Let
$$X= \biggl\{ u: u\in C[0,+\infty), u(0)=0, \sup_{t\in[0,+\infty)} \frac{\vert u(t)\vert }{1+t}< \infty \biggr\} , $$
with the norm \(\Vert u\Vert =\sup_{t\in[0,+\infty)}\frac{\vert u(t)\vert }{1+t}\), and
$$Y= \Bigl\{ y:y\in C[0,+\infty)\cap L[0,+\infty), \sup_{t\in[0,+\infty)} \bigl\vert y(t)\bigr\vert < +\infty \Bigr\} , $$
with the norm \(\Vert y\Vert _{1}=\int_{0}^{+\infty} \vert y(t)\vert \,dt+\sup_{t\in [0,+\infty)}\vert y(t)\vert \).
It is easy to prove that \((X, \Vert \cdot \Vert )\) and \((Y, \Vert \cdot \Vert _{1})\) are Banach spaces.
Define \(L:\operatorname {dom}L\subset X\rightarrow Y\) and \(N:X\rightarrow Y\) as follows:
$$(Lu) (t)=-u{''}(t),\qquad (Nu) (t)=f\bigl(t,u(t) \bigr),\qquad u(t)\in X, t\in[0,+\infty), $$
where
$$\operatorname {dom}L= \Biggl\{ u(t)\in X| u{''}(t)\in Y, u'(+\infty)=\sum_{i=1}^{m-1} \alpha _{i}u'(\xi_{i}) \Biggr\} . $$
Then the boundary value problem (1.1) can be written
$$(Lu) (t)=(Nu) (t),\quad u(t)\in \operatorname {dom}L. $$
For convenience, denote the function \(G(t,s)\) as follows:
$$G(t,s)=\textstyle\begin{cases} 0, & t=0,\\ \frac{(1-\sum_{i=k}^{m-1}\alpha_{i})(\frac {3}{2}t+e^{-t}-1)}{t\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}-\frac {t-s}{t}+e^{-s},& s\leq t, \xi_{k-1}\leq s < \xi_{k}, k=2,\ldots,m-1,\\ \frac{(1-\sum_{i=k}^{m-1}\alpha_{i})(\frac {3}{2}t+e^{-t}-1)}{t\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}+e^{-s}, & 0< t< s, \xi_{k-1}\leq s < \xi_{k}, k=2,\ldots,m-1,\\ \frac{\frac{3}{2}t+e^{-t}-1}{t\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}}-\frac{t-s}{t}+e^{-s}, & \xi_{m-1}\leq s\leq t,\\ \frac{\frac{3}{2}t+e^{-t}-1}{t\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}}+e^{-s}, & 0< t< s, \xi_{m-1}\leq s. \end{cases} $$
Clearly, \(G(t,s)\leq\frac{3}{2\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}}+1\), for \(t,s\in[0,+\infty)\).
Lemma 3.1
L
is a Fredholm operator of index zero.
Proof
It is easy to get
$$ \operatorname {Ker}L=\bigl\{ u\in \operatorname {dom}L| u(t)=ct,t\geq0, c\in\mathbf{R}\bigr\} , $$
(3.1)
and
$$ \operatorname {Im}L= \Biggl\{ y\in Y \Big| \sum_{i=1}^{m-1} \alpha_{i} \int_{\xi_{i}}^{+\infty }y(s)\,ds=0 \Biggr\} . $$
(3.2)
Define \(Q:Y\rightarrow Y\) by
$$ (Qy) (t)=\frac{e^{-t}}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi _{i}}}\sum_{i=1}^{m-1} \alpha_{i} \int_{\xi_{i}}^{+\infty }y(s)\,ds,\quad y\in Y. $$
(3.3)
Clearly, \(\operatorname {Ker}Q=\operatorname {Im}L\), \(\operatorname {Im}Q=\{y\mid y=ce^{-t},t\geq0, c\in\textbf{R}\}\), and \(Q:Y\rightarrow Y\) is a linear projector. In fact, for \(y(t)\in Y\), we have
$$\bigl(Q^{2}y\bigr) (t)=\bigl(Q(Qy)\bigr) (t)=Q\bigl(e^{-t} \bigr)\frac{1}{\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}}\sum_{i=1}^{m-1} \alpha_{i} \int_{\xi _{i}}^{+\infty}y(s)\,ds=(Qy) (t). $$
For \(y\in Y\), we have \(y=(y-Qy)+Qy\), \(Qy\in \operatorname {Im}Q\), \((I-Q)y\in \operatorname {Ker}Q=\operatorname {Im}L\). So we obtain \(Y=\operatorname {Im}Q+\operatorname {Im}L\). Take \(y_{0}\in \operatorname {Im}Q\cap \operatorname {Im}L\). \(y_{0}\in \operatorname {Im}Q\) means that \(y_{0}\) can be written \(y_{0}=ce^{-t}\), \(c\in\textbf{R}\). At the same time, by \(y_{0}\in \operatorname {Im}L\) and (3.2), we get
$$\sum_{i=1}^{m-1}\alpha_{i} \int_{\xi_{i}}^{+\infty}ce^{-s}\,ds=0, $$
i.e.
