Yu-Hao Liang Department of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan

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1 Detail proofs of some results in the article: Classification and evolution of bifurcation curves for the one-dimensional perturbed Gelfand equation with mixed boundary conditions Yu-Hao Liang Department of Applied Mathematics, National Chiao Tung University, Hsinchu 3, Taiwan Shin-Hwa Wang Department of Mathematics, National Tsing Hua University, Hsinchu 3, Taiwan In this article, we give detail proofs of Lemma 3.1(I)(ii) (I)(iv), (I)(vi) (I)(viii), (I)(x) (I)(xv) and (III)(v) (III)(ix) in the article Classification and evolution of bifurcation curves for the one-dimensional perturbed Gelfand equation with mixed boundary conditions 1], where these proofs are not given. To start with, we recall some definitions of functions and Lemma 3.1 therein. Remark that following labels (3.5) and (3.7) (3.19) are consistent with those in 1]. Define f(s) = exp ( ) as s a+s, F(s) = f(t)dt, θ(s) = 2F(s) sf(s) and P 1 (ρ,s) = ρf(ρ) sf(s) F(ρ) F(s), P 2(ρ,s) = ρ2 f (ρ) s 2 f (s) ρf(ρ) sf(s). (3.5) Lemma 3.1. Define M 1 (ρ) = ρf(ρ) F(ρ) for ρ >, (3.7) M 2 (ρ) = ρf (ρ) f(ρ) = a2 ρ for ρ, (3.8) (a+ρ) 2 Q(ρ,s) = θ(ρ) θ(s)+m 1 (ρ)f(ρ) F(s)] for s ρ, (3.9) This work is partially supported by the Ministry of Science and Technology of the Republic of China. Corresponding author. addresses: moonsea.am96g@g2.nctu.edu.tw (Y.-H. Liang), shwang@math.nthu.edu.tw (S.-H. Wang). 1

2 November 4, 215 R(ρ,s) = 2θ(ρ) θ(s)] f(ρ) F(ρ) ρ F(ρ) F(s)] 3/2 ρ s for s < ρ, (3.1) 8 3 K 1 (ρ) = ρ2 + 2ρ ρ (16ρ 3) for ρ, (3.11) ρ ( 8 K 2 (ρ) = 3 132a+165 ) ρ ρ+1 for ρ, (3.12) 5a+25 3 ] K 3 (ρ) = M 1 (ρ)+2ρ f(ρ) f (ρ) for ρ >, (3.13) f(ρ) L 1 (ρ,s) = 8 25 s M 2(s) 8 F(ρ) F(s) +M 1 (ρ) 1 for s ρ, (3.14) 25 f(s) L 2 (ρ,s) = P 1 (ρ,s) 8 25 s M 1(ρ) for s < ρ, (3.15) L 3 (ρ,s) = P 2 (ρ,s) 3 2 P 1(ρ,s) Then the following assertions (I) (III) hold: (I) For a >, the following assertions (i) (xv) hold: for s < ρ. (3.16) (i) f(ρ,a)(= f(ρ)) is a strictly increasing function of ρ on, ) and of a on (, ) when the other variable is fixed. Moreover, 1 < f(ρ) < min{e ρ,e a for ρ,a >. (ii) M 2 (ρ) < M 1 (ρ) for ρ >. (iii) 1 < 1 2 M 2(ρ)+2] < M 1 (ρ) for ρ >. (iv) M 1 (ρ) < M 2 (ρ)+1 for < ρ a. (v) M 1 (ρ) is strictly increasing on (,a]. (vi) M 2 (ρ) satisfies M 2(ρ) = a2 (a ρ) (a+ρ) 3 and M 2 (ρ) M 2 (a)(= a/4) for ρ >. > when ρ (,a), = when ρ = a, < when ρ (a, ), (3.17) (vii) P 1 (ρ,) = M 1 (ρ), lim s ρ P 1 (ρ,s) = M 2 (ρ)+1 for ρ > and lim ρ s +P 1 (ρ,s) = M 2 (s)+1 for s. (viii) P 1 (ρ,s) > 1 for s < ρ. (ix) For < ρ a, P 1 (ρ,s) is a strictly increasing function of s on,ρ). Moreover, M 1 (ρ) P 1 (ρ,s) < M 2 (ρ)+1 for s < ρ a, where the equality holds if and only if s =. (x) For s < a, P 1 (ρ,s) is a strictly increasing function of ρ on (s,a]. Moreover, P 1 (ρ,s) > M 2 (s)+1 for s < ρ a. (xi) P 2 (ρ,) = M 2 (ρ) and lim s ρ P 2 (ρ,s) = 2ρf (ρ)+ρ 2 f (ρ) f(ρ)+ρf (ρ) for ρ >. 2

3 November 4, 215 (xii) M 2 (ρ) < P 2 (ρ,s) for < s < ρ a. (xiii) Q(ρ,s) > for s < ρ. ] f(ρ) (xiv) + M 2(ρ)+1 f(ρ) > for < ρ a. F(ρ) F(ρ) 2ρ (xv) For ρ >, M 1 (ρ,a) and M 2 (ρ,a) are both strictly increasing functions of a on (, ). (II) For < a a (.51), R(ρ,s) > for s < ρ. Here a is the unique positive zero of a/2 e a a/ (3.18) (III) For a 4, the following assertions (i) (ix) hold: (i) R(ρ,s) < for ρ a and s < ρ. (ii) For < ρ < a, R(ρ,s) > for < s < ρ if ρ satisfies 21 M 2 (ρ)] f(ρ) M 1 (ρ) > ; while R(ρ,s) < for < s < ρ if ρ satisfies 22 M 1 (ρ)] f(ρ) <. (iii) R(ρ,s) > for s < ρ ρ 1. Here ρ 1 (.286) > 7/25 =.28 is the unique positive zero of 2(1 ρ) ρe 3ρ /(e ρ 1). (iv) R(1/2,s) < for s < 1/2. (v) K 1 (ρ) < M 1 (ρ) < K 2 (ρ) for 1/4 ρ 1/2. (vi) K 3 (ρ) < M 1 (ρ)+ 132a+165 5a+25 ρ2 for 1/4 ρ 1/2. (vii) For any ρ 1/4,1/2], there exists a unique positive s (= s (ρ,a)) on (,ρ) such that > when s (,s ), L 1 (ρ,s) = when s = s, < when s (s,ρ]. (viii) L 2 (ρ,s) for 1/4 ρ 1/2 and s < ρ. (ix) L 3 (ρ,s) for 7/25 ρ 1/2 and s < ρ. (3.19) Proof of Lemma 3.1. In the following, we would give detail proofs of Lemma 3.1 for parts (I)(ii) (I)(iv), (I)(vi) (I)(viii), (I)(x) (I)(xv) and (III)(v) (III)(ix). Proof of Lemma 3.1(I)(ii). Let n 1 (ρ) (a+ρ) 2F(ρ) ρ = (a+ρ) 2F(ρ) ρ = (a+ρ) 2 f(ρ) a 2 F(ρ) M 1 (ρ) M 2 (ρ)] ] ρf(ρ) F(ρ) a2 ρ (a+ρ) 2 (by (3.7) and (3.8)) 3

