10 HENRY C. WENTE
where a is a constant. Furthermore
(2.19) 4oo = -(4A + a )e - (4B + ft )e - 4a'e + 4/?'e +6^
where 6y = 6a/? - 4a. The system (2.18) is an algebraic completely
integrable Hamiltonian system with
Sf(a» /?»Pa»P/?) = PaP^ - V(*fft) = h
y {ait ft)
=
aa/?
- a
ft2
- A/9 -
Ba2.
Conversely, let (a(u),/?(u)} be a solution to(2.18) with the
property that the right side of (2.19) is positive for u = u and
o
some o o = o o . One may use (2.19) to solve for oo(u ,v) and then
o o
(2.17) to find oo(u,v) .
Proof: Let oo(u,v) be a solution to (2.16) with o o 2 0 and
v
such that there are functions {a(u), /?(u)} satisfying (2.17). We
first derive (2.19). Compute o o by differentiating (2.17) and
uu
substitute this into (2.16) to obtain an expression for 4oo
vv
Multiply this by 2oo and integrate in v to obtain (2.19) where
v
y(u) is some function of u. Now compute 8oo o o in two ways, once
v uv
by differentiating (2.19) with respect to u and also by
differentiating (2.17) with respect to v and multiplying the
result by 4oo . We obtain a polynomial expression in X = e with
coefficients being functions of u which is equal to zero. This
gives us the system (2.18) and the condition that 6^(u) = 6a/? - 4a.
The converse is just a reversal of the procedure. Q.E.D.
The solutions oo(u,v) obtained in this manner have some
specific properties which we now list.
oo
1) By setting X = e we may rewrite (2.19) in the form
(2.20) 4X
2
= p(u,X)
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