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

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