Fractional differential equations arise in many field, such as physics, mechanics, chemistry, engineering and biological sciences, etc. In recent years, many authors have investigated the existence of positive solutions for nonlinear fractional differential equation boundary value problems (see [18–21]). However, there are few papers concerned with the uniqueness of positive solutions. In this section, we only apply the results in Section 3 to study nonlinear fractional differential equation boundary value problems. We study the existence and uniqueness of positive solutions for the following nonlinear fractional differential equation boundary value problem:
$$ \textstyle\begin{cases} D^{\nu}_{0^{+}}u(t)=f(t,u(t),u(t))+g(t,u(t))+q(t,u(t)), \quad 0< t< 1, 3< \nu \leq4, \\ u(0)=u'(0)=u''(0)=u''(1)=0. \end{cases} $$
(4.1)
Here \(D^{\alpha}_{0^{+}}\) is the RiemannLiouville fractional derivative of order \(\nu>0\), defined by
$$ D^{\nu}_{0^{+}}u(t)=\frac{1}{\Gamma(n\nu)}\biggl(\frac{d}{dt} \biggr)^{n}\int^{t}_{0}(t \tau)^{n\nu1}u(\tau)\, d\tau, $$
where \(n=[\nu]+1\). \([\nu]\) denotes the integer part of the number ν; see [22]. \(f(t,u,v):[0,1]\times[0,+\infty)\times[0,+\infty )\rightarrow[0,+\infty)\) is continuous, and \(g(t,u), q(t,v):[0,1]\times[0,+\infty)\rightarrow[0,+\infty)\) are continuous.
In our considerations we will work in the Banach apace \(E=C[0,1]=\{ x:[0,1] \rightarrow\mathbb{R} \text{ is continuous}\}\) with the standard norm \(\x\=\sup\{x(t):t\in[0,1]\}\). Notice that this space can be equipped with a partial order given by
$$ x,y\in C[0,1],\quad x\leq y \quad \Leftrightarrow\quad x(t) \leq y(t) \quad \text{for all } t\in[0,1]. $$
Set \(P=\{x\in C[0,1]\mid x(t)\geq0,t\in[0,1]\}\), the standard cone. It is clear that P is a normal cone in \(C[0,1]\) and the normality constant is 1.
Definition 4.1
([23])
The integral
$$ I^{\nu}_{0^{+}}f(x)=\frac{1}{\Gamma(\nu)}\int^{x}_{0} \frac {f(t)}{(xt)^{1\nu}}\, dt,\quad x>0, $$
where \(\nu>0\), is called the RiemannLiouville fractional integral of order ν and \(\Gamma(\nu)\) is the Euler gamma function defined by
$$ \Gamma(\nu)=\int_{0}^{+\infty}t^{\nu1}e^{t} \, dt,\quad \nu>0. $$