4 November 4, 215 for ρ >. Then we compute that lim ρ + n 1 (ρ) = a 2 > and n 1(ρ) = 2(a+ρ)f(ρ) > for ρ >. It implies that, for ρ >, n 1 (ρ) > and hence M 1 (ρ) > M 2 (ρ). Hence the proof of Lemma 3.1(I)(ii) is complete. Proof of Lemma 3.1(I)(iii). First, for ρ >, 1 2 M 2(ρ)+2] > 1 since M 2 (ρ) > by (3.8). Next, we show that 1 2 M 2(ρ)+2] < M 1 (ρ) for ρ >. Let n 2 (ρ) for ρ >. Then we compute that n 2 () = and ρf(ρ) 1 M 2 2(ρ)+2] F(ρ) = ρf(ρ) 1 2 ρf (ρ)/f(ρ)+2] F(ρ) n 2 (ρ) = a 2 ρ 2 (4ρ+a 2 +4a) (2ρ 2 +a 2 ρ+4aρ+2a 2 ) 2f(ρ) > forρ >. Itimpliesthat,forρ >,n 2 (ρ) > andhence 1 2 M 2(ρ)+2] < M 1 (ρ) (= ρf(ρ)/f(ρ)) by (3.7). Hence the proof of Lemma 3.1(I)(iii) is complete. Proof of Lemma 3.1(I)(iv). Let n 3 (ρ) ρf(ρ) M 2 (ρ)+1 F(ρ) = for ρ >. Then we compute that n 3 () = and ρf(ρ) ρf (ρ)/f(ρ)+1 F(ρ) n 3 (ρ) = a 2 ρ(a 2 ρ 2 ) (ρ 2 +a 2 ρ+2aρ+a 2 ) 2f(ρ) < for < ρ < a. It implies that, for < ρ a, n 3 (ρ) < and hence M 1 (ρ) (= ρf(ρ)/f(ρ)) < M 2 (ρ)+1 by (3.7). Hence the proof of Lemma 3.1(I)(iv) is complete. Proof of Lemma 3.1(I)(vi). The proof is trivial and hence it is omitted. Proof of Lemma 3.1(I)(vii). The first equality in the statement of Lemma 3.1(I)(vii) is clear while the second and the third ones follow from the L Hôspital s rule directly. Proof of Lemma 3.1(I)(viii). By integration by parts, we have that, for s < ρ, F(ρ) F(s) = ρ f(t)dt = ρf(ρ) sf(s) ρ s s tf (t)dt < ρf(ρ) sf(s)sincef (t) = a2 f(t) > (a+t) 2 for t >. Hence P 1 (ρ,s) > 1 by (3.5), which completes the proof of Lemma 3.1(I)(viii). Proof of Lemma 3.1(I)(x). Fix s on,a). Then we compute that ρ P 1(ρ,s) = f(ρ){m 2(ρ)+1] P 1 (ρ,s) > F(ρ) F(s) for s < ρ a by Lemma 3.1(I)(ix). Hence, for s < a, P 1 (ρ,s) is a strictly increasing function of ρ on (s,a]. It implies that P 1 (ρ,s) > lim ρ s +P 1 (ρ,s) = M 2 (s) + 1 for s < ρ a by Lemma 3.1(I)(vii). Hence the proof of Lemma 3.1(I)(x) is complete. Proof of Lemma 3.1(I)(xi). The first equality in the statement of Lemma 3.1(I)(xi) is clear while the second one follows from the L Hôspital s rule directly. Proof of Lemma 3.1(I)(xii). By (3.5) and (3.8), for < s < ρ a, we compute that P 2 (ρ,s) M 2 (ρ) = ρ2 f (ρ) s 2 f (s) ρf(ρ) sf(s) ρf (ρ) f(ρ) = sf(s)m 2(ρ) M 2 (s)] >, ρf(ρ) sf(s)] 4

