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Oscillatory behavior of third-order neutral delay differential equations with distributed deviating arguments
Journal of Inequalities and Applications volume 2019, Article number: 207 (2019)
Abstract
The main contribution of this paper is to establish some new criteria, which ensure that every solution of third-order neutral delay differential equations with distributed deviating arguments is either oscillatory or tends to zero. The obtained theorems extend and improve several known results in the literature. Two examples are provided to illustrate the main results.
1 Introduction
Over the last several years, an increasing attention has been given to the oscillation theory and asymptotic behavior of various classes of second-order and high-order differential equations and dynamic equations on time scales [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17]. So far, much research activity concerns the oscillation problem of the third-order (TO) neutral differential and dynamic equations [18,19,20,21,22,23,24,25,26]. Recently, the research focus has been shifted to the study of the TO differential equations (DE) with distributed deviating arguments (DDA), and some results can be found in [27,28,29,30,31,32,33,34,35].
Li et al. [3] investigated
where \(z(t)=x(t)+p(t)x(\tau (t))\), \(0\leq p(t)\leq p_{0}<\infty \) and \(\alpha \geq 1\) is a constant. Under the methods proposed by Li et al. [3, 20], Jiang and Li [30] studied the following equation with DDA:
where \(\alpha >0\) is a constant, and obtained several theorems for (1.1) whenever
Furthermore, Elabbasy and Moaaz [31], and Wang et al. [32] examined a TODE with DDA under the assumption \(0\leq p(t) \leq P<1\). However, the obtained oscillation theorems cannot be applied when \(p(t)\geq 1\). Then Tunç [33] utilized a new technique, different from the existing methods, to give some criteria for a TODE with DDA, when \(p(t)\geq 1\).
The main objective here is to establish several oscillation criteria for the TO neutral delay (ND) DE with DDA
where \(t\geq t_{0}>0\),
and \(\alpha _{1}\), \(\alpha _{2}\), and \(\alpha _{3}\) are positive constants. We assume that:
- (A1)
\(r_{i}(t)\in C([t_{0},\infty ),(0,\infty ))\), \(\int ^{ \infty }_{t_{0}}r^{-\frac{1}{\alpha _{i}}}_{i}(t)\,dt=\infty \), \(i=1, 2\);
- (A2)
\(p(t)\in C([t_{0},\infty ),[1,\infty ))\) with \(p(t)\not \equiv 1\), and \(q(t,\xi )\in C([t_{0},\infty )\times [a,b],[0,\infty ))\) with \(q(t,\xi )\not \equiv 0\) eventually;
- (A3)
\(\tau (t)\in C^{1}([t_{0},\infty ),\mathbb{R})\), \(\tau (t)\leq t\), \(\tau '(t)>0\) and \(\lim_{t\rightarrow \infty } \tau (t)=\infty \);
- (A4)
\(g(t,\xi )\in C([t_{0},\infty )\times [a,b],\mathbb{R})\) is a nondecreasing function for ξ, which satisfies \(\liminf_{t\rightarrow \infty }g(t,\xi )=\infty \) for \(\xi \in [a,b]\);
- (A5)
\(\sigma (\xi )\in C([a,b],\mathbb{R})\) is nondecreasing and the integral of (1.2) is taken in the Riemann–Stieltjies sense.
This article is organized in the following manner. Section 2 presents three lemmas to prove our results. Section 3 establishes some new oscillation criteria for (1.2). Two examples finalize this article.
2 Some lemmas
For simplicity, we use some notations for sufficiently large \(t_{1}\) with \(t_{1}\geq t_{0}\) as below:
Furthermore, assume that
Lemma 2.1
Assume that (A1)–(A5) hold. Furthermore, let \(x(t)\) be an eventually positive solution of (1.2). Then \(z(t)\) satisfies either
or
Proof
From the condition of Lemma 2.1, there exists a \(t_{1}\geq t_{0}\) such that
for \(t\geq t_{1}\). Then (1.2) implies that
which means that \(E_{2}(t)\) is nonincreasing and of constant sign, and \(E'_{1}(t)\) is also of constant sign. We claim that \(E'_{1}(t)>0\). To prove this, assume on the contrary that \(E'_{1}(t)<0\) and there exists \(M_{1}\), s.t. for \(t\geq t_{2} \geq t_{1}\),
Then
We integrate (2.5) to get
Letting \(t\rightarrow \infty \) and from (A1), we obtain \(\lim_{t\rightarrow \infty }E_{1}(t)=-\infty \). Then there exist constants \(M_{2}\) and \(t_{3}\geq t_{2}\) such that
which yields that
Integrating the latter inequality from \(t_{3}\) to t, we have
We use (A1) again to have \(\lim_{t\rightarrow \infty }z(t)=-\infty \), which contradicts \(z(t)>0\). Hence, \(E'_{1}(t)>0\) for \(t\geq t_{1}\), and \(z(t)\) has properties (I) and (II). □
Lemma 2.2
Assume that (A1)–(A5) and (2.1) hold. Furthermore, suppose that \(x(t)\) is an eventually positive solution of (1.2) and \(z(t)\) has property (II) of Lemma 2.1. If
then \(\lim_{t\rightarrow \infty }x(t)=0\).
