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§4. Solution of the Scattering Problem 31
As we mentioned in Section 1 the flow is particuarly simple in this case
since no periodic or recurrent solutions are present. In the more interesting case
of the periodic boundary condition one has quasiperiodic solutions and the task
of finding coordinates for the integral surfaces which are tori, is more difficult.
The main problem is the «inverse problem» which consists in recovering L —
which is then a cyclic matrix — from its eigenvalues and appropriately chosen
quantities. This problem seems unsolved as yet.
§ 4. Solution of the Scattering Problem
From the results of Section 2 it follows that the asymptotic behavior of the
solutions of our problem (1.2
)isgivenby
x
(t)=α
k
x
(−t)=−α
k
+
k
t + β
−
t + β
k
+
k
+ o(e
−
+ o(e
k
−δt
−δt
),
(4.1)
)
for t → +∞ with some δ>0. Moreover, we found
+
α
k
= lim
k→k+∞
yk= −2λ
n−k+1
and α
−
= −2λ
k
k
i. e.
+
α
n−k+1
which expresses that the (n − k +1)
which the k
th
particle had in the past.
st
Our goal is to determine the relation between the phases β
−
= α
k
(4.2)
particle has then for t → +∞ velocity
+
−
, β
which can
k
k
be given explicitly too. This remarkable fact is also a consequence of the inte-
xk−x
k+1
grable character of the system and the representation of e
functions of λ
−λjt
, e
j
given by (3.10), (3.11). An explicit calculation seems
, ykas rational
prohibitive; nevertheless the following argument, which uses just rudimentary
properties of rational functions will lead to the goal. The result is
+
β
n−k+1
= β
−
+
k
ϕjk(α−)(4.3)
j=k
where
ϕ
(α)=
jk
−log(α
log(α
−
−
− α
j
−
j
)2for j<k,
k
−
− α
)2for j>k.
k
(4.4)
32 Finitely Many Mass Points on the Line
For n =2this amounts to
+
Thus ϕ
β
β
represents the phase shift between two particles with velocities α
jk
−
= β
= β
− log(α
1
−
+ log(α
2
2
+
1
−
−
− α
− α
)2,
2
−
)2.
2
(4.5)
−
, α
j
1
−
1
at t = −∞. The result (4.3) can therefore be interpreted as follows: The
particles are scattered just as if their interaction takes place two at a time! This
was suggested to me by M. Kruskal who described an analogous phenomenon
for solutions of the Korteweg – de Vries equation (see [6], Theorem 3.7) and
by P. D. Lax. This phenomenon which had been discovered by Zakharov et al.
(see [6] for references) is obviously intimately related to our result and it is
conceivable that one can be derived from the other — but we have not pursued
this point.
We illustrate the statement in Figures 1, 2. Figure 1 illustrates the
case n =2, which is given explicitly in terms of cosh(λ
totic behavior can be interpreted as the elastic reflection of two rods of length
ϕ
= log(α
21
of ϕ
21
−
−
− α
2
)2, provided this number is positive. For negative values
1
the particles reflect only after passing each other. However, this inter-
− λ1)t. The asymp-
2
pretation is somewhat misleading, especially if n>2, since the length of the
rods depends on their velocity, not on the label. We indicate schematically the
construction of the scattering for n =3in Figure 2.
−
k
Fig. 1 Fig. 2
§4. Solution of the Scattering Problem 33
To prove (4.3) first translate it into an asymptotic statement for (2.2). For
this purpose we note that on account of the linear t-dependence of the center of
mass we have
and therefore it suffices to prove (4.3) for the differences β
n
k=1
n
+
β
=
k
k=1
−
β
k
−
k+1
− β
−
,i.e. it
k
suffices to establish
β
+
n−k
− β
+
n−k+1
= β
−
k+1
− β
−
−
k
ϕjk+
j=k
j=k+1
ϕ
j, k+1
.
Using (2.1), (4.1) this amounts to
lim
t→+∞
a
n−k
(t)ak(−t)e
2(λ
k+1−λk
)t
= Ck(λ),k=1, 2, ..., n− 1(4.6)
with
Finally, with α
log(4C
−
= −2λkand (4.4) this gives for Ckthe expression
k
)2= −
k
j>k
C
=
k
j<k
j=k
(λj− λk)
(λk− λj)
ϕ
j, k
j<k+1
·
j>k+1
+
j=k+1
(λ
k+1
(λj− λ
ϕ
j, k+1
− λj)
k+1
.
