# Initial Boundary Value Problem of the General Three-Component Camassa-Holm Shallow Water System on an Interval

Authors: Lixin Tian; Qingwen Yuan; Lizhen Wang
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We study the initial boundary value problem of the general three-component
Camassa-Holm shallow water system on an interval subject to inhomogeneous
boundary conditions. First we prove a local in time existence theorem and
present a weak-strong uniqueness result. Then, we establish a asymptotic stabilization
of this system by a boundary feedback. Finally, we obtain a result of blow-up
solution with certain initial data and boundary profiles.


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## Body

## 1. Introduction

It is well known that the Camassa-Holm equation has attracted much attention in the past decade. It is a nonlinear dispersive wave equation that models the propagation of unidirectional irrotational shallow water waves over a flat bed, as well as water waves moving over an underlying shear flow. It was first introduced by Fo kas and Fuchssteiner as a bi-Hamiltonian model. Cauchy problem and initial boundary value problem for Camassa-Holm equation have been studied extensively in a number of papers (see [1–15] and the references within).Fu and Qu in [16] proposed a coupled Camassa-Holm equation,
(1)mt=2mux+mxu+(mv)x+nvx,nt=2nvx+nxv+(nu)x+mux;
with m=u-uxx,n=v-vxx, which has peakon solitons in the form of a superposition of multipeakons and may as well be integrable. They investigated the local well-posedness and blow-up solutions of system (1) by means of Kato’s semigroup approach to nonlinear hyperbolic evolution equation and obtained a criterion and condition on the initial data guaranteeing the development of singularities in finite time for strong solutions of system (1) by energy estimates. Recently the initial boundary value problem for the system (1) has been established in [17]; moreover, the local well-posedness and blow-up phenomena for the coupled Camassa-Holm equation were also established in [16, 18–32]. In [33], Tian and Xu obtained the compact and bounded absorbing set and the existence of the global attractor for viscous system (1) with the periodic boundary condition in by uniform prior estimate.Recently, Fu and Qu in [34] introduced a general three-component Camassa-Holm equation as follows:
(2)mt=2mux+mxu+(mv+mw)x+nvx+lwx,nt=2nvx+nxv+(nu+nw)x+mux+lwx,lt=2lwx+lxw+(lu+lv)x+mux+nvx,
where m=u-uxx, n=v-vxx, and l=w-wxx. Equation (2) also has peakon solitons in the form of a superposition of multipeakons. Such system also conserves the H1-norm conservation law. Moreover, the well-posedness and blow-up phenomena for system (2) with peakons have been established in [35]. To our knowledge, the initial boundary value problem of (2) has not been studied yet. The first aim of this paper is to consider an initial boundary value problem of the following
(3)mt=2mux+mxu+(mv+mw)x+nvx+lwx,nt=2nvx+nxv+(nu+nw)x+mux+lwx,lt=2lwx+lxw+(lu+lv)x+mux+nvx,m(0,·)=m0,n(0,·)=n0,l(0,·)=l0,m(·,0)|Γl=ml,n(·,0)|Γl=nl,l(·,0)|Γl=ll,m(·,1)|Γr=mr,n(·,1)|Γr=nr,l(·,1)|Γr=lr,
where Γl={t∈[0,T]∣(u+v+w)(t,x)<0},Γr={t∈[0,T]∣(u+v+w)(t,x)>0}.Then, we will consider the asymptotic stabilization of (3) by means of a stationary feedback law acting on the inhomogeneous boundary condition. Following the step in [11], we convert the initial boundary value problem of (3) on the interval into an ODE system and two PDE systems. Then, we can consider the system (3) easily. Consequently, we obtain a local in time existence theorem, a weak-strong uniqueness result, asymptotic stabilization result on the interval, and a result of blow-up solution, respectively.Our paper is organized as follows. In Section2, we will consider an initial boundary value problem and the uniqueness of the solution to (3). By using the feedback law enjoyed by (3), the asymptotic stabilization on an interval is considered in Section 3. Finally, in Section 4, a result of blow-up solution with certain initial data and boundary profiles will be established.First, we begin with a general remark that will be used many times later.Remark 1.
LetT be a positive number and ΩT=[0,T]×[0,1]. Changing u(t,x) in-u(t,1-x), v(t,x) in-v(t,1-x),  w(t,x) in-w(t,1-x), and t in T-t, and it will be more convenient for us to analysis the system, if we define the following sets
(4)Pl={t∈[0,T]∣(u+v+w)(t,0)=0},Pr={t∈[0,T]∣(u+v+w)(t,1)=0}.
LetΛ=(1-∂x2)1/2, then the operator Λ-2 can be expressed as
(5)Λ-2f(x)=G*f(x)=12∫01e-|x-y|f(y)dy,
where G=(1/2)e-|x|. Now, let Ai=Λ-2Bi=G*Bi,  i=1,2,3, where Bi is an auxiliary function which lifts the boundary values ml, mr, nl, nr, and ll defined by
(6)Bi(t,x)=0,(t,x)∈[0,T]×[0,1],G*Bi(t,0)=vli(t),G*Bi(t,1)=vri(t),∀t∈[0,T],
where i=1,2,3.
Settingu=p+G*B1, v=q+G*B2, and w=r+G*B3, we can further rewrite the system (3) as
(7)p=G*m,q=G*n,r=G*l,p(t,0)=p(t,1)=0,q(t,0)=q(t,1)=0,r(t,0)=r(t,1)=0,(8)mt=(p+G*B1+v+w)mx+[2(p+G*B1)x+vx+wx]m+nvx+lwx,nt=(q+G*B2+u+w)nx+[2(q+G*B2)x+ux+wx]n+mux+lwx,lt=(r+G*B3+u+v)lx+[2(r+G*B3)x+ux+vx]l+mux+nvx,m(0,·)=m0,m(·,0)|Γl=ml,m(·,1)|Γr=mr,n(0,·)=n0,n(·,0)|Γl=nl,n(·,1)|Γr=nr,l(0,·)=l0,l(·,0)|Γl=ll,l(·,1)|Γr=lr,
where functions ml,mr,nl,nr,ll, and lr in C0([0,1],R) are the boundary values and m0,n0, and l0 in L∞(0,1) are the initial datum.Lemma 2.
We haveAi=G*Bi∈C0([0,T];C∞[0,1])∈C0([0,T];C∞[0,1]) and p~,q~,r~∈L∞((0,T),C1,1([0,1]))∩Lip((0,T),H01(0,1)), i=1,2,3,m,n,l∈L∞(ΩT)∩Lip((0,T),H-1(0,1)). Moreover, we also have the bounds
(9)∥Ai∥L∞((0,T);C1,1[0,1])≤cosh(1)sinh(1)(∥vri∥L∞(0,T)+∥vli∥L∞(0,T)),(i=1,2,3),(10)∥p~∥L∞((0,T);C1,1[0,1])≤2(1+cosh(1))∥m∥L∞(ΩT),∥∂tp~∥L∞((0,T);H01[0,1])≤∥∂tm∥L∞((0,T);H-1[0,1]),∥q~∥L∞((0,T);C1,1[0,1])≤2(1+cosh(1))∥n∥L∞(ΩT),∥∂tq~∥L∞((0,T);H01[0,1])≤∥∂tn∥L∞((0,T);H-1[0,1]),∥r~∥L∞((0,T);C1,1[0,1])≤2(1+cosh(1))∥l∥L∞(ΩT),∥∂tr~∥L∞((0,T);H01[0,1])≤∥∂tl∥L∞((0,T);H-1[0,1]).Proof.
A
i
,
p
~
,
q
~
,
r
~
,
(
i
=
1,2
,
3
) can be expressed, respectively, as
(11)Ai(t,x)=G*Bi(t,x)=1sinh(1)(sinh(x)vri(t)+sinh(1-x)vli(t)),(i=1,2,3),p~(t,x)=-∫0xsinh(x-x~)m(t,x~)dx~+sinh(x)sinh(1)∫01sinh(1-x~)m(t,x~)dx~,q~(t,x)=-∫0xsinh(x-x~)n(t,x~)dx~+sinh(x)sinh(1)∫01sinh(1-x~)n(t,x~)dx~,r~(t,x)=-∫0xsinh(x-x~)l(t,x~)dx~+sinh(x)sinh(1)∫01sinh(1-x~)l(t,x~)dx~.
Estimates (9) and (10) can be easily obtained from the above expressions.


