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ANNALES
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L’INSTITUT FOURIER
Leonid DICKEY
Proof of the Treves theorem on the KdV hierarchy
Tome 55, no 6 (2005), p. 2015-2023.
<http://aif.cedram.org/item?id=AIF_2005__55_6_2015_0>
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Ann. Inst. Fourier, Grenoble
55, 6 (2005), 2015–2023
PROOF OF THE TREVES THEOREM
ON THE KdV HIERARCHY
by Leonid A. DICKEY
1. Necessity of the Treves condition for KdV.
Here we give a shorter proof of the Treves theorem [1] and some
addition to the theorem (Theorem 2 below). A discussion of the significance
of the theorem, and a part of the present proof (necessity) one can find in
[2] along with an attempt to generalize the theorem.
Theorem 1 (Treves). — A differential polynomial of u: P [u] =
P (u, u′, u′′, . . .) is, up to an exact derivative, a linear combination of
res∂ Lm/2 where L = ∂2 + u if and only if
(1) resx P (ũ(x), ũ′(x), ũ′′(x), . . .) = 0,
where ũ(x) is an arbitrary formal Laurent series of the form
(2) ũ(x) = −2x−2 +
∞∑
0
ui(xi/i!), u1 = 0.
(The following notations are used: res∂ symbolizes the coefficient in
∂−1, and resx the coefficient in x−1).
We also prove the following addition to the Treves theorem:
Theorem 2. — A differential polynomial P [u] = P (u, u′, u′′, . . .)
is exactly a linear combination of res∂ Lm/2 (without additional derivative
Keywords: Kdv hierarchy, first integrals, Treves’ criterion.
Math. classification: 35Q53.
2016 Leonid A. DICKEY
terms) if and only if P (ũ(x), ũ′(x), ũ′′(x), . . .) is a Laurent series with only
one singular term const ·x−2.
The beginning of the proof of the theorem 1. — In this section we
prove the necessity of the Treves condition (1).
Let us try to “undress” the operator L = ∂2 + u:
∂2+u = w∂2w−1 = (1+w1∂−1+w2∂−2+· · ·)∂2(1+w1∂−1+w2∂−2+· · ·)−1.
Rewrite this as
(1 + w1∂−1 + w2∂−2 + · · ·)∂2 = (∂2 + u)(1 + w1∂−1 + w2∂−2 + · · ·)
which yields the recurrence relations
2w′1 + u = 0
2w′k+1 + w
′′
k + uwk = 0, k > 0.
First, a lemma will be proven:
Lemma 1. — If a formal Laurent series ũ = −2/x2 +
∑∞
0 ui(x
i/i!),
u1 = 0, is taken for u, then all wk can be found in the form of Laurent
series.
The necessity of the Treves condition immediately follows from this
lemma. Indeed,
res∂ Lm/2 = res∂ w∂mw−1 = res∂ [w∂m, w−1] + res∂ w−1w∂m
= res∂ [w∂m, w−1] = ∂( )
since the residue of the commutator of any two operators is an exact
derivative. In this case this is an exact derivative of a Laurent series.
Therefore, it cannot contain a term with x−1, i.e., its residue with respect
to the variable x is zero.
Proof of the lemma 1. — We have 2w′1 + ũ = 0 whence
w1 = −
1
x
− 1
2
∞∑
0
ui
xi+1
(i+ 1)!
= − 1
x
−
∞∑
1
bi
xi
i!
, (b2 = 0).
Further, −w′2 = w′′1 − 2w′1w1 and −w2 = w′1 − w21. It is easy to calculate
(taking into account that b2 = 0) that w2 = (5b3/6 + b21)x
2 + O(x3) =
Ax2 +O(x3) = O(x2). Here, O(xn) means a power series starting with the
ANNALES DE L’INSTITUT FOURIER
PROOF OF THE TREVES THEOREM ON THE KdV HIERARCHY 2017
term involving xn. Using the recurrence formula, it is not difficult to show
by induction that all the next terms have the same form:
−w′k+1 = w′′k − 2w′1wk = 2A+O(x)− 2(x−2 + b1 +O(x2))(Ax2 +O(x3))
= O(x)
whence wk+1 = O(x2). 
2. Proof of the sufficiency of the Treves condition.
There is a grading in the differential algebra A of polynomials in
symbols u(k), P [u] = P (u, u′, u′′, . . .): w(u(n)) = n + 2, w(∂) = 1. If all
terms of a polynomial P have the same weight k then
P (λ2u, λ3u′, λ4u′′ . . .) = λkP (u, u′, u′′, . . .).
Lemma 2. — If a differential polynomial P satisfies the Treves
condition (1) then so does each homogeneous in weight component of this
polynomial.
Proof of the lemma 2. — Let P =
∑
Pκ where Pκ a homogeneous
polynomial of weight κ. Since {un} are arbitrary, we can replace them by
unλ
n+2. Now,
resx
∑
Pκ
[
−2/x2 + λ2u0 +
∞∑
2
λn+2un
xn
n!
]
= resx
∑
λκPκ
[
−2/(λx)2 + u0 +
∞∑
2
un
(λx)n
n!
