Abstract
In this paper, we obtain necessary and sufficient conditions for oscillation of a fourth order dynamic equation on time scales with deviating arguments. We discuss the oscillation behavior of solutions for strongly superlinear and strongly sublinear cases of the dynamic equation at hand. Our results unify and improve some known results for dynamic equations on time scales.
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1 Introduction
The topic of oscillation and stability of dynamic equations on time scales has been developed very rapidly in the past two decades. There are some excellent monographs [1,2,3,4] and papers [5,6,7,8,9,10] containing some interesting works in the field.
The oscillatory behavior of solutions for nonlinear fourth order functional differential equations of the form
has been discussed by Onose [11], where g is superlinear (sublinear) and strongly superlinear (strongly sublinear); he has extended and improved some interesting results of Kusano and Naito [12]. Furthermore, Gopalsamy et al. [13] obtained the sufficient and necessary conditions for oscillation of a fourth order differential equation with multiple deviating arguments given by
where g is strongly superlinear and strongly sublinear.
For some more results on oscillation of solutions for different kinds of fourth order equations on time scales, see [14,15,16,17,18,19,20,21] and the references cited therein. However, it has been observed that there is no work in the related literature concerning the sufficient and necessary conditions for oscillation of fourth order dynamic equations on time scales. Motivated by the aforementioned works, in this paper, we consider the following fourth order dynamic equation with deviating arguments:
where \(y^{\Delta}(t)\) is the delta (or Hilger) derivative of y at t, \(r\in C_{rd}(\mathbb{T}_{0}, \mathbb{R}^{+})\), \(\eta\in C_{rd}(\mathbb{T}_{0}, \mathbb{T})\), \(g: \mathbb {T}_{0}\times\mathbb{R}\to\mathbb{R}\) is a nonlinear continuous function, and \(\operatorname{sgn}g(t,y)=\operatorname{sgn} y\) for \(t\in\mathbb{T}_{0}\). In relation to (1), it is also assumed that \(\int_{t_{0}}^{\infty}t/r(t)\Delta t=\infty\).
The paper is organized as follows. In Sect. 2, we recall some basic concepts of dynamic equations on time scales. In Sect. 3, we establish necessary and sufficient criteria for oscillation of (1) when g is strongly superlinear as well as strongly sublinear.
2 Preliminaries
A time scale \(\mathbb{T}\) is a nonempty closed subset of the real numbers \(\mathbb{R}\) with \(\sup\mathbb{T}=\infty\). For example, \(\mathbb{R}\), \(h\mathbb{Z}\) for \(h>0\) and \(q^{\mathbb{N}}:=\{ q^{k},k\in\mathbb{N}\}\) for \(q>1\) are time scales. In the forthcoming analysis, we assume that \(\mathbb{T}\) has the topology that it inherits from the standard topology on \(\mathbb{R}\). Let the closed interval in \(\mathbb{T}\) be defined by \([c,d]:= \{t\in\mathbb {T}, c\leq t \leq d\}\). In a similar manner, one can define open intervals and half-open intervals, etc.
Definition 2.1
For \(t\in\mathbb{T}\) we define the forward jump operator \(\sigma:\mathbb{T}\rightarrow\mathbb{T}\) by \(\sigma (t):=\inf\{s\in\mathbb{T}:s > t\}\); the backward jump operator \(\rho: \mathbb{T}\rightarrow\mathbb{T}\) is defined by \(\rho(t):=\sup\{s\in\mathbb{T}:s < t\}\). If \(\sigma (t)>t\), then t is called right-scattered, while if \(\rho (t)< t\), it is called left-scattered. Also, if \(t<\sup\mathbb {T}\) and \(\sigma(t)=t\), then t is called right-dense, and if \(t>\inf\mathbb{T}\) and \(\rho(t)=t\), then t is called left-dense. The graininess function \(\mu(t):\mathbb{T}\to[0,\infty)\) is defined by \(\mu(t):=\sigma(t)-t\).
Definition 2.2
A function \(f:\mathbb {T}\rightarrow \mathbb{R}\) is rd-continuous if it is continuous at all right-dense points and its left-sided limit exists (and is finite) at a left-dense point. We denote the set of rd-continuous functions by \(C_{rd}(\mathbb{T}, \mathbb{R})\).
