EXPANDING IMMERSED CONVEX PLANE CURVES
∗Tai-Chia Lin, Chi-Cheung Poon, and Dong-Ho Tsai January 12, 2007
Abstract
We study the evolution driven by curvature of a given immersed convex closed plane curve. We show that it will converge to a self-similar solution eventually. This self-similar solution may or may not contain singularities. In case it does, we also have estimate on the curvature blow-up rate.
1
Introduction.
Let γ0 be an immersed convex closed plane curve with rotation index (number of times its tangent
vector winds around as one goes along the curve) m ∈ N. We can parametrize it by a smooth immersion X0 := X(·, 0) : S1 → R2, where S1 is a reference manifold. If we expand γ0 according
to the equation
∂X ∂t =
1
αkα · N, α > 1 (1)
where α > 1 is a constant, k(·, t) is the curvature of the curve γt, and N (·, t) is the outward unit
normal vector field to γt. It is well-known that under the flow (1), the function H = 1/k, which is
the radius of curvature of γt, will satisfy the equation with periodic boundary condition
αHt = (Hα)xx(x, t) + Hα(x, t) , H(x, t) = H(x + 2mπ, t) (2)
for all x ∈ R, t ∈ [0, T ). Here x ∈ R is the variable representing the outward normal angle of N, i.e., N (x, t) = (cos x, sin x).
For more information on the expansion and contraction of convex closed curves in R2, see [AN1],
[ANG2], [CT], [CZ], [GH], [T1], [U1], ... etc. In particular, the asymptotic behavior of the flow (1) for the case 0 < α ≤ 1 had been studied in a nice paper by Urbas [U1]. His result is that the flow will converge to m-fold cover of the unit circle S1 eventually.
If we let ¯H = Hα, then ¯ Ht = ¯H α−1 α ( ¯Hxx+ ¯H), H(x, t) = ¯¯ H(x + 2mπ, t), α − 1 α ∈ (0, 1) . (3)
Hence we consider the more general equation with periodic boundary condition ∂v
∂t = v
p(v
xx+ v) , v(x, t) = v(x + 2Lπ, t), p ∈ (0, 1) (4)
for all x ∈ R, t > 0. Here 0 < p < 1 and L > 0 are two fixed constants. We assume that the initial data v0(x) = v (x, 0) is an arbitrary smooth positive periodic function on R with period 2Lπ. In
case L ∈ N, v01−p(x) may not, in general, represent the radius of curvature of some immersed
convex closed curve.
Remark 1 The behavior of solutions v (x, t) to equation (4) for p ∈ (−∞, 0], p ∈ (0, 1) , p = 1, p ∈ (1, 2) , p = 2, and p > 2 are all quite different. This means that for p ∈ (−∞, ∞) , equation (4) has at least three critical values: p = 0, p = 1, p = 2. In this paper, we focus on the case p ∈ (0, 1) . Results in [U1] correspond to the case p ∈ (−∞, 0] with L ∈ N and v01−p(x) representing
the radius of curvature. Results in [T3] correspond to the case p = 1. The case p ∈ (1, 2) will be discussed in [PT].
It is well-known that the smooth solution v (x, t) to equation (4) exists until its space supremum blows up. Suppose that vmax(t) := supx∈Rv(x, t) → ∞ as t → Tmax. Then v (x, t) is defined on
R× [0, Tmax). If we let R (t) denote the unique solution to the ode dR
dt = R
p+1(t) , R (T
max) = ∞ (5)
then R(t) = [p (Tmax− t)]−1/pand we have the following comparison between R(t), vmin(t) and vmax(t) :
0 < vmin(t) ≤
1 p (Tmax− t)
1p
≤ vmax(t) for all t ∈ [0, Tmax). (6)
If we do the rescaling u (x, t) := v (x, t) /R (t) and let τ ∈ [0, ∞) be the new time defined by the relation t = Tmax(1 − e−pτ) , t ∈ [0, Tmax), which is motivated by the requirement dτ /dt =
Rp(t) , then the function
u(x, τ ) = p1/pTmax1/pe−τ · v x, T
max 1 − e−pτ , τ ∈ [0, ∞) (7)
will be a positive, bounded solution (see Proposition 8 below) of the rescaled equation ∂u ∂τ = u p u xx+ u − u1−p , 0 < p < 1 u(x, τ ) = u(x + 2Lπ, τ ) (8)
for all (x, τ ) ∈ R × [0, ∞), with the positive smooth initial condition u (x, 0) := u0(x) = p1/pTmax1/p ·
v0(x) . Moreover, we have
umax(τ ) ≥ 1 and umin(τ ) ≤ 1 (9)
for all τ ∈ [0, ∞) due to (6).
We also have the following estimate on Tmax. Let G (θ) = θ
1
1−p, 0 < p < 1, which is a convex
function on θ ∈ (0, ∞) and let s (t) = 1
2Lπ Z Lπ
−Lπ
z (x, t) dx, where z = v1−p, t ∈ [0, Tmax).
Jensen inequality implies ds dt = (1 − p) · 1 2Lπ Z Lπ −Lπ G (z (x, t)) dx ≥ (1 − p) G 1 2Lπ Z Lπ −Lπ z (x, t) dx = (1 − p) G (s) and by Proposition 7 below, we know that s (Tmax) = ∞. Since we also have
db
dt = (1 − p) G (b) , b (t) := R1−p(t) , b (Tmax) = ∞ comparison gives
s (t) ≤ R1−p(t) ≤ vmax1−p(t) , t ∈ [0, Tmax). (10)
In particular we have the estimate on the blow up time Tmax:
G (s (0)) = 1 2Lπ Z Lπ −Lπ v01−p(x) dx 1 1−p ≤ R (0) = 1 pTmax 1p ≤ vmax(0) . (11)
¿From now on, we always assume p ∈ (0, 1) and focus mainly on equation (8) in this paper. For notational convenience, we rewrite τ as t again. The main theorem is
Theorem 2 Let 0 < p < 1 and u0(x) > 0 be a smooth periodic function on R with period 2Lπ and
assume that u(x, t) is a positive bounded solution of (
ut= up uxx+ u − u1−p
u(x, t) = u(x + 2Lπ, t), u (x, 0) = u0(x) > 0
(12)
on R × [0, ∞). Then there is a function 0 ≤ w (x) ∈ C2(R) which is a non-negative entire solution of the ode
(
wxx(x) + w (x) − w1−p(x) = 0, x ∈ R
w(x) = w(x + 2Lπ), x ∈ R
(13) so that u (x, t) converges uniformly to w(x) in [−Lπ, Lπ] as t → ∞.
In the following discussions, the positive bounded solution u (x, t) of (12) may or may not come from the rescaling of v (x, t) . In the second case, if maxx∈[−Lπ,Lπ]u0(x) < 1, then u (x, t) converges
uniformly to w(x) ≡ 0 as t → ∞, and if minx∈[−Lπ,Lπ]u0(x) > 1, u (x, t) will blow up in finite
time. Hence a significant initial data must satisfy max
x∈[−Lπ,Lπ]u0(x) ≥ 1 and x∈[−Lπ,Lπ]min u0(x) ≤ 1. (14)
Another observation is: if u (x, t) stays positive and bounded, then the energy functional E (u (·, t)) = Z Lπ −Lπ u2x(x, t) − u2(x, t) + 2 2 − pu 2−p(x, t) dx, t ∈ [0, ∞) (15)
where u2x means (∂u/∂x)2, will be shown to be non-increasing in t ∈ [0, ∞) and it converges to a non-negative number (see Corollary 19). Hence for u (x, t) to stay positive and bounded, u0(x)
must satisfy E (u0) ≥ 0. However, E (u0) ≥ 0 is not sufficient. For example, one may take u0(x) to
be any constant λ ∈ (1, γ) , where γ = [2/ (2 − p)]1/p.
