aa r X i v : . [ m a t h . SP ] J un INVERSE PROBLEM FOR SINGULAR DIFFUSIONOPERATOR
ABDULLAH ERG ¨UN
Abstract.
In this study, singular diffusion operator with jump condi-tions is considered. Integral representations have been derived for solutionsthat satisfy boundary conditions and jump conditions. Some properties ofeigenvalues and eigenfunctions are investigated. Asymtotic representationof eigenvalues and eigenfunction have been obtained. Reconstruction ofthe singular diffusion operator have been shown by the Weyl function. Introduction.
Let’s define the following boundary value problem which will be denoted by L in the sequel all the paper l ( y ) := − y ′′ + [2 λp ( x ) + q ( x )] y = λ δ ( x ) y, x ∈ [0 , π ] / { p , p } (1.1)with the boundary conditions y ′ (0) = 0 , y ( π ) = 0 (1.2)and the jump conditions y ( p + 0) = α y ( p −
0) (1.3) y ′ ( p + 0) = β y ′ ( p −
0) + iλγ y ( p −
0) (1.4) y ( p + 0) = α y ( p −
0) (1.5) y ′ ( p + 0) = β y ′ ( p −
0) + iλγ y ( p −
0) (1.6)where λ is a spectral parameter, q ( x ) ∈ L [0 , π ], p ( x ) ∈ W [0 , π ], p , p ∈ (0 , π ), p < p , | α − | + γ = 0, | α − | + γ = 0, (cid:16) β i = α i ( i = 1 , (cid:17) and δ ( x ) = , x ∈ (0 , p ) α , x ∈ ( p , p ) β , x ∈ ( p , π ) to be α > , α = 1, β > , β = 1 real numbers.Direct and inverse problems are important in mathematics, physics and engi-neering. The inverse problem is called the reconstruction of the operator whose Mathematics Subject Classification.
Key words and phrases.
Inverse problem, Sturm-Liouville, Diffusion operator, Integralrepresentation. spectral characteristics are given in sequences. For example; to learn the dis-tribution of density in the nonhomogeneous arc according to the wave lengthsin mechanics and finding the field potentials according to scattering data inthe quantum physics are examples of inverse problems. The first study oninverse problems for differential equations was made by Ambartsumyan [25].A significant study in the spectral theory of the singular differential operatorswas carried out by Levitan in [4]. An important method in the solution ofinverse problems is the transformation operators. Guseinov [14] studied theregular differential equation and the direct spectral problem of the operatorunder certain initial conditions. In recent years, Weyl function has frequentlybeen used to solve inverse problems. The Weyl function was introduced byH. Weyl(1910) in the literature. Many studies have been made on direct orinverse problems [1–29] . The solution of discontinuous boundary value prob-lem can be given as an example of concrete problem of mathematical physics.Boundary value problems with discontinuous coefficients are important forapplied mathematics and applied sciences.H. Koyunbakan, E. S. Panakhov [17] proved that the potential function canbe determined on (cid:2) π , π (cid:3) while it is known on (cid:2) , π (cid:3) by single spectrum in[11].C. F. Yang [27] showed that can be determined uniquely diffusion operatorfrom nodal data. 2. Preliminaries.
Let φ ( x, λ ), ψ ( x, λ ) be solutions of (1 .
1) respectively under the boundaryconditions φ (0 , λ ) = 1 , φ ′ (0 , λ ) = 0 ψ ( π, λ ) = 0 , ψ ′ ( π, λ ) = 1and discontinuity conditions (1 . − (1 . Q ( t ) = 2 λp ( t ) + q ( t ) . It is obvious that the function φ ( x, λ ) is similar to [8] satisfies the followingintegral equationsif 0 ≤ x < p ; φ ( x, λ ) = e iλx + 1 λ Z x sin λ ( x − t ) Q ( t ) y ( t, λ ) dt (2.1)if p < x < p ; φ ( x, λ ) = β +1 e iλς + ( x ) + β − e iλς − ( x ) + γ α e iλς + ( x ) − γ α e iλς − ( x ) + β +1 R p λ ( ς + ( x ) − t ) λ J ( t ) y ( t, λ ) dt + β − R p λ ( ς − ( x ) − t ) λ J ( t ) y ( t, λ ) dt − i γ α R p λ ( ς + ( x ) − t ) λ J ( t ) y ( t, λ ) dt + + i γ α R p λ ( ς − ( x ) − t ) λ J ( t ) y ( t, λ ) dt + R xp sin λ ( x − t ) λ J ( t ) y ( t, λ ) dt (2.2)if p < x ≤ π ; NVERSE PROBLEM FOR SINGULAR DIFFUSION OPERATOR 3 φ ( x, λ ) = ξ + e iλb + ( x ) + ξ − e iλb − ( x ) + ϑ + e iλs + ( x ) + ϑ − e iλs − ( x ) + (cid:16) β +1 β +2 + γ γ αβ (cid:17) R p λ ( b + ( x ) − t ) λ J ( t ) y ( t, λ ) dt + (cid:16) β +1 β − − γ γ αβ (cid:17) R p λ ( s + ( x ) − t ) λ J ( t ) y ( t, λ ) dt + (cid:16) β − β − − γ γ αβ (cid:17) R p λ ( b − ( x ) − t ) λ J ( t ) y ( t, λ ) dt + (cid:16) β − β +2 + γ γ αβ (cid:17) R p p sin λ ( s − ( x ) − t ) λ J ( t ) y ( t, λ ) dt − i (cid:16) γ β +2 α + γ β +1 β (cid:17) R p λ ( b + ( x ) − t ) λ J ( t ) y ( t, λ ) dt − i (cid:16) γ β − α − γ β +1 β (cid:17) R p λ ( s + ( x ) − t ) λ J ( t ) y ( t, λ ) dt + i (cid:16) γ β − α − γ β − β (cid:17) R p λ ( b − ( x ) − t ) λ J ( t ) y ( t, λ ) dt + i (cid:16) γ β +2 α + γ β − β (cid:17) R p p cos λ ( s − ( x ) − t ) λ J ( t ) y ( t, λ ) dt + β +2 R p p sin λ ( βx − βp + αp − αt ) λ J ( t ) y ( t, λ ) dt − β − R p p sin λ ( βx − βp − αp + αt ) λ J ( t ) y ( t, λ ) dt − i γ β R p p cos λ ( βx − βp + αp − αt ) λ J ( t ) y ( t, λ ) dt + i γ β R p p cos λ ( βx − βp − αp + αt ) λ J ( t ) y ( t, λ ) dt + R xp sin λ ( x − t ) λ J ( t ) y ( t, λ ) dt (2.3)and it is obvious that the function ψ ( x, λ ) satisfies the following integral equa-tions;if p < x ≤ π , ψ ( x, λ ) = sin λβ ( x − π ) λβ + Z πx sin λβ ( x − t ) λβ Q ( t ) y ( t, λ ) dt (2.4) ABDULLAH ERG ¨UN if p < x < p ; ψ ( x, λ ) = (cid:16) αβ − γ αβ λα β − αβ λ (cid:17) e − iλ ( β ( p − π )+ α ( p − x )) + (cid:16) αβ + γ αβ λα β + αβ λ (cid:17) e − iλ ( β ( p − π ) − α ( p − x )) − (cid:16) αβ − γ αβ − (cid:17) R p p sin λ ( x − p + αt − αp ) λα Q ( t ) y ( t, λ ) dt + (cid:16) αβ − γ αβ + (cid:17) R p p sin λ ( x − p − αt + αp ) λα Q ( t ) y ( t, λ ) dt + (cid:16) αβ − γ αβ α β − αβ (cid:17) R πp sin λ ( x − p + β ( t − p )) λβ Q ( t ) y ( t, λ ) dt − (cid:16) αβ − γ αβ α β − αβ (cid:17) R πp sin λ ( x − p − β ( t − p )) λβ Q ( t ) y ( t, λ ) dt + γ αβ λ R p p cos λ ( x − p + αt − αp ) λα Q ( t ) y ( t, λ ) dt − γ αβ λ R p p cos λ ( x − p − αt + αp ) λα Q ( t ) y ( t, λ ) dt + R xp sin λα ( x − t ) λα Q ( t ) y ( t, λ ) dt (2.5)if 0 ≤ x < p ; ψ ( x, λ ) = (cid:16) ξ + + α β (cid:17) η − e − iλ ( b − ( π )+ x ) + (cid:16) ξ − − α β (cid:17) η + e − iλ ( b + ( π )+ x )+ (cid:16) ξ − + α β (cid:17) η − e − iλ ( s + ( π )+ x ) + (cid:16) ξ − − α β (cid:17) η + e − iλ ( s − ( π )+ x )+ (cid:16) α − µ + β (cid:17) R πa sin λ ( x − p − βt + βp ) λ Q ( t ) y ( t, λ ) dt − (cid:16) α + µ + β (cid:17) R πp sin λ ( x − p + p + βt − βp ) λ Q ( t ) y ( t, λ ) dt + (cid:16) α + µ − β (cid:17) R πp sin λ ( x − p − βt + βp ) λ Q ( t ) y ( t, λ ) dt − (cid:16) α − µ − β (cid:17) R πp sin λ ( x − p + p − βt + βp ) λ Q ( t ) y ( t, λ ) dt + iγ α β R πp cos λ ( x − p + βt − βp ) λ Q ( t ) y ( t, λ ) dt − iγ α β R πp cos λ ( x − p + p − βt + βp ) λ Q ( t ) y ( t, λ ) dt − iγ α β R πp cos λ ( x − p − βt + βp ) λ Q ( t ) y ( t, λ ) dt + iγ α β R πp cos λ ( x − p + p + βt − βp ) λ Q ( t ) y ( t, λ ) dt + A R p p sin λ ( x − p + αt − αp ) λα Q ( t ) y ( t, λ ) dt + A R p p sin λ ( x − p + p − αt + αp ) λα Q ( t ) y ( t, λ ) dt + B R p p cos λ ( x − p + αt − αp ) λα Q ( t ) y ( t, λ ) dt + B R p p cos λ ( x − p + p − αt + αp ) λα Q ( t ) y ( t, λ ) dt + C R p p sin λ ( x − p − αt + αp ) λα Q ( t ) y ( t, λ ) dt + C R p p sin λ ( x − p + p + αt − αp ) λα Q ( t ) y ( t, λ ) dt + D R p p cos λ ( x − p − αt + αp ) λα Q ( t ) y ( t, λ ) dt + D R p p cos λ ( x − p + p + αt − αp ) λα Q ( t ) y ( t, λ ) dt + R x λ ( x − t ) λ Q ( t ) y ( t, λ ) dt (2.6) NVERSE PROBLEM FOR SINGULAR DIFFUSION OPERATOR 5 where ς ± ( x ) = ± αx ∓ αp + p , β ± = (cid:16) α ± β α (cid:17) ,b ± ( x ) = βx − βp + µ ± ( p ) , s ± ( x ) = − βx + βp + µ ± ( p ) ,β ∓ = 12 (cid:18) α ∓ αβ β (cid:19) , ξ ∓ = 12 (cid:16) β ∓ ∓ γ α (cid:17) (cid:18) α ∓ αβ β + γ β (cid:19) ,ϑ ∓ = 12 (cid:16) β ∓ ∓ γ α (cid:17) (cid:18) α ± αβ β − γ β (cid:19) , µ ± = (cid:18) αβ ± γ αβ λα β ± αβ λ (cid:19) ,A = h(cid:16) iγ γ λαα β β + (cid:16) − α − β (cid:17) (cid:16) αβ − γ αβ − (cid:17)(cid:17)i ,B = h − iγ α β (cid:16) αβ − γ αβ − (cid:17) + α γ αβ λ i ,C = h(cid:16) α + β (cid:17) (cid:16) αβ − γ αβ + (cid:17) + iγ γ λαα β β i ,D = (cid:20) iγ α β (cid:16) αβ − γ αβ + (cid:17) − γ ( − α ) α αβ λ (cid:21) . Theorem 1. If p ( x ) ∈ W (0 , π ) and q ( x ) ∈ L (0 , π ) ; y υ ( x, λ ) be solutionsof (1 . , that satisfies conditions (1 . − (1 . , has the form y υ ( x, λ ) = y υ ( x, λ ) + Z x − x K υ ( x, t ) e iλt dt (cid:0) υ = 1 , (cid:1) where y υ ( x, λ ) = R ( x ) e iλx ; 0 ≤ x < p R ( x ) e iλς + ( x ) + R ( x ) e iλς − ( x ) ; p < x < p R ( x ) e iλb + ( x ) + R ( x ) e iλb − ( x ) + R ( x ) e iλs + ( x ) + R ( x ) e iλs − ( x ) ; p < x ≤ πR ( x ) = e − i R x p ( x ) dx , R ( x ) = (cid:16) β +1 + γ α (cid:17) R ( p ) e − iα R xp p ( t ) dt ,R ( x ) = (cid:16) β − − γ α (cid:17) R ( p ) e iα R xp p ( t ) dt , R ( x ) = (cid:18) β +2 + γ β (cid:19) R ( p ) e − iβ R xp p ( t ) dt ,R ( x ) = (cid:18) β − + γ β (cid:19) R ( p ) e − iβ R xp p ( t ) dt , R ( x ) = (cid:18) β − − γ β (cid:19) R ( p ) e iβ R xp p ( t ) dt ,R ( x ) = (cid:18) β +2 − γ β (cid:19) R ( p ) e iβ R xp p ( t ) dt and ̟ ( x ) = R x (2 | p ( t ) | + ( x − t ) | q ( t ) | ) dt and the functions K υ ( x, t ) satisfiesthe inequality Z x − x | K υ ( x, λ ) | dt ≤ e c υ ̟ ( x ) − with c = 1 , c = (cid:18) β +1 + (cid:12)(cid:12) β − (cid:12)(cid:12) + γ α + 2 α (cid:19) , c = (cid:18) α (cid:0) β +1 + (cid:12)(cid:12) β − (cid:12)(cid:12)(cid:1) + 1 α (cid:0) β +2 + (cid:12)(cid:12) β − (cid:12)(cid:12)(cid:1) + β + β + γ β (cid:19) ABDULLAH ERG ¨UN where ς ± ( x ) = ± αx ∓ αp + p , β ± = (cid:16) α ± β α (cid:17) , b ± ( x ) = βx − βp + ς ± ( p ) , s ± ( x ) = − βx + βp + ς ± ( p ) , β ∓ = (cid:16) α ∓ αβ β (cid:17) , ξ ∓ = (cid:0) β ∓ ∓ γ α (cid:1) (cid:16) α ∓ αβ β + γ β (cid:17) , ϑ ∓ = (cid:0) β ∓ ∓ γ α (cid:1) (cid:16) α ± αβ β − γ β (cid:17) , β ± = (cid:16) ± β (cid:17) . The proof is done as in [8].
Theorem 2.
Let p ( x ) ∈ W (0 , π ) and q ( x ) ∈ L (0 , π ) . The functions A ( x, t ) , B ( x, t ) , whose first order partial derivatives, are summable on [0 , π ] ,for each x ∈ [0 , π ] such that representation ϕ ( x, λ ) = ϕ ( x, λ ) + Z x A ( x, t ) cos λtdt + Z x B ( x, t ) sin λtdt is satisfied.If p < x < p ; ϕ ( x, λ ) = (cid:0) β +1 + γ α (cid:1) R ( p ) cos h λς + ( x ) − α R xp p ( t ) dt i + (cid:0) β − − γ α (cid:1) R ( p ) cos h λς − ( x ) + α R xp p ( t ) dt i + R ς + ( x )0 A ( x, t ) cos λtdt + R ς + ( x )0 B ( x, t ) sin λtdt (2.7)where β ± = (cid:16) α ± β α (cid:17) . If p < x ≤ π , ϕ ( x, λ ) = (cid:16) β +2 + γ β (cid:17) R ( p ) cos h λb + ( x ) − β R xp p ( t ) dt i + (cid:16) β − + γ β (cid:17) R ( p ) cos h λb − ( x ) − β R xp p ( t ) dt i + (cid:16) β − − γ β (cid:17) R ( p ) cos h λs + ( x ) + β R xp p ( t ) dt i + (cid:16) β +2 − γ β (cid:17) R ( p ) cos h λs − ( x ) + β R xp p ( t ) dt i + R xp A ( x, t ) cos λtdt + R xp B ( x, t ) sin λtdt (2.8)where β ∓ = (cid:16) α ∓ αβ β (cid:17) . Moreover, the equations A ( x, ς + ( x )) cos β ( x ) α + B ( x, ς + ( x )) sin β ( x ) α = (cid:0) β +1 + γ α (cid:1) R ( p )2 α R x (cid:16) q ( t ) + p ( t ) α (cid:17) dt (2.9) A ( x, ς + ( x )) sin β ( x ) α − B ( x, ς + ( x )) cos β ( x ) α = (cid:0) β +1 + γ α (cid:1) R ( p )2 α ( p ( x ) − p (0)) (2.10) NVERSE PROBLEM FOR SINGULAR DIFFUSION OPERATOR 7 A ( x, ς − ( x ) + 0) − A ( x, ς − ( x ) −
0) = (cid:0) β − − γ α (cid:1) R ( p )2 α sin β ( x ) α ( p ( x ) − p (0)) + (cid:0) β − − γ α (cid:1) R ( p )2 α cos β ( x ) α R x (cid:16) q ( t ) + p ( t ) α (cid:17) dt (2.11) B ( x, ς − ( x ) + 0) − B ( x, ς − ( x ) −
0) = (cid:0) β − − γ α (cid:1) R ( p )2 α cos β ( x ) α ( p ( x ) − p (0)) − (cid:0) β − − γ α (cid:1) R ( p )2 α sin β ( x ) α R x (cid:16) q ( t ) + p ( t ) α (cid:17) dt (2.12) B ( x,
0) = ∂A ( x, t ) ∂t (cid:12)(cid:12)(cid:12)(cid:12) t =0 = 0 (2.13) A ( x, s − ( x ) + 0) − A ( x, s − ( x ) −
0) = − (cid:16) β − − γ β (cid:17) R ( p )2 β ( p ( x ) − p (0)) sin ω ( x ) β − (cid:16) β − − γ β (cid:17) R ( p )2 β R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt cos ω ( x ) β (2.14) B ( x, s − ( x ) + 0) − B ( x, s − ( x ) −
