Huygens’ Principle, Integrable PDEs, and Solitons · Huygens’ Principle, Integrable PDEs, and Solitons Jorge P. Zubelli July 20, 2004 zubelli/ download:zubelli

Huygens’ Principle, Integrable PDEs, and Solitons

Jorge P. Zubelli

July 20, 2004

http://www.impa.br/ zubelli/ download:http://www.impa.br/˜zubelli/

Thanks: Prof. Marco Calahorrano

Joint work with Fabio Chalub - U. Lisboa

http://www.impa.br/~zubelli/

Overview

• Free Space Waves in (2+1) Dimensions × (3+1) Dimensions

Overview


• Huygens’ principle in Hadamard’s sense

Overview



• The Hadamard conjecture

Overview




• Calogero-Moser Systems

Overview





• Solitons - KdV, KP, and Friends

Overview






• Huygens for Dirac Systems

Overview






• Huygens for Dirac Systems

Cauchy Problem for the Wave Equation

ψdef= (∂2

t −∆)ψ = 0 (1)

ψ

∣∣∣∣t=0

= f (2)

∂tψ

∣∣∣∣t=0

= g . (3)

Domains of Dependence



Huygens Construction



Hadamard’s Approach Huygens’ Property

(A) “Major Premise”

The action of phenomena produced at the instant t = 0 on

the state of matter at the later time t = t0 takes place by the

mediation of every intermediary instant t = t′, i.e., (assuming

0 < t′ < t0), . . .

Hadamard’s Approach Huygens’ Property

(A) “Major Premise”

The action of phenomena produced at the instant t = 0 on

the state of matter at the later time t = t0 takes place by the

mediation of every intermediary instant t = t′, i.e., (assuming

0 < t′ < t0), . . .

(B) “Minor Premise”

If we produce a luminous disturbance localized in a

neighborhood of 0, its effect after an elapsed time t0 will

be localized in a neighborhood of the sphere centered at 0 with

radius ct0.


radius ct0.

(C) “Conclusion”

In order to calculate the effect of our initial luminous

phenomenon produced at 0 at t = 0, we may replace it

by a proper system of disturbances taking place at t = t′ and

distributed over the surface of the sphere with center 0 and

radius ct′.


radius ct0.






radius ct′.

(A) & (B) =⇒ (C)


radius ct0.






radius ct′.

(A) & (B) =⇒ (C)

Free Wave Equation

Notation: (n+ 1) dimensions, n = number of space dimensions and

1 indicates time dimension. space-time: (x0, x1, . . . , xn) ∈ Rn+1

where x0 = t is the time variable. wave operator:

ψdef= (∂2

t − c2∆)ψ ,

where

∆ψ =n∑i=1

∂2xiψ .

Free Wave Equation

Notation: (n+ 1) dimensions, n = number of space dimensions and

1 indicates time dimension. space-time: (x0, x1, . . . , xn) ∈ Rn+1

where x0 = t is the time variable. wave operator:

ψdef= (∂2

t − c2∆)ψ ,

where

∆ψ =n∑i=1

∂2xiψ .

Cauchy problem:

ψ = 0 (4)

ψ

∣∣∣∣t=0

= f (5)

∂tψ

∣∣∣∣t=0

= g . (6)

Method of Spherical Means

focus first: n = 3.

Method of Spherical Means

focus first: n = 3.

Mr [u] def=1

4πr2

∫|x−x′|=r

u(x′) dS(x′) . (7)

Exercise: Solution to the Cauchy problem given by

ψ =d

dt(tMct [f ]) + tMct [g] .

Consequence: At the point

(t, x) ∈ R4

solution ψ depends only on the initial data at the points on the

sphere in R3 of radius ct centered at x.

Hadamard’s Method of Descent:

Use the sol. of the given equation in a higher dimensional space to

produce the sol. in a lower dimensional space by introducing extra

“dummy variables.”

Using solutions of the wave eq for n = 3 one obtains a solution of

the wave equation for n = 2.

For n = 2 the dependence is on the interior and the surface of the

ball.

Theorem: For n > 1 odd: Solution of the wave equation depends

on the initial data on the intersection of the light cone with the

initial data manifold t = 0.

For n even: Solution depends on the values of the data on the

closure intersection of the interior part of the light cone and the

initial data manifold t = 0.

