Our starting point of this section is the following result about Green's function appearing in Section 2.
Lemma 4.
Proof. In fact,
By an elemental calculation, it can be proved that the maximum of is reached at , thus,
□
In the sequel, we present the main result of this paper.
For convenience, we put .
Theorem 3. Our Problem (2) has a unique nonnegative solution u(t) if the following conditions are satisfied:
(H1) f : [0, 1] × [0, ∞) → [0, ∞) is continuous and nondecreasing respect to the second argument.
(H2) There exists such that, for x, y ∈ [0, ∞) with y ≥ x and t ∈ [0, 1],
where.
Proof. Consider the cone
Obviously, (P, d) with d(x, y) = sup{|x(t) - y(t)|: t ∈ [0, 1]} is a complete metric space satisfying conditions (4) and (5).
Consider the operator defined by
where G(t, s) is the Green's function appearing in Section 2. Obviously, T applies P into itself since f(t, x) and G(t, s) are nonnegative continuous functions.
In what follows we check that assumptions in Theorem 2 are satisfied.
Firstly, the operator T is nondecreasing.
Indeed, by (H1), for u, v ∈ P, u ≥ v, and t ∈ [0, 1], we have
Now, we prove that T satisfies the contractive condition appearing in Theorem 1.
In fact, for u, v ∈ P and u ≥ v and, taking into account assumption (H2), we get
As , φ is nondecreasing, and, taking into account (H2) and Lemma 4, we obtain
Put ψ(x) = x - φ(x). As , this means that and from the last inequality
This proves that T satisfies the contractive condition of Theorem 1.
Finally, the nonnegative character of the function G(t, s) and f(t, x) [assumption (H1)] gives us
where 0 denotes the zero function.
Therefore, Theorem 2 says us that Problem (2) has a unique nonnegative solution.
□
In the sequel, we present a sufficient condition for the existence and uniqueness of positive solutions for Problem (2) (positive solution means x(t) > 0 for t ∈ (0, 1)). The proof of this condition is similar to the proof of Theorem 2.3 of [23]. We present this proof for completeness.
Theorem 4. Under assumptions of Theorem 3 and suppose that f(t0, 0) ≠ 0 for certain t0 ∈ [0, 1]. Then, Problem (2) has a unique positive solution.
Proof. Consider the nonnegative solution x(t) for Problem (2) whose existence is guaranteed by Theorem 3.
In the sequel, we will prove that x(t) is a positive solution.
Firstly, notice that x(t) is a fixed point of the operator and, consequently,
Now, suppose that there exists 0 < t* < 1 such that x(t*) = 0. This means that
Using that x(t) is a nonnegative function, f(t, y) is nondecreasing with respect to the second argument and the nonnegative character of G(t, s), we get
This gives us .
As G(t, s) ≥ 0 and f(s, 0) ≥ 0, the last expression implies
As G(t*, s) ≠ 0 a.e (s) (because G(t*, s) is given by a polynomial), we can obtain
(6)
On the other hand, as f(t0, 0) ≠ 0 for certain t0 ∈ [0, 1], the nonnegative character of f(t, y) gives us f(t0, 0) > 0. As f(t, y) is a continuous function, we can find a set A ⊂ [0, 1] with t0 ∈ A, μ(A) > 0, where μ is the Lebesgue measure and f(t, 0) > 0 for any t ∈ A. This contradicts (6).
Therefore, x(t) > 0 for t ∈ (0, 1). This finishes the proof. □
Remark 3. In Theorem 4, the condition f(t0, 0) ≠ 0 for certain t0 ∈ [0, 1] seems to be a strong condition in order to obtain a positive solution for Problem (2), but when the solution is unique, we will see that this condition is very adjusted one. In fact, suppose that Problem (2) has a unique nonnegative solution x(t) then
In fact, if f(t, 0) = 0 for each t ∈ [0, 1], it is easily seen that the zero function satisfies Problem (2) and the uniqueness of the solution gives us x(t) = 0. The reverse implication is obvious.
Remark 4. Notice that the hypotheses in Theorem 3 are invariant by continuous perturbation. More precisely, if f(t, 0) = 0 for any t ∈ [0, 1] and f satisfies (H1) and (H2) of Theorem 3 then g(t, x) = a(t) + f(t, x) with a : [0, 1] → [0, ∞) continuous and a ≠ 0, satisfies assumptions of Theorem 4, and this means that the following boundary value problem
has a unique positive solution.
Now, we present an example that illustrates our results.
Example 1. Consider the boundary value problem
(7)
In this case, and f(t, u) = c + λ · arctg u. It is easily seen that f(t, u) satisfies (H1) of Theorem 3.
In the sequel, we prove that f(t, u) satisfies (H2) of Theorem 3.
Previously, we consider the function ϕ : [0, ∞) → [0, ∞) given by ϕ(u) = arctg u and we will see that ϕ satisfies
In fact, put ϕ(u) = arctag u = α and ϕ(v) = arctg v = β (notice that, as u ≥ v and ϕ is nondecreasing, α ≥ β).
Then, from
and, as , then tgα, tgβ ∈ [0, ∞), we can obtain
Applying ϕ to the last inequality and taking into account the nondecreasing character of ϕ, we obtain
or, equivalently,
This proof our previous claim.
Now, for u ≥ v and t ∈ [0, 1], we have,
Now, we prove that ϕ(u) = arctg u belongs to . Obviously, ϕ : [0, ∞) → [0, ∞) is a continuous and nondecreasing function. Moreover, ψ(u) = u - ϕ(u) = u - arctg u is also continuous and nondecreasing and satisfies ψ(u) > 0 for u > 0 and ψ(0) = 0. Consequently, .
Finally, as f(t, 0) = c + arctg 0 = c > 0, by Theorem 4, Problem (7) has a unique positive solution for