Our starting point of this section is the following result about Green's function appearing in Section 2.

**Lemma 4.**
${max}_{t\in \left[0,1\right]}\underset{0}{\overset{1}{\int}}G\left(t,s\right)ds=\frac{1}{\Gamma \left(\alpha +1\right)}\left[{\left(\frac{\alpha -1}{\alpha}\right)}^{\alpha -1}-\left(\frac{\alpha -1}{\alpha}\right)\right]$

*Proof*. In fact,

$\begin{array}{lll}\hfill \underset{0}{\overset{1}{\int}}G\left(t,s\right)\mathsf{\text{d}}s& =\underset{0}{\overset{t}{\int}}G\left(t,s\right)\mathsf{\text{d}}s+\underset{t}{\overset{1}{\int}}G\left(t,s\right)\mathsf{\text{d}}s\phantom{\rule{2em}{0ex}}& \hfill \\ =\underset{0}{\overset{t}{\int}}\frac{{t}^{\alpha -1}{\left(1-s\right)}^{\alpha -1}-{\left(t-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}\mathsf{\text{d}}s+\underset{t}{\overset{1}{\int}}\frac{{t}^{\alpha -1}{\left(1-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}\mathsf{\text{d}}s\phantom{\rule{2em}{0ex}}& \hfill \\ =\underset{0}{\overset{1}{\int}}\frac{{t}^{\alpha -1}{\left(1-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}\mathsf{\text{d}}s-\underset{0}{\overset{t}{\int}}\frac{{\left(t-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}\mathsf{\text{d}}s\phantom{\rule{2em}{0ex}}& \hfill \\ =\frac{1}{\Gamma \left(\alpha \right)}\left[\frac{{t}^{\alpha -1}}{\alpha}-\frac{{t}^{\alpha}}{\alpha}\right]=\frac{1}{\Gamma \left(\alpha +1\right)}\left({t}^{\alpha -1}-{t}^{\alpha}\right)\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \end{array}$

By an elemental calculation, it can be proved that the maximum of

$h\left(t\right)={\int}_{0}^{1}G\left(t,s\right)\mathsf{\text{d}}s=\frac{1}{\Gamma \left(\alpha +1\right)}\left({t}^{\alpha -1}-{t}^{\alpha}\right)$ is reached at

${t}_{0}=\frac{\alpha -1}{\alpha}$, thus,

$\underset{0\le t\le 1}{max}\underset{0}{\overset{1}{\int}}G\left(t,s\right)\mathsf{\text{d}}s=\frac{1}{\Gamma \left(\alpha +1\right)}\left[{\left(\frac{\alpha -1}{\alpha}\right)}^{\alpha -1}-{\left(\frac{\alpha -1}{\alpha}\right)}^{\alpha}\right]$

□

In the sequel, we present the main result of this paper.

For convenience, we put $A=\frac{1}{\Gamma \left(\alpha +1\right)}\left[{\left(\frac{\alpha -1}{\alpha}\right)}^{\alpha -1}-{\left(\frac{\alpha -1}{\alpha}\right)}^{\alpha}\right]$.

**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*$0<\lambda \le \frac{1}{A}$ *such that, for x*,

*y* ∈ [0, ∞)

*with y* ≥

*x and t* ∈ [0, 1],

$f\left(t,y\right)-f\left(t,x\right)\le \lambda \phi \left(y-x\right),$

*where*$\phi \in \mathcal{J}$.

*Proof*. Consider the cone

$P=\left\{u\in C\left[0,1\right]:u\left(t\right)\ge 0\right\}.$

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

$\left(Tx\right)\left(t\right)=\underset{0}{\overset{1}{\int}}G\left(t,s\right)f\left(s,x\left(s\right)\right)\mathsf{\text{d}}s,\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}x\in P,$

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

$\left(Tu\right)\left(t\right)={\int}_{0}^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)\mathsf{\text{d}}s\ge {\int}_{0}^{1}G\left(t,s\right)f\left(s,v\left(s\right)\right)\mathsf{\text{d}}s=\left(Tv\right)\left(t\right).$