)
where empty products are to be set equal to 1.
Thus it suffices to prove (4.6) with (4.7). For n =2this is easily verified.
In that case our transformation (3.10) takes the explicit form
Since r
a
(t)=rk(0)e
k
b
=
1
λ
2
=
1
(λ
− λ1)r1r
2
2
r
1
−λkt
we find
a
(t)a1(−t)e
1
2
r
1
2
r
1
+ r
+ λ1r
2
+ r
2
2
2
2(λ2−λ1)t
2
2
,b2=
2
=(λ2− λ1)
2
λ
r
1
1
2
r
1
+ λ1r
+ r
r
1
r
2
2
2
,
2
2
−1
r
2
+
.
r
1
→ (λ2− λ1)2, (4.9)
which corresponds to (4.6) for k =1, n =2.
(4.7)
(4.8)
34 Finitely Many Mass Points on the Line
For n =3one can still, with some effort, verify the above statement by an
explicit calculation but for general n this seems a hopeless approach. Therefore
we proceed as follows. We know that for every solution
+
, c
n−k
−
e
k
−
k
−(λ
.
+
n−k
=
e
−(λ
r
, c
k+1
k+1−λk
rk(0)
a
(t) ∼ c
n−k
ak(−t) ∼ c
with positive constants c
+
n−k
(i) First we establish that c
(0), bj(0) of the solution. (ii) Second, we show that
data a
j
+
c
n−k
)t
k+1−λk
as t → +∞ (4.10)
)t
−
depend real analytically on the initial
k
(0)
+
(λ),
C
n−k
(4.11)
r
(0)
k
=
r
k+1
n−k
−
(λ)
C
k
(0)
+
−
(λ)C
(λ), (4.11)
k
with C
+
n−k
−
c
k
−
, C
depending on λ only. This shows that the limit (4.6) is equal to
k
C
(λ)=C
k
and therefore independent of the initial condition of r
determination of C
(λ) easy if we consider various limit situations for the initial
k
. This makes the actual
j
conditions, which will be the third step (iii).
+
−
To begin with the analytic dependence of the constants c
initial conditions we fix a solution a
(t), bj(t) of (2.2) and describe a nearby
j
n−k
, c
k
one by
a
= aje
j
,bj= bj+ v
j
u
j
where uj(0), vj(0) are small. The differential equations for u, v are then
˙u
˙v
k
=2a
= v
k
2
2u
k
(e
− 1) − a
k
k+1
− vk,
2
k−1
(e
2u
k−1
− 1).
The asymptotic behavior of the solutions is given by
u
=(v
k
v
= vk(+∞)+O(e
k
(∞) − vk(∞))t + γk+ O(e
k+1
−δt
)
−δt
)
for t → +∞ (4.12)
on the
§4. Solution of the Scattering Problem 35
if the initial data are small enough. It is sufficient to show that v
real analytically on u
(0), vj(0) if these are close to zero. For this purpose we
j
(∞), γkdepend
k
permit complex initial values and show that the above asymptotic description
holds for a complex neighborhood of the origin. This requires some simple a
priori estimates:
Obviously it suffices to establish the analytic dependence on the initial
values u
an estimate
with some positive constants δ, c
(τ), vk(τ) for some fixed positive τ . The fixed real solution satisfies
k
0 <a
(t) <c1e
k
−δt
for 0 t<∞
; we may assume δ<1 <c1. With
1
0 <η<
δ
8
and some τ, to be determined later, we consider complex initial values in
|u
(τ)| <η, |vk(τ)| <η. (4.13)
k
Let M (t)=max
which M(t) 2η. Then we get from the differential equations for τ t<τ
Using the inequality
and setting s = t − τ we get for τ t<τ
|vk(t)| and consider this function in an interval τ t<τin
k
t
|uk(t)| η +2
|vk(t)| η +8c
2
1
M(t) dt η(1 + 4(t − τ )).
τ
2u
|e
− 1| 2|u|e
t
−2δt
η
e
τ
(1 + 4(t −τ ))e
2 |u|
2η(1+4(t−τ ))
dt
hence
with c
2
M(t) η1+c
2
=8c
e2.Since2δ −8η>δwe get
1
M(t) η(1 + c
∞
−2δτ
e
2
e
0
e
2
−(2δ−8η)s
−2δτ5δ−2
(1 + 4s)ds
).
