Taking the derivative of the expressions (5.65) with respect to variables T and ζ$_{2}$, one can obtain the following To determine the above determinant, the monodromy matrix of (3.13) with the jump matrix D (2) i can be evaluated as For µ = 0 , with the values ω and ζ 0 2 the determinant (5.66) can be determined as This verifies condition ( A 6) . So, By Theorem 5.1, we can conclude that the perturbed system (5.60) admits a non-trivial T ( µ ) − periodic solution which converges in the B − topology to the non-trivial ω -periodic solution of (3.14) as µ tends to zero such that T (0) = ω. In Fig. 7, some numerical results are provided to show the solutions of system (5.60) with µ = 0 . 05 . The periodic solutions for µ = 0 are not grazing. For µ = 0 , we have one periodic solution which is orbitally stable, and for µ < 0 , there exist two periodic solutions. One of them has one discontinuity moment in each period, in other words, the cycle does not intersect the surface of discontinuity around grazing point and it is orbitally stable and the other one has two discontinuity moments in each period. This means, the number of periodic solutions increases by variation of µ, around µ = 0 . So, we will call that bifurcation of periodic solution from a grazing cycle. Example 5.2 Let us consider the following system with variable moments of impulses and a small parameter x ′ 1 = x$_{2}$, x ′ 2 = − 0 . 0001[ x 2 $_{2}$+ ( x$_{1}$ − 1) 2 − (1 + µ ) $^{2}$] x$_{2}$ − x$_{1}$ + 1 , Δ x$_{2}$ | $^{x}$∈ Γ = − (1 + Rx$_{2}$ + µx 3 $_{2}$) x$_{2}$ + µ $^{2}$, (5.69) 34 S $_{1}$( T , 0 , ζ$_{2}$, µ ) = x$_{1}$ ( T , 0 , ζ$_{2}$, µ ) = 0 , S $_{2}$( T , 0 , ζ$_{2}$, µ ) = x$_{2}$ ( T , 0 , ζ$_{2}$, µ ) − ζ$_{2}$ = 0 . (5.65) ∣ ∂ ( S$_{2}$ ( $^{T}$, 0 ,ζ$_{2}$,µ )) ∂ T ∂ ( S$_{1}$ ( $^{T}$, 0 ,ζ$_{2}$,µ )) ∂ζ$_{2}$ ∂ ( S$_{2}$ ( $^{T}$, 0 ,ζ$_{2}$,µ )) ∂ T ∂ ( S$_{2}$ ( $^{T}$, 0 ,ζ$_{2}$,µ )) ∂ζ$_{2}$ ∣ = ∣ ∂x$_{1}$ ( ω, 0 ,ζ 0 $_{2}$, 0) ∂ T ∂x$_{1}$ ( ω, 0 ,ζ 0 $_{2}$, 0) ∂ζ$_{2}$ ∂x$_{2}$ ( ω, 0 ,ζ 0 $_{2}$, 0) ∂ T ∂x$_{2}$ ( ω, 0 ,ζ 0 $_{2}$, 0) ∂ζ$_{2}$ − $_{1}$∣ ∣ . (5.66)    1 0 . 01 0 0 . $_{704}$   . (5.67) ∣ ∂ ( S$_{1}$ ( ω, 0 ,ζ 0 $_{2}$, 0)) ∂ T ∂ ( S$_{1}$ ( ω, 0 ,ζ 0 $_{2}$, 0)) ∂ζ$_{2}$ ∂ ( S$_{2}$ ( ω, 0 ,ζ 0 $_{2}$, 0)) ∂ T ∂ ( S$_{2}$ ( ω, 0 ,ζ 0 $_{2}$, 0)) ∂ζ$_{2}$ ∣ = ∣ 0 0 . 01 − exp(0 . 00025 π ) − 0 . $_{296}$∣ ∣ = 0 . 01 exp(0 . 00025 π ) = 0 . (5.68)