Abstract Elliptic Equations with Integral Boundary Conditons

1 Introduction

Boundary value problems(for short BVPs)for differential-operator equations(for short DOEs)have been studied extensively by many researchers(see[1–3,8,10–11,14–23,25–32]and the references therein).A comprehensive introduction to the DOEs and historical references may be found in[16,28].The maximal regularity properties for differential-operator equations have been investigated,e.g.,in[1,8,10–11,17–22,26,28,30–32].

In recent years,integral boundary conditions(for short IBC)for evolution problems have found many applications in various disciplines such as chemical engineering,thermoplasticity,underground water flow and population dynamics(see[4,7,9,13,24]and the references therein).

The main purpose aim of the present paper,is to show the separability properties of the integral boundary value problem(for short IBVP)for the following DOE:

and maximal regularity of Cauchy problem for the following abstract parabolic equation:

with integral boundary conditions,where A is a linear operator in the Banach space E and λ is a complex parameter.

Unlike the previous results,the boundary conditions here contain nonlocal integral terms.The maximal Lp-regularity and the Fredholmness are obtained.Moreover,it is proven that the corresponding elliptic operator is R-positive and is a generator of an analytic semigroup.These results are applied to nonlocal BVPs for partial differential equations and it’s finite or in finite systems on cylindrical domains.

We shall prove the separability of the problem(1.1),i.e.,we show that for each f∈Lp(0,1;E),there exists a unique strong solution u of the problem(1.1)and the following uniform coercive estimate holds:

Let E be a Banach space and Lp(Ω;E)denotes the space of strongly measurable E-valued functions that are defined on the measurable subset Ω⊂Rnwith the norm

The Banach space E is called a UMD-space if the Hilbert operator

is bounded in Lp(R,E),p ∈ (1,∞)(see[6]),where R=(−∞,∞).UMD spaces include,e.g.,Lp,lpspaces and Lorentz spaces Lpq,p,q∈(1,∞).

Let C be the set of the complex numbers and

A linear operator A is said to be ϕ-positive in a Banach space E with bound M>0 if D(A)is dense on E and

for any λ ∈ Sϕ,0 ≤ ϕ < π,where I is the identity operator in E,and B(E)is the space of bounded linear operators in E.Sometimes A+λI will be written as A+λ and denoted by Aλ.It is known in[27,§1.15.1]that there exist the fractional powers Aθof a positive operator A.Let E(Aθ)denote the space D(Aθ)with norm

Let E1and E2be two Banach spaces.By(E1,E2)θ,p,0< θ<1,1≤ p≤ ∞,we denote the interpolation spaces obtained from{E1,E2}by the K-method(see[27,§1.3.2]).

Let S(Rn;E)denote the Schwartz class,i.e.,the space of all E-valued rapidly decreasing smooth functions on Rn.Let F denote the Fourier transformation.A function Ψ∈C(Rn;B(E1,E2))is called a Fourier multiplier from Lp(Rn;E1)to Lp(Rn;E2)if the map u → Λu=F−1Ψ(ξ)Fu,u ∈ S(Rn;E1)is well defined and extends to a bounded linear operator

The set of all multipliers from Lp(Rn;E1)to Lp(Rn;E2)will be denoted by(E1,E2).For E1=E2=E,it will be denoted by(E).The most important facts about Fourier multipliers and some related references,can be found in[11–12,27,30].

Let Φh={Ψh∈(E1,E2),h ∈ Q}be a collection of multipliers.We say that Whis a uniform collection of multipliers if there exists a positive constant M independent of h∈Q such that

for all h∈Q and u∈S(Rn;E1).

Let N denote the set of natural numbers.A set G⊂B(E1,E2)is called R-bounded(see[10])if there is a positive constant C such that for all T1,T2,···,Tm∈ G and u1,u2,···,um∈ E1,m∈N,

where{rj}is a sequence of independent symmetric{−1,1}-valued random variables on Ω (see[11]).The smallest C for which the above estimate holds is called an R-bound of the collection G and denoted by R(G).

A set Gh⊂B(E1,E2)depending on parameter h∈Q is called uniformly R-bounded with respect to h if there exists a constant C,independent of h∈Q,such that for all T1(h),T2(h),···,Tm(h)∈ Ghand u1,u2,···,um∈ E1,m ∈ N,

It implies thatLet

Definition 1.1A Banach spaceEis said to be a space satisfying a multiplier condition if,for anyΨ∈C(n)(Rn;B(E)),theR-boundedness of theimplies thatΨis a Fourier multiplier,i.e.,for anyp∈(1,∞).

