# First-Order Regular and Degenerate Identification Differential Problems

Authors: A. Favini; A. Lorenzi; H. Tanabe
Journal: Abstract and Applied Analysis
 (2015)

---

## Abstract

We are concerned with both regular and degenerate first-order identification problems related to systems of differential equations of weakly parabolic type in Banach spaces. Several applications to partial differential equations and systems will be given in a subsequent paper to show the fullness of our abstract results.


---

## Body

## 1. Introduction

The basic aim of this paper consists in extending the results in [1] and in solving some identification problems in a product space X×X, where X is a Banach space X, endowed with a norm ·, related to systems of two possiblydegenerate first-order differential equations in time. More precisely, we will consider the differential problem:determine a pair of functionsy=(y1,y2):[0,τ]→X×Xandf:[0,τ]→R, τ∈R+,such that(1)M1,1y1′t+M1,2y2′t+L1,1y1t+L1,2y2t=ftz1,0≤t≤τ,M2,1y1′t+M2,2y2′t+L2,1y1t+L2,2y2t=f(t)z2,0≤t≤τ,M1,1y10+M1,2y20=M1,1y1,0+M1,2y2,0,M2,1y1(0)+M2,2y2(0)=M2,1y1,0+M2,2y2,0,Ψ1M1,1y1t+M1,2y2t+Ψ2M2,1y1t+M2,2y2t=gt,0≤t≤τ.
In particular, the regular choice corresponds to the case where
(2)M1,1=M2,1=I,M1,2=M2,2=0.
The starting point for this paper is provided by [1] (cf. also [2]) where the identification degenerate problem
(3)DtMyt+Lyt=ftz,0≤t≤τ,My0=My0,Φyt=gt,0≤t≤τ,
is studied in the Banach space X under assumptions of weak parabolicity (cf. [3]) on the linear closed linear operators L and M.Existence of solutions to evolution equations with matrix-valued operator coefficients has been considered very recently by Engel [4] whereno degeneration is involved.Abstract systems of parabolic equations of relevant importance in applications are described in the monograph by Yagi [5].Here we study both the regular and the degenerate cases using the basic results in [1] and the different methods introduced therein. This will allow us to handle, in particular, systems of elliptic-parabolic equations and systems of degenerate parabolic equations in different functional spaces.We indicate now the plan of the paper.In Section2, we will consider an identification problem for first-order regular systems of differential equations.In Section3, we will deal with identification problems for first-order in time systems of PDE’s, treating in particular some nonstandard boundary conditions.Section4 contains the main contributions to the degenerate case. The first two results—Theorems 14 and 18—can be easily extended to matrix-valued operators of the form
(4)ABCD,
where operators A and C map from X to X and Y, respectively, while operators B and D map from Y to X and Y, respectively, and Y being a suitable Banach space.Different levels of degeneration for operator matrices not entering the time derivative will be considered andcorresponding conditions for the solvability of the related identification problems will be given.Section5 contains some extensions of the basic results in [1], which will play a key role for the present paper.Section6 contains a number of applications to systems of PDE’s enlightening the concrete applications and the strict conditions to be satisfied by the single equations to guarantee both existence, uniqueness, and the regularity of solutions.The Appendix section describes a general approach to inverse problems using a perturbation theory of generators. In such a way, the inverse problem under consideration is reduced to a direct problem with a new generator.A number of specific examples illustrate the extension and the strength of this method.


## 2. Identification Problems for First-Order Regular Differential Systems in Banach Spaces

LetX be a complex Banach space with norm · and let
(5)Σα=λ∈C:Reλ≥-c01+Im⁡λα,c0>0.
The space X×X is endowed with the product norm.LetA:D(A)⊂X→X, B:D(B)⊂X→X, C:D(C)⊂X→X, D:D(D)⊂X→X be four closed linear operators satisfying the following properties, where ρ(L) denotes the resolvent set of a linear closed operator L:L(B)⊂X×X:(H1)
D(A)⊂D(C) and D(D)⊂D(B);(H2)
(λ+A)-1LX≤C1(1+|λ|)-β1 for all λ∈Σα;(H3)
(λ+D)-1LX≤C2(1+|λ|)-β2 for all λ∈Σα;(H4)
C(λ+A)-1LX≤C3(1+|λ|)-σ1 for all λ∈Σα;(H5)
B(λ+D)-1LX≤C4(1+|λ|)-σ2 for all λ∈Σα;(H6)
βj∈(0,1], σj∈R, j=1,2, σ1+σ2∈R+;(H7)
Φj∈X*, j=1,2.We consider here the problem consisting inrecovering the unknown scalar functionf:[0,T]→Rin the following differential system inX×X:
(6)Dtxtyt+ABCDxtyt=ftz1z2,0≤t≤τ,subject to the initial conditions(7)x0=x0,y0=y0,and to the additional information(8)Φ1xt+Φ2yt=gt,0≤t≤τ.
First we consider the following resolvent system in X×X, where λ∈Σα and w,z∈X:
(9)λ+Ax+By=z,Cx+λ+Dy=w
that we rewrite in the vector form:
(10)λ+ABCλ+Dxy=zw.
Note that according to our assumption (H3) we get
(11)y=-λ+D-1Cx+λ+D-1w,I-Bλ+D-1Cλ+A-1λ+Ax=z-Bλ+D-1w.
Observe now that
(12)Bλ+D-1Cλ+A-1LX≤Bλ+D-1LXC(λ+A)-1LX≤C3C41+λ-σ1-σ2<12,
if
(13)λ>2C3C41/(σ1+σ2)-1.
Then for all λ∈Σα∩B(0,r0)¯c=:S(α,r0) the linear operator I-B(λ+D)-1C(λ+A)-1 admits an inverse R(λ) in L(X) satisfying the estimate
(14)I-Bλ+D-1Cλ+A-1-1LX≤2.
Hence the solution to system (9) is given by
(15)x=λ+A-1Rλz-λ+A-1RλBλ+D-1w,y=-λ+D-1Cλ+A-1Rλzy=+λ+D-1I+Cλ+A-1RλBλ+D-1w.
Therefore, for all λ∈S(α,r0), we deduce the estimates
(16)x≤C51+λ-β1z+C61+λ-β1-σ2w,y≤C71+λ-β2-σ1zy≤+C81+λ-β21+1+λ-σ1-σ2w.
Since σ1+σ2>0, we conclude that the operator matrix
(17)A=ABCD,DA=DA×DD,
satisfies the following bound in the product space X×X:
(18)λ+A-1LX×X≤C9λ-β,λ∈Sα,r0,
where
(19)β=min⁡β1,β2,β1+σ2,β2+σ1.Remark 1.
If operatorsB and C are bounded, then σj=βj,j=1,2; thus, σ1+σ2>0 is trivially verified.Remark 2.
IfD(B)≃D(D), then σ2=β2-1≤0. Indeed, for any λ0∈-ρ(D) it follows
(20)λ0+Dλ+D-1L(X)≤λ0λ+D-1+I-λλ+D-1LX≤C2λ01+λ-β2+1+C2λ1+λ-β2≤C101+λ1-β2.
Moreover, the closed graph theorem implies B(λ0+D)-1∈L(X). This and the previous estimate imply
(21)Bλ+D-1LX≤C111+λ1-β2.
The relation σ2≤0, along with H9, implies σ1>-σ2=1-β2≥0. Consequently, the set relations D(C)≃D(A) and D(B)≃D(D) cannot occur simultaneously.Taking the results in [1] into account (reported and improved a bit in Section 5), we deduce the following Theorem 3.Theorem 3.
Let-A and -D be the generators of two differentiable (not necessarily densely defined) semigroups of parabolic type in the complex Banach spaceX satisfying (H1)–(H3). Let B and C be two linear closed operators satisfying properties (H4)–(H6) and let Φj, j=1,2, be two linear functionals with properties (H7). Let (α,β,θ) be a triplet of real numbers such that (cf. (19))
(22)α+β>32,2-α-β<θ<α+β-1.
Let the data (g,z1,z2) satisfy the properties
(23)g∈C1+θ([0,τ];***C***),z1∈D(A),z2∈D(D),***Φ1[z1]+Φ2[z2]≠0,***(24)Au0+Bv0,Cu0+Dv0∈DA×DD,g0=***Φ1u0+Φ2v0.***
Then the identification problem
(25)u′t=-Aut-Bvt+ftz1,0≤t≤τ,v′t=-Cut-Dvt+ftz2,0≤t≤τ,u0=u0,v0=v0Φ1ut+Φ2vt=gt,0≤t≤τ
admits a unique global solution (u,v,f)∈[C1+θ([0,τ];X)∩Cθ([0,τ],D(A))] × [C1+θ0,τ;X∩Cθ([0,τ],D(D))]×Cθ([0,τ];C).Remark 4.
IfC=kA for some k∈R+ and B∈L(X), then σ1=β1-1, σ2=β2. Thus σ1+σ2=β1+β2-1>0 implies β1+β2>1. In this case, owing to (19), we have β=β1+β2-1, so that we need to require α+β1+β2>5/2.In most applications,α=1, so that this relation reduces to β1+β2>3/2.


