- REDUCED VOLUME AT THE SINGULAR TIME FOR TYPE I SOLUTIONS 147
Therefore, applying this and the Type A assumption to (30.50), we obtain
(30.51)
IK (r)I
< i w(w _ i) ~ ( Cn,c,M + Cn,c,M 1 1' (i) I + nM Ir' (i) 1
2
+ n
2
M )di
- t (w-i)^2 r (w-i)~r (w-ir (w-ir+^1
i
w ( ' ) ~ ( 2Cn,c,M (nM + Cn,c,M) 1 1' (i) 1
2
n^2 M ) d '
< W - t 2 + , r + +1 t
- t (w-i) r (w-t) (w-ir
< i w , ~ (nM + Cn,c,M) ( R + 1 1' (i) i2) ,
_ ( w _ t ) ( , r dt
t w - t
+iw (w-i)~ (2Cn,c,M+(nM+Cn,c,M)n
2
M + n
2
M )di
t (w-i)2r (w-ir+1
~ (nM + Cn,c,M) i w (w - i) ~-r ( R +Ir' (i) i2) di
2CncM+(nM+CncM)n^2 M( ) 2-2r n^2 M( )2-r
+ " 5 " w-t2 +-3-w-t2'
2 - 2r 2 - r
provided r < ~.
Now we consider the specia l case where r = 1 (i.e., we have a Type I singular
solution) since this gives us a clear way to estimate the penultimate line of (30.51).
In this case we have
(30.52)
1
+ 2 (nM + Cn,c,M) (w - t)^2 R. (z, t).
STEP 2. Bounding IV'R.I. Again using r = 1, by (30.49) and (30.52), we have
(30.53) IV'R.^12 (z, t) ~ IRI (z, t) + IK (r)I + R. (z, t)
(w - t)3/2 w - t
< A+ BR. (z, t)
- w-t
on M x [a+ c,w), where
(30.54)
(30.55)
A~ 4Cn,c,M + 2 (nM + Cn,c,M + ~) n^2 M,
B ~ 2 ( nM + Cn,c ,M) + l.
That is, we have the key inequality:
(30.56)
on M x [a+c,w).
l\7£1 (z, t) ~ J A+ BR. (z, t)
w-t