Uniqueness results
Theorem 3.1
Suppose that
\(f(t,0)\not \equiv 0\)
on
\([0,1]\), and there exists
\(\lambda \in C(0,1)\cap L[0,1]\)
satisfying
$$ 0< \int_{0}^{1}a(s)\lambda (s)\,ds< +\infty $$
such that
$$ \bigl\vert f(t,x)-f(t,y) \bigr\vert \leq \lambda (t)\vert x-y \vert , \quad t \in [0,1], x, y \in [0,\infty ). $$
Then (1.1) has a unique positive solution if the spectral radius
\(r(L_{\lambda })\in (0,1)\), where
$$ L_{\lambda }u(t)= \int_{0}^{1}G(t,s) a(s)\lambda (s)u(s)\,ds. $$
Proof
It follows from \(f(t,0)\not \equiv 0\) that θ is not a fixed point of A. Then we only need to prove that A has a unique fixed point in Q.
Firstly, we will prove that A has a fixed point in Q.
Set
$$ Q_{1}= \bigl\{ u\in P: \exists l_{1}, l_{2}>0\mbox{ such that }l_{2}t^{ \alpha -1} \leq u(t)\leq l_{1}t^{\alpha -1} \bigr\} . $$
For any \(u \in Q\setminus \{\theta \}\), let
$$ l_{i}(u)= \int_{0}^{1} \Phi_{i}(s)a(s)\lambda (s) u(s)\,ds,\quad i=1,2. $$
It follows from Lemma 2.3 that \(l_{1}(u)\), \(l_{2}(u)> 0\) and
$$ l_{2}(u)t^{\alpha -1} \leq (L_{\lambda }u) (t)\leq l_{1}(u)t^{\alpha -1}. $$
(3.1)
Therefore,
$$ L_{\lambda }:Q\setminus \{\theta \} \rightarrow Q_{1}. $$
Similar to Lemma 2.5, we have that the spectral radius \(r(L_{\lambda })>0\) and \(L_{\lambda }\) has a positive eigenfunction \(\psi_{1}\), that is, \(L_{\lambda }\psi_{1}=r(L_{\lambda })\psi_{1}\). It is easy to see that
$$ \bigl(r(L_{\lambda }) \bigr)^{-1}l_{2}( \psi_{1})t^{\alpha -1} \leq \psi_{1}(t) \leq \bigl(r(L_{\lambda }) \bigr)^{-1}l_{1}( \psi_{1})t^{\alpha -1}. $$
(3.2)
For any \(u_{0}\in Q\setminus \{\theta \}\), let
$$ u_{n}=A(u_{n-1}),\quad n=1,2,\ldots . $$
We may suppose that \(u_{1}-u_{0} \neq \theta \) (otherwise, the proof is finished). It follows from (3.1) and (3.2) that
$$ L_{\lambda } \bigl(\vert u_{1}-u_{0} \vert \bigr) \leq \frac{r(L_{\lambda })l_{1}(\vert u_{1}-u_{0} \vert )}{l_{2}(\psi_{1})}\psi_{1}. $$
(3.3)
Then
$$\begin{aligned}& \vert u_{2}-u_{1} \vert = \biggl\vert \int_{0}^{1}G(t,s)a(s) \bigl[ f \bigl(s,u_{1}(s) \bigr)-f \bigl(s,u _{0}(s) \bigr) \bigr] \,ds \biggr\vert \\& \hphantom{\vert u_{2}-u_{1} \vert } \leq \int_{0}^{1}G(t,s)a(s)\lambda (s) \bigl\vert u_{1}(s)-u_{0}(s)\bigr\vert \,ds \\& \hphantom{\vert u_{2}-u_{1} \vert } =L_{\lambda } \bigl( \vert u_{1}-u_{0} \vert \bigr) \\& \hphantom{\vert u_{2}-u_{1} \vert } \leq \frac{r(L _{\lambda })l_{1}(\vert u_{1}-u_{0} \vert )}{l_{2}(\psi_{1})}\psi_{1}, \end{aligned}$$
