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UNFOLDING OF A CLASS OF SINGULAR FREE
BOUNDARIES FOR THE FOUR DIMENSIONAL AXI-SYMMETRIC
OBSTACLE PROBLEM

BY

Ahmad Feyzi Dizaji

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1983

/

GLVQOS

ABSTRACT

UNFOLDINGS OF A CLASS OF SINGULAR FREE
BOUNDARIES FOR THE FOUR DIMENSIONAL AXI—SYMMETRIC
OBSTACLE PROBLEM

BY

Ahmad Feyzi Dizaji

In this thesis we shall consider an elliptic free
boundary value problem, We shall deal with the four
dimensional aXi—symmetric obstacle problem and shall
consider a class of singularities of free boundary.
Specifically, we study generic perturbations of the
singularity x = H(y)-—% y log y, where H(t) is a
real analytic function about t = O with H(O) = O and
H'(O) # 0 near the origin. This locus described by
this formula represents the boundary of the region of

contact between a membrane and a smooth rigid obstacle.

The point of view taken is that of generic bifurcation,
where one scalar parameter (2K, the height of the membrane
above the obstacle at the origin) and one functional
parameter (¢) is present. Our prime interest is a
description of the unfolding of such singularities, their

normal forms, and generic conditions for unfoldings.

ACKNOWLEDGMENTS

I want to express my deep gratitude to Professor
Shui—Nee Chow and Professor John Mallet-Paret, my
research advisors, for all their patience, expert
guidance, and encouragement during the course of my
research. I am especially indebted to Professor
Mallet—Paret for his stimulating discussions and assistance

during the preparation of this thesis.

I am also grateful to my wife, Nasrin Habibi, for
her patience and cooperation displayed throughout my

graduate studies.

I am grateful to the members of my thesis committee as
well as guidance committee. I extend my appreciation to
all my instructors at Michigan State University, Tehran
University, high school, elementary school, and to my
parents from whom I learned a great deal of wisdom and

faith.

I also want to thank all the peOple who want to use

the science for justice and peace only.

Finally, I would like to thank Mrs, Cindy Lou Smith

for her patient typing of this manuscript.

TABLE OF CONTENTS

Chapter

1.

INTRODUCTION . . . . . . . . . . . . . . .
The minimal surface problem . . . . .

The obstacle problem . . . . . .

Some examples . . . . . . . . . . . . . . .

Local solution to the obstacle problem .

FORMULATION OF A PROBLEM . . . . . . . . .

Local solution to the axi- -symmetric
obstacle problem in R4 . . . . . . . .

A class of examples . . . . . . . . . . . .
Perturbation of Example 2.5: Heuristics .

Description of a problem . . . . . . o . .

A NECESSARY AND SUFFICIENT CONDITION FOR A
LOCAL SOLUTION . . . . . . o . . . . . .

An equivalent definition of a local
solution of axi-symmetric problem in R4 .

Proposition 3.9° Hypothesis 3.7 holds if
and only if Hypothesis 3.8 holds . . . .

Theorem 3.11. (u,O) in Definition 3.10
is a local solution to the obstacle
problem in the sense of Definition 3.1
if and only if . . . . . . . . . . . . .

Proposition 3.12° The local solution

(u,G) is symmetric with respect to
y-aXiS if and only if a o o o o o a a o 0

AN ANALYTIC CONTINUATION TO THE FREE
BOUNDARY . . . o . . o a o . . o . ._o .

Theorem 4.6. (The main Theorem) . . . . 0

Remark 4.7. (free boundary equation) ,

28

28

35

47

56

59

59

68

77

83

86
9O
91

Chapter

50

A CLASS OF UNFOLDING OF THE SINGULARITY
Problem 5.1. . . . . . . . . . . . . .

Necessary conditions in Problem 5.1 .

Remark 5.20. (a list of necessary

conditions for Problem 5.1) . . .

Sufficient conditions in Problem 5.1 continue

Conclusions . . . . . . . . . . . . .

i v

Theorem 5.30. (solving the Problem 5.1).

128
132
137
148

Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
~Figure
Figure
Figure

Figure

LIST

0 o
o o
n o
o o
o
e o
o n
o o
a o
o n
a o
o o
o a
o o
o o
a
a a
a

OF FIGURES

Figure
Figure
Figure
Figure
Figure
.Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure

Figure

5.9
5.10
5.11
5.12
5.13
5.14
5.15
5.16
5.17
5.18
5.19—a
5.19-b
5.20-a

5.20—b

vi

61

64

89

93

95

96

98
100
104
105
107
109
109
111
113
115
122
123
126
130
134
141
142
145

145

GLOSSARY OF NOTATIONS

Rn: Euclidean n-dimensional space, the product of

n copies of the real line R.
0: An open, generally bounded and connected subset of Rn.

an: The boundary of Q.

Bu bu Bu)

I [can]
8x1 5x2 axn

partial derivatives.

vu: The vector if u :Rn 4 R has

n 2u n
Au: Z) ——E if u :R 4 R
=1 6X.
Y J
u ,u : The partial derivatives 22 . i3 if u :R2 a R.
x y BX by
Cé(Q): The set of continuous functions with compact
support in Q.
Cm’l(5): The functions m times continuously differentiable
in 6 whose mth derivatives satisfy a Halder
condition with exponent X.
‘ Cm(5): The functions m times continuously differentiable
in 5
2,5 1 1
H (Q), H (0), HO(Q): The usual Sobolev spaces of
completion of Cm(5), 02(6), CS(Q), respectively,

each one with the corresponding norm.

1. INTRODUCTION

The minimal surface problem

1.1. Consider the classical minimal surface problem
of finding a surface in three—dimensional space, spanning
a fixed boundary loop, in such a way as to minimize the

area of the surface (Plateau's problem) [l8,16,6].

‘ X

7

./®

Figure 1.1

In particular in the non-parametric minimal surface problem
[18,6,8], we suppose the surface and its boundary are
graphs of functions u and g defined in subsets of

R2 as follows:

Given 0 C R2 a bounded domain with smooth
boundary BC, and g :53 4 R, find
1

u :5 4 R such that u = g on 30 and u

minimizes the area integral

H

A(u) = H‘ (1+ \vu‘2)2 dxdy (1.1)
0

among such functions.

Suppose this problem has a solution u, sufficiently

smooth. Let us derive the Euler equation satisfied by

this solution. Fix v E Cé(0)7 then for any t E R
A(u4—tv) 2 A(u) , (u-+tv)| = g .
6.
Hence
_1
0=-d—A(u+tv)| =r'f (1+)vu12) 2 (u v +u v )dxdy
dt 6:0 Juq xx yy

(Fréchet derivative)

Integrating by parts and assuming u twice differentiable

in 0 we get

uX u
o = - P ————-————— + -————X———— vcb<dy

I
J O J L+|vu|2 X J L+|vu\2

Y

for any such v. Hence u(x,y) satisfies the partial

Differential Equation

u u

a_aX_ —X_.___ + a;oy——L = O, for (XIY) E Q I
/l+\Vu12 ../1+\Vu|2

V

(1.2)

u = g on an

The nonlinear equation (1.2) is called the minimal surface

equation.

Existence, uniqueness, and regularity of the solution

It is known [18] that the above problem, in general,
has no solution. The methods developed for the solution
of the problem [18] are based on some further assumptions
concerning the prescribed boundary conditions. One of these

assumptions is introduced by the following definition.

*
Definition 1.1.1. Let C be the Jordan curve

defined in the xyz-Space by the equation 2 = g(P).
*

* *
1’ P2, P3 be three distinct points on

*
C , and denote by 9 the positive acute angle between

P E 50. Let P

* *
1' P2, P3.

We shall say that the boundary condition g(P) satisfies

the xy—plane and the plane passing through P

the three-point condition with the constant a if for all
‘ * * *
possible positions of the points P1’ P2, P3 we have

tan 9 g a for some fixed finite constant a.

Theorem 1.1.2. (existence [18]) Let there be
given, on a convex Jordan curve BC in the xy—plane,

a function g(P) of the point P varying on an which

4

satisfies the three-point condition (Definition 1.1.1)
with some constant a. Consider all the functions

2 = u(x,y) which satisfy, in the Jordan region Q, the
Lipschitz condition and u(P) = g(P) for any P 6 50.
Then there exists in this class a function uO(x,y) which

minimizes the integral

1
A(u) = ((0 [1+ ($9 + (:73)le dxdy 1+

Theorem 1.1.3 (regularity [18]). If 2 = uo(x,y)
is a solution of the problem under the conditions of
Theorem 1.1.2, then uo(x,y) is analytic and satisfies

the minimal surface equation (1.2) [18]. #

Theorem 1.1.4 (uniqueness). The solution stated

above is unique.

Proof: Suppose there are two such solutions w,v.
The class of functions considered in Theorem 1.1.2 is
clearly convex. The operator

1

A(u) = f)” [1+ (vu|2]2 dxdy
*2

k

. . . 2 .
15 also convex because if, in general f :R 4 R 1s

convex, then

is a convex function of u. Here f is the composition

of two convex functions |( :R2 4 R and 1+x2 from

R to R. Let's define B(t) = A(tw4—(l-—t)v), where
w, v are the solutions. Since w, v are both minimizer

of A(u) and A is convex thus
A(w) = A(v) = A(tw4-(l—t)v) constant for 0 g t g_1 .

Therefore B(t) is constant for 0 g t g 1. B is also

a convex function:

B(itl+ U.—k)t2)
= A(xtlw+ (1 ->.)t2w+ (1 —>.tl— (l ->.)t2)v)

==AM(Hw+(1—tflv)+(l—k)fliw+(l-tflv)

g AB(tl)4-(1-X)B(t2) for o.g i g_1.

So B :R 4 R is a convex and therefore continuous

function which is constant on [0,1]. Thus

B'(t) = B”(t) = o for t 6 [0,1].
B(t) = HQ f(tvw+ (l-t)vv) dxdy ,
B'(t) = 55A f'(tvw4—(l-—t)vv) ~v(w-—v)<b<dy
B”(t) = (5 [v(w-v)]T ~f”(tvw-+(1-t)VV)

'v(w-v)<h<dy (f” is the Hessian matrix)

6

Since f was convex, thus the integrand in B” is
positive semi—definite. Thus
T
B”(t) = o = [v(w—V)] ~f”<tvw+ (l-t)Vv):v(w-v) = o

for 0 g t g 1 .

The computation also shows that the integrand is nonnegative

as follows

2 2
(WX—VX) +—(wy-vy) +—[(wX-vx)(tw +(1—t)vy)

 

 

Bl/(t) = If y
0 (1+ |v(tw+(l-t)v)|2)3/2
+ (w -v )(th+(1-t)vx)]2
y V <h<dy = O .
Putting the integrand zero we get wX--vX = O = wy-—vy

on the connected domain 0. Thus w(x,y)-—u(x,y) =
constant on 0. Since w = u = g on 60 and the
solutions are continuous on 5, thus w = v on Q.

This completes the proof of uniqueness. #

The obstacle problem

1.2. An "obstacle problem” is obtained by introducing
a further constraint on the minimal surface, namely that,
it lies above a given fixed surface (the obstacle)

represented by the graph of some N :0 4 R, _where N < g

on 53. In other words, given g and m, the solution

u minimizes the area (1.1) subject to the boundary

condition u] = g and the constraint u 2 N in
50

6 (see Figure 1.2).

2
the surface u
.\3

the obstacle @

the graph of g

 

 

the contact
set I ' 5 the free

, boundary
an mm;

Figure 1.2

2,5

The existence of a solution u E H (0) fl Cl’t(6),

1 g s < m and 0 < t < 1, to this problem was proved by
H. Lewy and G. Stampacchia [13] subject to the following

conditions:

Q is a convex domain with smooth boundary,

9 E O, max N > O
Q

The proof in fact covers the higher dimensional case

where 3 : Rn. It was shown in [l] with the above

1,1

conditions that u E C (3).

 

8

The most interesting questions here focus on the

contact set and its boundary:

I

[(X,y) e 0 |u(X.y) = w(x,y)l .

T

BI = the free boundary .

The boundary al of the region of contact is called a
free boundary as it is not given apriori, but rather
depends on the solution, that is, it is a part of the
solution to be found. Observe that the derivation of (1.2)
is still valid (at least formally) in the complement

Q-—I of the contact set, namely where u > t.

A fundamental result on the geometry of T, in the
two dimensional cases was proved by Kinderlehrer [11]:
if Q c R2 is strictly convex with smooth boundary, if
g E O, and if N is strictly concave and analytic,
then T is an analytic Jordan curve enclosing the contact
set I. More generally if the smoothness of the obstacle

C2,s —

is relaxed to assume only N E (Q) (keeping all

other assumptions), then T is a Jordan curve admitting

Cl,t

a parametrization for any t < 5 (see [8]).

A main purpose of this thesis is to study different
~possible geometrical shapes T of free boundaries, for a
related class of problems. In particular, more complicated
shapes (cusps) will arise. We shall also study how the

set T varies as the data of the problem (such as m and

g) are allowed to change.

1.2.1. The Dirichlet Integral

 

If g, N and its gradiant VN are sufficiently small,
we might expect the solution u and its gradient also to
be sufficiently small. If this is so, then using the

approximation ./1+—x2 = 14“; x2+-0(x4), near x = 0,

leads us to consider the Dirichlet integral
J(u) = % j lVUl2
Q

as an approximation for A(u)-A(O). We may therefore
consider the obstacle problem obtained from the one

above by minimizing J(u) instead of A(u). The Euler
equation of J(u) is Laplace's equation vu = O, in
contrast to the (nonlinear) minimal surface equation (1.2)
associated with A(u). In general we may consider such

a problem in a space of any dimension, as follows:

Given 0 C Rn a bounded domain with smooth

boundary 50,

g :80 4 R sufficiently smooth

H

N :u 4 R sufficiently smooth
with w < g on 80.
find (*) u :5 a R, u = g on 50, u 2 N on 6,
such that u minimizes J(u) = % j lvu|2 among

those functions satisfying (*).

10

We stress that the relation between the minimal surface
problem (which minimizes A(u)) and the above Dirichlet
problem which minimizes J(u) (or energY) is purely
formal: they are two different problems. We may consider
the Dirichlet problem above as a model problem as it has
many features in common with the minimal surface problem,
yet is simpler in many respects. In particular the

Euler equation vu = 0 associated with the functional
J(u) is linear. In this thesis we study the Dirichlet

integral.

1.3. Existepce, uniqueness, regularity

 

More precisely, J(u) = % I [Vulzch<, X E Q C Rn,
Q

is minimized over the closed convex subset K of the

Sobolev space H1(Q):
r 1
K = (u E H (0) :u = g on 60, u 2 W a. e.}

Since K is a convex closed set and J is a convex

continuous functional, it is easy to show [8]

u minimizes J over K e DJ(u)(v-—u)

= JP. Vu°v(V~u) 20, VV 6 K ,

where DJ(u) denotes the Fréchet derivative.

Therefore, an equivalent formulation of the problem

seeks u E K so that

11

J0 vu .v(v-u)dx'2 0, VV 6 K

This new formulation is called a variational inequality.

If g and an are sufficiently smooth, then without

loss of generality we can assume g e O in the problem

as follows: simply replace u with v = u--ul and N
with a = w-ul, where ul is the solution to the
boundary value problem vu = 0 in Q, u l = g.

1 1 80

(It is known [12,6] that the solution u exists if g

1
and 60 are smooth enough). Since w E K implies

w-u = 0 on 80. we have

DJ(u)(w-—u) = f vu °v(w-u)dx = f v(u-—ul) ov(w.-u)dx
Q

L

=]‘ v(u-ul) -v((W-ul)- (u-ul))dX
o

f vv -v(wl-v)dx ,
Q

where w = (w-u

1 ) E K = [v E Hé(0) :v 2 a a.e.].

1 1
Thus u solves the minimizing problem with data

(g,¢) if and only if v solves it with (0,a).
Remark: Unless otherwise stated, from now on we
shall assume g E O on 50. #

We may summarize the above results in the following

precisely stated problems.

12

(1) Minimization problem: Given 1 E H (0),
with N g 0 on 50, minimize
J(u) pg; [vulzdx
Q
1 .
among u E HO(0) w1th u 2 N a.e.
(2) Variational problem: Given w E Hl(0), with

N.§ O on 30. find u E K
1 .
K = [v E HO(0) :VIZ w a.e. 1n 0]

so that f vu «u(v-u)dx 2 0, VV 6 K .
Q

Theorem 1.3.1. These two problems have unique

solutions, and they are the same solutions. #

For the proof and more detail in this context see
[8].
A basic regularity result is the following (we state

without proof):

Theorem 1.3.2 (Brezis and Kinderlehrer [1]). If

u E C2(5), and N < 0 on 50, then the unique solution
' u to the above problems satisfies: u E Cl(5) and
Du is uniformly Lipschitz in a. #

As before the contact set and free boundary

13

are compact subsets of 3 under the assumptions of

this theorem. On Q-—I the derivation of the Euler
equation is valid and so Au = 0 there. It is also the
case that I E {x E Q :A¢ g 0] because Au 3 0 in Q
(by applying the variational inequality, see [8,20]).
Finally, because u is C1, it follows immediately that

v(u-—¢) = O on T. See [8] for details. Conversely we

have the following result (see [2]).

Proposition 1.3.3. Suppose I C Q is a compact set

 

with boundary T which is piecewise smooth in the following

sense: there is a finite subset F E T such that if

x E T-—F, then near x, T is an embedded Cn"l

manifold. Suppose u E Cl(5) and satisfies:

Au = O and u 2 W on 0-—I
u = 0 on 50

u = N and AN 3 O on I

Then u is the solution of the obstacle problem;

Outline of Proof: We show 5 vu -v(v-—u)dx 2 0
Q

'for every v E K. It is enough to consider

v E Cl(5) 0 K, as this set is dense in K. Let

Br = U B(x,r), where B(x,r) is the r—ball about X.
xEF -
Let also E = 1-—I and set C = Q-—B , E = E-—B ,
r r r r

and I = I—IB.
r r

14

By Sard's Theorem [ 4], for almost every r, the
(n-1)-manifold aBr and T-F intersect transversally;
we choose r so that this is the case. We may apply
Green's Theorem (since aBr and T-—F are transverse)

to obtain

j Vu -v(v-u)dx
Q
r

[ vu-Wv—qu+j vw-WV—UNX

Er Ir

=§ % (v-u) + (BI g—r'l‘l (v-u) — j (v¢)(v—u)dx

BEr r Ir

As r 4 O, the integrals over Or and Ir tend to those

over 0 and I.

The integrals over arr and aIr can be divided
into two parts: those over F-—Br and those over the
spheres BBr. The integrals over F-—Br cancel as
%% = %% there, and the integrals over aEr and aIr
are taken with opposite orientation. The contribution
of the integrals over aBr tends to zero as r 4 0,

since the integrands are bounded and measure of aBr

tends to zero. Hence we get the result. #

Some Examples

Example 1. n = 1: A string fixed at endpoints

(a,0), (b,0), where Q = (alb).

15

 

Y
u(x)
\
w(x)
or a b >X
Figure 1.3

It is clear that on the intervals, where u(x) > u(x),

we have u”(x) = 0. So u(x) is linear. It is also
clear that u”(x) = w”(x) g 0 on intervals where

u(x) = u(x). In general u E C2((a,b)) as u” typically
has jump discontinuities: at the endpoints of the

intervals where u(x) = u(x), u”(x) jumps from $"(x)

to zero. See Figure 1.3.

Example 2. n = 2: Suppose Q = [(x,y) E R2 |x2+-y2< l],
g E O on an, and u(x,y) = -%(x24-y2)4-c, where
O < c < %. So t < 0 on 80. The radial symmetry of
the obstacle and the invariance of the Dirichlet integral
-under rigid rotations implies the unique solution u(x)

also is radially symmetric.

 

16

   
  

obstacle

2 2 » ~ .7 the free
boundary

Figure 1.4

As a candidate for the contact set, consider the

disk
I = {(X.y) e R2 |X2+y2 grg< 1]

of radius r for some r . If this is the contact set,

0’ 0

then in Q-—I, u(x) satisfies the Dirichlet problem

2 2 2
Au = O for r0 3 x +—y g 1
u = 0 for x2+y2 = 1
l 2 2 _
u — c-§ rO for x 4-y — r
and moreover Vu = vw on T, hence
ill_ 2_ 2
Br — -rO for x + v — rO , (l)

17

“U . . . . .
where g; 15 the outward radial der1vat1ve. The unique

solution of this Dirichlet problem is (with r2 = x2+y2

)
u(x,y) = A log r, rO g r g 1, where

_ .1. 2
A — (c--2 rO)/log r0 (2)
In addition (1) holds also if and only if ii = -r0, or
O
A: a?) (3)

There exists rO and A satisfying (2) and (3) if and

only if
r2 r2
_ O O
c — -7f log (7;)

It is an easy exercise to show that for any c E (O ,%),

there is a unique rO E (0,1) satisfying this relation.

We claim this unique rO and A = —r8 give the solution

w(x,y) . o g r 3 r0
U(X.y) =
A log r , r0 3 r g 1

of the problem. This follows from Preposition 1.3.3
, once we show u > ¢ on Q-—I, that is,
f(r) = -r3 log r4—% r2-c > O, for r E (rO,l). But

this is easy as f(ro) = O and f'(r) > 0 on (ro,1).