]
=
∑
λκ−1 resx Pκ
[
−2/x2 + u0 +
∞∑
2
un
xn
n!
]
.
If this is zero, then each term is zero since λ is arbitrary. 
Therefore, we can consider each component of weight κ separately.
The first integral res∂ Lm/2 where m = 2k − 1 has the weight 2k. It is
possible to prove that it contains a term Cuk with a non-zero coefficient C.
Indeed, dealing with the terms in res∂ Lm/2 where u is not differenti-
ated, one can consider ∂ and u as commuting. Then
res∂(∂2 + u)m/2 = res∂ ∂m(1 + u∂−2)m/2 = res∂ ∂m
∞∑
0
(
m/2
k
)
(u∂−2)k
=
(
m/2
(m+ 1)/2
)
u(m+1)/2.
TOME 55 (2005), FASCICULE 6
2018 Leonid A. DICKEY
If P is a differential polynomial satisfying the Treves condition (1),
then subtracting from it a linear combination of first integrals res∂ Lm/2,
one can achieve that it does not contain terms Cuk preserving the property
to satisfy the condition (1). Then we reduce this polynomial. Namely, we
reduce the order of the highest derivative involved in a differential monomial
by “integration by parts” as much as possible: if a differential monomial
(u(i1))p1(u(i2))p2 · · · (u(ik))pk where i1 < i2 < · · · < ik has pk = 1 then the
highest order of the derivative, ik, can be reduced if ik = 0 by addition of
an exact derivative, for example (u′)2u′′′ = −2u′(u′′)2 + ∂((u′)2u′′). The
second term is an exact derivative and the highest derivative involved in
the first term is the second one. Another example: uu′u′′ = u(u′2)′/2 =
∂(uu′2/2)− u′3/2. One can proceed doing this until all the monomials will
contain their highest derivatives in power > 1 (with a possible exception:
a term Cu). We call this the reduced form of a differential polynomial. It
is unique.
The reduced polynomial preserves the property (1) and does not
contain the terms Cuk. It remains to prove the following:
Lemma 3. — A reduced differential polynomial homogeneous with
respect to the weight which satisfies the condition (1) and does not contain
the term Cuk is zero.
Suppose that it is not zero. Let us write Eq. (1) in more detail:
resxQ
(
−2/x2 + u0 +
∞∑
2
ui(xi/i!), 4/x3 +
∞∑
2
ui∂(xi/i!),
−12/x4 +
∞∑
2
ui∂
2(xi/i!), . . .
)
= 0.
This equality can be differentiated with respect to u0 which is the same as
resx ∂Q[ũ]/∂u = 0. This operation can be repeated until there will be no
factor u at all, and, nevertheless, the polynomial is not zero since the term
Cuk is absent. More than that, the polynomial preserves all the properties
assumed in Lemma 3. We have
resxQ
(
4/x3 +
∞∑
2
ui∂(xi/i!),−12/x4 +
∞∑
2
ui∂
2(xi/i!), . . .
)
= 0.
Now let us take the derivative with respect to an arbitrary uk, k = 2, 3, . . .:
resx
∞∑
1
∂Q
(
4/x3 +
∑∞
2 ui∂(x
i/i!),−12/x4 +
∑∞
2 ui∂
2(xi/i!), . . .
)
∂u(n)
×∂n(xk/k!) = 0.
ANNALES DE L’INSTITUT FOURIER
PROOF OF THE TREVES THEOREM ON THE KdV HIERARCHY 2019
Integrating by parts, we get:
resx
∞∑
1
(−∂)n−1
×∂Q
(
4/x3 +
∑∞
2 ui∂(x
i/i!),−12/x4 +
∑∞
2 ui∂
2(xi/i!), . . .
)
∂u(n)
· x
k−1
(k − 1)! = 0
or
resx
δQ(4/x3+
∑∞
2 ui∂(x
i/i!),−12/x4+
∑∞
2 ui∂
2(xi/i!), . . .)
δu′
· x
k−1
(k−1)! = 0
where k = 2, 3, . . . . Denoting the variational derivative δQ/δu′ as R, we
have
resx xR(4/x3 +
∞∑
2
ui∂(xi/i!),−12/x4 +
∞∑
2
ui∂
2(xi/i!), . . .)xk−2 = 0
whence
(3)
(
xR(4/x3 +
∞∑
2
ui∂(xi/i!),−12/x4 +
∞∑
2
ui∂
2(xi/i!), . . .)
)
−
= 0.
Thus, the problem is now the following: to show that if R[v] (v = u′) is a
homogeneous polynomial satisfying Eq. (3), then R = 0. If this is proven,
then δQ/δu′ = 0 and Q is an exact derivative which is incompatible with
the fact that it is a non-zero reduced polynomial.
We shall prove a slightly more general lemma, the generalization is
needed in the proof of the theorem 2.
Lemma 4. — Let R(u, u′, u′′, . . .) be a homogeneous polynomial
satisfying(
xR(−2/x2 +
∞∑
0
ui(xi/i!), 4/x3 +
∞∑
2
ui∂(xi/i!), . . .)
)
−
= 0,
where u1 = 0 and other ui are arbitrary. Then R = 0.