5 November 4, 215 since M 2 (ρ) is strictly increasing on (,a] by (3.17). Hence the proof of Lemma 3.1(I)(xii) is complete. Proof of Lemma 3.1(I)(xiii). Weprovethat Q(ρ,s) = θ(ρ) θ(s)+m 1 (ρ)f(ρ) F(s)] > for s < ρ in the following three cases. Case 1: s < ρ a. We compute that Q(ρ,s) = F(ρ) F(s)]M 1 (ρ)+2 P 1 (ρ,s)] F(ρ) F(s)]{M 1 (ρ)+2 M 1 (ρ)+1] (by Lemma 3.1(I)(ii) and (I)(ix)) = F(ρ) F(s) >. (3.2) Case 2: s a < ρ. Note first that, since Q(a,s) > for s < a by (3.2) and Q(a,a) = by (3.9), we have that Q(a,s) for s a. Moreover, we compute that ρ F(ρ)Q(ρ,s)] = f(ρ)4f(ρ) 3F(s)+sf(s) M 2(ρ)F(s)] f(ρ)4f(a) 3F(s)+sf(s) M 2 (a)f(s)] (by (3.17)) f(ρ)n 4 (s), wheren 4 (s) = 4F(a) 3F(s)+sf(s) M 2 (a)f(s). Sincen 4 (a) = F(a)1+M 1 (a) M 2 (a)] > by (3.7) and Lemma 3.1(I)(ii) and since n 4 (s) = f(s) 2+M 2(s) M 2 (a)] 2f(s) < for < s < a by (3.17), we have that n 4 (s) > for s a. It follows that F(ρ)Q(ρ,s)] > for s a < ρ. Combining the fact with F(a)Q(a,s) for ρ s a as claimed above, we have that Q(ρ,s) > for s a < ρ. Case 3: a < s < ρ. Fix ρ > a. Suppose it is not true that Q(ρ,s) >, then there exists some s on (a,ρ) such that Q(ρ,s ). Note first that, by the result from Case 2 with s = a, we have that Q(ρ,a) >. On the other hand, since Q(ρ,ρ) = by (3.9) and lim s ρ s Q(ρ,s) = f(ρ)1+m 1(ρ) M 2 (ρ)] < by Lemma 3.1(I)(ii), there exists some s 1 on (s,ρ) such that Q(ρ,s 1 ) >. Hence Q(ρ,s ) < min{q(ρ,a), Q(ρ,s 1 ).It implies that Q(ρ,s) on a,s 1 ] has its global minimum at some point s 2 on (a,s 1 ). However, we compute that, for a < s < ρ, s Q(ρ,s) = f(s)1+m 1(ρ) M 2 (s)], 2 s 2Q(ρ,s) = f (s)1+m 1 (ρ) M 2 (s)]+f(s)m 2(s) = a2 (s+a) 2f(s)1+M 1(ρ) M 2 (s)] f(s) a2 (s a) (by (3.17)) (a+s) 3 a 2 (s a) = (s+a) 2 s Q(ρ,s) f(s)a2 (a+s), 3 which implies that 2 s 2 Q(ρ,s 2 ) = f(s 2 ) a2 (s 2 a) (s 2 +a) 3 < since s Q(ρ,s 2) =. Consequently, s 2 can not be a local minimum point, and we get a contradiction. Hence Q(ρ,s) > for a < s < ρ. 5

6 November 4, 215 By Cases 1 3, we complete the proof of Lemma 3.1(I)(xiii). Proof of Lemma 3.1(I)(xiv). For < ρ a, we compute that, by (3.7) and (3.8), ] f(ρ) + M 2(ρ)+1 F(ρ) 2ρ Hence the proof of Lemma 3.1(I)(xiv) is complete. f(ρ) = f(ρ) F(ρ) 2ρ F(ρ) 3M 2(ρ)+1 M 1 (ρ)] > f(ρ) 2ρ F(ρ) {3M 2(ρ)+1 M 2 (ρ)+1] (by Lemma 3.1(I)(iv)) = f(ρ) ρ F(ρ) M 2(ρ) >. Proof of Lemma 3.1(I)(xv). For ρ >, we compute that ρ2 f(ρ,a) = f(ρ,a) and a (ρ+a) 2 a M 1(ρ,a) = ρf(ρ,a) a ρ 2 f(ρ,a) (ρ+a) = ρ 2 F(ρ,a) = ρ ] ρ = ρf(ρ,a) F(ρ,a)] 2 = aρf(ρ,a) F(ρ,a)] 2 ρ ρ a M 2(ρ,a) = 2aρ2 (ρ+a) 3 >. f(ρ,a)] F(ρ,a) f(ρ,a) F(ρ,a)] a a F(ρ,a)] 2 f(s,a)ds f(ρ) ρ F(ρ,a)] 2 ] ρ 2 (ρ+a) 2 s 2 (s+a) 2 f(s,a)ds s 2 (s+a) 2 f(s,a)ds (ρ s)(aρ+2sρ+as) (ρ+a) 2 (s+a) 2 f(s,a)ds >, Hence, for ρ >, M 1 (ρ,a) and M 2 (ρ,a) are both strictly increasing functions of a on (, ), which completes the proof of Lemma 3.1(I)(xv). Proof of Lemma 3.1(III)(v). (I). We prove that K 1 (ρ) < M 1 (ρ) for 1/4 ρ 1/2 and a 4. Note first that, for 1/4 ρ 1/2, ] K 1 (ρ) = 32ρ3 +198ρ ρ (2ρ 2 +5ρ+153) > ρ2 ( 32/2+198) 5(2ρ 2 +5ρ+153) >. Hence, for 1/4 ρ 1/2 and a 4, to prove that K 1 (ρ) < M 1 (ρ,a)(= ρf(ρ,a)/f(ρ,a)) is equivalent to prove that Indeed, we compute that, ρ n 5(ρ,a) = n 5 (ρ,a) ρf(ρ,a) K 1 (ρ) F(ρ,a) >. (3.21) n 6 (ρ,a) (ρ+a) 2 (32ρ 3 198ρ 2 272ρ 7959) 2f(ρ,a), (3.22) where n 6 (ρ,a) n 6,2 (ρ)a 2 +n 6,1 (ρ)a+n 6, (ρ) 6

7 November 4, 215 and n 6,2 (ρ) = 4224ρ ρ ρ ρ ρ ρ , n 6,1 (ρ) = 248ρ ρ ρ ρ ρ ρ ρ, n 6, (ρ) = 124ρ ρ ρ ρ ρ ρ ρ 2. Then it can be verified easily (but tediously and hence omitted) that, for 1/4 ρ 1/2 and a 4, n 6 (ρ,4) = 124ρ ρ ρ ρ ρ ρ ρ ρ n 7 (ρ) >, (See Fig. 1(a).) (3.23) a n 6(ρ,4) = 248ρ ρ ρ ρ ρ ρ ρ n 8 (ρ) >, (See Fig. 1(b).) (3.24) 2 a 2n 6(ρ,a) = 8448ρ ρ ρ ρ ρ ρ n 9 (ρ) >. (See Fig. 1(c).) (3.25) It implies that, for 1/4 ρ 1/2 and a 4, n 6 (ρ,a) > and hence, by (3.22), ρ n 5(ρ,a) >. (3.26) n 7 (ρ) n 8 (ρ) n 9 (ρ) (a) ρ (b) ρ.25.5 (c) ρ Fig. 1. Function n 7 (ρ) (resp., n 8 (ρ) and n 9 (ρ)) defined in (3.23) (resp., (3.24) and (3.25)) is positive on the interval 1/4, 1/2]. 7