Remark 3 (private communication with Ben Andrews) According to Ben Andrews’s result in a preprint, the convergence of u(x, t) to w (x) as t → ∞ is more than uniform. It converges in higher regularity. This is established via nontrivial higher derivatives estimate of u.
Remark 4 If u0(x) > 0 on [−Lπ, Lπ] and the solution u (x, t) stays bounded (as long as it exists),
it is impossible for u (x, t) to attain 0 in finite time. This is clear by looking at the behavior of umin(t) .
Part of our arguments resembles the methods used by Matano, see [CM], [M]. In [M], Matano studied the semilinear parabolic equation of the form
ut = uxx+ f (u, t)
with x ∈ [0, L] and t > 0. He proved that any bounded solution converges for t → ∞ to a single solution of the corresponding stationary problem. We will adapt some of his methods to the present situation. Similar technique is also used by Feireisl and Simondon, [FS]. They considered the equation
ut = φ(u)xx+ f (u)
with Dirichlet boundary condition. Under certain conditions on φ and f , Feireisl and Simondon proved that any bounded solution converges for t → ∞ to a single solution of the corresponding stationary problem.
2
Preparations.
We first look at v (x, t) , where t ∈ [0, Tmax).
Proposition 5 Let v(x, t) be a solution of (4) with positive initial data and periodic boundary condition and let
A = max
x∈R v
2(x, 0) + v2 x(x, 0)
12 . (16)
Then at any point (x, t) ∈ R × [0, Tmax), we have either
(vxx+ v) (x, t) > 0 (17)
or
v2x(x, t) + v2(x, t) ≤ A2. (18)
In particular, if v(x, t) > A at some (x, t), then we must have vt(x, t) > 0.
Proposition 6 For any (x, t) ∈ R × [0, Tmax), one has
|vx(x, t)| ≤ A + 2Lπ · vmax(t) (19)
where A is the constant in (16).
The proofs of Propositions 5 and 6 are identical to the proofs in [ANG2], Lemma 4.1 and 4.3. We do not repeat them here. We also observe the following
Proposition 7 There exists a constant C depending only on the initial data such that Z Lπ −Lπ vx2(x, t)dx ≤ Z Lπ −Lπ v2(x, t) dx + C (20)
for all t ∈ [0, Tmax). Moreover, if vmax(t) = v (pt, t) , pt ∈ [−Lπ, Lπ] , then for any small ε > 0, there
exists a number δ > 0, depending only on ε, such that (1 − ε) vmax(t) ≤ v (x, t) +
√
2LπC (21)
for all x ∈ pt− δ2, pt+ δ2 and all t ∈ [0, Tmax). In particular, the blow-up set for the solution
v (x, t) must contain at least some open interval I ⊂ [−Lπ, Lπ] . Proof. (20) is clear due to the following
∂ ∂t Z Lπ −Lπ vx2− v2 dx = − Z Lπ −Lπ 2vp(vxx+ v)2dx ≤ 0.
For any fixed small ε > 0, we have by (20) vmax(t) = v (x, t) + Z pt x vx(x, t) dx ≤ v (x, t) + |pt− x|1/2 Z Lπ −Lπ v2(x, t) dx + C 1/2 ≤ v (x, t) + |pt− x|1/22Lπvmax2 (t) + C 1/2 ≤ |pt− x|1/2 √ 2Lπvmax(t) + v (x, t) + √ 2LπC. Now if |pt− x|1/2≤ δ := ε/ √
2Lπ, then the proposition follows.
Proposition 8 If v(x, t) blows up at time Tmax, then there is some constant C > 0, independent
of time, such that
vmax(t) ≤ C|Tmax− t|−1/p (22)
for all t ∈ [0, Tmax). That is, in terms of Hamilton’s terminology, there is no type 2 blow-up.
Proof. For each 0 < t < Tmax, let
I(t) = Z Lπ
−Lπ
v2−p(x, t)dx. (23)
We note that by (21), when t is large enough, there is a constant C > 0 so that
vmax(t) ≤ CI(t)1/(2−p). (24)
Also, by straightforward computations, we have I0(t) = (2 − p) Z Lπ −Lπ v1−pvtdx = (2 − p) Z Lπ −Lπ (v2− v2x)dx (25) and I00(t) = 2(2 − p) Z Lπ −Lπ vp(vxx+ v)2dx ≥ 0. (26)
We claim that there is a t1 > 0 so that I0(t) > 0 for all t ∈ [t1, Tmax). In fact, if I0(t) ≤ 0 for
all t, then I(t) is an decreasing function. This implies that I(t) ≤ I(0), and from (24), vmax(t) is
bounded. It contradicts our assumption that v(x, t) blows up at time Tmax. Thus, there is a t1 > 0
so that I0(t1) > 0 and by (26) we know that I0(t) > 0 on [t1, Tmax).
By H¨older inequality Z Lπ −Lπ v (vxx+ v) dx 2 ≤ Z Lπ −Lπ v2−pdx Z Lπ −Lπ vp(vxx+ v)2dx
when t > t1, from (23), (25) and (26), we get
I00(t) 2I0(t) − I0(t) (2 − p) I (t) ≥ 0, t ∈ [t1, Tmax). It follows that I0(t) I0(t1) ≥ I (t) I (t1) 2/(2−p) , t ∈ [t1, Tmax)
and there is a constant C1 > 0 so that
I(t)−2/(2−p)I0(t) ≥ C
1, t ∈ [t1, Tmax). (27)
We may integrate (27) from t to Tmax and obtain
I(t) ≤ C2(Tmax− t)−(2−p)/p, t ∈ [t1, Tmax).
¿From (24), we conclude that
vmax(t) ≤ C3(Tmax− t)−1/p, t ∈ [0, Tmax)
for some finite constant C3 > 0, which can be taken to be independent of time t ∈ [0, Tmax). The
proof is done.
By Propositions 6, 8, and (7), we know that both u and ux are uniformly bounded if it comes
from the rescaling of v (x, t). On the other hand, if we start with equation (12) alone, we also have the following
Lemma 10 Assume u (x, t) is a positive bounded solution of equation (12) on R × [0, ∞). Then there exists a constant C > 0, depending only on the initial data u0(x) , such that
|ux(x, t)| ≤ C (28)
for all (x, t) ∈ R × [0, ∞). Proof. Let ˜u = etu. Then
˜
ut = etu + et[up(uxx+ u) − u] = up(˜uxx+ ˜u) .
For the above equation, we have the known estimate (see [T2], p. 263) |˜ux(x, t)| ≤ max {C, ˜umax(t)}
for all (x, t) , where the constant C depends only on the initial data. (28) then follows. We now can draw the following conclusion:
Corollary 11 Let u (x, t) be a positive bounded solution of equation (12) on R × [0, ∞). For each sequence tn, tn → ∞ as n → ∞, there is a subsequence, also called tn, so that u(x, tn) converges
uniformly over R to a Lipschitz continuous function w(x) ≥ 0, which is 2Lπ-periodic over R. Proposition 12 Suppose that u(x, tn) converges uniformly to a Lipschitz function w(x) ≥ 0 for
x ∈ [−Lπ, Lπ]. For any sequence snwith 0 ≤ sn≤ 1 for all n, u(x, tn+sn) also converges uniformly
to w(x) in [−Lπ, Lπ].
