Oscillation criteria for second order Emden-Fowler functional differential equations of neutral type

Suppose that \((A_{1})\)-\((A_{4})\) and (3) hold. If there exists a function \(\rho(t)\in C^{1}([t_{0},\infty), (0,\infty))\) such that where Then equation (1) is oscillatory for all \(\alpha> 0\) and \(\beta> 0\).

Suppose that equation (1) has a nonoscillatory solution \(x(t)\). Without loss of generality, we assume that \(x(t)>0\) for all large t. The case of \(x(t)<0\) can be treated by the same method. In view of \((A_{3})\) and \((A_{4})\), there exists \(t_{1}\geq t_{0}\) such that \(x(t)>0\), \(x(\tau(t))>0\), \(x(\delta(t))>0\) on \([t_{1},\infty)\). It follows that \(z(t)=x(t)+p(t)x(\tau(t))\geq x(t)>0\). It follows from (1) that Hence, \(r(t)\vert z^{\prime}(t)\vert ^{\alpha-1}z^{\prime}(t)\) is nonincreasing on \([t_{1},\infty)\).

We now claim that If not, then there exists \(t_{3}\in[t_{2},\infty)\) such that \(z^{\prime}(t_{3})<0\). Hence which implies that Integrating (11) from \(t_{3}\) to t, we find from (3) that which implies that \(z(t)\) is eventually negative. This contradicts \(z(t)>0\). Hence our claim is true.

Now we have This inequality together with (1) and (10) suggest

We set Then \(w(t)>0\). By (13) and (14) we have In the following we consider three cases for (15):

Case (i): \(\alpha=\beta\). In view of the inequality \(r^{\frac{1}{\alpha }}(t)z^{\prime}(t)\leq r^{\frac{1}{\alpha}} (\sigma(t) )z^{\prime } (\sigma(t) )\) and (15) we see that

Case (ii): \(\alpha<\beta\). Noting that \(z(\sigma(t))\) is increasing on \([T,\infty)\), then there exists a constant \(m_{1}>0\) such that

Case (iii): \(\alpha>\beta\). From \((r(t) (z^{\prime}(t) )^{\alpha} )^{\prime}\leq0\) and \(r^{\prime}(t)\geq0\), we get \(z^{\prime\prime}(t)\leq0\), then \(z^{\prime}(t)\) is nonincreasing. Thus, there exists a positive constant \(m_{2}\), such that Combining (16)-(18), we obtain for any \(\alpha> 0\), \(\beta>0\), Multiplying (19) by \(\rho(t)\) and integrating it from T to t, we obtain By the inequality where \(A\geq0\), \(B>0\), \(w\geq0\), and \(\lambda>0\), we now can rewrite inequality (20) as Letting \(t\rightarrow\infty\) in the above inequality, we get a contradiction with (6). Hence the theorem is proved. □

Theorems 1-5 of [1], Theorem 1 of [2] and [7] hold only for equation (1) with \(p(t)=0\) and \(\alpha =\beta\). Theorem 2.1 of [5] (or [6]) holds only for equation (1) with \(\alpha\geq\beta\), and Theorem 3.1 of [6] holds only for equation (1) with \(\beta\geq\alpha\). Hence our theorem improves and unifies the above results.

Suppose that \((A_{1})\)-\((A_{4})\) and (3) hold. Suppose there exist functions H, h, and ρ, such that \((H_{1})\), \((H_{2})\), and \((H_{3})\) hold. Further assume for all sufficiently large T, where λ, m, \(Q_{1}(t)\), and \(\lambda(t)\) are given in (7) and (8). Then equation (1) is oscillatory for all \(\alpha >0\) and \(\beta>0\).

Similar to Theorem 1, we assume that there exists a solution x of equation (1) such that \(x(t)>0\) on \([t_{1},\infty )\) for some \(t_{1}\geq t_{0}\). Multiplying both sides of (19) by \(H(t,s)\rho(s)\) and integrating from T to t, we have, for all \(t\geq T\geq t_{1}\), where w is defined by (14) and Applying integration by parts, from (\(H_{3}\)) and (24) we have Using the inequality (21), combining (26) and (25), we get It follows that which contradicts the assumption (23). Therefore, equation (1) is oscillatory. Now we finish the proof of this theorem. □

Theorem  2 remains true if the condition (23) is replaced by and

Theorem 2.1 of [3] holds only for equation (1) with \(\alpha=1\) and \(\beta>1\), Theorem 2.2 of [5] holds only for equation (1) with \(\alpha\geq\beta\), Theorem 5 of [7] holds only for equation (1) with \(\beta\geq\alpha\). Hence, Theorem 2 improves and unifies above oscillation criteria.

Suppose that \((A_{1})\)-\((A_{4})\) and (4) hold. Suppose and (6) are satisfied. Further assume there exists a constant \(K>0\) such that where \(\mu=\max\{\alpha,\beta\}\). Then equation (1) is oscillatory for all \(\alpha>0\) and \(\beta>0\).