Definition 2.3
For a function \(f: \mathbb{T}\to \mathbb{R}\), let \(F^{\Delta}(t)\) represent the Hilger derivative of f at t. Assume that \(t_{0}\in\mathbb{T}\) and \(f\in C_{rd}(\mathbb {T}_{0}, \mathbb{R})\). If \(F^{\Delta}(t)=f(t)\), then we define
Definition 2.4
We say g is strongly superlinear if there exists a constant \(\alpha>1\) such that
while g is strongly sublinear if there exists a constant \(\alpha\in (0,1)\) such that
3 Main results
In the sequel, we use the following notations:
Lemma 3.1
If \(y(t)\) is a nonoscillatory solution of (1), then there are only four cases for all sufficiently large \(t\geq t_{0}\):
Proof
Without loss of generality, let \(y(t)\) be an eventually positive solution of (1), that is, there exists \(t_{1}\geq t_{0}\) such that \(y(t) > 0\) for \(t\geq t_{1}\). Then \(y(\eta(t)) > 0\) for \(t\geq t_{1}\). From (1), it yields that \([r(t)y^{\Delta \Delta}(t)]^{\Delta\Delta}<0\) for \(t\geq t_{1}\). Therefore, \([r(t)y^{\Delta\Delta}(t)]^{\Delta}\) is eventually of constant sign. Next we suppose that \([r(t)y^{\Delta\Delta }(t)]^{\Delta}<0\) at some \(t=t_{2}\geq t_{1}\). Then, integrating \([r(t)y^{\Delta\Delta}(t)]^{\Delta\Delta}<0\) twice from \(t_{2}\) to t, and multiplying the resulting inequality by \(1/r(t)\) and integrating again from \(t_{2}\) to t, we get
where \(\bar{a}=[r(t_{2})y^{\Delta\Delta}(t_{2})]^{\Delta}<0\), \(\bar {b}=r(t_{2})y^{\Delta\Delta}(t_{2})\), and \(\bar{c}=y^{\Delta}(t_{2})\). In consequence, it follows from the assumption \(\int_{t_{0}}^{\infty}s/r(s)\Delta s=\infty\) that \(\lim_{t\to\infty}y^{\Delta}(t)=-\infty \), which contradicts the positivity of \(y(t)\). Therefore, we have \([r(t)y^{\Delta\Delta}(t)]^{\Delta}>0\) for all \(t\geq t_{1}\). It means that \(r(t)y^{\Delta\Delta}(t)\) eventually keeps the same sign. On the other hand, let \(r(t)y^{\Delta\Delta}(t)<0\) for \(t \geq t_{1}\). Then it can easily be shown that \(y^{\Delta}(t)\) is eventually positive. This completes the proof of (i). If there exists \(t_{2}\geq t_{1}\) such that \(r(t)y^{\Delta\Delta}(t)>0\) for \(t\geq t_{2}\), then \(r(t)y^{\Delta\Delta}(t)\geq c\) for \(t\geq t_{2}\), where \(c= r(t_{2})y^{\Delta}(t_{2})\). Multiplying this inequality by \(t/r(t)\) and integrating from \(t_{2}\) to t, by using the integration by parts formula on time scales, we get
which, together with \(\int_{t_{0}}^{\infty}s/r(s)\Delta s =\infty\), implies that \(\lim_{t\to\infty}ty^{\Delta}(t) = \infty\). Thus \(y^{\Delta}(t) > 0\) for all large \(t\geq t_{0}\). The proof is completed. □
Lemma 3.2
If \(t_{1}\geq t_{0}\) and \(t>u\), then \(\lim_{t\to\infty }\frac{R_{t_{1}}(t)}{R(t)}=1\) and \(\widetilde{\mathbf{R}}_{u}(t)\) is nonincreasing for u, where \(R_{t_{1}}(t)\), \(R(t)\), \(\widetilde{\mathbf{R}}_{u}(t)\) are given in (2).
