A half-inverse problem for the singular diffusion operator with jump conditions
aa r X i v : . [ m a t h . C A ] J un A HALF-INVERSE PROBLEM FOR THE SINGULARDIFFUSION OPERATOR WITH JUMP CONDITIONS
ABDULLAH ERG ¨UN
Abstract.
In this paper, half inverse spectral problem for diffusion op-erator with jump conditions dependent on the spectral parameter and dis-continuoty coefficient is considered. The half inverse problems is studiedof determining the coefficient and two potential functions of the boundaryvalue problem its spectrum by Hocstadt- Lieberman and Yang-Zettl meth-ods. We show that two potential functions on the whole interval and theparameters in the boundary and jump conditions can be determined fromspectrum. Introduction and preliminaries.
We consider the boundary value problem of the form l ( y ) := − y ′′ + [2 λp ( x ) + q ( x )] y = λ δ ( x ) y, x ∈ [0 , π ] / { a , a } (1.1)with the boundary conditions y ′ (0) = 0 , y ( π ) = 0 (1.2)and the jump conditions y ( a + 0) = α y ( a −
0) (1.3) y ′ ( a + 0) = β y ′ ( a −
0) + iλγ y ( a −
0) (1.4) y ( a + 0) = α y ( a −
0) (1.5) y ′ ( a + 0) = β y ′ ( a −
0) + iλγ y ( a −
0) (1.6)Where λ is a spectral parameter, p ( x ) ∈ W [0 , π ], q ( x ) ∈ L [0 , π ] are realvalued functions, a ∈ (cid:2) , π (cid:3) , a ∈ (cid:2) π , π (cid:3) , α , α , γ , γ are real numbers, | α i − | + γ i = 0 ( α i > i = 1 , β i = α i ( i = 1 ,
2) and δ ( x ) = (cid:26) α , x ∈ (cid:0) , π (cid:1) β , x ∈ (cid:0) π , π (cid:1) where 0 < α < β < α + β > Mathematics Subject Classification.
Key words and phrases.
Differential equations, Discontinuous function, Singular Diffusionoperator. problem for Sturm-Liouville operators and diffusion operators [1–25]. Thefirst results an inverse problems theory of Sturm-Liouville operators wheregiven by Ambarzumyan [2]. The half inverse problems for Sturm-Liouvilleequations; the known potential in half interval is determined by the help ofa one spectrum over the interval. First the obtained results the half inverseproblem by Hochstadt and Lieberman [10]. They proved that spectrum of theproblem − y ′′ + q ( x ) y = λy, x ∈ [0 , y ′ (0) − hy (0) = 0 y ′ (1) + Hy (1) = 0and potential q ( x ) on the (cid:0) , (cid:1) uniquely determine the potential q ( x ) on thewhole interval [0 ,
1] almost everywhere. Hald [9] proved similar results in thecase when there exists a impulsive conditions inside the interval. Many stud-ies have been done by different authours for half invers problems using thismethods [13, 18]. In the work [18] studied the existence of the solution for hehalf-inverse problem of Sturm-Liouville problems and gave method of recon-structing this solution under same conditions by Sakhnovich [16]. Recently,same new uniqueness results on the inverse or half inverse spectral analysis ofdifferential operators have been given. Koyunbakan and Panakhov [13] provedthe half inverse problem for diffusion operator on the finite interval [0 , π ]. RanZhang, Xiao-Chuan Xu, Chuan-Fu Yang and Natalia Pavlovna Bondarenko,proved the determination of the impulsive Sturm-Liouville operator from a setof eigenvalues [25] .Purpose of this study is to prove half inverse problem by using the Hocstadt-Lieberman and Yang-Zettl methods for the following equations˜ l ( y ) := − y ′′ + [2 λ ˜ p ( x ) + ˜ q ( x )] y = λ ˜ δ ( x ) y, x ∈ [0 , π ] / { a , a } (1.7) y ′ (0) = 0 , y ( π ) = 0 (1.8) y ( a + 0) = ˜ α y ( a −
0) (1.9) y ′ ( a + 0) = ˜ β y ′ ( a −
0) + iλ ˜ γ y ( a −
0) (1.10) y ( a + 0) = ˜ α y ( a −
0) (1.11) y ′ ( a + 0) = ˜ β y ′ ( a −
0) + iλ ˜ γ y ( a − . (1.12) Lemma 1.