8 November 4, 215 Next, we show that n 5 (1/4,a) > for a 4. It is easy to see that 1/4 (6b1 48b +192)(s 1/4) 2 +b 1 (s 1/4)+b ] ds = 1 for any b,b 1 >. Thus, if we let b 9822 and b s f(1/4,a), then 2225 f(1/4,a) n 5 (1/4,a) = 1/ f(1/4,a) F(ρ,a) = K 1 (1/4) 9822 f(1/4,a) F(1/4,a) 1/4 { 2225 = 9822 f(1/4,a) (6b 1 48b +192)(s 1/4) 2 ] +b 1 (s 1/4)+b f(s,a) ds 1/4 n 1 (s,a)ds. (3.27) Note that n 1 (1/4,a) =, s n 1(1/4,a) = and we compute that, for < s < 1/4 and a 4, 2 s 2n 1(s,a) = (6b 1 48b +192)f(1/4,a) 2 s 2f(s,a) = a a 461 f(1/4,a) (4a+1) 2 s 2f(s,a) > 32 ] 2446a a 461 a 2 (a 2 2a 2s) 1637 (4a+1) 2 (a+s) 4 e 1/4 s= (since 2 s 2f(s,a) = a2 (a 2 2a 2s) f(s,a) < a2 (a 2 2a 2s) e 1/4 by Lemma 3.1(I)(i) (a+s) 4 (a+s) 4 and s a 2 (a 2 2a 2s) (a+s) 4 = 2a2 (2a 2 3a 3s) (a+s) 5 2a2 (2a 2 3a 3/4) (a+s) 5 < ) = ( e1/4 )a 3 ( e 1/4 )a 2 +(24555e 1/ )a+3274e 1/4 1637a(4a+1) 2 > a2 ( e 1/4 )a ( e 1/4 ) ] +3274e 1/4 1637a(4a+1) 2 (since 24555e 1/ ( 16777) > ) a2 4( e 1/4 ) ( e 1/4 ) ] +3274e 1/4 1637a(4a+1) 2 > >. (since e 1/4 ( 4464) > ) 3274e 1/4 1637a(4a+1) 2 (since 4( e 1/4 ) ( e 1/4 ) = e 1/4 ( 11994) > ) So n 1 (s,a) > for < s < 1/4 and a 4 and hence n 5 (1/4,a) > for a 4 by (3.27). 8

9 November 4, 215 Combining this fact with (3.26), we conclude that, for 1/4 ρ 1/2 and a 4, n 5 (ρ,a) > and hence K 1 (ρ) < M 1 (ρ,a) by (3.21). (II). We prove that M 1 (ρ) < K 2 (ρ) for < ρ 1/2 and a 4. Note first that, for ρ 1/2, K 2 (ρ,a) a+165 ] ρ ρ+1 = 31 5a+25 a= ρ2 + 2 ρ+1 >, 3 since ρ2 + 2 ρ+1] = 123/1 >. Hence, for < ρ 1/2 and a 4, to prove that 3 ρ=1/2 M 1 (ρ,a) (= ρf(ρ,a)/f(ρ,a)) < K 2 (ρ,a) is equivalent to prove that n 11 (ρ,a) F(ρ,a) ρf(ρ,a) >. (3.28) K 2 (ρ,a) By direct computations, we have that n 11 (,a) =, n ρ 11(,a) = and 2 n ρ 2 11 (,a) = 1/3 >. Hence, if we can prove that 3 n ρ 3 11 (ρ,a) < for < ρ < 1/2 and a 4 and that n 11 (1/2,a) > for a 4, then n 11 (ρ,a), a 4, is either strictly increasing or first strictly increasing and then strictly decreasing on,1/2]. Consequently, n 11 (ρ,a) > min{n 11 (,a),n 11 (1/2,a) = for < ρ < 1/2 and a 4, which completes the proof. Indeed, we compute that, for < ρ < 1/2 and a 4, where and 3 n 12 (ρ,a) ρ 3n 11(ρ,a) = (a+ρ) 6 (4ρ 2 +1ρ+15)a (295ρ 2 5ρ 75)] 4f(ρ,a), (3.29) n 12 (ρ,a) n 12,1 (ρ)a 1 +n 12,9 (ρ)a 9 +n 12,8 (ρ)a 8 +n 12,7 (ρ)a 7 +n 12,6 (ρ)a 6 +n 12,5 (ρ)a 5 +n 12,4 (ρ)a 4 +n 12,3 (ρ)a 3 +n 12,2 (ρ)a 2 +n 12,1 (ρ)a+n 12, (ρ) n 12,1 (ρ) = 256ρ 8 16ρ 7 372ρ ρ ρ ρ ρ ρ+2565 ρ 4( 256ρ 4 16ρ 3 372ρ ρ ) +( 19575ρ ) ρ 4 256ρ 4 16ρ 3 372ρ ρ ] ρ=1/2 +( 19575/ ) = ρ , n 12,9 (ρ) = 512ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ 8 ( 512ρ+24832)+ρ 4 ( ρ ) +( ρ ) ρ 8 ( 512/ )+ρ 4 ( / ) +( / ) = 24576ρ ρ , 9

10 November 4, 215 n 12,8 (ρ) = 256ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ 9 ( 256ρ ) ρ 9 ( 256/ ) = ρ 9, n 12,7 (ρ) = 7756ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ] ρ=1/2 = ρ 9 /4, n 12,6 (ρ) = 512ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ 9 ( ρ )+ρ 5 ( ρ ) ρ 9 ( / )+ρ 5 ( / ) = ρ ρ 5 / , n 12,5 (ρ) = 1514ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ 1 ( 1514ρ )+ρ 8 ( ρ ) + ( ρ ) ρ 1 ( 1514/ )+ρ 8 ( / ) +ρ 6 ( / ) = ρ ρ ρ , n 12,4 (ρ) = 16788ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ] ρ=1/2 = /124 > 12511, 1

11 November 4, 215 n 12,3 (ρ) = ρ ρ ρ ρ ρ ρ ρ ρ ρ 3 ρ 1 ( ρ )+ρ 8 ( ρ ) ρ 1 ( / )+ρ 8 ( / ) ρ ρ 8, n 12,2 (ρ) = ρ ρ ρ ρ ρ ρ ρ ρ ρ 3 ρ 9 ( ρ ) ρ 9 ( / ) = ρ 9, n 12,1 (ρ) = ρ ρ ρ ρ ρ ρ 5 ρ 5 ( ρ ) ρ 5 ( / ) = ρ 6, n 12, (ρ) = ρ ρ ρ ρ 6 ρ 6( ρ ρ ) ρ ρ ρ ] ρ=199125/( ) = ρ 6. Above inequalities imply that, for a 4 and ρ 1/2, n 12 (ρ,a) > ( a a a a 4) = a ( a 9 a 6) ( a 9 a 5) (a 9 a 4 ) > and hence, by (3.29), 3 ρ 3n 11(ρ,a) <. (3.3) Next, we show that n 11 (1/2,a) > for a 4. It is easy to see that 1/2 (3b1 12b +24)(s 1/2) 2 +b 1 (s 1/2)+b ] ds = 1 11

