Page 1

GENERALIZED HAMILTON-JACOBI EQUATION AND HEAT KERNEL

ON STEP TWO NILPOTENT LIE GROUPS

OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Abstract. We study geometrically invariant formulas for heat kernels of sub-elliptic differ-

ential operators on two step nilpotent Lie groups and for the Grusin operator in R2. We

deduce a general form of the solution to the Hamilton-Jacobi equation and its generalized

form in Rn× Rm. Using our results, we obtain explicit formulas of the heat kernels for these

differential operators.

1. Introduction

Let us start with the Laplace operator on Rn,

∆ =1

2

n

?

j=1

∂2

∂x2

j

.

It is well-known that the heat kernel for ∆ is the Gaussian:

Pt(x,x0) =

1

(2πt)

n

2e−|x−x0|2

2t

.

Given a general second order elliptic operator in n dimensional Euclidean space,

∆X=1

2

n

?

j=1

X2

j+ lower order term,

where the {X1,...,Xn} is a linearly independent set of vector fields, the heat kernel takes the

form

1

(2πt)

Pt(x,x0) =

n

2e−d2(x,x0)

2t

?a0+ a1t + a2t2+ ···?.

Here d(x,x0) stands for the Riemannian distance between x and x0if the metric is induced

by the orthonormal basis {X1,...,Xn}. The aj’s are functions of x and x0. Note that

∂

∂t

?d2

2t

?

+1

2

n

?

j=1

?

Xjd2

2t

?2

= 0,

i.e.,d2

2tis a solution of the Hamilton-Jacobi equation.

2000 Mathematics Subject Classification. 53C17,53C22, 35H20.

Key words and phrases. Sub-Laplacian, Heat operator, H-type groups, action function, volume element.

The first author is partially supported by the NSF grant #0631541.

The second author is partially supported by a Hong Kong RGC competitive earmarked research grant

#600607, a competitive research grant at Georgetown University, and NFR grant #180275/D15.

The third author is supported by NFR grants # 177355/V30, #180275/D15, and ESF Networking Pro-

gramme HCAA.

1

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2OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Now let us move to subelliptic operators. We first consider the famous example: Heisenberg

sub-Laplacian on H1

∆X=1

2

∂x1

∂y

We shall try for a heat kernel in the form

1

tqe−f

where h =f

tis a solution of the Hamilton-Jacobi equation

∂h

∂t+1

2

∂x1

∂y

In other words,

∂h

∂t+ H(x,∇h) = 0,

where

H =1

2

is the Hamilton function associated with the sub-elliptic operator (1.1) and ξ1, ξ2and η are

dual variable to x1, x2and y respectively. Using the Lagrange-Chapit method, let us look at

the following equation:

(1.1)

?∂

+ 2x2

∂

?2+1

2

?∂

∂x2

− 2x1

∂

∂y

?2.

t···

?∂h

+ 2x2∂h

?2

+1

2

?∂h

∂x2

− 2x1∂h

∂y

?2

= 0.

(1.2)

(1.3)

??ξ1+ 2x2η?2+?ξ2− 2x1η?2?

=1

2

?ζ2

1+ ζ2

2

?

F(x,y,t,h,ξ,η,γ) = γ + H(x,y,ξ,η) = 0.

We shall find the bicharacteristic curves which are solutions to the following Hamilton system:

˙ x1=Fξ1= ξ1+ 2x2η = ζ1,

˙ x2=Fξ2= ξ2− 2x1η = ζ2,

˙ y =Fη= 2˙ x1x2− 2x1˙ x2,

˙t =Fγ= 1,

˙ξ1= − Fx1− ξ1Fh= 2η ˙ x2,

˙ξ2= − Fx2− ξ2Fh= −2η ˙ x1,

˙ η = − Fy− γFh= 0,

˙ γ = − Ft− γFh= 0,

˙h =ξ · ∇ξF + ηFη+ γFγ= ξ · ˙ x + η ˙ y − H

since˙t = 1 and γ = −H. With 0 ≤ s ≤ t, one has

γ(s) =γ = constant,

η(s) =η = constant,

t(s) =s.

Here “constant” means “constant along the bicharacteristic curve”. Furthermore,

H =1

2˙ x2

Another way to see that E is constant along the bicharacteristic, note that

¨ x1=˙ξ1+ 2η ˙ x2= +4η ˙ x2,

¨ x2=˙ξ2− 2η ˙ x1= −4η ˙ x1.

1+1

2˙ x2

2= E = energy.

(1.4)

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HAMILTON-JACOBI EQUATION AND HEAT KERNEL3

Therefore, ¨ x1˙ x1+ ¨ x2˙ x2= 0, and E =constant.