Proof
By applying L’Hôpital’s rule [1, Theorem 1.120], we find that
On the other hand, let
Then, using [1, Theorem 1.117] for all u, we obtain
This completes the proof. □
Lemma 3.3
If \(y(t)\) is a nonoscillatory solution of (1), then there exist \(T> t_{0}\) and a constant \(\tilde{c} > 0\) such that
Proof
Without loss of generality, we suppose that \(y(t)\) is eventually positive. Then, in view of Lemma 3.1, there exists \(t_{1}\geq t_{0}\) such that
Integrating \([r(t)y^{\Delta\Delta}(t)]^{\Delta\Delta}<0\) twice from \(t_{1}\) to t, we have
Multiplying (4) by \(1/r(t)\) and integrating twice from \(t_{1}\) to t yields
where \(a_{2}=y^{\Delta}(t_{1})\) and \(a_{3}=y(t_{1})\) are constants. Noting that \(\int_{t_{0}}^{\infty}s/r(s)\Delta s=\infty\), we deduce that there exist \(t_{2}\geq t_{1}\) and \(\tilde{c} > 0\) such that \(y(t)< \tilde{c} R(t)\) for \(t\geq t_{2}\).
Now let us prove the left-sided inequality in the lemma. In view of (3), observe that \([r(t)y^{\Delta\Delta}(t)]^{\Delta}\) is nonincreasing, and hence
From Lemma 3.2, there exists \(t_{3}\geq t_{1}\) such that \(R_{t_{1}}(t)\geq\frac{1}{2} R(t)\) for \(t\geq t_{3}\), and hence
Letting \(T=\max(t_{2}, t_{3})\), the proof is complete. □
Lemma 3.4
If \(y(t)\) is a nonoscillatory solution of (1), then there exists \(t^{*}\geq t_{0}\) such that, for any \(T\geq t^{*}\),
Proof
Without loss of generality, we suppose that \(y(t)\) is eventually positive. Firstly, if \(y(t)\) is a solution of type-(i), then there exists \(t_{1}\geq t_{0}\) such that
Let
Obviously, \(h^{\Delta\Delta}(t)>0\) for \(t\geq t_{1}\). Indeed, differentiating the above equation twice, we get
In view of (1), we obtain \([r(t)h^{\Delta\Delta}(t)]^{\Delta \Delta}=0\) and hence \([r(t)h^{\Delta\Delta}(t)]^{\Delta}=c\). Integrating (1) from t to T, we have
In the limit \(T\to\infty\), we note that \([r(t)h^{\Delta\Delta }(t)]^{\Delta}>0\) for \(t\geq t_{1}\). Then there exists \(c>0\) such that \(h^{\Delta\Delta}(t)=ct/r(t)>0\) for \(t\geq t_{1}\), which, on integrating from \(t_{1}\) to t, yields
Taking the limit \(t\to\infty\) and using the assumption \(\int _{t_{0}}^{\infty}s/r(s)\Delta s=\infty\) in (6), we get \(h^{\Delta }(t)>0\) for all large t. Therefore, there exists \(t_{2}\geq t_{1}\) such that \(h^{\Delta}(t)>c\int_{t_{1}}^{t} s/r(s)\Delta s\) for \(t\geq t_{2}\). Next, integrating (6) from \(t_{1}\) to t, we get
which implies that \(h(t)>0\) for large values of t (i.e., \(t\to\infty\)). Thus, there exists \(T\geq t_{2}\) such that
Now, by interchanging the order of integration, we get
On the other hand, if \(y(t)\) is a solution of type-(ii), then there exists \(t_{3}\geq t_{0}\) such that
Integrating (1) from t to T, we get
which takes the following form after taking the limit \(T\to\infty\):
Integrating the above inequality from t to T, we have
Multiplying the above inequality by \(1/r(t)\), and then integrating from t to T, we get
which, on taking the limit \(T\to\infty\), yields
Integrating the above inequality from \(t_{1}\) to t, we get
Using the argument employed in (7) and defining \(t^{*}=\max (t_{2},t_{3})\), we deduce that the conclusion of the lemma holds. □
Lemma 3.5
Let f and g be Δ-differentiable on \(\mathbb{T}\). Assume that \(g(t)\), \(g^{\Delta}(t)\) are not equal to zero for all \(t\in\mathbb{T}\) and have the same sign. Then
if
Proof
Since \(g(t)\) and \(g^{\Delta}(t)\) are not equal to zero for all \(t\in \mathbb{T}\), it follows from the identity \(g(\sigma(t))=\mu (t)g^{\Delta}(t)+g(t)\) that \(g(\sigma(t))\) is not equal to zero for all \(t\in\mathbb{T}\). Hence, for any \(\varepsilon>0\), there exists \(T>t_{0}\) such that, for \(t\geq T\), we have
If \(\operatorname{sgn}g(t)=\operatorname{sgn} g^{\Delta}(t)>0\), then
Hence, noticing that \(f(\sigma(t))=\mu(t)f^{\Delta}(t)+f(t)\) and \(g(\sigma(t))>0\), we get
Consequently, we have
Also one can observe that the above expression holds for \(\operatorname {sgn}g(t)=\operatorname{sgn} g^{\Delta}(t)<0\). Thus, in view of the arbitrariness of ε, we obtain the desired result. □
Theorem 3.6
Assume that g is strongly superlinear and \(\eta(t)\geq\sigma(t)\). Then every solution of (1) is oscillatory if and only if
Proof
We first prove the necessity by contradiction. Let us suppose that condition (8) does not hold true. Then there exists a positive constant c such that
So we can choose a sufficiently large \(T>t_{0}\) such that
Let
It is clear that U is a Banach space with the norm \(\|y\|=\sup_{t\in \mathbb{T}_{0}}|y(t)|\). Let us introduce a closed, bounded, and convex subset of U defined by
Define a map \(\mathcal{P}\) on Ω as follows:
In the sequel, we will show that \(\mathcal{P}\) has a fixed point in Ω.
