Figure 6: The red orbit of system (4.50) non-axially grazes the surface Γ . The magenta trajectory with initial point (0 , 1 . 32) approaches the cycle as time increases. The green cycle with initial point (0 , 0 . 96) demonstrates the inside stability of the grazing orbit. 5 Small parameter analysis and grazing bifurcation In this part, we will discuss existence and bifurcation of cycles for perturbed systems, if the generating one admits a grazing periodic solution. In continuous dynamical systems, a small parameter may cause a change in the number of periodic solutions in critical cases. In the present analysis, we will demonstrate that the change may happen in non-critical cases, since of the non-transversality. That is why, one can say that grazing bifurcation is under discussion. Let us deal with the following system where x ∈ R $^{n}$, t ∈ $^{R}$, Γ( µ ) = { x | Φ( x )+ µφ ( x, µ ) = 0 } , µ ∈ ( − µ$_{0}$, µ$_{0}$ ) , and µ$_{0}$ is a sufficiently small positive number. Functions f ( x ) , I ( x ) and Φ( x ) are continuously differentiable up to second order, g ( x, µ ) , K ( x, µ ) are continuously differentiable in x and µ. The function φ ( x, µ ) is continuously differentiable in x up to second order and to first order in µ. We assume that the generating system for (5.57) is the system (2.1) with all conditions assumed for the system, earlier. The main assumption of this section is that (2.1) admits a ω − periodic solution, Ψ( t ) . Let Ψ(0) = ( ζ 0 $_{1}$, ζ 0 $_{2}$, . . . , ζ 0 $_{n}$) be the initial value of the solution. Our aim is to find conditions that verify the existence of periodic solutions of (5.57) with a period T such that for µ = 0 , the periodic solutions of (5.57) are turned down to Ψ( t ) . It is common for the autonomous systems that the period T does not coincide with ω. Thus, in the remaining part of the paper, we will consider the period T as an unknown variable. Since Ψ(0) is not an equilibrium, there is a number j = 1 , 2 , . . . , n, such that f$_{j}$ ( ζ 0 $_{1}$, ζ 0 $_{1}$, . . . , ζ 0 $_{n}$) = 0 . In other words, the vector field is transversal to line x$_{j}$ = ζ 0 j near the point. Hence, to try points near to −1.5 −1 −0.5 0 0.5 1 1.5 −1 −0.5 0 0.5 1 1.5 x$_{1}$(t) x 2 (t) x ′ = f ( x ) + µg ( x, µ ) , Δ x | $^{x}$∈ Γ( µ ) = I ( x ) + µK ( x, µ ) , (5.57) 31