We need to find the classical action integral

S(t) =

?t

0

?ξ · ˙ x + η ˙ y − H?ds.

...

x2+ 16η2˙ x2= 0

Let find ξ and x from the Hamilton system. We obtain

...

x1+ 16η2˙ x1= 0,

from (1.4). Hence

˙ x1(s) = ˙ x1(0)cos(4ηs) +¨ x1(0)

4η

sin(4ηs)

= ˙ x1(0)cos(4ηs) + ˙ x2(0)sin(4ηs)

=ζ1(0)cos(4ηs) + ζ2(0)sin(4ηs)

(1.5)

and

˙ x2(s) =˙ x2(0)cos(4ηs) +¨ x2(0)

4η

sin(4ηs)

=˙ x2(0)cos(4ηs) − ˙ x1(0)sin(4ηs)

= − ζ1(0)sin(4ηs) + ζ2(0)cos(4ηs),

(1.6)

which yields

(1.7)

x1(s) = x1(0) + ζ1(0)sin(4ηs)

4η

+ ζ2(0)1 − cos(4ηs)

4η

and

(1.8)

x2(s) = x2(0) − ζ1(0)1 − cos(4ηs)

4η

+ ζ2(0)sin(4ηs)

4η

.

At s = t one has x1(t) = x1and x2(t) = x2, so

1

2ζ1(0)sin(4ηt) +1

−1

2ζ2(0)?1 − cos(4ηt)?=2η?x1− x1(0)?,

2ζ1(0)?1 − cos(4ηt)?+1

+ζ1(0)cos(2ηt) + ζ2(0)sin(2ηt) =2η?x1− x1(0)?

−ζ1(0)sin(2ηt) + ζ2(0)cos(2ηt) =2η?x2− x2(0)?

Hamilton’s equations give

ξ2(s) = − 2ηx1(s) +?ξ2(0) + 2ηx1(0)?

= − 2ηx1(0) −1

?

and

ξ1(s) = −2ηx2(0) +1

2

2ζ2(0)sin(4ηt) =2η?x2− x2(0)?,

or,

sin(2ηt)

,

sin(2ηt)

.

(1.9)

2ζ1(0)sin(4ηs) −1

ζ1(0)sin(4ηs) − ζ2(0)?1 + cos(4ηs)??

?

2ζ2(0)?1 − cos(4ηs)?+ ζ2(0) + 4ηx1(0)

,

=2ηx1(0) −1

2

ζ1(0)?1 + cos(4ηs)?+ ζ2(0)sin(4ηs)

?

.

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4 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

The above calculations imply

ξ1˙ x1+ ξ2˙ x2= − 2η ˙ x1(s)x2(0) + 2ηx1(0)˙ x2(s) +1

= − 2η?˙ x1(s)x2(0) − x1(0)˙ x2(s)?+?1 + cos(4ηs)?E,

?t

To find E we square and add the two equations in (1.9),

2

?ζ2

1(0) + ζ2

2(0)??1 + cos(4ηs)?

and

0

?ξ · ˙ x + η ˙ y − H?ds = η

?

y − y(0) + 2?x1(0)x2− x1x2(0)?+sin(4ηt)

4η2

E

?

.

E =1

2ζ2

1(0) +1

2ζ2

2(0) = 2η2|x − x0|2

sin2(2ηt).

Hence,

S(t) =

?t

?

0

?ξ · ˙ x + η ˙ y − H?ds

y − y(0) + 2?x1(0)x2− x1x2(0)?+ |x − x0|2cot(2ηt)

We note that x, y, t, x0and η = η(0) are free parameters while y(0) = y(0;x,x0,y,η;t) is not.

Therefore, we need to introduce one more free variable h(0) such that h(t) = h(0) + S(t) is a

solution of the Hamilton-Jacobi equation (1.2).

It reduces to find h(0). To find it we shall substitute S into (1.2). Straightforward compu-

tation shows that

∂h

∂t+ H(x,y,ξ(t),η(t)) = 0

=η

?

.

where

(1.10)

h(t) = η(0)y(0) + S(t),

i.e.,

h(0) = η(0)y(0).

This yields

∂h

∂t+ H

?

x,y,∇xh,∂h

∂y

?

= 0.

We have the following theorem.

Theorem 1.1. We have shown that

h =η(0)y(0) +

=ηy + 2η?x1(0)x2− x1x2(0)?+ η|x − x0|2cot(2ηt)

is a “complete integral” of (1.2) and (1.3), i.e., a solution of (1.2) and (1.3) which depends

on 3 free parameters x1(0), x2(0) and η.

?t

0

?ξ · ˙ x + η ˙ y − H?ds

(1.11)

Before we move further, let us consider a more general situation.