Proof
One can see that (1.3) yields (see [33])
which can be rewritten as
Combining (2.4) and (2.8), we get
based on the fact that \(z'(t)<0\) for \(t\geq t_{1}\). Since \(z(t)\) has property (II) of Lemma 2.1, one gets \(\lim_{t\rightarrow \infty }z(t)=l\geq 0\). We claim that \(l=0\). Indeed, if we assume on the contrary that \(l>0\), then there exists \(t_{2}\geq t_{1}\) s.t. \(\tau ^{-1}(g_{2}(t))\geq t_{2}\) and \(z(\tau ^{-1}(g_{2}(t)))\geq l\), \(t\geq t_{2}\). Inequality (2.9) then yields
We integrate (2.10) to get
which indicates that
Integrating (2.11) from t to ∞, we have
and then
since \(z'(t)<0\) for \(t\geq t_{2}\). Integrating the latter inequality from \(t_{2}\) to ∞, we get
which contradicts (2.6). Thus, we obtain \(l=0\) and \(\lim_{t\rightarrow \infty }x(t)=0\), since \(0< x(t)\leq z(t)\). □
Lemma 2.3
Assume that (A1)–(A5) and (2.2) hold. Furthermore, suppose that \(x(t)\) is an eventually positive solution of (1.2) and \(z(t)\) has property (I) of Lemma 2.1. Then for \(t\geq t_{1} \geq t_{0}\),
Proof
Property (I) of \(z(t)\) yields
Applying the monotonicity of \(E^{\frac{1}{\alpha _{2}}}_{2}(t)\) gives
It can be seen from (2.13) that
which, together with \(r_{1}(t)(z'(t))^{\alpha _{1}}>0\), yields that \(z'(t)/\delta _{2}(t,t_{1})\) is nonincreasing for \(t\geq t_{1}\). Therefore,
which leads to
Then for \(t\geq t_{2}\geq t_{1}\),
due to \(\tau ^{-1}(t)\leq \tau ^{-1}(\tau ^{-1}(t))\). Substituting (2.16) into (2.7), one has
which leads to
3 Main results
We respectively consider two cases \(g_{1}(t)\leq \tau (t)\) and \(g_{1}(t)\geq \tau (t)\) for \(t\geq t_{0}\). Now, we begin with the first case.
Theorem 3.1
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold, and \(g_{1}(t)\leq \tau (t)\). If there exists \(\rho (t)\in C^{1}([t _{0},\infty ),(0,\infty ))\) s.t.
for \(t_{1}\) and \(t_{*}\) with \(t_{*}\geq t_{1}\geq t_{0}\), where
then every solution of (1.2) is either oscillatory or tends to zero.
Proof
Suppose that (1.2) has a nonoscillatory solution \(x(t)\). We may assume that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). So we have that \(z(t)\) is positive and satisfies the two properties for \(t\geq t_{1}\).
We first consider property (I). Define \(\omega (t)\) by
Then \(\omega (t)>0\) and
Based on (2.12), we have
Since \(g_{1}(t)\leq \tau (t)\), we get \(\tau ^{-1}(g_{1}(t))\leq t\). Applying (2.15), we obtain
From (2.13), we have
We combine (3.4)–(3.7) to conclude that
Next, we will compute \(z^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}(t)\) and consider the following two cases:
Case (i). Assume that \(\alpha _{1}\alpha _{2}>\alpha _{3}\). Since \(z(t)/\delta _{3}(t,t_{1})\) is nonincreasing, due to (2.15), there exist constants \(h_{1}>0\) and \(t_{2}\geq t_{1}\) such that
and
where \(m_{1}=h^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}_{1}\).