Idea of the proof: Use the method of spherical means indicated

above for n odd, and descending to even n.

Remarks on the Strict Huygens Property:

• Fascinating property of the solutions to the wave equation in 3

spatial dimensions

• meaningful transmission of information

• “instantaneous” signal in 3 spatial dimensions remains

“instantaneous”

“Thus our actual physical world, in which acoustic and

electromagnetic signals are the basis of communication, seems

to be singled out among other mathematically conceivable

models by intrinsic simplicity and harmony” (Courant & Hilbert,

Methods of Math. Phys.)

• meaningful transmission of information

• “instantaneous” signal in 3 spatial dimensions remains

“instantaneous”

“Thus our actual physical world, in which acoustic and

electromagnetic signals are the basis of communication, seems

to be singled out among other mathematically conceivable

models by intrinsic simplicity and harmony” (Courant & Hilbert,

Methods of Math. Phys.)

Strict Huygens

Let L be a strictly hyperbolic (linear) second order operator defined

on a “causal domain” (e.g d’Alembertian + lower order terms)

Let S be a space-like surface

Consider the Cauchy-problemLψ = 0ψ|S = f

∂νψ|S = g

(8)

Strict Huygens

Let L be a strictly hyperbolic (linear) second order operator defined

on a “causal domain” (e.g d’Alembertian + lower order terms)

Let S be a space-like surface

Consider the Cauchy-problemLψ = 0ψ|S = f

∂νψ|S = g

(8)

Definition: L satisfies a strict Huygens’ principle if the solution to

every (well posed) Cauchy problem (8) depends on the initial data

f, g only in the intersection of the characteristic conoid and the

space-like manifold S.

Remark: Equivalent definition in terms of the support of the

fundamental solutions

LΨ± = δξΨ|±t>0 = 0

every (well posed) Cauchy problem (8) depends on the initial data

f, g only in the intersection of the characteristic conoid and the

space-like manifold S.

Remark: Equivalent definition in terms of the support of the

fundamental solutions

LΨ± = δξΨ|±t>0 = 0

Trivial Symmetries

1. Nonsingular coordinate transformations:

x 7→ x = x(x)

det(Dx) 6= 0

2. Gauge transformations:

ψ 7→ ψ = λ−1ψ

L 7→ L = λ−1Lλ ,

where λ = λ(x) 6= 0.

3. Multiplication by a scalar function:

L 7→ L = µL ,

where µ = µ(x) 6= 0.

Hadamard’s Question:

Determine all the hyperbolic operators that satisfy a strict Huygens’

Property (HP).

Hadamard’s Question:

Determine all the hyperbolic operators that satisfy a strict Huygens’

Property (HP). HARD!!!

Facts:

• The usual wave operator satisfies a strict HP iff n is odd and n > 1.

• If n = 3 the op + u satisfies strict HP iff u = 0.

Hadamard’s Conjecture

Conjecture: The only potentials u that can be added to so that

L = + u satisfies a strict HP is (mod the trivial symmetries)

u = 0.

Hadamard’s Conjecture

Conjecture: The only potentials u that can be added to so that

L = + u satisfies a strict HP is (mod the trivial symmetries)

u = 0.

FALSE!

Historical Background

• Hadamard’s conjecture – Courant & Hilbert

• Asgeirsson

• Stellmacher (counter-example to Hadamard’s conjec)

• Stellmacher and Lagnese: DARBOUX TRANSFORMATIONS

RATIONAL SOLUTIONS OF THE KdV

• Gunther - Ibragimov

• Berest: Iso-Huygens’ deformations (JMP ‘92) – Relation to

Virasoro flows and the bispectral problem

“The symmetries that preserve HP for operators of the form

+ u(x0) are related to the symmetries that preserve the bispectral

property for u(x0), i.e., the positive Virasoro flows”

“The symmetries that preserve HP for operators of the form

+ u(x0) are related to the symmetries that preserve the bispectral

property for u(x0), i.e., the positive Virasoro flows” related to joint

work w/ F. Magri. CMP 1992

Lagnese & Stellmacher

Notation: x0 = t, x = (x0, x1, · · ·xn)

The simplest example discovered by Lagnese & Stellmacher

(dimension 5 + 1):

L = +2x2

0

, (9)

More generally

Lagnese & Stellmacher

Notation: x0 = t, x = (x0, x1, · · ·xn)

The simplest example discovered by Lagnese & Stellmacher

(dimension 5 + 1):

L = +2x2

0

, (9)

More generally If L = + u(x0) where u rational solution of the

KdV, i.e. u is of the form

uk(x0) = 2∂2x0

log ϑk(x0) , (10)

where ϑk is the k-th Adler-Moser polynomial. defined by the

relations

ϑ0 = 1

ϑ1 = x

ϑ′k+1ϑk−1 − ϑ′k−1ϑk+1 = (2k + 1)ϑ2k .

