# Life-Span of Classical Solutions to Hyperbolic Inverse Mean Curvature Flow

Authors: Zenggui Wang
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, we investigate the life-span of classical solutions to hyperbolic inverse mean curvature flow. Under the condition that the curve can be expressed in the form of a graph, we derive a hyperbolic Monge–Ampère equation which can be reduced to a quasilinear hyperbolic system in terms of Riemann invariants. By the theory on the local solution for the Cauchy problem of the quasilinear hyperbolic system, we discuss life-span of classical solutions to the Cauchy problem of hyperbolic inverse mean curvature.


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## Body

## 1. Introduction

In this paper, we study hyperbolic inverse mean curvature flow (HIMCF):(1)∂2γ∂t2=−k−1N⟶+k−1k−1sT⟶,∀z,t∈S1×0,T,γz,0=γ0z,∂γ∂tz,0=γ1z,γ1z,N⟶0=0,where k denotes the mean curvature of the curve γ and N⟶ and T⟶ are, respectively, the unit inner normal and tangent vectors of the curve γ (·, t). γ0 stands for a smooth strictly convex closed curve, γ1 denotes the initial velocity of γ0, and N⟶0 is the unit inner normal vector of γ0.Definition 1.
(see [1]). A flow is called a normal preserving flow if the normal vector of the curve is independent of time dN⟶/dt=0, i.e., γts,N⟶=0 holds for all time.
It is well known that inverse mean curvature flow is an important method to derive the energy estimates in general relativity; for example, Huisken and Ilmanen developed a theory of weak solutions of the inverse mean curvature flow and used it to prove successfully Riemannian Penrose inequality which plays an important role in general relativity (see [2]). In [3], Urbas proved that, for inverse mean curvature flow, the surfaces stay strictly convex and smooth for all time. Furthermore, the surfaces become more and more spherical in the process. Similar results have also been obtained for star-shaped initial data with the positive mean curvature of the surface, see Urbas [3] and Gerhardt [4]. In [2], Huisken and Ilmanen proved a sharp lower bound of mean curvature, from which they proved that if the initial surface is the boundary of a strictly star-shaped domain and has nonnegative mean curvature, a smooth solution of the inverse mean flow will exist for all time and converge to a manifold.
The hyperbolic mean curvature flow, i.e., the hyperbolic version of mean curvature flow, has been introduced by some authors, see Gurtin and Podio-Guidugli [5], He et al. [6], Kong et al. [7, 8], Lefloch and Smoczyk [9], Notz [10], Rotstein et al. [11], and Wang [12, 13]. In fact, Yau in [14] has suggested the following equation related to a vibrating membrane or the motion of a surface:(2)∂2X∂t2=HN⟶,where H is the mean curvature and N⟶ is the unit inner normal vector of the surface and pointed out that very little is known about the global time behavior of the hypersurfaces. Recently, in [1], Chou and Wo proposed a new hyperbolic curvature flow for convex hypersurfaces. This flow is most suited when the Gauss curvature is involved. The equation satisfied by the graph of the hypersurface under this flow gives rise to a new class of fully nonlinear Euclidean invariant hyperbolic equations. Finally, they consider the expanding Gauss curvature flow driving by negative powers −k−β of the Gauss curvature. In the special case β = 1, the expanding flow becomes, once written in terms of the support function S(θ, t), a linear problem:(3)Stt=Sθθ+S,S0=ϕ,St0=ψ,where S(θ, t) = {γ(θ, t), (cos θ, sin θ)}, and the normal angle θ is determined modulo 2π. They get the following result.Proposition 1.
(see [1]). Consider (3) where the initial values are smooth and satisfy ϕθθ + ϕ > 0 and ψθθ + ψ > 0. Then, the flow remains smooth and expands to infinity like a circle.
Motivated by the inverse mean curvature flow and the local solution theory of the Cauchy problem for the quasilinear hyperbolic system (see [15]), we will focus on life span (the maximum existence time of unique local classical solutions) of classical solutions to the Cauchy problem for the hyperbolic inverse mean curvature flow. Our first result is the following local existence theorem for initial value problem (1).Theorem 1.
(local existences and uniqueness). Suppose thatγ0 is a smooth, strictly convex curve, and γ1:S1⟶ℝ2 is a smooth vector function. Then, there exist positive T and a family of strictly convex closed curves γ(·, t) with t ∈ [0, T) such that γ(·, t) satisfies (1).
The following theorem concerns the life-span of local (in space) smooth solutions of flow (1) that can be written as convex graphs over an interval R⊂ℝ in the form(4)γt,z=x,ut,x,for someu:0,T×R⟶ℝ.G.
Writing the initial conditions as(5)u0,x=fx,∂u∂t=gx,for some f∈C2R¯ and g∈C1R¯, we obtain the following result.Theorem 2.