As in Theorem 1 we assume that there exists a solution x of equation (1) such that \(x(t)>0\) on \([t_{1},\infty)\) for some \(t_{1}\geq t_{0}\). Then we have from which we see that there exist two possible cases of the sign of \(z^{\prime}(t)\). If \(z^{\prime}(t)>0\), then we come back to the proof of Theorem 1, and we can get a contradiction with (6). If \(z^{\prime}(t)<0\), we have Therefore, \(x^{\prime}(t)<0\) and from which together with (32) we have Then equation (1) becomes Now we define a function v by Obviously, \(v(t)>0\) for \(t\geq t_{1}\). It follows from (34) that \(r(t)\vert z^{\prime}(t)\vert ^{\alpha-1}z^{\prime}(t)\) is nonincreasing. Hence we get Dividing the above inequality by \(r^{\frac{1}{\alpha}}(t)\) and integrating it from t to l, we have It follows that Moreover, we have By (38) and the fact \(z^{\prime}(t)<0\) we find that there exists a constant \(c_{1}>0\) such that On the other hand, from (39) we get Hence we have Since \(r^{\frac{1}{\alpha}}(t)(-z^{\prime}(t))\) is nondecreasing, then there exists a constant \(c_{2}>0\) such that Next, differentiating (38) yields We consider the following three cases:

Case (i): \(\alpha> \beta\). In this case, since \(z(t)\) is decreasing, it follows from (43) that where \(c_{1}= \beta [z(t_{1}) ]^{\frac{\beta-\alpha}{\alpha}}\).

Case (ii): \(\alpha=\beta\). In this case, we see that \([z(t)]^{\frac{\beta-\alpha}{\alpha}}=1\), then (43) becomes

Case (iii): \(\alpha<\beta\). By the inequality (37) we have \((r(t)(-z^{\prime}(t))^{\alpha})^{\prime}\geq0\), from which together with \(r^{\prime}(t)\geq0\) we find that \(z^{\prime\prime}(t)\leq0\). Hence we get \(z^{\prime}(t)\leq z^{\prime}(t_{2})\) for \(t\geq t_{2}\). Now the inequality (43) suggests that where \(c_{2}=\beta[-z^{\prime}(t_{2})]^{\frac{\beta-\alpha}{\beta}}\).

Combining (44)-(46), we obtain where \(\mu=\max\{{\alpha,\beta\}}\) and Multiplying (47) by \(\pi^{\mu}(t)\) and integrating it from \(t_{2}\) to t, we have Using integration by parts, the inequality (48) yields By the inequality (21), we get Substituting in (49), we obtain where \(K=\frac{\mu^{2\mu+1} }{c^{\mu}(\mu+1)^{\mu+1}}\).

In view of (41) and (42), we have which contradicts condition (33). Then equation (1) is oscillatory for all \(\alpha>0\) and \(\beta>0\). Hence the theorem is proved. □

Theorem 2.2 of [2] holds only for equation (1) with \(p(t)=0\) and \(\alpha=\beta\), Theorem 2.1-2.3 of [4] hold only for \(\alpha=1\) and \(\beta\geq1\), Theorem 2.5 of [5] and Theorem 2.3 of [6] hold only for \(\alpha\geq\beta\). Our Theorem 3 holds for equation (1) with all \(\alpha>0\) and \(\beta>0\).

Consider the differential equation where \(z(t)=x(t)+\frac{1}{2}x(t-1)\), \(\alpha>0\), \(\beta>0\), and \(\lambda=\min \{\alpha,\beta\}\). Noticing that \(r(t)=1\), \(p(t)=\frac{1}{2}\), \(q(t)=\frac{1}{t^{1+\frac {\lambda}{2}}}\), \(\tau(t)=t-1\), \(\sigma(t)=t-2\) and then (3) is satisfied. Since \(Q_{1}(t)=(\frac{1}{2})^{\beta}\frac {1}{t^{1+\frac{\lambda}{2}}}\), \(\lambda>0\), then \(\int^{\infty }_{t_{0}}Q_{1}(t)\,dt<\infty\). In order to apply Theorem 1, it remains to discuss condition (6). If we choose \(\rho(t)=t^{\frac{\lambda}{2}}\), we have Then by Theorem 1, every solution of (52) is oscillatory for all \(\alpha>0\) and \(\beta>0\).

Consider the differential equation where \(z(t)=x(t)+\frac{1}{3}x(t-2)\), \(\alpha>0\), \(\beta>0\). Observe \(r(t)=t^{2\alpha}\), \(p(t)=\frac{1}{3}\), \(q(t)=t^{2\alpha+\beta}\), \(\tau(t)=t-2\), \(\sigma(t)=t-3\) and Then (4) is satisfied. It is clear that (32) is satisfied. Since \(p(t)=\frac{1}{3}\), we have If we choose \(\rho(t)=1\), then condition (6) is satisfied. To apply Theorem 3, it remains to discuss the condition (33); in view of \(\pi(t)=\frac{1}{t}\), we have Then by Theorem 3, (53) is oscillatory for all \(\alpha >0\) and \(\beta>0\).

We note that the results obtained for those equations in [1‐20] cannot deal with (52) and (53).