## 2. Initial Boundary Value Problem

First, we define what we mean by a weak solution to (8). Our test functions will be in the space:
(12)Adm(ΩT)={{⋃{T}}φ∈C1(ΩT)∣φ(t,x)=0on[0,T]∖Γl×{0}∪[0,T]∖Γr×{0}∪{T}×[0,1]}.Definition 3.
When(p,q,r)∈L∞((0,T);Lip[0,1])×L∞((0,T);Lip[0,1])×L∞((0,T);Lip[0,1]), the function (m,n,l)∈L∞(ΩT)×L∞(ΩT)×L∞(ΩT) is the weak solution to (8) if for all φ∈Adm(ΩT):
(13)∬ΩT‍m(φt-(u+v+w)φx+(px+∂xG*B1)φ)dtdx=-∬ΩT‍(nvx+lwx)φ(t,x)dtdx-∫01m0(x)φ(0,x)dt+∫0T((u+v+w)(t,0)φ(t,0)m(t,0)-(u+v+w)(t,1)φ(t,1)m(t,1))dt,∬ΩT‍n(φt-(u+v+w)φx+(qx+∂xG*B2)φ)dtdx=-∬ΩT‍(mux+lwx)φ(t,x)dtdx-∫01n0(x)φ(0,x)dx+∫0T((u+v+w)(t,0)φ(t,0)n(t,0)-(u+v+w)(t,1)φ(t,1)n(t,1))dt,∬ΩT‍l(φt-(u+v+w)φx+(rx+∂xG*B3)φ)dtdx=-∬ΩT‍(mux+nvx)φ(t,x)dtdx-∫01l0(x)φ(0,x)dx+∫0T((u+v+w)(t,0)φ(t,0)l(t,0)-(u+v+w)(t,1)φ(t,1)l(t,1))dt.
It is obvious thatC01(ΩT)⊂Adm(ΩT); therefore, a weak solution to system (8) is also a solution to (8) in the distribution sense. And it is clear that a regular weak solution is a classical solution.Definition 4.
For(t,x)∈ΩT, we consider ω(·,t,x) the maximal solution satisfying
(14)ωt=-(u+v+w)(t,ω(t,x)),ω(0,x)=x.
We consider thatω is the flow of (u(t,x),v(t,x),w(t,x)). For (t,x)∈ΩT, ω(·,t,x) is defined on a set [e(t,x),h(t,x)]. Here e(t,x) is basically the entrance time in ΩT of the characteristic curve going through (t,x).Remark 5.
Obviouslye(t,x)>0 implies that ω(e(t,x),t,x)∈{0,1}.In the following, we consider a partition ofΩT, which allows us to distinguish the different influence zones in ΩT.Definition 6.
LetP={(t,x)∈ΩT|∃s∈[e(t,x),h(t,x)] such that ω∈{0,1} and (u+v+w)(s,ω(s,t,x))=0}∪{(s,ω(s,0,0))∣foralls∈[0,T]}∪{(s,ω(s,0,1))∣foralls∈[0,T]},
(15)I={(t,x)∈ΩT∖p∣e(t,x)=0},L={(t,x)∈ΩT∖p∣ω(e(t,x),t,x)=0},R={(t,x)∈ΩT∖p∣ω(e(t,x),t,x)=1}.Those points of the setP are tangent to the boundary, which are precisely the singular points of e and h. It's obviously that the sets P,I,L, and R constitute a partition of ΩT. Furthermore, if (t,x)∈L, then e(t,x)∈Γl, and if (t,x)∈R, then e(t,x)∈Γr.Definition 7.
Here, we consider the case of data(u,v,w)∈L∞([0,T];C1([0,1]))×L∞([0,T];C1([0,1]))×L∞([0,T];C1([0,1])),(ml,nl,ll)∈Cc1(Γl)×Cc1(Γl)×Cc1(Γl); (mr,nr,lr)∈Cc1(Γr)×Cc1(Γr)×Cc1(Γr),(m0,n0,l0)∈Cc1(0,1)×Cc1(0,1)×Cc1(0,1). We define the functions m,n,  and l in the following way.
When(t,x)∈P,  m(t,x)=0,  n(t,x)=0, and  l(t,x)=0, when (t,x)∈I,
(16)m(t,x)=m0(ω(0,t,x))×exp(∫0t[2(px+∂xG*B1)+vx+wx]×(s,ω(s,t,x))ds{∫})+∫0t(nvx+lwx)(s,ω(s,t,x))×exp(∫st[2(px+∂xG*B1)+vx+wx]×(s′,ω(s′,t,x))ds′{∫})ds,n(t,x)=n0(ω(0,t,x))×exp(∫0t[2(qx+∂xG*B2)+ux+wx]×(s,ω(s,t,x))ds{∫0t})+∫0t(mux+lwx)(s,ω(s,t,x))×exp({(s′,ω(s′,t,x))}[2(qx+∂xG*B2)+ux+wx]×(s′,ω(s′,t,x))ds′)ds,l(t,x)=l0(ω(0,t,x))×exp(∫0t[2(rx+∂xG*B3)+ux+vx]×(s,ω(s,t,x))ds{∫0t})+∫0t(mux+nvx)(s,ω(s,t,x))×exp(∫st[2(rx+∂xG*B3)+ux+vx]×(s′,ω(s′,t,x))ds′{∫st})ds,
when (t,x)∈L,
(17)m(t,x)=ml(e(t,x))×exp(∫e(t,x)t[2(px+∂xG*B1)+vx+wx]×(s,ω(s,t,x))ds{∫th})+∫e(t,x)t(nvx+lwx)(r,ω(r,t,x))×exp(∫st[2(px+∂xG*B1)+vx+wx]×(s′,ω(s′,t,x))ds′{∫st})dr,n(t,x)=nl(e(t,x))×exp(∫e(t,x)t[2(qx+∂xG*B2)+ux+wx]×(s,ω(s,t,x))ds{∫e(t,x)t})+∫e(t,x)t(mux+lwx)(s,ω(s,t,x))×exp(∫st[2(qx+∂xG*B2)+ux+wx]×(s′,ω(s′,t,x))ds′{∫st})ds,l(t,x)=ll(e(t,x))×exp(∫e(t,x)t[2(rx+∂xG*B3)+ux+vx]×(s,ω(s,t,x))ds{∫e(t,x)t})+∫e(t,x)t(mux+nvx)(s,ω(s,t,x))×exp(∫st[2(rx+∂xG*B3)+ux+vx]×(s′,ω(s′,t,x))ds′{∫st})ds,
when (t,x)∈R,
(18)m(t,x)=ml(e(t,x))×exp(∫e(t,x)t[2(px+∂xG*B1)+vx+wx]×(s,ω(s,t,x))ds{∫e(t,x)t})+∫e(t,x)t(nvx+lwx)(r,ω(r,t,x))×exp(∫st[2(px+∂xG*B1)+vx+wx]×(s′,ω(s′,t,x))ds′{∫st})dr,n(t,x)=nl(e(t,x))×exp(∫e(t,x)t[2(qx+∂xG*B2)+ux+wx]×(s,ω(s,t,x))ds{∫e(t,x)t})+∫e(t,x)t(mux+lwx)(s,ω(s,t,x))×exp(∫st[2(qx+∂xG*B2)+ux+wx]×(s′,ω(s′,t,x))ds′{∫st})ds,l(t,x)=ll(e(t,x))×exp(∫e(t,x)t[2(rx+∂xG*B3)+ux+vx]×(s,ω(s,t,x))ds{∫e(t,x)t})+∫e(t,x)t(mux+nvx)(s,ω(s,t,x))×exp(∫st[2(rx+∂xG*B3)+ux+vx]×(s′,ω(s′,t,x))ds′{∫st})ds.Lemma 8.
Since(m,n,l)∈L∞(ΩT)×L∞(ΩT)×L∞(ΩT) and satisfies (8), we immediately get that (m,n,l) is the weak solution of (8) and (m,n,l)∈W1,∞(0,T,H-1(0,1))×W1,∞(0,T,H-1(0,1))×W1,∞(0,T,H-1(0,1)). However, the functions m,n, and l satisfy the following estimates:
(19)∥m∥C0(ΩT)≤[max(∥m0∥L∞,∥ml∥L∞,∥mr∥L∞)+∥nvx+lwx∥L∞(0,1)T]×eT∥2(p+G*B1)x+vx+wx∥C0(ΩT),∥n∥C0(ΩT)≤[max(∥n0∥L∞,∥nl∥L∞,∥nr∥L∞)+∥mux+lwx∥L∞(0,1)T]×eT∥2(q+G*B2)x+ux+wx∥C0(ΩT),∥l∥C0(ΩT)≤[max(∥m0∥L∞,∥ml∥L∞,∥mr∥L∞)+∥mux+nvx∥L∞(0,1)T]×eT∥2(r+G*B3)x+ux+vx∥C0(ΩT),∥∂tm∥C0(ΩT)≤{2max(∥m0∥L∞,∥ml∥L∞,∥mr∥L∞)×(∥p+G*B1∥L∞((0,T);Lip[0,1])+∥v∥L∞((0,T);Lip[0,1])+∥w∥L∞((0,T);Lip[0,1]))+[(∥u∥L∞(ΩT)+∥∂xx2u∥L∞(ΩT))2+(∥v∥L∞(ΩT)+∥∂xx2v∥L∞(ΩT))2+(∥w∥L∞(ΩT)+∥∂xx2w∥L∞(ΩT))2]×(1+2(∥p+G*B1∥L∞((0,T);Lip[0,1])+∥v∥L∞((0,T);Lip[0,1])+∥w∥L∞((0,T);Lip[0,1])))T}×exp(2T(∥(p+G*B1)x∥L∞(ΩT)+∥vx∥L∞(ΩT)+∥wx∥L∞(ΩT))),∥∂tn∥C0(ΩT)≤{2max(∥n0∥L∞,∥nl∥L∞,∥nr∥L∞)×(∥q+G*B2∥L∞((0,T);Lip[0,1])+∥u∥L∞((0,T);Lip[0,1])+∥w∥L∞((0,T);Lip[0,1]))+[(∥u∥L∞(ΩT)+∥∂xx2u∥L∞(ΩT))2+(∥v∥L∞(ΩT)+∥∂xx2v∥L∞(ΩT))2+(∥w∥L∞(ΩT)+∥∂xx2w∥L∞(ΩT))2]×(1+2(∥q+G*B2∥L∞((0,T);Lip[0,1])+∥u∥L∞((0,T);Lip[0,1])+∥w∥L∞((0,T);Lip[0,1])))T}×exp(2T(∥(q+G*B2)x∥L∞(ΩT)+∥px∥L∞(ΩT)+∥wx∥L∞(ΩT))),∥∂tl∥C0(ΩT)≤{2max(∥l0∥L∞,∥ll∥L∞,∥lr∥L∞)×(∥r+G*B3∥L∞((0,T);Lip[0,1])+∥u∥L∞((0,T);Lip[0,1])+∥v∥L∞((0,T);Lip[0,1]))+[(∥u∥L∞(ΩT)+∥∂xx2u∥L∞(ΩT))2+(∥v∥L∞(ΩT)+∥∂xx2v∥L∞(ΩT))2+(∥w∥L∞(ΩT)+∥∂xx2w∥L∞(ΩT))2]×(1+2(∥r+G*B3∥L∞((0,T);Lip[0,1])+∥p∥L∞((0,T);Lip[0,1])+∥q∥L∞((0,T);Lip[0,1])))T}×exp(2T(∥(r+G*B3)x∥L∞(ΩT)+∥ux∥L∞(ΩT)+∥vx∥L∞(ΩT))).Definition 9.
We can define operatorE and a domain for the system (8) by:  forallp,q,r∈L∞((0,T);C1,1([0,1]))∩Lip([0,T];H01(0,1)),
(20)E(p)=p~∈L∞((0,T);C1,1([0,1]))∩Lip([0,T];H01(0,1)),E(q)=q~∈L∞((0,T);C1,1([0,1]))∩Lip([0,T];H01(0,1)),E(r)=r~∈L∞((0,T);C1,1([0,1]))∩Lip([0,T];H01(0,1)),CM0,M1,T={p,q,r∈L∞((0,T);C1,1([0,1]))∩Lip([0,T];H01(0,1))∣∥d∥L∞((0,T);C1,1[0,1])≤M0,∥d∥Lip((0,T);H01(0,1))≤M1},
where
(21)d(t,x)=max(p(t,x),q(t,x),r(t,x)),(t,x)∈[0,T]×[0,1].ObviouslyCM0,M1,T is convex and CM0,M1,T is compact with respect to the norm ∥·∥L∞((0,T);Lip([0,1])). We will endow CM0,M1,T with the norm ∥·∥L∞((0,T);Lip([0,1])). There exist positive numbers M0,M1, and T such that E maps CM0,M1,T into itself.Theorem 10.
There existsT>0, and (m,n,l) is a weak solution of (8) with p,q,r∈L∞((0,T);C1,1([0,1]))∩Lip([0,T];H01(0,1)) and m,n,l∈L∞(ΩT). Moreover, any such solution (p,q,r) is in fact in C0([0,T];W2,P(0,1))∩C1([0,T];W01,p(0,1)),forallp<+∞. Furthermore, the existence time of a maximal solution T≥min(T*,T~), with
(22)T*=maxα>0,β>0(16αln(|α-C1|4(1+cosh(1))α)),T~=maxα>0,β>0(16αln(|β|12α2)),C1=max[cosh(1)sinh(1)(∥vri∥L∞(0,T)+∥vli∥L∞(0,T))],i=1,2,3.Proof.
ForT~>0, we consider ml,mr,nl,nr,ll, and lr in C0([0,T~]) such that the sets Pl and Pr have only a finite number of connected components.
LetC0=max(∥mi∥L∞(0,1),∥ni∥L∞(Γl),∥li∥L∞(Γr)), where i=0,r,l and
(23)C1=max[cosh(1)sinh(1)(∥vri∥L∞(0,T)+∥vli∥L∞(0,T))],i=1,2,3.
Now, if u,v,w∈CM0,M1,T (see (21)), we have
(24)∥2(p+G*B1)x+vx+wx∥C∞(ΩT)≤2(∥∂xu∥C∞(ΩT)+∥∂xv∥C∞(ΩT)+∥∂xw∥C∞(ΩT)),∥2(q+G*B2)x+ux+wx∥C∞(ΩT)≤2(∥∂xu∥C∞(ΩT)+∥∂xv∥C∞(ΩT)+∥∂xw∥C∞(ΩT)),∥2(r+G*B3)x+ux+vx∥C∞(ΩT)≤2(∥∂xu∥C∞(ΩT)+∥∂xv∥C∞(ΩT)+∥∂xw∥C∞(ΩT)).
For all u,v,w∈L∞((0,T);W2,∞(0,1)), we have
(25)∥∂xu∥L∞(ΩT)≤2∥u∥L∞(ΩT)∥∂xx2u∥L∞(ΩT)≤(∥u∥L∞(ΩT)+∥∂xx2u∥L∞(ΩT)),∥∂xv∥L∞(ΩT)≤2∥v∥L∞(ΩT)∥∂xx2v∥L∞(ΩT)≤(∥v∥L∞(ΩT)+∥∂xx2v∥L∞(ΩT)),∥∂xw∥L∞(ΩT)≤2∥w∥L∞(ΩT)∥∂xx2w∥L∞(ΩT)≤(∥w∥L∞(ΩT)+∥∂xx2w∥L∞(ΩT)),∥nvx+lwx∥L∞(0,1)≤[(∥u∥L∞(ΩT)+∥∂xx2u∥L∞(ΩT))2+(∥v∥L∞(ΩT)+∥∂xx2v∥L∞(ΩT))2+(∥w∥L∞(ΩT)+∥∂xx2w∥L∞(ΩT))2],∥mux+lwx∥L∞(0,1)≤[(∥u∥L∞(ΩT)+∥∂xx2u∥L∞(ΩT))2+(∥v∥L∞(ΩT)+∥∂xx2v∥L∞(ΩT))2+(∥w∥L∞(ΩT)+∥∂xx2w∥L∞(ΩT))2],∥mux+nvx∥L∞(0,1)≤[(∥u∥L∞(ΩT)+∥∂xx2u∥L∞(ΩT))2+(∥v∥L∞(ΩT)+∥∂xx2v∥L∞(ΩT))2+(∥w∥L∞(ΩT)+∥∂xx2w∥L∞(ΩT))2].
We also define that d~(t,x)=max(p~(t,x),q~(t,x),r~(t,x)),(t,x)∈ΩT. If p,q,r∈CM0,M1,T, then from Lemmas 2 and 8, we derive that
(26)∥d~∥L∞((0,T);C1,1[0,1])≤2(1+cosh(1))(C0+12T*(M0+C1)2)×exp(6T*(M0+C1)),∥∂td~∥C0(ΩT)≤[6C0(M0+C1)+12(M0+2C1)2×(1+6(M0+C1)T~)]exp(6T~(M0+C1)).
Finally, to obtainp~,q~,r~∈CM0,M1,T, it is sufficient to show that
(27)2(1+cosh(1))(C0+12T*(M0+C1)2)×exp(6T*(M0+C1))≤M0,M0+[6C0(M0+C1)+12(M0+C1)2×(1+6T~(M0+C1))]×exp(6T~(M0+C1))≤M1,
if we have chosen T and M0; it is easy to choose M1 to satisfy the second inequality. For the above two inequalities, we just choose M0 and M1 sufficiently large and then T close to 0. More precisely:
(28)M0>2(1+cosh(1))C0,T*≤16(M0+C1)ln(M04(1+cosh(1))(M0+C1)),T~≤16(M0+C1)ln(M1-M012(M0+C1)2).
Maximizing the bound ofT, we can get minimum existence. Then, we get the result announced, where α=M0+C1, β=M1-M0.Lemma 11.
The operatorE:CM0,M1,T→CM0,M1,T is continuous with respect to ∥·∥L∞((0,T);Lip[0,1]).Proof.
The proof is omitted here; one can see a similar proof in [8, Proposition 2.4].
Now, we can apply Shauder's fixed point theorem to the operatorE, and we get the result that there exist fixed points p,q,r such that E(p)=p, E(q)=q, and E(r)=r, so we know that there exists a wake solution of (9).
(29)p,q,r∈L∞((0,T);C1,1([0,1]))∩Lip([0,T];H01(0,1)).