## 3. Identification Problems for Regular First-Order in Time Systems of PDE’s

In this section, we will deal with some identification problems related to systems of PDE’s.Problem 5.
LetΩ be a bounded domain in Rn with a C∞-boundary ∂Ω.
We want to recover the scalar functionf:[0,τ]→R in the initial-boundary value problem:
(26)Dtut,x+Ax,Dxut,x+Bx,Dxvt,x=ftz1x,t,x∈0,τ×Ω,Dtvt,x+Cx,Dxut,x+Dx,Dxvt,x=ftz2x,t,x∈0,τ×Ω,ux,0=u0x,v(x,0)=v0x,x∈Ω,axDνut,x+αx·∇ut,x+bxut,x=0,axDνut,x+αx·∇ut,x+t,x∈0,τ×∂Ω,cxDνvt,x+βx·∇vt,x+dxut,x=0,cxDνvt,x+βx·∇vt,x+t,x∈0,τ×∂Ω,
under the additional information
(27)∫Ω‍η1xux,t+η2xvx,tdx=gt,t∈0,τ.
Here
(28)-Ax,Dx=∑i,j=1n‍ai,jxDxiDxj+∑i=1n‍aixDxi+a0x,-Dx,Dx=∑i,j=1n‍a~i,jxDxiDxj+∑i=1n‍a~ixDxi+a~0x
are two second-order linear elliptic differential operators with real-valued C∞-coefficients on Ω¯ such that
(29)ai,jx=aj,ix,∑i,j=1n‍ai,jxξiξj≥c0ξ2,00000000000000000000000x,ξ∈Ω¯×Rn,a~i,jx=a~j,ix,∑i,j=1n‍a~i,jxξiξj≥c0ξ2,00000000000000000000000x,ξ∈Ω¯×Rn,c0 being a positive constant.Assume for the time being thatB(x,Dx) and C(x,Dx) are multiplication operators defined, respectively, by two C∞(Ω¯)-functions r and s. In this case we have σj=βj.Concerning the linear boundary differential operators defined by(30)A^u=aDyu+α·∇u+bu,D^u=a~Dyu+α~·∇u+b~u,
assume that a, a~, b, b~ and α, α~ are real-valued C∞ functions and vector fields on ∂Ω such that Tu=α·∇u. and T~v=α~·∇v are real C∞-tangential operators on ∂Ω, Dν and Dν~ standing for the conormal derivatives associated with the matrices (ai,j(x)) and (α~i,j(x)), respectively; that is,
(31)Dν=∑i,j=1n‍ai,jxnixnjx-1∑i,j=1n‍ai,jxnixDxi,Dν~=∑i,j=1n‍a~i,jxnixnjx-1∑i,j=1n‍a~i,jxnixDxi,n(x) denoting the unit outward normal vector to ∂Ω at x.Assume further (cf. page 515 in [6]) that the vector field α (resp., α~) does not vanish on Γ0={x∈∂Ω:a(x)=0} (resp., Γ~0={x∈∂Ω:a~(x)=0}) and the function t→a(x(t,x0)) (resp., t→a~(x(t,x0))) has zeros of even order not greater than 2κ1 (resp., 2κ2) along the integral curve x(t,x0) of α (resp., α~) passing through x0∈Γ0 (resp., x0∈Γ~0) at t=0. In other words, the so-called (H)δ-condition holds with δ=δ1=1/(1+2κ1) and δ=δ2=1/(1+2κ2), respectively.It is shown on page 516 in [6] that the operator L1 and L2 defined by
(32)DL1=u∈L2Ω:Ax,Dxu∈L2Ω,A^u=0,L1u=-Ax,Dxu,u∈DL1,D(L2)={v∈L2(Ω):D(x,Dx)v∈L2(Ω),D^v=0},L2v=-Dx,Dxu,v∈DL2,
satisfy in L2(Ω) the resolvent estimates
(33)λ+L1-1LL2Ω≤Cλ-1+δ1/2,λ+L2-1LL2Ω≤Cλ-1+δ2/2
for all complex λ in a sector |argλ|≤φ, with φ∈(π/2,π) and |λ| suitable large.We are in the subelliptic case:(34)β=min⁡β1,β2=min⁡1+δ12,1+δ22>12.
Choose now θ∈(1-β,β) and make the following assumptions:
(35)zj∈DLj,ηj∈L2Ω,j=1,2,u0∈DL1,v0∈DL2,A·,Du0+r·v0∈DL1,s·u0+D·,Dv0∈DL2,g∈C1+θ0,τ;R,g0=∫Ω‍η1xu0x+η2xv0xdx≠0,∫Ω‍η1xz1x+η2xz2xdx≠0.
Then our identification problem admits a unique global solution (u,v,f) ∈ [C1+θ([0,τ];L2(Ω))∩Cθ([0,τ],D(L1))]×[C1+θ([0,τ];L2(Ω))∩Cθ([0,τ],D(L2))]×Cθ([0,τ];R).We consider now the more general case whereC(x,Dx)=kA(x,Dx), k∈R+, and B is the multiplication operator by s∈C∞(Ω¯).From Remark4 we get β=β1+β2-1, so that α+β>3/2 reduces to β1+β2>3/2, that is, to δ1+δ2>1. Since
(36)δ1+δ2=11+2κ1+11+2κ2<1,ifκ1,κ2∈N∖0,
we are compelled to require that either of δ1 or δ2 must coincide with 1.Note that, ifδ2=1, then the corresponding boundary value problem iselliptic and this holds if and only if a~(x)≠0 for all x∈∂Ω (cf. [7], on page 515). In this case β=β1=(1+δ1)/2>1/2 so that 1-β=(1-δ1)/2.Choose nowθ∈((1-δ1)/2,(1+δ1)/2) and make the same assumptions as in the previous case, except for (u0,v0). The related condition has to be changed to the following:
(37)A·,Du0+s·v0∈DL1,kA·,D·u0+D·,Dv0∈DL2.
Then the given identification problem admits a unique global solution (u,v,f) ∈ [C1+θ([0,τ];L2(Ω))∩Cθ([0,τ],D(L1))]×[C1+θ([0,τ];L2(Ω))∩Cθ([0,τ],D(L2))]×Cθ([0,τ];R).A corresponding result holds whenδ2=1.Problem 6.
We note that in Problem5 the domain of the operator-matrix is a product of domains. However, Corollary 31 allows to handle also decoupled domains. For this purpose, we will consider a problem related to a reaction diffusion model describing a man-environment epidemic system investigated in [8]. Such a model consists in a parabolic equation coupled with an ordinary differential equation via a boundary feedback operator (cf. also [7]).
In order to obtain stability results the authors linearize the model and arrive at the following evolution system, whereu(t,x) and v(t,x) stand, respectively, for the concentration of the infection agent and the density of the infective population at time t and point x:
(38)Dtut,x=Δut,x-axut,x+ftz1,Dtut,x=Δut,x-at,x∈0,+∞×Ω,Dtvt,x=cxut,x-dxvt,x+ftz2,Dtvt,x=cxut,x-dxt,x∈0,τ×Ω,u0,x=u0x,v0,x=v0x,x∈Ω,Dνut,x+βxut,x=∫Ω‍kx,yvt,ydy,Dνut,x+βxut,x=t,x∈0,+∞×∂Ω,
where Ω is a bounded domain in Rn with a smooth boundary ∂Ω, Δ is the Laplacian, a,c,d∈C(Ω¯), β∈C(∂Ω), k∈L∞(∂Ω×Ω) are nonnegative functions, and Dν denoted the outward normal derivative on ∂Ω.We defineE=C(Ω¯), X=E×E and denote by Mh the multiplication operator induced by the function h. Moreover, we introduce the operator matrix
(39)A=Δ-MaOMc-Md,DA=u,v∈X:u∈H2Ω,Δu∈E,(40)Dνux+βxux=∫Ω‍kx,yvydyon⁡∂Ω.It can be proved (cf. [7, page 126]) that A generates an analytic semigroup on X.Consider then the identification problem consisting in finding a triplet(u,v,f), f being a scalar function satisfying the direct Problem (38) as well as the additional condition:
(41)σut,x¯+μvt,x~=gt,t∈0,τ,
where z1,z2,u0,v0∈C(Ω¯), g∈C([0,τ];R), x¯, x~ are fixed points in Ω,σ,μ∈R2∖{0}. As a consequence of [2, Theorem 3.2] we getProposition 7.
Letθ∈(0,1), g∈C1+θ([0,τ];R), (u0,v0)∈D-A(1+θ,∞), (z1,z2)∈D-A(θ0,∞), σu0(x¯)+μv0(x~)=g(0), σz1(x¯)+μz2(x~)≠0. Then the identification Problem (38), (41) admits a unique global strict solution ((u,v),f)∈[C1+θ([0,τ];X)∩Cθ(0,τ;DA)]×Cθ([0,τ];R).Notice that the interpolation spacesD-A(θ,∞) are well characterized both in an abstract form and for many boundary conditions (cf. [4, 9]), but in this concrete case it seems to be difficult to translate them as on page 321 in [4]. Therefore, one can use the more restrictive assumptions (u0,v0),(z1,z2)∈D(A) that can be easily checked.Problem 8.
We solve here an identification problem in Hölder spaces.LetΩ be a bounded domain in Rn with a smooth boundary ∂Ω. Then Cν+ωΩ¯, ν∈N∖{0}, and ω∈(0,1) denote the Banach space of all functions in Cν(Ω¯) whose derivatives of order ν are all Hölder continuous with exponent ω. Such a space will be endowed with the natural norm ·ν+ω.We introduce now some notation and assumption. As usual,α~=(α1,…,αn)∈Nn denotes a multi-index and we associate with it the monomial differential operator Dα~=∏j=1n‍Djαj, Dj=i-1∂/∂xj. Finally, for any (fixed) m∈N and ω∈0,1/3 let the functions α~:Ω¯→C, |α~|≤2m, satisfy, for some positive constant M>1, the relations:
(42)000000aα~∈C1+ωΩ¯;C,α~≤2m,aα~x∈R,α~∈0,2m,∀x∈Ω¯,M-1ξ2m≤∑α~=2maα~xξα~≤Mξ2m,M-1ξ2m∑α~=2m‍aα~x∀x∈Ω¯,ξ∈Rn.
Introduce now the linear operator A in X=C1+ω(Ω¯) defined by
(43)DA=u∈C2m+1+ωΩ¯:Dα~u=0DA=00on⁡∂Ω,α~≤m-1u∈C2m+1+ωDα~uΩ¯:,Aux=∑α~≤2maα~xDα~ux,x∈Ω¯,u∈DA.
A remarkable result by von Wahl (cf. [10] on page 241) establishes the resolvent bound
(44)z+A-1u1+ω≤Cz-1+1+ω/2mu1+ω,∀u∈C1+ωΩ¯,z∈Σ1∖B0,R,
for a large R>0, where Σ1={z∈C:Rez≥-c0(1+Im⁡z)},c0>0.For our application is not restrictive to assume that such a condition is satisfied for allz∈Σ1. In this case the resolvent estimate
(45)λ+A-1LC1+ωΩ¯≤C1+λ-β1,λ∈Σ1
holds with β1=1-(1+ω)/(2m).Let nowD be another linear differential operator of order 2r, with properties similar to those of A, defined by
(46)DD=v∈C2r+1+ωΩ¯:Dα~v=0on⁡∂Ω,α~≤r-1,Dvx=∑α~≤2rdα~xDα~vx,x∈Ω¯,v∈D(D),
where ω∈(0,1/3).Likewise as above, the spectral estimate(47)λ+D-1LC1+ωΩ¯≤C1+λ-β2,λ∈Σ1
holds with β2=1-(1+ω)/(2r).Notice thatβ1>1/2 and β2>1/2 imply m≥2 and r≥2.Let nowB and C be linear differential operators, with smooth coefficients (e.g., in C1+ω(Ω¯)), defined by
(48)DB=Cp+1+ωΩ¯,DC=Cq+1+ωΩ¯,B=∑β~≤pbβ~xDβ~,C=∑γ~≤qcγ~xDγ~,
where 0≤p<2r and 0≤q<2m.In view of Satz II on page 239 in [10] we have
(49)B(z+D)-1L(C1+w(Ω¯))≤C1+z-σ2,z∈Σ1,zlarge,Cz+A-1LC1+ωΩ¯≤C1+z-σ1,z∈Σ1,zlarge,
where
(50)σ1=1-q+1+ω2m,σ2=1-p+1+ω2r.
We introduce now the operator matrix
(51)DA=DA×DD,A=ABCD.
We can now apply Theorem 3 with X=C1+ω(Ω¯), α=1 and
(52)ω∈0,13,rq+mp+1+ωr+m<4mr,(53)β=min⁡1-1+ω2m,1-1+ω2r,2-1+ω2m-p+1+ω2r,β=min⁡-2-1+ω2r-q+1+ω2m>12.
Under such hypotheses the identification problem
(54)Dtut,x+Aut,x+Bvt,x=ftz1x,Dtut,x+A-ut,x+Bvt,x∈0,τ×Ω,Dtvt,x+Cut,x+Dvt,x=ftz2x,Dtut,x+Aut,-x+Bvt,x∈0,τ×Ω,u0,x=u0x,v0,x=v0(x),x∈Ω,Dα~ut,x=0,α~≤m-1,Dδ~vt,x=0,Dtut,x+Avδ~≤r-1,t,x∈0,τ×∂Ω,σut,x¯+μvt,x~=gt,t∈0,τ,x¯ and x~ being two fixed elements in Ω, admits a unique solution ((u,v),f)∈[C1+θ0,τ;C1+ωΩ¯×C1+ωΩ¯∩Cθ(0,τ;DA×DD)]×Cθ([0,τ];R) provided that the following conditions are satisfied: (H1)
u0∈D(A), v0∈D(D), Au0+Bv0∈D(A), Cu0+Dv0∈D(D);(H2)
z1∈D(A), z2∈D(D), σz1(x¯)+μz2(x~)≠0;(H3)
g∈C1+θ([0,τ];R), 1-β<θ<β, g(0)=σu0(x¯)+μv0(x~).Notice that, if p≤2r-2 and q≤2m-2, then β=min⁡{β1,β2} and condition rq+mp+(1+ω)(r+m)<4mr is satisfied.Problem 9.
Here we solve an identification problem related to the Ornstein-Uhlenbeck operator inRn, n≥1. For this purpose, we refer to the monograph [11]. Such an operator is the prototype of an elliptic operator with unbounded coefficients and is defined on smooth functions φ by
(55)Aφx=12Tr⁡QD2φx+Bx,Dφ,x∈Rn,
where Dφ and D2φ denote, respectively, the gradient and the Hessian matrix of φ, while Q and B≠O are n×n constant matrices, Q being strictly positive definite and the spectrum σ(B) of B being contained in the left complex half-plane Rez<0.It is well known that the realizationLp of A in the weighted space Lμp(Rn), where
(56)μdx=2π-n/2det⁡Q∞-1/2e-Q∞-1x,x/2dx,Q∞=∫0+∞esBQesB*ds,
generates an analytic semigroup with domain Wμ2,p(Rn), where
(57)Wμk,pRn=u∈LμpRn:Dαu∈LμpRn,α≤k
and is endowed with the norm uWμk,p(Rn)=∑|α|≤k‍DαuLμp(Rn), where fLμp(Rn)=(∫Rn‍|f(x)|pdμ(x))1/p.Moreover, the spectrum ofLp is the discrete set
(58)σLp=λ=∑j=1rnjλj:nj∈N∪0,j=1,…,r,λ1,…,λr denoting the (distinct) eigenvalues of B.We note also that (cf. [11, Theorem 9.3.4])
(59)∇uLμpRn≤C1uLμpRn1/2Lpu-uLμpRn1/2,∇uLμpRn≤C1uLμpRn1/2-u∈Wμ2,pRn,xuLμpRn≤C0uWμ1,pRn,u∈Wμ2,pRn,x=∑j=1nxj21/2.
Whence, since the operator Lp-1 is continuously invertible, we easily deduce the estimates
(60)xuLμp(Rn)≤C0uLμpRn+C1uLμpRn1/2Lpu-uLμpRn1/2≤C0C2uLμp(Rn)1/2Lpu-uLμp(Rn)1/2+C1uLμp(Rn)1/2Lpu-uLμp(Rn)1/2=C3uLμpRn1/2Lpu-uLμpRn1/2,u∈Wμ2,pRn.
Likewise we get the estimates
(61)∑i,j=1nxiDxjuLμpRn≤C4∇uLμpRn+D2uLμpRn≤C5Lpu-uLμpRn,u∈Wμ2,pRn.
Finally, recall that the dual space to Lμp(Rn) can be identified with Lμq(Rn), 1/p+1/q=1.Consider now the linear differential operatorB defined by
(62)Bu=∑j=1najxDxju+∑j=1nbjxxju,u∈Wμ1,pRn,
where aj, bj are uniformly continuous and bounded functions in Rn.The previous estimates yield(63)BuLμp(Rn)≤C7uLμp(Rn)1/2Lpu-uLμp(Rn)1/2,λ∈ρ(Lp).
Consider then the linear operator
(64)B1u=∑i,j=1n‍ai,jxxiDxju,u∈Wμ1,pRn,
where ai,j are uniformly continuous and bounded functions in Rn.Reasoning as above, we conclude thatB1 satisfies the estimate
(65)B1(Lp-λ)-1fLμp(Rn)≤C8fLμpRn,λ∈ρLp.
Consider now the identification problem:
(66)Dtut,x=Lp-λ0ut,x+∑j=1n‍ajxDxjvt,xDtut,x=+∑j=1n‍bjxxjvt,x+ftz1x,Dtut,x===========t,x∈0,τ×Rn,(67)Dtvt,x=∑j=1n‍a¯jxDxjut,x+∑j=1n‍b¯jxxjut,xDtvt,x=+Lp-λ0vt,x+ftz2x,Dtvt,x=====2=====t,x∈0,τ×Rn,(68)u0,x=u0x,v0,x=v0x,x∈Rn,(69)∫Rn‍η1xut,x+η2xvt,xdμx=gt,∫Rn‍η1xut,x+η2xvt,xd==t∈0,τ,
where η1,η2∈Lq(Rn), 1/p+1/q=1, g∈C1+θ([0,τ];R), θ∈(0,1), and
(70)g0=∫Rn‍η1xu0x+η2xv0xμdx.
Theorem 3 applies with X=Lμp(Rn), p∈(1,+∞). Here α=1, β1=β2=1, σ1=σ2=1/2, so that β=1. Assume further that z1,z2,u0,v0∈Wμ2,p(Rn) possess the additional properties:
(71)Lp-λ0u0+∑j‍ajxDxjv0+∑jbjxxjv0∈Wμ2,pRn,∑j‍a¯jxDxju0+∑j‍b¯jxxju0+Lp-λ0v0∈Wμ2,pRn,∫Rn‍η1xz1x+η2xz2xdμ(x)≠0,t∈0,τ.
Then the identification Problems (66)–(321) admit a unique strict global solution (u,v,f) with u,v∈C1+θ([0,τ];Lμp(Rn))∩Cθ([0,τ];Wμ2,p(Rn)), f∈Cθ([0,τ];R).We note that a corresponding result still holds if either of the lower order operators is replaced withB1u=∑i,j=1nai,j(x)xiDxju.SinceRn is an unbounded domain of cone type, the real interpolation spaces (Lμp(Rn),Wμ2,p(Rn))θ,p, θ∈(0,1) are well characterized. Exactly we have (cf. [9, Theorem 3.3.1]):
(72)LμpRn,Wμ2,pRnθ,∞=Bp,∞2θRn;μ.
Assume now that our data (u0,v0,z1,z2) possess the following properties:
(73)g∈C1+θ0,τ;R,0<θ<θ0<1,u0,v0∈Wμ2,p(Rn),Lp-λ0u0+∑j=1n‍ajxDxjv0+∑j=1n‍bjxxjv0∈Bp,∞2θRn;μ,∑j=1n‍a¯jxDxju0+∑j=1n‍b¯jxxju0+Lp-λ0v0∈Bp,∞2θRn;μ,g0=∫Rn‍η1xz1x+η2xz2xdμx≠0,g0=∫Rn‍η1xz1xη2xz2x==t∈0,τ,z1,z2∈Bp,∞2θ0Rn;μ.
Therefore, we can apply Theorem 3 or Proposition 5.1 in [1] and deduce the same conclusion as above.We can analogously deal with the case when either of the lower order operators is replaced withB1u=∑i,j‍ai,j(x)xjDxju.Problem 10.
In [12] the following remarkable result is proved. Let X be a Hilbert space, let A:D(A) be a densely defined, strictly positive self-adjoint operator, and let M=[mi,j]i,j=1n be a constant complex-valued matrix. Then the operator A, defined by
(74)DA=DAn,A=AM=mn,1A,…,mn,nAmn,1A,…,mn,nA,
generates a strongly continuous analytical semigroup in H=Xn if and only if (cf. [12, page 311]) the eigenvalues λi of the matrix M satisfy (i) Reλi<0 if λi is complex and (ii) λi≤0 if λi is real.This result allows to deal with a lot of very important problems related to evolution PDE’s. We confine ourselves to describe an identification problem related to one of such models.Suppose thatA=AM generates an analytical semigroup in H=Xn. Then owing to [1, Theorem 2.1, page 45] (as improved in a following paper for Al Horani and Favini) the identification problem
(75)y′t-Ayt=ftz,0≤t≤τ,y0=y0,Φyt=gt,0≤t≤τ,
admits a unique global solution (y,f)∈[Cθ([0,τ];D(A))∩C1+θ([0,τ];X)]×Cθ([0,τ];R), provided that
(76)y0∈DA,Ay0∈DAθ,∞=DAθ,∞,g∈C1+θ0,τ;R,Φy0=g0z∈DAθ,∞.
Let us apply this result to the following thermoelastic PDE problem with simplified “hinged” homogeneous boundary conditions and Dirichlet thermal boundary conditions (cf. [12, page 317]), where the reference bounded domain Ω has a smooth boundary ∂Ω:determine a triplet of functionsw,v:(0,τ)×Ω→Randf:[0,τ]→Rsuch that(P)Dt2wt,x+Δ2wt,x+αΔvt,x=ftz1x,Dt2wt,x+Δ2wt,x+αΔvt,x==t,x∈Q,Dtvt,x-ηΔvt,x-αΔDtwt,x=ftz2x,Dtvt,x-ηΔvt,x-αΔDtwtx==5t,x∈Q,w0,x=w0x,Dtw0,x=w1x,v0,x=v0x,w0,x=w0x,000Dtw0,x=w1x,w1x,1-x∈Ω,wt,x=0,Δwt,x=0,vt,x=0,wt,x=0,wt,x=0,t,x∈0,τ×∂Ω,η1,wt,·+∫Ω‍η2xvt,xdx=gt,η1,wt,·+++∫Ω‍η2xvt,xt∈0,τ,
where Q:=(0,τ)×Ω and 〈·,·〉 denotes the duality pairing between H2(Ω)∩H01(Ω) and its dual space.Introduce now the positive self-adjoint operatorA defined by
(77)DA=H2Ω∩H01Ω⊂X=L2Ω,A=-Δ.
This definition implies
(78)DA2=h∈H4Ω:u∂Ω=Δu∂Ω=0,A=Δ2.
The previous differential equations read now
(*)Dt2wt,x+A2wt,x-αAvt,x=ftz1x,0000000000000000000000t,x∈Q:=0,τ×Ω,Dtvt,x+ηAvt,x+αADtwt,x=ftz2x,t,x∈Q.
Introduce then the Banach space Y, the new variable u, the vector y, and the operator Al defined by
(79)Y=DA×L2Ω×L2Ω,u=Dtw,y=w,u,vT,DAe=DA2×DA×DA,Ae=OIO-A2OαAO-αA-ηA.
Then system (*) can be rewritten as the single equation:
(80)y′t-Aeyt=ftz,0≤t≤τ.
In [12, page 318] it is shown that operator Ae generates an analytic semigroup on Y if and only if the operator
(81)A=OAO-AOαAO-αA-ηA
generates a strongly continuous analytic semigroup in the space H=L2(Ω)3.Indeed, in this case we haveA=AM, where the constant matrix M is defined by
(82)M=010-10α0-α-η.
Consider then the eigenvalue equation m(λ)=:det⁡(λI-M)=λ3+ηλ2+(α2+1)λ+η=0 and observe that (i) since m(-η)=-α2η<0 and m(0)=η>0, the matrix M admits a real negative eigenvalue λ1=-a(α,η)∈(-η,0); (ii) all the real eigenvalues of M are strictly negative; (iii) all the nonreal eigenvalues λ have negative real parts Reλ=[a(α,η)-η]/2, since the sum of is -η. Consequently, all the eigenvalues of M have negative real parts so that A generates an analytic semigroup in H (resp., Y).Let us consider first our identification problem inY. Since
(83)z=0,z1,z2T,Φy=∫Ω‍η1xy1x+η2xy3xdx,Φy=∫Ω‍Φy=∫Ω‍Φy=η1,η2∈L2Ω
we must require
(84)Φz=∫Ω‍η2xz2xdx≠0.
Moreover, we assume
(85)g∈C1+θ0,τ;R,g0=∫Ωη1xw0x+η2xv0xdx,000000000000000000000(w0,w1,v0)∈Y,(86)OIO-A2OαAO-αA-ηAw0w1v0=w1-A2w0+αAv0-αAw1-ηAv0∈Y;DAlθ,∞=D(A)×L2(Ω)×L2(Ω);DA2×DA×DAθ,∞=DA;DA2θ,∞×L2Ω;DAθ,∞×L2Ω;DAθ,∞,0,z1,z2∈DA;DA2θ0,∞×L2Ω;DAθ0,∞×L2Ω;DAθ0,∞⟺z1,z2∈L2Ω;DAθ0,∞.
Note that the previous spaces are well characterised as Besov spaces (cf. [9]). More precisely, since A is a positive operator we have (cf. [9, page 105])
(87)DA;DA2θ,∞=L2Ω;DA21+θ/2,∞.
On the other hand, if Λ:DA⊂X→X is a positive operator, from [9, Theorem 1.14.3(b)] it follows that (X,D(Λ2))σ,∞, σ∈(1/2,1), coincides with {x∈DA:t2σ-1[Λ(Λ+1)-1]ΛxL∞*X}<∞ = {x∈D(Λ):Λx∈(X,D(Λ))2σ-1,∞}. Therefore,
(88)L2Ω,DA2(1+θ)/2,∞=u∈DA:Au∈L2Ω,DAθ,∞.
Using now the notation in [9, page 321],
(89)DA=Hp,B12,B1u=u∂Ω,
we deduce the following characterization in terms of Besov spaces:
(90)L2Ω,DAθ,∞=B2,∞,B12θΩ,ifθ≠14,=f∈B2,2,B11/2Ω:B1f=f∂Ω,∈B2,2,B11/2Ω,ifθ=14,
cf. [9, Definition 4.3.2, page 317].Therefore, we assume that our data satisfy the following properties:(91)w0,w1,v0∈DA2×DA×DA,Aw1,-A2w0+αAv0,-αAw1-ηAv0∈L2Ω,DAθ,∞⟺A2w0,Aw1,Av0∈L2Ω,DAθ,∞,z1,z2∈L2Ω,DAθ0,∞,0<θ<θ0<1,∫Ω‍η2xz2xdx≠0,g∈C1+θ0,τ;R,g0=∫Ω‍η1xw0x+η2xv0xdx.
Under these assumptions the identification Problem (P) admits a unique solution (w,v,f) such that w∈[C2+θ([0,τ];L2(Ω))×C1+θ([0,τ];D(A))×Cθ([0,τ];D(A2))], v∈[C1+θ([0,τ];L2(Ω)) × Cθ([0,τ];D(A))] and f∈Cθ([0,τ];R).Remark 11.
SinceA is a strictly positive and self-adjoint operator in the Hilbert space X, we could treat in the same way the case when X=Lμ2(RN), as in Problem 9. Indeed, the realization is that A of -L2+I, with BQ=QB*, B=-Q∞-1 is strictly positive and self-adjoint [11, Proposition 9.3.10, page 251].Remark 12.
If we consider the identification Problem(P) with the additional information
(92)∫Ω‍η3xDtw(t,x)dx=gt,t∈0,τ,
then, setting
(93)Φy=0,∫Ω‍η3xy2xdx,0,
under the assumptions
(94)∫Ω‍η3xz1xdx≠0,z1,z2∈L2Ω,DAAθ0,∞,g∈C1+θ0,τ;R,g0=∫Ω‍η3xw1xdx,g∈C1+θ0,τ;R,g∈C1+θ0,τ;R,η3∈L2Ω,A2w0,Aw1,Av0∈L2Ω,DAθ,∞.
Problem (P), admits a unique solution (w,v,f).


