“…ThusF u,s p (s) = 1/(2s), and evaluating both parts of (27) gives R p (t) = e t/2 /2 for t = 0. Therefore, from (25), Some sample calculations for the specific choice of n = 2, t 1 = 0, t 2 = 1, g 1,2 (x 1 , x 2 ) = e t 1 x 1 and g 2,2 (x 1 , x 2 ) = The above can be explicitly integrated, leading to a not particularly illuminating lengthy expression. Its behaviour with p and t is shown in Figure 11, bearing in mind that positive M relates to flow across the heteroclinic from the lower to the upper strip.…”