Proof. ¿From (12), it is easy to see that 0 ≤ Z Lπ −Lπ u2t updx = − 1 2 d dt Z Lπ −Lπ u2x− u2+ 2 2 − pu 2−p dx, t ∈ [0, ∞) (29)
which means that the non-increasing quantity E (u (·, t)) := Z Lπ −Lπ u2x− u2+ 2 2 − pu2−p dx (30)
is a Lyapunov functional for equation (12). Thus, for any S > 0, we have Z S 0 Z Lπ −Lπ u2t updxdt = 1 2E (u (·, 0)) − 1 2E (u (·, S)) . Since both u and ux are uniformly bounded, when S → ∞, we have
0 ≤ Z ∞ 0 Z Lπ −Lπ u2t updxdt = 1 2E (u (·, 0)) − 1 2S→∞lim E (u (·, S)) < ∞. (31)
Also note that for any x ∈ [−Lπ, Lπ], we have
u(2−p)/2(x, tn+ sn) − u(2−p)/2(x, tn) = 2 − p 2 Z tn+sn tn ut(x, t) up/2(x, t)dt.
It follows from the Cauchy-Schwarz inequality, and 0 ≤ sn≤ 1, that Z Lπ −Lπ u (2−p)/2(x, t n+ sn) − u(2−p)/2(x, tn) dx ≤ 2 − p 2 Z Lπ −Lπ Z tn+sn tn ut up/2 dtdx ≤ 2 − p2 (2Lπ)12 Z tn+1 tn Z Lπ −Lπ u2t updtdx 1 2 . (32) By (31), we have Z Lπ −Lπ u (2−p)/2(x, t n+ sn) − u(2−p)/2(x, tn) dx → 0 as n → ∞. (33)
If u(x, tn+ sn) does not converge uniformly to w(x) as n → ∞, then there exists ε > 0 and a
subsequence of tn+sn, still denoted as tn+sn, such that maxx∈[−Lπ,Lπ]|u(x, tn+ sn) − w (x)| ≥ ε for
all n, and moreover, u(x, tn+sn) converges uniformly on [−Lπ, Lπ] to another Lipschitz continuous
function ˜w ≥ 0. By (33), we must have w = ˜w everywhere, which gives a contradiction. Remark 13 From the above proof we actually see that as long as sn∈ R is a bounded sequence, then
u(x, tn+ sn) converges uniformly to w(x) in [−Lπ, Lπ] as n → ∞.
Remark 14 Since we upper bound estimate 0 < up ≤ c for some constant c > 0, we have u1p ≥
1 c > 0, which implies Z ∞ 0 Z Lπ −Lπ u2tdxdt < ∞.
If in addition we also have a positive lower bound on up (i.e., up ≥ const. > 0 for all time) (this will imply, by standard parabolic theory, that all higher derivatives of u stay uniformly bounded), we then can follow similar proof as in [ANG2], p. 612, to see that ut(x, t) → 0 in C∞[−Lπ, Lπ] as
t → ∞. In particular, it will imply that u (x, t) converges in C∞[−Lπ, Lπ] to a positive function
w (x) ∈ C∞(R) , which is a 2Lπ-periodic entire solution of the ode
wxx(x) + w (x) − w1−p(x) = 0, x ∈ R. (34)
Proposition 15 Suppose that u (x, tn) converges uniformly on R to a Lipschitz function w(x) ≥ 0.
Then w(x) ∈ C2(R) and is an entire solution of the ode (13). Proof. Let φ(x) ∈ C∞
0 (R) and h(t) ∈ C0∞(0, 1). We may rewrite (12) in the form
1
1 − p u1−p
t = uxx+ u − u1−p. (35)
Hence if [a, b] is any finite interval containing the support of φ(x), by integration by parts, we would have − 1 1 − p Z 1 0 Z b a u1−p(x, tn+ t) h0(t) φ (x) dxdt = Z 1 0 Z b a uxx(x, tn+ t) + u (x, tn+ t) − u1−p(x, tn+ t) h (t) φ (x) dxdt = Z 1 0 Z b a u (x, tn+ t) φxx(x) + u (x, tn+ t) φ (x) − u1−p(x, tn+ t) φ (x) h (t) dxdt.
If we let n → ∞ in above, by the Bounded Convergence Theorem, we would have
− 1 1 − p Z 1 0 Z b a w1−p(x) h0(t) φ (x) dxdt = Z 1 0 Z b a w (x) φxx(x) + w (x) φ (x) − w1−p(x) φ (x) h (t) dxdt = 0
since h(t) ∈ C∞ 0 (0, 1). As h ∈ C0∞(0, 1) is arbitrary, we conclude Z b a w (x) φxx(x) + w (x) φ (x) − w1−p(x) φ (x) dx = 0 (36) which, as φ ∈ C∞
0 (R) is also arbitrary, implies that w (x) solves the ode (13) in weak sense over R.
By regularity theory, since w (x) ≥ 0 and is Lipschitz continuous over R, we have w (x) ∈ C2(R) at
least.
Remark 16 Since u (x, tn) may approach zero somewhere as n → ∞, in applying the bounded
convergence theorem, the fact p ∈ (0, 1) is essential. In the case p ≥ 1, the above method of proof has to be confined to compact subset of the region {x : w (x) > 0} .
Remark 17 By regularity theory, u (x, tn) actually converges in C∞(I) to w (x) as n → ∞, where
I is any compact subset of the open set
Ω = {x ∈ [−Lπ, Lπ] : w (x) > 0} .
The reason is that on I, along the sequence of time tn, equation (12) is uniformly parabolic.
As for the convergence of the gradient ux, we have
Lemma 18 Assume u(x, tn) converges uniformly to a non-negative Lipschitz continuous function
w(x) ∈ C2(R) which is an entire solution of the ode. Then there exists some sequence sn with
0 ≤ sn≤ 1 for all n, such that
lim n→∞ Z Lπ −Lπ u2x(x, tn+ sn)dx = Z Lπ −Lπ wx2(x)dx. (37)
Proof. We first show that lim n→∞ Z Lπ −Lπ ux(x, tn) wx(x) dx = Z Lπ −Lπ w2x(x) dx (38)
as long as u(x, tn) converges uniformly to w(x). To see this, by (29) we know that E (u (·, t)) is
bounded below and non-increasing in t ∈ [0, ∞). Hence limt→∞E (u (·, t)) exists. In particular,
limn→∞RLπ
−Lπu 2
x(x, tn) dx exists also.
Along the sequence tnwe have
Z Lπ −Lπ u2 x(x, tn) − w2x(x) dx = Z Lπ −Lπ ux(x, tn) [ux(x, tn) − wx(x)] dx − Z Lπ −Lπ wxx(x) [u (x, tn) − w (x)] dx (39)
where by Bounded Convergence Theorem again, the second integral in (39) converges to 0 as n → ∞. Hence (38) holds.
Multiply the equation 1
1 − p u1−p− w1−p
t = (u − w)xx+ u − w − u1−p+ w1−p
by h(t)u(x, tn+ t), h(t) ∈ C0∞(0, 1), and integrate, using integration by parts, to get
1 1 − p Z 1 0 Z Lπ −Lπ u(x, tn+ t)u1−p(x, tn+ t) − w1−p(x)th(t)dxdt = − 1 1 − p Z 1 0 Z Lπ −Lπ u(x, tn+ t)u1−p(x, tn+ t) − w1−p(x) h0(t) +ut(x, tn+ t)u1−p(x, tn+ t) − w1−p(x) h (t) dxdt.
By (31), we have Z ∞ 0 Z Lπ −Lπ u2tdxdt ≤ (sup u)p Z ∞ 0 Z Lπ −Lπ u2t updxdt < ∞.
Thus, when t → ∞, by Proposition 12, Bounded Convergence Theorem, and H¨older inequality we know that lim n→∞ Z 1 0 Z Lπ −Lπ u(x, tn+ t)u1−p(x, tn+ t) − w1−p(x) th(t)dxdt = 0.