Assume thatγ0 is a strictly convex closed curve and γ1:S1⟶ℝ2 is a smooth vector function on S1. Then, a lower bound δ∗ for the maximum time T of existence of a solution of (1) in the form G is given by(6)δ∗=min‖ϕ‖A0−‖ϕ‖,A0+A1‖ϕ˙‖B0+B1‖ϕ˙‖+B2‖ϕ˙‖2,1C0+C1‖ϕ˙‖,where ϕ=ϕ1,…,ϕ5T=1+f′2−g′/f″,−1−f′2−g′/f″,f,g,f′T, ϕ˙ix=dϕix/dx,(7)A0ϕ=8ϕ3+8ϕ2+3ϕ+1,A1ϕ=2ϕ+1,A2ϕ=A0−ϕ,A3ϕ=48ϕ2+32ϕ+4,B0ϕ=2A0A1A3+A2,B1ϕ=6A0A1+2A12A3,B2ϕ=6A12,C0ϕ=41+2ϕ248ϕ2+32ϕ+512ϕ2+8ϕ+1,C1ϕ=161+2ϕ212ϕ2+8ϕ+1.
The paper is organized as follows. In Section2, we derive a hyperbolic Monge–Ampère equation by the hyperbolic inverse mean curvature flow and give the short-time existence theorem, i.e., Theorem 1. In Section 3, we reduce the hyperbolic Monge–Ampère equation to a quasilinear hyperbolic system in Riemann invariants. Section 4 is devoted to prove the main result, i.e., Theorem 2, by the local solution of theory for the Cauchy problem of the quasilinear hyperbolic system.


## 2. Hyperbolic Monge–Ampère Equation

Suppose forz ∈ (a, b) and t ∈ (t0, t1), the curve γ(z, t) can be expressed in the form of a graph (x, u(x, t)), and we have(8)γt=xt1,ux+0,ut,γtt=xtt1,ux+0,uxxxt2+2uxtxt+utt.Taking the inner product with the choiceN⟶=−ux,1/1+ux2 and T⟶=1,ux/1+ux2, respectively, yields(9)utt+2xtuxt+xt2uxx=1+ux2−k−1,(10)xtt=k−1s+uxk−11+ux2.It would be convenient to write the expression of the curvaturek for graphs:(11)k=ux1+ux2x=uxx1+ux23/2.Flow (1) is a normal preserving flow. First, we need to compute ∂tγtz,N⟶. Using the identities(12)N⟶t=N⟶t,T⟶T⟶=−N⟶,T⟶tT⟶=−<N⟶,∂tγzγz>T⟶=−γz−1N⟶,γtz+∂tγz−1N⟶,γz⟶T⟶=−γz−1N⟶,γtzT⟶,γttz,N⟶=−k−1z+k−1k−1sT⟶z,N⟶=−k−1z+k−1k−1skγz=0,one computes(13)∂∂tγtz,N⟶=γttz⟶,N⟶+γtz,N⟶t=−γz−1γtz,T⟶γtz,N⟶=−γts,T⟶γtz,N⟶.This is an ODE of the formdy/dt=by. Clearly, (1) is a normal preserving flow. Then, having observed that also the tangent vector is a constant, by using it, we can easily to show the identity(14)uxxxt+uxt=0.Byxt=−uxt/uxx, the normal preserving flow is reducible, and the curvature k for graphs uxx/1+ux23/2, graph form of flow (1), satisfies the following equation:(15)uxxutt−uxt2=−1+ux22.This is a fully nonlinear hyperbolic Monge–Ampère equation as long as the curve is uniformly convex, i.e., flow (1) reduces to(16)uxxutt−uxt2=−1+ux22,ux,0=fx,utx,0=gx.Remark 1.
We have expressed flow (1) as the equation of support function in (3). When the curve γ is represented as a graph and described by the support function simultaneously, the following relations hold:(17)S=u−xux1+ux2,tanθ=−1ux,Sθ=−x−uux1+ux2.
For an unknown functionh = h(x, t) defined on x,t∈ℝ2, Monge–Ampère equation is(18)A+Bhtt+Chxt+Dhxx+Ehxxhtt−hxt2=0,where the coefficients A, B, C, D, and E depend on t, x, h, ht, hx. We say that (18) is a hyperbolic Monge–Ampère equation if(19)Δ2t,x,h,ht,hx≜C2−4BD+4AE>0,hxx+Bt,x,h,ht,hx≠0.
Initial conditions for the Cauchy problem on theOx axis are(20)h0,x=h0x,ht0,x=h1x,where h0∈C3ℝ1 and h1∈C2ℝ1. Suppose h0 and h1 satisfy the next two conditions. First, the axis Ox is free, i.e.,(21)h0″+B0,x,h0x,h1x,h0′y≠0.
Secondly, on the axisOx, the hyperbolic condition(22)Δ20,x,h0y,h1y,h0′y>0,holds.