2 .
Again note u is not c2, as .2_% undergoes a
dr

jump at r = r from

18

——— = -l to Ji- (A log r) I = 1 .

2
dr dr r-rO

The anolog of this problem in dimensions n 2 3 may
be considered in a similar fashion. The contact set is
again a disk of radius r0,

In Q-—l, u has the form u = Ar

where rO depends on n.

—n+2__A°

Example 3. Lewy and Stampacchia [12] show that if
0 C R2 is convex and N is strictly concave and analytic
with N < 0 on 80, then T is an analytic Jordan curve,
Q-—I is homeomorphic to an annulus, and I is simply
connected. This result can be generalized for the case

of obstacle u E C2(a). See [8].

Some additional properties of the solution

 

1. If N < O in 50 and At < 0 in Q, then

Q-—I is connected.

Proof: This can be shown by strong maximum and
minimum principle: (Let Aw 2 0 (g 0) in A, a bounded
domain, and suppose there exists a point y E A for

which w(y) = sup w(inf w). Then w is constant in A.)
A

Suppose C-—I is not connected, then it has some component
6 whose boundary is part of the boundary of I. Then
11— t is zero on 56 and u-—¢ > 0, A(u-—$) > O in '3,

which contradicts the strong maximum principle.

19

2. Under quite general conditions [3,9,10,12]
F = bl consists of smoothly parametrized arcs, possibly
with cusps (in fact analytically parametrized arcs if N
is analytic). In particular if G is a neighborhood of

2 and (3 0 T is a Jordan are

a p01nt (XO’YO) E Q C R
through (xo,yo), and N is analytic, then the arc
0 0 F admits an analytic parametrization (possibly with

cusps). See [8] for more details.

3. If )1 < $2, then 0 < u2--ul < SUP(N2'-Wl)-

Proof: Let Kl = {w E Hé(0) :w 2 $1} and
1 .
= r . . =
K2 Kw E HO(0) .w 2 $2], and put u3 m1n(ul,u2),
v = max(ul,u2). Then by the variational inequality

formulation I vul -v(u3-ul)dx 2 O, f vu2 :v(v-u2)dx 2 O,
Q Q

fi ' ._ = _
where u3 - K1 and v E K2. Since v u2 (u

3 1)!
thus fqvu2 ~v(v-—u2)dx 2 O a vauz «v(u3-ul)dx g 0,
Suntracting the first inequality above from the last one

yields

f v(u2-ul) -v(u -ul)dx S O or

3

")

J{u <u }v(u2-ul) °v(u2‘-ul)dx g 0
2 1

The last inequality is obtained by Theorem A.1. in [8].

Then Cl property of u2--ul with
P lv(u2-u 2dx g 0 imply that u , 1.

. )]
. 1
(u2<ul}

20

To establish the second part of the inequality put
sup(¢2-wl) = c, and $1 3 $2 3 $14-c. Also put

u = min(ul+-c,u2), v = max(ul+-c,u

3 We want to show

2).:

u2 g ul4-c. We have [Qvul :v(v-ul-—c)dx

[Ovul »v(v-ul)dx 2 0. Since v = u +—c+—u thus

1 2‘u3'
IQ vul -v(u3-u2)dx g 0. Subtracting fovu2 -v(u3-u2) 2 0

from the last inequality we get

fl

J{u +c<u }V(ul-u2) °v(ul-u2)dx S O
1 2

1

C property of u14-c and u implies u2 g u14-c

2

on {2.

Remark: If $1 3 $2, with the same boundary
condition, then I1 is not necessarily contained in I2.
Consider the following example with n = 1: See Figure 1.5.

4’T\ Old u
new u V / u. \

    

,. ‘51d 1' f“
/ x .
new Y
Figure 1.5
Here I -I or I — I may be non—empty sets.

2 1 1 2

21

Example 4. Schaeffer gave [20] two examples,
for n = 2, for which the contact sets Ij (j = 1,2)
are bounded by curves with cusps as in Figures 1.6, 1.7

respectively.

Figure 1.6

/ i...— H“ -.‘H— T2

Figure 1.7

The curve F1 is the analytic image of a circle, with
non-vanishing derivative everywhere. In particular, near
the double point, Fl consists of two analytic arcs,

tangent to each other without crossing, and with different

curvatures at the point of tangency. The curve F2 is
analytic away from the cusp, near which it has the form
y2 ~ Kx5 for some K # 0.

Consider first the analytic function

1 -1

B(t) = (t+t— ),/'2+st(t)(t-t )/2, where s is a

real parameter and Pj(t) is chosen as

 

22

In example 1, Pl(t) = t24-24-t-2
-1 2
In example 2, P2(t) = (t-24-t ) .
For 5 = O, observe that B is a conformal map of

{t :|t| > 1] onto the plane cut along the real axis from
—1 to 1. By choosing s > 0 small, we may arrange
that, in either example 1 or example 2, B maps the
circles [t] = l and [t] = 2 onto Tj and a Jordan
curve Cj which contains Tj in its interior,
respectively. Let Ij be the (closed) region inside

Tj and let Qj be the (open) region inside Cj°
Schaeffer shows that B maps the annulus

A = [t :1 < [t] < 2] onto 0.-—I. in a one-to—one

3 J
manner. He proves that with the obstacle

111(XIY): _% (X2+Y2) I

The obstacle problem has its unique solution in Gj with
contact set Ij and the free boundary Fj (for sufficiently
small s > O). This is accomplished by considering an
equivalent (non—singular) problem in the disk \tl < 2

via the mapping B.

Example 5. In [20] Schaeffer gives an example with
a C” obstacle for which I has infinitely many connected

components in the neighborhood of some point.

 

23
Local solution to the obstacle problem

1.4. The obstacle problem discussed in last
sections has a local formulation as a differential

equation which we define it more precisely now.

Definition 1.4.1. A local solution in an open set

0‘9 0 C 191 is a 01 function u :0 4 I? satisfying

u(x) 2 u(x) on O
Au(x) = 0 if u(x) > w(x)

A¢(x) go if u(x) = w(x) . #

If u is a local solution in G, then the contact set
in 0 is the set I = [X E O |u(x) = u(x)], and the
free boundary in O is F = BI n 0. Observe that

wu-w)=0 on L

Now assume in the neighborhood of some point, P
is an (n-1)-dimensional manifold (arc if n = 2,
surface if n = 3) which is analytic with I lying on
one side of P. Then on a one sided neighborhood V of
f, contained in O-—I, we have Au = O, u > w, and on
T we have u = N, v(u-—¢) = 0. In fact the initial

boundary value problem

24

has a unique solution u near T in O -1 by the
Cauchy—Kowalewski Theorem. In fact an explicit formula
for u can be given when n = 2 (see [20]) and we shall

present the method later on.

The higher dimensional cases are much more difficult.
However, for n even, if the solution has enough symmetry,
results can be obtained. We shall study the case n = 4

for a class of axisymmetric problems.

Example 6. Kinderlehrer and Nirenberg [10,8] gave
an example of a local solution in which the free boundary
has a cusp and away from it the free boundary is analytic.
They showed for m 2 1 odd the free boundary can never

have a cusp of the form

2 2m+l
(y-yy Km-XO) , I<#O
near some (xo,yo) in two dimensional case (0 C R2),
whereas for m even such a cusp can occur. (In fact, the

case m = l was first noted by Schaeffer: Example 4).
Their proof involves first straightening out the cusp to
a line segment by means of a conformal mapping; then an
analysis of several terms of the Taylor series of u
near (xo,yo), based on the equation Au = O governing

u in C-—I, gives a contradiction to u 2 w.

 

 

25
Further specialization

We shall study local solutions in the neighborhood
of a point, say the origin. We assume the nondegeneracy
condition A¢(O) < 0 (if A¢(O) > 0, then T = ¢ near
0). Then, as we see from the Definition 1.4.1, upon
adding the same harmonic function h to a local solution
u and obstacle W and multiplying by a positive constant
c we get another local solution u1 = c(u—+h) for the

new obstacle W1 = c(¢+—h). In particular a quadratic

polynomial can always be chosen for h(x) to give
1
‘(J(X) = -§ IX]2+O(IX\3) . x E Rn .

We choose u(x) = -% )X)2 again as a model problem near
the origin and consider only this obstacle from now on.
Deleting the higher-order terms in N should not affect
the resulting theory significantly, although this has yet

to be established.

Changing Boundary conditions

222. Schaeffer [20,21] has studied how the set I
changes as the data t and g vary. He proved a theorem
that if g and g are C” with t < g on 50, T a
Ca curve, and Au < 0 on I, then if ml, gl are
sufficiently near 3, g in the C: topology, the free

7‘

boundary of .1

of the corresponding problem will be a

26

Cf curve near I in the C0° topology (in normal

coordinates about T).

'f : 44:74 solution u

1, - , obstacle

Figure 1.8

Schaeffer has also pointed out the need for a
generic theory of such variations of I and BI.
Mallet—Paret and Chow declared that such a theory presumably
could take the form of a bifurcation theory or unfolding
theory for the singularities of al. Such a theory was
described in the case n = 2 [14,5,15]. A significant
point here is that the unfoldings one encounters are not
generic in the sense of singularity theory as developed by
Thom Mather, Arnold and others; only very special types
of singularities can occur. This is seen for example
from the result of Kinderlehrer and Nirenberg [10,8]

described above in Example 6.

Mallet—Paret made a detailed study of a class of
singularities of a free boundary, and their bifurcation,
in the two dimensional case, as the data of the system
varies parametrically. Typically cusps on the free
boundary become smooth, and islands may appear as in

Figure 1.9.

27

singular free boundary after perturbation

Figure 1.9

In this thesis we shall study the unfoldings
(bifurcations) of a class of singularities of the four
dimensional axi-symmetric obstacle problem. This case
is the next simplest after the two-dimensional case, as
harmonic functions admit a particularly nice representation.

2

Let u(x) = V(X1' xg4---:4-xn), where v is defined

in a region in R2. Putting y = ,/x§+ o°°+-x: , for

n = 2 and 4 we have locally

Au = 0 if and only if

v(xl,y) = Imf(xl+-iy), f(z) analytic (n = 2)
v(xl,y) = % Imf(xl+-iy). f(z) analytic (n = 4)
There is no such formula in three dimensional space. In

general there are anologs of the above (n = 2,4) cases

for even (but not odd) dimensions.

 

2. FORMULATION OF A PROBLEM

Local solution to the axi—symmetric obstacle

 

problem in R4

2.1. In this chapter we consider the aXi—symmetric
obstacle problem in R4 (n = 4) with the obstacle
142
e(xl,x2,x3,x4) = --3 jg: xj , and we study the local

solutions w defined near the origin. By aXi—symmetric

we mean
W(xl.X2.x3,x4) = V(X,y) .
where x = x y = ,/X2+-x24-x Note then
1’ 2 3 4 ' ‘
6(X1.X2,X3,X4) = -% (X24-y2) = w(x,y). We shall formulate

a problem of unfolding a singular free boundary passing
through 0 in this chapter. But we need to state and

prove a prOposition and a lemma first.

Proposition 2.2. w is a local solution of the
axi-symmetric obstacle problem with obstacle e in a
region for which y > 0 if and only if u = yv is a
local solution to the obstacle problem (for »n = 2) with

obstacle &(x,y) = ym(x,y) there.

28

 

29

3599;: We check the conditions of definition of
local solution stated in Chapter 1. Clearly w_2 e in
the region if and only if yv‘2 t there. Also it is
easy to check that AR4 w = 0 when w > e if and only

if A 2(yv) = 0 when yv > ym (of course in the region
R

y > 0) because

2 2
8 v a v 2 5v 1
A W = —-—'+ ——- + -'-— = - A (YV)
R4 ax2 ayz y By y R2
Now, it is only left to check the C1 condition.
Assume V(X,y) is C1 in a region for which y > 0,
then its composition with y = x34—xi4—x: , i.e.,
. 1
w(xl,x2,x3,x4) Is also C there.
. l .
Conversely assume w(xl,x2,x3,x4) is C 1n the
region, then w(xl,x2,O,O) = V(X,y), where x = x1 and
y = x2, is also Cl there. Here is the proof complete
because v is C1 in the region if and only if u = yv
is C1 there. #

The last proposition shows that in a simply connected
domain in the upper half—plane in the noncontact set
V(z) = % Imf(z), where z = X4—iy, for an analytic

function f.

Remark 1: If w is an axi—symmetric local solution
with the above obstacle, then we can extend the function

v to be even in y. We shall use the extended even

 

30

function v and the odd one u = yv defined in a whole

neighborhood of the origin in (x,y)—plane later on.

Remark 2: Following Remark 1, sometimes, we shall
need to use a modification of Proposition 2.2 concerning

the x-axis (y = O) entering the region (recall y > O

in the Proposition). We claim here that w is C1 in a

region for which y 2 0 if and only if V(X,y) is C1

1

there. To show this let w(x1,x2,x3,x ) be C in the

4
region y 2 0. Then w(xl,x2,0,0) = V(X,y), where x = x1
and y = x2, is C1 there.
Conversely if V(X,y) is C1 in the region, then

w(xl,x2,x3,x4)

composite function of

2 2 2
w(xl,x2,x3,x4) — V(xl,y) , y — ./x24—x3+-x4 .

is continuous there by simply being a

 

. x.
Moreover since .VV = 3! —1 , j = 2,3,4, clearly these
oXj Y Y

partial derivatives of w are continuous when y # O.

. . . 8v .
Since v is an even function of y, thus 3? is an

odd function of y. If we now let y approach to zero,

 

then lim g2? = 0 because %z(x,0) = lim %X(X,y)
(x,y)4(0.0) ‘ j Y ..o vY
x.
and 7% remains bounded (between zero and one). Hence
w(xl,x2,x3,x4) is C1 in the region for which y 2 O. #

31

2.3. Local solution formula

Let us consider an analytic arc Fl :x = a(y)
through (xo,yo), where (x0,yo) is fixed near the
origin with yO > O and x0 = a(yo). We shall construct
a local solution in a neighborhood of (xo,yo), in the
region y > 0, for the above axi—symmetric problem,
with free boundary T1 and contact set I lying on one
side of F1. Let us define the conformal mapping Ll for
t complex near yO as 2 = L1(t) = a(t)4—it, where

20 = xO-i-iyO E Tl. See Figure 2.1.

t—plane

 

 

 

Figure 2.1

Lemma 2.4. For a sufficiently small neighborhood

V of 2 L-1 exists as a conformal mapping and

0' l
Lil(x+—iy) = y for X+-iy E El. Let VO be a
sufficiently small one—sided neighborhood of F1 about

32

0' and let u(x,y) = -% y(x24-y2). Then u = yv is

a local solution to the obstacle problem, in the region

Z

y > O in a neighborhood of z with obstacle t and the

Q!
noncontact set in V0, the free boundary T1, and the
contact set in V--VO if and only if
u(x,y) = (u(x,y), for (X.y) E (V-VO)
Z i 2 —1 2
u(x,y) = u(Z) = 1(zO)+Re f §[ +4(L1 (q)) ]dq ,
z
0
for z E VO U 11 .
Proof: Since Li(t) = a’(t)+—i # O for t (as

complex) near yO in the t—plane, then L1 is a conformal

mapping near yO taking the real axis (near yo) to the

1' Therefore L11 exists and it is conformal in
a small enough neighborhood V of 20 and LIl(x+-iy) = y
1' L1 also maps the upper or lower half (the other

half) of a neighborhood of y0 conformally onto a one—

arc T

on T

sided neighborhood V (the other one—sided neighborhood)

0

of F about 2 The proof of the first part is complete

‘1

here.

00

Necessary part: With obstacle N we have

. -- _ - 1 2._§ _ i _ - _E. _'
wX-iuy — -Xy+1(2 x 1-2 y ) — X(x 1y) 2 y(x 1y)

= g(x~ iy) (x+ 3 iy) = %(z- 21y)(Z+ 21y)

= é—(z2+4y2)

33

For 2 near 20 function f(z) = %[z2+-4(Lil(z))2]

is holomorphic. Note uX--iuy is holomorphic in V0

because u is harmonic there, and it is also continuous

on V6 because u is C1 in V. For 2 E T
. _ 2 2 2 _
Vu — VN- Therefore uX-1uy _ 2[z *‘4Y ] _
2] = f(z). for z E Fl. Hence

1 we have

—§[zz+4(L11<z))

uX—iuy = f(z), for z E VO [8,19] because both analytic
functions on V agree on F . Thus u -iu is well—
0 1 x y
defined and unique. Moreover uX-iuy = f(z) implies
u(z) = Re g(z) [19] for an appropriate analytic function
on V0 with g'(z) = f(z) by the Cauchy-Riemann equations.
2
This means u(z) = C-+Re j f(q)dq. Then u = w on T1
20
2
implies u(zo) = C = w(zo). Hence u(z) = N(20)*'Re f f(q)dq
2
O

for z E V U T1; and u = N on (V-V ). Here is the

O

necessary part proved.

O

Sufficient part: Conversely assume u is given by
the above formula on V. Then clearly u is harmonic in

and it is C1 in V . Also for z E T

V0' 0 1’ if we
choose the path of integration on F1 and q = §+-in, we
shall have
“2 i 2 2
u(z) = w(zo)+-Re J -§[§ + 3n + Zign](d§+ idn)
2o
z
1 . 2 2 _ ,
= u(ZO)-§ J (g +-3n )dn+ 23ndg
Zo

= 1(2)

 

34

Thus u is continuous on V. Now we want to show

Vu = vu on T1. For 2 E Tl we have

 

 

—— d z i 2 -1 2
Vu = uX-i-iuy = uX--1uy = 3; f2 §[q +4(Ll (Q)) ]dq
O
= %[224-4y2] = €(x24-3y24-21xy)
=v¢
Thus u is C1 in V, in particular along Tl. To
complete the proof we need to show u > N in V0. To
show this observe that on F1 we have u-¢ = O,
. 2 1
en
where n is the unit (outer) normal to F1. Thus u > N
near T1 in V0 in the region y > O. #
Remark. Instead of defining u = N on V-V = I,

0
if we extend the integral formula for u to be defined

in a full neighborhood of 20, then u > N holds on

both sides of T1 with u = N on T1. We also observe
in the proof that v(u-—¢) = 0 in the closure of the
noncontact set if and only if (L—l(z))2 = (Im z)2. #

Now we are going to start the main purpose of this
chapter by constructing a local solution near 0 with a
singular free boundary in Example 2.5. In Section 2.7 later
on we shall discuss heuristically unfolding of the above

free boundary by some perturbation of Example 2.5.

A class of examples

2.5. We want to construct a local solution to the
above axi-symmetric problem whose free boundary possesses
a cusp at the origin. Singular free boundaries for

axi-symmetric obstacle problem are only in the form

y2 = KX4n+l or z = H(t)-—% t2n+llog|t|jzit2n+l
, t real near zero
Figure 2.2—a
or isolated points [17], where n 2 O is an integer,
H(t) is a real analytic function near t = O with
H(O) = 0, H'(0) > 0; and let T = T1 U T2 be the union of

the four arcs, near the origin in the z—plane given
. 2 , .
parametrically by z = H(t)"? t log It] :lt, 0 g lt‘ < r0.
See Figure 2.2. We assume symmetry of v = - u as
before about the X—axis, namely that v is even in y.

In this example the contact set I will be the shaded

region as shown in the Figure 2.2

   

"—1

H
-"'"‘ “‘-“9'

"'1

Figure 2.2—b

 

36

Following the local solution formula given in the

Lemma 2.4, let us first define
L(t) = H(t)—gt log t+it, t g [io:o< o}
(i.e. Jig arg t < 11
I 2 2 ,

where log t has the branch which is real for t > O;
and L(t) for t > 0 represents the right hand part
of T1. We proceed in five steps as follows:
(i) 'We are going to show that L maps the real

axis near 0 onto T and an upper one—sided

1
neighborhood of the real axis about 0 conformally onto
an upper one-sided neighborhood of F1 about 0. To

see this let t move along the real axis from O to

r > 0. By the branch we chose L maps [O,r] into the

right hand part of P1 in a one—to—one manner. Let t
move from t = r along \t] = r in the upper half plane
to t = -r, then back again to t = 0 along the

negative axis (see Figure 2.3).

 

37
Continue the function L analytically along this closed
path Yr and its interior Gr° Clearly L is continuous
on Er and analytic on Gr-[O]. Along the negative
part of the real axis in the t—plane L has the image:

L(t) = H(t)-gt log ltl—it, -rgth. Argt=7r.

This image is exactly the left hand part of T1 (above

the x—axis) near the origin. We now want to show that

the image of Yr 0 {t :It] = r], for r small enough,

stays away from the origin in the upper half-plane. Let

z = L(t), t = rele, 0 g e g N, for r small enough. Then

Im z = Im[t(H’(0)-—% log r4—i(l.—%§)]+-O(r2)

r[(l--27T—a-)cos 9+ (H'(O) —% log r)sin e] +0(r2)

- f(67r) > O .

This shows that for r small enough, the image of

Yr 0 (t :[t] = r] stays away from the origin in the upper

half-plane. Observe also that

é%(Re z) = r[(—H’(O)+g log er)sin 6

”IT

2

+ ( 9-—l)cos 9]4-O(r ) < O .