On the other hand we have 1 1 − p Z 1 0 Z Lπ −Lπ u(x, tn+ t)u1−p(x, tn+ t) − w1−p(x) th(t)dxdt = Z 1 0 Z Lπ −Lπ u(x, tn+ t)(u − w)xx+ u − w − u1−p+ w1−p h(t)dxdt = Z 1 0 Z Lπ −Lπ −ux(x, tn+ t) [ux(x, tn+ t) − wx(x)] h (t) +u (x, tn+ t)u (x, tn+ t) − w (x) − u1−p(x, tn+ t) + w1−p(x) h (t) dxdt which implies that
lim n→∞ Z 1 0 h (t) Z Lπ −Lπ ux(x, tn+ t) [ux(x, tn+ t) − wx(x)] dx dt = 0. Since h(t) ∈ C∞
0 (0, 1) in above is arbitrary, one can find some sequence sn∈ [0, 1] such that
lim
n→∞
Z Lπ
−Lπ
ux(x, tn+ sn) [ux(x, tn+ sn) − wx(x)] dx = 0
and since u (x, tn+ sn) also converges uniformly to w (x) , by (38) we conclude
lim n→∞ Z Lπ −Lπ u2x(x, tn+ sn) dx = Z Lπ −Lπ wx2(x) dx.
The proof is done.
Corollary 19 Assume u(x, tn) converges uniformly to a non-negative function w(x) ∈ C2(R)
which is an entire solution of the ode (13). Then lim
t→∞E (u (·, t)) = E (w) ≥ 0. (40)
Proof. By Lemma 18, there exists sn with 0 ≤ sn≤ 1 so that
lim
n→∞E (u (·, tn+ sn)) = E (w) .
Since E(u(·, t)) is a non-increasing function, we must have limt→∞E(u(·, t)) = E (w) .
Finally, since 0 < p < 1, we have E (w) = Z Lπ −Lπ −w (x)wxx(x) + w (x) − w1−p(x) dx + 2 2 − p − 1 Z Lπ −Lπ w2−p(x) dx = 2 2 − p − 1 Z Lπ −Lπ w2−p(x) dx ≥ 0. (41)
The proof is done.
Corollary 20 Assume u(x, tn) converges uniformly to a non-negative function w(x) ∈ C2(R)
which is a solution of the ode (13) in R. Then for any bounded sequence sn∈ R we also have
lim n→∞ Z Lπ −Lπ u2x(x, tn+ sn)dx = Z Lπ −Lπ wx2(x)dx. (42)
In particular, we also have lim n→∞ Z Lπ −Lπ |ux(x, tn+ sn) − wx(x)|2dx = 0. (43)
3
Proof of Theorem 2.
The solutions of the equation
wxx(x) + w (x) − w1−p(x) = 0, x ∈ R, 0 < p < 1 (44)
has been studied by J. Urbas, [U2]. Let w(x; a), a > 0, be a solution of the initial value problem (here a represents the maximum value of w)
( w xx(x) + w (x) − w1−p(x) = 0 w(0; a) = a and wx(0; a) = 0. (45) Let γ = 2 2 − p 1/p > 1. (46)
If γ < a < ∞, there is a number R(a) so that, w(x; a) is a strictly decreasing function for x ∈ (0, R (a)) and w(R (a) ; a) = 0 and wx(R (a) ; a) < 0. The function R(a) is given by the improper
integral R(a) = Z a 0 ds pF (a) − F (s), where F (s) = s 2− 2 2 − ps2−p. (47)
The function R(a) is strictly decreasing with respect to a, and lim
a→γ+R(a) =
π
p and a→∞lim R(a) =
π 2.
When a = γ, we have R(γ) = π/p. Thus w(π/p; γ) = 0 and wx(π/p; γ) = 0, and w(x; γ)
can be defined as a C2, non-negative function and is an entire solution of (44). This does not contradict the uniqueness theorem for solutions to the initial value problem (45) since the exponent 1 − p ∈ (0, 1) . We denote w∗(x) = w(x; γ) for x ∈ [−π/p, π/p] and w∗(x) = 0 for |x| ≥ π/p. The
explicit formula for w∗ on [−π/p, π/p] is well-known: w∗(x) = 2 2 − pcos 2p 2x 1/p , x ∈ [−π/p, π/p] . (48)
For convenience, we shall call w∗(x), with x ∈ [π/p, π/p] , a bump. Note in particular that several bumps joined together (connected by zero functions) is also a non-negative C2 entire solution of the ode.
When 1 < a < γ, w(x; a) is positive over R and is a periodic entire solution of (44) with period 2R(a), where R(a) is now given by the formula
R (a) = Z a
b
ds
Here b ∈ (0, 1) is the unique number satisfying F (a) = F (b). For 0 < x < R(a), w(x; a) is decreasing, and w (R(a); a) = b, wx(R(a); a) = 0. Moreover, w(x; a) is symmetric about any local maximum
or minimum. When a = 1, w(x; 1) = 1 for all x ∈ R. The function R(a) is continuous in a for 1 < a < γ.
The monotonicity of R (a) is not easy to derive. Using some further transformations, B. Andrews [AN2] was able to show that R(a) is strictly increasing with respect to a ∈ (1, γ). Note that (49) is an improper integral near both a and b, which is more complicated than the one-sided improper integral (47). Urbas also proved that
lim
a→γ−
R(a) = π
p and a→1lim+R(a) =
π
√p. (50)
Hence the image of R(a) for a ∈ (1, γ] is given by (π/√p, π/p].
By Proposition 15, if u(x, tn) converges uniformly to a non-negative function w(x) ≥ 0, then
w(x) = w(x; a) for some a ∈ (1, γ) , or w(x) ≡ 1, or
w(x) ≡ one or several bumps joined together. (51)
or finally w(x) ≡ 0 (if u(x, t) does not come from the rescaling of v (x, t) in (4)). Moreover, the solution of the ode w(x; a) has different period for different a ∈ (1, γ].
Remark 21 In case u (x, tn) converges to w(x) of the form (51), one can infer some interesting
observation. Since We know that the optimal regularity of the bump function w∗(x) over R is w∗(x) ∈ C2p(R) . Hence it is impossible for ∂mu
∂xm(x, tn) to be uniformly bounded (independent of
time) if m >h2pi. In case u (x, t) comes from the rescaling of v (x, t) of equation (4), then ∂∂xmmv x, ˜tn
grows faster than R(˜tn) =p Tmax− ˜tn−1/p as n → ∞, where ˜tn= Tmax 1 − e−ptn .
To go further, we need the following theorem of S. Angenent [ANG1] (see Chen & Matano [CM] also):
Theorem 22 Let u : [−Lπ, Lπ] × [0, ∞) be a non-trivial classical solution of ut = a(x, t)uxx+ b(x, t)ux+ c(x, t)u
with the periodic boundary condition. Assume that a, b, c satisfy the condition a, a−1, a
t, axx, b, bt, bx, c ∈ L∞loc([−Lπ, Lπ] × [0, ∞)) .
Let z(t) denote the number of zeros of u (·, t) in (−Lπ, Lπ], i.e., z(t) is the number of points x ∈ (−Lπ, Lπ] such that u (x, t) = 0. Then
1. For all t > 0, z(t) is finite.
2. z(t) is non-increasing in t ∈ [0, ∞).
