UNIQUENESS OF POSITIVE SOLUTIONS OF A CLASS OF ODE WITH NONLINEAR BOUNDARY CONDITIONS RUYUN MA AND

UNIQUENESS OF POSITIVE SOLUTIONS OF A CLASS OF ODE WITH NONLINEAR BOUNDARY CONDITIONS

RUYUN MA AND YULIAN AN

Received 19 August 2004 and in revised form 27 January 2005

We study the uniqueness of positive solutions of the boundary value problem u(cid:1)(cid:1) + a(t)u(cid:1) + f (u) = 0, t ∈ (0,b), B1(u(0)) − u(cid:1)(0) = 0, B2(u(b)) + u(cid:1)(b) = 0, where 0 < b < ∞, B1 and B2 ∈ C1(R), a ∈ C[0, ∞) with a ≤ 0 on [0, ∞) and f ∈ C[0, ∞) ∩ C1(0, ∞) satisfy suitable conditions. The proof of our main result is based upon the shooting method and the Sturm comparison theorem.

1. Introduction The existence of positive solutions of second order ordinary diﬀerential equations (ODEs) with linear boundary conditions has been extensively studied in the literature, see Coﬀ- man [1], Henderson and Wang [7], Lan and Webb [8] and the references therein. Also the existence of positive solutions of second order ODEs with nonlinear boundary conditions has been studied by several authors, see Dunninger and Wang [2], Wang [11] and Wang and Jiang [12] for some references along this line. However for the uniqueness problem of second order ODEs, even in the linear boundary conditions case, very little was known, see Ni and Nussbaum [9], Fu and Lin [6] and Peletier and Serrin [10]. To the best of our knowledge, no uniqueness results of positive solutions were established for second order ODEs subject to nonlinear boundary conditions. In this paper, we attempt to prove some uniqueness results in this direction. More precisely, we consider the uniqueness of positive solutions of the boundary value problem

(cid:2)

− u(cid:1)(0) = 0,

(cid:1) u(0)

t ∈ (0,b) (1.1)

+ u(cid:1)(b) = 0, (1.2) u(cid:1)(cid:1) + a(t)u(cid:1) + f (u) = 0, (cid:1) (cid:2) u(b) B2 B1

where 0 < b < ∞. We make the following assumptions: (C1) f ∈ C[0, ∞) ∩ C1(0, ∞) with f (0) = 0,

f (u) > 0, u f (cid:1)(u) < f (u), for u > 0; (1.3)

290 Uniqueness of positive solutions of a class of ODE

i (x) is nondecreasing on

(C2) a ∈ C[0, ∞) with a(t) ≤ 0 for t ≥ 0; (C3) Bi ∈ C1[0, ∞) satisﬁes Bi(0) = 0, Bi(x) > 0 for x > 0, B(cid:1) (0, ∞) (i = 1,2).

i (x) ≥ 0 for x ≥ 0 (i = 1,2).

Remark 1.1. Condition (C3) implies that B(cid:1) In fact, we have from Bi(0) = 0 and Bi(x) > 0 for x > 0 that

i (x) is nondecreasing on (0, ∞) implies that B(cid:1)

i (x) ≥ 0

B(cid:1) i (0) ≥ 0. (1.4)

This together with the assumption B(cid:1) for x ≥ 0.

The main result of this paper is the following.

Theorem 1.2. Let (C1)–(C3) hold. Then problem (1.1), (1.2) has at most one positive so- lution.

Here we say u(t) is a positive solution of (1.1), (1.2), if that u(t) > 0 on [0,b] and satisﬁes the diﬀerential equation (1.1) as well as the boundary conditions (1.2).

Remark 1.3. As an application of Theorem 1.2, we consider the nonlinear problem

− u(cid:1)(0) = 0,

(cid:1) u(0)

(1.5) t ∈ (0,b), (cid:2)l u(cid:1)(cid:1) + a(t)u(cid:1) + up = 0, (cid:1) (cid:2)k u(b) + u(cid:1)(b) = 0,

where p ∈ (0,1), k, l ∈ (1, ∞) are given, a ∈ C[0, ∞) with a ≤ 0 on [0, ∞). Clearly all of the conditions of Theorem 1.2 are satisﬁed. Therefore by Theorem 1.2, (1.5) has at most a positive for any b ∈ (0, ∞).

The proof of the main result is motivated by the work of Erbe and Tang [3, 4, 5] and is based on the shooting method and the Sturm comparison theorem. The rest of the paper is organized as follows. In Section 2, we state and prove some preliminary lemmas. The proof of Theorem 1.2 will be given in Section 3.