Lemma 4.2
([19, 24])
Let
\(\nu>0\)
and
\(u\in C(0,1)\cap L(0,1)\). The fractional differential equation
$$ D^{\nu}_{0^{+}}u(t)=0 $$
has
$$ u(t)=c_{1}t^{\nu1}+c_{2}t^{\nu2}+ \cdots+c_{n}t^{\nun},\quad c_{i}\in R, i=0,1, \ldots,n, n=[\nu]+1, $$
as unique solution.
Lemma 4.3
([19, 24])
Assume that
\(u\in C(0,1)\cap L(0,1)\)
with a fractional derivative of order
\(\nu>0\)
that belongs to
\(C(0,1)\cap L(0,1)\). Then
$$ I^{\nu}_{0^{+}}D^{\nu}_{0^{+}}u(t)=u(t)+c_{1}t^{\nu1}+c_{2}t^{\nu 2}+ \cdots+c_{n}t^{\nun} $$
for some
\(c_{i}\in R\), \(i=0,1,\ldots,n\), \(n=[\nu]+1\).
Lemma 4.4
If
\(f(t,u(t),u(t))+g(t,u(t))+q(t,u(t))\geq0\), then the fractional boundary value problem (4.1) has a unique positive solution
$$ u(t)=\int_{0}^{1}G(t,s)\bigl[f\bigl(s,u(s),u(s) \bigr)+g\bigl(s,u(s)\bigr)+q\bigl(s,u(s)\bigr)\bigr]\, ds, $$
where
$$ G(t,s)=\frac{1}{\Gamma(\nu)} \textstyle\begin{cases} t^{\nu1}(1s)^{\nu3}(ts)^{\nu1}, &0\leq s\leq t\leq1, \\ t^{\nu1}(1s)^{\nu3}, &0\leq t\leq s\leq1. \end{cases} $$
(4.2)
Proof
Lemma 4.3 and Definition 4.1 imply that
$$\begin{aligned} u(t) =&c_{1}t^{\nu1}+c_{2}t^{\nu2}+c_{3}t^{\nu3}+c_{4}t^{\nu4} \\ &{}\int_{0}^{t}\frac{(ts)^{\nu1}}{\Gamma(\nu )}\bigl[f \bigl(s,u(s),u(s)\bigr)+g\bigl(s,u(s)\bigr)+q\bigl(s,u(s)\bigr)\bigr]\, ds. \end{aligned}$$
From (4.1), we know that \(c_{2}=c_{3}=c_{4}=0\) and
$$ c_{1}=\frac{1}{\Gamma(\nu)}\int_{0}^{t}(1s)^{\nu 3} \bigl[f\bigl(s,u(s),u(s)\bigr)+g\bigl(s,u(s)\bigr)+q\bigl(s,u(s)\bigr)\bigr]\, ds. $$
Then the unique solution of (4.1) is given by
$$\begin{aligned} \begin{aligned} u(t)={}&\int_{0}^{1}\frac{t^{\nu1}(1s)^{\nu3}}{\Gamma(\nu )}\bigl[f \bigl(s,u(s),u(s)\bigr)+g\bigl(s,u(s)\bigr)+q\bigl(s,u(s)\bigr)\bigr]\, ds \\ &{}\int_{0}^{t}\frac{(ts)^{\nu1}}{\Gamma(\nu )}\bigl[f \bigl(s,u(s),u(s)\bigr)+g\bigl(s,u(s)\bigr)+q\bigl(s,u(s)\bigr)\bigr]\, ds \\ ={}&\int_{0}^{1}G(t,s)\bigl[f\bigl(s,u(s),u(s) \bigr)+g\bigl(s,u(s)\bigr)+q\bigl(s,u(s)\bigr)\bigr]\, ds. \end{aligned} \end{aligned}$$
This completes the proof of Lemma 4.4. □
Lemma 4.5
Let
\(3<\nu\leq4\). Then the function
\(G(t,s)\)
defined by (4.2) satisfies the following conditions:

