Difference-differential Equations with Fredholm Operator in the Main ...

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3 The Reduction of Equation (1) to Regular Problems Suppose: 2. The operators A 2 , . . . , A q (P, Q)- commute. Then there are matrices A i , i = 2, q, such that A i Φ = A i Z, A ∗ i Ψ = A′ i Υ Following formulas (2), (3) we introduce the projection operators P, Q, which generate the direct decompositions E 1 = E 1k ⊕ E 1∞−k , E 2 = E 2k ⊕ E 2∞−k . Note that ΓE 2∞−k ⊂ E 1∞−k We look for the solution of equation (1) in the following form u(x) = Γv(x) + (C(x), Φ), (4) where Γ = (B + ∑ n i=1 < ., γ (1) i > z (1) i ) −1 is a bounded operator , v ∈ E 2∞−k , C(x) = (C 1 (x), . . . , C n (x)) ′ , C i (x) = (C i1 (x), . . . , C ipi (x)), Φ = (Φ 1 , . . . , Φ n ) ′ , Φ i = (φ 1 i , . . . , φ (p i) i ), i = 1, n. Substituting expression (4) into equation (1) and noting that BΓv = v, since BΓ = I − ∑ n i=1 < ., ψ (1) i > z (1) i , < v, ψ (1) i >= 0, we obtain q∑ q∑ L 0 v + L i A i Γv + L 0 B(C, Φ) + L i A i (C, Φ) = f(x). (5) i=1 i=1 The operator Γ (Q, P )-commutes, so from condition 2 and corollary 1 it follows that QA i Γ(I − Q) = 0, (I − Q)A i ΓQ = 0. Hence, QA i Γv = 0, ∀v ∈ E 2∞−k . According to corollary 2 BΦ = A B Z, where A B = (B 1 , . . . , B m ) is a symmetrical cell-diagonal matrix. Consequently, (I − Q)BΦ = 0, (I − Q)A i Φ = 0, i = 1, q, (6) because (I − Q)Z = 0. The following equalities hold: (A i (C, Φ), Ψ) = A ′ iC, (B(C, Φ), Ψ) = A B C. (7) Projecting equation (5) onto E 2∞−k using (6) we obtain the regular equation (solved according to operator L 0 ): ˜Lv = (I − Q)f(x), (8) where q∑ ˜L = L 0 + L i A i Γ. (9) i=1 In order to determine the vector-function C(x) : R r → R k , we project equation (5) onto E 2k and using (7) we obtain q∑ L 0 A B C + L i A ′ iC =< f(x), Ψ > . (10) i=1 4
So it is proved Theorem 1. Suppose conditions 1 and 2 are satisfied, f : Ω ⊂ R r → E 2 - sufficiently smooth function. Then any solution of equation (1) can be represented in the form u = Γv + (C, Φ) where v satisfies regular equation (8), and the vector C(x) is defined from system (10). Since detA B = 0, then system (10) is degenerated in the general case. Let us specify the sufficient conditions on which the systems (10) is recurrent sequence of regular linear differencedifferential equations (solved according to operator L 1 ). Introduce the following condition: 3 The operators A 2 , . . . , A q commute quasitriangularly (see Def.2), or k = n. Lemma 1. Suppose conditions 1, 2, 3 are satisfied. Then system (10) is a recurrent sequence of linear differencedifferential equations of order q 1 with the operators of the form ˜ L ks = L 1 + q∑ i=2 a ik p k −s+1,sL i . In particular, if condition 1 is satisfied and A 2 = . . . = A q = 0, system (10) takes the form L 1 C ipi (x) =< f(x), ψ (1) i >, L 1 C ipi −s(x) =< f(x), ψ (s+1) i > −L 0 C ipi −s+1(x), s = 1, p i − 1, i = 1, n. Proof follows from lemma 2 [6] or lemma [10]. So if the conditions of theorem 1 and lemma 1 are satisfied then equation (1) can be reduced to regular equation (8) and to sequence of regular linear difference-differential equations (10) with order q 1 . Thus under the condition 3 system (10) be pseudosingular system. Remark. Based on lemma 1 the vector components c i in the case p i ≥ 2 are determined from the recurrent sequence of linear equations of order q 1 . If p i ∗ = . . . = p i ∗ +s = 1 then the corresponding elements c i ∗(x), . . . , c i ∗ +s(x) are determined from the regular system of s + 1 equations of order q 1 . In particular if p 1 = · · · = p n = 1, then A B become zero. 4 The Choice of Boundary Conditions In this section conditions 1,2,3 are satisfied. In the preceding section for the definition of projections (I − P )u, P u the solution u of equation (1) we constructed equation (8) with regular operator q∑ ˜L = L 0 + L i A i Γ, i=1 and system (10), splitting into recurrent sequence of linear equations with regular differencedifferential operators q∑ L˜ ks = L 1 + a ik p k −s+1,s L i. i=2 5
Page 1 and 2: Difference-differential Equations w
Page 3: Definition 1. If u ∈ D(A), P u
Page 7 and 8: 5 Examples Theorem 3 can be used fo
Page 9 and 10: ✻ ✬ ✗ ✖ ✫ x 0 1 ω ∂
Page 11 and 12: [6] N.A.Sidorov and E.B. Blagodatsk

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