2. The preliminary results

To apply the shooting method, we need some properties of the solutions of the initial value problem

(2.1)

u(cid:1)(cid:1) + ¯a(t)u(cid:1) + ¯f (u) = 0, u(cid:1)(0) = λ. u(0) = δ, (2.2)

Lemma 2.1. Let ¯a ∈ C[0, ∞), ¯f ∈ C[0, ∞) ∩ C1(0, ∞) with ¯f (0) = 0 and ¯f (s) > 0 for s > 0. Let δ ∈ (0, ∞) and λ ∈ R be two given constants. Then (2.1), (2.2) has a unique solution u satisfying either

(I) u(t) > 0 for t ∈ [0, ∞); or (II) there exists ρ ∈ (0, ∞) such that

u(t) > 0 on t ∈ [0,ρ), u(ρ) = 0, u(cid:1)(ρ) < 0. (2.3)

R. Ma and Y. An 291

(cid:4)

Proof. For any r ∈ (0, ∞), let

(cid:3) (t,u, p) | t ∈ [0,r], u > 0

. (2.4) Ωr :=

Then the function

(2.5) F(t,u, p) := ¯a(t)p + ¯f (u)

satisﬁes locally Lipschitz condition in Ωr, and consequently (2.1), (2.2) has a unique so- lution u(t) such that one of the following cases must occur

(i) u > 0 on [0, ∞); (ii) there exists ρ ∈ (0, ∞) such that u > 0 on [0,ρ), and limt→ρ− u(t) = 0; (iii) there exists T ∈ (0, ∞) such that u > 0 on [0,T) and limsupt→T − u(t) = ∞. We claim that (iii) can not occur. Assume on the contrary that (iii) occurs, then

(2.6) u(cid:1)(t) = ∞. limsup t→T −

(cid:7)(cid:7)(cid:1)

(cid:5) (cid:6) t

(cid:5) u(cid:1)(t)exp

On the other hand, we have from (2.1) that

(cid:7) ¯a(s)ds

+ exp ¯a(s)ds ¯f (u) = 0, t ∈ [0,T) (2.7)

(cid:5) (cid:6) t

which together with the condition ¯f (s) > 0 for s > 0 implies that

(cid:7) ¯a(s)ds

u(cid:1)(t)exp is strictly decreasing on [0, T). (2.8)

However this contradicts the fact (2.6).

Therefore either (i) or (ii) must occur. Suppose on the contrary that (ii) occurs and u(cid:1)(ρ) = 0. Using the similar argument of (cid:8) t 0 ¯a(s)ds) is strictly decreasing on [0,ρ). Thus proving (2.8), we conclude that u(cid:1)(t)exp( (cid:8) t 0 ¯a(s)ds) > 0 on [0,ρ), and accordingly u(cid:1)(t) > 0 on [0,ρ). However this contra- u(cid:1)(t)exp( dicts the fact δ = u(0) > u(ρ) = 0. Therefore u(cid:1)(ρ) < 0 if (ii) occurs. This completes the (cid:1) proof.

In order to prove Theorem 1.2, we introduce an initial value problem

u(cid:1)(cid:1) + a(t)u(cid:1) + f (u) = 0, (2.9)

u(0) = α > 0, (2.10) u(cid:1)(0) = B1(α).

For any α > 0, we know from Lemma 2.1 that (2.9), (2.10) has a unique solution u such that one of the cases occurs: (i) u > 0 in [0, ∞); (ii) there exists a unique ρ = ρ(α) ∈ (0, ∞) such that u(t) > 0 on [0, ρ), u(ρ) = 0 and u(cid:1)(ρ) < 0.

292 Uniqueness of positive solutions of a class of ODE

 

Let



∞, ρ(α),

Tα = (2.11) if (i) occurs if (ii) occurs.

(cid:2)

(cid:1) u(0,α)

From α > 0, we have that u(0,α) = α > 0 and u(cid:1)(0,α) = B1(α) > 0, and consequently

(2.12) B2 + u(cid:1)(0,α) = B2(α) + B1(α) > 0.