Let p ( x ) ∈ W (0 , π ) , q ( x ) ∈ L (0 , π ) . M ( x, t ) , N ( x, t ) are sum-mable functions on [0 , π ] such that the representation for each x ∈ [0 , π ] / { a , a } . ϕ ( x, λ ) solution of the equations (1 . , providing boundary conditions (1 . and discontinuity conditions (1 . − (1 . ϕ ( x, λ ) = ϕ ( x, λ ) + Z x M ( x, t ) cos λtdt + Z x N ( x, t ) sin λtdt is satisfied, ALF-INVERSE PROBLEM 3 for < x < π , ϕ ( x, λ ) = (cid:0) β +1 + γ α (cid:1) cos h λξ + ( x ) − α R xa p ( t ) dt i + (cid:0) β − − γ α (cid:1) cos h λξ − ( x ) + α R xa p ( t ) dt i (1.13) for π < x ≤ π , ϕ ( x, λ ) = (cid:16) β +2 + γ β (cid:17) cos h λk + ( π ) − β R πa p ( t ) dt i + (cid:16) β − + γ β (cid:17) cos h λk − ( π ) − β R πa p ( t ) dt i + (cid:16) β − − γ β (cid:17) cos h λs + ( π ) + β R πa p ( t ) dt i + (cid:16) β +2 − γ β (cid:17) cos h λs − ( π ) + β R πa p ( t ) dt i (1.14) where ξ ± ( x ) = ± αx ∓ αa + a , k ± ( x ) = ξ + ( a ) ± βx ∓ βa , s ± ( x ) = ξ − ( a ) ± βx ∓ βa , β ∓ = (cid:16) α ∓ β α (cid:17) , β ∓ = (cid:16) α ∓ αβ β (cid:17) .Thus, following the relations hold;If p ( x ) ∈ W (0 , π ) , q ( x ) ∈ W (0 , π ) ( ∂ M ( x,t ) ∂x − ρ ( x ) ∂ M ( x,t ) ∂t = 2 p ( x ) ∂N ( x,t ) ∂t + q ( x ) M ( x, t ) ∂ N ( x,t ) ∂x − ρ ( x ) ∂ N ( x,t ) ∂t = − p ( x ) ∂M ( x,t ) ∂t + q ( x ) N ( x, t ) M (cid:0) x, ς + ( x ) (cid:1) cos β ( x ) α + N (cid:0) x, ς + ( x ) (cid:1) sin β ( x ) α = (cid:16) β +1 + γ α (cid:17) Z x (cid:18) q ( t ) + p ( t ) α (cid:19) dtM (cid:0) x, ς + ( x ) (cid:1) sin β ( x ) α − N (cid:0) x, ς + ( x ) (cid:1) cos β ( x ) α = (cid:16) β +1 + γ α (cid:17) ( p ( x ) − p (0)) M ( x, k + ( x ) + 0) − M ( x, k + ( x ) −
0) = − (cid:16) β +2 + γ β (cid:17) ( p ( x ) − p (0)) sin ω ( x ) β − (cid:16) β +2 + γ β (cid:17) R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt cos ω ( x ) β N ( x, k + ( x ) + 0) − N ( x, k + ( x ) −
0) = (cid:16) β +2 + γ β (cid:17) ( p ( x ) − p (0)) cos ω ( x ) β − (cid:16) β +2 + γ β (cid:17) R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt sin ω ( x ) β ∂M ( x, t ) ∂t (cid:12)(cid:12)(cid:12)(cid:12) t =0 = N ( x,
0) = 0 where β ( x ) = R x p ( t ) dt , ω ( x ) = R xa p ( t ) dt + R a p ( t ) dt .The proof is done as in [5]. Definition.
The function ∆ ( λ ) is called the characteristic function of theeigenvalues { λ n } of the problem (1 . − (1 . λ ) is called the characteristicfunction of the eigenvalues n ˜ λ n o of the problem (1 . − (1 . λ = s , s = σ + iτ , σ, τ ∈ R. The solution ϕ ( x, λ ) of (1 . − (1 .
6) havethe following asymptotic formulas hold on for | λ | → ∞ , ABDULLAH ERG ¨UN for 0 < x < π , ϕ ( x, λ ) = 12 (cid:18) α ∓ β α + γ α (cid:19) exp (cid:18) − i (cid:18) λξ + ( x ) − v ( x ) α (cid:19)(cid:19) (cid:18) O (cid:18) λ (cid:19)(cid:19) for π < x ≤ π , ϕ ( x, λ ) = 12 (cid:18) α αβ β + γ β (cid:19) exp (cid:18) − i (cid:18) λk + ( x ) − t ( x ) β (cid:19)(cid:19) (cid:18) O (cid:18) λ (cid:19)(cid:19) . where v ( x ) = R xa p ( t ) dt , t ( x ) = R xa p ( t ) dt .In this study, if q ( x ) and p ( x ) to be known almost everywhere (cid:0) π , π (cid:1) , sufficientto determine uniquely p ( x ) and q ( x ) whole interval (0 , π ) .2. main result If ϕ ( x, λ ) a nontrivial solution of equation (1 .
1) with conditions (1 . . λ is called eigenvalue. Additionally, ϕ ( x, λ ) is called the eigenfunctionof the problem corresponding to the eigenvalue λ . { λ n } are eigenvalues ofthe problem. Lemma 2. If λ n = ˜ λ n , α ˜ α = β ˜ β then α = ˜ α and β = ˜ β for all n ∈ N .Proof. Since λ n = ˜ λ n and ∆ ( λ ) , ˜∆ ( λ )are entire functions in λ of order oneby Hadamard factorization theorem for λ ∈ C∆ ( λ ) ≡ C ˜∆ ( λ )On the other hand, (1 .
1) can be written as∆ ( λ ) − C ˜∆ ( λ ) = C h ˜∆ ( λ ) − ˜∆ ( λ ) i − [∆ ( λ ) − ∆ ( λ )]Hence C h ˜∆ ( λ ) − ˜∆ ( λ ) i − [∆ ( λ ) − ∆ ( λ )] = (cid:16) β +2 + γ β (cid:17) cos h λk + ( π ) − w ( π ) β i + (cid:16) β − + γ β (cid:17) cos h λk − ( π ) − w ( π ) β i + (cid:16) β − − γ β (cid:17) cos h λs + ( π ) + w ( π ) β i + (cid:16) β +2 − γ β (cid:17) cos h λs − ( π ) + w ( π ) β i − C (cid:16) ˜ β +2 + ˜ γ β (cid:17) cos h λk + ( π ) − ˜ w ( π )˜ β i − C (cid:16) ˜ β − + ˜ γ β (cid:17) cos h λk − ( π ) − ˜ w ( π )˜ β i − C (cid:16) ˜ β − − ˜ γ β (cid:17) cos h λs + ( π ) + ˜ w ( π )˜ β i − C (cid:16) ˜ β +2 − ˜ γ β (cid:17) cos h λs − ( π ) + ˜ w ( π ) β i (2.1)If we multiply both sides of (2 .
