By Humbataliyev R.

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**Extra info for On the existence of solution of boundary value problems**

**Example text**

Further, we seek a general regular solution of the equation in the form m−1 u(z) = u0(z) + (zA)p e−zA Cp , (57) p=0 where Cp ∈ H2m− 1 , and e−zA is a holomorphic in S(α,β) group of bounded 2 operators generated by the operator (−A). Now, let’s deﬁne the vectors Cp (p = 0, m − 1) from condition (55). Then, obviously, for the vectors Cp (p = 0, m − 1) we get the following system of equations: ⎧ ⎪ c0 = −u0 (0), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −c0 + c1 = −A−1 u (0), ⎪ ⎪ ⎪ ⎨ c − 2c + 2c = −A−2 u (0), 0 1 2 ⎪ ····················· ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ (−1)m−1 c0 + (−1)m−2 c1 + · · · + cm−1 = −A−m+1 u(m−1) (0).

14. S. On correct solvability conditions of boundary value problems for operator – diﬀerential equations. Dokl. 273, No 2. (in Russian). 15. , Majenes E. Inhomogenous boundary value problems and their applications. , ”Mir”, 1971. (in Russian). 16. : Boundary value problems for diﬀerential – operator equations. Kiev, “Naukova Dumka”, 1984. (Russian). 17. : On generalized solutions for a class of fourth order operator-diﬀerential equations. J. Izv. , ser. -tech. math. 18-21 (Russian) 18. , Schechter M.

2 be fulﬁlled. Then elementary solutions of problem (32), (33) is complete in the space of generalized solutions. Proof. It is easy to see that if there exists a generalized solution, then m−1 u W2m (R+ ;H) ≤ const ϕj m−j−1/2 . j=0 Then it follows from the trace theorem [17] and these inequalities that m−1 Ck m−1 ϕν m−ν−1/2 ≤ u W2m (R+ ;H) = Ck ν=0 ϕν m−ν−1/2 . (37) ν=0 Further, from the theorem on the completeness of the system K(Π− ) it follows that for any collection {ϕν }m−1 ν=0 and ϕν ∈ Hm−ν−1/2 there is such a number N and Ck (ε, N) that N ϕν − (ν) Ck ϕi,j,h < ε/m, ν = 0, m − 1.