It is easy to prove that (15) is a hyperbolic Monge–Ampère equation, where(23)A=1+ux22,B=0,C=0,D=0,E=1.
In fact,(24)Δ2t,x,u,ut,ux=C2−4BD+4AE=02−4⋅02+41+ux22=41+ux22>0,uxx+Bt,x,u,ut,ux=uxx+0=k1+ux23≠0.
Furthermore, we suppose thatgx is the third continuous differential function and f(x) is second continuous differential function; then, initial conditions satisfy(25)Δ20,x,f,g,f′=41+f′22>0,uxxx,0+B0,x,f,g,f′=f′′+0=k01+f′23≠0,in which f′=df/dx and f″=d2f/dx2.
Hence, by the standard theory of hyperbolic equations (see [16]), we have local existences and uniqueness theorem (i.e., Theorem 1).Remark 2.
In fact, the derivation of (16) and Theorem 1 can also be obtained by a direct application of the arguments therein (cf. Section 1 in [1]). When establishing the local solvability for the normal preserving flow, the authors in [1] suppose the initial curve γ0 ∈ Hk(S1) and γt(0) ∈ Hk−1(S1), k > 5/2; however, we just provide that γ0 and γt(0) are smooth.


## 3. Systems in Riemann Invariants

This section is concerned with the reduction of (15). Let u(x, t) be a C3 solution of equation (15) in some domain D˜. Suppose(26)uxx≠0,∀x,t∈D˜.Condition (26) means that vertical lines t = const are free. By definition, put(27)r=C+Δ−2uxt2uxx=1+ux2−uxtuxx,s=C−Δ−2uxt2uxx=−1−ux2−uxtuxx.The functionsr and s are tangents of angles of inclinations of characteristics of equation (15), and we always call them Riemann invariants of (15). Let p = ut and q = ux; then,(28)rt+srx=r−sq,st+rsx=r−sq,ut+rux=p+rq,pt+spx=s1+q2,qt+rqx=1+q2.Theorem 3.
Letu(x, t) be a C3- solution of (15). Suppose (26) and (15) are satisfied by u. Then, the set of functions (r, s, u, p, q), where r and s are obtained from (27), p = ut, and q = ux, is a C1 solution of the system of five equations (28) in D˜.Theorem 4.
Let (r, s, u, p, q) be a C1–solution of (28) in the domain D˜, satisfying the initial value condition(29)u0,x=fx,p0,x=gx,q0,x=f′x,r0,x=1+f′2x−g′xf″x,s0,x=−1−f′2x−g′xf″x.
LetD˜ be a domain in which this solution is defined. Suppose(30)Δt,x,ut,x,utt,x,uxt,x>0,holds. Then, u(x, t) be a C3 solution of (15) in the domain D˜, and furthermore, ut = p, ux = q, and (26) holds.
Letu=u1,u2,u3,u4,u5T=r,s,u,p,qT,(31)Λ=λ100000λ200000λ300000λ400000λ5=s00000r00000r00000s00000r,μ=μ1,…,μ5T=r−sq,r−sq,p+rq,s1+q2,1+q2T.
Then, we reduce equation (28) to the following Cauchy problem for a quasilinear hyperbolic system in terms of Riemann invariants:(32)∑j=15δij∂uj∂t+λi∂uj∂x=μi,i=1,…,5,ui0,x=ϕix,i=1,…,5,where ϕ=ϕ1,…,ϕ5T=1+f′2−g′/f″,−1−f′2−g′/f″,f,g,f′T.
Throughout this paper, we shall use the following notation: the absolute value of any vectoru = (u1, …, u5) is defined as(33)u=maxiui,and the C0 norm of a vector function u = u(t, x) = (u1(t, x), …, un(t, x)) on a bounded domain R is defined as(34)u=supt,x∈ℝ2ut,x.
Firstly, supposeϕ=ϕ1,…,ϕ5T is bounded on R and ϕ∈C1ℝ. Take a suitable positive constant Ω such that(35)ϕ<Ω.
For the continuous differentiable vector functionu(t, x), we may define(36)u1=u+∂u∂t+∂u∂x.
Furthermore, for any finite set of functions setΓ⊂C1R¯ or Γ⊂C1R¯×0,δ, and we can similarly define its norm ‖Γ‖, ‖Γ‖1.
DenoteR(δ0) = {(t, x)∣t ∈ [0, δ0], x ∈ R}, where R is a bounded domain, and define its exterior domain Eδ0=t,x,ut,x∈Rδ0,u≤Ω. Furthermore, suppose λi,∂λi/∂x,∂λi/∂uj,μi,∂μi/∂x,∂μi/∂uj are C0 functions defined on E(δ0), by the standard hyperbolic theory, there exists a constant 0 < δ∗ ≤ δ0 such that the Cauchy problem of systems (28) and (29) has an unique C1 solution on R(δ∗).


