FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER
FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER
FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER
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24 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangThen denotingg(x) = dαdx α f(x),in the future we will mean byd αdx αthe fractional integral when α < 0 and the fractional derivative when α > 0.The fractional derivatived αd(1 − x) αof order α > 0 of a function f(x) ∈ L 1 (0, 1) , with ending point x = 1 isdefined in the similar way.Let {γ k } n 0 be any set of real numbers satisfying the condition 0 < γ j ≤ 1(0 ≤ j ≤ n). We denoteσ k =k∑γ j − 1; µ k = σ k + 1 =j=0and we assume that1ρ =n ∑j=0k∑j=0γ jγ j − 1 = σ n = µ n − 1 > 0.(0 ≤ k ≤ n),Following M. M. Dzhrbashyan (see [6]), we consider the integro-differentialoperatorsD (σ0) f(x) ≡ d−(1−γ0)dx f(x),−(1−γ 0)D (σ1) f(x) ≡ d−(1−γ 1)d γ 0dx −(1−γ 1) dx γ f(x),0D (σ2) f(x) ≡ d−(1−γ 2)d γ 1d γ 0f(x),dx −(1−γ2) dx γ1 γ0dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .D (σn) f(x) ≡ d−(1−γn)dx −(1−γ n)d γn−1dx γ · · · d γ0n−1 dx γ f(x).0Here we note that if γ 0 = γ 1 = · · · = γ n = 1, then obviouslyD (σ k) f(x) = f (k) (x) (k = 0, 1, 2, . . . , n).The objects of our investigation are boundary value problems for the followingequations:D (σ n) u − [ λ + q(x) ] u = 0, 0 < σ n < ∞, (1.1)u ′′ + D α 0xu + q(x)u = λu, 0 < α < 1. (1.2)
Boundary Value Problems for Differential Equations of Fractional Order 25We will consider various versions of the equation (1.1). For γ 0 = γ 1 = 1,γ 3 = γ 4 = · · · = γ n = 0, the equation (1.1) turns into the equation∫1xΓ(1 − γ 2 )0u ′′ (t)(x − t) γ dt − [λ + q(x)]u(x) = 0, (1.3)2which is called a fractional oscillating equation [5], and the operator D (σ 2)is called the operator of fractional differentiation in Caputo sense [5]. Forγ 0 = γ 2 = 1, γ 3 = γ 4 = · · · = γ n = 0, the equation (1.1) turns into theequation1Γ(1 − γ 1 )∫dxdx0u ′ (t)(x − t) γ dt − [λ + q(x)]u(x) = 0. (1.4)1The equation (1.4) has been investigated as a model equation of fractionalorder 1 < σ < 2 (see [3] and references therein). Further, if γ 0 = γ 2 = · · · =γ n = 1, then the equation (1.1) will be written asD (σ n) u =1Γ(1 − γ 1 )d n−1dx n−1∫x0d(t)(x − t) γ 1 dt − [λ + q(x)]u(x) = 0. (1′ )The two-point boundary value problem of Dirichlet u(0) = 0, u(1) = 0for the fractional oscillatory equation was studied by one of the authorsof this paper in [5]. Therein for the first time Green’s function for similarboundary value problems has been constructed. In particular, it was provedthat the two-point problem of Dirichlet u(0) = 0, u(1) = 0 for the fractionaloscillatory equation with q(x) = 0 is equivalent to the equationu(x) =[ ∫xλ(1 − t) 1−γ 2u(t) dt −Γ(1 − γ 2 )0∫ 10]x 1−γ 2(1 − t) 1−γ 2u(t) dt .The same problem for the model fractional differential equation of order1 < σ < 2 is equivalent to the equation [3]u(x) =[ ∫xλ(1 − t) γ 1u(t) dt −Γ(1 + γ 1 )0∫ 10]x(1 − t) γ 1u(t) dt .The operator A inverse to the operator B induced by the differential expression(1 ′ ) and natural boundary conditionsu(0) = 0; D (σ 1) u ∣ ∣x=0= 0, . . . , D (σ n−2) u ∣ ∣x=0= 0, u(1) = 0looks like (see [3] and [6])[ ∫x1Au =Γ(ρ −1 x(1 − t) 1 ρ −1 u(t) dt −)0∫ 10]x 1 ρ −1 (1 − t) 1 ρ −1 u(t) dt .
28 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangwhereM ε u =K ε (x, t) =N ε =˜K ε (x, t) =∫ x0K ε (x, t)u(t) dt, (1.6){(x − t) 1+ε , t < x,0, t ≥ x∫ 10˜K ε (x; t)u(t) dt,{x 1+ε (1 − t) 1+ε , t ≠ 1,0, t = 1.(1.7)(1.8)Considering the disturbance of the operator A ε , we write(A − A ε )u = (M − M ε )u − (N − N ε )u.First, we deal with (M − M ε )u. Clearly,(M − M ε )u =∫ 10[K(x, t) − Kε (x, t) ] u(t) dt.Since{(x − t)[1 − (x − t) ε ], t < xK(x, t) − K ε (x, t) =0, t ≥ x =⎧⎨[(x−t) ε ln(x−t) +ε=2 ln2 (x−t)+ · · · +ε n lnn (x−t)]+ · · · , t < x,⎩1!2!n!0, t ≥ x,we havewhereThusM ε u =(M − M ε )u = ε∫ 10∫ 10∫1K 1 (x, t)u(t) dt + · · · + ε n⎧⎨(x − t) ln n (x − t), t < x,K n (x, t) = n!⎩0, t ≥ x.K(x, t)u(t) dt−ε∫ 100K n (x, t)u(t) dt + · · · ,∫1K 1 (x, t)u(t) dt−· · ·−ε n K n (x, t)u(t) dt−· · · .0
Boundary Value Problems for Differential Equations of Fractional Order 29Similarly,N ε u =where∫ 10˜K(x, t)u(t) dt − εwhich proves Theorem 1.2.∫ 10∫1˜K 1 (x, t)u(t) dt − · · · − ε n⎧⎨x(1 − t) ln n (x − xt), t < 1,˜K n (x, t) =n!⎩0, t = 1,0˜K n (x, t)u(t) dt + · · · ,Theorem 1.3. All eigenvalues λ n (ε) of the operator A(ε) are real.Proof. We havewhereλ n (ε) = πn 2 + ελ 1 + ε 2 λ 2 + · · · , (1.9)ϕ n (ε) = sin nx + εϕ 1 + ε 2 ϕ 2 + · · · , (1.10)λ n =n∑(A k ϕ n−k , sin nx), (1.11)k=1ϕ n = Rn∑(λ k − A k )ϕ n−k . (1.12)k=1Here R is the resolvent of the operator A, corresponding to the eigenvalueπn 2 . This resolvent is an integral operator with the kernel[S(x, y)= − y 1]cos ny sin nx+1−x sin ny cos nx+n n 2n 2 sin ny sin nx , y ≤x(if y > x it is necessary to interchange y and x in the right part of thisformula).Clearly R transforms H 0 (H 0 is the orthogonal complement of the functionsin πnx) into itself and cancels sin πnx.From (1.11) it follows that λ 1 = (A 1 sin nx, sin x). As the kernel of theoperator A 1 assumes real values, we have sin λ 1 = 0. From (1.12) it followsthat ϕ 1 = R(nk 2 − A 1 ) sin nx and since the kernels of the operators R andA 1 assume real values, we have Im ϕ 1 = 0. So, successively it is possible toestablish that all λ i are real. Since ε is real, λu(ε) is real too. □Theorem 1.4. For the eigenvalues λ n (ε) and eigenfunctions ϕ n (ε) ofthe operator A(ε) there hold the estimates|λ n (ε) − πn 2 |
30 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangProof. From (1.11) and (1.12), assuming thatwhere‖A n u‖ ≤ p n−1{ u‖u‖ + b‖A 0 u‖ } , m = ‖A 0 ‖;{ 8(a + mb)c = maxd, 8p + 4 a + mb },d1d = ‖R‖, |ε| < 1 c ,we obtain the simple formulas [8]∣∣λ(ε) − λ 0 − ελ 1 − · · · − ε n λ n∣ ∣ ≤ d 2 (|ε|c)n+1 , (1.13)∣ ϕ(ε) − ϕ0 − εϕ 1 − · · · − ε n ϕ n∣ ∣ ≤12 (|ε|c)n+1 . (1.14)Let’s calculate the values of the parameters a, b, c, d, m. First we find m,m = ‖A‖ = sup A, where sup A is the spectral radius of the operator A.As sup A = π −1 , we have m = π −1 . Further, d = dist (πn 2 ; Σ ′′ ) (d is anisolating distance), where Σ ′′ is the spectrum of the operator A −1 with theexcluded point πn 2 . Clearly d = π(2n − 1). To find other parameters a, b,c, we will obtain an estimate of the norm of the operator A n :‖A n ϕ‖ L(0,1) ≤∫ 1 ∫ 1Here K n (x, t) is the kernel of the operator A n∫ 1 ∫ 100|K n (x, t)| · |ϕ(t)| dt dx ==∫ 10|ϕ(t)|∫1−t000∫ 1 ∫ x0z| ln n z|n!0|K n (x, t)| |ϕ(t)| dt dx.∣ (x − t) lnn (x − t)∣ |ϕ(t)| dt dx =n!dz dt ≤ 1 n!Let’s calculate the integral ∫ z ln z n dz.∫z ln z n dz =∫ 10z| ln n z| dz · ‖ϕ‖ L1(0,1).= z2 (ln z) n− nx2 (ln x) n−122 2 + · · · + (−1)n−1 n(n−1)(n−2) · · · 2( x22 n−1 2 − 1 )2 2 .Hence ‖A n ‖ ≤ 12. Now we will take a = 1/4, p = 1/2, and b = 0. Sincen+1{c = max 8 a + mb + 8p + 4 a + mb },ddwe haveThen{ 2c = maxd d}; 4 + 1 = 4 + 1 d = 5.|λ(ε) − λ| ≤ 1 2π(2n − 1)2.
Boundary Value Problems for Differential Equations of Fractional Order 31Thus we have found |ϕ ε − ϕ 0 | ≤ 1/2, and proved Theorem 1.4.□Theorem 1.5. Let u 0 (x), u 1 (x), . . . , u n (x), . . . be the eigenfunctionsof the operator A ρ , and λ 0 , λ 1 , . . . , λ n , . . . be the corresponding eigenvalues.Then the least frequency self-oscillation, i.e. ϕ 0 (x), has no units, i.e.u 0 (x) ≠ 0, 0 < x < 1.Proof. Let λ 0 be the least eigenvalue of the operator A ρ . Thenu 0 (x) = E ρ (−λ 0 x; ρ −1 ) =∞∑k=0(−λ 0 x) kΓ(ρ −1 + ρ −1k ) .Show that the function u 0 (x) has no zero in the interval (0, 1). Supposethat the function u 0 (x) in a point x 0 ∈ (0, 1) equals to zero, that is,u 0 (x 0 ) = E ρ (−λ 0 x 0 ; ρ −1 ) =∞∑k=0(−λ 0 x 0 ) kΓ(ρ −1 + ρ −1k ) = 0.That’s to say, the number −λ 0 x 0 is a zero point of E ρ (z, ρ −1 ). However,−λ 0 x 0 < λ 0 as x 0 ∈ (0, 1) while we have assumed that the least zero is λ 0 .The obtained contradiction proves Theorem 1.5.□Remark 1. It is also possible to show analogously that u 1 (x) in theinterval has exactly one zero, etc.Theorem of existence of the basis made of root spaces of the operatorA ρ (0 < ρ < 1/2) is connected with the problem of completeness of systemsof eigenfunctions of the operator induced by the differential expression⎧⎪⎨ 1 d n−1 ∫xu ′ (t)l(D) n = Γ(1 − γ 1 ) dx n−1dt − (λ + q(x))u,γ1(x − t)0⎪⎩u(0) = 0, D σ 1u ∣ x=0= 0, . . . , D σ n−2u ∣ x=0= 0, u(1) = 0which is studied in case where q(x) is a semi-function (see [3] and referencestherein). If completeness of system of eigenfunctions bounded is proved,then the question is: is it possible to make basis with the eigenfunctions ofthis operator. Let us give an answer to this question.Lemma 1.1. The operator A is dissipativeProof. We will consider the operatorAu =∫ x0(x − t) 1+ε u(t) dt −∫ 10x 1+ε (1 − t) 1+ε u(t) dt.
