We solve the inverse spectral problem for a class of Sturm - Liouville operators with singular nonlocal potentials and nonlocal boundary conditions

Nizhnik, L.P.

Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)

Methods of Functional Analysis and TopologyVol. 15 (2009), no. 1, pp. 41–47INVERSE EIGENVALUE PROBLEMS FOR NONLOCALSTURM–LIOUVILLE OPERATORSL. P. NIZHNIKTo the memory of A. Ya. PovznerAbstract. We solve the inverse spectral problem for a class of Sturm–Liouvilleoperators with singular nonlocal potentials and nonlocal boundary conditions.1. IntroductionThis paper is dedicated to the memory of A. Ya. Povzner whose outstanding workshave opened for me the area of spectral theory of Schrödinger operators. My personalacquaintance with A. Ya. Povzner took place in 1961 when I was presenting my PhDThesis and he was an opponent. Talks with him in that time and the scientific discussionswere very instructive and left an unforgettable trace in all my scientific life.A one-dimensional Schrödinger operator with nonlocal potential has the formLy ≡ − d2dx2y(x) +∫K(x, s)y(s) ds,where K(x, s) = K̄(x, s) is a Hermitian-symmetric kernel. If K(x, s) = v(x)δ(s − x0) +δ(x− x0)v̄(s), where δ is Dirac’s function, we have a Schrödinger operator with nonlocalpoint potential, Ly ≡ −d2y(x)dx2+v(x)y(x0)+ δ(x−x0)(y, v)L2 [1]. When considering suchan operator, one can avoid using Dirac’s δ-function if the differential expression ly(x) =d2y(x)dx2+ v(x)y(x0), x = x0, in the point x = x0 is supplemented with the boundary-valueconditions y(x0 − 0) = y(x0 + 0) = y(x0), y(x0 − 0)− y(x0 + 0) = (y, v)L2 [1]. Note thatsuch nonlocal operators appear not only in quantum mechanics but in other areas suchas the theory of diffusion processes, see the related references in [2].In this paper, we study a one-dimensional Schrödinger operator with nonlocal pointpotential on a bounded interval with periodic boundary-value conditions. Because of theperiodicity, we can limit the considerations to only the case where the point nonlocalpotential has its support at an endpoint of the interval. Then we have the followingnonlocal Sturm–Liouville eigenvalue problem:(1) ℓ(y) := −y′′(x) + v(x)y(1) = λy(x), 0 ≤ x ≤ 1,subject to the boundary-value condition(2) y(0) = y(1), y′(1) − y′(0) + (y, v)L2 = 0.Here v ∈ L2(0, 1) is the nonlocal “potential” and λ ∈ C is a spectral parameter. Thisproblem is close to the problem studied in [2] for equation (1) with the boundary-valueconditions y(0) = y′(1) + (y, v)L2 = 0. However, this unperturbed problem (1)–(2), forv = 0, has a simple eigenvalue, λ = 0, and double eigenvalues λ = (2nπ)2, n ∈ N ,2000 Mathematics Subject Classification. Primary 47A55, 47B25, 35J10.Key words and phrases. Inverse spectral problems, Sturm–Liouville operators, nonlocal potentials,nonlocal boundary conditions.4142 L. P. NIZHNIKwhereas the unperturbed problem considered in [2] has all simple eigenvalues λ = (nπ)2,n ∈ N . This influences the way for studying problem (1)–(2) and solving the inverseeigenvalue problem, that is, the problem of recovering the function v in equation (1) froma known collection of all eigenvalues of problem (1)–(2).For solving the inverse eigenvalue problem, as in the case of the usual potential, inaddition to eigenvalues it is also necessary to have some additional information [3]. Thesame is true for the real nonlocal potential, — we need to have signs of its even Fouriersine coefficients.2. Direct spectral analysisOn the space L2(0, 1), problem (1)–(2) is naturally related to an operator Tv that hasdomaindomT = {y ∈ W 22 (0, 1) | y(0) = y′(1) − y′(0) + (y, v)L2 = 0}and is defined by Tvy = ℓ(y). The operator Tv