# Almost Periodic (Type) Solutions to Parabolic Cauchy Inverse Problems

Authors: Fenglin Yang; Chuanyi Zhang
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We first show the existence and uniqueness of (pseudo) almost periodic solutions of some types of parabolic equations. Then, we apply the results to a type of Cauchy parabolic inverse problems and show the existence, uniqueness, and stability.


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## Body

## 1. Introduction

Zhang in [1, 2] defined pseudo almost periodic functions. As almost periodic functions, pseudo almost periodic functions are applied to many mathematical areas, particular to the theory of ordinary differential equations. (e.g., see [3–26] and references therein). However, the study of the related topic on partial differential equations has only a few important developments. On the other hand, almost periodic functions to various problems have been investigated (e.g., see [27–32] and references therein), but little has been done about the inverse problems except for our work in [33–36]. In [36], we study pseudo almost periodic solutions to parabolic boundary value inverse problems. In this paper, we devote such solutions to cauchy problems.To this end, we need first to define the spaces in a more general setting. LetJ∈{R,Rn}. Let 𝒞(J) (resp., 𝒞(J×Ω), where Ω⊂Rm) denote the C*-algebra of bounded continuous complex-valued functions on J (resp. J×Ω) with the supremum norm. For f∈𝒞(J) (resp., 𝒞(J×Ω)) and s∈J, the translation of f by s is the function Rsf(t)=f(t+s) (resp., Rsf(t,Z)=f(t+s,Z),(t,Z)∈J×Ω).Definition 1.1.
(1) A functionf∈𝒞(J) is called almost periodic if for every ϵ>0 the set
(1.1)T(f,ϵ)={τ∈J:‖Rτf-f‖<ϵ}
is relatively dense in J. Denote by 𝒜𝒫(J) the set of all such functions. The number (vector) τ is called ϵ-translation number (vector) of f.
(2) A functionf∈𝒞(J×Ω) is said to be almost periodic in t∈J and uniform on compact subsets of Ω if f(·,Z)∈𝒜𝒫(J) for each Z∈Ω and is uniformly continuous on J×K for any compact subset K⊂Ω. Denote by 𝒜𝒫(J×Ω) the set of all such functions. For convenience, such functions are also called uniformly almost periodic.
(3) A functionf∈𝒞(J)(𝒞(J×Ω)) is called pseudo almost periodic if
(1.2)f=g+φ,
where g∈𝒜𝒫(J)(𝒜𝒫(J×Ω)) and φ∈𝒫𝒜𝒫0(J)(𝒫𝒜𝒫0(J×Ω)),
(1.3)PAP0(J)={φ∈C(J):lim⁡r→∞1(2r)n∫[-r,r]n|φ(x)|dx=0},PAP0(J×Ω)={φ∈C(J×Ω):lim⁡r→∞1(2r)n∫[-r,r]n|φ(x,Z)|dx=0},
uniformly with respect to Z∈K, where K is any compact subset of Ω. Denote by 𝒫𝒜𝒫(J)(𝒫𝒜𝒫(J×Ω)) the set of all such functions.
Set(1.4)APT(J)∈{AP(J),PAP(J)},APT(J×Ω)∈{AP(J×Ω),PAP(J×Ω)}.Members of𝒜𝒫𝒯(J)(𝒜𝒫𝒯(J×Ω)) are called  almost periodic type.We will use the notations  throughout the paper:RTm=Rm×(0,T),∥F∥T=sup{|F(x,t)|:x∈Rn,0≤t≤T}. F∈𝒜𝒫𝒯(Rn×RTm) means that F(x(1),x(2),t) is almost periodic type in x(1)∈Rn and uniformly for (x(2),t)∈RTm;F∈𝒜𝒫𝒯(Rn×Rm) means that F(x(1),x(2)) is almost periodic type in x(1)∈Rn and uniformly for x(2)∈Rm.Let(1.5)Z(x,t;ξ,s)=1(2π(t-s))n+mexp⁡{-∑(xi-ξi)24(t-s)},(x,ξ∈Rn+m),
be the fundamental solution of the heat equation [37].In the next section, we will show the existence and uniqueness of some type of parabolic equations. Sections3 is devoted to a type of Cauchy Problem respectively.


