Interval of equilibrium for the constrained energy problem in the presence of an external field and a Continuum limit of the Toda lattice
Preprint, Inst. Appl. Math., the Russian Academy of Science

M.A.Lapik


Russian Academy of Science, Keldysh Institute of Applied Mathematics

Moscow, 2004

Abstract

We present the derivation of equations for the interval of equilibrium for the constrained energy problem in the presence of an external field and show that for some time dependence of the external field Q(y,t) = Q(y,0)-yt/2 the endpoints of interval of equilibrium are the solution of Continuum limit of the Toda lattice






 a

t
= -  b-a

4
 a

x
 b

t
=  b-a

4
 b

x
Also here we investigate the continuity of the set of equilibrium in general case.

1 Introduction.
We shall investigate the interval of equilibrium for the constrained energy problem in the presence of an external field. In this section we briefly recall the classical theorems dealing with weighted energy problems and constrained energy problems of potential theory. Let M be the collection of finite Borel measures, M S x M be measures with support in S and total mass x. Let MSx, s be the subset of MSx
MSx,s = :0 s, (R) = x , S S.
where Ss=S; s means that s - is a measure. Let Q:S R {+} be a continuous function on the compact S R and cap({x R: Q(x) < }) > 0. The logarithmic potential of the measure from M is defined by
U(z) =
log
1
|z-y|
d(y)
and weighted energy integral is defined by
IQ() =

log
1
|z-y|
d(y) d(z) + 2
Q d
and Q is called an external field. If the weighted energy functionals considered on unbounded S R, then the external field Q has to satisfy the additional condition
 
lim
|z| → , z S
Q(z)
ln|z|
= .
That provides existence of energy integral


[log
1
|z-t|
+Q(t)+Q(z)]
d(t) d(z).
Let us consider an extremal energy problem, i.e. the problem of finding a measure Q x such that
IQ (m Q x)=WQx =
inf
m M x 
IQ (m).
If in addition s has finite logarithmic energy (energy when Q = 0) over all compact sets and s({x R: Q(x) < }) > x then s is called a constraint. We can define a constrained extremal energy problem, i.e. the problem of finding a measure Qx,s such that
IQ(Qx,s)=WQx,s = inf MSx,sIQ().
Introduction in the theory of constrained extremal problems is in the paper by Dragnev and Saff [DS] and in the case without constraint there is the monograph by Saff and Totik [ST]. We introduce here some known facts. If WQx then there exists unique measure Qx Mx such that IQ(Qx) = WQx. Also if WQx,s then there exists unique measure Qx,s Mx,s such that IQ(Qx,s) = WQx,s. The extremal measures are called the equilibrium measures because they have the following properties (when Us and Q are continuous):
U Q x + Q =

FQx on S
=F Q x on SQx
    (1)
for some constant FQx, and correspondingly
UQx,s+ Q =

FQx,s on SQx,s
FQx,s on Ss -Qx,s
    (2)
for some constant FQx,s. Here S denotes the support of measure . Properties (1) and (2) are sufficient conditions for being equilibrium measures, namely if there is a constant such that the first or the second relation above holds for some measure with total mass x then the measure is solution of the extremal problem and it is called respectivelyequilibrium measure associated with Q with total mass x or s-constrained equilibrium measure with total mass x associated with Q. In the paper we use continuous external fields and potentials of constraints, but they may be lower semi-continuous (see [BR],[ST] and [DS]) and there exists similar equilibrium conditions. There is a functional introduced in [BR]
FQx(K):= -xlog cap(K)+
Q dwK,
where wK denotes the equilibrium measure associated with the regular compact set K. For the first time the functional for x=1 was considered in [MS] The support of the equilibrium measure in R minimizes the functional, more precisely
FQx (K) =

=FQx if SQx K SQx
> FQx   otherwise
were SQx={y:(UQx+ Q)(y) =FQx }. One can derive this by integration of above equilibrium conditions with respect to measure wK. Also V.S.Buyarov and E.A.Rahmanov have proved that SQx increase with respect to x and have represented the sets SQx and SQx for all x R+ in the following way
SQx=
 