is a closed symmetric operator, sinceintegration by part formula easily yields that the quadratic form (Tvy, y)L2 is real,(3) (Tvy, y) = (y′, y′)L2 + 2Re [y(1)(v, y)L2 ].For v = 0, the operator T0 is self-adjoint on the space L2(0, 1), with domain consisting offunctions periodic on (0, 1) and belonging to the Sobolev space W 22 (0, 1). Eigenfunctionsof the operator T0 are the functions 1, cos 2nπx, sin 2nπx, n ∈ N , and eigenvalues areλ0 = 0, λn = (2nπ)2. The operators Tv and T0 can be considered as extensions of thesymmetric operator Tmin that has domain(4) domTmin = {y ∈ W 22 (0, 1) | y(0) = y(1) = 0, y′(0) = y′(1), (y, v)L2 = 0}and is defined by Tminy = Tvy = T0y = −y′′. Since dim(domTv/domTmin) ≤ 2, theself-adjoint operator Tv is a rank r ≤ 2 perturbation of the self-adjoint operator T0. Sincethe operator T0 has discrete spectrum, which consists of the numbers λn = (2nπ)2, theoperator Tv has discrete spectrum consisting of real numbers λn → +∞ for n → ∞.Theorem 1. 1. All eigenvalues of problem (1), (2), distinct from (nπ)2, n ∈ N ,are simple.2. The number λ = (2nπ)2 is an eigenvalue of problem (1), (2) if and only if(5) vn =12∫ 10v(x) sin nπxdx ={0, if n even,4nπ, if n odd.3. The number λn = (2nπ)2 is a double eigenvalue of problem (1), (2) if and onlyif, in addition to (5), we have(6)∑k =n[1 + (−1)k+1]kπ[vk + v̄k] − 12 |vk|2(kπ)2 − (nπ)2 = 0.Problem (1), (2) has no eigenvalues with multiplicity exceeding 2.4. The number λ = 0 is an eigenvalue of problem (1), (2) if and only if (6) holdsfor n = 0.Proof. The eigenvalues of problem (1), (2) coincide with the eigenvalues of the operatorTv. Let y(x; z) be an eigenfunction of the operator Tv corresponding to an eigenvalueλ = z2. If y(0; z) = y(1; z) = 0, then it follows from equation −y′′ = z2 that y = C sinπxand z2 = (nπ)2, n ∈ N . Hence, for λ = (nπ)2, eigenfunctions take nonzero equal values atthe endpoints of the interval (0, 1). This leads to simplicity of the eigenvalues λ = (nπ)2,n ∈ N . Indeed, if an eigenvalue were multiple, there would exist at least two linearlyindependent eigenfunctions yi(x; z), i = 1, 2, corresponding to the same eigenvalue. Butthen y(x; z) = y1(0; z)y2(x; z)−y2(0; z)y1(x; z) would be a nontrivial eigenfunction whichINVERSE EIGENVALUE PROBLEMS FOR NONLOCAL STURM–LIOUVILLE OPERATORS 43vanishes at the endpoints of the interval (0, 1). This is impossible for λ = z2 = (nπ)2.Claim 1 of the theorem is proved.Let now λ = (nπ)2, n ∈ N , be an eigenvalue of the operator Tv, that is, there existsa nontrivial solution y(x; nπ) of problem (1), (2) for λ = (nπ)2. Since the nonlocalpotential satisfies v ∈ L2, this function can be represented by a Fourier sine series,(7) v(x) =∞∑k=1vk sinkπxdx, vk = 2∫ 10v(x) sin kπxdx.By substituting (7) into equation (1) and solving it, we get(8) y(x; nπ) = A cosnπx + B sin nπx − A2nπvnx cosnπx − A∑k =nvk sin kπx(kπ)2 − (nπ)2 .The boundary-value condition y(0) = y(1) leads to the identity(9) A[1 − (−1)n + vn2nπ(−1)n] = 0.Hence, if A = 0, then (5) holds. If A = 0 and the solution y is nontrivial, then it hasthe form y = B sin nπx and B = 0. Substituting this solution into the boundary-valuecondition y′(1) − y′(0) + (y, v) = 0 we are again led to (5). Hence, problem (1)–(2) hasa nontrivial solution for λ = (nπ)2 only if condition (5) is satisfied. Also, condition (5)is sufficient for the function y = sin nπx to be an eigenfunction. Claim 2 is proved.To prove Claim 3, note that λ = (nπ)2 is a double eigenvalue if and only if condition (5)is satisfied and the function in (8), with A = 1 and B = 0, satisfies the boundary-valuecondition y′(1) − y′(0) + (y, v) = 0. Substituting (8) into this boundary-value