(cid:2)

(cid:1) u(t,α)

Therefore, there exists (cid:1) ∈ (0,Tα) such that

+ u(cid:1)(t,α) > 0, t ∈ [0, (cid:1)). (2.13) B2

(cid:2)

(cid:1) u(t,α)

Denote

+ u(cid:1)(t,α). (2.14) B(t,α) := B2

(cid:12)

When B(t,α) vanishes at some t0 ∈ (0,Tα), we deﬁne b(α) to be the ﬁrst zero of B(t,α) in (0,Tα). More precisely, b(α) is a function of α which has the properties

(cid:2) .

(cid:2) (cid:1) b(α),α

= 0,

B t ∈ B(t,α) > 0, 0, b(α) (2.15)

(cid:4)

If B(t,α) is positive in [0,Tα), then we deﬁne b(α) = Tα. Let

(cid:3) α | α > 0, b(α) < Tα

. N := (2.16)

(cid:2)

It is obvious that (1.1), (1.2) has no positive solution if N is an empty set. (We recall that u is a positive solution means u(t) > 0 in [0,b]. So in the case B(Tα,α) = 0, u(t,α) is not a positive solution of (1.1), (1.2) since u(Tα,α) = 0). Hence we suppose N (cid:8)= ∅. Remark 2.2. It is worth remarking here that if (ii) occurs, and accordingly u(ρ(α),α) = 0, then b(α) ∈ (0,ρ(α)),

(cid:2) (cid:1) b(α),α

= 0,

(cid:12) 0,b(α)

(cid:2)

B . B(t,α) > 0 on (2.17)

(cid:2) (cid:1) ρ(α),α

= B2

B In fact, we have from Lemma 2.1 that (cid:1) u(ρ,α) + u(cid:1) < 0, (2.18)

(cid:12)

(cid:13)

which together with the fact B(0,α) > 0 yields the existence of zero of B(t,α) in (0,ρ(α)). Lemma 2.3. Let (C1)–(C3) hold and let α ∈ N. Let u(t,α) be the unique solution of (2.9), (2.10) on [0,Tα). Then

(cid:1)

, (2.19) u(t,α) > 0, u(cid:1) t ∈ (cid:2) b(α),α 0,b(α) < 0.

(cid:12)

(cid:13)

Proof. By Remark 2.2, b(α) ∈ (0,ρ(α)). Applying Lemma 2.1, we get that

. t ∈ u(t,α) > 0, 0,b(α) (2.20)

R. Ma and Y. An 293

(cid:2)(cid:2)

The second inequality in (2.19) can be easily deduced from (2.20) and (C3) and the fact

(cid:1) u

(cid:2) (cid:1) b(α),α

(cid:1) b(α),α

(cid:2) (cid:1) b(α),α

= 0.

= B2

B + u(cid:1) (2.21) (cid:1)

Lemma 2.4. Let (C1)–(C3) hold. Let u(t,α) be the unique solution of (2.9), (2.10) on [0,Tα). If η ∈ (0,Tα) is such that

B(η,α) = 0, (2.22)

then

B(t,α) > 0, t ∈ [0,η). (2.23)

(cid:7)(cid:7)(cid:1)

(cid:5) (cid:6) t

(cid:5) u(cid:1) exp

Proof. From (2.9), we conclude that

(cid:7) a(s)ds

+ exp a(s)ds f (u) = 0. (2.24)

(cid:7)(cid:7)(cid:1)

(cid:5) (cid:6) t

(cid:2)

Since u(t,α) > 0 for all t ∈ [0,η], we have

(cid:5) u(cid:1)(t,α)exp

= −exp

(cid:7) a(s)ds

(cid:1) u(t,α)

(cid:2)(cid:2)

(cid:1) u

f a(s)ds < 0, ∀t ∈ [0,η]. (2.25)

= 0.

(cid:2) (cid:1) τ2,α

(cid:1) τ2,α

= B2

(cid:2)(cid:2)

(cid:1) u

B + u(cid:1) (2.26) Suppose on the contrary that there exists τ2 ∈ [0,η) such that (cid:2) (cid:1) τ2,α

(cid:1) τ2,α

= −B2

u(cid:1) < 0 (2.27) Then we have from condition (C3) and the fact u(τ2,α) > 0 that (cid:2) (cid:1) τ2,α

(cid:5) (cid:6) τ2

and accordingly

(cid:7) a(s)ds

(cid:2) (cid:1) τ2,α

u(cid:1) exp < 0. (2.28)

(cid:13)

(cid:7) a(s)ds

(cid:12) τ2,η

This together with (2.25) implies that (cid:5) (cid:6) t t ∈ u(cid:1)(t,α)exp < 0, , (2.29)

(cid:13)

and consequently

(cid:12) τ2,η

t ∈ . u(cid:1)(t,α) < 0, (2.30)

This implies

(cid:2) (cid:1) τ2,α

u > u(η,α). (2.31)

294 Uniqueness of positive solutions of a class of ODE

(cid:2)(cid:2)

(cid:1) u

(cid:2) .

