Citation: Vadim S. Anishchenko, Tatjana E. Vadivasova, Galina I. Strelkova, George A. Okrokvertskhov. Statistical properties of dynamical chaos[J]. Mathematical Biosciences and Engineering, 2004, 1(1): 161-184. doi: 10.3934/mbe.2004.1.161
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Dedicated to Professor Neil Trudinger on the occasion of his 80th birthday.
Let M0 be a smooth, closed, uniformly convex hypersurface in Rn+1 enclosing the origin. In [20], the authors studied the following anisotropic Gauss curvature flow:
{∂tX(x,t)=−f(ν)rαK(x,t)ν,X(x,0)=X0(x), | (1.1) |
where K(⋅,t) is the Gauss curvature of the hypersurface Mt, parametrised by X(⋅,t): Sn→Rn+1, ν(⋅,t) is the unit outer normal at X(⋅,t), f is a given positive smooth function on Sn, and r=|X(x,t)| is the distance from the origin to the point X(x,t).
The Gauss curvature flow (1.1) was introduced to study the existence of solutions to the dual Minkowski problem proposed in [16]. It can be formulated as solving the following Monge-Ampère equation on the unit sphere Sn,
det(∇2u+uI)(x)=f(x)u(|∇u|2+u2)α2,x∈Sn, | (1.2) |
where u denotes the support function of a hypersurface solution M. By establishing the a priori estimates and studying the convergence of the normalized flows of (1.1), the following results were proved in [20].
Theorem 1.1. ([20]) Let f be a smooth and positive function on the sphere Sn.
(i) If α>n+1, there is a unique smooth, uniformly convex solution to (1.2).
(ii) If α=n+1, assume that f satisfies the condition (1.3) below, then there is a smooth, uniformly convex solution to (1.2). The solution is unique up to dilation.
(iii) If α<n+1 and f is even, there is an origin-symmetric, smooth and uniformly convex solution to (1.2).
(iv) If f≡1, then the solution must be a sphere when α≥n+1, and the origin-symmetric solution must be a sphere when 0≤α<n+1.
When α=n+1, Eq (1.2) is the Aleksandrov problem. It is known that there is a necessary and sufficient condition for the existence of solutions, namely
∫Snf=|Sn|,∫ωf<|Sn|−|ω∗|, ∀ convex domain ω⊂Sn, | (1.3) |
where ω∗ is the dual set of ω.
In this paper, as in [5,10,20] we employ a Gauss curvature flow to study the existence of solutions to the Lp dual Minkowski problem introduced in [23]
which extends the dual Minkowski problem (1.2). Let f and g be positive functions on Sn, and p,q∈R. We study the existence of solutions to the following equation,
det(∇2u+uI)(x)=f(x)up−1(|∇u|2+u2)n+1−q2g(∇u(x)+ux√|∇u|2+u2),x∈Sn. | (1.4) |
Equation (1.4) contains the Lp dual Minkowski problem as a special case (namely when g≡1). It extends the Lp-Minkowski problem (when q=n+1 and g≡1) and the dual Minkowski problem (when p=0 and g≡1).
In particular, when g≡1, p=1,q=n+1, Eq (1.4) is the classical Minkowski problem, which asks for the existence of closed convex hypersurfaces with prescribed surface area measure. It is a major impetus for the development of fully nonlinear PDEs. The Lp-Minkowski problem, introduced in [22], concerns the existence of closed convex hypersurfaces with prescribed p-area measures. It extends the classical Minkowski problem and includes the logarithmic Minkowski problem (p=0), and the centro-affine Minkowski problem (p=−n−1) as special cases [2,11]. In the last two decades, great progress has been made in the study of the Lp-Minkowski problem. There is a rich phenomena on the existence and multiplicity of solutions, depending on the range of the exponent p (see e.g., [1,9,12,15,18,19]).
For general exponents p and q, Eq (1.4) has been studied in [4,6,7,8,17]. Suppose that g≡1. The existence of smooth solutions to (1.4) was proved in [17] for p>q, and in [8] for p=q. When p<q, the solution may not be smooth in general. Weak solutions were obtained in [4] when p>1 and q>0, and later on in [8] for all p>0 and q∈R. If only origin-symmetric solutions are concerned, Eq (1.4) was solved in [6] when pq≥0, and in [7] when q>0 and −q∗<p<0 where q∗ is defined as
q∗={qq−n if q≥n+1,nqq−1 if 1<q<n+1,+∞ if 0<q≤1. |
Suppose now g≢1. Equation (1.4) with p=q=0 characterises the Gauss image problem proposed by [3], which extends the classical Aleksandrov problem. It was also considered in [21] from the optimal transportation viewpoint.
The main result of this paper is the following.
Theorem 1.2. Let f,g∈C1,1(Sn) be positive functionssatisfying c−10≤f,g≤c0 for some constant c0>1. Suppose that q>n and
p<{−nqq−1, if q≥n+1,−qq−n, if n<q<n+1. | (1.5) |
Then there is a uniformly convex and C3,α-smooth positive solution to (1.4), where α∈(0,1).
When q=n+1 and g≡1, Theorem (1.2) recovers the main result in [13]. The range of p and q in Theorem 1.2 has no overlap with that in [7], and to the best of our knowledge, has not been studied in other papers.
