1547845439-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_I__Chow_

4. CONSTRUCTING THE BRYANT STEADY SOLITON 21

the differential equation for y in (1.48) shows that there exists to such that

y :::; 0 for t 2:: to. Therefore,

dx 2

->x dt - +n-l

and limt/T x = +oo for some finite T >to.

PROPOSITION 1.32. The metric associated to this trajectory is incom-

plete.

PROOF. We will show that there exists a constant a E (0, 1) such that

(1.51)

a

x<--- T-t

for t sufficiently close to T. From this inequality, it follows· by integration

that w :::; C.(T-t)-a for some constant C and hence that r = J w dt is finite

as t /' T.

To establish the upper bound on x, suppose we knew that there is a

positive constant 5 and a t 1 E ( t, T) such that

(1.52) y + 5x:::; 0 for all t 2:: ti.

Then dx/dt 2:: (1+5)x^2 + n - 1, and this implies that

-arctan d ( x ) > f{!--1 --

dt .J a ( n - 1) - a '

a~ 1/(1+5).

Integrating from t to T on both sides and solving for x give the inequality

(1.51).

To establish (1.52), note that

d

dt (y + 5x) = x(y - nx) + 5(x^2 - xy + n -1).

When y + 5x = 0, the right-hand side equals (5^2 - n)x^2 + (n - 1)5. Let x 0

be the x-intercept of the trajectory, and let 5o > 0 be small enough so that

(1.53) (55 - n)x^2 + (n - 1)5o < 0 for all x > xo.

Let (x1, Y1) be any point on the trajectory such that x1 > xo and Yl < 0,
and choose 5 :::; 5o sufficiently small so that Y1+5x1 :::; 0. Then (1.53) shows
that this inequality persists for all later values oft. D

4.3. Phase plane analysis of the left-hand trajectory. We now

turn to the analysis of the other trajectory emerging from saddle point (1, n).
By the same reasoning as before, we see that x is decreasing, y is increasing,

and y - nx is positive along this trajectory. Moreover,

x -+ O+ and y -+ +oo
as t increases. In order to prove that the corresponding metric is complete,
we need to know the limiting behavior of wand r. To do this, we introduce