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By Julii A. Dubinskii

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Moreover, degP( z) 2: degQ( z). Example 2. Let A«() = 1/(, ( En = Cl\R~, where R~ = {x E Rl : x 2: O}. Evidently, n is a Runge domain. z(h(z). (z) are entire functions satisfying the inequality where M).. > 0, r).. < R('x) = dist(A, R~). ) such that I w(z) = D u(z). ) and denote it by w(z) =nat J u(z)dz. If, for example, u(z) is a quasipolynomial, then the "natural" inverse w(z) may be obtained by usual integration. Here we must "forget" to write down the additive constants during the process of integration.

For the proof that the values A(D)u(z) are well-defined, we use the Borel inversion formula. J (a+1' lal=O where a+ 1 = (a1 + 1, ... ,an + 1). If rl > O, ... ,rn > 0 are the types of exponential growth of u(z), then the Borel function Bu«() is analytic for I(j I > rj, j = 1, ... 12) we obtain u(z) = (2:i)n ;;= ir>. 14) where Bu),«() is the Borel function associated with u),(z) = exp AZ' 0 may be chosen so small that all the polycylinders U), will lie strictly inside rl.

Let A. ((), s = 0,1, ... , m - 1, be the roots of the polynomial L(A, e). (e) = 0(1(1) (1(1---+ (0). 16) is true if and only if polynomial L(A, e) is a K ovalevskaya polynomial. Proof The necessity is obvious since the coefficients of the polynomials Ak (A, e) are symmetric functions of order m-k of the roots AO((), ... 17) This means that L( A, () is a Kovalevskaya polynomial. Sufficiency. 16). (e)1 > IlKI. Then for such e if Il > 0 is large enough. 17). The lemma is proved. We now turn to the question of the local solvability of the Cauchy problem itself.

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