32 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangDenotev(t) =∫ x0(x − t) 1+ε u(t) dt −∫ 10x 1+ε (1 − t) 1+ε u(t) dt.From this expression it follows that u(x) = Dv ∈ L 2 (0, 1). Consider againthe product[ ∫x(Au, u) (x − t) 1+ε u(t) dt −=∫ 10= v(x)00∫ 1∫ 10]x 1+ε (1 − t) 1+ε u(t) dt u(x) =[ ∫x∫](x − t) 1+ε u(t) dt − x 1+ε (1 − t) 1+ε u(t) dt u(x) dx ==∫ 100v(x)D x v dt =0=∫ 10v(x)∫ 1v(x) d ( ∫xdx0∫0( x( ∫x)∣v ′ (t) ∣∣∣1v(x)(x − t) ε dt −= −∫ 10( ∫x00)v(t)(x − t) ε dt dx =0∫ 10( ∫x)v ′ (t)(x − t) ε dt v ′ (t) dx,0)v ′ (t)(x − t) ε dt dx =)v ′ (t)(x − t) ε dt v ′ (x) dx =as v(0) = v(1) = 0. Define v ′ (x) = z(x). Then (Au, u) = −(J ε z, z). Now,by virtue of a theorem of Matsaev–Polant, it follows that the values of theform (J ε z, z) lay in the angle |argz| < πε2, which proves Lemma 1.1. □In papers [1] and [23], the dissipativity of the operator l(D) is proved.4. The Basis Problem for Systems of EigenfunctionsTheorem 1.6. The system of eigenfunctions of the operator A formsa basis of the closed linear hull.Proof. Remind that the system of vectors {u 1 , u 2 , . . . , u n , . . . } forms a basisof the closed linear hull G ⊂ H if the inequalityn∑m |c k | 2 n∑ ∥≤ ∥ c k ϕ 2 ∥∥2 ∑ n≤ M |c k | 2k=1k=1holds, where m, M are positive constants independent of c 1 , c 2 , c 3 , . . . , c n(n = 1, 2, 3, . . . ). Any vector f (f ∈ G) in that case uniquely expands intok=1
Boundary Value Problems for Differential Equations of Fractional Order 33∑a series f = ∞ ( ∞∑c k u k |c k | 2 < ∞ ) . It is known that if the spectrum ofn=1n=1a dissipative operator with completely continuous imaginary component “ispressed enough” to the real axis, then it is possible to form a basis of thelinear closed hull of eigenvectors of this operator.We need a theorem of Glazman.Theorem of Glazman. Let A be a linear bounded dissipative operatorwith completely continuous imaginary component, possessing an infinitesystem of eigenvectors {u k } ∣ ∞ k=1 , normalized by the condition (u k, u u ) = 1(k = 1, 2, 3, . . . ), and {λ k } ∣ ∞ be the corresponding sequence of differentk=1eigenvalues. If the condition∑ Im λj Im λ k< ∞|λ j − λ k |is satisfied, then the system {ϕ k } ∣ ∞ is a Bari–Riesz’s basis of the closedk=1linear hull. Since for 0 < ρ < 1/2 all eigenvalues of the operator A ρ are real,due to theorem of Glazman the proof of Theorem 1.6 follows. Analogously,it is possible to prove similar statements for more general problems. □5. Partial Problem of Eigenvalues for the Operator A ρ(0 < ρ < 1/2)In [1], [3], there have been determined areas in the complex plane wherethere are no eigenvalues of the operator A ρ for any ρ. Here, as above, wewill assume that 0 < ρ < 1/2. In applied problems the greatest interestrepresents usually determination of first eigenvalues, therefore we will solvethis problem for eigenvalues of the operator A ρ .Let’s consider the operatorA= 1 [ ∫xΓ(ρ −1 (x−t) 1/ρ u(t) dt−)0∫ 10]x 1/ρ−1 (1−t) 1/ρ−1 u(t) dt =A 0 u+A 1 u.As 0 < ρ < 1/ρ, the operator A 0 and A 1 are invertible. Therefore sp(A) =spA 0 + spA 1 .Clearly1spA 0 = 0, spA 1 =Γ(2 − 1/ρ) .Therefore the determinant∞∑D A (λ) = (1 − λµ j ), µ k = 1 ,λ kj=1is meaningful.Earlier it has been established that the number λ is an eigenvalue of theoperator A only if λ −1 is a zero of the function E ρ (λ; ρ −1 ). Since the entire
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),
36 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangwhereG(x, x) = 0,ddx G(x, ξ) ∣∣∣∣ξ=x= 1.From Marchenko’s results it follows that the operator A is linearly equivalentto the operator of repeated integration [9]J 2 y =∫ x0(x − ξ)y(ξ) dξ,where y(ξ) ∈ L 2 [0, 1].In the present paper, similar results we will be obtained for the differentialequations of the fractional order σ (1 < σ ≤ 2). In what follows, wewill use the operators of fractional differentiation of M. M. Dzhrbashyan [9],which are defined as follows.Consider a problem of Cauchy typeJ{y; γ 0 , γ 1 , γ 2 } = D σ 2y − {λ + q(x)}y = 0, x ∈ (0, 1], (1.17)dd∣ = C 0 , ∣ = C 1 ,x=0 x=0(1.18)dx Dσ 0dx Dσ 1where q(x) ∈ C[0, 1].Note that some results of this paper have been published earlier in [1],[7]. In those works, operator of transformation translating a solution ofthe equation (1.18) to a solution of the equation D σ 2y − λy = 0 has beenconstructed. Let y(x, λ) be a solution of a problem of Cauchy type. Thenit is known [2] that the identityy(x; λ) = C 0 x µ0−1 E ρ (λx 1/ρ ; µ 0 ) + C 1 x µ1−1 E ρ (λx 1/ρ ; µ 1 )++∫ x0((x − τ) 1/ρ E ρ λ(x − τ) 1/ρ ; 1 )q(τ)y(τ; λ) dτ (1.19)ρholds, where∞∑ z kE ρ (x, µ) =Γ ( )µ + k (1.20)k=0 ρis a function of Mittag–Leffler type.In what follows, we will consider the case where C 0 = 0, C 1 = 1, λ = iz,Im z = 0, µ 1 = 1, |q(x)| < 1. As µ 1 = 1, we have(see [5]). In [7],y(x; λ) = x µ 1−1 E ρ (λx 1/ρ ; µ 1 )++|E ρ (λx 1/ρ ; µ 1 )| ≤ Cx −1/ρ (1 ≤ x < ∞)∫ x0((x − τ) 1/ρ−1 E ρ λ(x − τ) 1/ρ ; 1 )q(τ)y(τ; λ) dτ, (1.21)ρ
Boundary Value Problems for Differential Equations of Fractional Order 37From the general Volterra theory it follows that the equation (1.21) can besolved by a method of iterations. Since the function q(x) is continuous, wehavewhereϕ n (λ, x) =y(x, λ) = E ρ (λx 1/ρ ; µ 1 ) + ϕ 1 (λ, x) + ϕ 2 (λ, x) + · · · , (1.22)∫ x0dτ 1 · · ·∫τ n−10dτ 1 · · · q(τ n )E ρ (λ(x − τ 1 ) 1/ρ ; µ 1 ) · · · ×× E ρ (λ(τ n−1 − τ n ) 1/ρ ; µ 1 )E ρ (λτ 1/ρn ; µ 1 ).Obviously the series (1.22) converges in regular intervals in each interval(0; l), l < 1.Theorem 1.7. There is an operator of transformation V , transforminga solution E ρ (λx 1/ρ ; µ 1 ) of the equation (1.19) into a solution y(x; λ) of theequation (1.17).holds.Lemma 1.2. The equality(J − λJ 1/ρ ) −1 1Γ(ρ −1 ) = E ρ(λx 1/ρ ; 1) (1.23)Proof. Let f(x) ∈ L 1 (0, 1), ρ > 0 and λ be any complex parameter. Thenthe integral equationhas a unique solutionu(x) = f(x) + λu(x) = f(x) +λΓ(ρ −1 )belonging to the class L 1 (0, 1).In (1.24), let f(x) = 1Γ(ρ −1 ). We obtain∫ x0u(x) =∫ x0(x − 1) 1/ρ u(t) dt (1.24)((x − t) 1/ρ−1 E ρ λ(x − t) 1/ρ ; x )f(t) dt (1.25)ρ1Γ(ρ −1 ) + 1Γ(ρ −1 )Due to (1.25) and (1.26), we haveu(x) =1Γ(ρ −1 ) + 1Γ(ρ −1 )Using the known formula of M. M. Dzhrbashyan∫ x0∫ x0(x − t)u(t) dt. (1.26)((x − t) 1/ρ−1 E ρ λ(x − t) 1/ρ ; x )dt. (1.27)ρ
38 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tang1Γ(α)∫ z0(z − t) α−1 E ρ (λt 1/ρ ; µ)t µ−1 dt =we calculate the integral1Γ(ρ −1 )∫ x0= z µ+α−1 E ρ (λz 1/ρ ; µ + α) (µ > 0, α > 0),((x−t) 1/ρ−1 E ρ λ(x−t) 1/ρ ; x ) (dt=λx 1/ρ E ρ λx 1/ρ ; 1+ 1 ). (1.28)ρρFrom (1.27) and (1.28), it follows that1u(x) =Γ(ρ −1 ) + λx1/ρ E ρ(λx 1/ρ ; 1 + 1 )ρThus Lemma 1.2 is proved.= E ρ (λx 1/ρ ; 1).Lemma 1.3. Let y(x, λ) be a solution of a problem of Cauchy type(1.17)–(1.18). Theny(x, λ) = (J + λ)KA −1 f,whereAu =∫ x0(x − t) 1/ρ−1 u(t) dt, K = (J + Aq(x)), f = 1.□Proof. Obviously the solution of the equation⎧⎪⎨ D σ y + q(x)y − λy = 0,D⎪⎩σ0 y ∣ x=0= 0,D σ 1y ∣ x=0= −1coincides with the solution of the equationy + A −1 q(x)y + λAy = f.Denote K = (J + A −1 q(x)). Then we will obtainy = (J + λxA −1 ) −1 f,which proves Lemma 1.3.□Lemma 1.4. The operator B is linearly equivalent to the operator J 1/ρ .Proof. From Theorem 1.7 and Lemma 1.3 it follows thatFrom (1.23) and (1.29) it follows thatV E ρ (λx 1/ρ ; 1) = (I + λKA −1 ) −1 f. (1.29)(J + λB) −1 f = V (J + J 1/ρ ) −1 +1Γ(ρ −1 ) . (1.30)
Boundary Value Problems for Differential Equations of Fractional Order 39Dividing both parts of the equality (1.30) successively by degrees of λ, weobtainλ 0 f = λ 0 1VΓ(ρ −1 ) , Kf = V J 1/ρ 1Γ(ρ −1 ) .From these equalities it follows that K = V J 1/ρ V −1 , which proves Lemma1.4.□Theorem 1.8. The operator B is a monocell Volterra operator.Proof. As the operator J 1/ρ is monocell, then by virtue of a Lemma 1.4,the operator B is monocell too.□Theorem 1.8 can be used for solution of inverse problems for the differentialequations of the fractional order.
CHAPTER 2The Sturm–Liouville Problems for a SecondOrder Ordinary Differential Equation withFractional Derivatives in the Lower Terms1. On Some Problems from the Theory of Equations of MixedType Leading to Boundary Problems for the DifferentialEquations of the Second Order with Fractional DerivativesConsider the equation⎧m∑⎪⎨ u ′′ + a 0 (x)u ′ + a i (x)D αi0x ω i(x)u + u m+1 (x)u = f(x),i=1(2.1)u(0) cos α + u⎪⎩′ (0) sin α = 0,u(1) cos β + u ′ (1) sin β = 0,where 0 < α m < · · · < α 1 < 1, D α is the operator of fractional differentiationof order αD σ u =1Γ(1 − α)∫dxdx0u(t)dt, 0 < σ < 1.(x − t)σMany direct and inverse problems associated with a degenerating hyperbolicequation and equation of the mixed hyperbolic-parabolic type arereduced to equations of the type (2.1). In [3] it is shown that to a problem(2.1) reduces an analogue of a problem of Tricomi type for the hyperbolicparabolicequation with Gellerstendt operator in the body.On Euclid planes with the cartesian orthogonal coordinates x and ywe will consider the model equation in partial derivatives of the mixed(parabolic-hyperbolic) type|y| mH(−1) ∂2 u∂x 2 = ∂1+H(−y) u, (2.2)∂y1+H(−y) where m = const > 0, H(y) is the Heavyside function, u = u(x, y). Theequation (2.1) in the upper half-plane coincides with Fourier’s equation∂ 2 u∂x 2 = ∂u∂y(2.3)40
Boundary Value Problems for Differential Equations of Fractional Order 41and in the lower half-plane it coincides with(−y) m ∂2 u∂x 2 = ∂u∂ywhich for y = 0 transforms to an equation of parabolic type.Let Ω be an area bounded by the segments of straight linesand the characteristicsAA 0 : x = 0,A 0 B 0 : y = y 0 ,B 0 B : x = r,(2.4)AC : x − 2m+2(−y) 2 = 0m + 2andBC : x + 2m+2(−y) 2 = rm + 2of the equation (2.4); Ω + = {(x, y) : 0 < x < r, 0 < y < y 0 } be theparabolic part of the mixed area Ω; Ω − be the part of the area Ω, laying inthe lower half-plane y < 0 and bounded by the characteristics AC, 0 ≤ x ≤r2 , BC, r 2≤ x ≤ r and the segment AB; ]0, r[= {(x, 0) : 0 < x < r};{u + (x, y), ∀ (x, y) ∈ Ω + ,u =u − (x, y), ∀ (x, y) ∈ Ω − (2.5),where D l 0x is the operator of fractional integrodifferention of Sturm–Liouvilleof order |l| starting at the point 0.Consider a problem of Tricomi type for the equation (2.1) in Ω, withnonlocal condition of linear conjugation.Problem 1. Find a regular solution of the equation (2.1) in the areasΩ + , Ω − , with the following propertiesu + ∈ C(Ω + ) ∩ C 1( Ω + ∪]0, r[ ) , u − ∈ C(Ω − ) ∩ C 1( Ω − ∪]0, r[ ) , (2.6)u + (x, 0) = u − (x, 0) − λD0x −2 u− (t, 0), 0 ≤ x ≤ r, (2.7)∂(u + − u − )∂y ∣ = 0, 0 < x < r, (2.8)y=0u + (0, y) = ϕ 0 (y), u − (r, y) = ϕ r (y), 0 ≤ y ≤ y 0 , (2.9)u − ∣∣∣AC= ψ(0), 0 < x < r, (2.10)where Ω is the closure of Ω, λ is the spectral parameter, ϕ 0 (y) and ϕ r (y)are given functions of the class C 1 [0, r], and[ x( m + 2) 2/(m+2) ]ψ(x) = u2 , − x4is a given function of the class C 3 [0, r].
