where R = 0 . 9 and Γ = { x | x$_{1}$ = 0 , x$_{2}$ ≤ 0 } . It is easy to see that system (5.69) is of the form (5.57) and Φ( x$_{1}$, x$_{2}$ ) = x$_{1}$ = 0 . The system has a periodic solution where t ∈ $^{R}$for µ ∈ ( − 2 , 0] . The generating system of (5.69) has the following form and admits the periodic solution Ψ$_{0}$( t ) = (1+cos( t ) , − sin( t )) . By means of the equality 〈∇ Φ( x ∗ ) , f ( x ∗ ) 〉 = 〈 (1 , 0) , (0 , 1) 〉 = 0 with x ∗ = (0 , 0) ∈ ∂ Γ , it is easy to say that x ∗ is a grazing point of Ψ$_{0}$( t ) . Let us start with the linearization of system (5.71) around the periodic solution Ψ$_{0}$( t ) . Consider a near solution y ( t ) = y ( t, 0 , y ∗ +Δ y ) , where Δ y = (Δ y$_{1}$, Δ y$_{2}$ ) , to the periodic solution Ψ$_{0}$( t ) . Assume that y ( t ) satisfies condition ( N 1) , and it meets the surface of discontinuity Γ at the moment t = ξ and at the point ¯ y = y ( ξ, 0 , y ∗ + Δ y ) . Considering the formula (3.10) for the transversal point ¯ y = (¯ y$_{1}$, ¯ y$_{2}$ ) , the first component ∂τ (¯ y ) ∂y 0 1 can be evaluated as ∂τ (¯ y ) ∂y 0 1 = − 1 ¯ y$_{2}$ . From the last equality, the singularity is seen at the grazing point. By taking into account (3.17) with (5.71) and ∂τ (¯ y ) ∂y 0 1 , we obtain that 35 Figure 7: The red arcs are the trajectory of the system (5.60) with initial value (0 , 1 . 2) and the blue arcs are the orbit with initial value (0 , 1 . 5) . Through simulation, we observe that the trajectories approach to the periodic solution of (5.60) as time increases. Ψ$_{µ}$( t ) = (1 + (1 + µ ) cos( t ) , − (1 + µ ) sin( t )) , (5.70) x ′ 1 = x$_{2}$, x ′ 2 = − 0 . 0001[ x 2 + ( x$_{1}$ − 1) 2 − 1] x$_{2}$ − x$_{1}$ + 1 , Δ x$_{2}$ | $^{x}$∈ Γ = − (1 + Rx$_{2}$ ) x$_{2}$, (5.71) −0.2 0 0.2 0.4 0.6 0.8 1 1.2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 y$_{1}$ y 2