On mean curvature flow with driving force starting as singular initial hypersurface
Ryunosuke Mori Longjie Zhang
math.DG












On mean curvature flow with driving force starting as singular initial hypersurface
    Ryunosuke Mori ^1, Longjie ZHANG ^2
    September, 2016

^1 
Corresponding author University:Graduate School of Mathematical Sciences, The University of Tokyo. Address:3-8-1 Komaba Meguro-ku Tokyo 153-8914, Japan. Email:moriryu@ms.u-tokyo.ac.jp, 45c136045@gmail.com

^2 
Corresponding author University:Graduate School of Mathematical Sciences, The University of Tokyo. Address:3-8-1 Komaba Meguro-ku Tokyo 153-8914, Japan. Email:zhanglj@ms.u-tokyo.ac.jp, zhanglj919@gmail.com


The second author is the Research Fellow of Japan Society for the Promotion of Science, Number: 17J05160.
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140mm

Abstract: We consider an axisymmetric closed hypersurface evolving by its mean curvature with driving force under singular initial hypersurface. We study this problem by level set method. We give some criteria to judge whether the interface evolution is fattening or non-fattening. 

Keywords and phrases: mean curvature flow; driving force; level set method; singularity; fattening. 

2010MSC: 35A01, 35A02, 35K55, 53C44.




    






§ INTRODUCTION




 This paper studies the mean curvature flow with driving force under singular initial data

    V=-κ+A  on Γ(t)⊂ℝ^n+1 ,


    Γ(0)=Γ_0,

where Γ(t) is a smooth family of hypersurfaces embedded in ℝ^n+1, V is the outer normal velocity of Γ(t), κ is the mean curvature of Γ(t) and A>0, called driving force, is a constant. Here the sign of κ is taken so that the problem is parabolic. For example, under the definition, the mean curvature of unit sphere in ℝ^n+1 is n.



In this paper we consider that the initial data Γ_0 is smooth except for origin and can be wrriten into 

    Γ_0={(x,y)∈ℝ×ℝ^n| |y|=u_0(x), -b_0≤ x≤ b_0},

where u_0(x) is even and satisfies u_0(x)>0, for x∈ (-b_0,0)∪ (0,b_0), u_0(0)=u_0(-b_0)=u_0(b_0)=0. Since Γ_0 is smooth at (-b,0,⋯,0) and (b,0,⋯,0), it is easy to see that u_0^'(-b_0)=-u_0^'(b_0)=+∞.

Let 

    γ:=lim_x→ 0^+arctan u_0^'(x)∈[0,π/2],

seeing Figure <ref>.

Main assumptions for γ=π/2.
Under the condition γ=π/2, let Λ_0=Γ_0∩{(x,y)∈ℝ×ℝ^n| x≥0}.

We consider another problem.

    V=-κ+A  on Λ^+(t)⊂ℝ^n+1,<ref>*


    Λ^+(0)=Λ_0<ref>*

(Figure <ref>).

We consider this problem by level set method. By the theory in <cit.>, there exists a unique viscosity solution ϕ of the following level set equation

    {[ ϕ_t=|∇ϕ|(∇ϕ/|∇ϕ|)+A|∇ϕ| in ℝ^n+1×(0,T),;                      ϕ(x,y,0)=a_1(x,y), ].

where a_1(x,y) satisfies Λ_0={(x,y)| a_1(x,y)=0} and {(x,y)| a_1(x,y)>0} is bounded. The results in appendix show that the zero set of ϕ is not fattening in a short time. Indeed, thanks to Theorem <ref>, the zero set of ϕ can be written into

    Λ^+(t)={(x,y)|ϕ(x,y,t)=0}={(x,y)∈ℝ^n+1| |y|=v(x,t), a_*(t)≤ x≤ b_*(t)}, 0<t<T_*.
 
Moreover, (v,a_*,b_*) is the solution of the following free boundary problem

    {[ u_t=u_xx/1+u_x^2-n-1/u+A√(1+u_x^2), x∈(a_*(t),b_*(t)), 0<t< T_*,;                         u(a_*(t),t)=0, u(b_*(t),t)=0, 0≤ t< T_*,;                     u_x(a_*(t),t)=∞, u_x(b_*(t),t)=-∞, 0≤ t<T_*,;                                        u(x,0)=u_0(x), 0≤ x≤ b_0,;                           u(x,t)>0, x∈(a_*(t),b_*(t)), 0<t< T_*. ].*

In this paper, a_* and b_* are called the end points of Λ^+(t).

Assumption (A+): There exists δ>0 such that a_*(t)≥0 for 0≤ t<δ.

Assumption (A-): There exists δ>0 such that a_*(t)<0 for 0<t<δ.

Here we give some sufficient conditions such that main assumptions hold.

If κ(O)<A, then, since 

    a_*^'(0)=κ(O)-A<0,

a_*(t)<0 for any small t>0. Similarly, if κ(O)>A, a(t)>0 for any small t>0. Here 

    κ(O)=lim_x→ 0^+(-u_xx/(1+u_x^2)^3/2+n-1/u√(1+u_x^2))

denotes the mean curvature of Λ_0 at the origin O. 



Here we present our main results.


Let Γ_0 and γ be defined by (<ref>) and (<ref>).


Assume γ=π/2, n≥2.


(1) If the assumption (A-) holds, then there exists T>0 such that the interface evolution Γ(t) for (<ref>) with initial curve Γ_0 is not fattening for 0≤ t<T.

(2) If the assumption (A+) holds, then the interface evolution Γ(t) for (<ref>) with initial data Γ_0 is fattening.



Let Γ_0 and γ be defined by (<ref>) and (<ref>).

Then there exist α_n∈(0,π/2) (n≥2) such that α_n→π/2, as n→∞ and if 0≤γ<α_n, then there exists T_γ such that the interface evolution Γ(t) for (<ref>) with initial curve Γ_0 is not fattening for 0≤ t<T_γ.



Let Γ_0 and γ be defined by (<ref>) and (<ref>).

Assume 0≤γ<π/2, for n=1.

The interface evolution Γ(t) for (<ref>) in ℝ^2 with initial data Γ_0 is fattening.


The definitions of fattening, non-fattening, outer-evolution, inner-evolution and interface evolution are given in section 2.

Theorem <ref> can be explained by Figure <ref> and Figure <ref>. φ in Figure <ref> and Figure <ref> is the unique viscosity solution of 

    {[ φ_t=|∇φ|(∇φ/|∇φ|)+A|∇φ| in ℝ^n+1×(0,T),;       φ(x,y,0)=a_2(x,y) in ℝ^n+1×(0,T), ].

where a_2(x,y) satisfies Γ_0={(x,y)∈ℝ^n+1| a_2(x,y)=0} and {(x,y)∈ℝ^n+1| a_2(x,y)>0} is bounded. Let Γ(t)={(x,y)∈ℝ^n+1|φ(x,y,t)=0}.








Motivation. Recently, <cit.> has considered the mean curvature flow with driving force starting as singular initial curve in the plane. The author gets the same results as in Theorem <ref> under the condition n=1 (see <cit.>). In this paper, we give some criteria to judge whether the interface evolution starting as singular hypersurface is fattening or non-fattening in higher dimension. Combining the results in <cit.>, we can conclude the results as the following tables. In the following tables, “Connected” means that the evolution is a connected set and “Separated” means that the evolution consists of two disjoint components.

  

  

  


The role of a_*(t) in Theorem <ref>. In Section 5, we can prove that there exists a unique solution (u,b)  of the following free boundary problem 

    u_t=u_xx/1+u_x^2-n-1/u+A√(1+u_x^2), -b(t)<x<b(t), 0<t<T_1,


    u(b(t),t)=u(-b(t),t)=0, u_x(-b(t),t)=-u_x(b(t),t)=∞, 0<t<T_1,


    u(x,0)=u_0(x), -b_0≤ x≤ b_0,


    u(x,t)>0, -b(t)< x< b(t), 0<t<T_1.

Precisely, we say (u,b) is the solution of (<ref>), (<ref>), (<ref>) and (<ref>), if

(1) b(t) is a positive function and b∈ C([0,T_1))∩ C^1((0,T_1)).

(2) u∈ C(D_T_1)∩ C^2,1(D_T_1), where D_T_1=∪_0≤ t<T_1([-b(t),b(t)]×{t}) and D_T_1=∪_0<t<T_1((-b(t),b(t))×{t}) (We must note that D_T_1≠D_T_1).

(3) (u,b) satisfies (<ref>), (<ref>), (<ref>) and (<ref>).

 Obviously, the flow Γ^*(t)={(x,y)| |y|=u(x,t), -b(t)≤ x≤ b(t)} satisfies (<ref>), (<ref>) naturally.

Let (v,a_*,b_*) be the solution of the problem (*). If the assumption (A+) holds, the flow

    Λ^+(t)={(x,y)||y|=v(x,t), a_*(t)≤ x≤ b_*(t)}
 
does not intersect the flow

    Λ^-(t)={(-x,y)| (x,y)∈Λ^+(t)},
  
for 0<t<δ. Denote Λ(t)=Λ^+(t)∪Λ^-(t). Obviously, Λ(t) also satisfies (<ref>). Seeing Γ^*(0)=Λ(0)=Γ_0, this means that there exist two types of flows Γ^*(t) and Λ(t) evolving by V=-κ+A with the same initial curve Γ_0. Therefore under this condition, the solution of the original problem (<ref>), (<ref>) is not unique. Indeed, from the proof of Theorem <ref>, we see that the flow Γ^*(t) is the boundary of the closed evolution and the flow Λ(t) is the boundary of open evolution.

If a_*(t)<0 for 0<t<δ, Λ^+(t)∩Λ^-(t)≠∅. Obviously, Λ(t)=Λ^+(t)∪Λ^-(t) does not satisfy (<ref>). But Λ^+(t) plays the role of a sub-solution (in the proof of Lemma <ref>). Using this sub-solution, the boundaries of the open evolution and closed evolution are away from the x-axis. Moreover, it can be proved that the derivatives and the second fundamental forms of them are uniformly bounded. By the uniqueness result  (Proposition <ref>), we can prove they are coincide. 

For classical mean curvature flow i.e. A=0 and under the condition γ=π/2, since a_*(t)≥0 always holds, the interface evolution is fattening.  

Background. 
In 1995, <cit.> considered the classical mean curvature flow in dimension n, n≥2. They proved that the singular formations for axisymmetric flow can only be shrinking or pinching. Moreover, they used level set method to show that after pinching, the interface evolution is non-fattening and separated into some disjoint connected components. Indeed, this result can be seen as a special condition A=0 and γ=0 in this paper. 
  
