POSITIVE SOLUTIONS OF THE (n − 1,1) CONJUGATE BOUNDARY VALUE PROBLEM

We consider the (n−1, 1) conjugate boundary value problem. Some upper estimates to positive solutions for the problem are obtained. We also establish some explicit sufficient conditions for the existence and nonexistence of positive solutions of the problem.

In 2009, Webb [4] considered a nonlocal version of the (n − 1, 1) conjugate problem.He obtained a lower estimate for the Green's function G(t, s), based on which a lower estimate to positive solutions for the problem (1.1)-(1.2) can be proved (see Lemmas 2.8 and 2.9 below).However, to our knowledge, no satisfactory upper estimates to positive solutions for the problem (1.1)-(1.2) have been obtained in the literature.We know that upper and lower estimates for positive solutions of boundary value problems have important applications.For example, once we find some a priori upper and lower estimates for positive solutions of a certain boundary value problem, we can use them together with the Krasnosell'skii fixed point theorem to derive a set of existence and nonexistence conditions for positive solutions of the problem (See [5] for a paper taking this approach).With this motivation, we in this paper make a further study of positive solutions to the problem (1.1)-(1.2).Our main goal is to develop some new upper estimates for positive solutions of the problem (1.1)-(1.2).Here, by a positive solution, we mean a solution u(t) such that u(t) > 0 on (0, 1).Since the case n = 2 is a well-studied case, we in this paper assume that n ≥ 3.
Throughout the paper, we let X = C[0, 1] be equipped with the supremum norm Obviously X is a Banach space.Also, we define x .
These constants will be used later in the statements of our existence and nonexistence theorems.This paper is organized as follows.In Section 2, we obtain some new upper estimates to positive solutions to the (n − 1, 1) conjugate problem, and discuss some lower estimates from the literature.In Section 3, we establish some explicit sufficient conditions for the existence and nonexistence of positive solutions to the problem.

Upper and Lower Estimates for Positive Solutions
Throughout the paper we define the constants We also define the functions w 1 : [0, 1] → [0, +∞) and 1, if t ≥ q.The functions w 1 (t) and w 2 (t) will be used to estimate positive solutions of the problem (1.1)-(1.2).It is easy to see that w 2 = 1 and that both w 1 (t) and w 2 (t) are continuous functions.An equivalent definition for the function w 1 (t) is The function w 1 (t) first appeared in [4].It can be shown that (2.1) The verification of (2.1) is straightforward and is therefore left to the reader.
The proof of the lemma is complete.
The next lemma was proved by Eloe and Henderson in [1].
2) and (2.2), and u(t 0 ) > 0 for some t 0 ∈ (0, 1), then and there exists c ∈ (0, 1) such that u ′ (c) = 0, u ′ (t) > 0 for 0 < t < c, and u ′ (t) < 0 for c < t < 1. (2.4) In other words, c is the unique zero of u ′ in (0, 1 where Proof.We see from Lemma 2.2 that u(c) = u .The inequality in (2.5) is trivial for t = 0, t = c, and t = 1.We need only to show that the inequality holds for 0 < t < c and c < t < 1.Let x ∈ (0, c) ∪ (c, 1) be a fixed number.Define It is easy to verify the following facts: We take two cases to continue the proof.
Case I: 0 < x < c.In this case, because h(0) = h(x) = h(c) = 0, there exist t 1 ∈ (0, x) and If we continue this procedure, we can show that for each i = 2, 3, • • • , n − 2, there exist t i , s i , and r i such that 0 In particular, we have The above equation implies that In summary, if 0 < x < c then u(x) ≤ β(x)u(c).Hence, the inequality in (2.5) holds for 0 < t < c.
Case II: c < x < 1.In this case, because h(0) = h(c) = h(x) = 0, there exist t 1 ∈ (0, c) and If we continue from here and follow the same lines as in Case I, we can show that u(x) ≤ β(x)u(c) for c < x < 1.The proof in Case II is now complete.
Both Lemmas 2.5 and 2.6 provide a lower estimate to functions satisfying (1.2), (2.2), and (2.3).The difference is that the lower estimate in Lemma 2.5 depends on the unique zero c of u ′ in (0, 1), while the lower estimate in Lemma 2.6 does not.If we know where c is, then (2.6) is a better estimate that (2.7).However, if we don't know where c is, then we have to do with (2.7).
If we continue this procedure, then finally we can show that there exist t n−2 , s n−2 , and r n−2 such that 0 < t n−2 < s n−2 < r n−2 < q and Therefore, there exists t n ∈ (t n−1 , s n−1 ) such that h (n) (t n ) < 0, which contradicts (2.9).The proof is complete.
Both Lemmas 2.3 and 2.7 provide an upper estimate to functions satisfying (1.2), (2.2), and (2.3).The difference is that the upper estimate in Lemma 2.3 depends on the unique zero c of u ′ in (0, 1), while the upper estimate in Lemma 2.7 does not.
The next lemma was proved by Webb in [4].
Lemma 2.8.For t, s ∈ [0, 1] we have where Using Lemma 2.8, we can easily establish the following lower estimate.Lemma 2.9.If u ∈ C n [0, 1] satisfies (1.2), (2.2), and (2.3), then (2.10) Proof.On one hand, for 0 ≤ t ≤ 1 we have This means that On the other hand, for 0 ≤ t ≤ 1, we have The proof is complete.EJQTDE, 2010 No. 53, p. 7 The estimates (2.7) and (2.10) are of the same type, that is, both are independent of c.The two estimates are similar but (2.10) is a little better than (2.7).Both estimates are listed here because they are obtained by different methods, and both methods are useful in finding estimates.
We have shown that, for functions u satisfying (1.2), (2.2), and (2.3), there are several upper and lower estimates of different types -the lower estimates (2.6), (2.7), (2.10), the upper estimates (2.5), (2.8), (2.12), and the "natural" upper estimate These upper and lower estimates can be used in different situations.In the next section, we will show how to use the upper estimate (2.10) and the lower estimate (2.8) to establish some explicit existence and nonexistence conditions for positive solution of the problem (1.1)-(1.2).