3. If (x0, t0), t0 > 0, is a multiple zero of u, then for all t1< t0< t2, we have z(t1) > z(t2).
One can apply Angenent’s theorem to ux and get the following
Lemma 23 Assume ux is not identically zero. Let z∗(t) denote the number of zeros of ux(·, t)
in (−Lπ, Lπ], then it satisfies the conclusion of Theorem 22. Proof. By equation (12), letting Q = ux, we have
∂tQ = up· Qxx+ pup−1Q · Qx+ [(p + 1) up− 1] · Q
where the coefficients satisfy the assumption of Theorem 22. Hence the conclusion follows. We also need the following
Proposition 24 For any x0 ∈ [−Lπ, Lπ], there is a t0 > 0 so that ux(x0, t) does not change sign
for t > t0.
Proof. This is a consequence of Theorem C in p. 267 of D. H. Tsai [T2]. Its proof is also based
on Theorem 22.
Remark 25 By Proposition 24, if u (x, tn) converges uniformly to a non-negative function w (x) ∈
C2(R), and also converges in C1[a, b] to w (x) on some interval [a, b] and if wx(x) > 0 on [a, b] ,
then for fixed x0∈ [a, b] we have ux(x0, t) > 0 as t is large enough. As a consequence, if there exists
another sequence sn→ ∞ such that u (x, sn) converges in C1[a, b] to another ˜w (x) ∈ C2(R) , we
must have
˜
wx(x) ≥ 0 for all x ∈ [a, b] . (52)
Proof of Theorem 2: By Proposition 15, there is a sequence tn → ∞ so that u(x, tn)
con-verges uniformly to a function 0 ≤ w (x) ∈ C2(R) which is an entire solution of (13). Suppose that there is another sequence sn → ∞ so that u(x, sn) converges uniformly to a function ¯w (x). We
wish to prove that w (x) is identical to ¯w (x) . We first claim that
max
x∈R w (x) = maxx∈R w (x) .¯ (53)
This is based on a continuity argument. Suppose not and say α := max w < β := max ¯w. We know that α = lim n→∞ max x∈R u (x, tn) and β = lim n→∞ max x∈R u (x, sn) . (54)
Assume first that β > 1. Let τ (t) = maxx∈Ru (x, t) , which is a continuous function of t ∈ [0, ∞). For any A > 0, the image of τ (t) for t ∈ [A, ∞) must be a bounded interval IA in R. By (54), α, β are
limit points of IA, which implies that (α, β) ⊆ IA for any A > 0. In particular, for any ξ ∈ (α, β),
there must be a sequence rn → ∞ so that maxx∈Ru(x, rn) → ξ. Moreover, u(x, rn) converges
(passing to a subsequence if necessary) to some function ˜w (x) ≥ 0, which is also a C2entire solution of (13), with maxx∈Rw (x) = ξ ∈ (α, β) .˜
With the above observation, we can find at least (in fact infinitely many) two different positive entire solutions w1(x) , w2(x) of the ode (13) with different minimal periods (since their maxima
are different) such that u(x, tn) converges to w1(x) and u(x, sn) converges to w2(x) . By parabolic
regularity theory, both convergences are also in C1. This will contradict Proposition 24 as there exists some x0 ∈ R such that w01(x0) > 0 but w20 (x0) < 0 (see Remark 25 also). Hence (53) is
proved for the case β > 1.
If β = 1, then α must be α = 0. In this case, we have ¯w (x) ≡ 1, w (x) ≡ 0. By considering mini-mum instead of maximini-mum, one can also find two different positive entire solutions w3(x) , w4(x) of
the ode (13) with different minimal periods (since their minima are different) such that u(x, ˜tn)
converges to w3(x) and u(x, ˜sn) converges to w4(x) . Again we will get the same contradiction and
(53) is proved for the case β = 1 also.
In conclusion, we have α = β and maxx∈Ru(x, t) has a unique limit as t → ∞.
If α = β = 1, then clearly w (x) = ¯w (x) ≡ 1. Also if α = β = 0, then w (x) = ¯w (x) ≡ 0. If α = β ∈ (1, γ) , γ = [2/ (2 − p)]1/p > 1, then by uniqueness theorem in ode, we now must have
¯
w (x) = w (x + x0) for some x0 ∈ [−Lπ, Lπ] (55)
which means that these two solutions only differ by a translation in space. We want to claim that x0 = 0. Assuming x0 6= 0, then similar to the above continuity argument, we may find a sequence
rn→ ∞ so that u(x, rn) converges uniformly to a function ˜w(x) = w(x + ˜x0) with 0 < |˜x0| < 0.1.
Then, there is a point θ so that both w(θ) and ˜w(θ) are positive, and both wx(θ) and ˜wx(θ) are
The case α = β = γ = [2/ (2 − p)]1/p is argued differently. The uniqueness theorem of ode does not apply here and we do not have (55) in general. The number of bumps in w (x) may be different from the number of bumps in ¯w (x) , so their difference may not be just a translation. However, by energy consideration, we know that E (w) = E ( ¯w) , which implies that both solutions have the same number of bumps.
If w (x) is different from ¯w(x), then by Proposition 24 there must exist an interval I, assumed to be I = [−π/p, π/p] , over which w (x) is a bump, but ¯w (x) is identically equal to 0. Let I (t) = u (0, t) , t ∈ [0, ∞), we have limn→∞I (tn) = γ and limn→∞I (sn) = 0. By continuity argument, for
any ξ ∈ (0, γ), there exists a sequence rn so that limn→∞I (rn) = ξ. Moreover, u(x, rn) converges
uniformly to a function ˜w(x) with ˜ w(x) = 2 2 − pcos 2p 2(x − ˜x0) 1/p , x ∈ −πp + ˜x0, π p + ˜x0
for some small |˜x0|. Similar to the above discussion, we will get a contradiction.
The proof of Theorem 2 is complete.
We now can state the following results regarding the possible limits of w (x) :
Corollary 26 Assume u (x, t) is a positive bounded solution of (12) on R × [0, ∞), converging uniformly to w(x) ∈ C2(R) which is an entire solution of the ode (44). Then
(1). If 0 < L ≤ √p1 , then w (x) ≡ 0 or w (x) ≡ 1.
(2). If √p1 < L < 1p, then w (x) ≡ 0 or w (x) ≡ 1 or w (x) is a positive entire solution of the ode (44) whose period, after multiplying by some m ∈ N, is equal to 2Lπ.
(3). If L ≥ 1p, then in addition to the possibilities in (2), w (x) can be composed of one or several
bumps (joined together by zero function).
Remark 27 In case w (x) > 0 everywhere, then as t → ∞ equation (12) remains uniformly parabolic. Hence the convergence for t → ∞ is in C∞[−Lπ, Lπ] due to the parabolic regularity
theory.
The following result says that when L is large enough, the bump solution w∗(x) in (48) will be the asymptotic limit if one choose the initial data u0(x) suitably.
Proposition 28 If L is large enough, then there exists a symmetric initial data u0(x) such that
u(x, t) stays positive and bounded, and it converges to the following
w∗(x) = 2 2−pcos2 p 2x 1/p for x ∈h−πp,πpi 0 for x ∈h−Lπ, −πpi∪hπp, Lπi. (56)
In particular, the zero set Z = {x ∈ [−Lπ, Lπ] : w (x) = 0} is not empty. Remark 29 We conjecture that Proposition 28 holds as long as L > 1p.
Proof. Let v0(x) be a positive symmetric function satisfying v0(x) = v0(−x), x ∈ [−Lπ, Lπ]
and v0
0(x) < 0 for x ∈ (0, Lπ) , v00(0) = v00 (Lπ) = 0 (both are simple zeros of v00(x)). We
also require v0(x) and |v00 (x)| to be sufficiently small outside the interval [−3Lπ/4, 3Lπ/4] ⊂
rescaling u (x, t) := v (x, t) /R (t) , umin(t) ≤ 1, umax(t) ≥ 1. We know that u (x, t) will be a
positive, bounded solution of (12) with initial condition u0(x) = v0(x) /R (0) , where R (0) satisfies
1 2Lπ Z Lπ −Lπ v1−p0 (x) dx 1 1−p ≤ R (0) ≤ vmax(0) due to (11).