12 November 4, 215 for any b,b 1 >. If we let b 268a+35 and b 1a a+35 s f(1/2), then 1a+5 f(1/2) n 11 (1/2,a) 1 1a+5 = F(1/2,a) f(1/2,a) = F(1/2,a) 2K 2 (1/2,a) 268a+35 f(1/2,a) 1/2 { = f(s,a) 1a+5 268a+35 f(1/2,a) (3b 1 12b +24)(s 1/2) 2 ] +b 1 (s 1/2)+b 1/2 n 13 (s,a)ds. ds (3.31) Note that n 13 (1/2,a) =, s n 13(1/2,a) = and we compute that, for < s < 1/2 and a 4, 2 s 2n 13(s,a) = 2 2a+1 s2f(s,a) 268a+35 (3b 1 12b +24)f(1/2,a) = 2 24(4a3 23a 2 192a 65) s 2f(s,a) f(1/2,a) (268a+35)(2a+1) ] 2 a 2 (a 2 2a 2s) 24(4a 3 23a 2 192a 65) > f(s,a)] (a+s) 4 a=4 + a=4 (268a+35)(2a+1) 2 (by Lemma 3.1(I)(i) and since 2 a2 (a 2 2a 2s) s 2f(s,a)= (a+s) 4 a 2 (a 2 2a 2s) = 2a(1+2s)a2 sa 2s 2 ] a (a+s) 4 (a+s) ] 5 d 24(4a 3 23a 2 192a 65) da (268a+35)(2a+1) 2 = 16( 2s+8) e 4s 28 4+s (s+4) 4 ] 16( 2s+8) e 4s (s+4) 4 4+s 28 s=1/2 117 e1/2 (since d ] 16( 2s+8) e 4s ds (s+4) 4 4+s = e4/ e1/2 (.9) >. 117 e1/2 ] e 1/2 a=4 f(s,a), 2a ( a 2 1 a 1 2 2) >, (a+s) 5 = a3 +14a a+99 (268a+35) 2 (2a+1) 3 > ) = 32(3s2 24s 16) e 4s (s+4) 6 32(3/4 16) 4+s < (s+4) 6 e 4s 4+s < ) So n 13 (s,a) > for < s < 1/2 and a 4 and hence n 13 (1/2,a) > by (3.31). Combining this fact with (3.3), we conclude that, for 1/4 ρ 1/2 and a 4, n 11 (ρ,a) > and hence M 1 (ρ,a) < K 2 (ρ,a) by (3.28). Hence the proof of Lemma 3.1(III)(v) is complete. Proof of Lemma 3.1(III)(vi). Let n 14 (ρ,a) 1 { M 1 (ρ,a)+ 132a+165 ] ρ 5a+25 ρ2 K 3 (ρ,a) = 132a+165 5a+25 ρ+2 a 2 aρ 2ea+ρ (a+ρ) 2 12

13 November 4, 215 for < ρ 1/2 and a 4. Then lim ρ +n 14(ρ,a) =, (3.32) and n 14 (1/2,a) = (664a2 +462a+165) 1(4a 2 +4a+1)e a 2a+1 n 15 (α), = 8α (2a+1) 2 25 α eα whereα a 2a+1 satisfies4/9 α 1/2fora 4. Wehavethatn 15(α) > for4/9 α 1/2 since n 15 (4/9) = 12637/45 2e 4/9 (.1) >, n 15 (4/9) = 79/225 2e4/9 (.32) > and n 15(α) = 16 2e α > 16 2e 1/2 ( 12.73) > for 4/9 α 1/2. Hence n 14 (1/2,a) > (3.33) for a 4. Moreover, we compute that, for < ρ 1/2 and a 4, 2 ρ 2n 14(ρ,a) = 2a2 (a+ρ) 4 (a 2 2a 2ρ)e aρ a+ρ 6 ] 2a2 (a+ρ) 4 { a 2 2a 2/2 ] a=4 6 = 2a2 (a+ρ) 4 <. Combing this fact with (3.32) and (3.33), we conclude that, for < ρ 1/2 and a 4, n 14 (ρ,a) > and hence K 3 (ρ) < M 1 (ρ)+ 132a+165 5a+25 ρ2. Hence the proof of Lemma 3.1(III)(vi) is complete. Proof of Lemma 3.1(III)(vii). We first show that (3.19) holds for function L 1 (ρ,s) at the two endpoints s = and ρ when ρ is fixed on 1/4,1/2] and a 4. That is, L 1 (ρ,) > and L 1 (ρ,ρ) < for 1/4 ρ 1/2 and a 4. Indeed, we compute that, for 1/4 ρ 1/2 and a 4, L 1 (ρ,) = 8 25 F(ρ,a)+M 1(ρ,a) 1 > 8 ρ e s ds+k 1 (ρ) 1 (by Lemma 3.1(I)(i) and (III)(v)) 25 = (eρ 3 1)+ ρ2 + 2ρ (16ρ 3) 1 (by (3.11)) ρ ρ n 16 (ρ) >. (see Fig. 2.) (3.34) Remark that the proof of (3.34) is easy but tedious and hence we omit it here. On the other hand, for 1/4 ρ 1/2 and a 4, since L 1 (ρ,ρ) = 8 25 ρ M 2(ρ,a) + M 1 (ρ,a) 1 = ρf(ρ,a) F(ρ,a) M 2 (ρ,a) ρ] by (3.7) and M 2 (ρ,a) ρ M 2(ρ,a) = M 2(ρ,a) >, we have that L 1(ρ,ρ) < if and only if F(ρ,a) ρf(ρ,a) M 2 (ρ,a)+1 8 >. (3.35) ρ 25 13