(cid:1) u(η,α)

By Remark 1.1 and (2.31), we get

(cid:1) τ2,α

≥ B2

(2.32) B2

(cid:12)

(cid:13)

From (2.30) and (C1)–(C2) and the fact u(cid:1)(cid:1)(t,α) = −a(t)u(cid:1)(t,α) − f (u(t,α)), it follows that

t ∈ u(cid:1)(cid:1)(t,α) < 0, (2.33) τ2,η

and consequently

(cid:2) (cid:1) τ2,α

(cid:2)(cid:2)

(cid:1)

(cid:2)

(cid:1) u

u(cid:1) > u(cid:1)(η,α), (2.34)

(cid:1) u(η,α)

(cid:1) τ2,α

= B2

(cid:2) (cid:1) τ2,α

B + u(cid:1) + u(cid:1)(η,α) = 0. (2.35) which together with (2.32) implies that (cid:2) τ2,α > B2

(cid:1)

However this contradicts (2.26). Remark 2.5. From Lemmas 2.3 and 2.4, we have that if η ∈ (0,Tα) satisﬁes

B(η,α) = 0. (2.36)

Then

η = b(α). (2.37)

In other words, if α ∈ N, then b(α) is the unique zero of B(t,α) = 0 in [0,ρ(α)). Therefore to prove that (1.1), (1.2) has at most one positive solution, it is suﬃcient to show that for any l > 0, there exists at most one α ∈ N such that b(α) = l.

Now we denote the variation of u(t,α) by φ(t,α) = ∂u(t,α)/∂α. Then, φ(t,α) satisﬁes

(2.38)

1(α).

(cid:2)(cid:2)

(cid:1) u

φ(cid:1)(cid:1) + a(t)φ(cid:1) + f (cid:1)(u)φ = 0, φ(cid:1)(0,α) = B(cid:1) φ(0,α) = 1, (2.39)

(cid:2) (cid:1) b(α),α

(cid:8)= 0,

φ α ∈ N. Lemma 2.6. Suppose that (cid:1) b(α),α + φ(cid:1) (2.40) B(cid:1) 2

Then one of the following cases must occur

(cid:2)(cid:2)

= −a

(cid:2) u(cid:1)

− f

(i) N is an open interval; (ii) N = (0, j1) ∪ ( j2, ∞) with 0 < j1 < j2 < +∞. Moreover, b(cid:1)(α) > 0 for all (0, j1); b(cid:1)(α) < 0 for all ( j2, ∞).

(cid:1) u

(cid:2) (cid:1) b(α),α

(cid:1) b(α)

(cid:1) b(α),α

u(cid:1)(cid:1) Proof. We ﬁrstly show that b(α) ∈ C1(N) and b(cid:1)(α) (cid:8)= 0. From Lemma 2.3, (C1)–(C2), we conclude that (cid:2) (cid:1) b(α),α < 0. (2.41)

R. Ma and Y. An 295

This together with

(cid:2) (cid:1) b(α),α

= 0

B (2.42)

(cid:2)(cid:2)

(cid:14) (cid:14) (cid:14)

(cid:1) u

and (C3) and (2.19) implies that

(cid:1) b(α),α

(cid:2) (cid:1) b(α),α

= B(cid:1) 2

t=b(α)

u(cid:1) B(t,α) + u(cid:1)(cid:1) < 0. (2.43) ∂ ∂t

(cid:2)(cid:2)(cid:12)

(cid:2)(cid:13)

(cid:1)

(cid:1) u

So by Implicit Function theorem, b(α) is well-deﬁned as a function of α in N and b(α) ∈ C1(N). Furthermore, it follows from (2.43) that N is an open set. Diﬀerentiating both sides of (2.42) with respect to α, we obtain

(cid:1) b(α),α

(cid:1) b(α),α)b(cid:1)(α) + φ

(cid:1) b(α),α

(cid:2) (cid:1) b(α),α

(cid:2) b(α),α

= 0, (2.44)

u(cid:1) + u(cid:1)(cid:1) b(cid:1)(α) + φ(cid:1) B(cid:1) 2

(cid:12)

(cid:2)(cid:2)

(cid:2)(cid:13)

that is,

(cid:2) (cid:1) + u(cid:1)(cid:1) b(α),α (cid:2) (cid:1) (cid:2)(cid:2) b(α),α φ

(cid:1) b(cid:1)(α) b(α),α (cid:2) (cid:1) + φ(cid:1) b(α),α

= 0.