## 2. Solutions of Parabolic Equations

Lemma 2.1.
LetT>0. If  φ∈𝒜𝒫𝒯(Rn×Rm) and
(2.1)u(x,t;s)=∫Rn+mφ(ξ)Z(x,t;ξ,s)dξ,
then for each fixed s∈[0,T)u∈𝒜𝒫𝒯(Rn×Rm×[s,T]).Proof.
First consider the case thatφ∈𝒜𝒫(Rn×Rm). Let τ∈Rn be an ϵ-translation vector of φ:
(2.2)u(x(1)+τ,x(2),t;s)-u(x(1),x(2),t;s)=∫Rn+mφ(ξ(1),ξ(2))[Z(x(1)+τ,x(2),t;ξ(1),ξ(2),s)-Z(x(1),x(2),t;ξ(1),ξ(2),s)]dξ(1)dξ(2)=∫Rn+m[φ(x(1)+τ+ξ(1),x(2)+ξ(2))-φ(x(1)+ξ(1),x(2)+ξ(2))]Z(0,t;ξs)dξ,
where 0∈Rn+m is the zero vector. Note that∫Rn+mZ(0,t;ξ,s)dξ=1, we get
(2.3)‖Rτu-u‖≤‖Rτφ-φ‖∫Rn+mZ(0,t;ξ,s)dξ<ϵ,
where t∈[s,T] and x(2)∈B with B a bounded subset of Rm. This shows that u∈𝒜𝒫(Rn×Rm×[s,T]).
To show thatu∈𝒫𝒜𝒫(Rn×Rm×[s,T]) if φ∈𝒫𝒜𝒫(Rn×Rm), we only need to show that u∈𝒫𝒜𝒫0(Rn×Rm×[s,T]) if φ∈𝒫𝒜𝒫0(Rn×Rm). That is,
(2.4)lim⁡r→∞1(2r)n∫[-r,r]n|u(x(1),x(2),t;s)|ds(1)=0,
uniformly with respect to (x(2),t)∈Ω, here Ω is any compact subset of Rm×[s,T].
Sinceφ∈𝒫𝒜𝒫0(Rn×Rm), for ϵ>0 there exist positive numbers A and r0 such that, when r≥r0 for all ξ(1)∈[-A,A]n and ξ(2)∈Ω∩Rm, one has
(2.5)1(2r)n∫[-r,r]n|φ(x(1)+ξ(1),ξ(2))|dx(1)<ϵ3,∫-∞-A+∫A∞|φ(x(1)+ξ(1),ξ(2))|Z(0,x(2),t;ξ,s)dξ<2ϵ3.
Therefore,
(2.6)1(2r)n∫[-r,r]n|u(x(1),x(2),t;s)|dx(1)≤1(2r)n∫[-r,r]n∫Rn+m|φ(x(1)+ξ(1),ξ(2))|Z(0,x(2),t;ξ,s)dξdx(1)=∫-∞-A+∫-AA+∫A∞Z(0,x(2),t;ξ,s)dξ1(2r)n∫[-r,r]n|φ(x(1)+ξ(1),ξ(2))|dx(1)<ϵ
uniformly with respect to (x(2),t)∈Ω, where by ∫abF(ξ)dξ we mean that
(2.7)∫abF(ξ)ξ=∫[a,b]n+mF(ξ)dξ=∫ab⋯∫abF(ξ)dξ1dξ2⋯dξn+m.
This shows that u∈𝒫𝒜𝒫0(Rn×Rm). The proof is complete.Corollary 2.2.
Letφ,∂φ/∂xi∈𝒜𝒫𝒯(Rn×Rm), and let  u be as in Lemma 2.1. Then,  ∂u/∂xi∈𝒜𝒫𝒯(Rn×Rm×[s,T]).Proof.