Case (ii). Assume that \(\alpha _{1}\alpha _{2}\leq \alpha _{3}\). Since \(z'(t)>0\), there exists \(h_{2}>0\) such that
and
where \(m_{2}=h^{\frac{\alpha _{3}}{\alpha _{1}\alpha _{2}}-1}_{2}\). We combine (3.8) with (3.9) and (3.10) to have
By using the inequality (see [18])
where
we obtain
We combine (3.11) and (3.13) to conclude that
We integrate the latter inequality to make
which contradicts (3.1).
Secondly, we investigate property (II) and deduce \(\lim_{t\rightarrow \infty }x(t)=0\). □
Theorem 3.2
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that \(g_{1}(t)\leq \tau (t)\) and \(\alpha _{1} \alpha _{2}=\alpha _{3}\). If there exists \(\rho (t)\) s.t.
for \(t_{1}\) and \(t_{*}\) with \(t_{*}\geq t_{1}\geq t_{0}\), then we get the same conclusion as in Theorem 3.1.
Proof
Suppose that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). Then \(z(t)\) satisfies the two properties.
We first consider property (I). From (2.13) and (2.14), we have
and
Define \(\omega (t)\) by (3.2). As in the proof of Theorem 3.1, we get (3.3). Since \(E_{2}(t)>0\), (3.3) indicates
Combining the latter inequality with (3.5), (3.6), and (3.15), we see that
An integration of (3.16) from \(t_{2}\) (\(t_{2}\geq t_{1}\)) to t leads to
for all sufficiently large t, which contradicts (3.14).
Secondly, if property (II) holds, then \(\lim_{t\rightarrow \infty }x(t)=0\). □
Theorem 3.3
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that \(g_{1}(t)\leq \tau (t)\) and \(\alpha _{1} \alpha _{2}=\alpha _{3}\geq 1\). If there exists \(\rho (t)\) s.t.
for \(t_{1}\) and \(t_{*}\) with \(t_{*}\geq t_{1}\geq t_{0}\), where \(\gamma (t)\) is given in Theorem 3.1, then we get the same conclusion as in Theorem 3.1.
Proof
Suppose that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). Then \(z(t)\) satisfies the two properties.
We first consider property (I). Proceeding as in the proof of Theorem 3.1, we get (3.11), and
From (3.2) and (3.15), one has
We substitute (3.19) into (3.18) to see that
from which one gets
by completing the square with respect to \(\omega (t)\). We integrate the latter inequality from \(t_{2}\) (\(t_{2}\geq t_{1}\)) to t to obtain
for all sufficiently large t, which contradicts (3.17).
Secondly, if property (II) holds, then \(\lim_{t\rightarrow \infty }x(t)=0\). □
Next, we consider \(g_{1}(t)\geq \tau (t)\) for \(t\geq t_{0}\).
Theorem 3.4
Assume that conditions (A1)–(A5), (2.1), (2.2), and (2.6) hold, and \(g_{1}(t)\geq \tau (t)\). If there exists \(\rho (t)\) s.t.
for \(t_{1}\) and \(t_{*}\) with \(t_{*}\geq t_{1}\geq t_{0}\), then we get the same conclusion as in Theorem 3.1.
Proof
Suppose that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). Then \(z(t)\) satisfies the two properties.
We first consider property (I). Define \(\nu (t)\) by
Then \(\nu (t)>0\) and
Since \(\tau ^{-1}(g_{1}(t))\geq t\geq \tau (t)\) and \(z'(t)>0\), we have
which indicates that
due to (2.12). Based on (3.7), \(E'_{2}(t)\leq 0\) and \(\tau (t)\leq t\), so one has \(\tau (t)\geq t_{1}\) and
for \(t\geq t_{2}>t_{1}\). Combining (3.9), (3.10), (3.23)–(3.25), we conclude that
one gets
Integrating the latter inequality from \(t_{2}\) to t, we have
for all sufficiently large t, which contradicts (3.20).
Secondly, if property (II) holds, then \(\lim_{t\rightarrow \infty }x(t)=0\). □
Theorem 3.5
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that \(g_{1}(t)\geq \tau (t)\) and \(\alpha _{1} \alpha _{2}=\alpha _{3}\). If there exists \(\rho (t)\) s.t.
for \(t_{1}\) and \(t_{*}\) with \(t_{*}\geq t_{1}\geq t_{0}\), then we get the same conclusion as in Theorem 3.1.
