Hua C., Wong R. Differential Equations and Asymptotic Theory in Mathematical Physics
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12
n
asymptotic behavior of
pn(x).
One can also derive uniform asymptotic
expansions. For details see
R.
Wong
's
article in these proceedings
65.
Example
1.
From
(2.14)-(2.16)
we see that the orthonormal Laguerre
polynomials are
=
qa
+
iyX.
Similarly, integration by parts and the orthogonality relation yield
'Yn
'Yn-
1
-
-n
-
-
Therefore
A,(x)
=
d=/x,
Bn(z)
=
-n/z.
In this case
(3.4)
leads
to
d
dx
The differential equation
(3.8)
becomes
d2 d
dx2 dx
-X-LP)(X)
=
(n
+
a)Lpl(x)
-
nL?)(z).
z-Lp(x)
+
(1
+a
-
X)-LP)(X)
+
nL?)(x)
=
0,
(3.9)
which is of Sturm-Liouville type.
13
Example
2.
Consider the weight function
CeFX4
on
W,
so
u
=
x4
-In
C.
The boundary terms in
(3.2)-(3.3)
vanish and we have
Since
w
is an even function,
p,(-x)
=
(-l),p,(x)
so
pi(x)
is an even
function. Moreover
(2.2)
implies
b,
=
J,xpi(z)w(x)dx
=
0.
Thus the
right-hand side of the above equation is
and we find
&(x)
=
4a,(x2
+
a:
+a;+,).
(3.10)
Similarly
B,(x)
=
4a;x.
(3.11)
Therefore
pL(x)
=
4a,(x2
+ +
ai+l)p,-l(x)
-
4aixp,(~),
(3.12)
and
=
16ai(x2
+
+
~i-~)p,(~).
(3.13)
The differential equation
(3.13)
was first derived by Shohat
54
through a
complicated procedure.
For
applications see Nevai
47.
One can also derive
nonlinear relations among the recursion coefficients
{a,}
from
(3.12)
by
using the recurrence relation
xpn(x)
=
an+lpn+l(x)
+
anPn-l(X).
(3.14)
Clearly
(3.14)
yields
(3.15)
xn
+
C,Z,-~
+
lower order terms.
pQ(x)
=
a1ag..
.a,
Equating the coefficients of
x*-l
in
(3.12)
proves
-
44ai
+
d+l)
+
4a,(c,-1
-
a,c,).
(3.16)
n
al..
.a,
--
a1
.
.
.
a,-1
14
On the other hand, substituting from
(3.15)
into
(3.14)
and equating the
coefficients of
xn-'
we find
Therefore
(3.16)
becomes
n
=
4~2(ai-~
+
a:
+
a2+,).
(3.17)
Many additional nonlinear relations follow from equating coefficients of
other powers of
x.
Freud seems to have been the first to discover such
nonlinear equations
for
recursion coefficients of polynomials orthogonal with
respect to exponential weights,
W(Z)
=
exp(-z2"). He only studied the
case
m
=
2.
Formulas like
(3.17)
are instances of the string equation
in Physics.
It
is clear that one can derive nonlinear relations similar to
(3.17)
when
v
=
x6+
constant using the same technique. References to the
literature on this are in
47.
The treatment presented here
is
from
15.
Chen and Ismail
l5
analyzed the case when
v
is
a
polynomial and de-
scribed the corresponding
A,(x)
and
Bn(x)
functions.
Qiu
and Wong
49
used the differential equation and the Chen-Ismail analysis to derive large
n
uniform isymptotics for
p,(x).
The operators
L1,n
and
Lz,,
generate a Lie algebra where the product
of two operators
A
and
B
is the Lie bracket
[A,B]
=
AB
-
BA.
Finite
dimensional Lie algebras are of interest. When
u
=
x2+
constant,
L1,n
and
Lz,,
generate a three-dimensional Lie algebra called the harmonic oscillator
algebra, Miller
45.
Miller
44
characterized all finite dimensional Lie algebras
that are generated by first order differential operators. Chen and Ismail
l5
proved that when
u(x)
is a polynomial, ~(x)
=
e-"("),
then
LI,~
and
Lz,~
generate
a
Lie algebra of dimension
2m
+
1,
2m
being the degree of
u(x).
The converse is not known
so
we state it as
a
conjecture.
Conjecture.
If
the Lie algebra generated
by
LI,~
and
Lz,,
is
finite
dimensional and the
support of
v
is
I%,
then the Lie algebra has dimension
2m
+
1
and
u
must
be
a
polynomial
of
even
degree.