### 2.1. Uniqueness

We will prove the weak-strong uniqueness of weak solution of (8) in the following.Theorem 12.
Let(p,m),(q,n),(r,l)∈L∞((0,T);C1,1([0,1]))∩Lip([0,T];H01(0,1)×L∞((0,T);Lip([0,1])) be the weak solution of (7)-(8), then it is unique in L∞((0,T);C1,1([0,1]))×L∞(ΩT).Proof.
DefineΦ=m-m~,  Ψ=n-n~,  Υ=l-l~ and P=p-p~,  Q=q-q~,  H=r-r~, then we have
(30)P(t,·)=G*Φ(t,·),Q(t,·)=G*Ψ(t,·),H(t,·)=G*Υ(t,·),
where P,Q,H∈Lip([0,T];H01(0,1)), and (Φ,Ψ,Υ)∈L∞(ΩT)×L∞(ΩT)×L∞(ΩT) is the unique weak solution of
(31)Φt=(p+G*B+v+w)Φx+[2(p+G*B1)x+vx+wx]Φ+(P+Q+H)m~x+(2Px+Qx+Hx)m~+Ψvx+n~Qx+Υwx+l~Hx,Ψt=(q+G*B+u+w)Ψx+[2(q+G*B2)x+ux+wx]Ψ+(P+Q+H)n~x+(Px+2Qx+Hx)n~+Φux+m~Φx+Υwx+l~Hx,Υt=(r+G*B+u+v)Υx+[2(r+G*B3)x+ux+vx]Υ+(P+Q+H)l~x+(Px+Qx+2Hx)l~+Φux+m~Φx+Ψvx+n~Qx.
Let(32)b1=2(p+G*B1)x+vx+wx,b2=2(q+G*B2)x+ux+wx,b3=2(r+G*B3)x+ux+vx,f1=(P+Q+H)m~x+(2Px+Qx+Hx)m~+Ψvx+n~Qx+Υwx+l~Hx,f2=(P+Q+H)n~x+(Px+2Qx+Hx)n~+Φux+m~Φx+Υwx+l~Hx,f3=(P+Q+H)l~x+(Px+Qx+2Hx)l~+Φux+m~Φx+Ψvx+n~Qx,
with i0=0,  il=0, and  ir=0, where i=Φ,Ψ,Υ.
For(t,x)∈P, we have Φ(t,x)=0,Ψ(t,x)=0, and Υ(t,x)=0.
Then, we get the uniqueness result.
For(t,x)∈I, we have
(33)Φ(t,x)=∫0tf1(s,ω(s,t,x))×exp(∫rtb1(s′,ω(s′,t,x))ds′)ds,Ψ(t,x)=∫0tf2(s,ω(s,t,x))×exp(∫rtb2(s′,ω(s′,t,x))ds′)ds,Υ(t,x)=∫0tf3(s,ω(s,t,x))×exp(∫rtb3(s′,ω(s′,t,x))ds′)ds.
For(t,x)∈L, we have
(34)Φ(t,x)=∫e(t,x)tf1(s,ω(s,t,x))×exp(∫rtb1(s′,ω(s′,t,x))ds′)ds,Ψ(t,x)=∫e(t,x)tf2(s,ω(s,t,x))×exp(∫rtb2(s′,ω(s′,t,x))ds′)ds,Υ(t,x)=∫e(t,x)tf3(s,ω(s,t,x))×exp(∫rtb3(s′,ω(s′,t,x))ds′)ds.
For(t,x)∈R, we have
(35)Φ(t,x)=∫e(t,x)tf1(s,ω(s,t,x))×exp(∫rtb1(s′,ω(s′,t,x))ds′)ds,Ψ(t,x)=∫e(t,x)tf2(s,ω(s,t,x))×exp(∫rtb2(s′,ω(s′,t,x))ds′)ds,Υ(t,x)=∫e(t,x)tf3(s,ω(s,t,x))×exp(∫rtb3(s′,ω(s′,t,x))ds′)ds.
Now since∥P(t,·)∥L∞(0,1)≤5∥Φ(t,·)∥L∞(0,1), ∥Q(t,·)∥L∞(0,1)≤5∥Ψ(t,·)∥L∞(0,1), ∥H(t,·)∥L∞(0,1)≤5∥Υ(t,·)∥L∞(0,1) and m~,∂xm~,n~,∂xn~,l~,∂xl~ bounded, we see that for some λ1>0, λ2>0,  λ3>0,
(36)∥f1(t,·)∥L∞(0,1)≤λ1(∥Φ(t,·)∥L∞(0,1)+∥Ψ(t,·)∥L∞(0,1)+∥Υ(t,·)∥L∞(0,1)),∥f2(t,·)∥L∞(0,1)≤λ2(∥Φ(t,·)∥L∞(0,1)+∥Ψ(t,·)∥L∞(0,1)+∥Υ(t,·)∥L∞(0,1)),∥f3(t,·)∥L∞(0,1)≤λ3(∥Φ(t,·)∥L∞(0,1)+∥Ψ(t,·)∥L∞(0,1)+∥Υ(t,·)∥L∞(0,1)),
since p,q,r, and Bi,(i=1,2,3) are bounded, we get that for some λ1′>0, λ2′>0, λ3′>0,
(37)∥Φ(t,·)∥L∞(0,1)≤λ1′∫0t(∥Φ(s,·)∥L∞(0,1)+∥Ψ(s,·)∥L∞(0,1)+∥Υ(s,·)∥L∞(0,1))ds,∥Ψ(t,·)∥L∞(0,1)≤λ2′∫0t(∥Φ(s,·)∥L∞(0,1)+∥Ψ(s,·)∥L∞(0,1)+∥Υ(s,·)∥L∞(0,1))ds,∥Υ(t,·)∥L∞(0,1)≤λ3′∫0t(∥Φ(s,·)∥L∞(0,1)+∥Ψ(s,·)∥L∞(0,1)+∥Υ(s,·)∥L∞(0,1))ds.
We can obtain that
(38)∥Φ(t,·)∥L∞(0,1)+∥Ψ(t,·)∥L∞(0,1)+∥Υ(t,·)∥L∞(0,1)≤(λ1′+λ2′+λ3′)×∫0t(∥Φ(s,·)∥L∞(0,1)+∥Ψ(s,·)∥L∞(0,1)+∥Υ(s,·)∥L∞(0,1))ds.
As a result, we get the result of the uniqueness by Gronwall's inequality when(t,x)∈I, (t,x)∈L, (t,x)∈R. Then, we complete the proof of the uniqueness results.