The idea next is to adjust v0(x) such that u0(x) satisfies the energy condition
E (u0) < E (1) = 2Lπ · 2 2 − p − 1 (57) which will exclude the convergence of u (x, t) to 1. In doing so we need to assume L is large.
First we require v0(x) to be almost like a constant on [−Lπ/2, Lπ/2] . More precisely, let
ε ∈ (0, 1) be small and set v0 = 21/(1−p) + ε at x = 0, v0 = 21/(1−p) at x = Lπ/2, where
in-between we require v0 to be strictly decreasing with almost constant slope. The behavior of v0 on
[−Lπ/2, 0] can be obtained by symmetry. Then 1 2Lπ Z Lπ −Lπ v01−p(x) dx ≥ 1 2Lπ Z Lπ/2 −Lπ/2 v1−p0 (x) dx ≥ 1 2Lπ Z Lπ/2 −Lπ/2 2dx = 1 and so 1 ≤ R (0) ≤ vmax(0) . The value of |v00| over [0, Lπ/2] is about 2ε/ (Lπ) .
Next we set v0 on [Lπ/2, 3Lπ/4] to be v0 = 21/(1−p) at x = Lπ/2, v0= ε at x = 3Lπ/4, strictly
decreasing in-between with almost constant slope. The value of |v0
0| over the interval [Lπ/2, 3Lπ/4] is
about 421−p1 − ε
/ (Lπ) .
Finally set v0 = ε at x = 3Lπ/4, v0 = ε/2 at x = Lπ, strictly decreasing in-between with
almost constant slope. The value of |v0
0| over [3Lπ/4, Lπ] is about 2ε/ (Lπ) .
The corresponding u0(x) = v0(x) /R (0) has energy E (u0) = A + B, where
B = 2 Z Lπ2 0 + Z 3Lπ4 Lπ 2 + Z Lπ 3Lπ 4 ! −u20(x) + 2 2 − pu 2−p 0 (x) dx := B1+ B2+ B3 and A = 2 R2(0) Z Lπ2 0 + Z 3Lπ4 Lπ 2 + Z Lπ 3Lπ 4 ! (v0)2x(x) dx ≈ R22(0) (2ε/ (Lπ))2· Lπ 2 + 421−p1 − ε / (Lπ)2· Lπ 4 + (2ε/ (Lπ)) 2 · Lπ4 . Hence (since 1 ≤ R (0) ≤ 21/(1−p)+ ε) A ≤ 2 Lπ 2ε2+ 4 · 21−p2 + ε2 . (58)
To estimate B, note that the function
F (s) = 2
2 − ps2−p− s
2
satisfies F (s) = 0 at s = 0 and s = γ := (2/ (2 − p))1/p> 1; F > 0 on the interval (0, γ) and F < 0 on (γ, ∞). Moreover, F is increasing on (0, 1) , decreasing on (1, ∞) , and attains its maximum value F (1) = 2/ (2 − p) − 1 ∈ (0, 1) at s = 1. Hence we have
B1 = 2 Z Lπ2 0 −u20(x) + 2 2 − pu 2−p 0 (x) dx ≤ Lπ · F (1) (59)
and similarly B2 = 2 Z 3Lπ4 Lπ 2 −u20(x) + 2 2 − pu 2−p 0 (x) dx ≤ Lπ 2 · F (1) . (60)
For B3, since R (0) ≥ 1, we have u0(x) ∈ (0, ε) on [3Lπ/4, Lπ] . Hence
B3= 2 Z Lπ 3Lπ 4 −u20(x) + 2 2 − pu 2−p 0 (x) dx ≤ Lπ 2 · F (ε) . (61)
Combining (58), (59), (60), and (61), we conclude that 0 ≤ E (u0) = A + B ≤ 2 Lπ 3ε2+ 24−2p1−p +Lπ 2 (3F (1) + F (ε)) < 2Lπ · F (1) = E (1) (62) for L > 0 large enough and ε > 0 small enough.
By the maximum principle, reflection principle, and Lemma 23, the symmetry of the initial data is preserved with
(
u (x, t) = u (−x, t) , x ∈ [−Lπ, Lπ]
ux(x, t) < 0 for x ∈ (0, Lπ) ; ux(0, t) = ux(Lπ, t) = 0
for all t > 0. Let w∗(x) be the limit function of u(x, t) as t → ∞. Then w∗(x) = w∗(−x) and w0
∗(x) ≤ 0 when x ∈ [0, Lπ]. Note that those positive entire period solutions of the ode (44) have
bounded periods with the maximum period given by 2π/p. Thus, the function w∗(x) obtained in the above manner cannot be positive, entire, and periodic with period 2Lπ if L is large.
Finally we see that w∗(x) can not be either w∗(x) ≡ 0 or w∗(x) ≡ 1. The first case is clear since umax(t) ≥ 1 for all t > 0. For the second case, as the initial energy of u0(x) satisfies E (u0) < E (1)
and E (u (·, t)) is non-increasing and converges to E (w∗) ≥ 0, w∗(x) ≡ 1 can not happen. Hence it
must be given by (56).
Remark 30 We can also find a non-symmetric initial data u0(x) such that if L > 0 is large enough,
then the conclusion of Proposition 28 holds. One simply choose the positive periodic function v0(x) on
[−Lπ, Lπ] satisfying v0
0(x) > 0 for x ∈ (−Lπ, 0) , v00 (x) < 0 for x ∈ (0, Lπ) , v00 (x) = 0 at x =
0, Lπ (both are simple zeros of v0
0(x)). We also require v0(x) satisfying the energy condition E (u0) <
E (1) . Under the evolution, u (x, t) is positive and bounded. By Lemma 23, ux(x, t) = 0 at exactly
two points in [−Lπ, Lπ] (one at global maximum point, the other at global minimum point). Again, based on exactly the same reason, the limit function w∗(x) of u(x, t) must be of the single-bump form (56), which may center at some x 6= 0.
Remark 31 When L = m ∈ N, the symmetric initial data v0(x) clearly satisfies R−mπmπ v01−p(x) ·
sin xdx = 0. Since we have some freedom in choosing the symmetric initial data v0(x) (we only
require it to satisfy the energy condition (57)), we can also adjust it to satisfy Rmπ
−mπv 1−p 0 (x) ·
cos xdx = 0. In requiring so, v1−p0 (x) can be the radius of curvature of some locally convex immersed closed curve in R2.
4
A Non-blow-up Result.
Let w∗(x) be the function in (48) with w∗(x) = 0 outside [−π/p, π/p]. Note that w∗(x) is decreasing on [0, π/p] with w∗(π/p) = 0, w0
∗(π/p) = 0, and w∗(x) = 0 for x > π/p. We have the following
Lemma 32 There exist small positive number ρ > 0 such that for any 0 < α < w∗(π/p − ρ/2), the problem ( qt = qp qxx+ q − q1−p in (−ρ, ρ) × [0, ∞) q (−ρ, t) = α = q (ρ, t) for t > 0; q (x, 0) = α for x ∈ (−ρ, ρ) (63) has a unique positive solution which satisfies
0 < q (x, t) ≤ Ce−t in (−ρ/8, ρ/8) × [0, ∞) (64)
where C is a constant depending on p, α and ρ.