## 4. The Proof of Theorem2

In this section, we consider the life-span of the solution by the construct method of the local solution for the Cauchy problem of the quasilinear hyperbolic system. To prove this theorem, we first consider the initial value problem of the equation for the graph (x, u(x, t)) of the curve. Since the initial curve γ0 is a smooth curve, we can find open intervals Jα, α = 1, …, N, in different coordinates and smooth functions u0α defined on Jα such that γ0 can be described as the union of the graphs x,u0αx:x∈Jα, α = 1, …, N. Assume uα is the solution with initial value u0α for each α. Note that u(x, t) ≡ uα satisfies (16).The following is about the proof of Theorem2.First, we consider quasilinear hyperbolic system (28) to be linear. Take a suitable positive constant Ω1 such that (37)Ω1>ϕ+1+Ωϕ˙+Ω3+2Ω2+Ω+1.Here, Ω1 is a positive constant derived from second a priori estimate, ϕ˙=dϕ/dx. Suppose the following function set(38)∑δΩ,Ω1=νt,xν=ν1,…,ν5T∈C1Rδ,ν≤Ω,ν1≤Ω1,and for each ν ∈ ∑(δ∣Ω, Ω1), by the linear hyperbolic system,(39)∑j=15δlj∂uj∂t+λ˜lt,x∂uj∂x=μ˜lt,x,l=1,…,5,and the initial conditions ul(0, x) = ϕl(x), x ∈ R1, l = 1, …, 5, in which λ˜lt,x=λlt,x,νt,x, i.e., λ˜=ν2,ν1,ν1,ν2,ν1T and μ˜lt,x=μlt,x,νt,x, i.e.,(40)μ˜=μt,x,νt,x=ν1−ν2ν5,ν1−ν2ν5,ν4+ν1ν5,ν21+ν52,1+ν52T.Defining the function setΓ∗=λl,∂λl/∂x,∂λl/∂uj,μl,∂μl/∂x,∂μl/∂uj,l,j=1,…,5, we get the C0 norm estimates of the set Γ∗:(41)maxl=1,…,5λlt,x,ν=maxν1,ν2≤Ω,maxl=1,…,5∂λl∂xt,x,ν=0,maxl,j=1,…,5∂λl∂ujt,x,ν=1,maxl=1,…,5μlt,x,ν≤max2Ω2,2Ω2,Ω+Ω2,1+Ω2Ω,1+Ω2≤Ω3+2Ω2+Ω+1,maxl=1,…,5∂μl∂xt,x,ν=0,∂μl∂ujt,x,ν=ν5ν5ν500−ν5−ν501+ν5200000000100ν1−ν2ν1−ν2ν12ν1ν52ν5,maxl,j=1,…,5∂μl∂ujt,x,ν≤max2Ω2,1+Ω2,2Ω,1≤2Ω2+2Ω+1.Hence, theC0 norm of the set Γ∗ can be bounded as follows:(42)Γ∗=maxΩ,0,1,Ω3+2Ω2+Ω+1,2Ω2+2Ω+1.LetΓ2ν=λ˜lt,x,∂λ˜l/∂xt,x,,μ˜lt,x,∂μ˜l/∂xt,x; by(43)∂λ˜l∂xt,x=∂λl∂xt,x,ν+∑j=15∂λl∂ujt,x,ν∂νj∂x,l=1,…,5,(44)∂μ˜l∂xt,x=∂μl∂xt,x,ν+∑j=15∂μl∂ujt,x,ν∂νj∂x,l=1,…,5,we have(45)∂λ˜l∂x≤Ω1,∂μ˜l∂x≤Ω1max4Ω,2Ω+1,3Ω2+1≤3Ω2+4Ω+1Ω1.Therefore, we can bound theC0 norm of the set Γ2[ν] as follows:(46)Γ2ν≤maxΩ,Ω1,Ω3+2Ω2+Ω+1,3Ω2+4Ω+1Ω1.In order to prove Theorem2, we will get some priori estimates for the solutions of Cauchy problems for linear hyperbolic systems. These estimates will be useful in proving the existence and uniqueness of solutions for the initial value problem of quasilinear hyperbolic systems.Firstly, by the first associated integral relations of initial value problem (39),(47)ukt,x=uk0t,x+∫0tμkτ,ωkτ;t,xdτ,uk0,x=ϕkωk0;t,x,k=1,…,5.Hereafter, the argument of each function in thel-th integrand is (τ, ξ) = (τ, ωl(τ; t, x)),(48)dωldlτ=λlτ,ωlτ,x,ωlt;t,x=x,and d/dlτ means the directional derivative along the l-th