32 Ψ(0) for the periodicity, it is sufficient to consider those with j − th coordinate is equal to ζ 0 j , [31]. For the discontinuous dynamics, the choice of the fixed coordinate can be made easier if the surface of disconti- nuity is provided with a constant coordinate. We will demonstrate this in examples. Denote the initial values of the intended periodic solution by ζ$_{1}$, ζ$_{2}$, . . . , ζ$_{n}$. Assume that one initial value ζ$_{j}$ is known, i.e. ζ 0 j . Thus, the problem contains n − many unknowns, they can be presented as ζ$_{1}$, ζ$_{2}$$^{, . . . , ζ$_{j}$}$− $_{1}$, ζ$_{j}$$_{+1}$ , . . . , ζ$_{n}$, T . Denote the solution of (5.57) by x$_{s}$ ( t, ζ$_{1}$, ζ$_{2}$, . . . , ζ$_{n}$, µ ) with initial conditions x$_{s}$ (0 , ζ$_{1}$, ζ$_{2}$, . . . , ζ$_{n}$, µ ) = ζ$_{s}$. To determine the unknowns, we will consider the Poincar ´ e criterion, which can be written as where ζ$_{j}$ = ζ 0 j . The equations (5.58) are satisfied with µ = 0 , T = ω, ζ$_{i}$ = ζ 0 i , i = 1 , 2 , . . . , n, since Ψ( t ) is the periodic solution. The following condition for the determinant is also needed in the remaining part paper. (A6) Theorem 5.1 Assume that condition ( A 6) is valid. Then, (5.57) admits a non-trivial periodic solution, which converges in the B − topology to the non-trivial ω -periodic solution of (5.57) as µ tends to zero. We will present the following examples to realize our theoretical results. Example 5.1 In this example, we will consider the perturbed system in case the generating system has a graziness. To show that, let us take into account the following perturbed system It is easy to see that the system (5.60) is of the form (5.57). For µ = 0 , the generating system became (3.14). For the perturbed system (5.60), we will investigate existence of the periodic solution around the grazing periodic solution of (3.14) with the help of Theorem 5.1. There are two sorts of possible periodic solutions of (5.60) around the grazing one. One of them has two impulse moments during the period since it crosses both lines of discontinuity, i.e. x$_{1}$ = 0 and x$_{1}$ = exp(0 . 00025 π ) . The other sort is the periodic solution which does not intersect the line x$_{1}$ = ∣ ∣ ∂ ( S$_{1}$ ( ω,ζ 0 $_{1}$,ζ 0 $_{2}$,...,ζ 0 j$_{−}$ $_{1}$,ζ 0 j $_{+1}$,...,ζ 0 $_{n}$, 0)) ∂ T . . . ∂ ( S$_{1}$ ( ω,ζ 0 $_{1}$,ζ 0 $_{2}$,...,ζ 0 j$_{−}$ $_{1}$,ζ 0 j $_{+1}$,...,ζ 0 $_{n}$, 0)) ∂ζ$_{n}$ ∂ ( S$_{2}$ ( ω,ζ 0 $_{1}$,ζ 0 $_{2}$,...,ζ 0 j$_{−}$ $_{1}$,ζ 0 j $_{+1}$,...,ζ 0 $_{n}$, 0)) ∂ T . . . ∂ ( S$_{2}$ ( ω,ζ 0 $_{1}$,ζ 0 $_{2}$,...,ζ 0 j$_{−}$ $_{1}$,ζ 0 j $_{+1}$,...,ζ 0 $_{n}$, 0)) ∂ζ$_{n}$ . . . . . ∂ ( S$_{n}$ ( ω,ζ 0 $_{1}$,ζ 0 $_{2}$,...,ζ 0 j$_{−}$ $_{1}$,ζ 0 j $_{+1}$,...,ζ 0 $_{n}$, 0)) ∂ T . . . ∂ ( S$_{n}$ ( ω,ζ 0 $_{1}$,ζ 0 $_{2}$,...,ζ 0 j$_{−}$ $_{1}$,ζ 0 j $_{+1}$,...,ζ 0 $_{n}$, 0)) ∂ζ$_{n}$ ∣ ∣ = 0 (5.59) x ′ 1 = x$_{2}$, x ′ 2 = − 0 . 001 x$_{2}$ − x$_{1}$, Δ x$_{2}$ | $^{x}$∈ Γ$_{1}$ = − (1 + R$_{1}$x$_{2}$ + µx$_{2}$ ) x$_{2}$, Δ x$_{2}$ | $^{x}$∈ Γ$_{2}$ = − (1 + R$_{2}$ + µ ( x$_{2}$ − exp(0 . 001 π/ 2)) x$_{2}$. (5.60) S$_{k}$ ( T , ζ$_{1}$, ζ$_{2}$, . . . , ζ$_{n}$, µ ) ≡ x$_{k}$ ( T , ζ$_{1}$, ζ$_{2}$, . . . , ζ$_{n}$, µ ) − ζ$_{k}$ = 0 , k = 1 , 2 , . . . , n, (5.58)