**4. Criterion of an Iterative Nature**

In this section, we create a criterion of an iterative nature that ensures that the solutions to Equation (4) oscillate, when *F*(*x*) = *x*.

**Lemma 8.** *Suppose that x* ∈ P*<sup>s</sup> and ω satisfies case* (D3) *in Lemma 1. If (8) holds, then the function <sup>ω</sup>*/<sup>E</sup> *<sup>α</sup> <sup>n</sup>*−<sup>2</sup> *is decreasing and also converges to zero as u* <sup>→</sup> <sup>∞</sup>*.*

**Proof.** From Lemma 4, we have that (7) holds for *u* ≥ *u*<sup>1</sup> ≥ *u*0. Integrating (7) from *u*<sup>1</sup> to *u* and using (8), we get

$$\begin{split} a(u)\omega^{(n-1)}(u) &\quad \leq &\quad a(u\_1)\omega^{(n-1)}(u\_1) - \int\_{u\_1}^{u} \mathcal{Q}(\upsilon)\omega(\mathcal{J}(\upsilon))d\upsilon\\ &\quad \leq &\quad a(u\_1)\omega^{(n-1)}(u\_1) - a\omega(u)\int\_{u\_1}^{u} \frac{\mathcal{E}\_{n-3}(\upsilon)}{\mathcal{E}\_{n-2}^2(\upsilon)}d\upsilon\\ &=&\quad a(u\_1)\omega^{(n-1)}(u\_1) - a\omega(u)\left(\frac{1}{\mathcal{E}\_{n-2}(u)} - \frac{1}{\mathcal{E}\_{n-2}(u\_1)}\right). \end{split} \tag{23}$$

It follows from Lemma 5 that *ω*(*u*) → 0 as *u* → ∞. Thus, there is a *u*<sup>2</sup> ≥ *u*<sup>1</sup> such that *<sup>a</sup>*(*u*1)*ω*(*n*−<sup>1</sup>)(*u*1) <sup>+</sup> *αω*(*u*) <sup>E</sup>*n*−2(*u*1) <sup>≤</sup> 0 for *<sup>u</sup>* <sup>≥</sup> *<sup>u</sup>*2. Hence, (23) becomes

$$
\omega(u)\omega^{(n-1)}(u) \le -\frac{\mathfrak{a}}{\mathcal{E}\_{n-2}(u)}\omega(u). \tag{24}
$$

Using Lemma <sup>2</sup> with *<sup>s</sup>* <sup>=</sup> 1, we get *<sup>ω</sup>* <sup>≤</sup> *<sup>a</sup>* · *<sup>ω</sup>*(*n*−1) · E*n*−3, which with (24) yields

$$\frac{\omega^{\prime}(u)}{\mathcal{E}\_{n-3}(u)} \le a(u)\omega^{(n-1)}(u) \le -\frac{\alpha}{\mathcal{E}\_{n-2}(u)}\omega(u). \tag{25}$$

Hence,

$$\frac{\mathbf{d}}{\mathbf{d}\boldsymbol{\omega}} \left( \frac{\boldsymbol{\omega}}{\mathcal{E}\_{n-2}^{\boldsymbol{\alpha}}} \right) = \frac{1}{\mathcal{E}\_{n-2}^{\boldsymbol{\alpha}+1}} \left[ \mathcal{E}\_{n-2} \boldsymbol{\omega}^{\boldsymbol{\prime}} + \boldsymbol{\kappa} \mathcal{E}\_{n-3} \boldsymbol{\omega} \right] \le 0.1$$

Now, we have that *<sup>ω</sup>*/<sup>E</sup> *<sup>α</sup> <sup>n</sup>*−<sup>2</sup> is positive and decreasing. Then, *<sup>ω</sup>*/<sup>E</sup> *<sup>α</sup> <sup>n</sup>*−<sup>2</sup> <sup>→</sup> as *u* → ∞, where -≥ 0.