The uniform R-boundedness of the

implies that Ψhis a uniform collection of Fourier multipliers.

Remark 1.1Definition 1.1 is a restriction on the Banach spaces E.If E is a UMD space,then this space satisfies the multiplier condition.All UMD spaces satisfy the multiplier condition(see[12]).

The ϕ-positive operator A is said to be R-positive in a Banach space E if the set

is R-bounded.

Note that in Hilbert spaces,all norm-bounded sets are R-bounded.Therefore,in Hilbert spaces,all positive operators are R-positive.If A is the generator of a contraction semigroup on Lq,1≤ q ≤ ∞ or A has the bounded imaginary powers with(see[11])in E∈UMD,then those operators are R-positive.

Let D(Ω;E)denote the class of all E-valued in finite differentiable functions on domain Ω with compact supports.For E=C,denotes it by D(Ω).

Let σ∞(E)denote the space of all compact operators in E.Let E0and E be two Banach spaces and E0be continuously and densely embedded into E.Let m be a positive integer.Let us consider the spaceconsisting of all functions u ∈ Lp(Ω;E0)that have the generalized derivativeswith the norm

For Ω =(a,b),a,b ∈ (−∞,∞),the spacewill be denoted byand for E0=E,denotes it by

By using the techniques of[12,Theorem 3.7]we obtain the following proposition.

Proposition 1.1LetE1andE2be two UMD spaces and

Suppose that there exists a positive constantKsuch that

ThenΨhis a uniform collection of multipliers fromLp(Rn;E1)toLp(Rn;E2)forp ∈ (1,∞).

From[21,Theorem 4]we obtain the following theorem.

Theorem 1.1Let the following conditions be satisfied:

(1)Eis a Banach space satisfying the uniform multiplier conditionp∈(1,∞)and0

(2)mis a positive integer andα =(α1,α2,···,αn)aren-tuples of nonnegative integer numbers such that

(3)Ais anR-positive operator inEwith0≤ϕ<π;

(4)Ω∈Rnis a region such that there exists a bounded linear extension operator betweenfromand fromLp(Ω;E)toLp(Rn;E).

Then the embeddingis continuous and there ex-ists a positive constantCμsuch thatfor alland0

Remark 1.2If Ω ⊂ Rnis a region satisfying m-horn condition(see[5,§7]),E=R,A=I,then for p∈ (1,∞)there exists a bounded linear extension operator from

From[21,Theorem 6]we obtain the following theorem.

Theorem 1.2Suppose that all conditions of Theorem1.1are satisfied.Let0<μ≤1−κ.Then the embedding

is continuous and there exists a positive constantCμsuch that for alltheuniform estimate holds,i.e.,

Remark 1.3Note that the constant Cμin the above estimate depends only on h0,μ and T.Since these numbers are bounded and fixed,this dependence does not in fluence the further results.

Theorem 1.3(see[22])LetEbe a Banach space andAbe aϕ-positive operator inEwith boundM,0≤ ϕ < π.Letmbe a positive integer,Then forn operatorgenerates a semigroupwhichis holomorphic forx>0.Moreover,there exists a positive constantC(depending only onM,ϕ,m,αandp)such that for everyandλ ∈Sϕ,

From[28,§1.7.7,Theorem 2]we obtain the following theorem.

Theorem 1.4Letmandjbe integer numbers,Then,forthe transformationsare bounded linearly fromand the following inequality holds:

By using the integral representation formula,in a similar way as in[5,§10.1]we have the following theorem.

Theorem 1.5Letmandjbe integer numbers,Then,forthe transformationbounded linearly fromontoEand the following inequality holds:

Let A be a positive operator in a Banach space E.Consider the differential-operator equation

Let ω1,ω2,···,ωmbe the roots of the equation

and

where q is some integer number from(1,m).

A system of complex numbers ω1,ω2,···,ωmis called q-separated if there exists a straight line P passing through 0 such that no value of the numbers ωjlies on it,and ω1,ω2,···,ωqare on one side of P while ωq+1,···,ωmare on the other side.

Lemma 1.1(see[20])Let the following conditions be satisfied:

(1)andωj,j=1,···,m,areq-separated;

(2)Eis a Banach space satisfying the multiplier condition forp∈(1,∞);

(3)Ais anR-positive operator inE.