To prove Theorem 1.2, we will employ the following Gauss curvature flow,
∂X∂t(x,t)=−η(t)f(ν)g(ξ)⟨X,ν⟩p|X|n+1−qK(x,t)ν+X(x,t), | (1.6) |
where X(⋅,t):Sn→Rn+1 is a parametrisation of the evolving convex hypersurface Mt, ξ=X/|X|, ν and K are respectively the unit outward normal and the Gauss curvature of Mt. The multiplier η(t) is given by
η(t)=[∫Snf(x)up(x,t)dx]1−p−1[∫Sng(ξ)rq(ξ,t)dξ]1−1q, | (1.7) |
and as before, u(x,t) and r(ξ,t) are the support and radial functions of Mt.
Denote by Ko the set of convex bodies Ω⊂Rn+1 containing the origin in its interior. We will show that (1.6) is a gradient flow to the following functional:
Jp,q(Ω)=[∫Sng(ξ)rq(ξ)dξ]1q+[∫Snup(x)f(x)dx]1−p, | (1.8) |
where Ω∈Ko, u and r are the support function and the radial function of Ω, respectively. If Ω∈Ko is a critical point of the functional (1.8), then the support function of Ω satisfies Eq (1.4).
For the Gauss curvature flow (1.6), a main issue is the lack of uniform estimates, namely the control of the eccentricity of Ω (i.e., the eccentricity of the minimum ellipsoid of Ω). Our strategy is to use a topological method to find a special initial condition such that the evolving hypersurfaces Mt=∂Ωt satisfy
Br(0)⊂Ωt⊂BR(0), | (1.9) |
for some positive constants R≥r>0 independent of t. Once the solution satisfies such a C0-estimates, one can establish higher order a priori estimates for the flow (1.6). Hence by the monotonicity of the functional (1.8), the flow converges to a solution of (1.4).
To find the special initial hypersurface, we assume that q>n and (1.5) such that the functional Jp,q(Ω) will become very large if either the volume of Ω is sufficiently large or small, or the eccentricity of Ω is sufficiently large. This property enables us to find the special initial hypersurface by using the topological method and the variational structure of the equation as in [13], where we proved the existence of solutions to the Lp-Minkowski problem in the super-critical case (namely when q=n+1 and p<−n−1). Although the approach is similar to that in [13], equation (1.4) and the associated flow (1.6) are more complicated than the corresponding ones in [13]. Therefore, we need to present sufficient details of the argument for the approach.
We will consider in Section 2 the a priori estimates for the flow (1.6). In Section 3, we combine the a priori estimates and the topological method to find a special initial condition such that the flow converges to a solution of (1.4), and thus prove Theorem 1.2. Section 4 contains further remarks on some variants of Theorem 1.2.
For a closed convex hypersurface M⊂Rn+1, the support function of M is given by
u(x)=⟨x,ν−1M(x)⟩, ∀ x∈Sn, |
where νM:M→Sn is the Gauss map and ν−1M is its inverse. It is well known that ν−1M(x)=u(x)x+∇u(x), and the Gauss curvature of M at ν−1M(x) is given by
K=1/det(uij+uδij), | (2.1) |
where uij:=∇2iju. Assume that Cl(M)∈Ko, where Cl(M) denotes the convex body enclosed by M. Recall that the radial function of M, denoted by r, is given by
r(ξ)=max{λ:λξ∈Cl(M)} ∀ ξ∈Sn. | (2.2) |
Denote →r(ξ)=r(ξ)ξ. We also define the radial Gauss mapping by
AM(ξ)=νM(→r(ξ)) ∀ ξ∈Sn. |
It is easy to verify that
r∘A−1M(x)=|ν−1M(x)|=(u2(x)+|∇u(x)|2)12. |
For any convex body Ω∈Ko, its polar dual Ω∗ is given by
Ω∗={x∈Rn+1:x⋅y≤1,∀y∈Ω}. |
Let Mt be a solution to the flow (1.6) and X(⋅,t) be its parametrisation. Consider the new parametrisation
¯X(x,t)=X(ν−1Mt(x),t). |
It is straightforward to compute
∂¯X∂t=∑i∂X∂zi∂(ν−1Mt)i∂t+∂X∂t. |
Since the first term on the right hand side is tangential, taking inner product with the unit outer normal of Mt gives that
∂tu(x,t)=⟨x,∂t¯X(x,t)⟩=⟨x,∂tX(ν−1Mt(x),t)⟩. |
Hence by (2.1), the flow (1.6) can be expressed as
∂tu(x,t)=−η(t)f(x)g(∇u+ux√u2+|∇u|2)up(u2+|∇u|2)n+1−q2det(∇2u+uI)+u. | (2.3) |
Theorem 2.1. Suppose that both f and g are positive and C1,1-smooth. Let u(⋅,t) be a positive, smooth and uniformly convex solution to (2.3), t∈[0,T). Assume that
1/C0≤u(x,t)≤C0 | (2.4) |
for all (x,t)∈Sn×[0,T). Then
C−1I≤(∇2u+uI)(x,t)≤CI ∀ (x,t)∈Sn×[0,T), | (2.5) |
where C is a positive constant depending only on n,p,q,C0, minSnf, minSng, ‖f‖C1,1(Sn), ‖g‖C1,1(Sn), and the initial condition u(⋅,0), but is independent of T.
Proof. We first observe that, by the convexity
|∇u(x,t)|≤maxSnu(⋅,t)≤C0, ∀ (x,t)∈Sn×[0,T). | (2.6) |
It also yields the bound of η(t) defined by (1.7):
1/C1≤η(t)≤C1, for all t∈[0,T), | (2.7) |
where C1 depends only on n,p,q,minSnf,minSng,maxSnf,maxSng and C0.
