(s))g0(s)ds

 pmM Z v

01f(g(s))g0(s)ds

= pmM Z g(v)

01f(v)dv

= pmMG(g(v)) and Ru 0h 2(v)h

1(v)dvG

1(u)h1(u)=pMpm

uG

1(u)1pM

f(g(u)) pM m uG(g(u))f(g(u))  Mm g(u)G(g(u))f(g(u)) 60
where we used g(u)1pM u which follows fromg0(u)1pM andg(0) = 0. Hence, for any >0 suciently small, Z  0R u 0h 2(v)h

1(v)dvG

1(u)h1(u)

Z  0Mm g(u)G(g(u))f(g(u))du = Z  0Mm g(u)G(g(u))f(g(u))pa(g)g0(u)du  MpM m Z g()

0vG(v)f(v)dv:

So the growth condition in (

4:1:3) implies that for some ~ >0,

R u 0h 2(v)h

1(v)dvG

1(u)h1(u)2L1[0;~]:

Applying Theorem

4.2 , we have the existence of rupture solutionv0to (4:2:2). Then u

0=g(v0)

is a rupture solution to (

4:1:2). Moreover,u0is a weak solution to (4:1:2) follows from the

fact thatv0is a weak solution to (4:2:2). 61

4.3 POINT RUPTURE SOLUTION WITH NON-MONOTONICH

We prove Theorem

4 :2in this section. F oran y2(0;), we useuto denote the unique

solution to the initial value problem 8< :u rr+1r ur=h(u); u(0) =;u0(0) = 0:(4.3.1) Lemma 4.3.There existsr>0such thatuis dened on[0;r]withu(r) =.

Moreover,u0(r)>0on(0;r]and

G 114 r2 u(r)+H G 1114 r2 on[0;r]:(4.3.2) where

H(w) =Z

w 0R u 0h 2(v)h

1(v)dvG

1(u)h1(u)du:

Proof.For simplicity, we suppress thesubscript in this proof. We write u rr+1r ur=h(u) in the form of (rur)r=rh(u)0; so we have ru r=Z r 0 sh(u(s))ds0: In particular,uis monotone increasing anducan be extended wheneverf(u) is dened and bounded. Hence, there existsr>0 such thatuis dened on [0;r] withu(r) =. Sinceuis monotone increasing andh1is monotone decreasing, we have ru r=Z r 0 sh(u(s))dsZ r 0 sh

1(u(s))ds

h1(u(r))Z r 0 sds=12 r2h1(u(r)); hence, u rh

1(u)12

r: 62

Integrating again, we have

G

1(u(r))G1() +14

r214 r2: where G

1(v) =Z

v 01h

1(s)ds:

SinceG1is continuous and strictly monotone increasing,G11is well dened and we have u(r)G1114 r2 :

On the other hand, sinceh2is monotone increasing,

ru r=Z r 0 sh(u(s))dsZ r 0 sh

2(u(s))dsZ

r 0 h 2 G 1114 s2 sds:

Letv=G1114

s2
, we haveG1(v) =14 s2, and 1h

1(v)dv=12

sds:

Hence,

Zr 0 h 2 G 1114 s2 sds= 2Z G1 1(14 r2) 0h 2(v)h

1(v)dv:

Hence,

u r2r Z G1 1(14 r2) 0h 2(v)h

1(v)dv

which yields u(r)+Z r 02s " ZG1 1(14 s2) 0h 2(v)h

1(v)dv#

ds =+Z G1 1(14 r2) 02s  Zw 0h 2(v)h

1(v)dv2sh

1(w)dw

=+Z G1 1(14 r2) 0R w 0h 2(v)h

1(v)dvG

1(w)h1(w)dw

=+H G 1114 r2 where

H(w) =Z

w 0R u 0h 2(v)h

1(v)dvG

1(u)h1(u)du

63
and we used substitution w=G1114 s2 :The bounds onuimplies: Corollary 4.4.There existsr>0such that for any20;2 , r r:

We can take

r = 2sG 1 H 12  :

Proof.For any20;2

,  =u(r)+H G 1114 r2  2 +H G 1114 r2 :

Hence,

H G 1114 r2 2 : Since the functionHis strictly monotone increasing, we have r 2sG 1 H 12  :The point rupture solution can be constructed as the limit ofuas!0. 64

Proposition 4.5.There exists a sequencefkg1

k=10;2 satisfying lim k!1k= 0; such thatuk!u0uniformly inB r(0)ask! 1, for some function u