Suppose that -> 0. Hence,

$$\frac{\omega(\mu)}{\mathcal{E}\_{n-2}^a(\mu)} \ge \ell. \tag{26}$$

Next, we define

$$\phi := \frac{a \cdot \omega^{(n-1)} \cdot \mathcal{E}\_{n-2} + \omega}{\mathcal{E}\_{n-2}^a}, 0$$

and so

$$\begin{split} \boldsymbol{\phi}^{\prime} &= \quad \frac{\left(\boldsymbol{a}\cdot\boldsymbol{\omega}^{(n-1)}\right)^{\prime}}{\mathcal{E}\_{n-2}^{n-1}} - \frac{1}{\mathcal{E}\_{n-2}^{a}} \Big(\boldsymbol{a}\cdot\boldsymbol{\omega}^{(n-1)}\cdot\mathcal{E}\_{n-3} - \boldsymbol{\omega}^{\prime}\Big) + \boldsymbol{a}\frac{\boldsymbol{a}\cdot\boldsymbol{\omega}^{(n-1)}\cdot\mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}^{a}} + \boldsymbol{a}\frac{\mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}^{a+1}}\cdot\boldsymbol{\omega}^{\prime} \\ &\leq \quad -\frac{1}{\mathcal{E}\_{n-2}^{a-1}}\cdot\mathcal{Q}\cdot\left[\boldsymbol{\omega}\circ\boldsymbol{g}\right] + \boldsymbol{a}\frac{\boldsymbol{a}\cdot\boldsymbol{\omega}^{(n-1)}\cdot\mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}^{a}} + \boldsymbol{a}\frac{\mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}^{a+1}}\cdot\boldsymbol{\omega} .\end{split}$$

Thus, from (8), we get

$$\begin{split} \phi' &\leq \quad -a \frac{\mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}^{a+1}} \cdot [\omega \circ \emptyset] + a \frac{a \cdot \omega^{(n-1)} \cdot \mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}^{a}} + a \frac{\mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}^{a+1}} \cdot \omega^{n} \\ &\leq \quad a \frac{a \cdot \omega^{(n-1)} \cdot \mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}^{a}} .\end{split}$$

It follows from (25) and (26) that

$$\begin{array}{rcl} \phi' & \leq & -\alpha^2 \frac{\mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}^{n+1}} \cdot \omega(\mu), \\ & \leq & -\ell \alpha^2 \frac{\mathcal{E}\_{n-3}}{\mathcal{E}\_{n-2}} < 0. \end{array}$$

Integrating this inequality from *u*<sup>1</sup> to *u*, we find

$$\phi(u\_1) \ge \ell a^2 \ln \frac{\mathcal{E}\_{n-2}(u\_1)}{\mathcal{E}\_{n-2}(u)} \nu$$

which leads to a contradiction. Then, -= 0. The proof is complete. **Lemma 9.** *Assume that (8) holds. If*

$$\begin{split} \limsup\_{u \to \infty} \left[ \mathcal{E}\_{n-2}(\mathcal{g}(u)) F(\mathcal{E}\_{n-2}^{-\
u}(\mathcal{g}(u))) \int\_{u\_1}^{\mathcal{G}(u)} \mathcal{Q}(\upsilon) F(\mathcal{E}\_{n-2}^{\
u}(\mathcal{g}(\upsilon))) \mathrm{d}\upsilon \\ + F(\mathcal{E}\_{n-2}^{-\
u}(\mathcal{g}(u))) \int\_{\mathcal{G}(u)}^{u} \mathcal{E}\_{n-2}(\upsilon) \mathcal{Q}(\upsilon) F(\mathcal{E}\_{n-2}^{\
u}(\mathcal{g}(\upsilon))) \mathrm{d}\upsilon \\ + F(\mathcal{E}\_{n-2}^{-1}(\mathcal{g}(u))) \int\_{u}^{\infty} \mathcal{E}\_{n-2}(\upsilon) \mathcal{Q}(\upsilon) F(\mathcal{E}\_{n-2}(\mathcal{g}(\upsilon))) \mathrm{d}\upsilon \right] > K, \end{split} \tag{27}$$

*then ω satisfies case* (D2) *in Lemma 1.*

**Proof.** Proceeding as in the proof of Lemma 6, we get that (16) and (17) hold. Using Lemma 8, we get

$$
\omega(\mathcal{g}(\upsilon)) \ge \frac{\mathcal{E}\_{n-2}^{\mathfrak{a}}(\mathcal{g}(\upsilon))}{\mathcal{E}\_{n-2}^{\mathfrak{a}}(\mathcal{g}(\mathfrak{u}))} \omega(\mathcal{g}(\mathfrak{u})) \text{ for } \upsilon \le \mathfrak{u}. \tag{28}
$$