=1|N

Thus Re 2 decreases as Q varies from O -to F.

Finally

 

38

2
£L§(Im z) = r[-(l-—%§)cos 9
d9

+ (-H’(O)4—%~log re2)sin a]+-O(r2) < O.

This shows Im 2 has only one critical point, a maximum

point, at a point 9 near %, and Im z is increasing

0
on [0,90] and decreasing on [60,v]. Thus Im 2 takes
its minimum values at e = 0,w. Hence L1 is one—to-one

on Yr' and L(yr) is a Jordan curve.

Consider the topological degree (or winding number)

d(L,Gr,z) for any 2 E L(yr). This equals one in the
region bounded by L(Yr) (the Jordan curve), and is
zero in the unbounded component of ¢-—L(yr). Because

L is analytic, it follows that L maps Gr conformally
onto the region bounded by the Jordan curve L(yr).

1

Hence L— conformally maps L(Gr) onto Gr‘
Moreover L"1 is analytic at each point of the closure
L(Gr) except 0, where it is continuous

(ii) Near any 2 E T

O -(O] we have, by Lemma 2.4,

l
a local solution

2
u = w(zO)-% Im j [q2+4(L_l(q))2

Zo

]dq,

2 above -{0]

T1

u = u(z), z beneath T1-—[O] .

 

39

The overlapping solutions along Tl-[O] agree
and they are independent of 20 on each side of Tl-{O].
are two such initial points

This is because if z

21' 2
on the right hand part, for example, then the difference

of

Z
u(z) = ((21) -% 1m] [q2+4<L‘l(q)>21dq
Z
1

Z
u(z) = 1(zz>-%Im]‘z [q2+4(L'1<q)>21dq
2

Z
is (:(zl) -1(22)-% 1m 5 2 [q2+4<L‘1<q))21dq = o, by
Z
1

taking the last integration path on Tl-[O]. Therefore
by taking zero as the limit case of 20 in the formula

we get

2
u(z) = _% Im f [q2+4(L—l(q))2]dq.
0
This formula can be used for u in the noncontact set
corresponding to both right hand and left hand part of

Tl-{O]. See Figure 2.4 (shaded region). For emphasis

—-—--->

   

Figure 2.4

40

we may here check that u = N (by taking integration

path along Tl-{O]) and vu = vN on Tl-{O] by

observing the fact that [L_l(z)]2 = (Im 2)2 there.

2
Also 8 u; )
an

where n is the unit outer normal to T-{0] at any

= A(u-N) = -AN = Ny > O on Fl-{O],

z E 1‘1-{0}.

This shows u-N > O on the noncontact sets
corresponding to the left hand and right hand parts of
Tl-[O]. Note u given above is defined in 375;? and
it is the imaginary part of an analytic function in
fTG;T-—[O}. Thus Au = 0 in L(Gr). Define u = N on
T1 and between T1 and real axis. To show now that u
is a local solution in the upper half of z-plane, near
the origin, we need to show u is C1 there and u > N

in L(Gr).

(iii) At this step we want to show u > N in L(Gr)

Z

11(2) ~N(z) = -% Im j [q

2+4<If1<q>>21dq+% Y(X2+y2)

1

_E-Im yO[L2(s)+-4SZ]L'(s)dS

+% (Im L(t))[(Im L(t))2+ (Re L(t))2]
1 1 3 2 .t
= --3 Im[§(L(t)) +4t L(t)-8 J sL(s)ds]
o
+%(1m L(t))[(Im L(t))2+ (Re L(t))z]

 

41

2 3 2 t
g(Im L(t)) —2 Im(t L(t))+4 Im J‘ sL(s)ds
o

= --l£% r3sin3 e log3r

37

+ 3% r3sin2 9 log2r((l-%§)cos 9+ H'(0)sine)

w
+ 0(r3 log r sin2 e)
= F(r.e) ,
where t = re16 and capital "0" is used here uniformly

near (r,9) = (0,0).

For 9 away from zero or w, certainly
u(z)-N(z) > O for r sufficiently small. For 9 near

0 or v, we divide by sin29:

F(r,e) = E r3(log r)2((1_fi)cos 6+H’(0)sine)
. 2 2 v
Sln 9 F
- -l—6-3- r3(log r)3sin9+0(r3 logr)
3W
This shows that for 9 near 0, v we have EigLfll > 0
sin e

if r is small enough. But we claim that there is a
neighborhood of (r,9) = (0,0) in which _——7?_ > 0. To

show this we do as follows

F(r.a)

= (l-Ei)cos 5+-H’(0)sin9
3 2-- r
r (logr) sin 9

. I

1
log r

— (log r)sin e+—O( )

 

42

2 . .
G(r,e) = (l-7§)cos 64-H'(0)51n 9-—(log r)Sin e 2 c > 0
0 g e g F
for , for some fixed (constant) rO .
O < r 3 r0
. l c

We can assume rO is so small that 0(IBE_?) 2 -§-. Then
-?r——J:L£L%L-—§— > g-. This shows that u-N > O in the
r (log r) sin a
whole region L(Gr). In the next step we want to discuss

Cl condition of the local solution.

(iv) In order to show that u is a local solution
in the upper half of z—plane near the origin, it is now
left only to show that u is C1 there. By what we
explained in the first two steps it is clear that u is
C1 in 2 # 0 there. It is also clear that u is
continuous at the origin. So the only thing left is to

show that the partial derivatives of u are continuous at

z = 0. To do this it is enough to show that

 

lim :13 = a—‘1J(o,o) , lim 33 = :11 (0,0).
(x,y)4(O.O) °X 5" (x,y)4(O.O) Y Y
(XIY)EL(Gr) (le)€L(Gr)

By looking at the formula for u defined on L(G ), we

have

43

Z
2 —1 2
$1: "23—25 m f0 [q +4(L (q)) ]dq
= _% Im[22+4(L-1(2))2] .
a_u= -11 I... (2 [q2+4(L‘1(q))21dq
BY 2 5y 0
= ..% Re[zz+-4(L—l(z))2] .
Thus
lim a—u=O=M(OIO) I
(x,y)4(O.O) 5X ax
11m g—‘l o = SJ (0,0)
(X,Y)"(OIO) y y

In the next step we complete the process of constructing
a local solution to the axi—symmetric problem promised at

the beginning of Section 2.5.

(v) In order to show that v = % u is a local
solution to the axi—symmetric problem it is only left
to show that v is C1 at the origin. Since u is C1

at the origin by step (iv), thus v = % u is C0 there.

Thus we only need to show the partial derivatives of v
are continuous at the origin. As we argued in last

step we only need to show:

 

n- 1
a _ _2 2
y Y
t
y Re t2-Im f 52L'(s)ds
O
Y
Hence
2 I
2 Im J s L (s)ds
limfl=—2 lim Ret
240 By t4O Im L(t) [Im L(t)]2
. r cos 29
'2 1““ 2 . , . 29
r40 "F logr Sin 94—H (0)51n 9+—(l-—7?)cos 94-O(r)

because we've shown in step (i) that the minimum of the

function

(2 O)

44

1im 5—V = 549 (0,0) ,
(x,y)4(0,0) ex ex

lim EX =‘§£(0,0)
(x,y)4(0.0) BY BY

Z
y Retzz+4(L‘l(z))21—Im ] [q2+4(L‘1<q>)21dq
O

 

 

 

 

-—;-r(sin 39)log r-iL r sin 3e-—3%

3W 9W

 

(-% logr sin 94-H'(0)sin e

+ 5 cos 394—% H'(O)sin 3e+—0(r2)

3

 

+ (l-%)cos 9+0(r))2

£13) = -—%

occurs at 9 = 0 or T with f(O)

. . 2”
logr Sin 9+—H’(O)51n 9+—(l-—1S)cos E
i

45

and therefore for r g r0, r0 small enough,

f(e)+0(r) >%.

31

Thus lim
(x,y)4(0.0) BY

= o = 2&3, (0,0). Also

2
_5_V_ i 2 -1 2 _ 2 Im(t )
8X — -2y Im[z +4(L (Z)) ] — X-——ImL(t) .
Similarly we can show
6V —2 Im(t2) 8
lim 5;: 11mm: O=-O—:% (0:0) .
(XIY)"(OIO) r40

This completes step (v), hence v is a local solution to
the axi-symmetric problem near the origin of the z—plane.
The shaded region is the contact set; T1, T2 are the
free boundaries, and the rest of the neighborhood of the
origin is the noncontact set in the z—plane. See

Figure 2.5.

 

Figure 2.5

46

Remark 2.6. Simple calculation shows that F1 is
symmetric with respect to the y-axis if and only if H
is odd. In particular then L(io) is pure imaginary
for 0 > 0. Putting a(t) = L(t)-it = H(t)-—% t log t
and assuming H odd, we get L(t) = a(t)4—it, which
for t > 0 parametrizes the right hand part of F1. The
analytic continuation of this equation along Yr yields

at -r < O the equation
L(-r) = a(-r)-ir = -H(r)+—r log r4-ir = -a(r)+—ir ,

which shows that following the analytic continuation of

a along Yr from r to —r gives
a(ar) = -a(r) -21(-r) a (*)

Moreover the equation of the right hand part of the free

boundary Fl

x = a(y) = H(y) -% y log y

extended for complex variable implies

,, _ ~ 2- . ..
a (t) — H (Hm?t — L (t) ( )
is single—valued, in fact a”(t) is meromorphic in a

neighborhood of the origin. See (**).

 

47

We shall show later on that the analogue of (*),
(**) will also happen for the perturbed problem where

the cusp is unfolded to a smooth free boundary.

Perturbation of Example 2.5: Heuristics

 

221. Let us now physically pull the obstacle down
a little bit, keeping symmetry of the problem with respect
to both axes. We then expect a new local solution surface
v over m near (0,0) and a new free boundary. The
new contact set should be near the old one. We might
anticipate, for example, that the new free boundary would
consist of two analytic curves T1 and T2,

and left half plane, as shown in Figure 2.6 (recall

in the right

that the axi—symmetric solution V(X,y) is extended to

be even in y). We want the right hand part of the free

 

 

Figure 2.6

 

48

boundary T1 of this local solution to be represented
by x = a(y), an analytic even function near zero,

passing through 20 = x0 = a(O) with the region to the
right of F1 lying in the contact set. Fl has the

equation 2 = L(t) = a(t)+-it, for t real.

 

 

t-plane

Figure 2.7

Let us now first construct the local solution near

20 = a(O). We extend L again on the complex numbers.
Since L’(O) # 0, L_l(z) is defined as a conformal map

from a neighborhood of 20 to a neighborhood of t = 0,

taking the contact set (to the right of T ) to Im t < O,

l
and the non—contact set (to the left of Tl)_ to Im t > 0.
By the formula given in Section 2.3 for u = yv,

on the left hand side of Tl we have (note N(zo) = 0)

 

49

Z
V(X,y) = v<z> = —i Im [q2+4(L‘l(q)>2]dq.
2y 2
O

for y = Im z # 0.

Observe that the above formula for V(X,y) defines a

real analytic function, even when y = 0. Moreover this
function is even in y. This is because L-l(q) is
imaginary for q real, the integral above is analytic

function which is real for 2 real. Therefore
22 —1 2 .

Im I [q +-4(L (q)) ]dq is real analytic and is zero
2
0

when y = Im z = 0, and in fact is odd in y by the

reflection principle. When y = 0 this formula yields

V(Xl,O) = 113 V(Xl.y)

= -%[.— my [q2+4<L‘l<q>>21dq1
y 20 Z=Xl

= w(X110)-2(L_l(xl)) . (1)

On the right hand side of 1‘1 we have V(z) = :p(z).

From the past theory, 2.4, 2.5, we have that V(X,y)

30 defined is C1 across the free boundary (even at y = O),

 

50

with A(yv) = O to the left of T1. To check that v

is a local solution near all there remains is to show

v > w near 2 , to the left of F and that A W = O
O 1 R4
2
in the non—contact set. We show §_A§L%gfl_= 4 at any
6n
point of T1 where n the normal to T1 directed away
from the contact set. We have
32(v— ) 32 u-N 152(u—N) .
d—=——2( )=———2 (Slnce Vu=VNIU=1W
an an Y Y an
= l A(u..N) = -l-AN = 4 , for y # O .
Y Y '

This is also true for y = O by continuity. To see

A 4 W = O observe that w is analytic in noncontact set

R

since V(X,y) is analytic and even in y, and hence v
is an analytic function to x and y2. Also observe that
A W = 0 if y # 0. Hence by continuity A W = 0 if
R4 R4

y = 0 also. Here is the construction of the local solution
near 20 complete. Recall that in 2.7 we assumed the

local solution to the axi-symmetric problem exists near

the origin of the z-plane. This local solution agrees

with the one constructed near 2

0°

2.7.1. Further Assumptions. We are assuming the

 

 

perturbed local solution has a free boundary as shown
in Figure 2.5. In particular assume v-¢ >.0 on the
interval (—zO,zO) on the real axis (i.e., this interval

is in the non-contact set). Assume further that the only

 

51

local maximum of v(x,0)-m(x,0) in this interval is

x = O, that JL (v(x,0)-—m(x,0)) # O for x # O,

 

at , 8X 2
x E (—zo,zo), and $13 (v(0,0)-—m(0,0)) < 0. See
6X
Figure 2.8, where —K = -% (V(0.0)-$(0:0)):
y = —% <v<x.0) -cp(X.0))o
Y
—zO 20 ‘Y
-K
Figure 2.8

From (1) it follows that (L_l(x))2 has an analytic

continuation as an even function of X throughout

[-zo,zo]. Thus Y = (L_l(x))z in Figure 2.8.

2.7.2. Heuristic Derivation of Explicit Formula

For Function a.
First Step. We see from 2.7.1 that L"1 maps
[O,zo] monotonically onto the segment [in,0] in the
complex plane and is analytic there with nonzero

derivative in (0,20]. Thus L(t) (and a(t) = L(t)-it)

 

52

have analytic continuations along [O,iJR). Indeed,

the part of the graph in Figure 2.7 for x > O is

described by

The graph f

= a

(A/Y)+i(/Y, where fl 6 [o,i./1'<] .

or x < O is, by symmetry x = -L(JY). At

t = iJK, the Riemann surface has a branch point with

two sheets,

(t-i./I_<)2
As L(f’?)
each other
analytic on

at iJR.

Consid

so a(t),

i.e.

L

is an analytic function of

2
there. This is because §_iY§21 (0,0) # 0.
8X

and -L(jY) are analytic continuations of
near i/K, it follows that (L(t))2 is
[O,iJR]. In fact (L(t))2 has a simple zero

er t E [—iJK,O]. a(t) is real analytic,

L(t)

ha

closed segment and

a(t)

L(t)

ve analytic continuation along this

 

a(I) .

 

a(t)4—it = a(E)4-i€4—2 it

 

Thus (L(t) -2 it)2

simple zero

at

L(E) + 2 it

is analytic on [-iJR,0] with a
2

—i(/E. Consider (Ll/(t)) =((L(t)—21t)”)

(a”(t))2 which is analytic on (—iJK,iJR) as L(t) # O

2

53

H

there. Near iJR, L(t) = (t-in)2M(t), where
M(1JR) # O, M is analytic. Hence (L”(t))2 is
meromorphic near id? with a pole of order exactly

three. The same thing is true at —iJK with L(t)-—2 it.

Hence
// 2
(a (t)) = ——3‘t)3
(t +K)
where h is real analytic, non-zero on [-iJK,iJK], and

even in t.

For clarity we display the analytic continuations

of a(jY) and L(jY). See Figure 2.9. The symmetric

 

 

 

Figure 2.9

path passing through (O,—K) is the graph A of

 

54

x = 1mm. 7? 6 [0.1.7121 (*)

shown in Figure 2.7. Near Y = O we have

L(JY) = a(JY)+—ij§ , where a(JY) is analytic function

of Y because a(t) is an even function of t. Therefore
the graph of x = a(jY) = L(JY)-1JY cross the horizontal
axis at x = a(O), as shown by the solid line B in
Figure 2.8. Indeed, the dashed graph A (*) may be

displaced to the right to obtain the solid graph B

X: sL(fi)—ifi> sL(fi) .

Near (a(O),0) the graph A has the form
x = L(j?) = a(JY)+ LfY, so may be continued as a smooth

curve (as a(jY) is analytic in Y) to the graph of

x = a(fi) _ 1,)?

denoted C. Similarly, near (—a(0),O) the graph B

has the form

x = _wfi) — ifi = _a(fi) - 21.5

so may be continued to D: x = —a(JY)+-21JY. Continuing

in this fashion yields all graphs

(4) X: iamY)+NwY= tam)+Nit,

w
10

H
1“

- O,

2.:
ll

 

55

as shown in Figure 2.8. We may regard x as a function
of t, where t lies on a Riemann surface with infinitely
many sheets (#).
d2X 2 //
However, from (#) we see (——§) = (a (t)
dt

single valued as shown above. This last result and the

)2 is

equation of path D are analogue of the Remark 2.6.

Second step. Let us calculate the above h(t)
for the unperturbed problem of Example 2.5, where K = O.
2
) =

We have t2(a”(t) (tH”(t)-—%)2. Hence

I 4
h(t) = t6(a’(t>>2 = (%)2t +O(t6) (**)
is real analytic and even with a fourth order zero at
t = O.

In the perturbed problem we have (t2+K)3(a”(t))2 =

h(t,K), where the left hand side is some special
perturbation of the left hand side of (**). Since the
current problem is a perturbation of 2.5, we expect

that h: R2 -4 It must, near (0,0), be a real analytic
function even in t near the function

h(t,O) = (%)2t4+-O(t6) in some sense. Motivated by
the Weierstrass Preparation Theorem [7] we might expect
a family h(t,K) = (%)2(t4+2A(K)t2+B(K))¢l(t2,K) of
perturbations, where 01, A, B are real functions with

$1 analytic in t2 satisfying A(O) = O = 8(0),

¢l(0,0) = 1. Thus

 

56

4 2
(aI/(t))2 = (3)2 t +2At +B ¢2< 2

t IK) I
w (t2+K)3

where K, A, B E I! are near zero, K > O, B > O and

¢(O,K) is near 1. Let

fl.) -2@ NW2.) , p...)

(t +K)

Description of a problem

 

2.8. In the heuristic derivation of the above
formula we regarded A(KL B(K), and ¢(t2,K) as
depending on the perturbation parameter K. For K = 0,
an explicit local solution with A(O) = B(O) = O,

2 _ 2 _
¢(t ,0) ~ ¢O(t ) was constructed, where ¢O(0) — 1

(Example 2.5).

Now we adopt a somewhat different view, that of
multi—parameter bifurcation theory. We regard A, B, K
and the function ¢(-) themselves as independent
parameters. They vary in the parameter space as
(K,A,B,¢) E R3XZ, in a neighborhood of (O,O,0,¢O)

where Z is a Banach space of functions ¢ defined as

follows.

Definition 2.8.1. Fix p > O and let Z be the

Banach space of functions @(w) analytic on the open disk

2 . . .. 2
[ml < p , continuous on the closed disk )4) g p , such

57

that @(T) is real for T real, with the norm

HNH = supidpml, m 3 pz. .1

Remark: Fix $0 E Z 2(for some p) w1th ¢O(O) = l.

2 ¢O(y )
_%.__c;__
K = A = B = O and ¢O given in (***)). This gives a

Consider fo(y) = (the f corresponding to
family of curves Fl : x = a(y), a”(y) = fo(y). In

¢O(y2)—1

, _ 2 1. ________ _ 2. 1 .
fact a (y) _ -W (y+ y ) —- —W (Y+-No(y)). where

NO is odd. Integrating twice we get
2 2 .

a(y) = D4—Cy-—F (y log y-y)-; Nl(y), where N1 is

odd with cubic leading term. Comparing with Example 2.5

we get D = 0, H(t) = C-F%)y-—% Nl(y). Hence given

¢O we have a typical local solution as constructed in

Example 2.5, which is not unique by presence of the

constant C in the formula H above.

Problem 2.8.2. Fix 0, an open disk about the

origin in z = x4—iy plane, small enough (say 5 E
domain of the local solution in Example 2.5). Fix

20 = xo+~1yo, x0 > 0, yO > 0, on the unperturbed curve
(F1 in Example 2.5). Consider A, B, K near 0,

¢ near $0 in Z, and f(y) with the branch of f

for which f(yO) < O, and (a(yo),x ) near 2

O 0'

a’(yo) near ii (H(Y)-% Y 109 Y)) °
Y " y=yo

 

58

(1) Find necessary and sufficient condition that
there exists a symmetric local solution in G with part
of the free boundary

X = a(y)

a”(y) = f(y)

near yO with a'(yo), a(yo) as stated above.

(2) Are the constants of integration unique? In
other words, is the local solution unique for the given

f(y) above? #

We will devote chapter 5 to handle this problem.
But first on the third and fourth chapters we develop
some theory needed to go over chapter five. For this
purpose we put an end to the chapter 2 and we start

chapter 3 now.