conditionleads to identity (6). Since problem (1)–(2) deals with a second order equation onthe interval (0, 1), it can not have more than two linearly independent solutions and,consequently, there are no eigenvalues of multiplicity exceeding 2.Consider Claim 4. It follows from (1) for λ = 0 that(10) y(x; 0) = A + Bx − y(1)∞∑k=1vk sinkπx(kπ)2.The boundary-value condition y(1) = y(0), since the solution y(x; 0) is nontrivial, leadsto the condition B = 0, A = 0, y(1) = A. Substituting (10) into the boundary-valuecondition (2) we get identity (6) with n = 0. To give an exact description of the distribution of eigenvalues of problem (1), (2), it isconvenient to show that the eigenvalues are connected with zeros of an analytic function,which is a characteristic function of problem (1), (2).To this end, consider a special solution of equation (1) with λ = z2, satisfying thecondition y(0) = y(1),(11) ϕ(x; z) =sin zx + sin z(1 − x)z− sin zxz∞∑k=1vk sin kπx(kπ)2 − z2 .The function ϕ is an eigenfunction of problem (1), (2) if it satisfies the boundary-valuecondition (2). This gives the characteristic equation χ(z) = 0, where the characteristicfunction χ(z) is defined by χ(z) = ϕ′(1) − ϕ′(0) + (ϕ, v) and has the form(12) χ(z) = 2(cos z − 1) + sin zz∞∑k=1(−1)kkπαk(kπ)2 − z2 ,where(13) (−1)kαk = [1 + (−1)k+1](vk + v̄k) −12kπ|vk|2.44 L. P. NIZHNIKLemma 1. The characteristic function χ(z) of the form (12) is an entire function of zand, for z = nπ, n ∈ N , takes the values(14)χ(2mπ) =|v2m|2(4mπ)2,χ((2m − 1)π) = −|2 − v2m−12(2m − 1)π) |2.Proof. The proof is carried out by direct computations using the explicit form (12) ofthe characteristic function. Theorem 2. The number λ = z2 is an eigenvalue of problem (1), (2) if and only if z isa zero of the characteristic function χ(z). The number λ = z2 = 0 is a double eigenvalueof the problem (1), (2) if and only if z is a double zero of the characteristic function.All zeros z = nπ, n ∈ N , of the characteristic function are simple. The characteristicfunction does not have zeros of multiplicities greater than 2.Proof. It follows that the squares z2 = (nπ)2 of zeros of the characteristic function χ(z)are eigenvalues from the fact that the special solution (11) is an eigenfunction and viceversa. For z2 = (nπ)2 = λ, the number λ = z2 is an eigenvalue if and only if (5) issatisfied which, by (14), is equivalent to χ(nπ) = 0. It is easy to check that condition (6)is equivalent to χ̇(nπ) = 0 which, in its turn, together with χ(nπ) = 0, is equivalentto that nπ is a double eigenvalue. Condition (6), for n = 0, is equivalent to χ(0) = 0.Hence, the eigenvalues, counting multiplicities, coincide with squares of zeros of χ(z),counting multiplicities.If the function χ(z) had a multiple root z0 = nπ, this would imply that ∂∂z ϕ(x; z)|z=z0were a generalized eigenfunction, which is impossible since the operator Tv is self-adjoint.In the same way, we can prove that there are no zeros of χ(z) that have multiplicitiesgreater than 2 for z = nπ. Theorem 3. The increasingly ordered sequence λ1 ≤ λ2 ≤ . . . ≤ λn ≤ λn+1 ≤ . . . of alleigenvalues of problem (1), (2), counting multiplicities, has the following properties:1. the sequence weakly alternates with the sequence (nπ)2,(15) λn ≤ (nπ)2 ≤ λn+1, n ∈ N ;2. there is an asymptotic representation,(16)√λ2n = 2nπ − β2nn ,√λ2n+1 = 2nπ +β2n+1n,where βj ≥ 0 and∑∞j=1 β2j < +∞.Proof. An upper estimate for λn is easily obtained from the variation minimax princi-ple [4],λn = supϕ1,...,ϕn−1 inf ||ψ||=1, ψ⊥ϕk,ψ(0)=ψ(1)(Tvψ, ψ)≤ supϕ1,...,ϕn−1 inf ||ψ||=1, ψ⊥ϕk,ψ(0)=ψ(1)=0(ψ′, ψ′) = (nπ)2.For large n, we have the strict inequality(17) [(2n − 1)π]2 < λ2n ≤ λ2n+1 < [(2n + 