1) by cos h λk + ( π ) − w ( π ) β i and integrate withrespect to λ in ( ε, T ), ( ε is sufficiently small positive number) for any positivereal number T , then we get ALF-INVERSE PROBLEM 5 R Tε (cid:16) C h ˜∆ ( λ ) − ˜∆ ( λ ) i − [∆ ( λ ) − ∆ ( λ )] (cid:17) cos h λk + ( π ) − w ( π ) β i dλ =+ R Tε n(cid:16) β +2 + γ β (cid:17) cos h λk + ( π ) − w ( π ) β i + (cid:16) β − + γ β (cid:17) cos h λk − ( π ) − w ( π ) β i + (cid:16) β − − γ β (cid:17) cos h λs + ( π ) + w ( π ) β i + (cid:16) β +2 − γ β (cid:17) cos h λs − ( π ) + w ( π ) β i − C (cid:16) ˜ β +2 + ˜ γ β (cid:17) cos h λk + ( π ) − ˜ w ( π )˜ β i − C (cid:16) ˜ β − + ˜ γ β (cid:17) cos h λk − ( π ) − ˜ w ( π )˜ β i − C (cid:16) ˜ β − − ˜ γ β (cid:17) cos h λs + ( π ) + ˜ w ( π )˜ β i − C (cid:16) ˜ β +2 − ˜ γ β (cid:17) cos h λs − ( π ) + ˜ w ( π ) β io dλ And so R Tε (cid:16) C h ˜∆ ( λ ) − ˜∆ ( λ ) i − [∆ ( λ ) − ∆ ( λ )] (cid:17) cos h λk + ( π ) − w ( π ) β i dλ = R Tε (cid:16) β +2 + γ β (cid:17) cos h λk + ( π ) − w ( π ) β i dλ − C R Tε (cid:16) ˜ β +2 + ˜ γ β (cid:17) cos h λk + ( π ) − w ( π ) β i cos h λk + ( π ) − ˜ w ( π )˜ β i dλ = R Tε (cid:16) β +2 + γ β (cid:17) + (cid:16) β +2 + γ β (cid:17) cos h λk + ( π ) − w ( π ) β i dλ − C R Tε (cid:16) ˜ β +2 + ˜ γ β (cid:17) (cid:16) cos h λk + ( π ) − ˜ w ( π )+ w ( π ) β i + cos h w ( π ) − ˜ w ( π )˜ β i(cid:17) dλ ∆ ( λ ) − ∆ ( λ ) = O (cid:16) | λ | e | Imλ | k + ( π ) (cid:17) , ˜∆ ( λ ) − ˜∆ ( λ ) = O (cid:16) | λ | e | Imλ | k + ( π ) (cid:17) forall λ in ( ε, T ). C (cid:18) ˜ β +2 + ˜ γ β (cid:19) − (cid:18) β +2 + γ β (cid:19) = O (cid:18) T (cid:19) By letting T tend to infinity we see that C = ˜ β +2 + ˜ γ β β +2 + γ β Similarly, if we multiply both side of (2 .
1) cos h λk − ( π ) − w ( π ) β i and integrateagain with respect to λ in ( ε, T ) and by letting T tend to infinity, then we get C = ˜ β − + ˜ γ β β − + γ β But since α, β and ˜ α, ˜ β are positive, since w + ( π ) − ˜ w + ( π ) = w − ( π ) − ˜ w − ( π )we conclude that C = 1. Hence ˜ β +2 β +2 = ˜ β − β − is obtained. We have thereforeproved since α = ˜ α that β = ˜ β .The proof is completed. (cid:3) Lemma 3. If λ n = ˜ λ n then α i = ˜ α i and γ i = ˜ γ i ( i = 1 , for all n ∈ N .The proof is done as in [5]. ABDULLAH ERG ¨UN
Theorem 1.
Let { λ n } a eigenvalues of both problem (1 . − (1 . and (1 . − (1 . . If p ( x ) = ˜ p ( x ) and q ( x ) = ˜ q ( x ) on (cid:2) π , π (cid:3) , then p ( x ) = ˜ p ( x ) and q ( x ) = ˜ q ( x ) almost everywhere on [0 , π ] .Proof of Theorem 1. Let function ϕ ( x, λ ) the solution of equation (1 .
1) underthe conditions (1 . − (1 .
6) and the function ˜ ϕ ( x, λ ) the solution of equation(1 .
7) under the conditions (1 . − (1 . (cid:2) , π (cid:3) . The integral forms of thefunctions ϕ ( x, λ ) and ˜ ϕ ( x, λ ) can be obtained as follows ϕ ( x, λ ) = (cid:0) β +1 + γ α (cid:1) cos h λξ + ( x ) − α R xa p ( t ) dt i + (cid:0) β − − γ α (cid:1) cos h λξ − ( x ) + α R xa p ( t ) dt i + R x M ( x, t ) cos λtdt + R x N ( x, t ) sin λtdt (2.2)and˜ ϕ ( x, λ ) = (cid:16) ˜ β +1 + ˜ γ α (cid:17) cos h λξ + ( x ) − α R xa ˜ p ( t ) dt i + (cid:16) ˜ β − − ˜ γ α (cid:17) cos h λξ − ( x ) + α R xa ˜ p ( t ) dt i + R x ˜ M ( x, t ) cos λtdt + R x ˜ N ( x, t ) sin λtdt (2.3)If we multiply equations (2 .