Mean curvature with driving force under the condition γ=π/2 and n=1, the curve in plane, has been considered in <cit.> recently. The same results as in Theorem <ref> are given in <cit.>. In this paper, we give more general criteria to judge whether the interface evolution starting as singular initial hypersurface is fattening or non-fattening. 

A short review for mean curvature flow. For the classical mean curvature flow: A=0 in (<ref>), there are many results. Concerning this problem, Huisken <cit.> showed that any solution that starts out as a compact, smooth and convex surface preserves so until it shrinks to a "round point", its asymptotic shape is a sphere just before it disappears. He prove this result for hypersurfaces in ℝ^n+1 with n≥2, while Gage and Hamilton <cit.> showed that it still holds when n=1, the curves in the plane. Gage and Hamilton also showed that an embedded curve remains embedded, i.e. the curve will not intersect itself. Grayson <cit.> proved the remarkable fact that such family must become convex eventually. Thus, any embedded curve in the plane will shrink to "round point" under mean curvature flow. However in higher dimensions this is not true. Grayson <cit.> also showed that there exist smooth solutions that become singular before they shrink to a point. His example consists of a barbell: two spherical surfaces connected by a sufficiently thin "neck". In this example, the inward curvature of the neck is so large that it will force the neck to pinch before shrinking. In 1995, A. Altschuler, S. B. Angenent and Y. Giga <cit.> studied the global-in-time solutions whose initial value is a compact, smooth, rotationally symmetric hypersurface given by rotating a graph around an axis by level set method. They proved the hypersurface will separate into two smooth hypersurfaces after pinching. Indeed, the condition in <cit.> is the condition A=0 and γ=0 in this paper.

Main method.  Since the initial hypersurface is singular at the origin, we use the level set method to study the problem. The level set method will be introduced in Section 2. For the level set method, <cit.> gives the existence and uniqueness for the viscosity solution of the level set equation. However, in spite of the development of the level set method, it is still difficult to determine whether fattening occurs or not.

Another important tool is the intersection number principle. It was also used in <cit.> that the intersection number between two families evolving by mean curvature flow is non-increase. But for the problem with driving force this number may increase in time. The intersection number between two flows evolving by V=-κ+A has been investigated by <cit.>. We will give the references in Section 3.



The rest of this paper is organized as follows. In Section 2, we refer to the level set method by using viscosity solution which is established by <cit.>. We also refer to the definitions of open evolution, closed evolution and the basic knowledge concluding comparison principle, monotone convergence theorem and so on. In Section 3, we give some preliminary results including interior gradient estimate, modifying intersection number principle and its application. In Section 4, we prove the main results. In Section 5, we give an example for the phenomenon “second fattening”. In Section 6, we give the local existence and uniqueness of the problem (<ref>) with some initial hypersurfaces.










§ VISCOSITY SOLUTION AND LEVEL SET METHOD


Since the initial curve Γ_0 has singularity at (0,0), the equation V=-κ+A does not make sense at t=0. Therefore, we apply the level set method to our problem. In this section, we refer to the level set method in ℝ^N. Let Γ(t) be a smooth family of smooth, closed, compact hypersurfaces in ℝ^N given by Γ(t)={x|ψ(x,t)=0,x∈ℝ^N} for some ψ and {x|ψ(x,t)>0} is bounded. If Γ(t) evolves by (<ref>), we can see that ψ(x,t) satisfies 

    ψ_t=|∇ψ |(∇ψ/|∇ψ|)+A|∇ψ| on {(x,t)|ψ(x,t)=0}.

Next we consider the equation in whole space

    ψ_t=|∇ψ |(∇ψ/|∇ψ|)+A|∇ψ| in ℝ^N×(0,T).

Equation (<ref>) is called the level set equation of (<ref>). Theorem 4.3.1 in <cit.> gives the existence and uniqueness of the viscosity solution for (<ref>) with ψ(x,0)=ψ_0(x). Here ψ_0(x) is a bounded and uniform continuous function.


Level set method. Using the solution of level set equation, we introduce the level set method. 
 (1) Let D_0 be a bounded open set in ℝ^N. A family of  open sets {D(t)}_0<t<T in ℝ^N is called an (generlized) “open” or “inner” evolution of (<ref>) with initial data D_0 if there exists a viscosity solution ψ of (<ref>) that satisfies

    D(t)={x∈ℝ^N |ψ(x,t)>0}, D_0={x∈ℝ^N|ψ(x,0)>0}.


(2) Let E_0 be a bounded closed set in ℝ^N. A family of closed sets {E(t)}_0<t<T in ℝ^N is called a (generlized) “closed” or “outer” evolution of (<ref>) with initial data E_0 if there exists a viscosity solution ψ of  (<ref>) that satisfies

    E(t)={x∈ℝ^N |ψ(x,t)≥0}, E_0={x∈ℝ^N|ψ(x,0)≥0}.


The set Γ(t)=E(t)∖ D(t) is called an (generalized) interface evolution of  (<ref>) with initial data Γ_0=E_0∖ D_0.


 (1) For an open set D_0 and E_0=D_0, we often choose

    ψ(x,0)=max{sd(x,∂ D_0),-1}

where

    sd(x,∂ D_0)={[  dist(x,∂ D_0),  x∈ D_0,; -dist(x,∂ D_0),  x∉ D_0. ].


(2) Although the choice of ψ(x,0) is not unique, Theorem 4.2.8 in <cit.> implies that both the open evolution D(t) and closed evolution E(t) are independent of the choice of ψ(x,0).

(3) In general, even if E_0=D_0, we can not guarantee E(t)=D(t). If E(t)∖ D(t) has interior points for some t, we call the interface evolution is fattening. Respectively, if E(t)=D(t) for all 0<t<T, we say the interface evolution is not fattening. Therefore, for proving non-fattening, it is sufficient to show ∂ D(t)=∂ E(t). For proving fattening, it is sufficient to show that there exists a ball B such that B⊂ E(t)∖ D(t).

(4) If D_0 and D_0 are symmetric to x_i-axis, then it is also true for D(t) and E(t). Since level set equation (<ref>) is invariance under orthogonal transformation.


We now list some fundamental properties of open evolution and closed evolution of  (<ref>). (All the results listed below can be found in Chapter 4 of <cit.>)


(Semigroups <cit.>). Let N(t) and M(t) be the operators such that
N(t)D_0=D(t) and M(t)E_0=E(t) for t>0, where {D(t)}, {E(t)} are the open evolution and closed evolution with initial data D_0 and E_0,respectively. Then we have N(t)D(s)=D(t+s) and M(t)E(s)=E(t+s), for any t>0, s>0. 


(Order preserving property <cit.>). Let D_0, D_0^' be two open sets in ℝ^N and let E_0, E_0^' be two closed sets in ℝ^N and let N(t), M(t) be the operators defined in Theorem <ref>. Then

(1) N(t)D_0⊂ U(t)D_0^', if D_0⊂ D_0^';

(2) M(t)E_0⊂ M(t)E_0^', if E_0⊂ E_0^';

(3) N(t)D_0⊂ M(t)E_0^', if D_0⊂ E_0^';

(4) E_0⊂ D_0 and dist(E_0,∂ D_0)>0, then M(t)E_0⊂ N(t)D_0.


(Monotone convergence <cit.>).

(1) Let D(t) and {D_j(t)} be open evolutions with initial data D_0 and D_j0 respectively. If D_j0↑ D_0, then D_j(t)↑ D(t) for t>0, i.e., ⋃_j≥1D_j(t)=D(t);

(2) Let E(t) and {E_j(t)} be closed evolutions with initial data E_0 and E_j0 respectively. If E_j0↓ E_0, then E_j(t)↓ E(t) for t>0, i.e., ⋂_j≥1E_j(t)=E(t).


 Let {D_1(t)} and {D_2(t)} be the open evolution of V=-κ+A.  For A>0, even if D_1(0) and D_2(0) are disjoint, D_1(t) and D_2(t) may have intersections. The basic reason is that the level set equation (<ref>) is not orientation free (It is not true that -u is also a solution for (<ref>) when u is a solution for (<ref>)). 


(Separate)
Let {D_1(t)}_0≤ t<T be the open evolution of V=-κ+A and {D_2(t)}_0≤ t<T be the open evolution of V=-κ-A. If D_1(0)∩ D_2(0)=∅, then D_1(t)∩ D_2(t)=∅ for 0≤ t<T. 


The proof of this theorem is similar to that of Theorem 3.5 in <cit.>. We omit it.

 In order to prove the fattening result, we need the following lemma. 

(Lemma 2.8 in <cit.>) Assume that D_1(t) and D_2(t) are the open evolution of (<ref>) with D_1(0)=U_1 and D_2(0)=U_2, respectively. And D(t) is denoted as the open evolution of (<ref>) with D(0)=U_1∪ U_2. If D_1(t)∩ D_2(t)=∅ for 0≤ t≤ T, then  D(t)=D_1(t)∪ D_2(t) for 0≤ t≤ T.


(Local smoothness for graphs) Suppose that ψ is a viscosity solution of (<ref>). Assume in an open region U×(t_1,t_2),

    {(x,t)|ψ=0}∩ U={(x,t)| x_N=g(x^',t), x^'∈ U^'},

where x^'=(x_1,⋯,x_N-1), U^'=U∩{x_N=0} and g is continuous in U^'×(t_1,t_2). Then the function g is a viscosity solution of 

    g_t=(δ_ij-g_x_ig_x_j/1+|∇ g|^2)g_x_ix_j+ A√(1+|∇ g|^2),

or 

    g_t=(δ_ij-g_x_ig_x_j/1+|∇ g|^2)g_x_ix_j-A√(1+|∇ g|^2).

If the direction of the driving force of {(x,t)|ψ=0}∩ U is upward (downward), we choose “+” (“-”) in above graph equation.

Moreover, g is C^∞ in the region U^'×(t_1,t_2).

The proof is similar to that in <cit.> (Theorem 5.1 and Theorem 5.4), and we omit it.



§ PRELIMINARIES


In this section, we prepare some preliminaries for future purposes. The results in this section can be found in <cit.>.

Graph equation. Let u(x,t) be a function on an open subset of ℝ^n×ℝ, then the graph of u(x,t) is a family of hypersurfaces in ℝ^n+1. The family of hypersurfaces moves by V=-κ+A if and only if 

    u_t=(δ_ij-u_x_iu_x_j/1+|∇ u|^2)u_x_ix_j± A√(1+|∇ u|^2),

where the signs of the last terms are determined by the direction of the driving force.

