FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER
FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER
34 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tangfunctions D a (A) and E ρ (λ; ρ −1 ) are of genus zero, we will notice thatD A (λ) = cE ρ (λ; 2),where c is a constant unknown as yet. Consider the logarithmic derivative[ln DA (λ ] ′ D ′=A (λ)∞D A (λ) = − ∑ λ j∞∑ ∞∑= (λ k j k)λ k−1 (|λ| < λ −111 − λλ ),jj=1k=1 j=1that is,[ln DA (λ) ] ′=D ′ A (λ)D A (λ) = −sp( A(J − λA) −1) = − ∑ χ n+1λ n ,where χ n= spA n .Let D A (A) = ∞ ∑0=1a k λ k be the representation by Taylor’s series of thefunction D A (A). Establish the interrelation between χ kand a k . AsD A (λ) = E ρ (λ; ρ −1 ) = ∑ λ kΓ(ρ −1 + kρ −1 ) ,1we have a n =Γ(ρ −1 +kρ −1 ). Further, sincewe obtain the recurrent formulaD ′ A (λ)D A (λ) = ∑ (nan λ n )∑an λ n= − ∑ χ n+1λ n ,χ n+1+ ∑ a s χ n+1−s= −(n + 1)a n+1 ,χ n+1=− ∑ [a n−k χ k−na n = − ∑ (−1) n−k]Γ(2−β(n − k)) χ −n (−1)n ,kΓ(2−nβ)1χ 1=Γ(2 + 1/ρ) , 2a 2 + a 1 χ 1= −χ 2, χ 2= a 2 1 − 2a 2 ,2∑−χ 3= 3a 2 3χ 3−s, χ 3= a 2 a 1 + a 1 (2a 2 − a 2 1) − 3a 2 .s=1As 1/χ 1< λ 1 < χ 1/χ 2, we have1Γ(2 + 1/ρ) < λ 21 0.j=0
Boundary Value Problems for Differential Equations of Fractional Order 35Since 1 < σ 2 < 2, we have 1 < 1 λ< 2, i.e. 1/2 < ρ < 1. The last assumptionis very important.Introduce into consideration the differential operators [5]k∑˜µ k = ˜σ k + 1 = γ 2−j (k = 0, 1, 2),j=0D (˜σ0) d −(1−γ2)1 f(x) ≡d(1 − x) f(x),−(1−γ 2)D (˜σ 1)d −(1−γ 1)d γ 21 f(x) ≡ −d(1 − x) −(1−γ1) d(1 − x) γ f(x),2D (˜σ2)1 f(x) ≡d −(1−γ0)d(1 − x) −(1−γ 0)d γ1 d γ2f(x).d(1 − x) γ1 γ2d(1 − γ)Now the problem (Ã) may be formulated as follows. In the class L 2(0, 1)(or L 1 (0, 1)), find a nontrivial solution of the equationD (˜σ2) z − {λ + q(x)}z = 0, x ∈ [0, 1),satisfying the boundary conditionsD (˜σ0)1 z∣ cos α + D (˜σ1)1 z∣ sin α = 0,x=0 x=0D (˜σ0)∣ cos β + D (˜σ1)∣ sin β = 0.x=0 x=01 z1 zThe associated problem gives essentially new results, in the case wherethe order of the fractional differential equation is less than one. We willdevote a separate paper to this case. To show how to transfer the obtainedresults to the case of the differential equations of order higher than two, weconsider the following problem.6. Operators of TransformationV. A. Marchenko [7] when solving a reverse problem for the equationy ′′ − q(x)y + λy = 0 (0 ≤ x ≤ 1) (1.15)builds an operator of transformation, transforming a solution of the equation(1.15) into a solution of the equationy ′′ + λy = 0. (1.16)By means of Green’s function, with the initial data y(0) = 0, (1.16) correspondsto the differential operatorand the integral operatorAy =∫ x0l(y) = y ′′ − q(x)yG(x, ξ)y(ξ) dξ (0 ≤ x ≤ 1),
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Boundary Value Problems for Differential Equations of Fractional Order 35Since 1 < σ 2 < 2, we have 1 < 1 λ< 2, i.e. 1/2 < ρ < 1. The last assumptionis very important.Introduce into consideration the differential operators [5]k∑˜µ k = ˜σ k + 1 = γ 2−j (k = 0, 1, 2),j=0D (˜σ0) d −(1−γ2)1 f(x) ≡d(1 − x) f(x),−(1−γ 2)D (˜σ 1)d −(1−γ 1)d γ 21 f(x) ≡ −d(1 − x) −(1−γ1) d(1 − x) γ f(x),2D (˜σ2)1 f(x) ≡d −(1−γ0)d(1 − x) −(1−γ 0)d γ1 d γ2f(x).d(1 − x) γ1 γ2d(1 − γ)Now the problem (Ã) may be formulated as follows. In the class L 2(0, 1)(or L 1 (0, 1)), find a nontrivial solution of the equationD (˜σ2) z − {λ + q(x)}z = 0, x ∈ [0, 1),satisfying the boundary conditionsD (˜σ0)1 z∣ cos α + D (˜σ1)1 z∣ sin α = 0,x=0 x=0D (˜σ0)∣ cos β + D (˜σ1)∣ sin β = 0.x=0 x=01 z1 zThe associated problem gives essentially new results, in the case wherethe order of the fractional differential equation is less than one. We willdevote a separate paper to this case. To show how to transfer the obtainedresults to the case of the differential equations of order higher than two, weconsider the following problem.6. Operators of TransformationV. A. Marchenko [7] when solving a reverse problem for the equationy ′′ − q(x)y + λy = 0 (0 ≤ x ≤ 1) (1.15)builds an operator of transformation, transforming a solution of the equation(1.15) into a solution of the equationy ′′ + λy = 0. (1.16)By means of Green’s function, with the initial data y(0) = 0, (1.16) correspondsto the differential operatorand the integral operatorAy =∫ x0l(y) = y ′′ − q(x)yG(x, ξ)y(ξ) dξ (0 ≤ x ≤ 1),
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