42 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangNonlocal condition of conjugation set by the equation (2.5), and localboundary conditions (2.6), (2.7) coincide with Tricomi conditions. For λ =0, the problem (2.1) is an analogue of Tricomi type problem.From (2.5) and (2.8) for x = 0 and (2.7) for y = 0, it follows thatthe equality ϕ 0 (0) = ψ(0) is a necessary condition for the consistency ofboundary data.We have the following lemmaThenLemma 2.1. Let (2.3) be a solution of Problem 1τ(x) = u − (x, 0), ν(x) = ∂u−dy ∣ = 0; x −β ν(x) ∈ L[0, r].y=0τ ′′ (x) − λτ(x) = ν(x), (2.11)Γ(2β)D 1−2β0x τ(t) = [ β m ν(x) + x β D 1−β0x ψ(t)] Γ(β), (2.12)τ(0) = ϕ 0 (0),τ(r) − λwhere Γ(z) is Eulers gamma-function, andβ =m2m + 4 ,β m =∫ r0(r − t)τ(t) dt = ϕ r (0), (2.13)Γ(2 − 2βΓ(1 − β) (2 − 4β)2β−1 .In fact, as u = u + (x, y) is a solution of the problem (2.1) in the area Ω + ,from (2.2) by (2.4) we have u +′′ (x, 0) = ν(x). On the other hand, accordingto the nonlocal condition of conjunction (2.5) u +′′ (x, 0) = r ′′ (x) − λτ(x).From this equality the equality (2.10) follows between τ(x) and ν(x). Theequation (2.11) represents another form of the known equation of theoriesof the relation between τ(x) and ν(x) known from the theory of mixed typeequations, brought from the area Ω − to the line of parabolic degenerationfor Cellersted’s equations. The condition (2.13) is a consequence of (2.5)and (2.7).Excluding from the system (2.10)–(2.11), we havewhereL βτ(x) (x) = λτ + ψ β (x), (2.14)L βτ(x) (x) = τ ′′ (x) − µ β D 1−2β0x τ(t), (2.15)µ β = Γ(2β)β m Γ(β) , ψ β = − 1 x β D 1−β0x ψ(t). (2.16)β mHence, due to the condition of Lemma 2.1 the function τ(x) = u − (x, 0)should be a solution of the following nonlocal problem.Problem 2. Find a solution τ(x) of the equation (2.13) of classC 2 ]0, r[ ∩C 1 [0, r[ , satisfying the condition (2.12).
Boundary Value Problems for Differential Equations of Fractional Order 43Passing to investigation of structural and qualitative properties of thesolution of the Problem 2, we will notice that any solution of the equation(2.13) of the class C 2 ]0, r[ ∩C 1 [0, r[ ∩[0, r[ will be a solution of the equationτ(x) = τ ′ (0)x+τ(0)+µ β D −20x D1−2β 0tτ(ξ)+λD −20x τ(t)+D1−2β 0t ψ β (t). (2.17)The equation (2.16) is deduced from the equality (2.13) after applicationof the operator D 1−2β0t to both its parts.It is easy to see thatD0x −2 D1−2β 0tτ(ξ)=∫ x0(x−t) ∂ ∂t D−2β 0t τ(ξ) dt=D −10x D−2β 0tτ(ξ)=D −1In view of the latter from (2.16) by virtue of (2.12) we have0x D−2β 0tτ(t).ϕ r (0) − ϕ 0 (0) = rr ′ (0) + µ β D −1−2β0r τ(t) + D0r −2 ψ β(t). (2.18)Substituting the value of r ′ (0) (2.17) into the equation (2.16), we haveτ(x) = µ β D0x −α τ(t) + a 1xD0x −α τ(t) + λD−2 0x τ(t) + f 0(x), (2.19)where a 1 = −µ βr, α = 1 + 2β = 2(m + 1)/(m + 2),f β (x) = D −20x ψ β(t) + ϕ 0 (0) + x r[ϕr (0) − ϕ 0 (0) − D −20x ψ β(t) ] . (2.20)The equation (2.18) is an integral Fredholm’s equation of the secondkind and it is equivalent to Problem 1.2. Formulas for Calculation of Eigenvalues of a Boundary ValueProblemIntroduce some formulas from the theory of disturbance which we needin the further. Let T (χ, ε) be a linear operator bunch,T (χ, ε) = T + χT ′ + εT ′′ ,where T is a complete self-adjoined operator all eigenvalues of which areisolated and have multiplicity equal to 1; T ′ and T ′′ are defined in the sameHilbert space as T ; T is bounded.T (χ, ε) − ζ = T − ζ + χT ′ + εT ′′ = [ J + (χT ′ + εT ′′ )R(ζ, T ) ] (T − ζ)if ζ is not anigenvalue of T (χ, ε). Then, the resolventexists if the termR(ζ, x, ε) = R(ζ, T ) [ J − (χT ′ − εT ′′ )R(ζ, T ) ] −1[J − (χT ′ − εT ′′ )R(ζ, T ) ] −1may be determined as a Neumann’s series[J − (χT ′ − εT ′′ )R(ζ, T ) ] −1∑∞ [= − (χT ′ − εT ′′ )R(ζ, T ) ] n,n=0
44 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tangi.e. the operator A, inverse to the operator [ J + (χT ′ + εT ′′ )R(ζ, T ) ] −1,exists and equals to∑ [− (χT ′ + εT ′′ )R(ζ, T ) ] n,and for this a sufficient condition is the condition∥∥χT ′ + εT ′′ ‖ ≤ ‖R(ζ, T ) ∥ ∥.So, if the equality (∗) is fulfilled, then(∗)R(ζ, T ) [ J + (χT ′ + εT ′′ )R(ζ, T ) ] −1=∞∑ (= R(ζ, T ) − (χT ′ + εT ′′ )R(ζ, T ) ) p= R(ζ, χ, ε). (2.21)p=0Let λ be an eigenvalue of the operator T = T (0, 0) of multiplicity m = 1,Γ be a closed positively oriented contour contained in the resolvent set andcontaining only eigenvalues λ of T . Consider the projectorP (χ, ε) = − 12iπ∫ΓR(ζ, χ, ε).Assertion 2.1.dim P (χ, ε) = dim P (0, 0) = 1.Proof of Assertion 2.1 follows from Lemma 1.4.10 [16].Assertion 2.2. Let λ be an eigenvalue of the operator T (χ, ε) correspondingto the eigenvalue λ of the operator T . Thenλ(χ, ε) = tr ( T (χ, ε)P (χ, ε) ) = λ + tr ( T (χ, ε) − λ ) P (χ, ε).Proof of this assertion follows from the definition and fromtrP (χ, ε) = dim P (χ, ε) = 1.This formula gives a complete solution of the eigenvalues problem.Assertion 2.3.λ(χ, ε) = − 12iπ tr ∫Γ(ζ − λ)R(ζ, χ, ε) dζ. (2.22)In fact,(T (χ, ε) − λ)R(ζ, χ, ε) =(T (χ, ε) − ζ + ζ)R(ζ, χ, ε) == [ (T (χ, ε) − ζ) + (ζ − λ) ] R(ζ, χ, ε) = J + (ζ − λ)R(ζ, χ, ε).Integrating both parts of the last inequality over ζ, we obtain− 1 ∫R(ζ, χ, ε) dζ = − 1 ∫dζ − 1 ∫(ζ − λ)R(ζ, χ, ε) dζ.2iπ2iπ 2iπΓΓΓ
Boundary Value Problems for Differential Equations of Fractional Order 45As∫Assertion 2.3 is proved.ΓR(ζ, χ, ε) dζ = P (χ, ε),∫Γdζ = 0,Assertion 2.4. The following equality holdsλ(χ, ε) − λ = − 1 ∫log [ 1 + (χT ′ + εT ′′ ) ] R(ζ, T ) dζ.2iπΓProof. From the formula (2.22) we haveλ(χ, ε) − λ = − 1 ∫ ∞∑2iπ tr 1p (ζ − λ) d [− (χT ′ + εT ′′ )R(ζ, T ) ] pdζ.dζΓp=0Substitution of (2.21) instead of the resolvent R(ζ, χ, ε) gives usλ(χ, ε) − λ = − 1 ∫2iπ tr (ζ − λ)R(ζ, χ, ε) dζ =Γ= 1 ∫2iπΓlog (1 + [ (χT ′ + εT ′′ )R(ζ, T ) ] ) dζ.Until now we have considered various power series of χ and ε and did notspecify explicitly a condition of their convergence. Now we investigate suchconditions.The series (2.21) obviously converges if the equality∥∥(χT ′ + εT ′′ ) ∥ ∥ < 1holds and this condition takes place if|χ| < 1 (‖T ′ R(ζ, T )‖ ) , |ε| < 1 (‖T ′′ ‖R(ζ, T ) ) .22Denote by r 0 (χ) and r 1 (ε) the values of χ and ε, respectively for which wehave(‖T ′ R(ζ, T )‖ ) −1 (= 2|χ|, ‖T ′′ R(ζ, T )‖ ) −1= 2ε.Clearly the series (2.21) converges in regular intervals on ζ ∈ Γ if|χ| ≤ r 0 = minζ∈Γ r 0(ζ),|ε| < r 1 = minζ∈Γ r 1(ζ). (2.23)Hence it is clear that the radia of convergence r 0 and r 1 depend on thecontour Γ.□Assertion 2.5. Let ρ = maxζ∈Γ r 1|ζ − λ|. Then|λ(χ, ε) − λ| < β 2π .
46 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangThenIn fact, from (2.22) we haveλ(χ, ε) − λ = tr [ T (χ, ε)P (χ, ε) ] = − 1 ∫2πi tr (ζ − λ)R(ζ, χ, ε) dζ.|λ(χ, ε) − λ| = 12π tr ∫Sincewe haveΓ(ζ − λ)R(ζ, χ, ε) dζ ≤≤ ρ2π tr ∫ρtrP (χ, ε) = dim P (χ, ε) = 1,2πΓΓR(ζ, χ, ε) dζ = ρ trP (χ, ε).2π|λ(χ, ε) − λ| < ρ(2π) −1 . (2.24)A simple analysis of the formula (2.24) shows that the less is ρ, the closergets λ(χ, ε) to λ as expected. Moreover, the less are χ, ε, the less should be|χ|, |ε| in general, for λ(χ, ε) to get within Γ.Assertion 2.6. Let T ′ R(ζ, T ) ∈ G F :∥∥(χT ′ + εT ′′ )R(ζ, T ) ∥ ∥ < 1, (2.25){T ′ R(ζ, T ) + T ′′ R(ζ, T ) } v = Σ [ a k (v i ϕ i )ϕ i + b k (v i ϕ i )ψ i], (2.26)where (ϕ i ϕ j ) = 0, (ϕ i ϕ i ) = 1, (ϕ i ψ j ) = 0. Then(T ′ R + T ′′ R) 2n = (T ′ R) 2n−1 + (T ′′ R) 2n−1 ,(T ′ R + T ′′ R) 2n+1 = (T ′ R) 2n+1 + T ′ (T ′′ ) 2n R 2n+1 .Proof. Consider the double power series1+ ( −χT ′ R(ζ, T ) ) + ( −χT ′ R(ζ, T ) ) + ( −χT ′ R(ζ, T ) ) 2+(−χT ′ R(ζ, T ) ) 3++ · · · + ( − εT ′′ R(ζ, T ) ) + ( − χT ′ R(ζ, T ) )( − εT ′′ R(ζ, T ) ) ++ ( − εT ′′ R(ζ, T ) )( − χT ′ R(ζ, T ) ) ++ · · · + ( − χT ′′ R(ζ, T ) ) + ( − εT ′′ R(ζ, T ) ) 2+ · · · .The general member of this series can be rewritten as[(χT ′ + εT ′′ )(R(ζ, T )) ] p.Rewrite the formula (2.26):[χT ′ R(ζ, T ) + εT ′′ R(ζ, T ) ] v 2 ={ [χT= ′ R(ζ, T ) ] }2+ χεT ′ R(ζ, T )T ′′ R(ζ, T ) + χεT ′′ R(ζ, T )T ′ R(ζ, T )}={χ 2[ T ′ R(ζ, T ) ] 2χε[T ′′ R(ζ, T )T ′ R(ζ, T ) ] v,v =
Boundary Value Problems for Differential Equations of Fractional Order 47[χT ′ R(ζ, T ) + εT ′′ R(ζ, T ) ] 2nv =[χ 2n (T ′ R) 2n + εχ 2n−1 T ′′ R(T ′ R) 2n−1] v,(T ′ R(ζ, T ) + T ′′ R(ζ, T ) ) 3v == ( T ′ R(ζ, T ) + T ′′ R(ζ, T ) ) 2(T ′ R(ζ, T ) + T ′′ R(ζ, T ) ) v == [ (T ′ R) 3 + T ′′ R(T ′ R) 2] v,[T ′ R + T ′′ R ] 2n+1v =[(T ′ R) 2n + T ′′ (T ′ ) 2n (R(ζ, T )) 2n+1] vwhich proves Assertion 2.6.□As a consequence, we note the following interesting fact.Corollary 2.1. Let T (x) = χT ′ + T , λ n (x) be the eigenvalues of T (x),λ n be the eigenvalues of T , where T ′ R(ζ, T ) and R(ζ, T ) are completelycontinuous operators, χ be a complex parameter, andR(ζ, T )T ′ = T ′ R(ζ, T ).Thenλ n (χ) − λ n = χtrT ′ P n , (2.27)where [16]P n = − 1 ∫R(ζ, T ) dζ.2iπΓ nProof. From the equalityλ n (χ) − λ n = 1 ∫ ( ∑ (−1)ptr(χT ′ R) p) dζ,2iπp!consideringandwe obtainΓ nd n+1(R(ζ, T )) = n!R(ζ, T )dζn+1 n! [ T ′ R(ζ, T ) ] n=[R(ζ, T )] nn!(T ′ ) n ,λ n (x) − λ n = 12iπ tr ∑ (−χ) n(T ′ ) n dn+1R(ζ, T ) dζ.nn! dζn+1 Calculate the integrals∫∫ddζ (R(ζ, T )) dζ =ThereforeΓ nCorollary 2.1 is proved.λ n (x) − λ n = − 12iπ tr ∫Γ nd(R(ζ, T )) = 0, n > 1.Γ nχT ′ R(ζ, T ) dζ = χtrT ′ p n .□