\(c=0\). This implies that \(y_{0}=0\). Thus, \(Y=\operatorname {Im}Q\oplus \operatorname {Im}L\) and \(\dim \operatorname {Ker}L=\operatorname {codim}\operatorname {Im}L=1<+\infty\). Observing that ImL is closed in Y, L is a Fredholm operator of index zero. □
Define \(P:X\rightarrow X\) as
$$ (Pu) (t)=t \int_{0}^{+\infty}e^{-t}u(t)\,dt,\quad u(t)\in X. $$
(3.4)
Clearly, \(P:X\rightarrow X\) is a linear continuous projector and
$$\operatorname {Im}P=\bigl\{ u\mid u(t)=ct, t\geq0, c\in\textbf{R}\bigr\} =\operatorname {Ker}L. $$
Thus, \(X=\operatorname {Im}P\oplus \operatorname {Ker}P=\operatorname {Ker}L\oplus \operatorname {Ker}P\).
Define \(K_{P}:\operatorname {Im}L\rightarrow \operatorname {Ker}P\cap \operatorname {dom}L\) by
$$ (K_{P}y) (t)=- \int_{0}^{t}(t-s)y(s)\,ds+t \int_{0}^{+\infty }e^{-s}y(s)\,ds,\quad y\in \operatorname {Im}L. $$
(3.5)
By simple calculations, we have \((K_{P}L_{P})u=u\), \(\forall u\in \operatorname {dom}L\cap \operatorname {Ker}P\), and \((L_{P}K_{P})y=y\), \(\forall y\in \operatorname {Im}L\). So \(K_{P}=(L_{P})^{-1}\), where \(L_{P}=L\mid_{\operatorname {dom}L\cap \operatorname {Ker}P}:\operatorname {dom}L\cap \operatorname {Ker}P\rightarrow \operatorname {Im}L\).
Define the linear isomorphism \(J:\operatorname {Im}Q\rightarrow \operatorname {Ker}L\) as
$$J\bigl(ce^{-t}\bigr)=ct, \quad t\geq0, c\in\textbf{R}. $$
Thus, \(JQN+K_{P}(I-Q)N:X\rightarrow X\) is given by
$$ \bigl[JQN+K_{P}(I-Q)N\bigr]u(t)=t \int_{0}^{+\infty}G(t,s)f\bigl(s,u(s)\bigr)\,ds. $$
(3.6)
Lemma 3.2
\(QN:X\rightarrow Y\)
is continuous and bounded and
\(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\)
is compact, where
\(\Omega\subset X\)
is bounded.
Proof
For convenience, denote \(M_{r}:=\frac{\int_{0}^{+\infty}h_{r}(\tau)\,d\tau}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\).
We will prove that \(QN:X\rightarrow Y\) is continuous and bounded.
Since \(\Omega\subset X\) is bounded, for \(u\in\overline{\Omega}\), there exists a constant \(r>0\), such that \(\Vert u\Vert < r\). By the condition (A1), we have
$$\begin{aligned} \Vert QNu\Vert _{1}={}& \int_{0}^{+\infty} \Biggl\vert \frac{e^{-s}}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\sum _{i=1}^{m-1}\alpha_{i} \int _{\xi_{i}}^{+\infty}f\bigl(\tau, u(\tau)\bigr)\,d\tau \Biggr\vert \,ds \\ &{}+\sup_{t\in [0,+\infty)} \Biggl\vert \frac{e^{-t}}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi _{i}}}\sum _{i=1}^{m-1}\alpha_{i} \int_{\xi_{i}}^{+\infty}f\bigl(\tau , u(\tau)\bigr)\,d\tau \Biggr\vert \\ \leq{}& \int_{0}^{+\infty}\frac{e^{-s}}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\sum _{i=1}^{m-1}\alpha _{i} \int_{\xi_{i}}^{+\infty} \bigl\vert f\bigl(\tau, u(\tau)\bigr) \bigr\vert \,d\tau \,ds \\ &{}+\sup_{t\in[0,+\infty)}\frac{e^{-t}}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\sum _{i=1}^{m-1}\alpha_{i} \int _{\xi_{i}}^{+\infty} \bigl\vert f\bigl(\tau, u(\tau) \bigr) \bigr\vert \,d\tau \\ \leq{}&\frac{\int_{0}^{+\infty}h_{r}(\tau)\,d\tau }{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}} +\frac{\int_{0}^{+\infty}h_{r}(\tau)\,d\tau}{\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}}=2M_{r}. \end{aligned}$$
So \(QN:X\rightarrow Y\) is bounded. By (A1) and the Lebesgue dominated convergence theorem, we see that \(QN:X\rightarrow Y\) is continuous.
