θ$_{i}$, i ∈ $^{Z}$, then the linearization system for the equation around the equilibrium is Indeed, if u$_{1}$ ( t ) , u$_{1}$ (0) = e$_{1}$, u$_{2}$ ( t ) , u$_{2}$ (0) = e$_{2}$, are solutions of (3.28) , then one can see that x ( t ) − (0 , 0) = x 0 $_{1}$u$_{1}$ ( t ) + x 0 $_{2}$u$_{2}$ ( t ) , for all t ∈ $^{R}$. We have obtained that linearization exists for both grazing solutions Ψ( t ) , and the equilibrium at the origin. Moreover, conditions ( C 1) − ( C 10) are valid and all other solutions are B-differentiable in parameters [1]. Thus, the system (3.14) defines a K − smooth discontinuous flow in the plane. In the next example, we will finalize the linearization around the grazing solution Ψ( t ) . Example 3.2 (Linearization around the grazing discontinuous cycle). We continue analysis of the last example, and complete the variational system for Ψ( t ) . Let us consider this time, the linearization at the non-grazing moment ω = π. The discontinuity point is c = (0 , − exp(0 . 0005 π )) and it is of ( α ) − type, since By using formula (3.10), one can compute the gradient as ∇ τ ( c ) = (exp( − 0 . 0005 π ) , 0) . Then, utilizing ∇ τ ( c ) and formula (3.11), one can determine that the matrix of linearization at the moment π is From the monotonicity of the jump function, − R$_{1}$y 2 $_{2}$, it follows that the the yellow and blue subregions of G are invariant. Consequently, for each solution near to Ψ( t ) , the sequences B$_{i}$ is of two types B$_{i}$ = D ( j ) i , i ∈ Z and j = 1 , 2 , where D (1) 2 $^{i}$− 1 = O$_{2}$, D (2) 2 $^{i}$− 1 = D (2) 1 =    − 1 0 0 . 001 − 1 . 8 exp(0 . 00025 π ) $_{0}$   , D (1) 2 i = D (2) 2 i =    exp( − 0 . 0005 π ) 0 0 . 001 $_{0}$   , i ∈ $^{Z}$. That is, the condition ( A 4) is valid and the linearization around the periodic solution (3.15) on R is of two subsystems: $_{u}$′ 1 = u$_{2}$, $_{u}$′ 2 = − u$_{1}$ + 0 . 001 u$_{2}$, Δ u$_{2}$ | t = θ$_{i}$ = − (1 + R$_{2}$ ) u$_{2}$. (3.28) 〈∇ Φ( c ) , f ( c ) 〉 = 〈 (1 , 0)( − exp(0 . 0005 π ) , − 0 . 001 exp(0 . 0005 π )) 〉 = − exp(0 . 0005 π ) = 0 . B$_{2}$ =    exp( − 0 . 0005 π ) 0 0 . 001 $_{0}$   . 19