A Discontinuous Sturm-Liouville Operator With Indefinite Weight

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1 Journal of Mathematics Research Vol., No. 3; August A Discontinuous Sturm-Liouville Operator With Indefinite Weight Qiuxia Yang (Corresponding author) Mathematics Science college, Inner Mongolia University 35 University of West Street, Huhhot, China & Department of Computer, Dezhou University University Street, Dezhou 53, China dlzxyanglixia@63.com Wanyi Wang Mathematics Science college, Inner Mongolia University 35 University of West Street, Huhhot, China Tel: wwy@imu.edu.cn The research is financed by the Natural Science Foundation of China and the National Natural Science Foundation of Neimongo. No. 668 and 7 (Sponsoring information) Abstract In this paper, we consider an indefinite Sturm-Liouville operator with eigenparameter-dependent boundary conditions and transmission conditions. In an appropriate space K, we define a new self-adjoint operator A such that the eigenvalues of A coincide with those of such a problem and obtain asymptotic approximation for its eigenvalues and eigenfunctions. Keywords: Indefinite Sturm-Liouville operator, Eigenvalue, Eigenfunction, Transmission condition. Introduction In recent years, more and more researchers are interested in the discontinuous Sturm-Liouville problem for its application in physics (Demirci M.,, p.-3 and Buschmann D., 995, p.69-86). The various physics applications of this kind of problem are found in many literature, including some boundary value with transmission conditions that arise in the theory of heat and mass transfer (Aiping W., 6, p.66-7 and Akdoğan Z., 7, p ). Here we consider a discontinuous Sturm-Liouville problem with the indefinite weight function r(x). By using the technics of (Kadakal M., 5, p.9-5 and Kadakal M., 6, p.59-58) and some new approaches, we define a new linear operator A associated with the problem on an appropriate Krein space K. We discuss its eigenvalues and eigenfunctions, and derive asymptotic approximation formulas for eigenvalues and eigenfunctions. In this study, we consider a discontinuous eigenvalue problem consisting of indefinite Sturm-Liouville equation lu := (a(x)u (x)) + q(x)u(x) = λr(x)u(x), x I () where I = [, ) (, ], a(x) = a for x [, ) and a(x) = a for x (, ],, a are positive real constants, xr(x) > a.e., r(x), q(x) L [I, R], and λ C is a complex eigenparameter; with the eigenparameter-dependent boundary conditions at the endpoints and the transmission conditions at the point of discontinuity l u := λ(α u() α u ()) (α u() α u ()) = () l u := λ(β u() β u ()) + (β u() β u ()) = (3) l 3 u := u(+) α 3 u( ) β 3 u ( ) = () l u := u (+) α u( ) β u ( ) = (5) where the coefficients α i, β i, α j and β j (i =, and j =, ) are real numbers. Throughout this paper, we assume that = α 3 β 3 α β >, ρ = α α α α, ρ = β β β β Published by Canadian Center of Science and Education 6

