## 2 Preliminary Results
Integrating the last inequality from \(u\) to \(v\), we obtain  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]6From (1), (A1) and (A5), we have  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]7  
and so,  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]8  
Combining (1) with (12), we get  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]9  
Hence,  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]0  
From (11) and (13) becomes  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]1  
that is,  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]2  
The proof of the Lemma is complete. **3. Criteria for Nonexistence of Non-Kneser Solutions**  
For simplicity, we use the following notations:  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]3  
In the following, we establish a Hille and Nehari type criterion for nonexistence of non-Kneser solutions. **Lemma 6**. : _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 \(t\geq t_{1}\geq t_{0}\). If \(P<\infty\) and \(D<\infty\), then_  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]4  
_and_  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]5  
_where_  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]6 \[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]7_and_  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]8  
Proof. : Assume that \(x\) is a positive solution of (1) and \(z\in\widetilde{\aleph}\). By Lemma 4, we get that (5)-(8) hold. Now, we define the function  
\[z^{\prime}(t)<-K_{2}(r_{1}(t))^{-1/a_{1}},\]9  
Then \(\omega\) is positive for \(t\geq t_{1}\), and satisfies  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]0  
Thus, from (6) and (7), there exists a \(T\geq t_{1}\) such that  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]1  
for \(t\geq T\). This implies that  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]2  
Using (8), we get  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]3  
which with (2), gives  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]4  
On the other hand, we define the function  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]5  
From the definitions of \(\omega\left(t\right),L\) and \(U\), we see that  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]6  
Now, let \(\varepsilon>0\), then from the definition of \(P\) and \(L\), we can pick \(t_{3}\geq T\) sufficiently large such that  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]7  
By integrating (16) from \(t\) to \(\infty\) and using (17), we have  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]8  
Multiplying the latter inequality by \(\left(\widetilde{R}\left(g\left(t\right),T\right)\right)^{\beta}\), we obtain  
Taking the limit superior on both sides as \(t\rightarrow\infty\) and using (18), we get  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]9  
Thus, from (19), we arrive at  
\[z\left(t\right)\geq\frac{1}{2}\left(t-t_{1}\right)z^{\prime}\left(t\right). \tag{4}\]0  
which completes the proof. **Theorem 1**. : _Assume that \(r_{i}^{\prime}\left(t\right)>0\) for \(i=1,2\), and \(x\) is a positive solution of (1). If_  
\[z\left(t\right)\geq\frac{1}{2}\left(t-t_{1}\right)z^{\prime}\left(t\right). \tag{4}\]1  
_then the class \(\widetilde{\aleph}\) is empty._  
Proof. : Let \(x\) be a positive solution of (1) and \(z\in\widetilde{\aleph}\). First, let \(P=\infty\). As in the proof of Lemma 6, we obtain that (19) and (20). Then, from (20), we have  
\[z\left(t\right)\geq\frac{1}{2}\left(t-t_{1}\right)z^{\prime}\left(t\right). \tag{4}\]2  
Taking the limit inferior as \(t\rightarrow\infty\) and using (19), we get  
\[z\left(t\right)\geq\frac{1}{2}\left(t-t_{1}\right)z^{\prime}\left(t\right). \tag{4}\]3  
this is a contradiction. On the other hand, let \(P<\infty\). From Lemma 6, we have \(P\leq L-L^{\frac{1+\beta}{P}}\).