By A. V. Babin

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**Extra resources for Iterations of Differential Operators**

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Since p >2, then m/p - m/2 <0, and we arrive at a contradiction for fairly large N. 1 is proved. 1. 1, it is easy to note that the polynomial solvability of the equation Bu=f is only used to establish the "locality" of the dependence of u on F, ie. the fact that u(x0 ) is uniquely defined using the values At the same time, the way in which u(x0 ) is expressed in terms of B"Rx0 ), k = 0, 1, ... is completely unimportant. A. V. g. Malliavin Ul, Lyubich and Tkachenko CU, Khryptun CU and Chemyavskii [1]).

Firstly, therefore, u (t) is continuous with respect to t, and secondly, u(O) = f. 10). 2. 11) has a unique solution. 13) Proof. 11) is unique. 11). Below we shall take as the operator L a self-adjoint extension of the differential operator B. 14) where acx(x) are real functions belonging to A(O). 15) where (u ,v) is a scalar product in (L2 (0) = H. We shall require that the condition of lower boundedness of the operator B holds: (Bu, u)~bllull 2 , Vue~. 16). 1. e. 17) 28 A. V. l). 1. 3) holds.

E. 12) holds. Then for any function x. (B)f, x). N). Note that since lf(x) = 0 when S, then Pn(]Jjf(x) = 0 for these = 1 when Ix! ~~ Ix! ~~