Now, we will prove that \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\) is compact.
First of all, by the condition (A1) and \(u\in\overline{\Omega}\), we have
$$\begin{aligned} &\biggl\vert \frac{K_{P}(I-Q)Nu(t)}{1+t}\biggr\vert \\ &\quad =\biggl\vert - \int_{0}^{t}\frac {t-s}{1+t}(I-Q)Nu(s)\,ds+ \frac{t}{1+t} \int_{0}^{+\infty }e^{-s}(I-Q)Nu(s)\,ds\biggr\vert \\ &\quad \leq \int_{0}^{t}\bigl\vert (I-Q)Nu(s)\bigr\vert \,ds+ \int _{0}^{+\infty}\bigl\vert (I-Q)Nu(s)\bigr\vert \,ds \\ &\quad \leq2 \int_{0}^{+\infty}\bigl\vert Nu(s)\bigr\vert \,ds+2 \int_{0}^{+\infty}\bigl\vert QNu(s)\bigr\vert \,ds \\ &\quad \leq2 \int_{0}^{\infty}h_{r}(s) \,ds+2M_{r}\leq 4M_{r}< +\infty, \end{aligned}$$
i.e.
\(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\) is bounded.
Second, for \(u\in\overline{\Omega}\), \(0< t_{1}< t_{2}< T<\infty\),
$$\begin{aligned} &\biggl\vert \frac{K_{P}(I-Q)Nu(t_{2})}{1+t_{2}}-\frac {K_{P}(I-Q)Nu(t_{1})}{1+t_{1}}\biggr\vert \\ &\quad = \biggl\vert - \int_{0}^{t_{2}}\frac {t_{2}-s}{1+t_{2}}(I-Q)Nu(s)\,ds+ \frac{t_{2}}{1+t_{2}} \int_{0}^{+\infty }e^{-s}(I-Q)Nu(s)\,ds \\ &\qquad {} - \biggl(- \int_{0}^{t_{1}}\frac {t_{1}-s}{1+t_{1}}(I-Q)Nu(s)\,ds+ \frac{t_{1}}{1+t_{1}} \int_{0}^{+\infty }e^{-s}(I-Q)Nu(s)\,ds \biggr) \biggr\vert \\ &\quad \leq \int_{0}^{t_{1}}\biggl\vert \frac{t_{2}-s}{1+t_{2}}- \frac {t_{1}-s}{1+t_{1}}\biggr\vert \bigl\vert (I-Q)Nu(s)\bigr\vert \,ds+ \int_{t_{1}}^{t_{2}}\frac {t_{2}-s}{1+t_{2}}\bigl\vert (I-Q)Nu(s)\bigr\vert \,ds \\ &\qquad {}+\biggl\vert \frac{t_{2}}{1+t_{2}}-\frac {t_{1}}{1+t_{1}}\biggr\vert \int_{0}^{+\infty}e^{-s}\bigl\vert (I-Q)Nu(s) \bigr\vert \,ds \\ &\quad \leq \int_{0}^{t_{1}}\biggl\vert \frac{t_{2}-s}{1+t_{2}}- \frac {t_{1}-s}{1+t_{1}}\biggr\vert h_{r}(s)\,ds+ \int_{0}^{t_{1}}\biggl\vert \frac {t_{2}-s}{1+t_{2}}- \frac{t_{1}-s}{1+t_{1}}\biggr\vert \bigl\vert QNu(s)\bigr\vert \,ds \\ &\qquad {}+ \int_{t_{1}}^{t_{2}}\frac {t_{2}-s}{1+t_{2}}h_{r}(s) \,ds+ \int_{t_{1}}^{t_{2}}\frac {t_{2}-s}{1+t_{2}}\bigl\vert QNu(s) \bigr\vert \,ds \\ &\qquad {}+\biggl\vert \frac{t_{2}}{1+t_{2}}-\frac {t_{1}}{1+t_{1}}\biggr\vert \int_{0}^{+\infty}h_{r}(s)\,ds+\biggl\vert \frac {t_{2}}{1+t_{2}}-\frac{t_{1}}{1+t_{1}}\biggr\vert \int_{0}^{+\infty}\bigl\vert QNu(s)\bigr\vert \,ds \\ &\quad \leq \int_{0}^{t_{1}}\biggl\vert \frac {t_{2}-s}{1+t_{2}}- \frac{t_{1}-s}{1+t_{1}}\biggr\vert h_{r}(s)\,ds+M_{r} \int _{0}^{t_{1}}\biggl\vert \frac{t_{2}-s}{1+t_{2}}- \frac{t_{1}-s}{1+t_{1}}\biggr\vert e^{-s}\,ds \\ &\qquad {}+ \int_{t_{1}}^{t_{2}}h_{r}(s) \,ds+M_{r} \int _{t_{1}}^{t_{2}}e^{-s}\,ds+2\biggl\vert \frac{t_{2}}{1+t_{2}}-\frac {t_{1}}{1+t_{1}}\biggr\vert M_{r}. \end{aligned}$$