## 4. First-Order Systems of Singular Differential Equations in Banach Spaces and Identification Problems

Here we face identification problems for systems of singular first-order differential equations in the Banach spaceX×X, both applying the general results described in Section 5 and developingad hoc methods in order to improve the corresponding consequences in some cases. For this purpose, we need some preliminary lemmas on the resolvent estimates.Theorem 13.
Suppose that the closed linear operatorsL1, L2, L4, M1, M2, M4 in X satisfy the conditions:
(95)M1(zM1+L1)-1L(X)≤C1+z-β1,∀z∈Σα,(96)M4(zM4+L4)-1L(X)≤C1+z-β4,∀z∈Σα,(97)M2x≤CM4xδx1-δ,0<δ≤1,x∈D(M4),(98)DL4⊆DL2continuously.
Then the matrix operators A and B defined by
(99)A=L1L20L4,B=M1M20M4
with D(A)=D(L1)×D(L4), D(B)=D(M1)×D(M4), satisfy the estimate
(100)00BzB+A-1L(X×X)≤C1+z-β,∀z∈Σα∩B0,R¯c,forlargeenoughR,
where β=β1+β4+δ-2.Proof.
First of all, we need to verify thatA and B are closed linear operators.
Letxn∈D(M1), yn∈D(M4), xn→x, yn→y, M1xn+M2yn→ξ, M4yn→η as n→+∞. We immediately deduce that y∈D(M4) ad M4y=η. Moreover,
(101)M2yn-M2ym≤CM4ym-M4ymδyn-ym1-δ
implies that M2yn→σ as n→+∞ for some σ∈X. Since M2 is closed δ=M2y. But M1 is closed too, so that x∈D(M1) and M1x+M2y=ξ. Analogously, if xn∈D(L1), yn∈D(L4), xn→x, yn→y as n→+∞, L1xn+L2yn→ξ, L4yn→η, then y∈D(L4) and L4y=η. Since D(L4)⊆D(L2)(102)yn-ym+L2yn-L2ym≤Cyn-ym+L4yn-L4ym
and thus L2yn has a limit σ as n→+∞. It follows that L2yn tends to σ=L2y so that we can conclude as the above.
We easily see that(103)zB+A-1=zM1+L1-1-zM1+L1-1zM2+L2zM4+L4-10zM4+L4-1
so that(104)BzB+A-1=M1zM1+L1-1-M1zM1+L1-1zM2+L2zM4+L4-1+M2zM4+L4-10M4zM4+L4-1.
Notice that ifz≠0, then
(105)-M1zM1+L1-1zM2+L2zM4+L4-1+M2zM4+L4-1=-1zI-L1zM1+L1-1×zM2+L2zM4+L4-1+M2zM4+L4-1=-I-L1zM1+L1-1M2zM4+L4-1-1zI-L1zM1+L1-1L2zM4+L4-1+M2zM4+L4-1=L1zM1+L1-1M2zM4+L4-1-z-1I-L1zM1+L1-1L2zM4+L4-1.
Therefore, from (97)(106)M2zM4+L4-1LX≤CM4zM4+L4-1LXδzM4+L4-1L(X)1-δ≤C1+z-β4δ1+z1-β41-δ≤C1+z1-β4-δ,
we deduce the bound
(107)-M1zM1+L1-1zM2+L2zM4+L4-1+M2zM4+L4-1LX≤c11+z2-β1-β4-δ+c21+z-β1-β4+1≤c1+z2-β1-β4-δ.
Since β1+β4+δ-2≤min⁡{β1,β4}, this completes the proof.We can now apply Proposition29 to the identification problem:
(108)ddtM1M20M4x(t)y(t)+L1L20L4x(t)y(t)=f(t)z1z2,∀t∈[0,τ],M1x+M2y0=M1x0+M2y0,M4y0=M4y0,Φ1M1xt+M2yt+Φ2M4yt=gt,Φ1M1xt+M2ytΦ2M4yt∀t∈0,τ.
Suppose that assumptions (95)–(98) are satisfied, with α+β1+β4+δ>7/2 and let 4-α-β1-β4-δ<θ<α+β1+β4+δ-3. Further, assume
(109)z1z2=M1z1*+M2z2*M4z2*,L1x0+L2y0L4y0=M1x0*+M2y0*M4y0*
with
(110)z1*∈D(L1),z2*∈D(L4),x0*∈D(L1),y0*∈D(L4).We have the following.Theorem 14.
Under assumptions (95)–(98), if α+β1+β4+δ>7/2, θ∈(4-α-β1-β4 - δ,α + β1+β4+δ-3), (109) and (110) hold together with g∈C1+θ([0,τ];R), Φ1,Φ2∈X*,
(111)Φ1z1+Φ2z2≠0,(112)Φ1M1x0+M2y0+Φ2M4y0=g0,
then the identification Problem (108) admits a unique global solution x,y,f∈Cθ0,τ;DL1 × Cθ0,τ;DL4×Cθ0,τ;R, M1x+M2y∈C1+θ0,τ;X, M4y∈C1+θ([0,τ];X).Remark 15.
WhenM2=0, we can take δ=1 in (97).Particular attention deserves the case whenα=β1=β4=δ=1, X being a reflexive Banach space. Then
(113)BzB+A-1LX×X≤C1+z-1,∀z∈k0+Σ1.
Let T=B(k0B+A)-1; that is,(114)T:=M1k0M1+L1-1L1k0M1+L1-1M2k0M4+L4-1-k0-1I-L1k0M1+L1-1L2k0M4+L4-10M4k0M4+L4-1.ThenX×X=N(T)⊕R(T)¯. Denote by P the projection operator onto N(T) along R(T)¯ and suppose 0<θ<θ0<1, g∈C1+θ([0,τ];R), Φ1,Φ2∈X*, (112) holds, if (I-P)[z1,z2]T=[z¯1,z¯2]T,
(115)Φ1z¯1+Φ2z¯2≠0,(116)sup⁡t>0tθk0M1+L1k0M2+L20k0M4+L4x0y0tT+I-1I-P×k0M1+L1k0M2+L20k0M4+L4x0y0R(T)¯<+∞,(117)sup⁡t>0tθ0tT+I-1z¯1z¯2RT¯<+∞.
After applying Proposition 32, we get the following result.Proposition 16.
LetX be a reflexive Banach space, suppose (95)–(98) hold with α=β1=β4=δ=1, T, P being as above, g∈C1+θ([0,τ];R), 0<θ<θ0<1, (x0,y0)∈D(L1)×D(L4), Φ1,Φ2∈X*. If (112)–(117) hold, then the identification Problem (108) admits a unique global solution (x,y,f)∈Cθ([0,τ];D(L1))×Cθ([0,τ];D(L4))×Cθ([0,τ];R), such that M1x+M2y∈C1+θ([0,τ];X), M4y∈C1+θ([0,τ];X). Moreover, if T has a closed range, then (116)-(117) can be dropped out.Next we extend Theorem13 to nontriangular operator matrices. Precisely, we consider the system (A,B) where
(118)A=L1L2L3L4,B=M1M20M4.We have the following.Lemma 17.
Assume (95)–(98) hold and D(M1)⊆D(L3), with
(119)L3x≤CM1xδ1x1-δ1,x∈D(M1),0<δ1≤1.
If β1+β4+δ+δ1>3, then B(zB+A)-1L(X×X)≤C(1+|z|)-β,∀z∈Σα,|z| large, when β=β1+β4 + δ-2.Proof.
We proceed by a perturbation argument. Write(120)zM1+L1zM2+L2L3zM4+L4=00L30+zM1+L1zM2+L20zM4+L4=I00I+00L30×zM1+L1-1-zM1+L1-1zM2+L2zM4+L4-10zM4+L4-1×zM1+L1zM2+L20zM4+L4=I00I+00L3zM1+L1-1-L3zM1+L1-1zM2+L2zM4+L4-1×zM1+L1zM2+L20zM4+L4.
In view of (119), we get first
(121)L3zM1+L1-1LX≤C1+z1-β1-δ1,L3zM1+L1-1LX0≤Ccf.Lemma17zM2+L2zM4+L4-1L(X)≤zM2zM4+L4-1L(X)+L2zM4+L4-1L(X)≤C1+z2-β4-δ+1+z1-β4≤C1+z2-β4-δ,
and then
(122)L3zM1+L1-1zM2+L2zM4+L4-1≤C1+z3-β1-β4-δ-δ1.
Notice now that 1-β1-δ1≤3-β1-β4-δ-δ1 implies that the linear operator
(123)I00I+00L3zM1+L1-1-L3zM1+L1-1zM2+L2zM4+L4-1
has a bounded inverse for |z| large and its inverse can be estimated in the norm of L(X×X) by 1/2, for example. It follows that B(zB+A)-1 has precisely the same bound as in Lemma 17.Theorem 18.
Under assumptions (95)–(98) and (119), if α+β1+β4+δ>7/2 and β1+β4+δ+δ1>3, θ∈(4-α-β1-β4-δ,α+β1+β4+δ-3),
(124)z1z2=M1z1*+M2z2*M4z2*,L1x0+L2y0L3x0+L4y0=M1x0*+M2y0*M4y0*
for some z1*,x0*∈D(L1), z2*,y0*∈D(L4), g∈C1+θ([0,τ];R), Φ1,Φ2∈X* and (111)-(112) hold, then the identification problem
(125)ddtM1M20M4x(t)y(t)+L1L2L3L4x(t)y(t)=ftz1z2,ddtM1M20M4x(t)y(t)+L1L2L3L4x(t)y(t)=∀t∈0,τ,M1x+M2y0=M1x0+M2y0,M4y0=M4y0,Φ1M1xt+M2yt+Φ2M4yt=g(t),Φ1M1xt+M2ytΦ2M4yt∀t∈0,τ,
admits a unique global solution (x,y,f)∈Cθ([0,τ];D(L1))×Cθ([0,τ];D(L4))×Cθ([0,τ];R), such that M1x+M2y∈C1+θ([0,τ];X), M4y∈C1+θ([0,τ];X).Remark 19.
WhenM2=0, we can choose δ=1; thus, the first two conditions in the statement of Theorem 18 involving α, β1, β4, β5, δ reduce to α+β1+β5>5/2 and β1+β4+δ>2.Remark 20.
We could easily establish a result at all analogous to the one in Proposition16, concerning this more general case.We are now in a position to face the general identification problem:(126)ddtM1M2M3M4xtyt+L1L2L3L4xtyt=ftz1z2,∀t∈0,τ,(127)M1M2M3M4x0y0=M1M2M3M4x0y0,(128)Φ1M1xt+M2yt+Φ2M3xt+M4yt=gt,Φ1M1xt+M2yt+Φ2M3t+M4t∀t∈0,τ.
To this end, we will assume that
(129)M1,M3haveboundedinverses,withDM1⊆DM3.
Multiply the second equation in the system by M1M3-1 and substract the obtained equation from the first one. We obtain the following system equivalent to (126):
(130)ddtM1M20M2-M1M3-1M4xtyt+L1L2L1-M1M3-1L3L2-M1M3-1L4xtyt=ftz1z1-M1M3-1z2,∀t∈0,τ,
together with
(131)M1M20M2-M1M3-1M4x(0)y(0)=M1M20M2-M1M3-1M4x0y0
and the additional information
(132)Ψ1M2-M1M3-1M4yM1xt+M2yt+Ψ2M2-M1M3-1M4yt=gt,+Ψ2M2-M1M3-1M4yt∀t∈0,τ,
where Ψ=(Ψ1,Ψ2) is related to Φ=(Φ1,Φ2) by
(133)Ψ=Φ10M3M1-1-M3M1-1,
so that(134)ΨM1xt+M2yt,M2-M1M3-1M4yt=Ψ1M1xt+M2yt+Ψ2M2-M1M3-1M4yt=Φ10M3M1-1-M3M1-1M1x(t)+M2y(t)M2-M1M3-1M4y(t)=ΦM1x(t)+M2ytM3M1-1M1x(t)+M3M1-1M2y(t)-M3M1-1M2y(t)+M4y(t)=ΦM1x(t)+M2y(t)M3x(t)+M4y(t)=Φ1[M1x(t)+M2y(t)]+Φ2M3xt+M4yt.Notice that the operatorM2-M1M3-1M4 may have no bounded inverse. Our next step is to translate to this case the conditions established in Theorem 14. Assume, in addition to (95),
(135)×zM2-M1M3-1M4+L2-M1M3-1L4-1M2-M1M3-1M4×zM2-M1M3-1M4+L2-M1M3-1L4-1L(X)≤C1+z-β4,∀z∈Σα,(136)M2x≤CM2-M1M3-1M4xδx1-δ,0<δ≤1,M2xCM1M3-1M4δx1-δ,x∈D(M2-M1M3-1M4),(137)L1-M1M3-1L3hasaboundedextensiontoX,
and thus δ1=1. Compute now
(138)Ψz1z1-M1M3-1z2=Φ10M3M1-1-M3M1-1z1z1-M1M3-1z2=Φz1M3M1-1z1-M3M1-1z1+z2=Φ1[z1]+Φ2[z2],
as desired.Therefore, we are in a position to establish the following result.Theorem 21.
Suppose that operatorsLi, Mi satisfy (95), (135), (136), and (137) and that M1,M3∈L(X) have bounded inverses, β1+β4+δ>2, and α+β1+β4+δ>7/2, θ∈(4-α-β1-β4-δ,α+β1 + β4+δ-3),
(139)z1z2=M1z1*+M2z2*M3z1*+M4z2*,L1L2L3L4x0y0=M1M2M3M4x0*y0*,
where z1*,x0*∈D(L1), z2*,y0*∈D(L4), g∈C1+θ([0,τ];R), and (111) holds together with the compatibility relation
(140)Φ1M1x0+M2y0+Φ2M3x0+M4y0=g0.
Then the identification Problems (126)–(128) admit a unique strict global solution (x,y,f)∈Cθ([0,τ];D(L1))×Cθ([0,τ];D(L4))×Cθ([0,τ];R), with M1x+M2y∈C1+θ([0,τ];X), M3x+M4y∈C1+θ([0,τ];X).Proof.
It is a simple rewriting of the result in Theorem18 as applied to Problems (130)–(132), taking into account the relation to the starting Problems (126)–(128).Of course, the preceding results apply to the abstract strongly degenerate elliptic-parabolic system(141)ddtM1M200x(t)y(t)+L1L2L3L4x(t)y(t)=ftz1z2,∀t∈0,τ,(142)M1x+M2y0=M1x0+M2y0,(143)ΦM1xt+M2yt=gt,∀t∈[0,τ].
However, in view of this generality, the corresponding assumptions would yield the restrictions z2=0 and M2=0. To overcome this difficulty we will make suitable assumptions on the operators involved.Clearly, ifz2=0 and L4 has a bounded inverse, the second equation in (141) gives y(t)=-L4-1L3x(t), so that (141)–(143) reduces to the identification problem:
(144)ddtM1-M2L4-1L3xt+L1-L2L4-1L3xt=ftz1,0≤t≤τ,M1-M2L4-1L3x0=M1x0+M2y0,Φ(M1-M2L4-1L3)x(t)=gt,0≤t≤τ.
Observe that the compatibility relation y0=L4-1L3x0 must hold. On the other hand, all the results in Propositions 29 and 32 apply provided that operators A=L1-L2L4-1L3 and B=M1-M2L4-1L3 satisfy the assumptions described there.Ifz2≠0, so that y4(t)=f(t)L4-1z2-L4-1L3x(t), we arrive at the following identification problem:
(145)ddtM1xt-M2L4-1L3xt+ftM2L4-1z2+L1-L2L4-1L3xt=ftz1-L2L4-1z2,0≤t≤τ,M1-M2L4-1L3x+ftM2L4-1z20=M1x0+M2y0,Φ(M1-M2L4-1L3)x(t)+f(t)M2L4-1z2=gt,0≤t≤τ.