The differential equation
(3.8)
when expanded out becomes
P;(X)
-
[~'(x)
+
AL(x)/An(x)l~L(x)
+
sn(x)Pn(x)
=
0,
(3.18)
where
(3.19)
15
4.
Electrostatic Equilibrium Problems
We will study Coulomb interactions in the plane where the force is
2elea/r,
el,
and
e2
being the charges
of
two particles and
r
the distance between
them. This arises in
R3
if we have infinite uniformly charged wires perpen-
dicular to the plane under consideration and
el,
and
e2
are the charges per
unit length. The potential energy is
-2ele2
In
r.
Now consider
n
movable charged particles each carrying a unit charge.
The particles are restricted to
a
subset
E
of
R
in the presence
of
an external
field with potential energy
V(x).
Let
21,
.
. .
,
x,
denote the positions of the
particles and assume
x1
>
x2
>
. . .
>
2,.
In what follows
x
will denote
the vector
(XI, 22,.
.
.
,xn).
The total energy
E(x)
of this system is
n
k=l
l<i<zj<n
The question
is
to find the equilibrium position of the movable charges,
so
we must find the stationary points of
E(x)
and among them choose the
points that give a minimum. The equation
VE(x)
=
0
is equivalent to
(4.2)
For example when
V(x)
=
x2+
constant,
(4.2)
becomes a nonlinear system
of
algebraic equations. In general
(4.2)
is
a
nonlinear system.
Stieltjes
56
considered the case when
V(x)
=
-aln(l
-
x)
-
,81n(l
+x),
and
E
=
(-1,l).
(4.3)
In other words we have two fixed charges at
fl
and the movable particles
are restricted to
(-1,l).
Stieltjes' idea was to change the nonlinear problem
(4.2)
to
a
linear problem. He observed that, for
a
fixed
j,
x-xi x-xzj
1
i=l
"1
c
-
24Zj
i=l
i#i
=
Z+Zj
lim
(z
--),
x
-
xj
where
16
It is a calculus exercise to evaluate the above limit and show that (4.2) is
This transforms the system (4.2) to
f”(x)
-
V’(x)
f’(x)
=
0,
for
x
=
x1,.
.
.
,xn.
The idea now is to choose a suitable function
h(z)
such that the validity of
f”(x)
-
V’(x)
f’(z)
+
h(z)
f
(z)
=
0,
at
z
=
XI,.
. .
,x,,
(4.5)
implies the validity of (4.5) at all
x.
Of course (4.5) holds for any
h(x)
as
long as
h(xj)
is finite,
1
5
j
5
n.
Observe that (4.5) may have more
solutions than the polynomial solutions to
y”(x)
-
V’(z)y‘(rc)
+
h(x)y(x)
=
0,
for all
x
E
E,
(4.6)
E
is
the set to which the changes are restricted.
To
connect (4.6) to. (3.14),
we need to choose
v
so
that
V’(x)
=
v’(x)
+
AL(x)/An
and choose
h
as
S,
of (3.14). This will work if the electrostatic equilibrium problem has
a
unique solution and (4.6) has a unique polynomial solution for the chosen
h.
Stieltjes proved that this holds in the special case (4.3).
A
more general
theorem is the following.
Theorem
4.1.
(Ismail
29).
Assume that
w
=
e-”,
v(x)
is convex on
(a,
b)
and that the
n
movable charges are restricted to
(a,
b).
If
the external
field
V(z)
is
v(x)
+
ln(An(x)/un)
is convex on (a, b) then the electrostatic
equilibrium problem has a unique solution and
is
given by the zeros
ofpn(x).
To
find the value
of
E(x)
at equilibrium we need the concept of dis-
criminants.
Definition
4.1.
The discriminant
D(g)
of a poIynomial
g,
n
j=1
is
2
&7)
=
Y2n-2
J-J
(Xi
-
Zj)
.
17
Stieltjes
56
and Hilbert
26
proved that the discriminant of a Jacobi
polynomial is given by
~(pp,P)(~))
=
2-4,-1)
fiJ
.j-2n+2
(j
+a)j-l(j
+p)j-l(n+j
+a+p)"-j.
j=1
(4.9)
In
28
we generalized (4.9) to polynomials orthogonal with respect to a
weight function
w
=
e-".
The key was to combine our (3.4) with an idea
of
Schur, see
6.71
in
58.
The result is the following theorem.
Theorem
4.2.
The discriminant
of
pn is given
by
where
x1,52,.