14 November 4, n 16 (ρ) Fig. 2. Function n 16 (ρ) defined in (3.34) is positive on the interval 1/4,1/2]. ρ By direct computation, we have that F(ρ,a) a = ρ ρf(ρ,a) M 2 (ρ,a) ρ ] ( 17ρ+25)a 2 +16ρ 2 a+(8ρ 3 25ρ 2 ) a f(ρ,a)ds+25ρ3 f(ρ,a) (17ρ+25)a 2 (16ρ 5)ρa (8ρ 25)ρ 2 ] >, 2 since a f(ρ,a) by Lemma 3.1(I)(i) and since ( 17ρ+25)a2 + 16ρ 2 a + (8ρ 3 25ρ 2 ) 33 2 a2 + a 21 4 > for 1/4 ρ 1/2 and a 4. Hence L 1(ρ,ρ) < for 1/4 ρ 1/2 and a 4 if (3.35) holds for 1/4 ρ 1/2 and a = 4 or equivalently, L 1 (ρ,ρ)] a=4 < for 1/4 ρ 1/2. Indeed, by Lemma 3.1(III)(v), we have that L 1 (ρ,ρ)] a=4 = 8 25 ρ M 2(ρ)+M 1 (ρ) ρ M 2(ρ)+K 2 (ρ) 1 ρ ( = 31ρ ρ 2 96ρ+16 ) ρ ( < 174ρ 2 96ρ+16 ) 75(4+ρ) 2 75(4+ρ) 2 ρ 174ρ 2 96ρ+16 ] 75(4+ρ) = 8ρ 2 ρ=96/(2 174) 2175(4+ρ) < 2 for 1/4 ρ 1/2. Hence we have that for 1/4 ρ 1/2 and a 4. L 1 (ρ,ρ) < (3.36) Now we show that (3.19) holds in the following two cases, which completes the proof. 14

15 November 4, 215 Case 1: 4 a 5. Fix 1/4 ρ 1/2. Then we compute that, for < s < ρ, 2 s 2L 1(ρ,s) = 8 { a 2 (a 2 +2a+2s) 4a 2 +( 8s+5)a (4s 2 +25s) 25 (a+s) 4 f(s,a) 4(a 2 +2a+2s) 8 { ] a 2 (a 2 +2a+2s) 4a 2 +( 8s+5)a (4s 2 +25s) 25 (a+s) 4 f(s,a) 4(a 2 +2a+2s) (since ] 4a 2 +( 8s+5)a (4s 2 +25s) a 4(a 2 +2a+2s) f(s,a) F(ρ,a) F(s,a)] f(s,a) f(s,a)(ρ s) a=5 = (29 4s)a2 (4s 2 +17s)a+(4s 2 75s) 27a2 19a (a 2 +2a+2s) 2 2(a 2 +2a+2s) < and 2 by the Mean Value Theorem for some s s,ρ],1/2]) 8 ] a 2 (a 2 +2a+2s) 4s 2 65s+15 e 1/2 (1/2 s) 25 (a+s) 4 f(s) 8s+14 (since 4s 2 65s+15 4s 2 65s+15 ] = 233/2 > and s=1/2 1 f(s,a) < e 1/2 by Lemma 3.1(I)(i)) 2a 2 (a 2 +2a+2s) = (8e 1/2 4)s 2 +(136e 1/2 65)s+(15 7e 1/2 ) ] >, 25(2s+35)(a+s) 4 f(s) since 8e 1/2 4( 9) >, 136e 1/2 65( 159) > and 15 7e 1/2 ( 35) >. Combining this fact with (3.34) and (3.36), we conclude that (3.19) holds for 4 a 5. Case 2: a > 5. Fix 1/4 ρ 1/2. Then we compute that, for < s < ρ, s L 1(ρ,s) { = 8 25 a2 (a s) (a+s) 8 1 f(s,a) s 3 25 f(s,a)] 2 F(ρ,a) F(s,a)] = 8 a { 2 9a3 73sa 2 48s 2 a 16s 3 f(s,a)+f(ρ,a) F(s,a)] 25(a+s) 2 f(s,a) 8a 2 (a+s) 8 ] a { 2 9a3 73sa 2 48s 2 a 16s 3 f(s,a)+f(s 25(a+s) 2 f(s,a) 8a 2,a)(ρ s) (a+s) a=5 (since ] 9a3 73sa 2 48s 2 a 16s 3 a 8a 2 (a+s) = s(41a3 +48sa 2 +48s 2 a+16s 3 ) 4a 3 (a+s) 2 < and by the Mean Value Theorem for some s s,ρ],1/2]) 8 ] a 2 16s3 24s s e 1/2 (1/2 s) 25(a+s) 2 f(s,a) 2(5+s) (since 16s 3 24s s s 3 24s s+1125 ] s=1/2 = 31/2 > and 1 f(s,a) < e 1/2 by Lemma 3.1(I)(i)) = a2 16s 3 (2e 1/2 24)s 2 +(1825 9e 1/2 )s (1125 5e 1/2 ) ] 625(5+s)(a+s) 2 f(s,a) 15

