226
III,§3,3-III,§3,4
1°) If problem (2.31)25 has a solution g, then, for all r _< 2s,
Q3i94, R) = (-Q)3'g.(-Q, R)
is the value of a junction of R that is holomorphic in w. In particular, 92,
is a function of R that is meromorphic in w, with poles R1 and R2.
2°) Problem (2.31)25+1 is solvable [and thus problem (2.31)25+2 also]
if and only if the primitive of (Gg2i)dR is defined (that is, uniform) in
w\[R1, R2].
Proof.
Suppose 1°) is true. [1°) is evident for s = 0]. Then, by (3.14),
the function g25+1: I'[L0, Mo]\t - C is obviously the restriction of the
primitive of (Gg2s)dR to the edges of the cut [R1, R2] in C, which is
defined on the universal covering space of w\[R1, R2]. Obviously, 925+1
is defined on I'[L0, Mo]\E if and only if this primitive is defined on
w\[R1, R2)
Assume that this condition is satisfied.
With a convenient choice of the constant of integration, this primitive
925+1 is an odd meromorphic function of Q in a neighborhood of R1
and R2. Like Q, it takes opposite values on the edges of the cut [RI, R2].
For R near R1 or R2i Q is near 0 and g25 is an even function of Q having
a pole of order 6s at Q = 0. From (3.3), (3.5), (3.6), and (3.14),
d 1 = G925 =
2-93-
n 925
la
-
4d
RKR d 2s
,
where KR
0.
dQ dQ
Q[ Q
dQ
Thus g25+1 is locally an odd function of Q having a pole of order 3(2s + 1)
at Q = 0.
Consequently Q3c2s+ng25+1 is holomorphic in w.
g25+2 is defined by (3.11). Then
Q3(2,+2
'92,+2 is holomorphic in a).
Consequently 1°) holds when s is replaced by s + 1.
4. The SchrSdinger-Klein-Gordon Case
In order to establish, for any r, the solvability of problem (2.1) which
we have reduced to problem (2.31) an appropriate assumption is clearly
necessary.
Assume that K is a second-degree polynomial:
K[R,M] = -R2A(M) + 2RB(M) - C(M).