Let R be of weight κ. From the homogeneity, it follows that the given
equality can be written as
(4)
(
x1−κR(−2 +
∞∑
0
uix
i+2/i!, 4 +
∞∑
2
uix
i+2/(i− 1)!, . . .)
)
−
= 0.
Now, one must expand R(−2 +
∑∞
0 uix
i+2/i!, 4 +
∑∞
2 uix
i+2/
(i−1)!, . . .) in powers of x and write that all terms of power less than κ− 1
TOME 55 (2005), FASCICULE 6
2020 Leonid A. DICKEY
vanish. For that, it is more convenient to expand this expressions in powers
of ui, it automatically will be an expansion in powers of x.
The arguments of R: u, u′, . . . we denote as ξ0, ξ1, . . ., thus, R =
R(ξ0, ξ1, . . . , ξµ). We have (denoting ∂uj = ∂/∂uj etc)
∂ujR
(
−2 +
∞∑
0
ui
xi+2
i!
, 4 +
∞∑
2
ui
xi+2
(i− 1)! , . . .
)
= ∂ujR(ξ1, ξ2, . . .)
= DjR(ξ0, ξ1, . . .)xj+2 where Dj =
1
j!
∂ξ0 +
1
(j − 1)!∂ξ1 + · · ·+ ∂ξj
for j = 0, 2, 3, . . . . Now:
x1−κR
(
−2 +
∞∑
0
xi+2
i!
, 4 +
∞∑
2
ui
xi+2
(i− 1)! ,−12 +
∞∑
2
ui
xi+2
(i− 2)! , . . .
)
(5) =
∑
q0,q2,...
Dq00 D
q2
2 · · ·R(θ)
q0!q2! · · ·xκ−1−λ(q)
uq00 u
q2
2 · · ·
where
(θ) = (−2 · 1!, 2 · 2!,−2 · 3!, . . . , (−1)µ+12 · (µ+ 1)!), λ(q) =
∑
(j + 2)qj .
The condition that the expression (5) does not contain negative
powers of x becomes
(6) Dq00 D
q2
2 · · ·R(θ) = 0 when λ(q) < κ− 1.
Lemma 5 on homogeneous polynomials. — Let
R(ξ0, ξ1, ξ2, . . . , ξµ) =
∑
(p)
a(p0p1p2···pµ)ξ
p0
0 ξ
p1
1 ξ
p2
2 · · · ξpµµ
where 2p0 + 3p1 + 4p2 + · · · + (µ + 2)pµ = κ  2. Suppose the equation
(6) where (θ) = (ξ∗0 , ξ
∗
1 , ξ
∗
2 , . . .) is a fixed set of nonzero values of the
corresponding variables is satisfied for all sets of integers {q0, q2, q3, , , , }
such that λ(q) ≡
∑
(2 + j)qj  κ− 2. If κ  4, we consider only sets {qi}
such that not all of qi are zero. Then all coefficients a(p) are zero.
Notice that if κ < 4, the polynomial R contains only one term, and
the proof is obvious.
ANNALES DE L’INSTITUT FOURIER
PROOF OF THE TREVES THEOREM ON THE KdV HIERARCHY 2021
3. Proof of the Lemma 5 and the end of the proof
of the sufficiency of the Treves condition.
We use the induction with respect to the number of variables
ξ0, ξ1, . . . , ξµ. For µ = 0 the statement is trivial. Let it be proven for µ− 1.
The Euler formula for the weight-homogeneous polynomial R reads
2ξ0∂ξ0R+ 3ξ1∂ξ1R+ 4ξ2∂ξ2R+ · · ·+ (µ+ 2)ξµ∂ξµR = κR.
Solving µ equations with µ unknowns, one can express ∂ξ0 , . . . , ∂ξµ−1 in
terms of D0, D2, . . . , Dµ and ∂ξµ :
∂ξj = τj0D0 +
µ∑
k=2
τjkDk + σj∂ξµ , j = 0, . . . , µ− 1
with constant coefficients. Substituting this for ∂ξi in the Euler equation,
we get (
a0(ξ)D0 +
µ∑
2
αi(ξ)Di + β(ξ)∂ξµ
)
R = κR
where coefficients linearly depend on {ξi}. Hence,
∂ξµR =
(
b(ξ) + a0(ξ)D0 +
µ∑
2
ai(ξ)Di
)
R
with coefficients which are rational functions of {ξi}. Iterating this formula,
we have
(7) ∂mξµR =
∑
j0+j2+···+jµm
αj0,j2,...,jµ(ξ)D
j0
0 D
j2
2 · · ·Djµµ R.
One can write R =
∑d
0 aj(ξ0, . . . , ξµ−1)ξ
j
µ/j! where ad is not the identical
zero. We consider two cases: when ad is not a constant and when it is a
constant. If ad is not a constant, its weight is at least 2 (since this is the
smallest weight of variables ξi).
Now we can prove that the weight-homogeneous differential polyno-
mial ad(ξ0, . . . , ξµ−1) of weight κ1 = κ−(µ+2)d, if κ1  2, has the property
Dq00 D
q2
2 · · ·D
qµ−1
µ−1 ad(θ) = 0 if λ(q) = 2q0 +4q2 + · · ·+(µ+1)qµ−1  κ1− 2.