(1)
\(G(t,s)\geq0\), \((t,s)\in[0,1]\times[0,1]\);

(2)
\(\frac{1}{\Gamma(\nu)}s(2s)(1s)^{\nu3}t^{\nu1}\leq G(t,s)\leq \frac{1}{\Gamma(\nu)}(1s)^{\nu3}t^{\nu1}\)
for
\(t,s\in[0,1]\).
Proof
For the condition (1), when \(0\leq t\leq s\leq1\) it is obvious that
$$ G(t,s)=\frac{ t^{\nu1}(1s)^{\nu3}}{\Gamma(\nu)}\geq0. $$
In the case \(0\leq s\leq t\leq1\) (\(s\neq1\)), we have
$$\begin{aligned} G(t,s)&=\frac{1}{\Gamma(\nu)} \biggl[\frac{t^{\nu1}(1s)^{\nu 1}}{(1s)^{2}}(ts)^{\nu1} \biggr] \\ &\geq\frac{1}{\Gamma(\nu)} \bigl[t^{\nu1}(1s)^{\nu1}(ts)^{\nu 1} \bigr] \\ &=\frac{1}{\Gamma(\nu)} \bigl[(tts)^{\nu1}(ts)^{\nu1} \bigr] \geq0. \end{aligned}$$
Moreover, as \(G(t,1)=0\), then we conclude that \(G(t,s)\geq0\) for all \((t,s)\in[0,1]\times[0,1]\). So the condition (1) is true.
For the condition (2), first we prove the left inequality. If \(0\leq s\leq t\leq1\), then we have \(0\leq ts\leq tts=(1s)t\), and thus
$$ (ts)^{\nu1}\leq(1s)^{\nu1}t^{\nu1}. $$
Hence,
$$\begin{aligned} G(t,s) =&\frac{1}{\Gamma(\nu)}\bigl[(1s)^{\nu3}t^{\nu1}(ts)^{\nu1} \bigr]\geq \frac{1}{\Gamma(\nu)}\bigl[(1s)^{\nu3}t^{\nu1}(1s)^{\nu1}t^{\nu1} \bigr] \\ =&\frac{1}{\Gamma(\nu)}\bigl[(1s)^{\nu3}(1s)^{\nu1} \bigr]t^{\nu1}=\frac {1}{\Gamma(\nu)}s(2s) (1s)^{\nu3}t^{\nu1}. \end{aligned}$$
If \(0\leq t\leq s\leq1\), then we have
$$ G(t,s)=\frac{1}{\Gamma(\nu)}(1s)^{\nu3}t^{\nu1}\geq \frac{1}{\Gamma (\nu)}s(2s) (1s)^{\nu3}t^{\nu1}. $$
So the left inequality holds. Evidently, the right inequality also holds. The proof is completed. □
Theorem 4.6
Let
\(3<\nu\leq4\). Assume that
 (L_{1}):

\(f: [0,1]\times[0,+\infty)\times[0,+\infty)\rightarrow [0,+\infty)\)
is continuous, and
\(g, q: [0,1]\times[0,+\infty )\rightarrow[0,+\infty)\)
are continuous with
\(g(t,0)\not\equiv0\), \(q(t,1)\not\equiv0\), and
\(f(t,0,1)\not\equiv0\);
 (L_{2}):

\(f(t,u,v)\)
is increasing in
\(u\in[0,+\infty)\)
for fixed
\(t\in[0,1]\)
and
\(v\in[0,+\infty)\), decreasing in
\(v\in[0,+\infty)\)
for fixed
\(t\in[0,1]\)
and
\(u\in[0,+\infty)\), and
\(g(t,u)\)
is increasing in
\(u\in[0,+\infty)\)
for fixed
\(t\in[0,1]\), and
\(q(t,v)\)
is decreasing in
\(v\in[0,+\infty)\)
for fixed
\(t\in[0,1]\);
 (L_{3}):

\(g(t,\lambda u)\geq\lambda g(t,u)\)
for
\(\lambda\in(0,1)\), \(t\in[0,1]\), \(u\in[0,+\infty)\), and
\(q(t,\lambda^{1}v)\geq\lambda q(t,v)\)
for
\(\lambda\in(0,1)\), \(t\in[0,1]\), \(v\in[0,+\infty)\), and there exists a constant
\(\alpha\in(0,1)\)
such that
\(f(t,\lambda u,\lambda ^{1} v)\geq\lambda^{\alpha}f(t,u,v)\), \(\forall t\in[0,1]\), \(\lambda\in (0,1)\), \(u,v\in[0,+\infty)\);
 (L_{4}):

there exists a constant
\(\delta>0\)
such that
\(f(t,u,v)\geq \delta(g(t,u)+q(t,v))\), \(t\in[0,1]\), \(u,v\geq0\).
Then

(1)
there exists
\(u_{0},v_{0}\in P_{h}\)
and
\(r\in(0,1)\)
such that
\(rv_{0}\leq u_{0}< v_{0}\)
and
$$ \textstyle\begin{cases} u_{0}(t)\leq\int_{0}^{1}G(t,s)[f(s,u_{0}(s),v_{0}(s))+g(s,u_{0}(s))+q(s,v_{0}(s))]\, ds,\quad t\in[0,1], \\ v_{0}(t)\geq\int_{0}^{1}G(t,s)[f(s,v_{0}(s),u_{0}(s))+g(s,v_{0}(s))+q(s,u_{0}(s))]\, ds,\quad t\in[0,1], \end{cases} $$
where
\(h(t)=t^{\nu1}\), \(t\in[0,1]\);