Proof. The existence of a unique solution is standard. Since ¯q(x, t) = αe−t/2 is a sub-solution, by
the maximum principle we have q(x, t) ≥ ¯q(x, t) for all (x, t). Thus q(x, t) exists for all t ∈ [0, ∞) and is positive.
Let ρ ∈ (0, 2π/p) be a fixed small number such that the number α0 := w∗(π/p − ρ/2) satisfies
0 < αp0 ≤ p
2(1 − p) ∈ (0, 1) . (65)
If 0 < α < α0, there is π/p − ρ/2 < R1 < π/p so that w∗(R1) = α. We define the function
h(x) = w∗(x + R1+ ρ) for x ∈ [−ρ, 0], and do the extension h(x) = h(−x) for x ∈ [0, ρ], and note
that
h(x) = 0 for x ∈ [−ρ/2, ρ/2] . (66)
The function h(x) is a solution of hxx+ h − h1−p= 0 with boundary condition h(−ρ) = α = h(ρ).
Let q (x, t) be the solution of (63) with the above α ∈ (0, α0) as initial-boundary value. By the
reflection principle, we see that (since 0 < α < 1, q (x, t) decreases initially for x ∈ (−ρ, ρ)) (
q (−x, t) = q (x, t) for x ∈ (−ρ, ρ) , t > 0 qx(x, t) ≥ 0 for x ≥ 0, t > 0
(67) and by the maximum principle, we also have
q (x, t) ≥ h(x) for all x ∈ [−ρ, ρ] , t > 0. (68) The difference q − h satisfies the equation
1 1 − p q 1−p− h1−p t = (q − h)xx+ (q − h) − (q1−p− h1−p) (69) with ( q (−ρ, t) − h (−ρ, t) = 0 = q (ρ, t) − h (ρ, t) for t > 0 q (x, 0) − h (x, 0) > 0 for x ∈ (−ρ, ρ) . (70)
By integrating (69), we see that 1 1 − p d dt Z ρ −ρ q1−p− h1−p dx = (q − h)x(ρ, t) − (q − h)x(−ρ, t) + Z ρ −ρ (q − h) − q1−p− h1−p dx.
In view of (68) and (70), we have (q − h)x(ρ, t) ≤ 0 and (q − h)x(−ρ, t) ≥ 0. Also, since 0 ≤
h(x) ≤ q(x, t) ≤ α, we have the inequality q − h ≤ α
p
To see (71), we consider the function g (x) = A − x
A1−p− x1−p, x ∈ [0, A] , A > 0 is a constant
and note that g (x) increases from g (0) = Ap to g (A) = (1 − p)−1Apon [0, A] . Now let b = 1−p/2,
b ∈ (1 − p, 1). For any 0 < α < α0, by (65) and (71), we have
1 1 − p d dt Z ρ −ρ q1−p− h1−p dx ≤ −b Z ρ −ρ q1−p− h1−p dx for all t > 0. This implies that for all t > 0
Z ρ −ρ q1−p(x, t) − h1−p(x) dx ≤ Z ρ −ρ α1−p− h1−p(x) dx · e−(1−p)bt ≤ 2ρα1−p· e−(1−p)bt.
By (67) and (66), for any 0 ≤ x ≤ ρ/4 we have (note that the condition (66) is essential) ρ 4q 1−p(x, t) ≤Z ρ/2 x q1−p(y, t) dy ≤ Z ρ −ρ q1−p− h1−p dx ≤ 2ρα1−p· e−(1−p)bt. Thus q(x, t) ≤ 81−p1 α · e−bt for 0 ≤ x ≤ ρ/4, t > 0
and similar estimate also holds on −ρ/4 ≤ x ≤ 0. In conclusion we have
q(x, t) ≤ 81−p1 α · e−bt for − ρ/4 ≤ x ≤ ρ/4, t > 0 (72) where b = 1 − p/2. Also set S (t) = Z ρ/4 −ρ/4 ρ2− 16x2 q1−p(x, t) dx, S (0) = 1 3α 1−pρ3.
By direct differentiation and integration by parts, we get d dt h e(1−p)tS (t)i= (1 − p)e(1−p)t Z ρ/4 −ρ/4 ρ2− 16x2 (qxx+ q) dx = (1 − p)e(1−p)t ( Z ρ/4 −ρ/4 32x · qx(x, t) dx + Z ρ/4 −ρ/4 ρ2− 16x2 qdx ) = (1 − p)e(1−p)t ( 16ρq (ρ/4, t) − 32 Z ρ/4 −ρ/4 qdx + Z ρ/4 −ρ/4 ρ2− 16x2 qdx ) . By (72) we can derive d dt h e(1−p)tS (t)i≤ 81−p1 α (1 − p) 16ρ + ρ 3 3 e−pt/2 for t > 0.
After an integration, we obtain
e(1−p)tS (t) − S (0) ≤ 81−p1 α(1 − p) 16ρ + ρ 3 3 Z ∞ 0 e−pt/2dt and so e(1−p)tS (t) ≤ 81−p1 α(1 − p)2 p 16ρ +ρ 3 3 +1 3α1−pρ 3.
Hence we have
S(t) ≤ C∗e−(1−p)t for t > 0
where the constant C∗ can be given by C∗ = α1−p 81−p1 (1 − p)2 p 16ρ + ρ 3 3 +ρ 3 3 = α1−pC1(p) ρ + C2(p) ρ3
where C1(p) and C2(p) are positive constants depending only on p.
For any 0 ≤ x ≤ ρ/√32 we have ρ2 2 Z ρ/√32 −ρ/√32 q1−p(x, t) dx ≤ Z ρ/√32 −ρ/√32 ρ2− 16x2 q1−p(x, t) dx ≤ C ∗e−(1−p)t for t > 0 Hence for 0 ≤ x ≤ ρ/8 q1−p(x, t) ρ √ 32 − ρ 8 ≤ Z ρ/√32 x q1−p(y, t) dy ≤ 2C∗ ρ2 e−(1−p)t which implies q (x, t) ≤ Ce−t for 0 ≤ x ≤ ρ/8, t > 0 (73)
where C is given by C = αC1(p) ρ−2+ C2(p)1/(1−p). Similarly, we also have
q(x, t) ≤ Ce−t for − ρ/8 ≤ x ≤ 0, t > 0. (74)
The proof is done.
Remark 33 ¿From the proof we see that if Lemma 32 is valid for some ρ > 0 and α ∈ (0, w∗(π/p − ρ/2)) ,
then for any 0 < ˜ρ < ρ, ˜α ∈ (0, w∗(π/p − ˜ρ/2)) , Lemma 32 is still valid.
A higher dimension version of Lemma 32 is given in [CPE]. From Lemma 32, equation (7), and the maximum principle, we obtain the following non-blow up result (in below we assume u (x, t) comes from the rescaling of v (x, t)):
Theorem 34 Suppose that u (x, t) converges to w(x) uniformly in [−Lπ, Lπ] and w(x) = 0 for x ∈ [a, b] ⊂ [−Lπ, Lπ], w(x) > 0 outside [a, b]. Then for any ε > 0, there is a constant M > 0 depending on ε so that 0 < v (x, t) < M for all x ∈ [a + ε, b − ε] , t > 0. Here v(x, t) is the solution of (4).