Proof
Suppose that (2.3) holds for \(t\geq t_{1}\geq t _{0}\). Then \(z(t)\) satisfies the two properties.
We first consider property (I). Proceeding as in the proof of Theorem 3.4, we get (3.22), which implies
Applying (3.15), the monotonicity of \(E_{2}(t)\) and the fact that \(\tau (t)\leq t\), one has \(\tau (t)\geq t_{1}\) and
for \(t\geq t_{2}>t_{1}\). Combining (3.24), (3.28), and (3.29), one gets
Upon integrating (3.30) from \(t_{2}\) to t, we obtain a contradiction to (3.27).
Secondly, if property (II) holds, then \(\lim_{t\rightarrow \infty }x(t)=0\). □
Theorem 3.6
Assume that (A1)–(A5), (2.1), (2.2), and (2.6) hold. Furthermore, suppose that \(g_{1}(t)\geq \tau (t)\) and \(\alpha _{1} \alpha _{2}=\alpha _{3}\geq 1\). If there exists \(\rho (t)\) s.t.
for \(t_{1}\) and \(t_{*}\) with \(t_{*}\geq t_{1}\geq t_{0}\), then we get the same conclusion as in Theorem 3.1.
We omit the proof of Theorem 3.6 here, since it is similar to that of Theorem 3.3.
4 Examples
The following examples are given to show the applications of Theorems 3.1 and 3.5.
Example 4.1
For \(t>k_{1}\geq 1\), consider a TONDDE with DDA
where
Let \(\alpha _{1}=1/3\), \(\alpha _{2}=5\), \(\alpha _{3}=7/3\), \(a=k_{1}\), \(b=k_{1}+1\), \(r_{1}(t)=t-k_{1}\), \(r_{2}(t)=1\), \(\tau (t)=t-k_{1}\), \(g(t,\xi )=t-k_{1}-1/\xi \), \(\sigma (\xi )=\xi \), \(p(t)=(4t+5)/(t+1)\), \(q(t,\xi )=10(t+\xi )\). Choose \(t_{0}=t_{1}=k_{1}\). Then we obtain \(\alpha _{1}\alpha _{2}<\alpha _{3}\), \(4\leq p(t)<5\),
Furthermore, we deduce that
It is easy to verify that
Therefore, conditions (A1)–(A5), (2.1), (2.2), and (2.6) hold, and \(g_{1}(t)\leq \tau (t)\). We choose \(\rho (t)=t\) and \(t_{*}=k_{1}+2\). Applying Theorem 3.1, it remains to check (3.1), where
Then we get
as \(t\rightarrow \infty \), since \(\int ^{t}_{k_{1}+2}s^{-\frac{5}{3}}\,ds< \infty \). Hence, we get the same conclusion as in Theorem 3.1.
Example 4.2
For \(t>k_{1}\geq 1\), consider a TONDDE with DDA
where l is a positive integer,
Let \(\alpha _{1}=9\), \(\alpha _{2}=1/3\), \(\alpha _{3}=3\), \(a=k_{1}\), \(b=k_{1}+l\), \(r_{1}(t)=t-k_{1}\), \(r_{2}(t)=1\), \(\sigma (\xi )=\xi \),
Choose \(t_{0}=t_{1}=k_{1}\). Then we have \(\alpha _{1}\alpha _{2}=\alpha _{3}\), \(4< p(t)<5\),
where \(t\geq t_{2}>4k_{1}\), \(\delta _{1}(t,t_{1})\), \(\delta _{2}(t,t _{1})\), and \(\delta _{3}(t,t_{1})\) are the same as in Example 4.1. Furthermore, we deduce that
Clearly, (2.6) holds. Choosing \(\rho (t)=t^{2}\) and \(t_{*}=t _{2}\), one has
as \(t\rightarrow \infty \), which means that (3.27) holds, and all conditions of Theorem 3.5 are satisfied. Hence, we get the same conclusion as in Theorem 3.1.
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Funding
This research was supported by the National Natural Science Foundation of China (Grant No. 61803176, 61703180) and A Project of Shandong Province Higher Educational Science and Technology Program (Grant No. J18KA230).
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Sun, Y., Zhao, Y. Oscillatory behavior of third-order neutral delay differential equations with distributed deviating arguments. J Inequal Appl 2019, 207 (2019). https://doi.org/10.1186/s13660-019-2161-0
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DOI: https://doi.org/10.1186/s13660-019-2161-0