(2)
the problem (4.1) has a unique positive solution
\(u^{\ast}\)
in
\(P_{h}\);

(3)
for any
\(x_{0},y_{0}\in P_{h}\), constructing successively the sequences
$$ \textstyle\begin{cases} x_{n+1}(t)= \int_{0}^{1}G(t,s)[f(s,x_{n}(s),y_{n}(s))+g(s,x_{n}(s))+q(s,y_{n}(s))]\, ds,\quad n=0,1,2,\ldots, \\ y_{n+1}(t)= \int_{0}^{1}G(t,s)[f(s,y_{n}(s),x_{n}(s))+g(s,y_{n}(s))+q(s,x_{n}(s))]\, ds,\quad n=0,1,2,\ldots, \end{cases} $$
we have
\(\x_{n}u^{\ast}\\rightarrow0\)
and
\(\y_{n}u^{\ast }\\rightarrow0\)
as
\(n\rightarrow\infty\).
Proof
To begin with, from Lemma 4.4, problem (4.1) has an integral formation given by
$$ u(t)=\int_{0}^{1}G(t,s)\bigl[f\bigl(s,u(s),u(s) \bigr)+g\bigl(s,u(s)\bigr)+q\bigl(s,u(s)\bigr)\bigr]\, ds. $$
Define three operators \(A:P\rightarrow E\); \(B:P\rightarrow E\); \(C:P\rightarrow E\) by
$$\begin{aligned}& (Au) (t)=\int_{0}^{1}G(t,s)g\bigl(s,u(s)\bigr) \,ds,\qquad (Bv) (t)=\int_{0}^{1}G(t,s)q \bigl(s,v(s)\bigr)\,ds, \\& C(u,v) (t)=\int_{0}^{1}G(t,s)f\bigl(s,u(s),v(s) \bigr)\,ds. \end{aligned}$$
It is easy to prove that u is the solution of problem (4.1) if and only if \(u=Au+Bu+C(u,u)\). From (L_{1}), we know that \(A:P\rightarrow P\), \(B:P\rightarrow P\), \(C:P\times P\rightarrow P\). In the sequel, we check that A, B, C satisfy all the assumptions of Theorem 3.1.
First, we prove that C is a mixed monotone operator, A is increasing and B is decreasing.
In fact, for \(u_{i},v_{i}\in P\), \(i=1,2\) with \(u_{1}\geq u_{2}\), \(v_{1}\leq v_{2}\), we know that \(u_{1}(t)\geq u_{2}(t)\), \(v_{1}(t)\leq v_{2}(t)\), \(t\in[0,1]\), and by (L_{2}) and Lemma 4.5
$$\begin{aligned} C(u_{1},v_{1}) (t) =&\int_{0}^{1}G(t,s)f \bigl(s,u_{1}(s),v_{1}(s)\bigr)\,ds \\ \geq&\int _{0}^{1}G(t,s)f\bigl(s,u_{2}(s),v_{2}(s) \bigr)\,ds=C(u_{2},v_{2}) (t). \end{aligned}$$
That is, \(C(u_{1},v_{1})\geq C(u_{2},v_{2})\). Similarly, it follows from (L_{2}) and Lemma 4.5 that A is increasing and B is decreasing.
Second, we show that B, C satisfies the condition (3.1) and A is subhomogeneous operator.
For any \(\lambda\in(0,1)\) and \(u,v \in P\), by (L_{3}) we have
$$\begin{aligned}& B\bigl(\lambda^{1}v\bigr) (t)=\int_{0}^{1}G(t,s)q \bigl(s,\lambda^{1}v(s)\bigr)\,ds \\& \hphantom{B\bigl(\lambda^{1}v\bigr) (t)}\geq\lambda\int_{0}^{1}G(t,s)q \bigl(t,v(s)\bigr)\,ds =\lambda Bv(t), \\& C\bigl(\lambda u,\lambda^{1}v\bigr) (t)=\int_{0}^{1}G(t,s)f \bigl(s,\lambda u(s),\lambda^{1}v(s)\bigr)\,ds \\& \hphantom{C\bigl(\lambda u,\lambda^{1}v\bigr) (t)}\geq\lambda^{\alpha} \int_{0}^{1}G(t,s)f\bigl(s,u(s),v(s)\bigr)\,ds= \lambda^{\alpha }C(u,v) (t), \end{aligned}$$
that is, \(B(\lambda^{1}v)\geq\lambda Bv\) for \(\lambda\in(0,1)\), \(u \in P\), \(C(\lambda u,\lambda^{1}v)\geq\lambda^{\alpha}C(u,v)\) for \(\lambda\in(0,1)\), \(u,v \in P\).