Note that(2.8)∂u(x,t;s)∂xi=∫Rn+mφ(ξ)∂Z(x,t;ξ,s)∂xidξ=-∫Rn+mφ(ξ)∂Z(x,t;ξ,s)∂ξidξ=∫Rn+m∂φ(ξ)∂ξiZ(x,t;ξ,s)dξ.By Lemma2.1 we get the conclusion.Lemma 2.3.
Iff(x,t)∈𝒜𝒫𝒯(Rn×RTm¯) and
(2.9)u(x,t)=∫0tds∫Rn+mf(ξ,s)Z(x,t;ξ,s)dξ,
then u and ∂u(x,t)/∂xi(i=1,2,…,n+m) are all in 𝒜𝒫𝒯(Rn×RTm¯).The proof is similar to that of Lemma2.1, so we omit it.Theorem 2.4.
Consider the heat problem(2.10)∂u∂t-∑i=1n+m[∂2u∂xi2+bi(x,t)∂u∂xi]-c(x,t)u=f(x,t),(x,t)∈RTn+m,u(x,0)=φ(x),x∈Rn+m.
If f(x,t),bi(x,t),∂bi/∂xj(i,j=1,2,…,n+m),c(x,t) are in 𝒜𝒫𝒯(Rn×RTm¯) and φ,∂φ/∂xi(i=1,2,…,n+m) are in 𝒜𝒫𝒯(Rn×Rm), then (2.10) has a unique solution u∈𝒜𝒫𝒯(Rn×RTm¯).Proof.
Problem (2.10) has the standard solution (see [37]):
(2.11)u(x,t)=∫Rn+mφ(ξ)Γ(x,t;ξ,0)dξ+∫0tds∫Rn+mf(ξ,s)Γ(x,t;ξ,s)dξ=I1+I2,
where
(2.12)Γ(x,t;ξ,s)=Z(x,t;ξ,s)+∫st∫Rn+mZ(x,t;y,η)⋅Φ(y,η;ξ,s)dydη,Φ(y,η;ξ,s)=∑l=1∞(LZ)l(y,η;ξ,s),(LZ)1=LZ,(LZ)l+1(y,η;ξ,s)=∫st∫Rn+m[LZ(y,η;v,σ)](LZ)l(v,σ;ξ,s)dvdσ,
and L is the parabolic operator
(2.13)L=∑i=1n+m[∂2∂xi2+bi(x,t)∂∂xi]+c(x,t)-∂∂t.
Now, we show that u∈𝒜𝒫𝒯(Rn×RTm¯):(2.14)I1=∫Rn+mφ(ξ)Z(x,t;ξ,0)dξ+∫Rn+mφ(ξ)dξ∫0tdη∫Rn+mZ(x,t;y,η)∑l=1∞(LZ)l(y,η;ξ,0)dy=u1(x,t)+∑l=1∞∫0tdη∫Rn+mZ(x,t;y,η)dy∫Rn+mφ(ξ)(LZ)l(y,η;ξ,0)dξ,I2=∫0tds∫Rn+mf(ξ,s)Z(x,t;ξ,s)dξ+∫0tds∫Rn+mf(ξ,s)dξ∫stdη∫Rn+mZ(x,t;y,η)∑l=1∞(LZ)lZ(y,η;ξ,s)dy=u2(x,t)+∑l=1∞∫0tdη∫Rn+mZ(x,t;y,η)dy∫0ηds∫Rn+mf(ξ,s)(LZ)l(y,η;ξ,s)dξ.
By Lemmas 2.1 and 2.3, ui∈𝒜𝒫𝒯(Rn×RTm¯)(i=1,2,).
Obviously,(2.15)LZ(x,t;ξ,s)=∑i=1n+mbi(x,t)∂Z(x,t;ξ,s)∂xi+c(x,t)Z(x,t;ξ,s).