Nonexistence and Existence Results
We begin by fixing some notations.First, we define EJQTDE, 2010 No. 53, p. 8 Clearly, P is a positive cone of the Banach space X.Define the operator T : P → X and its associated linear operator L : X → X by It is well known that T : P → X and L : X → X are completely continuous operators.It is easy to see that T (P ) ⊂ P and L(P ) ⊂ P .Now the integral equation (1.3) is equivalent to the equality In order to solve the problem (1.1)-(1.2),we only need to find a fixed point u of T in P such that u = 0. We also define the constants Now we give some explicit sufficient conditions for the nonexistence of positive solutions.The proof of Theorem 3.2 is similar to that of Theorem 3.1 and is therefore omitted.The next theorem is from [4].
If either 2) has at least one positive solution.
Here comes the natural question -how can we find the value of µ 1 ?In general, there is no explicit formula for finding µ 1 , and some kind of approximation has to be made.One approach is to use a numerical method to find an approximation for µ 1 .Another approach is to develop some theoretic upper and lower bounds for µ 1 .Both approaches are interesting.The following upper and lower bounds for µ 1 were given in [4].With the newly found upper estimate (2.8) from Section 2, we can now improve the lower bound m for µ 1 .First, we introduce some notations.For each n ≥ 1, we let θ n = T n w 2 and σ n = T n w 1 .In other words, we define Proof.Let φ ∈ P \ {0} be a positive eigenfunction corresponding to the principal eigenvalue r(L) of L. By Lemma 2.2, we have φ(t) > 0 for 0 < t < 1.Let n ≥ 1.We have For each 0 ≤ t ≤ 1 we have This implies that φ ≤ φ µ n 1 θ n .Thus we have By Lemma 2.4, the maximum of θ n (t) must occur at a point in the interval [(n − 2)/(n − 1), 1].Therefore we have Thus we have proved that m n ≤ µ 1 .In a similar fashion, we can show that µ 1 ≤ M n .The proof is complete.
The next example was first considered in [4].We now reconsider it to illustrate some of our results.

.Theorem 3 . 5 .
No. 53, p. 10       Next, for each n ≥ 1, we define the constants Let r(L) be the radius of the spectrum of L and let µ 1 = 1/r(L).For each n ≥ 1, we have m n ≤ µ 1 ≤ M n .