Notice the extra difficulty arising from the fact that the unknown term f(t) is not supposed to be differentiable. However, if M1-M2L4-1L3 has a bounded inverse, we can introduce the new unknown:
(146)M1-M2L4-1L3xt+ftM2L4-1z2=ξt,
so that
(147)xt=M1-M2L4-1L3-1ξtxt=-ftM1-M2L4-1L3-1M2L4-1z2.Then (145) reads equivalently
(148)ddtξt+L1-L2L4-1L3M1-M2L4-1L3-1ξt=ftz¯,0≤t≤τ,ξ(0)=M1x0+M2y0,Φξ(t)=g(t),0≤t≤τ,
where
(149)z¯=L1-L2L4-1L3M1-M2L4-1L3-1M2L4-1z2z1-L2L4-1z2+L1-L2L4-1L3M1-M2L4-1L3-1M2L4-1z2
This is a regular identification problem if, for example, (L1-L2L4-1L3)(M1-M2L4-1L3)-1 is a bounded operator. Indeed, we have the following.Corollary 22.
LetM1-M2L4-1L3 have a bounded inverse and let (L1-L2L4-1L3)(M1-M2L4-1L3)-1 be bounded. Let the compatibility relation Φ[M1x0+M2y0]=g(0) hold together with Φ[z¯]≠0. Then Problems (141)–(143) admit a unique strict solution (x,y,f)∈Cθ([0,τ];D(L1)) × Cθ([0,τ];D(L4))×Cθ([0,τ];R), 0<θ<1.Corollary22 can, in fact, be refined weakening the assumption on (L1-L2L4-1L3)(M1-M2L4-1L3)-1. For this purpose consider the system
(150)M1x+M2y=ξ,L3x+L4y=ftz2.
If operator M1-M2L4-1L3:=S has an inverse, then
(151)x=S-1ξ-S-1M2L4-1z2ft,y=-L4-1L3S-1ξ+ftL4-1+L4-1L3S-1M2L4-1z2.
Therefore, the pair (ξ,f) satisfies the problem
(152)ξ′t+L1-L2L4-1L3S-1ξt=ftz¯,0≤t≤τ,ξ(0)=M1x0+M2y0,Φξt=gt,0≤t≤τ,
where
(153)Φz¯≠0,z¯∶=z1-L2L4-1z2+L1-L2L4-1L3S-1M2L4-1z2.
That is, (ξ,f) satisfies (148), as desired.LetA=(L1-L2L4-1L3)S-1. Then, from Theorem 2.1 in [1] we deduce the following.Theorem 23.
Let-A be the generator of an analytic semigroup of negative type in X, 0<θ<θ0<1, ξ0=M1x0+M2y0∈DA(θ+1,+∞), z¯∈DA(θ0,+∞), Φ[z¯]≠0, g∈C1+θ([0,τ];R), Φ[ξ0]=g(0). Then Problems (141)–(143) admit a unique strict global solution (x,y,f)∈Cθ([0,τ];D(L1))×Cθ([0,τ];D(L4)) × Cθ([0,τ];R), such that M1x+M2y∈C1+θ([0,τ];X).Applying Corollary31, we also obtain the following.Corollary 24.
If(154)z+A-1L(X)≤C1+z-β,∀z∈Σ1,
where β∈(1/2,1), z¯∈D(A), Φ[z¯]≠0, ξ0=M1x0+M2y0∈D(A2), Φξ0=g(0), g∈C1+θ([0,τ];R), 1-β<θ<β, then Problems (141)–(143) admit a unique strict global solution (x,y,f)∈Cθ([0,τ];D(L1))×Cθ([0,τ];D(L4))×Cθ([0,τ];R), with M1x+M2y∈C1+θ([0,τ];X).Example 25.
Consider the identification problem:(155)ddtII00x(t)y(t)+αLβLγLδLx(t)y(t)=ftz1z2,∀t∈0,τ,x+y0=ξ0,Φx+yt=gt,∀t∈0,τ,Φξ0=g0,
where L is a closed linear operator in the Banach space X and α,β,γ,δ∈C.To apply Theorem23 we set S=((δ-γ)/δ)I and we note that L1-L2L4-1L3=(αδ-βγ/δ)L, so that (L1-L2L4-1L3)S-1=((αδ-βγ)/(δ-γ))L. Moreover,
(156)z¯=δ-γz1+α-βz2δ-γ.
It follows that if ((αδ-βγ)/(-δ+γ))L generates an analytic semigroup of negative type and (δ-γ)z1+(α-β)z2∈DL(θ0,+∞), Φ[(δ-γ)z1+(α-β)z2]≠0, g∈C1+θ([0,τ];R), 0<θ<θ0<1, ξ0∈DL(θ+1,∞), then the identification Problem (155) admits a unique global strict solution (x,y,f)∈Cθ([0,τ];D(L))×Cθ([0,τ];D(L))×Cθ([0,τ];R), x+y∈C1+θ([0,τ];X).We stress that we have been able to determine a triplet of functionsu1,u2:[0,τ]×Ω→R and f:[0,τ]→R in the following parabolic-elliptic identification problem:
(157)Dtu1x,t+u2x,t+αLx,Dxu1x,t+βLx,Dxu2x,t=ftz1x,t∈0,τ,x∈Ω,(158)γLx,Dxu1x,t+δLx,Dxu2x,t=ftz2x,t∈0,τ,x∈Ω,(159)u1x,t+u2x,t=u1,0x+u2,0x,t∈0,τ,x∈∂Ω,(160)u1x,t=u2x,t=0,x∈∂Ω,(161)∫Ω‍ψxu1x,t+u2x,tdt=gt,t∈[0,τ].Here operatorL(x,Dx) is defined by
(162)DL=W2,pΩ∩W01,pΩ,p∈(1,+∞),Lux=-∑i,j=1n‍Dxiai,jxDxjLux=+∑i=1n‍aixDxi+a0x,x∈Ω,
where (αδ-βγ)(γ-δ)-1<0 and the coefficients ai,j, aj, a0 satisfy the following assumptions:
(163)ai,j,Dxiai,j,ai,Dxiai,a0∈CΩ¯,ai,jx=aj,ix,ai,j,Dxiai,j,ai,Dxiai,00∈∈CΩ¯,x∈Ω,i,j=1,…,n,∑i,j=1nai,jxξiξj≥c0ξ2,x∈Ω¯,ξ∈Rn,a0x-1p∑i=1n‍Dxiaix≥c1,x∈Ω¯.
The same argument applies when L(x,D) is the opposite of the realization in X=Lμp(Rn), p∈(1,+∞), of the Ornstein-Uhlenbeck operator Lp, precisely
(164)DL=Wμ2,pRn,p∈(1,+∞),L=-Lp+kI,DL=Wμ2,pRn,p∈(1,+∞),LLp0=k∈R+
(for the properties in Lμp, p∈(1,+∞), cf. Problem 9).In this case (161) changes to
(165)∫Ω‍ψxu1x,t+u2x,tμdx=gt,t∈0,τ.Example 26.
Consider the identification problem:(166)ddtIIIIx(t)y(t)+αLβLγLδLx(t)y(t)=ftz1z2,∀t∈0,τ,x+y0=ξ0,Φ1x+yt+Φ2x+yt=Φ1+Φ2x+yt=gt,000000000000000000000∀t∈0,τ,
with (Φ1+Φ2)(ξ0)=g(0).Such a problem is easily reduced to an equivalent problem related to the integral differential equation:(167)ddtII00xtyt+αLβLα-γLβ-δLxtyt=ftz1z1-z2,∀t∈0,τ.
Consequently, we have come back to Example 25 with γ, δ, z2 being replaced by α-γ, β-δ, z1-z2, respectively.Taking this into account, the same conclusions as Example25 can be obtained, provided that αδ-βγ≠0, α-β-γ+δ≠0, -(αδ-βγ)(α-β-γ+δ)-1L generates a holomorphic semigroup in X, (γ-δ)z1-(α-β)z2∈DL(θ0,+∞), (Φ1+Φ2)[(γ-δ)z1-(α-β)z2]≠0, ξ0∈DL(θ+1,+∞). Then the identification problem above admits a unique global strict solution (x,y,f)∈Cθ([0,τ];D(L))×Cθ([0,τ];D(L))×Cθ([0,τ];R), x+y∈C1+θ([0,τ];X).Consider now the identification problem:(168)ddt00M30x(t)y(t)+L1L2L3L4x(t)y(t)=ftz1z2,ddt00M30x(t)y(t)+=L1L2L3L4x(t)y(t)=∀t∈0,τ,M3x0=M3x0,ΦM3xt=gt,∀t∈[0,τ],Φ[M3x0]=g(0).
Under the assumption that L2 have a bounded inverse we get y=f(t)L2-1z1-L2-1L1x. Therefore we obtain the following differential equation for the single unknown x:
(169)ddtM3xt+L3-L4L2-1L1x=ftz2-L4L2-1z1.
Now it suffices to suppose that the pair M3,L3-L4L2-1L1 satisfies the properties described in [1] or in Section 5 in order to obtain existence and uniqueness for the given identification problem.IfM3 has an inverse M3-1, introduce function ξ defined by M3x=ξ, so that
(170)ξ′t+L3M3-1ξt+L4yt=ftz2.
Assuming that L4 has a bounded inverse, we have
(171)yt=ftL4-1z2-L4-1ξ′t-L4-1L3M3-1ξt.
Substituting in the first equation of the system, we get
(172)L1M3-1ξt=L2L4-1ξ′t-L2L4-1L3M3-1ξt=ftz1-L2L4-1z2.
That is, we have obtained the following differential identification problem for ξ:
(173)L2L4-1ξ′(t)+L2L4-1L3-L1M3-1ξt=ftL2L4-1z2-z1,0≤t≤τ,ξ0=M3x0,Φξt=gt,0≤t≤τ.
This identification problem might be treated similarly under the obvious hypothesis L2L4-1 to have a bounded inverse. But if L2 is invertible we come back to the first case.The assumption requiringL2 to be invertible seems really essential in some sense. As an example, take
(174)ddt00M30xtyt+L100L4xtyt=ftz1z2,ddt00M30xtyt+L100L4xtyt=f∀t∈0,τ,M3x0=ξ0,ΦM3xt=gt,∀t∈[0,τ],Φ[ξ0]=g(0),
where L1 and L2 admit bounded inverses. Then x(t)=f(t)L1-1z1 implies Φ[M3x(t)] = Φ[f(t)M3L1-1z1]=g(t). Hence, if Φ[M3L1-1z1]≠0, then we necessarily have
(175)ft=gtΦM3L1-1z1,xt=gtΦM3L1-1z1L1-1z1.
Moreover, the second equation furnishes uniquely y(t). This shows the importance of the invertibility of M3, as expected. Notice too that no assumption like semigroup generation is required to operators L1 and L2.Example 27.
LetA, B be two bounded linear operators in X. Observe first that the identification problem
(176)ddtABABx(t)y(t)+L1L2L3L4x(t)y(t)=ftz1z2,∀t∈0,τ,(Ax+By)(0)=ξ0,ΦAxt+Byt=gt,∀t∈0,τ,
is equivalent to the problem
(177)ddtAB00xtyt+L1L2L1-L3L2-L4xtyt=ftz1z1-z2,∀t∈[0,τ],Ax+By0=ξ0,ΦAxt+Byt=gt,∀t∈0,τ.
Thus we can apply both Theorem 23 and Corollary 22. Operator A is now given by (L1-L2(L2 - L4)-1(L1-L3))(A-B(L2-L4)-1(L1-L3))-1, while z¯ is given by z¯=z1-L2(L2 - L4)-1(z1-z2)+AB(L2-L4)-1(z1-z2).Example 28.
LetA, B be two bounded linear operators in X, with 0∈ρ(B), 0∈ρ(A+kB) for some k≠0.Consider the identification problem:(178)ddtAABBxtyt+L1L2L3L4xtyt=ftz1z1,∀t∈0,τ,A(x+y)(0)=Aξ0,B(x+y)(0)=Bη0,Φ1Ax+yt+Φ2Bx+yt=gt,∀t∈0,τ,
with a compatibility relation g(0)=Φ1[Aξ0]+Φ2[Bη0], Aξ0=Aη0.Note that, under our assumptions onA and B, such a problem is equivalent to the following:
(179)ddtIIIIx(t)y(t)+A+kB-1L1+kL3A+kB-1L2+kL4B-1L3B-1L4=ftA+kB-1z1+kz2B-1z2,∀t∈[0,τ],x+y0=A+kB-1Aξ0+kBη0=η0,Ψ1xt+yt+Ψ2xt+yt=gt,∀t∈0,τ,
where Ψ1=Φ1A, Ψ2=Φ2B.Since, in turn, this identification problem is equivalent to(180)ddtII00xtyt+A+kB-1L1+kL3A+kB-1L2+kL4A+kB-1L1+kL3-B-1L3A+kB-1L2+kL4-B-1L4xtyt=ftA+kB-1z1+kz2A+kB-1z1+kz2-B-1z2,∀t∈0,τ,x+y0=A+kB-1Aξ0+kBη0=η0,Ψ1+Ψ2x+yt=gt,∀t∈0,τ.Theorem23 and Corollary 24 run as well.As an example, letX=Lp(Ω), 1<p<+∞, Ω being a domain in Rn of class C2. Let L=-Δ be the laplacian in X endowed with Dirichlet boundary conditions, L1=αL, and let L2=βL, L3=γL, L4=δL, α,β,γ,δ∈C. Let m and n be two real-valued continuous functions on Ω¯, n(x)≥c>0, m being possibly negative. Let k∈R be such that m(x)+kn(x)>0 for all x∈Ω¯. Given z1,z2∈Lp(Ω), our identification problem consists in finding a triplet (u,v,f) such that
(181)Dtmxux,t+vx,t-αΔux,t-βΔvx,t=ftz1x,t∈0,τ,x∈Ω,Dtnxux,t+vx,t-γΔux,t-δΔvx,t=ftz2x,t∈0,τ,x∈Ω,ux,t=vx,t=0,t∈0,τ,x∈∂Ω,mxux,0+vx,0=mxξ0x,x∈Ω,nxux,0+vx,0=nxη0x,x∈Ω,∫Ω‍η1xmx+η2xnxux,t+vx,tdt=gt,t∈0,τ,η1, η2 being given functions in Lq(Ω),1/p+1/q=1, with ∫Ω‍[η1(x)m(x)+η2(x)n(x)]ξ0(x)dx=g(0), m(x)[ξ0(x)-η0(x)]≡0. If ν(x):=m(x)+kn(x), let
(182)ax=α+kγγx,bx=β+kδνx,cx=α+kγνx-γnx,d(x)=β+kδνx-γnx.
Then S-1=L-1(d/d-c)L,L1-L2L4-1L3=((ad-bc)/d)L, so that -A=-((ad-bc)/d-c)L = ((ad-bc)/d-c)Δ is assumed to generate an analytic semigroup in Lp(Ω).Since(183)z¯x=z1x+kz2xνx-bxdxz1x+kz2xνx-z2xnx+ad-bcd-cβ+kδrx-δnx-1×z1(x)+kz2xνx-z2xnx,
our previous abstract assumptions read as
(184)z¯∈DAθ0,∞=DΔθ0,∞,∫Ω‍η1xmx+η2xnxη0xdx=g0,∫Ω‍η1xmx+η2xnxz¯(x)dx≠0.
Suppose α,β,γ,δ∈R. Trivial computations show that
(185)ad-bcd-c=-α+kγnxνx+γν2xα+kγβ+kδnx-α+kγδνx-α+kγnxνx+γν2x×νxnxβ-α+kδ-γ+νxγ-δ-1=ωx.
We need ω(x)≥c>0 for all x∈Ω¯. Therefore, all the required elements are determined.Clearly, if-A generates an infinitely differentiable semigroup
(186)(z+A)-1L(X)≤C1+z-β,∀z∈Σ1,β∈12,1,
we are compelled to require z¯∈D(A), as in Corollary 24, (cf. [10, 13, 14]).