t<x
SQt
,
SQx=
 
t>x
SQt.
They have proven for unconstrained case that almost everywhere with respect to x we have cap(SQx\ SQx)=0. There is an analogical results for the support of the measure in the constrained case in [K]. In the paper we investigate continuity with respect to x of the set S Ss -. The paper deals with properties of endpoints of interval S Ss -=[a(x),b(x)], were = Qs, x is an equilibrium measure with total mass (R) = x s(R) for an external field Q and constraint s (see section 4). Also we reproduce a derivation of equations for the support SQx = [a(x),b(x)] of equilibrium measure with total mass x corresponding to external field Q . In some sense support defines the equilibrium measure and to find the support is a very essential task in solving the extremal problem in the potential theory. The set S Ss - is very interesting too, it is the set where the constraint is not hit and also it is the set of equilibrium. Here we restrict ourselves by the case were this set is an interval. To justify our interest to the equations for endpoints S Ss -=[a(x),b(x)] we mention that there are applications of the subject to theory of partial differential equations. Namely endpoint functions a(x,t) and b(x,t) of these extremal problems for some special time dependence t of the external field Q(y,t) are the solutions of known system of PDE called continuum limit of Toda lattice (see [AV], [DM]). In the section 5 we derive these equations.
2 The continuity of the set of equilibrium.
In this section we consider of continuity of the set of equilibrium which is SQs, x Ss -Qs, x. It is a compact set in the metric space R and we represent the definition of the notion of continuity for the general compact sets which one can find in the Bible for the set theory by K.Kuratowski "Topology". And after the definition we prove a lemma describing the breaking points of intersection of continuous families of sets and apply the lemma to the set of equilibrium. Let S(x) be the family of compact sets with parameter x. The set liminfxx0+0S(x) is called limit interior of the S(x) in x0 from above. Point p belongs to liminfxx0+0S(x) if any neighborhood of p have nonempty intersection with each S(x) where x0<x<x0+e for some e. By the same way we define the limit superior. Point p belongs to limsupxx0+0S(x) if any neighborhood of p have nonempty intersection with infinite number of S(x) where x0<x<x0+e for some e.
 
limsup
xx0+0
S(x):=
 
e >0
 
x0<x<x0+e
S(x)
.
Evident that
 
liminf
xx0+0
S(x)
 
limsup
xx0+0
S(x),
 
limsup
xx0+0
S1(x) S2(x)
 
limsup
xx0+0
S1(x)
 
limsup
xx0+0
S2(x).
The family of sets S(x) converges in x0 from above or has a limit in x0 from above if
 
limsup
xx0+0
S(x)=
 
liminf
xx0+0
S(x)=:
 
lim
xx0+0
S(x).
The family of compacts is continuous in x0 from above if
 
lim
xx0+0
S(x)
= S(x0).
The converging family of compacts S(x) has jump J in x0+0 if
JS(x0):=
 
e > 0
 
0<x x0 < e
S(x) S(x0)
.
There is the parallel with number theory: if J S (x0 )= then S(x) converges and limxx0 0S(x)=S(x0), but the converse is false. Evidently, that for increasing family of compact set these definitions have more simple representation
 
limsup
xx0+0
S(x)=
 
liminf
xx0+0
S(x) =
 
x>x0
S(x),
and for decreasing family of compact set respectively
 
limsup
xx0+0
S(x)=
 
liminf
xx0+0
S(x) =
 
x>x0
S(x)
.
The situation when x tends to x0 from below is defined in analogous way.
Theorem 1   Let Q and s be an external field and constraint respectively on S R. Let = Qx,s be the s-constrained equilibrium measure with total mass x associated with Q. (i)If S Ss-=[a(x),b(x)], a(x)<b(x) then the functions a(x) and b(x) are continuous almost everywhere. (ii) If S [A,B] R then the set of equilibrium S Ss- is continuous almost everywhere.
Lemma 1   (i) Let K(x) and L(x) be increasing and decreasing respectively families of compact sets with parameter x in the compact S R, namely if x1<x2 then K(x1) K(x2) and L(x1) L(x2). In a neighborhood of x0
K(x) L(x) = M(x).
K(x) and L(x) are continuous in x0, i.e.,
K(x0) =
 
x<x0
K(x)
=
 
x>x0
K(x),
L(x0)=
 
x<x0
L(x) =
 
x>x0
L(x)
.
Then the jump M(x) in x0, which we defines
JM(x0):=
 
e > 0
 
0<x x0 < e
M(x) M(x0)
,
is contained in M(x0) and, in particular, the jump does not contain an interior point. (ii) Let K(x) and L(x) be continuous in x0 families of compact sets with parameter x in [A,B] R. In a neighborhood of x0
K(x) L(x) = M(x).
If for any x holds K(x) L(x) [A,B] R then M(x) is continuous in x0.
Remark. The point (ii) does not hold for any compact in R. Let S = {0, 1, 1/2, 1/3, ...} and L(x) = {1, 1/2, 1/3, ..., [1/x]-1}, K(x) = S \ L(x) = {1/[1/x]+1, 1/[1/x]+2, ..., 0} for x (0,1]. L(0) = {0} = K(0). Then non trivial M(x) is only in 0.

