“…T d ds |s − x| d+2α−2 (∆ s g)(t, x, s, r, v, r, ṽ) = c β,d g(t, x, r, v) dr dṽ (r − r)(−∆ x ) α g(t, x, r, ṽ) =: −1 d,α Σ g (t, x, r) g(t, x, r, v) , and deduce from (39) the Vlasov-type equation ∂ t g(t, x, r, v) + v∂ r g(t, x, r, v) + C 1 [g, g](t, x, r, v) = 0, (…”