Recently in [14], we studied the centro-affine Minkowski problem. We established in [14] the a priori estimates for a more general equation of the form
∂tu(x,t)=−η(t)Φ(x,u,∇u)[det(∇2u+uI)]−1+u(x,t), | (2.8) |
so that when
Φ(x,u,∇u)=f(x)g(∇u+ux√u2+|∇u|2)up(u2+|∇u|2)n+1−q2, |
the Eq (2.8) becomes (2.3).
By virtue of Lemma 6.1 and Lemma 6.2 in [14], and using (2.6), (2.7) and (2.8), we conclude (2.5) as desired and hence complete the proof.
By the second derivative estimates (2.5), Eq (2.3) becomes uniformly parabolic. Hence, by Krylov's regularity theory, we have the following C3,α estimate,
‖u(⋅,t)‖C3,α(Sn)≤C ∀ (x,t)∈Sn×[0,T), | (2.9) |
for any given α∈(0,1), where the constant C depends only on n,p,q,C0, minSnf, minSng, ‖f‖C1,1(Sn), ‖g‖C1,1(Sn), and the initial condition u(⋅,0). By the a priori estimates (2.9), we have the longtime existence of solutions to the flow (1.6), provided that u satisfies (2.4).
Theorem 2.2. Assume the conditions in Theorem 2.1. Let Tmax be the maximal time such that the solution u(⋅,t) exists on [0,Tmax). Then Tmax=∞ and u satisfies the estimates(2.5) and (2.9).
In this section, we use a topological method to select an initial hypersurface N0, such that the flow (1.6) deforms N0 to a solution of (1.4).
We first prove the monotonicity of J:=Jp,q under the flow (1.6). Recall that
J(Ω)=[∫Sng(ξ)rq(ξ)dξ]1q+[∫Snup(x)f(x)dx]1−p. |
Lemma 3.1. Suppose Mt, t∈[0,T), is a solution to the flow (1.6) in Ko. Then
ddtJ(Ωt)≥0, |
where Ωt=Cl(Mt). Equality holds if and only if the support function of Mt satisfies
det(∇2u+uI)=fup−1(|∇u|2+u2)n+1−q2g(∇u+ux√u2+|∇u|2)η(t), | (3.1) |
where η is given in (1.7).
Proof. The following formulas can be found in [20]:
∂trr(ξ,t)=∂tuu(AMt(ξ),t),|JacA|(ξ)=rn+1K(→r(ξ,t))u(AMt(ξ)), | (3.2) |
where JacA is the Jacobian of the radial Gauss mapping.
By virtue of (2.1), (2.3) and (3.2), we obtain
ddtJ(Ωt)=[∫Sngrqdξ]1q−1∫Sngrq∂trrdξ−[∫Snupfdx]1−p−1∫Snfup∂tuudx=[∫Sngrqdξ]1q−1∫Sngrq−n−1K∂tudx−[∫Snupfdx]1−p−1∫Snfup−1∂tudx=[∫Sngrqdξ]1q−1∫Sngrq−n−1uK(u−η(t)fgrn+1−qupK)∂tudx≥0. |
Clearly, equality ddtJ(Ωt)=0 holds if and only if u(⋅,t) satisfies (3.1).
The proof of Lemma 3.1 also verifies that (1.4) is the Euler-Lagrange equation of the functional (1.8) up to a constant. Note that once we have a solution to (3.1), by a proper rescaling, we can obtain a solution to (1.4).
In the following, we aim to find an initial condition u(⋅,0) such that (2.4) is satisfied.
For any convex body Ω in Rn+1, let E(Ω) denote John's minimum ellipsoid of Ω. We have
1n+1E(Ω)⊂Ω⊂E(Ω). |
Let a1(Ω)≤a2(Ω)≤⋯≤an+1(Ω) be the lengths of semi-axes of E(Ω). Denote eM=eΩ=an+1(Ω)a1(Ω) the eccentricity of M:=∂Ω (or the eccentricity of Ω).
In this subsection, we will show Proposition 3.2 by assuming that Lemma 3.3 and Lemma 3.4 hold. The proofs of Lemma 3.3 and Lemma 3.4 will be presented after the proof of Proposition 3.2.
Proposition 3.2. Suppose that q>n and p satisfies (1.5). Suppose that f,g satisfy c−10≤f,g≤c0 for some constant c0>1. For any given constant A>J(B1(0)), if one of the quantities eΩ, Vol(Ω), [Vol(Ω)]−1, and [dist(O,∂Ω)]−1 is sufficiently large, then J(Ω)≥A.
Proof. We divide the proof into three steps.
Step 1: If eΩ≥e for a large constant e>1, we have J(Ω)>A.
By Lemmas 3.3 and 3.4, we have
−1p∫Snupdx≥Cdn−|p|n+1∏j=2aj, | (3.3) |
1q∫Snrqdξ≥Caq−nn+1n∏j=1aj. | (3.4) |
From (3.3) and (3.4), there exists a constant C>0 depending only on n,p,q,c0 such that
[J(Ω)]2≥C[∫Snupdx]−1p[∫Snrqdξ]1q≥Cdn|p|−1a1−nqn+1[n+1∏j=2a−1|p|j][n∏j=1a1qj]=C[a1d]1−n|p|[a2a1]1−n|p|−1q[a3a2]1−n−1|p|−2q⋯[an+1an]1−1|p|−nq. |
That is
[J(Ω)]2≥C[a1d]1−n|p|n∏j=1[aj+1aj]1−n+1−j|p|−jq. | (3.5) |
Condition (1.5) yields that
1−n|p|>0, and 1−n+1−j|p|−jq>0, ∀ j=1,⋯,n. |
To see this, if q≥n+1, then |p|>nqq−1 and
1−n+1−j|p|−jq>1−(n+1−j)(q−1)nq−jq=(q−n−1)(j−1)nq≥0. |
While if n<q<n+1, then |p|>qq−n and
1−n+1−j|p|−jq>1−(n+1−j)(q−n)+jq=(n−j)(n+1−q)q≥0. |
Note that a1≥d. If eΩ is large, there is a j such that aj+1/aj is large. We see from (3.5) that J(Ω)≥A.