Indeed, using (7), we obtain
Dq00 D
q2
2 · · ·D
qµ−1
µ−1 ad(θ) = D
q0
0 D
q2
2 · · ·D
qµ−1
µ−1 ∂
d
ξµR(θ)
=
∑
j0+j2+···+jµd
αj0,j2,...,jµ(ξ)D
j0
0 D
j2
2 · · ·Djµµ Dq00 Dq22 · · ·D
qµ−1
µ−1 R(θ).
TOME 55 (2005), FASCICULE 6
2022 Leonid A. DICKEY
Since
(2j0 + 4j2 + · · ·+ (µ+ 2)jµ)+(2q0 + 4q2 + · · ·+ (µ+ 1)qµ−1)
 (µ+ 2)(j0 + j2 + · · ·+ jµ) + κ1 − 2
 (µ+ 2)d+ κ− (µ+ 2)d− 2 = κ− 2,
all terms of the sum vanish. Since ad depends on µ−1 variables, the lemma
5 is supposed to be true for ad, and ad is identically zero. Therefore, R = 0.
Now, we discuss the second alternative: ad is a constant. Then, instead
of ad = ∂dξµR, we consider
ãd = ∂d−1ξµ R = ad−1(ξ0, . . . , ξµ−1) + cξµ, c = const .
We have κ2 = w(ãd) = µ + 2  2. Moreover, if ad−1 = 0, κ2  4
since ad−1 cannot be linear. Just as in the first case, we can prove that
Dq00 D
q2
2 · · · ãd(θ) = 0 if λ(q)  κ2 − 2. If not all of qi are zero, then
Dq00 D
q2
2 · · · ãd(θ) = Dq00 Dq22 · · · ad−1(θ) = 0, and we conclude that ad−1 = 0.
Then c = 0 since ∂d−1ξµ R(θ) = 0, and ad = 0. Thus, R = 0 and the lemmas
5, 4, and also the theorem 1 are proven. 
4. Proof of the Theorem 2.
In the same way as before, one can restrict himself to homogeneous
polynomials. Let resxR(ũ(x), ũ′(x), ũ′′(x), . . .) = 0 for all series (2),R being
a differential polynomial. One can differentiate this equality with respect
to u0, u2, u3, . . .:
resx
∂R
∂u
= 0, resx
(
∂R
∂u
· x
2
2
+
∂R
∂u′
· ∂ x
2
2
+
∂R
∂u′′
· ∂2x
2
2
)
= 0, . . . ,
or
resx
δR
δu
= 0, resx
δR
δu
· x
2
2!
= 0, resx
δR
δu
· x
3
3!
, . . .
which means that P = δR/δu, after the substitution (2), can have, as a
Laurent polynomial, only one singular term ax−2. It is well-known that
any polynomial res∂ Lm/2 is the variational derivative of the next one,
res∂ L(m+2)/2 (see [3], Proposition 3.5.2.). The theorem is proven one way.
Conversely, let P be a homogeneous differential polynomial such
that after the substitution (2) there is only one singular term, ax−2. In
particular, resx P (ũ, ũ′, . . .) = 0. Then P is a sum of c · res∂ Lm/2 and
a derivative, ∂S(u, u′, . . .). Unless S is identically zero, it must contain,
ANNALES DE L’INSTITUT FOURIER
PROOF OF THE TREVES THEOREM ON THE KdV HIERARCHY 2023
after the substitution (2), singular terms of order x−2 or higher since if it
were not so, i.e., (xS(ũ, ũ′, . . .))− = 0, then, according to the Lemma 4, S
would be zero. Then ∂S has nonzero singular terms of order at least x−3
in contradiction to the assumption. Thus, S = 0, and the given differential
polynomial is just c · res∂ Lm/2. 
BIBLIOGRAPHY
[1] F. TREVES, An algebraic characterization of the Korteweg - de Vries hierarchy.
Duke Math. J., 108-2 (2001), 251–294.
[2] L. A. DICKEY, On a generalization of the Treves criterion for the first integrals
of the KdV hierarchy to higher GD hierarchies. Lett. Math. Phys., 65 (2003),
187–197.
[3] L. A. DICKEY, Soliton Equations and Hamiltonian Systems, 2nd Ed., Advanced
Series in Mathematical Physics, 26, World Scientific (2003).
Leonid A. DICKEY,
University of Oklahoma
Department of Mathematics
Physical Sciences Center
601 Elm Avenue
Norman, Oklahoma 73019 (USA)
ldickey@ou.edu
TOME 55 (2005), FASCICULE 6



Proof of the Treves theorem on the KdV hierarchy

Iliev, Plamen

ANNA
LE
S
 D
E
 L’INSTIT
UT
 F
O
U
RI
ER
ANNALES
DE
L’INSTITUT FOURIER
Plamen ILIEV
On the heat kernel and the Korteweg–de Vries hierarchy
Tome 55, no 6 (2005), p. 2117-2127.
