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§2. Flaschka’s Form of the Differential Equation and Asymptotic Behavior 21
where
n
2
λkr
k
n
2
2
r
k
k=1
(1.4)
V =
k=1
and the variables are restricted to the (2n − 1) dimensional domain
n
λ
1<λ2
<... <λn;
Clearly, the solutions run from the maximum of V at r
imum of V of r
integrals of the motion, while r
surfaces λ
k
is up to translation of the x
= δk1as t runs from −∞ to ∞,andλ1, λ2, ..., λnare
k
, ..., r
1
=const. The mapping taking x, y into the variables λk, rkon (1.5)
one to one and will be given explicitly. The inverse
k
2
r
=1,rk> 0. (1.5)
k
k=1
= δknto the min-
k
can be viewed as parameters on the
n−1
mapping illustrates the inverse method of spectral theory.
Thus this note does not claim any new idea and should be considered as
providing a simple model illustrating the construction of integrals and its connection with the inverse method of spectral theory in extreme simplicity, yet with
all rigor. On the other hand, it leads immediately to an unsolved problem if one
wants to carry out this approach for the periodic boundary condition (1.3
though the integrals I
on the level surfaces I
for this problem are well known, no parameters are known
k
= ckwhich determine the xk(mod 1), ykuniquely. This
k
is related to the lack of an inverse theory for the Hill’s equation −u
). Al-
+ q(x)u =
= λu, q(x +1) = q(x) under periodic boundary conditions u(x +1) =u(x)
where the problem consists in finding a set of quantities which together with the
eigenvalues allow one to determine q(x). One can hope to shed some light on
this question if one could solve the above finite dimensional problem.
§ 2. Flaschka’s Form of the Differential Equation and
Asymptotic Behavior
We set, with Flaschka,
1
a
k
(xk−x
e
=
2
k+1
)/2
,bk= −
1
y
k
2
(2.1)
22 Finitely Many Mass Points on the Line
so that the differential equations (1.2) go into
˙a
k
˙
b
k
= ak(b
=2(a
− bk),k=1, 2, ..., n− 1
k+1
2
2
− a
k
),k=1, 2, ..., n
k−1
(2.2)
with the boundary conditions (1.3) being
=0,an=0. (2.3)
a
0
Observe that (2.1) provides a transformation of the (x, y) variables into
the (a, b)-variables. We identify points (x, y), (x, y) if x
− xkis independent
k
of k, and call the equivalence class a «configuration». It is characterized by the
(2n −1) numbers x
− xn, k =1, ..., n− 1,andyk, k =1, ..., n. Thus
k
(2.1) defines an invertible transformation of the (2n −1)-dimensional space of
configurations into the domain
D =a, b | a
> 0,k=1, ..., n− 1,
k
and it remains to study the flow given by the quadratic differential equation (2.2)
in D. The energy is given by
n
k=1
2
b
. (2.4)
k
H =4
n−1
k=1
1
2
a
+
k
2
We show first that for any solution in D
(t) → 0 for t →±∞ and k =1, ..., n− 1, (2.5)
a
k
which amounts to the assertion that x
k+1−xk
we consider the system (2.2) with prescribed a
+∞
2
that
(a
−∞
2
+ a
) dt < ∞ and prove the
0
n
→∞as t →±∞. To prove this
(t), an(t) ∈ L2(−∞, +∞) so
0
Lemma. For any solution of (2.2) with this modified boundary condition
we have
−∞
∞
2
1
+ a
2
) dt < ∞.
k−1
(a
§2. Flaschka’s Form of the Differential Equation and Asymptotic Behavior 23
ROOF.
P
Consider the function
ϕ(t)=b
− b
1
n
for which
Thus
dϕ
dt
=˙b
1
−˙bn=2(a
ψ =
1
2
ϕ −
2
0
−∞
+ a
t
(a
2
) − 2(a
n
2
+ a
0
2
n
) dt
2
1
+ a
2
n−1
).
satisfies
dψ
dt
= −(a
2
1
+ a
2
n−1
).