2) and (2 . ϕ ( x, λ ) · ˜ ϕ ( x, λ ) = S + ˜ S + [cos (2 λξ + ( x ) − K ( x )) + cos L ( x )]+ S + ˜ S − [cos (2 λa t − L ( x )) + cos (2 λα ( x − a ) − K ( x ))]+ S − ˜ S + [cos (2 λa + L ( x )) + cos (2 λα ( x − a ) + K ( x ))]+ S − ˜ S − [cos (2 λξ − ( x ) + L ( x )) + cos K ( x )]+ S + R x ˜ M ( x, t ) cos h λξ + ( x ) − t ( x ) α i cos λtdt + S + R x ˜ N ( x, t ) cos h λξ + ( x ) − t ( x ) α i sin λtdt + S − R x ˜ M ( x, t ) cos h λξ − ( x ) + t ( x ) α i cos λtdt + S − R x ˜ N ( x, t ) cos h λξ − ( x ) + t ( x ) α i sin λtdt + ˜ S + R x M ( x, t ) cos h λξ + ( x ) − ˜ t ( x ) α i cos λtdt + ˜ S + R x N ( x, t ) cos h λξ + ( x ) − ˜ t ( x ) α i sin λtdt + ˜ S − R x M ( x, t ) cos h λξ − ( x ) + ˜ t ( x ) α i cos λtdt + ˜ S − R x N ( x, t ) cos h λξ − ( x ) + ˜ t ( x ) α i sin λtdt + (cid:0)R x M ( x, t ) cos λtdt (cid:1) (cid:16)R x ˜ M ( x, t ) cos λtdt (cid:17) + (cid:0)R x N ( x, t ) sin λtdt (cid:1) (cid:16)R x ˜ N ( x, t ) sin λtdt (cid:17) + (cid:0)R x M ( x, t ) cos λtdt (cid:1) (cid:16)R x ˜ N ( x, t ) sin λtdt (cid:17) + (cid:16)R x ˜ M ( x, t ) cos λtdt (cid:17) (cid:0)R x N ( x, t ) sin λtdt (cid:1) ALF-INVERSE PROBLEM 7 ϕ ( x, λ ) · ˜ ϕ ( x, λ ) = S + ˜ S + [cos (2 λξ + ( x ) − K ( x )) + cos L ( x )]+ S + ˜ S − [cos (2 λa t − L ( x )) + cos (2 λα ( x − a ) − K ( x ))]+ S − ˜ S + [cos (2 λa + L ( x )) + cos (2 λα ( x − a ) + K ( x ))]+ S − ˜ S − [cos (2 λξ − ( x ) + L ( x )) + cos K ( x )]+ (cid:8)R x U c ( x, t ) cos (2 λt − K ( t )) dt − R x U s ( x, t ) sin (2 λt − K ( t )) dt (cid:9) (2.4)is obtained, being S ± = (cid:0) β ± ∓ γ α (cid:1) , ˜ S ± = (cid:16) ˜ β ± ∓ ˜ γ α (cid:17) , K ( x ) = t ( x )+˜ t ( x )2 , L ( x ) = t ( x ) − ˜ t ( x )2 , U c ( x, t ) = S + ˜ M ( x, ξ + ( x ) − t ) cos (cid:16) K ( t ) − t ( x ) α (cid:17) + S − ˜ M ( x, ξ − ( x ) − t ) cos (cid:16) K ( t ) − t ( x ) α (cid:17) + ˜ S + M ( x, ξ + ( x ) − t ) cos (cid:16) K ( t ) − ˜ t ( x ) α (cid:17) + ˜ S − M ( x, ξ − ( x ) − t ) sin (cid:16) K ( t ) − ˜ t ( x ) α (cid:17) − S − ˜ N ( x, ξ + ( x ) − t ) sin (cid:16) K ( t ) − t ( x ) α (cid:17) − S − ˜ N ( x, ξ − ( x ) − t ) sin (cid:16) K ( t ) − t ( x ) α (cid:17) − ˜ S + N ( x, ξ + ( x ) − t ) sin (cid:16) K ( t ) − ˜ t ( x ) α (cid:17) − ˜ S − N ( x, ξ − ( x ) − t ) sin (cid:16) K ( t ) − ˜ t ( x ) α (cid:17) + K ( x, t ) cos K ( t ) + K ( x, t ) cos K ( t )+ M ( x, t ) sin K ( t ) + M ( x, t ) sin K ( t ) U s ( x, t ) = S + ˜ M ( x, ξ + ( x ) − t ) sin (cid:16) K ( t ) − t ( x ) α (cid:17) + S − ˜ M ( x, ξ − ( x ) − t ) sin (cid:16) K ( t ) − t ( x ) α (cid:17) + ˜ S + M ( x, ξ + ( x ) − t ) sin (cid:16) K ( t ) − ˜ t ( x ) α (cid:17) + ˜ S − M ( x, ξ − ( x ) − t ) sin (cid:16) K ( t ) − ˜ t ( x ) α (cid:17) + S + ˜ N ( x, ξ + ( x ) − t ) cos (cid:16) K ( t ) − t ( x ) α (cid:17) + S − ˜ N ( x, ξ − ( x ) − t ) cos (cid:16) K ( t ) − t ( x ) α (cid:17) + ˜ S + N ( x, ξ + ( x ) − t ) cos (cid:16) K ( t ) − ˜ t ( x ) α (cid:17) + ˜ S − N ( x, ξ − ( x ) − t ) cos (cid:16) K ( t ) − ˜ t ( x ) α (cid:17) + K ( x, t ) sin K ( t ) + K ( x, t ) sin K ( t ) − M ( x, t ) cos K ( t ) − M ( x, t ) cos K ( t ) K ( x, t ) = Z x − t − x M ( x, s ) ˜ M ( x, s + 2 t ) ds + Z x t − x M ( x, s ) ˜ M ( x, s + 2 t ) ds ABDULLAH ERG ¨UN K ( x, t ) = Z x − t − x N ( x, s ) ˜ N ( x, s + 2 t ) ds + Z x t − x n ( x, s ) ˜ N ( x, s + 2 t ) dsM ( x, t ) = Z x − t − x M ( x, s ) ˜ N ( x, s + 2 t ) ds − Z x t − x M ( x, s ) ˜ N ( x, s + 2 t ) dsM ( x, t ) = − Z x − t − x N ( x, s ) ˜ M ( x, s + 2 t ) ds + Z x t − x N ( x, s ) ˜ M ( x, s + 2 t ) ds Let ϕ ( x, λ ) and ˜ ϕ ( x, λ ) are substituted into (1 .
1) and (1 . − ϕ ′′ ( x, λ ) + (2 λp ( x ) + q ( x )) ϕ ( x, λ ) = λ ρ ( x ) ϕ ( x, λ ) (2.5) − ˜ ϕ ′′ ( x, λ ) + (2 λp ( x ) + q ( x )) ˜ ϕ ( x, λ ) = λ ρ ( x ) ˜ ϕ ( x, λ ) (2.6)The following equations is obtained (2 .
5) and (2 . R π ϕ ( x, λ ) ˜ ϕ ( x, λ ) [2 λ ( p ( x ) − ˜ p ( x )) + ( q ( x ) − ˜ q ( x ))] dx = [ ˜ ϕ ′ ( x, λ ) ϕ ( x, λ ) − ϕ ′ ( x, λ ) ˜ ϕ ( x, λ )] π + | π π R π ϕ ( x, λ ) ˜ ϕ ( x, λ ) [2 λ ( p ( x ) − ˜ p ( x )) + ( q ( x ) − ˜ q ( x ))] dx + ˜ ϕ ′ ( π, λ ) ϕ ( π, λ ) − ϕ ′ ( π, λ ) ˜ ϕ ( π, λ ) = 0 (2.7)Let Q ( x ) = q ( x ) − ˜ q ( x ) and P ( x ) = p ( x ) − ˜ p ( x ) U ( λ ) = Z π [2 λP ( x ) + Q ( x )] ϕ ( x, λ ) ˜ ϕ ( x, λ ) dx It is obvious that the functions ϕ ( x, λ ) and ˜ ϕ ( x, λ )are the solutions whichsatisfy boundary value conditions of (1 .
