FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER
FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER
62 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangProof. Studying the spectrum of the problem is reduced to studying thespectrum of the linear operator bunchClearly,L(λ) = J + M −1{ n ∑M −1( n ∑i=1i=1}a i (x)D αi0x ω i − λM −1 .)a i (x)D α i0x ω i(x)u ∈ G 1 .M −1 is a positive operator. If for the function of distribution rn(r) it ispossible choose a nondecreasing function ϕ(r) (0 ≤ r ≤ ∞) possessing theproperties1. lim ϕ(r) = ∞, ϕ(r) ↑ (r → ∞);r→∞[ ] ′ ϕ ′ (r)2. lim ln(ϕ(r)) = limr→∞r→∞ ϕ(r) < ∞;n(r)3. limr→∞ ϕ(r) = 1,then by theorem of Keldysh,limr→∞n(r)n(r, M −1 ) = 1.As ϕ(r), in our case, obviously it is possible to take the function ϕ(r) =r 1/2 . This proves Theorem 2.9.□Consider the differential expressionLy = −y ′′ + D α 0xyon the finite interval [0, 1].Let T be an operator defined in the Hilbert space H = L 2 (0, 1) bythe operator L with boundary conditions y(0) = y(1) = 0. Certainly, theoperator T is a weak disturbance of the operator{u ′′ ,Au =u(0) = 0, u(1) = 0.Formulate a theorem which is quite important in studying the operatorsof the type T.Theorem 2.10. The operator T is dissipative.Proof.( )(T y, y) = − d2dx 2 y, y + (D0xy, α y), (D0xy, α y) ≥ 0.Thus the operator T is dissipative.□
Boundary Value Problems for Differential Equations of Fractional Order 637. Construction of a Biorthogonal SystemConsider the Sturm–Liouville problem for a fractional differential equation⎧⎪⎨ u ′′ + a m (x)u ′ + a i (x)D α i0x ω i(x)u + λu = 0,u(0) cos(α) + u⎪ ′ (0) sin(α) = 0,(2.47)⎩u(1) cos(β) + u ′ (1) sin(β) = 0and the problem⎧⎪⎨ z − [a m (x)z(x)] ′ − ω i (x)D αix1 a i(t)z(t) + λz = 0,z(0) cos(α) + z⎪⎩′ (0) sin(α) = 0,z(1) cos(β) + z ′ (1) sin(β) = 0.We will call the problems (2.47) and (2.48) mutually adjoint.Following M. M. Dzhrbashyan [10], we introduce the functionsω(λ) = u(1, λ) + u ′ (1, λ) sin(β),˜ω(λ ∗ ) = z(0, λ ∗ ) + z ′ (0, λ ∗ ) sin(α),where u(x, λ) is a solution of the following Cauchy problem(2.48)u ′′ + a m (x)u ′ + a i (x)D α i0x ω i(x)u + λu = 0, (2.49)u(0) = sin(α), u ′ (0) = − cos(α) (2.50)and z(x, λ ∗ ) is a solution of the problemz ′′ (x) + a m (x)z ′ (x) + ω i (x)D α ix1 a i(x)z + λ ∗ z = 0, (2.51)z(1) = sin(β), z ′ = − cos(β). (2.52)The existence and uniqueness of solutions of mutually adjoint problems(2.47) and (2.48) are already proved [3].Clearly if u(x, λ) is a solution of the problem (2.49)–(2.50), then anecessary and sufficient condition for it to be also a solution of the problem(2.47) isω(λ) = u(1, λ) cos(β) + u ′ (1, λ) sin(β).There is a similar statement for the solutions of the problems (2.51)–(2.52).For construction of an orthogonal system of eigenfunctions and associatedfunctions of mutually adjoint problems by method of M. M. Dzhrbashyanwe need the followingTheorem 2.11. Let a m (0) = a m (1) = 0. Then for any values of theparameters λ, λ ∗ the identityholds.(λ − λ ∗ )∫ 10u(x, λ)z(x, λ ∗ ) dx = ω(λ) − ˜ω(λ ∗ ) (2.53)
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62 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangProof. Studying the spectrum of the problem is reduced to studying thespectrum of the linear operator bunchClearly,L(λ) = J + M −1{ n ∑M −1( n ∑i=1i=1}a i (x)D αi0x ω i − λM −1 .)a i (x)D α i0x ω i(x)u ∈ G 1 .M −1 is a positive operator. If for the function of distribution rn(r) it ispossible choose a nondecreasing function ϕ(r) (0 ≤ r ≤ ∞) possessing theproperties1. lim ϕ(r) = ∞, ϕ(r) ↑ (r → ∞);r→∞[ ] ′ ϕ ′ (r)2. lim ln(ϕ(r)) = limr→∞r→∞ ϕ(r) < ∞;n(r)3. limr→∞ ϕ(r) = 1,then by theorem of Keldysh,limr→∞n(r)n(r, M −1 ) = 1.As ϕ(r), in our case, obviously it is possible to take the function ϕ(r) =r 1/2 . This proves Theorem 2.9.□Consider the differential expressionLy = −y ′′ + D α 0xyon the finite interval [0, 1].Let T be an operator defined in the Hilbert space H = L 2 (0, 1) bythe operator L with boundary conditions y(0) = y(1) = 0. Certainly, theoperator T is a weak disturbance of the operator{u ′′ ,Au =u(0) = 0, u(1) = 0.Formulate a theorem which is quite important in studying the operatorsof the type T.Theorem 2.10. The operator T is dissipative.Proof.( )(T y, y) = − d2dx 2 y, y + (D0xy, α y), (D0xy, α y) ≥ 0.Thus the operator T is dissipative.□