Horizontal and vertical graph equation. If Γ(t) is a family of rotationally symmetric hypersurfaces in ℝ^n+1, then for each t>0, a part of Γ(t) may be represented either as a horizontal graph, r=u(x,t), or a vertical graph, x=v(r,t), where r=√(y_1^2+y_2^2+⋯+y_n^2), (x,y)∈ℝ×ℝ^n.

If Γ(t) is given as a horizontal graph, then Γ(t) evolves by V=-κ+A in ℝ^n+1 and the direction of the driving force points to the positive direction of r=|y| axis if and only if u satisfies the horizontal graph equation 

    ∂ u/∂ t=u_xx/1+u_x^2-n-1/u+A√(1+u_x^2).

If Γ(t) is given as a vertical graph, then Γ(t) evolves by V=-κ+A in ℝ^n+1 if and only if v satisfies the vertical graph equation

    ∂ v/∂ t=v_rr/1+v_r^2+n-1/rv_r+ A√(1+v_r^2),

or

    ∂ v/∂ t=v_rr/1+v_r^2+n-1/rv_r-A√(1+v_r^2),

where the sign of the last term is determined by the direction of the driving force (We choose “+(-)” when the direction of the driving force is the positive (negative) direction of x axis).

(Theorem 3.1 in <cit.>) For u∈ C^3(Ω_T)∩ C^0 (Ω_T), u satisfies

    u_t=(δ_ij-u_x_iu_x_j/1+|∇ u|^2)u_x_ix_j± A√(1+|∇ u|^2),

For the condition “+”(“-”), we assume u<0 (u>0) in Ω_T, u(0,T)=-v_0 (u(0,T)=v_0). Then

    |∇ u(0,T)|≤ (3+16v_0)e^2K,

where K=20v_0^2(4n+1/T+4A+A/2v_0)+2, Ω_T=B_1(0)× (0, 2T)⊂ℝ^n×(0,∞) and

    δ_ij={[  1, i=j,; 0, i≠ j. ].



 (1) In Theorem <ref>, Ω_T can be replaced by Ω_T=B_R(x_0)×(0,2T) and v_0=u(x_0,T). Then the conclusion becomes

    |∇ u(x_0,T)|≤ e^2K(3+16v_0/R),

where K=20v_0^2/R^2(4n+R^2/T+4A/R+A/2v_0)+2. Here We set v(x,t)=u(Rx+x_0,R^2t)/R, and apply Theorem <ref> to v(x,t).

(2) When u is the solution of (<ref>) for “+” without the assumption “u<0”, if we set

    v=u-M-ϵ

where M=sup_Ω_T |u| and ϵ>0. and apply (1) in Remark <ref> to v, we get

    |∇ u(0,T)|≤(3+16M-u(0,T)+ϵ/R)e^2K_ϵ,

where K_ϵ=20(M-u(0,T)+ϵ)^2/R^2(4n+R^2/T+4A/R
+A/2(M+ϵ-u(0,T)))+2.

As ϵ→0, we have

    |∇ u(0,T)|≤(3+32M/R)e^2K,

where K=80M^2/R^2(4n+R^2/T+4A/R)
+20AM/R^2+2.


Then we use the (2) in Remark <ref> and the same method as in <cit.> to prove the next corollary.

 For s_1<s_2, ρ>0 and x_0∈ℝ^n we set

    Ω=B_ρ(x_0)×(s_1,s_2).

Suppose that u∈ C^3(Ω) solves the equation (<ref>) in Ω with M=sup_Ω|u|<∞. For any ϵ>0 there is a constant C=C(M,ϵ,n) such that

    |∇ u|≤ C  on Ω_ϵ=B_ρ-ϵ(x_0)×(s_1+ϵ^2,s_2).



(1) From Corollary <ref> and <cit.>, there exist C_k(M,ϵ,n) such that 

    |∇^k u|≤ C_k, (x,t)∈ B_ρ-2ϵ(x_0)×(s_1+2ϵ^2,s_2).

(2) Noting that C and C_k are all independent of s_2, if the solution u exists for all t>s_1, then s_2 can be taken as ∞ in Corollary <ref>.






§ INTERSECTION NUMBER PRINCIPLE


Next, we begin to introduce the modifying intersection number principle.

Intersection number for rotationally symmetric hypersurfaces. Let r=|y|, y∈ℝ^n. For two rotationally symmetric hypersurfaces Γ_1(t) and Γ_2(t), given by Γ_1(t)={(x,y)∈ℝ×ℝ^n| r=u_1(x,t)} and Γ_2(t)={(x,y)∈ℝ×ℝ^n| r=u_2(x,t)}, define the number of intersections between u_1(·,t) and u_2(·,t) as the intersection number between Γ_1(t) and Γ_2(t) denoted by 𝒵[Γ_1(t),Γ_2(t)]. 

It is well known that the intersection number between two families of rotationally symmetric hypersurfaces Γ_1(t) and Γ_2(t) evolving by V=-κ is not increasing (<cit.>). But it is not true for V=-κ+A. To conquer this difficulty, in <cit.>, they give some results for the intersection number in the case V=-κ+A. 

(Remark 4.4 in <cit.>)
Let Γ_1(t) and Γ_2(t) be two closed, compact, rotationally symmetric hypersurfaces given by Γ_1(t)={(x,y)∈ℝ×ℝ^n| r=u_1(x,t),a_1(t)≤ x≤ b_1(t)}, Γ_2(t)={(x,y)∈ℝ×ℝ^n| r=u_2(x,t), a_2(t)≤ x≤ b_2(t)}. If Γ_i(t) evolve by V=-κ+A in ℝ^n+1, i=1,2. Then 


(a). 𝒵(Γ_1(t),Γ_2(t))>0 does not increase in t when t satisfies 𝒵(Γ_1(t),Γ_2(t))>0.


(b). If 𝒵(Γ_1(t_0),Γ_2(t_0))=0, then 𝒵(Γ_1(t),Γ_2(t))≤ 1 for t_0<t<T. 


Using Theorem <ref>, we can prove following Theorem <ref> and Theorem <ref>.

(Theorem 4.2 in <cit.>) Γ(t)={(x,y)∈ℝ^n+1| r=u(x,t),a_2(t)≤ x≤ b_2(t)} is a smooth family of closed, smooth hypersurfaces in ℝ^n+1 for 0<t<T. If Γ(t) evolves by V=-κ+A in ℝ^n+1,  there is a function σ: ℝ_+×ℝ_+→ℝ such that

    |u_x(x,t)|≤σ(t,u(x,t))

holds for 0<t<T, a_2(t)<x<b_2(t). The function σ depends only on M=max_a_2(0)<x<b_2(0) u(x,0) and T.


(Theorem 4.5 in <cit.>) Let Γ(t), t∈ [0,T), be a family of smooth hypersurfaces evolving by V=-κ+A in ℝ^n+1. If Γ(0) is obtained by rotating the graph of a function around the x-axis, Γ(t) also has the same symmetry as Γ(0) for t∈[0,T).



In our problem, the hypersurface evolving by V=-κ+A maybe intersect itself at x-axis. To conquer this difficulty, we refer to the definition of α-domain in <cit.>.

 We say a domain U is an α-domain if

(1). U⊂ℝ^n+1 is an open set of the form

    U={(x,y)∈ℝ×ℝ^n| r<u(x)}.


(2). I={x∈ℝ| u(x)>0 } is a bounded, connected interval. Let the endpoints of I be a_1<a_2.

(3). u is smooth on I;

(4). ∂ U intersects each cylinder ∂ C_ρ with 0<ρ≤α twice and these intersections are transverse, where C_ρ={(x,y)∈ℝ^n+1| r<ρ}.



We observe that the boundary ∂ U of an α-domain U does not intersect itself at y=0. The condition (3) implies ∂ U is a smooth curve, except possibly at its endpoints (a_1,0),(a_2,0). The condition (4) implies that there exist δ_1,δ_2>0 such that

    u(a_1+δ_1)=u(a_2-δ_2)=α,

and

    u^'(x)={[ >0, x∈(a_1,a_1+δ_1],; <0, x∈[a_2-δ_2,a_2). ].

Therefore, the inverse of u|_[a_1,a_1+δ_1] and u|_[a_2-δ_2,a_2] exist, denoted by v_1, v_2:[0,α]→ℝ. By the implicit function theorem, they are smooth in (0,α]. Moreover, v_1^'(r)>0, v_2^'(r)<0, (0<r≤α) and

    ∂ U∩ C_α={(x,y)∈ℝ^n+1|0≤ r≤α, x=v_i(r), i=1,2}.

The two components of ∂ U∩ C_α are called the left and right caps of ∂ U.

Since U is an α-domain, by Theorem <ref>, there exists T_U such that ∂ D(t) is smooth and D(t)={(x,y)∈ℝ^2| |y|<u(x,t), a(t)< x < b(t)} for 0<t<T_U. Here (u,a,b) satisfies

    u_t=u_xx/1+u_x^2-n-1/u+A√(1+u_x^2), a(t)<x<b(t), 0<t<T_U,


    u(b(t),t)=u(a(t),t)=0, u_x(a(t),t)=-u_x(b(t),t)=∞, 0<t<T_U,


    u(x,t)>0, a(t)< x< b(t), 0<t<T_U.
 


For n≥2, there exists a t_U^α>0 such that D(t) is an α(t)-domain for all 0<t<min{t_U^α,T_U}, where α(t) is the solution of the following equation

    α^'(t)=A-n-1/α(t)

with initial data α(0)=α.


For the problem,

    {[ α^'(t)=A-n-1/α(t), t>0,;                 α(0)=α, ].

the following hold.

(1) when α<(n-1)/A, there exists T_α<∞ such that α(t)↓ 0 as t→ T_α and lim_t→ -∞α(t)=(n-1)/A; 

(2) when α=(n-1)/A, α(t)=(n-1)/A for 0≤ t<∞;

(3) when α>(n-1)/A, α(t)↑∞ as t→∞ and lim_t→ -∞α(t)=(n-1)/A. 

 Since U is not contained in the cylinder C_α, there is a small ball B_ϵ(P)⊂ U∖C_α. By (1) in Theorem <ref>, D(t) contains the ball B_ϵ(t)(P) for 0<t<δ_1, where ϵ(t) is the solution of the following equation

    ϵ^'(t)=-A-n/ϵ(t), 0<t<δ_1

with initial data ϵ(0)=ϵ and δ_1 is the maximal time for the solution existing.