(cid:1) (cid:1) b(α),α u (cid:1) (cid:1) + B(cid:1) b(α),α u 2

u(cid:1) B(cid:1) 2 (2.45)

which together with (2.40) implies that

b(cid:1)(α) (cid:8)= 0. (2.46)

Next we show that if ¯α ∈ (0, ∞) \ N is such that there is a sequence {αn} ⊂ N and αn → ¯α as n → ∞, then b(αn) → +∞.

(cid:2)

Suppose on the contrary that b(αn) (cid:1) +∞, then there exists a subsequence of {b(αn)} which converges to a limit number t∗. Without loss of generality, we may suppose that b(αn) → t∗ as n → ∞, and consequently

(cid:1) b

(cid:1) αn

(cid:2) (cid:1) t∗, ¯α

(cid:2) ,αn

= 0.

= limn→∞

B B (2.47)

However this contradicts ¯α /∈ N.

Finally we show that if N is not an open interval, then (ii) must occur. Suppose J1 = ( j0, j1) and J2 = ( j2, j3) are two distinct components of N with 0 < j1 < j2 < ∞. Then

b(α) = +∞. (2.48) lim α→ j− 1 b(α) = lim α→ j+ 2

Since b(α) is strictly monotonic in each component of N, we have that b(cid:1)(α) > 0 in J1, and b(cid:1)(α) < 0 in J2. Meanwhile

b(α) < +∞, b(α) < +∞. (2.49) lim α→ j− 3 lim α→ j+ 0

It follows that j0 = 0 and j3 = +∞, and accordingly N = (0, j1) ∪ ( j2, ∞) with b(cid:1)(α) > 0 in (0, j1), and b(cid:1)(α) < 0 in ( j2, ∞). (cid:1)

296 Uniqueness of positive solutions of a class of ODE

(cid:2)(cid:2)

(cid:1) u

3. Proof of Theorem 1.2 By Remark 2.5, we only need to show that for any l > 0, there exists at most one α ∈ N such that b(α) = l. Recall that for any given α ∈ N, (2.43), (2.45) hold. If we can show that

(cid:1) b(α),α

(cid:2) (cid:1) b(α),α

φ α ∈ N + φ(cid:1) > 0, (3.1) B(cid:1) 2

then it follows from (2.43) and (2.45) that

α ∈ N. b(cid:1)(α) > 0, (3.2)

Thus by Lemma 2.6, N must be an open interval. Moreover we know from (3.2) that b(α) is a strictly increasing function on N. Thus, for any given l > 0, there is at most one α ∈ N such that b(α) = l, and consequently, (1.1), (1.2) has at most one positive solution.

(cid:12)

(cid:13)

Proof of Theorem 1.2. Now we prove (3.1). First we claim that

t ∈ . φ(t,α) > 0, 0,b(α) (3.3)

(cid:14) (cid:14)

Suppose on the contrary that φ(t,α) has a zero in (0,b(α)]. We denote the ﬁrst zero of

= −u

≥ 0

(cid:2) (cid:1) t3,α

t=t3

u(cid:1)φ − uφ(cid:1) φ(cid:1) (3.4) φ(t,α) in (0,b(α)] by t3, then 0 < t3 ≤ b(α) and (cid:2) (cid:1) t3,α

since φ(t3,α) = 0 and φ(t,α) > 0 on (0,t3) implies φ(cid:1)(t3,α) ≤ 0. Notice that

(cid:15)

(cid:16)(cid:1)

(cid:5) (cid:6) t

(cid:12)

φ(cid:1)(cid:1) + a(t)φ(cid:1) + f (cid:1)(u)φ = 0 (3.5)

= exp

(cid:7) a(s)ds

(cid:13) φ < 0

so that using (C1) and (1.1) we can compute (cid:7) a(s)ds (u(cid:1)φ − uφ(cid:1)) exp f (cid:1)(u)u − f (u) (3.6)

(cid:5) (cid:6) t

(cid:2)

(cid:14) (cid:14)

− B(cid:1)

(cid:7) a(s)ds

for t ∈ (0,t3). Next we compute from (C3) and (2.39) and (2.10)