0) = − (cid:16) β − − γ β (cid:17) R ( p )2 β ( p ( x ) − p (0)) cos ω ( x ) β + (cid:16) β − − γ β (cid:17) R ( p )2 β R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt sin ω ( x ) β (2.15) A ( x, s + ( x ) + 0) − A ( x, s + ( x ) −
0) = − (cid:16) β − − γ β (cid:17) R ( p )2 β ( p ( x ) − p (0)) sin ω ( x ) β − (cid:16) β − − γ β (cid:17) R ( p )2 β R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt cos ω ( x ) β (2.16) B ( x, s + ( x ) + 0) − B ( x, s + ( x ) −
0) = − (cid:16) β − − γ β (cid:17) R ( p )2 β ( p ( x ) − p (0)) cos ω ( x ) β + (cid:16) β − − γ β (cid:17) R ( p )2 β R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt sin ω ( x ) β (2.17) A ( x, b − ( x ) + 0) − A ( x, b − ( x ) −
0) = − (cid:16) β − + γ β (cid:17) R ( p )2 β ( p ( x ) − p (0)) sin ω ( x ) β − (cid:16) β − − γ β (cid:17) R ( p )2 β R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt cos ω ( x ) β (2.18) B ( x, b − ( x ) + 0) − B ( x, b − ( x ) −
0) = (cid:16) β − + γ β (cid:17) R ( p )2 β ( p ( x ) − p (0)) cos ω ( x ) β − (cid:16) β − − γ β (cid:17) R ( p )2 β R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt sin ω ( x ) β (2.19) A ( x, b + ( x ) + 0) − A ( x, b + ( x ) −
0) = − (cid:16) β +2 + γ β (cid:17) R ( p )2 β ( p ( x ) − p (0)) sin ω ( x ) β − (cid:16) β +2 + γ β (cid:17) R ( p )2 β R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt cos ω ( x ) β (2.20) B ( x, b + ( x ) + 0) − B ( x, b + ( x ) −
0) = (cid:16) β +2 + γ β (cid:17) R ( p )2 β ( p ( x ) − p (0)) cos ω ( x ) β − (cid:16) β +2 + γ β (cid:17) R ( p )2 β R x (cid:16) q ( t ) + p ( t ) β (cid:17) dt sin ω ( x ) β (2.21)are held. ABDULLAH ERG ¨UN
If in addition we suppose that p ( x ) ∈ W (0 , π ) , q ( x ) ∈ W (0 , π ), the func-tions A ( x, t ), B ( x, t ) the following system are provided. ( ∂ A ( x,t ) ∂x − q ( x ) A ( x, t ) − p ( x ) ∂B ( x,t ) ∂t = η ∂ A ( x,t ) ∂t ∂ B ( x,t ) ∂x − q ( x ) B ( x, t ) + 2 p ( x ) ∂A ( x,t ) ∂t = η ∂ B ( x,t ) ∂t (2.22)where η = (cid:26) α ; p < x < p β ; p < x < π .The proof is done as in [7].Conversely, if the second order derivatives of functions A ( x, t ), B ( x, t ) aresummable on [0 , π ] and A ( x, t ), B ( x, t ) provides (2 .
22) system and equations(2 . . ϕ ( x, λ ) which is defined by (1 . .
6) is asolution of (1 .
1) satisfying boundary conditions (1 . Definition 1.
If there is a nontrivial solution y ( x ) that provides the (1 . .
1) problem, then λ is called eigenvalue. Additionally, y ( x ) is called the eigenfunction of the problem corresponding to the eigen-value λ .Let us assume that q ( x ) satisfies the following conditions. Z π n(cid:12)(cid:12) y ′ ( x ) (cid:12)(cid:12) + q ( x ) | y ( x ) | o dx > y ( x ) ∈ W [0 , π ] such that y ( x ) = 0 and y ′ (0) · y (0) − y ′ ( π ) · y ( π ) = 0. Lemma 1.
The eigenvalues of the baundary value problem L are real.Proof. We set l ( y ) := [ − y ′′ + q ( x ) y ]. Integration by part yields( l ( y ) , y ) = R π l ( y ) · y ( x ) dx = R π n | y ′ ( x ) | + q ( x ) | y ( x ) | o dx Since condition (2 .
23) holds, it follow that ( l ( y ) , y ) > (cid:3) Lemma 2.
Eigenfunction corresponding to different eigenvalues of problem L are orthogonal in the sense of the equality ( λ n + λ k ) Z π δ ( x ) y ( x, λ n ) y ( x, λ k ) dx − Z π p ( x ) y ( x, λ n ) y ( x, λ k ) dx = 0(2.24) The proof of Lemma 2 carried out as claim [14].
NVERSE PROBLEM FOR SINGULAR DIFFUSION OPERATOR 9 Properties of the spectrum
Let ψ ( x, λ ) and ϕ ( x, λ ) be any two solutions of equation (1 . W [ ψ ( x, λ ) , ϕ ( x, λ )] = ψ ( x, λ ) ϕ ′ ( x, λ ) − ψ ′ ( x, λ ) ϕ ( x, λ ), Wronskian dosen’tdepend on x . In this case, it depends only on the λ parameter. Although itis shown as W [ ψ, ϕ ] = ∆ ( λ ). ∆ ( λ ) is called the characteristic function of L . Clearly, the function ∆ ( λ ) is entire in λ . It follows that, ∆ ( λ ) has at mosta countable set of zeros { λ n } . Lemma 3.
