By A.V.; Kalinichenko, D.F. Bitsadze

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In both expressions the right-hand side is a strictly monotone function of the September 7, 2010 16:7 WSPC - Proceedings Trim Size: 9in x 6in 02 38 variable, so f is one-to-one, and therefore a homeomorphism onto image, on both I and J. It remains to check that f (I) and f (J) are disjoint except for the common endpoint. Suppose that f (x, 0) = f (1, y). Then by the formulas above, −x = 1 − 2/(1 + αy). Comparing x we get 1/(2 + αx) = (1 + y)/(2 + α(1 + y)). Eliminating x from this system of two equations we get 1 + αy 1+y = .

All points below the anti-diagonal are attracted to (0, 0) and all points above√ the anti-diagonal are attracted to (1, 1). (g) For α > 2(1 + 2), (1/2, 1/2) is an orientation reversing repelling fixed point. There is an attracting orbit of period two on the anti-diagonal. There are two saddle period two orbits that follow the orbit of period two on the anti-diagonal. One saddle orbit lies below the anti-diagonal and one above. There are no other periodic points. Every other point is attracted to one of the periodic orbits mentioned (including (0, 0) and (1, 1)).

Then: (a) f (C) is a Jordan curve and f (D ◦ ) is the region bounded by f (C); (b) f is a homeomorphism of D onto its image. 7. If α ≤ 1/2 then n is a homeomorphism onto its image. Proof. Assume that α ≤ 1/2. 1 it follows that it is enough to prove that n restricted to the triangle T with the vertices (0, 1), (1, 1) and (1/2, 1/2) is a homeomorphism onto its image and that n(T ) ⊆ T . From the formulas for the partial derivatives of n we see that the Jacobian of n in the N W quadrant is a fraction with a positive denominator and the numerator equal to 1−4α2 (1−x)2 y 2 .

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