48 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangIf T coincides with a projector, the statement of Corollary 2.1 is obviouswithout planimetric integration. Asλ n (x) − λ n = − 1 ∫tr [ (ζ − λ)R(ζ, χ, ε) ] dζ,2iπwe haveλ n (x) − λ n = 1 ∫ ( ∑ [tr (χT ′ + εT ′′ )R(ζ, T ) ] P 1)dζ =2iπP !Γ n= 1 ∫ [tr χT ′ R(ζ, T ) + εT ′′ R(ζ, T ) + χ 2 (T ′ ) 2 (R(ζ, T )) 2 +2iπΓ nΓ n+χεT ′ R(ζ, T )T ′′ R(ζ, T )+χεT ′′ R(ζ, T )T ′ R(ζ, T )+ε 2 (T ′′ R(ζ, T )) 2 +· · ·]dζ.Theorem 2.1. Let the operator T ′ be permutable with the operatorT , and the operator T ′′ R(ζ, T ) be a nilpotent operator with parameter ofnilpotence equal to 2, T ′ R, T ′′ R ∈ G F . Thenλ n (x) − λ n = χspT ′ P n + εspT ′′ P n ,where P n is a Riesz’s projector of the operator T corresponding to λ n :P n = 1 ∫R(ζ, T ) dζ2iπΓ nProof. From (2.27),wherêλ nm = (−1)m+n+12iπλ n (χ, ε) − λ n =∫Γ n∞∑m=0 n=0∞∑χ m ε n̂λnm ,a nm tr(T ′ R(ζ, T )) m (T ′′ R(ζ, T )) n dζ.As the operator T ′′ R(ζ, T ) is nilpotent, we have∫̂λ 0m = (−1)m+1 Cmkk (T ′ ) m (R(ζ, T )) m ,2iπΓ n∫̂λ 1m = (−1)m+2 Cmkm (T ′′ ) m R(ζ, T )(T ′ )R m ,2iπΓ n∫̂λ jm = (−1)m+1 a jm (T ′′ R(ζ, T )) m T ′ R.2iπΓ n
Boundary Value Problems for Differential Equations of Fractional Order 49We obtain ̂λ 0m = χtrT ′ P n . Now as in proof of Corollary 2.1 we haveλ n (χ, ε) = χ̂λ 0m + ε̂λ 1m . For calculation of ̂λ 1m we will use∫Γ nd(R 2 (ζ, T )) = 0.Thus∫̂λ = (−1)m+2 a 1m T ′ (R(ζ, T ))(T ′′ R(ζ, T )) dζ =2iπΓ∫n= R(ζ, T )T ′ T ′′ R(ζ, T ) dζ = 0.Γ nTaking into account that( d) nR(ζ,T ) = n! [R(ζ, T )] n+1 (n = 1, 2, 3, . . . ),dζwe obtain∫Γ nT ′ T ′′ R k+1 (ζ, T ) dζ ={T ′ P n , k = 0,0, k ≠ 0.Thus ̂λ 0m = χtrT ′ P n , ̂λ 1m = εtrT ′′ P n , which proves Theorem 2.1.□Now using Theorem 2.1 we will calculate the eigenvalues of the problemu ′′ + cD 1−α u + λu = 0, u(0) = 0, u(π) = 0, 0 < α < 1, (2.28)where D 1−α is the operator of fractional order 1 − α in Weil sense, i.e.D 1−α u = 1 ∫2πdu(x − t)ψ 1−α (t) dt,2π dxψ 1−α (t) = 2∞∑k=10cos [kt − (1 − α)π/2]k 1−α .Suppose first that c(x) ≡ const. Represent the operator D 1−α R(ζ, T ) asthe sumD 1−α R(ζ, T ) = χT ′ R(ζ, T ) + εT ′′ R(ζ, T )so that T ′ and T ′′ satisfy the conditions of Theorem 2.1. Since the operator{−u ′′ ,T u =u(0) = 0, u(π) = 0,is a full self-adjoined operator,R(ζ, T )v =∞∑k=12 (v, sin kx)π k 2 − ζsin kx.
50 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangThusAsD 1−α R(ζ, T )v = 2 π D1−α( ∞ ∑k=12π(v, sin kx))k 2 sin kx.− ζD 1−α sin kx = d [D 1−α sin kx ] =dx= d [ (n −α sin nx − απ )]= n 1−α[ cos nx cos απdx22we obtainD 1−α R(ζ, T )v = 2 ∞∑ n 1−α (v, sin kx)[cos kx cos αππk2k=1= 2 απ∞∑ (v, sin kx)cosπ 2 k 2 cos kx + 2 απ∞∑sin− ζπ 2ThuswhereIndeed,k=1AR(ζ, T )v = 2 πk=1D 1−α R(ζ, T )v = A ˜R(ζ, T )v + B ˜R(ζ, T )v,Av = sin απ2Bv = cos απ2sinαπ2k=1∞∑(v, sin kx) sin kx,k=1∞∑(v, sin kx) cos kx.k=1∞∑k 1−α[ ∑∞ (v, sin kx)k 2 − ζk=1= 2 απ∞∑sinπ 2k=1BR(ζ, T )v = 2 απ∞∑cos k 1−α( ∑∞π 2k=11−α (v, sin kx)kk 2 ,− ζk=1(v, sin jx)k 2 − ζ= 2 απcosπ 2 ,[AR+BR] 2n+1 v = [ (AR) 2n+1 + BA 2n R 2n+1] v.+ sinαπ2 sin nx ],απ]+ sin =21−α (v, sin kx)kk 2 .− ζ]sin jx sin kx sin kx =)sin jx, sin jx cos kx =The fact that the operators A, R(ζ, T ) are permutable, is checked directly:AR(ζ, T )v = 2 απ∞∑cos k 1−α( ∑∞ (v, sin jx))π 2j 2 sin jx, sin kx sin kx =− ζ= 2 πsinαπ2k=1∞∑k=1j=11−α (v, sin kx)kk 2 − ζsin kx,
Boundary Value Problems for Differential Equations of Fractional Order 51R(ζ, T )Av = 2 απ∞∑ (Av, sin nx)cosπ 2 n 2 sin nx =− ζn=1∞∑= 2 ∞ π sin ∑n 1−α (v, sin kx) sin nxk=1n 2 sin nx =− ζn=1= 2 sin απ∞∑ n 1−α (v, sin kx)π 2 n 2 sin nx =− ζ= AR(ζ, T )v,n=1i.e.,AR(ζ, T )v = R(ζ, T )Av.Now show the nilpotence of the operator BR(ζ, T ):[ ] 2v BR(ζ, T ) = BR(ζ, T )BR(ζ, T )v == 2 π[ 2=π= 2 πcosαπ2cosαπ2∞∑k=1cosαπ2( 2π∞∑k=1] 2 ∞ ∑k=11−α (BR(ζ, T )v, sin kx)kk 2 − ζcosαπ2∞∑k=1cos kx =(v,sin jx)j 2 −ζcos jx, sin kx )k 2 − ζ( ∞∑ (v,sin jx)k 1−α j 2 −ζk=1cos kx =cos jx, sin kx ) cos kxk 2 = 0.− ζAccording to Theorem 1.4.1 [16], for the eigenvalues of the problem we haveµ k = n 2 + ̂λ 0m + ̂λ 1m .Calculate ̂λ∫0m = trAp n . As P n = 12iπR(ζ, T ) dζ, p n is an integral operatorwith the kernelΓ nP n (x, y) = 2 sin nx sin ny,πi.e.P n v = 2 π∫ π0sin nx sin nyv(y) dy = 2 (v, sin nx) sin nx.πWe obtainAp n v = 2 απ ∑sin k 1−α (v, sin nx)(sin nx, sin kx) sin kx =π 2= 2 απsinπ 2 (v, sin nx)n1−α sin nx.Find the eigenvalues of the operator AP n . AP n v = λv, that is,2 απsinπ 2 n1−α (v, sin nx) = vλ.
52 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangThus λ 1 = n 1−α 2 πi.e.,BP n = 2 π= 2 π= 2 πsinαπ2sinαπ2απsin2 , then A P n = λ 1 = 2 π n1−α sin απ2 . Further∑(pn v, sin nx) sin nx =∞∑k=1n 1−α( 2π (v, sin kx) sin kx, sin nx )cos nx =sinαπ2 (v, sin kx) cos nx = 2 πBP n v = 2 πCalculate the trace of the operator BP n∫παπsin2 cos nx v(t) sin nt dt,∫παπsin2 cos nx v(t) sin nt dt.BP n = ∑ λ n (BP n ).To find the eigenvalues of the operator BP n , we will solve the equationthat is,2πClearly B P n = 0. Thus0BP n v = vλ,∫παπsin2 cos nx v(t) sin nt dt = λv(t).0µ n = n 2 + n 1−α sin πα 2 .For solution of the problem (2.28) in case c(x) = const we use the followingtheorem being certainly of an independent interest as well.3. Estimation of EigenvaluesTheorem 2.2. Let the eigenvalues of the self-adjoint operator A 0 withdiscrete spectrum be λ (0)n = n q (n = 1, 2, 3, . . . , q), the normed eigenvectorsbe ϕ n and let B be a closed A 0 -bounded operator withD A0 ⊂ D B , ‖Bϕ n ‖ = O(n ρ )Then for the eigenvalues λ n of the operator A = A 0 + B we haveλ n − n q = O(n ρ ).0Proof. Let Γ n be the circle with radius 1 and with the center at the pointn q . Let n be so enough that inside of Γ n there are no eigennumbers exceptλ n . Let the operator A be defined by the formula A = A 0 + B. We needthe following known result.□
Boundary Value Problems for Differential Equations of Fractional Order 53Lemma 2.2. Let A 0 be an operator with compact resolvent R(ζ, A 0 ),operator B be compact with respect to A 0 . If ∆ is a bounded area withrectifiable boundary Γ and σ(A 0 + tB) ∩ Γ = ∅ for all t ∈ [0, 1], then theoperator A also has a compact resolvent andN(∆, A 0 + B) = N(∆, A 0 ),where N(A) is the number of eigenvalues of the operator laying inside thearea.Proof. Let R(ζ, A 0 ) be the resolvent of the operator A 0 . Then∞∑ (v, ϕ n )ϕ nR(A 0 , ζ) =n q .− ζn=1By Lemma 2.1 it follows that if ‖BR(ζ, A 0 )‖ Γn < 1, then σ(A 0 +tB)∩Γ = ∅for all t ∈ [0, 1] and‖BR(ζ, A 0 )‖ Γn =n=1∥∞∑n=1(v, ϕ n )ϕ nk q − ζ∥ ∥∥ΓnBϕ k ≤∞∑≤ ∥ (v, ϕ n)ϕ nk q ∥ ‖Bϕ k ‖ ≤ ‖(v, ϕ n)ϕ n ‖∞∑− ζ Γn |k q n ρ k ρ≤− ζ|k ρ − (n q − 1/3) .Further,=n−1∑n=1∞∑n=1k ρk ρ − (n q − 1/3) +k ρk ρ − (n q − 1/3) =n ρn−1n ρ − n q + 1/3 + ∑k=n+1n=1k ρk ρ − (n q − 1/3) .The last sum will be less than 1 if by means of finite disturbance the operatorA 0 is replaced by the operator Ã0v = A 0 v − 3n ρ (v, ϕ n )ϕ n .□Theorem 2.3 essentially strengthens and generalizes a known result ofM. K. Gavurin.Theorem of Gavurin. Let the eigenvalues of a self-adjoint operatorwith discrete spectrum be λ (0)n = n q (n = 1, 2, 3, . . . ; q > 1), the normedeigenfunctions ϕ n and B be a closed symmetric operator withD A0 ⊂ D B , ‖Bϕ n ‖ = O(n α ).Then, for the eigenvalues of the operator Av = A 0 v + Bv we have∑λ n = n q + ∑ k≠n(Bϕ k , ϕ n )|(Bϕ k , ϕ n )| 2|k q − n q | 2 −k≠n1 + ∑k≠n|(Bϕ k ,ϕ n )| 2|k q −n q | 2|(Bϕ k ,ϕ n )| 2|k q −n q | 2 +
54 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tang+∑k,l≠n1 + ∑(Bϕ k ,ϕ n )(Bϕ n ,ϕ l )(k q −n q )(l q −n q )k≠n|(Bϕ k ,ϕ n )| 2|k q −n q | 2 + O(n q−5 ).Here we will note that from Theorem 2.2 follows the known formula ofI. M. Lifshits, having greater applications in quantum statistics.Theorem of I. M. Lifshits. Let H be a self-adjoint hypermaximaloperator, T be a self-adjoint finite operator. Then∑(µn − λ n ) = spT,where µ n are the eigenvalues of the operator H + T , λ n = λ n (H).From Theorem 2.2 it follows a theorem generalizing Lifshits’s theorem.Theorem 2.4. Under the conditions of Theorem 2.2 the formula∞∑ ( )λn (χ, ε) − λ n = εspT ′ + χspT ′′n=1holds.Now in the case c(x) ≠ const from Theorem 2.2 followsTheorem 2.5. For the eigenvalues λ n of the problem the followingestimates holdλ n − n 2 = O(n α ).Proof of this theorem follows from the proof of Theorem 2.3 for q = 2.□4. Cauchy Problem for Differential Equations with FractionalDerivativesConsider the problem−u ′′ + D α 0xu + λu = 0, (2.29)u(0) = 0, u ′ (0) = 1. (2.30)We know [63] that the problem (2.29)–(2.30) has a unique solution u(x, λ)for any λ. For the further it will be important to know whether the functionu(x, λ) will be an entire function of genus zero.Theorem 2.6. The solution u(x, λ) of the problem (2.29)–(2.30) is anentire function of genus zero of the parameter λ.Proof. Assume that u 0 = x andu n (x, λ) = u 0 (x, λ) +∫ x0{}(x − t) 1−α u n−1 (t, λ) + λ(x − t)u n−1 (t, λ) dt.