Since by the energy relation ϕ and hence ψ is bounded, also
T
2
1
+ a
2
) dt = ψ(−T ) − ψ(T )
n−1
(a
−T
is bounded for T →±∞, proving the lemma.
We can apply this argument, in particular, to a
= an=0. Applying this
0
lemma to the reduced system where the first and last equations in the first and
+∞
2
second line of (2.2) are cancelled we conclude that
(a
−∞
2
+ a
n
) dt < ∞ and
n−2
inductively that
+∞
n−1
k=1
Since on the other hand |˙p| 2a
n−1
that p =
quence t
is so selected that |t
intervals |t − t
2
a
→ 0 for t →±∞. Indeed, otherwise there would exist a se-
k
1
→∞with p(tk) δ>0. We may assume that the sequence
k
− tk| δ/M.Sincep(t) δ/2 in the disjoint
k+1
δ
1
k
| <
it cannot be integrable, contradicting (2.6). This
2
M
proves (2.5). Moreover, we conclude from (2.2) that b
2
a
dt < ∞. (2.6)
k
−∞
2
|bk− b
k
| M is bounded it follows
k+1
tends to a limit bk(∞)
k
as t → +∞.
24 Finitely Many Mass Points on the Line
Flaschka [1, 2] noted that the above system (2.2) can be expressed in matrix
form
d
L = BL − LB (2.7)
dt
where
L =
⎛
b
1a1
⎜
a
1b2
⎜
⎜
⎜
⎜
⎝
0 a
.
.
.
b
n−1an−1
n−1bn
0
⎞
⎟
⎟
⎟
; B =
⎟
⎟
⎠
⎛
0 a
⎜
⎜
⎜
⎜
⎜
⎝
−a
1
0
1
0 −a
.
.
.
0 a
n−1
0
n−1
0
⎞
⎟
⎟
⎟
⎟
⎟
⎠
Thus if U = U(t) is the orthogonal matrix satisfying
dU
= BU; U(0) = I
dt
then by (2.7)
d
−1
(U
dt
hence
U
Thus, L(t) is similar to L(0) and the eigenvalues λ
LU)=0
−1
LU = L(0).
of the Jacobi matrix L,
k
which are real and distinct, are independent of t. This description of the integrals
as eigenvalues of a linear operator is due to Lax [5] and Flaschka’s derivation
was based on his approach.
Thus the characteristic polynomial
(λ)=det(λI −L)=
Δ
n
as well as the coefficients I
n
(λ − λk)=
k=1
, ..., Inare constants of the motion (2.2). For
1
n
k=0
Ikλ
n−k
(2.8)
definiteness we order the eigenvalues according to their size,
.
λ
1<λ2
<... <λn.
Notice that L(t) → L(∞) as t → +∞ where L(∞) is a diagonal matrix whose
diagonal elements must be the eigenvalues λ
˙a
k
∼ b
(∞) − bk(∞)
k+1
a
k
in appropriate order. From
k
§3. Partial Fractions and Continued Fractions 25
and (2.5) we conclude that b
(∞) <bk(∞) or
k+1
b
(∞)=λ
k
n−k+1
i. e.
L(∞)=diag(λ
n,λn−1
, ..., λ1).
Using that the t-reversing substitution
t →−t; a
k
→ a
; bk→ b
n−k
n+1−k
leaves the system invariant, we conclude that
L(−∞)=diag(λ
1,λ2
, ..., λn)
i. e. L(∞), L(−∞) differ just in the order of the diagonal elements. The
physical interpretation of this result is: If for t →−∞the particles x
the velocities y
the particles x
= −2λkwhere y1<y2< ... < ynthen for t → +∞
k
have the velocities y
k
n−k+1
so that the particles exchange their
approach
k
velocities.