Then for a functionu(x)to be a solution of Equation(1.4),which belongs to the spaceE(A),E),it is necessary thatwhere

2 Statement of the Problem

In a Banach space E,consider the integral boundary value problem

where f ∈ Lp(0,1;E),fk∈ Ek=(E(A),E)αk,p,p∈ (1,∞);A and Bkare linear operators in E and λ is a complex parameter.

3 Homogeneous Equations

Let us first consider the following problem:

where λ is a complex parameter,A and Bkare linear operators in E,Ek=(E(A),E)αk,p.

Remark 3.1Let A be a positive operator in E.By definition,the operator Aλ=A+λ is ϕ1-positive in E for|argλ| ≤ ϕ and ϕ + ϕ1< π.From the beginning of the proof of[28,Lemma 5.4.2/4],for|argλ|≤ ϕ and|argμ|≤ ϕ1,ϕ,ϕ1∈ (0,π),we have the estimate with M0depending on ϕ only.Then in view of[8,Lemma 2.6],there exist semigroups U1λ(x)=which are holomorphic for x>0 and strongly continuous for x≥0.

Let

Condition 3.1Assume that

(1)E is a Banach space satisfying the multiplier condition and η0;

(2)A is an R-positive operator in E;

(3)D(Bk) ⊂ (E(A),E)κ+ε,pforand

(4)the operators u→Lku are bounded from(0,1;E(A),E)into Ek.

Theorem 3.1Suppose that Condition3.1is satisfied.Then,the problem(3.1)–(3.2)forfk∈ Ek,p ∈ (1,∞)andλ ∈ Sϕwith sufficiently large|λ|has a unique solutionu ∈(0,1;E(A),E)and the following coercive uniform estimate holds:

ProofBy virtue of Lemma 1.1 and Remark 1.3,an arbitrary solution of Equation(3.2)belonging to the space(0,1;E(A),E),has the form

where

Now taking into account boundary conditions(3.3)we obtain the algebraic linear equations with respect to g1,g2:

Since the operator determinant η0,the system(3.5)has the solutionwhere

Hence,the problem(3.1)–(3.2)has a solution given bellow

By virtue of[8,Lemma 2.6],we have

In view of(3.6),by estimate(3.7),the condition 3.1(2)and in view of the positivity of A,we obtain

Hence,from Theorem 1.3 we obtain the assertion

4 Non-homogenous Equations

Now consider IBVPs for non-homogenous equation

Theorem 4.1Suppose that Condition3.1is satisfied.Then the operatoru→{(L+λ)u,L1u,L2u}for|argλ|≤ ϕ,0 ≤ ϕ < πand sufficiently large|λ|,is an isomorphism from(0,1;E(A),E)ontoLp(0,1;E)× E1× E2.Moreover,for theseλ,the following uniform coercive estimate holds:

ProofWe have proved the uniqueness of solution of the problem(4.1)–(4.2)in Theorem 3.1.By definition of(0,1;E(A),E)and by(4)of Condition 3.1,it is easy to see that the operator u→{(L+λ)u,L1u,L2u}is bounded from(0,1;E(A),E)into Lp(0,1;E)×E1×E2.Hence,by the Banach theorem,it is sufficient to show that the operator u→{(L+λ)u,L1u,L2u}is surjective from(0,1;E(A),E)into Lp(0,1;E)×E1×E2.Let us define

We now show that the problem(4.1)–(4.2)has a solution u ∈(0,1;E(A),E)for all f∈Lp(0,1;E),fk∈Ekand u=u1+u2,where u1is the restriction on[0,1]of the solution of the equation

and u2is a solution of the problem

By virtue of Theorem 3.1,the problem(4.4)has a unique solution

A solution of Equation(4.4)is given by the formula

where L(λ,ξ)=A+ ξ2+ λ.It follows from the above expression that

Let us show that operator-functions

are Fourier multipliers in Lp(R;E)uniformly with respect to λ.In fact,due to positivity of A and by virtue of[8,Lemma 2.3]we have the following uniform estimates:

It is clear to see that

Due to R-positivity of the operator A,the sets

are R-bounded.Then in view of the Kahane’s contraction principle and from the product properties of the collection of R-bounded operators(see[11,Lemma 3.5,Proposition 3.4])we obtain

That is to say,the R-bound of the set

is independent of λ.Next,let us consider σλ(ξ).It is clear to see that

Then by using the well-known inequality yj≤ C(1+ym),y ≥ 0,j ≤ m for y=(|λ|−1l|ξ|)jand m=2,we get the uniform estimate

