260 9. TYPE I SINGULARITIES
PROOF. By (6.32), we calculate
v^2 dt d D (M) = -v dt d (tr M) + (tr M) dt d v - v dv dt
d d
= -v dt (>. + μ + v) + (>. + μ) dt v
= -1/3 - I/ (>.2 + μ2 + >.μ) + (>. + μ) >.μ.
Thus it suffices to prove that
7f ~ -1/ (>.2 + μ2 + >.μ) + (>. + μ) >.μ 2: 0
whenever v < 0. There are two cases: ifμ< 0, then we write
7f = (μ v) (>.2 + μ2 + >.μ) μ3 2: -μ3 > O,
while if μ 2: 0, we have
7f = >.^2 (μ - v) - vμ^2 + >.μ (μ-v) 2: 0.
0
PROOF OF THEOREM 9.4. Fix any x E M^3 with v (x, 0) < 0. Let M be
the quadratic form corresponding to Rm [g], and let
/Cc (A^2 TM^3 &Jsf\^2 TM^3 )x
be the time-dependent set defined in Lemma 9.5. Then since
JC (O) = {JP I tr JP 2: -3 and }
trJP2: lv(JP)I (logjv(JP)j-3) if v(JP)::; -1 '
we have M (0) E JC by the hypothesis inf v (-, 0) 2: -1. We claim M (t)
remains in JC as long as v < 0. The theorem follows directly from the claim,
which implies that
=-R -log ( -v) 2: -i+3t (1)
I/ l+t^1 - - log -- 1 + t = log (^1 + t) -^3
when 0 < -v < 1/ (1 + t), and
R
- 2: log (- v) +log (1 + t) - 3
- 1/
when - v 2: 1/ (1 + t).
The proof that M (t) E JC uses the maximum principle for systems. The
inequality tr M 2: -3/ (1 + t) is preserved, because
d 1 1
dt (trM) - 3 (trM)2 2: 3 (>.2 + μ2 + v2) 2: 0.
- 1/
And if lvl 2: 1/ (1 + t), then Lemma 9.6 implies that ftn (M) 2: 1/ (1 + t),
which is equivalent to the inequality
:t [ tlv7 - log lvl - log (1 + t)] 2: 0.
0