(3.4)
$$\begin{aligned}& \vert u_{3}-u_{2} \vert = \biggl\vert \int_{0}^{1}G(t,s)a(s) \bigl[ f \bigl(s,u_{2}(s) \bigr)-f \bigl(s,u _{1}(s) \bigr) \bigr] \,ds \biggr\vert \\& \hphantom{\vert u_{2}-u_{1} \vert }\leq \int_{0}^{1}G(t,s)a(s)\lambda (s) \bigl\vert u_{2}(s)-u_{1}(s)\bigr\vert \,ds \\& \hphantom{\vert u_{2}-u_{1} \vert } =L_{\lambda } \bigl( \vert u_{2}-u_{1} \vert \bigr) \\& \hphantom{\vert u_{2}-u_{1} \vert } \leq \frac{r(L _{\lambda })l_{1}(\vert u_{1}-u_{0} \vert )}{l_{2}(\psi_{1})}L_{\lambda }\psi _{1} \\& \hphantom{\vert u_{2}-u_{1} \vert } =\frac{[r(L_{\lambda })]^{2}l_{1}(\vert u_{1}-u_{0} \vert )}{l_{2}(\psi _{1})}\psi_{1}, \\& \hphantom{\vert u-u_{m} \vert } \cdots \end{aligned}$$
(3.5)
$$\begin{aligned}& \vert u_{m+1}-u_{m} \vert = \biggl\vert \int_{0}^{1}G(t,s)a(s) \bigl[ f \bigl(s,u_{m}(s) \bigr)-f \bigl(s,u _{m-1}(s) \bigr) \bigr] \,ds \biggr\vert \\& \hphantom{\vert u_{m+1}-u_{m} \vert } \leq \int_{0}^{1}G(t,s)a(s)\lambda (s) \bigl\vert u_{m}(s)-u_{m-1}(s) \bigr\vert \,ds \\& \hphantom{\vert u_{m+1}-u_{m} \vert } =L_{\lambda } \bigl(\vert u_{m}-u_{m-1} \vert \bigr) \\& \hphantom{\vert u_{m+1}-u_{m} \vert } \leq \frac{[r(L _{\lambda })]^{m-1}l_{1}(\vert u_{1}-u_{0} \vert )}{l_{2}(\psi_{1})}L_{\lambda } \psi_{1} \\& \hphantom{\vert u_{m+1}-u_{m} \vert } =\frac{[r(L_{\lambda })]^{m}l_{1}(\vert u_{1}-u_{0} \vert )}{l_{2}( \psi_{1})}\psi_{1}. \end{aligned}$$
(3.6)
By induction, we can get
$$ \vert u_{n+1}-u_{n} \vert \leq \frac{[r(L_{\lambda })]^{n}l_{1}(\vert u_{1}-u_{0} \vert )}{l _{2}(\psi_{1})} \psi_{1}, \quad n=1,2,\ldots . $$
(3.7)
Then, for any \(n, m \in \mathbb{N}\), we have
$$\begin{aligned} \vert u_{n+m}-u_{n} \vert &\leq \vert u_{n+m}-u_{n+m-1} \vert +\cdots +\vert u_{n+1}-u_{n} \vert \\ &\leq \bigl( \bigl[r(L_{\lambda }) \bigr]^{n+m-1}+\cdots + \bigl[r(L_{\lambda }) \bigr]^{n} \bigr) \frac{l _{1}(\vert u_{1}-u_{0} \vert )}{l_{2}(\psi_{1})} \psi_{1} \\ & \leq \frac{[r(L_{ \lambda })]^{n}l_{1}(\vert u_{1}-u_{0} \vert )}{[1-r(L_{\lambda })]l_{2}(\psi _{1})}\psi_{1}. \end{aligned}$$
It follows from \(r(L_{\lambda })<1\) that
$$ \Vert u_{n+m}-u_{m} \Vert \rightarrow 0\quad (n \rightarrow \infty ), $$
which implies \(\{u_{n}\}\) is a Cauchy sequence. Therefore, there exists \(u^{\ast }\in Q\) such that \(\{u_{n}\}\) converges to \(u^{\ast }\). Clearly \(u^{\ast }\) is a fixed point of A.
In the following, we will prove that the fixed point of A is unique.
