S. B. Damelin, Peter Dragnev*, and A. B. J. Kuijlaars

The Support of the Equilibrium Measure for a Class of External Fields on a Finite Interval

• Introduction and Results• Potential Theory Preliminaries• Elasticity Interpretation• Proofs

1. Introduction and ResultsExternal field Let Q C( ), say =[-1,1].

Weighted energy Define IQ():

)()(2)()(||

1log)( tdtQtdsdts

IQ

Extremal measure Q is defined as:

IQ(Q)=min{IQ(): ( )},

where ( ) are all unit Borel measures supported on .

weighted approximation orthogonal polynomials integrable systemsrandom matrix theory

Applications of extremal measures:

Remark: The support of the extremal measure is a main ingredient in the solution. If we know that the support consists of N intervals, then we set up a system of equations for the endpoints; once we know the endpoints, we find the extremal measure from a Riemann-Hilbert problem.

Two results on the nature of the support:

• (Mhaskar-Saff, ‘85) If Q is convex, then supp(Q ) is one interval.

• (Deift-Kriecherbauer-McLaughlin, ‘98) If Q is real analytic, then supp(Q ) consists of a finite number of intervals.

The determination of the number of intervals is a

nontrivial problem.

We shall study nonconvex and non real analytic external fields on = [-1,1] of the type:

Q,c(x)= -c sign(x) |x| (1.1)

• For integer - considered in {DKM} and {DaK};• For = [0,1] - considered in {KDr};

Main result: The support of Q for Q = Q,c consists of at most two intervals.

Definition: Let Q be -Hölder continuous on [-1,1], i.e.

Q C1+([-1,1]) |Q´(x)-Q´(y)|C|x-y|,for some >0 and some positive C.

Theorem 1. Let Q C1+([-1,1]). Suppose there exists a1 [-1,1], such that

a) Q is convex on [-1, a1];

b) for every a [-1, a1], there is t0 [a1 ,1], such that

,

dsa)s)(s(1ts

(s)Qπ1t

1

a

decreases on [a1 , t0] and increases on [t0 ,1].

Then supp(Q ) is the union of at most two intervals.

(1.2)

Theorem 2. For 1 and c>0, let Q,c be given by (1.1).

Then for every a[-1,0], there is t0 [0 ,1], such that

,

, dsa)s)(s(1ts(s)Q

π1t

1

a

c

decreases on [0, t0] and increases on [t0 ,1]. As a result,

the support of Q,c consists of at most two intervals.

(1.3)

Remark: The result for any 1 is harder than the integer cases and required a new approach (Th1). This simplified the proofs in the other cases as well.

2. Potential Theory Preliminaries

• Frostman characterization of Q for Q C1+([-1,1]):

U (x)+Q(x)=F for xsupp()

U (x)+Q(x)F for x[-1,1], (2.1)

where U (x)= - log|x-t| d(t).Let =supp(Q ) and dQ = v(t) dt, then we have

log|x-t| v(t)dt = Q(x)-F, x (2.2)

v(t)dt = 1.

If =k [ak ,bk] we get a singular integral equation

int ),( )( xxQdt

txtv

The solution v(t) of the SIE in (2.3) has N (the number of intervals in ) free parameters, determined from the fact that the constant F is unique for all intervals and the total mass of v is one. In general, the solution is not positive. Denote the signed measure =v(t)dt. For =[a,1]

(2.3)

] 11[))(1(

1

dsa)s)(s(1ts

(s)Qattdt

d1

a

(2.4)

Balayage of a measure:

The balayage onto of a nonnegative measure is a

measure that is supported on and

U (x)= U (x)+c, for q. e. x

We write also =Bal(; ). For signed measure = +--

Bal(;) = Bal(+; )-Bal(- ;).

From the definition of it is clear that if 12 then

1 = Bal( 2

; 1)

Two Lemmas on balayage and convexity:

Lemma 1. Let and n be finite unions of intervals with

limn n= . Then limn n = (in weak* sense).

Lemma 2. Let be a finite union of closed intervals and be the associated signed measure, and let v be its density. Suppose that [a,b] and:

(a) Q is convex on [a,b]; (b) v(a)0 and v(b)0; (c) v(t)0 on \ [a,b];

Then v(t)>0 for all t[a,b];

3. Elasticity interpretation [KV]Let the lower half-plane be elastic and Q be the profile of a rigid punch (up to a constant). Suppose a force f is applied to Q. Let D be the displacement of Q. Denote the region of

contact S and the pressure p(x)dx. For small D the profile of the plane is given by -U= log|x-y|p(y)dy. Thus

log|x-y|p(y)dy= Q(x)+F on S

log|x-y|p(y)dy Q(x)+F on [-1,1]. (3.1)

Then =p(t)dt/f is an extremal measure with external field

Q/f (compare (3.1) with Frostman conditions (2.1)).

Now it is easy to illustrate our result. The rigid punch has a profile -x3. Applying force f corresponds to an external field Qc = -cx3 with c=1/f. Then we expect to have three

critical numbers 0<c1<c2 <c3, such that the support is

[-1,1] for c<c1; for c1<c<c2 it is [a,1] with a<0; for c2

<c<c3 it is [a,p][q,1]; and for c>c3 the support is [a,1]

with a<0.