Step I. \(\mathcal{P}\) maps Ω into Ω. Let \(y\in \varOmega\). Then \(c/2\leq|y(t)|\leq c\) for \(t\geq t_{0}\). In view of Lemma 3.2, we have that \(\widetilde{\mathbf{R}}_{T}(t)\leq \widetilde{\mathbf{R}}(t)\) for \(T\geq t_{0}\). Then, for \(t\geq T\), we obtain
Then it follows from the strong superlinearity of g that
which implies that \(c/2\leq(\mathcal{P}y)(t)\leq c\) for \(t\in\mathbb {T}_{0}\). This shows that \(\mathcal{P}\varOmega\subseteq\varOmega\).
Step II. \(\mathcal{P}\) is completely continuous.
We first show that \(\mathcal{P}\) is continuous. Let \(y_{n}\in\varOmega\) (\(n=1,2,\ldots\)) such that \(\|y_{n}-y\|\rightarrow0\) as \(n\rightarrow \infty\). Hence we get \(y\in\varOmega\) since Ω is a closed set. Then
which, by the strong superlinearity of g, yields
In consequence, we get
Since \(|g(s,y_{n}(\eta(s)))-g(s,y(\eta(s)))|\to0\) as \(n\to\infty\), the Lebesgue dominated convergence theorem implies that \(\lim_{n \rightarrow\infty}\|\mathcal{P}y_{n}-\mathcal{P}y\|=0\), and hence we obtain that \(\mathcal{P}\) is continuous in Ω.
Next, we show that \(\mathcal{P}\varOmega\) is relatively compact. According to the Arzela–Ascoli theorem on time scales (see [6]), we just need to verify that the family of functions \(\{\mathcal{P}y: y\in\varOmega\}\) is bounded and uniformly Cauchy, and \(\{\mathcal{P}y: y\in\varOmega\}\) is equi-continuous on \([t_{0}, T_{1}]\) for any \(T_{1}\in\mathbb{T}_{0}\). Firstly, the boundedness is obvious. Secondly, in view of (9), for any \(\varepsilon>0\), we can choose a sufficiently large number \(T^{*}\geq T\) so that
Hence, for \(y\in\varOmega\), \(t_{2}>t_{1}\geq T^{*}\), we get
which implies that \(\{\mathcal{P}y: y\in\varOmega\}\) is uniformly Cauchy.
For any \(T_{1}\in\mathbb{T}_{0}\) and \(y\in \varOmega\), if \(T\leq t_{1}< t_{2}\leq T_{1}\), then
where
On the other hand, by L’Hôpital’s rule, we get
which, in view of Lemma 3.5, implies that \(\lim_{t\to\infty}\frac{K_{t_{1}}(\sigma(t))}{\widetilde{\mathbf {K}}_{t_{1}}(\sigma(t))}=0\), where
Furthermore, by using the earlier argument, we find that \(\lim_{t\to\infty}\frac{\overline{R}_{t_{1}}(\sigma(t))}{ \widetilde{\mathbf{R}}_{t_{1}}(\sigma(t))}=0\). Hence, for any \(\epsilon>0\), there exists \(T_{1}^{*}\geq t_{1}\) such that \(\overline {R}_{t_{1}}(\sigma(t))<\epsilon\widetilde{\mathbf{R}}_{t_{1}}(\sigma (t))\) for \(t\geq T_{1}^{*}\). This means that
Therefore, there exists \(\delta>0\) such that
Moreover, we have
From the preceding arguments, we conclude that \(\{\mathcal{P}y: y\in \varOmega\}\) is equi-continuous on \([t_{0}, T_{1}]\). Hence, \(\mathcal{P}\varOmega\) is relatively compact. Thus, \(\mathcal{P}\) is completely continuous. Hence, by Schauder’s fixed point theorem, \(\mathcal{P}\) has a fixed point \(y_{0}\in\varOmega\), which is a nonoscillatory solution of (1). This is a contradiction.