Let δ_2 denote the maximal existence time of the solution for equation (<ref>) with initial data α(0)=α.

Theorem <ref> implies B_ϵ(t)(P)∩C_α(t)=∅ for 0<t<t_U^α. Here

    t_U^α=min{δ_1,δ_2}.


Fix 0<ρ<α(t_0) and 0<t_0<min{t_U^α,T_U}, and let ρ(t) be the solution of the equation (<ref>) with initial data ρ(0)=ρ. Next we prove that ∂ C_ρ intersects ∂ D(t_0) at most twice.

By comparison principle for ordinary differential equation, ρ(-t_0)<α holds. Therefore, y=ρ(-t_0) intersects y=u(x,0) only twice. Noting ρ(t-t_0)>0 for t≥ 0, 

    y=ρ(t-t_0)>u(a(t),t)=0, y=ρ(t-t_0)>u(b(t),t)=0 for t>0.

By Theorem D in <cit.>, 𝒵_[a(t),b(t)](ρ(t-t_0),u(·,t)) is not increasing for t>0. Here 𝒵_I(u_1(·,t),u_2(·,t)) denotes the intersection number between u_1 and u_2 in I.  

Therefore,  

    𝒵_[a(t),b(t)](ρ(t-t_0),u(·,t))≤𝒵_[a(0),b(0)](ρ(-t_0),u(·,0))=2.


This means that ∂ C_ρ intersects ∂ D(t_0) at most twice.

On the other hand, by the choice of t_U^α, D(t) contains the ball B_ϵ(t)(P). Since B_ϵ(t)(P) lies outside of the cylinder C_α(t), each ∂ C_ρ must intersect ∂ D(t) at least twice for any 0<t<min{t_U^α,T_U}.

Therefore ∂ C_ρ intersects ∂ D(t_0) exactly twice. Consequently, D(t) is an α(t)-domain for all 0<t<min{t_U^α,T_U}.


 For t_U^α and T_U given in Lemma <ref> holds, t_U^α≤ T_U.


To prove this proposition, we need the following lemma.

 Assume that D(t)={(x,y)| |y|<u(x,t), a(t)≤ x≤ b(t)} is a ρ-domain for 0<t<T. Let  w_1<w_2 such that

    C_ρ∩∂ D(t)={(x,y)| x=w_1(y,t) or x=w_2(y,t)}.

 Then

    lim_t→ Tw_1(y,t)=w_1(y,T)     and    lim_t→ Tw_2(y,t)=w_2(y,T)

exist and these convergences are uniformly convergent for |y|≤ρ/2. Moreover, v_1(r_1,T)<v_1(r_2,T) and v_2(r_1,T)>v_2(r_2,T) for 0<r_1<r_2<ρ/2,
where v_1(r,t)=w_1(y,t) and v_2(r,t)=w_2(y,t).

 w_1(y,t) and w_2(y,t) satisfy the equation (<ref>), respectively for "∓". We only prove the result for w_1(y,t). Since w_1 is uniformly bounded, Corollary <ref> and Remark <ref> imply that derivatives ∇^j_y w_1, j=1,2, are uniformly bounded for 0≤|y|≤ρ/2, T/2≤ t<T. Consequently, ∂ w_1/∂ t is bounded for 0≤|y|≤ρ/2, T/2≤ t<T. So there exists w_1(y,T) such that w_1(y,t) converges to w_1(y,T) uniformly for 0≤|y|≤ρ/2, as t→ T.

Note that the following hold

    ∂ v_1/∂ r(ρ/2,t)>0, ∂ v_1/∂ r(0,t)=0 for 0<t<T
 
and 

    ∂ v_1/∂ r(r,0)>0 for 0< r<ρ/2.

In fact, differentiating (<ref>) in r, 

    p_t=a(r,t)p_rr+b(r,t)p_r+c(r,t)p,

where p=w_r, a(r,t)=1/(1+w_r^2), b(r,t)=-2w_rw_rr/(1+w_r^2)^2+(n-1)/r-Aw_r/√(1+w_r^2), c(r,t)=-(n-1)/r^2. Strong maximum principle implies
 
    ∂ v_1/∂ r>0 for  0<r<ρ/2, 0<t≤ T.
 
Therefore v_1(r_1,T)<v_1(r_2,T) for 0<r_1<r_2<ρ/2. Similarly, we can prove the conclusion for v_2.


 If T_U<t_U^α. By Lemma <ref>, there exists ρ>0 such that D(t) is a ρ-domain for 0<t<T_U. 

We divide ∂ D(t) into two parts: ∂ D(t)=(∂ D(t)∩{r< ρ/2})∪(∂ D(t)∩{r≥ρ/2}).


Step 1. ∂ D(t)∩{r< ρ/2}

 Since ∂ D(t) is a ρ-domain, there exist w_1<w_2 such that ∂ D(t)∩{r<ρ}={(x,y)| x=w_1(y,t), |y|<ρ}∪{(x,y)| x=w_2(y,t), |y|<ρ}. By the same argument as in Lemma <ref>, ∇^j_y w_i, j=1,2, i=1,2, are uniformly bounded for 0≤|y|≤ρ/2, T_U/2≤ t<T_U. Therefore, the mean curvature of ∂ D(t)∩{r< ρ/2} is uniformly bounded for T_U/2≤ t<T_U. 


Step 2. ∂ D(t)∩{r≥ρ/2}

Recalling ∂ D(t)={(x,y)| |y|=u(x,t),a(t)≤ x≤ b(t)}, by Lemma <ref>, v_1(ρ/4,T_U)<v_1(ρ/2,T_U) and v_2(ρ/4,T_U)>v_2(ρ/2,T_U). Then for any sufficiently small ϵ>0 for all t close to T_U there holds

    [v_1(ρ/2,t),v_2(ρ/2,t)]⊂ (v_1(ρ/4,t)+ϵ,v_2(ρ/4,t)-ϵ).

Theorem <ref> shows that u_x is bounded for |y|≥ρ/4. i.e. u_x is bounded in [v_1(ρ/4,t),v_2(ρ/4,t)], t close to T_U. Remark <ref> implies that u_xx is uniformly bounded in (v_1(ρ/4,t)+ϵ,v_2(ρ/4,t)-ϵ), t close to T_U. 

Therefore, (<ref>) shows that u_x and u_xx are uniformly bounded for x∈[v_1(ρ/2,t),v_2(ρ/2,t)], t close to T_U. Consequently, the curvature of ∂ D(t)∩{r≥ρ/2} is bounded for t close to T_U. Here we show that the curvature of ∂ D(t) is uniformly bounded as t↑ T_U. It contradicts that ∂ D(t) become singular at T_U.


 In Lemma <ref>, 0<t<min{t_U^α,T_U} is equivalent to 0<t<t_U^α. By the choice of t_U^α, if U⊂ W, t_U^α≤ t_W^α.






§ PROOF OF THEOREM <REF>


Denote U={(x,y)∈ℝ^n+1| |y|<u_0(x),-b_0≤ x≤ b_0}. By assumption of u_0 in Section 1, we know that U∩{x≥0} is an α-domain with smooth boundary, for some α>0. 

We choose vector field X∈ C^1(ℝ^n+1∖{O}→ℝ^n+1) such that

(i) At any P∈∂ U not on the x-axis has ⟨ X,n(P)⟩>0, n(P) is the inward unit normal vector to ∂ U at P.

(ii) We set X((x,y))=(0,-y/|y|), near O and set X=(-1,0,⋯,0) near (b,0,⋯,0), X=(1,0,⋯,0) near (-b,0,⋯,0). 


We note that X has no definition at O.

Since X≠0 on ∂ U∖{O} and |X|=1 near O, by continuity, there exists a neighbourhood V⊃∂ U such that |X|≥δ>0 for some δ>0 in V∖{(0,0)}. 

 For ρ small enough, there exists a smooth hypersurface Σ⊂ V∖{O} with

(i) X(P)∉ T_PΣ at all P∈Σ, i.e., Σ is transverse to the vector field X;

(ii) Σ=∂ U in {(x,y)||y|≥2ρ};

(iii) Σ∩{(x,y)||y|≤ρ} consists of discs Δ_± c={(± c,y)||y|≤ρ} and pipe B_d={(x,y)|-d ≤ x≤ d,|y|=ρ}.



 Because U∩{x≥0} is an α-domain, there exist δ_j, γ_j and 0<δ_j<γ_j such that

    u_0(δ_j)=u_0(γ_j)=u_0(-δ_j)=u_0(-γ_j)=α/2^j

and

    ∂ U∩ C_α={(x,y)| x=± v(y), |y|<α}∪{(x,y)| x=± w(y),|y|<α},

where v,w∈ C^∞((-α,α)) and 0<v(y)<w(y) for |y|<α. Here δ_j is decreasing and γ_j is increasing in j, respectively.

We let w_j∈ C^∞((-α/2^j-1,α/2^j-1)) be a function satisfying

    w_j(y)={[    γ_j+2, 0≤|y|<α/2^j+1,; w(y), α/2^j<|y|<α/2^j-1, ].

and let u_j∈ C^∞((-δ_j-1,δ_j-1)) be a function satisfying

    u_j(x)={[   α/2^j+1, x∈[0,δ_j+2],; u_0(x),  x∈[δ_j,δ_j-1). ].


Let Σ_j consist of three parts: {(x,y)| |y|=u_j(x), x∈(-δ_j,δ_j)}, {(x,y)| x=± w_j(y), |y|<α/2^j} and ∂ U∩{|y|≥α/2^j}. It is easy to see that for sufficiently large j, Σ_j⊂ V∖{(0,0)} satisfies (i), (ii), (iii) for c=γ_j+2, ρ=α/2^j+1 and d=δ_j+2.


Denote σ(P,α):Σ×(-δ,δ)→ V (V is given at the begining of this section and Σ is given by Proposition <ref>) the flow generated by vector field X in ℝ^n+1. Precisely, σ(P,α) is defined as following:

    {[ dσ(P,α)/dα=X(σ(P,α)), P∈Σ,;         σ(P,0)=P,     P∈Σ. ].