Proof. Let ρ > 0 be small enough such that Lemma 32 holds and ρ < ε. Hence for any θ ∈ [a + ε, b − ε] the interval (θ − ρ, θ + ρ) is contained in [a, b] . Let α ∈ (0, w∗(π/p − ρ/2)) . There
exists some large time T∗ such that |u (x, t)| < α on [a, b] × [T∗, ∞). For fixed θ ∈ [a + ε, b − ε] , consider the equation
(
qt= qp qxx+ q − q1−p
in (θ − ρ, θ + ρ) × [0, ∞)
q (θ − ρ, t) = α = q (θ + ρ, t) for t > 0; q (x, 0) = α for x ∈ (θ − ρ, θ + ρ) . By the maximum principle we have
0 < u (x, t + T∗) ≤ q (x, t) in (θ − ρ, θ + ρ) × [0, ∞) where by Lemma 32, q (x, t) satisfies
which implies
0 < u (x, t + T∗) ≤ Ce−t in (θ − ρ/8, θ + ρ/8) × [0, ∞).
Also from (7) we know that
u(x, t) = p1/pTmax1/pe−t· v x, T max 1 − e−pt , t ∈ [0, ∞) and so 0 < p1/pTmax1/pe−(t+T∗)· v x, Tmax 1 − e−p(t+T∗) ≤ Ce−t in (θ − ρ/8, θ + ρ/8) × [0, ∞) where C = αC1(p) ρ−2+ C2(p) 1/(1−p) . Hence we derive vx, Tmax 1 − e−p(t+T∗) ≤ Ce T∗ p1/pT1/p max = αC1(p) ρ −2+ C2(p)1/(1−p) p1/pT1/p max · eT∗ in (x, t) ∈ (θ − ρ/8, θ + ρ/8) × [0, ∞). In particular we have vθ, Tmax 1 − e−p(t+T∗) ≤ αC1(p) ρ −2+ C 2(p) 1/(1−p) p1/pT1/p max · eT∗ (75)
for all θ ∈ [a + ε, b − ε] and t ∈ [0, ∞). The proof is done.
5
Application to Plane Curves Expansion.
As mentioned in the introduction section, if γt, t ∈ [0, T ), is a family of locally convex immersed
closed curves in R2 (with rotation index m ∈ N) given by smooth immersions Xt := X (·, t) :
S1 → R2, and satisfies the curvature flow equation (1), then the radius of curvature H of γ t will
satisfy equation (2) for all x ∈ R, t ∈ [0, T ). It is well-known that the smooth solution H (x, t) to equation (2) exists on R × [0, Tmax) with Hmax(t) → ∞ as t → Tmax. We also have Hmin(t) is
non-decreasing on [0, Tmax). Moreover, γt expands to infinity (perhaps non-uniformly if m > 1) also
as t → Tmax, i.e., limt→Tmax(supS1kX (·, t)k) = ∞. See [U1], [T1] for details.
Now the corresponding equation for v (x, t) = Hα(x, t) is given by equation (4) with v(x, t) =
v(x + 2mπ, t), t ∈ [0, Tmax). If the rescaled curve ˜γt := γt/R (t) converges, as t → Tmax, to a smooth
locally convex non-circular (i.e., w (x) 6≡ 1) immersed homothetic solution ˜γ∞, then all of this type
of solutions have been classified according to results in the nice papers by Urbas [U2] and Andrews [AN2]. Thus we only mention the case when ˜γ∞ has singularity formation. The basic concept is quite easy: when we rescale back the curve γt by R (t) , the part of γt whose radius of curvature does
not expand to infinity collapses into singularity.
¿From Remark 31, if m ∈ N is large enough, there exists a symmetric locally convex immersed closed curve γ0 such that its v0(x) satisfies all of the conditions required in the proof of Proposition
28. Hence the rescaled function u (x, t) = v (x, t) R (t) = Hα(x, t) R (t) , R 0(t) = Rp+1(t) , R (T max) = ∞, p = 1 − 1 α
will converge to w∗(x) of (56) on [−mπ, mπ] as t → Tmax. Thus the rescaled radius of curvature,
denoted as ˜H, will converges to the limit function ˜H∞(x) , given by ˜
H∞(x) =
w1−p∗ (x) for x ∈ [−π/p, π/p] 0 for x ∈ [−mπ, −π/p] ∪ [π/p, mπ]
From ˜H∞(x) one can construct the symmetric limiting curve ˜γ∞up to a translation by the following formula (recall that the variable x represents the outward normal angle)
P (x) = Z x −mπ w1−p∗(θ) · − sin θ cos θ dθ, x ∈ [−mπ, mπ] (76)
2 1 0 -1 -2 2 1 0 -1 -2 3e-05 1.5e-05 0 -1.5e-05 -3e-05 3e-05 1.5e-05 0 -1.5e-05 -3e-05
Figure 1: p = 0.1187, m > 1p, the right hand side is a zoom-in picture of the singularity.
2 1 0 -1 -2 2 1 0 -1 -2 0.001 0.0005 0 -0.0005 -0.001 0.001 0.0005 0 -0.0005 -0.001
Figure 2: p = 0.279, m > 1p, the right hand side is a zoom-in picture of the singularity.
2 1 0 -1 -2 2 1 0 -1 -2 0.003 0.0015 0 -0.0015 -0.003 0.003 0.0015 0 -0.0015 -0.003
Figure 3: p = 0.531, m > 1p, the right hand side is a zoom-in picture of the singularity.
2 1 0 -1 -2 2 1 0 -1 -2 Figure 4: p = 0.8, m > 1p.
where w∗(x) is given by (56). It is a curve symmetric to the x-axis and has a cusp-like singularity at the origin. Some of the curves are pictured in Figures 1 - 4.
Remark 35 Since w∗(x) = 0 on x ∈ h−mπ, −πpi∪hπp, mπi, the integral in (76) is completely determined by w∗(x) over the interval h−πp,πpi. Hence as long as p ∈ (0, 1) is chosen, the picture will be determined regardless of the value of m ∈ N, where m > 1p.
The curvature blow-up rate for singularity in each of the above pictures (single bump case) can be determined using Theorem 34. Let ε > 0 be small. There exists some constant M > 0 depending on ε such that
0 < Hmin(0) ≤ H (x, t) =
1
k (x, t) ≤ M (77)
for all (x, t) ∈ ([−mπ, −π/p − ε] ∪ [π/p + ε, mπ]) × [0, Tmax). Hence for the rescaled radius of
cur-vature ˜H we have 0 < Hmin(0) R1/α(t) ≤ ˜H (x, t) = H (x, t) R1/α(t) = 1 R1/α(t) k (x, t) ≤ M R1/α(t)
which implies the following blow-up rate for the rescaled curvature ˜k (x, t) = 1/ ˜H (x, t) : 1 M · R 1 α (t) ≤ ˜k (x, t) ≤ 1 Hmin(0)· R 1 α(t) (78)
for all (x, t) ∈ ([−mπ, −π/p − ε] ∪ [π/p + ε, mπ]) × [0, Tmax), where
Rα1 (t) = α α − 1· 1 Tmax− t α−1 , α > 1.
For the rescaled maximum curvature ˜kmax(t) , it also satisfies estimate (78). Finally we see that as
ε → 0, M may tend to infinity, which only affects the lower bound of ˜k (x, t) in (78).
The curvature blow-up rate for the case of several bumps is similar. However, we need to restrict to compact subsets strictly contained inside the region {x ∈ [−mπ, mπ] : w (x) = 0} .
Acknowledgments. The first and third authors would like to thank the support of the National Sciences Council of Taiwan. Thanks also go to Professor Ben Andrews of the Australian National University for some useful discussions, and to Dr. S. M. Chang and C. H. Nien for helping us with the pictures.
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Asian J. Math., Vol. 3, No. 3, 635-658.
Department of Mathematics, National Taiwan University, Taipei 106, TAIWAN. Email: [email protected]
Department of Mathematics, National Chung Cheng University, Chiayi 621, TAIWAN. Email: [email protected]
Department of Mathematics, National Tsing Hua University, Hsinchu 300, TAIWAN. Email: [email protected]