The variable x0 is the time variable

THEOREM: (Lagnese and Stelmacher) Let ϑk be the k-th

Adler-Moser polynomial and uk as in equation (10). If the number n

of spatial dimensions is odd and n ≥ 3 + 2k, then, the operator

L = + uk(x0) , (11)

is a strict Huygens operator.

THEOREM: (Lagnese) If L is a strict Huygens operator of the

form (11), and u(x0) is an analytic potential, then (mod trivial

symmetries) u(·) is a rational solution of KdV.

KdV, KP, and Friends

Korteweg-de Vries equation:

ψt + cψx + αψxxx + βψψx = 0 . (12)

The Kadmontsev-Petviashvilli equation

∂

∂x(ut − 6uux − uxxx) = κuyy (13)

are archetypical examples of infinite dimensional completely

integrable systems.

They exhibit a tremendous amount of structure in terms of

conservation laws and symmetries.

ut = −uxxx + 6uux . (14)

solitary wave solutions, i.e.

u(x, t) = f(x− ct) . (15)

exercise: Assuming that f is sufficiently smooth, and decays at ∞together with its derivatives, it can be written as

f(x) = −12c sech2

(12√c(x− x0)

)(16)

The J. S. Russel Story

Figure 1: The term soliton was coined by M. Kruskal to denote the

aspect of a solitary wave and a particle, whence the greek suffix “ton”

Solutions of the KP

Soliton Equations are Full of Structure



It can be shown, that the KdV admits solutions of the form

u(x, t) = −2∂2x log det[A(x, t)] ,

where A is the N ×N matrix whose entries are

Aij = δij +βi

κi + κjexp(−(κi + κj)x+ 8κ3

i t) .

The asymptotic behavior along lines of slope 4κ2n on the (t, x)-plane

displays interesting particle like interactions, between the different

solitons.

Furthermore, the KdV comes together with a full hierarchy of

commuting Hamiltonian flows

It can be shown, that the KdV admits solutions of the form

u(x, t) = −2∂2x log det[A(x, t)] ,

where A is the N ×N matrix whose entries are

Aij = δij +βi

κi + κjexp(−(κi + κj)x+ 8κ3

i t) .

The asymptotic behavior along lines of slope 4κ2n on the (t, x)-plane

displays interesting particle like interactions, between the different

solitons.

Furthermore, the KdV comes together with a full hierarchy of

commuting Hamiltonian flows

The KdV hierarchy Xkk=1,3,5,···

X1(u) = ux

X3(u) = −uxxx + 6uux...

Rational Solutions of KdV

Airault, McKean, Moser, Adler, Ablowitz, Choodnovsky2

The Adler-Moser Polynomials:

If one iterates the Darboux process starting from u0 = 0 one obtains

from the results of the previous section that un = −2∂2x logWn. The

polynomial Wn, after a suitable choice of parameters and

normalization, is nothing more than the n-th Adler-Moser

polynomial. After a suitable choice of parameters in the sequence of

Darboux transformations can be made so that

un = −2∂2x log ϑn(x+ t1, t3, ..., t2n−1) ,

satisfies the flows of the KdV hierarchy, i.e.,

∂tkun = Xk[un] .

Furthermore:

satisfies the flows of the KdV hierarchy, i.e.,

∂tkun = Xk[un] .