 

3. A NECESSARY AND SUFFICIENT CONDITION

FOR A LOCAL SOLUTION NEAR THE ORIGIN

The main purpose of this chapter is stated in
problem 3.2 below. But first, using Proposition 2.2 and
the Remarks 1, 2 after, we can restate the definition of
local solution of the axi—symmetric obstacle problem as

follows:

An equivalent definition of a local solution of

axi-symmetric problem in R4

Definition 3.1. By a local solution (u,O) of
the 4-dimensional axi—symmetric obstacle problem, symmetric
with reSpect to x—axis, on a fixed open disk 0:: R2

centered at the origin, we mean u :0 4 R such that

(i) u(X.-y) = —u(X.y)
.. 1 . 1 .
(11) § u(x,y) = V(X,y) is C in 0
(iii) Ell-u(x,y) = V(X,y) 2 u(x,y) in o,
where m(x,y) = -% (Xzi-yz)
. \ . 2, 1
(1V; if I = [(x,y) E J I? u(x,y) = w(X,Y)]I

then Au = 0 in G —I. 4

59

60

Remark. In part (iv), Au = 0 in (0-—I) is

equivalent to A 4 W = 0 there. To see this let
R

A 4 W = O in (O-—I), then u(x,y) = yw(x,y,0,0)
R
implies

azu _ 62w azu _ aw 82w

——y—2', —ZE+Y'—‘§.

ax ax By BY

_ _ ii = 2E -

If y — 0, then Au — 2 8y 0 because BY 18 an odd

real analytic function of y. If y # 0, then

 

2 2
v a v 2 Av
Au - y(0 + -§ + - :—) = O .
3X2 y y 01/
Conversely if Au = 0 in (O-I), then as in (the end of)

2.7 we can show A 4 W = O.
R

Let us now consider (as before) an analytic arc
:x = al(y), near (xO,yO) E 0, where yO > O,

X =a

0 l(yo) and al(y) is real analytic near yo.

Problem 3.2. What are the necessary and sufficient
conditions for (u,0) to be a local solution to the obstacle
problem (in the sense of Definition 3.1) with El as a
part of the free boundary and the contact set I to one
side of T1? #

We shall obtain the solution to this problem step

by step throughout this Chapter. Let us start by considering

the complex form of

 

61

I -z = Ll(t) = al(t)+ it, t E R, near yO .

Extend L1 to be defined for complex t near yO as
in 2.3, 2.4. As in Lemma 2.4 we can construct a local

solution uO in a neighborhood of (xo,yo), with free

boundary T1, and noncontact set lying on one side of

P Let v Cio be a one-sided neighborhood of z on

1' O O

the side of F1 in which the noncontact set lies. For

 

 

 

 

Figure 3.1

the remainder of this chapter, we deal with this local

solution (u ) arising from the given are Fl as

O'Vo
above. We shall try to extend this to a local solution
on the disk 9. Before doing this, we need to state and

prove the following lemma.

 

62

Lemma 3.3. Assume V(E 0) containing V is an open

0
connected set. Then Au = O in V with u = uO in V0
if and only if (Lil)2 defined in V0 has a (single-valued)

analytic continuation in V and I (L11(q))2dq is real for
c

every closed path c C V. Moreover we have then

i 2

z
u(z) = N(zo)+-Re I2 2[q +—4(Lil(q))2]dq, z E V

O

which is independent of the path of integration in V.

Proof: First assume u is harmonic in V, and

equals 110 in V0. Then uX--1uy is holomorphic in V.
In VO we have that

—— _ . _ i 2 -l 2 _

vu — uX-—1uy — 2[ +4(Ll (2)) ] — f(z)
is analytic, so f and (L_l)2 have analytic continuation

1

in V. Hence the integral representation of u remains

valid:

Z
u(Z) = (h(zo)+Ref f(q)dq
Z
0
Z
_ l p 2 —l 2
— (:(zo)—2 ImuzZ [q +4(Ll (q)) ]dq, 2 e V
o

as both sides of this equation are harmonic functions

agreeing on V0. If 2 can be reached from 20 by
different paths 11’ v2 in V, then
Re " f(q)dq = Re 5 f(q)dq because u is single—valued.

"‘ 1 "'2

 

63

Thus Re f f(q)dq = O, which implies

Y1'Y2

. -1 2 _ .
Re § 21(Ll (q)) dq — O, or equivalently

Yl'Yz
-l 2 . . . . -1 2 .
§ [Ll (q)) dq is real. This implies I (Ll (q)) dq ls
Yi'Yz C

real for any closed path C C V. Conversely let (Lil)2

have an analytic continuation from V0 to V, and put

d
Z

Z
u(z) = N(20)-+Re V f(q)dq .
O .

where the integral is along some path from 20 to z

in V. We assume § (L11(q))2dq is real, and therefore
C

f f(q)dq is imaginary for any closed path C C V. This
C

2

guarantees u(z) = N(zO)+-Re E f(q)dq to be single—
2
O

valued in V and independent of the path of integration

there. Hence u is harmonic in V and u = 110 in V0.

This completes the proof. 4

Let us now begin extending u to a local solution

0

of the obstacle problem in the sense of Definition 3.1.

The first step is defining a set containing VO on which

uO is extended as a harmonic function ul with the

property ul > N.

Definition 3.4. Consider (continuous) paths

v: [0,1] 4 0' with V(O) E VO such that

 

64

(l) uO can be harmonically extended along the
path, i.e., there exist 0 = t0 < tl < ... < tN = l, and
harmonic functions uO k defined on some open disks

Dk E O centered at y(tk) with uO,k{Z) = uo,k+l(z) on

Dk fl Dk+l # ¢ and u0 = u0,0 on DO. See Figure 3.2

 

Figure 3.2

(Note this extension may not be single—valued on U Dk);

and

(2) uO,k(z) > w(z) on Dk’ where
w(z) = -% y(x24-y2). Let Vi = {y(l) ly is a path as
above with properties stated in (l) and (2)}. #

Remark 1. V1 is clearly an open connected set
containing V0. Following the proof of Lemma 3.3, we
see that condition (1) above is equivalent to analytically
continuing (Lil)2 from y(O) to V(l) (along v) with
the disks Dk' Let ul(z) denote the corresponding harmonic

continuation of uolz).

65
—l 2
Remark 2. ul as well as (Ll ) extended above
may not be single—valued° By Lemma 3.3 ul is single-
valued harmonic in Vi if and only if (Lil)2 has a

single—valued analytic continuation (from V0) to
Vi with § [LIl(q)]2dq = real for any closed path
C

(: C IV. .
1
Remark 3 o O V] is a dis .IOint union Of tWO SUbsetS

P and Q defined as follows:

P = [z 6 avi [there is no y :[O,l] 4 5 with

y<1)= z. y(O) 6 v0, and wow) co.

having properties (1) and (2) as in

Definition 3.4} , Q = (avi) —P

observe Q E (avi) D 50. We call P the Natural Boundary

of Vi and it is closed. #
Let us now extend (uO’VO) anti-symmetrically to
its reflection in the lower half plane.

Definition 3.5. Let V5 = {z :E 6 Vi},

_ . _ I , x w
OO—[Z‘ZEVO}I Wl—Vln{(XIY)‘YZO/‘I

v
w2=v2’l: {(x,y)zy<o}, w=w uw

l 2’
J=Wln{(x,y):y=o}=filnwz, r2={z-.Eerl},
uOO\X,y) = —uO(x,-y) for (x,y) é V00.

 

 

 

66

Remark l° W E G is open, and for (TZ’VOO) the

—l . -1
maps L2, L2 can be defined as L1’ L1 for (T1,VO),
namely, L2(t) = a(t)-—it for t near yo. Moreover
on VOO uOO can be constructed similarly as we did for
uO and a modification of Lemma 2.4 (uOO < w instead of

1.10 > W) is valid.

Remark 2. Note even if W1 is not connected,
ul(z) can still be defined as before (by means of
integration along the paths which extend into the lower
half plane). However we easily see that
W1 = Vi C {(x,y) :y > 0], W2 = V5 C [(x,y) :y < O},

and W W are connected if u extends to a local

1’ 2 0
solution u in 0, since u(x,O) = w(X,O) = 0.
Remark 3. awl is a disjoint union of three subsets:
= P '
v1 fl 8W1, V1 0 5W1 C J, and Q N 5W1. Set
v2 = [z :z 6 v1}, and v = v1 U v2. If we have
Wl = Vi C {(x,y) :y > 0}, then the second subset of

5W1 above is empty.

Lemma 3.6. For 2 = L (t) E V we have

1 O
—l - _ -l —l — 2 _ -l 2
L2 (2) — —L1 (2) and [L2 (2)] — [L1 (2)] .

Proof: T1, T2 have equations 2 = Ll(t) = a(t)+ it
for t real near yO > O and z = L2(t) = a(—t)+—it for
t real near -yO < 0, respectively. Therefore, for
t e L'lw)

l O

 

67

 

 

Ll(t) = a(t) a? = a(t) —i't‘ = L2(_E)
. -l — — -1
Putting z = L1(t) above we get L2 (z) = -t = -L1 (2)

for z 6 V0. Hence

[L;l(E§]2 = [Lil(z)]2 for z 6 VO .

We are now going to state two hypotheses and then prove

they are equivalent.

Hypothesis 3.7.

The functions uo and uOO in V0 and VOO
respectively have a common single—valued harmonic extension
u to w, which is C1 in W; and u = w and vu = vw
on (v-J). Observe that v-—J C aW is the closure of

the set of Natural Boundary points which do not lie on

the x—axis. Note by this Hypothesis we have

D
u2(X,y) = u(x,y) = -u(X.-y) = —ul(x,-y) for (x,y) 6 W2
well-defined as a harmonic extension of uOO in W2 for
which u2 < $2. #

Hypothesis 3.8.

1) (Lil)2 defined in V0 has a single-valued
analytic continuation in W1 and § (Lil(q))2dq is
C
real for any closed path C C Wl (which imply similar
properties for (Lgl)2 in W2)°

68

2) (Lil)2 is continuous on W1, [L11(z)]2 = y2
for z = X+—iy 6 (v--J),[Lil(z)]2 = [LIl(z)]2 = [L;l(z)]2
«Z -1 2
is real for z E J, and lim Im J [L1 (q)] dq exists
z4z* zO
ZEWl
for an 2 5 BW and equals l'-(y3--y3) for any
Y * P l 3 * Q
2* 6 (vl-J), where the integration path is in Wl U {20}. #

Proposition 3.9. Hypothesis 3.7 holds if and only
if Hypothesis 3.8 holds, and in such a case we have
vi = wl c ((x,y) =y > o]. v; = W2 C (my) =Y< 0}:

with W1 and W2 open and connected.

Note: Under either one of the above Hypotheses if

J-—(vl-J) # ¢, which is a countable union of disjoint

Open intervals on the real axis, then along any path

from 20 to 20 in W U (J-(v-—J)) passing through
a point z in J-(vl-J) we shall have
2 3 E0 —1 2
—3 Yo = Im f (L (q)) dq.
2o
—l 2 . . —l 2
where (L ) shows, in this case, both (L ) and

(L31)2 because these two are the analytic continuation
of each other in W U (J-(v-—J)) by the Reflection
principle. Instead of the latter integral condition we

can impose an equivalent condition:

69

Proof: Assume Hypothesis 3.7 hold. Let us first
show the second part (the set equation). If J = ¢,

then clearly we have

wl=vic ((x,y):y>0}. W2=V5C {(x,y)zy<0}-

1

If J # ¢, then u being C in W and u(z) 2 w(z)

in Wl imply “.2 O on J (continuity of u is enough

here). Similarly u(z) g w(z) in W2 imply u S O on
J. Thus u E O on J. Hence Vi = Wl C {(x,y) :y > O],

Vé = W2 C ((x,y) :y < 0}. These show W1 and W2 are

open and connected.

Let us now show the first part. The Hypothesis 3.7

2

is assumed, thus, by Lemma 3.3, (Lil) has a single

valued analytic continuation in W1 and § (Lil(q))2dq
C

is real for any closed path C C W1. Since u is C1

in W, thus

 

 

u(z) = @(zo)+Re J." f(q)dq
= vu = f(z) = i-ifgz4-4(L-l(z))2], for z E W ,
2 l l
_ z
u(z) = ((zOHRe j_ f(q)dq
2
O —
= vu = f(z) = -i[324-4(L_l(z))2], for z E W ,
2 2 2
imply (LZ1)2, (L31)2 are continuous on W1, W2

respectively.

70

If J # ¢, then putting the above values of vu

equal on J we get:

(Lflznz = (L;1(z))2 for z 6 J = wl n 562 .

—1

Using Lemma 3.6 we can conclude [L;1(E)]2 = [L1 (2)]2

for z E J because [L;l(E)]2-[Lil(z)]2is an analytic
function which is identically zero on V0, and therefore
it is zero on W1. Comparing the last two equations we
get [L;l(z)]2 = [L11(z)]2 for z 6 J. Hence

2

[Lil(z)]2 = [L;l(z)] is real for z E J. For any

2 E W we have
_ = ' _ 1 = "TT*__— 1
vu vm ux+iuy vu uX iuy VU
' — - 2
= —%[ZZ+4(le(Z)) ]—V1L’

= 21[y2-<Lgl<z))21

Since u is C1 on W and consequently (L31)2 is
continuous on Wj' thus
. 2 _l 2 r
vu(2) -vw(2) = 21[y --(Lj (2)) ], for z : aw.
In particular for z = X+—iy E (V-J), where vu-—vw = O,
we have

71
For any z* E 8W1,

u(zQ = lim u(z)

242*
zEWl
Z
= ((20) _% lim my [q2+4(LIl(Q))2]dq
z4z* zO
ZEWl

z
exists, that is, lim Im I (LIl(q))2dq exists. In
2
O

z»z*
ZEWl
particular for 2* E (v-—J), where u(z*) = w(z*), we
have then
u(z*) = lim u(z) = w(z*)
z»z*
26Wl

Z
a w(zO)-¢(z*) = % lim Im j [q2-+4(L11(q))2]dq
Z"Z Z

26W: O
1 3 3 z -1 2
a 3(y*-yo) = lim Im f (Ll (q)) dq
242* 20
zewl

z
Note that the last limit equals Im I * (L11(q))2dq if
z
0

z 2 can be joined in Wl U {20,z*} by a path with

*' 0

only endpoints 20' 2* in 5W1. In this case then for
c w ,
Z ‘ 1

72

2

Z
U(z) = ((20) -—§- Imf [q +4(Lil(q))2]dq

20

Z
= 1<zO)-—§- Im ( *[q2+4<L11<q))21dq
Z
0
1 z 2 -1 2
-3 Im j‘ [q +4(Ll (q)) ]dq
2*
1 ,2 2 —1 2
= w(z*)-—-2- ImJ [q +4(Ll (q)) ]dq .
2*

which will be used later on. Note W1 may not be path
connected even if W1 is. If X E J- (vl-J) , then there

exists a disk D(x;r) with D(X;r) fl (vl-J) = (1). Thus

(D(x;r) — [x-r,x+r]) C W. In fact 2 can be joined to

Z0 by a path in W1. For any 2 6 [J- (vl-JH (which

is a countable union of disjoint open intervals) we have

1 z 2 —1 2
ul(z) = u2(z) = O a1‘1(zO)—§ Im (‘2 [q +r(Ll (q)) ]dq
O
— ((3 we Im (21 +r<L'l< ))2]d
‘ o 2 .1— q 2 q q
z
0
— (‘EO 2 -l 2
a ((20) -11:(zo) — 2 ImJ [q +4(L (q)) ]dq
z
0
EO —l 2 2 3
= Im f (L (q)) dq = "3 yO I
Zo
_ _1 2 — 2
where (L 1)2 represents both (Ll‘) , (L21) because

these are analytic continuation of one another by the

reflection principle (recall they are real—valued on x—axis).

73

Observe u(z) = 0 implies (as above)
z
-l 2 _ l 3 -————-
Im I (Ll (q)) dq — -3 yo, V z E J-(vl-J). Here is
z
0
the proof of the note also complete.
Conversely assume Hypothesis 3.8 holds. (1) in

3.8, by Lemma 3.3, implies u is single—valued harmonic

extension of uO (defined in V0) at least in the

connected component of W containing V

1 0°

If J = ¢, then clearly Wl = Vi C {(le) =Y > O}

and W2 = Vé C [(x,y) :y < O}. In this case (1) in 3.8

implies that u (defined on W) is single-valued harmonic
z
in W by Lemma 3.3. Since lim Im j (L11(q))2dq,
242* 20
26W
if

z 3)

* 6 BW exists and equals to ‘%(Yi"YO

l

2* E (vl-J), thus u is continuous on W1 and

u = W on (vl-J). For a detail see the proof of
"only if" part. Since (L11)2 is continuous on W,

thus u is C1 on W. [Lil(z)]2 = y2 for z E (vl-J)

implies vu = vw there (again for a detail look at the

"only if" part). Similarly we can argue in W2, (L51)2,

.in the lower half of z—plane.

If J # ¢, then we have three subcases

a"?

 

74

Consider the subcase J-(vl-J) = ¢. Then all points

of J are the limit points of -J. In this subcase

V1

W1 = Vi C [(x,y) :y > 0}, W2 = Vé C [(X.y) zy < 0}
(l) in 3.8 again implies, by Lemma 3.3, that u is a
single-valued harmonic extension of uO defined in V

The first integral condition in (2) of 3.8 implies

O.

u(z*) = lim u(z) = w(z*) = O , V 2* E J .
z z*
ZEW

The rest of the proof goes as in the case J = ¢.

Let us now conSider the subcase ¢.§ J"(Vl“J).§ J.
Then (1) in 3.8 again implies, by Lemma 3.3, that u is

the single—valued harmonic extension of uO at least in

the component of W containing V (similarly we

1 O
can argue about the lower half of the z—plane). Note

J-(vl-J) are those points of J which are not limit

points of the part of v above (outside of) the real

1

axis. Thus for any small enough disk, D(z*;r),
neighborhood of any 2* E J-(vl-J) we have

D(2*,r) n{(x,y):y(_1)3 < o} c w.. In fact J— (V

3 l"j)
is open and it is a countable union of disjoint Open
intervals on the real axis. Let (a,b) be the maximal

 

interval containing 2* and contained in J-—(vl-—J).

Any such point 2* in this open interval can be joined

75

to 20

6 BW

by a path in W with only endpoints z*,

1

z 1° Clearly a or b e (v --J)° By the first

0 l

integral condition in (2) of (3.8) we get

lim u(z) = w(a) (or w(b)) = O (for detail look
z-0a(or b)
zEWl

u ' u I .
at only if part). Suppose now (al'bl) C Wl‘g Vl With

u(al) = w(al) = O or u(bl) = w(b ) = 0. Then

1
2 _
u(2*) = ((20 )-— my: OIq +4(L11<q))21dq
1 2* 2 —1
_ 3 Im f2 [q +4(Ll (q)) 2]dq, for 2* E (al,bl), where
z = a1 or z = bl according to whether w(al) = u(al)

or w(bl) = u(b ) respectively. Thus

1

z
u(z*) = u(z)-—% Im f *[q24-4(Lil(q))2]dq. Therefore
2
2
u(z*) = --Im f: *[q2 4—4( l(q)) ]dq .
On the other hand (L'l'l)2 is real valued on J, thus

u(z*) = O, V 2* E (al,bl). This contradicts Wl-; V’
V

Hence such an (al,bl) does not exist and W1 = i. Hence

we conclude W = V’ C {(x,y) :y > O}, and by symmetry

1 1

W2 = Vé C {(x,y) :y < 0}. Since the limit in (2) of 3.8
. l 3 3 —————

eXists, and equals to Ԥ(y*-yo) for z* E (vl-J),
thus u(z*) = lim u(z), 2* E 5W, exists and can be

zaz

'k

defined, and equals to $(2*) for 2* E (vlfl-J). Since
(L_l)2 is continuous on W , thus vu is continuous

l

76

on W1 (for detail look at "only if" part) and
therefore u is C1 there. Moreover u s O on J.
Also [Lil(z)]2 = y2 for z E (187:7?) implies vu = v
on (Vl’J)°

Let us finally consider the subcase .I- 63;:73) =
where (31::3) O J = ¢. Then any point z 6 J can be
joined to 20 (and EC) byapath in W1 (and in w2
with only endpoints 20’ z perhaps in awl (and E0'
2 E 5W2 perhaps). Let us show now J C awl. Since
(L_]‘)2 is continuous on W, thus for z 6 J we have

Z
ul(z) = (<20) ——§- Im “r [q2+4(L11<q>)21dq 2 ((2) = o

2o
u (z) = w("z‘ )—l Im (2 [ 2+4(L‘1(q))2]d w(z) — o
2 o 2 9— q 2 ‘13- ‘
Zo
that is -i 3 Im (2 (L_l( ))2d and
, 3yo.>. . i q q
z
0
z
i y3 2 Im F (L_l(q))2dq. Since (L—l)2, (L-l)2 are
3 O 02' 2 l 2
O

analytic continuation of one another by the Reflection

principle (use Lemma 3.6 and (2) in 3.8), thus

2 z ————————— 2
Im Ur; (L;l(Q))2dq = Im j’_ (L;l<q))2dq = Imf (Lglmnz
20 20 20
Z 2
-Im f (L;l(q))2dq. Hence -% y: 2 Im f (L_l(q)) dq
z Z
O O-
1 3 .z —1 2
and -g-yo 3 Im 5 (L1 (q)) dq which imply ul(z) = O

W

J,

)

 

77

and u (z) = 0. That is u(J) = 0 and

2
_ I . = I .