2) and (1 . . U ( λ n ) = 0 (2.8)for each eigenvalue λ n . Let us marked U ( λ ) = Z π P ( x ) ϕ ( x, λ ) ˜ ϕ ( x, λ ) dx, U ( λ ) = Z π Q ( x ) ϕ ( x, λ ) ˜ ϕ ( x, λ ) dx Then equations (2 .
7) can be rewritten as2 λ n U ( λ n ) + U ( λ n ) = 0 . From (2 .
4) ve (2 .
7) we obtain | U ( λ ) | ≤ ( C + C | λ | ) exp ( τ π ) (2.9) C , C > λ . Because λ n = ˜ λ n , ∆ ( λ ) = ϕ ( π, λ ) = ˜ ϕ ( π, λ ). Thus, U ( λ ) = Z π [2 λP ( x ) + Q ( x )] ϕ ( x, λ ) ˜ ϕ ( x, λ ) dx = ∆ ( λ ) [ ϕ ( π, λ ) − ˜ ϕ ( π, λ )] . The function φ ( λ ) = U ( λ )∆( λ ) is an entire function with respect to λ . ALF-INVERSE PROBLEM 9
It follows from ∆ ( λ ) ≥ ( | λβ | − C ) exp ( τ ξ + ( x )) and (2 . φ ( λ ) = O (1) forsufficient large | λ | . We obtain φ ( λ ) = C , for all λ by Liouville’s Theorem. U ( λ ) = C ∆ ( λ ) R π ϕ ( x, λ ) ˜ ϕ ( x, λ ) [2 λP ( x ) + Q ( x )] dx == C h(cid:16) β +2 + γ β (cid:17) R ( a ) cos h λk + ( π ) − β R πa p ( t ) dt i + (cid:16) β − + γ β (cid:17) R ( a ) cos h λk − ( π ) − β R πa p ( t ) dt i + (cid:16) β − − γ β (cid:17) R ( a ) cos h λs + ( π ) + β R πa p ( t ) dt i + (cid:16) β +2 − γ β (cid:17) R ( a ) cos h λs − ( π ) + β R πa p ( t ) dt ii + O (exp ( τ k + ( π )))By the Riemann-Lebesgue lemma, for λ → ∞ , λ ∈ R we get C = 0. Then,2 U ( λ ) = S + ˜ S + R π P ( x ) cos (2 λξ + ( x ) − K ( x )) dx + S + ˜ S + R π P ( x ) cos L ( x ) dx + S + ˜ S − R π P ( x ) cos (2 λa t − L ( x )) dx + S + ˜ S − R π P ( x ) cos (2 λα ( x − a ) − K ( x )) dx + S − ˜ S + R π P ( x ) cos (2 λa + L ( x )) dx + S − ˜ S + R π P ( x ) cos cos (2 λα ( x − a ) + K ( x )) dx + S − ˜ S − R π P ( x ) cos (2 λξ − ( x ) + L ( x )) dx + S − ˜ S − R π P ( x ) cos K ( x ) dx + R π P ( x ) (cid:0)R x U c ( x, t ) cos (2 λt − K ( t )) dt (cid:1) dx − R π P ( x ) (cid:0)R x U s ( x, t ) sin (2 λt − K ( t )) dt (cid:1) dx. where ξ ± ( x ) = ± αx ∓ αa + a , k ± ( x ) = µ + ( a ) ± βx ∓ βa , s ± ( x ) = µ − ( a ) ± βx ∓ βa , β ∓ = (cid:16) α ∓ β α (cid:17) , β ∓ = (cid:16) α ∓ αβ β (cid:17) . U ( λ ) = S + ˜ S + R π P ( t ) e − i ( K ( t )) e i ( λξ + ( t ) ) dt + S + ˜ S + R π P ( t ) e i ( K ( t )) e − i ( λξ + ( t ) ) dt + S + ˜ S − R π P ( t ) e − i ( L ( t )) e i (2 λa t ) dt + S + ˜ S − R π P ( t ) e i ( L ( t )) e − i (2 λa t ) dt + S + ˜ S − R π P ( t ) e − i ( K ( t )) e i (2 λα ( t − a )) dt + + S + ˜ S − R π P ( t ) e i ( K ( t )) e − i (2 λα ( t − a )) dt + S − ˜ S + R π P ( t ) e i ( L ( t )) e i (2 λa t ) dt + S − ˜ S + R π P ( t ) e − i ( L ( t )) e i (2 λa t ) dt + S − ˜ S + R π P ( t ) e i ( K ( t )) e i (2 λα ( t − a )) dt + S − ˜ S + R π P ( t ) e − i ( K ( t )) e i (2 λα ( t − a )) dt + S − ˜ S − R π P ( t ) e i ( L ( t )) e − i ( λξ − ( t ) ) dt + S − ˜ S − R π P ( t ) e − i ( L ( t )) e i ( λξ − ( t ) ) dt + S + ˜ S + R π P ( x ) cos L ( x ) dx + S − ˜ S − R π P ( x ) cos K ( x ) dx + R π P ( x ) (cid:0)R x U c ( x, t ) cos (2 λt − K ( t )) dt (cid:1) dx − R π P ( x ) (cid:0)R x U s ( x, t ) sin (2 λt − K ( t )) dt (cid:1) dx if necessary operations are performed and integrals are calculated U ( λ ) = S + ˜ S + (cid:20) T ( π / iλα e i ( λξ + ( π )) − T (0)2 iλα e iλ ( αa + a ) − iλα R π T ′ ( t ) e i ( λξ + ( t ) ) dt (cid:21) + S + ˜ S + (cid:20) − T ( π / iλα e − i ( λξ + ( π )) + T (0)2 iλα e − iλ ( αa + a ) + iλα R π T ′ ( t ) e − i ( λξ + ( t ) ) dt (cid:21) + S + ˜ S − (cid:20) T ( π / iλα e iλa − T (0)2 iλα − iλα R π T ′ ( t ) e ia t dt (cid:21) + S + ˜ S − (cid:20) − T ( π / iλα e iλa + T (0)2 iλα + iλα R π T ′ ( t ) e − ia t dt (cid:21) + S + S − (cid:20) T ( π / iλα e iλα ( π − a ) − T (0)2 iλα e − iλαa − iλα R π T ′ ( t ) e iλα ( t − a ) dt (cid:21) + S + S − (cid:20) − T ( π / iλα e − iλα ( π − a ) + T (0)2 iλα e iλαa + iλα R π T ′ ( t ) e − iλα ( t − a ) dt (cid:21) + S − ˜ S + (cid:20) − T ( π / iλα e − iλa π + T (0)2 iλα + iλα R π T ′ ( t ) e − ia t dt (cid:21) + S − ˜ S + (cid:20) T ( π / iλα e iλa π − T (0)2 iλα − iλα R π T ′ ( t ) e ia t dt (cid:21) + S − ˜ S + (cid:20) − T ( π / iλα e − iλα ( π − a ) + T (0)2 iλα e iλαa + iλα R π T ′ ( t ) e − iλα ( t − a ) dt (cid:21) + S − ˜ S + (cid:20) T ( π / iλα e iλα ( π − a ) − T (0)2 iλα e − iλαa − iλα R π T ′ ( t ) e iλα ( t − a ) dt (cid:21) + S − ˜ S − (cid:20) − T ( π / iλα e i ( λξ − ( π )) + T (0)2 iλα e iλ ( αa + a ) + iλα R π T ′ ( t ) e i ( λξ − ( t ) ) dt (cid:21) + S − ˜ S − (cid:20) T ( π / iλα e − i ( λξ − ( π )) − T (0)2 iλα e iλ ( αa − a ) − iλα R π T ′ ( t ) e − i ( λξ + ( t ) ) dt (cid:21) + S + ˜ S + R π P ( x ) cos L ( x ) dx + S − ˜ S − R π P ( x ) cos K ( x ) dx + (cid:20) T ( π / iλ e iπλ − T (0)2 iλ − iλ R π T ′ ( t ) e iλt dt (cid:21) + (cid:20) − T ( π / iλ e − iπλ + T (0)2 iλ + iλ R π T ′ ( t ) e − iλt dt (cid:21) where T ( t ) = P ( t ) e − i ( K ( t )) , T ( t ) = P ( t ) e i ( K ( t )) , T ( t ) = P ( t ) e − i ( L ( t )) , T ( t ) = P ( t ) e i ( L ( t )) , P ( t ) = R π t P ( x ) U c ( x, t ) dx, P ( t ) = R π t P ( x ) U s ( x, t ) dx, T ( t ) = P ( t )+ iP ( t )2 e − iK ( t ) , T ( t ) = P ( t ) − iP ( t )2 e iK ( t ) By the Riemann-Lebesgue lemma R π P ( x ) cos L ( x ) dx = 0 , R π P ( x ) cos K ( x ) dx = 0 and P (cid:0) π (cid:1) = 0 for λ → ∞ . ALF-INVERSE PROBLEM 11
Thus,2 U ( λ ) = − S + ˜ S + iλα R π T ′ ( t ) e i ( λξ + ( t ) ) dt + S + ˜ S + iλα R π T ′ ( t ) e − i ( λξ + ( t ) ) dt − S + ˜ S − iλα R π T ′ ( t ) e ia t dt + S + ˜ S − iλα R π T ′ ( t ) e − ia t dt − S + S − iλα R π T ′ ( t ) e iλα ( t − a ) dt + S + S − iλα R π T ′ ( t ) e − iλα ( t − a ) dt + S − ˜ S + iλα R π T ′ ( t ) e − ia t dt − S − ˜ S + iλα R π T ′ ( t ) e ia t dt + S − ˜ S + iλα R π T ′ ( t ) e − iλα ( t − a ) dt + S − ˜ S + iλα R π T ′ ( t ) e iλα ( t − a ) dt + S − ˜ S − iλα R π T ′ ( t ) e i ( λξ − ( t ) ) dt − S − ˜ S − iλα R π T ′ ( t ) e − i ( λξ − ( t ) ) dt + i λ R π T ′ ( t ) e iλt dt − i λ R π T ′ ( t ) e − iλt dt (2.10)2 U ( λ ) = S + ˜ S + R π Q ( x ) (cid:18) e i ( λξ +( x ) − K ( x ) ) + e − i ( λξ +( x ) − K ( x ) ) (cid:19) dx + S + ˜ S − R π Q ( x ) (cid:16) e i (2 λa t − L ( x )) + e − i (2 λa t − L ( x )) (cid:17) dx + S + ˜ S − R π Q ( x ) (cid:16) e i (2 λα ( x − a − K ( x )) + e − i (2 λα ( x − a − K ( x )) (cid:17) dx + S − ˜ S + R π Q ( x ) (cid:16) e i (2 λa t + L ( x )) + e − i (2 λa t + L ( x )) (cid:17) dx + S − ˜ S + R π Q ( x ) (cid:16) e i (2 λα ( x − a K ( x )) + e − i (2 λα ( x − a K ( x )) (cid:17) dx + S − ˜ S − R π Q ( x ) (cid:18) e i ( λξ − ( x )+ L ( x ) ) + e − i ( λξ − ( x )+ L ( x ) ) (cid:19) dx + S + ˜ S + R π Q ( x ) cos L ( x ) dx + S − ˜ S − R π Q ( x ) cos K ( x ) dx + R π Q ( x ) (cid:0)R x U c ( x, t ) cos (2 λt − K ( t )) dt (cid:1) dx − R π Q ( x ) (cid:0)R x U s ( x, t ) sin (2 λt − K ( t )) dt (cid:1) dx where R ( t ) = Q ( t ) e − i ( K ( t )) , R ( t ) = Q ( t ) e i ( K ( t )) , R ( t ) = Q ( t ) e − i ( L ( t )) , R ( t ) = Q ( t ) e i ( L ( t )) , Q ( t ) = R π t P ( x ) U c ( x, t ) dx,Q ( t ) = R π t P ( x ) U s ( x, t ) dx, R ( t ) = Q ( t )+ iQ ( t )2 e − iK ( t ) , R ( t ) = Q ( t ) − iQ ( t )2 e iK ( t ) By the Riemann-Lebesgue lemma R π Q ( x ) cos L ( x ) dx = 0 , R π Q ( x ) cos K ( x ) dx = 0. Thus,2 U ( λ ) = S + ˜ S + R π R ( t ) e i ( λξ + ( t ) ) dt + S + ˜ S + R π R ( t ) e − i ( λξ + ( t ) ) dt + S + ˜ S − R π R ( t ) e ia t dt + S + ˜ S − R π R ( t ) e − ia t dt + S + S − R π R ( t ) e iλα ( t − a ) dt + S + S − R π R ( t ) e − iλα ( t − a ) dt + S − ˜ S + R π R ( t ) e − ia t dt + S − ˜ S + R π R ( t ) e ia t dt + S − ˜ S + R π R ( t ) e − iλα ( t − a ) dt + S − ˜ S + R π R ( t ) e iλα ( t − a ) dt + S − ˜ S − R π R ( t ) e i ( λξ − ( t ) ) dt + S − ˜ S − R π R ( t ) e − i ( λξ − ( t ) ) dt + i λ R π R ( t ) e iλt dt + i λ R π R ( t ) e − iλt dt (2.11) λU ( λ ) + U ( λ ) = 0 . (2.12)If (2 .