From (17) and (28), (16) becomes

$$\begin{split} \frac{\omega(\mathcal{g}(u))}{F(\omega(\mathcal{g}(u)))} &\geq \ \mathcal{E}\_{n-2}(\mathcal{g}(u))F(\mathcal{E}\_{n-2}^{-\mathfrak{a}}(\mathcal{g}(u))) \int\_{u\_{1}}^{\mathcal{G}(u)} \mathcal{Q}(\upsilon)F(\mathcal{E}\_{n-2}^{\mathfrak{a}}(\mathcal{g}(\upsilon))) \mathrm{d}\upsilon \\ &\quad + F(\mathcal{E}\_{n-2}^{-\mathfrak{a}}(\mathcal{g}(u))) \int\_{\mathcal{G}(u)}^{u} \mathcal{E}\_{n-2}(\upsilon)\,\mathcal{Q}(\upsilon)F(\mathcal{E}\_{n-2}^{\mathfrak{a}}(\mathcal{g}(\upsilon))) \mathrm{d}\upsilon \\ &\quad + F(\mathcal{E}\_{n-2}^{-1}(\mathcal{g}(\mu))) \int\_{u}^{\infty} \mathcal{E}\_{n-2}(\upsilon)\,\mathcal{Q}(\upsilon)F(\mathcal{E}\_{n-2}(\mathcal{g}(\upsilon))) \mathrm{d}\upsilon, \end{split}$$

which contradicts (27).

**Theorem 4.** *Assume that (8), (22) and (27) hold. Then, Equation (4) is oscillatory.*

It is also possible to continue to improve the monotonic property of the function *ω*/E*n*−<sup>2</sup> and then use it in the oscillation criteria.

**Notation 3.** *Since* E*n*−2(*u*) *is decreasing, there is a* > 1 *such that*

$$\frac{\left[\mathcal{E}\_{n-2}\circ\mathcal{g}\right]}{\mathcal{E}\_{n-2}}\geq\epsilon.$$

*Let α*<sup>0</sup> ∈ (0, 1)*, we define α*<sup>0</sup> = *α*

$$\mathfrak{a}\_{k+1} := \mathfrak{a}\_0 \frac{\epsilon^{a\_k}}{1 - a\_k} \prime$$

*for k* = 0, 1, . . . *.*

**Lemma 10.** *Suppose that x* ∈ P*<sup>s</sup> and ω satisfies case* (D3) *in Lemma 1. Suppose also that there is <sup>m</sup>* <sup>∈</sup> <sup>N</sup> *such that <sup>α</sup><sup>k</sup>* <sup>∈</sup> (0, 1) *and <sup>α</sup>k*−<sup>1</sup> <sup>&</sup>lt; *<sup>α</sup><sup>k</sup> for <sup>k</sup>* <sup>=</sup> 0, 1, ... , *m. If (8) holds, then the function <sup>ω</sup>*/E*<sup>m</sup> <sup>n</sup>*−<sup>2</sup> *is decreasing, and also converges to zero as u* <sup>→</sup> <sup>∞</sup>*.*

**Proof.** From Lemma 8, we have that *<sup>ω</sup>*/<sup>E</sup> *<sup>α</sup> <sup>n</sup>*−<sup>2</sup> is decreasing and also converges to zero as *u* → ∞. We will prove the required when *m* = 1.

From Lemma 4, we have that (7) holds for *u* ≥ *u*<sup>1</sup> ≥ *u*0. Integrating (7) from *u*<sup>1</sup> to *u* and using the fact that *<sup>ω</sup>*/<sup>E</sup> *<sup>α</sup> <sup>n</sup>*−<sup>2</sup> is decreasing, we get