02C0B

r(0) \C2B r(0)nf0g : Moreover,u0is a classical solution to(4:2:5)inBr(0)nf0gand G 1114 r2 u0(r)H G 1114 r2 on[0;r]:

Proof.For any" >0,u,20;2

is a family of uniformly bounded classical solutions to
u=h(u) inB r(0)nB"(0); hence by a diagonal argument, there exists a sequencefkg1 k=10;2 satisfying limk!1k=

0, such thatuk!u0locally uniformly inB

r(0)nf0gask! 1. Now (4:3:2) implies G 1114 r2 u0(r)H G 1114 r2 on [0;r]: Since lim r!0H G 1114 r2 = 0; it is not diculty to see, from the bounds ofuandu0, thatuk!u0uniformly inB r(0) ask! 1.Remark4.6.The above limit should be independent of the choice offkg1 k=1. Actually, we expect thatu!u0uniformly on [0;r] as!0. Unfortunately, we are unable to provide a proof here. In order to show thatu0is a weak solution. We need the following lemma:

Lemma 4.7.

lim r!0+ru00(r) = 0:(4.3.3) 65

Proof.For anyr2(0;r), we have

(ru00(r))0=rf(u0)>0: Hence,ru00(r) is monotone increasing in (0;r). Sinceru00(r)0 in (0;r), = lim r!0+ru00(r)0 is well dened. If >0, we have forrsuciently small, sayr2(0;~r], ru

00(r)2

hence, for anyr2(0;~r], u

0(r) =u0(~r)Z

~r r u00(r)dru0(~r)Z ~r r2rdr: which contradicts to the fact thatu0is continuous at 0 if we letr!0+. Hence= 0 and (

4:3:3) holds.Proposition 4.8.h(u0)2L1(Br(0))andu0is a weak solution to(4:2:5)inBr(0).

Proof.For any test function'2C1c(Br(0)), we have

Z B r(0)u

0
'dx= lim

"!0+Z B r(0)nB "(0)u 0
'dx = lim "!0+ Z B r(0)nB "(0)
u0'dxZ @B "(0) u 0@'@n '@u0@n  ds x = lim "!0+ Z B r(0)nB "(0)h(u0)'dxZ @B "(0)u 0@'@n dsx+Z @B "(0)'@u0@n dsx : Now for any"2(0;r), sinceu0(")u0(r), we have Z @B "(0)u 0@'@n dsxu0(")kr'kL1(Br(0))j@B"(0)j 2"u0(")kr'kL1(Br(0))!0 as"!0+. On the other hand, (4:3:3) implies that Z @B "(0)'@u0@n dsx2"u00(")k'kL1(Br(0))!0 66
as"!0+. Hence, we have for any'2C1c(Br(0)), Z B r(0)u

0
'dx= lim

"!0+Z B r(0)nB "(0)f(u0)'dx: Choosing'such that'1 near the origin, the above limit implies thatf(u0) is integrable near the origin. Sincef(u0) is a positive continuous function inBr(0)nf0g, we conclude f(u0)2L1(Br(0)). So we have for any test function'2C1c(Br(0)) Z B r(0)u

0
'dx= lim

"!0+Z B r(0)nB "(0)h(u0)'dx=Z B r(0)h(u0)'dx;

i.e.,u0is a weak solution to (4:2:5) inBr(0).Theorem4.2 is a com binationof Prop osition4.5 and Prop osition4.8 .

Finally, we discuss several examples in this section to get a better understanding of the technical assumption on the growth rate ofh.

Example 4.9.

h(u) =b(u)u where >0 is a constant andb(u) satises B

1b(u)B2

for some constants 0< B1< B2. If we take h

1=B1uandh2=B2u;

we have Ru 0h 2(v)h

1(v)dvG

1(u)h1(u)=(1 +)B2B

12L1[0;1]:

67

Example 4.10.For some 0< p <1,

h(u) =b(u)vp+1e1v p andb(u) satises B

1b(u)B2

for some constants 0< B1< B2. If we take h

1=B1vp+1e1v

pandh2=B2vp+1e1v p; we have Ru 0h 2(v)h

1(v)dvG

1(u)h1(u)=B2pB

1vp2L1[0;1]:

Example 4.11.

f(u) =12 

1 + sin1u

 u +

1sin1u

 u  where

0<  <  < + 1:

We take

f

1(u) =u;f2(u) =u;

then we have for anyu2(0;1], h

1(u)h(u)h2(u):

