2152 Meccanica(2013)48:2149-215s Table 1 Comparison results obtained for y1:(.results reported by [1]:[.results reported by [5] -8'(0 f"(0 P(0) Present results Ramanaiah et al.[1 Present results Ramanaiah et al.10]Present results Ramanaiah et al.1 0.357406 0.3574 0.978400 0.9784 1.734930 1.7349 (0.358) (0.971 (1.73 [1.73492 0.36022 0.976978 0.977 1.708942 1.7089 0.425113 0.4251 1.247154 1.2472 1.39001 1.3900 10 1.134254 1.3432 19.732022 19.7320 0.51230 0.5123 1003.583404 3.5834 621.332230 621.3332 0.162144 0.1621 pmlc es f"(0)= 0.8646,90 The highe for =1. 09 f"(0)0.86446-0.20256v-+.. (17) 0.3 00)~1-0.39054y-1+0.18302y-2+. asyoo.Asymptotic expressions(17)are shown in Fig.1 by broken lines.Both expressions in(17)are in very good agreement with the numerically determined values even at relatively small values of y.Numeri- cal solutions have been obtained for Eqs.(10).(11). (13).(14)both for the free convection,M=0.limit and for other values of M.namely M=0.1.1.10 and 100 when is large.i.e.applying the boundary condi- tion(0)=1.The results iven in Table 1 which shows 牌cgk的oa川a时 3.1.2 Small y Fig1 Plots of (a))and (b)0)againstyfor For this case we have to scale the variables by writing with M=0subject to oundary conditions (1).Asy f=y/67 0=y5/6a. P=y2/P. n=yl/6n (18) 3.1.1 Large y aveE.u0.B0wi地Mo pt that now For this case we leave the equations unscaled with respect to元.The only change is in the boundary con- ditions(11)which is now the leading-order problem still given by (10).(13). (14)but now subject to the condition that 0(0)=1. 日=-1+y5/6aom万=0 (19) Springer 2152 Meccanica (2013) 48:2149–2158 Table 1 Comparison results obtained for γ  1; (.) results reported by [1]; [.] results reported by [5] M −θ (0) f (0) P(0) Present results Ramanaiah et al. [10] Present results Ramanaiah et al. [10] Present results Ramanaiah et al. [10] 0 0.357406 0.3574 0.978400 0.9784 1.734930 1.7349 (0.358) (0.971) (1.73) [0.35741] [0.97840] [1.73492] 0.1 0.360227 0.3602 0.976978 0.9770 1.708942 1.7089 1 0.425113 0.4251 1.247154 1.2472 1.390015 1.3900 10 1.134254 1.3432 19.732022 19.7320 0.512309 0.5123 100 3.583404 3.5834 621.332230 621.3332 0.162144 0.1621 Fig. 1 Plots of (a) f (0) and (b) θ(0) against γ for σ = 1.0 and obtained from the numerical solution to Eqs. (10), (13), (14) with M = 0 subject to boundary conditions (11). Asymptotic expressions (17), (20) for large and small γ are shown by broken lines 3.1.1 Large γ For this case we leave the equations unscaled with the leading-order problem still given by (10), (13), (14) but now subject to the condition that θ(0) = 1. Our numerical solution to this problem gives f (0) = 0.86446, θ (0) = −0.39054 for σ = 1.0. The higher￾order terms are found by expanding in inverse powers of γ . The details are straightforward and we find that, for σ = 1, f (0) ∼ 0.86446 − 0.20256γ −1 +··· θ(0) ∼ 1 − 0.39054γ −1 + 0.18302γ −2 +··· (17) as γ → ∞. Asymptotic expressions (17) are shown in Fig. 1 by broken lines. Both expressions in (17) are in very good agreement with the numerically determined values even at relatively small values of γ . Numeri￾cal solutions have been obtained for Eqs. (10), (11), (13), (14) both for the free convection, M = 0, limit and for other values of M, namely M = 0.1, 1, 10 and 100 when γ is large, i.e. applying the boundary condi￾tion θ(0) = 1. The results are given in Table 1 which shows a very good agreement with the previous results reported by Ramanaiah et al. [10], Stewartson [1] and Jones [5]. 3.1.2 Small γ For this case we have to scale the variables by writing f = γ 1/6f, θ = γ 5/6θ, P = γ 2/3P, η = γ 1/6η (18) This leaves Eqs. (10), (13), (14) with M = 0 essen￾tially unaltered except that now differentiation is with respect to η. The only change is in the boundary con￾ditions (11) which is now θ  = −1 + γ 5/6θ on η = 0 (19)