By Kravchenko V. V.

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From Theorem 10 we see immediately that the solution of Problem 1 does not always exist because not all functions g are -extendable into + . 33). If this is the case then the solution of Problem 1, according to the Cauchy integral formula, is obtained from the Cauchy integral of g: f = K g. Let us consider another boundary value problem for the operator D , the so-called jump problem. 5. THE OPERATOR D + I 39 Problem 3. 35) is ful…lled: f+ f = K [g]+ K [g] = P [g] + Q [g] = g: A much more di¢ cult problem is the analogue of the famous Riemann boundary value problem.

Dg(x) = D[eh ; x i ] f (x) + eh ; x i D[f ](x) !! eh ; x i f (x) !! eh ; x i f (x) = 0 g(x): Of course, in the opposite direction the proof is similar. More details can be found in [8]. Thus the operator D+ I, when being reduced to D + 0I is a constant complex quaternion, represents less interest compared to the operator D + M . 1) where D := D + M ; 2 H(C). 1) we have to distinguish di¤erent cases depending on the algebraic properties of . The following observations will help us to understand the structure of the integral operators.

N (y)f (y)d y Z R K (x y) R := B R n y f (y)d jyj R y: Now let us consider the limit of this equality when R ! 1. We have the following asymptotic relation Z Z y R #(y)( K (x y) f (y)d y jyj R R y y )f (y)d 2 +i jyj jyj R y; R ! 1: Using the radiation condition we obtain that this integral tends to zero when R ! 1. Thus, f (x) = Z K (x y)! n (y)f (y)d y which gives us the statement of the theorem. 22) allowed us to obtain a very simple proof of the Cauchy integral formula for the exterior domain.

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