.
.
,
x,
are
the zeros
of
p,.
It is clear from (4.1) that at equilibrium the term
-2
C15i<j5n
In
Izi
-
zjl
is -1nD(f) with
f
as in (4.4).
Theorem
4.3.
(Ismail
2g).
Let
{p,} be orthonormal with respect to
w
and define an external field with potential
V(x),
V(z)
=
~(z)
+
lnA,(z)/a,.
(4.11)
Under the assumptions
of
Theorem
4.1,
the total energy at equilibrium
is
given by
n
n
~(51,
...,
2,)
=Cv(zj)
-2Cj1naj.
(4.12)
j=1
j=1
In
other words
n
n
exp(-E(zl,.
. . ,
2,))
=
n
w(zj)
n
a?.
(4.13)
j=1
j=1
The standard electrostatic equilibrium problem uses
v
as the potential
of the external field and the equilibrium occurs at the zeros of a Fekete
polynomial, see Saff and Totik 51. In general the Fekete polynomials are
not orthogonal,
so
we cannot generate them through
a
convenient recursion
relation. Our thesis is that the introduction of the term ln(A,(z)/a,) does
not change the energy at equilibrium in any significant way but has the
added advantage of having a closed form expression (4.12) for the energy
18
at equilibrium. In fact when we take
~(z)
=
z4+
constant, we proved in
29
that the energy at equilibrium is
qn2 Inn
+
czn2
+
csnlnn
+
.
.
.
,
as
n
+
00,
(4.14)
where c1 and
cz
are the same whether the external potential is
~(z)
or
V(z).
Moreover our formulation leads to closed form expressions for the energy
and to the identification of the positions of the particles. Computationally
the zeros of orthogonal polynomials are stable under slight variations of the
recursion coefficients.
It is our opinion that Theorems
4.1
and
4.3
are in the same spirit as
the work of Stieltjes
56,
57.
To explain this latter point consider the Jacobi
polynomials
{Pp’p’(z)}.
In this case the weight function is
wa,p(z)
=
C(1
-
z),(l+
z)P,
(4.15)
u
=
v,,p(z)
=
-aln(l
-
z)
-
pln(l+
z)
-
lnC1.
Here
C1
is a constant. In this case
(4.16)
(4.17)
2
a,
=
Therefore
V(z)
=
V,,p(z)
is
v,,p(z)
=
w,+l,p+l(z)
+constant,
and Stieltjes proved Theorem
4.3
in this case. Moreover it is clear that
computing the second product on the right-hand side of
(4.10)
is straight-
forward. To compute the first product in
(4.10),
we only need to compute
n:=,(l
-
zj).
On the other hand
hence
n
It is known that
PP”’(1)
=
(-l)nP~p’a’(-l)
=
(a
+
l),/n!.
Therefore
the discriminant of the orthonormal Jacobi polynomials can be found in
closed form. The interested reader is encouraged to carry out the details
and verify
(4.9).
19
To state a recent result of Ercoloni and McLaughlin
2o
we need to remind
the reader of the Laplace asymptotic method. Let
[u,b]
c
R
where a
function
f
has unique local minimum at
a
with
f”(u)
>
0
and assume
that
f
is real analytic in a neighborhood of
a
and continuous on
S.
Thus
f(z)
=
f(a)
+
(z
-
~)~f”(a)/2!
+
.
. . .
We wish to determine the large
n
behavior of the integral
b
I,
=
Jd
e-nf(x)
g(z)
dz.
The idea is that as
n
4
00,
e-nf(x) can be well-approximated by
e-”f@)exp(-n(z
-
f”(a)/2).
After the change of variable
z
=
a
+
fi
t/fl(a)
,
we see that the integral over
[a,
b]
can be replaced by
sw
and the result is that
provided that
g(a)
#
0.
more
40
who considered the integral
A
multidimensional analogue of
I,
arose in the work of Lax and Liver-
Assuming that
N
=
cn, c
E
(0,
l),
the problem is to determine the asymp-
totics of
J,,N
as
n
-+
co
and fixed
c.
Over the years this problem has
attracted the attention of many mathematicians who contributed to its
so-
lution. The final step was taken by Ercolani and McLaughlin
20.
They
wrote the integrand
as
exp(-E), and minimize
E,
then apply the nonlin-
ear steepest descent technique
of
Deift and Zhou
la.
It will be interesting
to consider the integral
In
(4.19),
{pn(z;
N)}
is
a
sequence of polynomials orthonormal with respect
to exp(-Nv(z)), with
XI,.