By the uniform continuity of \(\frac{t-s}{1+t}\) in \([0,T]\times[0,T]\) and \(\frac{t}{1+t}\) in \([0,T]\), and the absolute continuity of the integral, we see that \(K_{P}(I-Q)N:\overline {\Omega}\rightarrow X\) is equicontinuous on \([0,T]\), \(\forall T>0\).
Third, for \(\varepsilon>0\), there exists a constant \(l>0\), such that
$$\int_{l}^{+\infty}h_{r}(s)\,ds< \frac{\varepsilon}{12}, \qquad \int_{l}^{+\infty }e^{-s}\,ds< \frac{\varepsilon}{12M_{r}}. $$
Since \(\lim_{t\rightarrow+\infty}\frac{t-l}{1+t}=1\), \(\lim_{t\rightarrow+\infty}\frac{t}{1+t}=1\), there exists a constant \(T>l\) such that
$$\biggl\vert 1-\frac{t-l}{1+t}\biggr\vert < \frac{\varepsilon}{12M_{r}},\qquad \biggl\vert 1-\frac{t}{1+t}\biggr\vert < \frac{\varepsilon}{12M_{r}} ,\quad t\geq T. $$
For \(u\in\overline{\Omega}\), \(T\leq t_{1}< t_{2}\), we have
$$\begin{aligned} &\biggl\vert \frac{K_{P}(I-Q)Nu(t_{2})}{1+t_{2}}-\frac {K_{P}(I-Q)Nu(t_{1})}{1+t_{1}}\biggr\vert \\ &\quad = \biggl\vert - \int_{0}^{t_{2}}\frac {t_{2}-s}{1+t_{2}}(I-Q)Nu(s)\,ds+ \frac{t_{2}}{1+t_{2}} \int_{0}^{+\infty }e^{-s}(I-Q)Nu(s)\,ds \\ &\qquad {}- \biggl(- \int_{0}^{t_{1}}\frac {t_{1}-s}{1+t_{1}}(I-Q)Nu(s)\,ds+ \frac{t_{1}}{1+t_{1}} \int_{0}^{+\infty }e^{-s}(I-Q)Nu(s)\,ds \biggr) \biggr\vert \\ &\quad \leq \int_{0}^{l}\biggl\vert \frac{t_{2}-s}{1+t_{2}}- \frac {t_{1}-s}{1+t_{1}}\biggr\vert \bigl\vert (I-Q)Nu(s)\bigr\vert \,ds+ \int_{l}^{t_{2}}\frac {t_{2}-s}{1+t_{2}}\bigl\vert (I-Q)Nu(s)\bigr\vert \,ds \\ &\qquad {}+ \int_{l}^{t_{1}}\frac {t_{1}-s}{1+t_{1}}\bigl\vert (I-Q)Nu(s)\bigr\vert \,ds+\biggl\vert \frac{t_{2}}{1+t_{2}}-\frac {t_{1}}{1+t_{1}} \biggr\vert \int_{0}^{+\infty}e^{-s}\bigl\vert (I-Q)Nu(s) \bigr\vert \,ds \\ &\quad \leq \biggl[ \biggl(1-\frac{t_{2}-l}{1+t_{2}} \biggr)+ \biggl(1- \frac{t_{1}-l}{1+t_{1}} \biggr) \biggr] \int_{0}^{+\infty }\bigl\vert (I-Q)Nu(s)\bigr\vert \,ds+2 \int_{l}^{+\infty}\bigl\vert (I-Q)Nu(s)\bigr\vert \,ds \\ &\qquad {} + \biggl[ \biggl(1-\frac{t_{2}}{1+t_{2}} \biggr)+ \biggl(1- \frac {t_{1}}{1+t_{1}} \biggr) \biggr] \int_{0}^{+\infty}\bigl\vert (I-Q)Nu(s)\bigr\vert \,ds \\ &\quad \leq \biggl[ \biggl(1-\frac{t_{2}-l}{1+t_{2}} \biggr)+ \biggl(1- \frac{t_{1}-l}{1+t_{1}} \biggr) \biggr]2M_{r}+2 \int_{l}^{+\infty }h_{r}(s) \,ds+2M_{r} \int_{l}^{+\infty}e^{-s}\,ds \\ &\qquad {} + \biggl[ \biggl(1-\frac{t_{2}}{1+t_{2}} \biggr)+ \biggl(1- \frac {t_{1}}{1+t_{1}} \biggr) \biggr]2M_{r}< \varepsilon. \end{aligned}$$
