On solvability of a partial integral equation in the space ${L_2(OmegatimesOmega)}$

a rXiv:084.472v1[math-ph]3Apr28On solvability of a partial integral equation in the space L 2(Ω×Ω)Eshkabilov Yu.Kh.National University of Uzbekistan e-mail:yusup62@rambler.ru Abstract In this paper we investigate solvability of a partial integral equation in the space L 2(Ω×Ω),where Ω=[a,b ]ν.We define a determinant for the partial integral equation as a continuous function on Ωand for a continu-ous kernels of the partial integral equation we give explicit description of the solution.Key words:partial integral operator,partial integral equation,the Fredholm integral equation.2000MSC Subject Classification :45A05,45B05,45C05,45P05In the models of solid state physics [1]and also in the lattice field theory [2],there appear so called discrete Schrodinger operators,which are lattice analogies of usual Schrodinger operators in continuous space.The study of spectra of lattice Hamiltonians (that is discrete Schrodinger operators)is an important matter of mathematical physics.Nevertheless,on studying spectral properties of discrete Schrodinger operators three appear partial integral equations in a Hilbert space of multi-variable functions [1,3].Therefore,on the investigation of spectra of Hamiltionians considered on a lattice,the study of a solvability problem for partial integral equations in L 2is essential (and even interesting from the point of view of functional analysis).A question on the existence of a solution of partial integral equations for functions of two variables were considered in [4-8]and others.In this paperwe consider an integral equation on the space of functions of two variables L 2(Ω×Ω),where Ω=[a,b ]ν⊂R ν,with one partial integral operator.We define a determinant for the partial intergal equation (PIE)as a continuous function on Ω,which helps to obtain the classical Fredholm theorems for a PIE,and for a continuous kernels of the PIE we give explicit description of the solution.Let H =L 2(Ω×Ω)(H 0=L 2(Ω))be a Hilbert space of measurable and quadratic integrable functions on Ω×Ω(on Ω),where Ω=[a,b ]ν.We denote by µthe Lebesgue measure on Ωand define the measure µon Ω×Ωby µ=µ⊗µ.1In the space H,we consider a partial integral operator(PIO)T1defined byT1f= Ωk(x,s,y)f(s,y)ds,f∈Hwhere k(x,s,y)∈L2(Ω3).The function k(x,s,y)is called kernel of the PIO T1.If there exists a number M such thatb(t)≤M for almost all t∈Ω,(I) then the operator T1is a linear bounded operator on H and it is uniquely defined by its kernel k(x,s,y),whereb(t)= Ω Ω|k(x,s,t)|2dxds.A kernelk(s,x,y)f(s,y)ds,f∈H.Consider a family of operators{Kα}α∈Ωin H0associated with T1by the following formulaKαϕ= Ωk(x,s,α)ϕ(s)ds,ϕ∈H0,where k(x,s,y)is the kernel of T1.Further,if a set of integralabity in the integral is absent,then we mean integralabity by the setΩ.First,we consider certain properties of PIO T1with the kernel k(x,s,y)∈L2(Ω3)satisfying the condition(I)and then we study solvability of the PIE with the kernel k(x,s,y)∈C(Ω3).Lemma1.Let f∈H andϕy(x)=f(x,y),where y∈Ωisfixed.Then for an arbitraryε>0,there exists a subsetΩε⊂Ωsuch thatµ(Ωε)≥µ(Ω)−εand ϕα∈H0,α∈Ωε.Moreover, ϕα ≤C,α∈Ωεfor some C>0.Proof.Let f∈H and d= f 2=0.Define two sequences of measurable subsets inΩby the following equalities:A n= y: |f(x,y)|2dx<n,y∈Ω ,n∈N,B n= y: |f(x,y)|2dx≥n,y∈Ω ,n∈N.The sequences of subsets{A n}and{B n}hold the following properties:1o.A1⊂A2⊂...