Boundary Value Problems for Differential Equations of Fractional Order 55Let |λ| ≤ N. ThenFor n ≥ 2 we have≤Hence≤∫ x0∣∣u 1 (x, λ) − u 0 (x, λ) ∣ ≤ x2−α2 − α + N x22 .∣∣u 2 (x, λ) − u 0 (x, λ) ∣ ∣ ≤{}(x − t) 1−α u(t) + λ(x − t)u 1 (t) − (x − t) 1−α t − λ(x − t)t dt.x 3−α(x3+N(2−α)(3−α) 3! +N∣∣u 2 (x, λ) − u 1 (x, λ) ∣ ∣ ≤x 3−α(2−α)(3−α)and in general∣∣u n (x, λ) − u n−1 (x, λ) ∣ ≤ (N + M) n[Hence the seriesu(x, λ) = u 0 (x, λ) +) [=(N +1)x n−αx 3−α(2−α)(3−α)],(2 − α)(3 − α) · · · (n − α) + N xnn!].∞∑ {un (x, λ) − u n−1 (x, λ) } (2.31)n=1converges in regular intervals of λ, for |λ| ≤ N and for 0 ≤ x ≤ 1. As forany n ≥ 2=∫ x0u ′ n(x, λ) − u ′ n−1(x, λ) ={(x − t) −α (u n−1 (t, λ) − u n−2 (t, λ)) + λ ( u n−1 (t, λ) − u n−2 (t, λ) )} dt,u ′′ n(x, λ) − u ′′ n−1(x, λ) = {D α 0x + λ} ( u n−1 (x, λ) − u n−2 (x, λ) ) ,we have that the series obtained by once and twice differentiation of theseries (2.31) also converge in regular intervals of x. Thus∞∑u ′′ {(x, λ) = u′′n(x, λ) − u ′′ n−1(x, λ) } =n=1= u ′′1(x, λ) − u ′′0(x, λ) += {D α 0x + λ}{u 0 (x, λ)} +∞∑n=2{u′′n(x, λ) − u ′′ n−1(x, λ) } =∞∑ {un (x, λ) − u n−1 (x, λ) } =n=2= {D α 0x + λ}{u 0 (x, λ)} + λu 0 (x, λ),and u(x, λ) satisfies the equation (2.29). Clearly u(x, λ) satisfies the conditions(2.30).
56 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangNow, along with the problem (2.29)–(2.30), we will consider the followingproblemv ′′ − v ′ (x) = λv(x), v(0) = 0, v ′ (0) = 1.It is known that the solution v(x, λ) of this problem is an entire function ofgenus zero. Take a v 0 (x, λ) and letv n (x, λ) = v 0 (x, λ) +Write u n (x, λ) and v n (x, λ) as∫ x0{1 + λ(x − t)vn−1 (t, λ) } dt.u n (x, λ) = a 0 (x) + λa 1 (x) + · · · + λ n a n (x),v n (x, λ) = b 0 (x) + λb 1 (x) + · · · + λ n b n (x).Clearly |a i (x)| < b i (x), x ∈ (0, 1). Then, from the theorem of Hadamardfollows that u(x, λ) is an entire function of genus zero.□5. On a Method of Estimation of First EigenvaluesConsider the problem−u ′′ + D α 0xu + λu = 0, (2.32)u(0) = 0, u ′ (0) = 1. (2.33)Let u(x, λ) be a solution of the problem (2.32)–(2.33). We have alreadyestablished that u(x, λ) is an entire function of genus zero. Hence it ispossible to represent it as an infinite product∞∏u(x, λ) = c(1 − λ ),λ jj=1where c is a constant unknown as yet, λ j are zeros of the function u(1, λ).Since the zeros of the function u(1, λ) coincide with the eigenvalues of theproblem (2.32)–(2.33)−u ′′ + D α 0xu = λu, (2.32 ′ )u(0) = 0, u(1) = 0, (2.33 ′ )the investigation of the eigenvalues of the problem (2.32 ′ )–(2.33 ′ ) is reducedto the investigation of the zeros of the function u(1, λ). First of all we needthe following interesting statement.Lemma 2.3. All eigenvalues of the problem (2.32 ′ )–(2.33 ′ ) are positive.Proof. For proof of the given statement we will consider the following problem−u ′′ + D α 0xu = λu, u(0) = 0, u(0) = 0.
Boundary Value Problems for Differential Equations of Fractional Order 57This problem is equivalent to the equationDenoteu(x) = x +K 1 (x, t) =∫ x0[(x − t) 1−α − λ(x − t) ] u(t) dt. (2.34){[(x − t)1−α ] , 0 ≤ t < x < 1,0, x < t ≤ 1.We will define the further sequence of kernels {K n (x, t)} ∞ 1recurrent equalitiesK n+1 (x, t) =∫ xElementary calculations show thatK 2 (x, t) = λΓ(1 − α)Γ(2 − α)Γ(4 − 2α)Using induction in n we obtaint− 2λK n (x, t 1 )K 1 (x, t 1 ) dt 1 .(x − t) 3−2α −Γ(2)Γ(2 − α)Γ(4 − α)by means of the(x − t) 3−α + 1 3! (x − t)3 .K n+1 (x, t) = K 0 n+1(x, t) − λK 1 n+1(x, t) + λK 2 n+1(x, t) + · · · , x ≥ t.Elementary calculations show that Kn+1(x, i t) ≥ 0 for x ≥ t for any n. Fromhere for the resolvent of the equation (2.34) we have the formula∞∑R(x, λ, t) = K n+1 (x, t, λ) = a 0 (x, t) − λa 0 (x, t) + λ 2 a 2 (x, t) + · · · ,n=0where a i (x, t) ≥ 0 for x ≥ t. Since the solution (2.34) is represented aswe haveu(1, λ)1 +∫ 10u(x) = x +∫ x0R(x, λ, t)t dt,R(1, λ, t)t dt = ã 0 − λã 1 + λ 2 ã 2 − λ 3 ã 3 + · · · ,where ã 0 , ã 1 , . . . , ã n , . . . are nonnegative numbers. Thus λ is an eigenvalueof the problem (2.32)–(2.33), iff λ is a zero of the function w(λ) = u(1, λ) =ã 0 − λã 1 + λ 2 ã 2 − · · · . But it is obvious that w(λ) hasn’t negative zeros.From the estimates obtained in (2.32) it follows that the eigenvalues{λ j } of the problem satisfy the equality j=1 ∑∞λ −1j< ∞. Therefore for the
58 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tangoperator A induced by the differential equation (2.32 ′ ) and the boundaryconditions (2.33 ′ ) we can introduce the determinant∞∏ (D a (λ) = 1 − λλ−1j (A) ) .j=1Clearly D a (λ) = u(1, λ). Further, as in [3], we will consider the logarithmicderivative of the function[ln (DA (λ)) ] ′ D ′=A (λ)∞D A (λ) = − ∑χ n+1λ n , (2.35)n=1where χ n= spurA n . By virtue of the equality (2.35) we haveu ′ (λ, 1)u(λ, 1) = − ∑ χ n+1λ n .Let Taylor’s series u(x, λ) look likeu(x, λ) =∞∑S n (x)λ n ,n=0where u(x, λ) is the solution of the problem (2.32 ′ )–(2.33 ′ ). Then∑ nSn (1)λ n−1S n λ n = ∑ χ n+1λ n .From this equality we haven∑χ n+1+ S m (1)χ n+1−m= −(n + 1)λ n .m=1These equalities form a system of non-homogeneous linear equations for χ n+1S 0 (1)χ 1+ 0 ∗ χ 2+ 0 ∗ χ 3+ · · · + 0 ∗ χ n= S 1 ,S 1 (1)χ 1+ S 0χ 2+ 0 ∗ χ 3+ · · · + 0 ∗ χ n= S 2 ,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .S n−1 (1)χ 1+ S n−2 χ 2+ · · · + S 0 (1)χ n= −nS n .Hence for determining χ jwe have the recurrent formulas[n−1∑= − nS n (1) + S p−1 (1)χ n−p]. (2.36)χ nFrom (2.36) it follows that for determining χ n(n = 1, 2, 3, . . . ) it isenough to know that S n (1) (n = 1, 2, 3, . . . ). We know that the problemis equivalent to the equationp=1−u ′′ + D α 0xu = λu, u(0) = 0, u(1) = 1u(x) = x +∫ x0{(x − t) 1−α − λ(x − t) } dt. (2.37)
Boundary Value Problems for Differential Equations of Fractional Order 59Since the solution (2.37) is an entire function of the parameter λ, we have∞∑u(x, λ) = S n (x)λ n . (2.38)Substituting (2.38) in (2.37) we have= x +∫ x0From (2.39) it followsn=0S 0 (x) + λS 1 (x) + λ 2 S 2 (x) + · · · =[(x − t) 1−α − λ(x − t) ][ S 0 (t) + · · · + λ n S n (t) + · · · ] dt. (2.39)S 0 (x) = x +S 1 (x) = −S 2 (x) =∫ x0∫ x0∫ x0(x − t) 1−α u(t) dt, (2.40)(x − t)S 0 (t) dt +∫ x(x − t) 1−α S 2 (x) dt +0∫ x(x − t) 1−α S 1 (x) dt, (2.41)0(x − t) 1−α S 1 (x) dt, (2.42). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Solving the equation (2.40), we obtainS 0 (x) = xE ρ (x 1/ρ ; 2); S 0 (1) = 1.For solution of the equation (2.41) we will calculateThen∫ x0(x − t)S 0 (t) dt =a 0 (x)=x 3 E ρ (x 1/ρ ; 4)+∫ x0∫ x0tE ρ (t 1/ρ ; 2)(x − t) dt = x 3 E ρ (x 1/ρ ; 4).((x − t) 1/ρ−1 E ρ (x−t) 1/ρ ; 1 )t 3 E ρ (t 1/ρ ; 4) dt =ρ=−c 1(x 3 E ρ (x 1/ρ ; 4)+c 0 ρx 3+1/ρ[ E ρ(x 1/ρ ; 3+ 1 ρIt is likewise possible to show that(− 3+ 1 ( )E ρ x 1/ρ ; 4+ 1 ρρ)] ) .a 2 (x) = −cx 5 E ρ (x 1/ρ ; 6) + cρx 5+1/ρ E ρ(x 1/ρ ; 5 + 1 ρ)−(− cρ 5 + 1 ()x 5+1/ρ E ρ x 1/ρ ; 6 + 1 )−ρρ)−
60 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tang(− c[ρx 5+1/ρ E ρ(+ cρ{ρx 2/ρ+5 E ρx 1/ρ ; 5 + 1 ρ) (− ρ 5 + 1 ()x 5+1/ρ E ρ x 1/ρ ; 6 + 1 ρρ) ] +()x 2/ρ+5 E ρ x 1/ρ ; 6 + 1 )−ρx 1/ρ ; 5 + 1 ρ(− cρ 5 + 1 ()[x 2/ρ+5 E ρ x 1/ρ ; 6 + 1 ) (− 5 + 2 ( )E ρ x 1/ρ ; 6 + 1 )x 2/ρ+5 +ρρ ρρ∞∑+ 12ρk=0Now note that from (2.36) it follows thatx (k+2)/ρ−5 k 2)] }Γ ( kρ + 6 + 2 .ρS 1 (1) = −χ 1, S 2 (1) = 1 2 (χ2 1 − χ 2).As all eigenvalues of the problem (2.27 ′ )–(2.28 ′ ) are positive, obviouslyλ 1 > 1 χ 1= − 1S 1 (1) .The estimate from below for λ 1 looks like λ 1 < χ 1/χ 2.Now taking into account that it is possible to calculate S 1 and S 2 towithin 10 −2 , we will obtainTheorem 2.7. For the first eigenvalue λ 1 of the problem (2.32 ′ )–(2.33 ′ )we have the relation(1.85) −1 < λ 1 < 3.86.Note that it is likewise possible to find estimates for the eigenvalues ofthe problem−u ′′ + D α 00x u + Dα 10x u = λu,u(0) = 0, u(1) = 0.6. Mutually Adjoint Problems and Problem of Completeness ofEigenfunctionsFor the equationu ′′ +consider the problemn∑i=1a i (x)D α i0x ω j(x)u = λu, 0 < α i < 1, (2.43)u(0) = 0, u(1) = 0. (2.44)Along with the problem (2.43)–(2.44), we consider the problemn∑z ′′ + ω j (x)(D α i0x )∗ a i z + λz = 0, (2.45)i=1z(0) = 0, z(1) = 0. (2.46)□
Boundary Value Problems for Differential Equations of Fractional Order 61Here (D α i0x )∗ is the adjoint operator to the operator D α i0x i.e.(D α 1i0x )∗ u = −Γ(1 − α)∫d1dxxu(t)(t − x) α dt,in a certain sense associated with the problem (2.43)–(2.44). We will callthese problems mutually adjoint.Let {v n (x)} ∞ n=1 be a system of eigenfunctions of the problem (2.43)–(2.44), and {z n (x)} ∞ n=1 be a system of eigenfunctions of the problem (2.45)–(2.46).We will establish that the systems of functions {v n (x)} ∞ n=1, {z n (x)} ∞ n=1are biorthogonal on [0, 1].Theorem 2.8. The system of eigenfunctions {v n (x)} ∞ n=1 is completein L 2 (0, 1).Proof. Introduce the operator{−u ′′ ,Mu =u(0) = 0, u(1) = 0.Then the problem (2.43)–(2.44) is equivalent to the equationu + M −1{ n∑i=1}a i (x)D αi0x ω i(x)u − λM −1 u = 0i.e., the problem is reduced to investigation of operators of Keldysh type,where∫ 1{M −1 t(x − 1), t ≤ x,u = G T (x, t)u dt, G T (x, t) =x(t − 1), t ≥ 1.0Clearly M −1 is a complete self-adjoint operator,M −1( n ∑i=1)a i (x)D α i0x ω i(x)u ∈ G 1 ,M −1 is a kernel operator. Then, from a theorem of Keldysh [3], it followsthe corresponding completeness. Denote by n(r) the exact number ofeigenvalues of the problem (2.43)–(2.44) laying in the circle |λ| < r.Problems on distribution of eigenvalues consist in studying asymptoticproperties of n(r) as r → ∞.□Theorem 2.9. The following equality holdslimr−→∞ n(r)/r1/2 = 1.