characteristic curve, i.e.,(49)ddlτ=∂∂τ+λl∂∂ξ.Moreover,(50)uk0=∑l,j=1nδklt,xδlj0,ξlt,xϕjξlt,x,k=1,2,…,5,where (δkl(x, t)) is the inverse matrix of (δlj(t, x)), and each δkl(x, t) is also a C1 function of (t, x). By(51)maxk=1,…,5uk0t,x≤ϕ,maxl=1,…,5μlt,x,ν≤Ω3+2Ω2+Ω+1,if 0 ≤ t ≤ δ, we have(52)u≤ϕ+Ω3+2Ω2+Ω+1δ.This is first a priori estimate for the solution of Cauchy problem.We are going to construct second a priori estimate for the first derivatives of the solutions of Cauchy problem. Define the function set(53)p=pit,x=∂uit,x∂t,w=wit,x=∂uit,x∂x,i=1,…,5,μ¯lt,x=∂μlt,x∂x−∂λl∂xwl,l=1,…,5.By the second associated integral,(54)wkt,x=wk0t,x+∫0tμ¯kτ,ωkτ;t,xdτ,k=1,…,5,(55)wk0t,x=ϕ˙kωk0;t,x,k=1,…,5,(56)pkt,x=pk0t,x−λk∫0tμ¯kτ,ωkτ;t,xdτ,k=1,…,5,(57)pk0t,x=μkt,x−λlt,xϕ˙kωk0;t,x,k=1,…,5.OnR(δ), we have(58)maxk=1,…,5wk0t,x≤ϕ.,maxk=1,…,5pk0t,x≤Ωϕ.+Ω3+2Ω2+Ω+1.Let(59)wτ=supk=1,…,5wkt,x,t,x∈Rτ,pτ=supk=1,…,5pkt,x,t,x∈Rτ,obviously; we have(60)w=sup0≤τ≤δ=wδ,p=sup0≤τ≤δ=pδ.By (54), if 0 ≤ t ≤ δ,(61)wt≤ϕ˙+3Ω2+4Ω+1Ω1δ+Ω1∫0twτdτ.By Gronwall inequality,(62)wt≤ϕ˙+3Ω2+4Ω+1Ω1δeΩ1t,we have(63)w≤ϕ˙+3Ω2+4Ω+1Ω1δeΩ1δ.By (62), we have(64)pt≤Ωϕ˙+Ω3+2Ω2+Ω+1+3Ω2+4Ω+1ΩΩ1δ+ΩΩ1∫0tϕ˙+3Ω2+4Ω+1Ω1δeΩ1τdτ=Ω3+2Ω2+Ω+1+Ωϕ˙+3Ω2+4Ω+1Ω1δeΩ1t,i.e.,(65)p≤Ω3+2Ω2+Ω+1+Ωϕ˙+3Ω2+4Ω+1Ω1δeΩ1δ.By the definition of ‖u‖1,(66)u1≤ϕ+Ω3+2Ω2+Ω+11+δ+Ω+1ϕ˙+3Ω2+4Ω+1Ω1δeΩ1δ.Without loss generality, we assumeδ0≤1/Ω1, and we restrict to δ ∈ (0, δ0); then,(67)eΩ1δ≤e1<3.By Taylor’s expansion,(68)eΩ1δ<1+Ω1δeΩ1δ<1+3Ω1δ.Plugging (68) into (66) and simplifying the second order term of δ,(69)u1≤ϕ+Ω+1ϕ˙+Ω3+2Ω2+Ω+11+δ+Ω+112Ω2+16Ω+4+3ϕ˙Ω1δ.This is second estimate for the first derivatives of the solution of the Cauchy problem.Next, we are going to estimate the modulus of continuity of the first-order derivatives of the solution. Letψ(t, x) be a function on the bounded domain R. The module of continuity of ψ(t, x) is defined by the following nonnegative function:(70)ρη=ρηψ=supt′−t″≤ηx′−x″≤ηt′,x′,t″,x″∈Rψt′,x′−ψt″,x″,0≤η<∞.Similarly, we can define the module of continuity of a function with any number of independent variables. Also, the modulus of continuity of a vector functionψ = (ψi(t, x))(i = 1, …, n) or a set of functions Γ = {ψi} is defined as follows:(71)ρηψ=maxi=1,…,nρηψi,ρηΓ=maxψi∈Γρηψi.It is easy to see that the modulus of continuity possesses the properties stated in Lemmas1 and 2.Lemma 1.
(see [15]). It holds that(72)ρη2≥ρη1≥0;ifη2≥η1≥0,ρη1+η2≤ρη1+ρη2.
f is a continuous function on t1,t2⊂ℝ iff(73)ρη⟶0as,η⟶0.
f is Hölder continuous ont1,t2⊂ℝ with exponent α(0 < α ≤ 1) iff there exists a constant L such that(74)ρη≤Lηα,for any natural number N and any positive constant C, we have(75)ρNη≤Nρη,ρCη≤C+1ρη,where [C] denotes the integer part of C.Lemma 2.