Wl — Vl C {(x,y) .y > 0}, W2 V2 C [(x,y) .y < 0}.

Condition (1) of 3.8 again implies of course, by Lemma 3.3,

that u is a single-valued harmonic extension of uO

and in W U J. The rest of the proof

in W1, W2,

can be stated as before. #

Definition 3.10. Assume either one of Hypothesis
3.7 or 3.8. Now let us extend the function u from the
set W to the set 0' as

u(z), if z 6 W
u(z) = .
w(z), if z E 0-—W

Theorem 3.11. (u,@) in Definition 3.10 is a local
solution to the obstacle problem in the sense of

Definition 3.1 if and only if

z
1) 1im if; y3+-Im f (L-l(q))2dq] exists and
y 3 O 1
z-vxl zO
ZEWl
equals to (L11(xl))2 for any x1 6 J,
~—---—- 2
_ — n — 2

2) 11m -J3[-y(Lll(Z))2+“% ygi-Im J (L l(q)) ]

24x1 y 20

zEWl

exists and equals to i[(L11(xl))2]’, V X1 6 u.

2

l (x)> 20,

Note: Since (— u-w)( = -2(L-l
x

Y

-1 ,
] at any X E J, thus any

possible real zero of (L ) in J-(V-J) is not

 

78

simple. These zeros are isolated in J- (v -J) and

they form a set S so that S U (v-J) establishes the

free boundary aI for the obstacle problem.

Proof: Assume u satisfies the requirement of
Definition 3.1. Then v = % u is C1 in O, in
particular, it is C1 on the real axis. If J = O,

then nothing needs to be proved. Assume therefore J # O.

For any z 6 W1 we have then

Z
u(z) =1<zO)-%Im (z [q2+4<L11(q>)21dq

0
Hence
1 _ l 2 _1_ 2 _2; 3 z —1 2
y u(z) - —2 x +6 y —y[3 yO+Im “(2 (L1 (q)) dq],
O
for z 6 W1 . (*)
Furthermore for any 2 6 W1 we have
1 -1 -_ A __l. - -
v(§u(z)) — yuX+i( u — 2 u) — (uX+iu )— 2 11
Y Y
_ i _i 2_i —l 2 __i_ 2 i
— -x+2y-2yx — y[Ll (2)] +2yx -6y
2
2i —1 2
+ 713 yO+Im ( (L1 (q)) dq]
Y 20
= —x+—— y+-2%[—y(L'l'l(z))2+-§ yg
Y
z
. _ 2
+ ImJ (L.l(q)) dq] (**)

   

79

J may consist of two classes of points, one which are
limit points of v-—J (the part of v outside of J)
and the others are the interior points of J-(v-—J).

We treat them in different ways.

First let Xl be a limit point of v-—J, i.e., in

the first class. Since % u(z) is continuous on W

1!
' 11 3 .z —1 2
thus by (*) we have lim '—[§ yoi-Im J (Ll (q)) dq]
24x y z
__1 o
ZEWl

exists, where z = xi-iy. To find this limit let

z* 4 x1, where 2* E (v-J), then
.13 z-12_i3133_13
11m {-3- y0+1mf (Ll (q)) dq] — 3 yo+3(y*-yo) — 3 y.
zaz* zO
zEWl
and
1 1 3 “2* —1 2 . 1 2
lim —— 3 yoi-Im 3 (L1 (q)) dq] = 1im 3-y* = 0
2*7X1 y* 20 Yieqo
2*6wl
(there are such 2*) .
Or we can argue as follows: 1im % u(z) =
z—vx1 -
‘ 2
x
. 1 _ . 1 _ 1
lim ——-u(z*) — lim —— w(z*) — "7f° Thus by (*)
z*+x1y* 2*4xly*
2
lim l[l y34-Im F (L_l(q))2dq] = O, which proves
y 3 0 . l
24xl 20

C"
Z‘Wl

 

80

condition (1). (Note for such an xl we have
[Lil(xl)]2 = 0). Since u(z) is C1 on the real axis

also, thus by (**)

, 1 —1 2 1 3 Z -l 2
lim —2[_Y(Ll (z)) +w§ y04-Im f (Ll (Q)) dq]
27:1 Y 2O
ZEWl

l 2

is continuous on W and

exists. Recall that (L1 1

)

2
equals to y2 on (v-—J), and lim Im I (Lil(q))2dq =
z

 

242*
zEW1 O
Y3-Y3
*
3 O for 2* 6 vl-J. To compute the above limit we

can make these substitutions to get it equal to

. 2 . . .
11m --3 y* = O (which proves condition (2))

Y*40
Or we can argue as saying 1im v(£ u(z)) = lim VO = -x1
24xl 24x
ZEWl zEWl

and consequently by (**):

—— Z

. 1 -1 2 1 3 —1 2 _
ii: 2[-y(Ll (2)) +3 yO+Im (‘2 (L1 (q)) dq] - o ,
zEWl

which proves (2) .

Now let xl be an interior point of J-—(v-—J),

i.e., in the second class. Then any point z c D(x1;r),

 

81

for sufficiently small radius r, can be joined to

20 or 20 in W, where only endpoints may not be in
-l 2 -l 2
L2 )

W. Recall that in this case (Ll ) , ( are

analytic continuation of one another across the x-axis

'1>2.

and we may call both as (L Again since % u(z)

is continuous on W1, thus by (*)

z
_ - 2
lim %[% ygi-Im f (L l(q)) dQ]
22:1 z0
zEWl

exists for any such x1. To compute this limit we can do

as follows

Z
lim hi y3+1m§ (L‘1(q))2dq1
y 3 0
24X 2
O
ZEW

x z
= lim i[% yg-i- Im i (L'l(q))2dq+ Imj‘ (L‘l(q))2dq]
zax y z x
_l O
26W
= lim — Im j:( 1(q)) exists,
z+x
l
zEWl
where z = Xi'iY: % yg+-Im j (L—l(q))qu = O.

(The latter equation was shown at the end of first

part, ”only if”, of the proof of 3.9.) Moreover

 

82
z -1— 2—

Imf:(Lc1))2dq=Im‘((L(q))dq=Im§:(Ll(q))2dq=
X

z -1 2 z —1 2
Im f (L (q)) dq, which shows Im I (L (q)) dq is
x x

an odd function, real analytic vanishing for y = O.
1 z -1 2
Thus § Im I (L (q)) dq is a real analytic function even

in y. Hence

Z
1% yg+ 1erz (L'l<q)>2dq1

0

lim l
Z-’X y
zEWl

Z
= — Im 5 (L'lmnqut
X

II
721
(D
b

l
N

Since % u(z) is C , i.e., v(% u) is C

in W, thus by (**) we have

11 %[-y(L_l(z))2+% Im J“ (L'l(q))2dq]
24:1 y 20
26Wl
—"'_‘ Z
= lim —l§[—y(LZl(Z))2+ Im j‘ (L_l(q))2dq]
11 X1

_1
W1

exists. To compute the value of this limit, since we have

real analytic function (which vanishes at y = 0) divided

 

83

by y2 involved, thus it is enough to consider z a Xl

vertically, i.e. z = xli-iy. Then by L'Hopital's Rule

 

 

— 2
lim -J§[-y(L-l(z))2+-Im ( (L_l(q))2dQ]
z-vxl y xl
zer
-(L‘1<z>)2 —y ga—uL‘Hz) >21 +Re(L_l(Z) )2
= lim y 2y
Z"X
26Wl
i Im(L-l(z))2 —y Big/(film?
= lim 2
Y
24X]-

ZEW:L

= gnaw-1n. ))2]'+-i- Re1<L'l(xl)>21'

l 2
. - 2
= 1[(L 1(x1))]’
where the last equality is because (L—l)2 is real on
J. This proves condition (2). We discover also that
(% u(z)-—m(z)| = -2(L—l(x))2, for any x E J, including
x

. ———— —l 2 2

the pOints of J 0 (v-J) because (L (z)) = (Im z)

for these points. The rest of the note is clear now.

Conversely assume conditions (1), (2) hold. If
J = O, then we are done. So let J # O. Using
conditions (1), (2), equations (*), (**) imply that u

is C1 on the real axis.

Proposition 3.12. The local solution (u,e) is

symmetric with respect to y—axis if and only if

 

84
[L‘l<z>12 = [L‘H—En

Proof: Assume (u,0) is symmetric with respect

.. _ Lu - .53.-
to y—aXis. u(-X,y) — u(x,y). Then ax(x,y) — _BX( X,y),
i.e., uX is odd with respect to X while u is even.

2E _ fiE._ _ - _ _
Moreover 5y(x,y) — ay( X,y). Hence ux( X,y)4-iuy( X,y) —
—ux(x,y)4-iuy(x,y), that is, uX(-z)-—iuy(—z) =

-(uX(z)-iuy(z)); where u(z) =

z .
- 2 .
((zo)4-Re f -%[q24-4(L l(q)) ]dq. Applying Cauchy—
z
0

 

Riemann equations we get %[(—E)24-4(L-l(—E))2] =

-—§-[22+4(L'l(z))2]. Thus —2i(1.'l(—E))2 = -2i(L—l(z))2,

that is, [L-l(-E)]2 = [L-l(z)]2° In particular if z

is imaginary, then —E = z and therefore [L'l(z)]2

is real. Conversely assume now [L-l(—E)]2 [L_l(z)]2.

First notice that u(JE) = u(z) if and only if

-% Im ”(2 [q2+4(L—l(q))2]dq = _% Im d(-z[q2+4(L-l(q))2]dq.
Z0 Z0

N I

This is true if and only if Im f [q2+-4(L-l(q))2]dq = 0,
2

-Z

if and only if Im f (L-l(q))2dq = 0. Let us show now
2
"z -1 2
the last equation: Im F (L (q)) dq =
”2
C -1 2 ‘E -1 2
Im[f (L (q)) dq-E-r (L (q)) dq], where the path from
2 c

z to —z is symmetric with respect to y—axis crossing

85

y—axis at c. Hence

_]_(

_E 2
Im I (L q)) dq
2

C Z _ __
— Im[f (L‘l(q>>2dq+-j (L'1(-q>)2<—dq)1
Z C

C C—
Im[f (L'1<q>)2dqi-§ (L‘l<q))2d61
Z Z

C c —_..__
( Im[(L‘1<q))2dq1+-5 Im[(L‘l(q))2dE1
Z Z

This proves therefore u(JE) = u(z). Here is the proof

complete.

 

4. AN ANALYTIC CONTINUATION

TO THE FREE BOUNDARY

Introduction

In this chapter we want to make some observations

on the conditions stated in Chapter three as necessary

and sufficient for a local solution u to the obstacle
problem to exist. In particular we want to see what we
can conclude about the maps L and a from the conditions
imposed on (L_l)2. We shall discover, roughly speaking,
that the free boundary of the local solution everywhere

has the equation X = a(y) or x = a(-y)-—2iy, where

a here is some analytic continuation of the original a
inside a: = w u [J—(U—_JH = (0—1) (1 5. Since (III)2
is analytic single—valued in W, thus in a neighborhood

of any point 21 in this set we have

on .
Ufl(zH2 = (z—zl)m 23 d.(z—zl)j,
j=o 3
where do # O, m 2 O (4.1)
. . 2 —l 2
PrODOSition 4.1. If we let t = [L (2)] for

z in the above neighborhood, then 2 can be written as

86

 

87
_1_
. . 2 -l 2 p .
a power series in [t -(L (zl)) ] , for some integer

p‘Z l, t2 in a small enough neighborhood of (LIl(zl))2.

Proof: By (4.1) above we have

t2 = (z--zl)m Z) d.(z-z )3. Consider two cases.
i=0 3 l

The case m 2 1: Taking m—th root of both sides we
2

get tm = (z-—zl)g(z), where g is analytic near 21

with g(zl) # 0. By invoking the inverse function Theorem
2

we get z-—zl = a(tm), where B(T) is analytic in a

small enough neighborhood of zero. Here is the proof of
this case complete.

a v
The case m = 0: Let t2--cxO = Z) d.(z-—zl)j =
j=r 3
co .
(zuzl)r Z) Gj(Z-Zl)j_r, where Or is the first non-

j=r
zero coefficient among O1, d2, G3,... Taking r—th
1
root of both sides we get (t2-Go)r = (z-zl)g(z),

where g is analytic near 21 with g(zl) # 0. Again

by invoking the inverse function Theorem we get

z-—z = w((t2-do)r), where w(T) is analytic in a

Isufficiently small neighborhood of zero. Thus

2 = 214-w((t2-oo)r), and the proof is complete. #
. . -1 2

Corollary 4.2. If 2]— is a Simple zero of (L )
in W, then 21 is a branch point of L_l, and L is

a single valued analytic function in a neighborhood of

 

L_l(zl) = O (z = L(t) = a(tz) by the notation of the
proposition). If 21 is a second order zero of (L—l)2
in W, then (L_l)2 has two analytic square roots in a

sufficiently small neighborhood of z]- each of which is
invertible with analytic inverse L (z = L(t) = a(t) by

the proposition notation).

Proof: It is an immediate consequence of m = l,

m = 2, respectively, in the Proposition 4.1.

Definition 4.3. By (4.1) (power series) we can write

E
L—1(z) = (z--zl)2 ’

j
aj (2‘21) I

W

m 2 0, d6 # o
3

which is double valued with a branch point at 21 if

m is odd, and consists of two analytic functions (with
opposite signs) if m is even. Define

A = (z E W [2 is a zero of (L“l)'],

B = [z E W [2 is a simple zero of (L-l)2}. Clearly
point of A U B are isolated in W and 20 is not a
limit point of A U B. In particular W U {20} is

path connected.

Lemma 4.4. Along any path y :[0,1] 4 W U {z

},

Q
where y(0) = 20 E V0 U T1 and y(t) E A U B, for any
t, L“1 has an analytic continuation and is invertible

locally with analytic inverse L. Moreover there exists

a path c :[o,1] 4 ¢ such that c(o) = L-l(zo) = Yo

 

89

and L(O(t)) = y(t); further L is analytic

continuation along 0, of the original L.

Proof: Consider 21 = y(tl

Because 21 z A U B, either [L_l(zl)]2 # 0 and
-l

) for some t1 6 [0,1].

(L )’(zl) # O, or else [L-l(zl)]2 = O and m = 2
in (4.1). In either case, (L-l)2 has two square roots
l L_1 which are analytic near z1 and which satisfy
(L'l)’<zl) a! o.

The fact that i L_1 are analytic (locally) implies
the original L-l has an analytic continuation along

the path y. This is locally invertible because

(L-l)'(y(t)) # 0 for this analytic continuation. The

path 0 is given by o(t) = L_l(y(t)). #

Remark 4.5. Note that because L—1 is double valued

in W, there may be two paths yl(t) and Y2(t) with

the same endpoint yl(l) = y2(1), which gives rise to
different continuations of L_l: L_l(yl(l))=-L-l(y2(l)).
Given a path Y1 there exists such a path y2 if
2

and only if (L‘l) has an odd order zero in W. For

example we can modify Y1 as shown in Figure 4.1, where
z is an odd order zero of (L_l)2.
='\ =‘/ Y
21 (1(1) {2 l) 1

Figure 4.1

 

90

Theorem 4.6. For any 2* E (v-—J) U S = T, the

free boundary (established in Theorem 3.11) we have

z* = x*+iy* = L( :‘:y*) = a( ty*)+i( ty*) in the
following sense: For any zj 4 2* in W-—(A U B) there
exists sequence t. 4 y* or -y*, and branches (i.e.

J
analytic continuations) Lj of L near t. such that

. = L. t. 4
23 3(1) 2*

Proof: Let 2* belong to the free boundary F.
Since W-—(A U B) is open and path connected, thus for
any sequence {zj] C W-—(A U B) with zj 4 2*, each
0 (or 20) by a path Yj in
W-(A U B). By Lemma 4.4 the original map L"l has a

zj can be joined to 2

(unique) analytic continuation L31 along Yj which

is locally invertible with analytic inverse, say Lj’

along oj, where Lj(Oj(s)) = yj(s). On the other hand

zj 4 2* implies (L_l(z.))2 4 yi, where y: = [L-l(z*)]2,
by continuity of (L‘l)2 in W. Putting tj = L;l(zj),

2 2 .

w t z. = L. t. and t. 4 , where t. is th

ege J 3( J) 3 y* 3 e
endpoint of Cj in t-plane. If there is no odd order
zero of (L-l)2 in W, then L51 = L_l, V j, and

either tj 4 y* or tj 4 —y* (one of these cases only).
If there is an odd order zero of (L_l)2 in W, then

by Remark 4.5 we can assume either tj 4 y* -or t. 4 -V

(each one is possible).

 

91

This shows that for any sequence 2. 4 2* E T in
J

W, there exists {tj] tj 4 y* or —y* and branches

of L near tj such that

.=L.t. '* .
23 3( 3) 2* #

Remark 4.7. In the cases we will consider the

branches Lj are in fact all the same branch of L,

in a neighborhood of :ty*. Thus we can write

2* = IA iy*), or equivalently
x* = a(y*) or X* = a(—y*)-2iy*
-l 2 . t
In case (L ) has an odd order zero in w, we may
assume in fact t3. 4 y*, so we can write

X* = a(y*) for X*4-iy* E T . #

5. A CLASS OF UNFOLDING OF THE SINGULARITY

In this chapter we use the theory developed in
Chapters 3 and 4 to study a class of problems derived
at the end of Chapter 2 (in 2.8.2). As before, we
consider here free boundary curves x = a(y) near
a point 20 = xo+iyO fixed on the unperturbed free
boundary curve in C} with x0 > O, yO > 0, where
f(yo) < O and a”(y) = f(y) near yO as stated in
Problem 2.8.2. Following Section 2.7 we want to treat a

special case of Problem 2.8.2 here:

Problem 5.1. Study Problem 2.8.2 in the special

case when K > O. 4

.1

Lemma 5.2. Let IO be a compact segment (e.g.

interval), and L be analytic in an open neighborhood

of IO. Let L’(z) # O, V 2 6 IO, and also let L be
one-to—one on IO. Then there exists a neighborhood of
IO on which L is a conformal map.

Proof: Suppose not. Then there are sequences

1’ r ' =
‘an}’ (hm), an 4 IO, bn 4 IO, w1th L(an) _L(bn). Then

there are subsequences an 4 a, bn 4 b, where a, b 6 IO.

92

 

93

‘-~-—-_—..—

Figure 5.1

If a # b, then we get a contradiction to one—to-one

assumption because L(a) = L(b). If a = b, then
L(bnk)-L(ank)
L’(a) = lim ———————-————-—- = 0 because L(b ) = L(a ),
k4m bn -a nk nk
k “k
V k 5 EL Here we get a contradiction to the assumption
f'(t) # O on IO. This completes the proof. #

Necessary conditions in Problem 5.1

5.3. We want to find necessary conditions that
there exist a local solution symmetric with respect to

both axes in 6, with part of the free boundary

Tl :x = a(y), a”(y) = f(y), near yO
Here (a(yo),yo) 18 near 20 = (XO’YO) fixed on the
unperturbed free boundary curve XO > O, yO > O, and
al(yo) is near é§(H(y)-% y log y)| , f(yo) < O,
y=YO
/ 4; ,2
f(y) = -%-4Jéfi§§3:% @(yZ), ¢ near $0 in Z;
M (y +K) /

A,B,K near 0, K > O

 

94

We begin by building up the set v on one side

0
of T1 (over) as in Lemma 2.4° Consider the complex
form of Fl:
2 = L(t) = a(t)+ it, t E R near yo .

Extend L to be defined for complex t near yO again

as in the past. On an interval about yO (on the real

axis) L is one—to—one, L' # 0, thus by Lemma 5.2. L-1

exists and is a conformal map in a neighborhood of L(yo).

As long as f remains real on the real axis to the left

of yo, or equivalently y4+ 2Ay2+-B remains non-

negative there, we can extend F L analytically and the

1!

local solution also nearby T as in Section 2.7 by

1
Lemma 5.2.
Lemma 5.4° The case where t44-2At24-B has a simple
root t2 = 32 E (O,yg) cannot happen.

Proof: Suppose it can. The proof proceeds in

three steps.

First step: Without loss of generality we can

assume t44-2At24—B > O for t E (B,yo). The relation

“y 4_ 2
” yl (t 4—K)

a’(y2)-a'(yl) =

’ ( ) I

B S y2 < y1 g'yo shows that a (y2 > a (yl) for the
above yl, y2. This shows a’(y) > O for B < y < yO

and a(B) < X We now want to show a(B) 2 O and

O.

 

95

conclude that a(B)4—iB 6(3. Suppose a(fi) < 0. Then
a(B)+-iB is in the second quadrant. We can extend

T as the graph of function a as long as it remains

1
inside 0. We can also extend L nearby (B,yO]. The
local solution built up near a(yo)+iyO can then be

extended near the new Fl as shown in Figure 5.2 with

 

 

Figure 5.2

the shaded region as noncontact set. By symmetry with
respect to the y-axis we have the same construction

starting in second quadrant with a free boundary F

L2
intersecting Tl at 2 on the y—axis. Here we get a
contradiction because above T1 near 2 we have u > Q,
and on T in the noncontact set to the left of 2 we

2

have u = 9. This shows a(B) x O.