## 2.1. Uniqueness

We will prove the weak-strong uniqueness of weak solution of (8) in the following.Theorem 12.
Let(p,m),(q,n),(r,l)∈L∞((0,T);C1,1([0,1]))∩Lip([0,T];H01(0,1)×L∞((0,T);Lip([0,1])) be the weak solution of (7)-(8), then it is unique in L∞((0,T);C1,1([0,1]))×L∞(ΩT).Proof.
DefineΦ=m-m~,  Ψ=n-n~,  Υ=l-l~ and P=p-p~,  Q=q-q~,  H=r-r~, then we have
(30)P(t,·)=G*Φ(t,·),Q(t,·)=G*Ψ(t,·),H(t,·)=G*Υ(t,·),
where P,Q,H∈Lip([0,T];H01(0,1)), and (Φ,Ψ,Υ)∈L∞(ΩT)×L∞(ΩT)×L∞(ΩT) is the unique weak solution of
(31)Φt=(p+G*B+v+w)Φx+[2(p+G*B1)x+vx+wx]Φ+(P+Q+H)m~x+(2Px+Qx+Hx)m~+Ψvx+n~Qx+Υwx+l~Hx,Ψt=(q+G*B+u+w)Ψx+[2(q+G*B2)x+ux+wx]Ψ+(P+Q+H)n~x+(Px+2Qx+Hx)n~+Φux+m~Φx+Υwx+l~Hx,Υt=(r+G*B+u+v)Υx+[2(r+G*B3)x+ux+vx]Υ+(P+Q+H)l~x+(Px+Qx+2Hx)l~+Φux+m~Φx+Ψvx+n~Qx.
Let(32)b1=2(p+G*B1)x+vx+wx,b2=2(q+G*B2)x+ux+wx,b3=2(r+G*B3)x+ux+vx,f1=(P+Q+H)m~x+(2Px+Qx+Hx)m~+Ψvx+n~Qx+Υwx+l~Hx,f2=(P+Q+H)n~x+(Px+2Qx+Hx)n~+Φux+m~Φx+Υwx+l~Hx,f3=(P+Q+H)l~x+(Px+Qx+2Hx)l~+Φux+m~Φx+Ψvx+n~Qx,
with i0=0,  il=0, and  ir=0, where i=Φ,Ψ,Υ.
For(t,x)∈P, we have Φ(t,x)=0,Ψ(t,x)=0, and Υ(t,x)=0.
Then, we get the uniqueness result.
For(t,x)∈I, we have
(33)Φ(t,x)=∫0tf1(s,ω(s,t,x))×exp(∫rtb1(s′,ω(s′,t,x))ds′)ds,Ψ(t,x)=∫0tf2(s,ω(s,t,x))×exp(∫rtb2(s′,ω(s′,t,x))ds′)ds,Υ(t,x)=∫0tf3(s,ω(s,t,x))×exp(∫rtb3(s′,ω(s′,t,x))ds′)ds.
For(t,x)∈L, we have
(34)Φ(t,x)=∫e(t,x)tf1(s,ω(s,t,x))×exp(∫rtb1(s′,ω(s′,t,x))ds′)ds,Ψ(t,x)=∫e(t,x)tf2(s,ω(s,t,x))×exp(∫rtb2(s′,ω(s′,t,x))ds′)ds,Υ(t,x)=∫e(t,x)tf3(s,ω(s,t,x))×exp(∫rtb3(s′,ω(s′,t,x))ds′)ds.
For(t,x)∈R, we have
(35)Φ(t,x)=∫e(t,x)tf1(s,ω(s,t,x))×exp(∫rtb1(s′,ω(s′,t,x))ds′)ds,Ψ(t,x)=∫e(t,x)tf2(s,ω(s,t,x))×exp(∫rtb2(s′,ω(s′,t,x))ds′)ds,Υ(t,x)=∫e(t,x)tf3(s,ω(s,t,x))×exp(∫rtb3(s′,ω(s′,t,x))ds′)ds.
Now since∥P(t,·)∥L∞(0,1)≤5∥Φ(t,·)∥L∞(0,1), ∥Q(t,·)∥L∞(0,1)≤5∥Ψ(t,·)∥L∞(0,1), ∥H(t,·)∥L∞(0,1)≤5∥Υ(t,·)∥L∞(0,1) and m~,∂xm~,n~,∂xn~,l~,∂xl~ bounded, we see that for some λ1>0, λ2>0,  λ3>0,
(36)∥f1(t,·)∥L∞(0,1)≤λ1(∥Φ(t,·)∥L∞(0,1)+∥Ψ(t,·)∥L∞(0,1)+∥Υ(t,·)∥L∞(0,1)),∥f2(t,·)∥L∞(0,1)≤λ2(∥Φ(t,·)∥L∞(0,1)+∥Ψ(t,·)∥L∞(0,1)+∥Υ(t,·)∥L∞(0,1)),∥f3(t,·)∥L∞(0,1)≤λ3(∥Φ(t,·)∥L∞(0,1)+∥Ψ(t,·)∥L∞(0,1)+∥Υ(t,·)∥L∞(0,1)),
since p,q,r, and Bi,(i=1,2,3) are bounded, we get that for some λ1′>0, λ2′>0, λ3′>0,
(37)∥Φ(t,·)∥L∞(0,1)≤λ1′∫0t(∥Φ(s,·)∥L∞(0,1)+∥Ψ(s,·)∥L∞(0,1)+∥Υ(s,·)∥L∞(0,1))ds,∥Ψ(t,·)∥L∞(0,1)≤λ2′∫0t(∥Φ(s,·)∥L∞(0,1)+∥Ψ(s,·)∥L∞(0,1)+∥Υ(s,·)∥L∞(0,1))ds,∥Υ(t,·)∥L∞(0,1)≤λ3′∫0t(∥Φ(s,·)∥L∞(0,1)+∥Ψ(s,·)∥L∞(0,1)+∥Υ(s,·)∥L∞(0,1))ds.
We can obtain that
(38)∥Φ(t,·)∥L∞(0,1)+∥Ψ(t,·)∥L∞(0,1)+∥Υ(t,·)∥L∞(0,1)≤(λ1′+λ2′+λ3′)×∫0t(∥Φ(s,·)∥L∞(0,1)+∥Ψ(s,·)∥L∞(0,1)+∥Υ(s,·)∥L∞(0,1))ds.
As a result, we get the result of the uniqueness by Gronwall's inequality when(t,x)∈I, (t,x)∈L, (t,x)∈R. Then, we complete the proof of the uniqueness results.


## 3. Asymptotic Stabilization

### 3.1. Preliminary Results

The equilibrium state that we want to stabilize ism=n=l=0, p=q=r=G*Bi=0, and i=1,2,3. A natural idea is using Lyapunov indirection method to investigate whether the linearized system around the equilibrium state is stabilizable or not. Its stabilization would provide a local stabilization result on the nonlinear system. However, there is a difficulty in the stabilization problem. We have to prescribe yl, and we just need to make a continuous transition at (t,x)=(0,0), and that yl asymptotically converge in time. For convenience, the system (6)–(8) can rewrite in the following
(39)∂ty-a˘(t,x)∂xy=b˘(t,x)y+f˘(t,x),y(0,·)=y0,y(·,0)=yl,y(1,·)=yr,g˘(t,x)=G*y(t,x),g˘(t,0)=g˘(t,1)=0,B˘(t,x)=0,G*B˘(t,0)=vl(t),G*B˘(t,1)=vr(t),
where
(40)b˘(t,x)=(2(p+G*B˘)x+vx+wx00002(q+G*B˘)x+ux+wx002(r+G*B˘)x+ux+vx),a˘(t,x)=((p+G*B˘+v+w)0000(u+q+G*B˘+w)00(u+v+r+G*B˘)),y=(mnl),y0=(m0n0l0),f˘(t,x)=(nvx+lwxmux+lwxmux+nvx)=(0nlm0lmn0)(uxvxwx),g˘(t,x)=(pqr),yl=(mlnlll),yr=(mrnrlr).Our feedback law for (3) reads
(41)y∈C0([0,1])↦{vl(y)=Al∥y∥C0([0,1]),vr(y)=Ar∥y∥C0([0,1]),∂tyl=Myl,
where Al>2sinh(1), Ar>Alcosh(1)+sinh(2), M>0, T>0, and M=(μ000μ000μ),μ<0, a symmetric matrix, is the unique matrix solution to the matrix function:
(42)PM+MTP=-Z,
for some symmetric positive-definite matrices P and Z. Indeed, let V(t,yl)=ylTPyl be the Lyapunov candidate, and that yl asymptotically converges in time is equivalent to that the time derivative of the V,V˙(t,yl)=-ylTZyl is strictly negative. A fixed-point strategy will be used again to prove the existence of a solution to the closed-loop system, we begin by defining the domain of the operator.Definition 13.
LetX be the space of (g,N)∈C0([0,T]×[0,1])×C0([0,1]) satisfying(1)
forall(t,x)∈[0,T]×[0,1],  g(0,x)=y0(x),  g(t,0)=y0(0)eMt,(2)
forallt∈[0,T],  ∥g(t,·)∥C0([0,1])≤N(t),(3)
N is nonincreasing, and N(0)≤∥y0∥C0([0,1])×C0([0,1])×C0([0,1]).Lemma 14.
The domainX is nonempty, convex, bounded, and closed with respect to the uniform topology.
The proof is elementary and one notices that(y0(x)eMt,∥y0∥C0([0,1])×C0([0,1])×C0([0,1])eMt)∈X, so X is nonempty.
Now for(y,N)∈X, we define g∨ and G*B∨ as the solutions of
(43)∀(t,x)∈ΩT,g∨(t,x)=G*y(t,x),g∨(t,0)=g∨(t,1)=0,B∨(t,x)=0,G*B∨(t,0)=AlN(t),G*B∨(t,1)=ArN(t).
One has the following exact formulas:
(44)∀(t,x)∈ΩT,g∨(t,x)=-∫0xsinh(x-x˘)y(t,x˘)dx˘-sinh(x)sinh(1)∫01sinh(x˘-1)y(t,x˘)dx˘,G*B∨(t,x)=N(t)sinh(1)(Arsinh(x)+Alsinh(1-x)).
Therefore, we have the following inequalities:
(45)∀(t,x)∈[0,T]×[0,1],|g∨(t,x)|≤2(1+cosh(1))∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),|∂xg∨(t,x)|≤2cosh(1)∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),|∂xx2g∨(t,x)|≤[2(cosh(1)+1)+1]×∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),|∂x(G*B∨)(t,x)|≥Ar-2cosh(1)Alsinh(1)N(t),|G*B∨(t,x)|≥AlN(t).
Letc∨(t,x)=G*B∨(t,x)+g∨(t,x), where c∨(t,x)=(uvw)(t,x), and in turn those provide
(46)c∨(t,x)≤[2(1+cosh(1))+cosh(1)sinh(1)(Ar+Al)]×∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),∂xc∨(t,x)≤sinh(2)+2Alcosh(1)-Arsinh(1)×∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),∂xx2c∨(t,x)≤[2(1+cosh(1))+1+cosh(1)sinh(1)(Ar+Al)]×∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]).
Now, if ω is the flow of c∨, ω is C1, and since c∨≥0, ω(·,t,x) is nondecreasing. This allows us to define the entrance time and then the operator S as follows. Let e(t,x)=min{s∈[0,t]∣ω(s,t,x)=0}.
Now, forforall(t,x)∈[0,T]×[0,1], S(y,N)=(y~,N~) with the following:(1)
ifx≥ω(t,0,0),
(47)y(t,x)=y0(ω(0,t,x))exp(∫0tb∨(r,ω(r,t,x))dr)+∫0tf∨(r,ω(r,t,x))×exp(∫rtb∨(r′,ω(r′,t,x))dr′)dr,(2)
ifx≤ω(t,0,0),
(48)y(t,x)=y0(0)eMe(t,x)exp(∫0tb∨(r,ω(r,t,x))dr)+∫e(t,x)tf∨(r,ω(r,t,x))×exp(∫rtb∨(r′,ω(r′,t,x))dr′)dr,(3)
N(t)=∥y~(t,·)∥C0([0,1]).Lemma 15.
(1) The operatorS maps X to X.
(2) The familyS(X) is uniformly bounded and equicontinuous.
(3)S is continuous w.r.t. the uniform topology.
The proof is very similar to [10], except for the state y here is a three-component vector and the proof is omitted.
Now, we can apply Schauder’s fixed point theorem toS and get (y,N) fixed point of S.