Thus, \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\) is equiconvergent at infinity.
By Lemma 2.1, we see that \(K_{P}(I-Q)N:\overline{\Omega}\rightarrow X\) is compact. □
Theorem 3.1
Assume that (A1) and the following conditions hold.
(A2) For
\(u\geq0\), there exist three nonnegative functions
\(\mu (t)\), \(\beta_{i}(t)\), \(i=1,2\), such that
$$-\mu(t)ue^{-t}\leq f(t,u)\leq-\beta_{1}(t)ue^{-t}+ \beta _{2}(t), \qquad G(t,s)f(s,u)\geq-e^{-s}u,\quad t,s\in[0,+\infty), $$
where
\(\mu(t)ue^{-t}, \beta_{2}(t), \beta_{1}(t)ue^{-t}\in L[0,+\infty )\), \(\inf_{t\in[0,+\infty)}\beta_{1}(t) :=\beta_{0}>0\)
and
\(\mu(t)\)
satisfying
-
(i)
\(\sup_{t\in[0,+\infty)}\mu(t):=\mu _{1}<\frac{2\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi _{i}}}{3+2\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\),
-
(ii)
there exists
\(t_{0}\in[0,+\infty)\), such that
\(d_{0}:=\frac {t_{0}}{1+t_{0}}\int_{0}^{+\infty}[1-G(t_{0},s)\mu (s)](1+s)e^{-s}\,ds>1\), \(G(t_{0},s)\geq0\).
(A3) There exists
\(R>\frac{\mu_{1}+\alpha_{1}\beta_{0}+1}{\alpha_{1}\beta _{0}}\int_{0}^{\infty}\beta_{2}(s)\,ds\), such that
\(f(t,Rt)<0\), \(t\in[0,+\infty)\).
Then the problem (1.1) has at least one positive solution.
Proof
Take a cone
$$C=\bigl\{ u(t)\in X|u(t)\geq0, t\in[0,+\infty)\bigr\} . $$
Set
$$\Omega_{1}= \biggl\{ u\in X\Big|\frac{1}{d_{0}}\Vert u\Vert < \frac{\vert u(t)\vert }{1+t}< r< R, t\in[0,+\infty) \biggr\} , \qquad \Omega_{2}= \bigl\{ u\in X\mid \Vert u\Vert < R\bigr\} , $$
where \(d_{0}\) is given by the condition (A2) and \(R>\frac{\mu_{1}+\alpha_{1}\beta_{0}+1}{\alpha_{1}\beta _{0}}\int_{0}^{\infty}\beta_{2}(s)\,ds\). Clearly, \(\Omega_{1}\), \(\Omega_{2}\) are open and bounded sets of X, \(\overline{\Omega}_{1}= \{u\in X |\frac{1}{d_{0}}\Vert u\Vert \leq\frac{\vert u(t)\vert }{1+t}\leq r< R \}\subset\Omega_{2}\), and \({C\cap(\overline{\Omega}_{2}\setminus\Omega_{1})\neq\emptyset}\).
In view of Lemmas 3.1 and 3.2, L is a Fredholm operator of index zero and the condition (C1) of Theorem 2.1 is fulfilled.