By Lemmas 2.1 and 2.3 we only need to show that the functions
(2.16)wl(x,t)=∫Rn+mφ(ξ)(LZ)l(x,t;ξ,0)dξ,vl(x,t)=∫0tds∫Rn+mf(ξ,s)(LZ)l(x,t;ξ,s)dξ,
are in 𝒜𝒫𝒯(Rn×RTm¯). We do this by induction. By Lemmas 2.1 and 2.3 and Corollary 2.2, it is true for the case l=1. Suppose that wl(x,t),vl(x,t)∈𝒜𝒫𝒯(Rn×RTm). Then,
(2.17)wl+1(x,t)=∫Rn+mφ(ξ)dξ∫0tdη∫Rn+mLZ(x,t;y,η)(LZ)l(y,η;ξ,0)dy=∫0tdη∫Rn+mLZ(x,t;y,η)dy∫Rn+mφ(ξ)(LZ)l(y,η;ξ,0)dξ=∫0tdη∫Rn+mwl(y,η)LZ(x,t;y,η)dy.
By the induction assumption and Lemma 2.3, we have wl+1(x,t)∈𝒜𝒫𝒯(Rn×RTm¯). Similarly one shows that vl+1(x,t)∈𝒜𝒫𝒯(Rn×RTm¯). The proof is complete.


## 3. Cauchy Problem

Starting this section we will apply the results of the last section to inverse problems of partial differential equations. We will investigate two types of initial value problems in this and the next sections, respectively. We will keep the notation in Section2 and, at the same time, introduce the following new notation:(3.1)x=(x1,x2,…,xn-1),ξ=(ξ1,ξ2,…,ξn-1),X=(x,xn),ζ=(ξ,ξn).The following estimates are easily obtained:(3.2)‖∫0tds∫RnZ(X,t;ζ,s)dζ‖T≤m(T),
where m(T)are positive and increasing for T≥0 and m(T)→0 as T→0.To show the main results of this and the next sections, the following lemmas are needed. The first lemma is the Gronwall-Bellman lemma; the convenient reference should be an ODE text, for instance, it is proved on page 15 of [38].Lemma 3.1.
Letφ,  ϕ, and  χ be real, continuous functions on  [0,T] with  χ≥0. If
(3.3)φ(t)≤ϕ(t)+∫0tχ(s)φ(s)ds(t∈[0,T]),
then
(3.4)φ(t)≤ϕ(t)+∫0tχ(s)ϕ(s)exp⁡{∫stχ(ρ)dρ}ds(t∈[0,T]).Lemma 3.2.
Letφ be a continuous function on  [0,T]. If  ϕ,  χ1, and  χ2 are nondecreasing and nonnegative on  [0,T] and
*φ(t)≤ϕ(t)+χ1(t)∫0tφ(s)ds+χ2(t)∫0tφ(s)t-sds(t∈[0,T]),
then
(3.5)φ(t)≤ϕ(t)[1+tχ1(t)+2tχ2(t)]etχ(t),
where
(3.6)χ(t)=tχ12(t)+4tχ1(t)χ2(t)+πχ22(t).Proof.
Replacingφ(s) in the two integrals of (*) by the expression on the right-hand side in (26), changing the integral order of the resulting inequality, and making use of the monotonicity of ϕ, χ1, and χ2, one gets
(3.7)φ(t)≤ϕ(t)[1+tχ1(t)+2tχ2(t)]+[tχ12(t)+4tχ1(t)χ2(t)+πχ22(t)]∫0tφ(s)Ds.
Using Lemma 3.1 leads to the conclusion.Lemma 3.3.
LetF(X,t)∈𝒞(RTn¯) and  φ∈𝒞(Rn). If  u(X,t) is a solution of the problem
(3.8)ut-Δu+qu=F(X,t),(X,t)∈RTn,u(X,0)=φ(X),X∈Rn,
then
(3.9)‖u‖T≤K(T)(T‖F‖T+‖φ‖),
where K(T)=1+T∥q∥TeT∥q∥T.One sees thatK(T) depends on ∥q∥T only and is bounded near zero.Proof.
The solutionu can been written as
(3.10)u(X,t)=∫Rnφ(ζ)Z(X,t;ζ,0)dζ+∫0tds∫RnF(ζ,s)Z(X,t;ζ,s)dζ-∫0tds∫Rnq(ξ,s)u(ζ,s)Z(X,t;ζ,s)dζ,
so,
(3.11)‖u‖t≤‖φ‖+∫0t‖F‖sds+∫0t‖q‖s‖u‖sds.