### 3.2. Stabilization and Global Existence

Theorem 16.
For anyy0∈C0([0,1])×C0([0,1])×C0([0,1]), there exists y∈C0(ΩT)×C0([0,T],C2([0,1])) a weak solution of (39) satisfying
(49)∀x∈[0,1]y(0,x)=y0(x).
Furthermore, any maximal solution of (39) and (41) is global, and if we let
(50)k=max(2(1+cosh(1))+1+cosh(1)sinh(1)(Ar+Al),sinh(2)+2Alcosh(1)-Arsinh(1)),τ=1∥M∥3ln(∥M∥328k3∥y0∥C0([0,1])×C0([0,1])×C0([0,1])2),
then we have
(51)∀t≥τ∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥M∥32k(1+2k∥M∥3)e∥M∥3(t-x).To finish the proof of Theorem (39), we have to prove the global existence of a maximal solution and the estimate (51).Proof .
First, we rewrite (46) as the following:
(52)∀(t,x)∈ΩTc∨(t,x)≤k∥y(t,·)∥C0([0,1]),∂xc∨(t,x)≤k∥y(t,·)∥C0([0,1]),∂xxc∨(t,x)≤k∥y(t,·)∥C0([0,1]),
where k=max(2(1+cosh(1))+1+(cosh(1)/sinh(1))(Ar+Al),(sinh(2)+2Alcosh(1)-Ar)/sinh(1)).
Fory is the solution of the transport (39) and it satisfies
(53)y(t,x)=y(s,ω(s,t,x))exp(∫0tb∨(r,ω(r,t,x))dr)+∫0tf∨(r,ω(r,t,x))×exp(∫rtb∨(r′,ω(r′,t,x))dr′)dr.
Combining those facts, we get for t≥s the following:
(54)|y(t,x)|≤|y(s,ω(s,t,x))|(1+4k2|y(s,ω(s,t,x))|t)×exp(2∫stk∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1])dr).
We have also imposed y(t,0)=y(s,0)eM(t-s) and thanks to the existence theorem that a maximal solution of the closed loop system is global. To get a more precise statement, we consider all the between time t and s, and we obtain.
For0≤s≤t,
(55)∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥y(s,·)∥C0([0,1])×C0([0,1])×C0([0,1])×max[e∥M∥3(r-x)(1+4k2∥y(s,·)∥C0([0,1])×C0([0,1])×C0([0,1])r)]×exp(2k∫rt∥y(α,·)∥C0([0,1])×C0([0,1])×C0([0,1])dα).
We define(56)g(r)=[e∥M∥3(r-s)(1+4k2∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1])r)]×exp(2k∫st∥y(α,·)∥C0([0,1])×C0([0,1])×C0([0,1])dα),
and we set g(r)=g1(r)+g2(r), where
(57)g1(r)=e∥M∥3(r-s)×exp(2k∫rt∥y(α,·)∥C0([0,1])×C0([0,1])×C0([0,1])),g2(r)=e∥M∥3(r-s)4k2∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1])r×exp(2k∫rt∥y(α,·)∥C0([0,1])×C0([0,1])×C0([0,1])).
Then, we have(58)g′(r)=(∥M∥3-2k∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1]))g1(r)+(1r+∥M∥3-2k∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1]))×g2(r),
as long as the quantity ∥y(r,·)∥C0([0,1]) is not equal to zero, it strictly decreases, so if ∥y0∥C0([0,1])×C0([0,1])×C0([0,1])>∥M∥3/2k, for t small enough ∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≥∥M∥3/2k, and we have the following.
(59)∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥y0∥C0([0,1])×C0([0,1])×C0([0,1])×(1+4k2∥y0∥C0([0,1])×C0([0,1])×C0([0,1])t)e∥M∥3t.
If we defineτ=(1/∥M∥3)            ln(∥M∥32/8k3∥y0∥C0([0,1])×C0([0,1])×C0([0,1])2), we get that
(60)∥y(τ,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥M∥32k.
This providesτ≤s≤t, the inequality (which was clear when ∥y0∥C0([0,1])×C0([0,1])×C0([0,1])≤∥M∥3/2k)(61)∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥y(τ,·)∥C0([0,1])×C0([0,1])×C0([0,1])×(1+4k2∥y(τ,·)∥C0([0,1])×C0([0,1])×C0([0,1])t)×exp(2k∫xt∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1])dr)≤∥M∥32k(1+2k∥M∥3)e∥M∥3(t-x).


