74 29. COMPACT 2-DIMENSIONAL ANCIENT SOLUTIONS
Let 9KR ~ (vKR)-^1 gcyl ~ (vKR)-^1 9euc· The ansatz (29.8) is equivalent to(29.10) f)KR (s, B, t) = 2a (t) cosh^2 s + /3 (t)
in cylindrical coordinates on S^1 x IR and is also equivalent to
(29.11) vKR (r, B, t) = a ~t) +(a (t) + /3 (t)) r^2 + a ~t) r^4
in polar coordinates on IR^2.
Substituting (29.8) and the scalar curvature formula
2asin^2 'I/;
(29.12) RKR = 4a + 4/3
2
f3 2 'lj;
a+ cos
into a~;R = RKRVKR, we see that vKR is a solution to the Ricci flow if and only if
(29.13) a'= 4a (a+ /3) and /3' = -4a/3,
where ' denotes a time derivative. Let
()
_,_a(t)
a t --:-
2
, b(t)~/3(t)+a(t),so that
(29.14) vKR(r, B,t) = a(t) +b(t)r^2 +a(t)r^4.
Then (29.13) is equivalent to
(29.15) a' = 4ab and b' = 16a^2.Since a and /3 are positive, we have that a > 0 and b > 2a. Hence a' > 8a^2 , which
implies that there exists a time t at which both a(t) and b(t) tend to infinity; by
translating time we may assume that this time is 0.
We compute that (b^2 - 4a^2 )' = 0. Define μ > 0 by b(t)^2 - 4a(t)^2 = μ^2.
Hence b' = 4(b^2 - μ^2 ). The general solution to this ODE is b(t) = μ~:g::~:. Since
limvo b(t) = oo, we have C = 1 and hence
(29.16) b(t) = -μcoth(4μt), a(t) = -~csch(4μt), whereμ> O;
that is,
(29.17) a(t) = -μcsch(4μt), /3(t) = -μtanh(2μt).
In conclusion, we haveLEMMA 29.3 (Characterization of the King- Rosenau solution). Let (S^2 , 9KR(t)),
t E (-oo, 0), be an ancient solution to th e Ricci flow of the form (29.8), where
a, /3 > 0 and limt/'O a(t) = oo. Then 9KR(t) is the King-Rosenau solution, where
for some μ > 0 we have that
9KR(t) = -~ (2 csch( 4μt) + tanh(2μt) COS^2 'lj;f1 g52
μ
1 ( 1 1 )-l
= -μ 2 csch(4μt) + coth(4μt)r^2 + '2 csch(4μt)r^4 geuc= -~ (csch(4μt) cosh(2s) + coth(4μt))-
1
9cyl·
μ