Furthermore: If we look at the dynamics of the poles of the

rational solutions of KdV we get Calogero-Moser systems

Calogero-Moser Systems

x = ∂H/∂y (17)

y = −∂H/∂x (18)

H =12

∑y2j +

∑j<k

(xj − xk)−2 (19)

Integrability - Introduce (with Moser)

A(x, y) def= diag[y1, · · · , yn] +((√−1/(xj − xk)

))j 6=k

Eigenvalues of A(x, y) are constants of motion & in involution

Fjdef= Tr(Aj/j)

The Fj Poisson commute

Taking j = 2 and H = F2 the locus defined by

gradF2 = 0

x|∑j 6=k

(xj − xk)−3 = 0 ∀j = 1, · · · , n

× y = 0 (20)

is invariant by the flow of

xj = 6∑j 6=k

(xj − xk)−2 j =, · · · , n (21)

Miracle: If we set

vdef= 2

∑j

(x− xj)−2 v = v(x, t) xj = xj(t)

The requirement that v satisfies the KdV

vt = 3vvx − vxxx

is equivalent to (20) and (21).

Only the tip of the iceberg:

Miracle: If we set

vdef= 2

∑j

(x− xj)−2 v = v(x, t) xj = xj(t)

The requirement that v satisfies the KdV

vt = 3vvx − vxxx

is equivalent to (20) and (21).

Only the tip of the iceberg: Extended in many directions...

• Rational solutions of KP (Krichever)

• Rational solutions of AKNS (Sachs)

AKNS: Ablowitz-Kaup-Newell-Segur-Zakharov-Shabat Obtained a

way of constructing certain hierarchies of integrable evolution

equations.



equations. Integrability of the nonlinear-Schroredinger equation, the

mKdV, Sine-Gordon, ...

Set

L =[∂x −qr −∂x

]. (22)



equations. Integrability of the nonlinear-Schroredinger equation, the

mKdV, Sine-Gordon, ...

Set

L =[∂x −qr −∂x

]. (22)

The AKNS is obtained by constructing Lax pairs of the form

Ltk =[P(k) , L

]

This induces flows on the fields (q, r)qt` = A`(q, ..., ∂`xq; r, ..., ∂

`xr) ,

rt` = B`(q, ..., ∂`xq; r, ..., ∂`xr) .

(23)

flow generated by the hierarchy for ` = 3qt3 = 1

4(qxxx − 6qrqx) ,

rt3 = 14(rxxx − 6qrrx) .

Rational Solutions of AKNS

Set for p & q non-negative.

σ =p∏j=1

(x− xj(t)) (24)

τ =p+q∏j=p+1

(x− xj(t)) (25)

qdef= (log σ)xxσ/τ (26)

rdef= τ/σ (27)

Dynamics of the poles of (q, r)

The dynamics of (q, r) under the first equation of the AKNS

hierarchy is essentially equivalent to the dynamics of the roots xj of

σ and τ in two Calogero-Moser systems as follows: Set

al =

+1 if 1 ≤ l ≤ p−1 if p+ 1 ≤ l ≤ p+ q

and

xj = −8∑

k 6=j ak=aj

(xj − xk)−3 j = 1, · · · , p+ q

coupled through the constraints∑k 6=j

(aj − ak)(xj − xk)−2 = 0 j = 1, 2, · · · , p+ q

Dirac operators

Similar problem for Dirac operators ought to be studied.

D = γµ∂µ + v

The Dirac matrices γµ obey the anti-commutation rule

γµ, γν def= γµγν + γνγµ = 2gµν ,

where

((gµν)) = diag[1,−1, · · · ,−1]

is the Minkowski tensor,

and v is a matrix potential

Notation:6∂ = γµ∂µ .

Hence

6∂2 =

From now on: xj denotes the j-th coordinate.

Huygens’ principle for Dirac operators

Joint work F. Chalub (Lisboa)

Fundamental Solutions for the free Dirac operator:

(6∂ + v)Φ = δξ ,

where δξ denotes Dirac-delta at ξ = (t0, y)

Notation: λ is the geodesic distance from (t, x) to the point

ξ = (t0, y):

λdef=√

(t− t0)2 − ||x− y||2 =√

(xµ − ξµ)(xµ − ξµ) .

C(ξ) def= (t, x)|λ((t, x), ξ) = 0 ,

is the light cone with vertex in ξ

Definition: Φ = Φ(x, ξ) satisfies Huygens’ principle iff

supp Φ ⊂ C(ξ) ,

∀ξ ∈ Ω, (Ω a causal domain).

Theorem 1. The free Dirac operator in dimension n obeys theHuygens’ principle if, and only if, n is odd.

