15. The quantity Q must be identically zero
and
(29.199)
8
at l~eucvil
2
= V~euc l~eucvil
2
- 2v l~eucval
2- 4~eucV l~eucvil
2
- 4~eucV l~eucvil
+ 2vi~eucVi~~uc V - 8~eucViV£jiVej.
From (29.198), (29.199), and dropping the vanishing terms
~eucv(lvijk/
2
- l~eucvi) - 2V£iVijkV£jk + 2VjkVijk~eucVi = 0,
we compute using (29.175) that
(29.200)
8
at laijk 1
2
= V~euc laijk 1
2
- 4R laijk /
2
2
- ::(l8e (vi.iijk)l
2- l"Vv l
2
li.iijkl
2- 3vi.iejkVe~eucVjk)·
v
- 3vi.iejkVe~eucVjk)·
- l"Vv l
To analyze the RHS of (29.200), let Zjk = ~eucVjk - ~~;ucVbjk· Since
I--^1
(^2 1) l"-1 (^2) 1- 12
veaejk = - v v aijk
2
125
(which is analogous to (29.161)) and since 15ei8e (vaijk) = vei.iejk + ~Zjk, by (29.157)
we have
/Be (vaijk)l^2 - I TF(8 (va))eijkl^2 = l8i (vaijk)l^2
1 -2
= 2 l"Vvl
2
laijkl
2
+ viaijkVZjk + v
4
lzjkl^2.
Applying
2 l"Vvl
2
laijkl
2
+ vvt /aijkl; = 2veaijk TF(a(va))eijk + 2 lvtiitjkl
2
+ i!ti.itjkVZjk
to (29.200), we calculate t hat
BQ - - -
Dt = V~eucQ - 4RQ
- 2 (lae (vaijk)l
2
+ ~ l"Vv l
2
li.iijkl
2
- 3vei.iejkV~eucVjk + i!i!t laijkl;)
= v~eucCJ - 4RQ - 2(1 TF(a(va))eijk + i!ti.iijkl^2 - li!tiitjkl
2
)
- 2
- 21 ~zjk - vei.iejk I
Again applying (29.157) yields (29.196). This completes the proof of Lemma 29.48.
- The quantity Q must be identically zero
Let Q = v /TF(B)l^2 , as defined in (29.134). In this section we shall prove the
following. Since we have shown in Proposition 29.46 that Q being zero implies that
g(t) is either a shrinking round 2-sphere or the King- Rosenau solution, this will
complete the proof of the main Theorem 29.l.
PROPOSITION 29.49. Ifμ> 0 in Proposition 29.36, then Qmax(t) ~ maxx Q(x, t)
is nonincreasing and satisfies
(29.201) lim Qmax(t) = 0.
t-+-oo
Hence Q(x, t) = 0 on S^2 x (-oo, 0).