16 a2 (2e 1/2 24)s 2 +(1825 9e 1/2 )s ( 16/ e 1/2 ) ] 625(5+s)(a+s) 2 f(s,a) a2 (1825 9e 1/2 )/2 (1123 5e 1/2 ) ] 625(5+s)(a+s) 2 f(s,a) (since 2e 1/2 24 ( 9) > and (1825 9e 1/2 ) ( 341) > ) = a2 421/2 5e 1/2] 625(5+s)(a+s) 2 f(s,a) <, November 4, 215 since 421/2 5e 1/2 ( 128) >. Combining this fact with (3.34) and (3.36), we conclude that (3.19) holds for a > 5. The proof of Lemma 3.1(III)(vii) is now complete. Proof of Lemma 3.1(III)(viii). Fix a 4 and 1/4 ρ 1/2. Let s on (,ρ) be defined in (3.19). We claim first that L 2 (ρ,s) > for < s < s. Suppose it is not true. Then there exists some s on (,s ) such that L 2 (ρ,s ) =. Consequently, since L 2 (ρ,) = M 1 (ρ) M 1 (ρ) = and lim s + s L 2(ρ,s) = M 2 (ρ) ρ M 1(ρ) = L 1 (ρ,ρ) > by Lemma 3.1(I)(vii), we have that there exists some t on (,s ] such that L 2 (ρ,t ) = and L s 2(ρ,t ). Note that, by the definition of L 2 (ρ,s) in (3.15), we have that P 1 (ρ,t ) = 8 t 25 +M 1 (ρ). Moreover, we compute that s L 2(ρ,t ) = f(t ) P 1 (ρ,t ) 1 M 2 (t ) 8 ] F(ρ) F(t ) F(ρ) F(t ) 25 f(t ) f(t ) 8 = F(ρ) F(t ) 25 t +M 1 (ρ) 1 M 2 (t ) 8 25 f(t ) = F(ρ) F(t ) L 1(ρ,t ) >, F(ρ) F(t ) f(t ) by (3.14) and (3.19), which makes a contradiction with s L 2(ρ,t ). Hence L 2 (ρ,s) > for < s < s. By similar arguments, we can prove that L 2 (ρ,s) > for s < s < ρ. Hence L 2 (ρ,s) for s < ρ, which completes the proof of Lemma 3.1(III)(viii). Proof of Lemma 3.1(III)(ix). We would show that L 3 (ρ,s) > for 7/25 ρ 1/2, < s < ρ and a 4 by the following strategy. For fixed ρ on (7/25,1/2) and a 4, suppose that there exists some s on (,ρ) such that L 3 (ρ,s ) =. Then we claim that it must hold that s L 3(ρ,s ) <. However, it makes a contradiction with lim s ρ L 3 (ρ,s) > as claimed below. Claim 1: lim s ρ L 3 (ρ,s) > for 7/25 ρ 1/2 and a 4. Proof of Claim 1. By Lemma 3.1(I)(vii) and (I)(xi), for 7/25 ρ 1/2 and a 4, we ] 16

17 November 4, 215 have that lim L 3(ρ,s) = ( 25ρ2 +24ρ 1)a 4 (52ρ 2 +4ρ)a 3 (76ρ 3 +6ρ 2 )a 2 (4ρ 3 )a ρ 4 s ρ 5(a+ρ) 2 (ρ+1)a 2 +2ρa+ρ 2 ] > ( 25ρ2 +24ρ 1)a 4 (52ρ 2 +4ρ)a 3 11a 2 a 1 5(a+ρ) 2 (ρ+1)a 2 +2ρa+ρ 2 ] n 17 (ρ,a) 5(a+ρ) 2 (ρ+1)a 2 +2ρa+ρ 2 ] >, since n 17 (ρ,a) > which follows from the facts that, for 7/25 ρ 1/2, n 17 (ρ,4) = 9728ρ ρ 449 a n 17(ρ,4) = 8896ρ ρ a 2n 17(ρ,4) = 648ρ ρ a 3n 17(ρ,4) = 2712ρ ρ 96 4 a 4n 17(ρ,a) = 6ρ ρ 24 So Claim 1 holds. ( min 9728ρ ρ 449 ) = 63 >, ρ {7/25,1/2 ( 8896ρ ρ 348 ) = 44 >, min ρ {7/25,1/2 min ρ {7/25,1/2 min ρ {7/25,1/2 min ρ {7/25,1/2 ( 648ρ ρ 214 ) = 53 >, ( 2712ρ ρ 96 ) = 26112/625 >, ( 6ρ ρ 24 ) = 2256/25 >. Claim 2: For 7/25 ρ 1/2 and a 4, if there exists some s on (,ρ) such that L 3 (ρ,s ) =, then s L 3(ρ,s ) >. Proof of Claim 2. Fix ρ on (7/25,1/2) and a 4. Suppose s be an arbitrary point on (,ρ) such that L 3 (ρ,s ) =. Then, by (3.5) and (3.16), we have that ρ 2 f (ρ) s 2 f (s ) = 3 ρf(ρ) s f(s )] F(ρ) F(s ) 25 ρf(ρ) s f(s )]. 17

18 November 4, 215 Moreover, where s L 3(ρ,s ) = 3 f(s )ρf(ρ) s f(s )] 2 F(ρ) F(s )] f(s )+s f (s ) 2s f (s )+s 2 f (s ) 2 F(ρ) F(s ) ρf(ρ) s f(s ) + f(s )+s f (s )]ρ 2 f (ρ) s 2 f (s )] ρf(ρ) s f(s )] 2 = 3 f(s )ρf(ρ) s f(s )] 2 F(ρ) F(s )] { + f(s )+s f (s ) ρf(ρ) s f(s )] f(s )+s f (s ) F(ρ) F(s ) ρf(ρ) s f(s )] 2 F(ρ) F(s ) = 3 f(s )ρf(ρ) s f(s )] 2 F(ρ) F(s )] 2 +3 f(s )+s f (s ) F(ρ) F(s ) = 2s f (s )+s 2 f (s ) ρf(ρ) s f(s ) ρf(ρ) s f(s )] 2s f (s )+s 2 f (s )+ 37 f(s 25 )+s f (s )] ρf(ρ) s f(s ) { f(s ) 3 ρf(ρ) s f(s ) 2 P 1(ρ,s )] 2 3M 2 (s )+1]P 1 (ρ,s )+ s 2 f (s )+ 87s 25 f (s )+ 37f(s 25 ) (by (3.5) and (3.8)) f(s ) f(s ) ρf(ρ) s f(s ) n 18(ρ,s ), (3.37) n 18 (ρ,s) 3 2 P 1(ρ,s)] 2 3M 2 (s)+1]p 1 (ρ,s)+ s2 f (s) sf (s)+ 37 f(s) 25. (3.38) f(s) Hence to prove that s L 3(ρ,s ) <, it suffices to prove n 18 (ρ,s) > for < s < ρ, 7/25 ρ 1/2, and a 4. We show it by the following Claims 2(a) (c), which completes the proof of Claim 2. Claim 2(a): For < s < 1/2 and a 4, n 18 (ρ,s) is a strictly increasing function of ρ on (s,1/2). Proof of Claim 2(a). Fix s on (,1/2) and a 4. Then we compute that, for s < ρ < 1/2(< a), ρ n 18(ρ,s) = 3{P 1 (ρ,s)] M 2 (s)+1] ρ P 1(ρ,s) = 3{P 1 (ρ,s)] M 2 (s)+1] f(ρ){m 2(ρ)+1] P 1 (ρ,s) F(ρ) F(s) >, by Lemma 3.1(I)(ix) and (I)(x). So Claim 2(a) holds. By Claim 2(a), for a 4, to prove n 18 (ρ,s) > for < s < ρ and 7/25 ρ 1/2, it suffices to prove that lim ρ s +n 18 (ρ,s) > for 7/25 s < 1/2, and n 18 (7/25,s) > for < s < 7/25. We show them by the following Claims 2(b) and 2(c), respectively. Claim 2(b): lim ρ s +n 18 (ρ,s) > for 7/25 s < 1/2 and a 4. 18