(see [15]). Assuming that the right hand side in each of the following formulas makes sense, then(76)ρηφ±ψ≤ρηφ+ρηψ;ρηφψ≤φρηψ+ψρηφ;if(77)ψ≥a=constant>0,then(78)ρηφψ≤φa2ρηψ+1aρηφ;if φ = φ(t), then(79)ρηφ≤φ.η;φ.=dφdt,ρηφψt≤ρρηψφ;letting(80)ψt=∫atbtφτ,tdτ,we have(81)ρηψt≤φρηa+ρηb+supt∫atbtρηφτ,tdτ,where the integrand ρ(η∣φ(τ, t)) is the modulus of continuity of φ with respect to t (τ is regarded as a parameter).
By the ordinary equation whichξ = ωl(τ; t, x) satisfies(82)∂ξ∂τ=λlτ,ξτ;t,x,and the initial conditions(85)τ=t:ωlt;t,x=x.
It is easy to see that(86)∂ωl∂xτ;t,x=exp∫tτ∂λl∂ξτ1,ωlτ1;t,xdτ1,∂ωl∂tτ;t,x=−λlexp∫tτ∂λl∂ξτ1,ωlτ1;t,xdτ1.
Hence,(87)∂ωl∂x≤exp∂λl∂xt≤1+∂λl∂xδexp∂λl∂xδ,l=1,…,5,∂ωl∂t≤λlexp∂λl∂xt≤‖λl‖1+∂λl∂xδexp∂λl∂xδ,l=1,…,5.
Sinceδ<δ0≤1/Ω1,(88)∂ω∂x≤1+3Ω1δ,∂ω∂t≤1+3Ω1δΩ.
By (55) and (57) and Lemma 1, on R(δ)(0 ≤ δ ≤ δ0), we have(89)ρηwk0t,x=ρηϕ˙ωk0;t,x≤1+Ω1+3Ω1δρηϕ˙k,ρηpk0t,x=ρημk−λkϕ˙ωk0;t,x≤ρημk+λkρηϕ˙ωk0;t,x+ϕ˙kρηλk.
Then,(90)ρηpk0t,x≤Ω1+Ω1+3Ω1δρηϕ˙k+1+ϕ˙χη,where(91)χη=maxρημk,ρηλk.
Hence, we have(92)ρηw0≤1+Ω1+3Ω1δρηϕ˙,(93)ρηp0≤Ω1+Ω1+3Ω1δρηϕ˙+1+ϕ˙χη.
Furthermore,(94)ρηw0+ρηp0≤1+Ω21+3Ω1δρηϕ˙+1+ϕ˙χη.
Define(95)ρτ,ηw=supk=1,…,5t′,x′,t″,x″∈Rτt′−t″|≤η,x′−x″≤ηwkt′,x′−wkt″,x″,ρτ,ηp=supk=1,…,5t′,x′,t″,x″∈Rτt′−t″|≤η,x′−x″≤ηpkt′,x′−pkt″,x″.
Then, it is obvious that, onR(δ)(0 ≤ δ ≤ δ0),(96)ρηw=sup0≤τ≤δρτ,ηw=ρδ,ηw,ρηp=sup0≤τ≤δρτ,ηp=ρδ,ηp.
By the second integral relations (54) and (56), we will estimate ρt,ηw and ρ(t, η∣ p) for 0 ≤ t ≤ δ. By Lemmas 1 and 2 and (94), for 0≤t¯≤δ (note that here and in the following, τ (τ ≤ t) is regarded as a parameter), we get(97)ρt¯,ημ¯kτ,ωkτ;t,x≤41+Ωρτ,ημ¯kτ,ξ,and as τ ≤ t ≤ δ, we have(98)ρτ,ημ¯kτ,ξ=ρτ,η∂μkτ,ξ∂x−∂λkτ,ξ∂xwkτ,ξ≤ρτ,η∂μk∂x+∂λk∂xρτ,ηwkτ,ξ+wkρτ,η∂λk∂xτ,ξ≤Ω1ρτ,ηw+1+wχ1η≤Ω1ρτ,ηw+4ϕ˙+12Ω2+16Ω+5χ1η,where(99)χ1η=maxk=1,…,5ρτ,η∂μk∂x,ρτ,η∂λk∂x.
Hence, we have(100)ρt¯,ημ¯kτ,ωkτ;t,x≤41+ΩΩ1ρτ,ηw+4ϕ˙+12Ω2+16Ω+5χ1η.
By(101)μ¯k=∂μ˜k∂x−∂λ˜k∂xwk,k=1,…,5,we obtain(102)μ¯k≤∂μ˜k∂x+∂λ˜k∂xwk≤15Ω2+20Ω+5+4ϕ˙Ω1,i.e.,(103)μ¯≤15Ω2+20Ω+5+4ϕ˙Ω1.