Step 2: If either Vol(Ω)≤v0 or Vol(Ω)≥v−10, for a small constant v0>0, then J(Ω)>A.
We have
J(Ω)≥[∫Snrqgdξ]1q≥Cd≥Cdan+1[Vol(Ω)]1n+1, | (3.6) |
and
J(Ω)≥[∫Snupfdx]1|p|≥Can+1≥a1an+1C[Vol(Ω)]1n+1. | (3.7) |
If a1/d is large, then (3.5) implies that J(Ω)≥A. Therefore we may assume a1≤Cd for some C, and so a1/an+1≤Cd/an+1. Hence if either d/an+1 or a1/an+1 is sufficiently close to 0, then (3.5) again shows that J(Ω)≥A. Hence we assume that d/an+1 or a1/an+1 are away from 0. By (3.6) and (3.7), if either Vol(Ω) or [Vol(Ω)]−1 is large, then J(Ω)≥A.
Step 3: If dist(O,∂Ω)≤d0 for a sufficiently small d0>0, then J(Ω)>A.
In this case, we may assume that aj≤Cd for all 1≤j≤n+1 for some C≥1. As discussed in Step 2, a1≤Cd, otherwise we are done. If aj/d is sufficiently large for some j, then eΩ is also huge as eΩ≥aj/a1≥aj/(Cd). Step 1 shows that J(Ω)≥A.
Under the above assumption, if d is sufficiently small, then Vol(Ω) becomes very small. By Step 2, we have J(Ω)>A.
Lemma 3.3. Let Ω∈Ko. Suppose q>0. There exists a constant C>0 depending only on n and q such that
1q∫Snrq(ξ)dξ≥Caq−nn+1n∏i=1ai. |
Here, r is the radial function of Ω, and a1≤⋯≤an+1 are the lengths of semi-axes of E(Ω).
Proof. By a proper rotation of coordinates, we assume that E=E(Ω) is given by
E−ζE={z∈Rn+1:n+1∑i=1z2ia2i≤1}, |
where ζE=(ζ1,⋯,ζn+1) is the center of E. We can further assume that ζn+1≥0.
Since 1n+1E⊂Ω, we have
u(en+1)≥ζn+1+1n+1an+1. | (3.8) |
Hence, there exists a point p0∈Ω such that
p0⋅en+1=u(en+1)≥ζn+1+1n+1an+1. |
Consider the hyperplane L which is orthogonal to en+1 and passes through ζE:
L={z∈Rn+1:(z−ζE)⋅en+1=0}. |
Let P=L∩1n+1E be the intersection of L with the ellipsoid 1n+1E, and V be the cone in Rn+1 with base P and the vertex p0. Clearly V⊂Ω.
Case 1: q>n+1. Let us consider the following subset of V:
V′={z∈V:zn+1−ζn+1≥12(p0⋅en+1−ζn+1)}. |
This together with (3.8) implies that
|z|≥an+12(n+1), ∀z∈V′. | (3.9) |
Using V′⊂Ω, q>n+1 and (3.9), we have
1q∫Snrqdξ=∫Ω|z|q−n−1dz≥∫V′|z|q−n−1dz≥Caq−n−1n+1Vol(V′), | (3.10) |
where C is a constant depending only on n and q.
It is easy to see that
Vol(V′)≥cnan+1n∏i=1ai. | (3.11) |
Combining (3.10) and (3.11), we conclude that
∫Snrqdξ≥Caq−nn+1n∏i=1ai. |
Case 2: 0<q≤n+1. Since Ω contains the origin and Ω⊂E, we have
|z|≤cnan+1,∀z∈Ω. | (3.12) |
Using 1n+1E⊂Ω, n<q≤n+1 and (3.12), we derive that
1q∫Snrqdξ=∫Ω|z|q−n−1dz≥Caq−n−1n+1Vol(Ω)≥Caq−nn+1n∏i=1ai. |
This completes the proof.
Lemma 3.4. Let Ω∈Ko. Suppose p<0. There exists a constant C>0 depending only on n and p such that
−1p∫Snupdx≥Cdn−|p|n+1∏j=2aj. |
Here, u is the support function of Ω, d=dist(O,∂Ω) and a1≤⋯≤an+1 are the lengths of semi-axes of E(Ω).
Proof. By a proper rotation of coordinates, we assume that E=E(Ω) is given by
E−ζE={z∈Rn+1:n+1∑i=1z2ia2i≤1}, |
where ζE=(ζ1,⋯,ζn+1) is the center of E. Let x0∈Sn be a point such that u(x0)=minSnu=d, where u is the support function of Ω and d=dist(O,∂Ω). We choose i# and switch ei# and −ei# if necessary such that
x0⋅ei#=max{|x0⋅ei|:1≤i≤n+1}. |
This implies that x0⋅ei#≥cn. We use cn to denote a constant which depends only on n but may change from line to line.