2 Journal of Mathematics Research Vol., No. 3; August We define the inner product in L r (I)as [ f, g] = f a g rdx + f a g rdx, f, g L r (I) where f (x) = f (x) [,) and f (x) = f (x) (,]. Obviously (Lr (I), [, ] ) is a Krein space.. An operator formulation in the adequate Krein space In this section, we introduce the special inner product in Krein space K := Lr (I) C ρ C ρ and a symmetric operator A defined on K such that ()-(5). Namely, we define an inner product on K by [F, G] := f a g rdx + f a g rdx + h, k + q, s (6) ρ ρ for F := ( f, h, q), G := (g, k, s) K. Then (Lr (I) C ρ C ρ, [, ]) is a Krein space, which we denote by Lr (I) C ρ C ρ. A fundamental symmetry on the Krein space is given by J J := sgnρ sgnρ where sgnρ i {, }, (i =, ) and J : L r (I) L r (I) is defined by (J f )(x) = f (x)sgn(r(x)), x [, ] Let, = [J, ], then, is a positive definite inner product which turns K into a Hilbert space H = (L r (I) C ρ C ρ, [J, ]). We define the operator A in K as follows: D(A) = {( f (x), h, q) K f, f AC loc((, )), f, f AC loc((, )), lf L r (I) l 3 f = l f =, h = α f () α f (), q = β f () β f ()} AF = (lf,α f () α f (), (β f () β f ())) for F = ( f,α f () α f (),β f () β f ()) D(A) For convenience, for ( f, h, q) D(A), set N ( f ) = α f () α f (), N ( f ) = α f () α f () N ( f ) = β f () β f (), N ( f ) = β f () β f () Now we can rewrite the considered problem ()-(5) in the operator form AF = λrf. Lemma. The eigenvalues and eigenfunctions of the problem ()-(5) are defined as the eigenvalues and the first components of the corresponding eigenelements of the operator A, respectively. Lemma. The domain D(A) is dense in H. Proof: Let F = ( f (x), h, q) H, F D(A) and C be a functional set such that { ϕ (x), x [, ) ϕ(x) = ϕ (x), x (, ] where ϕ (x) C [, ) and ϕ (x) C (, ]. Since C D(A) ( C), any U = (u(x),, ) C is orthogonal to F, namely F, U = f a u r dx + f a u r dx = f, u This implies that f (x) is orthogonal to C in L r (I) and hence f (x) =. Next suppose that G = (g(x), k, ) D(A), then F, G = ρ hk,soh =. Similarly we have q =. So F = (,, ). Hence, D(A) is dense in H. Theorem.3 The linear operator A is self-adjoint in K. Proof: The operator A is self-adjoint in Krein space K if and only if the operator JA is self-adjoint in Hilbert space H. For all F, G D(A). By two partial integrations we obtain [AF, G] = [F, AG] + W( f, g; ) W( f, g; ) + W( f, g;) W( f, g;+) 6 ISSN E-ISSN

3 Journal of Mathematics Research Vol., No. 3; August + (N ( f )N ρ (g) N ( f )N (g)) + (N ρ ( f )N (g) N ( f )N (g)) where, as usual, by W( f, g; x) we denote the Wronskians f (x)g (x) f (x)g(x). Since f and g satisfy the boundary conditions ()-(3) and transmission conditions ()-(5), we get W( f, g; ) = (N ( f )N ρ (g) N ( f )N (g)) W( f, g;)= ρ (N ( f )N (g) N ( f )N (g)) W( f, g;+) = W( f, g; ) Then we have [AF, G] = [F, AG](F, G D(A)), so A is symmetric, JA is also symmetric. Let JA = B, J l = L. Ifρ <, then If ρ >, then L u = l u. Similarly, if ρ <, then L u := λ(α u() α u ()) + (α u() α u ()) = L u := λ(β u() β u ()) (β u() β u ()) = If ρ >, then L u := l u. And L 3 u := l 3 u, L u := l u. For simplicity, we set ρ > and ρ >. In the following, we show that for all F = ( f, N ( f ), N ( f )) D(A), s.t. BF, W = F, U. Then W D(A) and BW = U, here W = (w(x), h, q), U = (u(x), k, s), i.e. (i) w, w AC loc((, )), w, w AC loc((, )), lw L r (I); (ii) h = α w() α w (), q = β w() β w (); (iii) l 3 w = l w = ; (iv)u(x) = lw; (v) k = α w() α w (), s = β w() β w (). For all F C D(A), we obtain + a (lf)w r dx (lf)w r dx = f u r dx + f u r dx a a a namely, lf, w = f, u. Hence, by standard Sturm-Liouville theory, (i) and (iv) hold. By (iv), the equation AF, W = F, U becomes a Then However (lf)w r dx + a (lf)w r dx + N ( f )h ρ N ( f )r ρ = a f u r dx + a lf, w = f, lw + N ( f )k ρ N ( f )h ρ + N ( f )r ρ N ( f )s ρ lf, w = f, lw + W( f, w; ) W( f, w; ) + W( f, w;) W( f, w;+) f u r dx + N ( f )k ρ N ( f )s ρ So N ( f )k N ( f )h + N ( f )r N ( f )s = ( f ( )w ( ) f ( )w( )) ρ ρ ρ ρ ( f ()w () f ()w()) + ( f ()w () f ()w()) ( f (+)w (+) f (+)w(+)) (7) By Naimark, s Patching Lemma (Naimark M.A. (968)) that there exists an F D(A) such that f ( ) = f ( ) = f (+) = f (+) =, f () = α, f () = α, f () = β, f () = β Thus N ( f ) =, N ( f ) =. Then from (7), (ii) be true. Similarly (v) is proved. Next choose function F D(A) and satisfies f () = f () = f () = f () = f (+) =, f ( ) = β 3, f ( ) = α 3, f (+) = Thus N ( f ) = N ( f ) = N ( f ) = N ( f ) =. Then from (7), we can have w(+) = α 3 w( ) + β 3 w ( ) Published by Canadian Center of Science and Education 63

4 Journal of Mathematics Research Vol., No. 3; August Similarly we can have w (+) = α w( ) + β w ( ) Corollary. All eigenvalues of the operator JA are real, and if λ and λ be the two different eigenvalues of the problem ()-(5), then the corresponding eigenfunctions f (x) and g(x) are orthogonal in the sense of a f g r + a f g r + ρ (α f () + α f ())(α g() α g ()) + ρ (β f () β f ())(β g() β g ()) = 3. Asymptotic approximations of fundamental solutions Let ϕ λ (x) be the solution of equation () on the interval [, ), satisfying the initial conditions ϕ λ () = α + λα,ϕ λ () = α + λα (8) After defining this solution we can define the solution ϕ λ (x) of equation () on the interval (, ] by the initial conditions ϕ λ () = α 3 ϕ λ () + β 3 ϕ λ (), ϕ λ () = α ϕ λ () + β ϕ λ () (9) Analogously we shall define the solutions χ λ (x), χ λ (x) by initial conditions χ λ () = β + λβ,χ λ () = β + λβ () χ λ () = β χ λ () β 3 χ λ (),χ λ () = α χ λ () α 3 χ λ () () Let us consider the Wronskians ω i (λ) := W λ (ϕ i,χ i ; x) (i =, ) which are independent of x Ω i and are entire functions, where Ω = [, ) and Ω = (, ]. This sort of calculation gives ω (λ) = ω (λ). Now we introduce the characteristic function ω(λ)asω(λ) := ω (λ). Theorem 3. The eigenvalues of the problem ()-(5) consist of the zeros of function ω(λ). Proof: Let u (x) be any eigenfunction corresponding to eigenvalue λ. Then the function u (x) may be represented in the form { C ϕ u (x) = λ (x) + C χ λ (x), x [, ) C 3 ϕ λ (x) + C χ λ (x), x (, ] where at least one of the constants c i (i =, ) is not zero. Consider the true function l v (u (x)) =, v =, as the homogenous system of linear equations in the variables c i (i =, ) and talking into account (8)-(), it follows that the determinant of this system is ω (λ ) ω (λ ) ϕ λ () χ λ () ϕ λ () χ λ () = ω(λ ) 3 = ϕ λ () χ λ () ϕ λ () χ λ () Lemma 3. Let λsgnx = s, s = σ + it. Then the following integral equations hold for k =, dx ϕ λ(x) = ( α k s α ) dk s(x + ) cos + dxk s ( α s α ) dk s(x + ) sin + dxk s dx ϕ λ(x) = (α k 3 ϕ λ () + β 3 ϕ dk sx λ ()) cos + a dxk a s (α ϕ λ () + β ϕ dk sx λ ()) sin dxk + a s s(x y) sin q(y)ϕ dxk λ (y)dy () s(x y) sin q(y)ϕ dxk λ (y)dy (3) a Proof: Regard ϕ λ (x) as the solution of the following non-homogeneous Cauchy problem { a u (x) + s u(x) = q(x)ϕ λ (x) a Using the method of constant changing, ϕ λ (x) satisfies ϕ λ () = α s α,ϕ λ () = α s α ϕ λ (x) = ( α s α s(x + ) ) cos + s ( α s α s(x + ) ) sin + s sin s(x y) (y)ϕ λ (x)dy 6 ISSN E-ISSN