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)
64 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangProof. Due to the definition of the functions u(x, λ) and z(x, λ ∗ ) as solutionsof problems of Cauchy type, we have (2.49)-(2.50) andz ′′ (x, λ ∗ ) [ a m z(x, λ ′′ ) ] ′+ ωi (x)D αi0x a i(x) + λ ∗ z = 0,z(1, λ ∗ ) = sin β, z ′ (1, λ ∗ ) = − cos β.Multiplying both parts of the first equation of (2.47) by z(x, λ ∗ ) andintegrating it from 0 to 1, we have+∫ x0∫ x0= ω(λ) − ˜ω(λ ∗ )+∫ 10z(x, λ ∗ )u ′′ (x, λ) dx +∫ 10a m (x)z(x, λ ∗ )u ′ (x, λ) dx+∫xa i (x)z(x, λ ∗ )D αi0x u(x, λ) dx + λu(x, λ)z(x, λ ∗ ) dx =∫ x0u(x, λ)z ′′ (x, λ) dx −0∫ 1ω i (x)D α ix1 a i(t)z(x, λ ∗ )u(x, λ) dx + λ0∫ 10[am (x)z(x, λ ∗ ) ] ′u(x, λ)+u(x, λ)z(x, λ ∗ ) dx. (2.54)At the same time, we multiply both parts of the first equation of (2.48) byu(x, λ) and integrate it from 0 up to 1. We have+∫ 10∫ 10z(x, λ ∗ )u(x, λ) dx −∫ 1ω i (x)D αix1 a i(t)z(x, λ ∗ )u(x, λ) dx + λSubtracting (2.55) from (2.56), we obtain∫ 100[am (x)z(x, λ ∗ ) ] ′u(x, λ) dx+∫ 1z(x, λ ∗ ) dx = ˜ω(λ∗ ) − ω(λ)λ − λ ∗ .0u(x, λ)z(x, λ ∗ ) dx = 0. (2.55)So we obtain an analogue of identity of M. M. Dzhrbashyan.Theorem 2.12. The formulaholds if λ = λ ∗ .ω(λ) = ˜ω(λ) (2.56)The equality (2.56) follows from the identity (2.53).We needed to construct a biorthogonal system of eigenfunctions andassociated functions of mutually adjoint problems (2.47).□
Boundary Value Problems for Differential Equations of Fractional Order 65Let {λ n } be a sequence of all eigenvalues of the problem (2.47) arrangedin the non-decreasing order of moduli 0 ≤ |λ 1 | ≤ |λ 2 | ≤ · · · ≤ |λ n | ≤ · · ·Note that a same eigenvalue may appear repeatedly in the above sequence.According to M. M. Dzhrbashyan [1], for each natural number n ≥ 1,we denote by p n the multiplicity of occurrence of the number λ n in thesequence { λ 1 , |λ 2 |, . . . , |λ n |, . . . } .Let {u n (x)} ∞ n=1, {z n (x)} ∞ n=1 be sequences of eigenfunctions of the problems(2.47) and (2.48), respectively. As in [7] we prove that the systems offunctions {u n (x)} ∞ n=1, {u ∗ n(x)} ∞ n=1 are continuous on [0, 1), and the systemsof functions {z n (x)} ∞ n=1, {z ∗ n(x)} ∞ n=1 are continuous on (0, 1].We will call a system of functions {u n (x)} ∞ n=1 a normal system of eigenfunctionsand associated functions of problem (2.47), and system of functions{z n (x)} ∞ n=1 a normal system of eigenfunctions and associated functionsof problem (2.48).Then we have the following important theorem of biorthogonality ofthese constructed systems.Theorem 2.13. The system of functions {u n (x), z n (x)} is biorthogonalon [0, 1] i.e.∫ 1∫ 1{δ mn 1, for n ≠ m,u n (x)z n (x) dx = z n (x)u n (x) dx =0, for n = m.00Proof of this theorem is bulky and mainly does not differ from the proof ofthe corresponding theorem of M. M. Dzhrbashyan [1] if there is the identity∫ 10u n (x)z n (x) dx = ω(λ) − ˜ω(λ∗ )λ − λ ∗ .Systems of eigenfunctions and associated functions (3.19)–(3.22) areconstructed only on the basis of the identity (3.23) without reference tothe problem (2.47). Naturally from the identity it is possible to obtain thesufficient information on the spectrum of the problem (2.47).
CHAPTER 3Solving Two-Point Boundary Value Problemsfor Fractional Differential Equations (FBVPs)1. The Basic ConceptsIn [8] and [33], the existence and uniqueness of solutions, and numericalmethods for FBVPs with Caputo’s derivatives and with Riemann–Liouvillederivatives are studied, respectively.FBVPs with Caputo’s derivatives:andCa D γ t y(t) + f (t, y(t)) = 0, a < t < b, 1 < γ ≤ 2, (3.1)y(a) = α, y(b) = β (3.2)Ca D γ t y(t) + g ( t, y(t), C a D θ t y(t) ) = 0, a < t < b, (3.3)y(a) = α, y(b) = β, 1 < γ ≤ 2, 0 < θ ≤ 1, (3.4)where y : [a, b] ↦→ R, f : [a, b] × R ↦→ R, g : [a, b] × R × R ↦→ R are continuousand satisfy Lipschitz conditions|f(t, x) − f(t, y)| ≤ K f |x − y|, (3.5)∣∣g(t, x 2 , y 2 ) − g(t, x 1 , y 1 ) ∣ ≤ K g |x 2 − x 1 | + L g |y 2 − y 1 | (3.6)with Lipschitz constants K f , K g , L g > 0.FBVPs with Riemann–Liouville derivatives:andRLa D γ t y(t) + f (t, y(t)) = 0, a < t < b, 1 < γ ≤ 2, (3.7)y(a) = 0, y(b) = β, (3.8)RLa D γ t y(t) + g ( t, y(t), RLa Dt θ y(t) ) = 0, a < t < b, (3.9)y(a) = 0, y(b) = β, 1 < γ ≤ 2, 0 < θ ≤ γ − 1, (3.10)where y : [a, b] ↦→ R, f : [a, b] × R ↦→ R, g : [a, b] × R × R ↦→ R are continuousand satisfy Lipschitz conditions (3.5) and (3.6), respectively.In order to state the problems in concern, we introduce the definitionsand properties of Caputo’s derivatives and Riemann–Liouville fractionalintegrals in [8].Definition 3.1 (see [29]). Let γ > 0, n−1 < γ ≤ n and let C n [a, b] :={y(t) : [a, b] → R; y(t) has a continuous n-th derivative}.66
Boundary Value Problems for Differential Equations of Fractional Order 67(1) The operator RLa D γ t defined byRLa D γ t y(t) = dn 1dt n Γ(n − γ)∫ ta(t − τ) n−γ−1 y(τ) dτ (3.11)for t ∈ [a, b] and y(t) ∈ C n [a, b], is called the Riemann–Liouville differentialoperator of order γ.(2) The operator C a D γ t defined byCa D γ 1t y(t) =Γ(n − γ)∫ ta(t − τ) n−γ−1( d) ny(τ)dτ (3.12)dτfor t ∈ [a, b] and y(t) ∈ C n [a, b], is called the Caputo differential operator oforder γ.Definition 3.2 (see [29]). Provided γ > 0, the operator Ja γ , definedon L 1 [a, b] byJ γ a y(t) = 1Γ(γ)∫ ta(t − τ) γ−1 y(τ) dτ (3.13)for t ∈ [a, b], is called the Riemann–Liouville fractional integral operator oforder γ, where L 1 [a, b] := { y(t) : [a, b] → R; y(t) is measurable on [a, b] and∫ ba|y(t)| dt < ∞ } .Lemma 3.1 (see [31]). (1) Let γ > 0. Then, for every f ∈ L 1 [a, b],RLa D γ t J γ a f = f (3.14)almost everywhere.(2) Let γ > 0 and n − 1 < γ ≤ n. Assume that f is such that Jan−γ f ∈A n [a, b]. Then,n−1∑Ja γ RLa D γ (t − a) γ−k−1t f(t) = f(t) −Γ(γ − k)k=0limz→a +(3) Let γ > θ > 0 and f be continuous. ThenCa D γ t J γ a f = f,d n−k−1n−γJdzn−k−1 a f(z). (3.15)Ca Dt θ Ja γ f = Ja γ−θ f. (3.16)(4) Let γ ≥ 0, n − 1 < γ ≤ n and f ∈ A n [a, b]. Thenn−1∑Ja γ C a D γ t f(t) = f(t) −k=0D k f(a)k!(t − a) k , (3.17)where A n [a, b] is the set of functions with absolutely continuous derivativeof order n − 1.
68 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangLet S be a Banach space, T : S ↦→ S be a mapping, and ‖ · ‖ denote thenorm of S.Definition 3.3 (see [25]). If there exists a constant ρ (0 ≤ ρ < 1)such that‖T x − T y‖ ≤ ρ‖x − y‖ (3.18)for any x, y ∈ S, then T is said to be a contractive mapping of S.Lemma 3.2 (Contractive Mapping Principle) (see [25]). If T isa contractive mapping of a Banach space S, then there exists a unique fixedpoint y ∈ S satisfying y = T y.2. Existence and Uniqueness of the Solutions for FBVPsIn this section, the existence and uniqueness of the solutions for FBVPs(3.7)–(3.8), (3.9)–(3.10), (3.1)–(3.2) and (3.3)–(3.4) are studied. Withoutloss of generality, only the case of homogeneous boundary conditions α =β = 0 in the above four kinds of FBVPs are considered.Lemma 3.3. (1) FBVP (3.1)–(3.2) is equivalent toy(t) =∫ baG(t, s)f(s, y(s)) ds, (3.19)where G(t, s) is called the fractional Green function defined as follows:⎧(t − a)(b − s)⎪⎨γ−1 (t − s)γ−1− , a ≤ s ≤ t ≤ b,(b − a)Γ(γ) Γ(γ)G(t, s) =(t − a)(b − s) ⎪⎩γ−1(3.20), a ≤ t ≤ s ≤ b.(b − a)Γ(γ)(2) FBVP (3.7)–(3.8) is equivalent toy(t) =∫ baĜ(t, s)f(s, y(s)) ds, (3.21)where Ĝ(t, s) is the fractional Green function defined as follows:⎧(t − a)⎪⎨γ−1 ( b − s) γ−1 (t − s) γ−1− , a ≤ s ≤ t ≤ b,Ĝ(t, s) =Γ(γ) b − a Γ(γ)(t − a) ⎪⎩γ−1 ( b − s) γ−1(3.22), a ≤ t ≤ s ≤ b.Γ(γ) b − a(3) FBVP (3.3)–(3.4) is equivalent toy(t) =∫ baG(t, s)g ( s, y(s), C a D θ sy(s) ) ds, (3.23)where G(t, s) is the fractional Green function defined in (3.20).
Boundary Value Problems for Differential Equations of Fractional Order 69(4) FBVP (3.9)–(3.10) is equivalent toy(t) =∫ baĜ(t, s)g ( s, y(s), RLa D θ sy(s) ) ds, (3.24)where Ĝ(t, s) is the fractional Green function defined in (3.22).Proof. We only give the proof for (1). The case for (3) is similar to that for(1). The proofs of (2) and (4) are referred to [33].According to Lemma 3.1(4), applying the operator J γ a to both sides ofthe equation in (3.1) yieldsy(t) − y(a) − y ′ (a)(t − a) + J γ a f(t, y(t)) = 0. (3.25)Since y(a) = y(b) = 0, from (3.25) one easily obtainsand theny ′ (a) =∫ by(t) = y ′ (a)(t − a) − J γ a f(t, y(t)) ===∫ ba∫ baa(b − s) γ−1f(s, y(s))ds, (3.26)(b − a)Γ(γ)(t − a)(b − s) γ−1∫ tf(s, y(s)) ds −(b − a)Γ(γ)aG(t, s)f(s, y(s)) ds.Conversely, applying the operator C a D γ t∫ bCa D γ t y(t) = C a D γ t=∫ baaG(t, s)f(s, y(s)) ds =(t − s) γ−1f(s, y(s)) ds =Γ(γ)to both sides of (3.19) yields(b − s) γ−1(b − a)Γ(γ) f(s, y(s)) dsC a D γ t (t − a) − C a D γ t J γ a f(t, y(t)) == −f(t, y(t)),and the homogeneous boundary condition is verified easily.LetP = C 0 [a, b],P 1 = C 1 [a, b] := { y(t) : y(t), y ′ (t) ∈ C 0 [a, b] } ,{}P 2 = C θ [a, b] := y(t) : y(t) ∈ C 0 [a, b], RLa Dt θ y(t) ∈ C 0 [a, b] .□
70 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tangwhere C 0 [a, b] denotes the space of all continuous functions on [a, b]. Wedefine the norms ‖·‖ f , ‖·‖ g , ‖·‖ĝ and the operator T f , ̂T f , T g , ̂T g as follows:‖y‖ f := K fT f y(t) :=̂T f y(t) :=∫ ba∫ bamax |y(t)|,a≤t≤bG(t, s)f(s, y(s)) ds,Ĝ(t, s)f(s, y(s)) ds, ∀ y(t) ∈ P;[‖y‖ g := max Kg |y(t)| + L g | C a Dt θ y(t)| ] ,a≤t≤bT g y(t) :=∫ baG(t, s)g ( s, y(s), C a D θ sy(s) ) ds, ∀ y(t) ∈ P 1 ,[‖y‖ĝ := max Kg |y(t)| + L g | RLa Dt θ y(t)| ] ,a≤t≤b̂T g y(t) :=∫ baĜ(t, s)g ( s, y(s), RLa D θ sy(s) ) ds, ∀ y(t) ∈ P 2 .Obviously, P, P 1 , P 2 are complete normed spaces with respect to ‖ · ‖ f ,‖ · ‖ g and ‖ · ‖ĝ, respectively. The operators T f , ̂T f , T g , ̂T g are continuous.Now (3.19) and (3.21) can be rewritten as y = T y and y = ̂T f y, for y ∈ P;(3.23), (3.24) can be rewritten as y = T g y and y = ̂T g y, respectively. Accordingto the contractive mapping principle, “finding a sufficient conditionfor existence and uniqueness of the solution for FBVP (3.1)–(3.2) (or (3.7)–(3.8) or (3.3)–(3.4) or (3.9)–(3.10))” is equivalent to “finding a sufficientcondition under which T f (or ̂T f or T or ̂T g ) is a contractive mapping ofP(or P 1 or P 2 )”. Then, the following theorems are true.Theorem 3.1. (1) Let f be a continuous function on [a, b] × R andsatisfy the Lipschitz condition (3.5).(1.1) If2K f (b − a) γ< 1, (3.27)Γ(γ + 1)then there exists a unique solution for FBVP (3.1)–(3.2) in P.(1.2) If(γ − 1) γ−1 (b − a) γK fγ γ < 1, (3.28)Γ(γ + 1)then there exists a unique solution for FBVP (3.7)–(3.8) in the space P.(2) Let g be a continuous function on [a, b] × R × R and satisfy theLipschitz condition (3.6).
