1547845439-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_I__Chow_

22 1. RICCI SOLITONS

new phase plane coordinates

x

X~y'n-,

y

y ~ y'n(n - 1).

y

As x -----t O+ and y -----t +oo,

X -----t O+ and Y -----t O+.

(The constants are chosen so that the invariant hyperbola becomes the unit

circle in the XY-plane.) With a new independent variable s such that

ds ~ y dt, we have

(1.54) --dX -X3 - X +a y2

ds ' ~ =Y(X

(^2) -aX),
where a ~ 1/ y'n. In these coordinates, the first integral corresponds to a
Lyapunov function L ~ X^2 + Y^2 which satisfies dL/ds = 2X^2 (L-l). Since
L < 1 for this trajectory, L is strictly decreasing but dL/ds 2: 2L(L - 1).
So, the trajectory approaches the origin exponentially as s -----t +oo. Since
(1.55)

1 dY

d(logw) = xdt = y'ii,X _ 1 Y'

it suffices to find the limiting behavior of X in terms of Y. Note that for X

and Y small, (1.54) implies

dX 2

-R:!-X+aY

ds '

dY

ds R:! -aXY.

In particular, we expect -X + aY^2 -----t 0 as s -----t +oo. Indeed, we have the
following.

LEMMA 1.33.

and

where a = 1/ y'n.

. x

s->+oo hm y^2 =a

. X-aY^2 3
s->+oo hm y^4 = 2a,

PROOF. One might try to apply l'Hopital's Rule to find the limit of
X/Y^2 • However,

(1.56) 2..ik____ .!l. x - -x ( x

(^2) - 1 ) 1
.!l.y2 ds - y2 x2 - aX + ynX2 -X
shows that this is not straightforward, since the limit of X/Y^2 is also in-
volved in the right-hand side. Let g(s) = X and h(s) = Y^2 and apply the
Cauchy mean value formula (see, e.g., [10]):
g^1 (c)(h(a) - h(b)) = h'(c)(g(a) - g(b))