“…In order to apply the results obtained in the previous sections we consider the following change of variable. Let us set y − y =: z and use this equality in (2), where y solves (3). It is easy to verify that z satisfies z t + z xxx − ν(t)z xx + (zy) x + zz x = v1 ω in Q, z(0, t) = z(L, t) = 0 on (0, T ), z x (0, t) = z x (L, t) on (0, T ), z(•, 0) = y 0 − y 0 in (0, L).…”