=0.i 由(3) 7=l+i-(6 Ll′=ul+i-t(3) 将(1),(6)式代入(2)式 r(1 u+I-u l=l=h(5) u+ n'ur+n(l-r)u-n'ur=n(l-r)u lu=hl→∞时 r h u = 0, i = 由(3) i' = u +i −u' (6) 将(1)，(6)式代入(2)式 u r l r n n u i u − + − =  ' ' n'ur + n'(l − r)u − n'u'r = n(l − r)u lu = h h r n n n u nu −  − = ' ' ' u r l r n n u u r l r u − − =  −  + ' ' u r l r i − = (1) i n n i   = (2) u = u +i −i (3) (1 ) u i l r    = + (4) lu = lu = h (5) 