autonomous system with variable moments of impulses $_{x}$′$_{1}$ = x$_{2}$, $_{x}$′$_{2}$ = − x$_{1}$, Δ x$_{1}$ | $^{x}$∈ Γ = 1 √$_{2}$ − x$_{1}$ + K ( x$_{2}$ − x$_{1}$ ) $^{2}$, Δ x$_{2}$ | $^{x}$∈ Γ = 1 √$_{2}$ − x$_{2}$ + K ( x$_{2}$ − x$_{1}$ ) $^{2}$, (4.50) where Γ = { ( x$_{1}$, x$_{2}$ ) | x$_{1}$ + x$_{2}$ = √$_{2}$ } , ˜ Γ = { ( x$_{1}$, x$_{2}$ ) | x$_{1}$ = x$_{2}$ } and K = 0 . 11 . It is easy to verify that the point x ∗ = ( 1 2 , 1 2 ) is a grazing point because 〈∇ τ ( x ∗ ) , f ( x ∗ ) 〉 = 〈 (1 , 1) , ( 1 $_{2}$, − 1 $_{2}$) 〉 = 0 and the grazing is non-axial. We assume that the domain is the plane. The solution Ψ( t ) = (sin( t ) , cos( t )) , t ∈ R is a grazing one, since the point x ∗ = Ψ( π $_{4}$) is from its orbit. The cycle and the line of discontinuity are depicted in Figure 5. Let us consider the linearization at the grazing point x ∗ next. We will consider the near solution x ( t ) = x ( t, 0 , x ∗ + Δ x ) . Denote t = ξ, the moment when the solution meets the surface of discontinuity Γ at the point ¯ x = x ( ξ ) = x ( ξ, 0 , x ∗ + Δ x ) . Taking into account formulae (3.17) , (3.18) with (4.50) , one can obtain the following matrix Calculating the right hand side of the expression (4.51) , we obtain that ∂W$_{i}$ ( x ( ξ, 0 , x ∗ + Δ x )) ∂x 0 1 =    ¯ x$_{2}$ − ¯ $_{x$_{1}$}$   1 ¯ x$_{1}$ − ¯ x$_{2}$ +    − 2 K (¯ x$_{2}$ − ¯ x$_{1}$ ) − 2 K (¯ x$_{2}$ − ¯ x$_{1}$ ) − 2 K (¯ x$_{2}$ − ¯ x$_{1}$ ) − 2 K (¯ x$_{2}$ − ¯ x$_{1}$ $_{)}$   × ( e$_{1}$ +    ¯ x$_{2}$ − ¯ $_{x$_{1}$}$   1 ¯ $^{x$_{1}$}$− ¯ x$_{2}$ ) +    − 1 √$_{2}$ + K (¯ x$_{2}$ − ¯ x$_{1}$ ) 2 1 √$_{2}$ − K (¯ x$_{2}$ − ¯ x$_{1}$ ) 2    1 ¯ x$_{1}$ − ¯ x$_{2}$ . (4.51) 28 Figure 5: The red curve is the orbit of Ψ( t ) which grazes non-axially the line of discontinuity. −1 −0.5 0 0.5 1 1.5 −0.5 0 0.5 1 1.5 x$_{1}$ x 2