By Lemma 3.1 one gets the desired result. The proof is complete.Problem 1.
Find functionsu(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯) and q(x,t)∈𝒜𝒫𝒯(RTn-1¯) such that
(3.12)ut-Δu+qu=F(X,t),(X,t)∈RTn,(3.13)u(X,0)=φ(X),X∈Rn,(3.14)u(x,0,t)=h(x,t),(x,t)∈RTn-1¯,
where φ(x,xn),φxnxn(x,xn)∈𝒜𝒫𝒯(Rn-1×R),h(x,t)≥const>0,h(x,t),(Δh-ht)∈𝒜𝒫𝒯(RT0n-1¯), and F(x,xn,t),Fxnxn(x,xn,t)∈𝒜𝒫𝒯(Rn-1×RT0¯). T0>T>0 are constants.By (3.13) and (3.14), one sees that h(x,0)=φ(x,0).We have the following additional problem.Problem 2.
Find functionsW(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯) and q(x,t)∈𝒜𝒫𝒯(RTn-1¯) such that
(3.15)Wt-ΔW+qW=Fxnxn(X,t),(X,t)∈RTn,(3.16)W(X,0)=φxnxn(X),X∈Rn,(3.17)W(x,0,t)=ht-Δh+qh-F(x,0,t),(x,t)∈RTn-1¯.
The Cauchy problems with unknown coefficient belong to inverse problems [39]. “In the last two decades, the field of inverse problems has certainly been one of the fastest growing areas in applied mathematics. This growth has largely been driven by the needs of applications both in other sciences and in industry.” [40]. For the two problems above, we have the following.Lemma 3.4.
Problems1 and 2 are equivalent to each other.Proof.
LetV(X,t)=uxn(X,t). Then, V(X,t) satisfies
(3.18)Vt-ΔV+qV=Fxn(X,t),(X,t)∈RTn,(3.19)V(X,0)=φxn(X),X∈Rn,(3.20)Vxn(x,0,t)=ht-Δh+qh-F(x,0,t),φ(x,0)=h(x,0),(x,t)∈RTn-1¯.
So, if Problem 1 has a solution (u,q), then Problems (3.18)–(3.20) have the solution (V,q) with V(X,t)=uxn(X,t). Obviously V(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯) if u(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯).
On the other hand, ifV(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯) and q(x,t)∈𝒜𝒫𝒯(RTn-1¯) satisfy (3.18)–(3.20), then we will show that Problem 1 has a unique solution (u,q) and u(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯).
The uniqueness comes from the uniqueness of Cauchy Problem (1)-(2). For the existence, note the fact that if (u,q) is a solution of (3.12)–(3.14), then V=uxn. Thus, we define
(3.21)u(X,t)=∫0xnV(x,y,t)dy+Φ(x,t).
It follows from (3.14) that Φ=h. Now, u satisfies (3.13) because
(3.22)u(X,0)=∫0xnV(x,y,0)dy+h(x,0)=φ(X)-φ(x,0)+h(x,0)=φ(X).
It follows from (3.18), (3.20), and (3.21) that
(3.23)ut-Δu+qu=∫0xn(Vt-ΔV+qV)dy+∫0xn∂2V(x,y,t)∂y2dy-∂2∂xn2∫0xnV(x,y,t)dy+ht-∇h+qh=F(x,xn,t)-F(x,0,t)+Vxn(x,xn,t)-Vxn(x,0,t)-Vxn(x,xn,t)+ht-Δh+qh=F(X,t).
Thus, u satisfies (3.12) and (u,q) is a unique solution of Problem 1.
It follows fromV(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯,h(x,t)∈𝒜𝒫𝒯(RTn-1¯), and (3.21) that u(X,t) in (3.21) is in 𝒜𝒫𝒯(Rn-1×RT¯).
Since we have shown that Problem1 is equivalent to (3.18)–(3.20), to show the lemma we only need to show that Problem 2, equivalent to (3.18)–(3.20) too.