## 5. Some Improvements of Known Results

For the reader’s convenience, we report here the main results in [1] with some minor improvements.Proposition 29.
Let0<β≤α≤1, α+β>3/2, 2-α-β<θ<α+β-1. Let X be a Banach space, let Φ∈X* and A:D(A)⊂X→X and B:D(B)⊂X→X be two closed linear operators such that (i) D(A)⊆D(B)j; (ii) A is invertible; (iii) B(λB+A)-1L(X)≤c(1+|λ|)-β for all λ∈Σα, where
(187)Σα=λ∈C:Reλ≥-c01+λα,c0>0.
Let g∈C1+θ([0,τ];R) and z∈X satisfy the following properties for some z*, v*∈D(A):
(188)Φz≠0,ΦBu0=g0,z=Bz*,Au0=Bv*.
Then the identification problem
(189)Bu′t+Aut=ftz,0≤t≤τ,Bu0=Bu0,ΦBut=gt,0≤t≤τ,
admits a unique global solution:
(190)u,f∈Cθ0,τ;DA×Cθ0,τ;R,Bu∈C1+θ0,τ;X.Remark 30.
Assumptions (ii) and (iii) can be weakened to(iibis)
k0B+A is invertible for some k0∈R;(iiibis)
B(λB+A)-1L(X)≤c(1+|λ|)-β for all λ∈k0+Σα.Indeed, let us introduce the new unknownv(t)=e-k0tu(t). Then Problem (189) is equivalent to the following:
(191)Bv′t+A+k0Bvt=e-k0tftz,0≤t≤τ,Bv0=Bu0,ΦBut=e-k0tgt,0≤t≤τ.
Proposition 29 applies immediately provided we replace the triplet (A,f,g) by ((A+k0B),e-k0tf,e-k0tg). Once v and f1(t)=e-k0tf(t) have been determined so are u and f, with the same regularity.As a consequence, we have the following result relative to the generators-A of infinitely differentiable semigroups of parabolic type with nonnecessarily dense domains, satisfying
(192)(λ+A)-1L(X)≤c1+λ-β,λ∈k0+Σα.Corollary 31.
Let0<β≤α≤1, α+β>3/2, and 2-α-β<θ<α+β-1. Let Ψ∈X* and the closed linear operator A satisfy (192) in X. Let g∈C1+θ([0,τ];R), Φ∈X*, z∈D(A), Φ[z]≠0, w0∈D(A2), Φ[w0]=g(0). Then the identification problem
(193)w′t+Awt=ftz,0≤t≤τ,w0=w0,Ψwt=gt,0≤t≤τ,
has a unique global solution (w,f)∈[C1+θ([0,τ];X)∩Cθ([0,τ],D(A))]×Cθ([0,τ];R).In particular, but very important case, whereα=β=1, and X is reflexive, it is possible to weaken the assumptions on the initial data and z. Notice that in the statement of Proposition 29z must belong to the range of BA-1 (or of B(A+k0B)-1)) and Φ[z]≠0.The following extension to Theorem 2.2 in [1] holds.Proposition 32.
LetA:D(A)⊂X→X and B:D(B)⊂X→X be two closed linear operators in the reflexive Banach space X, with D(A)⊆D(B), Φ∈X* and g∈C1+θ([0,τ];R), θ∈(0,1). Suppose
(194)ΦBu0=g0,BλB+A-1L(X)≤c1+λ-1,λ∈k0+Σα.
Let T=B(k0B+A)-1 and let P be the projection of X on the null space N(T) along R(T)¯. Suppose θ0∈(θ,1) and
(195)ΦI-Pz≠0,(196)sup⁡t>0tθA+k0Bt+k0B+A-1×I-PA+k0Bu0A+k0Bt+k0B+A-1R(T)¯<+∞,(197)sup⁡t>0tθ0A+k0Bt+k0B+A-1I-PzR(T)¯<+∞.
Then the identification Problem (189) admits a unique global solution
(198)u,f∈Cθ0,τ;DA×Cθ0,τ;R,Bu∈C1+θ0,τ;X.Remark 33.
IfT has aclosed range, conditions (196)-(197) can be dropped out. Indeed, it suffices to apply Theorem 2.2 in [1] to (191) and to observe that (D(T~-1)¯,D(T~-1))θ,∞ reduces to D(T~-1)=R(T~)=R(T), T~ denoting the restriction of T to R(T)¯=R(T).Remark 34.
Very recently in [8, Theorem 3.1] the above results have been improved to the case α=β=1 and B=I.Proposition 35.
LetA be the generator of an analytic semigroup in the complex Banach space X, let Φ∈X*, y0∈D(A), Ay0∈(X,D(A))θ,∞ for some θ∈(0,1), and let g∈C1+θ([0,τ];C), z∈(X,D(A))θ,∞, Φ[z]≠0, Φ[y0]=g(0). Then the inverse problem
(199)y′t=Ay(t)+f(t)z,0≤t≤τ,y0=y0,Φyt=gt,0≤t≤τ,
admits a unique solution (y,f) such that
(200)t∈C1+θ0,τ;X∩Cθ0,τ;DA,f∈Cθ0,τ;R.
As for Problem (199), also the hyperbolic case, corresponding to the case where A generates a C0-semigroup, has been dealt with in [8, Corollary 2.1].Proposition 36.
LetA be the generator of a C0-semigroup in the complex Banach space X, and let Φ∈X*, z∈D(A), Φ[z]≠0, y0∈D(A), g∈C1([0,τ];C), Φ[y0]=g(0). Then the inverse Problem (199) admits a unique solution (y,f)∈[C1([0,τ];X)∩C([0,τ];D(A))]×C([0,τ];R).In the next section, we are giving specific applications of the results listed in this section.