By (i) in Proposition <ref>, for any C^1 function u:Σ→ℝ, “the image of u under σ”:={σ(P,u(P))| P∈Σ} is a C^1 hypersurface. Conversely, for any curve Γ⊂ V which is C^1 close to Σ, there exists a unique C^1 function u:Σ→ℝ such that Γ={σ(P,u(P))| P∈Σ}. In other words, the map σ(·,t) defines new coordinates from Σ to V. Therefore, if Γ(t)⊂ V(0<t<T) is a smooth family of smooth hypersurfaces and C^1 close to Σ, there exists a unique function u ∈ C^∞(Σ×(0,T)) such that Γ(t)={σ(P,u(P,t))| P∈Σ}. Let z=(z_1,⋯,z_n) be the local coordinates on an open subset of Σ. If Γ(t) evolves by V=-κ+A, under these coordinates u satisfies the following equation

    ∂ u/∂ t=a_ij(z,u,∇_z u)∇^2_z_iz_ju+b(z,u,∇_z u).

Here {a_ij} is a smooth, positive matrix. Precisely, we can see Section 3 in <cit.>. Consequently, (<ref>) is a parabolic equation. 

For example, σ(·,α) is the flow defined as above.  We can easily deduce that 

    σ(P,α)={[  (x,ρ-α), P∈ B_d,; (-c+α,y), P∈Δ_-c,;   (c-α,y), P∈Δ_c, ].

where we choose the local coordinates:


(1). (x,ρ y) on B_d;


(2). (± c,y) on Δ_± c. 

Suppose that Γ(t) is symmetric to x-axis. Then, on B_d, u depends only on x, t and satisfies

    u_t=u_xx/1+u_x^2+n-1/ρ-u-A√(1+u_x^2).

In this case, b=n-1/ρ-u-A√(1+u_x^2). For n≥ 2, it is easy to see b is not smooth at u=ρ. This is the most significant difference between the condition n=1 and condition n≥ 2.

On Δ_± c, since u depends only on y=(y_1,⋯,y_n), u satisfies 

    u_t=(δ_ij-u_y_iu_y_j/1+|∇ u|^2)u_y_iy_j- A√(1+|∇ u|^2).





 For smooth function v(x,t) on V×(0,T), where V is a compact set, we denote m(t) by 
    m(t)=max{v(x,t)| x∈ V}.

Then there exists P_t∈ V such that v(P_t,t)=m(t) and m^'(t)=v_t(P_t,t) for t>0.

This is a well known result, for example, seeing <cit.>.

 For n≥2, let Γ_j(t), t∈[0,T] be two families of hypersurfaces with σ^-1(Γ_j(t)) the graph of u_j(·,t) for certain u_j∈ C(Σ×[0,T]), j=1,2. Let D_j(t) be bounded open domain with ∂ D_j(t)=Γ_j(t) and assume that D_j(t) are α(t)-domain, j=1,2. Moreover, assume that u_j are smooth on Σ×(0,T] and smooth on Σ∖(Δ_± c∪ B_d)×[0,T]. And suppose that ρ-u_j are bounded from below on Σ∖(Δ_± c∪ B_d)×[0,T]. If 

(1). Γ_j(t) evolves by (<ref>);

(2). Γ_1(0)=Γ_2(0); 

(3). ∫_0^T1/α^2(t)dt<∞, 


then Γ_1(t)=Γ_2(t) for 0<t≤ T.


 Consider function v(P,t)=u_1(P,t)-u_2(P,t). From our assumptions, there holds v∈ C(Σ×[0,T]) and v is smooth on (Σ∖Δ_± c∪ B_d)×[0,T] and smooth on Σ×(0,T]. Moreover v(P,0)≡0. We define M(t)=max{v(P,t)| P∈Σ}. Choose P_t as in Lemma <ref> such that M(t)=v(P_t,t) and M^'(t)=v_t(P_t,t).

Case 1. P_t∈ B_d, u_j satisfy

    u_t=u_xx/1+u_x^2+n-1/ρ-u-A√(1+u_x^2).<ref>

Obviously, v satisfies the following equation

    v_t=a^1(x,t)v_xx+b^1(x,t)v_x+c^1(x,t)v,

where a^1(x,t)>0 and c^1(x,t)=n-1/(ρ-u_1)(ρ-u_2). Since v attains its maximum at P_t, then
v_x(P_t,t)=0 and v_xx(P_t,t)≤0. So we have v_t(P_t,t)≤ c^1(x,t)v. By assumption that D_j are α(t)-domain, then ρ-u_j>α(t), j=1,2. Therefore, 

    v_t(P_t,t)≤n-1/α^2(t)v(P_t,t).

Consequently, M^'(t)≤n-1/α^2(t)M(t).

Case 2. P_t∈Σ∖(Δ_± c∪ B_d). Then we can choose coordinates z on some neighbourhood of P_t on Σ and u_j satisfy (<ref>). We may write this equation as u_t=F(z,t,u,∇ u, ∇^2 u). Then v satisfies

    v_t=a^2_ij(z,t)v_z_iz_j+b^2_i(z,t)v_z_i+c^2(z,t)v

where

    c^2(z,t)=∫_0^1F_u(z,t,u^θ,∇ u^θ,∇ ^2u^θ)dθ,

where u^θ=(1-θ)u_0+θ u_1 and {a^2_ij} is a positive definite.

Since v is smooth on Σ∖(Δ_± c∪ B_d)×[0,T] and ρ-u_j are bounded from below on Σ∖(Δ_± c∪ B_d), 0<t<T, then there exists a positive constant C such that |c^2(z,t)|≤ C. The constant C may depend on the choice of local coordinates z. By compactness of Σ, Σ has a finite covering consisting of neighborhoods of local coordinates, and we can choose C independent of the choice of local coordinates. Since ∇ v(P_t,t)=0, {v_z_iz_j(P_t,t)} is negative semi-definite, 

    v_t(P_t,t)≤ c(P_t,t)v(P_t,t)≤ Cv(P_t,t).

Consequently, M^'(t)≤ CM(t).

Case 3. P_t∈Δ_± c. We only consider P_t∈Δ_-c. Then in the z-coordinates of Δ_-c, u_j satisfy the full graph equation

    u_t=(δ_ij-u_y_iu_y_j/1+|∇ u|^2)u_y_iy_j- A√(1+|∇ u|^2).<ref>

 Hence v=u_1-u_2 satisfies a linear parabolic equation 

    v_t=a^3_ij(z,t)v_z_iz_j+b^3_i(z,t)v_z_i,

where {a^3_ij} is positive definite. Obviously, ∇ v(P_t,t)=0 and {v_z_iz_j(P_t,t)} is negative semi-definite. It follows that M^'(t)≤0.

From the three cases above, if we put

    r(t)=n-1/α^2(t)+C,

then there holds M^'(t)≤ r(t)M(t). Consequently, by the assumption of α(t), 
    M(t)≤ M(0)e^∫_0^tr(t)ds=0.


By considering m(t)=min{v(P,t)| P∈Σ}, we can similarly prove m(t)≥ 0. Therefore Γ_1(t)=Γ_2(t) for 0≤ t≤ T.


Note that the intial curve in our problem is singular at x-axis. The assumption that “D_j(t) are α(t) domain” in Proposition <ref> means that Γ_j(t) “escape” from origin with speed α(t). If the “escape speed” satisfies 

    ∫_0^T1/α^2(t)dt<∞,

we can get the uniqueness.

Following Lemmas <ref> and <ref> can be proved by similar argument to the argument in ℝ^2 given by <cit.>. For reader's convenience, we give the proof.

 There exists a sequence of closed sets E_j such that E_j^∘ are α/2^j-domains and E_j↓U. Here U is given at the beginning of the section and E^∘ denotes the interior of the set E.

 We choose δ_j as in the proof of Proposition <ref>. We can construct even functions v_j∈ C^∞((-b_0,b_0)) such that 

    v_j(x)={[         α/2^j, x∈ (-δ_j/2,δ_j/2),; u_0(x), x∈ [-b_0,-δ_j]∪[δ_j,b_0], ].

v_j(x)≥ u_0(x), x∈[-b_0,b_0] and v^'_j(x)>0, x∈(δ_j/2,δ_j). It is easy to see v_j↓ u_0 uniformly in [-b_0,b_0] as j→∞.

Let E_j={(x,y)||y|≤ v_j(x), -b_0≤ x≤ b_0}. Since v_j↓ u_0 uniformly in [-b_0,b_0], E_j↓U as j→∞. It is easy to check E_j^∘ are α/2^j-domain.


  Let the same assumption of (1) in Theorem <ref> be given. Then there exists t_1>0 such that, for all t_2 satisfying 0<t_2<t_1, the second fundamental forms and derivatives of ∂ E_j(t) are uniformly bounded for t_2≤ t≤ t_1,  where E_j(t) denote the closed evolution of V=-κ+A with E_j(0)=E_j and E_j are chosen as in Lemma <ref>.

 
Let E_j(t)={(x,y)||y|≤ v_j(x,t), -c_j(t)≤ x≤ c_j(t)}.

Step 1. For all t_2 satisfying 0<t_2<δ (δ given by assumption (A-)), there exists a constant c>0 such that 

    v_j(0,t)>c, t_2/2<t<δ.


Let U^+(t) denote the bounded set with ∂ U^+(t)=Λ^+(t). Since U^+(0)=U∩{x≥0}⊂ E_j=E_j(0), U^+(t)⊂ E_j(t). By our assumption that a_*(t)<0 for 0<t≤δ, O∈ U^+(t)⊂ E_j(t) for 0<t<δ. For all t_2∈(0,δ), there exists c>0 such that v_j(0,t)> c for t_2/2≤ t≤δ.

Step 2. Construction of four auxiliary balls.

Since U∩{x≥0} is an α-domain, there exist β_2>β_1>0 such that u_0(±β_1)=u_0(±β_2)=α and u_0^'(x)<0 for x>β_2, u_0^'(x)>0 for 0<x<β_1. There exist p>β_1 and 0<q<β_2 such that u_0(± q)=u_0(± p)=α/2. We consider the points

    Q=(-p,0),                  P=(p,0),


    Q^'=(-p,α),                 P^'=(p,α).




Since P∈ U and P^'∈U^c, there exists ϵ such that B_ϵ(P)⊂ U and B_ϵ(P^')⊂U^c. Consequently, B_ϵ(P)∪B_ϵ(Q)⊂ E^∘ and B_ϵ(P^')∪B_ϵ(Q^')⊂ E^c. Then for j large enough, B_ϵ(P)∪B_ϵ(Q)⊂ E_j^∘ and B_ϵ(P^')∪B_ϵ(Q^')⊂ E_j^c. Comparison principle shows that

    B_ϵ(t)(P)∪B_ϵ(t)(Q)⊂ E_j(t)^∘

for 0<t<δ_2. By Theorem <ref>,

    B_ϵ(t)(P^')∪B_ϵ(t)(Q^')⊂ E_j(t)^c

for 0< t<δ_2. Here ϵ(t) is the solution of (<ref>) with ϵ(0)=ϵ on the interval [0,δ_1). Take δ_2 independent of j such that ϵ(t)>ϵ/2 for 0<t<δ_2. 