Boundary Value Problems for Differential Equations of Fractional Order 71(2.1) IfK g2(b − a) γΓ(γ + 1) + L g[(b − a) γ−θΓ(γ + 1)Γ(2 − θ) + (b −]a)γ−θ< 1, (3.29)Γ(γ + 1 − θ)then there exists a unique solution for FBVP (3.3)–(3.4) in P 1 .(2.2) IfK g(γ − 1) γ−1 (b − a) γγ γ Γ(γ + 1)+ L g(2γ − θ)(b − a) γ−θγΓ(γ − θ + 1)< 1, (3.30)then there exists a unique solution for FBVP (3.9)–(3.10) in the space P 2 .Proof. The proof for (1.1): For u(t), v(t) ∈ P, and (t, s) ∈ [a, b] × [a, b],according to the definition for the operator “T f ”, we have∣ Tf u(t) − T f v(t) ∣ ∣ ≤∫ba{ ∫b≤ ‖u − v‖ fThus∥ Tf u − T f v ∥ ∥f= maxaa≤t≤b K f|G(t, s)| · |f(s, u(s)) − f(s, v(s))| ds ≤(t − a)(b − s) γ−1(b − a)Γ(γ)∫ tds +≤a(t − s) γ−1Γ(γ)2(b − a)γΓ(γ + 1) ‖u − v‖ f .}ds ≤∣ Tf u(t) − T f v(t) ∣ 2K f (b − a) γ≤ ‖u − v‖ f .Γ(γ + 1)Considering (3.27), we finish the proof according to Lemma 3.2.The proof for (2.1): On one hand, we have∣∣T g u(t) − T g v(t) ∣ 2(b − a)γ≤Γ(γ + 1) ‖u − v‖ gfor u(t), v(t) ∈ P 1 , and (t, s) ∈ [a, b] × [a, b], which is similar to (1.1). Onthe other hand, according to Lemma 3.1, we haveCa D θ t T g u(t) =∫ b= (t−a)1−θ (b−s) γ−1Γ(2−θ) Γ(γ)(b − a) g( s, u(s), C a Dsu(s) θ ) ds−Ja γ−θ g ( t, u(t), C a Dt θ u(t) ) .aThen∣∣Ca Dt θ T g u − C a Dt θ T g v ∣ [≤ ‖u − v‖ g ·(b − a) γ−θΓ(2 − θ)Γ(γ + 1)(b −]a)γ−θ+ .Γ(γ − θ + 1)Combined with the definition of ‖ · ‖ g and Lemma 3.2, (3.29) holds.The proof for (1.2) and (2.2) are referred to [33].□
72 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tang3. Shooting Methods for FBVPsIn this section, single shooting methods are applied to solve the FBVPs(3.1)–(3.2), (3.3)–(3.4), (3.7)–(3.8) and (3.9)–(3.10) numerically.According to the idea of the shooting method, a FBVP is turned into afractional initial value problem (FIVP) which can be solved by some suitablenumerical method (see [26]–[28], [31]). We write down the correspondingprocedure for FBVPs.Denote the corresponding initial value conditions of FBVPs (3.1)–(3.2)and (3.3)–(3.4) asy(a) = a 0 , y ′ (a) = a 1 , a 0 , a 1 ∈ R. (3.31)Then FBVPs (3.1)–(3.2) and (3.3)–(3.4) can turn into FIVPs (3.1), (3.31)and (3.3), (3.31), respectively.Usually, not all a k (k = 0, 1) are equal to zero, in other words, the initialvalue conditions (3.31) are inhomogeneous. Settingz(t) = y(t) − a 0 − a 1 (t − a), (3.32)(3.1), (3.31) and (3.3)–(3.31) can be transformed into another FIVPs withhomogeneous initial value conditions.For a given equispaced mesh a = t 0 < t 1 < · · · < t N = b with stepsizeh = (b − a)/N, we give a fractional backward difference scheme of order one(refer to [31]) to solve FIVPs (3.1) and (3.3) with homogeneous initial valueconditions y(a) = 0, y ′ (a) = 0:m∑z m = −h γ f(t m , z m ) − ω k z m−k , (3.33)andi=0k=1z m = −h γ g(t m , z m , 1 ∑m ) m∑h θ ˜ω i z m−i − ω k z m−k , (3.34)where(ω 0 = 1, ω k = 1 − γ + 1 )ω k−1 , k = 1, 2, . . . , N, (3.35)k(˜ω 0 = 1, ˜ω i = 1 − θ + 1 )˜ω i−1 , i = 1, 2, . . . , N. (3.36)iThe case for (3.7)–(3.8) and (3.9)–(3.10) is similar. The reader shouldnote that the initial values take the following formRLa D γ−1t y(a) = b 1 ∈ R, lim t→a + a y(t) = b 2 ∈ R. (3.37)It is easy to check thatUsually, b 1 ≠ 0 and letk=1b 2 = limt→a + J 2−γa y(t) = 0. (3.38)z(t) = y(t) − b 1(t − a) γ−1Γ(γ)(3.39)
Boundary Value Problems for Differential Equations of Fractional Order 73in (3.7), (3.37) and (3.9), (3.37). Then they are transformed into FIVPswith homogeneous initial conditions. (3.33) or (3.34) is also employed tosimulate them.3.1. Shooting method for linear problems. The linear case of fractionaltwo-point boundary value problem (3.1) with homogeneous boundaryvalue conditionsCa D γ t y(t) + c(t)y(t) + d(t) = 0, a < t < b, 1 < γ ≤ 2, (3.40)y(a) = 0, y(b) = 0 (3.41)where c(t), d(t) ∈ C 0 [a, b].According to Theorem 3.1, ifb − a
74 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangDue to (3.44)–(3.45), we consider {|y(t n ) − y n |} N n=0. Then we getmax |y(t n) − y n | = O(h r ). (3.47)0≤n≤NAs to the linear case of (3.7)–(3.8), we only need to replace (3.42) and(3.43) with the initial value conditionsRLa D γ−1t y(t) ∣ = ξ 1 , lim J 2−γt=a t→a + a y(t) = 0, (3.48)D γ−1y(t) ∣ = ξ 2 , ;t=aRLatlim J 2−γt→a + a y(t) = 0. (3.49)Again, the shooting procedures for linear FBVPs (3.3)–(3.4) and (3.9)–(3.10) are similar to that of (3.1)–(3.2) and (3.7)–(3.8), respectively.3.2. Shooting method for nonlinear problems. For nonlinearFBVP (3.1) and (3.7) with homogeneous boundary value conditions (3.41),we consider the following FIVP:Ca D γ t y(t) + f(t, y(t)) = 0, a ≤ t ≤ b, 1 < γ ≤ 2, (3.50)y(a) = 0, y ′ (a) = ξ, (3.51)The FIVPs (3.50) is corresponding to FBVP (3.1), (3.41) with analyticsolutions denoted as y(t; ξ). In general, y(b; ξ) ≠ 0. Once a zero point ξ ∗of φ(ξ) := y(b; ξ) is found, one obtains y(t; ξ ∗ ), the solution of FBVP (3.1),(3.41). When y(t; ξ), and hence φ(ξ) are continuously differentiable withrespect to ξ, Newton’s method can be employed to determine ξ ∗ . Startingwith an initial approximation ξ (0) , one gets ξ (k) as follows:ξ (k+1) = ξ (k) − φ(ξ(k) )φ ′ (ξ (k) ) . (3.52)On one hand, y(b; ξ), hence φ(ξ) can be determined by solving FIVP (3.50)numerically. On the other hand, it is easy to check that w(t; ξ) := ∂y(t;ξ)∂ξisa solution of the following FIVP (refer to [30]):Ca D γ ∂f(t, y(t; ξ))t w(t; ξ) + w(t; ξ) = 0, (3.53)∂yw(a; ξ) = 0, w ′ (a; ξ) = 1, a ≤ t ≤ b, 1 < γ ≤ 2. (3.54)So we can compute w(b; ξ), i.e., φ ′ (ξ) by solving FIVPs (3.53) numerically.However, considering the computation complexity for ∂f(t,y(t;ξ))∂yin practice,we usually calculate the difference quotient△φ(ξ (k) ) := φ(ξ(k) + △ξ (k) ) − φ(ξ (k) )△ξ (k)instead of the derivative φ ′ (ξ (k) ) itself, where △ξ (k) is a sufficiently smallnumber. And the formula (3.52) is replaced byξ (k+1) = ξ (k) − φ(ξ(k) )△φ(ξ (k) ) . (3.55)
Boundary Value Problems for Differential Equations of Fractional Order 75In short, the procedure of solving FBVP (3.1), (3.41) by shootingmethod is displayed as follows:step 1. choose a starting value ξ (0) and an iterative precision ɛ for Newton’smethod; then for k = 0, 1, 2, . . . ;step 2. obtain y(b; ξ (k) ) by solving FIVP (3.50) with y ′ (a) = ξ (k) andcompute φ(ξ (k) );step 3. choose a sufficiently small number △ξ (k) ≠ 0 and determiney(b; ξ (k) + △ξ (k) ) by solving the FIVP (3.50) with y ′ (a) = ξ (k) +△ξ (k) , and then compute φ(ξ (k) + △ξ (k) );step 4. compute △φ(ξ (k) ) and determine ξ (k+1) by the formula (3.55);repeat steps 2–4 until |ξ (k+1) − ξ (k) | ≤ ɛ and denote the finalξ (k+1) by ξ ∗ ;step 5. finally, we obtain the numerical solution of FBVP (3.1), (3.41) bysolving FIVP (3.50) with y ′ (a) = ξ ∗ .Next, we want to give the error analysis of shooting method for nonlinearFBVP (3.1), (3.7). The error consists of two parts. One is from the errorbetween |ξ ∗ − ˜ξ|. By shooting method, we can only get the approximation˜ξ of the exact initial value ξ ∗ (= y ′ (a)). That is to say, we solve FIVP{CaDt α y(t) + f(t, y(t)) = 0,y(a) = 0, y ′ (a) = ˜ξ(3.56)instead of FIVP {CaD α t y(t) + f(t, y(t)) = 0,y(a) = 0, y ′ (a) = ξ ∗ .(3.57)which is an initial perturbation problem in fact. Denote the exact solutionsof (3.56) and (3.57) as y(t; ˜ξ) and y(t; ξ ∗ ), respectively.Lemma 3.4 (see [32]). Suppose α > 0, A(t) is a nonnegative functionlocally integrable on [a, b] and B(t) is a nonnegative, nondecreasing continuousfunction defined on [a, b], and suppose ψ(t) is nonnegative and locallyintegrable on [a, b] withψ(t) ≤ A(t) + B(t)on this interval. Then∫ ta(t − τ) α−1 ψ(τ) dτ, t ∈ [a, b] (3.58)ψ(t) ≤ A(t)E α(B(t)Γ(α)(t − a)α ) , (3.59)where E α is the Mittag–Leffler function defined by∞∑ x kE α (x) :=Γ(kα + 1) . (3.60)k=0
76 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangAccording to the above lemma, we get the error estimate∣∣y(t; ξ ∗ ) − y(t; ˜ξ) ∣ ∣ ≤≤ |ξ ∗ − ˜ξ|(t − a)E α (K f (t − a) α ) ≤ C 1 |ξ ∗ − ˜ξ|, t ∈ [a, b], (3.61)where C 1 := max (t − a)E α(K f (t − a) α ).t∈[a,b]Another part of the error is from numerical solving procedure of FIVP(3.50) by scheme (3.33), (3.35). Suppose that the scheme of FIVP convergeswith order r, and denote the approximate solution of y(t n ; ˜ξ) as y n (˜ξ). Weget ∃ C 2 > 0, s.t.∣max ∣y(t n ; ˜ξ) − y n (˜ξ) ∣ ≤ C 2 h r . (3.62)0≤n≤NTo sum up, we have∣∣y(t n ; ξ ∗ ) − y n (˜ξ) ∣ ∣ = ∣ ∣y(t n ; ξ ∗ ) − y(t n ; ˜ξ) + y(t n ; ˜ξ) − y n (˜ξ) ∣ ∣ ≤that is,≤ ∣ ∣y(t n ; ξ ∗ ) − y(t n ; ˜ξ) ∣ ∣ + ∣ ∣y(t n ; ˜ξ) − y n (˜ξ) ∣ ∣max0≤n≤N∣∣y(t n ; ξ ∗ ) − y n (˜ξ) ∣ ≤ C 1 |ξ ∗ − ˜ξ| + C 2 h r . (3.63)As to (3.7), (3.41), we turn it into its corresponding FIVPRLa D γ t y(t) + f (t, y(t)) = 0, a ≤ t ≤ b, 1 < γ ≤ 2, (3.64)RLa D γ−1t y(t) ∣ = ξ, lim J 2−γt=a t→a + a y(t) = 0. (3.65)The numerical procedure of simulating (3.7), (3.41) is completely similar tothat of (3.1), (3.41).Again, the procedures of shooting method for (3.2) and (3.8) with homogenousboundary conditions (3.41) are similar to the case of (3.1), (3.41)and (3.7), (3.41), respectively.4. Numerical Experiments for Solving FBVPsIn this section, two numerical examples are given to show the feasibilityand validity of single shooting methods for the FBVPs.First, let us introduce the parameters. a = 0 = t 0 < t 1 < · · · < t n = bis a given equispaced mesh with stepsize h = (b − a)/n, n = 100, γ = 1.5,θ = 0.5. For the linear cases in Examples 1, the two “initial speeds” ξ 1 = 0.5,ξ 2 = 1.5. For the nonlinear cases in Examples 2, ɛ = 10 −12 , ξ (0) = 0,△ξ (k) ≡ 10 −8 .Example 3.1. Consider the following linear FBVPC0 D 1.5C0 D 0.5t y(t) + 1 3 y(t) + 1 4t y(t) = r(t), 0 < t < 1,y(0) = 0, y(1) = 0,
Boundary Value Problems for Differential Equations of Fractional Order 770 x 10−4−0.5−1error−1.5−2−2.5−30 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1tFigure 1. The shooting error for Example 3.10.35numerical solutionanalytical solution0.30.250.2y(t)0.150.10.0500 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1tFigure 2. The true solution and numerical solution of Example 3.1whereSincer(t) = − 7t0.52 √ π + t 3 − 2t1.53 √ π − t2 3 .∣∣g(t, u 2 , v 2 ) − g(t, u 1 , v 1 ) ∣ ∣ ≤ 1 3 |u 2 − u 1 | + 1 4 |v 2 − v 1 |,K g = 1 3 , L g = 1 , a = 0, b = 1, γ = 1.5, θ = 0.5,4