The zeros { λ n } of the characteristic function ∆ ( λ ) coincide withthe eigenvalues of the baundary value problem L . The functions ψ ( x, λ ) and ϕ ( x, λ ) are eigenfunctions corresponding to the eigenvalue λ n , and there exista sequence ( β n ) such that ψ ( x, λ n ) = β n ϕ ( x, λ n ) , β n = 0 (3.1) The proof of the Lemma 3 is done as in [28].
Let use denote α n = Z π δ ( x ) ϕ ( x, λ n ) dx − λ n Z π p ( x ) ϕ ( x, λ n ) dx, n = 1 , , , ... (3.2)The numbers { α n } are called normalized numbers of the problem L . Lemma 4.
The equality • ∆ ( λ n ) = 2 λ n β n α n is obtained. Here • ∆ = d ∆ dλ .Proof. Since ϕ ( x, λ ) and ψ ( x, λ ) are the solutions of (1 . − ϕ ′′ ( x, λ ) + [2 λp ( x ) + q ( x )] ϕ ( x, λ ) = λ δ ( x ) ϕ ( x, λ ) − ψ ′′ ( x, λ ) + [2 λp ( x ) + q ( x )] ψ ( x, λ ) = λ δ ( x ) ψ ( x, λ )equations are provided. Hence, we differentiate the equalities with respect to − • ϕ ′′ ( x, λ )+[2 λp ( x ) + q ( x )] • ϕ ( x, λ ) = λ δ ( x ) • ϕ ( x, λ )+[2 λδ ( x ) − p ( x )] ϕ ( x, λ ) − • ψ ′′ ( x, λ )+[2 λp ( x ) + q ( x )] • ψ ( x, λ ) = λ δ ( x ) • ψ ( x, λ )+[2 λδ ( x ) − p ( x )] ψ ( x, λ ) . Thanks to these equations ddx (cid:26) ϕ ( x, λ ) · • ψ ′ ( x, λ ) − ϕ ′ ( x, λ ) · • ψ ( x, λ ) (cid:27) = − [2 λδ ( x ) − p ( x )] ϕ ( x, λ ) ψ ( x, λ ) ddx (cid:26) • ϕ ( x, λ ) · ψ ′ ( x, λ ) − • ϕ ′ ( x, λ ) · ψ ( x, λ ) (cid:27) = [2 λδ ( x ) − p ( x )] ϕ ( x, λ ) ψ ( x, λ ) . If the last equations are integrated from x to π and from 0 to x , respectively,by the discontinuity conditions, we obtain − (cid:26) ϕ ( ξ, λ ) · • ψ ′ ( ξ, λ ) − ϕ ′ ( ξ, λ ) · • ψ ( ξ, λ ) (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) πx = Z πx [2 λδ ( ξ ) − p ( ξ )] ϕ ( ξ, λ ) ψ ( ξ, λ ) dξ and (cid:26) • ϕ ( ξ, λ ) · ψ ′ ( ξ, λ ) − • ϕ ′ ( ξ, λ ) · ψ ( ξ, λ ) (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) x = Z x [2 λδ ( ξ ) − p ( ξ )] ϕ ( ξ, λ ) ψ ( ξ, λ ) dξ. If we add the last equalities side by side, we get W h ϕ ( ξ, λ ) , . ψ ( ξ, λ ) i + W h . ϕ ( ξ, λ ) , ψ ( ξ, λ ) i = − . ∆ ( λ ) = R π [2 λδ ( ξ ) − p ( ξ )] ϕ ( ξ, λ ) ψ ( ξ, λ ) dξ for λ → λ n , this yields • ∆ ( λ n ) = − R π [2 λ n δ ( ξ ) − p ( ξ )] β n ϕ ( ξ, λ n ) dξ = 2 λ n β n nR π δ ( ξ ) ϕ ( ξ, λ n ) dξ − λ n R π p ( ξ ) ϕ ( ξ, λ n ) dξ o = 2 λ n β n α n . Denote,Γ n = (cid:8) λ : | λ | = (cid:12)(cid:12) λ n (cid:12)(cid:12) + δ, δ > , n = 0 , , , ... (cid:9) , G n = (cid:8) λ : (cid:12)(cid:12) λ − λ n (cid:12)(cid:12) ≥ δ, δ > , n = 0 , , , ... (cid:9) where δ is sufficiently small positive number.For sufficiently large values of n , one has | ∆ ( λ ) − ∆ ( λ ) | < C δ e | τ | ( βπ − βa + αp − αp + p ) , λ ∈ Γ n . (3.3)As it is shown in [19] , | ∆ ( λ ) | ≥ C δ e | Imλ | π for all λ ∈ ¯ G δ , where C δ > | λ |→∞ e −| Imλ | π (∆ ( λ ) − ∆ ( λ ))= lim | λ |→∞ e −| Imλ | π (cid:16)R π ˜ A ( π, t ) cos λtdt + R π ˜ B ( π, t ) sin λtdt (cid:17) = 0is constant. On the other hand, since for sufficiently large values of n (see [23])we get (3 . (cid:3) Lemma 5.
The problem L ( α, p , p ) has countable set of eigenvalues. Ifone denotes by λ , λ , ... the positive eigenvalues arranged in increasing or-der and by λ − , λ − , ... the negative eigenvalues arranged in decreasing order,then eigenvalues of the problem L ( α, p , p ) have the asymptotic behavior λ n = λ n + d n λ n + k n λ n , n → ∞ where k n ∈ l , d n is a bounded sequence and λ n = nπβπ − βp + αp − αp + p + ψ ( n ) ; sup n | ψ ( n ) | = c < + ∞ . NVERSE PROBLEM FOR SINGULAR DIFFUSION OPERATOR 11
Proof.