Step 3. Divide ∂ E_j(t) into two parts by auxiliary balls. 

Since for all ρ<α/2, C_ρ intersects ∂ E_j at most four times, by the intersection number argument as in the proof of Lemma <ref>, there exists t_0>0 such that C_ρ intersects ∂ E_j(t) at most four times for 0<t<t_0. By continuity, we can deduce that there exists δ_4 such that for all ρ<α, the equation v_j(x,t)=ρ has just one root for x>p for all t<δ_4. By symmetry, it also holds for x<-p. 

Put t_1=min{t_0,δ_2,δ_3,δ_4}. Then Step 1 and intersection argument show that E_j(t)^∘ are all c-domains for t_2/2<t<t_1.  Let d<min{c,ϵ/4}. By (<ref>) in Step 2, we have v_j(x,t)>d for t_2/2<t<t_1 and x with |x-p|<√(ϵ^2(t)-d^2) or |x+p|<√(ϵ^2(t)-d^2). By ϵ(t)>ϵ/2,

    v_j(x,t)≥ d in Ω=(-p-√(3)/4ϵ,p+√(3)/4ϵ)×(t_2/2,t_1).


For x≤-p, by (<ref>) in Step 2,

    v_j(x,t)<α/2-ϵ(t)<α/2-ϵ/2 for x≤-p, 0≤ t<t_1.

This is also true for x≥ p.

Step 4. The derivatives and second fundamental forms of ∂ E_j(t) are bounded in Ω^'=[-p,p]×(t_2,t_1).

Since v_j(x,t)≥ d in Ω=(-p-√(3)/4ϵ,p+√(3)/4ϵ)×(t_2/2,t_1), Theorem <ref> implies that v_jx are uniformly bounded in Ω. By Remark <ref>, v_jxx are uniformly bounded in Ω^'. 

Step 5. The derivatives and second fundamental forms of ∂ E_j(t) are bounded for x≤ -p and x≥ p, t_2<t<t_1.

We only consider for x≤-p. For 0<t<t_1, the part of ∂ E_j(t) on x≤-p can be represented by x=w_j(y,t) for |y|<α/2, t∈ (0,t_1), and w_j satisfy the equation (<ref>) in the condition “-” and n=1. Then Corollary <ref> and Remark <ref> imply that all ∂ ^k/∂ y^kw_j(y,t), k=1,2, are uniformly bounded for |y|≤α/2-ϵ/2, t_2<t<t_1 and for any t_2>0. Then the derivatives and second fundamental forms of ∂ E_j(t) are uniformly bounded for x≤-p, t_2<t<t_1. 

The proof of this lemma is completed.


 There exists a sequence of open sets {U_j} such that and U_j∩{x≥0} being an α-domain such that U_j↑ U as j→∞.


 Let the same assumption in Theorem <ref> hold. Then there exists t_1>0 such that for all t_2∈(0,t_1), the second fundamental forms and derivatives of ∂ U_j(t) is uniformly bounded for t_2<t<t_1, where U_j(t) is the open evolution of V=-κ+A with U_j(0)=U_j and {U_j} is chosen as in Lemma <ref>.


Lemmas <ref> and <ref> are all able to be proved as in <cit.>.

As mentioned in Proposition <ref>, in order to get the uniqueness, we must give the estimate of “escape speed”. Let R_0 be taken small enough such that B_R_0((R_0,0,⋯,0))∪ B_R_0((-R_0,0,⋯,0))⊂ U. In next lemma, we construct a sub-solution.

(Sub-solution)
Take R_0 as above. Function u is even and defined by

    u(x,t)={[ √((r(t)+R(t))^2-R_0^2)-√(r^2(t)-x^2), 0≤ x<R_0r(t)/R(t)+r(t),;          √(R^2(t)-(x-R_0)^2), R_0r(t)/R(t)+r(t)≤ x≤ R_0+R(t). ].

Here r(t)=t^3/4 and R(t) satisfies R^'=A-n/R, R(0)=R_0. Then there exists t_*>0 such that (u,R_0+R(t)) is a sub-solution of (<ref>), (<ref>), (<ref>), (<ref>) for 0<t<t_*.



We can easily deduce that |R(t)-R_0|=O(t), as t→0. Since r(t)=t^3/4>R(t), for sufficient small t, u is well-defined for small t.

1. Positive: Obviously, u_0(x,t)>0 for -R_0-R(t)<x<R_0+R(t).

2. Initial condition: By the choice of R_0, u(x,0)=√(R_0-(x+R_0)^2)≤ u_0(x) for 0≤ x≤ R_0 and R_0≤ b_0.

3. Boundary condition: Obviously, at boundary, 

    u(-R_0-R(t),t)=u(R_0+R(t),t)=0

and

    u_x(-R_0-R(t),t)=-u_x(R_0+R(t),t)=∞.

4. Interior: For R_0r(t)/R(t)+r(t)< x< R_0+R(t) or -R_0-R(t)<x<-R_0r(t)/R(t)+r(t), u(x,t)=√(R^2(t)-(x-R_0)^2) satisfies (<ref>). Next we only need prove u is a sub-solution of (<ref>) for -R_0r(t)/R(t)+r(t)<x<R_0r(t)/R(t)+r(t) and t small. By calculation,

    1+u_x^2=r^2/r^2-x^2;


    u_xx=r^2/(r^2-x^2)^3/2;


    u_t=(r+R)(r^'+R^')/√((r+R)^2-R_0^2)-rr^'/√(r^2-x^2).

Therefore, 

    u_t   -   u_xx/1+u_x^2+n-1/u-A√(1+u_x^2)=(r+R)(r^'+R^')/√((r+R)^2-R_0^2)-rr^'/√(r^2-x^2)-1/√(r^2-x^2)
       +   n-1/√((r+R)^2-R_0^2)-√(r^2-x^2)-Ar/√(r^2-x^2)≤(r+R)(r^'+R^')/√((r+R)^2-R_0^2)-r^'-1/r
       +   n-1/√((r+R)^2-R_0^2)-r.

Since |R^'(t)| is bounded, R(t) is bounded from above and below for small t, and r(t)=t^3/4, we can deduce that 

    1/√((r+R)^2-R_0^2)=1/√((r+R-R_0)(r+R+R_0))=O(t^-3/8);


    (r+R)(r^'+R^')=O(t^-1/4)

as t→0. Consequently, 

    u_t   -   u_xx/1+u_x^2+n-1/u-A√(1+u_x^2)≤ C_1 t^-5/8-C_2 t^-1/4-C_3 t^-3/4+C_4 t^-3/8
       =   t^-3/4(C_1t^1/8-C_2t^1/2-C_3+C_4t^3/8)<0

for any sufficient small t>0. It is easy to check that u is C^1 at x=R_0r(t)/R(t)+r(t). Then there exists t_*>0 such that u is a sub-solution of (<ref>) in viscosity sense for 0<t<t_*. 

We complete the proof.



Recall U={(x,y)∈ℝ^n+1| |y|<u_0(x), b_0<x<b_0} and E(t) be the outer evolution of U. Then there exists t^*>0 such that, for each t∈[0,t^*), E(t) can be described as follows,

    E(t)={(x,y)∈ℝ^n+1| |y|≤ v(x,t), -b_1(t)≤ x≤ b_1(t)}.

Here (v,b_1) is the uniqueness solution of (<ref>), (<ref>), (<ref>), (<ref>) on the interval [0,t^*). Moreover, there exists α(t) with ∫_0^t^*1/α^2(t)dt<∞ such that E(t)^∘ is α(t)-domain for t<t^*.


Let E_j(t) be given by Lemma <ref>. Since {(x,y)∈ℝ^n+1||y|≤u(x,0)}⊂ E_j(0)={(x,y)∈ℝ^n+1||y|≤ v_j(x,0), -c_j(0)≤ x≤ c_j(0)}), {(x,y)∈ℝ×ℝ^N||y|≤u(x,t)}⊂ E_j(t)={(x,y)∈ℝ^n+1||y|≤ v_j(x,t), -c_j(t)≤ x≤ c_j(t)} for 0<t<t_* and for all j. 

By r(t)=t^3/4, |R(t)-R_0|=O(t) and boundedness of R(t) from below as t→0, there exists t^*>0 such that

    u(0,t)   =   √((r(t)+R(t))^2-R_0^2)-√(r^2(t))= √(r(t))(√(r(t)+2R(t)+R(t)-R_0/r(t))-√(r(t)))
       ≥    C^'t^3/8-t^3/4≥ Ct^3/8,

for t<t^*. If we taken

    α(t)=Ct^3/8 for 0≤ t<t^*,

then ∫_0^t^*1/α^2(t)dt<∞.
Therefore, E_j(t)^∘ are all α(t)-domains. Moreover, (v_j,c_j) satisfy
(<ref>), (<ref>), (<ref>). By E_j(0)↓ E(0), Theorem <ref> and the same method as in Lemma <ref>, we can show that E_j(t)↓ E(t) and derivatives and second fundamental forms of E_j(t) are uniformly bounded. (v,b_1)=lim_j→∞(v_j,c_j) is the solution of (<ref>), (<ref>), (<ref>), (<ref>). Moreover, {(x,y)∈ℝ^n+1||y|≤u(x,t)}⊂ E(t) for 0<t<t^*. E(t)^∘ is also an α(t)-domain 0<t<t^*. The uniqueness of the solution follows from Proposition <ref>.


 Let E(t) and U(t) be the closed and open evolution of (<ref>) with E(0)=U and U(0)=U, respectively.

Let U^+(t) denote the bounded open set with ∂ U^+(t)=Λ^+(t). Since U^+(0)=U∩{x≥0}⊂ U=U(0), there holds U^+(t)⊂ U(t). By our assumption that a_*(t)<0 for 0<t≤δ, O∈ U^+(t)⊂ U(t) for 0<t<δ, where δ is given in assumption (A-). This means that ∂ U(t) escapes away from origin. Therefore, 

    {(x,y)∈ℝ^n+1||y|≤u(x,t)}⊂ U(t), 0<t<T,

where T=min{t^*,δ, t_1}. (Recall t_1 is given by Lemma <ref>) Consequently, U(t) is also an α(t)-domain. For small t, we can easily check that ∂ E(t) and ∂ U(t) satisfy the assumption in Proposition <ref>. We get ∂ E(t)=∂ U(t) for 0≤ t<T.