Theorem 1. Let Q C1+([-1,1]). Suppose there exists a1 [-1,1], such that a) Q is convex on [-1, a1]; b) for every a [-1, a1], there is t0 [a1 ,1], such that

,

dsa)s)(s(1ts

(s)Qπ1t

1

a

decreases on [a1 , t0] and increases on [t0 ,1].

Then supp(Q) is the union of at most two intervals.

(1.2)

4. Proofs

Proof: Let a=min {x: x supp()}.

If supp() [a1 ,1], then the problem reduces to

{KDr}. Therefore, assume a< a1.

For every pair (p,q), with a<pq1, let vp,q be the

density of the signed measure with =[a,p][q,1] (if q=1, then =[a,p]);

Introduce Z, consisting of all (p,q) such that a) a<pq1 and q a1;

b) supp() [a,p][q,1];

c) If q<1, then vp,q(t) increases on (q,1).

d) If p>a1, then vp,q(t) decreases on (a1,1).

))(1( att ))(1( att

First, Z. Indeed by formula (2.4) (recall) and condition b) of the theorem we have that (t0, t0) Z.

] 11[))(1(

1

dsa)s)(s(1ts

(s)Qattdt

d1

a

(2.4)

Next, we show that Z is closed. Let (pn,qn) Z and

pnp, qnq. We have to show that (p,q) Z. It is clear that (p,q) satisfies conditions a) and b). By Lemma 1 we obtain vpn ,qn

vp,q from which we derive c) and d).

Finally, we find a pair (p,q) in Z such that q-p

maximal. For this choice vp,q is positive. Since supp()

[a,p][q,1], it follows supp() = [a,p][q,1]. QED

Theorem 2. For 1 and c>0, let Q,c be given by (1.1). Then for every a[-1,0], there is t0 [0 ,1], such that

,

, dsa)s)(s(1ts(s)Q

π1t

1

a

c

decreases on [0, t0] and increases on [t0 ,1]. As a result, the support of Q,c

consists of at most two intervals.

Proof: Write Q=Q,c. Since Q is convex on [-1,1], it is left to show that

, )(

dsa)s)(s(1ts

(s)Qπc

1tG1

a

(4.1)

has the decreasing/increasing property for t[0,1].

21

1

0

1 || IIds

tsa)s)(s(1s

π1

dsts

a)s)(s(1sπ1tG

a || )(

0 1

where I2 is p.v. integral. We establish the properties

(i) G (0) 0;

(ii) G (1) > 0;

(iii) For every >1, there is t [0 ,1], such that

G´ (t)<0 on [0,t ], G´ (t)>0 on [t ,1], and

G´´ (t) 0 on [ t ,1].

(4.2)

Let

Suppose now 1< <2. Consider

f(z)=z -1[(z-1)(z-a)]1/2

defined for z(-,1]. Then I2 may be written as

R

dztz

zfi

dztz

zfi

I )(2

1)(2

12

where the contours and R are given below.

(4.3)

Properties (i) and (ii) are straight forward. We prove (iii) by induction on k=[]. For =1 we

explicitly find G1 (t)=t-(1+a)/2 and (iii) is true with t1=0.

Using contour integration we get the representation:

RCdz

tzzf

itG )(

21)(

dxtx

a)x)(x(1xπ

a

R ||sin 1

dxtx

a)x)(x(1xa

||cos1 0 1

(4.4)

dxtx

a)x)(x(1xπ

tGa

R

)(||sin2)( 3

1

dxtx

a)x)(x(1xπ a

)(

||)cos1(2 0

3

1

Differentiating (4.4) twice and letting R we obtain:

We now conclude that G´´ (t) > 0 for t (0 ,1), in the case

when 1< <2. Therefore G is strictly convex and (iii)

follows easily from G (0) < 0 and G (1) > 0. Thus we

have established (iii) for []=1.

(4.5)

dxa)x)(x(1tx

xtxtxπ

tGa

)(

||||||1)(1 221

Now let k 2, and suppose (iii) is true for all with [] = k-1.

where

(4.6)

)( )(

||||11

1 21ttGdxa)x)(x(1

txxtx

π a

)()(: 1 ttGtF

dxa)x)(x(1xtxtx

πtF

a ||||1)( 2

1

For the derivatives of F(t) we get for 0<t<1:

0 ||)(

21)( 20

2

dxa)x)(x(1xtxx

πtF

a

0 ||)(

41)( 20

3

dxa)x)(x(1xtxx

πtF

a

Differentiating (4.6) we get

)()()()( 11 tGttGtFtG

)()(2)()( 11 tGttGtFtG

(4.7)

(4.8)

By the inductive hypothesis there exists t -1, such that

G´ -1(t)<0 on (0,t -1), and G´ -1(t)>0, G´´ -1 (t) 0 on

(t -1,1). Consider first t -1 >0 (the other case is similar).

)()()()( 11 tGttGtFtG

)()(2)()( 11 tGttGtFtG

F´(t)<0F´´(t)>0 (4.7) (4.8)

• On (0,t -1); G (0) 0, G´ -1(t) <0 G -1(t)

• On (0,t -1); G -1(t) < 0 G´ (t) < 0 on (0,t -1)

• On (t -1,1); G´ -1(t)>0, G´´ -1(t)>0 G´´ (t)>0 (s.c.)

• Then t >t -1: G´ (t) is (+) on (0,t ) and (-) on (t ,1).

• Of course G´´ (t)>0 on (t ,1). This proves (iii). QED