By the properties of the continuity modulus,(104)ρη∫atbtψτ,tdτ≤ψρηa+ρηb+supt∫atbtρηψτ,tdτ,we have(105)ρt¯,η∫0tμ¯kτ,ωkτ;t,xdτ≤μ¯kη+∫0t¯ρτ1,ημ¯kτ,ωkτ;t,xdτ1≤15Ω2+20Ω+5+4ϕ˙Ω1η+41+ΩΩ1∫0t¯ρτ,ηwdτ+41+Ω4ϕ˙+12Ω2+16Ω+5δχ1η.
Hence, putting together (104) and (93),(106)ρt,ηpk+ρt,ηwk≤ρt,ηpk0+ρt,ηwk0+1+λρt,η∫0tμ¯kdτ+∫0tμ¯kdτχη≤1+ϕ˙+1+Ω1+ΩΩ115Ω2+20Ω+5+4ϕ˙χη+41+Ω24ϕ˙+12Ω2+16Ω+5δχ1η+41+Ω2ρηϕ˙+41+Ω2Ω1∫0tρτ,ηwdτ+∫0tρτ,ηpdτ.
Denote(107)ρt,η=ρt,ηp+ρt,ηw,and by Gronwall inequality and t≤δ<δ0<1/Ω1, we have(108)ρt,η≤e41+Ω21+ϕ˙+1+Ω1+ΩΩ115Ω2+20Ω+5+4ϕ˙χη+41+Ω2ρηϕ˙+4ϕ˙+12Ω2+16Ω+5δχ1η.
By Cauchy problems of the linear hyperbolic system, there exists a unique continuous differential solutionu = (ui(t, x)) of (39) on R(δ). Denote(109)u=Tv.
Obviously, the fixed point of the operatorT is the solution of the initial value problem. Therefore, it is sufficient to prove that the operator T has a unique fixed point, provided that δ > 0 is suitably small. We first prove that there exists a positive number δ1 ≤ δ0 depending only on the norm ‖Γ∗‖ and ‖ϕ‖1:(110)δ1=δ1Γ∗,ϕ1,such that, for any δ ≤ δ1, the operator T maps ∑(δ ∣ Ω, O1) to itself. To do so, we use first estimate (52) and second estimate (69) to bound ‖u‖ and ‖u‖1. It follows that, for u = Tv, we have on R(δ)(111)u≤ϕ+Ω3+Ω2+Ω+1δ.
Take Ω = 2‖ϕ‖,(112)δ=ϕ8ϕ3+8ϕ2+2ϕ+1,we have(113)u≤Ω.
By (69),(114)u1≤ϕ+Ω3+2Ω2+ϕ˙+1Ω+1+Ω3+2Ω2+Ω+1δ+Ω+112Ω2+16Ω+4+3ϕ˙Ω1δ.
We take(115)Ω1=2ϕ+2Ω3+4Ω2+2ϕ˙+1Ω+1=4ϕ+2ϕ˙+16ϕ3+16ϕ2+6ϕ+2.
LetA0(‖ϕ‖) = (8‖ϕ‖3 + 8‖ϕ‖2 + 3‖ϕ‖ + 1), A1(‖ϕ‖) = 2‖ϕ‖ + 1, A2(‖ϕ‖) = A0 − ‖ϕ‖, A3(‖ϕ‖) = 48‖ϕ‖2 + 32‖ϕ‖ + 4, B0(‖ϕ‖) = 2A0A1A3 + A2, B1ϕ=6A0A1+2A12A3, and B2ϕ=6A12. Then, we can choose(116)δ=A0+A1ϕ˙B0+B1ϕ˙+B2ϕ˙2.
Take(117)δ1=minϕA0−ϕ,A0+A1ϕ˙B0+B1ϕ+B2ϕ2.
It is easy to check that, for eachδ ∈ (0, δ1], on R(δ), we have(118)u≤Ω,u1≤Ω1,and therefore, the operator T maps ∑(δ ∣ Ω, O1) to itself.
Now, we will use third estimate (108) to prove the following Lemma.Lemma 3.
There exists a positive numberδ2=δ2ϕ,ϕ˙≤δ1 and a nonnegative function Ω2(η) (0 < η < ∞) such that(119)limη⟶0Ω2η=0,for any δ ≤ δ2, if v ∈∑(δ ∣ Ω, Ω1) and if(120)ρη∂v∂t+ρη∂v∂x≤Ω2η,then for u = Tv, we have ρ(η ∣ p) + ρηw ≤ Ω2(η).Proof.
Let(121)χ˜η=maxl=1,…,5ρηλ˜l,ρημ˜l,η.
By the definition of the modulus of continuity and the construction of Ω1, we have(122)χ˜η≤maxΩ1η,3Ω2+4Ω+1Ω1η,η=3Ω2+4Ω+1Ω1η,ρt,ηp+ρt,ηw≤e41+Ω21+ϕ˙+1+Ω1+ΩΩ115Ω2+20Ω+5+4ϕ˙χη+41+Ω2ρηϕ˙+41+Ω24ϕ˙+12Ω2+16Ω+5δχ1η,where(123)χ1η=maxk=1,…,5ρτ,η∂μ˜k∂x,ρτ,η∂λ˜k∂x.