We next prove the sufficiency by contradiction. Without loss of generality, let \(y(t)\) be an eventually positive solution of (1). Then, from Lemma 3.4, there exists \(t_{1}\geq t_{0}\) such that, for any \(T\geq t_{1}\), we have
It follows from \(y^{\Delta}(t)>0\) for \(t\geq t_{1}\) that there exists a constant \(c_{1}>0\) such that \(y(\eta(t))\geq c_{1}\) for \(t\geq t_{1}\). Then, by the strong superlinearity of g, we have
Hence
that is,
Notice that there exists \(\zeta\in[s,\sigma(s)]\) such that
Multiplying (10) by \(\widetilde{\mathbf{R}}_{t_{1}}(\sigma (t))y^{\alpha}(\sigma(t))g(t,c_{1})\) and then integrating from \(t_{2}\) (\(t_{2}>T\)) to t, we get
This means that
On the other hand, we have
Now, we show that \(\lim_{t\to\infty}\frac{\widetilde {\mathbf{R}}_{t_{1}}(\sigma(t))}{\widetilde{\mathbf{R}}(\sigma (t))}=1\). In fact, from L’Hôpital’s rule and Lemma 3.5, we need to prove that
i.e., it is sufficient to show that
Furthermore, one can see that expression (11) is true. Indeed, from L’Hôpital’s rule and Lemma 3.5 again, we just need to show that
Obviously, (12) is satisfied. Thus, there exists \(t_{3}\geq T\) such that \(\widetilde{\mathbf{R}}_{t_{1}}(\sigma(t))>\frac {1}{2}\widetilde{\mathbf{R}}(\sigma(t))\) for \(t\geq t_{3}\). It means that
which contradicts (3). The proof is complete. □
Theorem 3.7
Assume that g is strongly sublinear and \(\eta(t)\leq t\). Then every solution of (1) is oscillatory if and only if
Proof
We first prove the necessity by contradiction. Suppose that condition (13) does not hold true. Then there exists \(c>0\) such that
Let \(T>t_{0}\) be so large that
Let
Obviously, U is a Banach space with the norm \(\|y\|=\sup_{t\in \mathbb{T}_{0}}\frac{|y(t)|}{R^{2}(t)}\). Define a closed and convex subset of U as follows:
We define a map \(\mathcal{P}\) on Ω by
In order to show that \(\mathcal{P}\) has a fixed point in Ω, we proceed as follows.
Step I. \(\mathcal{P}\) maps Ω into Ω. Let \(y\in \varOmega\). Then \(cR(t)\leq|y(t)|\leq2cR(t)\) for \(t\geq t_{0}\). Furthermore, we have
which implies that \(cR(t)\leq|(\mathcal{P}y)(t)|\leq2cR(t)\) for \(t\in\mathbb{T}_{0}\). This shows that \(\mathcal{P}\varOmega\subseteq \varOmega\).
Step II. \(\mathcal{P}\) is completely continuous.
Firstly, we show that \(\mathcal{P}\) is continuous. Set \(y_{n}\in\varOmega \) and \(\|y_{n}-y\|\rightarrow0\) as \(n\rightarrow\infty\). Hence, we have \(y\in\varOmega\) since Ω is a closed set. Then
By the strong sublinearity of g, we have
Then
Since \(|g(s,y_{n}(\eta(s)))-g(s,y(\eta(s)))|\to0\) as \(n\to\infty\), the Lebesgue dominated convergence theorem implies that \(\lim_{n \rightarrow\infty}\|\mathcal{P}y_{n}-\mathcal{P}y\|=0\), and thus \(\mathcal{P}\) is continuous in Ω.