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HAMILTON-JACOBI EQUATION AND HEAT KERNEL5

2. Generalized Hamilton-Jacobi equations

In this section we study the Hamilton-Jacobi equation which is crucial in the construction

of the heat kernel associated with elliptic and sub-elliptic operators. We deduce a general

form of the solution to the Hamilton-Jacobi equation and its generalized form. We consider

an (n + m)-dimensional space Rn× Rm. The coordinates are denoted x = (x1,...,xn) ∈ Rn

and y = (y1,...,ym) ∈ Rmwith dual variables (ξ1,...,ξn) and (η1,...,ηm) respectively. The

roman indices i,j,k,... will vary from 1 to n and the Greek indices α,β,... will vary from 1

to m. As usual, the Hamiltonian function H(x,y,ξ,η) is a homogeneous polynomial of degree

2 in the variables (ξ,η) and has smooth coefficients in (x,y).

We have the following nice generalizaition of a result from [11].

Theorem 2.1. Set

(2.1)

h(t;x,y,ξ,η) =

m

?

α=1

ηα(0)yα(0) + S(t;x,y,ξ,η)

where

xj= xj(s;x,y,ξ,η;t),j = 1,...,n;

yα= yα(s;x,y,ξ,η;t),α = 1,...,m

and

S(t;x,y,ξ,η) =

?t

0

?

ξ(u) · ˙ x(u) + η(u) · ˙ y(u) − H(x(u),y(u),ξ(u),η(u))

?

du.

Then h satisfies the usual Hamilton-Jacobi equation:

∂h

∂t+ H

?

x,y,∇xh,∇yh

?

= 0.

Proof. In order to prove the theorem, we first calculate the partial derivatives of the function

S with respect to all variables explicitly. For j = 1,...,n,

∂S

∂xj(t;x,y,ξ,η)

?t

k=1

n

?

n

?

?t

n

?

It follows that

=

0

?

n

?

∂H

∂ξk

?∂ξk

∂ξk

∂xj

∂xj

dxj

ds

+ ξk

d

ds

∂xk(s;x,y,ξ,η;t)

∂xj

?+

m

?

α=1

?∂ηα

∂xj

dyα

ds

+ ηα

d

ds

∂yα(s;x,··· ;t)

∂xj

?

−

k=1

−

m

?

α=1

∂H

∂ηα

∂ηα

∂xj

−

k=1

∂H

∂xk

∂xk(s;x,y,ξ,η;t)

∂xj

−

m

?

α=1

∂H

∂yα

∂yα(s;x,y,ξ,η;t)

∂xj

?

ds

=

0

d

ds

?

n

?

k=1

ξk∂xk(s;x,y,ξ,η;t)

∂xj

+

m

?

m

?

α=1

ηα∂yα(s;x,y,ξ,η;t)

∂xj

?

ds

=

k=1

ξk(s)∂xk(s;x,y,ξ,η;t)

∂xj

???

s=t

s=0+

α=1

ηα(s)∂yα(s;x,y,ξ,η;t)

∂xj

???

s=t

s=0.

∂S

∂xj(t;x,y,ξ,η) = ξj(t) −

m

?

α=1

ηα(0)∂yα(0;x,y,ξ,η;t)

∂xj

.

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6 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Similarly, for β = 1,...,m,

∂S

∂yβ(t;x,y,ξ,η) = ηβ(t) −

m

?

α=1

ηα(0)∂yα(0;x,y,ξ,η;t)

∂yβ

.

Moreover,

∂S

∂t(t;···) =

n

?

k=1

ξk(t;···)˙ xk(t;···) +

m

?

α=1

ηα(t;···)˙ yα(t;···) − H?x,y,ξ(t;···),η(t;···)?

s=t

s=0+

α=1

∂t

+

n

?

k=1

ξk(s;···)∂xk(s;···)

∂t

???

m

?

ηα(s;···)∂yα(s;···)

???

s=t

s=0.

Differentiating x1= x1(t;x,y,ξ,η;t) yields

0 =d

dtx1(t;x,y,ξ,η;t) = ˙ x1(t;···) +∂x1(s;x,y,ξ,η;t)

On the other hand, one has

???

ηα(s;···)∂yα(s;···)

∂t

therefore,

∂S

∂t= −H(t;···) −

∂t

???s=t.

ξk(s;···)∂xk(s;···)

∂t

s=t

s=0= −ξk(t;···)˙ xk(t;···),k = 1,...,n,

and

???

s=t

s=0= −ηα(t;···)˙ yα(t;···) − ηα(0;···)∂yα(0;···)

∂t

,α = 1,...,n,

m

?