Proof. (i) Taking in to account that K(x) and L(x) are increasing and decreasing set families respectively we can easy see that for x > x
0
M(x) M(x0) = (K(x) L(x))(K(x0) L(x0))
(K(x) \ K(x0))(L(x0) \ L(x)).
Let us define the set C:
 
n
 
x0+
1
n
>x>x0
M(x) M(x0)
 
n
[
 
x0+
1
n
>x>x0
(K(x) \ K(x0))
 
x0+
1
n
>x>x0
(L(x0) \ L(x))
]=: C
and prove that C K(x0) L(x0). Consider any p from C. If p K(x0) then there exists K(x'), x'>x0 such that p K(x'). Let fix any n1 > 1/x'-x0.
 
x0+
1
n1
>x>x0
(K(x) \ K(x0))
K(x')
provide that p K(x0) or
p
 
x0+
1
n
>x>x0
(L(x0) \ L(x))
for any n>n1. If p int(L0) then there is L(x''), x''>x0 so that p L(x''), it is simple sequence of continuity of L(x) in x0. Let fix any n2 >max( 1/x''-x0, n1). So, in the case p int(L0) for any n>n2 we have
p
 
x0+
1
n
>x>x0
(L(x0) \ L(x))
and hence p K(x0). We have got p K(x0) L(x0) for any n>n2 and J+ K(x0) L(x0). The above consideration yields that
 
n
 
x0+
1
n
>x>x0
M(x)\ M(x0)
= ,
hence to complete the proof we have to investigate the set
 
n
[
 
x0+
1
n
>x>x0
M(x) M(x0)]
 
n
[
 
x0+
1
n
>x>x0
M(x)] M(x0).
Let
p
 
n
[
 
x0+
1
n
>x>x0
M(x) M(x0)],
this means that each neighborhood of y contains point from M(x) = L(x) K(x), hence p limsup L(x) limsupK(x) = M(x0). To prove the assertion "from below" one should apply above reasoning with the exchange of K(x) and L(x). (ii)Let p be any point from M(x0), then p is the limit of K(x) and L(x) in x0. Then there are points k(x) and l(x) from K(x) and L(x) for all x, |x-x0|<e in each neighborhood of p. By familiar Cantor method of bisecting (see proof of Lemma 4) there exists a point of M(x) in [k, l]. (or in [l, k]) for each x. Then p liminfxx0+0M(x) and
M(x0)
 
liminf
xx0+0
M(x)
 
limsup
xx0+0
M(x)=
=
 
limsup
xx0+0
(L(x) K(x))
 
limsup
xx0+0
L(x)
 
limsup
xx0+0
K(x) = M(x0)



Now we represent the proof of Theorem 1.

Proof. From [
K] it followers that for continuous external field the map x Qs, x is increasing for positive x s(R). That is , if 0<x1< x2 s(R), then the difference Qs, x2-Qs, x1 is a positive measure. Hence the supports S Qs, x form an increasing family with respect to x. Families of increasing compacts on R has at most countable number of breaking points. The decreasing family Ss - Qs, x has the same breaking points with respect to x. One can see it from (s -Qs, x1)-(s -Qs, x2)= Qs, x2-Qs, x1. So we are allowed to apply Lemma 1 (i) for S Qs, x =: K(x) and Ss - Qs, x =: L(x) for almost all x0. Hence the functions a(x) and b(x), that are defined from S Ss-=[a(x),b(x)], are continuous on R\ N were N is at most countable set, since non degenerated interval can have only jump that contain interior points. (ii) is evident from (i) and Lemma 1(ii).


3 The equations for the endpoints of equilibrium measure support in case of constrain absence.
We start with a modification of the theorem from ([ST],p.201). We deal with an equilibrium measure with total mass x and some special kind of external field that will be made use in the next section.
Theorem 2   Suppose that Q: S R, S is a compact in R, satisfies the following conditions: a) Q is continuous function on S , b) Q' is differentiated a. e. and Q' L(S). Let n be some signed measure, so that a) Un is continuous, b) R1/|y-g| dn(y) < + for an arbitrary g S . Let = Qs, x be the equilibrium measure with total mass x = x0 on S corresponding to the external field Q = Q - Un and the support of is a finite closed non degenerated interval SQx = [a(x0),b(x0)]=[a,b] S. Then we have
1
p

b


a
Q'(y)
y-a
b-y
 dy -
 


[a,b ]c
y-a
y-b
 dn(y)

= x0 -n(R) if b int(S)
x0 -n(R) if b (S)
1
p

b


a
Q'(y)
b-y
y-a
 dy +
 


[a,b ]c
y-b
y-a
 dn(y)

= x0 -n(R) if a int(S)
x0 -n(R) if a (S)
where [a,b ]c = R \ [a,b] is complement of [a,b] in R.