A few examples of Huygens ops are:

1. Depending only on t = x0, Huygens in three dimensions are

v =1t

v =(t t

t t

)2. An example of a potential depending on one space parameter.

v =

(0 1

x31x3

0

)is Huygens for n = 3.

Natural Question: Can one construct a family of (strict) Huygens

Dirac operators similar to the one obtained by Lagnese and

Stellmacher for the wave operator?




Theorem 2 (Chalub & JPZ). If q(x0) and r(x0) are solutions ofthe rational solutions of AKNS constructed in (24), then

6∂ − q + r

2I +

q − r2

γ (28)

satisfies Huygens’ property in dimension d+2, if d is odd, or d+3if d is even.




Theorem 2 (Chalub & JPZ). If q(x0) and r(x0) are solutions ofthe rational solutions of AKNS constructed in (24), then

6∂ − q + r

2I +

q − r2

γ (28)

satisfies Huygens’ property in dimension d+2, if d is odd, or d+3if d is even.

Remark: (JPZ, Jr. Diff. Eqs, 97(1):71–98, 1992) The

corresponding AKNS operators are bispectral.

Technical Details

Construction of Fundamental Solutions:

LF = δ

Technical Details

Construction of Fundamental Solutions:

LF = δ

consider the following problem for the wave eqE = δξ

E

∣∣∣∣t<t0

= 0 ,(29)

where ξ = (t0, ξ1, . . . , ξn) where ξ = (t0, ξ1, . . . , ξn)

admits the following solutions depending on the value of n.

E(x, ξ) =

cnδ

(n−3)/2(Υ) n odd

cnΥ−(n−1)/2+ n even,

(30)

where Υ denotes

Υ = (t− t0)2 −n∑i=1

(xi − ξi)2 ,

and δ(k) denotes the k-th derivative of Dirac’s delta function.

The distribution that appears for n even in equation (30)

Sn(Υ) =Υ−(n−1)/2

+

Γ((n− 3)/2)

plays an important role in what follows.

It should be interpreted by evaluating at x = Υ the function

Sν(x) =xν+

Γ(ν + 1)def=

xν

Γ(ν+1)x > 0

0 x < 0(31)

This function in turn has the special property of being homogeneous

of degree ν.

Seek solutions in the form:

∞∑ν=0

Uν(x, ξ)Sν−p(Υ) , (32)

where now Υ is the square of the geodesic in the metric defined by

the gij and

pdef=n− 1

2. (33)

For odd values of n: must search for E(x, ξ) of the form

E(x, ξ) = W1(x, ξ)Υ−p (34)

where

W1(x, ξ) =∞∑j=0

Uj(x, ξ)Υj .

where now Υ is the square of the geodesic in the metric defined by

the gij and

pdef=n− 1

2. (33)

For odd values of n: must search for E(x, ξ) of the form

E(x, ξ) = W1(x, ξ)Υ−p (34)

where

W1(x, ξ) =∞∑j=0

Uj(x, ξ)Υj .

For even n:

E(x, ξ) = V (x, ξ)Υ−p +W0(x, ξ) log Υ +R , (35)

where

W0(x, ξ) =∞∑j=p

Uj(x, ξ)Υj−p ,

V (x, ξ) =p−1∑j=0

Uj(x, ξ)Υj ,

and R is a smooth function.

where

W0(x, ξ) =∞∑j=p

Uj(x, ξ)Υj−p ,

V (x, ξ) =p−1∑j=0

Uj(x, ξ)Υj ,

and R is a smooth function.

Remark: The coefficients of the formal expansion (32) found by

substituting this ansatz into the equation

LE =∑

gijExixj +∑

biExi + uE = 0

and matching the behavior of the powers of Υ.

This yields for r = 0, 1, . . .

2∑

gijΥxi∂xjUr + 4((r − 1)− p)Ur + Ur(LΥ− uΥ) = −LUr−1 ,

(36)

where we set U−1def= 0.

Hadamard’s Criterion

The question of determining which hyperbolic operators satisfy a

strict Huygens’ principle received an indirect characterization in the

work of Hadamard.

Formal adjoint L† to the operator L.

The operator L† is defined as

L†ψ def=∑

∂xi∂xj(gijψ)−

∑∂xi(b

iψ) + uψ .