19 November 4, 215 Proof of Claim 2(b). By Lemma 3.1(I)(vii), we compute that lim ρ s +n 18(ρ,s) = 3 2 M 2(s)+1] 2 3M 2 (s)+1] 2 + s2 f (s) sf (s)+ 37f(s) 25 f(s) = ( 25s2 +24s 1)a 4 (52s 2 +4s)a 3 (76s 3 +6s 2 )a 2 4s 3 a s 4 5(a+s) 4 > for a 4. Above inequality holds by the exactly same arguments given in Claim 1. So Claim 2(b) holds. Claim 2(c): n 18 (7/25,s) > for < s < 7/25 and a 4. Proof of Claim 2(c). Let k = 3/25. Then, by (3.38), we have that { 3 n 18 (7/25,s) = 2 P 1(7/25,s)] 2 3(k +1)P 1 (7/25,s) + {3k M 2 (s)]p 1 (7/25,s)+ s2 f (s) sf (s)+ 37f(s) 25 f(s) 3 2 (1+k)2 + {3k M 2 (s)]p 1 (7/25,s)+ s2 f (s) sf (s)+ 37f(s) 25 f(s) (since 3 ] 3 2 x2 3(k +1)x 2 x2 3(k +1)x x=k (1+k)2 for any k,x R) { 3 = 3k M 2 (s)]p 1 (7/25,s) 2 (1+k)2 s2 f (s) sf (s)+ 37 f(s) 25 f(s) n 19 (s)p 1 (7/25,s) n 2 (s) n 21 (s). (3.39) Note first that n 19 () = 3k = 9/25 >, n 19 (7/25) = 3 25a2 15a 147 < for a 4, and 25(25a+7) 2 n 19 (s) is strictly decreasing on (,7/25) by Lemma 3.1(I)(vi). Thus there exists some s 1 on (, 7/25) such that > when s (,s 1 ), n 19 (s) = when s = s 1, < when s (s 1,7/25]. Moreover, since n 19 (s 1 ) = 3k M 2 (s 1 )] =, i.e., M 2 (s 1 ) = k, we have that, for a 4, n 2 (s 1 ) = s2 1f (s 1 ) f(s 1 ) = f (s 1 )f(s 1 ) f (s 1 )] 2 87 s 1 f (s 1 ) f(s 1 ) (1+k)2 s1 f (s 1 ) f(s 1 ) ] 2 87 s 1 f (s 1 ) f(s 1 ) (1+k)2 = a2 2a 2s 1 M a 2 2 (s 1 )] M 2(s 1 ) (1+k)2 (by (3.8)) ] a 2 2a 2s k 2 87 a 2 25 k (1+k)2 (since < s 1 < 7/25) s=7/25 = 475a2 45a a 2 475a2 45a 126] a= a 2 = a 2 <. 19 (3.4)

20 November 4, 215 It follows, by (3.39), that n 18 (7/25,s 1 ) = n 2 (s 1 ) >. Next, we show that n 18 (7/25,s) > for s (,s 1 ) (s 1,7/25) and a 4. By (3.39), it suffices to show that n 21 (s) = n 19 (s) P 1 (7/25,s) n ] 2(s) > (3.41) n 19 (s) for s (,s 1 ) (s 1,7/25) and a 4. We first compute that ] d n2 (s) 5a 2 n 22 (s,a) = ds n 19 (s) 3(a+s) 3 (25s 3)a 2 6sa 3s 2 ] 2, where n 22 (s,a) = ( 125s 2 +3s+2)a 5 +(125s 3 +28s 2 54s)a 4 +(22s 3 146s 2 )a 3 (3s 4 +94s 3 )a 2 +24s 4 a+28s 5 a 2 ( 125s 2 +3s+2)a 3 +(28s 2 54s)a 2 +( 146s 2 )a /78125 ] (since 28s 2 54s ] s=1/2 = /78125) a 2 n 23 (s,a) >, since, for < s < 7/25 and a 4, n 23 (s,4) = 414s s / s s /78125 ] = /78125 >, s=7/25 a n 23(s,4) = 396s 2 +18s s 2 +18s+96 ] = 456/625 >, s=7/25 2 a 2n 23(s,4) = 244s s s s+48 ] = 358/125 >, s=7/25 3 a 3n 23(s,a) = 75s 2 +18s+12 75s 2 +18s+12 ] = 18/5 >. s=7/25 ] Hence d n2 (s) ds n 19 < for s (,s (s) 1 ) (s 1,7/25) and a 4. Moreover, since P 1 (7/25,s) is strictly increasing on,7/25] by Lemma 3.1(I)(x), P 1 (7/25,s) n 2(s) n 19 is strictly increasing (s) on (,s 1 ) and (s 1,7/25). Hence by (3.4) and (3.39), n 18 (7/25,s) > for s (,s 1 ) if n 21 () >. Similarly, n 18 (7/25,s) > for s (s 1,7/25) if lim s (7/25) n 21 (s) >. Indeed, we 2

21 November 4, 215 compute that n 21 () = 3kM 1 (7/25) ] 2 (1+k)2 > 3kK 1 (7/25) ] 2 (1+k)2 (by Lemma 3.1(III)(v)) = 3k = >, and, by Lemma 3.1(I)(vii), (1+k)2 = ( 1+ 7 ) 2 25 lim s (7/25) n 21(s) = a4 175a a a (25a+7) 4 So Claim 2(c) holds. > a4 ( )a 3 625(25a+7) 4 ( a )a3 625(25a+7) 4 >. By Claims 1 and 2, we have that L 3 (ρ,s) > for < s < ρ, 7/25 ρ 1/2 and a 4, which completes the proof of Lemma 3.1(III)(ix). References 1] Y.-H. Liang, S.-H. Wang, Classification and evolution of bifurcation curves for the onedimensional perturbed Gelfand equation with mixed boundary conditions, submitted. 21