Indeed, one can easily show that the derivative at the grazing point y ∗ is Similarly, all other points of the grazing orbit can be discussed. Next, differentiating (2.3) with y ( t ) again we obtain that where e$_{2}$ = (0 , 1) $^{T}$. Calculate the right hand side of (3.22) at the point ¯ y = (¯ y$_{1}$, ¯ y$_{2}$ ) to obtain To calculate the fraction ∂τ ( y ( ξ, 0 , y ∗ + Δ y )) ∂y 0 2 in (3.23) , we apply formula (3.10) for the transversal point ¯ y = (¯ y$_{1}$, ¯ y$_{2}$ ) . The second component ∂τ (¯ y ) ∂y 0 2 takes the form ∂τ (¯ y ) ∂y 0 2 = 0 . This and formula (3.23) imply Similar to (3.21) , one can obtain that Joining (3.21) and (3.25) , it can be obtained that    − 1 0 . 001 − 1 . 8 exp(0 . 00025 π $_{)}$   . (3.21) ∂W$_{i}$ ( y ) ∂y 0 2 = τ ( y ) ∫ θ$_{i}$ ∂f ( y ) ∂y ∂y ( s ) ∂y 0 2 ds + f ( y ( s )) ∂τ ( y ) ∂y 0 2 + J$_{y}$ ( y )( e$_{2}$ + f ( y ( s )) ∂τ ( y ) ∂y 0 2 ) + f ( y + J ( y )) ∂τ ( y ) ∂y 0 2 + θ$_{i}$ ∫ τ ( y ) ∂f ( y ( s ) + J ( y ( s ))) ∂x ∂y ( s ) ∂y 0 2 ds, (3.22) ∂W$_{i}$ ( y ( ξ, 0 , y ∗ + Δ y )) ∂y 0 2 = f ( y ( ξ, 0 , y ∗ + Δ y )) ∂τ ( y ( ξ, 0 , y ∗ + Δ y )) ∂y 0 2 + J$_{y}$ ( y ( ξ, 0 $_{, y}$∗ + Δ y )) ( e$_{2}$ + f ( y ( ξ, 0 $_{, y}$∗ + Δ y ))) ∂τ ( y ( ξ, 0 , y ∗ + Δ y )) ∂y 0 2 ) + f ( y ( ξ, 0 , y ∗ + Δ y )) ∂τ ( y ( ξ, 0 , y ∗ + Δ y )) ∂y 0 2 . (3.23) ∂W$_{i}$ (¯ y ) ∂y 0 2 =    0 − 2 R ¯ $_{y$_{2}$}$   . (3.24) ∂W$_{i}$ ( y ∗ ) ∂y 0 2 =    0 $_{0}$   . (3.25) W$_{iy}$ ( $_{y}$∗$_{) =}$    − 1 0 0 . 001 − 1 . 8 exp(0 . 00025 π ) $_{0}$   . (3.26) 17