1)π]2.Indeed, by the Rouché theorem, the entire function χ(z) and the function 2(cos z − 1)have the same number of zeros, counting multiplicities, in the strip −(2n+1)π < Re z <(2n + 1)π for large n. Since, by Theorem 2, the eigenvalues λn of problem (1), (2)are squares of zeros of the function χ(z), the function χ(z) is even, and double zerosINVERSE EIGENVALUE PROBLEMS FOR NONLOCAL STURM–LIOUVILLE OPERATORS 45of the function 2(cos z − 1) are the numbers z = 2nπ, n = 0,±1, ..., which shows thatinequalities (17) hold.On the other hand, if(18) v2m = 0, v2m+1 = 4(2m + 1)π,then, by (14), χ(nπ) = 0 for any n and, moreover, χ(nπ)(−1)n > 0. Hence, in everyinterval (nπ, (n + 1)π), the characteristic function χ has at least one zero z0 and, con-sequently, there is one eigenvalue λ = z2 in the interval In = ((nπ)2, (n + 1)2π2). Theassumption that at least one interval In contains more than one eigenvalue leads to acontradiction with estimate (17). Hence, if conditions (18) are satisfied, inequality (15)holds,(19) λn < (nπ)2 < λn+1.Since condition (18) can be satisfied by an arbitrary small change of the potential, passingto the limit in (19) we get (15).To prove the asymptotic representations (16) let us first find an asymptotic represen-tation of zeros of the characteristic function χ(z) defined by (12), since√λn = zn.Write the characteristic function in (12), χ(z), as(20) χ(z) = 2(cos z − 1) − 1z∫ 10sin ztα(t) dt,where α(t) ∈ L2, and the numbers αk are coefficients in the Fourier sine expansion. Ifz = 2mπ + εm, then (20) for m → ∞ gives(21) −ε2m +|v2m|2(4mπ)2− εmw2m2mπ+ o(ε2m) = 0,where w2m =∫ 10cos 2mπtα(t)t dt and∑m w22m < ∞.Identity (21) gives two solutions for εm, one of which is nonpositive, ε+m = −β2m,β2m ≥ 0, and the other one is nonnegative, ε−m = β2m+1, β2m+1 ≥ 0. Here∑j |βj |2 <+∞. This gives representation (16). 3. Isospectral nonlocal potentialsLet Λ(v) = {λj}∞j=1 be an ordered sequence of all eigenvalues, counting multiplic-ities, of problem (1), (2) with a nonlocal potential v. Two potentials v1 and v2 arecalled isospectral if Λ(v1) = Λ(v2), that is, eigenvalues and their multiplicities of prob-lems (1), (2), coincide for the potentials v1 and v2. It follows from Theorem 2 thatpotentials v1 and v2 are isospectral if and only if the corresponding characteristic func-tions coincide. Using representations (12), (13) the for characteristic functions we obtaina criterion for two potentials v1 and v2 to be isospectral in terms of their Fourier sinecoefficients v(j)k , j = 1, 2, k ∈ N ,(22)|v(1)2m| = |v(2)2m|,∣∣∣1 − v(1)2m+14(2m+1)π∣∣∣=∣∣∣1 − v(2)2m+14(2m+1)π∣∣∣.On the space L2(0, 1) of real-valued functions, define nonlinear projections Πs thatdepend on a sequence of real numbers s = {s1, s2, ..} by(23) Πs(∞∑n=1vn sin nπx)=∞∑n=1v(s)n sin nπx,∞∑n=1v2n < +∞,46 L. P. NIZHNIKwhere(24) v(s)n =|vn|, if sn = 0,vn, if |vn| ≤ |2sn − vn|, sn = 0,2sn − vn, if |2sn − vn| < |vn|.Theorem 4. Two real nonlocal potentials v1, v2 ∈ L2(0, 1) are isospectral if and only ifthe corresponding nonlinear projections coincide,(25) Πsv1 = Πsv2.Here s = {sn}∞n=1 consists of the numbers s2m = 0 and s2m−1 = 2(2m − 1)π.Proof. The proof follows from the isospectral criterion (22) and definitions (23), (24) ofthe nonlinear projection Πs. 4. The inverse spectral analysisThe main tool for describing eigenvalues of problem (1), (2) is the characteristic func-tion χ(z) explicitly given by (12). This function can be written using the ordered set ofall eigenvalues, Λ = {λj}.Theorem 5. The characteristic function χ(z) can be expressed in terms of the orderedsequence {λj}, counting multiplicities, of problem (1), (2) as(26) χ(z) = (λ1 − z2)∞∏k=2λk − z24[k2 ]2π2,where [α] denotes the entire part of α.Proof. Since χ(z) is an even function of exponential type 1 andlimξ→∞χ(iξ)2(cos iξ − 1) = 1,it can uniquely be represented in terms of its zeros zk =√λk as the product in (26). Consider now the problem of recovering the nonlocal potential for problem (1), (2) interms of the set Λ = {λj} of all eigenvalues of the problem. To solve this problem, wecan propose the following algorithm:Step 1. Using the eigenvalues Λ = {λj}∞j=1 construct the characteristic function χ(z) byformula (26).Step 2. Calculate the values χ(nπ), n ∈ N .Step 3. Use formula (14) to find vn that have the minimal modulus.Step 4. Use the Fourier coefficients vn to construct the nonlocal potential(27) v(x) =∞∑n=1vn sin nπx.Example. Let, for problem (1), (2), we have λ1 = π2, λ2n = λ2n+1 = (2nπ)2, n ∈ N .Then χ(z) = z2−π2z22(cos z − 1). In this case, χ(2nπ) = 0, χ((2n − 1)π) = −4 + 4(2n−1)2 ,n ∈ N . It follows from formulas (14) that v2n = 0, v2n−1 = 4π 12n−1+√(2n−1)2−1, n ∈ N .Hence, the nonlocal potential v(x) has the following Fourier series:v(x) = 4π∞∑n=1sin(2n − 1)πx2n− 1 +√(2n − 1)2 − 1.INVERSE EIGENVALUE PROBLEMS FOR NONLOCAL STURM–LIOUVILLE OPERATORS 47Since∞∑n=1sin(2n − 1)πx2n− 1 =π4, we havev(x) =π22+ 2π∞∑n=1sin(2n − 1)πx(2n − 1)[2n − 1 +√(2n − 1)2 − 1]2.As was remarked in the previous section, the nonlocal potential v is defined by itseigenvalues Λ(v) non-uniquely. However, it is easy to give conditions such that thenonlocal potential can be found in a unique way, as it was done in [2]. In particular,the inverse problem for (1), (2) with nonlocal potential has a unique solution if weuse Λ = {λj}∞j=1 to find a nonlocal potential with a minimal norm and an additionalassignment of signs of all nonzero even Fourier coefficients.For the inverse problem, it is important to have a description of the initial data of theproblem, that is the set Λ = {λj}∞j=1 of all eigenvalues of problem (1), (2) with nonlocalpotential v ∈ L2.Theorem 6. Conditions 1 and 2 in Theorem 3 are necessary and sufficient for a sequenceΛ = {λj}∞j=1 to be an ordered sequence of all eigenvalues of problem (1), (2) with anonlocal potential v ∈ L2.Proof. Let conditions 1 and 2 of Theorem 3 be satisfied. Then the function χ(z) con-structed using (26) admits representation (19) and the value χ(nπ) can be written asin (14) with some vn,∑∞n=1 |vn|2 < ∞. The nonlocal potential v(x) =∑vk sin kπx ∈ L2gives rise to a characteristic function that coincides with χ(z) that is constructed usingformula (26). Hence, problem (1), (2) with the potential constructed as above has eigen-values that coincide with Λ = {λj}∞j=1. References1. S. Albeverio and L. P. Nizhnik, Schrödinger operators with nonlocal point interactions, J. Math.Anal. Appl. 332 (2007), 884–895.2. S. Albeverio, R. Hryniv and L. Nizhnik, Inverse spectral problems for nonlocal Sturm-Liouvilleoperators, Inverse Problems 23 (2007), 523-535.3. V. A. Marchenko, Sturm-Liouville Operators and Their Applications, Naukova Dumka Publ.,Kiev, 1977 (in Russian); Engl. transl.: Birkhäuser Verlag, Basel, 1986.4. M. Reed and B. Simon, Methods of Modern Mathematical Physics. IV. Analysis of Operators,Academic Press, New York, 1978.Institute of Mathematics, National Academy of Sciences of Ukraine, 3 Tereshchenkivs’ka,Kyiv, 01601, UkraineE-mail address: nizhnik@imath.kiev.uaReceived 01/11/2008

Inverse Eigenvalue Problems for Nonlocal Sturm-Liouville Operators

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Inverse Eigenvalue Problems for Nonlocal Sturm-Liouville Operators

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Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)