“…Since g (u) < 0 for (0, 1), g(u) is decreasing in (0, 1), and so t 2 > 7 4 . A simple exercise shows that (23) does not hold in this case for all values of u ∈ 7 18 , 1 ; thus, there are no critical points of G in (0, 2) × 7 18 , 1 × (0, 1). Suppose that there is a critical point ( t, ũ, ṽ) of G existing in the interior of cuboid S. Clearly, it must satisfy that ũ < 7 18 .…”