So, the operators B, C satisfy (3.1). Also, for any \(\lambda\in(0,1)\), \(u\in P\), from (L_{3}) we know that
$$ A(\lambda u) (t)=\int_{0}^{1}G(t,s)g\bigl(s, \lambda u(s)\bigr)\,ds\geq\lambda\int_{0}^{1}G(t,s)g \bigl(s,u(s)\bigr)\,ds=\lambda Au(t), $$
that is, \(A(\lambda u)\geq\lambda Au\) for \(\lambda\in(0,1)\), \(u\in P\). So, the operator A is subhomogeneous.
Third, we show that \(Ah\in P_{h}\), \(Bh\in P_{h}\), and \(C(h,h)\in P_{h}\).
In fact, from (L_{1}), (L_{2}) and Lemma 4.5, for any \(t\in[0,1]\), we have
$$\begin{aligned}& C(h,h) (t) =\int_{0}^{1}G(t,s)f\bigl(s,h(s),h(s) \bigr)\,ds=\int_{0}^{1}G(t,s)f \bigl(s,s^{\nu1},s^{\nu1}\bigr)\,ds \\& \hphantom{C(h,h)(t)} \geq\frac{1}{\Gamma(\nu)}h(t)\int_{0}^{1}s(2s) (1s)^{\nu3}f(s,0,1)\,ds, \\& C(h,h) (t) =\int_{0}^{1}G(t,s)f\bigl(s,h(s),h(s) \bigr)\,ds=\int_{0}^{1}G(t,s)f \bigl(s,s^{\nu1},s^{\nu1}\bigr)\,ds \\& \hphantom{C(h,h)(t)} \leq\frac{1}{\Gamma(\nu)}h(t)\int_{0}^{1}(1s)^{\nu3}f(s,1,0) \,ds, \end{aligned}$$
from (L_{2}), (L_{4}), we have
$$ f(s,1,0)\geq f(s,0,1)\geq0. $$
Since \(f(t,0,1)\not\equiv0\), we get
$$ \int_{0}^{1}f(s,1,0)\,ds\geq\int _{0}^{1}f(s,0,1)\,ds> 0, $$
and in consequence
$$\begin{aligned}& l_{1}=\frac{1}{\Gamma(\nu)}\int_{0}^{1}s(2s) (1s)^{\nu3}f(s,0,1)\,ds>0, \\& l_{2}=\frac{1}{\Gamma(\nu)}\int_{0}^{1}(1s)^{\nu3}f(s,1,0) \,ds>0. \end{aligned}$$
So, \(l_{1}h(t)\leq C(h,h)(t)\leq l_{2}h(t)\), \(t\in[0,1]\), and hence we have \(C(h,h)\in P_{h}\). Similarly,
$$\begin{aligned}& \frac{1}{\Gamma(\nu)}h(t)\int_{0}^{1}s(2s) (1s)^{\nu3}g(s,0)\,ds\leq Ah(t)\leq\frac{1}{\Gamma(\nu)}h(t)\int _{0}^{1}(1s)^{\nu3}g(s,1)\,ds, \\& \frac{1}{\Gamma(\nu)}h(t)\int_{0}^{1}s(2s) (1s)^{\nu3}q(s,1)\,ds\leq Bh(t)\leq\frac{1}{\Gamma(\nu)}h(t)\int _{0}^{1}(1s)^{\nu3}q(s,0)\,ds, \end{aligned}$$
from \(g(t,0)\not\equiv0\), \(q(t,1)\not\equiv0\), we easily prove \(Ah\in P_{h}\), \(Bh\in P_{h}\). Hence the condition (H_{1}) of Theorem 3.1 is satisfied.
Lastly, we show the condition (H_{2}) of Theorem 3.1 is satisfied.
For \(u,v\in P\) and any \(t\in[0,1]\). From (L_{4})
$$\begin{aligned} C(u,v) (t)&=\int_{0}^{1}G(t,s)f\bigl(s,u(s),v(s) \bigr)\,ds \\ &\geq\delta\int_{0}^{1}G(t,s) \bigl(g \bigl(s,u(s)\bigr)+q\bigl(s,v(s)\bigr)\bigr)\,ds \\ &=\delta \biggl[\int_{0}^{1}G(t,s)g\bigl(s,u(s) \bigr)\,ds+\int_{0}^{1}G(t,s)q\bigl(s,v(s)\bigr) \,ds \biggr] \\ &=\delta \bigl[(Au) (t)+(Bv) (t) \bigr], \end{aligned}$$
then we get \(C(u,v)\geq\delta(Au+Bv)\) for \(u,v\in P\).
So the condition of Theorem 4.6 follows from Theorem 3.1. □
By using Theorem 3.7, we can easily prove the following conclusion.
Theorem 4.7
Let
\(3<\nu\leq4\). Assume that (L_{1}) and (L_{2}) hold and satisfy the following conditions:
 (L_{5}):