Suppose that there exist \(u_{1}(t)\in C\cap\partial\Omega_{2}\cap \operatorname {dom}L\) and \(\lambda_{0}\in(0,1)\) such that \(Lu_{1}=\lambda_{0}Nu_{1}\), i.e.
\(u_{1}''(t)+\lambda_{0} f(t,u_{1}(t))=0\). By \(u_{1}(t)\in \operatorname {dom}L\), we have
$$\begin{aligned}& u_{1}'(+\infty)-\sum_{i=1}^{m-1} \alpha_{i}u_{1}'(\xi _{i})=0, \\& \quad \textit{i.e. } {-}\lambda_{0} \int_{0}^{+\infty }f\bigl(s,u_{1}(s)\bigr)\,ds+ \sum_{i=1}^{m-1}\alpha_{i} \lambda_{0} \int _{0}^{\xi_{i}}f\bigl(s,u_{1}(s)\bigr) \,ds=0. \end{aligned}$$
It follows from (A2) that
$$0= \sum_{i=1}^{m-1}\alpha_{i} \int_{\xi_{i}}^{+\infty }f\bigl(s,u_{1}(s)\bigr)\,ds \leq\sum_{i=1}^{m-1}\alpha_{i} \int_{\xi _{i}}^{+\infty}\bigl[-\beta_{1}(s)u_{1}(s)e^{-s}+ \beta_{2}(s)\bigr]\,ds. $$
So
$$ \alpha_{1} \int_{0}^{+\infty}\beta_{1}(s)u_{1}(s)e^{-s} \,ds\leq \int _{0}^{+\infty}\beta_{2}(s)\,ds. $$
(3.7)
Considering (A2), (3.7), and
$$\begin{aligned} u_{1}(t)&=(I-P)u_{1}(t)+Pu_{1}(t)=K_{P}L(I-P)u_{1}(t)+Pu_{1}(t)=K_{P}Lu_{1}(t)+Pu_{1}(t) \\ &=-\lambda_{0} \int _{0}^{t}(t-s)f\bigl(s,u_{1}(s) \bigr)\,ds+\lambda_{0}t \int_{0}^{+\infty }e^{-s}f\bigl(s,u_{1}(s) \bigr)\,ds+t \int_{0}^{+\infty}e^{-s}u_{1}(s)\,ds, \end{aligned}$$
we obtain
$$\begin{aligned} &\frac{u_{1}(t)}{1+t}=-\frac{\lambda_{0}}{1+t} \int _{0}^{t}(t-s)f\bigl(s,u_{1}(s) \bigr)\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}=}{}+\frac{\lambda_{0}t}{1+t} \int_{0}^{+\infty }e^{-s}f\bigl(s,u_{1}(s) \bigr)\,ds+\frac{t}{1+t} \int_{0}^{+\infty}e^{-s}u_{1}(s)\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}}\leq\lambda_{0} \int_{0}^{t}\frac{(t-s)}{1+t}\mu (s)u_{1}(s)e^{-s}\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}=}{}+\frac{\lambda_{0}t}{1+t} \int_{0}^{+\infty }e^{-s}\bigl[- \beta_{1}(s)u_{1}(s)e^{-s}+\beta_{2}(s) \bigr]\,ds+\frac{t}{1+t} \int _{0}^{+\infty}e^{-s}u_{1}(s)\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}}\leq \int_{0}^{t}\frac{\beta_{1}(s)e^{-s}\mu (s)u_{1}(s)}{\beta_{1}(s)}\,ds+ \int_{0}^{+\infty}\beta_{2}(s)\,ds+ \int _{0}^{+\infty}\frac{\beta_{1}(s)e^{-s}u_{1}(s)}{\beta_{1}(s)}\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}}\leq\frac{\mu_{1}+\alpha_{1}\beta_{0}+1}{\alpha _{1}\beta_{0}} \int_{0}^{+\infty}\beta_{2}(s)\,ds< R, \\ &\frac{u_{1}(t)}{1+t}=-\frac{\lambda_{0}}{1+t} \int _{0}^{t}(t-s)f\bigl(s,u_{1}(s) \bigr)\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}=}{}+\frac{\lambda_{0}t}{1+t} \int_{0}^{+\infty }e^{-s}f\bigl(s,u_{1}(s) \bigr)\,ds+\frac{t}{1+t} \int_{0}^{+\infty}e^{-s}u_{1}(s)\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}}\geq\lambda_{0} \int_{0}^{t}\frac{t-s}{1+t}\bigl[\beta _{1}(s)u_{1}(s)e^{-s}-\beta_{2}(s) \bigr]\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}=}{}+\frac{\lambda_{0}t}{1+t} \int _{0}^{+\infty}e^{-s}\bigl[- \mu(s)u_{1}(s)e^{-s}\bigr]\,ds+\frac{t}{1+t} \int _{0}^{+\infty}e^{-s}u_{1}(s)\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}}\geq- \int_{0}^{t}\beta_{2}(s)\,ds- \int_{0}^{+\infty}\frac {\beta_{1}(s)\mu(s)e^{-s}u_{1}(s)}{\beta_{1}(s)}\,ds \\ &\hphantom{\frac{u_{1}(t)}{1+t}}\geq-\frac{\alpha_{1}\beta_{0}+\mu_{1}}{\alpha_{1}\beta _{0}} \int_{0}^{+\infty}\beta_{2}(s)\,ds>-R. \end{aligned}$$
These contradict \(u_{1}(t)\in C\cap\partial\Omega_{2}\cap \operatorname {dom}L\). So (C2) is satisfied.