α=1

ηα(0;···)∂yα(0;···)

∂t

.

It follows that if we set as in the statement of the theorem

h(t;x,y,ξ,η) =

m

?

α=1

ηα(0)yα(0) + S(t;x,y,ξ,η),

then it satisfies

∂h

∂xk

∂h

∂yα

=ξk(t;x,y,ξ,η;t),k = 1,...,n

=ηα(t;x,y,ξ,η;t),α = 1,...,m,

and

∂h

∂t+ H

?

x,y,ξ(t),η(t)

?

= 0⇒

∂h

∂t+ H

?

x,y,∇xh,∇yh

?

= 0.

This completes the proof of the theorem.

?

We note that the derivation that (2.1) satisfies the Hamilton-Jacobi equation was complete

general, not restriction to H

x,y,∇xh,∇yh

that ηα(s) =constant for α = 1,...,m. The action integral S is not a solution of the Hamilton-

Jacobi equation because some of our free parameters are dual variables ηα(0) instead of yα(0).

For the Heisenberg sub-Laplacian or the Grusin operator, η(0) = η cannot be switched to y(0).

As we know, ˙ y = 2(˙ x1x2− x1˙ x2). From (1.5) – (1.8), one has

?

??

being (1.3). In particular we did not assume

˙ y = 2˙ x1x2(0) − x1(0)˙ x2+1

2

?ζ2

1(0) + ζ2

2(0)?1 − cos(4ηs)

2η

?

,

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HAMILTON-JACOBI EQUATION AND HEAT KERNEL7

and

y(s) = 2?x1(s)x2(0) − x1(0)x2(s)?+

At s = t, one has x1(t) = x1, x2(t) = x2and

E

4η2

?4ηs − sin(4ηs)?+ C.

y = 2?x1x2(0) − x1(0)x2

?+

E

4η2

?4ηt − sin(4ηt)?+ C.

Hence, one has

y(s) =y − 2

??x1− x1(s)?x2(0) − x1(0)?x2− x2(s)??

E

4η2

−

?4η(t − s) − (sin(4ηt) − sin(4ηs))?.

y(0) = y + 2?x1(0)x2− x1x2(0)?+ |x − x0|2µ(2ηt),

φ

sin2φ− cot φ.

At s = 0,

where we set

µ(φ) =

To replace η by y(0), one needs to invert µ,

µ(2ηt) =y − y(0) + 2?x1(0)x2− x1x2(0)?

|x − x0|2

.

This is impossible since for most of the values on the right hand side µ−1is a many valued

function [2]. Therefore we must leave η as one of the free parameters which does not permit S

to be a solution of the Hamilton-Jacobi equation.

Before we go further, we present a scaling property of the solution to the Hamiltonian system

dxj

ds

=∂H

∂ξj,

dyα

ds

=∂H

∂ηα,

dξj

ds= −∂H

∂xj,

dηα

ds

= −∂H

∂yα,

s ∈ [0,t] with the boundary conditions

x(0) = x0,

x(t) = x,

y(t) = y,η(0) = η(0).

Lemma 2.1. One has the following scaling property

xj(s;x,x0,y,ξ,η(0);t) =xj

?λs;x,x0,y,ξ,η(0)

?λs;x,x0,y,ξ,η(0)

?λs;x,x0,y,ξ,η(0)

?λs;x,x0,y,ξ,η(0)

λ

;λt?, j = 1,...,n

;λt?, α = 1,...,m

;λt?, j = 1,...,n

;λt?, α = 1,...,m

yα(s;x,x0,y,ξ,η(0);t) =yα

λ

ξj(s;x,x0,y,ξ,η(0);t) =λξj

λ

ηα(s;x,x0,y,ξ,η(0);t) =ληα

λ

(2.2)

for λ > 0, if the two sides of (2.2) stays in the domain of unique solvability of the Hamiltonian

system.

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8 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Proof. Denote the curve on the right-hand side of (2.2) by {˜ x(s), ˜ y(s),˜ξ(s), ˜ η(s)}. Note that

s ∈ (0,t). Then for j = 1,...,n

∂˜ xj

∂s

=λ˙ xj

?λs;x,x0,y,ξ,η(0)

=λ∂H

∂ξj

=∂H

∂ξj

λ

;λt?

λ

?x1

?˜ x(s), ˜ y(s),˜ξ(s), ˜ η(s)?,

?λs;x,x0,y,ξ,η(0)

;λt?,x2

?λs;x,x0,y,ξ,η(0)

λ

;λt?,...?

since

calculations and homogeneity of degree 2 of∂H

∂H

∂ξj, j = 1...,n, are homogeneous of degree 1 in ξ1,...,ξn and η1,...,ηm. Similar

∂xjand

∂H

∂yαin ξ1,...,ξnand η1,...,ηmyield

∂˜ yα

∂s

=∂H

∂ηα,

∂˜ξj

∂s= −∂H

∂xj,

∂˜ ηα

∂s

= −∂H

∂yα.