5 Journal of Mathematics Research Vol., No. 3; August Then differentiating it with respect to x, we have (). The proof for (3) is similar. Lemma 3.3 Let λsgnx = s, Ims = t. Then for α while if α = dx ϕ λ(x) = α dk s(x + ) k s cos + O( s k+ e t x+ ) () dxk dx ϕ λ(x) = β 3α s3 sin s sx cos + O( s k+ e t ( k a + x a ) ) (5) dxk a dx ϕ λ(x) = a k α s dk s(x + ) sin + O( s k e t x+ ) (6) dxk dx ϕ λ(x) = β k 3 α s cos s sx cos + O( s k+ e t ( a + x a ) ) (7) dxk a k =,. Each of this asymptotic equalities hold uniformly for x as λ. Theorem 3. Let λsgnx = s, Ims = t. Then the characteristic function ω(λ) has the following asymptotic representations: Case α, β Case α, β = Case 3 α =, β Case α =, β = ω(λ) = β 3β α s6 a ω(λ) = β 3β α s5 a ω(λ) = β 3β α s5 ω(λ) = β 3β α s sin s sin s a + O( s 5 e t ( a ) ) cos s sin s a + O( s e t ( a ) ) sin s cos s a + O( s e t ( a ) ) cos s cos s a + O( s 3 e t ( a ) ) Proof: The proof is obtained by substituting (5) into the representation ω(λ) = (β + λβ )ϕ λ() (β + λβ )ϕ λ () Corollary 3.5 The eigenvalues of the problem ()-(5) are semibounded below. Theorem 3.6 The following asymptotic formulae hold for the real eigenvalues of the problem ()-(5) with r(x) = sgnx: Case α, β λ n = (n )π + O( n ), λ n = a (n )π + O( n ) Case α =, β λ n = (n )π + O( n ), λ n = a (n )π + O( n ) Case 3 α, β = λ n = (n )π + O( n ), λ n = a (n )π + O( n ) Case α =, β = λ n = (n )π + O( n ), λ n = a (n )π + O( n ) Proof: By applying the known Rouche theorem, we can obtain these conclusions.. More accuracy asymptotic formulae for eigenvalues and eigenfunctions In this section, for the sake of simplicity we assume that = a =, α = β 3 =, α 3 = β and the weight function r(x) = sgnx. We will use the following method to obtain more accurate conclusions. Similarity with the third section, we can get the following three conclusions: Published by Canadian Center of Science and Education 65

6 Journal of Mathematics Research Vol., No. 3; August Lemma. Let λsgnx = s, Ims = t. Then α dx ϕ λ(x) = α dk k s dx cos s(x + ) k α s dk dx sin s(x + ) + k s dx sin s(x y)q(y)ϕ λ(y)dy + O( s k e t (x+) ) k while if α = dx ϕ λ(x) = α k α 3s dk dx cos s(x + ) k α α 3s dk dx sin s(x + ) + α 3 k s dx ϕ λ(x) = α k + s dx k sin s(x y)q(y)ϕ λ(y)dy + O( s k e t (x+) ) dx cos s(x + ) k α s dk dx sin s(x + ) + k s dx k ϕ(k) λ (x) = α α 3 dx cos s(x + ) k α α 3s dk dx sin s(x + ) + α 3 k s + s dx sin s(x y)q(y)ϕ λ(y)dy + O( s k e t (x+) ) k k =,. Each of this asymptotic equalities hold uniformly for x as λ. Theorem. The characteristic function ω(λ) has the following asymptotic representations: Case α, β dx k sin s(x y)q(y)ϕ λ(y)dy dx k sin s(x y)q(y)ϕ λ(y)dy + O( s k e t (x+) ) dx k sin s(x y)q(y)ϕ λ(y)dy ω(λ) = α α 3β s5 