## 6. Applications

In this section, we will give several concrete applications of our previous abstract results.Problem 37.
First, we recall some previous results from [15–17]. Let L and M be two linear differential operators with domains in Lp(Ω), p∈(1,+∞), Ω∈Rn being a bounded region with a boundary ∂Ω of class C2. M—a multiplication operator by a nonnegative function m∈L∞(Ω)—is defined by
(201)DM=LpΩ,Mux=mxux,x∈Ω.
Operator L is defined either by
(202)DL=W2,pΩ∩W01,pΩ,Lux=LuxLux∶=-∑i,j=1n‍Dxiai,jxDxjuxLux+∑j=1n‍ajxDxjux+a0xux,L+∑j=1n‍ajxDxjuxxx∈Ω,u∈DL,
or by
(203)DL=u∈W2,pΩ:∑i,j=1n‍ai,jxνixDxjux+bxux=0on⁡∂Ω∑i,j=1n‍,Lux=Lux,x∈Ω,u∈D(L).
We assume that the coefficients ai,j, aj, a0 enjoy the following properties:
(204)ai,j,Dxiai,j,ai,Dxiai,a0∈CΩ¯,i,j=1,…,n,ai,jx=aj,ix,x∈Ω¯,∑i,j=1n‍ai,jxξiξj≥c0ξ2,x∈Ω¯,ξ∈Rn,a0x-1p∑i=1n‍Dxiaix≥c1,x∈Ω¯,bx+1p∑i=1n‍aixνix≥0,x∈∂Ω,c0 and c1 being two positive constants.Then it is shown in Theorem 2.1 in [17] that the pair (L,M) satisfies in the sector
(205)Σc=z∈C:Rez≥c1+z,c∈R+
the following estimate with β=1/p:
(206)M(zM+L)-1L(X)≤C1+z-β.
Let us consider the following identification parabolic-elliptic problem:
(207)Dtm1xkm2xb0m2xaut,xvt,x+L1x,DxL2x,DxOL4x,Dxut,xvt,x=ftz1z1,∀t,x∈0,τ×Ω,m1(x)km2xb0m2xau0,xv0,x=m1xkm2xb0m2xau0xv0x,∀x∈Ω,∫Ω‍η1xm1xut,x+km2xbvt,x0000+η2xm2xavt,xm1xut,x+km2xbvt,xdx=g(t),∀t∈0,τ,
where the linear operators Li(x,Dx), i=1,2,4, and D(L4)⊂D(L2) enjoy the same properties as L(x,Dx) with β=1/p, m1,m2∈L∞(Ω) with m1≥0 and m2≥0 a.e. in Ω, k≠0 (if k=0 we can apply Remark 19), 0<a≤b, z1,z2,u0,v0∈Lp(Ω), η1,η2∈Lq(Ω), 1/p+1/q=1, g∈C([0,τ];R).We know that the pairs(L1(x,Dx),M1), (L2(x,Dx),M4), where M4u(x)=m2(x)au(x), satisfy (206) in Lp(Ω).In view of the moment inequality (cfr. [18, page 115]), assumption (97) holds with δ=b/a. Therefore, Theorem 14 applies provided that (2/p)+(b/a)>5/2; that is, p<((5a-2b)/4a)-1. So, we must necessarily have b≤a<2b and 1<p<4/3 respectively.Letθ∈(3-(2/p)-(b/a),(2/p)+(b/a)-2), 1<p<4a/(5a-2b),
(208)z1z2=m1z1*+km2bz2*m2az2*,L1·,Dxu0+L2·,Dxv0L4·,Dxv0=m1u0*+km2bv0*m2av0*,
where z1*,u0*∈D(L1), z2*,v0*∈D(L4),
(209)∫Ω‍η1xz1x+η2xz2xdx≠0,∫Ω‍η1xm1xu0x+km2xbv0x+η2xm2xav0xm1xu0x+km2xbv0xdx=g0.
Then Problem (207) admits a unique global solution (u,v,f)∈Cθ([0,τ];D(L1))×Cθ([0,τ];D(L4)) × Cθ([0,τ];R), m1u+km2bv, m2av∈C1+θ([0,τ];Lp(Ω)).Using the same scheme, one could handle the more general problem:(210)Dtm1(x)km2xb0m2xaut,xvt,x+L1x,DxL2x,Dxk1m1xcL4x,Dxut,xvt,x=ftz1z2,∀(t,x)∈[0,τ]×Ω,
where c∈(0,1], k1 is constant, provided that (2/p)+(b/a)+c>3 (cf. Theorem 14).As a particular case, we can also treat the problem(211)Dtm1(x)m2xb0m2xaut,xvt,x+L1x,Dxn1xkm1xcn2xut,xvt,x=ftz1z2,∀t,x∈0,τ×Ω,
where n1,n2∈C(Ω¯;R), n2(x)≥c~>0, c∈(0,1], k∈R.Problem37bis. We show here how some more regularity of functions m1 and m2 allow to choose a larger exponent p in the reference space Lp(Ω).We recall that a functionm∈C1(Ω¯), m(x)≥0, x∈Ω¯, is said to be ρ-regular for some ρ∈(0,1] [3, 15, 16] if there exists a positive constant C such that
(212)∇mx≤Cmxρ,x∈Ω¯.
If m0∈C1(Ω¯), m0(x)≥0, x∈Ω¯, and k>1, then m(x)=m0(x)k is ρ-regular with ρ=(k-1)/k.Ifm is ρ-regular and
(213)∑i=1n‍aixDximx≤0,x∈Ω¯,bx+1pmxp-1∑i=1n‍aixνix≥0,x∈Ω¯,
then (cfr. Theorem 3.3 in [17]) the pair (L,M) satisfies (206) with α=1, β=2[p(2-ρ)]-1. Note that β>1/2 if and only if p∈[2,4(2-ρ)-1).Let nowm1 be ρ1-regular, so that (L1(x,Dx),M1) satisfies (206) with β1=2[p(2-ρ1)]-1 according to [15]. Moreover, let m2∈C1(Ω¯), m2(x)≥0, x∈Ω¯, a>1. Then, β4=2[p(1+(1/a))]-1. Thus, we must have
(214)2p12-ρ1+a1+a+ba>52,ba>12
to find
(215)2≤p<4a5a-2b12-ρ1+a1+a2⟹2a5a-2b12-ρ1+a1+a>1.
Let b=νa, 1/2<ν<1. Then the right-hand side in (215) changes to
(216)12-ρ1+bb+ν+ν>52.
In particular, the last property is achieved if ρ1 tends to 1 and a is large enough.Therefore, Theorem14 applies in the reference space Lp(Ω) with b=νa and
(217)2≤p<45-2ν12-ρ1+a1+a,
whenever the following inequalities hold:
(218)12-ρ1+a1+a+ν>52,12<ν<1.Problem 38.
We are concerned with the initial and boundary value problem:(219)Dtm1xm2xm3xm4xut,xvt,x+α1xL1x,Dxα2xL2x,Dxα3xL3x,Dxα4xL2x,Dxut,xvt,x=ftz1z1,∀(t,x)∈0,τ×Ω,m1xm2xm3xm4xu0,xv0,x=m1xm2xm3xm4xu0xv0x,x∈Ω,∫Ω‍η1xm1xut,x+m2xvt,x+η2xm3xut,x+m4xvt,xdx=gt,0≤t≤τ,
where for the sake of simplicity, we have set
(220)Ljx,Dx=-Δ+ajx,j=1,2,3
and have endowed such operators with either the Dirichlet or the Robin boundary condition.We will assume thatmj∈L∞(Ω), j=1,…,4, are real-valued, while αj∈C(Ω¯), j=1,…,4, and ai∈C(Ω¯), j=1,2,3, are scalar functions, the ai’s being nonnegative, satisfying the following properties:
(221)mjx≥Cj>0,j=1,3,-Cm1xm3xm4x-m2x≤m2x≤m1xm3xm4x,x∈Ω,(222)α1≥c>0,α2x-m1xm3xα4(x)≤-c¯<0,α3x=m1xm3xα1x,x∈Ω.
We observe that properties (221) hold if m2(x)≤0 and m4(x)≥0 for all x∈Ω¯. This choice implies C=1 and δ=1 in (97). Moreover, the last condition in (222) implies (137) and δ1=1.Assume now that the matrix operator(223)α1xL1x,Dxα2xL2x,Dxα3xL3x,Dxα4xL2x,Dx
is invertible in Lp(Ω)×Lp(Ω), p∈(1,+∞). For this purpose, it is enough to assume
(224)α1xα4x-α2xα3x≥C0>0,α1xα4xa1x-α2xα3xa2xα1(x)α4(x)-α2xα3x≥0,
for all x∈Ω. Further, z1, z2, η1, η2 are given functions on Ω.Relying on the proof of Theorem18 we are led to the following equation:(225)Dtm1(x)m2(x)0-m1(x)m3(x)m4(x)+m2(x)u(t,x)v(t,x)+α1(x)L1(x,Dx)α2(x)L2(x,Dx)-m1(x)m3(x)α3(x)L3(x,Dx)+α1(x)L1(x,Dx)(α2(x)-m1(x)m3(x)α4(x))L2(x,Dx)×u(t,x)v(t,x)=f(t)z1xz1x-m1xm3xz2x,∀(t,x)∈[0,τ]×Ω.Under our assumptions (221)–(224) Theorem 18 applies with α=δ=β1=1, β4=1/p>1/2, that is, p∈(1,2), if the following additional properties are satisfied when z1*,u0*∈D(L1), z2*,v0*∈D(L2), g∈C1+θ([0,τ];R), with θ∈(1/p′,1/p), η1,η2∈Lp′(Ω):
(226)z1(x)z2(x)=m1(x)z1*(x)+m2(x)z2*(x)m3(x)z1*(x)+m4(x)z2*(x),α1(x)L1(x,Dx)u0(x)+α2(x)L2(x,Dx)v0(x)α3(x)L3(x,Dx)u0(x)+α4(x)L2(x,Dx)v0(x)=m1xu0*x+m2xv0*xm3xu0*x+m4xv0*x,∫Ω‍η1xz1x+η2xz2xdx≠0,∫Ω‍η1xm1xu0x+m2xv0x+η2xm3xu0x+m4xv0xdx=g(0).
Then the identification Problems (221)–(224) admit a unique global strict solution (u,v,f) ∈ Cθ([0,τ];D(L1))×Cθ([0,τ];D(L2))×Cθ([0,τ];R) such that m1+2ju+m2+2jv∈C1+θ([0,τ];Lp(Ω)), j=0,1 and p∈(1,2).We observe that ifmj,αj,j=1,…,4, and ai,i=1,2,3, are more regular, the previous result can be extended to the case where p∈[2,+∞) and β4 is larger. For this purpose, we assume that m:=m1m4-m2m3≥0 is ρ-regular; that is,
(227)m∈C1Ω¯,∇mx≤cmxρ,x∈Ω¯,ρ∈0,1.
Since n(x)=α4(x)(m1x/m3x)-α2(x)≥c¯>0 for all x∈Ω¯, we get
(228)Dxjmxnx=nx-2nxDxjmx-mxDxjnx≤Cmxρnx+mxnxDxjnxnx≤C′mxnxρ,≤≤≤Cmxρnx+mxnxDxjnxnx≤C′x∈Ω¯.
Therefore, owing to Theorem 3.3 in [15], we can choose β4=2[p(2-ρ)]-1, p∈[2,+∞). This choice implies β4>1/2 and p∈[2,4(2-ρ)-1), so that in this case θ runs in the interval (1-2[p(2-ρ)]-1,2[p(2-ρ)]-1).In the case whenp∈(1,2) and m is ρ-regular with ρ∈(2-p,1), then, owing to Theorem 4.2 in [15], we can choose β4=[p(2-ρ)]-1 so that θ∈((1-ρ)(2-ρ)-1,(2-ρ)-1).Problem38 bis. We could handle also the case where the determinant m1(x)m4(x)-m2(x)m3(x) vanishes everywhere in Ω¯. However, in this case, Theorem 18 forces us to assume m2(x)=0 for all x∈Ω¯.In view of assumptions (222) the second equation in system (225) becomes
(229)m1xm3xα4x-α1xL2x,Dvt,x=ftm1xm3xz2x-z1x-m1xm3xα3xa3x-α1xa1xut,x.
Thus all is reduced to aregular identification problem.However, using Theorem23, in some cases we can handle the situation when m2≢0. For the sake of simplicity we choose L1=L2=L3=L and assume α4/α3=α2/α1=ν=const., so that
(230)νm1m3α3-α1=m1m3α4-α2
for some constant ν≠0. Therefore our differential equation becomes(231)Dtm1(x)m2(x)00u(t,x)v(t,x)+α1(x)L(x,Dx)α2(x)L(x,Dx)(α1(x)-m1(x)m3(x)α3(x))L(x,Dx)α2x-m1xm3xα4xL(x,Dx)u(t,x)v(t,x)=f(t)z1(x)z1(x)-m1(x)m3(x)z2(x),∀(t,x)∈[0,τ]×Ω.Referring to the notation in Theorem23, we get
(232)S=M1-M2L4-1L3=m1-ν-1m2I,L1-L2L4-1L3=α1-ν-1α2L·,D,A=(α1-ν-1α2)L(·,D)m1-ν-1m2-1.
Consequently, the linear closed operator -A generates an analytic semigroup of linear bounded operators provided α1(x)-ν-1α2(x)>0 and m1(x)-ν-1m2(x)>0 for all x∈Ω¯. In this case the resolvent of -A admits the representation:
(233)λ+A-1=(m1-ν-1m2)λ+A-1=×λm1-ν-1m2α1-ν-1α2-1+A-1λ+A-1=×α1-να2-1.
The element z¯ in Theorem 23 can be easily described.Example 39.
LetΩ=B(0,1)={x∈Rn:x<1} and let operator L(x,D)=-Δ+a(x) be endowed with either Dirichlet or Robin boundary conditions.
Consider then the following identification problem, wherer and s are real positive numbers:
(234)Dt1-1-x2r11-x2su(t,x)v(t,x)+α1(x)L(x,Dx)α2(x)L(x,Dx)α1(x)L(x,Dx)α4(x)L(x,Dx)u(t,x)v(t,x)=f(t)z1(x)z2(x),∀(t,x)∈[0,τ]×Ω,1-1-x2r11-x2su(0,x)v(0,x)=1-1-x2r11-x2su0(x)v0(x),x∈Ω,∫Ω‍η1xut,x-1-x2rvt,x+η2xut,x+1-x2svt,xdx=gt,++η2xut,x+1-x2svt,xdx0≤t≤τ.
If we suppose that
(235)α1,α2,α4∈CΩ¯∩C1Ω,α1x≥C>0,α2x-α4x<0,x∈Ω¯,
all our conditions in the last lines in Problem 37 are verified for β4=1/p, p∈(1,2). Moreover, if r,s∈(1,+∞], then m(x)=(1-x2)r+(1-x2)s belongs to C1(Ω¯) and from the inequality ∇(1-x2)r≤C[(1-x2)r](r-1)/r we deduce
(236)∇mx=∇1-x2r+1-x2s∇mx≤C1-x2r(r-1)/r+1-x2s(s-1)/s∇mx≤C1-x2r+1-x2s(r-1)/r∇mx+1-x2r+1-x2s(s-1)/s∇mx≤mx(min⁡(r,s)-1)/min⁡(r,s),x∈Ω¯.
That is, m is ρ-regular with ρ=(min⁡(r,s)-1)/min⁡(r,s), and the arguments in the treatment of Problem 38 apply as well.Example 40.
Consider the identification problem:(237)Dtmx2m(x)n(x)m(x)n(x)nx2u(t,x)v(t,x)+α1(x)L(x,Dx)α2(x)L(x,Dx)α3(x)L(x,Dx)α4(x)L(x,Dx)u(t,x)v(t,x)=f(t)z1xz2x,∀(t,x)∈0,τ×Ω,mx2m(x)n(x)m(x)n(x)nx2u(0,x)v(0,x)=mx2m(x)n(x)mxnxnx2u0(x)v0(x),∫Ω‍ηxmx2ut,x+mxnxvt,xdx=gt,ηx0x2x2ut,x+mxnxvt,x00≤t≤τ,
where m(x)≥c>0 and n(x)≥c>0 for all x∈Ω¯. Observe that the determinant of the matrix inside the time derivative vanishes identically in Ω¯.As in the proof of Theorem18 we obtain the equivalent problem:(238)Dtmx2m(x)n(x)00u(t,x)v(t,x)+α1(x)L(x,Dx)α2(x)L(x,Dx)(α1(x)-m(x)n(x)α3(x))L(x,Dx)(α2(x)-m(x)n(x)α4(x))L(x,Dx)u(t,x)v(t,x)=f(t)z1(x)z1(x)-m(x)n(x)z2(x),∀(t,x)∈[0,τ]×Ω,(239)mx2m(x)n(x)00u(0,x)v(0,x)=mx2mxnx00u0xv0x,∀x∈Ω,(240)∫Ω‍ηxmx2ut,x+mxnxvt,xdx=gt,0≤t≤τ.If(241)mα3-nα1mα4-nα2=ν=const.,
then A in Theorem 23 specifies to
(242)A=α1-να2L·,Dm-1m-νn-1.
We observe that assumptions on A are satisfied if α1(x)-να2(x)≥c>0 and m(x)-νn(x)≥c>0 for all x∈Ω¯.We compute nowz¯:
(243)z¯x=z1x-α2xα2x-mxnxα4x-1×z1x-mxnxz2x+α1x-να2x×Lx,Dmx-νnx-1n(x)Lx,D-1×α2x-mxnxα4x-1z1x-mxnxz2x.
Finally, observe that in this case we have
(244)S=mm-νnI,L1-L2L4-1L3=(α1-να2)L(·,D),L2L4-1=α2α2-mnα4I,M2L4-1=mnL·,D-1α2-mnα4.
Assume now that the data satisfy
(245)m2u0+mnv0∈DAθ+1,∞,z¯∈DAθ0,∞,00m2u0+mnv0∈DAθ+1,∞,0<θ<θ0<1,∫Ω‍ηxz¯xdx≠0,g∈C1+θ([0,τ];R),g0=∫Ω‍ηxmx2u0x+mxnxv0xdx.
Then Problem (237) admits a unique strict global solution (u,v,f)∈Cθ([0,τ];D(L))×Cθ([0,τ];D(L)) × Cθ([0,τ];R) such that m2u+mnv∈Cθ([0,τ];Lp(Ω)), p∈(1,+∞).We now face the general case where the additional information is(246)∫Ω‍η1xmx2ut,x+mxnxvt,x+η2xmxnxut,x+nx2vt,xdx=g(t),0≤t≤τ.
Taking advantage of the identity
(247)m(x)n(x)u(t,x)+nx2vt,x=nxmxmx2ut,x+mxnxvt,x,
condition (246) takes the form
(248)∫Ω‍η1x+nxmxη2x×mx2ut,x+mxnxvt,xdx=g(t),0≤t≤τ.
Moreover, solving (238) for v, we get the following equation for all (t,x)∈[0,τ]×Ω:
(249)vt,x=-νu(t,x)+f(t)Lx,Dx-1×α2x-mxnxα4x-1z1x-mxnxz2x,
so that
(250)mx2ut,x+mxnxvt,x=mx2-mxnxut,x+f(t)m(x)n(x)Lx,Dx-1α2x-mxnxα4x-1×z1x-mxnxz2x,α1(x)L(x,Dx)u(t,x)+α2xLx,Dxvt,x=α1x-να2xLx,Dxut,x+f(t)α2(x)α2x-mxnxα4x-1×z1x-mxnxz2x.