If(V,q) is a solution of (3.18)–(3.20), let W(X,t)=Vxn(X,t). Then one can directly calculate that (W,q) is a solution of (3.15)–(3.17) and W(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯).
On the other hand, if(W,q) is a solution of (3.15)–(3.17), let
(3.24)V(X,t)=∫0xnW(x,y,t)dy+Φ(x,t),
where Φ is the solution of the Cauchy problem
(3.25)Φt-ΔΦ+qΦ=Wxn(x,0,t)+Fxn(x,0,t),(x,t)∈RTn-1Φ(x,0)=φxn(x,0),x∈Rn-1.
Since W(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯), Wxn(x,0,t)∈𝒜𝒫𝒯(RTn-1¯). By Theorem 2.4, Φ(x,t)∈𝒜𝒫𝒯(RTn-1¯) and so V(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯).
Obviously,Vxn(x,0,t)=W(x,0,t)=ht-Δh+qh-F(x,0,t), and this shows that V satisfies (3.20). V satisfies (3.19) because
(3.26)V(X,0)=∫0xnW(x,y,0)dy+Φ(x,0)=∫0xnφxnxn(x,y)dy+Φ(x,0)=φxn(x,xn)-φxn(x,0)+Φ(x,0)=φxn(X).
Finally,
(3.27)Vt-ΔV+qV=∫0xn(Wt-ΔW+qW)dy+∫0xn∂2W(x,y,t)∂y2dy-∂2∂xn2∫0xnW(x,η,t)dη+Φt-ΔΦ+qΦ=Fxn(x,xn,t)-Fxn(x,0,t)+Wxn(x,xn,t)-Wxn(x,0,t)-Wxn(x,xn,t)+Φt-ΔΦ+qΦ=Fxn(X,t).
This shows that V satisfies (3.18). The proof is complete.By (3.15)-(3.16) we have the integral equation about W(X,t):(3.28)W(X,t)=∫Rnφxnxn(ζ)Z(X,t;ζ,0)dζ+∫0tds∫RnFxnxn(ζ,s)Z(X,t;ζ,s)dζ-∫0tds∫Rnq(ξ,s)W(ζ,s)Z(X,t;ζ,s)dζ,(3.29)q=Lq=[Δh-ht+F(x,0,t)]h-1+h-1∫Rnφxnxn(ζ)Z(x,0,t;ζ,0)dζ+h-1∫0tds∫RnFxnxn(ζ,s)Z(x,0,t;ζ,s)dζ-h-1∫0tds∫Rnq(ξ,s)W(ζ,s)Z(x,0,t;ζ,s)dζ(x,t)∈RTn-1,
where W is determined by (3.28).It is readily to show that (3.15)–(3.17) are equivalent to (3.28)-(3.29).Note that, for a givenq(x,t)∈𝒜𝒫𝒯(RTn-1¯), Theorem 2.4 shows the Cauchy problem (3.15) and (3.16) (or equivalently (3.28)) has a unique solution W∈𝒜𝒫𝒯(Rn×RT¯). Thus, (3.29) does define an operator L.  To show that Problem 2 (and so Problem 1) has a unique solution, we only need to show that (3.29) has a solution q(x,t)∈𝒜𝒫𝒯(RTn-1¯). That is, L has a fixed point in 𝒜𝒫𝒯(RTn-1¯). To this end, let(3.30)‖[Δh-ht+F(x,0,t)]h-1+h-1∫Rnφxnxn(ζ)Z(x,0,t;ζ,0)dζ+h-1∫0Tds∫RnFxnxn(ζ,s)Z(x,0,t;ζ,s)dζ‖T0=M2.