From(4.7)–(4.8)we have

Due to the R-positivity of the operator A,the set

is R-bounded.Then from the equality

and by Kahane’s contraction principle we obtain the R-boundedness of set

Now by differentiating σλ(ξ)we have

Consider the set

Due to the R-positivity of the operator A,the sets

are R-bounded.Then from the above formula in view of estimate(4.7),by virtue of the Kahane’s contraction principle and from the product and additional properties of the collection of R-bounded operators,for all ξ1,ξ2,···,ξm∈ R, σk(ξ1,λ), σ(ξ2,λ),···,σ(ξm,λ),u1,u2,···,um∈ E and independent symmetric{−1,1}-valued random variables rj(y),j=1,2,···,m,m ∈ N,we obtain the uniform estimate

i.e.,

Then in view of Definition 1.1,it follows that Ψλ(ξ)and σλ(ξ)are the uniform collection of multipliers in Lp(R;E).Then,by using the equality(4.6)we obtain that the problem(4.4)has a solution u∈(R;E(A),E)and the uniform estimate holds,i.e.,

Let u1be the restriction of u on(0,1).Then the estimate(4.9)implies that

By virtue of Theorem 1.5,we get

Hence,Lku1∈Ek.Thus by virtue of Theorem 4.1,the problem(4.9)has a unique solution u2(x)that belongs to the space(0,1;E(A),E)and for sufficiently large|λ|we have

Moreover,from(4.10),for|argλ|≤ ϕ we obtain

By Theorem 4.1,we have

Finally,from(4.11)–(4.12)we obtain the estimate(4.3).

Consider the problem

Let B denote the operator in F=Lp(0,1;E)generated by the problem(4.13)–(4.14),i.e.,

Theorem 4.1 implies the following corollary.

Corollary 4.1Suppose that Condition 3.1 is satisfied.Then for sufficiently large k>0,there exist positive constants C1and C2so that

for u∈W(0,1,E(A),E).

Theorem 4.2Suppose that Condition3.1is satisfied.Then the operatorBis uniformlyR-positive inLp(0,1;E).

ProofThe estimate(4.3)implies that,for λ ∈ Sϕand enough large|λ|,B+ λ is invertible and the operator B is positive in Lp(0,1;E).By using a similar technique as in[28,Lemma 5.3.2/1],we obtain that,for f∈D(0,1;E(A)),the solution of the Equation(4.13)is represented as

where

andare analytic semigroups generated by operatorBy taking into account the boundary conditions(4.14),we obtain the following equation with respect to g1and g2:

By solving the above system and substituting it into(4.15),in a similar way as in Theorem 3.1,we obtain the representation of the solution for the problem(4.13)–(4.14):

where Ckjandare the same as in(3.6).By calculating Lk(Φλ)we obtain from the above

where Bkj(λ)are like.So are the uniformly bounded operators in E and

Let us first show that the set{G(λ,x,y);λ ∈ S(ϕ)}is uniformly R-bounded.Really,by using the generalized Minkowcki’s Young inequalities,by semigroup estimates(3.7),have the uniform estimate

Due to R-positivity of A,in view of properties of holomorphic semigroups Ujλ(x)and the uniform boundedness of operators Bkj(λ)and by using the Kahane’s contraction principle,we get that the sets

are uniformly R-bounded.Then by using the Kahane’s contraction principle,product and additional properties of the collection of R-bounded operators and the R-boundedness of the sets bkj,d0for all u1,u2,···,uμ∈ F, λ1,λ2,···,λμ∈ S(ϕ),for independent symmetric{−1,1}-valued random variables ri(y),i=1,2,···,μ, μ ∈ N,we have the estimate

uniformly in x and y.This implies that

In view of R-bondedness property of kernel operators(see[11,Proposition 4.12])and due to the density of D(0,1;E(A))in Lp(0,1;E)(see[23]),we obtain the assertion.

5 Cauchy Problem for Abstract Parabolic Equation

Consider the problem

In this section we obtain the well-posedeness of problem(4.14)in mixed Lpspace.

If G+=(0,∞)× (0,1),p=(p,p1),Lp(G+;E)will denote the space of all p-summable scalar-valued functions with mixed norm(see,[5,§1]for E=C),i.e.,the space of all measurable functions f defined on G,for which

Analogously,(G+;E)denotes the E-valued Sobolev space with corresponding mixed norm(see,[5,§10]for E=C).