Clearly,

˜ xj(0) = xj

?0;x,x0,y,ξ,η(0)

λ

;λt?= xj(0),

˜ xj(t) = xj

?λt;x,x0,y,ξ,η(0)

λ

;λt?= xj,

for j = 1,...,n and

˜ yα(t) =yα(λt;x,x0,y,ξ,η(0)

λ

;λt?= yα,

;λt?= ληα(0)

˜ ηα(0) =ληα(0;x,x0,y,ξ,η(0)

λλ

= ηα(0)

for α = 1,...,m. The bicharacteristic curves are unique, so the two sides of (2.2) agree.

?

Corollary 2.2. One has

h(x,x0,y,ξ,η(0);t) = λh?x,x0,y,ξ,η(0)

λ

;λt?.

Proof. In the case of Heisenberg group, the corollary is a direct consequence of the explicit

formula (1.11) and in this case, η(0) = η is a constant. Here we would like to give a proof

which applies in more general case. We know that for j = 1,...,m,

˙ xj(s;x,x0,y,ξ,η(0);t) =dxj

ds(s;x,x0,y,ξ,η(0);t)

=dxj

ds

?λs;x,x0,y,ξ,η(0)

?λs;x,x0,y,ξ,η(0)

λ

;λt?

;λt?.

=λ˙ xj

λ

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HAMILTON-JACOBI EQUATION AND HEAT KERNEL9

Similar result holds for ˙ yαfor α = 1,...,m. Therefore,

?t

?t

− λ2H(x(λs;...),...)?ds

=1

λ

0

k=1

?t

=λS?x,x0,y,ξ,η(0)

Also,

m

?

and the proof of the corollary is therefore complete.

0

?ξ(s) · ˙ x(s) + η(s) · ˙ y(s) − H(x(s;...),...)?ds

?λξ?λs;x,x0,y,ξ,η(0)

=

0

λ

;λt?· λ˙ x(λs;...) +

m

?

α=1

ληα(λs;...) · λ˙ yα(λs;...)

?t

?λ2

?ξ?s′;x,x0,y,η(0)

n

?

ξk(λs;...)˙ xk(λs;...) + λ2

m

?

α=1

ηα(λs;...)˙ yα(λs;...) − λ2H(x(λs;...),...)?d(λs)

,λt?· ˙ x(s′;...) + η(s′;...) · ˙ y(s′;...) − H(x(s′;...),...)?ds′

=λ

0

λ

λ

,λt?.

α=1

ηα(0)yα(0;x,x0,y,ξ,η(0);t) = λ

m

?

α=1

ηα(0)

λ

yα

?0;x,x0,y,ξ,η(0)

λ

;λt?

?

Set

f(x,x0,y,ξ,η(0)) = h(x,x0,y,ξ,η(0),t)

???t=1.

Then

Theorem 2.2. f is a solution of the generalized Hamilton-Jacobi equation

(2.3)

m

?

α=1

ηα(0)

∂f

∂ηα(0)+ H

?

x,y,∇xf,∇yf

?

= f.

Proof. By homogeneity property of the function h, one has

h(x,x0,y,ξ,η(0),t) =1

th(x,x0,y,ξ,tη(0),1) =1

tf(x,x0,y,ξ,tη(0)),

so,

(2.4)

∂h

∂t= −1

t2f +1

t

m

?

α=1

ηα(0)

∂f

∂ηα(0)

on one hand. On the other hand,

(2.5)

∂h

∂t= −H?x,y,∇xh,∇yh?

from Theorem 2.1. Since (2.4) agrees with (2.5) for all t so we may set t = 1 which yields the

proposition.

?

At the rest of the sectionwe present some examples that revealthe geometrical nature of

functions h and f.

Page 10

10 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

2.3. Laplace operator. We start from the Laplace operator ∆ =?n

Hamiltonian function H(ξ) is

k=1

∂2

∂x2

k

in Rn. The

H(ξ) =1

2

n

?

k=1

ξ2

k

and hence we need to deal with F(ξ,γ) = H + γ = 0. The Hamilton’s system is

˙ x = ξ,

˙ξ = 0,

˙ γ = 0.

with initial-boundary conditions x(0) = x0, x(t) = x. Since˙ξ = 0, it follows that ξ(s) = ξ(0) =

constants, is a constant vector. Then

¨ x =˙ξ = 0⇒

x(s) = ξ(0)s + x0.