⊂A n⊂...and B1⊃B2⊃...⊃B n⊃...;22o .lim n →∞A n = n ∈N A n and lim n →∞B n = n ∈N B n ;30.Ω=A n ∪B n and A n ∩B n =∅,n ∈N .Further,we define two bounded sequences of non-negative numbers a n and b n by a n =A n dy Ω|f (x,y )|2dx and b n =B n dy Ω|f (x,y )|2dx.The sequences of numbers a n and b n have the properties:4o .a n is increasing and b n is decreasing;5o .a n +b n =d,n ∈N .¿From the boundeness and monotonicity of the sequences a n and b n we infer that both of them have finite limit.By the property 5o and by the construction of the set B n we obtain that d −a n ≥0,n ∈N and d ≥a n +nµ(B n ),n ∈N .Then µ(B n )≤(d −a n )/n,n ∈N .Therefore lim n →∞µ(B n )=0.By the property 3o we have µ(A n )=µ(Ω)−µ(B n ),n ∈N .Hence,lim n →∞µ(A n )=µ(Ω),i.e.for an arbitrary small ε>0there exists a number n 0∈N such that µ(Ω)−ε≤µ(A n 0)≤µ(Ω)and 0≤µ(B n 0)<ε.Moreover,this means that|ϕα(x )|2dx = |f (x,α)|2dx <n 0,α∈A n 0.Then,for the set Ωε=A n 0we have ϕα∈H 0,α∈Ωεand ϕα ≤C,α∈Ωεfor all C ≥n 0.Corollary 1.Let f ∈H , f =1and ϕy (x )=f (x,y ),where y ∈Ωis fixed.Then there exists a measurable subset Ω0⊂Ωsuch that,µ(Ω0)>0and the family {ϕα}α∈Ωof functions on Ωhas the following property:ϕα∈H 0,α∈Ω0and 0< ϕα ≤C,α∈Ω0for some C >0.Corollary 2.Let f ∈H .Then there exists a decreasing sequence {εn }n ∈N of posity numbers such that lim n →∞εn =0and(a)for each n ∈N there exists a measurable subset Ωn ⊂Ωwith the propertieµ(Ωn )>µ(Ω)−εn such that Ω1⊂Ω2⊂...⊂Ωn ⊂...andn ∈NΩn =Ω;(b)for each n ∈N ,ϕ(n )α∈H 0,α∈Ωn and there exists a positive number C n such that ϕ(n )α ≤C n ,∀α∈Ωn ,where ϕ(n )α(x )=f (x,α),α∈Ωn ;(c)for any n ∈N ,the functionf n (x,y )= f (x,y ),if (x,y )∈Ω×Ωn ,0,otherwisebelongs to H and lim n →∞f n (x,y )=f (x,y ).Proposition 1.The following two conditions are equivalent:(i)A number λ∈C is an eigenvalue for the operator T 1;(ii)A number λ∈C is an eigenvalue for operators {K α}α∈Ω0,where Ω0is some subset of Ωsuch that µ(Ω0)>0.3Proof.We start with the implication(i)⇒(ii).Letλ∈C be an eigenvalue of operator T1,i.e.T1f0=λf0for some f0∈H, f0 =1.We defineϕα=ϕα(x)=f0(x,α),α∈Ω.Therefore,we have a family{ϕα}α∈Ωof functions on Ω.Then,by Corollary1,there exists a subsetΩ0⊂Ωsuch thatµ(Ω0)>0and ϕα∈H0,α∈Ω0, ϕα =0,∀α∈Ω0.For an arbitraryα∈Ω0we haveKαϕα= k(x,s,α)ϕα(s)ds= k(x,s,α)f0(s,α)ds=λf0(x,α)=λϕα(x),i.e.the numberλis an eigenvalue for Kα,α∈Ω0.Now,we prove the implication(ii)⇒(i).Suppose that there exists a subset Ω0inΩwithµ(Ω0)>0and a numberλ∈C is an eigenvalue for operators Kα,α∈Ω0.Since Kαis a compact operator for allα∈Ω,then there exists a function f0∈H,f0=0[9]such that T1f0=λf0.Proposition2.Ifλ∈C is an eigenvalue of the operator T1,then the numberλis an eigenvalue of every operator K∗α,α∈Ω0:K∗αϕ= λis