10) and (2 .
11) are substituted into (2 . S + ˜ S + α R π ( R ( t ) + iT ′ ( t )) e i ( λξ + ( t ) ) dt + S + ˜ S + α R π ( R ( t ) − iT ′ ( t )) e − i ( λξ + ( t ) ) dt + S + ˜ S − α R π ( R ( t ) + iT ′ ( t )) e ia t dt + S + ˜ S − α R π ( R ( t ) − iT ′ ( t )) e − ia t dt + S + S − α R π ( R ( t ) + iT ′ ( t )) e iλα ( t − a ) dt + S + S − α R π ( R ( t ) − iT ′ ( t )) e − iλα ( t − a ) dt + S − ˜ S + α R π ( R ( t ) + iT ′ ( t )) e − ia t dt + S − ˜ S + α R π ( R ( t ) − iT ′ ( t )) e ia t dt + S − ˜ S + R π ( R ( t ) + iT ′ ( t )) e iλα ( t − a ) dt + S − ˜ S + R π ( R ( t ) − iT ′ ( t )) e − iλα ( t − a ) dt + S − ˜ S − R π ( R ( t ) − iT ′ ( t )) e i ( λξ − ( t ) ) dt + S − ˜ S − R π ( R ( t ) + iT ′ ( t )) e − i ( λξ − ( t ) ) dt + R π ( R ( t ) + iT ′ ( t )) e iλt dt + R π ( R ( t ) − iT ′ ( t )) e − iλt dt = 0Since the systems n e ± iλξ + ( t ) : λ ∈ R o , (cid:8) e ± iλa t : λ ∈ R (cid:9) , (cid:8) e ± iλα ( t − a ) : λ ∈ R (cid:9) and (cid:8) e ± iλt : λ ∈ R (cid:9) are entire in L (cid:0) − π , π (cid:1) , it follows R ( t ) + iT ′ ( t ) = 0 , R ( t ) − iT ′ ( t ) = 0 , R ( t ) + iT ′ ( t ) = 0 R ( t ) − iT ′ ( t ) = 0 , R ( t ) + iT ′ ( t ) = 0 , R ( t ) − iT ′ ( t ) = 0 R ( t ) + iT ′ ( t ) = 0 , R ( t ) − iT ′ ( t ) = 0 , R ( t ) + iT ′ ( t ) = 0 R ( t ) − iT ′ ( t ) = 0 , R ( t ) − iT ′ ( t ) = 0 , R ( t ) + iT ′ ( t ) = 0 R ( t ) + iT ′ ( t ) = 0 , R ( t ) − iT ′ ( t ) = 0Then, we get the following system. R ( t ) + iT ′ ( t ) = 0 R ( t ) − iT ′ ( t ) = 0and hence, h Q ( t ) + P ( t ) K ′ ( t ) − P ′ ( t ) i + i h Q ( t ) + P ( t ) K ′ ( t ) + P ′ ( t ) i = 0 h Q ( t ) + P ( t ) K ′ ( t ) − P ′ ( t ) i − i h Q ( t ) + P ( t ) K ′ ( t ) + P ′ ( t ) i = 0and hence, (cid:26) Q ( t ) + P ( t ) K ′ ( t ) − P ′ ( t ) = 0 Q ( t ) + P ( t ) K ′ ( t ) + P ′ ( t ) = 0 P ′ ( t ) = U c ( t, t ) P ( t ) − R π t U s ( x, t ) Q ( x ) dx − R π t (cid:16) K ′ ( t ) U s ( x, t ) + ∂H s ( x,t ) ∂t (cid:17) P ( x ) dxP ( t ) = − R π t P ′ ( x ) dxQ ( t ) = − ( K ′ ( t ) + U s ( t, t )) P ( t ) − R π t U c ( x, t ) Q ( x ) dx − R π t (cid:16) K ′ ( t ) U c ( x, t ) − ∂H s ( x,t ) ∂t (cid:17) P ( x ) dx (2.13) ALF-INVERSE PROBLEM 13
If we mark this S ( t ) = (cid:0) Q ( t ) , P ( t ) , P ′ ( t ) (cid:1) T and K ( x, t ) = U c ( x, t ) K ′ ( t ) U c ( x, t ) − ∂U s ( x,t ) ∂t − ( K ′ ( t ) + U s ( t, t ))0 0 1 U s ( x, t ) K ′ ( t ) U s ( x, t ) + ∂U s ( x,t ) ∂t U c ( x, t ) Equations (2 .
13) can be reduced to a vector from S ( t ) + Z π t K ( x, t ) S ( x ) dx = 0 (2.14)for 0 < t < π .Since the equation (2 . .