Moreover,

x = x(t) = ξ(0)t + x0

⇒

ξ(0) =x − x0

t

and

∂h

∂t=1

2

n

?

k=1

ξ2

k=

n

?

k=1

(xk− x(0)

2t2

k)2

=|x − x0|2

2t2

or,

h(x,x0,t) = h(0) +|x − x0|2

2t2

t = h(0) +|x − x0|2

2t

.

Since this is a translation invariant case, we may assume that h(0) = 0. Therefore,

f(x,x0) = h(x,x0,t)

???t=1=|x − x0|2

2

gives us the Euclidean action function.

2.4. Grusin operator. We are in R2now and the horizontal vector fields X1, X2are given

by

∂

∂x,

X1=

and

X2= x∂

?

∂y.

?2

The Grusin operator is given as follows: ∆X =

1

2

∂

∂x

+1

2x2?

∂

∂y

?2.It is obvious that

∆X is elliptic away from the y-axis but degenerate on the y-axis.

hence {X1,X2,[X1,X2]} spanned the tangent bundle of R2everywhere.

theorem [12], ∆Xis hypoelliptic.

The Hamiltonian function H for the ∆Xis

H(x,y,ξ,η) =1

Since [X1,X2] =

By H¨ ormander’s

∂

∂y,

(2.6)

2ξ2+1

2x2η2.

The Hamilton system can be obtained as follows;

˙ x =Hξ= ξ,

˙ y =Hη= ηx2,

˙ξ = − Hx= −η2x,

˙ η = − Hy= 0,

˙S =ξ ˙ x + η ˙ y − H.

With 0 ≤ s ≤ t,

η(s) = η(0) = η0= constant,

Page 11

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 11

“constant” means “constant along the bicharacteristic curve”. Next,

¨ x =˙ξ = −xη2,

so

¨ x + η2x = 0.

It follows that

x(s) = Acos(ηs) + B sin(ηs) = x(0)cos(ηs) +ξ(0)

η

sin(ηs) = x0cos(ηs) +ξ(0)

η

sin(ηs).

Hence,

ξ(s) = ˙ x(s)

yields

ξ(s) = ξ(0)cos(ηs) − ηx0sin(ηs).

We also have

x = x(t) = x0cos(ηt) +ξ(0)

η

sin(ηt),

and

(2.7)

ξ(0)

η

=x − x0cos(ηt)

sin(ηt)

.

Consequently,

x(s) = x(0)cos(ηs) +x − x0cos(ηt)

sin(ηt)

sin(ηs).

The singularities occur at η = η0=kπ

η0=2kπ

t

if x = −x0. Next,

˙ y(s) =ηx2(s)

?

=d

ds

2

=d

ds

2

twhen x = ±x0; they are η =

(2k+1)π

t

if x = x0and

=ηx0

?1

η

2+1

?x2

?η

ξ(0)

η

by (2.7) and collect terms with x2

?

?

x2

0

2sin2(ηt)

2

x2

0

4sin2(ηt)

The terms containing x2are:

2cos(2ηs)?+ 2x0ξ(0)

?s +sin(2ηs)

0+?ξ(0)

η

sin(ηs)cos(ηs) +?ξ(0)

?+x0ξ(0)

s +1

4

η

η

?2?1

2−1

2cos(2ηs)??

???

?1 − cos(2ηs)??

?

0

2η

?2?

η2

0−?ξ(0)

sin2(ηs) +1

2

?ξ(0)

η

?2?s −sin(2ηs)

2

η

2η

?

x2

η

?

x2

?2?

0:

sin(2ηs) +x0

ξ(0)

.

We replace

x2

2

0

ηs +1

2sin(2ηs) + ηscos2(ηt)

cos2(ηt) − sin2(ηt)

sin2(ηt)

?

?

sin2(ηt)−1

2

cos2(ηt)

sin2(ηt)sin(2ηt) −cos(ηt)

sin(2ηs) −cos(ηt)

sin(ηt)

sin(ηt)

?1 − cos(2ηs)??

?1 − cos(2ηs)??

=x2

0

2

ηs

sin2(ηt)−1

2

=

ηs −1

?cos(2ηt)sin(2ηs) + sin(2ηt)?1 − cos(2ηs)???

2ηs −?sin(2ηt) − sin?2η(t − s)???

=

.

1

4

x2

sin2(ηt)

?2ηs − sin(2ηs)?,

Page 12

12 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

and the terms with x0x are the following:

1

2

2xx0

sin2(ηt)

?1

2

?sin?η(2s − t)?+ sin(ηt)?− ηscos(ηt)

?

.

So,

˙ y(s) =d

ds

?

x2

0

4sin2(ηt)

x2

4sin2(ηt)

2xx0

4sin2(ηt)

?