Proof. Functional F
Qx(K):= -xlogcap(K)+Q dwK satisfies
FQ x(SQx) = FQx([a,b]) =
 
min
a < b: a , b S
FQx([a,b ]),
where wK denotes the Robin distribution associated with the set K (the equilibrium measure for Q=0 on K). By known facts about capacity and equilibrium measure of the interval we have
FQx0([a,b ]):= -x0log


b -a
4



+
1
p

b


a
Q(y)
(y-a)(b-y)
 dy =
=-x0log


b -a
4



+
1
p

b


a
Q(y)
(y-a)(b-y)
 dy -
Un dw[a,b ]
In the last integral the both integrals exist and we are allowed to apply Fubini's theorem:
FQx0([a,b ])=
-x0 log


b -a
4



+
1
p

b


a
Q(y)
(y-a)(b-y)
 dy-
Uw[a,b ](ydn(y)=
= -x0log


b -a
4



+
1
p

b


a
Q(y)
(y-a)(b-y)
 dy +
+



log


b -a
4



+g[a,b ](y)


 dn(y)=
= log


b -a
4



( n (R)-x0) +
1
p

b


a
Q(y)
(y-a)(b-y)
 dy +
g[a,b ](ydn(y).
On making the change of variable
y =
b +a
2
+
b -a
2
cosq ,      0 q p
we can rewrite FQx([a,b ]) as
log


b -a
4



( n (R)-x0) +
1
p

p


0
Q(
b +a
2
+
b -a
2
cos q )  dq+
g[a,b ](ydn(y)
Now we consider the case when a(x0) or b(x0) lies in int(S). We take derivatives F/b and F/a which equal to
F
b
([a ,b]) =
=
n (R)-x0
b -a
+
1
2p

p


0
Q'(
b +a
2
+
b -a
2
cosq ) (1+cosq )  dq+
+
g
b
[a,b ](ydn(y) =
=
n (R)-x0
b -a
+
1
(b -a)p

b


a
Q'(y)
y-a
b-y
 dy-
1
b -a

 


[a,b ]c
y-a
y-b
 dn(y)
and correspondingly
F
a
([a ,b]) =
=
x0 - n (R)
b -a
+
1
2p

p


0
Q'(
b +a
2
+
b -a
2
cosq ) (1-cosq )  dq+
+
g
a
[a,b ](ydn(y) =
=
x0 - n (R)
b -a
+
1
(b -a)p

b


a
Q'(y)
b-y
y-a
 dy+
1
b -a

 


[a,b ]c
y-b
y-a
 dn(y)
were [a,b ]c is R\ [a,b ]. Possibility to differentiate under the integral sign is guaranteed by lemma 2 and lemma 3 which we present just after the proof. On taking a=a(x0) and b=b(x0), it follows that
F
b
([a ,b]) =
=
n (R)-x0
b -a
+
1
(b -a)p

b


a
Q'(y)
y-a
b-y
 dy -
1
b -a

 


[a,b ]c
y-a
y-b
 dn(y) = 0.
A similar argument yields the same formula for the derivative with respect to a
F
a
([a ,b]) =
=
x0 - n (R)
b-a
+
1
(b-a)p

b


a
Q'(y)
b-y
y-a
 dy+
1
b-a

 


[a,b]c
y-b
y-a
 dn(y) = 0
If b(x0) is not interior point of S, we can apply Saff Totik arguments ([ST],p.202) for our F-functional and will be able to write that F/+b ([a ,b]) tends to same limit when b tends to b-0 and minimizes the F-functional in point b. So we have inequalities in the theorem because the derivative of F-functional is not greater than zero
1
p

b


a
Q'(y)
y-a
b-y
 dy -
 


[a,b ]c
y-a
y-b
 dn(y) x0 -n(R).
By the same way we have F/-a ([a ,b]) 0
1
p

b


a
Q'(y)
b-y
y-a
 dy +
 


[a,b ]c
y-b
y-a
 dn(y) n(R)-x0.



To complete the section we prove the above mentioned technical lemmas which deal with changing of integration and differentiation.
Lemma 2   Suppose that Q is a continuous function on the compact S R, with Q' L(S). Furthermore, let I1,I2 be some compacts sets, and y:I1 I2 | S, with y C1(I1 I2). Then for all b I1 with (b-d,b+d) I1 for some d>0
b

 


I2
Q(y(b,q ))  dq =
 


I2
Q' (y(b, q ))
y
b
(b, q )  dq,
that is, we may exchange integration and differentiation with respect to the parameter b.


Proof. Since Q
' L1(S), we have for [b1,b2] I1
Q(y(b2,q)-Q(y(b1, q ) =
b2


b1
Q' (y(b, q ))
y
b
(b, q )  db.
Consequently,
|Q(y(b2,q )-Q(y(b1, q )| Q' L(S)
y
b
L( I1 I2) |b2-b1|,
and the claim follows by applying Lebesgue's Dominated Convergence Theorem to fn(q)= [Q(y(b 1/n, q ))-Q(y(b, q ))]/(1/n).


The next Lemma deals with the differentiation of a Green potential of a non degenerate interval [a,b] with respect to one of the endpoints a or b.
Lemma 3   Let n be some measure with compact support S(n), and a<b. Provided that

 


R
1
|y-b|
dn(y) < ,
the Green potential g[a,b](ydn(y) can be differentiated with respect to b, and
b

g[a,b](ydn(y)
=

b
g[a,b](y)  dn(y)
  =
-1
b - a

 


R\ [a,b]
y-a
y-b
  dn(y).
Similarly, provided that

 


R
1
|y-a|
dn(y) < .
the Green potential g[a,b](ydn(y) can be differentiated with respect to a, and
a

g[a,b](ydn(y)
=

a
g[a,b](ydn(y)
  =
1
b- a

 


R\ [a, b]
y-b
y-a
  dn(y).