14) only has the trivial solution. Thus, we obtain S ( t ) = 0 for 0 < t < π .This gives us Q ( t ) = P ( t ) = 0 for 0 < t < π .Thus, we obtain q ( x ) = ˜ q ( x ) and p ( x ) = ˜ p ( x ) on (0 , π ). The proof is com-leted. (cid:3) Acknowledgement
Not applicable.
References [1] O. Acan and D. Baleanu, “A new numerical technique for solving fractional partialdifferential equations,”
Miskolc Mathematical Notes , vol. 19, no. 1, pp. 3–18, 2018.[2] D. Alpay and I. Gohberg, “Inverse problems associated to a canonical differential sys-tem,” in
Recent Advances in Operator Theory and Related Topics . Birkhauser Basel,2001, pp. 1–27.[3] R. K. Amirov and A. A. Nabiev, “Inverse Problems for the Quadratic Pencil of theSturm-Liouville Equations with Impulse,”
Abstract and Applied Analysis , vol. 2013, pp.1–10, 2013, doi: 10.1155/2013/361989.[4] R. Carlson, “An inverse spectral problem for Sturm-Liouville operators with discontin-uous coefficients,”
Proceedings of the American Mathematical Society , vol. 120, no. 2,pp. 475–475, feb 1994, doi: 10.1090/s0002-9939-1994-1197532-5.[5] A. Ergun and R. Amirov, “Direct and inverse problem for diffusion operator with dis-continuity points,”
TWMS J. App. Eng. Math. , vol. (9), pp. 9–21, 2019.[6] S. Gala, Q. Liu, and M. Ragusa, “A new regularity criterion for the nematic liquidcrystal flows,”
Applicable Analysis , vol. 91, no. 9, pp. 1741–1747, 2012.[7] S. Gala and M. Ragusa, “Logarithmically improved regularity criterion for the boussi-nesq equations in besov spaces with negative indices,”
Applicable Analysis , vol. 95, no. 6,pp. 1271–1279, 2016. [8] F. Gesztesy and B. Simon, “Inverse spectral analysis with partial information on thepotential ii: The case of discrete spectrum,”
Transactions of the American MathematicalSociety , vol. 352, no. 06, pp. 2765–2787, jun 2000, doi: 10.1090/s0002-9947-99-02544-1.[9] O. H. Hald, “Discontinuous inverse eigenvalue problems,”
Communications onPure and Applied Mathematics , vol. 37, no. 5, pp. 539–577, sep 1984, doi:10.1002/cpa.3160370502.[10] H.Hochstadt and B. Lieberman, “An Inverse Sturm–Liouville Problem with MixedGiven Data,”
SIAM Journal on Applied Mathematics , vol. 34, no. 4, pp. 676–680, jun1978, doi: 10.1137/0134054.[11] R. O. Hryniv and Y. V. Mykytyuk, “Half-inverse spectral problems for Sturm-Liouvilleoperators with singular potentials,”
Inverse Problems , vol. 20, no. 5, pp. 1423–1444, jul2004, doi: 10.1088/0266-5611/20/5/006.[12] M. Keldysh, “On the eigenvalues and eigenfunctions of some classes of nonselfadjointequations,”
Dokl. Akad. Nauk. SSSR , vol. 77, pp. 11–14, 1951.[13] H. Koyunbakan and E. S. Panakhov, “Half-inverse problem for diffusion operators onthe finite interval,”
Journal of Mathematical Analysis and Applications , vol. 326, no. 2,pp. 1024–1030, feb 2007, doi: 10.1016/j.jmaa.2006.03.068.[14] B. I. Levin,
Distribution of zeros of Entire Functions . American Mathematical Society,1964, vol. 424-436.[15] B. Levitan,
Inverse Sturm-Liouville problems . Netherlands: VNU Science Press., 1987.[16] A. S. Markus,
Introduction to the Spectral Theory of Polynomial Op-erator Pencils
Inverse Problems in Science and En-gineering , vol. 22, no. 5, pp. 848–859, sep 2013, doi: 10.1080/17415977.2013.832241.[18] L. Sakhnovich, “Half-inverse problems on the finite interval,”
Inverse Problems , vol. 17,no. 3, pp. 527–532, may 2001, doi: 10.1088/0266-5611/17/3/311.[19] G. Wei and H.-K. Xu, “On the missing eigenvalue problem for an inverse sturm–liouvilleproblem,”
Journal de Math´ematiques Pures et Appliqu´ees , vol. 91, no. 5, pp. 468–475,may 2009, doi: 10.1016/j.matpur.2009.01.007.[20] C.-F. Yang, “Reconstruction of the diffusion operator from nodal data,”
Zeitschrift frNaturforschung A , vol. 65, no. 1-2, pp. 100–106, jan 2010, doi: 10.1515/zna-2010-1-211.[21] C.-F. Yang and Z.-Y. Huang, “A half-inverse problem with eigenparameter dependentboundary conditions,”
Numerical Functional Analysis and Optimization , vol. 31, no. 6,pp. 754–762, jul 2010, doi: 10.1080/01630563.2010.490934.[22] C.-F. Yang and X.-P. Yang, “An interior inverse problem for the Sturm–Liouville op-erator with discontinuous conditions,”
Applied Mathematics Letters , vol. 22, no. 9, pp.1315–1319, sep 2009, doi: 10.1016/j.aml.2008.12.001.[23] C.-F. Yang and A. Zettl, “Half Inverse Problems For Quadratic Pencils of Sturm-Liouville Operators,”
Taiwanese Journal of Mathematics , vol. 16, no. 5, pp. 1829–1846,sep 2012, doi: 10.11650/twjm/1500406800.[24] V. A. Yurko,
Inverse Spectral Problems for Linear Differen-tial Operators and Their Applications
ALF-INVERSE PROBLEM 15 [25] R. Zhang, X.-C. Xu, C.-F. Yang, and N. Bondarenko, “Determination of the impul-sive Sturm-Liouville operator from a set of eigenvalues,”
J.Inverse and III-Posed Probl ,vol. 28, pp. 341–348, 2019.
Cumhuriyet University, Vocational School of Sivas, Sivas., 58140, Turkey
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