Oscillatory behavior of third-order neutral delay differential equations with distributed deviating arguments

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.

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:

1. (A1)

   \(r_{i}(t)\in C([t_{0},\infty ),(0,\infty ))\), \(\int ^{ \infty }_{t_{0}}r^{-\frac{1}{\alpha _{i}}}_{i}(t)\,dt=\infty \), \(i=1, 2\);

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;

3. (A3)

   \(\tau (t)\in C^{1}([t_{0},\infty ),\mathbb{R})\), \(\tau (t)\leq t\), \(\tau '(t)>0\) and \(\lim_{t\rightarrow \infty } \tau (t)=\infty \);

4. (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 \(\xi \in [a,b]\);

5. (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.

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

Combining (1.2) and (2.17), we obtain (2.12). □

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

Using (3.12) and (3.26) with

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.

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.

Affiliations

1. School of Mathematical Sciences, University of Jinan, Jinan, P.R. China

   • Yibing Sun
   •  & Yige Zhao

Contributions

All authors contributed equally to this work. They all read and approved the final version of the manuscript.

Corresponding author

Correspondence to Yibing Sun.

Competing interests

The authors declare that they have no competing interests.

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