Suppose that \(v \neq u^{\ast }\) is a positive fixed point of A. Then there exists \(l_{1}(\vert u^{\ast }-v \vert )> 0\) such that
$$ L_{\lambda } \bigl( \bigl\vert u^{\ast }-v \bigr\vert \bigr)\leq \frac{r(L_{\lambda })l_{1}(\vert u^{\ast }-v \vert )}{l _{2}(\psi_{1})}\psi_{1}. $$
Therefore,
$$\begin{aligned} \bigl\vert Au^{\ast }-Av \bigr\vert &= \biggl\vert \int_{0}^{1}G(t,s)a(s) \bigl[f \bigl(s,u^{\ast }(s) \bigr)-f \bigl(s,v(s) \bigr) \bigr]\,ds \biggr\vert \\ & \leq \int_{0}^{1}G(t,s)a(s)\lambda (s) \bigl\vert u^{\ast }(s)-v(s) \bigr\vert \,ds \\ & \leq \frac{r(L_{\lambda })l_{1}(\vert u^{\ast }-v \vert )}{l_{2}(\psi_{1})}\psi _{1}. \end{aligned}$$
Similar to the proof of (3.7), we can get
$$ \bigl\vert A^{n}u^{\ast }-A^{n}v \bigr\vert \leq \frac{[r(L_{\lambda })]^{n}l_{1}(\vert u^{\ast }-v \vert )}{l_{2}(\psi_{1})}\psi_{1}. $$
(3.8)
It follows from \(r(L_{\lambda })<1\) that
$$ \bigl\Vert u^{\ast }-v \bigr\Vert = \bigl\Vert A^{n}u^{\ast }-A^{n}v \bigr\Vert \rightarrow 0\quad (n\rightarrow \infty ), $$
which implies that the positive fixed point of A is unique. □
Remark 3.1
The unique positive solution \(u^{\ast }\) of (1.1) can be approximated by the iterative schemes: for any \(u_{0}\in Q\setminus \{\theta \}\), let
$$ u_{n}=A(u_{n-1}),\quad n=1,2,\ldots , $$
then \(u_{n}\rightarrow u^{\ast }\). Furthermore, we have error estimation
$$ \bigl\vert u_{n}-u^{\ast } \bigr\vert \leq \frac{[r(L_{\lambda })]^{n}l_{1}(\vert u_{1}-u_{0} \vert )}{[1-r(L _{\lambda })]l_{2}(\psi_{1})}\psi_{1}, $$
and with the rate of convergence
$$ \bigl\Vert u_{n}-u^{\ast } \bigr\Vert =O \bigl( \bigl[r(L_{\lambda }) \bigr]^{n} \bigr). $$
Remark 3.2
The spectral radius satisfies \(r(L_{\lambda })= \lim_{n\rightarrow \infty }\Vert L_{\lambda }^{n} \Vert ^{\frac{1}{n}}\) and \(r(L_{\lambda })\leq \Vert L_{\lambda }^{n} \Vert ^{\frac{1}{n}}\). Particularly,
$$ r(L_{\lambda })\leq \Vert L_{\lambda } \Vert =\sup _{0\leq t \leq 1} \int_{0} ^{1}G(t,s) a(s)\lambda (s)\,ds. $$
Nonexistence results
Theorem 3.2
Suppose that there exists
\(b_{1} \in C(0,1)\cap L[0,1]\)
satisfying
$$ 0< \int_{0}^{1}a(s)b_{1}(s)\,ds< +\infty $$
such that
$$ f(t,x)\leq b_{1} (t)x, \quad t \in [0,1], x\in [0,\infty ). $$
Then (1.1) has no positive solution if the spectral radius
\(r(L_{b _{1}})\in (0,1)\), where
$$ L_{b_{1}}u(t)= \int_{0}^{1}G(t,s) a(s)b_{1} (s)u(s) \,ds. $$
Proof
We only need to prove that A has no fixed point in \(Q_{1}\). If otherwise, there exists \(v\in Q_{1}\) such that \(Av= v\). Similar to Lemma 2.5, we have that the spectral radius \(r(L_{b_{1}})>0\) and \(L_{b_{1}}\) has a positive eigenfunction \(\varphi_{b_{1}}\) satisfying
$$ L_{b_{1}}\varphi_{b_{1}}=r(L_{b_{1}})\varphi_{b_{1}}. $$
It is clear that \(v, \varphi_{b_{1}}\in Q_{1}\). Therefore, there exists \(c_{1}>0\) such that
$$ v\leq c_{1} \varphi_{b_{1}}. $$
By \(f(t,x)\leq b_{1} (t)x\), we have \(v=Av\leq L_{b_{1}}v\). It is obvious that \(L_{b_{1}}\) is increasing in \(Q_{1}\). By induction, we can get \(v\leq L^{n}_{b_{1}}v\), \(\forall n=1,2,3,\ldots \) . Then
$$ v\leq L^{n}_{b_{1}}v\leq L^{n}_{b_{1}}c_{1} \varphi_{b_{1}}= c_{1} \bigl[r(L _{b_{1}}) \bigr]^{n} \varphi_{b_{1}},\quad \forall n=1,2,3,\ldots . $$
Noticing \(r(L_{b_{1}})<1\), we have \(v=\theta \), which contradicts \(v\in Q_{1}\). □
Theorem 3.3
Suppose that there exists
\(b_{2}\in C(0,1)\cap L[0,1]\)
satisfying
$$ 0< \int_{0}^{1}a(s)b_{2}(s)\,ds< +\infty $$
such that
$$ f(t,x)\geq b_{2} (t)x, \quad t \in [0,1], x\in [0,\infty ). $$
Then (1.1) has no positive solution if the spectral radius
\(r(L_{b_{2}})> 1\), where
$$ L_{b_{2}}u(t)= \int_{0}^{1}G(t,s) a(s)b_{2} (s)u(s) \,ds. $$
Proof
The proof is just similar to Theorem 3.2, so we omit it. □