We next show that \(\mathcal{P}\varOmega\) is relatively compact. The boundedness is obvious. For any \(\varepsilon>0\), let \(T^{*}\geq T\) be so large that
Hence, for \(y\in\varOmega\), \(t_{2}>t_{1}\geq T^{*}\),
which implies that \(\{\mathcal{P}y: y\in\varOmega\}\) is uniformly Cauchy. Furthermore, for any \(y\in \varOmega\) and \(T_{1}\in\mathbb{T}_{0}\), if \(T\leq t_{1}< t_{2}\leq T_{1}\), then
Hence, there exists \(\delta>0\) such that
Moreover, we have
In consequence, \(\{\mathcal{P}y: y\in\varOmega\}\) is equi-continuous on \([t_{0}, T_{1}]\). According to the Arzela–Ascoli theorem on time scales, we know that \(\mathcal{P}\) is a compact operator. Hence \(\mathcal{P}\) is completely continuous. Therefore, \(\mathcal{P}\) has a fixed point \(y_{0}\in\varOmega\) according to Schauder’s fixed point theorem, which is a nonoscillatory solution of (1). This is a contradiction.
Now, we prove the sufficiency by contradiction. Without loss of generality, let \(y(t)\) be an eventually positive solution of (1). From Lemmas 3.1 and 3.3, there exist \(t_{1}\geq t_{0}\) and a positive constant \(c_{1}\) such that
and
Noting that \([r(t)y^{\Delta\Delta}(t)]^{\Delta\Delta}<0\), we have
From (14), (15), and the strong sublinearity of g, there exists \(\zeta\in[t,\sigma(t)]\) such that
Integrating the inequalities above from \(t_{2}\) to t, we get
and so
which contradicts (13). This completes the proof. □
Remark 3.8
It is noteworthy that the results given in the aforementioned theorems are the same as those in [13] where the dynamic equation on time scales is reduced to a differential equation when \(\mathbb{T}=\mathbb {R}_{+}\), \(\sigma(t)=t\), and \(x^{\Delta}=x'\). If further we set \(r(t)=1\) and \(\eta(t)=\sin t\) in (1), we conclude that \(R(t)=\widetilde{\mathbf{R}}(t)=t^{3}/6\), and every solution of (1) is oscillatory in view of Theorem 3.7 whenever (13) holds under a suitable strongly sublinear function g. However, for the case of \(\mathbb{T}=\mathbb{N}\), we know that \(\sigma (t)=t+1\) and \(x^{\Delta}(t)=\Delta x(t)=x(t+1)-x(t)\), there is no work concerning the sufficient and necessary conditions for the following corresponding difference equation:
Example 3.9
Let \(\mathbb{T}=\{t: t\in\mathbb{N}\}\), \(r(t)=\frac{t+1}{2t} \), \(\eta(t)=\lambda t\) for \(\lambda>2\), and \(g(t,x(t))=x^{3}(t)\) in (16). It is easy to see that \(\sum_{t=1}^{\infty}t/r(t)=\infty\) and \(\sum_{t=1}^{\infty}\widetilde{\mathbf{R}}(t+1)|c^{3}|=\infty\) for \(c\neq0\), and then all the conditions of Theorem 3.6 are satisfied. Thus every solution of (16) is oscillatory.
Example 3.10
Consider the fourth order dynamic equation
Here, \(\mathbb{T}=2^{\mathbb{N}}\), \(\eta(t)=4t\) and \(g(t,y(t))=\frac {1}{t}y^{5}(t)\). Hence we have \(\sigma(t)=2t\). In this case \(r(t)=t\), one can check that \(\int_{2}^{\infty}t/r(t) \Delta t=\infty\) and \(\widetilde{\mathbf{R}}(t)=\frac{4}{3}t^{2}-\frac{4}{3}-2t\frac{\ln t}{\ln2}\). Hence \(\int_{2}^{\infty}\widetilde{\mathbf {R}}(2t)|g(t,c)|\Delta t=\infty\). It means that all the conditions of Theorem 3.6 are satisfied. Then every solution of (17) is oscillatory.
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Zhou, Y., He, J.W., Ahmad, B. et al. Necessary and sufficient conditions for oscillation of fourth order dynamic equations on time scales. Adv Differ Equ 2019, 308 (2019). https://doi.org/10.1186/s13662-019-2245-7
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DOI: https://doi.org/10.1186/s13662-019-2245-7