Let w(x)=u(x)+u(−x), x∈Sn, be the width of Ω in x. Since 1n+1E(Ω)⊂Ω⊂E(Ω), we have
d≤minSnw≤cnai# |
and
2ain+1≤w(ei)≤2ai, ∀ i=1,⋯,n+1. |
By switching ei and −ei if necessary, we assume that u(ei)≤cnai for all i=1,⋯,n+1.
Consider the cone V in Rn+1 with the vertex p0=r∗(x0)x0 and the base
C:=convex hull of {O,r∗(ek)ek}k≠i#. |
Here, r∗ is the radial function of the polar dual Ω∗ of Ω:
Ω∗={y∈Rn+1:y⋅z≤1 ∀ z∈Ω}. |
Let V′ be a subset of V:
V′={z=(z1,⋯,zn+1)∈V:zi#≥r∗(x0)2x0⋅ei#}. |
Recall that r∗=1/u. So r∗(x0)=1d. Since x0⋅ei#≥cn, we see that
1cnd≤|z|≤1d for all z∈V′. |
The second inequality above follows by V′⊂Ω∗⊂B1/d(0) (as Bd(0)⊂Ω). Therefore
−1p∫Snupdx=1|p|∫Sn(r∗)|p|=∫Ω∗|z||p|−n−1dz≥∫V′|z||p|−n−1dz≥Cdn+1−|p|Vol(V′). | (3.13) |
Since r∗(ek)=1u(ek)≥cnak for all k≥1, we obtain
Vol(V′)≥cndai#n+1∏j=1a−1j. |
Plugging this into (3.13), we obtain
−1p∫Snupdx≥Cdn−|p|ai#n+1∏j=1a−1j≥Cdn−|p|n+1∏j=2aj. |
Remark 3.5. Let Mt, t∈[0,Tmax), be a solution to (1.6). By Proposition 3.2, if J(Mt)<A for a constant A independent of t, then there exist positive constants e0,v0,d0 depending on A, but independent of t, such that
eMt≤e0, v0≤Vol(Mt)≤v−10, and Bd0(0)⊂Ωt, | (3.14) |
where Ωt is the convex body enclosed by Mt. Note that (3.14) implies (2.4). Hence, the a priori estimates (2.5) and (2.9) hold, and one has the long-time existence of solution (Theorem 2.2). Therefore, all we need is to establish the condition J(Mt)<A for some constant A.
We introduce a modified flow (as in [13]) such that for any initial condition, the solution exists for all time t≥0. It is more convenient to work with a flow which exists for all t≥0.
Let us fix a constant
A0=10‖g‖L1(Sn)+10(n+1)‖f‖L1(Sn). | (3.15) |
If the minimum ellipsoid of Ω is B1(0), then 1n+1B1(0)⊂Ω⊂B1(0) and hence
J(Ω)≤12A0. | (3.16) |
For a closed, smooth and uniformly convex hypersurface N such that Ω0=Cl(N)∈Ko, we define ˉMN(t) as follows:
a): If J(MN(t))<A0 for all time t≥0, let ˉMN(t)=MN(t) for all t≥0, where MN(t) is the solution to (1.6).
b): If J(N)<A0, and J(MN(t)) reaches A0 at the first time t0>0, we define
ˉMN(t)={MN(t), if 0≤t<t0,MN(t0), if t≥t0. |
c): If J(N)≥A0, we let ˉMN(t)≡N for all t≥0.
For convenience, we call ˉMN a modified flow of (1.6). By the a priori estimates in Section 2, ˉMN(t) is smooth for any fixed time t, and Lipschitz continuous in time t. Moreover, we have the following properties.
i) ˉMN(t) is defined for all time t≥0, and by Lemma 3.1, J(ˉMN(t)) is non-decreasing. In particular, we have J(ˉMN(t))≤max{A0,J(N)} ∀ t≥0.
ii) If either dist(O,N) is very small, or Vol(Ω0) is sufficiently large or small, or eΩ0 is sufficiently large, by Proposition 3.2, we have ˉMN(t)≡N ∀ t≥0.
Here we recall the homology of a class of ellipsoids AI introduced in [13], such that an ellipsoid E with J(E)<A0 is contained in AI. By Proposition 3.2, we have
Corollary 3.6. For the constant A0 given by (3.15), there exist sufficiently small constants ˉd and ˉv, and sufficiently large constant ˉe, such that for any Ω∈Ko,
(i) if dist(O,∂Ω)≤ˉd, then J(Ω)>A0;
(ii) if eΩ≥ˉe, then J(Ω)>A0;
(iii) if Vol(Ω)≤ˉv or Vol(Ω)≥1/(n+1)n+1ˉv−1, then J(Ω)>A0.
Let K be the metric space consisting of non-empty, compact and convex sets in Rn+1, equipped with the Hausdorff distance. Denote by ˉKo the closure of Ko in K.
Fix the constants ˉd,ˉv,ˉe in Corollary 3.6. Let AI be the set of ellipsoids E∈ˉKo such that ˉv≤Vol(E)≤1/ˉv, and eE≤ˉe. Denote by A the following subset of AI
A={E∈AI:Vol(E)=ωn, and either eE=ˉe or dist(O,∂E)=0}. |
Here, ωn=|B1(0)| is the volume of B1(0), and eE is the eccentricity of E.
We also denote by EI the set of ellipsoids in AI centred at the origin, and by E the set of ellipsoids in A centred at the origin. These sets are all metric spaces by equipping the Hausdorff distance.