an eigenvalue of the adjoint operator T∗1.Proposition3.Every eigenvalue of the operator T1has infinite multiplicity. Proof.Letλ∈C be an eigenvalue of T1.Hence,there exists an element f0∈H, f0 =1such that T1f0=λf0.We consider a subspace L0⊂H:L0={ f∈H: f(x,y)=b(y)f0(x,y),where b=b(y)is an arbitrary bounded measurable function onΩ}.For every f∈L0we have T1 f=λ f,i.e.L0⊂Mλ,where Mλis the eigen-subspace corresponding toλ.But,the subspace L0is infinite dimensional,therefore,Mλis also infinite dimensional subspace of H.Now we consider the equationf−κT1f=g0,(1) in the space H,where f is an unknown function from H,g0∈H is given(known) function,κ∈C is a parameter of the equation,T1is PIO with a kernel k(x,s,y) continuous onΩ3.It is clear that,if k(x,s,y)∈C(Ω3)then for allα∈Ωthe integral operatorsKαon H0are compact.For eachα∈Ωwe denote by∆(1)α(κ)and M(1)α(x,s;κ), respectively,the Fredholm determinant and the Fredholm minor of the operator E−κKα,κ∈C[10],where E is the identity operator in H0.According to the continuity of the kernel k(x,s,y)and uniform convergence of the series for∆(1)α(κ)and M(1)α(x,s;κ)for everyκ∈C we obtain[10]that the function4D1(y)=D1(y;κ)onΩand the function M1(x,s,y)=M1(x,s,y;κ)onΩ3, which are given respectively by the equalitiesD1(y;κ)=∆(1)y(κ),y∈Ωand M1(x,s,y;κ)=M(1)y(x,s;κ),y∈Ω, are continuous functions onΩandΩ3for everyκ∈C.The continuous function D1(y)=D1(y;κ)(M1(x,s,y)=M1(x,s,y;κ))is called a determinant(a minor)of the operator E−κT1,κ∈C.Definition1.If for a numberκ0∈C D1(y;κ0)=0for all y∈Ω,thenκ0 is called a regular number of the PIE(1).A set of all regular numbers of the PIE(1)is denoted by R T1.Definition2.If for a numberκ0∈C there exists a point y0∈Ωsuch that D1(y0;κ0)=0,thenκ0is called a singular number of the PIE(1).A set of all singular numbers of the the PIE(1)is denoted by S T1.Definition3.If for a numberκ0∈C there exists a measurable subset Ω0⊂Ωwithµ(Ω0)>0such that D1(y;κ0)=0,∀y∈Ω0,thenκ0is called a characteristic number of the the PIE(1).A set of all characteristic numbers of the PIE(1)is denoted by X T1.Definition4.A numberκ0∈C is called an essential number of the PIE(1)ifκ0∈S T1\X T1.A set of all essential numbers of the PIE(1)is denotedby E T1.Thus,for a parameterκof the PIE(1),there exist subsets R T1,S T1,X T1,and E T1in C,which have the following relations:(i)R T1∪S T1=C and R T1∩S T1=∅;(ii)X T1∪E T1=S T1and X T1∩E T1=∅.¿From Definitions1,2,3and4one gets that for an arbitrary non-zero PIOT1sets R T1and E T1are non-empty,but X T1may be empty.For example,consider a PIE in the space L2([0,1]2):f(x,y)−κ1e x−s e y f(s,y)ds=g0(x,y),where f is an unknown function in L2([0,1]2),g0∈L2([0,1]2)is an arbitrary given function.For this PIE,the determinant has a simple form D1(y;κ)= 1−κe y,y∈[0,1].Therefore S T1= e−1,1 and X T1=∅.