Set B(M,T)={q(x,t)∈𝒜𝒫𝒯(RTn-1¯):∥q∥T≤M}. Now, we show that for small T the operator L in (3.29) is a contraction from B(M,T) into itself.Ifq∈B(M,T), then, according to Theorem 2.4, the function W determined by (3.15)-(3.16) and therefore by (3.28) belongs to 𝒜𝒫𝒯(Rn-1×RT¯). Note that φxnxn∈𝒜𝒫𝒯(Rn-1×R), (Δh-ht)∈𝒜𝒫𝒯(RTn-1¯), and Fxnxn∈𝒜𝒫𝒯(Rn-1×RT¯). It follows from Lemma 2.3, Theorem 2.4, and (3.29) that Lq∈𝒜𝒫𝒯(RTn-1¯) and(3.31)‖Lq‖T≤‖[Δh-tt+F(x,0,t)]h-1+h-1∫Rnφxnxn(ζ)Z(x,0,t;ζ,0)dζ+h-1∫0tds∫RnFxnxn(ζ,s)Z(x,0,t;ζ,s)dζ‖T+‖h-1‖T‖q‖T‖W‖Tm(T)≤12M+‖h-1‖TMK0(T)(‖φxnxn‖+T‖Fxnxn‖T)m(T),
where K0 comes from Lemma 3.3. Noting that m(T)→0 ad T→0, we choose T1≤T0 such that when T<T1 one has(3.32)‖h-1‖TK0(T)(‖φxnxn‖+T‖Fxnxn‖T)m(T)≤12.
So, Lq∈B(M,T). For q1,q2∈B(M,T), by (3.29)(3.33)‖Lq1-Lq2‖≤‖h-1‖T‖W1q1-W2q2‖Tm(T)≤m(T)‖h-1‖T[‖W1‖T‖q1-q2‖T+‖q2‖T‖W1-W2‖T].
The function V=W1-W2 is a solution of the Cauchy problem(3.34)Vt-ΔV+q1V=W2(q2-q1),(X,t)∈RTn,V(X,0)=0,X∈Rn.
Thus, by Lemma 3.3(3.35)‖V‖T≤K1(T)‖W2‖T‖q2-q1‖.
Applying Lemma 3.3 to (W1,q1) and (W2,q2), respectively, one gets(3.36)‖W1‖≤K1(T)(‖φxnxn‖+T‖Fxnxn‖),‖W2‖≤K2(T)(‖φxnxn‖+T‖Fxnxn‖).
If we choose T2≤T1 so that when T≤T2(3.37)2m(T)‖h-1‖T(‖φxnxn‖+T‖Fxnxn‖)K1(T)[1+MK2(T)]≤1,
then∥Lq1-Lq2∥T≤m(T)∥h-1∥T[∥W1∥+MK1(T)∥W2∥]∥q2-q1∥T≤m(T)∥h-1∥TK1(T)(1+MK2(T))(∥φxnxn∥+T∥Fxnxn∥T)∥q2-q1∥T≤12∥q2-q1∥T.One sees that for suchT, the operator L is a contraction from B(M,T) into itself and, therefore, has a unique fixed point in B(M,T). Thus, we have shown.Theorem 3.5.
If functionsF,φ, and  h satisfy the conditions of  Problem 1,  M and  T are determined by (3.30) and (3.32), (3.37) respectively, then in  RTn,  Problem 1 has a unique solution  (u,q) with  u∈𝒜𝒫𝒯(Rn-1×RT¯) and  q∈𝒜𝒫𝒯(RTn-1¯).Furthermore, we have the following.Theorem 3.6.
LetF,φ,andh be as in Problem 1.  Then, there exists an almost periodic type solution for Problem 1 in  RT0n.Proof.
We show that the conclusion of Theorem3.5 can be extended to RT0n. Let T=sup{s: Problem 1 has solution in Rsn}. By Theorem 3.5, T>0. Suppose that T<T0. Consider the problem
(3.38)ut-Δu+qu=F(X,t),X∈Rn,T≤t≤T0,u(X,T)=f(X),X∈Rn,u(x,0,t)=h(x,t),(x,t)∈Rn-1×[T,T0].Forq we can write the integral equation similar to (3.29), but this time its domain is x∈Rn-1,t∈[T,T0]. As the proof above, define the ball B1(M,S) in 𝒜𝒫𝒯(Rn-1×[T,T0]); then there exists a t0>0 such that the operator Lt0 is a contraction from B1(M,S) into itself. So, (3.29) has a solution for the domain x∈Rn-1,t∈[T,T+t0]. This contradicts the definition of T. We must have T=T0.For the stability, we have the following.Theorem 3.7.