Let \((\gamma u)(t)=\vert u(t)\vert \), \(u(t)\in X\). Then \(\gamma:X\rightarrow C\) is a retraction and maps subsets of \(\overline{\Omega}_{2}\) into bounded subsets of C, i.e. (C3) holds.
Let \(u(t)\in \operatorname {Ker}L\cap\partial\Omega_{2}\), then \(u(t)=ct\), \(t\geq0\). Define
$$\begin{aligned} H(ct,\lambda)&= \bigl[I-\lambda(P+JQN)\gamma \bigr](ct) \\ &= ct-\lambda t \int_{0}^{+\infty}e^{-t}\vert c\vert t\,dt- \frac{\lambda t}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\sum_{i=1}^{m-1} \alpha_{i} \int _{\xi_{i}}^{+\infty}f\bigl(t,\vert c\vert t\bigr)\,dt, \end{aligned}$$
where \(c\in\{-R,R\}\) and \(\lambda\in[0,1]\). Suppose \(H(ct,\lambda )=0\), by (A2), we obtain
$$\begin{aligned} c&=\lambda \Biggl(\vert c\vert +\frac{1}{\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}}\sum _{i=1}^{m-1}\alpha_{i} \int_{\xi _{i}}^{+\infty}f\bigl(t,\vert c\vert t\bigr)\,dt \Biggr) \\ &\geq\lambda \vert c\vert \Biggl(1-\frac{1}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\sum _{i=1}^{m-1}\alpha_{i} \int _{\xi_{i}}^{+\infty}\mu(t)te^{-t}\,dt \Biggr) \\ & \geq\lambda \vert c\vert \Biggl(1-\frac{\mu_{1}}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\sum _{i=1}^{m-1}\alpha_{i} \int _{0}^{+\infty}te^{-t}\,dt \Biggr)\\ &=\lambda \vert c\vert \biggl(1-\frac{\mu _{1}}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}} \biggr)\geq0. \end{aligned}$$
Hence \(H(ct,\lambda)= 0\) implies \(c\geq0\). Furthermore, if \(H(Rt,\lambda)=0\), we have
$$R(1-\lambda)=\frac{\lambda}{\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}} \sum_{i=1}^{m-1} \alpha_{i} \int_{\xi _{i}}^{+\infty}f(t,Rt)\,dt\geq0, $$
which is a contradiction to the condition (A3).
Thus, \(H(u,\lambda)\neq0\), for \(u\in \operatorname {Ker}L\cap\partial\Omega_{2}\), and \(\lambda\in[0,1]\). Therefore
$$\begin{aligned} &d_{B}\bigl(\bigl[I-(P+JQN)\gamma\bigr]\mid_{\operatorname {Ker}L}, \operatorname {Ker}L \cap\Omega_{2}, 0\bigr) \\ &\quad =d_{B}\bigl(H(\cdot,1), \operatorname {Ker}L\cap\Omega_{2}, 0 \bigr) \\ &\quad =d_{B}\bigl(H(\cdot,0), \operatorname {Ker}L\cap\Omega_{2}, 0 \bigr)=d_{B}(I, \operatorname {Ker}L\cap\Omega _{2}, 0)=1\neq0. \end{aligned}$$
Thus, (C4) holds.