. .
,A,
denotes the zeros of
p,(z;
N).
Moreover
A,(z;
N)
is the
A,
function in Theorem
3.1.
As
in
(4.18),
we take
N
=
cn,
with
c
E
(0,l).
20
5.
Generating Functions
and
Asymptotics
In this section we discuss techniques of finding the large
n
behavior
of
orthogonal polynomials.
A
generating function of a sequence of polynomials
{p,(x)}
is a function
G(x,
t)
which has the expansion
m
G(x,
t>
=
c
wn(z)t",
(5.1)
0
where
{en}
are suitably chosen constants to make
G(x,t)
have
a
closed
form. The Poisson kernel is the bilinear generating function
*(5.2)
n=O
It
is not clear where the Poisson kernel converge, but its convergence or
divergence clearly depends on the large
n
behavior of
pn(z)pn(y).
The first asymptotic technique we treat is Darboux's asymptotic method
which depends on finding
a
reasonable generating function.
As
an example
we first derive a generating function for the ultraspherical (Gegenbauer)
polynomials
{C,(x)}.
The polynomials
{C,"(x)}
are generated by
C,.(x)
=
1)
C,"(Z)
=
2v2,
(5.3)
2(x
+
v)C,(x)
=
(n
+
l)c,+l(x)
+
(2v
+
n
-
l)c;-l(x),
(5.4)
for
n.
>
0.
The Legendre polynomials correspond to
v
=
1/2.
Let
M
_.
G(z,
t)
=
c
C;(x)tn
n=O
(5.5)
Multiply
(5.4)
by
tn
and add for
n
=
1,2,.
. . .
In view of
(5.3)
and after
replacing
atn
by
tatt",
we establish the differential equation
2vxG(x,
t)
+
2xt&G(x,
t)
=
&G(x,
t)
+
2vtG(~,
t)
+
t2&G(x,
t).
Therefore
dtG(x,t)
-
~V(Z
-
t)
-
G(x,
t)
1
-
2xt
+
t2
'
hence
G(x,t)
=
f(x)(l
-
2xt
+t2)+'.
Now
f(x)
=
1
because
1
=
Cg(x)
=
G(x,
0)
=
f(x).
This proves
00
c
c,(x)tn
=
(1
-
2xt
+
ty.
(5.6)
n=O
21
To justify the above formal steps we start with the answer and reverse the
steps until we reach
(5.3)-(5.4).
Theorem
5.1.
(Darboux's
Method).
xr
gntn
be
analytic in
It1
<
r
and
f
-
g
be
continuous
on
It1
5
r.
Then
Let
f(t)
=
xrfnt",
g(z)
=
f"
-
gn
=
o(?--").
(5.7)
This theorem follows from the Riemann-Lebesgue lemma, see Olver
48.
Given
f,
the function
g
is called a comparison function and it normally
consists of the singular part off.
Example.
Consider the Legendre Polynomials. Their generating func-
tion is
00
c
P,(z)t"
=
(1
-
2zt
+
ty2,
(5.8)
n=O
since the Legendre polynomials correspond to
v
=
1/2
in
{C,"(z)}.
The
notation
P,(z)
is standard in the literature but
Pn
is not monic. The
t-singularities of the generating function
(5.8)
are
at
t
=
a,
t
=
p,
where
a,p=zfdz2-1,
and
IPI5IaI.
Here
z
E
C.
Note that
a
=
,B
if and only if
z
=
fl.
Moreover
la1
=
IpI
if
and only if
z
E
[-1,1].
Case
I:
z
=
fl.
In this case the right-hand side of
(5.8)
is
(1
~t)-',
hence
Pn(fl)
=
(fl)"
and no asymptotic analysis is needed.
Case
11:
z
E
C
\
[-I,
11.
In this case
IPI
<
la[.
Let
where we used the binomial theorem in the last equation.
It
is easy to see
that
C,"
Pn(z)t"-g(t)
is continuous in
JtJ
5
and,
g(t)
and
C,"
Pn(z)tn
are analytic in
It1
<
IpI.
Therefore Darboux's method yield
(1
-
r(n
+
1/2)
P"
r(1/2p
.
Pn(x)
M
Using
nb-"r(a
+
n)/r(b
+
n)
4
1
as
n
4
00,
we see from the above
asymptotic formula that
P"(2)
M
(1
-
p/a)-1/2p-"/J..n,
(5.9)
since
r(1/2)
=
J;;.