This describes the flow for our problem (2.2), (2.3). Still we will find
another set of variables, r
> 0, k =1, ..., n− 1, which together with the λ
k
form a set of coordinates, and represent the differential equations in these new
variables.
k
§ 3. Partial Fractions and Continued Fractions
Let
R(λ)=(λI − L)
where we suppress the dependence in t.Thisisann by n matrix and we single
out the element in the last row and last column
R
(λ)=(R(λ)en,en)=f(λ) where en=(0, 0, ..., 0, 1), (3.1)
nn
and f (λ) is hereby defined. Since L is symmetric it follows that f (λ) is an
analytic function for Im λ =0and
Im f(λ) > 0 for Im λ>0.
−1
26 Finitely Many Mass Points on the Line
Moreover, it is rational with simple poles at the eigenvalues λkand so admits
the partial fraction expansion
n
n
k=1
2
r
k
,rk> 0, (3.2)
λ − λ
k
2
r
=1.
k
with positive residua r
f(λ)=
k=1
2
. Moreover, for |λ|→∞one has λf(λ) → 1 and
k
Thus we have a mapping ϕ associating with every point in
D = {a
, ..., a
1
, ..., bnwith ak> 0} (3.3)
n−1,b1
a point in
Λ={λ
, ..., λn,r1, ..., rnwith λ1<λ2<... <λn,
1
n
2
r
=1,rk> 0}.
k
k=1
(3.4)
We claim that this mapping ϕ: D → Λ is one to one and onto. We will view
it as a coordinate transformation and then describe the differential equations in
−1
the new variables. The fact that the mapping ϕ has an inverse ϕ
:Λ→ D
corresponds to the inverse method of spectral theory, which in the elementary
form described here goes back to Stieltjes [3]. It is based on the fact that f (λ)
admits a continued fraction expansion
where the entries a
f(λ)=
λ − bn−
, bkagree precisely with those of L.
k
a
λ − b
2
n−1
n−1
1
−
.
.
.
−
2
a
1
λ − b
1
(3.5)
To prove this we establish the identity
Δ
f(λ)=
n−1
Δ
n
(3.6)
§3. Partial Fractions and Continued Fractions 27
where Δ
is the characteristic polynomial of (λI − L), see (2.8), and Δ
n
the k by k subdeterminant obtained by canceling the last n−k rows and columns
of (λI −L). Expanding Δ
by the last row one finds
k
Δ
=(λ − bk)Δ
k
k−1
− a
2
k−1
Δ
k−2
(3.7)
for k =3, 4, ..., n; it holds also for k =1, 2 if we set
Δ
=0, Δ0=1.
−1
Thus the ratios s
which leads to a finite continued fraction for s
=Δk/Δ
k
s
k
k−1
= λ −bk−
satisfy the recursion formula
2
a
k−1
for k =2, 3, ..., n
s
k−1
Δ
n
=
n
Δ
n−1
= f−1(λ).
Thus the representation (3.5) follows from (3.6) which we prove now. For
this purpose we compute the last column
= z of R = R(λ).
Re
n
We fin d
⎧
k+1
z
Δ
k
=
a
... a
k+1
Δ
n
Δ
n−1
=
n
Δ
.
n
for k =0, 1, ..., n− 2,
n−1
(3.8)
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
z
k
Indeed z is the solution of
(λI −L)z = e
n
and using the recursion formula one readily verifies (3.8). Thus
Δ
f(λ)=R
nn
= zn=
n−1
Δ
n
as we wanted to show.
Thus for a given matrix L we can compute the rational function f(λ) which
has n simple real poles with positive residua, since Im f(λ) > 0 for Im λ>0.
28 Finitely Many Mass Points on the Line
Ordering these poles according to size we have defined the mapping ϕ taking D
into Λ (see (3.3), (3.4)).
We come to the «inverse problem» which requires that we determine ϕ
−1
With any point in Λ we associate f (λ) by (3.2). Then Im f>0 for Im λ>0
and λf(Λ) → 1 for |λ|→∞. Thus
1
= λ + A − g(λ)
f(λ)
where A is a real constant and g(λ) is a rational function which satisfies
Im f
Im g(λ)=Imλ +
> 0 for Im λ>0.
2
|f|
.