Theorem 5.1Suppose that Condition3.1is satisfied forϕ ∈ (,π).Then forf ∈ Lp(G+;E)and sufficiently largea>0,the problem(5.1)has a unique solution belonging toE(A),E)and the uniform estimate holds,i.e.,

ProofThe problem(4.14)can be express as the following Cauchy problem:

Theorem 4.2 implies that the operator O is uniformly R-positive and is a generator of analytic semigroups in F=Lp(0,1;E).Then by virtue of[30,Theorem 4.2],we obtain that for all f∈Lp1(R+;F),the problem(4.15)has a unique solution belonging to(R+;D(B),F)and the following estimate holds:

Since Lp1(R+;F)=Lp(G+;E),by Theorem 4.1 we have

These relations and the above estimate imply the assertion.

6 Nonlocal Boundary Value Problems for Elliptic Equations

The Fredholm property of BVPs for elliptic equations with parameters in smooth domains was studied in[1].In this section,the coercive estimate on the solution of integral boundary conditions for elliptic equations will be established in mixed Lpspaces.

Let G⊂Rm,m≥2,be a bounded domain with an(m−1)-dimensional boundary∂G∈C∞which locally admits rectification.Consider the following nonlocal BVP for the following anisotropic elliptic equation:

where Bk(x)are bounded operator fromto Lp1(G)fordenotes the Besov space[6,§18].

Dx=,Dj=,Dy=(D1,···,Dm),mk∈{0,1},αk,βkare complex numbers,r=0 or r=1,y=(y1,···,ym).Let Q denote the operator generated by problem(6.1)–(6.3)for λ=0.

If Ω =(0,1)× G,p=(p1,p),Lp(Ω)will denote the space of all p-summable scalar-valued functions with a mixed norm(see,[5,§1]),i.e.,the space of all measurable functions f defined on Ω,for which

Analogously,(Ω)denotes the Sobolev space with corresponding mixed norm(see[6,§10]).

We have the result as following theorem.

Theorem 6.1Let the following conditions be satisfied:

(1)Bkare bounded operators from

(2)aα∈ C(Ω)for each|α|=2mandaα∈ [L∞+Lrk](Ω)for each|α|=k<2mwithrk≥p1,p1∈(1,∞)and

(3)bjβ∈ C2m−mj(∂Ω)for eachj,β,mj<2m,p ∈ (1,∞);

(4)fory ∈ Ω,ξ∈ Rμ,η ∈ S(ϕ1),ϕ1∈ [0,),|ξ|+|η|0,let

(5)for eachy0∈ ∂Ω,the local BVP’s in local coordinates corresponding toy0

has a unique solutionϑ ∈ C0(R+)for allh=(h1,h2,···,hm) ∈ Rmand forξ'∈ Rμ−1with

Then

(a)for allf ∈ Lp(Ω),p=(p1,p),andsufficiently large|λ|,the problem(6.1)–(6.3)has a unique solutionu ∈and the following coercive uniform estimate holds:

(b)the operatoru → Qu={Lu,L1u,L2u}is Fredholm from(Ω)intoLp(Ω)×

ProofLet E=Lp1(G).Consider the operator A defined by

Then the problem(6.1)–(6.3)can be rewritten in the form of(2.1),(3.1),i.e.,

Let us apply Theorem 4.1 to problem(6.4).It is known that Lp1(G)∈ UMD for p1∈ (1,∞)(see[3]).Then in view of the multiplier theorems in E-valued Lpspaces(see[30]),the space Lp1(G)satisfies the multiplier condition.By virtue of[11,Theorem 8.2],the operator A is R-positive in Lp1(G)and has the fractional powers,i.e.,all conditions of the Theorem 4.1 hold and we obtain the assertion.

7 In finite Systems of Parabolic Equations

Consider the mixed problem for an in finite system of parabolic equations

where um=um(x,t)and Bk(x)are linear operators fromto lq(see[27,§1.18.2]for the definition of),and here

Theorem 7.1Suppose thatBkare bounded operators fromandThen,forand forsufficiently largea>0,the problem(7.1)has a unique solutionthat belongstoand the following coercive uniform estimate holds:

ProofReally,let E=lq,A be in finite matrices,defined by

It is easy to see that the operator A is R-positive in lq.Therefore,all conditions of Theorem 5.1 hold and we obtain the assertion.

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