JOURNAL Series A - Pure and Applied Mathematics

60 At. Georgieva, St. Kostadinov3.2 Let be again X = R. We will show that the evolutionary process induced by theordinary equationdxdt = x + √ x (12)does not satisfy the condition (1) and following is not k(t, s)-continuous.Indeed, in this case we haveandX(t, s)ξ = (e t−s √ t−s2 ξ + e 2 − 1) 2|X(t, s)ξ − X(t, s)η| = |(e t−s √ t−s2 ξ + e= |e t−s (ξ − η) + 2e t−s2 (e t−s2 − 1)( √ ξ − √ η)|Hence, there does not exist a function k(t, s) with2 − 1) 2 − (e t−s2|X(t, s)ξ − X(t, s)η| ≤ k(t, s)|ξ − η| (ξ, η ∈ R).√ η + et−s2 − 1) 2 | =We will show that there are not numbers q n and functions h n (.) : R → R suchthat the impulse equationdxdt = x + √ x for t ≠ t n ,x(t + n ) = q n x(t n ) + h n (x(t n )), (n = 0, ±1, ±2, . . .)does not satisfy the condition (1).Really, in this case for t 0 < s ≤ t 1 < t ≤ t 2 and ξ, η ∈ R we have|X(t, s)ξ − X(t, s)η| == |(e t−t √12 x(t+1 ) + e t−t 12 − 1) 2 − (e t−t 12√y(t+1 ) + e t−t 12 − 1) 2 | == |e t−t 1(x(t + 1 ) − y(t + 1 )) + 2e t−t 12 (e t−t 12 − 1)( √ x(t + 1 ) − √ y(t + 1 ))|,where x(t) = x(t; s, ξ), y(t) = x(t; s, η). It is clear, that even in this case thecondition (1) is not fulfilled.3.3 Analogously consideretions are also valid for the equation.dxdt = 2tx + t√ x (13)although the right-hand side is t-dependent. In this case for t ≥ s and ξ ∈ R wehaveX(t, s) = (e t2 −s 2 √ 12 ξ +2 e t2 −s 22 − 1 22 )