.1—

 

96

Since we assume the existence of a local solution
of the perturbed problem in 0, thus the shaded region
is in the set W defined in Chapter 3. We are first going
to show that any small circle about a(B)4-iB does not
lie entirely in W. Hence such a circle must intersect

the free boundary somewhere other than on F

1°

t-plane

 

 

 

 

Figure 5.3

Since B is a simple zero of t44-2At24-B, it

(t-B)2

; .

3

follows a”(t) = —

4IN

—' I)

aj(t--B)j near t = B.

O

 

97

Integrating twice we get a(t) = a(B)+—d:l(t-B)-

& .
2(t--B)2 27 d{(t-B)j, where o’l, of are real, and
Tr j=o j - j

L(t) = L(a)+ (a;1+i)(t-a) -%<t-B)2 Z a’.<t-B)j.
1 j=o J
for t near B° Putting (t--B)2 = g and z = L(t)

we get for an analytic U

2

z-L(B) = §2[(a_’l+i)—%§3u(§ )1,

u(o) = as > O, G:1+'i # O .

1
Then (z-—L(B))2 = g(C04-C3§3+O(§4)), where CO # 0,

C3 # O, and inverting this function gives

H
(.A.’

5 = (z-L(B))2[b0+b3(Z-L(B))2+0((Z-L(B))2)],
where bO # 0, b3 # 0. Hence

(t-B)

U.)

= <2-L<meg+ 2b3bO(Z-L(B))2+0((z—L(B))2)].
(*)

This shows upon calculating t2 = (L—l(z))2 that

(L‘l<z>)2 = 132+ 23b§<z-L<e)> +b§<z —L(e)>2

rolm

. 7
+ 4BbOb3(z—L(B)) +o<<z —L(B))2)

 

98

-l)2 must be continuous on W (by Theorem 3.11),

Since (L
and since BbO b3 # O, the formula above shows that no
closed path (e.g. circle) about L(B), close enough

to this point, can be entirely in W. Such a closed path
therefore must intersect the free boundary somewhere other

than on F1. See Figure 5.3. If we evaluate L at

y > B, y1 = B4—(y-—B)e2Wl, which are in two different
Riemann sheets, then L(yl)-L(y) = $(y-—B)2 Z) d3(y-—B)J;>O
j=0

for (y-B) small enough, which shows that L is
locally one-to—one from a cut neighborhood of B (the
interval (B,y) deleted) into a cut neighborhood of

L(B) (a cusp deleted) as shown in Figure 5.4. In

 

TVs

7”

t—plane

 

Figure 5.4

 

99

fact if y moves to B, then the image of y moves

to L(B) on T and the image of y1 moves to L(B)

l

on some path P The paths T1 and T' are tangent

I
1' l
—1 2 _ 2
at L(B) by (*). Note [L (2*)] — (Im z*) for a
point on the free boundary. Thus L-l(z*) = y* near

B here.

If there were a segment with left endpoint B

(on the real axis) mapped onto a new part of the free

I] ll

boundary passing through L(B), then L = a would
have to be real there which is not true. Thus Fl,
Ti establish the boundary of W1 near L(B)° Ti is

the new part of the free boundary, and it is easy to show
directly u-¢ 2 O in the cusp deleted neighborhood of
L(B) with equality occuring only on Tl, Ti (one can
also refer here to the Kinderlehrer and Nirenberg cusp
result [10]). This shows a(B) = 0 cannot happen by
symmetry with respect to the y—axis if we argue as

in Figure 5.2. Then we can extend the local solution

near Ti also as we did for T1.

Second Step. In this step we want to end up with
a contradiction which proves the lemma. Note this
Lemma involves two cases according to whether there
is another root, say a, O < a < B, of t44-2At24-B
or not. Because of these singularities we may consider

,—

cuts along the x-axis as in Figure 3.5. In the upper

 

lOO

 

Figure 5.5

figure there is another simple root d, where in the
lower one there is not any. Consider the analytic
continuation in the upper half plane of the branch of

f we were using for definition of T1 on the right hand

side of B. Note
t
a(t) = a(B)+—j a’(¢)d¢
B
t t
= a(B)+-[Ta'(T)] -f Tf(T)dT
B B
t
= a(B) +a'(B)(t-B) + f (t- g)f(g)dg
B
Since the zeros of L’ are isolated and

L’(t) = a’(B)+ i+‘F f(;)d§ # O for —B S t‘g B, thus

we can join B to d (or B to -G, or B to -B

 

101

in the second case in Figure 5.5) by a path y staying

away from the zeros of L' with

Im y(s) > O for O < s < l, y(0) = B.

y(1) = a (or y(1) = -a. or y(1) = —B) ,

so that L is invertible locally in a neighborhood of
any point y(s), O < s < l, with analytic inverse L-l.
So we have an open connected region containing y((O,l))
in the upper half plane in which L is locally invertible.
L is continuous on the closure of this region.

The part of y near B is mapped by L in W1.
Thus either there is a point on y, y(g), O < 9 < l,
which is mapped by L to a point on the free boundary
or all points y(s), O < s < l are mapped in W (assume
the image stays inside 0 in this case, we will show

A
this in the third step). If there is such a point y(s),

then

[L_l(L(Y(Q)))]2 = [Im L(Y(g))]2;
L-1(L(y(g))) = mg) ,
which is not true no matter which one of pairs [B,d},
{B,-o}, {B,—B) is joined by v. Therefore there is no
such a y(g), and all y(s), O < s < l, are mapped in

W. At the end of this step we will show that L(G)

 

102

(or L(—d), or L(-B) cannot be mapped inside W and
they must be mapped into the free boundary by L. Assume

this now. Hence

-1 1 a
a = L (L(a)) = iIm L(a) = i[c1+—if(C1-Q)f(§)d§
'13
(note a(B), OMB) > O) *
-a = L_l(L(—G)) = :tIm L(—d)
1 +0 '
= 1pm.: fa (_a_g)f(g)dg] *
-B = film-13)) = iIm L(—B)
= 1 [43»:— f (—B-g>f(g)dgl
B
(there is no a in this case) **

Positive sign case in any one of these leads to a

I

contradiction immediately (in *, * leads to o = B and

in ** leads to —B = B) because f has fixed sign

there. In *, *’ negative sign case implies
1 .B . 1 B
2a = 33 (d—g)f(g)dg, +2a = Ti (+a+g)f(g)dg,
d a
l . .
respectively, where I f(g) is negative. These two

results cannot be true simultaneously. Hence the case of
4 2

having two roots 0 < d < B for t +—2At 4—B is ruled out.

Now assume this polynomial does not have any zero in (O,B).

illnllllll

 

103

Then ** implies = — If B(B4—g f(g)d§ which is not

again possible because i f(g) g 0. This contradiction
proves the Lemma if we can resolve the two questions which
are left open. We conclude this step by answering the

second question:
Claim: L(a) gin L(-O() £177. L(—B) e 17v.

Proof of the claim: We only prove the first one

 

and one can argue similarly for the others. Suppose

L(a) fi W, then (L‘l)2 has to be analytic at L(a)

(note (L_l)2 here is defined in the noncontact set) by

results of Chapter 3. To check this observe that

l
2 2 2 2 —
L”(t) = ii: ‘5 )‘t/gc‘ ) ¢<t2) = -%<t-a)2gl(t),

v 2
./2o(a2_g2
and g(a) = —fi—¢(a2 ).

(t +K)3

where g1 is analytic near a

(q2+ K)3 2
2
Thus 2 = L(t) = A4—B(t-—c)-%(t-d)2g2(t), where
g2(o) = {L91 (d) and g2 is analytic at d. If B # O,
Q
then z—A = L(t) —L(G) = B(t—o)[l—%(t—d)2g3(t)].
1
Putting (t-—o)2 = g, taking square root from both

sides, and applying the Inverse mapping theorem we can
1 1
find (t--o)2 in terms of (z —A)2. Squaring both
3

 

(z-A) 2 z-A 2 ..
B (———4- +

. +_ _ = 1 _
Sides then we ge- t a [r4-T BO B

 

 

(higher order terms)] = B‘+ + '°°, where

 

104

)+ooo]

NW

2

50510. Then t = (171(2) 2

 

 

)2 z-A4_% B (z—A

=[G+B MOB

will not be analytic at L(a) which is a contradiction

to L(a) E Q. If B = 0, then (z-—A)5 = (t-a)g4(t),

where g4 is analytic at a with g4(d) # 0. Again

by applying the Inverse Function Theorem as above we get
.2 _

t-c= (z-A)5[YO+Y1(2-A)5+'°']. where YO#0- This

again shows that t2 = [L_l(z)]2 is not analytic at

A = L(a). This contradicts L(a) e w again. SO L(a) E %.

Third step. To conclude the proof of the Lemma we
only need to show that the image of the path y in the
above three situations (y joining pairs {B,d], {3,—a},
and {fi,-B}) remains inside 03 This was the first
question left open in the second step° To do this we
first show that Im L(y) g B, Re L(y) g a(B) if we let
y be the limit case of the paths (chosen before)

coincided with the real axis as shown in Figure 5.6.

m

Figure 596

 

105

t
Clearly Im L(t) = t+—Im f (t-g)f(§)d§ < t,
B
for t < B, for all three pairs and Re L(t) = a(B)+

G’(B)(t-B) < a(B). for the pairs {Ba}. {fi.—B}. and

t
Re L(t) = a(BHa’mHt-BHf (t-g)f(g>dg < ans). for
d
the pair {6,-a], where t < B in any case above. So
L(y) remains to the left of X = a(B) and beneath

y = B in any case.

Let us now show that L(y) does not intersect

the x—axis. Suppose it does, say at L(Y(Q)). Then by
Theorem 3.11 and the fact that (% u-¢)| = -2(L‘l(x))2210
x
_ /\
on J, we must have L l[L(y(s))] imaginary. This

cannot happen obviously if y is joining the pair [B.G}.
To show it cannot occur in the other two cases we do as

_ A
follows. If L l[L(y(s))] is imaginary we can assume

the path y intersects the imaginary axis only at 0.

See Figure 5.7. Then L(O) = a(B)-—Ba’(B)-—j g f(g)
3

Figure 5.7

 

 

106

which is supposed to be real. But this is not real if
Y is joining any one of the pairs [B,-o}, {B,-B}
shown in Figure 5.7. This contradiction results that

L(y) does not intersect x-axis in any case.

Now in order to show that L(y) remains inside
0, it is enough to show that it does not intersect
y-axis° Suppose it does. Then by symmetry with respect

Bu 2
B

to y~axis we have —;(O,y) = —2 Im(L_l(iy)) = 0

(on y-axis) by Proposition 3.12. This shows then

L_l(iy) is either real or imaginary. For the pair [B,o}
we can take the path y in such a way that it does

not intersect the X-axis except at endpoints. This

rules out the case {B,a]. For the other pairs {B,—a},

{B,—B] we can assume y intersects X—axis and y—axis

only at O as in Figure 5.7. Then L-l(iy) = O, and
3

therefore L(O) = a(B)-—Ba'(5)4-f Q f(Q) = iy, for
O

y > O. The latter result is not true because % f(g) < O
and a(B)-—Ba'(B) is real. Thus L(y) cannot intersect
y—axis either. Hereby we showed L(y) remains inside
0 and in the first quadrant. Here are the third step
and the proof of the Lemma both complete. é

u

Lemma 5.4 proved that t44-2At2+-B cannot have a

simple root 0 < B < yo. Thus it has either a double
root o, O < o < YC’ or no root in (O,yo). Next Lemma

shows that the latter case cannot happen either.

 

107

Lemma 5.5. The case t44—2At24-B > O, for

O < t2 < yg, cannot happen.

Y
Proof: Since a’(y)-a'(yo) = j f(§)d§ > O for
yo
0 g y < yo, thus a'(y) is increasing to the left of

y0 but it is always finite. Moreover a defined by

y
a(y) = a(yo)+a’(yo)(y-y0)+f (y-§)f(§)d§ ,
yo

for O S y < yO

is finite, decreasing as y moves to the left of yo.

Therefore F intersects x—axis transversally (or

1
y-axis non-horizontally). This provides a contradiction
as follows. The local solution built up near a(yo)4-iyo,

with T1 as a piece of free boundary, is continued to

the left with noncontact set over Fl. By symmetry

  

WWW
‘Q/ WL/fl/
Jb/Wqé%%@%z

O

 

I
..//,

I

[7/ ,
// ”7/7

 

/' WW a( )_ i
”WW yo Yo

Figure 5.8

 

108

with respect to x-axis (or y-axis) we must have u < w
on the part of T1 below X-axis (or u > w on the
right hand side of y—axis) as shown in Figure 5.8.

On the other hand we must have u = W there. Here is

the proof complete. #

Corollary 5.6. The only possible case for f is

2
2 a -t 2 2 2
f(t) = — (t ), where O < d < y . Moreover
W (t24-K)3/2 ¢ 0
yo
a'(O) = o, a(O) > o, a’(y ) = j f(g)dg.
0 0

Proof: The first part is an immediate conclusion
of Lemmas 5.4, 5.5. If a’(O) # 0 (note a’(O) is
finite), then we get a contradiction as in Figure 5.8 in

Y
Lemma 5.5. This result yields a'(yo) = I O f(§)dg. We
0

can now build up the local solution near a(O) with the

noncontact set to the left of F as in Section 2.7°

1

If a(O) < 0, since a'(y) = j f(g)dg shows that the
o

tangent to the curve Fl never becomes horizontal,

thus at the intersection point of F1 with y-axis we
get a contradiction, by symmetry with respect to y-axis,
similar to the one we obtained at a(O) in Figure 5.8.
See Figure 5.9. Hence a(O) 2 0. If a(O) = 0, then
similar argument leads to a contradiction. Nearby the

origin to the left of we have u > h: -on the other

i1

hand on the part of the free boundary symmetric to T1

 

109

  
  
 

// Y \ 1‘
\€%;/ 1
We) 1
/ .
-a(yo) + iyo WWW/W J a(Yo) + iyo
{é

 

Figure 5.9

with respect to y—axis we must have u = w. This
contradiction shows a(O) # 0. Hence we conclude

a(O) > O and the proof is complete. #

2%”

-a(yO )+iyO

 

 

Figure 5.10

llO

Remark 5.7. We can build up the local solution near
a(O) with the noncontact set on the left hand side of
T1 in the first and forth quadrants as in Section 2,7.
Then T1 :2 = L(t) = a(t)+-it, t near zero, is

symmetric with respect to x-axis, and

z

u(z) = -% Im f [g24-4(L_l(g))2]dg, for 2 near
a(0)

a(O) to the left of T1, where L_1 is a conformal

map in a uniform neighborhood of T1 near a(O), and

% u-—w = —2(L-1(X))2 2 O on the real axis to the left

of a(O), where is contained in J. #

Remark 5.8. Since u is a local solution to the
problem, thus by Theorem 3.11 the map (L-l)2 must have
analytic continuation to the left of a(O) on the real
axis and it must be negative—valued until it becomes
zero again (by symmetry with respect to y—axis) and we

get a new point of the free boundary on the X-axis.

Since (L-l)2 # 0 there, thus (L—l)2 has analytic
continuation if and only if L"1 has, as long as (L-l)2
is nonzero. 4

M

For the time being we want to check the necessary
conditions for L (obtained already) to fulfill these
requirements, so that by these conditions we can conversely
construct a local solution near a(O) continuable to

the left to get the local solution u eventuallyo

 

lll

Lemma 509° L"1 in 5.7 does have analytic continuation

to the left of a(O) on the real axis at least on
(L(iJE),a(O)) and (L_l)2 is negative-valued. L(iji)
is the first critical point of (L_l)2 to the left of
a(O). Moreover, the value of (L-l)2 is (i/E)2 = -K.
Further % L_l(z) is indeed strictly increasing as

2 < a(O) decreases.

t
Proof: Note a’(t) = j f(T)dT,
O
t
a(t) = a(o)+f (t-T)f(T)dT, L(t) = a(t)+it,
O
L'(O) = i # O and L is conformal in a neighborhood of

0 taking the real axis to Tl and the imaginary axis
to the real axis to the left of a(O) by z = L(t) = a(t)+-it
as shown in Figure 5.11 (one can check the formulas above

to see there).

t-plane

 

Figure 5.11

 

112

Furthermore L’(t) = i+ l f(g)dg shows that

l

i L’(iO) > O for O < C < J? and it becomes unbounded

as o 4 (JR)-. Hence L is real and strictly decreasing

on [O,i/R], which can be seen from
t

L(t) = a(O)+—f (t-g)f(g)dg+—it also. Applying Lemma
0

5.2 for L on a sequence of intervals [O,ib/R-en)],
where an 4 O, we get L a conformal mapping on a
neighborhood of each [O,i(Jfi-en)] not containing

iJE. Hence L"1 is a conformal mapping in a neighborhood

of (L(ijfi),a(0)] not containing L(L/R). Moreover

L-l takes this part of the real axis to the upper

part of the imaginary axis and imlL—l is strictly

increasing to the left a(O) on the closed interval

[L(iJE‘,a(O)] (because of having similar property for L)°

A _
Let x < a(O) be the first critical point of (L l)2
to the left of a(O). Since éi (L_1(X))2lA =
X
-l d -l d -l
2 L (x) d; L (x))A = O a a; L (X)\A = 0, thus
x x
_ A

L’(L 1(x)) is infinity and by definition of L’ we
_ /\ ._ ,_ A
conclude that L 1(X) = iJK. Thus L(iJK) = x. #

Definition 5.10. For the singularities t iJE,
we assume the imaginary axis in t-plane is cut from i]?
to m (upward) and from —ij? to m (downward). Define
m(t), for t near zero, in the lower Riemann sheet as

analytic continuation of L(t) along the paths of the

 

113

t
form a, where L(t) = a(t)+it = a(0)+i (t—g)f(g)dg+it.
- 0

Note a'(LfR) is not defined (infinity) but a(ifi) is
defined as a finite number. Also m(t) = b(t)4—it, where
b is the analytic continuation of a along the paths

of type 0 (see Figure 5.12). #

 

l t—plane

Figure 5.12

Remark 5.11. Recall that during move from a(O)
to the left, we are looking for necessary conditions for
having local solution u so that they are going to be
sufficient at least for constructing a local solution
near a(O) continuable to the left. At this stage in
order to have a local solution u it is necessary to

2

have (L_l) negative-valued on the real axis near

_ _1
L(L/K) and L ‘ must be analytic in a full neighborhood

 

114

of this point. Let us now check these requirements

for our map L_l.

Lemma 5.12. L‘1 is analytic in a full neighborhood
of L(ifio with (L'1)'(L(ifi<)) = o. L"1 is imaginary-
valued on the real axis part of the neighborhood and
3— L-1
l
the left of L(ifi), L‘

attains a local maximum j? at L(ijE). To

1 . . . . .
is invertible w1th inverse

m as defined in 5.10.

Proof: In a sufficiently small neighborhood of

ij? we have

3

2 2 ——— m

2 _

L”(t) = "777% ¢(t2) = (t—iwll—O 2
(t +-K)

A0 # 0, t near iji ,

i an .
and L(t)—L(ifii) =A_'l(t-i/E)+(t_t/E)2 '2 Ajf(t_iJE)3,

3:0 1
A6 # 0 (by integrating twice). Putting (t-iJE)2 = g,
z = L(t) we get z-L(iN/Iz) = A11g2+ E A]! §2j+l = §g(§).

3:0

where g(O) # 0 and g is analytic near zero. By
invoking the Inverse Function Theorem we get
i = n(z-—L(iV?)), where 9(T) is analytic near zero
with 5(0) = 0, SIM” 7’0. Hence t—zufik= 92(z-L(iJE)),
and t = L_l(z) = iAth92(z-L(i/EO) is analytic in
a full neighborhood of L(iJE). Since 9(0) = 0,

\

a’(O) i 0, thus the latter equation of L_l(z) shows

 

115

(L_l)'(L(iJ§)) = 0. Since L-1 is imaginary—valued

to the right of L(ijfi), thus in the expansion of L-1
in power series of (z —L(iJ§)) all coefficients are
imaginary and L_1 is imaginary—valued to the left of
L(iji) also. By looking at the formula for L"1 we
see that if z-L(ij§) changes its argument by N, then
t -i/§ will change its argument by almost 2w if
|t-iji| is small enough. In particular by starting
from 21 > L(ifiE) and ending with 22 < L(ifi)

(see Figure 5.13), t-i/E will change its argument

by exactly 2? by the above argument, that is,

t = L_l(z) starts from t1 = L-l(zl) on the imaginary
axis and ends up with t2 = L_l(zz) on the imaginary
axis again. Note tl is in the upper Riemann sheet

and t2 in the lower one. We observe that -% L."l
attains a local maximum J? at L(iji). The last

part of the Lemma is clear.

Figure 5.13

 

116

Lemma 5.13. % L_1 is strictly decreasing as
2 < L(iJE) decreases, with inverse m, and it vanishes
at m(O) = b(O) with (—i L'l)’(b(o)) a! o. b defined

in 5.10 has the equation b(t) = ui—Xt-—a(t), where

X = f f(g)dg is pure imaginary, u is real, and
C

l
I X4—l S 0.
Proof: If (% L_l)' vanishes at some point
E e (b(O),L(ijE)), then m’(t) = Li—i-—a’(t) becomes

infinity at some point i0, 0 < O < JR. This result
is not true because m’ becomes infinity only at i L/K.
l —l . . . .