Proof. We will only show the first assertion of the Lemma, the second part is similar. Notice first that, for y [a,b],
b
g[a,b](y) = -
1
b - a
y-a
y-b
.
In the case bS(n), we notice that this derivative is smooth for y S(n)\ [a,b], and clearly we may exchange the order of differentiation and integration, see for instance the argument used in the preceding Lemma. By splitting if necessary the Green potential into two parts, we see that it only remains to consider the case S(n) [a,3b-a/2]. For the second derivative we obtain for y [a,b]
[
b
]2 g[a,b](y) = -
b
1
b - a
y-a
y-b
=
1
(b - a)2
y-a
y-b
y-3b/2+a/2
y-b
,
which is negative when y S(n)\ [a,b]. Consider the two sequences of functions
hn(y)=
g[a,b 1/n](y) -g[a,b ](y)
1/n
,
converging point-wise to the derivative of g[a,b](y) with respect to b for almost all y. Then we know that the Green potential has left-hand or right hand derivatives with respect to b, with
b

g[a,b](ydn(y) =
 
lim
n

hn(y)  dn(y) ,
provided that the right-hand limits exist.For the right-hand and left-hand derivatives, consider the functions
fn+(y) :=







g[a,b +1/n](y) -g[a,b ](y)
1/n
,
y R\ [a, b +
1
n
],
g[a,y](y) -g[a,b](y)
y-b
,
y (b, b +
1
n
],
0, y [a, b ].
fn-(y) :=



-
g[a,b -1/n](y) -g[a,b ](y)
1/n
,
y R\ [a, b ],
0, y [a, b ].
Then by the above sign considerations we may conclude that fn+(y) and fn-(y) is increasing and decreasing respectively in n for y S(n)\ [a,b]. Moreover, fn(y) tends to the derivative of g[a,b](y) with respect to b for almost all y, and hence
 
lim
n

fn(y)  dn(y) =
b
g[a,b](y)  dn(y)
by Lebesgue's Monotone Convergence Theorem. Thus the assertion of the Lemma follows by showing that lim[ fn(y)-hn(y)]   dn(y) = 0. Indeed, by construction, fn+(y) hn+(y) and fn-(y) hn-(y), and, since g[a,b](y)=O(y-b)y b+, we get for some suitable constant C
   
0
 
limsup
n

[ hn+(y)-fn+(y)]   dn(y)
   
=
 
limsup
n

b+1/n


b
[hn+(y)- fn+(y)]   dn(y)
   
=
 
limsup
n

b+1/n


b
[
1
y-b
- n ] g[a,b](y)   dn(y)
   
C
 
limsup
n

b+1/n


b
[
1
y-b
- n ] y-b   dn(y)
   
C
b-a
 
limsup
n

b+1/n


b
y-a
y-b
  dn(y) = 0 ,
and
   
0
 
limsup
n

[ fn-(y)- hn-(y)]   dn(y)
   
=
 
limsup
n

b


b-1/n
[fn-(y)- hn-(y)]   dn(y)
   
=
 
limsup
n

b


b-1/n
n g[a,b-1/n](y)   dn(y)
   
C
 
limsup
n

b


b-1/n
n y-(b - 1/n)   dn(y)
   
C
 
limsup
n

b


b-1/n
1
b-y
  dn(y) = 0 ,
the last equalities following from the assumption on the finiteness of the above integral. Thus the assertion of the Lemma follows.


4 The equations for the endpoints of the interval of equilibrium with presence of constraint.
The main aim of the paper is to describe the interval of equilibrium, i. e. the set where constrain is not hit.
Theorem 3   Suppose that Q: [A,B] R, ( - < A < B < +) satisfies the following conditions: a) Q is continuous function on the interval [A,B], b) Q' is differentiated a. e. and Q' L([A,B]). Let s be a constraint on [A,B] ([A,B] = Ss ), so that a) Us is continuous, b) R1/|y-g| ds(y) < + for an arbitrary g [A, B] . Let = Qs, x be the equilibrium measure for external field Q and constraint s with total mass (R) = x = x0 s(R). If S Ss -=[a(x0),b(x0)] = [a,b], A a(x0) < b (x0) B then

 


[A,B] \ Ss -
l-a
l - b
 ds (l)+
1
p

b


a
Q'(l)
l-a
b-l
 dl=

= x0 if b < B
x0 if b=B
-
 


[A,B] \ Ss -
l - b
l-a
 ds (l)+
1
p

b


a
Q'(l)
b-l
l-a
 dl=

= -x0 if a > A
-x0 if a = A
where [A,B] \ Ss - = S \ Ss - is part of S where constraint is hit.