Consequently, our identification problem is reduced to the previous one with η being replaced by η1+nη2/m.Problem 41.
LetΩ be a bounded region in Rn with a smooth boundary ∂Ω. We are concerned with recovering function f in the following problem related to a degenerate parabolic weakly coupled linear system:
(251)Dtut,x=Δaxut,x+b(x)v(t,x)+ftz1x,Dtut,x=Δaxut,x+bxv00(t,x)∈0,τ×Ω,Dtvt,x=cxut,x+Δdxvt,x+ftz2x,Dtut,x=Δaxut,x+bxv0t,x∈0,τ×Ω,u0,x=u0(x),v(0,x)=v0(x),x∈Ω,axut,x=0,dxvt,x=0,Dtut,xΔax(t,x)∈(0,τ]×∂Ω,∫Ω‍η1xut,x+η2xvt,xdx=g(t),t∈[0,τ].
We assume that functions a,b,c,d∈C(Ω¯;R) and a(x)>0, d(x)>0 for a.e. x∈Ω, while z1,z2,η1,η2∈L2(Ω)=:X, u0,v0∈H01(Ω)∩H2(Ω), and g∈C([0,τ];R). Finally, we assume that the consistency condition
(252)∫Ω‍η1xu0x+η2xv0xdx=g0
is fulfilled.Recall now that if1/a∈Lr(n)(Ω), where
(253)r1≥2,r2>2,rn≥n,ifn≥3,
then [3, page 83] operator K defined by
(254)DK=u∈L2Ω:au∈H01Ω∩H2Ω,Ku=-Δau,u∈DK,
satisfies the resolvent bound:
(255)λ+K-1L(L2(Ω))≤cλ-(2r-n)/(2r)
for all λ’s in a sector containing the closed half complex plane Reλ≥0. This implies α=1 and β=(2r-n)/(2r).Suppose now that1/a∈Lr1(n)(Ω) and 1/d∈Lr4(n)(Ω), so that the corresponding differential operators A and D defined by
(256)DA=u∈L2Ω:au∈H01Ω∩H2Ω,Au=-Δau,u∈DA,DD=v∈L2Ω:dv∈H01Ω∩H2Ω,Dv=-Δdv,v∈DD,
satisfy (H1) and (H3) in Theorem 3, with B1=B4=I, α=1, β1=[2r1(n)-n]/[2r1(n)], β4=[2r4(n)-n]/[2r4(n)].LetB and C be the multiplication operators by b(x) and c(x) in L2(Ω), respectively.Theorem3 runs provided
(257)min⁡1-n2r1n,1-n2r4n>12⟺min⁡r1n,r4n>n,
when θ0=:n[2min⁡{r1(n),r4(n)}]-1.Whence it follows that ifθ∈(θ0,1-θ0), z1∈D(A), z4∈D(D), Δ(au0)+bv0∈D(L1), Δ(bv0)+cu0∈D(L4), g∈C1+θ([0,τ];R)(258)∫Ω‍η1xz1x+η2xz2xdx≠0,∫Ω‍η1xu0x+η2xv0xdx=g0,
then Problem (251) admits a unique strict global solution (u,v,f)∈Cθ([0,τ];D(L1)) × Cθ([0,τ];D(L4))×Cθ([0,τ];R) such that u,v∈C1+θ([0,τ];L2(Ω)).More generally, consider the degenerate parabolic identification problem:(259)Dtmxut,x=Δaxut,x+bxvt,x+ftz1x,=Δaxut,x+bxvt,xt,x∈0,τ×Ω,Dtnxvt,x=cxvt,x+Δdxvt,x+ftz2x,=Δaxut,x+bxv+0(t,x)∈0,τ×Ω,lim⁡t→0+mxut,x=mxu0x,lim⁡t→0+nxvt,x=nxv0x,0000000000000000000x∈Ω,a(x)u(t,x)=0,d(x)v(t,x)=0,0000000000000000(t,x)∈(0,τ]×∂Ω,∫Ω‍η1xmxut,x+η2xnxvt,xdx=gt,=Δaxut,x+bxvt,x+ftvt,xt∈0,τ,
where m,n∈C(Ω¯;R) and satisfy m(x)>0, n(x)>0 for all x∈Ω.The change of unknown functionsmu=u1 and nv=v1 leads to the following (equivalent) identification problem:
(260)Dtu1t,x=Δaxmxu1t,x+bxmxv1t,x+ftz1x,=Δaxut,x+bxvt,x+t,x∈0,τ×Ω,Dtv1t,x=cxnxu1t,xDtv1t,x=+Δdxnxv1t,x+ftz2x,Dtv1t,xt,xt,xt,x=t,x∈0,τ×Ω,u10,x=mxu0x,v10,x=nxv0x,0000000000000x∈Ω,a(x)m(x)u1(t,x)=0,d(x)n(x)v1(t,x)=0,0000000000000000(t,x)∈(0,τ]×∂Ω,∫Ω‍η1xu1t,x+η2xv1t,xdx=g(t),t,xt,xt,xt,xt,xt,xt,xtt∈[0,τ].
Suppose that a/m,d/n,b/m,c/n∈C(Ω¯;R)—so that b and c must vanish on ∂Ω if m and n do. Moreover, we assume m/a∈Lr1(n)(Ω)n/d∈Lr4(n)(Ω), r1(n) and r4(n) being defined as above and satisfy min⁡{r1(n),r4(n)}>n, as well as
(261)DL1=u∈L2Ω:amu∈H01Ω∩H2Ω,L1u=-Δamu,u∈DL1,DL4=v∈L2Ω:dnv∈H01Ω∩H2Ω,L4v=-Δdnv,v∈DL4,Δ(au0)+bv0∈D(L1),Δ(dv0)+cu0∈D(L4),g∈C1+θ([0,τ];R),θ∈(θ0,1-θ0),∫Ω‍η1xz1x+η2xz2xdx≠0,∫Ω‍η1xmxu0x+η2xnxv0xdx=g0.
Then Problem (259) admits a unique strict global solution (u,v,f)∈Cθ([0,τ];D(A))×Cθ([0,τ];D(D))×Cθ([0,τ];R) such that mu,nv∈C1+θ([0,τ];L2(Ω)).Problem 42.
Consider the following one-dimensional parabolic identification problem of Sobolev type:(262)Dtut,x+Dx2ut,x-Dx2ut,x+axDx2vt,x+bxDxvt,x+cxvt,x=ftz1x,t,x∈0,τ×Ω,Dtvt,x+Dx2vt,x-Dx2vt,x+dvt,x=f(t)z2(x),(t,x)∈(0,τ]×Ω,1+Dx2u0,x=1+Dx2u0x,1+Dx2v0,x=1+Dx2v0x,000000000000000000000000x∈Ω,ut,0=ut,π=0,Dxvt,0=Dxvt,π=0,0000000000000000000000000000000000t∈0,τ,∫Ω‍η1x1+Dx2ut,x+η2x1+Dx2vt,xdx=gt,t∈0,τ.
Here a,b,c∈C([0,π];R) and d is a positive constant.Introduce now the linear operatorsLi, i=1,2,4, and Mj, j=1,2,3,4, defined by
(263)DL1=H010,π∩H20,π,L1u=-Dx2u,000000000000000000000000000000000u∈DL1,DL2=H20,π,L2v=aDx2v+bDxv+cv,00000000000000000000000000000000v∈DL2,DL4=v∈H20,π:Dxv0=Dxvπ=0,L4v=-Dx2v+dv,000000v∈DL4,DM1+3j=DL1+3j,M1+3ju=1+Dx2u,000000000000000000u∈DL1+3j,j=0,1,M2=M3=O.
Note that the pairs (M1,L1) and (M4,L4) satisfy our spectral assumptions with α=β1=β4=1 so that we can apply Theorem 14 with δ=1 provided we make the following assumptions:
(264)zj=1+Dx2zj*,zj*∈D(L1+3j),j=0,1,-Dx2u0=1+Dx2u0*-aDx2v0-bDxv0-cv0,0000000000000000000000000000u0*∈DL1,-Dx2v0+dv0=(1+Dx2)v0*,v0*∈D(L4),g∈C1+θ0,τ;R,θ∈(0,1),η1,η2∈L2((0,π)),∫Ω‍η1xz1x+η2xz2xdx≠0,∫Ω‍η1x1+Dx2u0x+η2x1+Dx2v0xdx=g(0).
Then Problems (262)–(265) admit a unique strict global solution (u,v,f)∈Cθ([0,τ];D(L1)) × Cθ([0,τ];D(L4))×Cθ([0,τ];R) such that (1+Dx2)u,(1+Dx2)v∈C1+θ([0,τ];L2((0,π))).Problem 43.
We want to point out a different approach for solving the identification problem described in Example28. For the sake of simplicity, we confine ourselves to the problem:
(265)DtMMIIx(t)y(t)+L1OOL2xtyt=ftz1z2,00000000000000000000∀t∈0,τ,MMIIxy(0)=MMIIx0y0,Φ1Mx+yt+Φ2x+yt=gt,00000000000000000000000000∀t∈0,τ,
with a compatibility relation g(0)=Φ1[M(x0+y0)]+Φ2[x0+y0].HereM is a bounded linear operator in X, Lj:D(Lj)⊂X→X, j=1,2 are two densely defined linear operators in X such that L2, M+L1L2-1, and z(M+L1L2-1)+L1, Rez≥0, are invertible with inverses in L(X) and
(266)zM+L1L2-1+L1-1L(X)≤Cz+1-1,Rez≥0,L1 being a sectorial operator with D(L2)⊂D(L1).From the matrix equation in (265) we obtain the following equation for the new unknown u=x+y:
(267)ftz1=DtMut+L1ut-L1ytftz1=DtMut+L1ut-L1L2-1ftz2-Dtut.
Introduce now the linear functional defined by
(268)Ψw=Φ1MM+L1L2-1-1w+Φ2M+L1L2-1-1w.
Then the pair (u,f) solves the identification problem:
(269)DtM+L1L2-1ut+L1ut=ftz1+L1L2-1z2,t∈0,τ,M+L1L2-1u(0)=M+L1L2-1x0+y0,ΨM+L1L2-1ut=gt,t∈0,τ.
Finally, set
(270)ξt=M+L1L2-1u(t)⟺ut=M+L1L2-1-1ξt,0000000000000000000000000000000000000t∈0,τ.
It is immediate to check that function ξ solves thenondegenerate identification problem:
(271)Dtξt+L1M+L1L2-1-1ξ(t)=ftz1+L1L2-1z2,0000000000000000000000000000000000000t∈0,τ,ξ0=M+L1L2-1x0+y0,Ψξt=gt,t∈0,τ.
We can now apply [1, Theorem 2.1] if the following conditions are satisfied:
(272)0<θ<θ0<1,g∈C1+θ0,τ;R;sup⁡t>0tθL1tM+L1L2-1+L1-1L1(x0+y0)<+∞⟺L1(x0+y0)∈DL1M+L1L2-1-1(θ,∞);sup⁡t>0tθ0L1tM+L1L2-1+L1-1z1+L1L2-1z2<+∞⟺z1+L1L2-1z2∈DL1M+L1L2-1-1θ0,∞;Φ1Mx0+y0+Φ2x0+y0=g0;Ψz1+L1L2-1z2≠0.
Since
(273)tI+M+L1L2-1-1L1-1=tM+L1L2-1+L1-1M+L1L2-1=M+L1L2-1-1tI+L1M+L1L2-1-1×M+L1L2-1,
it follows that (M+L1L2-1)-1L1 is sectorial, too. Moreover, from
(274)tθL1M+L1L2-1-1tI+L1M+L1L2-1-1-1=tθL1tM+L1L2-1+L1-1=tθL1tI+M+L1L2-1-1L1-1M+L1L2-1-1=tθ(M+L1L2-1)M+L1L2-1-1×L1tI+M+L1L2-1-1L1-1M+L1L2-1-1,
we deduce that ξ∈DL1(M+L1L2-1)-1(θ,∞) if and only if (M+L1L2-1)-1ξ∈D(M+L1L2-1)-1L1(θ,∞).Since-(M+L1L2-1)-1L1 is sectorial and has domain D(L1), from [19] (cf. also [20]), it follows D(M+L1L2-1)-1L1(θ,∞)=DL1(θ,∞). Therefore ξ∈DL1(M+L1L2-1)-1(θ,∞) if and only if (M + L1L2-1)-1ξ∈DL1(θ,∞).Consequently, our condition on the data read equivalently(275)M+L1L2-1-1L1x0+y0∈DL1θ,∞,M+L1L2-1-1z1+L1L2-1z2∈DL1θ0,∞.
Then Problem (265) admits a unique solution (ξ,f)∈[C1+θ([0,τ];X) ∩ Cθ([0,τ];D(L1(M+L1L2-1)-1))]×Cθ([0,τ];R). Consequently, since u(t)=[M+L1L2-1]-1ξ(t) and ξ∈C1+θ0,τ;X∩Cθ([0,τ];D(M+L1L2-1]-1), then the solution (u,f) to Problem (269) has the properties (M+L1L2-1)-1u∈C1+θ([0,τ];X), (u,f)∈Cθ([0,τ];D(L1))×Cθ([0,τ];R).Finally, from the latter differential equation in (265) we easily deduce the following representation formula for the pair (x,y):
(276)yt=-L2-1ut+ftL2-1z2,xt=ut-yt=ut+L2-1ut-ftL2-1z2,00000000000000000000000000000000t∈0,τ.
We can conclude by stating that Problem (269) admits a unique solution (x,y,f)∈Cθ([0,τ];D(L1))×Cθ([0,τ];D(L2))×Cθ([0,τ];R) such that x+y∈C1+θ([0,τ];X).Remark 44.
Iff is not differentiable with respect to time, functions x and y need not to be differentiable, even though their sum is. This fact exhibits the degeneracy of our Problem (265).Remark 45.
Consider the identification problem:(277)Dt-I-IIIx(t)y(t)+LOOLx(t)y(t)=ftz1z2,000000000000000000000000000000000∀t∈0,τ,-I-IIIxy0=-I-IIIx0y0,-Φ1x+yt+Φ2x+yt=gt,000000000000000000000000000∀t∈0,τ,
with a compatibility relation g(0)=(Φ1-Φ2)[x0+y0].HereL:D(L)⊂X→X is a densely defined invertible linear operator in X such that L-1∈L(X), while x0,y0,z1,z2∈X and g∈C1([0,τ],R).First we note that our assumptions arenot satisfied, since z(M+I)+L=L for all z∈C. However, a trivial computation yields
(278)x+yt=ftL-1z1+z2,∀t∈0,τ.
Therefore, if
(279)Φ1L-1z1+z2≠Φ2L-1z1+z2,f is uniquely determined by
(280)f(t)=Φ2-Φ1L-1z1+z2-1g(t),∀t∈[0,τ].
Since
(281)Lx(t)=f(t)z1+f′tL-1z1+z2,∀t∈[0,τ],Ly(t)=f(t)z2-f′tL-1z1+z2,∀t∈0,τ,
then the assumptions on our data imply
(282)x+y∈C10,τ,X,f∈C1([0,τ],R),
without any resolvent estimate involving L, but 0∈ρ(L).Application 1. Let X=Lp(Ω), p∈(1,+∞), Ω being a domain in Rn with a boundary ∂Ω of class C2. Let ai,j∈C(Ω¯;R) with ai,j=aj,i, i,j=1,…,n, be given function satisfying the ellipticity condition:
(283)∑i,j=1n‍ai,jxξiξj≥c0ξ2,x,ξ∈Ω¯×Rn.
Define then the linear differential second-order operator by
(284)DL=W2,pΩ∩W01,pΩ,Lu=∑i,j=1n‍ai,jDxiDxj-a0u,
where a0∈C(Ω¯;R+).Introduce also two functionsm,a1∈C(Ω¯;R) such that a1(x)>0 and m(x)+a1(x)-1≥c1>0 for all x∈Ω¯ and denote by M the multiplication operator in Lp(Ω) corresponding to m.Following the same steps as in [3, pages 79-80], we can readily derive the resolvent estimate
(285)zM+a1-1I-L-1L(X)≤C1+z-1,ifRez≥0.
Consider then the identification problem:
(286)Dtm(x)m(x)11u(t,x)v(t,x)-LOOa1Lu(t,x)v(t,x)=f(t)z1(x)z2(x),∀t∈[0,τ]m(x)m(x)11u(0,x)v(0,x)=m(x)m(x)11u0(x)v0(x),∫Ω‍η1xmx+η2xut,x+vt,xdx=gt,00000000000000000000000000000000000∀t∈0,τ.
We assume that our data satisfy
(287)u0,v0∈Lp(Ω),u0+v0∈W2,p(Ω)∩W01,p(Ω),m+a1-1-1Lu0+v0∈DLθ,∞,m+a1-1-1z1+a1-1z2∈DL(θ0,∞),η1+η2∈Lp′Ω,1p+1p′=1,g∈C1+θ0,τ;R,g0=∫Ω‍η1xmx+η2xu0x+v0xdx.∫Ω‍η1xmx+η2xm(x)+a1x-1-1×z1x+a1x-1z2xdx≠0.
We note that the spaces DL(θ,∞) are well characterised in [9, page 321].By virtue of the previous result we deduce that our degenerate parabolic identification problem admits a unique global strict solution(u,v,f)∈Cθ([0,τ];D(L))×Cθ([0,τ];D(L))×Cθ([0,τ];R) such that u+v∈C1+θ([0,τ];Lp(Ω)) and m(u+v)∈C1+θ([0,τ];Lp(Ω)).Application 2. According to [21] the projection operator P onto N(T) relative to the direct sum representation N(T)⊕R(T)¯ is well characterised. Now we show how the corresponding projection in product space X×X can be characterised as well.For the sake of simplicity we confine ourselves to a bit less general case than the one discussed in Proposition16, choosing here M2=O. Hence the corresponding differential system becomes
(288)DtM1OOM4u(t)v(t)+L1L2OL4u(t)v(t)=ftz1z2,00000000000000000000000000000000000∀t∈0,τ.
Suppose that L1 and L4 admit bounded inverses satisfying
(289)D(L2)⊂D(L4),D(Lj)⊂D(Mj),j=1,4,zMizMi+Li-1D(X)≤c,ifRez≥0,j=1,4.
From our assumption we deduce the following formula:
(290)L1L2OL4-1=L1-1-L1-1L2L4-1OL4-1.
Setting Ti=MiLi-1,i=1,4, we get
(291)T=M1OOM4L1-1-L1-1L2L4-1OL4-1=T1-T1L2L4-1OT4.
Hence(292)zT+I-1=zT1+I-1L2L4-1zT4+I-1-zT1+I-1L2L4-1zT4+I-1OzT4+I-1.Whence it follows(293)zT+I-1L(X×X)≤C,ifRez≥0.
Note now that if X is areflexive Banach space, we have the decomposition:
(294)X×X=NT⊕RT¯.
Observe then that
(295)x,y∈NT⟺T4y=0,T1x-L2L4-1y=0.
Introduce finally the linear bounded operator:
(296)P=P1-P1L2L4-1+L2L4-1P2OP2,