<http://aif.cedram.org/item?id=AIF_2005__55_6_2117_0>
© Association des Annales de l’institut Fourier, 2005, tous droits
réservés.
L’accès aux articles de la revue « Annales de l’institut Fourier »
(http://aif.cedram.org/), implique l’accord avec les conditions
générales d’utilisation (http://aif.cedram.org/legal/). Toute re-
production en tout ou partie cet article sous quelque forme que ce
soit pour tout usage autre que l’utilisation à fin strictement per-
sonnelle du copiste est constitutive d’une infraction pénale. Toute
copie ou impression de ce fichier doit contenir la présente mention
de copyright.
cedram
Article mis en ligne dans le cadre du
Centre de diffusion des revues académiques de mathématiques
http://www.cedram.org/
Ann. Inst. Fourier, Grenoble
55, 6 (2005), 2117–2127
ON THE HEAT KERNEL AND
THE KORTEWEG-DE VRIES HIERARCHY
by Plamen ILIEV
1. Introduction and examples.
Consider the one-dimensional Schrödinger (or Sturm-Liouville) oper-
ator
(1.1) L =
∂2
∂x2
+ u(x).
Its heat kernel H(x, y, t) is the fundamental solution of the heat equation
(1.2)
(
∂
∂t
− L
)
f = 0.
It is well known that H(t, x, y) has an asymptotic expansion of the form
(1.3) H(x, y, t) ∼ e
− (x−y)
2
4t
√
4πt
(
1 +
∞∑
n=1
Hn(x, y)tn
)
as t→ 0 + .
The differential equation (1.2) forH(x, y, t) implies the recursion-differential
equations for the coefficients Hn = Hn(x, y):
H0 = 1(1.4)
(x− y)∂Hn
∂x
+ nHn = LHn−1 for n  1.(1.5)
This system is known to admit unique smooth solutions Hn = Hn(x, y) in
some neighborhood of the diagonal x = y. The coefficients Hn are named
after J. Hadamard [13], who constructed them for the first time.
Keywords: Heat kernel expansions, KdV hierarchy, tau functions.
Math. classification: 35Q53, 35K05, 37K10.
2118 Plamen ILIEV
Computation of heat invariants of self-adjoint elliptic operators is a
well known problem in spectral theory which has many applications, in
particular to geometry and theoretical physics [5, 6, 10, 11, 17, 18]. The
asymptotics of the one-dimensional Schrödinger operator are of particular
interest due to their relations to the Korteweg-de Vries (KdV) hierarchy.
More precisely, it is known that the restriction of the heat coefficients on
the diagonal gives the right-hand sides of the KdV hierarchy, see [19, 21].
In the present paper we show that there are simple formulas for the
Hadamard’s coefficients Hn(x, y) in terms of the τ -function of the KdV
hierarchy (see the next section for a precise definition of the τ -function).
Remarkable explicit formulas for the coefficients of the Taylor expan-
sion of Hn(x, y) around the diagonal x = y were previously constructed
in [4], Theorem 1.3. However, these formulas have a rather complicated
combinatorial structure and it is practically impossible to write a closed
formula for the coefficients even for simple potentials u. One advantage of
the formulas derived in this paper is that they give finite expressions for
the heat coefficient if the τ -function is known (e.g. the solitons, or the more
general algebro-geometric solutions of KdV).
To see the importance of the KdV equations, let us compute the first
few coefficients using the defining relations (1.4)–(1.5). Anticipating the
appearance of the τ -function, let us write u(x) as
u(x) = 2
∂2 log(τ(x))
∂x2
.
From (1.5) one can easily obtain simple formulas for H1 and H2
(1.6) H1(x, y) =
2
x− y
(
τ ′(x)
τ(x)
− τ
′(y)
τ(y)
)
and
(1.7) H2(x, y) =
2
(x−y)2
((
τ ′′(x)
τ(x)
+
τ ′′(y)
τ(y)
)
−H1(x, y)−2
τ ′(x)τ ′(y)
τ(x)τ(y)
)
.
For the third coefficient we have the following formula
(x− y)3H3(x, y) = −6(x− y)H2(x, y)2
(
τ ′′′(x)
τ(x)
− τ
′′′(y)
τ(y)
)
− 2
(
τ ′′(x)τ ′(x)
τ2(x)
− τ
′′(y)τ ′(y)
τ2(y)
)
(1.8)
+ 4
(
τ ′(x)
τ(x)
τ ′′(y)
τ(y)
− τ
′(y)
τ(y)
τ ′′(x)
τ(x)
)
+
4
3
((
τ ′(x)
τ(x)
)3
−
(
τ ′(y)
τ(y)
)3)
+
∫ x
y
u2(ξ)dξ.