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

$\begin{array}{lll}\hfill \mathsf{\text{d}}\left(Tu,Tv\right)& =\underset{t\in \left[0,1\right]}{sup}\left\{|Tu\left(t\right)-Tv\left(t\right)|\right\}\phantom{\rule{2em}{0ex}}& \hfill \\ =\underset{t\in \left[0,1\right]}{sup}\left\{\left(Tu\left(t\right)-Tv\left(t\right)\right)\right\}\phantom{\rule{2em}{0ex}}& \hfill \\ =\underset{t\in \left[0,1\right]}{sup}\underset{0}{\overset{1}{\int}}G\left(t,s\right)\left(f\left(s,u\left(s\right)\right)-f\left(s,v\left(s\right)\right)\right)\mathsf{\text{d}}s\phantom{\rule{2em}{0ex}}& \hfill \\ \le \underset{t\in \left[0,1\right]}{sup}\underset{0}{\overset{1}{\int}}G\left(t,s\right)\lambda \phi \left(u\left(s\right)-v\left(s\right)\right)\mathsf{\text{d}}s.\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \end{array}$

As

$\phi \in \mathcal{J}$,

*φ* is nondecreasing, and, taking into account (H2) and Lemma 4, we obtain

$\begin{array}{lll}\hfill \mathsf{\text{d}}\left(Tu,Tv\right)& \le \lambda \phi \left(\mathsf{\text{d}}\left(u,v\right)\right)\cdot \underset{t\in \left[0,1\right]}{sup}\underset{0}{\overset{1}{\int}}G\left(t,s\right)\mathsf{\text{d}}s\phantom{\rule{2em}{0ex}}& \hfill \\ =\lambda \phi \left(\mathsf{\text{d}}\left(u,v\right)\right)\cdot A\le \phi \left(\mathsf{\text{d}}\left(u,v\right)\right)=\mathsf{\text{d}}\left(u,v\right)-\left(\mathsf{\text{d}}\left(u,v\right)-\phi \left(\mathsf{\text{d}}\left(u,v\right)\right)\right).\phantom{\rule{2em}{0ex}}& \hfill \\ \hfill \end{array}$

Put

*ψ*(

*x*) =

*x* -

*φ*(

*x*). As

$\phi \in \mathcal{J}$, this means that

$\psi \in \mathcal{F}$ and from the last inequality

$\mathsf{\text{d}}\left(Tu,Tv\right)\le \mathsf{\text{d}}\left(u,v\right)-\psi \left(\mathsf{\text{d}}\left(u,v\right)\right).$

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

$\left(T0\right)\left(t\right)=\underset{0}{\overset{1}{\int}}G\left(t,s\right)f\left(s,0\right)\mathsf{\text{d}}s\ge 0,$

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*(*t*_{0}, 0) ≠ 0 *for certain t*_{0} ∈ [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

$\left(Tu\right)\left(t\right)={\int}_{0}^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)\mathsf{\text{d}}s$ and, consequently,

$x\left(t\right)=\underset{0}{\overset{1}{\int}}G\left(t,s\right)f\left(s,x\left(s\right)\right)\mathsf{\text{d}}s.$

Now, suppose that there exists 0 <

*t** < 1 such that

*x*(

*t**) = 0. This means that

$x\left({t}^{*}\right)=\underset{0}{\overset{1}{\int}}G\left({t}^{*},s\right)f\left(s,x\left(s\right)\right)\mathsf{\text{d}}s=0.$

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

$0=x\left({t}^{*}\right)=\underset{0}{\overset{1}{\int}}G\left({t}^{*},s\right)f\left(s,x\left(s\right)\right)\mathsf{\text{d}}s\ge \underset{0}{\overset{1}{\int}}G\left({t}^{*},s\right)f\left(s,0\right)\mathsf{\text{d}}s\ge 0.$

This gives us $x\left({t}^{*}\right)={\int}_{0}^{1}G\left({t}^{*},s\right)f\left(s,0\right)\mathsf{\text{d}}s=0$.

