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)