- 2-DIMENSIONAL ANCIENT SOLUTIONS WITH FINITE WIDTH 59
We conclude that we have the lower bound
u(s, B, t) ~ e-~W-(O,t) on [1, oo) x 51 x (-oo, 0). DNext, we turn to the upper bound for u. First, we have the following.LEMMA 28.49 (A bound for the circular averages of (lnu)+)- For each t E(-oo, 0) there exists C(t) < oo such that the circular averages of the positive part
of ln u satisfy(28.60) ( (lnu)+(s,B,t)dB:::; C(t) for alls E [O, oo).}5 1
PROOF. Fix t E (-oo, 0). DefineW(s, t) ~ ( ln u(s, e, t)dB,} 51
W+(s, t) ~ ( (lnu)+(s, B,t)dB,}5 1
so that W = W+ - W, where W is defined by (28.57). By (28.59) we h ave
that W ( s, t) ~ -C for s E [O, oo). Since .6. ln u < 0 and smooth, we have that
W 55 (s, t) < O; i.e., s H W 5 (s, t) is strictly decreasing. Hence
μ(t) ~ lim W 5 (s, t) E [O, oo)
s-+oo
exists (μ cannot be negative by the lower bound for W).Claim. μ(t) = 0 for each t < 0.
Proof of the claim. Suppose that μ(t) > 0 for some t. Since W 5 (s , t) > μ , we
then have that lims-+oo W(s, t) = oo. Hence, given any k EN, there exists Sk < oo
such that W (s, t) ~ k for alls~ Sk· We shall now derive a contradiction.
Denote 7r : [1, oo) x IR ~ [1, oo) x 51 to be the universal cover with the lifted
Euclidean metric on t he total space and define u to be the lift of u; i.e., u = u o 7r.
Let (s, B) E [sk + 7r, oo) x 51 and choose e E IR such that 7r(s, B) = (s , B). By
.6. ln u < 0 and the mean value inequality, we have
ln u(s, e , t) - = ln u(s, e, t) - ~ 23 1 1 ln u(s, e , t)dsde.
7r B(s,ii) ( ,/27r)Observe t hat (s - 7r, s + 7r) x 51 c 7r(B(s,o/-/27r)). Since lnu is bounded from
below by (28.56), there exists C < oo independent of k such that
( lnu(s,B,t)dsdB~ ( l s+"Trlnu(s, B,t)dsdB-C
JB(i<,iJ)(,/27r) ls^1 s-"Tr
= 1~:7r W(s, t)ds - C
~ 27rk - c.
Therefore ln u( s, B, t) ~ 'Tr\ - 2 ~ 3 for any ( s, {j) E [ Sk + 7r , oo) x 51. Finally, because
k is arbitrary, this contradicts the finite-width condition in the same way as in the
proof of Lemma 28.43. D