As

*G*(

*t*,

*s*) ≥ 0 and

*f*(

*s*, 0) ≥ 0, the last expression implies

$G\left({t}^{*},s\right)f\left(s,0\right)=0\phantom{\rule{1em}{0ex}}\mathsf{\text{a}}\mathsf{\text{.e}}\left(s\right).$

As

*G*(

*t**,

*s*) ≠ 0 a.e (

*s*) (because

*G*(

*t**,

*s*) is given by a polynomial), we can obtain

$f\left(s,0\right)=0\phantom{\rule{1em}{0ex}}\mathsf{\text{a}}\mathsf{\text{.e}}\left(s\right).$

(6)

On the other hand, as *f*(*t*_{0}, 0) ≠ 0 for certain *t*_{0} ∈ [0, 1], the nonnegative character of *f*(*t*, *y*) gives us *f*(*t*_{0}, 0) > 0. As *f*(*t*, *y*) is a continuous function, we can find a set *A* ⊂ [0, 1] with *t*_{0} ∈ *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*(

*t*_{0}, 0) ≠ 0 for certain

*t*_{0} ∈ [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

$f\left(t,0\right)=0\mathsf{\text{foreach}}t\in \left[0,1\right]\mathsf{\text{ifandonlyif}}x\left(t\right)\equiv 0.$

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

$\left.\begin{array}{cc}\hfill {D}_{{0}^{+}}^{\alpha}u\left(t\right)+g\left(t,u\left(t\right)\right)=0,\hfill & \hfill \phantom{\rule{1em}{0ex}}0<t<1\hfill \\ \hfill u\left(0\right)=u\left(1\right)={u}^{\prime}\left(0\right)=0\hfill \end{array}\right\}$

has a unique positive solution.

Now, we present an example that illustrates our results.

*Example* 1. Consider the boundary value problem

$\left.\begin{array}{cc}\hfill {D}_{{0}^{+}}^{\frac{5}{2}}u\left(t\right)+c+\lambda \cdot arctg\phantom{\rule{2.77695pt}{0ex}}u\left(t\right)=0,\hfill & \hfill \phantom{\rule{1em}{0ex}}0<t<1,\phantom{\rule{2.77695pt}{0ex}}c,\lambda >0\hfill \\ \hfill u\left(0\right)=u\left(1\right)={u}^{\prime}\left(0\right)=0\hfill \end{array}\right\}$

(7)

In this case, $\alpha =\frac{5}{2}$ 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

$\varphi \left(u\right)-\varphi \left(v\right)\le \varphi \left(u-v\right),\phantom{\rule{1em}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}u\ge v.$

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

$tg\left(\alpha -\beta \right)=\frac{tg\alpha -tg\beta}{1+tg\alpha \cdot tg\beta}$

and, as

$\alpha ,\beta \in \left[0,\frac{\pi}{2}\right)$, then

*tgα*,

*tgβ* ∈ [0, ∞), we can obtain

$tg\left(\alpha -\beta \right)\le tg\alpha -tg\beta .$

Applying

*ϕ* to the last inequality and taking into account the nondecreasing character of

*ϕ*, we obtain

$\alpha -\beta \le arctg\left(tg\alpha -tg\beta \right),$

or, equivalently,

$\varphi \left(u\right)-\varphi \left(v\right)=arctg\phantom{\rule{2.77695pt}{0ex}}u-arctg\phantom{\rule{2.77695pt}{0ex}}v=\alpha -\beta \le arctg\left(u-v\right)=\varphi \left(u-v\right).$

This proof our previous claim.

Now, for

*u* ≥

*v* and

*t* ∈ [0, 1], we have,

$f\left(t,u\right)-f\left(t,v\right)=\lambda \left(arctg\phantom{\rule{2.77695pt}{0ex}}u-arctg\phantom{\rule{2.77695pt}{0ex}}v\right)\le \lambda arctg\left(u-v\right).$

Now, we prove that *ϕ*(*u*) = *arctg u* belongs to $\mathcal{J}$. 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, $\varphi \in \mathcal{J}$.

Finally, as

*f*(

*t*, 0) =

*c* +

*arctg* 0 =

*c* > 0, by Theorem 4, Problem (7) has a unique positive solution for

$0<\lambda \le {\left(\frac{1}{\Gamma \left(5/2+1\right)}\left[{\left(\frac{3}{5}\right)}^{3/2}-{\left(\frac{3}{5}\right)}^{5/2}\right]\right)}^{-1}\approx 17.8682.$