ANNALES DE L’INSTITUT FOURIER
THE HEAT KERNEL AND THE KdV HIERARCHY 2119
Notice that the integral cannot be computed explicitly, unless some-
thing remarkable happens. This is the place where the KdV equation comes
in. Assume that u(x) depends on a additional parameter s3 and it satisfies
the KdV equation
(1.9) 4∂3u = u′′′ + 6uu′,
where ∂3 = ∂/∂s3 stands for the partial derivative with respect to s3, and
u′ is the derivative with respect to x. Then one can easily see that
∫ x
y
u2(ξ)dξ =−2
3
(
τ ′′′(x)
τ(x)
−τ
′′′(y)
τ(y)
)
+ 2
(
τ ′′(x)τ ′(x)
τ2(x)
−τ
′′(y)τ ′(y)
τ2(y)
)
(1.10)
−4
3
((
τ ′(x)
τ(x)
)3
−
(
τ ′(y)
τ(y)
)3)
+
8
3
∂3 log
τ(x)
τ(y)
,
which combined with (1.8) leads to a simple formula for H3(x, y).
We’ll extend these computations by showing that if u is a solution of
the KdV hierarchy and τ is the corresponding τ -function, then there are
simple explicit formulas for Hn(x, y) in terms of τ .
The paper is organized as follows. In the next section we recall some
basic facts about the KdV hierarchy and Sato theory, which are needed
for the formulation and the proof of the main result. In Section 3, we
prove a general formula for Hn(x, y). It is interesting that the smoothness
of the coefficient Hn(x, y) on the diagonal is related to the Gegenbauer
polynomials. As a corollary of the main theorem, we see the symmetry of
the coefficients about the diagonal x = y as well as the connection between
Hn(x, x) and KdV equations.
As another application of the explicit formula, we show in [16] that
the expansion is finite if and only if the potential u(x) is a rational solution
of the KdV hierarchy decaying at infinity studied in [1, 2]. Equivalently,
one can characterize the corresponding operators as the rank one bispectral
family in [9]. For related results concerning the finiteness property of the
heat kernel expansion on the integers and rational solutions of the Toda
lattice hierarchy see [12]. For solitons of the Toda lattice and purely discrete
versions of the heat kernel see [14].
TOME 55 (2005), FASCICULE 6
2120 Plamen ILIEV
2. Korteweg-de Vries hierarchy and Sato theory.
In this section we recall some basic facts about KdV hierarchy and
Sato theory. For more details on this and the more general Kadomtsev-
Petviashvili hierarchy we refer the reader to the papers [20, 7] or the more
detailed expositions [8, 22].
Let
L =
∂2
∂x2
+ u(x)
be a second order differential operator. The KdV hierarchy is defined by
the Lax equations
(2.1)
∂L
∂sj
=
[
(Lj/2)+ , L
]
,
where j = 1, 3, 5, . . . is an odd positive integer and (Lj/2)+ is the differential
part of the pseudo-differential operator Lj/2. The first equation (for j = 1)
simply means that u(x, s1, s3, s5, . . .) = u(x + s1, s3, s5, . . .), giving us the
convenience to occasionally identify x and s1. The next equation (for j = 3)
is exactly the KdV equation (1.9). Let us represent L in a dressing form
(2.2) L = W∂2W−1,
where W is a pseudo-differential operator of the form
(2.3) W =
∞∑
k=0
ψk∂
−k, ψ0 = 1.
The wave (Baker) function Ψ(x, s, z) and the adjoint wave function
Ψ∗(x, s, z) are defined as
Ψ(x, s, z) = W exp
(
xz +
∞∑
i=1
s2i−1z
2i−1
)
(2.4)
=
( ∞∑
k=0
ψkz
−k
)
exp
(
xz +
∞∑
i=1
s2i−1z
2i−1
)
and
Ψ∗(x, s, z) = (W ∗)−1 exp
(
−xz −
∞∑
i=1
s2i−1z
2i−1
)
(2.5)
=
( ∞∑
k=0
ψ∗kz
−k
)
exp
(
−xz −
∞∑
i=1
s2i−1z
2i−1
)
,
ANNALES DE L’INSTITUT FOURIER
THE HEAT KERNEL AND THE KdV HIERARCHY 2121
whereW ∗ is the formal adjoint to the pseudo-differential operatorW . Using
(2.2) one can easily see that
(2.6) LΨ(x, s, z) = z2Ψ(x, s, z) and LΨ∗(x, s, z) = z2Ψ∗(x, s, z).
We shall also use the reduced wave function Ψ̄ and the reduced adjoint
wave function Ψ̄∗ obtained from Ψ and Ψ∗, respectively, by omitting the
exponential factor, i.e.
(2.7) Ψ̄(x, s, z) =
∞∑
k=0
ψkz
−k
and
(2.8) Ψ̄∗(x, s, z) =
∞∑
k=0
ψ∗kz
−k.
Equations (2.6) imply
(2.9)
LΨ̄(x, s, z) + 2z∂xΨ̄(x, s, z) = 0 and LΨ̄∗(x, s, z)− 2z∂xΨ̄∗(x, s, z) = 0.
Using equations (2.2)–(2.5) one can show that the wave and the adjoint
wave function satisfy the following bilinear identities
(2.10) resz
(
z2nΨ(l)(x, s, z)Ψ∗(x, s, z)
)
= 0,
for all nonnegative integers n and l, where Ψ(l)(x, s, z) is the lth derivative
of Ψ with respect to x, and the residue is around z =∞.