36 Finitely Many Mass Points on the Line
Now we fix τ so that
so that
Thus we can take τ
M(t) < 2η for τ t<τ
= ∞ and have the estimate
(t)|−< 2η for all t τ,
|v
k
5c
−2δτδ−2
e
2
< 1
.
and all complex initial data in the polydisk (4.13).
Since v
real axis for t → +∞ the limit function v
(t) depends analytically on those initial data and converges on the
k
(∞) is analytic in (4.13). From the
k
differential equation we obtain
(t) − vk(∞)| c4e
|v
k
−δt
for t τ
and
u
(t) − (v
k
(∞) − vk(∞))t =
k+1
= u
k
(0) +
t
(v
(t) − v
k+1
0
(∞) − vk(t)+vk(∞)) dt
k+1
converges for t → +∞ with a uniform bound. Hence its limit γ
in (4.13), completing the proof of (i).
To prove (ii) we use the representation (3.10) of the solutions by
is analytic
k
2
a
Here A
(t)=A
n−k
is a rational function in r, λ,say,
n−k
(r, λ) with rj= rj(0)e
n−k
=
A
n−k
P
Q
with P , Q being polynomials in r, λ. They are homogeneous in the r
the same degree, since A
for t → +∞ we assume first that λ
is of degree 0. To study its asymptotic behavior
n−k
j+1/λj
2, ..., n− 1. Then the dominant term in P is the one which comes first in
lexicographical ordering of the exponents of r
−λjt
,λj= λj(0).
, both of
j
is sufficiently large for all j =1,
n
p
.LetP0=
j
j=1
j
r
be this term
j
n
in P and Q
=
0
j=1
q
r
j
in λ and
Since on the other hand
§4. Solution of the Scattering Problem 37
j
the dominant term in Q.ThenP0, Q0are polynomials
n
A
n−k
P
∼
Q
(pj−qj)
0
r
0
j=1
.
j
2
(t) ∼ const · e
a
n−k
we conclude that p
k
Here the coefficient P
− qk= −2, p
A
n−k
0/Q0
−2(λ
k+1
r
P
0
∼
Q
0
k+1−λk
− q
k+1
r
k
)t
,k=1, ..., n− 1,
=+2, pj= qjotherwise, and
k+1
2
for t → +∞.
is positive for λ1<λ2<... <λn. This proves the
first line of (4.11) with
P
0
Q
0
at least for large values of λ
real λ in λ
<λ2< ...< λnthis equation holds for all those λ1. The second
1
+
)2=
(c
n−k
. Since the coefficient is real analytic for all
j+1/λj
equation of (4.11) follows in just the same manner.
This shows that the coefficient C
2
that C
(λ) is a rational function of λ. We will determine Ck(λ) by induction
k
on n. Incidentally, this will show that even C
(λ) in (4.11) is independent of rj(0),and
k
(λ) is rational. For n =2the
k
formula (4.7) or equivalently (4.3), (4.4) was verified. We prefer to prove the
statement in the form (4.3). From our argument we know that
+
β
n−k+1
1
In general, the limit of such a rational function of exponentials may even be discontinuous, e. g.
for
Ar
Cr1r3+ Dr
but
1r3
−
= β
+Φk(α),k=1, 2, ...,
k
2
+ Br
2
→
2
2
λ
A
for
C
λ3− λ
− λ
2
1
> 1,
2
→
− λ
λ
2
λ3− λ
1
< 1.
