grazing orbit. Fix points y = ( y$_{1}$, y$_{2}$ ) ∈ H and ψ = ( ψ$_{1}$, ψ$_{2}$ ) of the orbits y ( t ) and Ψ$_{0}$( t ) , respectively such that 0 ≤ y$_{1}$ = ψ$_{1}$ ≤ 1 and ψ$_{2}$ < 0 . Since of the equation y ′ 1 = y$_{2}$, the speed of y$_{1}$ ( t ) at ( y$_{1}$, y$_{2}$ ) is larger than the speed of Ψ$_{1}$( t ) at ( ψ$_{1}$, ψ$_{2}$ ) . Consequently, one can find that τ ( y ) ≤ π 4 < 2 π for y ∈ H. Thus, the condition ( A 3) is valid and Lemma 3.1 verifies the condition ( C ) . It is easy to demonstrate that the condition ( A 4) is valid such that near solutions to the grazing one are either continuous or discontinuous. That is, they don’t intersect the line of discontinuity Γ or intersect it permanently near to the grazing point and by means of the formula (5.76) , the linearization system for (5.71) around the grazing cycle Ψ$_{0}$( t ) consists of the following two subsystems and The system (5.77) + (5.78) is (2 π, 1) periodic. The Floquet multipliers of system (5.77) + (5.78) are ρ (1) 1 = 1 , ρ (1) 2 = 0 . 939 , ρ (2) 1 = 1 , ρ (2) 2 = 0 . 912 . Thus, condition ( A 5) is validated. Moreover, the conditions ( C 1) − ( C 7) and ( A 1) , ( A 2) can be verified utilizing similar way presented in Example 3.1. Consequently, Theorem 4.1 authenticates that the grazing periodic solution (cycle), Ψ$_{0}$( t ) of the system (5.71) is orbitally stable. The simulation results demonstrating the orbital stability of Ψ$_{0}$( t ) are depicted in Figure 8. Next, we will investigate two sorts of periodic solutions of system (5.69) with a period T near to 2 π. The first one is continuous and the second admits discontinuities once on a period. For those solutions, corresponding linearization systems around the grazing cycle Ψ$_{0}$( t ) are (5.77) and (5.78) , respectively. Figure 8: The grazing cycle of system (5.71) is in red. The blue arcs are the trajectory of the system with initial point (0 . 5 , 1 . 2) and the green continuous orbit is with initial value (0 . 1 , 0) . They demonstrate stability of the grazing solution. $_{u}$′ 1 = u$_{2}$, $_{u}$′ 2 = − 0 . 0001 sin(2 t ) u$_{1}$ + 0 . 0002 sin $^{2}$( t ) u$_{2}$, Δ u | 2 πi =    − 1 0 0 . 0003 R $_{0}$   u. (5.78) u ′ 1 = u$_{2}$, u ′ 2 = − 0 . 0001 sin(2 t ) u$_{1}$ + 0 . 0002 sin $^{2}$( t ) u$_{2}$, (5.77) 37 −0.5 0 0.5 1 1.5 2 2.5 −1.5 −1 −0.5 0 0.5 1 1.5 x$_{1}$(t) x 2 (t)