*a*(*u*)*ω*(*n*−1) (*u*) <sup>≤</sup> *<sup>a</sup>*(*u*1)*ω*(*n*−1) (*u*1) − *u u*1 Q(*υ*)*ω*(*g*(*υ*))d*υ* <sup>≤</sup> *<sup>a</sup>*(*u*1)*ω*(*n*−1) (*u*1) − *u u*1 Q(*υ*) E *α <sup>n</sup>*−2(*g*(*υ*)) E *α <sup>n</sup>*−2(*υ*) *<sup>ω</sup>*(*υ*)d*<sup>υ</sup>* <sup>≤</sup> *<sup>a</sup>*(*u*1)*ω*(*n*−1) (*u*1) <sup>−</sup> *<sup>ω</sup>*(*u*) E *α <sup>n</sup>*−2(*u*) *u u*1 <sup>Q</sup>(*υ*)<sup>E</sup> *<sup>α</sup> <sup>n</sup>*−2(*g*(*υ*))d*<sup>υ</sup>* <sup>≤</sup> *<sup>a</sup>*(*u*1)*ω*(*n*−1) (*u*1) <sup>−</sup> *<sup>ω</sup>*(*u*) E *α <sup>n</sup>*−2(*u*) *u u*1 *α*E*n*−3(*υ*) <sup>E</sup>2−*<sup>α</sup> <sup>n</sup>*−<sup>2</sup> (*υ*) E *α <sup>n</sup>*−2(*g*(*υ*)) E *α <sup>n</sup>*−2(*υ*) <sup>d</sup>*<sup>υ</sup>* <sup>≤</sup> *<sup>a</sup>*(*u*1)*ω*(*n*−1) (*u*1) <sup>−</sup> *α<sup>α</sup> <sup>ω</sup>*(*u*) E *α <sup>n</sup>*−2(*u*) *u u*1 E*n*−3(*υ*) <sup>E</sup>2−*<sup>α</sup> <sup>n</sup>*−<sup>2</sup> (*υ*) d*υ* <sup>≤</sup> *<sup>a</sup>*(*u*1)*ω*(*n*−1) (*u*1) <sup>−</sup> *α<sup>α</sup>* 1 − *α ω*(*u*) E *α <sup>n</sup>*−2(*u*) 1 <sup>E</sup>1−*<sup>α</sup> <sup>n</sup>*−<sup>2</sup> (*u*) <sup>−</sup> <sup>1</sup> <sup>E</sup>1−*<sup>α</sup> <sup>n</sup>*−<sup>2</sup> (*u*1) . (29)

It follows from Lemma 5 that *<sup>ω</sup>*(*u*) E *α <sup>n</sup>*−2(*u*) <sup>→</sup> 0 as *<sup>u</sup>* <sup>→</sup> <sup>∞</sup>. Hence, (29) becomes

$$a(\mu)\omega^{(\mathfrak{n}-1)}(\mu) \le -\frac{\mathfrak{a}\mathfrak{e}^{\mathfrak{n}}}{1-\mathfrak{a}}\frac{\omega(\mathfrak{u})}{\mathcal{E}\_{\mathfrak{n}-2}(\mathfrak{u})}.$$

The remainder of the proof has not been considered because it is identical to the proof of Lemma 8.

**Theorem 5.** *Assume that (8) and (22) hold. If there is <sup>m</sup>* <sup>∈</sup> <sup>N</sup> *such that <sup>α</sup><sup>k</sup>* <sup>∈</sup> (0, 1) *and <sup>α</sup>k*−<sup>1</sup> <sup>&</sup>lt; *<sup>α</sup><sup>k</sup> for k* = 0, 1, . . . , *m, and*

$$\begin{split} \limsup\_{u \to \infty} \left[ \mathcal{E}\_{n-2}(\mathcal{g}(u)) F \left( \mathcal{E}\_{n-2}^{-a\_m}(\mathcal{g}(u)) \right) \int\_{u\_1}^{\mathcal{G}(u)} \mathcal{Q}(\upsilon) F \left( \mathcal{E}\_{n-2}^{a\_m}(\mathcal{g}(\upsilon)) \right) d\upsilon \\ &+ F \left( \mathcal{E}\_{n-2}^{-a\_m}(\mathcal{g}(u)) \right) \int\_{\mathcal{S}(u)}^u \mathcal{E}\_{n-2}(\upsilon) \mathcal{Q}(\upsilon) F \left( \mathcal{E}\_{n-2}^{a\_m}(\mathcal{g}(\upsilon)) \right) d\upsilon \\ &+ F \left( \mathcal{E}\_{n-2}^{-1}(\mathcal{g}(u)) \right) \int\_u^\infty \mathcal{E}\_{n-2}(\upsilon) \mathcal{Q}(\upsilon) F (\mathcal{E}\_{n-2}(\mathcal{g}(\upsilon))) d\upsilon \right] > K, \end{split}$$

*then Equation (4) is oscillatory.*