Proof. We consider two cases: 1)The case a=a(x
0) or b=b(x0) int(Ss -), 2)The case a=a(x0) or b=b(x0) int(S). 1) We can think about as = n + t, where t := Ss -c = s Ss -c. Let us define Q(l) := Q(l ) + Ut(l) and will write the equilibrium conditions for and after n.
Q(l )+U(l) = Q(l )+US Qs, x0\ [a, b](l)+U[a, b](l)
=Q (l) + U n (l) =

F on S
F on Ss - = Sn
= F on S Ss - = Sn = [a, b]
The last two expressions are the sufficient conditions for n being equilibrium measure on n with support [a, b]. Also we have
n (R) = (R) -t (R)= (R) - (Ss -c) = (R) -s Ss -c (R) = x0- s(Ss -c) 0.
Clear that if Us is continuous then Us K is continuous for any compact K Ss. It follows from Us K+Us Ss \ K=Us and from the well known fact that logarithmic potential is a harmonic function outside support of the measure. So Q is continuous. By means of Theorem 1 for Sn, n and Q we have
1
p

b


a
Q'(l)
l-a
b-l
 dl +
 


[a,b ]c
y-a
y-b
 dt(y) = n (R) +t (R)      b int(Ss -)
1
p

b


a
Q'(l)
b-l
l-a
 dl -
 


[a, b]c
y-b
y-a
 dt(y) = -t (R)-n (R)      a int(Ss -)
With substituting explicit expressions for t and n one should get for b int(Ss -)

 


Ss - c
l-a
l - b
 ds(l )+
1
p

b


a
Q'(l)
l-a
b-l
 dl = s(Ss -c) + x- s(Ss - c) = x0
and for a int(Ss -)
-
 


Ss - c
b-l
a -l
 ds(l )+
1
p

b


a
Q'(l)
b-l
l-a
 dl= -x0.
2)We present s - =: * as * = n* + t*, where t* := * Ss -*c = s Sc. Let us define Q*(l) := -Q(l ) - Us(l)+ Ut*(l)= -Q(l ) - UsS (l) and write the equilibrium conditions for and then for n* .
-Q(l )-U(l) = Q*(l) + Un*(l) =

-F on S = Ss - * = Sn*
-F on Ss -
= -F on S Ss - = Sn* = [a, b]
The first and the last expressions are the sufficient conditions for n* be to the equilibrium measure on Sn* with support [a,b]. Also we can write
n* (R) = s (R)- (R) - s Sc (R) = s(S ) -x0.
Take in mind that s is the constrain and n* (R)= s(S )- (S ) 0. By means of Theorem 2 for Q(l) = -Q(l) - UsS(l) and n* we have in the case b int(S).
-
1
p

b


a
Q'(y)
y-a
b-y
 dy-
 


[a,b]c
y-a
y-b
 dsS(y) = n* (R) - s (S) = -x0
and for a int(S)
-
1
p

b


a
Q'(y)
b-y
y-a
 dy+
 


[a, b]c
y-b
y-a
 dsS(y) = s (S) - n* (R) = x0
We remember S\ [a,b ] = Ss -c. Thus we have get equations for a, b int(S) int(Ss - ) int(S Ss - ) = int([A,B]) by Lemma 4 which we present after this proof. Inequalities in the theorem can be derived by the same way.


Now we prove above mentioned Lemma 4.
Lemma 4   Let A, B be compact subsets in R, A B = [a, b] R. If x int(A B) then x int(A) or x int(B). That is int(A B) int(A) int(B).


Proof. Suppose that x int(A B) but x int(A) and x int(B). Then there are three possibilities: 1)x A and x B. Since x int(A B) and x B then $ U(x) (A B) which is an open set containing x, so that U(x) B = (by definition of set interior and the fact that R is a Hausdorff topology space). Since x A then there exists consequence {x
n} in U (x) with limit x and not belonging to A. Since {xn} A and {xn} B then U (x) (A B) and we have a contradiction. 2)x A and x B. The proof is the same as for the case 1). 3)x A and x B. It means that x [a, b]. if x (a, b) then the assertion of the lemma is evident. We will consider the case when x = a, and the case x = b is similar. Evidently [x, x + b-a/2) A and [x, x + b-a/2) B and $ U (x) (A B) which is an open set containing x, then there is e > 0 such that (a-e,a) (A B). And (a-e,a) A B = because of lemma condition. if (a-e,a) A or (a-e,a) B then x is an interior point of A or B, and we have a contradiction. So there are x A and y B in the interval (a-e,a) , let x<y. Without the loss of generality we can consider that x+y/2 A and put x1 = x+y/2, y1 = y and then we will bisect the interval [xn, yn] and after will choose the interval [xn+1, yn+1] with endpoints belonging different sets. So we have the system of included intervals { [xn, yn] } n=1 with lengths tending to 0, and one endpoint of each interval belongs A, the other belongs B. The intersection of all this intervals is one point. It is boundary point of A and B, and belongs to A and B thanks to compactness. But this contradicts to (a-e,a) A B = .