It was proved in [13] that EI is contractible and so the homology Hk(EI)=0 for all k≥1. Moreover, AI is homeomorphic to EI×B1(0). Hence, AI is contractible and the homology
Hk(AI)=0 for all k≥1. | (3.17) |
Denote
P={E∈AI:either Vol(E)=ˉv, or Vol(E)=1/ˉv, or eE=ˉe, or O∈∂E}. | (3.18) |
It is the boundary of AI if we regard AI as a set in the topological space of all ellipsoids. Moreover, there is a retraction Ψ from AI∖{B1} to P. Namely, Ψ:AI∖{B1}→P is continuous and Ψ|P=id. The following two theorems were also proved in [13].
Proposition 3.7. We have the following results.
(i) Hk+1(P)=Hk(A) for all k≥1.
(ii) There is a long exact sequence
⋯→Hk+1(A)→Hk(E×Sn)→Hk(E)⊕Hk(Sn)→Hk(A)→⋯. |
Proposition 3.8. Let n∗=n(n+1)2. The homology group Hn∗+n−1(E)=Z.
In this subsection, we use Propositions 3.7 and 3.8 to select a special initial condition in AI such that the solution to the Gauss curvature flow (1.6) satisfies the uniform estimate. The idea is similar to that in [13].
For any ellipsoid N such that Cl(N)∈AI, let ˉMN(t) be the solution to the modified flow. We have the following properties:
1) If Cl(N) is close to P in Hausdorff distance or in P, we have J(N)≥A0 and so ˉMN(t)≡N for all t (see Corollary 3.6).
2) If Cl(N) is close to B1(0) in Hausdorff distance, then J(N)<A0.
3) By our definition of the modified flow, J(ˉMN(t))<max{A0,J(N)} for all t. Hence by Remark 3.5, if ˉMN(t) is not identical to ˉMN(0)=N, then
eˉMN(t)≤ˉe, ˉv≤Vol(ˉMN(t))≤1/ˉv, and Bˉd(0)⊂Cl(ˉMN(t)) ∀ t≥0. | (3.19) |
With these properties, we can prove the following key lemma.
Lemma 3.9. For every t>0, there exists N=Nt with Cl(N)∈AI, such that the minimum ellipsoid of ˉMN(t) is the unit ball B1(0).
Proof. Suppose by contradiction that there exists t′>0 such that, for any Ω∈AI, EN(t′)≠B1(0), where N=∂Ω and EN(t′) is the minimum ellipsoid of ΩN(t′):=Cl(ˉMN(t′)).
By Corollary 3.6, EN(t′)∈AI. Hence we can define a continuous map T:AI→P by
Ω∈AI↦EN(t′)∈AI∖{B1}↦Ψ(EN(t′))∈P, |
where Ψ is the retraction after (3.18), and B1=B1(0) for short. Note that when Ω∈P, we have J(Ω)≥A0 and thus EN(t′)=EN(0)=Ω. This implies that T|P=idP. Hence, T is a retraction from AI to P, and so there is an injection from H∗(P) to H∗(AI). By (3.17), we then have
Hk(P)=0 for all k≥1. |
It follows from Proposition 3.7 (ⅱ) that
Hk(E×Sn)=Hk(E)⊕Hk(Sn) for all k≥1. |
Computing the left-hand side by the Künneth formula and using the fact Hk(Sn)=Z if k=0 or k=n, and Hk(Sn)=0 otherwise, we further obtain
Hk(E)⊕Hk−n(E)=Hk(E)⊕Hk(Sn). |
However, this contradicts Proposition 3.8 by taking k=n∗+2n−1 in the above.
In the following we prove the convergence of the flow (1.6) with a specially chosen initial condition. Take a sequence tk→∞ and let Nk=Ntk be the initial data from Lemma 3.9. By our choice of A0 (see (3.15) and (3.16)), Lemma 3.9 implies that
J(ˉMNk(tk))≤12A0. | (3.20) |
Hence, by the monotonicity of the functional J, we have
ˉMNk(t)=MNk(t) ∀ t≤tk. |
Since Cl(Nk)∈AI and Bˉd(0)⊂Cl(Nk), by Blaschke's selection theorem, there is a subsequence of Nk which converges in Hausdorff distance to a limit N∗ such that Cl(N∗)∈AI and Bˉd⊂Cl(N∗).
Next, we show that the flow (1.6) starting from N∗ satisfying J(MN∗(t))<A0 for all t.
Lemma 3.10. For any t≥0, we have
J(ˉMN∗(t))≤34A0. |
Hence
ˉMN∗(t)=MN∗(t) ∀ t>0. |
Proof. For any given t>0, since Nk→N∗ and tk→∞, when k is sufficiently large such that tk>t, we have
J(ˉMN∗(t))−J(MNk(t))≤14A0. |
By the monotonicity of the functional J,
J(MNk(t))≤J(MNk(tk)). |
Combining above two inequalities with (3.20), we obtain that
J(ˉMN∗(t))=J(ˉMN∗(t))−J(MNk(t))+J(MNk(t))≤J(ˉMN∗(t))−J(MNk(t))+J(MNk(tk))≤14A0+12A0=34A0. |
This completes the proof.
Let ΩN∗(t)=Cl(MN∗(t)) and u(⋅,t) be its support function. By Lemma 3.10, MN∗(t) satisfies (3.19). Hence,
ˉd≤u(x,t)≤C, ∀ (x,t)∈Sn×[0,∞), |
where C=(n+1)/(ˉvωn−1ˉdn). Hence, condition (2.4) holds, and we obtain the existence of solutions to (1.4) as follows.
