1316 
Ê 
GES 
8 oD (S—1) = — 3 [Bo —dfp. 
dp 
0 
5 dh . Mi 
Since PT is supposed never to be positive, the integral also cannot 
; 
be positive, and we conclude 
age 
Similarly we find 
$2 6. 
Now differentiating (9) again, we find 
: dy “ s ; 
B 4759 dB 10; 
dp 
10 
| aes (0) 
A 7E L 50 + 4 — (1 + 0) =0. 
di 
For 8=0 we have 1j=—= 60 =0. For small values of @, 4 and 
dy ; 8 3 : 
on therefore necessarily of the same sign. It follows from (10) 
Bil 
ee . a dy 
that this is only possible when 4 is positive ; 7 and EE thus begin 
dj 
by being both positive, and * cannot become negative without 
passing through zero. But, for values of 2 larger than zero, we find 
Aga: jee om 
from (10) that, for 7 = 0, a is positive. It follows that 1 can never 
dj 
become negative. The same reasoning holds for 6. Collecting 
the different inequalities, which have been found we can write 
Cis GS 3; 
= | Ee 
0<A<5<3. 
From (10) we find 
d A 
BPF CO) ee ier ar) 
eh 
Putting now 
=e 
oa 
we find 
dy v 
B dp aa [6 + 1 + fed —— 26] y == 0 e ° e . . (12) 
The factor in square brackets is necessarily positive, and is equal 
to-6 for B=0O. Putting thus 