Let \(u_{0}(t)=t\), \(t \in[0,+\infty)\), then \(u_{0}\in C\setminus\{0\}\), \(C(u_{0})=\{u\in C| u(t)\geq\mu t \mbox{ for some } \mu>0, t \in[0,+\infty )\}\), and we take \(\sigma(u_{0})=1\). Let \(u \in C(u_{0})\cap\partial \Omega_{1}\), we have \(\frac{1}{d_{0}}\Vert u\Vert \leq\frac{\vert u(t)\vert }{1+t}\leq r\), \(t \in[0,+\infty)\).
For \(u\in C(u_{0})\cap\partial\Omega_{1}\), by (A2), we get
$$\begin{aligned} \frac{\Psi u(t_{0})}{1+t_{0}}&=\frac{t_{0}}{1+t_{0}} \int_{0}^{+\infty}e^{-s}u(s)\,ds+ \frac {t_{0}}{1+t_{0}} \int_{0}^{+\infty}G(t_{0},s)f\bigl(s,u(s)\bigr) \,ds \\ & \geq\frac{t_{0}}{1+t_{0}} \int_{0}^{+\infty} \bigl(e^{-s}u(s)-G(t_{0},s) \mu(s)u(s)e^{-s} \bigr)\,ds \\ & =\frac{t_{0}}{1+t_{0}} \int_{0}^{+\infty }\bigl[1-G(t_{0},s)\mu(s) \bigr](1+s)e^{-s}\frac{u(s)}{1+s}\,ds \\ & \geq\frac{t_{0}}{1+t_{0}} \int_{0}^{+\infty }\bigl[1-G(t_{0},s)\mu(s) \bigr](1+s)e^{-s}\,ds\frac{1}{d_{0}}\Vert u\Vert =\Vert u \Vert . \end{aligned}$$
Thus, \(\Vert u\Vert \leq\sigma(u_{0})\Vert \Psi u\Vert \), for \(u\in C(u_{0})\cap\partial \Omega_{1}\). So (C5) holds.
For \(u(t)\in\partial\Omega_{2}\), \(t\in[0,+\infty)\), by the condition (A2), we have
$$\begin{aligned} (P+JQN)\gamma(u)&=t \int_{0}^{+\infty}e^{-s}\bigl\vert u(s)\bigr\vert \,ds+\frac{t}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi_{i}}}\sum_{i=1}^{m-1} \alpha_{i} \int _{\xi_{i}}^{+\infty}f\bigl(s,\bigl\vert u(s)\bigr\vert \bigr)\,ds \\ & \geq t \int_{0}^{+\infty}e^{-s}\bigl\vert u(s)\bigr\vert \,ds-\frac{t}{\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}}\sum_{i=1}^{m-1} \alpha_{i} \int_{\xi _{i}}^{+\infty}\mu(s)\bigl\vert u(s)\bigr\vert e^{-s}\,ds \\ &\geq t \int_{0}^{+\infty}e^{-s}\bigl\vert u(s)\bigr\vert \,ds-\frac{t}{\sum_{i=1}^{m-1}\alpha _{i}e^{-\xi_{i}}} \int_{0}^{+\infty}\mu(s)\bigl\vert u(s)\bigr\vert e^{-s}\,ds \\ &= t \int_{0}^{+\infty }e^{-s}\bigl\vert u(s)\bigr\vert \biggl(1-\frac{\mu(s)}{\sum_{i=1}^{m-1}\alpha_{i}e^{-\xi _{i}}}\biggr)\,ds\geq0, \end{aligned}$$
which means that \((P+JQN)\gamma(\partial\Omega_{2})\subset C\). Hence, (C6) holds.
For \(u(t)\in\overline{\Omega}_{2}\setminus\Omega_{1}\), \(t\in[0,+\infty)\), by the condition (A2), we have
$$\begin{aligned} (\Psi_{\gamma}u) (t)&=t \int_{0}^{+\infty}e^{-s}\bigl\vert u(s)\bigr\vert \,ds+t \int_{0}^{+\infty }G(t,s)f\bigl(s,\bigl\vert u(s)\bigr\vert \bigr)\,ds \\ &\geq t \int_{0}^{+\infty}e^{-s}\bigl\vert u(s)\bigr\vert \,ds-t \int_{0}^{+\infty }e^{-s}\bigl\vert u(s)\bigr\vert \,ds \\ &=0. \end{aligned}$$
So \(\Psi_{\gamma}(\overline{\Omega}_{2}\setminus\Omega_{1})\subset C\), i.e. (C7) is satisfied.
By Theorem 2.1, we confirm that the equation \(Lu=Nu\) has a positive solution u, i.e. the problem (1.1) has at least one positive solution. □