¿From Proposition1and Definition3it followsTheorem1.A numberλ∈C,λ=0,is an eigenvalue of the operator T1if and only ifλ−1∈X T1.Theorem2.a)ifκ0∈E T1,thenκ0∈X T∗1.Proof.Letκ0∈E T1.Then there exists a point y0∈Ωwith D1(y0;κ0)=0 and we haveµ{y∈Ω:D1(y;κ0)=0}=0.But using a property of the determinant D1(y;κ0)we obtain thatκ0),where D1(y;is a determinant of the operator E −κ0)=0and µ y ∈Ω: D 1(y ;κ0is an essential number ofthe adjoint equation f −D 1(y ;κ0)g (s,y )ds,g ∈H ,(2)but the corresponding homogeneous equation f −κ0T 1f =0has only trivial solution (zero solution).Here D 1(y ;κ0)and M 1(x,s,y ;κ0)are the determinant and the minor of the operator E −κ0T 1,respectively.Proof.Let κ0∈R T 1and κ0=0.First,we prove that PIE (1)is solvable in H .By Corollary 2,for the function g 0there exists a decreasing sequence of non-negative numbers εn and a sequence of increasing measurable subsets Ωn ⊂Ω,which satisfy the properties (a),(b)and (c)with lim n →∞εn =0.For every Ωn wedefine a subspace L (n )2=L (n )2(Ω×Ω)as follows:a function f∈H belongs to the subspace L (n )2,if it satisfies the following conditions:(i)ϕ(n )α(x )= f(x,α)∈H 0,∀α∈Ωn ;(ii)there exists a positive number C n such that ϕ(n )α ≤C n ,∀α∈Ωn ;(iii) f(x,y )=0if (x,y )∈Ω×(Ω\Ωn ).For every f ∈H ,there exists a sequence f n ∈L (n )2,n ∈N ,such thatlim n →∞f n =f.Therefore,first we find a solution of the equation (1)in the space L (n )2andwe can find a solution of the equation (1)in the space H as the limit f (x,y )=lim n →∞f n (x,y ),where f n are solutions of the equation (1)in the space L (n )2.Thus,the equation (1)in L (n )2reduces to the following one:f n (x,y )−κ0T 1 f n (x,y )=g n (x,y ),(3)where g n is an element of L (n )2corresponding to the function g 0(x,y ).Hence,by the property (b)of Corollary 2,for each fixed y ∈Ω,the equation(3)reduces to the following second type Fredholm integral equation in H 0:ϕ(n )α(x )−κ0K αϕ(n )α(x )=h (n )α(x ),α∈Ω(3′)where ϕ(n )α(x )= f n (x,α)is an unknown function in H 0,h (n )α(x )=g n (x,α)is agiven function in H 0.By the first fundamental Fredholm theorem,the equation (3′)for every α∈Ωn has the only solutionϕ(n )α=ϕ(n )α(x )=h (n )α(x )+κ0B αh (n )α(x ),6where the operator Bα=Bα(κ0)acts in H0by the formulaBαϕ= M(1)α(x,s;κ0)dy<∞,(II)|D1(y;κ0)|2then PIE(1)has a unique solution on H and it has a form f=g0+κ0Bg0∈H, but corresponding homogeneous equation f−κ0T1f=0has only trivial solution, where the operator B is given by(2).Proof.Letκ0∈E T.PutΩ′={y∈Ω:D1(y;κ0)=0}.It is evident that1Ω′=∅andµ(Ω′)=0.However,for every y∈Ω\Ω′the function f0(x,y)= g0(x,y)+κ0Bg0(x,y)satisfies the equation(1).Now it is enough to show that f0∈H.Suppose that g0satisfies the condition(II).We have|Bg0(x,y)|2dxdy= M1(x,s,y;κ0)≤ |M1(x,s,y;κ0)|·|g0(s,y)|ds|D1(y;κ0)|2dxdy≤≤N20µ(Ω) |g0(s,y)|ds 2µ(Ω)·|D1(y;κ0)|2dy<∞,i.e.Bg0∈H,therefore f0=g0+κ0Bg0∈H and f0is a solution of the equation(1).Uniqueness of the solution follows from ing Proposition1one can also show that the homogeneous equation f−κ0T1f=0has only the trivial solution.Remark1.The condition(II)in Theorem4is natural.For example,for the equationf(x,y)−κ1e x−s yf(s,y)ds=e x y1/2(4)in the space L2([0,1]2),we have D1(y;κ)=1−κy,y∈[0,1]and M1(x,s,y;κ)=e x−s y.Hence,S T1=E T1=[1,∞).For eachκ∈[1,∞),the equation(4)hasthe solutionf0(x,y)=e x 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