(cid:2) α ≤ 0,

= B1(α) − αB(cid:1)

(cid:1) ξ1(α)

1(α) =

(cid:1) B(cid:1) 1

1(α)

t=0

exp (u(cid:1)φ − uφ(cid:1)) (3.7)

(cid:5) (cid:6) t

(cid:14) (cid:14)

(cid:7) a(s)ds

where ξ1(α) ∈ (0,α). This means that

t=t3

exp (u(cid:1)φ − uφ(cid:1)) < 0 (3.8)

(cid:14) (cid:14)

and accordingly

t=t3

u(cid:1)φ − uφ(cid:1) < 0. (3.9)

However this contradicts (3.4). Therefore (3.3) is true.

R. Ma and Y. An 297

(cid:15)

(cid:16)(cid:1)

(cid:5) (cid:6) t

(cid:12)

(cid:13)

Using (3.5), (1.1), (C1) and (3.3), we can conclude

(cid:7) a(s)ds

= exp

(cid:7) a(s)ds

(cid:13) φ < 0,

(cid:1) 0,b(α)

t ∈ exp (u(cid:1)φ − uφ(cid:1)) f (cid:1)(u)u − f (u)

(3.10)

(cid:14) (cid:14)

which together with (3.7) implies that

t=b(α)

(u(cid:1)φ − uφ(cid:1)) < 0. (3.11)

(cid:2)(cid:2)

Since

(cid:2) (cid:1) b(α),α (cid:2) (cid:1) + u(cid:1) b(α),α

(cid:2) (cid:1) b(α),α (cid:1) (cid:1) + u(cid:1) b(α),α u (cid:2) (cid:1) (cid:2) (cid:1) b(α),α u ξ2(α)

(3.12)

0 = B = B2 = B(cid:1) 2

(cid:2) u

(cid:2) (cid:1) b(α),α

= −B(cid:1) 2

u(cid:1) . (3.13) for some ξ2(α) ∈ (0,u(b(α),α)), we have that (cid:1) ξ2(α)

(cid:1)

(cid:2)(cid:12)

(cid:1)

(cid:2)(cid:13)

− u

(cid:2) φ

(cid:2) (cid:1) b(α),α

This together with (3.11) implies

− u

= u(cid:1)

(cid:1) B(cid:1) b(α),α 2 (cid:1) = −B(cid:1) ξ2(α) 2 (cid:2) (cid:1) b(α),α φ (cid:14) (cid:14) = u(cid:1)φ − uφ(cid:1)

(cid:1) (cid:2) b(α),α ξ2(α) (cid:1) (cid:2) (cid:1) (cid:2) b(α),α φ u (cid:2) (cid:1) b(α),α < 0

t=b(α)

(3.14) + φ(cid:1) b(α),α (cid:2) (cid:1) (cid:2) φ(cid:1) b(α),α b(α),α − u (cid:2) (cid:1) (cid:2) (cid:1) φ(cid:1) b(α),α b(α),α

(cid:2) φ

(cid:2) (cid:1) b(α),α

and consequently

(cid:1) ξ2(α)

(cid:2)(cid:2)

(cid:2) φ

(cid:1) u

+ φ(cid:1) > 0. (3.15) B(cid:1) 2

(cid:2) (cid:1) b(α),α

(cid:1) b(α),α

(cid:1) ξ2(α)

≥ B(cid:1) 2

φ Now we have from (C3) and the facts ξ2(α) ≤ u(b(α),α) and φ(b(α),α) > 0 that (cid:2) (cid:1) b(α),α + φ(cid:1) + φ(cid:1) > 0. (3.16) B(cid:1) 2

(cid:1)

Therefore (3.1) holds.

Acknowledgments

The authors are very grateful to the anonymous referee for his/her valuable suggestions. Supported by the NSFC (no. 10271095), GG-110-10736-1003, NSF of Gansu province (no. 3ZS051-A25-016), the Foundation of Excellent Young Teacher of the Chinese Edu- cation Ministry.

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Ruyun Ma: Department of Mathematics, Northwest Normal University, Lanzhou 730070, China E-mail address: mary@maths.uq.edu.au

Yulian An: Physical Software & Engineering, Lanzhou Jiaotong University, Lanzhou 730070,

Gansu, China

E-mail address: an yulian@tom.com

References

UNIQUENESS OF POSITIVE SOLUTIONS OF A CLASS OF ODE WITH NONLINEAR BOUNDARY CONDITIONS RUYUN MA AND