Thus g(λ) has only simple poles on the real axis and their number is n −1.One
2
λkr
computes easily −A =
g = Bf
with B>0,andλf
n−1
1
f(λ)=
k
n−1
, λg(λ) →λ
→ 1 for |λ|→∞. Thus
1
λ + A − Bf
2
k
n−1
2
r
−
k
2
2
r
k
> 0. Thus
k
λ
and by induction we get a unique continued fraction of the form (3.5) with
A = −b
, B = a
n
2
> 0, etc. This shows that ϕ maps D one to one onto Λ.
n−1
Finally we express the differential equation (2.2) in these new variables.
For this purpose we deduce from (2.7)
d
dt
and taking the last element R
df
=(e
, (BR − RB)en)=−2(en,Ra
dt
Since R
n
n, n−1
agrees with z
dL
R = R
nn
n−1
df
dt
R = BR − RB
dt
= f in R we find
in (3.8) we obtain
= −2a
2
n−1
n−1en−1
Δ
n−2
Δ
n
.
)=−2a
n−1Rn, n−1
.
§3. Partial Fractions and Continued Fractions 29
This formula allows us to determine the desired differential equations. Since
we established already that dλ
/dt =0we have
k
n
df
dt
2rk˙r
=
λ − λ
k=1
k
.
k
Comparing the residue of the last two expressions we get
Δ
n−2
2r
k˙rk
= −2a
2
n−1
Δ
.
n
λ=λ
k
By the recursion formula (3.7) we have
=(λ −bn)Δ
Δ
n
n−1
− a
2
n−1
Δ
n−2
or, since Δn(λk)=0,
− b
λ
k
Δ
n−2(λk
)=
a
2
n−1
n
Δ
n−1(λk
),
hence
2r
= −2(λk− bn)
k˙rk
A similar comparison of the residua of
f(λ)=
gives
Δ
hence
2r
k˙rk
n
k=1
2
r
=1we find
k
Since
0=r
Δ
Δ
n−1
n
λ=λ
=
k=1
= r
k
Δ
n−1
Δ
Δ
n
n
= −2(λk− bn)r
= −λkr
k˙rk
n−1
n
r
λ − λ
2
k
2
k
2
k
2
k
λ=λ
.
+ b
.
k
k
n
30 Finitely Many Mass Points on the Line
and so, bn=λkr
2
, as we had seen before, and
k
˙r
= −λk−λjr
k
2
r
k
j
which gives the differential equation (1.4) of Section 1. These differential
equations represent the vector field along the gradient of the function V (r)
(see (1.4
Thus every solution approaches for t → +∞ the minimum: r(t) → e
for t →−∞the maximum: r(t) → e
)) restricted to the part of the unit sphere lying in the positive quadrant.
and
1
. Of course, it is also possible to give an
n
analytical representation for the solutions, since they are obtained by projecting
the linear differential equations ˙r
= −λkrkon the unit sphere. Thus we find
k
⎧
λ
(t)=λk(0),
⎪
k
⎪
⎪
⎪
⎨
2
r
(t)=
k
⎪
⎪
⎪
⎪
⎩
To summarize our result we consider the r
n
j=1
2
−2λkt
r
(0)e
k
2
r
(0)e
j
k
.
−2λjt
as homogeneous variables, still
positive, and set, accordingly,
n
f(λ)=
k=1
λ − λ
From (3.5) and the calculation of continued fractions it is clear that the a
2
r
k
k
n
k=1
−1
2
r
. (3.9)
k
2
, b
k
are rational functions of rj, λjof degree 0 in the rj. One verifies that ak, bkare
of degree 1 in λ
. Thus we have the following rational transformation
j
2
a
= Ak(r, λ),k=1, 2, ..., n− 1,
k
2
= Bk(r, λ),k=1, 2, ..., n
b
k
(3.10)
k
of λ
< ... < λn; rk> 0 into the domain D. If we identify two proportional
1
vectors r the mapping is one to one.
In these homogeneous coordinates r
the differential equations become
k
linear
dλ
dt
k
=0;
dr
dt
k
= −λ
. (3.11)
krk
Thus the solutions of (2.2) can be represented as rational functions of the n
constants λ
and n exponential functions e
j
−λjt
.
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