\(f(t,\lambda u,\lambda^{1} v)\geq\lambda f(t,u,v)\), \(\forall t\in[0,1]\), \(\lambda\in(0,1)\), \(u,v\in[0,+\infty)\)
and
\(g(t,\lambda u)\geq\lambda g(t,u)\)
for
\(\lambda\in(0,1)\), \(t\in[0,1]\), \(u\in[0,+\infty)\), and there exists a constant
\(\alpha\in(0,1)\)
such that
\(q(t,\lambda^{1}v)\geq\lambda^{\alpha} q(t,v)\)
for
\(\lambda\in (0,1)\), \(t\in[0,1]\), \(v\in[0,+\infty)\);
 (L_{6}):

there exists a constant
\(\delta>0\)
such that
\(g(t,u)+f(t,u,v)\leq\delta q(t,v)\), \(t\in[0,1]\), \(u,v \geq0\).
Then the conclusions (1)(3) of Theorem
4.6
hold.
By using Theorem 3.8, we can easily prove the following conclusion.
Theorem 4.8
Let
\(3<\nu\leq4\). Assume that (L_{1}) and (L_{2}) hold and satisfy the following conditions:
 (L_{7}):

\(f(t,\lambda u,\lambda^{1} v)\geq\lambda f(t,u,v)\), \(\forall t\in[0,1]\), \(\lambda\in(0,1)\), \(u,v\in[0,+\infty)\)
and there exists a constant
\(\alpha\in(0,1)\)
such that
\(g(t,\lambda u)\geq \lambda^{\alpha} g(t,u)\)
for
\(\lambda\in(0,1)\), \(t\in[0,1]\), \(u\in [0,+\infty)\), and
\(q(t,\lambda^{1}v)\geq\lambda q(t,v)\)
for
\(\lambda \in(0,1)\), \(t\in[0,1]\), \(v\in[0,+\infty)\);
 (L_{8}):