According to previous lemma, if n is a sufficiently large natural numberand λ ∈ Γ n , we have | ∆ ( λ ) | ≥ C δ e | Imλ | π > C δ e | Imλ | π > | ∆ ( λ ) − ∆ ( λ ) | . Applying Rouche’s theorem, we conclude that for sufficiently large n inside thecontour Γ n the functions ∆ ( λ ) and ∆ ( λ ) + { ∆ ( λ ) − ∆ ( λ ) } = ∆ ( λ ) havethe same number of zeros. That is, there are exactly ( n + 1) zeros λ , λ , ..., λ n .Analogously, it is shown by Rouche’s theorem that, for sufficiently large valuesof n , the function ∆ ( λ ) has a unique zero inside each circle (cid:12)(cid:12) λ − λ n (cid:12)(cid:12) < δ . Since δ > λ n = λ n + ε n , where lim n →∞ ε n = 0 . If∆ ( λ n ) = 0, we have∆ (cid:0) λ n + ε n (cid:1) + Z π A ( π, t ) cos (cid:0) λ n + ε n (cid:1) tdt + Z π B ( π, t ) sin (cid:0) λ n + ε n (cid:1) tdt = 0(3.4)∆ (cid:0) λ n + ε n (cid:1) = (cid:16) β +2 + γ β (cid:17) R ( p ) cos h(cid:0) λ n + ε n (cid:1) b + ( π ) − β R πp p ( t ) dt i + (cid:16) β − + γ β (cid:17) R ( p ) cos h(cid:0) λ n + ε n (cid:1) b − ( π ) − β R πp p ( t ) dt i + (cid:16) β − − γ β (cid:17) R ( p ) cos h(cid:0) λ n + ε n (cid:1) s + ( π ) + β R πp p ( t ) dt i + (cid:16) β +2 − γ β (cid:17) R ( p ) cos h(cid:0) λ n + ε n (cid:1) s − ( π ) + β R πp p ( t ) dt i (3.5)Since ∆ ( λ ) is an analytical function,∆ (cid:0) λ n + ε n (cid:1) = ∆ (cid:0) λ n (cid:1) ε n + · ∆ (cid:0) λ n (cid:1) ε n + ·· ∆ ( λ n ) ε n + ... , lim n →∞ ε n = 0. λ n is the roots of the ∆ ( λ ) = 0 equation∆ (cid:0) λ n + ε n (cid:1) = (cid:20) . ∆ (cid:0) λ n (cid:1) + o (1) (cid:21) ε n , n → ∞ is provided. (cid:20) . ∆ (cid:0) λ n (cid:1) + o (1) (cid:21) ε n + R s − ( x ) − p A ( π, t ) cos (cid:0) λ n + ε n (cid:1) tdt + R s + ( x ) − s − ( x )+0 A ( π, t ) cos (cid:0) λ n + ε n (cid:1) tdt + R b − ( x ) − s + ( x )+0 A ( π, t ) cos (cid:0) λ n + ε n (cid:1) tdt + R b + ( x ) − b − ( x )+0 A ( π, t ) cos (cid:0) λ n + ε n (cid:1) tdt + R xb + ( x )+0 A ( π, t ) cos (cid:0) λ n + ε n (cid:1) tdt + R s − ( x ) − p B ( π, t ) sin (cid:0) λ n + ε n (cid:1) tdt + R s + ( x ) − s − ( x )+0 B ( π, t ) sin (cid:0) λ n + ε n (cid:1) tdt + R b − ( x ) − s + ( x )+0 B ( π, t ) sin (cid:0) λ n + ε n (cid:1) tdt + R b + ( x ) − b − ( x )+0 B ( π, t ) sin (cid:0) λ n + ε n (cid:1) tdt + R xb + ( x )+0 B ( π, t ) sin (cid:0) λ n + ε n (cid:1) tdt = 0It is easy to see that the function ∆ ( λ ) = 0 is type of [16], so there is a η δ > (cid:12)(cid:12)(cid:12)(cid:12) . ∆ (cid:0) λ n (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≥ η δ > n . We also have λ n = nπβπ − βp + αp − αp + p + ψ ( n ) (3.6) where sup n | ψ ( n ) | < M is for some constant M > .
6) into (3 .
5) after certain transformations, we reach ε n ∈ l . We can obtainmore preciselySince (cid:0)R π A t ( π, t ) sin (cid:0) λ n + ε n (cid:1) tdt (cid:1) ∈ l and (cid:0)R π B t ( π, t ) cos (cid:0) λ n + ε n (cid:1) tdt (cid:1) ∈ l , we have ε n = λ n ∆ ( λ n ) nh(cid:16) β − − γ β (cid:17) R ( p )2 β sin h λ n s − ( π ) + ω ( x ) β i + (cid:16) β − − γ β (cid:17) R ( p )2 β sin h λ n s + ( π ) + ω ( x ) β i + (cid:16) β − − γ β (cid:17) R ( p )2 β sin h λ n b − ( π ) − ω ( x ) β i + (cid:16) β +2 + γ β (cid:17) R ( p )2 β sin h λ n b + ( π ) − ω ( x ) β ii R π (cid:0) q ( t ) + p ( t ) (cid:1) dt + h − (cid:16) β − − γ β (cid:17) R ( p )2 β cos h λ n s − ( π ) + ω ( x ) β i − (cid:16) β − − γ β (cid:17) R ( p )2 β cos h λ n s + ( π ) + ω ( x ) β i + (cid:16) β − + γ β (cid:17) R ( p )2 β cos h λ n b − ( π ) − ω ( x ) β i + (cid:16) β +2 + γ β (cid:17) R ( p )2 β cos h λ n b + ( π ) − ω ( x ) β ii [ p ( π ) − p (0)] o + k n λ n where d n = ( λ n ) nh(cid:16) β − − γ β (cid:17) R ( p )2 β sin h λ n s − ( π ) + ω ( x ) β i + (cid:16) β − − γ β (cid:17) R ( p )2 β sin h λ n s + ( π ) + ω ( x ) β i + (cid:16) β − − γ β (cid:17) R ( p )2 β sin h λ n b − ( π ) − ω ( x ) β i + (cid:16) β +2 + γ β (cid:17) R ( p )2 β sin h λ n b + ( π ) − ω ( x ) β ii R π (cid:0) q ( t ) + p ( t ) (cid:1) dt + h − (cid:16) β − − γ β (cid:17) R ( p )2 β cos h λ n s − ( π ) + ω ( x ) β i − (cid:16) β − − γ β (cid:17) R ( p )2 β cos h λ n s + ( π ) + ω ( x ) β i + (cid:16) β − + γ β (cid:17) R ( p )2 β cos h λ n b − ( π ) − ω ( x ) β i + (cid:16) β +2 + γ β (cid:17) R ( p )2 β cos h λ n b + ( π ) − ω ( x ) β ii [ p ( π ) − p (0)] o is bounded sequence. The proof is completed. (cid:3) The ϕ ( x, λ ) function is | λ | → ∞ in the region D = { λ : arg λ ∈ [ ε, π − ε ] } for x > p , ϕ ( x, λ ) = 12 (cid:18) β +2 + γ β (cid:19) exp (cid:0) − i (cid:0) λb + ( x ) − w ( x ) (cid:1)(cid:1) (cid:18) O (cid:18) λ (cid:19)(cid:19) , | λ | → ∞ it has an asymptotic representation where w ( x ) = R xp p ( t ) dt and β ∓ = (cid:16) α ∓ αβ β (cid:17) . NVERSE PROBLEM FOR SINGULAR DIFFUSION OPERATOR 13 Inverse Problem.