Hence i L is strictly decreaSing as z < L(iJ?)
decreases with inverse m, strictly decreasing as
t falls below id? on the imaginary axis. By symmetry
with respect to y—axis L_l must vanish at some point
and it cannot have minimum before that. This implies

. . . . .f— . l —l
m is strictly increaSing on [O,lVK]. Again i L
cannot vanish at point m(0) with derivative zero
there by m’(0) being finite. To show the formula for
b observe that m” = b” is also an analytic continuation

II

of L” = a along the paths of type 6 (referring

to Definition 5.10). Since along 0 the argument of
3

(t-—i¢§)2 is changed by 3w, thus b"(t) = —a”(t)
for t near 0. By integrating twice we get

b(t) = u-tlt-—a(t) and m(t) = ui—(ki—i)t-ea(t), where

117

x = b'(O) =J‘ f(g)dg. m(O) = L(“21(0) and a(0) are
’a

real, therefore u is real. Since m is real-valued
on [O,iji], and a(t), u are real there, thus A
must be imaginary there. Moreover m being invertible

on this interval implies
l
i x+-1 g 0

We also have the following relations

m(ifi) - L(ifi) = b(ifi) = aw?) = kiffi

= 2am?) -u = km?

i K _
2a(o>+2j (iv'K-§)f(§)dC.-u
O

and b(O) = a(O) _f gf(g)d<;.
0,

Corollary 5.13. % u-w attains its absolute
maximum (2k) on [b(O),a(O)] at L(iji), whereas
it vanishes at a(O), b(O) and strictly monotone

on each side of L(iji).

Remark 5.14. By symmetry with respect to y—axis
we must have —a(0) g b(O) < a(O), and this condition
will be fulfilled if u (the constant of integration)
ischosen real and appropriately. We have thus far

imposed the following conditions also: k = j f(g)dg

ll8
iV _

imaginary, u = —>.iJIZ+2a(o)+2 J" (iV’K—g)f(g)dg,
O

and %(X4-i) S 0. In the next Lemma we show that

equality case is not necessary here.

Lemma 5.15. The point m(O) = b(O) = u-a(0)

belongs to the free boundary and this implies the

necessary condition k+—i # O, i.e., %(l+—i) < 0.
. l -l 2 .
Proof: Since y u-m = —2(L ) on J, vanishes
at m(O), thus m(O) E (v-J) U s, the free boundary.
To show l+—i # 0, we check v(% u-w) = 0 at
m(O) = b(O). Suppose x+i = 0. Then m(t) —u+a(0) =
2 4' ll .
-llt -k2t -..., where kl — a (O) > O, i.e.,
x2 3 2
z—u+a(0) = —)\ s(l+——- s+— s + --:) if we put
1 Al kl
2 . dz _
t = 5. Since a; I — —xl # 0, thus by the Inverse
mapping theorem we have -llt2 = (z-u4-a(0))+
k 2 2
—2- (z—u+a(o)) +0((z—u+a(0)) )= e(z-u+a(0)).
)‘1
where e is an analytic function with 9(0) = 0,
a'(O) # 0. We discover that (L_l)2 is analytic at
1.1—am) but [(L“l(x))2]’\ #0. Thus
u-a(0)

v(% u-$)(m(0)) # 0, which violates a necessary condition
of Theorem 3.11. Hence k—ti = 0 is impossible. #

 

119

Lemma 5.16. Either b(O) is an isolated point of

the free boundary or 1 = -2i.
Proof: We discovered so far that m(t)-(u-a(0)) =
z-b(O) = (1+-i)t-xlt2-12t4---- for t near zero,

where 1+ i # 0 by Lemma 5.15. This means that the
inverse of the function m exists and is analytic in a
full neighborhood of b(O). The inverse, which is the

analytic continuation of L'l, can still be called L_l:

1 1
_Z_:_b_(_0_l. [1+Tl— t+__2._ t3+ 090

 

 

t: >.+i +i 1+i
A 1
l 2 3 son 2 co.
+(>.—+i‘t+>.+it+ )+ 3'
i.e.,
1 1 t z—z
_ -l _ z-b(0) 1 z—bgo) l 0 ...
t _ L (2) _ xi-i [14'x+i ( li—i +x+i 1+1 + )
2
+ *1 (3:09....
(k+i)§ 1+1

where the rest of the terms are of order greater than three.

 

Hence
2 2 3
_ —l _ z-b(0) 11(z—b(0)) 211(z—b(0))
t—L(z)—>\+i+ 3 + 5
(k+-i) (Xi-i)
Let's investigate % u(z)-—m(z) near b(O) = m(O).
For every 2 near b(0) in W we have
1 222 r2 -1.2,
V u(z)-—w(z) = 3 y -— Im d [L (s)] ds ,
_ y a(o)

120

where the integration is along a path in W U {a(0)]
-l 2

from a(O) to 2. Since (L ) is real valued on
Z
[b(O),a(O)], thus l u(z)-—o<z) = 3 y2-3-Im j [L‘l(g)12dg.
Y 3 Y b(o)

where the path of integration is in W U [b(O)] (remember
W = W U [J-(v-J)]). Let's now consider a disk

neighborhood v of b(O) in which L”1

(z) is analytic
and can be expanned in power series of z-b(0). In the
connected component C of open set V 0 W, which
contains V D [b(O),a(O)], we can join every z to a
point in V n [b(O),a(O)] by a path entirely inside the
component. Thus every point z in the component can be

joined to b(O) by a path in C U [b(O)] C V. Since

(L‘l)2 is analytic in V, thus we can redefine
1 2 2 2 z -1 2
—u(z)-o<z) =—3- y -— Im f [L (QH dg,
Y Y b(O)

integration on the ray path. Now

 

3
Z - Z _ 2 21 (g-bmn
I [L 1<g>12dg=§ 1‘g b(°))2+ l 4 mm
b(O) b(O) (1+ 1) (x.+i)
=1(_z_-.1-;12>>_3+il (z—b(o>)4+u
3 (>~+i)2 2 (7\.+i)
and
Z
In“? (L‘l(;))2dg= l 2[(x_h(0))2.—3- Y3]
“20 (1+1)
1
+-+2[4(x-b(0))3y
2(x+i)

_ 4(x—b(0))y3]+ m,

 

121

where the higher order terms are ignored. Note all terms
on the right hand side of the latter equation have y
factor because we've taken the imaginary part of the right

hand side of the former equation. Thus for z E C we have

Z
--2- my (L‘l<g)>2d;
y Z
0
—2 2 1 2
=————'HX-MOH -*Y)
(>.+i)2 3
)‘1 3 2
-—4 [4(x-b<0)) —4(X-b(O))y 1+m.
(1+i)
1 —2 2 2 2 1
and — u(z) -o(z) = —— (x-b(0)) +— y (1+————)
y (X+i)2 3 (>.+i)2
411 3 2
- ‘——*_Z [(X-b(0)) —(x-—b(0))y ]+—"' , where the higher
(X+i)

order terms are ignored. The right hand sides are real

analytic.

Let‘s now define w(x,y) as a function of x with

y as a fixed parameter as follows:

(u(x,y) = —‘—27 (x-b<0))2+§y2(1+——l——2>
(1+i) (1+i)
At this point we are going to show l+~——£——§_2 0. Suppose
(Mi)
not. Then
2 2 —2 2
:h—§y(1+l—2)= ———2- (x-b<0)> .
(1+i) (k+i) -

for fixed small \yi # 0

 

122

This shows a parabola with negative minimum % y2(l+-—JL—3)

(1+i)

which intersects the horizontal x-axis transversally

(look at Figure 5.14). Since m(x,y) has the leading

Figure 5.14

terms of % u(x,y)-—w(x,y) near b(O) and
52 1
—§ (— U(x,y) —cp(x,y))l > O ,
ax y (b(O),O)

bifurcation theory in this case [4] says that

% u(x,y)-¢(x,y) behaves similarly as w(x,y) does, in
the following sense, for small fixed perturbation of y
from zero. % u(x,y)-—¢(x,y) as a function of X with
parameter y, for any small fixed ly) # 0, intersects
the xy-plane transversally and it has a negative minimum.
But we know that if % u(x,y)-—¢(X,y) vanishes for any
fixed small perturbation of y from zero, then it must

have even multiple root with reSpect to X, i.e.,

% u(x,y)-—;(x,y) must touch the horizontal plane

 

123

tangentially, not transversally. Hence 1+ 1 2 2 O,
(l+i)
. l . .-l 2 . .
i.e., l -—:1____2 , that is, (i 14-1) ‘2 l which is
(i 1+1)

equivalent to: i-lk-tl 2 l or i_lk-+1 g -1. We already

knew that i-ll < —l (by Lemma 5.15), thus the case
i-ll 2 0 can't happen. Therefore 1‘11 3 -2.

If -; l < -2, i.e. 1+-——L——- > 0, then we'll have
i . 2
(1+1)
an isolated point b(O) of the free boundary because

% u(z)-m(z) > O for z E V-[b(0)], where V is a

small enough neighborhood of b(O). We'll show that this

case cannot happen either. Finally if l+-——L—-§ = 0,
_ (1+i)
then i 11 = -2 will be the only possible case which

works.

Lemma 5.17. The case % 1 < -2, which says b(0)

is an isolated point of the free boundary, can't happen.

Figure 5.15

Proof: Suppose it can. Then (L—l)2 has to have

analytic continuation along the real axis to the left of

 

124

b(O) and it has to be real—valued on the real axis. Again

%u(z) —cp(Z) = -2[L_l(z)]2 > O '

for z = x < b(O) near b(O) .

Since l+—i # 0, thus L_l exists and it is analytic
in a neighborhood of b(O) and is continued analytically

2

to the left of b(O) because (L-l) is nonzero there.

Since L-1 is pure imaginary—valued on the real axis to
the left of b(O), thus L-l has the image on the lower
half of the pure imaginary axis because if it had the
image on the upper half of the pure imaginary axis, then
m would be doublevalued on the upper part of the pure
imaginary axis near zero which is not true because m is
one—to—one near 0. Thus % L_l(x) < 0 for x < b(O).

But then m(t) = ui—(li—i)t-—a(t) implies

m'(t) = (l+2i)—a’(t)—i = i(i'1x+2)—i(% a’(t)+l) ,

which shows %-m'(t) < 0 for % t < 0 and It)
sufficiently small, while t 4 —id§ on the pure imaginary
axis implies % m’(t) a 3. Thus m'(t) must vanish some—
» where between 0 and —iJE on the pure—imaginary axis.
But this implies, then, m is not invertible there which
is not true because If:L exists and is analytic on the

u '/_ n - _12 1 ~' .' '
segment [-i.K,0] and (L ) takes its minimum not

before -iJK. Hence we get contradiction and we conclude

 

125

that i-ll < -2 cannot happen. Here is the proof

complete. #

Our investigation and search on the real axis is still
going on in O as long as we haven't reached to the part
of the free boundary on the left symmetric to the earlier

one with respect to y-axis. At this stage we realize

that i_ll = —2 is the only possible case left. Then
1 = —2i and the midpoint of
L(t) = a(t)+-it, m(t) = u-—a(t)-it,

for t = i0, J? > O > 0

is %(L(t)+-m(t)) = % independent of G. In particular

as we've already also seen b(O) = u-—a(0), i.e.
% = %(a(0)+-b(0)). By the next theorem and symmetry with
respect to y—axis we'll discover the midpoint % = 0.

Theorem 5.18. There is a new part T2 of the free

boundary passing through b(O) which is the mirror

reflection of the old one (passing through a(0)) with
respect to the vertical line x = g = L(ijE). Thereby
u = O and b(O) = —a(O).

Proof: Since 1 = —2i, thus m(t) = n-—a(t)-—it.

This formula, for t real, determines a curve passing
through m(O) = b(O) = u-—a(0) (as a vertex which we
already know belongs to the free boundary) concave to the

left and symmetric with reSpect to X—axis because a is even.

 

126

m which was the analytic continuation of L along
the paths of type 0 determines the upper (lower) half of
the new curve for negative (positive) t, and clearly for

Itl small enough but % t > 0 we have

m(t) > u-—a(0) = b(0) = m(O) .

  

 

t—plane
x/‘>‘/(
;/\l:'/‘<
. M'fi-XX x mm o

__
a
v

I

I
1“

\l‘ \l‘ \/ \l
\ Ix 1‘ /\ z

......

Y
y i
#UY)
Xxxxk.xxxx

I

Figure 5.16

 

   

Let's try to show that the new curve is the left part

of the free boundary. We have

m’(t) = —a’(t)-—i # 0, m is one-to-one,

for t real

 

Thus by Lemma 5.2 there is a strip neighborhood of any
segment of the real axis, centered at O, in which m

is a conformal mapping taking the upper (lower) part of
the neighborhood to the right (left) hand side of the new
curve. The inverse of m is L_l. Along the new

curve (the image of the real-axis under m) we have

-l(z) = —Im z = -y. For 2 on the new curve and taking

the integration path on this curve we get

[L‘1(z)12-y2 = o a v6 I1(2) -cp(Z)) = o,
i u(z) —o(z) = g y2—3 Im drz [L‘l(g)]2dg
Y Y b(O)
2 2 2 z 2
=—y——Im (Img)dz=O,
3 Y jb(0)

where the first equality is shown by computations in

the proof of Theorem 3.9. Moreover

2 1
(-u—co) _ 2 _
2 = -—252(u—'-h-)=-l-—-L92—M=AA(u—¢)=4>0
an an y an y

on the new curve. The latter three results about the new
curve passing through b(O) = m(O) imply that this new
curve is part of the free boundary. This new free boundary
is symmetric to the earlier one with respect to the
vertical line X = %. Since we have symmetry with respect
to both axis, thus % = O, i.e., v = O, B(O) = —a(0),

u

and m(t) = —a(t)-it. Moreover

 

128

L(ifi) = mm) = a(ifi) + mm?)
= (ham/E) -i(i./—IE) => (1 = 2mm?) + 2i(ifi)

= L(iffi) = a(ifi) +i(L/—IE) = % = 0

which was expected by the symmetry with respect to y-axis.

Hence the result. #

Remark 5.19. If we started by considering a piece
of arc on the left (new) free boundary, which specifically
starts with the branch of a” negative at O;
m(t) = —a(t)-—it, and assumed the noncontact set on the
right, then we could make a similar study and by analytic
continuation of m—1 we could reach to the right free
boundary: L(t) = a(t)+-it. In both cases cross signs (x)
are mapped into cross signs and dot signs are mapped into
dot signs in Figure 5.16. The signed area in the z—plane
are contained in W. We've already seen that Q44-2Ag24-B

cannot have a simple root i0, 0 g 0 3 ji, and it can't

have a double root at O, iJE.

Remark 5.20. (a list of necessary conditions for
Problem 5.1). So far we have derived some necessary

conditions for problem 5.1 as follows:
2 2
d — 2 2
————————- ¢(t ), where A = —d , B = d
(t2+K)3/2

HM

 

129

(iii) f f(g)d; = —2i,
0
(iv) a(O) = 3 f gf(g)dg > o
G

Y
(v) a(O) = a(yO)-yo a’(yo>+-(OO cf<g)
(this is obtained from a(yo) = a(0)+

yo
j (yO-g>f<g>dg)
0

Note using (iii) we can show that (iv) is equivalent to

ifi
a0)=.fi—f (Mi-QH(OdL
0

We realized that there is only one constant of
integration arbitrary and the other one is imposed by the
symmetry (e.g. by a’(0) = 0). Recall that in the
unperturbed problem case also, given 00, the local
solution was not unique by depending on one parameter c

as stated in the Remark following Definition 2.8.1. #

Lemma 5.21. Except perhaps near the semicircle the
mapping L is conformal in a semi—circular domain T
bounded by v consisting of segments [O,yo], [O,iJE]
in the upper Riemann sheet, segments [-yO,0], [O,ijfi] in
Ithe lower one, and the semi—circle centered at O (in both

sheets) and radius yO started from yO in the upper

Riemann sheet and ended with -yO in the lower one. The
image will be bounded from below by [—a(0),a(0)], Fl,
12.

 

130

 

 

 

 

Figure 5.17

Proof: The mapping L is analytic in T, continuous

up to the boundary. L takes the lower part of the boundary

of T to [—a(0),a(0)] and (T1 U T2) 0 [(x,y) :y > 0}.
Recall that by assumptions of problem 5.1, a(yo), a’(yo)
are near x0, é§(H(y)-—% y log y)| (in unperturbed
Y
0

problem) respectively. For any t on the semicircle we

have

for unperturbed case:

t
_ - _ I r _, —
Lu(t) — it+-au(yo)+-(t yo)au(yO)-+J (t :)fo(g)dt
Y0
= it+x + (t—y )[i(H(t) -—2— t 1og't)]
o 0 dt‘ 1‘ - _
t—Y
<1 2> O
t (S
-%7 (t—;) 0, dc
1 V. :2

 

131

for perturbed case:

t
L(t) = it+a(yO)+ (t—yo)a’(yo)+f (t—g)f(g)dg
Yo

iti—a(yo)+-(t-yo)a'(yo)

t

+% (y (t—g) 7—953 @(gzmc.

0
where the path of integration can be chosen on y starting
from 0 on the positive direction. Since K, o2 are
small, in comparison with y0 = |t| = lg\, thus the
integrands are in both cases uniformly close; and therefore
Lu(t), L(t) are also uniformly close, for all t on the
semicircle. But the image of the semicircle is a simple
path in the unperturbed case as we showed in Example 2.5
(Figure 2.3) . Hence the image of the semicircle under

L is uniformly close to the above simple path.

Now consider the topological degree (or winding number)

d(L,T,Z) for any Z f L(Y). This is zero in the unbounded
component of ¢-—L(y) and it is one in the major bounded
component of ¢-—L(y) which is bounded by F1’ T2,

[-a(0),a(0)] from below (note there may be small bounded

components bounded only by the image of the semicircle

under the map L). Hence L is a conformal mapping in
the preimage T0 of the above major bounded component of
¢-—L(v). (One can show directly, as in Example 2.5,

that L(v) is a simple curve here also). #

 

132

We can argue similarly about the other semi-circular

domain which is mapped below T [—a(0),a(O)], F by

l’ 2
the analytic continuation of L. Hence we have the

following:

Theorem 5.22. L"1 is a conformal mapping in a
neighborhood of the origin, more Specifically in

L(TO) U [—a(0),a(0)] U L(TO) U T U T2, where by bar

1
here we mean the complex conjugate. We can assume the
disk 0 is inside this neighborhood. Moreover the free

boundary consists of T1 U T2 only.

Proof: The first part is clear from the above
argument. The last part is simply because L_l(z) = i Im 2
can happen only on T

l
l U T2. 7

Sufficient conditions in problem 5.1 continue

Remark 5.23. By using the necessary conditions
derived so far and stated in the Remark 5.20, we could
think of process after Corollary 5.6 as part of the

establishing the sufficient conditions for having a local

solution v = % u in the following sense. We could
establish a local solution near a(O), T1 as explained
in Remark 5.7, and we could continue it along a strip
neighborhood of [—a(O),a(0)], where L_1 is analytic

and imaginary valued and the local solution is bigger

than w except at : a(O). We could then continue the

 

133

construction of the local solution to agree with the
local solution near —a(O), T2 without any need to Lemmas
5.15. 5.16, 5.17, and Theorem 5.18 because we have now
the formula for T2 which is represented by m, the
analytic continuation of L.

We can then apply Lemma 5.21 and Theorem 5.22 to

1

show that L_ is a conformal mapping in the subset

of O bounded between T1 and T2. Thus the formula
for the local solution can be extended to the whole disk
0. In order to show i u = v is a local solution in

the sense stated in Problem 5.1, it is enough to show

% ui-m > 0 in the domain bounded between F1. T2, or

u-—¢ > 0 away from (—a(0),a(0)) there. #

Theorem 5.24. u-—w > 0 in the domain

2:
Y
L(TO) U [-a(0),a(o)] U L(TO), where by bar here we

mean the complex conjugation.

Proof: It is enough to show u-¢ > 0 away
from (—a(0),a(0)) in L(TO), because we can do

similarly to show u-—¢ < 0 away from this segment in

L(TO) by symmetry of L, L—l, and also because % u-m > 0
in a uniform strip around (-a(0),a(0)) bounded by
T1, T2. Let us first investigate the sign of u-v along
' i e A A
the image of the arc segment t = yOe , 0 g 7 g :0,

 

134

_-‘\~\\\\\\\\:;plane

T T

2 ./fi; 1
4’ ’1’

“

\\

\

\

\‘
d
~

\ ,/

 

 

 

\| _‘ "_T(,— x
O \
1
/ / ’
//l t—plane
/
I
f i J?
’l
0
yo

 

 

 

Figure 5.18

under the mapping L. Put 9 = §L . Let 1(s) = L(t(s)) =
s O
L(yoe yo) and F(s) = u(l(s)) —¢(l(s)). Then F(O) = O,
F'(O)=V(U-tl;)').'(s)l =0, and F”(O)=
S=O
= —Ale'(O)H2 which is near

A(u-'¢:)H7~'(S‘IHZ|
s=0

4yOHké(O)H2 where the index u indicates unperturbed

case. Thus u-w > 0 along 1(5) uniformly with respect

to perturbation parameters. Similarly we can argue and

 

 

135

show u-¢ > 0 along the image of the arc segment
ie .
t = yOe , v-eo g e g v under the mapping L.