there exists a constant
\(\delta>0\)
such that
\(q(t,v)+f(t,u,v)\leq\delta g(t,u)\), \(t\in[0,1]\), \(u,v\geq0\).
Then the conclusions (1)(3) of Theorem
4.6
hold.
Example 4.9
Consider the following problem:
$$ \textstyle\begin{cases} D^{\frac{10}{3}}_{0^{+}}u(t)=u^{\frac{1}{3}}(t)+u^{\frac {1}{3}}(t)+u^{1}(t)+\frac{u(t)}{1+u(t)}m(t)+a(t)+b,\quad 0< t< 1, \\ u(0)=u'(0)=u''(0)=u''(1)=0, \end{cases} $$
(4.3)
where \(b>0\) is a constant, \(a,m: [0,1]\rightarrow[0,+\infty]\) are continuous with \(m\neq0\).
In this example, we have \(\nu=\frac{10}{3}\). Take \(0< c< b\) and let
$$\begin{aligned}& f(t,x,y)=x^{\frac{1}{3}}+y^{\frac{1}{3}}+a(t)+c,\qquad g(t,x)= \frac{x}{1+x}m(t)+bc,\qquad q(t,y)=\frac{1}{y}+a(t)+c, \\& \alpha=\frac{1}{3},\qquad m_{\mathrm{max}}=\max\bigl\{ m(t):t\in[0,1]\bigr\} , \qquad a_{\mathrm {max}}=\max\bigl\{ a(t):t\in[0,1]\bigr\} . \end{aligned}$$
Obviously, \(m_{\mathrm{max}}>0\), \(a_{\mathrm{max}}>0\). \(f:[0,1] \times [0,+\infty]\times[0,+\infty]\rightarrow[0,+\infty]\), and \(g, q:[0,1]\times[0,+\infty]\rightarrow[0,+\infty]\) are continuous. \(f(t,x,y)\) is increasing in \(x\in[0,+\infty)\) for fixed \(t\in[0,1]\) and \(y\in[0,+\infty)\), decreasing in \(y\in[0,+\infty)\) for fixed \(t\in [0,1]\) and \(x\in[0,+\infty)\), and \(g(t,x)\) is increasing in \(x\in [0,+\infty)\) for fixed \(t\in[0,1]\), and \(q(t,y)\) is decreasing in \(y\in [0,+\infty)\) for fixed \(t\in[0,1]\). \(g(t,0)=bc>0\), \(q(t,1)=1+a(t)+c>0\), \(f(t,0,1)=1+a(t)+c>0\). Besides, for \(\lambda\in (0,1)\), \(t\in[0,1]\), \(x\in[0,+\infty)\), \(y \in[0,+\infty)\), we have
$$\begin{aligned}& f\bigl(t,\lambda x,\lambda^{1} y\bigr)=(\lambda x)^{\frac{1}{3}}+ \bigl(\lambda^{1} y\bigr)^{\frac{1}{3}}+a(t)+c \geq \lambda^{\frac{1}{3}}\bigl(x^{\frac {1}{3}}+y^{\frac{1}{3}}+a(t)+c\bigr)= \lambda^{\alpha}f(t,x,y), \\& g(t,\lambda x)=\frac{\lambda x}{1+\lambda x}m(t)+bc\geq\frac{\lambda x}{1+x}m(t)+ \lambda(bc)=\lambda g(t,x), \\& q\bigl(t,\lambda^{1}y\bigr)=\bigl(\lambda^{1}y \bigr)^{1}+a(t)+c\geq\lambda \bigl(y^{1}+a(t)+c\bigr)=\lambda q(t,y). \end{aligned}$$
Moreover, if we take \(\delta\in (0,\frac{c}{m_{\mathrm {max}}+b+a_{\mathrm{max}}} ]\), then we obtain
$$\begin{aligned} f(t,x,y)&=x^{\frac{1}{3}}+y^{\frac{1}{3}}+a(t)+c\geq c+y^{1}= \frac {c}{m_{\mathrm{max}}+b+a_{\mathrm{max}}}(m_{\mathrm{max}}+b+a_{\mathrm {max}})+y^{1} \\ &\geq\delta \biggl[\frac{x}{1+x}m(t)+bc+a(t)+c+\frac{1}{y} \biggr]= \delta\bigl[g(t,x)+q(t,y)\bigr]. \end{aligned}$$
Hence all the conditions of Theorem 4.6 are satisfied. An application of Theorem 4.6 implies that problem (4.3) has a unique positive solution in \(P_{h}\), where \(h(t)=t^{\nu1}=t^{\frac{7}{3}}\), \(t\in [0,1]\).