THEOREM: [(Hadamard’s Criterion)] The operator L satisfies a

strict HP iff n is odd, n > 1, and the elementary solution of the

adjoint operator L† contains no logarithmic term, i.e., W0(x, ξ) = 0for all ξ and all x in the internal part of the characteristic conoid.

adjoint operator L† contains no logarithmic term, i.e., W0(x, ξ) = 0for all ξ and all x in the internal part of the characteristic conoid.

Riesz Kernels

Marcel Riesz: unified treatment of the expression for the solution to

the wave equation for different n.

Basic Idea: Extending for λ ∈ C the expression

Ξ(x, t, λ) def=

(x20 −

∑ni x

2i )λ (

∑ni x

2i )

1/2 < x0

0 elsewhere.

This is defined for <[λ] > 0 and the meromorphic extension takes

values in an appropriate space of distributions.

Normalizing Ξ by a suitable meromorphic function (of λ) one gets

for non-positive integers derivatives of the δ function.

A Few Technicalities

g a Lorentzian metric of signature +,−,−, . . . ,−.

Square of the geodesic distance Υ(x, ξ),

Characteristic conoid C(ξ), by the equation Υ(x, ξ) = 0.

C(ξ) \ ξ: two connected components C+(ξ) and C−(ξ), which are

naturally associated with the forward and backward time.

Open subsets of Ω as D+(ξ) and D−(ξ).

working w/ an operator of the form

L = div grad + 〈a,∇〉+ u ,

all coefficients (real) analytic and defined in a causal domain Ω.

Causal Domain:

1. Any two pts x and ξ are joined by a unique geodesic.

2. D+(x) ∩D−(ξ) is either empty or compact in Ω.

DEFINITION

A forward Riesz kernel of L is a holomorphic mapping

C 3 λ 7→ ΦΩλ(·, ξ) ∈ D′(Ω) ,

s.t.

supp[ΦΩλ(·, ξ)] ⊆ D+(ξ) (37)

L[ΦΩλ(·, ξ)] = ΦΩ

λ−1(·, ξ) (38)

ΦΩ0 (·, ξ) = δξ . (39)

Remarks:

• λ = 1 the Riesz kernel gives a distribution E+(·, ξ) = ΦΩ1 (·, ξ) such

that

LE+(·, ξ) = δξ

• Causality

supp[E+(·, ξ)] ⊆ D+(ξ) .

• Modern treatment of the asymptotic behavior of solutions

to hyperbolic problems studies the so called “asymptotics in

smoothness”

• If the operator L is analytic, this asymptotic behavior can be

replaced by a convergent series.

• This type of argument has been used extensively in the recent work

of Berest.

Construction of Riesz kernels

Analytic continuation:

Let g ∈ D(Ω) and start w/

〈Rλ(x, ξ)|g〉 =∫D+(ξ)

Υ(x, ξ)λ−n+1

2

Hn+1(λ)g(x)dx ,

for

<[λ] >n− 1

2and

Hn+1(λ) def= 2πn−1

2 Γ(λ)Γ(λ− n− 12

) .

Then, extend for all λ ∈ C the above definition by analytic

continuation, using iterates of the (classical) formula

Rλ = Rλ−1 , (40)

which is valid for <[λ] > (n+ 3)/2.

Fact: The distribution Rλ(x, ξ) satisfies

supp[Rλ(x, ξ)] ⊆ D+(ξ) ,

Then, extend for all λ ∈ C the above definition by analytic

continuation, using iterates of the (classical) formula

Rλ = Rλ−1 , (40)

which is valid for <[λ] > (n+ 3)/2.

Fact: The distribution Rλ(x, ξ) satisfies

supp[Rλ(x, ξ)] ⊆ D+(ξ) ,

Exercise:Rλ ∗Rµ = Rλ+µ . (41)

(x− ξ|∂x)Rλ = (2λ− n+ 1)Rλ , (42)

R0(x, ξ) = δξ(x) , (43)

Proposition: For an odd n λ ∈ 1, 2, . . . , (n− 1)/2 we have

Rλ(x, ξ) =δ

(n+12 −λ)

+ (Υ)βn,λ

, (44)

where δ+(Υ) stands for the the Dirac’s delta measure concentrated

on C+(ξ) ∩ Υ = 0, and βn,λ a numerical constant.