Proof of Theorem 1.2. Denote M(t)=MN∗(t) and \mathcal{J}(t) = \mathcal{J}({\mathcal M}(t)) . By Lemma 3.1 and Lemma 3.10,
{\mathcal J}(t) < A_0\ \text{and }\ {\mathcal J}'(t)\geq 0\ \ \forall\ t\ge0. |
Therefore,
\int_0^\infty \mathcal{J}'(t)dt \le \limsup\limits_{T\to\infty} {\mathcal J}(T)-{\mathcal J}(0) \leq A_0. |
This implies that there exists a sequence t_i\to \infty such that
\begin{eqnarray*} {\mathcal J}'(t_i) = \Big[\int_{{\mathbb S}^n}g r^q d\xi\Big]^{\frac{1}{q}-1}\int_{{\mathbb S}^n}g\frac{r^{q-n-1}}{uK} \Big(u -\eta(t) \frac{f}{g}r^{n+1-q}u^{p}K\Big){\partial}_t u dx\Big|_{t = t_i} \to 0. \end{eqnarray*} |
Passing to a subsequence, we obtain by the a priori estimates (2.9) that u(\cdot, t_i)\to u_\infty in C^{3, \alpha}({\mathbb S}^{n}) -topology and u_\infty satisfies
\begin{eqnarray*} \det({\nabla}^2 u+uI) = \lambda\frac{fu^{p-1} (|{\nabla} u|^2+u^2)^{\frac{n+1-q}{2}}}{g\big(\frac{{\nabla} u+u x}{\sqrt{u^2+|{\nabla} u|^2}}\big)}, \end{eqnarray*} |
where \lambda = \lim_{t_i\to\infty}\eta(t_i) . As p\ne q , it can be seen that u_\infty^\lambda = \lambda^{\frac{1}{p-q}} u_\infty satisfies (1.4).
This section is devoted to some variants of Theorem 1.2. We show that when g is a positive function defined on {\mathbb R}^{n+1} instead of {\mathbb S}^n , our argument still works. We first show Eq (4.1) below admits a solution u up to some multiplier \lambda > 0 when p and q are in the same range as in Theorem 1.2. Since now g is a function on {\mathbb R}^{n+1} , it does not imply that the equation with \lambda = 1 has a solution by a scaling argument as mentioned in Section 3.6.
Theorem 4.1. Let f\in C^{1, 1}({\mathbb S}^n) and g\in C^{1, 1}({\mathbb R}^{n+1}) be two positive functions satisfying c_0^{-1}\le f, g\le c_0 for some c_0 > 1 . Suppose q > n and p satisfies (1.5). Then there is a constant \lambda > 0 , and a uniformly convex and C^{3, \alpha} function u , \alpha\in (0, 1) , such that
\begin{equation} \det({\nabla}^2 u+uI)(x) = \lambda \frac{f(x)u^{p-1}(|{\nabla} u|^2+u^2)^{\frac{n+1-q}{2}}}{g({\nabla} u(x)+ux)} , \quad x\in {\mathbb S}^n. \end{equation} | (4.1) |
Proof. Similarly to (1.6), we study the flow
\begin{equation} \frac{{\partial} X}{{\partial} t}(x, t) = -\eta(t) \frac{f(\nu)}{ g(X)}\langle{X, \nu}\rangle^p|X|^{n+1-q}K(x, t) \nu +X(x, t), \end{equation} | (4.2) |
where
\begin{eqnarray*} \eta(t) = \Big[ \int_{{\mathbb S}^n} f(x) u^{p}(x, t) dx \Big]^{\frac{1}{-p}-1} \Big[q \int_{{\mathbb S}^n} \int_0^{r(\xi, t)} g(\tau\xi)\tau^{q-1} d\tau d\xi \Big]^{1-\frac{1}{q}}. \end{eqnarray*} |
The same computation as in Lemma 3.1 implies that the functional
\begin{eqnarray*} {\mathcal J}^{(1)}(\Omega) = \Big[q\int_{{\mathbb S}^n} \int_0^{r(\xi)} g(\tau \xi) \tau^{q-1} d\tau d\xi \Big]^{\frac{1}{q}} +\Big[\int_{ \mathbb{S}^n} u^p(x)f(x)dx\Big]^{\frac{1}{-p}} \end{eqnarray*} |
is non-decreasing along (4.2).
Since g is bounded and positive, we have
\begin{eqnarray*} c_0^{-1}\Big[\int_{{\mathbb S}^n}r^qd\xi\Big]^{\frac1q}\le\Big[q\int_{{\mathbb S}^n} \int_0^{r(\xi)}g(\tau \xi) \tau^{q-1} d\xi d\tau\Big]^{\frac{1}{q}} \le c_0\Big[\int_{{\mathbb S}^n}r^qd\xi\Big]^{\frac1q}. \end{eqnarray*} |
Therefore, Proposition 3.2 is also valid for the functional {\mathcal J}^{(1)} .
