## 2 Preliminary Results  
We use the following notations for the simplicity:  
\[R_{i}\left(t,u\right) : =\int_{u}^{t}\frac{1}{r_{i}^{1/a_{i}}\left(s\right)}\mathrm{d}s,\quad i =1,2,\] \[\bar{R}\left(v,u\right) : =\int_{u}^{v}\left(\frac{R_{2}\left(v,s\right)}{r_{1}\left(s \right)}\right)^{1/a_{1}}\mathrm{d}s,\] \[G_{u}\left(t\right) : =\frac{1}{r_{1}^{\frac{1}{a_{1}}}\left(t\right)}\left(\int_{u}^{t }\frac{1}{r_{2}^{\frac{1}{a_{2}}}\left(s\right)}\mathrm{d}s\right)^{\frac{1}{ a_{1}}},\quad\widetilde{G}_{u}\left(t\right)=\int_{u}^{t}G_{u}\left(s\right)ds\]  
and  
\[Q\left(t\right)=\min\left\{kq\left(t\right),kq\left(\tau\left(t\right)\right) \right\},\]  
for \(t\geq u\geq t_{0}\). Next we present the following six Lemmas that will be used as tools to prove our main results in the next sections. **Lemma 1** ([23]). : _Assume that \(h_{1}\), \(h_{2}\in\left[0,\infty\right)\) and \(\gamma>0\). Then_  
\[(h_{1}+h_{2})^{\gamma}\leq\mu\left(h_{1}^{\gamma}+h_{2}^{\gamma}\right),\]  
_where_  
\[\mu=\left\{ \begin{array}{ll}1&\text{for}\quad 0<\gamma\leq 1,\\ 2^{\gamma-1}&\text{for}\quad\gamma>1.\end{array} \right.\]  
**Lemma 2**. : _Assume that \(x\) is a positive solution of (1). Then, \(L_{3}z\left(t\right)\leq 0\) and there are only two possible classes for the corresponding function \(z:\)_  
\[ \begin{array}{lll}\textbf{Case }\left(1\right)&:&z\left(t\right)>0,\text{ }L_{1}z\left(t\right)<0\text{ and }L_{2}z\left(t\right)>0;\\ \textbf{Case }\left(2\right)&:&z\left(t\right)>0,\text{ }L_{1}z\left(t\right)>0 \text{ and }L_{2}z\left(t\right)>0.\end{array} \]  
Proof. : Let \(x\) be a positive solution of (1). Then, there exists a \(t_{1}\geq t_{0}\) such that \(x(t)>0\), \(x\left(\tau(t)\right)>0\) and \(x\left(g(t)\right)>0\) for \(t\geq t_{1}\). Therefore, \(z(t)>0\) and (1) implies that  
\[L_{3}z\left(t\right)=-q\left(t\right)f\left(x\left(g\left(t\right)\right) \right)\leq 0.\]  
Hence, \(L_{2}z\left(t\right)\) is a non-increasing function and of one sign. We claim that \(L_{2}z\left(t\right)>0\) for \(t\geq t_{1}\). Suppose that \(L_{2}z\left(t\right)<0\) for \(t\geq t_{2}\geq t_{1}\), then there exists a \(t_{3}\geq t_{2}\) and constant \(K_{1}>0\) such that  
\[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]  
for \(t\geq t_{3}\). By integrating the last inequality from \(t_{3}\) to \(t\), we get  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]  
Letting \(t\rightarrow\infty\), we have \(\lim_{t\rightarrow\infty}E_{1}(t)=-\infty\). Then there exists a \(t_{4}\geq t_{3}\) and constant \(K_{2}>0\) such that  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]for \(t\geq t_{4}\). By integrating this inequality from \(t_{4}\) to \(\infty\), we get \(\lim_{t\rightarrow\infty}z(t)=-\infty\), which contradicts \(z(t)>0\). Now we have \(E_{2}(t)>0\) for \(t\geq t_{1}\). Therefore, \(E_{1}(t)\) is increasing function and of one sign. The proof is complete. The set of all functions \(z\) satisfy that Case (1) is denoted by \(\aleph\). The set of all functions \(z\) that satisfy Case (2) is denoted by \(\widetilde{\aleph}\). Solutions \(x\) whose corresponding function \(z\in\aleph\) are called Kneser-solutions. ([25] Lemmas 3-5). Assume that \(r_{i}^{\prime}\left(t\right)>0\) for \(i=1,2\), \(x\) is a positive solution of (1) with corresponding function \(z\in\widetilde{\aleph}\) for all \(t\geq t_{2}\geq t_{0}\). Then, the following results are achieved.