The remarkable discovery of the Kyoto school was that the KdV
hierarchy (2.1) could be described by a function τ(x, s). This goes back
to an earlier work of Hirota, see [15]. The reduced wave and the reduced
adjoint wave functions can be expressed in terms of τ(x, s) by the following
formulas
(2.11) Ψ̄(x, s, z) =
τ(x; s− [z−1])
τ(x, s)
and Ψ̄∗(x, s, z) =
τ(x; s+ [z−1])
τ(x, s)
,
where [z] = (z, z3/3, z5/5, . . .).
Finally, let us denote by Wn(x, y) the coefficients of the function1
Ψ̄(x, s, z)Ψ̄∗(y, s, z), i.e.
(2.12) Ψ̄(x, s, z)Ψ̄∗(y, s, z) =
∞∑
n=0
Wn(x, y)z−n.
1 This function is closely related to the Green function for L.
TOME 55 (2005), FASCICULE 6
2122 Plamen ILIEV
Using (2.11) we can easily write an explicit formula for Wn in terms of
the τ -function. If we denote by Sk(s) the elementary Schur polynomials
defined by
(2.13)
∞∑
k=0
Sk(s)zk = exp
( ∞∑
k=1
s2k−1z
2k−1
)
,
then we have
(2.14) Wn(x, y) =
∑n
k=0[Sk(−∂̃)τ(x, s)] [Sn−k(∂̃)τ(y, s)]
τ(x, s)τ(y, s)
,
where
(2.15) ∂̃ = (∂1, ∂3/3, . . . , ∂2k−1/(2k − 1), . . .).
3. Explicit formulas for Hadamard’s coefficients.
The main result of the paper is the following theorem.
Theorem 3.1. — The Hadamard’s coefficients can be computed
from the following relation
(3.1) Hn(x, y) = (−1)n
n−1∑
k=0
2n−k(n− k)2k
k!
Wn−k(x, y)
(x− y)n+k ,
where (α)k = α(α + 1) . . . (α + k − 1) denotes the Pochhammer symbol,
and Wn(x, y) are defined by (2.14).
Proof. — To prove that the Hadamard’s coefficients are given by
(3.1) we need to check that (1.5) holds and that Hn(x, y) are smooth on
the diagonal x = y.
To see that (1.5) holds, let us denote
(3.2) fn(x, y, z) = (−1)n
n−1∑
k=0
2n−k(n− k)2k
k!(x− y)n+k z
n−k−1.
Then (3.1) can be rewritten as
(3.3) Hn(x, y) = resz
[
fn(x, y, z)Ψ̄(x, s, z)Ψ̄∗(y, s, z)
]
.
ANNALES DE L’INSTITUT FOURIER
THE HEAT KERNEL AND THE KdV HIERARCHY 2123
Using the last equation together with (2.9) one can easily see that
[(x− y)∂x + n]Hn(x, y)− L(x, ∂x)Hn−1(x, y)
= resz
[(
(x− y)∂xfn(x, y, z) + nfn(x, y, z)
− ∂2xfn−1(x, y, z)
)
Ψ̄(x, s, z)Ψ̄∗(y, s, z)
+
(
(x− y)fn(x, y, z) + 2zfn−1(x, y, z)
− 2∂xfn−1(x, y, z)
)
∂xΨ̄(x, s, z)Ψ̄∗(y, s, z)
]
.
A direct computation now shows that
(x− y)∂xfn(x, y, z) + nfn(x, y, z)− ∂2xfn−1(x, y, z) = 0
(x− y)fn(x, y, z) + 2zfn−1(x, y, z)− 2∂xfn−1(x, y, z) = 0,
which proves (1.5).
Next we need to show that Hn(x, y) is well defined on the diagonal.
Writing Hn(x, y) as
(3.4) Hn(x, y) =
2(−1)n
(x− y)2n−1
n−1∑
k=0
2k(x− y)k(k + 1)2n−2k−2
(n− k − 1)! Wk+1(x, y),
and applying L’Hôpital’s rule we see that we need to prove that for
j = 0, 1, . . . , 2n− 2 we have
(3.5)
n−1∑
k=0
2k
(
j
k
)
(2n− 2− k)!
(n− k − 1)! ∂
j−k
x Wk+1(x, y)|x=y = 0.
Using (2.12) and
∂jxΨ̄(x, s, z) = exp
(
−xz −
∞∑
i=1
s2i−1z
2i−1
)
(∂x − z)jΨ(x, s, z)
we see that (3.5) is equivalent to the following identities
(3.6)
resz
[(
n−1∑
k=0
2k
(
j
k
)
(2n− 2− k)!
(n− k − 1)! z
k(∂x − z)j−kΨ(x, s, z)
)
Ψ∗(x, s, z)
]
= 0.
Equation (3.6) will follow from the bilinear identities (2.10) if we can
show that the polynomial
(3.7) Pn,j(w) =
n−1∑
k=0
2k
(
j
k
)
(2n− 2− k)!
(n− k − 1)! (w − 1)
j−k,
is an even/odd function when j is an even/odd number, respectively. It is
a pleasant surprise to see that these polynomials are closely related to very
TOME 55 (2005), FASCICULE 6
2124 Plamen ILIEV
well known classical orthogonal polynomials – the so called Gegenbauer
polynomials.