## 3.1. Preliminary Results

The equilibrium state that we want to stabilize ism=n=l=0, p=q=r=G*Bi=0, and i=1,2,3. A natural idea is using Lyapunov indirection method to investigate whether the linearized system around the equilibrium state is stabilizable or not. Its stabilization would provide a local stabilization result on the nonlinear system. However, there is a difficulty in the stabilization problem. We have to prescribe yl, and we just need to make a continuous transition at (t,x)=(0,0), and that yl asymptotically converge in time. For convenience, the system (6)–(8) can rewrite in the following
(39)∂ty-a˘(t,x)∂xy=b˘(t,x)y+f˘(t,x),y(0,·)=y0,y(·,0)=yl,y(1,·)=yr,g˘(t,x)=G*y(t,x),g˘(t,0)=g˘(t,1)=0,B˘(t,x)=0,G*B˘(t,0)=vl(t),G*B˘(t,1)=vr(t),
where
(40)b˘(t,x)=(2(p+G*B˘)x+vx+wx00002(q+G*B˘)x+ux+wx002(r+G*B˘)x+ux+vx),a˘(t,x)=((p+G*B˘+v+w)0000(u+q+G*B˘+w)00(u+v+r+G*B˘)),y=(mnl),y0=(m0n0l0),f˘(t,x)=(nvx+lwxmux+lwxmux+nvx)=(0nlm0lmn0)(uxvxwx),g˘(t,x)=(pqr),yl=(mlnlll),yr=(mrnrlr).Our feedback law for (3) reads
(41)y∈C0([0,1])↦{vl(y)=Al∥y∥C0([0,1]),vr(y)=Ar∥y∥C0([0,1]),∂tyl=Myl,
where Al>2sinh(1), Ar>Alcosh(1)+sinh(2), M>0, T>0, and M=(μ000μ000μ),μ<0, a symmetric matrix, is the unique matrix solution to the matrix function:
(42)PM+MTP=-Z,
for some symmetric positive-definite matrices P and Z. Indeed, let V(t,yl)=ylTPyl be the Lyapunov candidate, and that yl asymptotically converges in time is equivalent to that the time derivative of the V,V˙(t,yl)=-ylTZyl is strictly negative. A fixed-point strategy will be used again to prove the existence of a solution to the closed-loop system, we begin by defining the domain of the operator.Definition 13.
LetX be the space of (g,N)∈C0([0,T]×[0,1])×C0([0,1]) satisfying(1)
forall(t,x)∈[0,T]×[0,1],  g(0,x)=y0(x),  g(t,0)=y0(0)eMt,(2)
forallt∈[0,T],  ∥g(t,·)∥C0([0,1])≤N(t),(3)
N is nonincreasing, and N(0)≤∥y0∥C0([0,1])×C0([0,1])×C0([0,1]).Lemma 14.
The domainX is nonempty, convex, bounded, and closed with respect to the uniform topology.
The proof is elementary and one notices that(y0(x)eMt,∥y0∥C0([0,1])×C0([0,1])×C0([0,1])eMt)∈X, so X is nonempty.
Now for(y,N)∈X, we define g∨ and G*B∨ as the solutions of
(43)∀(t,x)∈ΩT,g∨(t,x)=G*y(t,x),g∨(t,0)=g∨(t,1)=0,B∨(t,x)=0,G*B∨(t,0)=AlN(t),G*B∨(t,1)=ArN(t).
One has the following exact formulas:
(44)∀(t,x)∈ΩT,g∨(t,x)=-∫0xsinh(x-x˘)y(t,x˘)dx˘-sinh(x)sinh(1)∫01sinh(x˘-1)y(t,x˘)dx˘,G*B∨(t,x)=N(t)sinh(1)(Arsinh(x)+Alsinh(1-x)).
Therefore, we have the following inequalities:
(45)∀(t,x)∈[0,T]×[0,1],|g∨(t,x)|≤2(1+cosh(1))∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),|∂xg∨(t,x)|≤2cosh(1)∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),|∂xx2g∨(t,x)|≤[2(cosh(1)+1)+1]×∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),|∂x(G*B∨)(t,x)|≥Ar-2cosh(1)Alsinh(1)N(t),|G*B∨(t,x)|≥AlN(t).
Letc∨(t,x)=G*B∨(t,x)+g∨(t,x), where c∨(t,x)=(uvw)(t,x), and in turn those provide
(46)c∨(t,x)≤[2(1+cosh(1))+cosh(1)sinh(1)(Ar+Al)]×∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),∂xc∨(t,x)≤sinh(2)+2Alcosh(1)-Arsinh(1)×∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]),∂xx2c∨(t,x)≤[2(1+cosh(1))+1+cosh(1)sinh(1)(Ar+Al)]×∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1]).
Now, if ω is the flow of c∨, ω is C1, and since c∨≥0, ω(·,t,x) is nondecreasing. This allows us to define the entrance time and then the operator S as follows. Let e(t,x)=min{s∈[0,t]∣ω(s,t,x)=0}.
Now, forforall(t,x)∈[0,T]×[0,1], S(y,N)=(y~,N~) with the following:(1)
ifx≥ω(t,0,0),
(47)y(t,x)=y0(ω(0,t,x))exp(∫0tb∨(r,ω(r,t,x))dr)+∫0tf∨(r,ω(r,t,x))×exp(∫rtb∨(r′,ω(r′,t,x))dr′)dr,(2)
ifx≤ω(t,0,0),
(48)y(t,x)=y0(0)eMe(t,x)exp(∫0tb∨(r,ω(r,t,x))dr)+∫e(t,x)tf∨(r,ω(r,t,x))×exp(∫rtb∨(r′,ω(r′,t,x))dr′)dr,(3)
N(t)=∥y~(t,·)∥C0([0,1]).Lemma 15.
(1) The operatorS maps X to X.
(2) The familyS(X) is uniformly bounded and equicontinuous.
(3)S is continuous w.r.t. the uniform topology.
The proof is very similar to [10], except for the state y here is a three-component vector and the proof is omitted.
Now, we can apply Schauder’s fixed point theorem toS and get (y,N) fixed point of S.


## 3.2. Stabilization and Global Existence

Theorem 16.
For anyy0∈C0([0,1])×C0([0,1])×C0([0,1]), there exists y∈C0(ΩT)×C0([0,T],C2([0,1])) a weak solution of (39) satisfying
(49)∀x∈[0,1]y(0,x)=y0(x).
Furthermore, any maximal solution of (39) and (41) is global, and if we let
(50)k=max(2(1+cosh(1))+1+cosh(1)sinh(1)(Ar+Al),sinh(2)+2Alcosh(1)-Arsinh(1)),τ=1∥M∥3ln(∥M∥328k3∥y0∥C0([0,1])×C0([0,1])×C0([0,1])2),
then we have
(51)∀t≥τ∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥M∥32k(1+2k∥M∥3)e∥M∥3(t-x).To finish the proof of Theorem (39), we have to prove the global existence of a maximal solution and the estimate (51).Proof .
First, we rewrite (46) as the following:
(52)∀(t,x)∈ΩTc∨(t,x)≤k∥y(t,·)∥C0([0,1]),∂xc∨(t,x)≤k∥y(t,·)∥C0([0,1]),∂xxc∨(t,x)≤k∥y(t,·)∥C0([0,1]),
where k=max(2(1+cosh(1))+1+(cosh(1)/sinh(1))(Ar+Al),(sinh(2)+2Alcosh(1)-Ar)/sinh(1)).
Fory is the solution of the transport (39) and it satisfies
(53)y(t,x)=y(s,ω(s,t,x))exp(∫0tb∨(r,ω(r,t,x))dr)+∫0tf∨(r,ω(r,t,x))×exp(∫rtb∨(r′,ω(r′,t,x))dr′)dr.
Combining those facts, we get for t≥s the following:
(54)|y(t,x)|≤|y(s,ω(s,t,x))|(1+4k2|y(s,ω(s,t,x))|t)×exp(2∫stk∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1])dr).
We have also imposed y(t,0)=y(s,0)eM(t-s) and thanks to the existence theorem that a maximal solution of the closed loop system is global. To get a more precise statement, we consider all the between time t and s, and we obtain.
For0≤s≤t,
(55)∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥y(s,·)∥C0([0,1])×C0([0,1])×C0([0,1])×max[e∥M∥3(r-x)(1+4k2∥y(s,·)∥C0([0,1])×C0([0,1])×C0([0,1])r)]×exp(2k∫rt∥y(α,·)∥C0([0,1])×C0([0,1])×C0([0,1])dα).
We define(56)g(r)=[e∥M∥3(r-s)(1+4k2∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1])r)]×exp(2k∫st∥y(α,·)∥C0([0,1])×C0([0,1])×C0([0,1])dα),
and we set g(r)=g1(r)+g2(r), where
(57)g1(r)=e∥M∥3(r-s)×exp(2k∫rt∥y(α,·)∥C0([0,1])×C0([0,1])×C0([0,1])),g2(r)=e∥M∥3(r-s)4k2∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1])r×exp(2k∫rt∥y(α,·)∥C0([0,1])×C0([0,1])×C0([0,1])).
Then, we have(58)g′(r)=(∥M∥3-2k∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1]))g1(r)+(1r+∥M∥3-2k∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1]))×g2(r),
as long as the quantity ∥y(r,·)∥C0([0,1]) is not equal to zero, it strictly decreases, so if ∥y0∥C0([0,1])×C0([0,1])×C0([0,1])>∥M∥3/2k, for t small enough ∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≥∥M∥3/2k, and we have the following.
(59)∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥y0∥C0([0,1])×C0([0,1])×C0([0,1])×(1+4k2∥y0∥C0([0,1])×C0([0,1])×C0([0,1])t)e∥M∥3t.
If we defineτ=(1/∥M∥3)            ln(∥M∥32/8k3∥y0∥C0([0,1])×C0([0,1])×C0([0,1])2), we get that
(60)∥y(τ,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥M∥32k.
This providesτ≤s≤t, the inequality (which was clear when ∥y0∥C0([0,1])×C0([0,1])×C0([0,1])≤∥M∥3/2k)(61)∥y(t,·)∥C0([0,1])×C0([0,1])×C0([0,1])≤∥y(τ,·)∥C0([0,1])×C0([0,1])×C0([0,1])×(1+4k2∥y(τ,·)∥C0([0,1])×C0([0,1])×C0([0,1])t)×exp(2k∫xt∥y(r,·)∥C0([0,1])×C0([0,1])×C0([0,1])dr)≤∥M∥32k(1+2k∥M∥3)e∥M∥3(t-x).