sin s + (α α 3β s α α 3β s ) cos s + α α 3β s cos s( y) cos s( + y)q(y)dy Case α =, β +α α 3β s cos s( y) cos s( + y)q(y)dy + O( s 3 e t ) ω(λ) = α α 3β s cos s (α α 3 β s3 + α α 3β s3 ) sin s + α α 3β s3 cos s( y) sin s( + y)q(y)dy Case 3 α, β = +α α 3β s3 cos s( y) sin s( + y)q(y)dy + O( s e t ) ω(λ) = α α 3β s cos s + (α α 3β s 3 α α 3β s3 ) sin s α α 3β s3 sin s( y) cos s( + y)q(y)dy Case α =, β = α α 3β s3 sin s( y) cos s( + y)q(y)dy + O( s e t ) ω(λ) = α α 3β s3 sin s (α α 3β s + α α 3 β s ) cos s α α 3 β s sin s( y) sin s( + y)q(y)dy α α 3β s sin s( y) sin s( + y)q(y)dy + O( s e t ) Theorem.3 The following asymptotic formulas hold for the real eigenvalues and eigenfunctions of the boundary value transmission problem ()-(5): Case α, β s n = (n )π + O ( ),ϕ(x,λn ) = n (n) π α (n)π(x+) cos + O(n), x [, ) (n) π α α 3 cos (n)π(x+) + O(n), x (, ] 66 ISSN E-ISSN

7 Journal of Mathematics Research Vol., No. 3; August Case α =, β Case 3 α, β = Case α =, β = s n = (n )π s n = (n )π + O ( ),ϕ(x,λn ) = n + O ( ),ϕ(x,λn ) = n (n ) π α cos (n )π(x+) + O(n), x [, ) (n ) π α α 3 cos (n )π(x+) + O(n), x (, ] (n ) π α cos (n )π(x+) + O(n), x [, ) (n ) π α α 3 cos (n )π(x+) + O(n), x (, ] s n = nπ + O( ) { π α,ϕ(x,λn ) = cos nπ(x+) + O( n ), x [, ) n n π α α 3 cos nπ(x+) + O( n ), x (, ] Theorem. The following more accuracy asymptotic formulas hold for the real eigenvalues of the boundary value transmission problem ()-(5): Case α, β sn = Case α =, β Case 3 α, β = (n )π s n = (n )π s n = (n )π + + (α (n )π α β β + Q ) + O( ) n ( α (n )π α + β β Q ) ( ) + O n Case α =, β = s n = nπ ( α nπ α + β β Q ) ( ) + O n where Q := q(y)dy + q(y)dy. Proof: Let us consider Case only. Putting s n = (n)π + δ n,δ n = O ( n) in the equality ω(λ). We find that (8) (9) (α (n )π α β β + Q ) + O( n ) () sin δ n = cos δ n (α s n α β β + Q + O( n )) + O ( s n ) () () where Q := q(y)dy + q(y)dy. Consequently, from () it follows that δ n = (α (n )π α β β + Q ) ( ) + O n (3) substituting (3) in s n = (n)π + δ n, we have (8). The proof of other cases are similar. References Aiping W. (6). Research on Weidmann conjecture and differential operators with transmission conditions [Ph D Thesis], Inner Mongolia University,66-7. [in Chinese] Akdoğan Z, Demirci M etal Mukhtarov O Sh. (7). Green function of discontinuous boundary-value problem with transmission conditions. Math. Meth. Appl. Sci, 3: Buschmann D., Stolz G., Weidmann J. (995). One-dimensional Schrödinger operators with local point interactions. J.Reine Angew. Math, 67: Demirci M., Akdoğan Z., Mukhtarov O.sH. (). Asymptotic behavior of eigenvalues and eigenfuctions of one discontinuous boundary-value problem. International Journal of Computational Cogonition, (3):-3. Published by Canadian Center of Science and Education 67

8 Journal of Mathematics Research Vol., No. 3; August Kadakal M., Mukhtarov O Sh. (6) Discontinuous Sturm-Liouville problems containing eigenparameter in the boundary condition. Acta Mathematical sinic, English Series Sep., (5): Naimark M.A. (968). Linear Differential operators, Part II. Ungar, New York. Zhijiang C. (986). Ordinary differential operator, Shanghai, (Chapter 3).[in Chinese] 68 ISSN E-ISSN