Since∣∣f(t, u 2 ) − f(t, u 1 ) ∣ ∣ = ∣ ∣ sin(u 2 ) − sin(u 1 ) ∣ ∣ ≤ |u 2 − u 1 |,78 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tang2(b − a) γ [K gΓ(γ + 1) + L g(b − a) γ−θΓ(2 − θ)Γ(γ + 1)(b −]a)γ−θ+ ≈ 0.96 < 1,Γ(γ − θ + 1)there exists a unique solution for this FBVP according to Theorem 3.1. Infact, one can easily check that y(t) = t(1 − t) is the analytical solution. Theerrors between the numerical solution (obtained by using shooting methodmentioned above) and the analytical solution at mesh points are plotted inFigure 1. The true solution (denoted by real line) and numerical solution(denoted by □) on equispaced mesh are plotted in Figure 2.whereExample 3.2. Consider the following nonlinear FBVPRL−1 Dt 1.5 y(t) + sin(y) + r(t) = 0, −1 < t < 1,y(−1) = 0, y(1) = 0,r(t) = sin(C t+1 (2, 2π)) + C t+1 (0.5, 2π),C t−a (α, ω) = (t − a) α∞∑j=0(−1) j (ω(t − a)) 2jΓ(α + 2j + 1).3 x 10−32.521.5error10.50−0.5−1−1.5−2−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1tFigure 3. the shooting error for Example 3.2K f = 1, a = −1, b = 1, γ = 1.5,K f(γ − 1) γ−1 (b − a) γγ γ Γ(γ + 1)≈ 0.82 < 1,
Boundary Value Problems for Differential Equations of Fractional Order 79there exists a unique solution for this FBVP according to Theorem 3.1. Infact, one can easily check that y(t) = C t+1 (2, 2π) is the analytical solution.The errors between the numerical solution (obtained by using shootingmethod mentioned above) and the analytical solution (approximatedby summarizing the first 21 terms of the infinite series C t+1 (2, 2π) becauseof (2π)40Γ(43)≈ 6.02 × 10 −20 ) at different points are plotted in Figure 3. Theanalytical solution (denoted by real line) and numerical solution (denotedby □) on equispaced mesh are plotted in Figure 4.0.06numerical solutionanalytical solution0.050.040.03y(t)0.020.010−0.01−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1tFigure 4. the true solution and numerical solution of Example3.2stepsize h = 1/nTable 1max |y(t i) − y n (t i )|1≤i≤n1/16 2.46e − 2rates of convergence1/32 1.06e − 2 1.21981/64 4.9e − 3 1.10131/128 2.34e − 3 1.07401/256 1.31e − 3 1.0471The rates of convergence and maximum errors for Example 3.2 betweenthe numerical solution and the analytical solution are given in Table 1.Table 1 shows that the shooting method is of order one, which is in good
agreement with the fact that (3.33), (3.35) or (3.34), (3.36) is a method oforder one.The above numerical results indicate that the single shooting methodis a successful tool to solve fractional boundary value problems (3.1), (3.2),(3.3), (3.4). The numerical experiments for FBVPs (3.7), (3.8) and (3.9),(3.10) can be referred to [33]. The results are very similar.Bibliography1. T. S. Aleroev, On a class of operators associated with differential equations offractional order. (Russian) Sibirsk. Mat. Zh. 46 (2005), No. 6, 1201–1207; Englishtransl.: Siberian Math. J. 46 (2005), No. 6, 963–968.2. G. M. Gubreev, Regular Mittag–Leffler kernels and spectral decomposition of a classof nonselfadjoint operators. (Russian) Izv. Ross. Akad. Nauk Ser. Mat. 69 (2005),No. 1, 17–60; English transl.: Izv. Math. 69 (2005), No. 1, 15–57.3. T. S. Aleroev, The boundary problems for differential equations with fractionalderivatives. Dissertation, doctor of Physical and Mathematical Sciences, MoscowState Univercity, 2000.4. M. S. Brodskiǐ, Triangular and Jordan representations of linear operators. Translatedfrom the Russian by J. M. Danskin. Translations of Mathematical Monographs, Vol.32. American Mathematical Society, Providence, R.I., 1971; Russian original: Nauka,Moscow, 1969.5. T. S. Aleroev, The Sturm–Liouville problem for a second-order differential equationwith fractional derivatives in the lower terms. (Russian) Differentsial’nye Uravneniya18 (1982), No. 2, 341–342.6. T. S. Aleroev, On a boundary value problem for a fractional-order differential operator.(Russian) Differ. Uravn. 34 (1998), No. 1, 123, 144; English transl.: DifferentialEquations 34 (1998), No. 1, 126.7. T. S. Aleroev, Aleroev, T. S. Some problems in the theory of linear differentialoperators of fractional order. (Russian) Dokl. Akad. Nauk 341 (1995), No. 1, 5–6.8. M. Li, N. M. Nie, S. Jiménez, Y. F. Tang, and L. Vázquez, Solving two-pointboundary value problems of fractional differetial equations,http://www.cc.ac.cn/2009research report/0902.pdf.9. M. M. Dzhrbashyan and A. B. Nersesyan, Fractional derivativer and Cauchy’sproblem for differential equations of fractional order. (Russian) Izv. Akad. NaukArmyan. SSR, Ser. Mat. 3 (1968), No. 1, 3–29.10. M. M. Dzhrbashyan, A boundary value problem for a Sturm-Liouville type differentialoperator of fractional order. (Russian) Izv. Akad. Nauk Armyan. SSR, Ser.Mat. 5 (1970), No. 2, 71–96.11. A. M. Gachaev, The boundary problems for differential equations of fractional order,Dissertation, Nalchik, Russia, 2005.12. T. S. Aleroev, The boundary problem for differential operator of fractional order,The reports of Circassian International Academy of Sciences (Doklady Adygskoy(Cherkesskoy) Mezjdunarodnoy Akademii Nauk 1 (1994), No. 1.13. M. M. Malamud and L. L. Oridoroga, On some questions of the spectral theoryof ordinary differential equations of fractional order. Dopov. Nats. Akad. Nauk Ukr.Mat. Prirodozn. Tekh. Nauki 1998, No. 9, 39–47.80
Boundary Value Problems for Differential Equations of Fractional Order 8114. M. M. Malamud and L. L. Oridoroga, Analog of the Birkhoff theorem and thecompleteness results for fractional order differential equations. Russ. J. Math. Phys.8 (2001), No. 3, 287–308.15. T. S. Aleroev and H. T. Aleroeva, Some applications of the theory of disturbancein fractional calculation. Mem. Differ. Equations Math. Phys. (to appear).16. T. Kato, Perturbation theory for linear operators. Die Grundlehren der mathematischenWissenschaften, Band 132. Springer-Verlag New York, Inc., New York, 1966;Russian transl.: Mir, Moscow, 1972.17. I. C. Gohberg and M. G. Krein, Theory of Volterra operators in Hilbert space andits applications. (Russian) Nauka, Moscow, 1967.18. M. M. Dzhrbashjan, Integral transformations and presentations of functions in complexspace. Nauka, Moscow, 1966.19. V. I. Matsaev and Y. A. Polant, Über die Potenzen eines beschränkten dissipativenOperators. (Russian) Ukr. Mat. Zh. 14 (1962), 329–337 (1962).20. F. R. Gantmaher and M. G. Krein, Oscillation matrices and kernels and smalloscillations of mechanical systems. (Russian) 2d ed. Gosudarstv. Isdat. Tehn.-Teor.Lit., Moscow–Leningrad, 1950.21. E. Hille and R. Fillips, Functional analysis and semi-groups. (Russian) Translatedfrom the English by D. A. Vasil’kov. Edited by V. M. Alekseev and S. V. FominIzdat. Inostran. Lit., Moscow, 1962.22. I. C. Gohberg and M. G. Krein, Introduction to the theory of linear non-selfadjointoperators in Hilbert space. (Russian) Izdat. Nauka, Moscow, 1965.23. T. S. Aleroev and H. T. Aleroeva, A problem on the zeros of the Mittag-Lefflerfunction and the spectrum of a fractional-order differential operator. Electron. J.Qual. Theory Differ. Equ. 2009, No. 25, 18 pp.24. V. B. Lidskiǐ, Conditions for completeness of a system of root subspaces for nonselfadjointoperators with discrete spectrum. (Russian) Trudy Moskov. Mat. Obsc. 8(1959), 83–120.25. P. B. Bailey, L. F. Shampine, and P. E. Waltman, Nonlinear two point boundaryvalue problems. Mathematics in Science and Engineering, Vol. 44 Academic Press,New York–London, 1968.26. K. Diethelm and N. J. Ford, Numerical solution of the Bagley–Torvik equation.BIT 42 (2002), No. 3, 490–507.27. K. Diethelm, N. J. Ford, and A. D. Freed, A predictor-corrector approach for thenumerical solution of fractional differential equations. Fractional order calculus andits applications. Nonlinear Dynam. 29 (2002), No. 1-4, 3–22.28. Ch. Lubich, Discretized fractional calculus. SIAM J. Math. Anal. 17 (1986), No. 3,704–719.29. I. Podlubny, Fractional differential equations. An introduction to fractional derivatives,fractional differential equations, to methods of their solution and some of theirapplications. Mathematics in Science and Engineering, 198. Academic Press, Inc.,San Diego, CA, 1999.30. J. Stoer and R. Bulirsch, Introduction to numerical analysis. Translated from theGerman by R. Bartels, W. Gautschi and C. Witzgall. Springer-Verlag, New York–Heidelberg, 1980.31. M. Weilbeer, Efficient numerical methods for fractional differential equations andtheir analytical background. Univ. Braunschweig, Fakultat fur Mathematik und Informatik(Diss.)., Braunschweig, 2005, 218 p.32. H. Ye, J. Gao, and Y. Ding, A generalized Gronwall inequality and its applicationto a fractional differential equation. J. Math. Anal. Appl. 328 (2007), No. 2, 1075–1081.33. N. M. Nie, Y. M. Zhao, S. Jiménez, M. Li, Y. F. Tang and L. Vázquez, Solvingtwo-point boundary value problems of fractional differential equations with Riemann–Liouville derivatives, J. Syst. Simul. 22 (2010), No. 1.
82 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangAuthors’ Addresses:(Received 3.09.2009)T. S. AleroevMoscow Institute of a Municipal Services and ConstructionFaculty of Higher MathematicsRussiaThe Academy of National Economy under the Government of theRussian FederationMoscow 119571, RussiaE-mail: aleroev@mail.ruH. T. AleroevaMoscow Technical University of Communications and InformaticsMoscow 111024, RussiaE-mail: BinSabanur@gmail.comNing-Ming NieLSEC, ICMSEC, Academy of Mathematics and Systems ScienceChinese Academy of SciencesBeijing 100190, P. R. ChinaGraduate School of the Chinese Academy of SciencesBeijing 100190, P. R. ChinaE-mail: nienm@lsec.cc.ac.cnYi-Fa TangLSEC, ICMSEC, Academy of Mathematics and Systems ScienceChinese Academy of SciencesBeijing 100190, P. R. ChinaE-mail: tyf@lsec.cc.ac.cn
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