## 4. Blow-Up Phenomena

In this section, we present a result with the initial data and boundary profiles under a special condition that ensure strong solutions to following system blow-up in finite time as follows:(62)∂tm-(u+v+w)(t,x)∂xm=(2(p+G*B)+vx+wx)(t,x)m+(nvx+lwx)(t,x),∂tn-(u+v+w)(t,x)∂xn=(2(q+G*B)+ux+wx)(t,x)n+(mux+lwx)(t,x),∂tl-(u+v+w)(t,x)∂xl=(2(r+G*B)+ux+vx)(t,x)l+(mux+nvx)(t,x),m(0,·)=m0,n(0,·)=n0,l(0,·)=l0,x∈[0,1],m(t,0)=ml=m(t,1)=mr,t∈[0,T],n(t,0)=nl=n(t,1)=nr,t∈[0,T],l(t,0)=ll=l(t,1)=lr,t∈[0,T],
where m=u-uxx,  n=v-vxx,  l=w-wxx. This imply that
(63)u(t,0)=u(t,1),v(t,0)=v(t,1),w(t,0)=w(t,1).From Definition3, we can also define
(64)ωt=-(u+v+w)(t,ω(t,y)),(t,x)∈[0,T]×[0,1],ω(0,x)1=x,x∈[0,1];
where u,v, and w denote the solution to (62). Applying classical results in the theory of ordinary differential equations, one can obtain a result on which is crucial in studying blow-up phenomena.From (62), we obtain that
(65)mt-∂x(mu+mv+mw)=mux+nvx+lwx,nt-∂x(mu+nv+mw)=mux+nvx+lwx,lt-∂x(lu+lv+lw)=mux+nvx+lwx.Lemma 17.
Letu,v,w∈C([0,T);Hs)∩C1([0,T);Hs-1),(s≥2), then (64) has a unique solution ω∈C([0,T)×[0,1]). Moreover, the map ω(t,·) is an increasing diffeomorphism with
(66)ωx(t,x)=exp{-∫0t(ux(s,ω(s,x))+vx(s,ω(s,x))+wx(s,ω(s,x))ds){∫0t}}>0,ωx(0,x)=1,∀x∈[0,1].Proof.
The proof is omitted here, one can see a similar proof in [12].Now, we have the following lemma that the potentialm-n,  n-l,  m-l with compactly supported initial datum m0-n0,  n0-l0,  m0-l0 also has compact x support as long as it exists.Lemma 18.
Assume thatu0(t,x),v0(t,x),w0(t,x)∈Hs×Hs with s>(3/2), (u,v,w) is the corresponding solution, if m0-n0,n0-l0,m0-l0 has compact support, then m-n,n-l,m-l also has compact support, moreover, we can obtain that
(67)∥m(t,·)-n(t,·)∥≤e3KT∥m0(·)-n0(·)∥,∥m(t,·)-l(t,·)∥≤e3KT∥m0(·)-l0(·)∥,∥n(t,·)-l(t,·)∥≤e3KT∥n0(·)-l0(·)∥.Proof.
Since(68)ddt(m(t,ω(t,x))ωx)=(mt+mxωt)ωx+mωxt=[mt-∂x(mu+mv+mw)]ωx=(mux+nvx+lwx)ωx.
Similarly,(69)ddt(n(t,ω(t,x))ωx)=(mux+nvx+lwx)ωx,ddt(l(t,ω(t,x))ωx)=(mux+nvx+lwx)ωx.
So it follows that(70)ddt((m(t,ω(t,x))-n(t,ω(t,x)))ωx)=0,ddt((m(t,ω(t,x))-l(t,ω(t,x)))ωx)=0,ddt((n(t,ω(t,x))-l(t,ω(t,x)))ωx)=0.
We obtain(71)[m(t,ω(t,x))-n(t,ω(t,x))]ωx=m0(x)-n0(x),[m(t,ω(t,x))-l(t,ω(t,x))]ωx=m0(x)-l0(x),[n(t,ω(t,x))-l(t,ω(t,x))]ωx=n0(x)-l0(x).
From Lemma17, we have
(72)ωx=exp(-∫0tux(s,ω(s,x))+vx(s,ω(s,x))+wx(s,ω(s,x))ds{∫}).
If there exist four constants K,K1,K2, and K3 such that ux≤K1, vx≤K2, wx≤K3, and K=max(K1,K2,K3), we can get that
(73)∥m(t,·)-n(t,·)∥L∞=∥m(t,ω(t,·))-n(t,ω(t,·))∥L∞=∥exp(∫0tux(s,ω(s,x))+vx(s,ω(s,x))+wx(s,ω(s,x))ds{∫})×(m0(x)-n0(x))ds∥L∞≤e3KT∥m0(·)-n0(·)∥.
Similarly,
(74)∥m(t,·)-l(t,·)∥≤e3KT∥m0(·)-l0(·)∥,∥n(t,·)-l(t,·)∥≤e3KT∥n0(·)-l0(·)∥.Theorem 19.
Letu0(x),v0(x), and w0(x), s>3/2 and u(t,x),v(t,x),w(t,x) be the solution in (62) with time T. Then, T is finite if and only if
(75)limt→Tinf{infx∈[0,1][ux(x,t)]}=-∞
or
(76)limt→Tinf{infx∈[0,1][vx(x,t)]}=-∞
or
(77)limt→Tinf{infx∈[0,1][wx(x,t)]}=-∞.Proof.
Letu0(x),v0(x), and w0(x), s>3/2 and u(t,x),v(t,x),w(t,x) be the solution (62) with time T. We know that u(t,0)=u(t,1),v(t,0)=v(t,1), and w(t,0)=w(t,1). By the definition of m,n, and l, we have
(78)∥m∥L22=∫01(u-uxx)2dx=∫01(u2+2ux2+uxx2)dx,∥n∥L22=∫01(v-vxx)2dx=∫01(v2+2vx2+vxx2)dx,∥l∥L22=∫01(w-wxx)2dx=∫01(w2+2wx2+wxx2)dx.
Hence,∥u∥H22≤∥m∥L22≤2∥u∥H22,∥v∥H22≤∥n∥L22≤2∥v∥H22,∥w∥H22≤∥l∥L22≤2∥w∥H22.
Multiplying the first equation bym, the second one by n, and the third one by l, after integration by parts and adding up the results, we see that
(79)12ddt∫01(m2+n2+l2)dx=∫01[(2ux+vx+wx)m2+(2vx+ux+wx)n2+(2wx+ux+vx)l2+(mmx+nnx+llx)×(u+v+w)+mn(ux+vx)+lm(ux+wx)+ln(vx+wx){(2ux+vx+wx)m2}]dx=∫01[(2ux+vx+wx-12ux-12vx-12wx)m2+(2vx+ux+wx-12ux-12vx-12wx)n2+(2wx+ux+vx-12ux-12vx-12wx)l2+mn(ux+vx)+lm(ux+wx)ln(vx+wx){(2ux+vx+wx-12ux-12vx-12wx)}]dx=∫01[(32ux+12vx+12wx)m2+(32vx+12ux+12wx)n2+(32wx+12ux+12vx)l2+mn(ux+vx)+lm(ux+wx)+ln(vx+wx){(32ux+12vx+12wx)}]dx=∫01[12(ux+vx+wx)(m2+n2+l2)+m2ux+n2vx+l2wx+mn(ux+vx)+lm(ux+wx)+ln(vx+wx){12}]dx=∫01[12(ux+vx)(m+n)2+12(ux+wx)(m+l)2+12(vx+wx)(l+n)2+12uxm2+12vxn2+12wxl2]dx
So, we have
(80)ddt∫01(m2+n2+l2)dx≤(K1+K2+K3)×∫01[(m+n)2+(m+l)2+(l+n)2+m2+n2+l2]dx≤5(K1+K2+K3)∫01(m2+n2+l2)dx.
By Gronwall's inequality, we get(81)∥u∥H22+∥v∥H22+∥w∥H22≤∫01(m2+n2+l2)dx≤exp[5T(K1+K2+K3)]×∫01(m02+n02+l02)dx≤2exp[5T(K1+K2+K3)]×(∥u0∥H22+∥v0∥H22+∥w0∥H22).
The above inequality, Soblev’s embedding theorem, ensure that the solution (u(t,x),v(t,x),w(t,x)) cannot blow up in finite time.
On the other hand, if(82)limt→Tinf{infx∈[0,1][ux(x,t)]}=-∞
or
(83)limt→Tinf{infx∈[0,1][vx(x,t)]}=-∞
or
(84)limt→Tinf{infx∈[0,1][wx(x,t)]}=-∞,
then the solution will blow up in finite time.



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*Source: 691731-2013-03-20.xml*