where P1 and P2 are the projections on N(T1) and N(T4), respectively.It is easily checked thatP2=P and
(297)Pxy=P1x-P1L2L4-1y+L2L4-1P2yP2y∈NT,
since T4P2y=0 and
(298)T1P1x-P1L2L4-1y+L2L4-1P2y-T1L2L4-1P2y=T1L2L4-1P2y-T1L2L4-1P2y=0.
Once we have characterised P, we obtain the basic formula R(T)¯=(I-P)(X×X). Consequently, we have at our disposal all the elements required in Proposition 16.As an application, we can handle the identification problem:(299)Dtm1(x)00m4(x)u(t,x)v(t,x)-L1(x,D)L2(x,D)OL4(x,D)u(t,x)v(t,x)=f(t)z1(x)z2(x),∀t∈[0,τ],m1(x)00m4(x)u(0,x)v(0,x)=m1(x)00m4(x)u0(x)v0(x),m1u,η1H-1Ω×H01Ω+m2v,η2H-1Ω×H01Ω=g(t),∀t∈0,τ.
Assume now that m1,m2 are nonnegative functions in L∞(Ω), z1,z2∈H-1(Ω), u0,v0,η1,η2∈H01(Ω), while L1(x,D), L4(x,D), and L2(x,D) are second-order linear operators such that the first two are isomorphisms from H01(Ω) into H-1(Ω), while the third one is bounded from H01(Ω) into H-1(Ω).Under such conditions, the conditions in Proposition16 ensuring the solvability of the identification Problem (299) are fulfilled. Such conditions, related to the projection P, are expressed by means of the projections P1 and P2.Application 3. We generalize here to a system Problem 9 related to the scalar Ornstein-Uhlenbeck equation and use the same notation.Letm~ be a uniformly continuous and bounded function on RN such that
(300)m~x≥c0>0,x∈RN.
Consider first, for all λ∈C+={z∈C:Rez≥0} and f∈Lμ2(RN), the resolvent equation:
(301)λm~u+-L2+Iu=f,
where
(302)BQ=QB*,B=-Q∞-1.
Then -L2 is self-adjoint in Lμ2(RN) and [19] (cf. [20, Section 9.3.2, page 251]):
(303)∫Rn‍g·L2fdμ=-∫RN‍QDf,Dgdμ.
Thus, taking the real and imaginary parts in the equality:
(304)λ∫RN‍m~u2dμ+∫RN‍QDu,Du¯dμ+∫RN‍u2dμ=∫RN‍fu¯dμ,
we get the relations as follows:
(305)Reλ∫RN‍m~u2dμ+∫RN‍〈QDu,Du¯〉dμ+∫RN‍u2dμ=Re∫RN‍fu¯dμ,Im⁡λ∫RN‍m~u2dμ=Im⁡∫RN‍fu¯dμ.
They, in turn, imply
(306)Reλ+Im⁡λ∫RN‍m~u2dμ+∫RN‍QDu,Du¯dμ+∫RN‍u2dμ≤2fLμ2RNuLμ2(RN).
Since Reλ+|Im⁡λ|≥0, we deduce
(307)∫RN‍QDu,Du¯dμ+∫RN‍u2dμ≤2fLμ2RNuLμ2(RN).
implying
(308)c0λ∫RN‍u2dμ≤λ∫RN‍m~u2dμ≤Reλ+Im⁡λ∫RN‍m~u2dμ≤4fLμ2RN2=4λm~u+-L2+IuLμ2(RN)2,λ∈C+.
Since [λm~I+(-L2+I)]*=λ¯m~I+(-L2+I), we can conclude that λm~I+(-L2+I) has a bounded inverse and
(309)λm~I+-L2+I-1L(Lμ2(RN))≤4c0-1λ-1,λ∈C+.
Let us now consider the following degenerate identification problem related to amatrix-valued Ornstein-Uhlenbeck operator:look for a triplet of functionsu,v:RN→Randf:[0,T]→Rsuch that(310)Dtm(x)m(x)IIu(t,x)v(t,x)-L2-IOOa1(x)(L2-I)u(t,x)v(t,x)=f(t)z1(x)z2(x),∀t∈[0,τ],x∈RN,m(x)m(x)IIu(0,x)v(0,x)=m(x)m(x)IIu0(x)v0(x),x∈RN,∫RN‍η1xmx+η2ut,x+vt,xdμ,=gt,∀t∈0,τ,
where
(311)u0,v0,η1,η2∈Lμ2RN.
We now assume that functions m and a1 are uniformly continuous, bounded, and strictly positive in RN. Moreover, we assume that function m+(1/a1) has the same properties and satisfies m(x)+(1/a1(x))≥c0>0 for all x∈RN. Then condition (309) is satisfied with m~=m+(1/a1). Further, the interpolation space is characterised by
(312)Lμ2RN,Wμ2,2RNθ,∞=B2,∞2θRN;μ,θ∈0,1.
Assume now that our data enjoy the following properties:
(313)u0,v0∈Lμ2RN,u0+v0∈Wμ2,2RN,[a1(ma1+1)](L2-I)(u0+v0)∈B2,∞2θ(RN;μ),z1,z2∈Lμ2RN,a1z1+z2ma1+1∈B2,∞2θ0RN;μ,z1,z2∈Lμ2RNz1,z2∈Lμ2RN,000,θ0∈θ,1,g∈C1+θ0,τ;R,g0=∫RN‍η1xmx+η2xu0x+v0xdμ.∫RN‍η1xmx+η2x×1mxa1x+1a1z1x+z2xdx≠0.
Consequently, our degenerate parabolic identification problem possesses a unique global strict solution (u,v,f)∈Cθ([0,τ];Wμ2,2(RN))×Cθ([0,τ];Wμ2,2(RN))×Cθ([0,τ];R) such that u+v, m(u+v) ∈ C1+θ([0,τ];Lμ2(RN)).Application 4. We are here concerned with hyperbolic systems.LetAj=(ah,kj)h,k=1N, j=1,…,n, N,n∈R∖{0}, be a N×N Hermitian matrix, that is, ah,kj=ak,hj, and let ah,kj∈Cb1(Rn;C), the space of all bounded functions along with all first-order derivatives. Let B=(bh,k)h,k=1N be a N×N matrix with bh,k∈Cb1(Rn;C).SetX=L2(Rn)N and for all u,v∈L2(Rn)N define the inner product:
(314)u,v=∑j=1N‍uj,vjL2(Rn)=∑j=1N‍∫Rn‍ujxv¯jxdx,
where u=(u1,…,uN) and v=(v1,…,vN), so that u2=∑j=1N‍ujL2(Rn)N2.IfT∈S'(Rn), DxjT is the temperate distribution defined by 〈DxjT,φ〉=-〈T,Dxjφ〉 for all φ∈S(Rn). If f∈L2(Rn) and a∈C1(Rn;C) one defines
(315)aDxjf=Dxjaf-fDxja.
Of course, f∈L2(Rn) is identified with the temperate distribution 〈f,φ〉=∫Rn‍f(x)φ(x)dx,φ∈S(Rn).Therefore the components of∑j=1n‍Aj(x)Dxju+B(x)u are given by
(316)∑j=1n‍∑l=1N‍ak,ljxDxjul+∑l=1N‍bk,lxulx,k=1,…,N,
where Dxjul is meant in the sense of distributions.We can now introduce the linear first-order differential operatorA defined by
(317)Au=∑j=1n‍AjxDxju+Bxu∈S′Rn,u∈X,(318)DA=u∈X:Au∈X,Au=Au,u∈DA.
It is well known (cf. [22]) that A generates a C0-semigroup in X. Consequently, our abstract identification problem reduces to the following:find a pair(u,f)such that(319)Dtuk=∑j=1n‍∑l=1N‍ak,ljxDxjul+∑l=1N‍bk,lxulx+ftzk,Dtukak,ljxDxjulbk,lt,x∈0,τ×Rn,k=1,…,N,(320)uk(0,x)=u0,k(x),x∈Rn,k=1,…,N,(321)η,ut,·=gt,t∈[0,τ],
where z=(z1,…,zN), u0=(u0,1,…,u0,N),η∈X and g∈C1([0,τ];R) are given.Theorem13 allows to conclude that if z,u0∈D(A) (in particular they belong to H1(Rn), 〈〈η,u0〉〉=g(0) and 〈〈η,z〉〉≠0) the problems (319)–(321) admit a unique solution (u,f)∈[C1([0,τ];X) ∩ C([0,τ];D(A))]×C([0,τ];R).As an example, consider the following identification problem related to Maxwell equations:(322)DtE=-crot⁡H+f(t)z,(t,x)∈[0,τ]×R3,(323)DtH=-crot⁡E,(t,x)∈[0,τ]×R3,(324)E0,x=E0x,H0,x=H0x,x∈R3,(325)η1,Et,·+η2,Ht,·=g(t),t∈0,τ,
where c∈R+,z,E0,H0,η1,η2∈L2(R3)3, div⁡E0=0, div⁡H0=0 and div⁡z=0 in R3, and g∈C1([0,τ];R) are given.We note thatdiv⁡E0=0, div⁡H0=0, and div⁡z=0 in R3 imply div⁡E(t,·)=0 and div⁡H(t,·)=0 in R3, via (322) and (323).We notice that our assumption concerningz and E0 ensure thatno electric charges occur in R3.Introduce now the vectoru=(E,H)∈L2(R3)6. Recalling that
(326)rot⁡K=Dx2K3-Dx3K2,Dx3K1-Dx1K3,Dx1K2-Dx2K1,
we observe that (322) and (323) can be rewritten in the unified form:
(327)Dtu=A1Dx1u+A2Dx2u+A3Dx3u+ftz0,Dtu=A1Dx1u+A2Dx2A3Dx3t,x∈0,τ×R3,
where all the elements in Aj=(ah,kj)h,k=16 vanish but the following:
(328)a2,61=a6,21=c,a3,51=a5,31=-c,a3,42=a4,32=c,a1,62=a6,12=-c,a1,53=a5,13=c,a2,43=a4,23=-c.Define now(329)X=u∈L2R36:div⁡u1,u2,u3=div⁡u4,u5,u6=0inR3,Y=u∈L2R36:rot⁡u1,u2,u300000=rot⁡u4,u5,u6=0inR3.We observe that it is easy to check thatX is a closed subspace in L2(R3)6 orthogonal to Y.We observe now that, according to our assumptions, the initial value(330)u0=E0,H0
belongs to X, while the additional condition can be expressed as
(331)η,ut,·=gt,t∈0,τ,
where η=(η1,η2)∈L2(R3)6.Since operatorA is defined by (318), with N=6, it generates a C0-semigroup in X. Moreover, D(A)=L2(R3;rot⁡)×L2(R3;rot⁡), where L2R3;rot⁡={k∈L2R33:rot⁡k∈L2R33}. Consequently, Corollary 2.1 in [8] applies provided z,η1,η2∈L2(R3;rot⁡), 〈〈η,u0〉〉=g(0), and 〈〈η1,z〉〉≠0. Then the identification Problems (322)–(325) admit a unique solution (E,H)∈[C1([0,τ];X) ∩ C([0,τ];L2(R3;rot⁡)×L2(R3;rot⁡))]×C([0,τ];R).Remark 46.
Likewise we could treat identification problems for the Dirac equation in nonrelativistic Quantum Mechanics (cf. [22, page 54]), making use of the function space L2(R3)4.Application 5. Let A be the linear operator defined in X=L2(Rn), n∈N∖{0} by
(332)DA=H2Rn,Aux=iΔux-Vxux,Dt00000u=A1Dx1u+A2Dx2u+A3Dx3u∈DA,
where the potential V belongs to Lp(Rn;R) for some p>n/2 and p≥2. Then it is well known (cf. [23]) that A generates a group of unitary operators on L2(Rn). Therefore, Proposition 35 yields the following result.Corollary 47.
Letz,u0∈H2(Rn), η∈L2(Rn), 〈η,z〉≠0, g∈C1([0,τ];C), 〈η,u0〉=g(0), V∈Lp(Rn;R) for some p>n/2 and p≥2. Then the identification problem
(333)Dtut,x=iΔut,x-Vxut,x+ftzx,Dtu=A1Dx1u+A2Dx2uDx3t,x∈0,τ×Rn,u0,x=u0x,x∈Rn,∫Rn‍ηxut,xdx=g(t),t∈0,τ,
admits a unique solution (u,f)∈[C1([0,τ];L2(Rn))∩C([0,τ];H2(Rn))]×C([0,τ];R).Application 6. In a very recent paper [24] Taira exhibits a functional approach to the existence of Markov processes endowed with both Dirichlet and oblique derivative and the so-calledfirst-order Wentzell boundary conditions for second-order uniformly elliptic differential equations with discontinuous coefficients. More exactly, the related differential operator is assumed to be of the form
(334)Au=∑i,jN‍ai,jDxiDxju+∑i,jN‍biDxiu+cxu
on a bounded domain Ω⊂RN, N≥3, with boundary ∂Ω of class C1,1.The discontinuous real-valued coefficients are assumed to satisfy the following properties:(i)
ai,j∈VMO(Ω)∩L∞(Ω), ai,j(x)=aj,i(x) for a.e. x∈Ω and λ-1|ξ|2≤∑i,jN‍ai,j(x)ξiξj≤λ|ξ|2 for a.e. x∈Ω, all ξ∈RN and some λ∈[1,+∞);(ii)
bi∈L∞(Ω);(iii)
c∈L∞(Ω) and c(x)≤0 for a.e. x∈Ω.The differential operatorA is a diffusion operator describing a strong Markov process with continuous paths in Ω such as Brownian motions.Consider also a boundary operator of the form(335)Lux=μ(x)Dnu(x)+β(x)·∇τu(x)+γ(x)u(x)-δ(x)Au∂Ω,Dtu=A1Dx1u+A2Dx2x∈∂Ω,
where ∇τ denotes the tangential component of the gradient along ∂Ω.The following assumptions on the coefficients ofL are made:(iv)
μ is Lipschitz continuous on ∂Ω and μ(x)≥0 for all x∈∂Ω;(v)
β is a Lipschitz continuous vector field on ∂Ω;(vi)
γ is a Lipschitz continuous on ∂Ω and γ(x)≤1 for all x∈∂Ω;(vii)
δ is a Lipschitz continuous on ∂Ω and δ(x)≥0 for all x∈∂Ω;(viii)
n(x)=(n1(x),…,nN(x)) is the unit interior normal at x to ∂Ω.The termsμDnu, β·∇τu, γu, δ(Au)∂Ω correspond, respectively, to the following phenomena: reflection, drift along the boundary, absorption, and sticking (or viscosity).LetC(Ω¯) be the Banach space of real-valued functions continuous on Ω¯ endowed with the sup-norm.Associated with the formal differential operatorA we introduce the realization A defined by
(336)DA=u∈CΩ¯:Au∈CΩ¯,Lu=0on⁡∂Ω.
Functions in D(A) are said to satisfyWentzell conditions of the first order.Finally, recall that a strongly continuous semigroupTtt≥0 on C(Ω¯) is called a Feller semigroup if it satisfies
(337)0000000000000f∈CΩ¯,0≤fx≤1,x∈Ω¯implies0≤Ttfx≤1,x∈Ω¯.
This implies that Ttt≥0 is nonnegative and contracting on C(Ω¯).Assume now thatp∈(N,+∞) and A:D(A)⊂C(Ω¯)→C(Ω¯) is defined by
(338)DA=u∈W2,pΩ:Au∈CΩ¯,Lu=0on⁡∂Ω,Au=Au,u∈DA.
Then Theorem 1.1 in [24, page 717] establishes that A generates a Feller semigroup on C(Ω¯). Therefore, Theorem 13 applies. Recall now that the dual C(Ω¯)* consists of real measure ν with finite variation coinciding with μC(Ω¯).Finally, consider the identification problem:(339)Dtu(t,x)=Au(t,x)+f(t)z(x),(t,x)∈[0,τ]×Ω,u0,x=u0x,x∈Rn,Lut,x=0,(t,x)∈[0,τ]×∂Ω,∫Ω‍ut,xνdx=gt,t∈0,τ.
If z,u0∈D(A), g∈C1([0,τ];R), ∫Ω‍u0(x)ν(dx)=g(0), ∫Ω‍z(x)ν(dx)≠0, then the previous Problems admit a unique solution (u,f)∈[C1([0,τ];C(Ω¯))∩C([0,τ];D(A))]×C([0,τ];R).Application 7. We are here again concerned with the Ornstein-Uhlenbeck operator, but in Lp(Rn). For this purpose denote by L~p, p∈(1,+∞), the realization of the the Ornstein-Uhlenbeck operator A. Consider now the identification problem:
(340)Dtut,x=L~put,x+ftzx,(t,x)∈[0,τ]×Rn,u0,x=u0x,x∈Rn,∫Ω‍ut,xdx=gt,t∈[0,τ],In view of [11, Proposition 9.4.2, page 280] operator L~ defined by
(341)DL~p=u∈W2,pΩ:Au∈LpRn,L~pu=Au,u∈DL~p,
generates a strongly continuous semigroup {Tp(t)}t≥0, which isnot analytic.In this case, we can apply Theorem13. Indeed, if z,u0∈D(L~p), η∈Lp′(Rn), (1/p)+(1/p′)=1, g∈C1([0,τ];R), ∫Ω‍η(x)u0(x)dx=g(0), ∫Ω‍η(x)z(x)dx≠0, then the previous Problems admit a unique solution (u,f)∈[C1([0,τ];Lp(Rn))∩C([0,τ];D(L~p))]×C([0,τ];R).



---

*Source: 393624-2015-01-28.xml*