We complete the proof.



 Choose T_1<min{t^*,δ}. Here δ is given in assumption (A+), and t^* is given in Corollary <ref>.

Let U^±(t) be the bounded open domain with ∂ U^±(t)=Λ^±(t). Thus the left end point of U^+(t) and the right end point of U^-(t) are (a_*(t),0,⋯,0) and (-a_*(t),0,·,0), respectively. By assumption (A+), -a_*(t)≤ a_*(t), 0≤ t<T_1. Therefore, U^+(t)∩ U^-(t)=∅, 0≤ t<δ. From Lemma <ref>, the inner evolution U(t) satisfies U(t)=U^+(t)∪ U^-(t), for 0≤ t<δ.

Corollary <ref> shows that E(t)^∘ is an α(t)-domain for 0<t<T_1. By α(t)= Ct^3/8, there exists small enough ρ such that the ball 

    B_ρ((0,C(T_1/4)^3/8,0,⋯,0))⊂ E(t), T_1/2<t<T_1

and 

    B_ρ((0,C(T_1/4)^3/8,0,⋯,0))∩ U(t)=∅, T_1/2<t<T_1.

This means that the interface evolution Γ(t)=E(t)∖ U(t) has interior.



(1) In the proof of (1) in Theorem <ref>, we get the closed evolution and the open evolution are all connected sets. Therefore, they are homeomorphic. Moreover, using the unique result, we can prove that they are coincide.

(2) In the proof of (2) in Theorem <ref>, we get the closed evolution is connected and the open evolution is separated. Therefore, they are not homeomorphic.




§ PROOF OF THEOREM <REF>

In this section, we give the proof of Theorem <ref>. First, we introduce the following similarity transformation: for T>0,

    z=x/√(2(T-t)), τ=-1/2ln(T-t)

and

    w(z,τ)=1/√(2)e^τu(√(2)e^-τz,T-e^-2τ).


Then u satisfies 

    u_t=u_xx/1+u_x^2-n-1/u+A√(1+u_x^2),<ref>

 if and only if w satisfies

    w_τ=w_zz/1+w_z^2-zw_z+w-n-1/w+√(2)Ae^-τ√(1+w_z^2).

In 1992, <cit.> shows that there is a torus shape self-similar solution of (<ref>) for A=0 called “Angenent shrinking doughnut”. The self-similar solution remaining the shape of doughnut shrinks to a point. Moreover, in <cit.>, using this self-similar solution, they prove that after a rotational hypersurface pinches, the hypersurface will be separated into two disjoint components. We also expect to prove Theorem 1.2 by using some self-similar solution of (<ref>), however, it is difficult to find such solution. Therefore, we construct a compact self-similar super-solution of (<ref>).

(Super-solution of equation (<ref>))
Denote w:=C-√(ρ^2-z^2), -ρ≤ z≤ρ. 

For n>2, for every C, ρ with C^2+ρ^2<n and C>ρ>1, there exists τ_0(C,ρ) such that w is a super-solution of (<ref>) for -ρ< z<ρ, τ>τ_0.

For n=2, Fix θ∈(0,1), ϵ_0∈(0,2/9θ) arbitrary. Then, for each 1+θϵ_0<ρ<C<1+ϵ_0, there exists τ_0(C,ρ) such that w is a super-solution of (<ref>) for -ρ< z<ρ, τ>τ_0.


By calculation,

    w_τ=0,


    w_z=z/√(ρ^2-z^2)

and

    w_zz=ρ^2/(ρ^2-z^2)^3/2.

For convenience, we put q=√(ρ^2-z^2) for -ρ≤ z≤ρ.

    w_τ   -   w_zz/1+w_z^2+zw_z-w+n-1/w-√(2)Ae^-τ√(1+w_z^2)
       =   -1/q+n-1/C-q-C+q+ρ^2-q^2/q-√(2)Ae^-τρ/q
       =   -1/q+n-1/C-q-C+ρ^2/q-√(2)Ae^-τρ/q
       =   1/q(C-q)(C q^2-(ρ^2+C^2-n)q-C+ρ^2C-√(2)Ae^-τρ(C-q)).


For n>2, if ρ^2+C^2<n and ρ<C, we can deduce C q^2-(ρ^2+C^2-n)q-C+ρ^2C of the right hand side of the formula above attains its minimum at q=0.
Consequently,

    w_τ-w_zz/1+w_z^2+zw_z-w+n-1/w-√(2)Ae^-τ√(1+w_z^2)≥1/q(C-q)(-C+ρ^2C-√(2)Ae^-τρ C).

Therefore, if ρ>1, then there exits τ_0(C,ρ) such that 

    w_τ-w_zz/1+w_z^2+zw_z-w+n-1/w-√(2)Ae^-τ√(1+w_z^2)>0

for τ>τ_0.

For n=2, C q^2-(ρ^2+C^2-2)q-C+ρ^2C attains its minimum at ρ^2+C^2-2/2C. Consequently,

    w_τ   -   w_zz/1+w_z^2+zw_z-w+1/w-√(2)Ae^-τ√(1+w_z^2)
       ≥   1/q(C-q)(-(C^2+ρ^2-2)^2/4C+(ρ^2-1)C-√(2)Ae^-τρ C).

Then fixed θ∈(0,1), for any ϵ_0∈(0,2/9θ) and 1+θϵ_0<ρ<C<1+ϵ_0,  

    -(C^2+ρ^2-2)^2/4C+(ρ^2-1)C≥8θϵ_0-36ϵ_0^2/4(1+θϵ_0)=2θϵ_0-9ϵ_0^2/(1+θϵ_0)>0.

Therefore, there exists τ_0(C,ρ)>0 such that 

    w_τ-w_zz/1+w_z^2+zw_z-w+1/w-√(2)Ae^-τ√(1+w_z^2)>0

for τ>τ_0.



Under the condition n>2, in the proof of Proposition <ref>, for convenience, we assume 

    ρ^2+C^2<n.

Indeed, it is not necessary. In the proof, we can use that C q^2-(ρ^2+C^2-n)q-C+ρ^2C attains its minimum at ρ^2+C^2-n/2C.



Let w and τ_0 be given by Proposition <ref>. Then for T<e^-τ_0, 

    u(x,t;T)=√(2(T-t))w(x/√(2(T-t)))

is a super-solution of equation (<ref>) for -ρ√(2(T-t))<x<ρ√(2(T-t)), 0<t<T.

This result is obvious by Proposition <ref>. Here we omit the proof.


For n=2, recall γ given in Section 1 and ϵ_0 given by Proposition <ref>. Then for all 0≤γ<arctan√((1+ϵ_0)^2-1), we can choose ρ and C satisfying 1<ρ<C<1+ϵ_0 such that

    γ<√(C^2-ρ^2)/ρ.


For n>2, if 0≤γ<arctan√(n-2), we can choose ρ, C satisfying 1<ρ<C and ρ^2+C^2<n such that

    γ<√(C^2-ρ^2)/ρ.

It is obvious that the cone 

    |y|=√(C^2-ρ^2)/ρ|x|

is the envelop of the family of hypersurfaces {|y|=λw(x/λ)}_λ>0.

By the property of u_0, we can get

    y=√(C^2-ρ^2)/ρ|x|> u_0(x),

for small |x|.

Therefore, there exists T(ρ,C) such that for all 0≤ T<T(ρ,C), 

    u(x,0;T)>u_0(x),

-ρ√(2T)<x<ρ√(2T).

 
As mentioned in Remark <ref>, we choose 

    α_n={[ arctan√((1+ϵ_0)^2-1), n=2,;         arctan√(n-2), n>2. ].
 
Obviously, α_n→π/2, as n→∞.

For every γ<α_n, we choose C and ρ as in Remark <ref>. Let T_γ<min{T(ρ,C),e^-τ_0(ρ,C)}, where τ_0(ρ,C) is given by Proposition <ref> and T(ρ,C) is given by Remark <ref>. Next we show that for every 0<t<T_γ, the origin O∈ E(t)^c. Let flow 

    Γ(t;T)={(x,y)||y|=u(x,t,T), -ρ√(2(T-t))≤ x≤ρ√(2(T-t))},

0≤ t<T. Remark <ref> shows that for every 0<t<T_γ,

    Γ(0;t)∩ E(0)=∅.
 
By comparison principle, we can easily show that 

    Γ(s;t)∩ E(s)=∅, 0≤ s≤ t.

(Noting that Γ(s,t) is a hypersurface with boundary, we must consider the comparison principle in interior and boundary separately. By comparison principle, ∂ E(s) can not touch Γ(s,t) interior. At the boundary, gradient of u(x,s,t) is infinity. Theorem <ref> and <ref> implies that ∂ E(s) can not touch Γ(s,t) at the boundary. The details are left to the reader) Especially, 

    {O}∩ E(t)=Γ(t;t)∩ E(t)=∅.

Therefore for every 0<t<T_γ, there holds O∈ E(t)^c. 

Here we show that E(t) is separated into two connected components, for 0<t<T_γ. Let E^+(t) (E^-(t)) and U^+(t) (U^-(t)) be the outer evolution and inner evolution of V=-κ+A with E^+(0)=U∩{x≥0} (E^-(0)=U∩{x≤ 0}) and U^+(0)=U∩{x≥0} (U^-(0)=U∩{x≤0}), respectively.

Since U∩{x≥0} and U∩{x≤ 0} are α-domains and E^+(t)∩ E^-(t)=∅, using Theorem <ref>, we obtain 

    ∂ E^+(t)=∂ U^+(t), ∂ E^-(t)=∂ U^-(t), 0<t<T_γ.
 
Therefore 

    ∂ E(t)=∂ E^+(t)∪∂ E^-(t)=∂ U^+(t)∪∂ U^-(t)=∂ U(t), 0<t<T_γ.

Here we complete the proof.