Let functionshi,φi, and  Fi(i=1,2) be as in  Problem 1. If  Wi(X,t)∈𝒜𝒫𝒯(Rn-1×RT¯) and  qi∈𝒜𝒫𝒯(RTn-1¯)(i=1,2) are solutions to (3.15)–(3.17), then
(3.39)‖q2-q1‖t≤c1‖h2-h1‖t+c2‖(∂∂t-Δ)(h2-h1)‖t+c3‖∂2∂xn2(φ2-φ1)‖+c4‖F2-F1‖t+c5‖∂2∂xn2(F2-F1)‖t,
where cj(1≤j≤5) depends on T,∥h1∥t,∥(∂2/∂xn2)φ1∥, ∥F1∥t,∥(∂2/∂xn2)F1∥t, and ∥qi∥t(i=1,2) only.Proof.
By (3.29),
(3.40)h1(q2-q1)=-q2(h2-h1)-(∂∂t-Δ)(h2-h1)+F2(x,0,t)-F1(x,0,t)+∫Rn∂2∂ξn2(φ2(ζ)-φ1(ζ))Z(x,0,t;ζ,0)dζ+∫0tds∫Rn∂2∂ξn2(F2-F1)Z(x,0,t;ζ,s)dζ-∫0tds∫Rn[(W2-W1)q2+W1(q2-q1)]Z(x,0,t;ζ,s)dζ.
By Lemma 3.3,
(3.41)‖W1‖t≤K1(t)[‖∂2φ1∂xn2‖+t‖∂2F1∂xn2‖t].
Since the function V=W2-W1 is the solution of the Cauchy problem
(3.42)Vt-ΔV+q2V=∂2∂xn2(F2-F1)-W1(q2-q1),(X,t)∈RTnV(X,0)=∂2∂xn2[φ2(X)-φ1(X)],X∈Rn,
one has
(3.43)‖V‖t≤K2(T)[‖∂2∂xn2(φ2-φ1)‖+t‖∂2∂xn2(F2-F1)‖t+K1(t)(‖∂2∂xn2φ1‖+t‖∂2∂xn2F1‖)‖q2-q1‖t].
Therefore,
(3.44)‖q2-q1‖t≤‖h1-1‖t{∂2∂t‖q2‖t‖h2-h1‖t+‖(∂∂t-Δ)(h2-h1)‖t+‖F2(x,0,t)-F1(x,0,t)‖t+‖∂2∂xn2(φ2-φ1)‖+T‖∂2∂xn2(F2-F1)‖t+∫0tK2(s)[‖∂2∂xn2(φ2-φ1)‖+s‖∂2∂xn2(F2-F1)‖s]‖q2‖sds+∫0tK1(s)[‖∂2∂xn2φ1‖+s‖∂2∂xn2F1‖s]‖q2-q1‖sds+∫0tK1(s)K2(s)[‖∂2∂xn2φ1‖+s‖∂2∂xn2F1‖s]‖q2-q1‖s‖q2‖sds}.
Using Lemma 3.1, we get the estimates desired if we let
(3.45)c1=‖h1-1‖t‖q2‖t[1+‖h1-1‖t∫0tχ(s)exp⁡{‖h1-1‖t∫stχ(ρ)dρ}ds],c2=‖h1-1‖t[1+‖h1-1‖t∫0tχ(s)exp⁡{‖h1-1‖t∫stχ(ρ)dρ}ds],c3=‖h1-1‖t[1+∫0t‖q2‖sK2(s)ds][1+‖h1-1‖t∫0tχ(s)exp⁡{‖h1-1‖t∫stχ(ρ)dρ}ds],c4=c2c5=c3,
where
(3.46)χ(t)=[‖∂2∂xn2φ1‖+t‖∂2∂xn2F1‖t]K1(t)(1+K2(t)‖q2‖t).
The proof is complete.Corollary 3.8.
If Problem1 has a solution in RT0n, then it has a unique one.



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*Source: 180894-2012-05-17.xml*