(−An2 + Kn + 2Ta) θk − Ta (αk+1 + αk) = 0,

(8)

Mn2xk + T (αk+1 − αk) = 0,

(9)

xk+1 − xk − a (θk+1 + θk) − 2lak = 0,

(10)

and the rest of the solution proceeds as before in § 14, putting

xk, θk, αk = (L, P, Q) exp cki.

(11)