2ηs −?(sin(2ηt) − sin?2η(t − s)??)Big]

?2ηs − sin(2ηs)?

?1

+

+

2

?sin?η(2s − t)?+ sin(ηt)?− η scos(ηt)

??

.

The action function has the form

S =

?t

0

(ξ ˙ x + η ˙ y − H)ds = η(y − y(0)) +

?t

0

(ξ2− H)ds.

We find ξ2as follows

ξ2(s) =ξ2(0)

2

?ξ2(0) + η2x2

=H(0)

?1 + cos(2ηs)?− ξ(0)ηx0sin(2ηs) +1

0

??

H = H(0) =1

2η2x2

0

?1 − cos(2ηs)

?

=1

2

?

?

?

+1

2

?ξ2(0) − η2x2

0

?cos(2ηs) − ηx0ξ(0)sin(2ηs).

Since H is constant along the bicharacteristic, one has

2

?ξ2(0) + η2x2

0

?.

Continuing, we obtain the action function

S =η(y − y(0)) +

?t

0

?ξ2(0) − η2x2

?(ξ2(0) − η2x2

0)cos(2ηs)

2

− ηx0ξ(0)sin(2ηs)?ds

+ ηx0ξ(0)cos(2ηt) − 1

=η(y − y(0)) +1

2

0

?sin(2ηt)

2η

2η

.

We simplify this

S − η(y − y(0))

=η2

2

=η

4

?x − x0cos(ηt)

??x − x0cos(ηt)

sin(ηt)

?2sin(2ηt)

?2sin(2ηt) − x2

2η

−1

2η2x2

0

sin(2ηt)

2η

+ η2x0x − x0cos(ηt)

sin(ηt)

cos(2ηt) − 1

2η

?1 − cos(2ηt)??

sin(ηt)

0sin(2ηt) − 2x0x − x0cos(ηt)

sin(ηt)

.

(2.8)

In the bracket {···} of (2.8), terms involved x2

??cos2(ηt)

=x2

0

0are

x2

0

sin2(ηt)− 1

?cos(2ηt)sin(2ηt)

=2x2

0cot(ηt),

?

sin(2ηt) + 2cos(ηt)

sin(ηt)(1 − cos(2ηt))

+ 2cos(ηt)

?

?

sin2(ηt)

sin(ηt)−cos(2ηt)sin(2ηt)

sin2(ηt)

Page 13

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 13

terms involved x2are

x2sin(2ηt)

sin2(ηt)= 2x2cot(ηt),

and terms containing x0x are

2xx0

?

−cos(ηt)

sin2(ηt)sin(2ηt) −1 − cos(2ηt)

?2cos2(ηt)

4xx0

sin(ηt).

sin(ηt)

?

?

= − 2xx0

sin(ηt)

+ 2sin(ηt)

= −

Hence,

{···} =2(x2+ x2

0)cot(ηt) −

4xx0

sin(ηt)

=?(x + x0)2+ (x − x0)2?cot(ηt) −(x + x0)2− (x − x0)2

=(x + x0)2?

=(x + x0)2cos(ηt) − 1

sin(ηt)

= − (x + x0)2tan?ηt

Thus S has the following form:

?

By Theorem 2.1, we know that

sin(ηt)

cot(ηt) −

1

sin(ηt)

+ (x − x0)2cos(ηt) + 1

?

+ (x − x0)2?

cot(ηt) +

1

sin(ηt)

?

sin(ηt)

2

?+ (x − x0)2cot?ηt

2

?.

S = η(y − y(0)) −η

4

(x + x0)2tan?ηt

2

?− (x − x0)2cot?ηt

2

??

.

h(t;x,x0,y,η) =ηy(0) + S(t;x,y,η)

=ηy(0) + η(y − y(0)) −η

4

?

(x + x0)2tan?η

?− B2cot?ηt

2

?− (x − x0)2cot?η

2

??

=ηy −η

4

?

A2tan?ηt

22

??

is a solution of the Hamilton-Jacobi equation. Here A = x + x0and B = x − x0. Now by

Theorem 2.2, the function

???t=1=1

η∂f

∂η+ H

f(x,x0,y,η) = h(t;x,x0,y,η)

2

η

2

?

4y − A2tan?η

2

?+ B2cot?η

2

??

is a solution of the generalized Hamilton-Jacobi equation

?

x,x0,y,∂xf,∂yf

?

= f.

We set

η

2= ? ητ,

where τ ∈ R yields the domain of integration and ? η is a fixed complex number.

Page 14

14 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Lemma 2.5. Suppose f is a smooth function of τ ∈ R and

lim

τ→±∞Re(f)(τ) = ∞

off the canonical curve x2

0+ x2= 0. Then ? η is pure imaginary.