There is essential question about uniqueness of the solutions of equations from theorems 2 and 3. The equations describe the set where derivative with respect to endpoints of interval of the functional FQx vanishes. FQx achieves its minimum and has vanish derivation on the sets K so that SQx K SQx. The next example shows that SQx\ SQx may be non trivial, and by the same way one can construct very various sets. In section 2 we have proved that set were constraint is not hit is continuous almost everywhere that provides SQx=SQx= SQx+e, i. e. the minimum of FQx is unique a. e. and endpoints of support or interval of equilibrium is isolated in the set of solutions. Continuity does not mean that solution is unique a. e., there may be points where derivative is zero. If the external field Q so that Q''>|s| then we may guaranty that there is unique solution of systems from Theorem 2 and 3. There are more complicated symptoms of uniqueness of the solutions of equations. Example. Let Q be
Q :=

= 0 on [-2,2]
= g[-2,2](y) on [2,3]
= + on (-,-2] [3,),
the extremal measure with unique total mass Q1 is the Chebyshev measure of [-2,2] because we know that Uw[-2,2](y)= -g[-2,2](y), it provide
Uw[-2,2](y)+Q(y) =

= 0 on [-2,2] =: SQ1
0 on R.
(It is the sufficient condition, see the section 1). SQ1=[-2,3]. Easy to see that
FQ1
b
([-2,b]) =
b
( ln
4
b+2
+
b


-2
g[-2,2](y)
p (y+2)(b-y)
 dy) 0      on     [2,3].
5 Interval of equilibrium and the continuum limit of the Toda lattice.
In this section we show that equations from theorem 3 locally satisfy the system of partial differential equations
a
t
= -
b -a
4
a
x
b
t
=
b -a
4
b
x
which supplemented with the initial boundary conditions is well-known Toda lattice.
Theorem 4  Let Q(l, t) := -l t/2 + Q(l,t0) for t from some neighborhood of t0 containing [t0-T1,t0+T2] satisfies a) Q(l,t0) is continuous function on the interval [A,B] with respect to l, ( - < A < B < +) b) Q'(l,t0) is differentiated a. e. with respect to l and Q'(l,t0) L([A,B]). Let s be a constraint on the interval [A,B] so that c) Us is continuous, d) R1/|y-g| ds(y) < + for an arbitrary g [A, B] . e) the set of equilibrium for the extremal problem (??) is an interval for (x,t) [x0-X1, x0+X2] [t0-T1,t0+T2] , to wit S Ss -=[a(x,t),b(x,t)] , A < a(x,t) < b (x,t) < B where = Qs, x is the equilibrium measure for external field Q(l, t) and constraint s with total mass x s(R). f) the map [x0-X1, x0+X2] [t0-T1,t0+T2] {( a(x,t),b (x,t))} so that Jacobi matrix is well-defined, continuous and nowhere degenerate in the neighborhood of (x0,t0). Then a(x,t) and b (x,t) satisfy PDE in some neighborhood of (x0,t0)
a
t
= -
b -a
4
a
x
b
t
=
b -a
4
b
x
    (3)


Proof. Since Jacobian in the condition f) is nowhere degenerate in the neighborhood of (x
0 ,t0 ) we can inverse the PDE (3) by
b
x
x
b
+
b
t
t
b
=
b
x
(
x
b
+
b -a
4
t
b
) =1
b
x
x
a
+
b
t
t
a
=
b
x
(
x
b
+
b -a
4
t
b
) =0
a
x
x
a
+
a
t
t
a
=
a
x
(
x
a
-
b -a
4
t
a
) =1
a
x
x
b
+
a
t
t
b
=
a
x
(
x
a
-
b -a
4
t
a
) =0
Then the inversed system for (3) is
x
b
=
b -a
4
t
b
x
a
= -
b -a
4
t
a
    (4)
We have got equations for endpoints of interval of equilibrium (Theorem 3)

 


Ss - c
l-a
l - b
 ds (l)+
1
p

b


a
Q'(l)
l-a
b-l
 dl= x+
t(b - a)
4

 


Ss -c
l - b
l-a
 ds (l)-
1
p

b


a
Q'(l)
b-l
l-a
 dl= x -
t(b - a)
4
    (5)
After addition and subtraction we arrive to

 


[b,B] \ Ss -
 ds (l)
(l-a)(l - b)
 ds (l) -
 


[A,a] \ Ss -
 ds (l)
(l-a)(l - b)
+
+
1
p

b


a
Q'(l)
 dl
(l-a)(b-l )
=
t
2

 


[b,B] \ Ss -
(2l-a-bds (l)
(l-a)(l - b)
-
 


[A,a] \ Ss -
(2l-a-bds (l)
(l-a)(l - b)
+
+
1
p

b


a
Q'(l)
2l-a-b
(l-a)(b-l )
 dl = 2x
    (6)
By differentiating the first and the second equations in (5) with respect to a and b respectively we have
2
x
a
=
=
 