2
B
for
D
38 Finitely Many Mass Points on the Line
where Φk(α) is a real analytic function of α =(α
−
of β
. This follows from the fact that Ck(λ) is independent of rj(0),and
j
depends on λ
function by ψ
j
k
= −
−
1
α
only. For n +1 particles we denote the corresponding
j
2
(α), α =(α
−
, ..., α
1
−
). The induction proof requires the
n+1
−
, ..., α
1
−
), independent
n
verification of
ψ
(α)=Φk(α)+ϕ
k
The determination of ψ
n+1,k
(α) follows then from
n+1
n+1
for k =1, 2, ..., n. (4.14)
ψk(α)=0
k=1
which is a consequence of the linear t-dependence of the center of mass. Thus
it suffices to prove (4.14). Since both sides are independent of the β
and will, choose β
−
very large positive, so that the n particles x1,x2, ..., x
n+1
−
we may,
j
have already undergone their mutual interaction and are very far apart by the
time x
force on x
and t is large positive. Thus the interaction of x
essentially place pairwise. Since before the interaction with x
interacts with any of them. In other words, when x
n+1
, ..., xnthere are already close to
1
+
x
k
∼ α
x
n−k+1
t + β
k
+
, where β
k
−
∼ α
t + β
k
+
−
= β
k
n−k+1
n+1
−
+Φk,k=1, 2, ..., n,
k
(t) exerts some
n+1
+Φ
n−k+1
with xk(t), k n takes
we have
n+1
n
we obtain after interaction
−
x
n−k+1
∼ α
which shows that ψk∼ Φk+ ϕ
−
independent of β
the assertions (4.14) follow. The situation is depicted in
t + β
k
n+1,k
−
+Φk+ ϕ
k
−
if β
n+1
n+1,k
→∞. But since ψk, Φkare
Figure 3.
§ 5. Associated Differential Equations
The above Hamiltonian system (1.2) possesses, according to the above,
n integrals I
, I2, ..., In, which incidentally are polynomials in y
1
k
§5. Associated Differential Equations 39
Fig. 3
xk−x
and e
k+1
. One may show, which we will not do here, that these
integrals are in involution, i. e. the Poisson bracket for any two of
these vanishes. Using the integrals as new Hamiltonians one can introduce n new vector fields which possess the same integrals and commute
with each other. This makes the manifolds I
=constinto commuta-
k
tive groups. These are well known facts for integrable Hamiltonian systems (see, for example, Appendix 26 of [7]) which we will verify here directly.
As a starting point we take the differential equation (2.7) which represents
a deformation of the Jacobi matrix L leaving the spectrum fixed. But there are
many such isospectral deformations corresponding to different choices of B.We
restrict ourselves to skew symmetric matrices B giving rise to orthogonal similarity transformations. But instead of permitting only one pair of off diagonals
we allow several. Let B
stand for a skew symmetric matrix with p off diago-
p
nals above and adjacent to the diagonal. Thus the matrix B defined below (2.7)
would be denoted by B
nontrivial matrices B
tial equation, that is that the commutator B
. We claim that for every p in 1 p<nwe can find
1
such that˙L = BpL −LBpdefines a meaningful differen-
p
L − LBphas only one off diagonal
p
above the diagonal, while the others all vanish. We will establish this assertion
below but point out first that all these differential equations have the eigenvalues
40 Finitely Many Mass Points on the Line
of L as integrals and can therefore be transformed into the variables rk, λkas
˙
λ
=0, ˙rk= fk(λ, r).Forp =2one finds the matrix
k
β
n−2
n−1
0
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
⎛
0 β
⎜
⎜
−β10 β
⎜
⎜
−γ
B
=
2
⎜
⎜
⎝
−β
1
.
γ
1
.
1
2
0 γ
2
.
−γ
n−2
−β
.
.
.
.
.
.
n−1
where
β
=(bk+ b
k
γ
= aka
k
k+1
and the differential equation˙L = B
˙a
= ak(a
k
˙
=2bk(a
b
k
where we set a
2
k+1
=0, an=0. Introducing r, λ again by the transformation (3.10)
0
2
k
− a
− a
2
k−1
2
k−1
)+2b
+ b
,k=1, 2, ..., n− 1,
k+1)ak
,k=1, 2, ..., n− 2
L − LB2takes the explicit form
2
2
k+1
k+1
2
− b
),k n − 1,
k
2
a
− b
a
k−1
k
we find the differential equation
dλ
dt
k
=0,
dr
dt
k
= −λ
We just indicate the calculation. First restricting r
df
dt
=
2r
k˙rk
λ − λ
k
On the other hand
df
=(˙R(λ)e
dt
= −2(RB
= −2(γ
)=((B2R − RB2)en,en)=
n,en
2en,en
n−2Rn, n−2
)=−2(R(γ
+ β
n−1Rn, n−1
n−2en−2
).
(5.1)
2
,k n,
k−1
2
rk. (5.3)
k
2
tor
k
=1we have
k
(5.2)
. (5.4)
+ β
n−1en−1
),en)=
With (5.1) and
R
n, n−1
Δ
n−2
=
Δ
a
,R
n−1
n
n, n−2
Δ
n−3
=
Δ
a
n
n−2an−1
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