By(124)ρτ,η∂λ˜k∂x≤maxρτ,η∂ν1∂x,ρτ,η∂ν2∂x≤ρτ,η∂ν∂x,ρτ,η∂μ˜k∂x≤max4Ωρτ,η∂ν∂x,2Ω+1ρτ,η∂ν∂x,1+3Ω2ρτ,η∂ν∂x≤3Ω2+4Ω+1ρη∂ν∂x,we have(125)χ1η≤3Ω2+4Ω+1ρη∂ν∂x.
Denote(126)Ω2η=2e41+Ω241+Ω2ρηϕ˙+3Ω2+4Ω+1×1+ϕ˙+1+Ω1+ΩΩ115Ω2+20Ω+5+4ϕ˙Ω1η,and then we take(127)δ2=141+Ω24‖ϕ˙‖+12Ω2+16Ω+53Ω2+4Ω+1.
Let(128)C0ϕ=41+2ϕ248ϕ2+32ϕ+512ϕ2+8ϕ+1,C1ϕ=161+2ϕ212ϕ2+8ϕ+1,then(129)δ2=1C0+C1ϕ˙.
Then, if 0 <δ2, Lemma 3, (1) and (2), holds. The proof is completed.
Takeδ3 = min{δ1, δ2}, and we introduce the function set(130)∑δ3=∑δ3Ω,Ω1,Ω2η=νt,xν∈∑δ3Ω,Ω1,ρη∂ν∂x+ρη∂ν∂t≤Ω2η,then the operator T maps ∑(δ3) to itself. It is easy to see that T is compact on C1[R(δ)] and closed with respect to the C0 norm. We now prove that there exists a positive number δ∗=δ∗ϕ,ϕ˙≤δ2 such that, on R(δ∗), the operator T is a contraction operator with respect to the C0 norm.
Firstly, under theC1 norm, the set ∑(δ3) is bounded and equicontinuous; hence, it is precompact. Assume(131)limn⟶∞νn=ν,then(132)ρην=limn⟶∞ρηνn≤Ω2η,ν≤Ω,ν1≤Ω1.
Hence, the set is compact inC1[R(δ)]. Moreover, ∑(δ3) is closed with respect to the C0 norm (because ‖ν‖ ≤ ‖ν‖1, the convergence under the C1 norm provides the convergence under C0 norm; hence, it is complete with respect to C0 norm). Hence, we only need to prove that there exists δ∗ ≤ δ3 such that the operator T is a contraction operator in R(δ∗) with respect to the C0 norm. If we can show this, the existence of a unique fixed point for T will be proved.
Letν(1), ν(2) ∈∑(δ3); then, u(1) = Tν(1), u(2) = Tν(2) ∈∑(δ3). Assume ν∗ = ν(1) − ν(2), u∗ = u(1) − u(2), and(133)λ˜lαt,x=λlαt,x,ναt,x,λ˜lαt,x=λlαt,x,ναt,x,α=1,2.
Then, we have(134)∂uk∗∂t+λ˜k1t,x∂uk∗∂x=h˜kt,x,t=0:uk∗=0,k=1,…,5,where h˜k=−λ˜k1−λ˜k2∂uk2/∂x+μ˜k1−μ˜k2, k = 1, …, 5.
Now, notice that, onR(δ), we have(135)h˜≤Ω1+3Ω2+4Ω+1ν∗.
By the estimate in (52), it follows from (134) that, on R(δ),(136)u∗≤Ω1+3Ω2+4Ω+1δν∗.
Take(137)δ4=13Ω1+3Ω2+4Ω+1,Ω1=2ϕ+2Ω3+4Ω2+2ϕ˙+1Ω+1=4ϕ+2ϕ˙+16ϕ3+16ϕ2+6ϕ+2.
LetD0(‖ϕ‖) = 48‖ϕ‖3 + 84‖ϕ‖2 + 42‖ϕ‖ + 9 and D1(‖ϕ‖) = 6(2‖ϕ‖ + 1); then, we have(138)δ4=1D0+D1ϕ˙,if we choose δ∗ = min{δ3, δ4}, and the operator T is a contraction operator on R(δ∗).
From the above, we haveδ∗ = min{δ3, δ4}, and by(139)D0ϕ<C0ϕ,D1ϕ<C1ϕ,we get δ2 < δ4. Hence, the lower bound of the local solution of the Cauchy problem δ∗ is(140)δ∗=minϕA0−ϕ,A0+A1ϕ˙B0+B1ϕ+B2ϕ2,1C0+C1ϕ.
The proof of Theorem2 is completed.



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*Source: 6905269-2020-03-19.xml*