Now, look for an expansion of the fundamental solution

E(x, ξ) ∼∞∑ν=0

Uν(x, ξ)Rλ+ν(x, ξ) . (45)

Substitute expression (45) into

LE = δξ

obtain a set of recursive transport relations

(x− ξ|∂x)Uν(x, ξ) + νUν(x, ξ) = −14L[Uν−1(x, ξ)] . (46)

Facts:

• The above recursive system has a unique solution provided one

normalizes

U0(x, ξ) = 1 , (47)

and requires for r ≥ 1

Ur(x, ξ) = O(1) , x→ ξ . (48)

• If the operator L has analytic coefficients, then the series (46) is

uniformly convergent in a sufficiently small neighborhood of Υ = 0.

Riesz kernel for the operator L can be expanded as

ΦΩλ(x, ξ) =

∞∑ν=0

4νΓ(λ+ ν)

Γ(λ)Uν(x, ξ)Rλ+ν(x, ξ) .

Main argument for the proof of Hadamard’s result:

If n is even then for ν = 0, 1, 2, . . . we have

supp[Rν+1(x, ξ)] = D+(ξ) ,

and so Huygens principle does not hold.

If n is odd then for ν = 0, 1, 2, . . . , (n− 3)/2 we have

supp[Rν+1(x, ξ)] = C+(ξ) .

Notice: Using equation (46), and p = (n− 1)/2 one gets

Hadamard’s classical formula for n odd

E+(x, ξ) =1

2πp(V (x, ξ)δ(p−1)

+ (Υ) +W (x, ξ)H+(Υ)) ,

where H+(γ) is the the Heaviside step distribution on the forward

region D+(ξ).

〈H+(γ)|ϕ〉 =∫D+(ξ)

ϕ(x)dx .

Furthermore, V (x, ξ) and W (x, ξ) are analytic in a neighborhood of

x = ξ with expansions

V (x, ξ) =p−1∑ν=0

sνUν(x, ξ)Υν ,

where sν = [(1− p) . . . (ν − p)]−1 and

W (x, ξ) =∞∑ν=p

1(ν − p)!

Uν(x, ξ)Υν−p .

i.e.: strict Huygens’ principle iff vanishing of the term W .

Lemma: The term W (x, ξ) vanishes iff Up(x, ξ) = 0 for x on the

surface of the forward light cone C+(ξ).

Idea:

The lemma follows from the fact that W (x, ξ) is a solution of the

characteristic Goursat problem

L[W (x, ξ)] = 0 ,

with a bdry value given on the cone surface C+(ξ).

This Goursat problem has a unique solution.

Hence, W (x, ξ) ≡ 0 iff W (x, ξ) = 0 for x ∈ C+(ξ).

DEFINITION: The Hadamard series is said to be truncated (or

terminated) at level ν0 if the sequence defined by the recursions

(46), (47), and (48) is zero for ν ≥ ν0.

Corollary of previous Lemma

THEOREM:

Let L be a real-analytic formally self-adjoint hyperbolic on a causal

domain Ω ⊂ Rn+1 with n > 1.

Then, L is strictly Huygens iff n is odd and the Hadamard series for

L is truncated at level p = (n− 1)/2.

References

[1] J. J. Duistermaat and F. A. Grunbaum. Differential equations in

the spectral parameter. Comm. Math. Phys., 103(2):177–240,

1986.

[2] J. P. Zubelli and F. Magri. Differential equations in the spectral

parameter, Darboux transformations, and a hierarchy of master

symmetries for KdV. Commun. Math. Phys., 141(2):329–351,

1991.

[3] Jorge P. Zubelli. Rational solutions of nonlinear evolution

equations, vertex operators, and bispectrality. J. Differential

Equations, 97(1):71–98, 1992.

[4] Jorge P. Zubelli and D.S. Valerio Silva. Rational solutions of

the master symmetries of the KdV equation. Commun. Math.

Phys., 211(1):85–109, 2000.

[5] George Wilson. Collisions of Calogero-Moser particles and an

adelic Grassmannian (with an appendix by I. G. Macdonald).

Invent. Math., 133(1):1–41, 1998.

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Huygens’ Principle, Integrable PDEs, and Solitons · Huygens’ Principle, Integrable PDEs, and Solitons Jorge P. Zubelli July 20, 2004 zubelli/ download:zubelli