Let u_0 be a function as in Section 1 and let γ be the constant in (<ref>). For n≥ 2 and 0≤γ<α_n, there is no solution of the following free boundary problem,

    u_t=u_xx/1+u_x^2-n-1/u+A√(1+u_x^2), -b(t)<x<b(t), 0<t<T_1,<ref>


    u(b(t),t)=u(-b(t),t)=0, u_x(-b(t),t)=-u_x(b(t),t)=∞, 0<t<T_1,<ref>


    u(x,0)=u_0(x), -b_0≤ x≤ b_0,<ref>


    u(x,t)>0, -b(t)< x< b(t), 0<t<T_1.<ref>



Assume that there exists the solution (u,b) of the free boundary problem. We can use the approximate argument similarly as in Lemma <ref> to prove that the outer evolution E(t) is written as follows

    E(t)={(x,y)|ℝ^n+1| |y|≤ u(x,t), -b(t)≤ x≤ b(t)}.

This contradicts that E(t) is separated into two connected components.





§ PROOF OF THEOREM <REF>

In this section, we give the proof of Theorem <ref>. 

(Connected Outer evolution) Let u_0 be a function as in Section 1. In the plane, there is T_1>0 such that (u,b) is a unique solution of the following free boundary problem,

    u_t=u_xx/1+u_x^2+A√(1+u_x^2), -b(t)<x<b(t), 0<t<T_1,<ref>*


    u(b(t),t)=u(-b(t),t)=0, u_x(-b(t),t)=-u_x(b(t),t)=∞, 0<t<T_1,<ref>


    u(x,0)=u_0(x), -b_0≤ x≤ b_0,<ref>


    u(x,t)>0, -b(t)< x< b(t), 0<t<T_1.<ref>


Moreover, the outer evolution

    E(t)={(x,y)∈ℝ^2||y|≤ u(x,t), -b(t)≤ x≤ b(t)}, 0<t<T_1.


To prove this proposition, we need the following lemma.

(Lemma 4.7 in <cit.>) Let U be an α-domain in ℝ^2 and let U(t) be the open evolution with U(0)=U. Then there exists t_U>0 such that U(t) is an (α+At)-domain, 0<t<t_U.



Using the approximate argument as in Lemma <ref>, we can prove that there exists T_1>0 such that E(t)^∘ is At-domain, 0<t<T_1. Moreover, 

    E(t)={(x,y)∈ℝ^2||y|≤ u(x,t), -b(t)≤ x≤ b(t)}, 0≤ t<T_1.

Here (u,b) is the unique solution of (<ref>*), (<ref>), (<ref>), (<ref>). For the precise proof, we can see <cit.> similarly. Here we omit the details.



Indeed, it is determined by the existence of the solution (u,b) of (<ref>*), (<ref>), (<ref>), (<ref>) whether the outer evolution is connected or separated. 

Under the condition n=1, saying roughly, if u satisfies 

    u_t=u_xx/1+u_x^2+A√(1+u_x^2), <ref>*
 
and u(x,0)=u_0(x)≥0, by comparison principle, u(x,t)>0, t>0. This means that the problem always has a “positive” solution in the plane. This can be explained precisely by Lemma <ref>

However, under the conditon n≥2, the equation

    u_t=u_xx/1+u_x^2-n-1/u+A√(1+u_x^2)<ref>

has the “contraction power -n-1/u”. We can not ensure that the problem has a “positive” solution with u(x,0)=u_0(x)≥0. Lemma <ref> shows that if γ=π/2, this problem has a unique “positive” solution. 


(Separated inner evolution)
Suppose 0≤γ<π/2. Let (u,a,b) be the solution of the following problem

    u_t=u_xx/1+u_x^2+A√(1+u_x^2), a(t)<x<b(t), 0<t<T_1,<ref>**


    u(b(t),t)=u(a(t),t)=0, u_x(a(t),t)=-u_x(b(t),t)=∞, 0<t<T_1,<ref>*


    u(x,0)=u_0(x), 0≤ x≤ b_0,<ref>*


    u(x,t)>0, a(t)< x< b(t), 0<t<T_1.<ref>*

Then there exists δ>0 such that a(t)>0, 0<t<δ. Moreover, the inner evolution U(t) can be written as follows,

    U(t)=U^+(t)∪ U^-(t), 0<t<δ,

where U^+(t)={(x,y)∈ℝ^2||y|<u(x,t), a(t)≤ x≤ b(t)} and U^-(t)={(-x,y)| (x,y)∈ U^+(t)}.



We claim that there exists δ>0 such that a(t)>0 for 0<t<δ. If the claim holds, U^+(t)∩ U^-(t)=∅, 0<t<δ. Using Lemma <ref>, U(t)=U^+(t)∩ U^-(t), 0<t<δ.

We give the sketch of the proof of the claim.  

Let γ<γ_1<π/2. Define a family of circles 

    v_λ(y)=λ C-√((λ Ccos(π/2-γ_1))^2-y^2).

It is easy to find that the envelop of {v_λ}_λ>0 is |y|=tanγ_1 x.

Let {(x,y)| x=v_0(y), -δ_0<y<δ_0} be the left cap of ∂ U∩{x≥0}.

By the choice of γ_1, if necessary, choose δ_0 smaller such that 

    v_0(y)≤tan(π/2-γ_1)|y|, -δ_0<y<δ_0.
 
Consider the following inverse equation

    v_t=v_yy/1+v_y^2-A√(1+v_y^2), -δ_0<y<δ_0,

with v(y,0)=tan(π/2-γ_1)|y|. Since the initial function is not smooth, we modify it by the family {v_λ}_λ>0 near the origin. Let v_λ(y,t) be the solution with the modified initial data. We calculate 

    ∂/∂ tv_λ(0,0)=1/λ Ccos(π/2-γ_1)-A→∞,

as λ→0. Therefore, there exists constant C>0 such that v_λ(0,t)>Ct, for t small. It is easy to see v_λ(0,t)→ a(t), for every t small. Then

    a(t)>Ct, 0<t<δ,

for some δ. Here we complete the proof.



This result is an easy consequence of Proposition <ref> and Proposition <ref>.




§ APPENDIX 

In this section, we prove there exists a unique smooth family of smooth hypersurfaces Γ(t) satisfying

    V=-κ+A, on Γ(t)⊂ℝ^n+1,

where Γ(0)=∂ U and U is an α-domain.

Since ∂ U is not necessary smooth, we also use the level set method and prove the interface evolution is not fattening.

For α-domain U, we choose a smooth vector field X:ℝ^n+1→ℝ^n+1 such that


(i) At any point P∈∂ U which is not on the x-axis ⟨ X(P), n(P)⟩>0, where n is the inward unit normal vector at P.


(ii) Near the two end points of ∂ U, X is constant vector with X≡± e_0=(±1,0,⋯,0).

Since X≠0 on the compact set ∂ U, there are a constant δ>0 and an open neighbourhood V⊃∂ U on which |X|≥δ>0.



For sufficiently small ρ>0, there exists a smooth hypersurface Σ⊂ V with

(i) X(P)∉ T_PΣ at all P∈Σ, i.e., Σ is transverse to the vector field X.

(ii) Σ=∂ U in {(x,y)∈ℝ×ℝ^n| |y|≥2ρ}.

(iii) Σ∩{(x,y)∈ℝ×ℝ^n||y|≤ρ} consists of two flat disks Δ_a={(a,y)∈ℝ×ℝ^n||y|≤ρ} and Δ_b={(b,y)∈ℝ×ℝ^n||y|≤ρ} for some a<b.


Seeing Figure <ref>, this proposition can be proved as in Proposition <ref>.




Let ϕ^α:ℝ^n+1→ℝ^n+1 (α∈ℝ), t∈(-δ,δ) be the flow generated by vector field X on ℝ^n+1, that is,

    {[ dϕ^α(P)/dα=X(ϕ^α), P∈Σ,;      ϕ^0(P)=P,     P∈Σ. ].


We denote σ(P,s):=ϕ^s(P). As in Section 5, suppose Γ(t)⊂ V (0<t<T) are smooth hypersurfaces such that σ^-1(Γ(t)) is the graph u(·,t) for u:Σ×[0,T)→ℝ. Let z_1,z_2,⋯,z_n be local coordinates on an open subset of Σ. If Γ(t) evolving by V=-κ+A, then in these coordinates u satisfies the following parabolic equation

    ∂ u/∂ t=a_ij(z,u,∇ u)∂^2u/∂ z_i∂ z_j+b(z,u,∇ u).



For example, on Δ_a, by calculation, σ(y_1,,y_2,⋯,y_n,s)=(a-s,y_1,y_2,⋯,y_n). Then u satisfies the "-" condition of (<ref>).
 For n≥1, let Γ_1(t), Γ_2(t) (0≤ t<T) be two families of smooth hypersurface smooth and σ^-1(Γ_j(t)) be the graph of u_j(·,t) for certain u_j∈ C(Σ×[0,T)). Assume that the u_j are smooth on Σ×(0,T) as well as on Σ∖(Δ_a∪Δ_b)×[0,T). Then if the Γ_j(t) evolve by V=-κ+A and if Γ_1(0)=Γ_2(0), then there holds Γ_1(t)=Γ_2(t) for 0<t<T.

We use the same method in <cit.>. 
The proof is similar as in Proposition <ref>. Here we omit it.



If U is an α-domain with a smooth boundary, let D(t) and E(t) be the open and closed evolutions of V=-κ+A with D(0)=U and E(0)=U. Then there exists T>0 such that ∂ D(t) and ∂ E(t) are smooth hypersurfaces for 0<t≤ T and ∂ D(t)=∂ E(t). Moreover, denoting Σ(t)=∂ D(t)=∂ E(t), Σ(t) can be written into Σ(t)={(x,y)∈ℝ×ℝ^n| |y|=u(x,t), a(t)≤ x≤ b(t)} and (u,a,b) satisfies

    {[ u_t=u_xx/1+u_x^2-n-1/u+A√(1+u_x^2), x∈(a(t),b(t)), 0<t< T,;                         u(a(t),t)=0, u(b(t),t)=0, 0≤ t< T,;                     u_x(a(t),t)=∞, u_x(b(t),t)=-∞, 0≤ t<T,;                                                            ].**



We only give the sketch of the proof. By approximate argument similarly as in Lemma 6.2 and Lemma 6.4, ∂ D(t) and ∂ E(t) are smooth hypersurfaces and can be represented by σ(P,u_j(P)), for some u_j, j=1,2. Then we can use Proposition <ref> to prove ∂ D(t)=∂ E(t). Therefore Γ(t)=∂ E(t) can be represented by Γ(t)={(x,y)∈ℝ×ℝ^n| |y|=u(x,t), a(t)≤ x≤ b(t)}. Using Theorem <ref>, (u,a,b) satisfies (**).


Conflict of interest. 
We declare that we do not have any commercial or associative interest that represents a conflict of interest in connection with the work submitted.

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