To show u-¢ > 0 elsewhere on L(t), ie.., for

t = yoele, 90 g e g W-eo, let zp = a(yo)+-iyo. Then

for any 2 on this arc we have
for unperturbed solution:

Z
2 _
uu(z) = $(20)-% Im (2 [g +—4(Lul<g))21dg
O

for perturbed candidate:

Z
u(z) = ((zp)-% Im 52 [@24-4(L‘l(g))21dg

p
Thus
1 Zp 2
uu(z)-u(z) = w(ZO)-_w(zp)-3 Im fzo g dg
Z
- 2 Im f (L;l(g>>2dc
Zo
Z
+ 2 Im in (L_l(§))2d<;
Zp
_ ‘ ‘ l rzp 2
—1(zO)—1(zp)—2 Im ‘20 g dC
Z
_ 2 Im r [<L;1<g>>2-<L‘1(g>>21dc
Zp
2 .2p —1 2
— Imv‘. (Lu (QH dg
Z

 

 

136
Since zp is near 20' thus w(zO)-¢(zp),
1 z 2 z -1 2
_—2— Im 5 p g dg, and —2 Im ( p(Lu (g)) dg are small.
. 2O 20
Putting Lglm = t we get (L;l<g>)2- (L'l(g>)2 =
[L;1(L(t))]z—t2. Since L(t) is uniformly (with respect to
t) close to Lu(t), thus L;l(L(t))-t is near zero
uniformly with respect to t. Hence uu(z)-u(z) is
near zero for all such 2. This implies (nu-1b)- hl-w)

is near zero for all 2 on the boundary. Since uu-¢ > 0
on the noncontact set of the unperturbed problem, thus

away from F1 U T2 on the arc we have u-—¢ > 0. Hence
u-—¢ 2 0 on the image of the semicircle under the map L
vanishing only at 2p and -Eé.

Now we want to show u-w > 0 everywhere in the

domain bounded by T1, F2, the segment [-a(0),a(0)],

and the image of the semicircle under L. Suppose not.
Then u-—m takes its minimum at some point in the interior

of the domain. At such a point V(u-—¢) = 0. Since

2
u(z)—(1(2) = 3 y3-2 Imj (151(0)de
a(O)

3
‘_u fl__ -1( 2: 5_u_fl_ 2_ —l 2-
thus 5X'-ax — 21m[L 2)] 0, By By — 2y 2Re[L (2)] —(L
This shows [L_l(z)]2 = y2, which means 2 on F1 U T2.

This contradiction completes the proof.

 

 

137

Conclusions

5.25. Summary: We want to put the answer we got for
the Problem 5.1 in a summary. Given 0 near 00 in Z,
and parameters K, A = -d2, B = A2 fixed, as stated in
the Remark 5.20, there is only one constant of integration
to be chosen for the equation a”(y) = f(y) because the
other one is determined by symmetry. More specifically
we observed that a'(O) = O and a(O) > 0 was to be
chosen. We need also to see this point by considering
the initial data at yO (as we did already) in the
following way. We shall see that choosing a'(yo) near
§§(H(y) _% y log y)‘ , then (ii) and (iii) in the

y=YO

Remark 5.20 will determine d2, ¢(O), with

___J§fllL:fL____ > O, in terms of other parameters, where

(_l)r¢(r+l) (O)

¢(r+l)

is the first derivative of 0 nonzero at 0 and
(iv) will give a(O). Then a(yO) will be completely

determined by (v), that is, we have only one constant

of integration, a'(yo), to be chosen. We stress that
we choose a'(yO)-é%(H(y)-% y log y)‘ as close to
y=Y
0

zero as we wish. We will show that a(yo) will then be
close to x0 and the answer to the Problem 5.1 will be

unique.

Lemma 5.26. From (ii) in the Remark 5.20 we can

. 2 . .
find a as a function of K, 0 (or as a function of

i J

 

138
2 °° 2j
K, a , d , d ,..., where ¢(t ) = Z d.t ). Moreover
0 l 2 j=0 j
o2 Yo + +
we have asymptotically 7? = (log fi:)0 (l), where 0 is
./K

the usual 0 with positive leading term.
y 2 2
.. 0 g @(g )
Proof: By (ii) we have a’(y )+ dg =
—— 0 i0 (§2+K)3/2

y 2
d2 I O __$U£L_L__ dg. Put g - éjK. Then

0 (C2+K)3/2

Yo Yo
My H(fi _ifiwsidheijfi Agra)— d. >0,
0 o (g +1)3/2 K o (§+1)3/2

(*)
where the integral on the right hand side stays away from
zero because ¢ is near $0 in Z. This shows d2 as
a function of K, 0 (or parameters K,do,dl,d2,...).

To get the asymptotic representation, observe that for
the integral on the left hand side we have
Yo

[m g2mg) dg

o (g2+1)3/2

Yo
1 2 2 K 2 2
a (K) J.“ K)
1M—7d§+ M7d§
IO (§2+1)3 2 1 (§2+1)3 2
{q

_ .1 gmegz) [12 1:
J 2 3/2 dg + (pm) g
o (g +1) 1

yo

JR -2 2 .d
+i (KEN—Oak:
1 E

(§2+1)3/2

 

139

V0

1%dwioneg—
(-3 +1) MK

where the first integral on the right is pOSitive and
K, and the integrand

bounded (away from zero) for small

in the second one is:

 

§31¢<K§2)-¢<o>1+—¢<0)[§3 -<§2+1)3/21
§<§2+1)3/2
But (§2+-1)2-§3§-——5—1%%%———— and therefore
(g 2+1) + g

3% 2 2 d

<1+§2)3

Yo
_J‘ffi .2IW52 4101...;
l

(2)g+13/2

*<
O

©(o)<3§4+-3§2+ 1)d g
y 2 3/2] '

+J‘
1 §<§2+1)3/2[§3 (é +1)

where the second integral on the right hand Side is pOSitive
For the first integral on the

<1
2:)

and bounded for small K.

 

140

p

right hand side, by going back to the t variable we have

y w
j 0 [ ( dg, which is bounded for small k
J? (g2 +K)3

(easy to check). Substituting these results back in (*)

yields
Y_o

./K 2
§(K§ )
zio (1+g2) 3 2 d3 (**)

Hence asymptotically (as K approaches zero) we have:

Yo
¢(o) 1og -:+ 0(1) =
./K

leNI

Y
%? = (log —9)0+(l), where 0+ is the usual 0 with
j}?

the leading term of 0+(l) positive (as we see above).

Remark. If we put ¢(t2) = ¢(0)e(t2), then (**)

will have the form

Y_o
2 ,,K y
3? 79—32%dgflog—91‘al-0‘1):
0 (1+: JR
co , G.
e(t2)= 213.133, B.=51.

Lemma 5.27. From (iii) in the Remark 5.20 we

can find do as a function of K, Bj’ for all integers

. \ ,. Li— 1 H _.J;

j 4 1, With B (_l)r+l 2 0, where a — GO and Br+l
r+l

is the first nonzero among Bl, B2, B3,... . Moreover

 

141

do = 1 if and only if Bj = O V j 2 l or K = O = d.
. 1 _ yo + 3
Asymptotically we have l-—a— — BlK[(log ——)O (l)+75+
o R

0(K log ZS2)] if B # O and (1.--l—-)(_1)r =
J? l 0‘o

y Y
Br+lKr+l[(l°g —9)0+(l)+-O+(l)*'O(K log -9)] if
jg J?

51 = B2 = on. = fir = O and Br‘l‘l 7'1 OI r 2 l“
2 (a2 2) 2
Proof: Since F j ‘——‘_:J;—_ ¢(C )dg =

clucz (g2+K)3/2

2 2 2 2 2 2
2 a — ( ) . 2 (a -C >9<g>
_ f dg T — (—dg), where
W val (€2+_K)3/2 w fol _(€2*_K)3/2

g is changed to -Q in the last integral, thus

2 2
2 (a —g ) 2 . .
f(g)dg = — ¢<g )dg = ~41 1f and
fa U0 7 ”Fa uoz (g2+K)3/2

1 2 1
only if (iii) in the Remark 5.20 holds. If we now add
the infinity point to each Riemann sheet in Figure 5.19—a,
then the double Riemann sheet will topologically be

equivalent to a sphere, as in the Figure 5.19—b, with
[N

|

1 C
1"“! l
t.‘
\l

 

 

Figure 5 l9-a

 

142

-iJE

Figure 5°19—b

two (spherical) poles lid? on it. Then, we can
integrate the above integral along the path y, rather

than 6 U 62, and the integrand has its singularity

l
at infinity which is reduced to a singularity at zero if
1

we make the change g Thus

2 2 2
2 d — ’
f f(§)dg = w ( 2 g 2) dg
clue/2 v (Km )
m 2
_ 2_ <2 O -‘ ) 2j
‘wFO‘jf 2:32ng
J=0 Y (K-C )
w l-2j
2 2 n
= — 2: c1. o r __ an
F i=0 3 ”513(0) (1+Kn2)3/2
_ l dn
513(0) (3L+an)°/2 12.23“
where 5B(O) is the boundary of a small enough disk

about 0 traversed in the positive direction Note

_; 3 5 3 5 7
2 4
(1+-Kw ) 2 = l- ‘2 + 2 2 2m. _ 2 2 2

 

K +‘-- . Hence

 

 

. _ 2 . . 2 3 .
—41 - v [e2w1d04-ol(2710 + 27T1K)+

3 . 2 Swi 2
+ d2(-2 KVlG "7ET K )+ ~oo+

 

 

 

3
a O(2 2m 2 2 (2+r'l) _l)rr
r+l (2r): r:
3 5 3
2wi 2"-'°°'.(24. ) r r+l .°.]
+ (2r+2).' (r+l)1 ('1) K + '
- 1 _ 2 2 2 2 _5_ 2
i.e., aS-l — -Bl(a 4-4 K)+B2(4 Kg 4'64 K )+ +
3 5 3 l 3
2‘2'” (Try-l) r+l (:2 2‘2"”)
(3m r'(2r)' (“K’ (‘r+——2‘——) +°°° ~
° ° (r+l) (2r+l)
Thinking of this equation as a relation among u = 3L ,
O
K, Bj, for all j.2 l, and recalling from the last
a'(y )
Remark, by (*) d2 has only one term K ——a—9— = ua'(yO)K
0

depending on n. This term has small coefficient Ka’(yo),

and therefore by applying the implicit function theorem

 

- _ 2 2 2 2 2. 2 -
in ue-l — -Bl(G 4-4 K)+Bz(4 Kc 4.64 r )+ to find
u as a function of K, Bl, 32’ B3,... . We see from
the equation
2 2
_ 2.2- 2.3.0__5_
3 3
B (_K)r+12°2°° (2+r‘l)
r+l r'(2r)'
3 _E
d2 11'2 ‘
(76+ 2 )+
(r+l) (2r+l)

144

that if Br+l = a<r+l>

(C) is the first nonzero coefficient
___2L;i___
among B1, 52, B3,..., then we must have =

r+l
Br+l(-l)

r+l 2 O with equality case only when K = O.

l--0(O

C1r+1(-l)
More precisely u = 1 if and only if all Bj = O or
K = o, i.e., if and only if ¢ 5 1 or K = o = a = A = B.

Moreover asymptotically we have (by (1) above)

3:; _ (O)—l

Y Y
K[(log -g)0+(l)-+%+'O(K 109 —%)].
JK

 

JK
if B1 2 o, (2)
and
3 5 3
‘_ -§ ._ (—~+r-—l) y
2: l)(_l)r+l 2 r1’2i)' Kr+l[(log -9)O+(1)
r+l ‘ ' JK
3 r
+ 42 2 +—O(K log —Q)]
(rfl)(2ml) ,R
if Br+l is the first nonzero among 82, B3, B4,°.. . If

all Bj' j 2 l, are zero, then u = 1, a0 = 1 = ¢(O)

(no matter what K is — of course small). If Bl # 0,
then by (2) we have asymptotically the graph 5.20—a.
If Bl = Q = 62 = ... = Br but Br+l ; O for some r 2 1,

then by (2) we have asymptotically the graph 5,20—b.

 

 

 

Figure 5.20-a

 

Figure 5,20—b

Corollary 5.28o Given K and e as in the last
Remark, Lemma 5.27 shows that u (i.e., G0 = ¢(O)) can
be found uniquely in terms of K, Bj = 9(3)(O), j 2 1.

By Lemma 5.26 then d2 can be found in terms of these

'7

parameters° 7
yo
Lemma 5929, a(O) = VK(log -:)0+ (1), Moreover
UK
a(y ) is really close to X

0 ‘ O'

 

146

Proof: By the formula given for a(O) in the

Remark 5.20 we have

N

a(O)

JR-(ijj K(i./I'<- mag) d;

J? _ 2 2 2
2mg (1K-..) —92—<a +0) 332mm

0 (K-02)

 

 

NE 2 2 2
= MK+v§ j (a 4l32)Q(_O 3/2 do
0 (JR—o) (K+o)
2
<9-—+ €2)¢(-K§2)d§
= f+%fI11/23/2
0 (l- E) (l+§)
— 2 d2 Q(- -K§2 )d§ 5 HQg—Kg )d§
= J —— +17
V f 0(1-;>1/2((31+§) J«o 214(2th

Recall that the leading term of ¢(g2) is do = i = ¢(O

Thus by substituting the value of d2 from Lemma 5.26

Y
we get JK(log jg)0+(l) as the leading term for a(O):

VK
a(O) = fi+fi<log —)o Ky iii—LL— d:
E o ./1— §(./ ()1+g )3
(negligible)

This shows the first part of the Lemma. For the second

part observe that the Remark following Definition 2.8.1

indicates for the unperturbed problem au(y) =

.-2-y log y-+(c-+2)y-3 1 (y) a a (Y ) = Y (C -
v 7 w l‘ u 0 O

2 2 _ r 4 _ 3 . .

v YO.” wl(yo), where 41(y) — O(y ) lS analytic at O.

7T

 

147

For the perturbed problem
Y
a(y) = a(0)+.§ (y-g)f(g)dg =
O
I yO
a(yO) = a(0)+yoa (yo) -§ gfmdg
0

By assumption of the original problem a'(yo) is close
2 _ , . . ..
to c -W_log yO — au(yo)° Since a(O) lS negligible
(by the above estimate), thus in order to show au(y) is
Y
close to a(y) we only need to compare -f ng(g)dg with
"O

% yO-% wl(yo). We proceed as follows
2 yO G2- 2 2
—- g—97- ¢<g M;
V I0 (g2+K)3 2
= _2. IYO _CKZ_Q7_ ¢(§2)d§
F O (K+g2)3 2

Y
— f O Cf(g)d§
0

Yo 3

+ % IO W ¢<g2)dg
= -3 (yo —-—-—2—°‘:' ¢<g2>d§
V o (K+g )3/2
+ 3 IYO {—61—- - 11¢<gz>dg
V o (K+g )3/2
+ % §:O(¢<g2)-1.+1)dg
yo

2 JR 2
_ %.g_ EQ(K§ ) dg

” J? o (1+g2)3/2

 

 

148
yo
3? 2 4 2
2 ,— (1+3g +32 )¢(Kg)
_ ; ¢KJ d;
" o <1+§2)3/2[§3+(1+22>3/21
2 Yo 2 2
+FJ‘O (¢(C )-l)dg+jfi_‘ YO .

The last term hereis % yo, therefore we only need to

compare the other terms with -% w(yo). The first and
the second terms on the right hand side above are

negligible because of coefficients and the bounded integrals

Y
for small K. For the last integral % f O(¢(g2)—¢(O))d§+
O

2 yo
HO

series expansion of ¢(g2). The first term is close

(¢(O)-—l)dg, where $(O) = d by our last power

0

enough to -% ¢l(yo) (both with leading qubic term,
at least, with respect to yo), and the second term
g(dO--l)yO is negligible by the results of the Lemma 5.2.7.

n

Here is the proof complete. f

Remark. As the above proof shows, the difference
_ . r 3
of a(yo), xO — au(yo) is O(K )4-O(yo) for any
0 < r < %. #
What we have proved so far can be summarized into the

following (main result) theorem:

.. o ' =:+'r ‘ I
TheOiem 5.30. Fix zO <0 iyo, A0 > O yO /

on the unperturbed free boundary (in Example 2.5).

L1

 

149

Consider A, B, K near 0, K > O, and ¢ near $0 in
z, and f(y) with the branch of f for which f(yo) < O,

u = n I
and a(yo)+iyo near 20 xO+-iyo, a (yo) near
d

dy(H(y)-—% y log y)| . Then a necessary and sufficient

y=YO
condition that there exists a symmetric local solution in
0, an open fixed disk (independent of parameters K,A,B)
about the origin contained in the domain of the local
solution in Example 2.5, with part of the free boundary

II

x = a(y), a (y) = f(y)

Tl:
a’(y)| = a’(yo). a(y)! = a(yo)
yo yo
near yO is
(i) f(t) = 3 q2"t2 ¢(t2) where A = —d2
w (t2+K)372
B = a4, a > o
. yo
(ii) a'(yo) = f0 f(g)dg
(iii) j f(g)dg = -21
fl
, nyo
(iv) a(yo) = a(O)4—yoa (yO)-J gf(g)dg near
0
X0, where
1 iv? _
a(O) = 3? gfmdg = fi-f (iv‘K—§)f(§)d<;
'40 0

Moreover these four conditions are consistent and given

K and 9 = $%67 ¢ fixed, do = ¢(O), d2 = -A, B = A2

 

 

150

can be found uniquely as a function of the given

parameters (K,a'(0),e”(0),...,S(r)(0),...). In particular
the set of admissible parameters for the above necessary
and sufficient conditions is not vacuous. Furthermore
given the above fixed parameters the local solution is

unique with only one constant of integration: a’(yo)

to be chosen near aé(yo). #

 

 

LIST OF REFERENCES

 

 

LIST OF REFERENCES

[1] H. Brezis — D. Kinderlehrer, The smoothness of
solutions to non-linear variational inequalities,
Indiana Math. J., 23 (1974), pp. 831-844.

[2] H. Brezis, Operateurs Maximaux Monotones, North
Holland, Amsterdam, 1973.

[3] L.A. Caffarelli and N.M. Riviere, Smoothness and
analyticity of free boundaries in variational
inequality, Ann. Scuda Norm. Sup. Pisa, Ser. IV 3,
289—310 (1976).

[4] S.N. Chow — J. Hale, Methods of Bifurcation theory,
Springer-Verlag 1982.

[5] S.—N. Chow and J. Mallet—Paret, The parameterized
obstacle problem, Nonlinear Analysis, Theory,
Methods and Applications, vol. 4, No. 1,
pp. 73-91.

[6] R. Courant — D. Hilbert, Methods of Mathematical
physics, volumes I, II.

[7] M. Golubitsky, Stable Mappings and their singularities,
Springer—Verlag, 1973.

[8] D. Kinderlehrer - G. Stampacchia, An Introduction
to variational Inequalities and their Applications,
Academic Press, 1980.

[9] D. Kinderlehrer, The free boundary determined
by the solution to a differential equation,
Indiana Univ. Math. J. 25, 195—208 (1976).

[10] D. Kinderlehrer and L. Nirenberg, Regularity in
free boundary problems, Ann. Scuola Norm. Sup.
Pisa, Ser. IV 4, 373—391 (1977).

[ll] Kinderlehrer, D., The coincidence set of solutions

of certain variational inequalities, Archs.
ration. Mech. Analysis 40, 231-250 (1971).

151

 

[121

[l3]

[14]

[15]

[l6]

[171

[18]

[19]

[20]

[21]

152

H. Lewy and G. Stampacchia, On the regularity of
the solution of variational inequality, Comm.
Pure Appl. Math. 22, 153-188 (1969).

H. Lewy and G. Stampacchia, On existence and smoothness
of solutions of some noncoercive variational
inequalities, Arch. Rational Mech. Anal. 41
(1971), 241—253.

J. Mallet-Paret, Generic unfoldings and normal
forms of some singularities arising in the
obstacle problem, Duke Math. J. 46, 645—683 (1979).

J. Mallet-Paret, Generic bifurcation in the obstacle
problem, Quarterly Appl. Math. Jan. 1980,
pp. 355—387.

C. Morrey, Multiple integrals in the calculus of
variations, N.Y., Springer—Verlag, 1966.

Private communication with Professor J. Mallet—Paret.

T. Rado, On the problem of plateau, Springer-Verlag,
1971.

W. Rudin, Real and complex Analysis, McGraw—Hill, 1974.
D.G. Schaeffer, Some examples of singularities in
a free boundary, Ann. Scuola Norm. Sup. Pisa,

Ser. IV, 4, 133—144 (1977).

D.G. Schaeffer, A stability theorem for the obstacle
problem, Adv. Math. 17, 34-47 (1975).

 

 

 

 

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