20 1. RJCCI SOLITONS
saddle points since the linearization of (1.48) at (1, n) is^8
du
dt = - (n - 2) u - v,
dv
- = -nu+v
dt '
and the negative of this at (-1, -n).^9 Since we want the metric to close up
smoothly as r--+ 0, by Lemma A.2, we will need
x --+ 1 and y --+ n.
Therefore, we will limit our attention to the two trajectories that emerge
from the saddle point (1, n) as t increases. (These, plus the point itself,
comprise the unstable manifold of the saddle point; the stable manifold
is the hyperbola.) Given either of these trajectories, we can reconstruct
functions w, r and f oft by successively integrating
dw
(1.50) - = x dt dr = w dt, df = (nx - y)dt.
w '
Furthermore, we can choose constants of integration such that w \,, 0, r \,, 0,
and df /dr --+ 0 as t --+ -oo. It then follows that f(r) and w(r) smoothly
extend to even and odd functions of r, respectively (cf. [217], Proposition
5.2).
4.2. Phase plane analysis of the right-hand trajectory. The lin-
earization of (1.48) at the saddle point shows that the right-hand trajec-
tory emerges from the saddle point into the region where dx / dt > 0 and
dy / dt < 0. Because trajectories along the boundary of this region only
point into the region, x continues to increase and y continues to decrease.
Linearization also shows that y-nx is initially negative along the trajectory,
and the equation
d dx
-(y-nx) = x(y-nx) - n-
dt dt
shows that y - nx is negative and monotone decreasing. Because x 2: 1
and y - nx < -C for some positive constant C when t is sufficiently large,
(^8) The linearization of the general system
dx
dt =f(x,y),
dy
dt = g(x,y)
at (x,y) is
dx af at
dt =ax (x,y)x+ ay (x,y)y,
djj ag - ag -
dt = ax (x, y) x + ay (x, y) y.
(^9) The eigenvalues of the matrix
( 2~nn ~1)
are 2 and 1-n, with corresponding eigenvectors (-1,n) and (1, 1), respectively.