The Gegenbauer (or ultraspherical) polynomials are defined by
(3.8) Cλn(w) =
n∑
k=0
2k
(
λ+ k − 1
k
)(
2λ+ n+ k − 1
n− k
)
(w − 1)k,
see for example [3], pages 302–303. Notice that this definition can be used
for arbitrary λ. If λ > −1/2 and λ = 0 these polynomials are orthogonal on
the interval (−1, 1) with respect to (1−x2)λ− 12 which, in particular, implies
that Cλn(w) is an even/odd function when n is even/odd, respectively.
However, we need these polynomials also for negative values of λ. In this
case, we can use the three term recurrence relation
(3.9) 2(n+ λ)wCλn(w) = (n+ 1)Cλn+1(w) + (n+ 2λ− 1)Cλn−1(w).
and Cλ0 = 1 and Cλ1 = 2λw to deduce that Cλn(w) is an even/odd polynomial
when n is even/odd, respectively.
Changing the summation index in (3.7) we can rewrite Pn,j(w) as
(3.10) Pn,j(w) =
j∑
k=max(0,j−n+1)
2j−k
(
j
k
)
(2n− j − 2 + k)!
(n− j + k − 1)! (w − 1)
k.
From the last equation and the defining relation (3.8) for the Gegenbauer
polynomials one can see that2
Pn,j(w) = j!2n−1(2n− 2j − 3)!! Cn−j−
1
2
j (w) for 0  j  n− 1,(3.11)
Pn,j(w) =
(−1)j−n+1j!2n−1
(2j − 2n+ 1)!! C
n−j− 12
j (w) for n  j  2n− 2,(3.12)
which completes the proof. 	

From (2.14) and (3.1) we obtain the following
Corollary 3.2. — The Hadamard’s coefficientsHn(x, y) are sym-
metric functions of x and y, i.e. we have
Hn(x, y) = Hn(y, x).
Finally, we show that the heat coefficients {Hn(x, x)} determine the
right-hand sides of KdV equations (2.1).
2 (−1)!! = 1 and (2k − 1)!! = 1 · 3 · · · (2k − 1) for k  1.
ANNALES DE L’INSTITUT FOURIER
THE HEAT KERNEL AND THE KdV HIERARCHY 2125
Corollary 3.3. — We have
(3.13) Hn(x, x) =
2n
(2n− 1)!!W2n(x, x)
and
(3.14)
[
(L
2n−1
2 )+, L
]
= 2∂xW2n(x, x).
Thus, the KdV hierarchy (2.1) is equivalent to the following equations
(3.15) ∂2n−1u =
(2n− 1)!!
2n−1
∂xHn(x, x), for n = 1, 2, . . . .
Proof. — Using (3.1) and applying L’Hôpital’s rule 2n− 1 times we
see that
Hn(x, x) = (−1)n
2
(2n− 1)!
×
n−1∑
k=0
2k
(
2n− 1
k
)
(2n− k − 2)!
(n− k − 1)! ∂
2n−1−k
x Wk+1(y, x)|y=x(3.16)
= (−1)n 2
(2n− 1)!
× resz
[
z2n−1
(
Pn,2n−1(z−1∂x)Ψ(x, s, z)
)
Ψ∗(x, s, z)
]
,
where Pn,2n−1(w) is the polynomial defined by (3.10) for j = 2n− 1
(3.17) Pn,2n−1(w) =
2n−1∑
k=n
22n−1−k
(
2n− 1
k
)
(k − 1)!
(k − n)! (w − 1)
k.
Notice that this time we have
(3.18) Pn,2n−1(w) = (−1)n2n−1(2n− 2)!!
(
C−n+
1
2
2n−1 (w) + 1
)
,
and using the same argument (the bilinear identity and the fact that
C−n+
1
2
2n−1 (w) is an odd polynomial) we obtain (3.13) from (3.16).
From (2.2), (2.4), (2.5) and (2.11) it follows that
L
2n−1
2 = W∂2n−1x W
−1 =
∞∑
i,j=0
Si(−∂̃)τ(x, s)
τ(x, s)
∂2n−1−i−jx ·
Sj(∂̃)τ(x, s)
τ(x, s)
.
Combining this formula with (2.14) we find that the coefficient of ∂−1x in
L
2n−1
2 is W2n(x, x). If we denote by (L
2n−1
2 )− the integral (Volterra) part
of the pseudo-differential operator L
2n−1
2 we obtain
[(L
2n−1
2 )+, L] = [(L
2n−1
2 )+, L]+
= [L
2n−1
2 − (L 2n−12 )−, L]+ = [L, (L
2n−1
2 )−]+
= [∂2x + u(x),W2n(x, x)∂
−1
x +O(∂
−2
x )]+ = 2∂x(W2n(x, x)),
which gives (3.14) and completes the proof. 	

TOME 55 (2005), FASCICULE 6
2126 Plamen ILIEV
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Plamen ILIEV,
Georgia Institute of Technology
School of Mathematics
Atlanta, GA 30332–0160 (USA)
iliev@math.gatech.edu
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