Hence,

Ru 0h 2(v)h

1(v)dvG

1(u)h1(u)=R

u

0udv1

1+u=1 +1 +u2L1[0;1]

since >1. In this example,hcan't be expressed as a product of a bounded function and a monotone function. Interested readers should verify this. 68

5.0 CONCLUSION

In this thesis, we focused on the dynamics of thin lm equation modeling the steady states solutions of van der Waals force driven thin lm for a viscous uid assuming that there is no ux across the boundary and ignoring wetting and non-wetting eect and that the uid surface is perpendicular to the boundary of the container. The use of radial solutions was the approach to study solutions near the origin, point rupture solutions must be discrete for nite energy solutions and we expect them to be important in the sense that they may serve as model and blow up prole for the solutions near the rupture point. In chapter two, we performed many numerical experiments to study the most physical case of thin lm equation that is the two dimensional type of thin lm and for the special equation
h=1
hpwhen= 3. RK4 numerical method was used and most of the properties of the radial solution were clearly revealed in particular, their uniform boundedness and there oscillatory behavior. Smooth radial solutions could be plotted easily for any initial data, however for larger values then the method starts taking considerable amount of time, therefore we expect to use higher techniques if one needs faster and more accurate numerical approximations to the radial solutions. The rupture solution is computed using RK4, however it starts from zero by the use of an expansion to reach the rst step, and then proceeds via RK4 with controllable tolerance. The plots of the energy function versus and the average volumehsuggest the existence of bifurcations and their number must increase withk. The behavior of the energy shows that it approaches1asget larger. Asymptotic analysis revealed properties for the limiting solution as the initial data ! 1and behavior of the averagehand the energy were computed in terms of the initial data. 69
In chapter three, we consider the elliptic equation
h=f(h) as we stated and proved the theorem for a sucient condition onffor radial solutions as we obtain the bounds for the weak and radial point rupture solution under a light condition onf, also the assumption onfto be monotone decreasing can easily be replaced by assuming thatfcan be written as the product of two functionsf1andf2wheref1is uniformly bounded andf2is monotone decreasing such hypothesis can be useful in proving a similar result in any space of dimen- sion greater than two. From the proof one sees that the rupture is constructed from an arbitrary uniformly bounded sequence of smooth radial solutions and therefore uniqueness of the point rupture solution may not be guaranteed, thus the question remains open. The rupture set  =fx2 :u(x) = 0gcorresponds to "dry spots" in the thin lms, which is of great signicance in the coatings industry where non uniformities are very undesirable. In the two dimensional case we proved that the point rupture solution is never zero away from the origin therefore the only point where the solution touches the x-axis is the origin and therefore the measure of the rupture set is indeed zero. In chapter four, we consider the elliptic equation
h=f(h) in a two dimensional space such that the functionfis uniformly bounded near zero by two functionsf1and f

2continuous, monotone decreasing and satisfying

lim v!0+f1(v) = lim v!0+f2(v) =1 and under the conditions, f 2f

12L1[0;] andR

u 0f 2(v)f

1(v)dvG

1(u)f1(u)2L1[0;]:

The local weak solutionu0was derived with its appropriate bounds. It is important to note that the solution is actually a global weak solution. Many examples were discussed to show how we can apply these results. In particular, an interesting example of a non monotone function was mentioned. Afterward, we showed how these results can be applied in the plane to a quasi-linear elliptic equation of the form, 70
div(a(u)
u) =a0(u)2 jruj2+g(u) Next, we consider the elliptic equation
h=f(h) in anyndimensional vector space with dimensionn3 and using a dierent approach as in [11] and using change of variable procedure which works in any space dimension, the results of the existence of weak rupture solution still hold with appropriate bounds and can be be proved under the sole condition offbeing decreasing. Then we adapted the method to the above quasi-linear equation and obtain similar results. Furthermore, in the near future I plan and I would like to invest more on thin lm equations in general and consider the question of analyzing the dynamic of the fourth order partial dierential equation more carefully since it is derived from the very important Navier Stokes Equation under some simplifying assumptions. Radial solutions are very special ones for this partial dierential equation and these solutions must be very important in studying and approaching the general solutions and they may serve as a model or a blow up prole. I also would like to consider not only the steady states solutions but solutions depending on time as well. Numerical experiments suggest bifurcations of solutions and the number must vary with the average volume of the container and withk, the order of the critical point. Better tools and numerical solvers could help to give a clearer picture and would help understanding better the behavior of solutions. In the future, I also would like to explore in more detail the question of estimating the Hausdor dimension of the rupture set. 71

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