Let us consider the baundary value problem ˜ L :˜ L := l ( y ) := − y ′′ + [2 λ ˜ p ( x ) + ˜ q ( x )] y = λ ˜ δ ( x ) y, x ∈ (0 , π ) y ′ (0) = 0 , y ( π ) = 0 y (˜ p + 0) = ˜ α y (˜ p − y ′ (˜ p + 0) = ˜ β y ′ ( ˜ p −
0) + iλ ˜ γ y (˜ p − y (˜ p + 0) = ˜ α y (˜ p − y ′ (˜ p + 0) = ˜ β y ′ ( ˜ p −
0) + iλ ˜ γ y (˜ p − x, λ ) denote solution of (1 .
1) that satisfy the conditionsΦ ′ (0) = 1 , Φ ( π ) = 0 respectively and jump conditions (1 . − (1 . M ( λ ) := Φ (0 , λ ).The Φ ( x, λ ) and M ( λ ) functions are called the Weyl solution and the Weylfunction, respectively.Φ ( x, λ ) = M ( λ ) .ϕ ( x, λ )+ S ( x, λ ) , ( λ = λ n ; n = 1 , , , ... ) is true. Becauseof W [ ϕ, S ] | x =0 = ϕ (0 , λ ) S ′ (0 , λ ) − ϕ ′ (0 , λ ) S (0 , λ ) = 1 = 0, ϕ ( x, λ ) and S ( x, λ ) solutions are linear independent. When ψ ( x, λ ) is solution (1 . ψ ( x, λ ) = A ( λ ) ϕ ( x, λ ) + B ( λ ) S ( x, λ ) ψ ′ ( x, λ ) = A ( λ ) ϕ ′ ( x, λ ) + B ( λ ) S ′ ( x, λ ) . Due to boundary conditions, A ( λ ) = ψ (0 , λ ) , B ( λ ) = ψ ′ (0 , λ ) = − ∆ ( λ ) .Then ψ ( x, λ ) = ψ (0 , λ ) ϕ ( x, λ ) − ∆ ( λ ) S ( x, λ ) is obtained. Hence,Φ ( x, λ ) := − ψ ( x,λ )∆( λ ) = S ( x, λ ) + M ( λ ) ϕ ( x, λ ) , M ( λ ) = − ψ (0 ,λ )∆( λ ) .The M ( λ ) function is a meromorphic function. Theorem 3. If M ( λ ) = ˜ M ( λ ) , then L = ˜ L .Proof. Let us define the matrix P ( x, λ ) = [ P j,k ( x, λ )] , ( j, k = 1 ,
2) by theformula P ( x, λ ) · (cid:18) e ϕ ( x, λ ) e Φ ( x, λ ) e ϕ ′ ( x, λ ) e Φ ′ ( x, λ ) (cid:19) = (cid:18) ϕ ( x, λ ) Φ ( x, λ ) ϕ ′ ( x, λ ) Φ ′ ( x, λ ) (cid:19) .In this case P ( x, λ ) = − ϕ ( x, λ ) e ψ ′ ( x,λ ) e ∆( λ ) + e ϕ ′ ( x, λ ) ψ ( x,λ )∆( λ ) P ( x, λ ) = − e ϕ ( x, λ ) ψ ( x,λ )∆( λ ) + ϕ ( x, λ ) e ψ ( x,λ ) e ∆( λ ) P ( x, λ ) = − ϕ ′ ( x, λ ) e ψ ′ ( x,λ ) e ∆( λ ) − e ϕ ′ ( x, λ ) ψ ′ ( x,λ )∆( λ ) P ( x, λ ) = − e ϕ ( x, λ ) ψ ′ ( x,λ )∆( λ ) + ϕ ′ ( x, λ ) e ψ ( x,λ ) e ∆( λ )4 ABDULLAH ERG ¨UN Hence, P ( x, λ ) = ϕ ( x, λ ) h e S ′ ( x, λ ) + f M ( λ ) · e ϕ ′ ( x, λ ) i − e ϕ ′ ( x, λ ) [ S ( x, λ ) + M ( λ ) · ϕ ( x, λ )]= ϕ ( x, λ ) e S ′ ( x, λ ) − e ϕ ′ ( x, λ ) S ( x, λ ) + h f M ( λ ) − M ( λ ) i ϕ ( x, λ ) e ϕ ′ ( x, λ ) P ( x, λ ) = e ϕ ( x, λ ) [ S ( x, λ ) + M ( λ ) · ϕ ( x, λ )] − ϕ ( x, λ ) h e S ( x, λ ) + f M ( λ ) · e ϕ ( x, λ ) i = e ϕ ( x, λ ) S ( x, λ ) − ϕ ( x, λ ) e S ( x, λ ) + h M ( λ ) − f M ( λ ) i ϕ ( x, λ ) e ϕ ( x, λ ) P ( x, λ ) = ϕ ′ ( x, λ ) h e S ′ ( x, λ ) + f M ( λ ) · e ϕ ′ ( x, λ ) i − e ϕ ′ ( x, λ ) [ S ′ ( x, λ ) + M ( λ ) · ϕ ′ ( x, λ )]= ϕ ′ ( x, λ ) e S ′ ( x, λ ) − e ϕ ′ ( x, λ ) S ′ ( x, λ ) + h f M ( λ ) − M ( λ ) i ϕ ′ ( x, λ ) e ϕ ′ ( x, λ ) P ( x, λ ) = e ϕ ( x, λ ) [ S ′ ( x, λ ) + M ( λ ) · ϕ ′ ( x, λ )] + ϕ ′ ( x, λ ) h e S ( x, λ ) + f M ( λ ) · e ϕ ( x, λ ) i = ϕ ′ ( x, λ ) S ′ ( x, λ ) − ϕ ′ ( x, λ ) e S ( x, λ ) + h M ( λ ) − f M ( λ ) i ϕ ′ ( x, λ ) e ϕ ( x, λ )from M ( λ ) ≡ f M ( λ ); P ( x, λ ) = ϕ ( x, λ ) e S ′ ( x, λ ) − e ϕ ′ ( x, λ ) S ( x, λ ) P ( x, λ ) = e ϕ ( x, λ ) S ( x, λ ) − ϕ ( x, λ ) e S ( x, λ ) P ( x, λ ) = ϕ ′ ( x, λ ) e S ′ ( x, λ ) − e ϕ ′ ( x, λ ) S ′ ( x, λ ) P ( x, λ ) = ϕ ′ ( x, λ ) S ′ ( x, λ ) − ϕ ′ ( x, λ ) e S ( x, λ )are obtained. When M ( λ ) ≡ f M ( λ ), it is clear that the P j,k ( x, λ ) , ( j, k = 1 , λ . From (3 . ∀ x ∈ [0 , π ] , c δ , C δ constants that provide | P ( x, λ ) | ≤ c δ and | P ( x, λ ) | ≤ C δ inequalities canbe shown. From the Liouville theorem P ( x, λ ) ≡ A ( x ) and P ( x, λ ) ≡ ϕ ( x, λ ) · Φ ′ ( x, λ ) − e ϕ ′ ( x, λ ) · Φ ( x, λ ) = A ( x ) e ϕ ( x, λ ) · Φ ( x, λ ) − ϕ ( x, λ ) · e Φ ( x, λ ) = 0 ϕ ( x, λ ) = e ϕ ( x, λ ) · A ( x )Φ ( x, λ ) = e Φ ( x, λ ) · A ( x ) (cid:27) (4.1)are obtained and W [ ϕ, Φ] = W h ϕ ( x, λ ) , − ψ ( x,λ )∆( λ ) i = λ ) W [ ϕ ( x, λ ) , − ψ (0 , λ ) ϕ ( x, λ ) + ∆ ( λ ) S ( x, λ )]= − ψ (0 ,λ )∆( λ ) W [ ϕ ( x, λ ) , ϕ ( x, λ )] + W [ ϕ ( x, λ ) , S ( x, λ )] = 1 NVERSE PROBLEM FOR SINGULAR DIFFUSION OPERATOR 15
And similarly W h e ϕ, e Φ i = 1 is obtained. If this equation is written in place of(4 . W [ ϕ ( x, λ ) , Φ ( x, λ )] = W h A ( x ) e ϕ ( x, λ ) , A ( x ) e Φ ( x, λ ) i = A ( x ) W h e ϕ ( x, λ ) , e Φ ( x, λ ) i = A ( x )is obtained.Therefore, (cid:16) β +2 + γ β (cid:17) = 1; p = ˜ p , p = ˜ p . We have A ( x ) = 1 from (4 . ϕ ( x, λ ) ≡ ˜ ϕ ( x, λ ) and Φ ( x, λ ) ≡ ˜Φ ( x, λ ).When ϕ ( x, λ ) ≡ ˜ ϕ ( x, λ ), − ϕ ′′ + [2 λp ( x ) + q ( x )] ϕ = λ δ ( x ) ϕ − ϕ ′′ + [2 λp ( x ) + q ( x )] ϕ = λ ˜ δ ( x ) ϕ are obtained. n λ (cid:16) δ ( x ) − ˜ δ ( x ) (cid:17) + 2 λ ( p ( x ) − ˜ p ( x )) + ( q ( x ) − ˜ q ( x )) o ϕ ≡ f or ∀ λ ) δ ( x ) = ˜ δ ( x ), p ( x ) = ˜ p ( x ) and q ( x ) = ˜ q ( x ) a.e. For every λ in discontinuityconditions, ( α − ˜ α ) ϕ ( p − , λ ) = 0 (cid:16) β − ˜ β (cid:17) ϕ ′ ( p − , λ ) + ( γ − ˜ γ ) ϕ ( p − , λ ) = 0( α − ˜ α ) ϕ ( p − , λ ) = 0 (cid:16) β − ˜ β (cid:17) ϕ ′ ( p − , λ ) + ( γ − ˜ γ ) ϕ ( p − , λ ) = 0 α = ˜ α , β = ˜ β , γ = ˜ γ ve α = ˜ α , β = ˜ β , γ = ˜ γ .Consequently L = ˜ L . The proof is completed. (cid:3) Acknowledgement
Not applicable.
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