?

− isinh(2η2τ)?(B2+ A2)cosh(2η2τ) + (B2− A2)cos(2η1τ)?

(i). η1= 0, i.e., η ∈ iR. When τ ≈ ±∞,

f ≈1

2iη2τ

and

Re(f) ≈1

4(x2

as τ → ±∞ as long as x2

(ii). η2= 0, that is η ∈ R. Then

f = 2η1τy +1

4 sin(2η1τ)

is singular in τ ∈ R when x2

Proof. Let ? η = η1+ iη2. An elementary calculation yields

2

f =1

?η1+ iη2

?τ

4y +sin(2η1τ)?(B2− A2)cosh(2η2τ) + (B2+ A2)cos(2η1τ)?

cosh2(2η2τ) − cos2(2η1τ)

cosh2(2η2τ) − cos2(2η1τ)

?

?

4y − i2(x2

0+ x2)tanh(2η2τ)

?

,

0+ x2)2η2τ tanh(2η2τ) → ±∞

0+ x2?= 0.

2η1τ

?

B2− A2+ (B2+ A2)cos(2η1τ)

?

0+ x2?= 0, otherwise

Re(f) = f = 2η1τy −→

????

τ→±∞

±(sgn(y))∞.

(iii). η1?= 0, η2?= 0. Here

f ≈1

2

?η1+ iη2

?τ

?

4y − i(A2+ B2)tanh(2η2τ)

?

as τ → ±∞, and

Re(f) ≈2η1τy + (x2

=|τ|?2(sgn(τ))η1y + (x2

2η1y > (x2

0+ x2)??η2τ??

0+ x2)|η2|?

and choosing x0,x,y so that

0+ x2)|η2|

we have

lim

τ→±∞Re(f) = ±∞

which we do not want. This complete the proof of Lemma (2.5).

?

Following the tradition, we shall choose

? η = −i

2(x2

2.

Then

f = −iτy +1

0+ x2)τ coth τ −

τx0x

sinh τ.

Page 15

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 15

2.6. Sub-Laplace operator on step 2 nilpotent Lie groups. Let M be a simply connected

2-step nilpotent Lie group G equipped with a left invariant metric. Let G be its Lie algebra

and it is identified with the group G by the exponential map:

exp : G → G.

We assume

G = [G,G] ⊕ [G,G]⊥= C ⊕ [G,G]⊥= C ⊕ H,

where H and C are vector spaces over R with an skew-symmetric bilinear form

B : H × H → C

such that B(H,H) = C. The group law is given by

(H ⊕ C) × (H ⊕ C) → H ⊕ C

with

(x,y) ∗ (x′,y′) =

?x + x′,y + y′+1

2B(x,x′)?

and then the exponential map is the identity map. Let {X1,...,Xn} be a basis of H and let

{Y1,...,Ym} be a basis of the center [G,G] = C. We assume {X1,...,Xn} and {Y1,...,Ym} are

orthonormal, and introduce a left invariant Riemannian metric on the group G in an obvious

way.

We write the vector fields Xj, j = 1,...,n by:

Xj=

∂

∂xj

+

n

?

k=1

m

?

α=1

aα

jkxk

∂

∂yα

where the aα

We are interested in the sub-Laplacian ∆Xwhich can be defined as follows:

jkare real numbers and form skew-symmetric matrices?aα

∆X=1

jk

?

j,k, i.e., aα

jk= −aα

kj.

2

n

?

j=1

X2

j

It is easy to see that

(2.9)

?Xj,Xk

?= 2

n

?

k=1

m

?

α=1

aα

jk

∂

∂yα.

Lemma 2.7. The operator ∆X is hypoelliptic if and only if the rectangular matrix of order

n(n−1)

2

jk

× m with element?aα

Proof. The operator ∆X is hypoelliptic when the vector fields {Xj}n

bracket generating condition. This implies that we can recover all the

relations (2.9). If we consider

jk

(j,k) where j < k, this means that this matrix should have rank m.

?

{(j<k),α}is of rank m (which implies that m ≤n(n−1)

2

).

j=1satisfy the “first”

∂

∂yαfrom the

n(n−1)

2

?aα

?

as a matrix with indices α = 1,...,m and the couples

?

We may define a Lie group structure on Rn× Rmwith the following group law:

(2.10)

?

(x,y) ◦ (x′,y′) =

x1+ x′

1,...,xn+ x′

n,y1+ y′

1+

n

?

j,k=1

a1

jkx′

jxk,...,ym+ y′

m+

n

?

j,k=1

am

jkx′

jxk

?

.

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