[A,a] \ Ss -
1
(l-a)(l - b)
 ds (l) -
 


[b,B] \ Ss -
1
(l-a)(l - b)
 ds (l)
-
1
p

b


a
Q'(l)
(l-a)(b-l )
 dl -
b -a
2
t
a
+
t
2
 
2
x
b
=
=
 


[A,a] \ Ss -
1
(l-a)(l - b)
 ds (l)-
 


[b,B] \ Ss -
1
(l-a)(l - b)
 ds (l)
-
1
p

b


a
Q'(l)
(l-a)(b-l )
 dl +
b -a
2
t
b
+
t
2
With the system (6) one can easily see that this is our PDE (4) because the sum of all rest members is equal to zero. To complete the proof we have to check possibility of differentiating in (5). We discuss only the first equation. There is problematic place in the first integral. Like in the section 3 we define functions
hn(l)=
l-(a 1/n)
l - b
-
l-a
l - b
1/n
,
and hn+(l) = 0 when a < l < a + 1/n. The functions hn(l) converge point-wise to the derivative of l-a/l - b with respect to a. The absolute value of hn+(l) is dominated by 1/(l-a)(l - b). By d) it is integrable function for the measure s. The claim follows by applying Lebesgue's Dominated Convergence Theorem. By convexity of l-a/l - b with respect to a the functions hn-(l) are decreasing and they absolute values are dominated by the absolute value of derivative of l-a/l - b with respect to a, which has a finite integral by d). The assertion follows by Lebesgue's Monotone Convergence Theorem. Now we consider derivative of F(a) = ab Q'(l)l-a/b-l  dl.
F(a+
1
n
)-F(a)
1
n
=
=

b


a+
1
n
Q'(l)
l-a-
1
n
b-l
 dl-
b


a+
1
n
Q'(l)
l-a
b-l
 dl
1
n
+

b


a+
1
n
Q'(l)
l-a
b-l
 dl-
b


a
Q'(l)
l-a
b-l
 dl
1
n
.
The limit of the last part when n is equal to zero because
|
a


a+
1
n
Q'(l)
l-a
b-l
 dl| ||Q'||L*
1
n
*
a+
1
n


a
l-a
b-l
 dl.
Now we define functions hn(l) and our task is to change integration and differentiation.
hn(l) :=





Q'(l)
l-a-
1
n
b-l
-
l-a
b-l
1/n
,
l a+
1
n
,
0,
l (a, a +
1
n
).
Then

b


a+
1
n
Q'(l)
l-a-
1
n
b-l
 dl-
b


a+
1
n
Q'(l)
l-a
b-l
 dl
1
n
=
b


a
hn(l)  dl.
The absolute values of the functions hn(l) are dominated by ||Q'||L/(l-a)(b-l) and by applying Lebesgue's Dominated Convergence Theorem we have considered the case of write-hand derivative. For the left-end derivative we can apply the above reasoning with the same dominator function for the
F(a-
1
n
)-F(a)
-
1
n
=
=
b


a
Q'(l)
l-a+
1
n
b-l
-
l-a
b-l
-
1
n
 dl+

a


a-
1
n
Q'(l)
l-a+
1
n
b-l
 dl
1
n



References
[R]
E. A. Rahmanov. Equilibrium measure and the distribution of zeros of the external polynomials of a discrete variables. Mat. Sb. 1996, 187, 109-124(in Russian);English transl. in Sb.Math. 1996, 187, 1213-1228.
[AV]
A.I.Aptekarev, W. Van Assche. Asymptotic of discrete orthogonal polynomials and the continuum limit of the Toda lattice. Journal of physics A: Mathematics and General, 34(48), 2001, 10627-10639.
[BR]
V. S.Buyarov, E. A. Rahmanov. Families of equilibrium measures in an external field on the real axis (in Russian),Mat.Sb. 190(1999), 11-22; English transl. in Russian Acad. Sci. Sb. Math. 190(1999),791-802.
[DS]
P. D. Dragnev, E. B. Saff. Constrained energy problems with applications to orthogonal polynomials of a discrete variable. J.Anal. Math. 72(1997),223-259.
[K]
A. B. J. Kuijlaars. On the finite-gap ansatz in the continuum limit of the Toda lattice. Duke Math. J. 104(2000), 3, 433-462.
[ST]
E. B. Saff, V. Totik. Logarithmic Potentials with External Fields. Grundlehren Math. Wiss. 316, Springer, Berlin,1997.
[MS]
H.N. Maskar, E. B. Saff. Where does the sup norm of a weighted polynomial live? Constr. Approx(1985),1,71-91.
[DM]
P. Deift, K. T-R McLaughlin. A Continuum Limit of the Toda Lattice. Memoirs of the American Mathematical Society. Number 624. January 1998, Volume 131.

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