As a result, we can define the modified flow for (4.2) as in Section 3.3 with A_0 = 10c_0|{\mathbb S}^n|+ 10(n+1)\|f\|_{L^{1}({\mathbb S}^n)} . By using the topological argument as in Section 3.5, we can show the existence of an initial hypersurface {\mathcal M}_0 with {\rm{Cl}}({\mathcal M}_0)\in{\mathcal K}_o such that the evolving hypersurface {\mathcal M}_t = \partial \Omega_t of the flow (4.2) satisfies
\begin{eqnarray*} B_r(0)\subset{\Omega}_t\subset B_{R}(0) \end{eqnarray*} |
for some constants R > r > 0 which are independent of time t . Equivalently, the support function of {\mathcal M}_t , which satisfies the parabolic equation
\begin{equation} {\partial}_tu(x, t) = -\frac{\eta(t)f(x)}{ g({\nabla} u+u x)} \frac{u^p(u^2+|{\nabla} u|^2)^{\frac{n+1-q}{2}}}{\det({\nabla}^2u+uI)}+u, \end{equation} | (4.3) |
enjoys the C^0 -estimates:
\begin{eqnarray*} C_0^{-1}\le u(\cdot, t)\le C_0 . \end{eqnarray*} |
Since (4.3) is of the form (2.8), Theorem 2.1 is valid for (4.3). Therefore, u(\cdot, t) exists for all time t\ge0 and is of C^{3, \alpha} -smooth.
The same argument as in Section 3.6 implies that u(\cdot, t_i) converges to a solution of (4.1) with \lambda = \lim_{t_i\to\infty}\eta(t_i) .
When q\ge n+1 and p < -q , we can show the existence of solutions to (4.1) with \lambda = 1 .
Theorem 4.2. Let f\in C^{1, 1}({\mathbb S}^n) and g\in C^{1, 1}({\mathbb R}^{n+1}) be two positive functions satisfying c_0^{-1}\le f, g\le c_0 for some c_0 > 1 . Suppose q\ge n+1 and p < -q . Then there is a uniformly convex and C^{3, \alpha} function to (4.1) with \lambda = 1 , where \alpha\in (0, 1) .
Proof. Consider the functional
\begin{eqnarray*} {\mathcal J}^{(2)}(\Omega) = -\frac{1}{p} \int_{{\mathbb S}^n} u^p(x)f(x)dx+\int_{{\mathbb S}^n} \int_0^{r(\xi)} g(\tau \xi)\tau^{q-1}d\tau\ d\xi, \end{eqnarray*} |
and the flow
\begin{equation} \frac{{\partial} X}{{\partial} t } (x, t) = -\frac{f(\nu)K(x, t)\langle{X, \nu}\rangle^p}{g(X)}|X|^{n+1-q} \nu+X(x, t). \end{equation} | (4.4) |
Similar calculations as in Lemma 3.1 show that {\mathcal J}^{(2)} is non-decreasing under (4.4).
One can verify that Proposition 3.2 holds for the functional {\mathcal J}^{(2)} .
Step 1: if d = \operatorname{dist}(O, {\partial}{\Omega}) is sufficiently small, then {\mathcal J}^{(2)}({\Omega}) is sufficiently large. We adopt the same notations as in Proposition 3.2. By virtue of (3.3) and (3.4), we obtain
\begin{eqnarray*} [{\mathcal J}^{(2)}({\Omega})]^2\ge C\Big(\int_{{\mathbb S}^n} u^pdx\Big)\Big(\int_{{\mathbb S}^n} r^qd\xi\Big) \ge Ca_1 a_{n+1}^{q-n-1}d^{n-|p|}, \end{eqnarray*} |
where C > 0 depends only on n, p, q and c_0 . As q\ge n+1 and a_1\geq d , we have
\begin{eqnarray*} [{\mathcal J}^{(2)}({\Omega})]^2\ge Cd^{q-|p|}. \end{eqnarray*} |
This shows that if d is tiny, then {\mathcal J}^{(2)}({\Omega}) is huge.
Step 2: if {\rm{Vol}}({\Omega}) or [{\rm{Vol}}({\Omega})]^{-1} is sufficiently small, then {\mathcal J}^{(2)}({\Omega}) is sufficiently large. Since {\rm{Vol}}({\Omega}) being small implies that d is tiny, we are done by Step 1 in this case. Suppose that {\rm{Vol}}({\Omega}) is huge. The conclusion then follows from
\begin{equation} {\mathcal J}^{(2)}({\Omega})\ge C \int_{{\mathbb S}^n}r^q d\xi\ge C[{\rm{Vol}}({\Omega})]^{\frac{q}{n+1}}. \end{equation} | (4.5) |
Step 3: if e_{\Omega} is sufficiently large, then so is {\mathcal J}^{(2)}({\Omega}) . By Step 1, we assume without loss of generality that d is bounded from below by a constant C > 0 depending on n, p, q and c_0 . Using (4.5),
\begin{eqnarray*} [{\mathcal J}^{(2)}({\Omega})]^{\frac{n+1}{q}}\ge C{\rm{Vol}}({\Omega})\ge C e_{\Omega} d^{n+1}\ge Ce_{\Omega}. \end{eqnarray*} |
This proves Step 3.
To complete the proof, we introduce a modified flow of (4.4) and use the topological argument as in Section 3 to find the needed initial hypersurface {\mathcal N} , such that the evolving hypersurfaces {\mathcal M}_t with {\mathcal M}_0 = {\mathcal N} satisfy B_r(0)\subset{\Omega}_t\subset B_{R}(0) for some constants R > r > 0 . Since the support function u(\cdot, t) of (4.4) satisfies a parabolic equation of the form (2.8), the higher order derivative estimates follow by Theorem 2.1. The remaining proof follows exactly as that of Theorem 1.2. Since (4.4) does not contain an integral term like \eta(t) in (4.2), along a sequence of times \{t_i\}_{i = 1}^\infty , u(\cdot, t_i) converges to a solution of (4.1) with \lambda = 1 .
The first author was supported by ARC DE210100535. The second author was supported by NSFC12031017 and the Fundamental Research Funds for the Central Universities (No. 226-2022- 00157). The third author was supported by ARC DP200101084.
The authors declare no conflict of interest.
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